AP Biology

Lesson 1.1 · Unit 1 · CED topic 1.1

Structure of water and hydrogen bonding

A water molecule is two hydrogens bonded to one oxygen at about 104.5°, and nearly every property that makes life possible falls out of that bent, lopsided shape. Oxygen pulls the shared electrons much harder than hydrogen does, so each O–H bond is polar covalent: a partial negative charge (δ−) on the oxygen, a partial positive (δ+) on each hydrogen.

Because the molecule is bent, those charges don't cancel, and one water can hydrogen-bond to four neighbors at once. Cohesion, surface tension, high specific heat, evaporative cooling, floating ice, and water's power as a solvent are that one fact stated six ways.

Lesson 1.2 · Unit 1 · CED topics 1.2–1.3

Elements of life, functional groups, and polymers

Six elements (carbon, hydrogen, nitrogen, oxygen, phosphorus, and sulfur (CHNOPS)) make up about 97% of the mass of a living organism. Carbon is the one the rest hang from: with four valence electrons it forms four covalent bonds, so a carbon skeleton can branch, ring, and run to thousands of atoms while staying stable.

A skeleton alone is inert, though. What a molecule does comes from the functional groups attached to it: small clusters of atoms that behave the same way wherever they appear.

Lesson 1.3 · Unit 1 · CED topic 1.4

Carbohydrates and lipids

Glucose, fructose, and galactose all share the formula C₆H₁₂O₆. Same atoms, different arrangement, and the cell treats them as different molecules. That is the theme of this whole lesson: in biology, shape is function, and a single bond flipped upside down decides whether you can digest something or not.

Lesson 1.4 · Unit 1 · CED topics 1.4–1.5

Proteins: structure and function

Proteins do almost everything: catalysis, transport, signaling, structure, defense. They manage it because a chain of twenty possible monomers folds into an essentially unlimited number of specific shapes. And the shape is everything: swap one amino acid out of the 146 in hemoglobin's β chain and you get sickle-cell disease.

Lesson 1.5 · Unit 1 · CED topic 1.6

Nucleic acids: structure and information

A protein's shape is its function. A nucleic acid's sequence is its function: the order of four bases is the message, and every feature of DNA's structure exists to store that message, copy it faithfully, and hand it out.

Unit 1 quiz · 15 multiple-choice · 5 free-response

Unit 1 quiz: Chemistry of Life

Fifteen multiple-choice items and five short free-response questions on water, functional groups and polymers, carbohydrates and lipids, protein structure, and nucleic acids: click an option to see why each choice is right or wrong, and write each FRQ out before opening the model response.

Multiple choice

1. Data: three 200.0 g samples, each in an insulated container, were given 4180 J of heat. Every sample started at 22.0 °C.

   Substance	Water	Ethanol	Dry sand
   Specific heat (J/g·°C)	4.18	2.44	0.80
   Temperature change (°C)	5.0	8.6	26.1

   Which claim is best supported by the data?

   The sand absorbed more energy than the water, which is why its temperature rose the most. Water stores the most energy per degree of temperature change, because much of the heat added to it goes into breaking hydrogen bonds rather than speeding molecules up. Ethanol and water differ because the ethanol sample contains a smaller mass of molecules. Sand would be the best coolant of the three, because it reaches a high temperature the fastest.

   Every sample received exactly 4180 J, so heat is the variable being held constant and cannot explain the difference. A larger temperature change for the same energy means the substance stores less energy per degree, which is the opposite of what this choice claims.

   Water's 5.0 °C rise is the smallest of the three for the same 4180 J, so it takes the most energy to warm each gram by one degree. A water molecule can hydrogen-bond to up to four neighbors, and energy spent pulling those bonds apart does not show up as faster motion, which is what temperature measures.

   Mass is held constant at 200.0 g in all three containers, so the number of grams is not the variable. What differs is specific heat, which traces back to how many hydrogen bonds each molecule can form: ethanol's single hydroxyl group offers far fewer than water's two hydrogens and two lone pairs.

   Reaching a high temperature quickly is the mark of a poor coolant, not a good one. A coolant has to carry heat away, and sand's 26.1 °C rise for the same 4180 J: about 5.2 times water's rise: shows it carries the least energy for a given temperature change.

2. Which of the following best explains why a sheet of ice forms on the surface of a temperate lake in winter rather than at the bottom?

   In ice each water molecule is locked into four hydrogen bonds in a rigid lattice that holds its neighbors farther apart than they sit in the liquid, so the same mass occupies more volume. Hydrogen bonds break when water freezes, which lets the molecules spread out and makes the solid less dense. The water at the bottom of the lake is warmer, so heat rising through the lake carries the ice upward. Freezing converts water's polar covalent O–H bonds into nonpolar bonds, which lowers the density of the solid.

   This is the mechanism: in liquid water hydrogen bonds constantly break and re-form, letting molecules crowd close, while the ice lattice fixes every molecule at four bonds and a set distance. Lower density means the ice floats, the lake freezes top-down, and organisms survive under an insulating lid.

   Freezing does the reverse: it maximizes hydrogen bonding, holding every molecule in four bonds at once. If hydrogen bonds simply broke, water would behave like most substances and contract as it solidified, and the ice would sink.

   Ice floats whether or not any heat is moving, because its density is lower than the liquid's; buoyancy, not convection, holds it up. A lake's bottom water does sit near 4 °C, but that is a consequence of water's density maximum, not the cause of floating ice.

   Covalent bonds inside a water molecule are untouched by freezing: only the weak attractions between molecules rearrange. Oxygen keeps pulling the shared electrons harder than hydrogen does in ice exactly as it does in liquid water, so the bonds stay polar.

3. A researcher coats the inner wall of one glass capillary tube with a nonpolar wax and leaves an identical tube untreated. Both are stood upright in a dish of water. Which statement most likely describes the result and its cause?

   Water rises higher in the waxed tube, because the coating reduces friction between the water and the wall. Water rises to the same height in both tubes, because cohesion among water molecules is what lifts the column. Water rises far less in the waxed tube, because capillary rise depends on adhesion, hydrogen bonds between water and a polar surface, and wax presents no partial charges to bond with. Water rises higher in the waxed tube, because the coating increases the surface tension of the water.

   Capillary rise is not limited by friction; it is driven by an attraction that pulls water up the wall. Removing that attraction removes the driving force, so the waxed tube performs worse rather than better.

   Cohesion holds the column together once it is climbing, but something has to grip the wall first. Without adhesion the water has nothing to climb, which is why the same cohesive liquid behaves differently in the two tubes, and why a tree needs both to move water up xylem.

   Glass exposes δ− oxygens that hydrogen-bond with water's δ+ hydrogens, and that grip on the wall is adhesion. A nonpolar wax has no partial charges, so no hydrogen bonds form, and the cohesive column has no anchor to pull itself up along.

   Surface tension comes from cohesion among the water molecules themselves and is a property of the liquid, not of the tube's coating. Even if it rose, higher surface tension alone would not lift a column up a wall the water cannot bond to.

4. Data: the pH of 0.1 M aqueous solutions of four small organic molecules at 25 °C, with the functional groups each molecule carries.

   Molecule	Ethanol	Acetic acid	Ethylamine	Glycine
   Functional group(s)	hydroxyl	carboxyl	amino	amino and carboxyl
   pH of 0.1 M solution	7.0	2.9	11.8	6.1

   Which claim about functional groups is best supported by the data?

   Any functional group containing oxygen makes an aqueous solution acidic. A carboxyl group releases H⁺ into solution and an amino group takes H⁺ out of it, so a molecule carrying one of each ends up near neutral. The more polar the molecule, the lower the pH of its solution. Glycine's solution is near neutral because glycine carries no charged groups at pH 6.1.

   Ethanol's hydroxyl group contains oxygen, yet its 0.1 M solution sits at pH 7.0: exactly neutral. The table separates "contains oxygen" from "donates H⁺": only the carboxyl group does the second, because losing H⁺ leaves a resonance-stabilized: COO⁻.

   Acetic acid's carboxyl drives the solution to pH 2.9 by donating H⁺, ethylamine's amino group drives it to pH 11.8 by accepting H⁺, and glycine, which has both, lands at 6.1: close to ethanol's neutral 7.0 and nowhere near either extreme. The two groups act in opposite directions and largely cancel.

   Polarity and acidity are different properties. Ethanol's hydroxyl is polar enough to hydrogen-bond freely with water, which is why ethanol dissolves without limit, yet its pH is 7.0: polarity governs solubility, while acidity depends on whether a group can actually release a proton.

   Glycine at pH 6.1 is a zwitterion: its carboxyl has donated H⁺ to become, COO⁻ and its amino group has accepted one to become, NH₃⁺. It carries two charges, not none; the solution is near neutral because those charges balance, not because they are absent.

5. A polypeptide is built by dehydration synthesis from 25 alanine monomers (M = 89.09 g/mol). Water has M = 18.02 g/mol. What is the molar mass of the finished polypeptide?

   1776.75 g/mol 1794.77 g/mol 2227.25 g/mol 2659.73 g/mol

   This subtracts 25 waters instead of 24: 2227.25 − 25(18.02) = 1776.75. A chain of n monomers has only n − 1 bonds between them, so 25 residues release 24 waters, not 25: count the links, not the beads.

   Twenty-five residues require 24 peptide bonds, and each bond releases one water: 25(89.09) − 24(18.02) = 2227.25 − 432.48 = 1794.77 g/mol. Equivalently, every residue after the first adds only 89.09 − 18.02 = 71.07 g/mol to the chain.

   This is the combined mass of 25 free alanine molecules with no water removed at all. The polymer must weigh less than its monomers did separately, and the missing 432.48 g/mol is exactly the water the cell released while building it.

   This adds the water instead of removing it: 2227.25 + 432.48 = 2659.73. Adding water is hydrolysis, the reaction that takes the chain apart; dehydration synthesis, which builds it, removes one H₂O per bond formed.

6. A polypeptide with the sequence Gly–Ala–Ser and one with the sequence Ser–Ala–Gly are different molecules. Which statement best explains why?

   The two chains are built from different amino acids. The two chains have different molar masses, because a different number of water molecules was removed in building each one. Each peptide bond joins the carboxyl group of one residue to the amino group of the next, so the chain runs from a free amino end to a free carboxyl end, and the order of R groups along that direction is reversed in the two molecules. One chain was assembled by dehydration synthesis and the other by hydrolysis.

   Both chains contain one glycine, one alanine, and one serine: the composition is identical. The exam is testing the difference between composition and sequence, and that difference is the whole reason twenty monomers can build an unlimited number of distinct proteins.

   Both chains have three residues and therefore two peptide bonds, so both released two waters and both have the same molar mass. Mass cannot distinguish them; only the order along the chain can.

   Dehydration synthesis always links a carboxyl to an amino group, which gives every polypeptide a direction: an N-terminus at one end and a C-terminus at the other. Reading Gly–Ala–Ser from the N-terminus puts glycine's hydrogen R group first and serine's hydroxyl last, while the reverse sequence puts them the other way around, producing a different fold and a different function.

   Hydrolysis breaks polymers apart; it cannot assemble one. Both of these chains were built the same way, by dehydration synthesis, and the difference between them lies in the order in which the monomers were added.

7. Data: melting points of four fatty acids at 1 atm.

   Fatty acid	Myristic	Palmitic	Stearic	Oleic
   Carbons in the tail	14	16	18	18
   C=C double bonds	0	0	0	1
   Melting point (°C)	54.4	62.9	69.6	13.4

   Which claim is best supported by the data?

   Melting point is determined only by the number of carbons in the tail. Unsaturated fatty acids melt at higher temperatures, because a double bond is stronger than a single bond. Tail length has no measurable effect on melting point. Introducing a single cis double bond lowers the melting point far more than removing four carbons from a saturated tail does.

   Stearic and oleic acid have the same 18 carbons yet melt 56.2 °C apart, so chain length cannot be the only factor. The table was built with that pair deliberately matched so that the effect of the double bond could be isolated.

   The data run the other way: oleic acid, the only unsaturated one, melts lowest at 13.4 °C. The strength of the C=C bond itself is irrelevant to melting, which depends on the weak attractions between neighboring tails, and a cis kink keeps those tails from stacking closely enough to form many.

   Among the three saturated acids the melting point climbs steadily with length, from 54.4 °C at 14 carbons to 69.6 °C at 18. Length matters; it simply matters less than unsaturation does.

   Compare the two controlled contrasts. Going from stearic (18:0) to oleic (18:1) holds length constant and drops the melting point 56.2 °C; going from stearic (18:0) to myristic (14:0) holds saturation constant and drops it only 15.2 °C. One kink outweighs four carbons by nearly four to one, which is why a membrane rich in unsaturated tails stays fluid in the cold.

8. Humans digest starch but pass cellulose through largely unchanged, even though both are polymers of glucose. Which of the following best explains the difference?

   Starch's α-1,4 linkages curl the chain into a helix that fits the active site of amylase, while cellulose's β-1,4 linkages flip every other glucose and give a straight chain that no human enzyme's active site is complementary to. Cellulose is built from a different monosaccharide than starch is. Cellulose molecules are too large to be absorbed, so they cannot be digested. Starch is held together by hydrogen bonds, which are easily broken, while cellulose is held together by covalent bonds.

   The monomer is the same; the geometry of the bond is not. An α-1,4 linkage produces a helical chain amylase can grip and cut, while a β-1,4 linkage rotates alternate residues 180°, giving flat chains that hydrogen-bond to each other into microfibrils. Enzyme–substrate specificity depends on shape, and humans make no enzyme shaped for β-1,4.

   Both polymers are made entirely of glucose. That is precisely what makes the comparison useful: with the monomer held constant, the only variable left is the linkage, so the linkage has to be the explanation.

   Size is not the barrier: starch molecules are also far too large to absorb, which is exactly why amylase digests them into maltose and glucose first. The problem with cellulose is that the digestive step never happens.

   Both starch and cellulose are held together by covalent glycosidic linkages. Hydrogen bonds do appear in cellulose, but between neighboring chains, where they add strength rather than weakness: the opposite of what this choice implies.

9. A migrating bird must store 1800 kcal. Fat yields about 9 kcal/g. Glycogen yields about 4 kcal/g and binds roughly 2 g of water for every gram stored. Approximately how many times greater is the total mass the bird must carry if it stores the energy as hydrated glycogen rather than as fat?

   2.3 times 4.5 times 6.8 times 13.5 times

   This is 450/200, the comparison of dry glycogen to fat. It leaves out the water, which is the whole reason glycogen is a poor long-term store: the stem gives you 2 g of water per gram of glycogen for a reason.

   This is 900/200, counting only the bound water and forgetting the glycogen itself. The carried mass is the polymer plus its water, not one or the other.

   Fat: 1800/9 = 200 g. Glycogen: 1800/4 = 450 g, plus 2(450) = 900 g of bound water, for 1350 g total. Then 1350/200 = 6.8. Fat wins twice over (its highly reduced, C) H-rich tails carry more energy per gram, and being hydrophobic it drags no water along.

   This is 2700/200, which doubles the hydrated glycogen mass once too often. Two grams of water per gram of glycogen gives 900 g of water on top of 450 g of polymer, for 1350 g, not 2700 g.

10. Data: initial rate of a purified intestinal enzyme at 37 °C. Row 1 gives the rate measured at the pH in that column. Row 2 gives the rate measured at pH 7 for a sample that was first held for 20 minutes at the pH in that column and then returned to pH 7.

    pH of treatment	3	5	7	9	11
    Rate at that pH (µmol/min)	4	22	55	11	1
    Rate after return to pH 7 (µmol/min)	3	54	55	53	2

    Which statement best explains why the pH 5 and pH 9 samples recover after being returned to pH 7 while the pH 3 and pH 11 samples do not?

    Extreme pH hydrolyzes the peptide bonds of the primary structure, so at pH 3 and pH 11 there is no enzyme left to assay. Mild pH shifts only change which R groups are protonated, and that reverses when the pH is restored; at pH 3 and pH 11 the ionic and hydrogen bonds holding the tertiary fold are disrupted so far that the active site does not reform. The enzyme functions best at pH 7, so any sample tested away from pH 7 is permanently slowed. H⁺ and OH⁻ ions are competitive inhibitors that stay bound in the active site after the pH is restored.

    If peptide bonds were broken there would be no chain left to recover, but the row-2 values of 3 and 2 µmol/min are not the issue: the pH 5 and pH 9 samples show the chain survives this treatment intact and simply needs its fold back. Denaturation rearranges weak interactions; it does not cut the covalent backbone.

    Row 2 is the decisive comparison. The pH 5 and pH 9 samples return to 54 and 53 µmol/min, essentially the untreated 55, so their loss of activity was nothing but a reversible change in the charge on acidic and basic R groups. At pH 3 and pH 11 the ionic and hydrogen bonds between R groups were disrupted enough that the chain settled into a misfolded state, and restoring the pH restores only 3 and 2 µmol/min.

    Row 1 alone would fit this idea, but row 2 refutes it: the pH 5 sample was tested away from pH 7 and came back to 54 µmol/min. Being off the optimum is not the same as being permanently damaged, and telling the two apart is exactly what the second row was designed to do.

    A competitive inhibitor is removed when it is washed out, so returning the sample to pH 7 would restore activity in every column, which is not what row 2 shows. H⁺ changes the enzyme by protonating its R groups rather than by occupying the active site.

11. Figure: hemoglobin from three individuals separated by gel electrophoresis at pH 8.6, with the sample wells at the top of the gel and the anode (+) at the bottom.

    Lane 1 shows a single band 62 mm below the wells. Lane 2 shows a single band 41 mm below the wells. Lane 3 shows two bands of roughly equal intensity, one at 62 mm and one at 41 mm. Hemoglobin A carries glutamate at position 6 of each β chain; hemoglobin S carries valine at that position. The two proteins have the same molar mass.

    Which statement best explains the position of the band in lane 2 and correctly identifies the heterozygous individual?

    Valine's R group is nonpolar and uncharged while glutamate's is negatively charged at pH 8.6, so hemoglobin S carries less net negative charge and is pulled less strongly toward the anode; lane 3 is the heterozygote. Lane 2's hemoglobin is a larger molecule and is slowed by the gel matrix; lane 1 is the heterozygote. Lane 2's hemoglobin is denatured by the substitution and cannot migrate normally; lane 2 is the heterozygote. Valine adds a positive charge that pushes the molecule back toward the well; lane 1 is the heterozygote.

    At pH 8.6 glutamate's carboxyl R group is deprotonated and negative, so each HbA tetramer carries two more negative charges than an HbS tetramer, whose valines are neutral. More negative charge means a stronger pull to the anode and a band farther down: 62 mm for HbA, 41 mm for HbS. Lane 3 shows both bands, so that person makes both proteins and carries one allele of each.

    The stimulus states that the two proteins have the same molar mass, which removes size as an explanation: the substitution swaps one residue for another of similar bulk. A single band also means a single kind of hemoglobin, so lane 1 cannot be the heterozygote.

    A denatured protein would still carry charge and still migrate; nothing in the figure indicates unfolding, and sickle hemoglobin is a correctly folded, functional oxygen carrier that simply polymerizes when deoxygenated. One band in lane 2 means one allele's product, so lane 2 is homozygous.

    Valine's R group is a nonpolar hydrocarbon; it carries no charge at any physiological pH. The substitution does not add positive charge, it removes negative charge, which is enough to slow migration without reversing it, and lane 1's single band rules it out as the heterozygote anyway.

12. A mutation replaces a valine buried in the hydrophobic core of a globular enzyme with a lysine. Which of the following is the most likely consequence, with the best reason?

    Activity is unchanged, because only the amino acids that make up the active site itself affect an enzyme's function. Activity rises, because a charged R group can form additional ionic bonds and hold the fold together more tightly. No protein is produced at all, because the mutation destroys the primary structure of the chain. Activity falls sharply, because lysine's R group is positively charged and is unstable in a water-excluded core, so the chain distorts to move that charge toward the surface and the active site's geometry changes with it.

    The active site is not an independent part; it is a shape produced by the whole fold, and the fold is held together by R-group interactions throughout the chain. Anfinsen's result that the sequence alone dictates the native structure is the reason a change far from the site can still abolish activity.

    Ionic bonds do stabilize tertiary structure, but only where a partner of opposite charge sits nearby and water can reach them. Inside a water-excluded core there is no hydration and usually no counter-ion, so the buried charge destabilizes the fold instead of reinforcing it.

    A substitution changes the primary structure; it does not destroy it. The ribosome still builds a full-length chain of the same number of residues: the problem arises afterward, when that chain folds into the wrong shape.

    Burying a hydrophobic R group in the core is energetically favorable because it lets water maximize its own hydrogen bonding. A charged lysine there is the opposite: the protein must partially unfold or rearrange to let the charge reach solvent, and the active site, whose geometry depends on that fold, no longer matches the substrate.

13. Data: base composition of three nucleic acid samples, in mole percent of total bases.

    Sample	A	T or U	G	C
    W	22.1	22.0 (T)	27.9	28.0
    X	29.8	30.1 (T)	20.2	19.9
    Y	24.0	19.5 (U)	31.2	25.3

    Which claim is best supported by the data?

    Sample X separates into single strands at a higher temperature than sample W does. Sample Y is double-stranded RNA, since its four bases still sum to 100%. All three samples obey Chargaff's rules, so all three are double-stranded. Samples W and X are double-stranded DNA, and of the two, W requires the higher temperature to separate its strands; sample Y is not double-stranded.

    This reverses the relationship. Sample X is 20.2 + 19.9 = 40.1% G + C, while W is 27.9 + 28.0 = 55.9%, and each G–C pair is held by three hydrogen bonds against A–T's two, so W is the sample that needs more thermal energy to come apart.

    Summing to 100% only means the four bases account for all the bases present; it says nothing about pairing. The test for a duplex is whether A matches its partner and G matches its partner, and Y fails both, which is why its composition is free to be lopsided.

    Y does not obey them: its A and U differ by 4.5 percentage points and its G and C by 5.9. Chargaff's rules hold only where every base has a complementary partner on a second strand, so a violation is evidence that no such strand exists.

    W and X both satisfy %A = %T and %G = %C within rounding, the signature of complementary base pairing in a duplex. W's 55.9% G + C against X's 40.1% means more three-bond pairs per base, hence the higher melting temperature. Y contains uracil and breaks both equalities, 24.0 ≠ 19.5 and 31.2 ≠ 25.3, so it is single-stranded RNA.

14. A double-stranded DNA fragment is 12 base pairs long and contains 7 G–C pairs. How many hydrogen bonds hold the two strands together?

    24 29 31 36

    This treats all 12 pairs as if each had two hydrogen bonds. Only A–T pairs have two; G–C pairs have three, and the extra bond per G–C pair is the reason GC-rich DNA melts at a higher temperature.

    This swaps the two counts, giving G–C pairs two bonds and A–T pairs three: 7(2) + 5(3) = 29. Remember it by ring count and letters: guanine and cytosine, the "three-letter" pair in the mnemonic sense, share three bonds.

    If 7 of the 12 pairs are G–C, the remaining 5 are A–T. Then 7(3) + 5(2) = 21 + 10 = 31 hydrogen bonds. Note that the purine–pyrimidine rule guarantees exactly 12 pairs total, so no base is left unpaired.

    This treats all 12 pairs as three-bond G–C pairs, which would require a fragment with no adenine or thymine at all. The stem specifies only 7 G–C pairs, and the other 5 must be A–T.

15. Figure: the 1952 X-ray diffraction photograph of the B form of DNA, taken by Rosalind Franklin and Raymond Gosling.

    Dark spots form a large X across the center of the film. The vertical spacing of the layer lines corresponds to a repeating distance of 3.4 nm. A strong reflection far out along the vertical axis corresponds to a spacing of 0.34 nm. The overall width of the pattern indicates a molecule whose diameter is a uniform 2.0 nm along its entire length.

    Which feature of the pattern provides the strongest evidence that every base pair joins a purine to a pyrimidine?

    The X-shaped array of spots across the center of the film. The 3.4 nm repeat given by the spacing of the layer lines. The uniform 2.0 nm diameter along the whole length of the molecule. The strong 0.34 nm reflection far out on the vertical axis.

    The X is the classic signature of a helix, and it is what told Watson and Crick the molecule coils, but a helix of any composition would produce it. It constrains the overall shape, not what sits at each rung.

    The 3.4 nm layer-line spacing gives the pitch, the distance along the axis for one complete turn. Combined with the 0.34 nm rise it yields about ten base pairs per turn, which is a count, not a statement about which bases pair with which.

    Constant width is the clinching argument. Purines have two fused rings and pyrimidines one, so a purine–purine rung would bulge wider than 2.0 nm and a pyrimidine–pyrimidine rung would pinch narrower. Only one large ring system paired with one small one keeps every rung the same length, and the film shows no variation in width.

    The 0.34 nm reflection gives the rise between stacked bases, showing how tightly the rungs are packed along the axis. It measures spacing between pairs rather than the composition of any one pair.

Free response

1. Investigation: the effect of heat on purified lactase, the enzyme that hydrolyzes lactose into glucose and galactose.

   A student divided one stock solution of purified lactase into five identical tubes. Tube 1 was held at 25 °C for the entire experiment. Tubes 2 through 5 were held for 10 minutes at 45 °C, 65 °C, 85 °C, and 100 °C respectively, then cooled back to 25 °C. The student then added the same volume of the same lactose solution to every tube and measured the initial rate at which glucose appeared, in µmol per minute, with every tube at 25 °C.

   Using the investigation described, answer (a) through (d).

   1. Identify the independent variable in the investigation, and describe the purpose of tube 1.
   2. Explain how holding lactase at 85 °C changes its ability to bind lactose.
   3. Construct a graph the student could use to display the results: state which variable belongs on the x-axis and which on the y-axis, give an appropriate scale and label for each axis, and describe the shape of the curve you would expect.
   4. The student claims that the loss of activity is caused by a change in the enzyme's three-dimensional shape rather than by destruction of its amino acid sequence. Justify this claim using evidence from Anfinsen's ribonuclease experiment.
   Your response
   Scoring notes

   • (a) Accept: the independent variable is the temperature at which each tube was held for the 10-minute pre-treatment (45, 65, 85, or 100 °C, with 25 °C as the untreated level). Tube 1 is the control: it receives every step except the heating, so it establishes the activity of undamaged enzyme against which the heated tubes are compared. Do not accept: "temperature" with no indication that it is the pre-treatment temperature rather than the assay temperature; naming tube 1 "the control" without describing what it establishes; identifying the rate of glucose production as the independent variable.
   • (b) Accept: heat raises molecular motion enough to break the hydrogen bonds, ionic bonds, and hydrophobic interactions among R groups that hold the tertiary fold; the active site loses the shape complementary to lactose, the enzyme–substrate complex no longer forms, and the rate of glucose production falls. Do not accept: "the enzyme is denatured" or "the enzyme is destroyed" with no mechanism; "the enzyme dies"; any claim that heat breaks peptide bonds or alters the amino acid sequence; a general statement about enzymes that is never tied to lactase and lactose.
   • (c) Accept: pre-treatment temperature in °C on the x-axis because it is the independent variable, with an even scale covering 25–100 °C; initial rate of glucose production in µmol/min on the y-axis, starting at 0 and extending just past the tube 1 rate; both axes labeled with the quantity and its unit. Expected shape: near the tube 1 rate at 25 °C and 45 °C, falling steeply between 45 °C and 85 °C, and near zero at 85 °C and 100 °C. Do not accept: the independent variable on the y-axis; axes named without units; a rate axis that does not begin at zero; "plot rate against temperature" with no scale or labels given; a described shape that rises with temperature throughout.
   • (d) Accept: Anfinsen treated ribonuclease A with urea and β-mercaptoethanol, which disrupted the weak interactions and reduced the disulfide bridges, and activity fell to nearly zero; when both reagents were dialyzed away, activity returned to about 95% of the untreated control with no cellular machinery present. Because the same chain regained function once the weak interactions could re-form, the sequence must have survived the treatment intact and only the folded shape had been lost. Do not accept: restating the claim in other words; naming Anfinsen without describing the refolding result; "primary structure determines tertiary structure" asserted with no experimental evidence attached; evidence cited with no statement of why it supports the claim.
   Show a 4/4 response

   a The independent variable is the temperature each tube was held at for the 10-minute pre-treatment, 25, 45, 65, 85, or 100 °C. Tube 1 is the control: it gets every step except the heating, so its rate shows what lactase does undamaged.

   b At 85 °C the extra molecular motion shakes apart the hydrogen bonds, ionic bonds, and hydrophobic clustering between R groups that hold lactase in its folded shape. The active site is made by that fold, so once it opens up lactose no longer fits, the enzyme–substrate complex does not form, and almost no glucose appears.

   c Pre-treatment temperature in °C goes on the x-axis, since that is what I changed, scaled 25 to 100 °C in steps of 25; initial rate of glucose production in µmol/min goes on the y-axis, starting at 0 and running just past tube 1's rate. Both axes get the quantity and the unit. The curve stays near tube 1's rate through 45 °C, drops steeply from 45 to 85 °C, and flattens near zero at 85 and 100 °C.

   d Anfinsen treated ribonuclease A with urea and β-mercaptoethanol until it lost nearly all activity, then dialyzed both reagents away, and activity returned to about 95% of the untreated control with no help from any cell. The same chain was there throughout, so the sequence was never destroyed: only the fold was, and it came back on its own.

2. A shallow pond and a bare granite outcrop sit side by side in a desert and receive the same sunlight through the day. Answer (a) through (d).

   1. Describe the feature of a water molecule's structure that allows one water molecule to hydrogen-bond to as many as four neighbors.
   2. Explain how hydrogen bonding between water molecules accounts for water's unusually high specific heat.
   3. Water has a specific heat of 4.18 J/g·°C and granite 0.79 J/g·°C. Calculate the temperature change of a 500.0 g sample of each when it absorbs 1.045 × 10⁴ J, showing your setup.
   4. Using your calculated values, justify a claim about which of the two habitats offers an aquatic insect a more thermally stable daytime environment.
   Your response
   Scoring notes

   • (a) Accept: the molecule is bent at about 104.5°, and each O–H bond is polar covalent because oxygen pulls the shared electrons harder, leaving a partial negative charge on the oxygen and a partial positive charge on each hydrogen; because the molecule is bent those charges do not cancel, so each molecule can donate two hydrogen bonds through its two hydrogens and accept two at the oxygen's lone pairs. Do not accept: "water is polar" with no structural feature named; "water forms hydrogen bonds," which restates the stem; describing a hydrogen bond as a covalent bond between two water molecules.
   • (b) Accept: temperature measures average kinetic energy, and a large share of the heat added to water is consumed breaking hydrogen bonds between molecules rather than speeding molecules up, so a great deal of energy must be added per gram to raise the temperature one degree. Do not accept: "water has a high specific heat because of hydrogen bonding" with no statement of where the energy goes; confusing specific heat with heat of vaporization without linking either to bond breaking; a correct general statement never connected to water.
   • (c) Accept: \(\Delta T = q/(mc)\). Water: 10450/[(500.0)(4.18)] = 5.00 °C. Granite: 10450/[(500.0)(0.79)] = 26.5 °C. Both values with units and the setup shown. Do not accept: an answer with no setup; values with no units; multiplying instead of dividing by mc; a single value when two were requested.
   • (d) Accept: the pond, justified by the calculated values; for the same 1.045 × 10⁴ J absorbed, the water warms only 5.00 °C while the granite warms 26.5 °C, about 5.3 times as much, because water's hydrogen-bond network absorbs energy the rock cannot. The insect in the pond therefore experiences a far smaller temperature swing over the day. Do not accept: naming the pond with no reference to the calculated values; citing the two numbers without saying which habitat they favor or why; "water has a high specific heat" alone with no evidence from (c).
   Show a 4/4 response

   a A water molecule is bent, with its two hydrogens about 104.5° apart, and oxygen pulls the shared electrons much harder than hydrogen does. That leaves a partial negative charge on the oxygen and a partial positive on each hydrogen, and because the shape is bent those charges do not cancel. One molecule can therefore donate two hydrogen bonds through its hydrogens and accept two at the oxygen's lone pairs.

   b Temperature measures how fast molecules move, but in water much of the heat I add goes into pulling hydrogen bonds apart instead of speeding molecules up. Energy spent that way does not raise the temperature, so warming a gram of water by one degree takes an unusually large amount of heat.

   c Rearranging q = mcΔT gives ΔT = q/(mc). Water: 10450/[(500.0)(4.18)] = 5.00 °C. Granite: 10450/[(500.0)(0.79)] = 26.5 °C.

   d The pond is the more stable habitat. Both absorb the same 1.045 × 10⁴ J, but the water warms only 5.00 °C while the rock warms 26.5 °C, about 5.3 times as much, because water's hydrogen bonds soak up energy granite cannot store. The insect in the pond faces a temperature swing a fifth the size of the one on the rock.

3. Model: a space-filling model of a 10 base-pair segment of B-form DNA, shown beside the X-ray diffraction pattern of the same form.

   In the model, two sugar–phosphate backbones spiral around the outside of the molecule and the paired bases lie flat in the interior, stacked 0.34 nm apart. The model's diameter is a uniform 2.0 nm along its entire length. In the diffraction pattern, dark spots form an X across the film and the layer-line spacing corresponds to a 3.4 nm repeat. One strand of the modeled segment reads 5'-GGATCCGTAC-3'.

   Using the model, answer (a) through (d).

   1. Identify the bond that joins one nucleotide to the next within a single strand, and describe how the two strands of the model are oriented relative to each other.
   2. Explain how the uniform 2.0 nm diameter shown in the model supports the conclusion that each base pair joins a purine to a pyrimidine.
   3. Represent the strand complementary to 5'-GGATCCGTAC-3' with both of its ends labeled, and calculate the number of hydrogen bonds holding the 10 base-pair segment together, showing your setup.
   4. A student claims that a 10 base-pair segment whose strand reads 5'-AATTAATTAA-3' would separate into single strands at a lower temperature than the modeled segment. Justify this claim using your calculation from (c).
   Your response
   Scoring notes

   • (a) Accept: a phosphodiester bond, joining the 3' carbon of one sugar to the phosphate on the 5' carbon of the next. The two strands are antiparallel: one runs 5' to 3' while the other runs 3' to 5', so each strand's free 5' phosphate lies opposite the other's free 3' hydroxyl. Do not accept: "hydrogen bond" or "peptide bond"; "the strands are opposite" with no statement of 5' and 3' orientation; describing the backbones as running in the same direction.
   • (b) Accept: a purine has two fused rings and a pyrimidine one, so a purine paired with a purine would make a rung wider than 2.0 nm and two pyrimidines would make one narrower; only a two-ring base opposite a one-ring base keeps every rung the same length, and the model shows no variation in diameter anywhere along the molecule. Do not accept: "A pairs with T and G with C" restated with no reference to ring number or to width; any explanation that appeals only to hydrogen-bond counts; a correct general statement about purines and pyrimidines that is never linked to the uniform diameter.
   • (c) Accept: 3'-CCTAGGCATG-5' (equivalently written 5'-GTACGGATCC-3'), with both ends labeled. Hydrogen bonds: the segment has 6 G–C pairs and 4 A–T pairs, so 6(3) + 4(2) = 18 + 8 = 26. Do not accept: an unlabeled sequence; a complement written parallel to the given strand; 20 or 30 (treating all pairs alike); a bare number with no setup; a sequence containing uracil.
   • (d) Accept: the all-A–T segment is held by 10 A–T pairs, or 10(2) = 20 hydrogen bonds, against the modeled segment's 26 (six fewer, about 23% fewer), because each G–C pair contributes three hydrogen bonds and each A–T pair only two. Fewer hydrogen bonds means less thermal energy is needed to pull the strands apart, so the A–T segment melts first. Do not accept: "A–T pairs are weaker" with no bond counts; repeating the claim; citing the 20 and 26 without saying what they imply about the energy required.
   Show a 4/4 response

   a Nucleotides within a strand are joined by phosphodiester bonds, linking the 3' carbon of one sugar to the phosphate on the 5' carbon of the next. The two strands are antiparallel: one runs 5' to 3' while the other runs 3' to 5', so a free 5' phosphate sits opposite a free 3' hydroxyl.

   b Purines have two fused rings and pyrimidines one. If two purines ever paired, that rung would be wider than the rest and the molecule would bulge; two pyrimidines would pinch it in. The model is a uniform 2.0 nm wide everywhere, so every rung must be the same length, which happens only when a two-ring base pairs with a one-ring base.

   c The complement is 3'-CCTAGGCATG-5'. The segment has 6 G–C pairs and 4 A–T pairs, so the hydrogen bonds come to 6(3) + 4(2) = 18 + 8 = 26.

   d The claim holds. A segment of 5'-AATTAATTAA-3' is 10 A–T pairs, or 10(2) = 20 hydrogen bonds, against the modeled segment's 26: six fewer, about 23% fewer, since G–C pairs bring three bonds each and A–T pairs two. Separating the strands means breaking all of them, so the segment with fewer needs less thermal energy and melts first.

4. Data: fatty acid composition of the membrane phospholipids of rainbow trout held for six weeks at two water temperatures, as a percent of total fatty acids.

   Fatty acids (% of total)	Trout at 5 °C	Trout at 20 °C
   Saturated	34	55
   Monounsaturated	30	27
   Polyunsaturated	36	18

   Using the data, answer (a) through (d).

   1. Identify the percent of fatty acids that are unsaturated in each group of trout, and describe the trend in the data.
   2. Explain how the number of cis double bonds in a fatty acid tail is related to the fluidity of the membrane that tail sits in.
   3. Calculate the ratio of unsaturated to saturated fatty acids for each acclimation temperature, showing your setup.
   4. A student claims that trout acclimated to 12 °C would have a percent of unsaturated fatty acids between the two values you found in (a). Justify this claim using the data and the relationship you described in (b).
   Your response
   Scoring notes

   • (a) Accept: unsaturated = monounsaturated + polyunsaturated, so 30 + 36 = 66% at 5 °C and 27 + 18 = 45% at 20 °C; the trend is that the percent of unsaturated fatty acids falls, by 21 percentage points, as acclimation temperature rises, with the polyunsaturated fraction accounting for most of the change (36% to 18%). Do not accept: reporting only the saturated percentages; giving the two values with no trend stated; "the fatty acids change" with no direction; adding the columns incorrectly.
   • (b) Accept: a cis double bond puts a permanent kink in the hydrocarbon tail, so neighboring tails cannot stack closely; fewer attractions form between them, the phospholipids move past one another more easily, and the bilayer stays fluid at lower temperatures. More double bonds means more kinks and a more fluid membrane at any given temperature. Do not accept: "unsaturated fats are liquid" with no mechanism; describing the double bond as making the tail itself more flexible without reference to packing between tails; any statement about fluidity that is never tied to the tails' arrangement.
   • (c) Accept: at 5 °C, 66/34 = 1.94; at 20 °C, 45/55 = 0.82. Setups shown; two significant figures or better acceptable. Do not accept: inverted ratios presented as unsaturated to saturated; one ratio when two were asked for; a ratio with no setup; use of only the polyunsaturated row.
   • (d) Accept: the claim is supported because 12 °C lies between the two tested temperatures and the data show a consistent inverse relationship, 66% unsaturated at 5 °C and 45% at 20 °C, so a fish held between those temperatures needs a membrane that is fluid enough for 12 °C but not so fluid as it would be at 5 °C, giving an intermediate value. Do not accept: "yes, because it is in the middle" with no reference to the data or the mechanism; asserting a specific number with no reasoning; restating the relationship from (b) with no connection to the claim about 12 °C.
   Show a 4/4 response

   a Adding the monounsaturated and polyunsaturated rows, the 5 °C trout are 30 + 36 = 66% unsaturated and the 20 °C trout are 27 + 18 = 45%. Unsaturated fatty acids drop by 21 percentage points as the water warms, and most of that comes from the polyunsaturated fraction falling from 36% to 18%.

   b Every cis double bond puts a fixed kink in the tail, and kinked tails cannot pack tightly against their neighbors. With fewer attractions holding them the phospholipids slide past each other more easily, so a membrane with more double bonds stays fluid at a lower temperature than one built from straight saturated tails.

   c At 5 °C the ratio of unsaturated to saturated is 66/34 = 1.94. At 20 °C it is 45/55 = 0.82.

   d I agree. The data show an inverse relationship between temperature and unsaturation, 66% at 5 °C against 45% at 20 °C, because cold would stiffen a saturated membrane into a gel, so a cold fish needs kinks to stay fluid. At 12 °C that pressure is real but milder than at 5 °C, which puts the trout's unsaturated percentage between the two measured values.

5. A cell assembles a polypeptide from 40 glycine monomers (M = 75.07 g/mol). Water has M = 18.02 g/mol. Answer (a) through (d).

   1. Identify the reaction that joins two glycine monomers, and describe what happens to the atoms that are removed from them.
   2. Explain why the molar mass of the finished polypeptide is lower than the combined molar mass of 40 free glycine molecules.
   3. Calculate the molar mass of the finished polypeptide, showing your setup.
   4. A protease completely hydrolyzes the polypeptide back to free glycine. Justify a claim about how the total mass of dissolved solute changes, using your calculation from (c).
   Your response
   Scoring notes

   • (a) Accept: dehydration synthesis (condensation), forming a peptide bond between the carboxyl group of one glycine and the amino group of the next; the, OH taken from one monomer and the, H taken from the other combine and leave as a molecule of water. Do not accept: "hydrolysis"; naming the reaction with no account of the removed atoms; saying atoms are "lost" or "destroyed" rather than released as water.
   • (b) Accept: each of the 39 bonds formed removes one whole water molecule from the chain, so the polymer's mass is the summed monomer mass minus the mass of the water released; the missing mass is not destroyed but has left the molecule as H₂O. Do not accept: "some mass is lost" with no identification of water as what left; an answer that describes the reaction but never addresses mass; claiming the bond itself has negative mass or that atoms disappear.
   • (c) Accept: 40 residues require 39 bonds, so 40(75.07) − 39(18.02) = 3002.80 − 702.78 = 2300.02 g/mol. Setup shown and units given. Do not accept: 40 waters subtracted (2282.00); no water subtracted (3002.80); water added (3705.58); a value with no setup or no units.
   • (d) Accept: the total solute mass rises, from 2300.02 g/mol of polypeptide to 40(75.07) = 3002.80 g/mol of free glycine, an increase of 702.78 g/mol: exactly the 39(18.02) g/mol of water subtracted in (c). The justification is that hydrolysis inserts one water molecule at each of the 39 bonds, transferring mass from the solvent into the solutes rather than creating it. Do not accept: "the mass increases" with no value or no mechanism; claiming mass is created; claiming the total is unchanged; citing the numbers without explaining where the added mass came from.
   Show a 4/4 response

   a The two glycines are joined by dehydration synthesis, which forms a peptide bond between the carboxyl group of one and the amino group of the other. An, OH comes off one monomer and an, H off the other, and those atoms leave together as one molecule of water.

   b Building a chain of 40 residues takes 39 bonds, and every bond costs one whole water molecule that used to be part of the monomers. The mass is not gone, it walked away as H₂O, but it is no longer counted in the polymer, so the finished chain weighs less than the 40 separate glycines did.

   c Forty residues means 39 peptide bonds, so the molar mass is 40(75.07) − 39(18.02) = 3002.80 − 702.78 = 2300.02 g/mol.

   d The total dissolved solute mass goes up, from 2300.02 g/mol of polypeptide to 40(75.07) = 3002.80 g/mol of free glycine, a gain of 702.78 g/mol. That figure is exactly the 39(18.02) g/mol I subtracted in (c), which is my evidence for what is happening: hydrolysis puts one water molecule back into each of the 39 bonds, so the mass is transferred out of the solvent and into the solutes instead of being created.

Lesson 2.1 · Unit 2 · CED topics 2.1–2.2

Subcellular components and their functions

A eukaryotic cell is not a bag of enzymes. It is a factory with rooms, and each room is a membrane-bounded compartment that holds one set of reactions at one pH with one crew of enzymes. Learn the rooms and you can predict what a cell does for a living just by looking at what it has a lot of.

That is the skill the exam tests. It rarely asks you to list organelles; it hands you a cell and asks you to explain why its structure fits its function.

Lesson 2.2 · Unit 2 · CED topic 2.3

Cell size and the surface area-to-volume ratio

Why is there no elephant-sized amoeba? Because everything a cell needs has to cross its surface, and everything it needs the material for is spread through its volume. Volume grows as the cube of length while surface grows only as the square, so a cell that doubles in width doubles its demand relative to its supply.

Lesson 2.3 · Unit 2 · CED topics 2.4–2.5

Plasma membrane structure and selective permeability

The membrane is the reason a cell can have an inside at all. It is about 7–8 nm thick, and its whole job is to be picky: some molecules stroll through, others cannot cross without help, and a few never cross at all. Which is which follows from two properties of the molecule: its size and whether it is charged or polar.

Lesson 2.4 · Unit 2 · CED topics 2.6–2.7

Passive transport: diffusion, osmosis, and facilitated diffusion

Nothing in this lesson costs the cell a molecule of ATP. A concentration gradient is stored free energy, and when particles spread out they are spending it: moving to the more probable, higher-entropy arrangement. The cell's only decision is whether to open a door.

Lesson 2.5 · Unit 2 · CED topic 2.8

Water potential, tonicity, and osmoregulation

"Water moves toward the saltier side" is fine until a plant cell pushes back with its wall. Water potential, written Ψ (psi), is the bookkeeping that handles both effects at once: it combines how much solute a compartment holds with how hard it is being squeezed, and water always moves from higher Ψ to lower Ψ.

Lesson 2.6 · Unit 2 · CED topics 2.9–2.11

Active transport, bulk transport, and compartmentalization

Every gradient in the last two lessons was something the cell spent down. This lesson is how it builds them back. Moving a solute against its gradient lowers entropy locally, so it cannot happen for free: the cell pays, directly with ATP or indirectly with a gradient it made earlier.

Unit 2 quiz · 15 multiple-choice · 5 free-response

Unit 2 quiz: Cell Structure and Function

Fifteen multiple-choice items and five short free-response questions on organelles and the endomembrane system, surface area-to-volume constraints, membrane structure, passive and active transport, water potential, and the endosymbiotic origin of compartments: click an option to see why each choice is right or wrong, and write each FRQ out before opening the model response.

Multiple choice

1. Data: the volume occupied by four structures in three mammalian cell types, as a percent of total cell volume.

   Structure (% of cell volume)	Cell D	Cell E	Cell F
   Rough ER	25	3	7
   Smooth ER	2	4	19
   Mitochondria	7	32	15
   Secretory vesicles	18	0	1

   Which claim is best supported by the data?

   Cell D is specialized for synthesizing and exporting protein, because rough ER and secretory vesicles together fill 43% of its volume. Cell E is specialized for secretion, because it holds no secretory vesicles waiting to be filled. Cell F is specialized for protein export, because it contains more rough ER than cell E does. All three cells must export protein at the same rate, because all three contain rough ER.

   Ribosomes on the rough ER deposit proteins destined for export into the ER lumen, and secretory vesicles carry the finished product to the plasma membrane. Cell D devotes 25% plus 18% of its volume to those two steps, more than five times cell F's combined 8%, which is what a plasma cell or a pancreatic acinar cell looks like inside.

   A secretory cell is full of loaded vesicles, not empty of them; cell E has none at all. Its 32% mitochondria, more than four times cell D's, points instead to a cell that spends ATP continuously, such as cardiac muscle.

   Cell F's 7% rough ER does exceed cell E's 3%, but that is not evidence of specialization when cell D's is 25%. Cell F's real signature is 19% smooth ER against only 1% secretory vesicles, which fits lipid synthesis and drug detoxification in a liver cell rather than protein export.

   Every cell needs some rough ER for membrane and lysosomal proteins, so its mere presence says nothing about rate. The data are given as percentages precisely so that the proportion of the cell devoted to a task can be compared, and those proportions differ by a factor of eight.

2. Data: a pulse–chase experiment. Cells were given radioactive leucine for 3 minutes, washed, and returned to unlabeled leucine. Three membrane fractions were separated at intervals and the label in each was measured as a percent of the total label recovered.

   Minutes after the pulse	2	8	25	50
   Fraction P (% of label)	88	40	14	5
   Fraction Q (% of label)	10	48	41	15
   Fraction R (% of label)	2	12	45	80

   Which statement is best supported by the data?

   Fraction R synthesizes the most protein, because it holds the most label at 50 minutes. The three fractions each make protein independently, because all three contain some label at every time point. The three fractions act in series, P before Q and Q before R, because each fraction peaks only after the one above it has peaked and every column sums to 100%. Label is destroyed in fraction P and manufactured again in fraction R, because P falls while R rises.

   Holding the most label late says where protein ends up, not where it was made. At 2 minutes, when the labeled amino acid had just been incorporated, fraction R held only 2%: synthesis happened in P, which held 88%.

   Independent synthesis would give three curves that rise together and stay up. Instead each fraction rises and then falls as the next one rises, and that hand-off pattern is what a shared assembly line looks like.

   P peaks at 2 minutes, Q at 8, and R at 50, and no fraction peaks before the one upstream of it does: the signature of compartments acting in order, matching ribosome and rough ER, then Golgi, then secretory vesicle. The columns summing to 100% shows the label is being relocated, not created or lost.

   The columns sum to 100% at every time point, so no label leaves the system and none is added. Radioactive leucine that disappears from P appears in Q and then R, which is transfer, not destruction and resynthesis.

3. A drug prevents vesicles from budding off the trans face of the Golgi in a pancreatic acinar cell. Which of the following is the most likely result 60 minutes after a pulse of radioactive leucine?

   Most of the label remains in the rough ER, because that is the compartment the drug acts on. Most of the label accumulates in the Golgi, because the compartments act in series and cargo piles up immediately before the block while the steps upstream keep running. No label ever leaves the ribosome, because the endomembrane system shuts down as a single unit. Protein is secreted at the normal rate, because vesicles can travel from the rough ER straight to the plasma membrane.

   The drug acts at the Golgi's exit, not at the ER. Transfer from the rough ER to the Golgi is upstream of the block and keeps working, so the ER empties on roughly its normal schedule.

   In a pulse–chase, blocking the last step in a series leaves every earlier transfer intact, so label moves out of the ER, into the Golgi, and then stops. The Golgi fraction climbs toward the value the secretory granules would otherwise have reached, which is the reverse of the untreated pattern.

   Translation on ribosomes and import into the ER lumen are unaffected by a drug acting at the Golgi's trans face; the compartments are linked in sequence, not wired to fail together. Label does leave the ribosome and does reach the Golgi.

   The secretory route in this cell runs rough ER to Golgi to secretory vesicle, and Palade's pulse–chase shows no fraction peaking before the one upstream of it. With the Golgi's exit blocked there is no path to the surface at anything like the normal rate.

4. Model: three model cells, each with a volume of 64 µm³ but a different shape. All dimensions are in micrometers.

   Model cell	Cube 4 × 4 × 4	Sheet 16 × 4 × 1	Rod 32 × 2 × 1
   Surface area (µm²)	96	168	196
   Volume (µm³)	64	64	64
   Surface area-to-volume ratio (µm−1)	1.50	2.63	3.06
   Greatest distance from a surface to the interior (µm)	2.0	0.5	0.5

   Which claim is best supported by the model?

   Surface area-to-volume ratio is set by volume alone, so the three cells must exchange materials with their surroundings at the same rate. A cell can raise its surface area-to-volume ratio without changing its volume by flattening or elongating, and doing so also shortens the longest diffusion path into its interior. The rod has the highest ratio because it contains the largest volume. The cube is best adapted for exchange, because a compact shape puts every part of the cell closest to the center.

   Volume is held constant at 64 µm³ across all three cells, yet the ratio doubles from 1.50 to 3.06 µm⁻¹. That is exactly the comparison the model was built to make: with volume fixed, any difference left has to come from shape.

   The sheet and the rod hold the same 64 µm³ as the cube but wrap it in 168 and 196 µm² of membrane instead of 96, raising the ratio to 2.63 and 3.06 µm⁻¹. Because both are only 1 µm thick, nothing inside them sits more than 0.5 µm from a surface, against 2.0 µm in the cube, and since diffusion time scales with the square of distance, that trip is about sixteen times faster.

   All three cells have the same volume, 64 µm³, so volume cannot explain the difference. The rod wins on ratio because stretching a fixed volume into a thin shape exposes more of it as surface.

   Being compact is the problem, not the solution. The cube has the lowest ratio, 1.50 µm⁻¹, and the longest distance from any surface to its interior, 2.0 µm, so it is the worst of the three at supplying its own inside.

5. A spherical cell doubles its radius while keeping its shape. Which of the following best explains why supplying its interior becomes harder?

   Cytoplasm becomes more viscous in larger cells, so diffusion slows down inside them. The plasma membrane stretches thinner as the cell grows, so fewer transport proteins fit into it. Surface area grows with the square of the radius while volume grows with the cube, so the surface area-to-volume ratio is halved, and because diffusion time scales with the square of distance, the trip to the center takes four times as long. A larger cell requires less material per unit volume, so its supply falls behind its demand.

   Nothing about being larger changes the cytoplasm's viscosity, and the diffusion coefficient of a solute is the same in a big cell and a small one. What changes is the distance that has to be covered and the area available to cover it.

   Membrane is manufactured as the cell grows rather than stretched like a balloon, and its thickness stays about 7–8 nm. Protein density per square micrometer is not the limiting factor here; total square micrometers per cubic micrometer is.

   For a sphere, SA = 4πr² and V = (4/3)πr³, so SA:V = 3/r; doubling r halves the ratio. At the same time the center moves twice as far from the surface, and because diffusion time goes as the square of distance, that doubles into a fourfold delay. Supply per unit of demand falls and delivery slows at once.

   A larger cell needs more material in total, and roughly the same amount per unit volume, since the same metabolic reactions run everywhere in the cytoplasm. The mismatch arises because the demand rises with volume while the supply rises only with surface.

6. Data: permeability coefficients measured for a pure phospholipid bilayer containing no proteins, at 25 °C.

   Solute	O2	H2O	Glycerol	Glucose	K+
   Molar mass (g/mol)	32	18	92	180	39
   Permeability (cm/s)	10−1	10−3	10−6	10−7	10−12

   Which claim is best supported by the data?

   Permeability falls steadily as molar mass rises. Water crosses this bilayer only through aquaporins, since its permeability is far below that of O₂. Charge is a stronger barrier than size, because K⁺ is far smaller than glucose yet is 10⁵ times less permeant. Glycerol must be carried across by a transport protein, since it is more permeant than glucose.

   The table breaks that pattern twice. Water is the lightest solute at 18 g/mol yet is 100 times less permeant than O₂ at 32 g/mol, and K⁺ at 39 g/mol is the least permeant of all: less than glucose, which is more than four times its mass.

   The bilayer in this experiment contains no proteins at all, and water still crosses it at 10⁻³ cm/s. Aquaporins raise that rate enormously in cells that need it, but the data show they are not required for water to get through.

   K⁺ has a molar mass of 39 g/mol against glucose's 180, so on size alone it should cross more easily; instead it is 10⁻¹² against 10⁻⁷ cm/s, a factor of 10⁵ the other way. An ion must shed the hydration shell of water molecules bound to its charge before it can enter the hydrocarbon core, and that costs more energy than it has.

   Being more permeant than glucose is evidence that glycerol crosses the bare lipid well, not that it needs help. Glycerol is small and uncharged, so like water and urea it trickles through the hydrophobic core unaided: this bilayer has no proteins in it at all.

7. Which statement best explains how cholesterol helps a mammalian cell survive both a cold night and a hot afternoon?

   Its rigid ring system wedges between neighboring phospholipid tails, blocking them from packing into an ordered gel in the cold and restraining their motion in the heat. It dissolves into the hydrophobic core when the cell is cold and moves out to the membrane surface when the cell is warm. It forms hydrogen bonds with the phosphate heads that hold the two layers of the bilayer together at every temperature. It converts saturated tails into unsaturated ones as the temperature falls and reverses the change as it rises.

   Cholesterol sits among the tails permanently and buffers fluidity in both directions. In the cold its bulky fused rings get in the way of tight packing, so the bilayer does not solidify; in the heat those same rings limit how freely the tails can move, so the bilayer does not become too fluid and leaky.

   Cholesterol does not migrate in and out with temperature; it is a permanent component of the animal cell membrane, anchored by its hydroxyl group near the phosphate heads with its rings among the tails. A molecule that left the core in the heat could not restrain the tails, which is half of what cholesterol does.

   Cholesterol's single hydroxyl group does interact with the head region, but that is an anchor, not the mechanism. The two leaflets are held together by the hydrophobic effect on the tails, and cholesterol's effect on fluidity comes from the rings physically wedged between those tails.

   Nothing converts one fatty acid into another inside a finished membrane; changing tail saturation requires enzymes that synthesize new phospholipids, which is what bacteria do over hours or days. Cholesterol works instantly and physically, without any chemical change to the phospholipids.

8. An integral membrane protein is extracted from a bilayer and its sequence examined. Which of the following would most likely be found in the stretch of the chain that spanned the hydrophobic core?

   A run of residues with acidic and basic R groups, stabilized by the charged phosphate heads. A chain of sugar residues, which anchors the protein among the fatty acid tails. Alternating cysteines forming disulfide bridges to the surrounding phospholipids. A run of residues with nonpolar R groups, which can interact with the fatty acid tails without the energetic cost of burying a charge.

   Phosphate heads line the two surfaces of the bilayer, not its core, so they cannot stabilize anything in the middle. Charged R groups do cluster where the protein meets the head region and the surrounding water, which is the opposite end of the span.

   Sugar chains mark glycoproteins, and they are always on the outer face where they serve cell–cell recognition. Sugars are highly polar and hydrogen-bond with water, so the hydrophobic core is the last place one would sit.

   Disulfide bridges form between two cysteine R groups within or between polypeptides, never between a protein and a phospholipid, which has no sulfhydryl group to bond with. This choice swaps a real protein-stabilizing interaction into a place it cannot occur.

   The membrane's interior is a hydrocarbon environment, so the residues facing it must be hydrocarbon-like too: leucine, isoleucine, valine, phenylalanine. Burying a charged or polar R group there would mean stripping away its hydration shell at a large energetic cost, which is why membrane-spanning helices read as long nonpolar stretches.

9. Data: rate of glucose uptake at 37 °C by human red blood cells and by artificial lipid vesicles that contain no proteins.

   [Glucose] outside (mM)	1	2	5	10	25	50
   Red blood cells (µmol/min)	20	36	66	84	96	100
   Protein-free vesicles (µmol/min)	0.8	1.6	4.0	8.0	20.0	40.0

   Which statement best explains why the red blood cell rate levels off while the vesicle rate does not?

   The red cells run out of ATP at high glucose concentrations, so transport stops. Glucose is consumed inside the red cell, so its concentration gradient disappears. The vesicle membrane is more permeable to glucose than the red cell membrane is. Every GLUT carrier protein is occupied at high concentration, so the number of carriers rather than the size of the gradient sets the maximum rate; the protein-free vesicles have no carriers and depend on the gradient alone.

   Facilitated diffusion spends no ATP: glucose is moving down its own gradient, and the carrier only provides a path. A plateau produced by an energy shortage would also appear in a red cell starved of glucose, which is the opposite of the condition here.

   Consumption inside the cell would keep the internal concentration low and so keep the gradient large, which would raise uptake rather than cap it. The ceiling appears at high external concentration, exactly where the gradient is biggest.

   The data say the reverse. At every concentration the red cells take up more, and at 1 mM they take up 20 µmol/min against the vesicles' 0.8, twenty-five times as much. What the vesicles have is not more permeability but a rate that never stops rising.

   Each carrier binds a glucose molecule, changes shape, releases it, and resets, and that cycle takes time. Once every carrier is working continuously, near 100 µmol/min here, adding substrate cannot add rate. The vesicle row has no such ceiling: it is a straight line at 0.8 µmol/min per mM through the whole range, because simple diffusion has nothing to saturate.

10. A kidney collecting-duct cell responds to a hormone by inserting additional aquaporins into its plasma membrane. Which of the following best describes the effect?

    Water is now pumped into the cell against its water potential gradient at a cost of ATP. The rate of osmosis rises sharply, but the direction of net water movement is still set by the water potential difference across the membrane. Water begins to cross the membrane for the first time, since a bilayer with no aquaporins is impermeable to water. The cell's internal solute concentration rises, which lowers its water potential.

    An aquaporin is a channel, not a pump: it forms an open hydrophilic pore that a single file of water molecules passes through, and it hydrolyzes no ATP. Nothing in the cell can move water against its own potential gradient directly.

    Adding channels changes how fast water can cross, not which way it goes. Water still moves from higher water potential to lower, so the hormone lets the kidney reabsorb water quickly when the interstitial fluid is concentrated, and stops the flow when the potentials equalize.

    Water is small and uncharged, so it trickles through a bare bilayer on its own at about 10⁻³ cm/s. That is slow for a kidney cell, which is why aquaporins matter, but it is not zero.

    Aquaporins pass water and exclude solutes, including H⁺, so inserting them moves no solute at all. The cell's solute potential is set by what is dissolved in it, and that is unchanged.

11. Two compartments separated by a membrane permeable to glucose have reached equal glucose concentrations. Which statement best describes what happens next?

    Glucose molecules stop crossing the membrane, because there is no longer a gradient to drive them. Glucose molecules keep crossing in both directions at equal rates, so there is no further net change: the system is at dynamic equilibrium. The system must now spend ATP to hold the two concentrations equal. Glucose begins moving back toward the compartment it came from, re-establishing the original gradient.

    Molecular motion does not switch off; every glucose molecule is still moving and still encountering the membrane. What disappears at equilibrium is the imbalance in crossings, not the crossings themselves.

    This is what "dynamic" means in dynamic equilibrium. With equal concentrations on both sides, the probability of a molecule crossing left to right equals the probability of crossing right to left, so the two flows cancel and the measured concentrations hold steady while traffic continues in both directions.

    Maintaining equal concentrations is the spontaneous, high-entropy arrangement and costs nothing. ATP is needed only to move a solute against its gradient, which lowers entropy locally: the situation in active transport, not here.

    Spontaneous movement back into a gradient would lower the system's entropy with no energy input, which does not happen. Re-establishing a gradient is precisely the job a cell has to pay for with ATP.

12. Data: six dialysis bags, all filled with the same sucrose solution of unknown molarity, were weighed, submerged for 30 minutes at 22 °C in the beakers shown, blotted, and reweighed.

    Beaker [sucrose] (M)	0.0	0.2	0.4	0.6	0.8	1.0
    % change in mass	+13.8	+8.4	+2.9	−2.9	−8.2	−13.6

    Which value best estimates the sucrose molarity of the solution inside the bags?

    0.30 M 0.40 M 0.50 M 0.60 M

    Bags in 0.2 M gained 8.4% and bags in 0.4 M gained 2.9%, so a bag in 0.30 M would gain mass too. Water entering means the bag contents have the lower water potential, so the bag must be more concentrated than 0.30 M.

    At 0.4 M the bags still gained 2.9% of their mass, and a gain means net water movement inward. The bags cannot be 0.40 M, because a bag matched to its bath shows no change at all.

    The percent change crosses zero exactly halfway between 0.4 M (+2.9%) and 0.6 M (−2.9%), at 0.50 M. Zero net water movement means the bath and the bag have the same water potential, and since both contain only sucrose in water, they must have the same molarity. Divide by initial mass rather than final, and read the crossing rather than any single point.

    In 0.6 M the bags lost 2.9%, so water left them and the bath had the lower water potential of the two. That places the bag's concentration below 0.60 M, not at it.

13. Calculate the solute potential of a 0.25 M CaCl₂ solution at 27 °C. Use \(\Psi_s = -iCRT\) with \(R = 0.0831\) L·bar/(mol·K).

    −1.68 bars −6.23 bars −12.46 bars −18.70 bars

    This leaves the temperature in degrees Celsius: −(3)(0.25)(0.0831)(27) = −1.68. The gas constant is defined per kelvin, so the temperature must be converted first, 27 + 273 = 300 K.

    This uses i = 1, as if CaCl₂ stayed intact in solution: −(1)(0.25)(0.0831)(300) = −6.23. Solute potential depends on the number of dissolved particles, and calcium chloride dissociates into three of them.

    This uses i = 2, the value for NaCl. Each CaCl₂ formula unit releases one Ca²⁺ and two Cl⁻, so the ionization constant is 3, not 2: count the ions, not the elements.

    With i = 3, C = 0.25 M, R = 0.0831, and T = 27 + 273 = 300 K, \(\Psi_s = -(3)(0.25)(0.0831)(300) = -18.70\) bars. The two places students lose this point are forgetting i and leaving T in Celsius; the negative sign is required because dissolved solute can only pull water potential down.

14. Data: glucose uptake across the apical membrane of cultured intestinal epithelial cells, measured over 10 minutes under four conditions. Ouabain inhibits the sodium–potassium pump; phlorizin blocks the SGLT1 sodium–glucose cotransporter.

    Condition	Complete medium	Na+ removed	Ouabain added	Phlorizin added
    Glucose uptake (nmol/min per mg protein)	62	6	9	5

    Which claim is best supported by the data?

    Glucose crosses the apical membrane by secondary active transport driven by the Na⁺ gradient, which the sodium–potassium pump maintains. Glucose crosses by simple diffusion through the bilayer, since uptake is high in the complete medium. Ouabain must bind SGLT1 directly, since it lowers uptake almost as much as phlorizin does. Glucose uptake requires ATP to be hydrolyzed at the SGLT1 protein itself.

    Three results converge on this. Removing Na⁺ drops uptake from 62 to 6, so glucose entry depends on Na⁺ moving in beside it; phlorizin's drop to 5 identifies SGLT1 as the protein doing it; and ouabain's drop to 9 shows the whole thing collapses once the pump can no longer rebuild the Na⁺ gradient that supplies the free energy.

    Simple diffusion through the bilayer would be indifferent to Na⁺ in the medium and to a drug that blocks one protein, yet both treatments cut uptake by about 90%. Glucose is a large polar molecule and crosses a bare bilayer at a negligible rate.

    The stimulus states that ouabain inhibits the sodium–potassium pump, a different protein in a different membrane. Its effect is indirect and delayed: with the pump stopped, Na⁺ leaks in until the gradient runs down, and with no gradient there is no energy source for the cotransporter.

    SGLT1 hydrolyzes no ATP; that is what makes the transport secondary. The ATP is spent earlier and elsewhere, by the sodium–potassium pump on the basolateral membrane, and stored in the Na⁺ gradient the cotransporter then spends.

15. A neuron's sodium–potassium pumps hydrolyze 4.5 × 10⁸ ATP molecules per second. How many Na⁺ ions leave the cell per second, and what net positive charge leaves per second?

    9.0 × 10⁸ Na⁺, and a net of 4.5 × 10⁸ positive charges 1.35 × 10⁹ Na⁺, and a net of 4.5 × 10⁸ positive charges 1.35 × 10⁹ Na⁺, and a net of 2.25 × 10⁹ positive charges 1.35 × 10⁹ Na⁺, and no net charge, because the pump exchanges one ion for another

    This uses the potassium stoichiometry for sodium. Each ATP drives one cycle that sends 3 Na⁺ out and brings 2 K⁺ in, so 9.0 × 10⁸ is the number of K⁺ imported, not the number of Na⁺ exported.

    One ATP per cycle gives 3(4.5 × 10⁸) = 1.35 × 10⁹ Na⁺ out and 2(4.5 × 10⁸) = 9.0 × 10⁸ K⁺ in. The charge balance per cycle is three positive out against two positive in, a net of one, so 4.5 × 10⁸ positive charges leave each second, which is why the pump is called electrogenic and adds a few millivolts to the resting potential.

    2.25 × 10⁹ is 5(4.5 × 10⁸), the total number of ions moved per second in both directions. Net charge is a difference, not a sum: the 9.0 × 10⁸ K⁺ entering cancel all but 4.5 × 10⁸ of the sodium's charge.

    The exchange is not one for one. Three sodium ions leave for every two potassium ions that enter, so each cycle leaves the interior one positive charge more negative: an imbalance that accumulates at 4.5 × 10⁸ charges per second.

Free response

1. Investigation: cubes of agar containing the indicator phenolphthalein were soaked in 0.1 M NaOH, which turns the agar bright pink wherever it penetrates.

   Four cubes with sides of 1.0, 2.0, 3.0, and 4.0 cm were cut from one slab of agar, soaked together in the same beaker of NaOH for 10 minutes at 22 °C, blotted, and cut open. In every cube the pink zone extended 0.5 cm inward from each face.

   Cube side (cm)	1.0	2.0	3.0	4.0
   Surface area (cm²)	6.0	24.0	54.0	96.0
   Volume (cm³)	1.0	8.0	27.0	64.0
   Surface area-to-volume ratio (cm−1)	6.0	3.0	2.0	1.5
   Percent of volume turned pink	100	87.5	70.4	57.8

   Using the investigation and the data, answer (a) through (d).

   1. Identify the independent variable in the investigation, and describe one variable that had to be held constant and why holding it constant mattered.
   2. Explain the relationship between a cube's surface area-to-volume ratio and the percent of its volume that turned pink.
   3. Construct a graph that displays the relationship between cube side length and percent of volume turned pink: state which variable belongs on each axis, give an appropriate scale and label for each, and describe the shape of the curve.
   4. A student claims that no living cell could function at the size of the 4.0 cm cube. Justify this claim using the data.
   Your response
   Scoring notes

   • (a) Accept: the independent variable is the cube's side length (its size). One constant, with a reason: soak time, because the NaOH front advances with time and a longer soak would deepen the pink zone for a reason unrelated to size; or NaOH concentration, temperature, or agar composition, each because it sets how fast the front advances. Do not accept: naming a constant with no reason attached; "all other variables were kept the same" with none named; identifying percent pink or penetration depth as the independent variable.
   • (b) Accept: as the ratio falls from 6.0 to 1.5 cm⁻¹, the percent pink falls from 100 to 57.8; because the NaOH travels the same 0.5 cm into every cube, the reacted shell scales with surface area while the interior needing to be reached scales with volume, so a cube with less surface per unit volume leaves a larger fraction untouched. Do not accept: "larger cubes take longer to soak" (penetration depth was the same in all four); restating both columns with no causal connection; a general statement about SA:V never applied to these cubes.
   • (c) Accept: cube side length in cm on the x-axis because it is the independent variable, with an even scale from 0 to 4 cm; percent of volume turned pink on the y-axis, scaled 0 to 100% in even intervals; both axes labeled with quantity and unit. Shape: starts at 100% at 1.0 cm and falls, steeply at first and then more gradually, to 57.8% at 4.0 cm. Do not accept: the independent variable on the y-axis; axes with no units; a percent axis that does not include 0 and 100; "plot the data" with no axes, scale, or labels specified; a straight line through the origin.
   • (d) Accept: the 4.0 cm cube has a surface area-to-volume ratio of only 1.5 cm⁻¹ and just 57.8% of its volume was reached in 10 minutes, leaving 42.2% untouched; a cell of that size would have more than two-fifths of its cytoplasm receiving no oxygen or nutrients and unable to clear waste, because the membrane area available for exchange is too small for the volume behind it. Do not accept: "it is too big" with no value cited; quoting 57.8% or 1.5 cm⁻¹ without saying what it means for a living cell; claiming diffusion slows down inside large cells.
   Show a 4/4 response

   a The independent variable is the side length of the cube: 1.0, 2.0, 3.0, or 4.0 cm. Soak time had to be held at 10 minutes for all four, because the NaOH front advances with time, so a cube left in longer would have a deeper pink zone for a reason unrelated to its size.

   b The NaOH moved the same 0.5 cm into every cube, so the pink shell tracks surface area while the pale core tracks volume. As side length grows, volume grows faster than surface does and the ratio drops from 6.0 to 1.5 cm⁻¹, so a shell of fixed thickness covers a smaller share of a bigger cube, which is why percent pink falls from 100 to 57.8.

   c Side length in cm goes on the x-axis because that is what I set, scaled 0 to 4 cm in 1 cm steps; percent of volume turned pink goes on the y-axis, scaled 0 to 100% in steps of 20. Both axes get the quantity and the unit. The curve starts at 100% at 1.0 cm and falls, steeply at first and then more gently, to 57.8% at 4.0 cm.

   d I agree. The 4.0 cm cube has a ratio of only 1.5 cm⁻¹ and only 57.8% of it turned pink, so 42.2% of the interior got nothing in 10 minutes. A cell that size would have more than two-fifths of its cytoplasm cut off from oxygen and nutrients and from waste removal, because its surface is far too small to serve the volume behind it.

2. A researcher prepares artificial vesicles from pure phospholipid and a second batch of vesicles from the same phospholipid plus cholesterol. Answer (a) through (d).

   1. Describe how the phospholipid molecules are arranged in the bilayer of one of these vesicles, and identify the region of the bilayer a solute must pass through to cross it.
   2. Explain why O₂ crosses a pure phospholipid bilayer readily while Na⁺ crosses at a negligible rate, even though a sodium ion is smaller than an oxygen molecule.
   3. Predict how the fluidity of the cholesterol-containing vesicles compares with that of the pure phospholipid vesicles at 5 °C and at 40 °C.
   4. Justify your prediction in (c) by describing how cholesterol interacts with the phospholipid tails.
   Your response
   Scoring notes

   • (a) Accept: two layers of phospholipids with their hydrophilic phosphate heads facing the water on the outside of the vesicle and on the inside, and their hydrophobic fatty acid tails pointing inward toward each other, away from water; a solute must pass through the hydrophobic core formed by those tails. Do not accept: "a phospholipid bilayer" restated with no description of head and tail orientation; heads and tails reversed; identifying the phosphate heads as the barrier.
   • (b) Accept: O₂ is small and nonpolar, so it dissolves directly into the hydrocarbon core and passes through. Na⁺ carries a full positive charge and holds a shell of water molecules oriented around it; entering the core would require stripping that hydration shell away, which costs far more energy than the ion has, so charge rather than size is the barrier. Do not accept: "ions are too big" with no mention of charge or hydration; "the membrane is selectively permeable," which restates the question; a correct general rule never applied to O₂ and Na⁺.
   • (c) Accept a two-part directional prediction: at 5 °C the cholesterol vesicles are more fluid than the pure phospholipid vesicles, and at 40 °C they are less fluid. Do not accept: a single prediction covering only one temperature; "cholesterol changes fluidity" with no direction; "cholesterol makes the membrane more fluid" at both temperatures.
   • (d) Accept: cholesterol's rigid fused-ring system sits wedged between neighboring phospholipid tails; in the cold it physically prevents the tails from packing into a tightly ordered gel, keeping the bilayer fluid, and in the heat the same rings restrict how far and how fast the tails can swing, holding the bilayer together and preventing it from becoming too fluid and leaky. Do not accept: calling cholesterol a "fluidity buffer" with no mechanism; describing only one of the two temperature effects; claiming cholesterol changes the saturation of the tails.
   Show a 4/4 response

   a The phospholipids sit in two layers with their phosphate heads facing water (the outside of the vesicle for one layer, the enclosed interior for the other), and their fatty acid tails pointing inward at each other, away from water. Anything crossing must pass through that hydrophobic core of tails.

   b O₂ is small and nonpolar, so it dissolves into the hydrocarbon core the way it would into oil and comes out the far side. Na⁺ carries a full positive charge, and in water it is wrapped in a shell of water molecules turned toward it. Entering the core would mean shedding that shell, which costs more energy than the ion has, so the barrier is charge, not size.

   c I predict the cholesterol vesicles are more fluid than the pure phospholipid vesicles at 5 °C and less fluid than them at 40 °C.

   d Cholesterol's stiff ring system wedges between neighboring tails. At 5 °C those rings block the tails from packing into a tight, ordered gel, so the cholesterol membrane stays fluid while the pure one starts to solidify. At 40 °C the same rings limit how far the tails can swing, so it holds together while the pure one turns too fluid and leaky.

3. Model: a cross-sectional diagram of a mitochondrion inside a cultured human cell.

   The diagram shows a smooth outer membrane and, within it, an inner membrane thrown into deep folds labeled cristae; the space enclosed by the inner membrane is labeled the matrix. In the matrix the diagram shows a small closed loop of DNA with no histone proteins attached and several ribosomes labeled 70S. Ribosomes drawn in the surrounding cytosol are labeled 80S. An arrow beside the organelle is labeled "divides by binary fission."

   Using the model, answer (a) through (d).

   1. Identify TWO features shown in the model that the endosymbiotic theory predicts, and describe what that theory proposes each of the two membranes was derived from.
   2. Explain how the cristae shown in the model increase the amount of ATP a single mitochondrion can produce.
   3. An antibiotic that binds 70S ribosomes and blocks translation is added to the culture medium. Predict its effect on protein synthesis in the cytosol and on protein synthesis inside the mitochondrion.
   4. Justify your prediction in (c) using specific features labeled in the model.
   Your response
   Scoring notes

   • (a) Accept any two of: the double membrane, the circular DNA lacking histones, the 70S (bacterial-size) ribosomes, and division by binary fission. Membranes: the inner membrane is proposed to be the engulfed bacterium's own plasma membrane and the outer membrane the host cell's engulfing vesicle. Do not accept: only one feature named; features listed with no account of where each membrane came from; naming the cristae as evidence of endosymbiosis, since folding is a surface-area adaptation rather than a prediction of the model.
   • (b) Accept: the folds pack a large area of inner membrane into a small volume, and the electron transport chain proteins and ATP synthase are embedded in that membrane, so more folding means more of those complexes per organelle, more H⁺ pumped into the intermembrane space, and more ATP made per unit of mitochondrial volume. Do not accept: "cristae increase surface area" with no statement of what the added area holds; placing the electron transport chain in the matrix or in the outer membrane; an answer about ATP that never mentions membrane area.
   • (c) Accept a two-part directional prediction: cytosolic protein synthesis continues at close to its normal rate, while protein synthesis inside the mitochondrion is inhibited, so mitochondrially encoded proteins are not made and ATP output falls over time. Do not accept: "protein synthesis stops" without distinguishing the two compartments; predicting an effect with no direction; "nothing changes."
   • (d) Accept: the model labels the matrix ribosomes 70S and the cytosolic ribosomes 80S, and the drug binds only 70S, so the mitochondrial ribosomes are targets and the cell's own are not. The 70S label is itself one of the endosymbiotic predictions, since 70S is the ribosome size found in bacteria. Do not accept: repeating the prediction in other words; asserting shared ancestry with no labeled feature named; citing the double membrane or circular DNA, neither of which determines what the drug binds.
   Show a 4/4 response

   a Two features the theory predicts are the circular DNA with no histones and the 70S ribosomes in the matrix. On this model the inner membrane is the plasma membrane of the bacterium that was engulfed long ago, and the outer membrane is the vesicle the host cell wrapped around it as it took the bacterium in.

   b The cristae fold the inner membrane back on itself so a large membrane area fits inside a small organelle. The electron transport chain proteins and ATP synthase are embedded in that membrane, so the more of it a mitochondrion packs in, the more of those complexes it holds, the more H⁺ it pumps into the intermembrane space, and the more ATP it makes per unit volume.

   c I predict cytosolic protein synthesis continues at close to its normal rate while translation inside the mitochondrion shuts down, so mitochondrial proteins stop being made and the cell's ATP output drops over the following hours.

   d The model labels the matrix ribosomes 70S and the cytosolic ones 80S, and the drug binds only 70S, so the mitochondrion's ribosomes are targets and the cell's own are not. That 70S label is itself one of the endosymbiotic predictions, since 70S is the bacterial ribosome size.

4. Data: potato cores of equal starting mass were blotted, weighed, submerged for 24 hours at 25 °C in the open beakers shown, blotted again, and reweighed.

   [Sucrose] in the beaker (M)	0.0	0.2	0.4	0.6	0.8
   % change in mass	+16.2	+8.7	+1.8	−5.4	−12.0

   Using the data, answer (a) through (d).

   1. Identify the dependent variable in the investigation, and describe the trend in the data.
   2. Explain, in terms of water potential, why the cores in the 0.0 M beaker gained mass while the cores in the 0.8 M beaker lost mass.
   3. Determine the sucrose molarity at which the cores would show no net change in mass, then calculate the solute potential of that solution at 25 °C, showing your setup. Use \(\Psi_s = -iCRT\) with \(R = 0.0831\) L·bar/(mol·K) and \(i = 1\) for sucrose.
   4. Justify the claim that the water potential of the potato cores is −11.14 bars, using your result from (c) and the conditions of the investigation.
   Your response
   Scoring notes

   • (a) Accept: the dependent variable is the percent change in mass of the potato cores. Trend: percent change falls steadily as beaker sucrose concentration rises, from +16.2% at 0.0 M to −12.0% at 0.8 M, changing from a gain to a loss between 0.4 M and 0.6 M. Do not accept: naming sucrose concentration as the dependent variable; "the mass changed" with no direction; listing the endpoints with no statement of the trend.
   • (b) Accept: pure water in an open beaker has Ψ = 0 bars, the highest value possible, which is higher than the cores' negative Ψ, so water moved into the cells and the cores gained mass. The 0.8 M beaker has a strongly negative solute potential and therefore a Ψ lower than the cores', so water moved out of the cells into the beaker and the cores lost mass. Do not accept: "water moves toward the higher solute concentration" with no reference to water potential; a reversed direction; a general rule never applied to these cores.
   • (c) Accept: interpolating between 0.4 M (+1.8%) and 0.6 M (−5.4%), the line falls 7.2 percentage points across 0.2 M, so zero lies 1.8/7.2 = 0.25 of the way, at 0.4 + 0.25(0.2) = 0.45 M. Then with T = 25 + 273 = 298 K, \(\Psi_s = -(1)(0.45)(0.0831)(298) = -11.14\) bars. Do not accept: reading the crossing as 0.4 M or 0.6 M with no interpolation; T left in °C; a missing negative sign; a value with no units; a value with no setup shown.
   • (d) Accept: at 0.45 M there is no net movement of water, so the cores and the bath are at equilibrium and their water potentials are equal. The beaker is open, so its pressure potential is 0 and its water potential equals its solute potential, −11.14 bars; therefore the cores' water potential is −11.14 bars as well. Do not accept: restating the value; asserting the equality with no mention of zero net water movement or of Ψ_p = 0 in an open beaker; using the 0.8 M beaker's value.
   Show a 4/4 response

   a The dependent variable is the percent change in mass of the potato cores. It falls steadily as the beaker gets more concentrated, from +16.2% in pure water to −12.0% at 0.8 M, switching from a gain to a loss between 0.4 M and 0.6 M.

   b Pure water in an open beaker has a water potential of 0 bars, the highest there is, while the cores' Ψ is negative because of the solutes in their cells. Water moves from higher Ψ to lower, so it entered the cells and those cores gained mass. The 0.8 M beaker has a strongly negative solute potential, putting its Ψ below the cores', so water left the cells instead.

   c Between 0.4 M (+1.8%) and 0.6 M (−5.4%) the line drops 7.2 percentage points, so zero sits 1.8/7.2 = 0.25 of the way across: 0.4 + 0.25(0.2) = 0.45 M. With T = 25 + 273 = 298 K, Ψs = −(1)(0.45)(0.0831)(298) = −11.14 bars.

   d At 0.45 M the cores neither gain nor lose mass, which means no net water movement and therefore equal water potentials on both sides. The beaker is open to the air, so its pressure potential is 0 and its water potential is nothing but its solute potential, −11.14 bars. Since the cores match it at equilibrium, their water potential is −11.14 bars too.

5. Data: glucose uptake across the apical membrane of cultured intestinal epithelial cells over 10 minutes under four conditions. Ouabain inhibits the sodium–potassium pump; phlorizin blocks the SGLT1 sodium–glucose cotransporter.

   Condition	Complete medium	Na+ removed	Ouabain added	Phlorizin added
   Glucose uptake (nmol/min per mg protein)	62	6	9	5

   Using the data, answer (a) through (d).

   1. Identify the transport mechanism that carries glucose across the apical membrane in the complete medium, and describe the immediate source of free energy that drives it.
   2. Explain why ouabain lowers glucose uptake even though it does not bind the SGLT1 cotransporter.
   3. Calculate the percent decrease in glucose uptake caused by removing Na⁺ and the percent decrease caused by ouabain, showing your setup.
   4. A student claims that cells in a complete medium whose ATP synthesis has been blocked for 30 minutes would take up glucose at a rate close to the ouabain value rather than the complete-medium value. Justify this claim using the data.
   Your response
   Scoring notes

   • (a) Accept: secondary active transport by the SGLT1 Na⁺–glucose cotransporter; the immediate energy source is the electrochemical gradient of Na⁺ across the apical membrane, whose free energy is released as Na⁺ moves down that gradient into the cell and is used to haul glucose in against its own gradient. Do not accept: "active transport" with no statement that it is secondary or that Na⁺ supplies the energy; naming ATP hydrolysis at SGLT1 as the immediate source; "facilitated diffusion" or "simple diffusion."
   • (b) Accept: ouabain inhibits the sodium–potassium pump, which is what continually exports Na⁺ and keeps intracellular Na⁺ low; with the pump stopped, Na⁺ leaks in until the gradient runs down, and once there is no gradient there is no free energy for the cotransporter to spend, so uptake falls to 9: close to the value with Na⁺ removed entirely. Do not accept: claiming ouabain blocks SGLT1; "ATP is needed for transport" with no reference to the pump or the gradient; naming the pump with no explanation of how its loss reaches glucose uptake.
   • (c) Accept: Na⁺ removed, (62 − 6)/62 × 100 = 90.3%; ouabain, (62 − 9)/62 × 100 = 85.5%. Setups shown; rounding to one decimal or to the nearest whole percent acceptable. Do not accept: dividing by the treated value instead of the control; reporting the remaining percentage (9.7% and 14.5%) as the decrease; one calculation when two were asked for; a bare number with no setup.
   • (d) Accept: the claim is supported because blocking ATP synthesis stops the sodium–potassium pump for the same reason ouabain does, and the data show that stopping that pump drops uptake from 62 to 9 even with Na⁺ still present in the medium. That result is close to the 6 obtained with no Na⁺ at all, which shows the gradient, not the presence of Na⁺, is what the cotransporter depends on. Do not accept: "no ATP means no transport" with no mention of the pump or the gradient; restating the claim; citing the ouabain value with no explanation of why blocking ATP synthesis produces the same outcome.
   Show a 4/4 response

   a Glucose crosses the apical membrane by secondary active transport through the SGLT1 cotransporter. The immediate energy source is the Na⁺ electrochemical gradient: sodium falls down its own gradient into the cell, and the free energy released as it does drags glucose in with it, uphill against glucose's gradient.

   b Ouabain blocks the sodium–potassium pump, which is the protein that keeps pushing Na⁺ back out and holds internal sodium low. Once the pump stops, Na⁺ leaks in until inside and outside are nearly equal, and a gradient that has run down releases no free energy. The cotransporter is untouched but has nothing left to spend, so uptake collapses to 9.

   c Removing Na⁺: (62 − 6)/62 × 100 = 90.3%. Ouabain: (62 − 9)/62 × 100 = 85.5%.

   d I agree with the student. Blocking ATP synthesis stops the sodium–potassium pump for the same reason ouabain does, just from the fuel side instead of the protein side, and the data show what happens when that pump stops: uptake fell from 62 to 9 even though Na⁺ was still in the medium the whole time. That 9 is close to the 6 measured with no Na⁺ at all, which tells me the cotransporter needs the gradient rather than just the ion.

Lesson 3.1 · Unit 3 · CED topics 3.1–3.2

Enzyme structure and catalysis

Almost every reaction a cell needs is thermodynamically allowed and kinetically hopeless. A jar of sucrose solution really will hydrolyze on its own: you would just have to wait years. A cell cannot wait, and it cannot boil itself to force the issue, so it does the only other thing available: it lowers the barrier.

That is the entire job of an enzyme. It does not make an unfavorable reaction favorable, change how much free energy the reaction releases, or shift the equilibrium position. It changes how fast the system gets to that equilibrium: routinely by a factor of a million.

Lesson 3.2 · Unit 3 · CED topic 3.3

Environmental effects on enzyme function

An enzyme's power comes entirely from the shape of its active site, so anything that alters that shape alters the rate. Heat, acid, a molecule that squats in the pocket, a molecule that binds elsewhere and bends the protein: all of them show up as a change in µmol of product per minute. The exam gives you this as a curve or a table and asks you to describe the pattern and explain the mechanism: two tasks, two points, so answer both.

Lesson 3.3 · Unit 3 · CED topic 3.4

Cellular energy, free energy, and ATP coupling

Staying alive is expensive. A cell holds ion gradients against leakage, builds polymers out of scattered monomers, and keeps its interior far more ordered than its surroundings: all of it uphill. The universe permits this only because the cell pays, continuously, by running downhill reactions alongside the uphill ones. ATP is the currency it pays in.

Lesson 3.4 · Unit 3 · CED topic 3.5

Photosynthesis I: capturing light

The light reactions solve one problem: turn photons into the two chemically useful things the Calvin cycle needs, ATP and NADPH. Every structure in a chloroplast exists to serve that conversion, and the by-product, the oxygen you are breathing, is simply what is left over when water is torn apart for its electrons.

Lesson 3.5 · Unit 3 · CED topic 3.5

Photosynthesis II: the Calvin cycle and carbon fixation

The light reactions hand the stroma a supply of ATP and NADPH that is useless as storage: neither lasts. The Calvin cycle spends both immediately to pull carbon out of the air and build it into sugar. It needs light only indirectly, which is why calling it the "dark reaction" is misleading: it runs in daylight, on the products of the light reactions.

Lesson 3.6 · Unit 3 · CED topic 3.6

Cellular respiration I: glucose to carbon dioxide

Respiration does the reverse of photosynthesis, and it does it in small steps on purpose. Burning glucose in one shot would release far more energy than any molecule could capture. Instead the cell strips the electrons off six carbons a pair at a time, loading them onto NAD⁺ and FAD, and saves the big payoff for Lesson 3.7. Everything before the electron transport chain is preparation: it disassembles the glucose and fills the carriers.

Lesson 3.7 · Unit 3 · CED topics 3.6–3.7

Cellular respiration II: chemiosmosis, fermentation, and fitness

Lesson 3.6 left the cell holding 10 NADH and 2 FADH₂ and only 4 ATP to show for it. Now those carriers get cashed. The surprising part is the mechanism: the cell does not hand energy from the carriers to ADP chemically. It uses the electrons to pump protons, stores the energy as a gradient, and then lets the gradient turn a rotary motor.

Unit 3 quiz · 15 multiple-choice · 5 free-response

Unit 3 quiz: Cellular Energetics

Fifteen multiple-choice questions and five short free-response questions across all seven lessons: enzymes and their inhibitors, free energy and ATP coupling, the light reactions and the Calvin cycle, and the whole path from glucose to ATP. Click an option to see why each choice is right or wrong, then write each free response on paper against the clock before you open the model.

Multiple choice

1. Model: a reaction-coordinate diagram for the conversion of substrate S to product P, drawn twice on the same axes.

   Free energy in kJ/mol runs up the y-axis; reaction progress runs along the x-axis. Both curves start on the same reactant shelf at 0 and end on the same product shelf at −32. Curve 1, for the uncatalyzed reaction, climbs to a peak of +84 before dropping to the product shelf. Curve 2, measured with enzyme present, follows the same shelves but peaks at only +35.

   Which statement is best supported by the two curves?

   The enzyme lowers the activation energy by 49 kJ/mol and leaves ΔG at −32 kJ/mol The enzyme lowers ΔG from −32 to −81 kJ/mol, making the reaction more exergonic The enzyme raises the free energy of the products, so more product accumulates at equilibrium The enzyme lowers the activation energy by 32 kJ/mol and shifts the equilibrium toward product

   Correct. The barrier falls from +84 to +35, a drop of \(84 - 35 = 49\) kJ/mol, and both curves end on the same product shelf, so \(\Delta G\) is −32 kJ/mol with and without enzyme. Naming the right quantity is the whole item: an enzyme changes \(E_a\), never \(\Delta G\).

   This is the arithmetic slip of subtracting the barrier reduction from \(\Delta G\): −32 − 49 gives −81. But the product shelf never moved on the diagram, so the free-energy change is identical on both curves. "The enzyme lowers the energy of the reaction" is the wrong quantity and earns nothing on an exam.

   Nothing in the figure shows the product shelf rising, and if it did the reaction would release less free energy, not more. Enzymes also cannot shift an equilibrium position: they speed the forward and reverse directions by the same factor, so the system simply arrives at the same equilibrium sooner.

   The 32 here is \(\Delta G\), not the change in the barrier; the barrier fell by 49 kJ/mol. The second half of the choice repeats the most common enzyme misconception: a catalyst changes the rate of approach to equilibrium, not the position of the equilibrium itself.

2. Which observation provides the strongest evidence for the induced-fit model of enzyme action rather than a rigid lock-and-key active site?

   The enzyme is not consumed and can catalyze thousands of reactions in succession Structures of the enzyme solved with and without substrate show the active site closing around the substrate, bringing two catalytic side chains into alignment only after binding The enzyme lowers the activation energy of the reaction without changing its ΔG Raising substrate concentration increases the reaction rate until every active site is occupied

   This is true and important, it is why a cell runs a high-throughput pathway on a tiny amount of protein, but a rigid lock-and-key enzyme would also be regenerated unchanged. An observation that both models predict cannot distinguish between them.

   Correct. Induced fit claims the active site is not pre-formed: contact with the substrate changes its shape, straining the bonds about to break and positioning the catalytic groups. A structure that differs before and after binding is direct evidence of that conformational change, which lock-and-key does not predict.

   Every catalyst, rigid or flexible, lowers \(E_a\) without touching \(\Delta G\). This statement describes what enzymes do, not how the active site behaves when substrate arrives, so it says nothing about which binding model is right.

   Saturation follows from having a finite number of active sites and is predicted equally well by both models. It is good evidence that catalysis happens at discrete sites, but it does not tell you whether those sites change shape on binding.

3. Data: initial reaction rate of one enzyme at 30 °C and pH 7.0 across a range of substrate concentrations, with no inhibitor and with each of two inhibitors held at a fixed concentration.

   [Substrate] (mM)	1	2	5	10	20
   No inhibitor (µmol/min)	8.0	14.0	25.0	32.0	36.0
   + Inhibitor X (µmol/min)	3.0	6.0	13.0	22.0	33.0
   + Inhibitor Y (µmol/min)	3.2	5.6	10.0	12.8	14.4

   Which conclusion about inhibitor Y is best supported by the data?

   Y is competitive, because it lowers the rate at every substrate concentration tested Y is noncompetitive, because at 20 mM substrate the rate is still only 14.4 µmol/min, 40% of the uninhibited value, so added substrate cannot restore the maximum Y is competitive, because its curve is displaced to the right of the uninhibited curve Y has denatured the enzyme, because the rate with Y never reaches the uninhibited maximum

   Both inhibitors lower the rate at every concentration in the table, so this feature cannot tell them apart. The diagnostic question is what happens as substrate is flooded in: X climbs to 33.0 µmol/min, nearly the uninhibited 36.0, while Y stalls at 14.4.

   Correct. \(14.4/36.0 = 0.40\), so Y caps the enzyme at 40% of its maximum however much substrate is added. That is the signature of a noncompetitive inhibitor: it binds an allosteric site and distorts the active site, so a fixed fraction of enzyme molecules is dead and substrate cannot compete it away.

   A right-shifted curve that still reaches the uninhibited plateau does describe a competitive inhibitor, but that is inhibitor X, not Y. Y's curve is not merely shifted; its ceiling is lower, which is a different effect with a different mechanism.

   A denatured enzyme has lost its tertiary fold and gives a rate near zero at every substrate concentration. Y-treated enzyme still rises from 3.2 to 14.4 µmol/min as substrate increases, so the remaining enzyme is folded and working; it is inhibited, not destroyed.

4. A lizard's oxygen consumption is 3.50 mL O₂ g⁻¹ h⁻¹ at 20 °C and 8.75 mL O₂ g⁻¹ h⁻¹ at 30 °C, both temperatures below the optimum for its enzymes. What is Q₁₀, and what does that value indicate?

   2.5: a 10 °C rise multiplies the rate by 2.5, which is the range expected for enzyme-catalyzed reactions below the optimum 5.25: the rate rose by 5.25 mL O₂ g⁻¹ h⁻¹ across the 10 °C interval 0.40: the rate fell by a factor of 0.40, showing that the enzymes have begun to denature 1.25: the rate rose by 25% per 10 °C, close to the value expected for a purely physical process

   Correct. \(Q_{10} = (8.75/3.50)^{10/10} = 2.50\). Most biological rates give 2 to 3 below the optimum, because warming both increases collision frequency and raises the fraction of collisions carrying enough energy to cross the activation barrier.

   This is the arithmetic difference, \(8.75 - 3.50\), not a ratio. \(Q_{10}\) is defined as the factor by which a rate is multiplied per 10 °C, so it is unitless; an answer carrying mL O₂ g⁻¹ h⁻¹ cannot be a \(Q_{10}\).

   0.40 is \(3.50/8.75\): the ratio taken upside down. A \(Q_{10}\) below 1 does signal denaturation, but only when the measured rate actually falls with warming, and here it more than doubled.

   1.25 would mean a 25% increase per 10 °C, which is roughly what simple diffusion does. Enzyme-catalyzed processes respond far more steeply, and 2.5 is exactly that steeper response.

5. Synthesis of a cellular intermediate has ΔG = +21.0 kJ/mol. In the cell this step is coupled to ATP hydrolysis, ΔG = −30.5 kJ/mol. What is ΔG for the coupled reaction, and will it proceed?

   +51.5 kJ/mol; it does not proceed, because the two energy costs add together −9.5 kJ/mol; it proceeds, because the sum of the two free-energy changes is negative −30.5 kJ/mol; it proceeds, because ATP hydrolysis sets the free-energy change of the coupled reaction +21.0 kJ/mol; it does not proceed, because ATP hydrolysis only lowers the activation energy of the synthesis

   Adding 21.0 to 30.5 ignores the sign on ATP hydrolysis, which releases free energy. Coupled reactions are summed with their signs: \(+21.0 + (-30.5)\). Losing the sign here turns a spontaneous process into an impossible one.

   Correct. \(+21.0 + (-30.5) = -9.5\) kJ/mol, and a negative sum means the coupled reaction is exergonic and proceeds. Mechanically, ATP phosphorylates an intermediate so the two steps share one pathway; the 9.5 kJ/mol left over is the margin that makes the pair spontaneous.

   This treats ATP hydrolysis as if the endergonic step were free. The synthesis still costs +21.0 kJ/mol, and that cost is paid out of the 30.5 released, leaving −9.5. Coupling is never free: that is why cells need so much ATP.

   ATP hydrolysis is a source of free energy, not a catalyst; it changes \(\Delta G\), while enzymes change \(E_a\). Keeping those two roles separate is the whole point of this topic, and confusing them is the most common error on coupling questions.

6. Data: daily energy budget of a 40 g mouse and a 40 g lizard, each measured in a 25 °C chamber over 24 hours.

   Measurement	Mouse (endotherm)	Lizard (ectotherm)
   Energy taken in (kJ/day)	120	20
   Energy spent on maintenance (kJ/day)	108	8

   Which statement is best supported by the data?

   The mouse has more energy available for growth and reproduction, because it takes in six times as much energy as the lizard Both animals have 12 kJ/day available for growth and reproduction, but that surplus is 10% of the mouse's intake and 60% of the lizard's, because endothermy raises the maintenance cost The lizard converts 60% of its intake into new tissue, so it is a more efficient feeder than the mouse Neither animal can grow, because in both cases maintenance consumes the whole intake

   Intake alone does not say what is left over. The energy available for growth and reproduction is intake minus maintenance, and both animals land on the same surplus: \(120 - 108 = 12\) kJ/day and \(20 - 8 = 12\) kJ/day.

   Correct. The surpluses are identical at 12 kJ/day, but \(12/120 = 10\%\) versus \(12/20 = 60\%\). The mouse holds 37 °C against a 25 °C room and burns most of its intake as heat; the lizard lets its body temperature follow the chamber, so almost nothing is spent on thermoregulation.

   The 60% is the share of intake left after maintenance, not the share converted into tissue: some of that surplus is lost in assimilation and in the cost of building. The table gives no measurement of tissue gained, so this claim goes beyond the data.

   Both maintenance figures are below the corresponding intake, so each animal has a positive surplus. An organism whose maintenance truly consumed its whole intake would lose mass and stop reproducing, which is the link between energy budgets and fitness.

7. Figure: two traces on the same axes; percent of each trace's own maximum on the y-axis (0 to 100), wavelength in nm on the x-axis (400 to 700).

   Trace 1 is the absorption spectrum of purified chlorophyll a: 100 at 430 nm, falling to 6 at 550 nm, rising again to 92 at 680 nm. Trace 2 is the action spectrum of an intact Elodea leaf, measured as oxygen released: 85 at 430 nm, 24 at 550 nm, and 100 at 680 nm.

   Which best explains why the two traces differ most at 550 nm?

   The leaf reflects green light, and reflected light raises the measured rate of oxygen release at 550 nm Extraction denatures chlorophyll a, so trace 1 underestimates absorption at every wavelength Chlorophyll b and the carotenoids in the intact leaf absorb wavelengths that chlorophyll a does not and pass the energy to the reaction center, so the leaf harvests a wider band than any single pigment At 550 nm the leaf photosynthesizes without absorbing light, using ATP and NADPH stored from earlier illumination

   Reflected light has by definition left the leaf without being absorbed, so it cannot drive photolysis. Reflection in the green band is why leaves look green; it is the reason the action spectrum dips at 550 nm, not the reason it is higher than chlorophyll a's absorption there.

   If extraction lowered absorption uniformly, trace 1 would sit below trace 2 everywhere. It does not, chlorophyll a reads 100 at 430 nm while the leaf reads 85, so a general underestimate cannot be the explanation for a gap that appears only in the green.

   Correct. Trace 1 is one purified pigment; trace 2 is a whole leaf containing chlorophyll b and carotenoids, which absorb in the band where chlorophyll a is nearly transparent and transfer that excitation energy to chlorophyll a in the reaction center. That is why the action spectrum reads 24 where absorption reads 6.

   Oxygen release requires photolysis of water at photosystem II, which needs photons arriving now; stored ATP and NADPH are products of the light reactions and cannot substitute for them. A leaf in the dark releases no oxygen at all.

8. DCMU blocks the transfer of electrons from photosystem II to plastoquinone. A suspension of isolated chloroplasts is illuminated in the presence of DCMU. Which result is predicted?

   Oxygen release continues but NADPH production stops, because photolysis is driven directly by light and does not depend on electron flow Oxygen release stops, because water is split only to replace the electrons photosystem II has donated to the chain Both oxygen release and NADPH production increase, because electrons back up and accumulate in photosystem II Oxygen release stops but the proton gradient is unaffected, because ATP synthase works independently of the electron transport chain

   This reverses the dependency. Photolysis is not a light-driven reaction that runs on its own: it is the replacement step, triggered when P680 has lost electrons. Block the downstream transfer and photosystem II stays reduced, so it stops pulling electrons off water.

   Correct. Water is the electron source, and it is split only to refill photosystem II. With the exit blocked, no photolysis occurs, so no O₂ is released; NADPH production also collapses, because photosystem I receives no electrons to pass to NADP⁺ reductase.

   Electrons that cannot leave photosystem II do not generate additional products: the excited electrons simply return to the ground state, releasing the energy as heat and fluorescence. Nothing in a blocked chain makes more oxygen or more NADPH.

   The proton gradient is built by the electron transport chain pumping H⁺ into the lumen and by photolysis releasing H⁺ there. Blocking electron flow removes both sources, so the gradient dissipates and ATP synthase runs down.

9. A leaf fixes 18 molecules of CO₂ through the Calvin cycle. How much ATP and NADPH are consumed, and how many molecules of glucose could be assembled from the G3P exported?

   18 ATP and 12 NADPH; 1 glucose 54 ATP and 36 NADPH; 3 glucose 108 ATP and 72 NADPH; 3 glucose 54 ATP and 36 NADPH; 6 glucose

   These are the per-G3P figures for three turns, not for eighteen. One turn fixes one CO₂ and costs 3 ATP and 2 NADPH, so eighteen fixations cost six times as much. Anchor every Calvin-cycle calculation on "one CO₂ per turn."

   Correct. Eighteen turns cost \(18 \times 3 = 54\) ATP and \(18 \times 2 = 36\) NADPH. Every three turns export one net G3P, giving \(18/3 = 6\) G3P, and two G3P build one glucose, so \(6/2 = 3\) glucose. Check it the other way: \(3 \times 18 = 54\) ATP and \(3 \times 12 = 36\) NADPH.

   Doubling the correct values usually comes from applying the per-glucose costs (18 ATP, 12 NADPH) to each of six turns. Those per-glucose figures already include all six turns, so using them again multiplies the cost twice.

   The ATP and NADPH are right, but six glucose would need \(6 \times 6 = 36\) carbons fixed, twice what the stem gives. Dividing 18 by 3 to get glucose confuses turns per exported G3P with turns per glucose, which is six.

10. Data: net photosynthesis of soybean (a C3 plant) and maize (a C4 plant), measured at a leaf temperature of 35 °C in full sun at three atmospheric CO₂ concentrations.

    Atmospheric CO2 (ppm)	200	400	800
    Soybean, C3 (µmol CO2 m−2 s−1)	8	18	30
    Maize, C4 (µmol CO2 m−2 s−1)	28	34	35

    Raising CO₂ from 200 to 800 ppm multiplies soybean's rate by 3.75 but raises maize's by only 25%. Which explanation is best supported?

    Maize's PEP carboxylase concentrates CO₂ at rubisco inside the bundle-sheath cells, so its rubisco is already near saturation and photorespiration is suppressed, while soybean's rubisco sees only leaf-air CO₂ and loses carbon to photorespiration at 200 ppm Maize has fewer stomata than soybean, so it cannot take up the extra CO₂ made available Soybean contains more chlorophyll per unit leaf area, so it can use the extra CO₂ in the light reactions Maize's rubisco has a much higher affinity for CO₂ than soybean's rubisco, so additional CO₂ has little effect on it

    Correct. C4 anatomy delivers CO₂ to rubisco at a high local concentration, so maize starts at 28 and has little room to gain: \(35/28 = 1.25\). Soybean's rubisco is CO₂-limited and competing with O₂ at 200 ppm, so extra CO₂ both feeds carboxylation and suppresses photorespiration: \(30/8 = 3.75\).

    Maize at 200 ppm already outperforms soybean at 400 ppm, which rules out a plant that cannot get CO₂ in. C4 plants in fact run with less stomatal opening for the same carbon gain, which is why they conserve water, but stomatal number is not what the data here distinguish.

    CO₂ is not used in the light reactions at all: it enters at the carbon-fixation step of the Calvin cycle. Chlorophyll content sets how much light energy is captured, which is not the limiting factor in full sun at 200 ppm CO₂.

    Rubisco is essentially the same enzyme in both plants; the C4 advantage is in delivery, not affinity. PEP carboxylase, the enzyme that first captures CO₂ in maize, is the one with no affinity for O₂: swapping the two enzymes is the standard error on this question.

11. A cell is supplied with glucose in which all six carbons are ¹⁴C and respires it completely in the presence of oxygen. Where does the labeled carbon leave, and in what amounts per glucose?

    2 ¹⁴CO₂ from glycolysis, 2 from pyruvate oxidation, and 2 from the citric acid cycle All 6 ¹⁴CO₂ from the citric acid cycle; the earlier stages only rearrange the carbon skeleton 2 ¹⁴CO₂ from pyruvate oxidation and 4 from the citric acid cycle; glycolysis releases none All 6 ¹⁴CO₂ at the electron transport chain, where oxygen accepts carbon and electrons together

    Glycolysis splits a 6-carbon glucose into two 3-carbon pyruvate, \(2 \times 3 = 6\), so every carbon is still in the cell when glycolysis ends. Expecting CO₂ from glycolysis is one of the most common errors on respiration items.

    The citric acid cycle contributes 4 of the 6, two per turn over two turns. The other two leave earlier, during pyruvate oxidation, when each pyruvate loses one carbon as it is converted to the 2-carbon acetyl group on coenzyme A.

    Correct. Pyruvate oxidation releases one CO₂ per pyruvate, so 2 per glucose; each of the two Krebs turns releases 2, so 4 more. \(2 + 4 = 6\), and the glucose skeleton is fully dismantled before the electron transport chain does anything.

    Oxygen's role at the end of the chain is to accept electrons and H⁺, forming water, no carbon is involved. Every CO₂ is released by a decarboxylation reaction upstream, in the matrix, before the electrons ever reach the chain.

12. Per molecule of glucose, how much ATP is produced before the electron transport chain, and by what mechanism?

    2 net ATP, by oxidative phosphorylation in the mitochondrial matrix 10 ATP, one for each NADH produced before the chain 32 ATP, by chemiosmosis across the inner mitochondrial membrane 4 net ATP, all by substrate-level phosphorylation: a net 2 from glycolysis and 2 from the two turns of the citric acid cycle

    The 2 is glycolysis alone, and the mechanism is wrong twice over: glycolysis happens in the cytosol, and oxidative phosphorylation is the chemiosmotic process that occurs only once electrons reach the chain. Substrate-level phosphorylation hands a phosphate straight from an intermediate to ADP.

    Ten is the number of NADH per glucose, not ATP. Those carriers hold most of the energy, but it is not released as ATP until the electron transport chain, and each NADH yields roughly 2.5 ATP there, not 1.

    Thirty-two is the approximate total for the whole of aerobic respiration, including the chain, and chemiosmosis happens after the stages in question. The point of this item is that the stages before the chain produce very little ATP directly.

    Correct. Glycolysis makes 4 ATP but invests 2, for a net 2, and each of the two Krebs turns yields 1 ATP (or GTP). All four come from substrate-level phosphorylation; the real product of these stages is the 10 NADH and 2 FADH₂.

13. Data: pipette readings (mL) from three respirometers held in a 22 °C water bath for 20 minutes. KOH in each vial absorbs the CO₂ released, and each seed vial contains 15.0 g of peas.

    Vial	Reading at 0 min	Reading at 20 min
    Germinating peas	1.00	0.55
    Dormant peas	0.95	0.92
    Glass beads only (control)	0.90	0.93

    What is the rate of oxygen consumption of the germinating peas, corrected using the control?

    1.5 × 10⁻³ mL O₂ g⁻¹ min⁻¹ 2.0 × 10⁻⁴ mL O₂ g⁻¹ min⁻¹ 1.6 × 10⁻³ mL O₂ g⁻¹ min⁻¹ 2.4 × 10⁻² mL O₂ min⁻¹

    This is the uncorrected value: \(0.45/(15.0 \times 20) = 1.5 \times 10^{-3}\). The bead vial drifted \(0.93 - 0.90 = +0.03\) mL from changes in room temperature and pressure, and that drift pushed every reading up, so it must be subtracted before the rate is computed.

    \(2.0 \times 10^{-4}\) is the dormant peas' rate: \(0.06/(15.0 \times 20)\) after the same correction. It is a useful comparison: germinating seeds respire \(0.48/0.06 = 8\) times as fast, but it is not what the question asks for.

    Correct. Drift \(= 0.93 - 0.90 = +0.03\) mL. Germinating: \((0.55 - 1.00) - 0.03 = -0.48\), so 0.48 mL of O₂ was consumed, and \(0.48/(15.0 \times 20) = 1.6 \times 10^{-3}\) mL O₂ g⁻¹ min⁻¹.

    This is \(0.48/20\): the time division only. Respirometer rates are normalized per gram as well as per minute so that vials with different amounts of tissue can be compared; dropping the mass term also drops g⁻¹ from the units, which is the clue that something is missing.

14. Data: isolated mitochondria supplied with substrate and treated with one of three compounds. All values are percentages of the untreated control.

    Treatment	O2 consumption	ATP synthesis	Heat released
    Control	100	100	100
    Compound M	165	12	230
    Compound N	8	6	20
    Compound P	22	9	35

    Compound P inhibits ATP synthase and does not act on the electron carriers directly. Which feature of the data is consistent with that action, and why?

    Oxygen consumption rises to 165% while ATP falls to 12%, because blocking ATP synthase forces electron transport to speed up Oxygen consumption falls to 22% while ATP falls to 9%, because protons cannot return through the blocked synthase, the gradient builds, and the back-pressure slows the pumps that build it Oxygen consumption falls to 8% and heat to 20%, because ATP synthase is the final electron acceptor of the chain Heat rises to 230% while ATP falls to 12%, because inhibiting ATP synthase releases the gradient's energy as heat

    That row is compound M, and the pattern is the signature of an uncoupler: protons leak back across the membrane without passing the synthase, so nothing opposes the pumps and electron transport accelerates. An ATP synthase inhibitor does the opposite: it leaves the gradient in place.

    Correct. With the return path shut, H⁺ accumulates in the intermembrane space until the pumps can no longer push against it, so electron transport slows and oxygen consumption falls to 22%. This coupling of ATP demand to oxygen use is called respiratory control.

    Oxygen, not ATP synthase, is the final electron acceptor: that is the vocabulary error the distractor is testing. The 8%/20% row is compound N, consistent with a chemical such as cyanide that blocks the last complex of the chain so oxygen cannot be reduced at all.

    Energy that cannot be captured as ATP is released as heat only when the protons are allowed to flow back, which is what an uncoupler permits and an ATP synthase inhibitor prevents. Compound P's heat is 35%, well below the control, exactly because the whole chain has slowed.

15. A yeast culture producing about 32 ATP per glucose aerobically is shifted to anaerobic conditions, where it produces 2 ATP per glucose. By what factor must glucose consumption rise to hold ATP production steady, and what does fermentation itself contribute?

    16-fold; fermentation makes no ATP of its own: it regenerates NAD⁺ so that glycolysis can keep running 16-fold; fermentation makes 2 additional ATP per glucose by substrate-level phosphorylation 30-fold; fermentation regenerates NADH from NAD⁺ so the citric acid cycle can continue No increase is needed; fermentation yields as much ATP as aerobic respiration but releases it more slowly

    Correct. \(32/2 = 16\), so the cell must burn sixteen times as much glucose per minute. Fermentation's own reactions (pyruvate to ethanol and CO₂, or pyruvate to lactate) make no ATP; they oxidize NADH back to NAD⁺, without which glycolysis would stall within seconds and even the 2 ATP would stop.

    The factor is right, but the 2 ATP are made in glycolysis, not in the fermentation step that follows. Attributing them to fermentation hides the real function of those reactions, which is recycling the electron carrier.

    Thirty is the difference \(32 - 2\), not the ratio, and the recycling runs the other way: NADH is oxidized to NAD⁺. The citric acid cycle is not running at all under anaerobic conditions, because the chain that reoxidizes its carriers has no final electron acceptor.

    Fermentation yields \(2/32 = 6.25\%\) of the aerobic ATP per glucose. It is actually faster per unit time than aerobic respiration, which is why sprinting muscle uses it, but it is far less thorough: most of the glucose's energy is left in ethanol or lactate.

Free response

1. Data: initial rate of O₂ production by catalase acting on 1.0% hydrogen peroxide at 25 °C. Each value is the mean of three trials. A second set of tubes received catalase that had been boiled for 10 minutes and cooled before use.

   pH of buffer	3	5	7	9	11
   Catalase (µmol O2/min)	4	22	48	20	3
   Boiled catalase (µmol O2/min)	0	0	0	0	0

   A student investigated how pH affects catalase activity. Using the investigation and the data, answer (a) through (d).

   1. Identify the independent variable in the investigation, and describe the purpose of the boiled-catalase treatment.
   2. Explain the relationship between pH and the initial reaction rate in terms of the structure of the enzyme.
   3. Construct a graph of the catalase data: place pH on the x-axis scaled from 3 to 11 and initial rate on the y-axis scaled from 0 to 50 µmol O₂/min, label both axes with units, plot the five points, and connect them.
   4. A classmate claims the data show that catalase is permanently denatured at pH 11. Justify whether the data provided are sufficient to support that claim.
   Your response
   Scoring notes

   • (a) Accept: the independent variable is the pH of the buffer the reaction is run in. The boiled tubes are a control showing that the oxygen measured depends on folded, functional enzyme, not on hydrogen peroxide decomposing on its own or on any other component of the mixture at that pH. Do not accept: naming the enzyme, the peroxide, or the reaction rate as the independent variable; "it is the control" with no statement of what it controls for; restating the stem.
   • (b) Accept: the rate rises from pH 3 to a maximum at pH 7 and falls on either side, and away from the optimum excess H⁺ or OH⁻ changes the charges on the R groups lining the active site, breaking the ionic and hydrogen bonds that hold its shape, so the site no longer complements hydrogen peroxide and fewer enzyme–substrate complexes form. Do not accept: "pH affects the enzyme"; "the enzyme dies"; a description of the trend with no structural mechanism: describing is not explaining.
   • (c) Accept: pH on the x-axis from 3 to 11 with an even scale, initial rate on the y-axis from 0 to 50, both axes labeled with units (pH; µmol O₂/min), the five points (3, 4), (5, 22), (7, 48), (9, 20), (11, 3) plotted and connected into a single peak at pH 7. Do not accept: axes reversed; units omitted from the axis labels; points that do not match the table.
   • (d) Accept: the data are not sufficient; they show only that activity is low at pH 11, which is equally consistent with reversible inhibition; supporting permanent denaturation requires returning the pH 11 enzyme to pH 7 and showing that activity does not recover. Do not accept: "yes, because the rate is only 3 µmol/min"; a claim with no reference to a missing recovery test; asserting denaturation from the boiled row, which was a different treatment.
   Show a 4/4 response

   a The independent variable is the pH of the buffer the reaction runs in. The boiled catalase is the control: the oxygen I measured came from folded, working enzyme and not from peroxide breaking down on its own, because those tubes gave 0 µmol O₂/min at every pH.

   b The rate climbs from 4 µmol/min at pH 3 to 48 at pH 7, then drops back to 3 at pH 11. Away from pH 7 the extra H⁺ or OH⁻ changes the charges on the R groups lining catalase's active site, breaking the ionic and hydrogen bonds that hold that site in shape. A misshapen site no longer fits hydrogen peroxide, so fewer enzyme–substrate complexes form each second and the rate falls.

   c I would put pH on the x-axis from 3 to 11 and initial rate on the y-axis from 0 to 50 µmol O₂/min, label both axes with those units, plot (3, 4), (5, 22), (7, 48), (9, 20) and (11, 3), and connect them into a single peak at pH 7.

   d The data are not enough. All the table shows is that activity is low at pH 11, and reversible inhibition would look exactly the same. To support "permanently denatured" I would have to take the enzyme out of pH 11 buffer, put it back at pH 7, and show that the rate does not come back.

2. Data: standard free-energy changes for three reactions in a plant cell.

   Reaction	ΔG (kJ/mol)
   1. glucose + fructose → sucrose + H2O	+27.0
   2. ATP + H2O → ADP + Pi	−30.5
   3. activation of one amino acid for protein synthesis	+34.0

   Using the values in the table, answer (a) through (d).

   1. Identify which reactions in the table are endergonic, and describe one structural feature of ATP that makes reaction 2 exergonic.
   2. Explain how coupling reaction 2 to reaction 1 allows a plant cell to make sucrose even though reaction 1 by itself will not proceed.
   3. Calculate ΔG for reaction 1 coupled to one ATP hydrolysis, and for reaction 3 coupled to one ATP hydrolysis. Show your setup and give units.
   4. Justify, using your calculated values, the claim that reaction 3 requires the hydrolysis of two ATP rather than one.
   Your response
   Scoring notes

   • (a) Accept: reactions 1 and 3 are endergonic (positive ΔG). Feature: the three phosphate groups sit in a row and each carries a negative charge, so they repel one another; removing the terminal phosphate relieves that repulsion, and ADP and inorganic phosphate are more stable in water than ATP was. Do not accept: "ATP's bonds store a lot of energy" with nothing structural; identifying reaction 2 as endergonic; naming the reactions with no described feature of ATP.
   • (b) Accept: ATP transfers its terminal phosphate onto one of the sugars, and that phosphorylated intermediate is reactive enough to proceed; because the two steps share the intermediate they occur as a single reaction, so only the sum of the two ΔG values determines whether it runs. Do not accept: "ATP gives the reaction energy" with no intermediate and no reference to summing; a correct general statement about coupling that is never applied to sucrose.
   • (c) Accept: reaction 1, \(+27.0 + (-30.5) = -3.5\) kJ/mol; reaction 3, \(+34.0 + (-30.5) = +3.5\) kJ/mol. Setup must be shown and units are required. Do not accept: values with no setup; a sign error that reports reaction 1 as +3.5; answers with no units.
   • (d) Accept: with one ATP, reaction 3 is still +3.5 kJ/mol: positive, so it remains endergonic and will not proceed; with two, \(+34.0 + 2(-30.5) = -27.0\) kJ/mol, which is negative and therefore spontaneous. The evidence cited must be the sign of a computed sum. Do not accept: "two ATP give more energy" with no value; restating part (c) without drawing the conclusion about spontaneity.
   Show a 4/4 response

   a Reactions 1 and 3 are endergonic: both have a positive ΔG. ATP's three phosphate groups sit in a row and each carries a negative charge, so they repel one another. Cutting off the terminal phosphate relieves that repulsion, and ADP plus free phosphate are more stable in water than ATP was, so the hydrolysis releases free energy.

   b The cell does not run the two reactions side by side and hope. ATP hands its terminal phosphate to one of the sugars first, and that phosphorylated sugar is unstable enough to react with the other one. Because both steps go through the same intermediate they are really one reaction, so what decides whether sucrose gets made is the sum of the two ΔG values, not reaction 1's value on its own.

   c Reaction 1 with one ATP: \(+27.0 + (-30.5) = -3.5\) kJ/mol. Reaction 3 with one ATP: \(+34.0 + (-30.5) = +3.5\) kJ/mol.

   d With one ATP, reaction 3 still comes out at +3.5 kJ/mol. That is positive, so the coupled reaction is still endergonic and will not run on its own. With two ATP, \(+34.0 + 2(-30.5) = -27.0\) kJ/mol, which is negative and therefore spontaneous. The sign of that sum is the evidence that one ATP is not enough and two are.

3. Model: a thylakoid membrane drawn in cross-section, with the stroma above the membrane and the thylakoid lumen below it.

   Embedded in the membrane from left to right are photosystem II, plastoquinone, the cytochrome complex, plastocyanin, photosystem I, and, at the far right, ATP synthase. An arrow labeled 2 H₂O → 4 H⁺ + 4 e⁻ + O₂ points from the lumen into photosystem II. Curved arrows at the cytochrome complex show H⁺ moving from the stroma into the lumen. On the stroma side, NADP⁺ reductase is drawn accepting electrons from photosystem I. A pair of labels gives the stroma as pH 8.0 and the lumen as pH 5.0 in a leaf that has been illuminated for several minutes.

   Using the model, answer (a) through (d).

   1. Identify the source of the electrons that replace those lost by photosystem II, and describe where in the model protons accumulate.
   2. Explain how electron flow through the model generates the pH difference shown between the stroma and the lumen.
   3. Calculate the ratio of the H⁺ concentration in the lumen to the H⁺ concentration in the stroma in the illuminated leaf. Show your setup.
   4. A chemical is added that makes the thylakoid membrane freely permeable to H⁺. Justify the claim that ATP synthesis will fall sharply even though electrons continue to flow from photosystem II to NADP⁺.
   Your response
   Scoring notes

   • (a) Accept: the electrons come from water, split by photolysis on the lumen side (2 H₂O → 4 H⁺ + 4 e⁻ + O₂); protons accumulate in the thylakoid lumen. Do not accept: "from the sun," "from NADPH," or "from chlorophyll"; naming water without saying which compartment the protons build up in.
   • (b) Accept: two sources with directions stated; photolysis releases H⁺ directly into the lumen, and the cytochrome complex uses the energy of electrons falling through the chain to pump H⁺ from the stroma into the lumen; both raise lumen H⁺ and lower stroma H⁺. Do not accept: "the chain builds a gradient" with no mechanism or no direction; reversing the two compartments.
   • (c) Accept: pH 5.0 gives [H⁺] = 1 × 10⁻⁵ M and pH 8.0 gives 1 × 10⁻⁸ M, so the ratio is \((1 \times 10^{-5})/(1 \times 10^{-8}) = 1 \times 10^{3}\), i.e. 1000 : 1 with the lumen higher. Do not accept: 8/5 or a difference of 3 reported as the ratio; an answer that does not say which compartment is more concentrated; no setup shown.
   • (d) Accept: ATP synthase is driven by H⁺ flowing down the gradient through it, so if H⁺ can cross the membrane anywhere the gradient dissipates as fast as it is built, few protons take the route through the synthase, and little ADP is phosphorylated; electron transport is a separate process and continues because no carrier has been blocked. Do not accept: "the chemical breaks ATP synthase"; a claim that never identifies the gradient as the energy source for ATP synthesis.
   Show a 4/4 response

   a The replacement electrons come from water. In the model, water is split on the lumen side into 4 H⁺, 4 e⁻, and O₂, and those electrons go into photosystem II. The protons build up inside the thylakoid lumen, which is the compartment labeled pH 5.0.

   b Two things load the lumen with H⁺. Splitting water dumps protons straight into it, and as electrons fall from photosystem II through plastoquinone to the cytochrome complex, that complex uses the released energy to pump more H⁺ out of the stroma and into the lumen. Both moves raise lumen H⁺ and lower stroma H⁺, which is why the lumen sits at pH 5.0 and the stroma at pH 8.0.

   c pH 5.0 means [H⁺] = 1 × 10⁻⁵ M and pH 8.0 means [H⁺] = 1 × 10⁻⁸ M, so the ratio is \((1 \times 10^{-5})/(1 \times 10^{-8}) = 1 \times 10^{3}\). The lumen is 1000 times more concentrated in H⁺ than the stroma.

   d ATP synthase only works because protons are forced back into the stroma through it. If the membrane leaks H⁺ everywhere, the gradient drains as fast as the chain builds it, so hardly any protons take the route through the synthase and ATP output collapses. Nothing has touched the electron carriers, so electrons still reach NADP⁺, which is what shows ATP synthesis depends on the gradient, not on electron flow alone.

4. Data: pipette readings (mL) from three respirometers in a 25 °C water bath, read at 0 and 15 minutes. Each vial contains KOH to absorb CO₂, and the two seed vials each hold 20.0 g of peas.

   Vial	Reading at 0 min	Reading at 15 min
   Germinating peas	0.80	0.35
   Dormant peas	0.82	0.79
   Glass beads only	0.78	0.80

   Using the data, answer (a) through (d).

   1. Identify the independent variable in the investigation, and describe the purpose of the vial containing only glass beads.
   2. Explain the relationship between the metabolic state of the peas and the volume of oxygen consumed.
   3. Calculate the corrected rate of oxygen consumption of the germinating peas in mL O₂ per gram per minute. Show your setup and give units.
   4. Justify the claim that the difference between the two seed vials is caused by respiration rather than by the amount of seed used.
   Your response
   Scoring notes

   • (a) Accept: the independent variable is the metabolic state of the vial's contents (germinating peas, dormant peas, or no living tissue). The bead vial is a control for volume changes caused by shifts in room temperature and barometric pressure over the 15 minutes; its drift is subtracted from the other vials' readings. Do not accept: naming oxygen consumption (the dependent variable) as the independent variable; "the beads are the control" with no statement of what they control for.
   • (b) Accept: germinating seeds are synthesizing proteins, building walls and dividing cells, so their ATP demand is high; most of that ATP comes from oxidative phosphorylation, and the electron transport chain runs only while oxygen accepts electrons at the end, so a high ATP demand produces a high rate of oxygen consumption. Dormant seeds have very low ATP demand and consume little. Do not accept: "germinating seeds respire more," which restates the data; a mechanism that never mentions ATP demand or oxygen's role as final electron acceptor.
   • (c) Accept: drift \(= 0.80 - 0.78 = +0.02\) mL; germinating change \(= 0.35 - 0.80 = -0.45\) mL; corrected \(= -0.45 - 0.02 = -0.47\) mL, so 0.47 mL of O₂ was consumed; rate \(= 0.47/(20.0 \times 15) = 1.57 \times 10^{-3}\) mL O₂ g⁻¹ min⁻¹. Accept 1.6 × 10⁻³ if the setup shows the correction. Units are required. Do not accept: 1.5 × 10⁻³ with no correction shown; dividing by time or by mass alone; a bare number.
   • (d) Accept: mass was a controlled variable, both seed vials held 20.0 g, so it cannot account for 0.47 mL versus 0.05 mL, a difference of about ninefold; the only variable that differed between those two vials is whether the seeds were germinating. Do not accept: repeating the two volumes with no reference to the controlled mass; "germinating seeds are more active" as the entire justification.
   Show a 4/4 response

   a The independent variable is what is in the vial: germinating peas, dormant peas, or just beads. The bead vial has no living tissue, so anything that changes its reading has to be the room warming or the pressure shifting rather than respiration. It drifted +0.02 mL in 15 minutes, and I subtract that drift from the other two vials.

   b Germinating seeds are building tissue: transcribing genes, making proteins, dividing cells, and all of that costs ATP. Most of that ATP comes from oxidative phosphorylation, and the electron transport chain can only keep running while oxygen accepts electrons at the end, so a high ATP demand means a high oxygen demand. Dormant seeds are barely metabolizing, so they take up almost none.

   c Drift \(= 0.80 - 0.78 = +0.02\) mL. Germinating peas: \(0.35 - 0.80 = -0.45\) mL, corrected to \(-0.45 - 0.02 = -0.47\) mL, so 0.47 mL of O₂ was consumed. Rate \(= 0.47/(20.0 \times 15) = 1.57 \times 10^{-3}\) mL O₂ g⁻¹ min⁻¹.

   d Mass was held constant at 20.0 g in both seed vials, so it cannot be why germinating peas used 0.47 mL and dormant peas only 0.05 mL: about nine times less. The one thing that differed between those two vials is whether the seeds were germinating, so the evidence supports metabolic state as the cause.

5. Data: relative concentrations of two Calvin-cycle intermediates in illuminated Chlorella. The light was switched off at time 0; CO₂ was held at 1% throughout.

   Time (s)	−30	0	15	30	60
   3-PGA (relative units)	100	100	145	168	175
   RuBP (relative units)	100	100	42	18	8

   Using the data, answer (a) through (d).

   1. Identify the two products of the light reactions that the Calvin cycle requires, and describe what happens to the concentration of RuBP after the light is switched off.
   2. Explain why 3-PGA accumulates over the same interval in which RuBP falls.
   3. Predict the direction of change in 3-PGA and in RuBP if instead the light were left on and CO₂ were removed from the chamber at time 0.
   4. Justify your prediction in (c) using the sequence of reactions in the Calvin cycle.
   Your response
   Scoring notes

   • (a) Accept: ATP and NADPH. RuBP falls steeply, from 100 relative units to 42 within 15 s and to 8 by 60 s, a 92% decrease. Do not accept: naming O₂, G3P, or glucose as a required product of the light reactions; "RuBP changes" with no direction; a restatement of the stem.
   • (b) Accept: with the light off no new ATP or NADPH is made, so the reduction of 3-PGA to G3P stalls and 3-PGA stops being consumed, while rubisco, which needs no light, keeps fixing the CO₂ still present onto the RuBP already in the stroma, making more 3-PGA and using up RuBP, which cannot be regenerated because regeneration also costs ATP. Do not accept: "the Calvin cycle stops," which the data contradict; an answer that never links ATP or NADPH to a specific step of the cycle.
   • (c) Accept: 3-PGA falls and RuBP rises. Both directions are required. Do not accept: "the cycle slows down"; only one of the two directions; "no change."
   • (d) Accept: CO₂ is rubisco's other substrate, so removing it stops carboxylation and no new 3-PGA is formed; because the light is still on, ATP and NADPH keep arriving and the existing 3-PGA continues to be reduced to G3P and drained away, while G3P still feeds regeneration, so RuBP is rebuilt with nothing to consume it and accumulates. Do not accept: restating the prediction in other words; a justification that does not name CO₂ as rubisco's substrate or does not use the continued supply of ATP and NADPH.
   Show a 4/4 response

   a The Calvin cycle needs ATP and NADPH from the light reactions. Once the light goes off, RuBP drops fast, from 100 relative units to 42 in the first 15 seconds and down to 8 by 60 seconds, a 92% fall.

   b In the dark there is no new ATP or NADPH, so the step that reduces 3-PGA to G3P stalls and 3-PGA stops getting used up. Rubisco does not need light, though, so it keeps attaching the CO₂ still in the chamber onto the RuBP already sitting in the stroma. That makes more 3-PGA while consuming RuBP, and rebuilding RuBP also costs ATP, so the RuBP pool drains and is never refilled. That is why one rises 75% while the other collapses.

   c 3-PGA would fall and RuBP would rise: the opposite of what the table shows.

   d CO₂ is rubisco's other substrate, so taking it away stops carboxylation and no new 3-PGA is made. The light is still on, so ATP and NADPH keep arriving and the 3-PGA already there keeps being reduced to G3P, draining that pool. The G3P still feeds the regeneration phase, so RuBP gets rebuilt with nothing to carboxylate it and piles up. That is the mirror image of the dark experiment, which is what makes the prediction consistent with the model.

Lesson 4.1 · Unit 4 · CED topic 4.1

Cell communication: contact, local, and long-distance signals

A cell is a sealed bag of chemistry. Nothing inside a liver cell can feel you start to sprint. Multicellular life works anyway because cells send chemical messages, and every message needs three things: a cell that releases a signal molecule (the ligand), a cell carrying a receptor shaped to bind it, and a distance the molecule can actually cross.

That last item sorts signaling into types. Touching cells hand molecules over directly; cells 20 nm apart let one diffuse across; cells a meter apart need the bloodstream. Even bacteria do this: they are unicellular, but they are not alone.

Lesson 4.2 · Unit 4 · CED topics 4.2–4.3

Signal transduction: reception, transduction, response

Binding is not responding. Epinephrine never enters a liver cell, yet within seconds that cell is dumping glucose into the blood. Something has to carry the news across the membrane, and along the way it has to get much louder: a hormone present at \(10^{-10}\) M has to move millions of molecules of sugar.

Every pathway solves this the same way, in three stages the exam expects you to name.

Lesson 4.3 · Unit 4 · CED topic 4.4

When a pathway changes: mutations, drugs, and toxins

A pathway with six steps has six places to break, and the exam likes this precisely because a student who has memorized the word "transduction" cannot answer, while a student who knows the order of the steps can. The reasoning is always the same: if you can supply something downstream of a suspected break and the cell responds, the break is upstream of what you supplied.

Lesson 4.4 · Unit 4 · CED topic 4.5

Feedback mechanisms and homeostasis

Homeostasis is not constancy. Your blood glucose is never at 90 mg/dL: it is always a little above or below and always being pushed back. Every one of these systems has the same four parts: a set point, a sensor that detects the deviation, a control center, and an effector that does something about it.

What makes the loop work is the sign of what the effector does to the original stimulus.

Lesson 4.5 · Unit 4 · CED topic 4.6

The cell cycle: interphase, mitosis, and cytokinesis

One cell becomes two, and each of the two must carry a complete and identical copy of the genome. Getting that right means copying every chromosome exactly once, holding the copies together until the whole set is attached to the spindle, and only then pulling them apart. Almost all of that work happens before mitosis starts, in a typical dividing cell roughly 90% of the cycle is interphase.

Lesson 4.6 · Unit 4 · CED topic 4.7

Regulation of the cell cycle: checkpoints, cyclins, and cancer

A cell that divides with damaged DNA, or with one chromosome unattached to the spindle, does not make two working cells: it makes two broken ones. So the cycle is not a clock that simply runs. It is a series of gates, and at each one the cell asks a specific question and refuses to proceed until the answer is yes.

Unit 4 quiz · 15 multiple-choice · 5 free-response

Unit 4 quiz: Cell Communication and Cell Cycle

Fifteen multiple-choice questions and five short free-response questions across all six lessons: signal types and receptors, the three stages of transduction, pathways broken by mutation or toxin, feedback and homeostasis, and the cell cycle with its checkpoints. Click an option to see why each choice is right or wrong, then time yourself on each free response before opening the model.

Multiple choice

1. Data: luminescence of Vibrio fischeri cultures grown in seawater medium and assayed at five cell densities.

   Cell density (cells/mL)	2×105	2×106	8×106	2×107	8×107
   Light output (relative units)	0.2	2.0	8.0	60	800
   Light output per 106 cells	1.0	1.0	1.0	3.0	10.0

   Which conclusion about the threshold for light production is best supported by the data?

   The threshold is 2×10⁵ cells/mL, because that is the lowest density at which any light is detected There is no threshold; total light output rises steadily with density across the whole range, so the relationship is simply proportional The threshold lies between 8×10⁶ and 2×10⁷ cells/mL, because light output per 10⁶ cells is flat at 1.0 below that range and rises above it, so individual cells are changing what they express The threshold is 8×10⁷ cells/mL, because that is where total light output is greatest

   Light at the lowest density is 0.2 units: exactly what 0.2 million cells glowing at the baseline rate of 1.0 per 10⁶ would give. A faint glow proportional to cell number is what an unregulated population looks like, so this density shows no switching on at all.

   Total output does rise across the whole range, but not proportionally. From 8×10⁶ to 8×10⁷ the density rises tenfold while light rises \(800/8.0 = 100\)-fold. A proportional relationship would hold the per-cell row constant, and it does not.

   Correct. Dividing light by density removes the effect of simply adding more cells. The per-cell row is 1.0, 1.0, 1.0, then 3.0 and 10.0, so somewhere between 8×10⁶ and 2×10⁷ cells/mL each cell begins producing more light: the autoinducer has reached a concentration that switches the luminescence genes on.

   8×10⁷ is where output is highest, not where it changes character. A threshold is the point at which the behavior of the system changes, and the per-cell row shows that change happening an order of magnitude earlier.

2. Epinephrine released into the bloodstream circulates past every cell in the body, yet only some cells respond. Which statement best accounts for this?

   Only cells carrying a receptor that binds epinephrine can transduce the signal, so the receptor rather than the hormone determines which cells respond Epinephrine is degraded before it reaches most tissues, so only cells near the adrenal glands receive enough to respond Epinephrine crosses the plasma membrane only in target cells, whose membranes contain more cholesterol Cells that do not respond lack the gene for epinephrine and so cannot produce the signal themselves

   Correct. Specificity comes from the complementary fit between ligand and receptor, exactly as it does between substrate and enzyme. One ligand can even produce different responses in different cell types, because what lies downstream of the receptor differs from tissue to tissue.

   Endocrine signals are built to travel: that is the point of releasing them into the blood. Epinephrine reaches the heart in about 20 seconds and acts on tissues throughout the body, so distance from the gland is not what sorts responders from non-responders.

   Epinephrine is hydrophilic and does not cross the bilayer in any cell; its receptor is a membrane protein. The signals that diffuse through membranes are small hydrophobic ones such as steroid and thyroid hormones, and cholesterol buffers fluidity rather than gating entry.

   Target cells do not make the hormone: the adrenal glands do. Responding to a signal requires a receptor, not the gene for the signal, and confusing the sending cell with the receiving cell is the misconception being tested here.

3. Testosterone's receptor is inside the cell, while insulin's receptor spans the plasma membrane. Which property of the two signals best explains the difference?

   Testosterone circulates at a much higher concentration, so it can saturate an intracellular receptor Testosterone is a small hydrophobic steroid and diffuses through the hydrophobic core of the bilayer, while insulin is a large hydrophilic protein that cannot Insulin is a steroid and testosterone is a peptide, so each receptor sits where its ligand can reach it Testosterone acts faster than insulin, and intracellular receptors give faster responses than membrane receptors

   Concentration determines how much receptor is occupied, not where the receptor can usefully sit. A hydrophilic hormone at any concentration still cannot cross the hydrophobic interior of the bilayer, which is what sets receptor location.

   Correct. The bilayer's interior is a hydrophobic barrier: small nonpolar molecules pass freely, large polar ones do not. Testosterone diffuses in and binds an intracellular receptor that acts as a transcription factor; insulin must be detected at the surface and have its message relayed inward by a phosphorylation cascade.

   The two categories are swapped: insulin is a protein hormone and testosterone is the steroid. The reasoning in the second half is right, which is what makes this a tempting choice: check the chemical class before you apply the rule.

   The speeds are reversed. Steroid responses require transcription and translation and take hours; membrane-receptor cascades change enzyme activity within seconds. Slow and long-lasting is the trade-off steroid signaling makes for changing gene expression.

4. Which feature of a phosphorylation cascade allows a signal both to be amplified and to be switched off quickly once the ligand is gone?

   Each kinase phosphorylates exactly one molecule of the next kinase, and the whole chain is degraded after a single use Each activated kinase phosphorylates many molecules of the next kinase, while protein phosphatases continuously remove those phosphates cAMP is a protein, so proteases destroy it as soon as the ligand leaves the receptor The receptor is internalized, which reverses the phosphorylations that have already occurred

   A one-to-one relay would transmit the signal without amplifying it, and destroying the chain after each use would make the response impossible to repeat. Cascades are built from reusable enzymes precisely so the cell can respond again a minute later.

   Correct. The multiplication at each step is the amplification, and the phosphatases are the off switch: they are always working, so as soon as new phosphorylation stops the existing phosphates are stripped and the relay proteins return to their inactive shapes.

   cAMP is a nucleotide derived from ATP, not a protein, so proteases have no effect on it. It is degraded to AMP by phosphodiesterase, which is a real off switch, but the question asks about the kinase cascade itself.

   Receptor internalization does end new signaling at the top of the pathway, but it does nothing to proteins already phosphorylated further down. Reversing those requires an enzyme that removes phosphate groups, which is what a phosphatase does.

5. Data: glycogen phosphorylase activity (relative units) measured after liver cells were broken open and separated into a membrane fraction and a soluble fraction containing the enzyme.

   Treatment	Activity
   Intact cells + epinephrine	100
   Soluble fraction alone	4
   Soluble fraction + epinephrine	5
   Membrane fraction + epinephrine; boiled supernatant then added to soluble fraction	96
   Membrane fraction, no epinephrine; boiled supernatant then added to soluble fraction	6

   Which conclusion is best supported by these results?

   Epinephrine activates glycogen phosphorylase directly, because adding it to the soluble fraction raises activity above the untreated soluble fraction Epinephrine acts at the membrane and causes release of a small, heat-stable soluble substance that activates the enzyme, because activity appears only when the membranes were exposed to the hormone first The membrane fraction contains the glycogen phosphorylase, because the transferred supernatant produced 96 units of activity The substance released from the membranes is a protein, because its activity survived boiling

   Four units becomes five: a change far smaller than the 96 obtained through the membrane route, and within the noise of the assay. If the hormone acted on the enzyme directly, hormone plus enzyme alone should have given close to the intact-cell value of 100.

   Correct. Activity appears only in the treatment where membranes met the hormone first and the supernatant was then transferred; membranes without hormone gave 6. That is the logic that identified cyclic AMP: the hormone never enters the cell, and something small and diffusible carries the message inward.

   The enzyme was in the soluble fraction, which is where the assay was performed in every row. What was transferred from the membrane treatment was the small molecule in the supernatant, and it produced activity only because the soluble fraction supplying the enzyme was already there.

   Boiling denatures proteins, so surviving it argues that the messenger is not a protein: it is a small, heat-stable molecule. Reading the evidence backwards is exactly the trap; ask what the treatment would have destroyed.

6. Data: glucose release (µmol/min per 10⁶ cells) from four liver-cell lines, each assayed untreated, with the hormone, with forskolin (which activates adenylyl cyclase directly), and with a membrane-permeable cAMP analog.

   Treatment	Wild type	Line A	Line B	Line C
   None	2.0	2.1	1.9	2.0
   + hormone	40.0	2.4	2.2	2.3
   + forskolin	39.0	38.5	2.5	2.4
   + cAMP analog	38.0	38.0	37.5	2.6

   In which line is the defect best placed at adenylyl cyclase, and on what evidence?

   Line A, because it fails to respond to the hormone Line C, because it fails to respond to every treatment tested Line B, because it fails with the hormone and with forskolin but responds fully to the cAMP analog: forskolin requires a working cyclase and the analog does not Wild type, because its forskolin response (39.0) is slightly lower than its hormone response (40.0)

   Line A does fail with the hormone, but it responds fully to forskolin (\(38.5/2.1 = 18\)-fold), which acts directly on adenylyl cyclase. A working cyclase places the break above it, at the receptor or the G protein.

   Line C fails with everything including the cAMP analog (\(2.6/2.0 = 1.3\)-fold), so the break is at or below protein kinase A. The cAMP analog enters the pathway below the cyclase, so a cell that cannot respond to it has a defect further downstream still.

   Correct. Line B gives \(2.5/1.9 = 1.3\)-fold with forskolin but \(37.5/1.9 = 19.7\)-fold with the analog. Forskolin needs a functional cyclase to make cAMP; the analog is cAMP and skips the cyclase entirely. The one component that separates the two treatments is adenylyl cyclase.

   Forty versus thirty-nine is assay-to-assay variation, not a defect; the wild type is the reference every line is compared against, and it responds about twentyfold to all three activating treatments.

7. Cholera toxin modifies the α subunit of a G_s protein so that it can no longer hydrolyze its bound GTP. Which description of the effect is correct?

   A loss of function: the G protein can no longer bind GTP, so the pathway cannot be switched on A gain of function at the receptor: the toxin acts as an agonist, binding the receptor and triggering the normal response A loss of function at adenylyl cyclase: cAMP falls, and the intestinal cells stop secreting chloride A gain of function: the G protein is trapped in its active, GTP-bound form and keeps stimulating adenylyl cyclase even with no ligand bound to the receptor

   The toxin blocks GTP hydrolysis, not GTP binding: the difference between failing to switch on and failing to switch off. Because the G protein's own GTPase activity is its timer, removing it leaves the protein permanently on.

   An agonist binds the receptor's ligand site and triggers the normal, reversible response. Cholera toxin acts one step below the receptor, on the G protein itself, which is why deleting the receptor would not protect the cell from it.

   The clinical effect runs the other way: cAMP rises sharply, chloride channels open, water follows the salt into the gut, and the result is severe diarrhea. A pathway locked on produces too much signal, not too little.

   Correct. A component stuck in its active state produces the response constitutively, no ligand required. That is the definition of a gain-of-function change, and it is why cholera's effects continue long after any hormone would have been cleared.

8. A patient taking a β-adrenergic antagonist shows a smaller rise in heart rate during exercise than an untreated person. Which additional result would best confirm that the drug acts at the receptor rather than further down the pathway?

   Cardiac cells from the treated patient still respond normally to a membrane-permeable cAMP analog The treated patient's blood epinephrine concentration rises normally during exercise The treated patient's resting heart rate is within the normal range Cardiac cells from the treated patient fail to respond to epinephrine and to the cAMP analog alike

   Correct. This is the bypass logic the exam rewards: the analog enters the pathway below the receptor, the G protein, and the cyclase. A full response to it shows everything from cAMP downward is intact, so the block must lie above, at the receptor the drug occupies.

   Normal epinephrine only shows that the signal is being released; it says nothing about where the receiving cell fails. Both a receptor block and a downstream block would leave circulating hormone untouched.

   Resting heart rate is set largely by parasympathetic input and pacemaker activity, so a normal resting value is compatible with a block at any step of the β-adrenergic pathway. It does not localize anything.

   Failing both would place the break at or below cAMP: the opposite of the conclusion being tested. That pattern would argue the drug acts on protein kinase A or downstream, not on the receptor.

9. Data: blood glucose (mg/dL) measured hourly after the same standard meal in two adults.

   Time after meal (h)	0	1	2	3	4
   Person X (mg/dL)	85	135	100	88	85
   Person Y (mg/dL)	118	215	195	178	165

   Between hour 1 and hour 3, at what average rate does Person Y's blood glucose fall, and what does the comparison with Person X indicate?

   18.5 mg/dL per hour: slower than X's 23.5 mg/dL per hour, indicating that Y's negative feedback loop corrects the excursion less effectively 37 mg/dL per hour: faster than X's 47 mg/dL per hour, indicating that Y clears glucose more effectively 16.7 mg/dL per hour: the slope from the peak to hour 4, which is the standard measure of insulin response 18.5 mg/dL per hour: the same rate as X once Y's higher baseline is taken into account, so the two loops are equally effective

   Correct. \((178 - 215)/(3 - 1) = -18.5\) mg/dL per hour for Y and \((88 - 135)/(3 - 1) = -23.5\) for X. Y also starts higher and peaks higher, and at hour 4 has corrected only \((97 - 47)/97 = 51.5\%\) of its excursion against X's 100%: the loop runs but its gain is low, which is what insulin resistance looks like.

   These are the total drops over the two-hour interval, not rates: \(215 - 178 = 37\) and \(135 - 88 = 47\). A rate requires dividing by the elapsed time, and note that the larger number here belongs to X, so even the comparison is stated backwards.

   \((165 - 215)/3 = -16.7\) mg/dL per hour is the slope from hour 1 to hour 4, a different interval from the one the question names. There is also no standard that privileges the four-hour slope; you must read which interval is asked for.

   The rates are genuinely different, 18.5 against 23.5, and a higher baseline does not make a slower correction equivalent. Measured against each person's own excursion the gap widens rather than closes, because Y rose 97 mg/dL and X only 50.

10. Which of the following is controlled by positive feedback, and what distinguishes it from the others?

    Shivering when body temperature falls, because the response drives the variable further from the set point Oxytocin release during labor: cervical stretch triggers oxytocin, oxytocin strengthens contractions, and contractions increase stretch, so the response amplifies its own stimulus until birth ends the loop ADH release when blood osmolarity rises, because water reabsorption drives osmolarity still higher Insulin release after a meal, because the response continues until blood glucose is exhausted

    Shivering is negative feedback: the muscle activity generates heat, which raises body temperature back toward the set point and therefore removes the stimulus that started it. The description of the effect here is reversed.

    Correct. The output of the loop, stronger contractions, increases the very stimulus that produced it, so the process accelerates instead of settling at a value. Positive loops have no stable point to sit at, which is why they are reserved for fast, committed, one-way transitions ended by an outside event.

    ADH inserts aquaporins into the collecting duct so more water is reabsorbed into the blood, which dilutes it and lowers osmolarity toward the set point. That is negative feedback, and the choice states the direction backwards.

    Insulin is negative feedback: glucose uptake and glycogen storage lower blood glucose, which removes the stimulus for further insulin release. The loop stops at the set point near 90 mg/dL, not at exhaustion: running glucose to zero would be lethal.

11. Data: a partly completed table for a somatic cell of an organism with 2n = 16 dividing by mitosis. One cell of the table has been left blank.

    Stage	G1	G2	Metaphase	Anaphase	Each daughter cell
    Chromosomes	16	16	16	?	16
    DNA molecules (chromatids)	16	32	32	32	16

    What value belongs in the cell marked ?, and why?

    16, because the chromosome number already doubled during S phase and cannot double again 32, because the centromeres split in anaphase and each separated sister chromatid is now counted as a full chromosome 8, because anaphase separates homologous chromosomes and halves the count 64, because both the DNA content and the chromosome number double as the chromatids move to opposite poles

    The chromosome number does not double in S phase: that is the single most common error on this topic, and the table itself shows it: the count stays 16 from G₁ through metaphase while DNA molecules go from 16 to 32, because the two chromatids share one centromere.

    Correct. Count centromeres for chromosomes and chromatids for DNA. When cohesin is cleaved in anaphase, each chromatid gains its own centromere and therefore counts as a chromosome, so 16 becomes 32 with no new DNA synthesized at all. Cytokinesis then returns each daughter to 16.

    Homologous chromosomes are separated in meiosis I, not in mitosis. Mitotic anaphase separates sister chromatids, and the halving to 16 per cell happens at cytokinesis when the cytoplasm divides, not by any event that reduces the count below the parental number.

    No DNA is replicated during anaphase; the DNA-molecule row correctly stays at 32 throughout M phase. Doubling both rows counts the same chromatids twice, once as DNA and once again as new chromosomes.

12. Which statement correctly compares cytokinesis in a plant cell with cytokinesis in an animal cell and explains the difference?

    The plant cell pinches in two with a cleavage furrow while the animal cell builds a cell plate, because animal membranes are more flexible The plant cell divides by binary fission because it has a cell wall, while the animal cell divides by mitosis The animal cell pinches in two using an actin–myosin contractile ring, while the plant cell builds a cell plate from Golgi vesicles at the midline, because a rigid cellulose wall cannot be drawn inward Both form a cleavage furrow, but the plant cell's furrow is reinforced with cellulose delivered by the Golgi

    The two mechanisms are swapped. Animal cells form the furrow; plant cells form the plate. Membrane flexibility is not the constraint either: the cellulose wall outside the plant membrane is what cannot be squeezed inward.

    Binary fission is prokaryotic: one circular chromosome, one origin, no spindle and no mitosis. Plant cells are eukaryotic and divide by mitosis followed by cell-plate formation; having a wall does not make a cell prokaryotic.

    Correct. The animal cell is pinched from the outside in by a contractile ring just inside the plasma membrane; the plant cell is divided from the inside out by Golgi vesicles that line up at the midline and fuse into a plate that joins the existing wall. The wall is the reason the new boundary must be built in place.

    A plant cell forms no furrow at all. The vesicles that carry wall material fuse into a disc that grows outward from the center: the opposite geometry from a furrow, which closes inward from the perimeter.

13. Figure: two traces on one set of axes for an early frog embryo; relative amount from 0 to 100 on the y-axis, time in minutes from 0 to 80 on the x-axis.

    The cyclin B trace rises in a straight line from 5 at 0 min to 95 at 36 min, then falls to 5 within about 5 minutes, and repeats the same pattern through a second cycle. The MPF activity trace is flat near 4 until 26 min, rises steeply to 94 by 32 min, holds through metaphase, and drops below 5 by 38 min.

    Which pair of rise rates is correct, and what does the comparison show about how MPF is controlled?

    Cyclin 2.5 and MPF 15.0 relative units per minute; MPF is synthesized six times faster than cyclin B is Cyclin 2.5 and MPF 2.8 relative units per minute; the two rise at nearly the same rate, so MPF simply tracks cyclin Cyclin 2.5 and MPF 15.0 relative units per minute; MPF activity is a threshold switch rather than a readout proportional to cyclin concentration Cyclin 95 and MPF 94 relative units per minute; both reach nearly the same maximum, so the controls are equivalent

    The rates are right, \((95-5)/36 = 2.5\) and \((94-4)/(32-26) = 15.0\), but the interpretation is not. MPF is not a molecule being synthesized; it is cyclin B bound to CDK1, and the trace shows its activity. CDK1 concentration is constant.

    2.8 comes from \((94-4)/32\), dividing by the time since 0 min rather than by the 6 minutes over which MPF actually rose. The trace is flat until 26 min, so those first 26 minutes are not part of the rise and must not be in the denominator.

    Correct. \((95-5)/36 = 2.5\) and \((94-4)/6 = 15.0\), a ratio of 6. Because activity climbs six times more steeply than the cyclin driving it, MPF cannot be a proportional readout: cyclin must accumulate past a threshold before enough CDK1 is bound, and then activation is nearly all-or-none.

    These are peak heights, not rates: a rate needs a change divided by an elapsed time. Matching maxima say nothing about control, and the interesting feature of the figure is precisely that the two traces have different shapes.

14. A tumor cell line carries two nonfunctional copies of the gene for p53. Which consequence follows most directly?

    Cells cannot assemble a mitotic spindle, so they arrest at the M checkpoint Cyclin B is no longer degraded, so cells are trapped in metaphase Cells can no longer enter G₀ and therefore divide only when growth factors are present Cells with damaged DNA are not arrested in G₁, so the damage is replicated in S phase and copied into both daughter cells

    p53 has no role in spindle assembly; the M checkpoint monitors kinetochore attachment and works independently of it. A p53-null cell builds a normal spindle, which is part of why the damage it carries is passed on so efficiently.

    Cyclin B degradation is triggered by the anaphase-promoting complex that MPF itself switches on, and it is what allows exit from mitosis. p53 acts much earlier, before S phase, and losing it does not trap cells in metaphase.

    Entering G₀ is what a cell does when it fails the G₁ checkpoint, so losing p53 makes cells less likely to stop, not more. Cancer cells are characteristically able to divide without the growth-factor signals normal cells require.

    Correct. DNA damage normally activates p53, which induces the CDK inhibitor p21 and halts the cycle in G₁ for repair or directs the cell to apoptosis. Without it a damaged cell proceeds into S phase, the damage becomes a fixed mutation, and every descendant inherits it.

15. Data: distribution of cells by stage in a human cell culture 6 hours after adding a drug that prevents microtubules from attaching to kinetochores, compared with an untreated culture. Values are percentages of 1,000 cells scored.

    Stage	Untreated (%)	Drug-treated (%)
    Interphase	88	34
    Prophase	6	8
    Metaphase	3	56
    Anaphase	1	1
    Telophase	2	1

    Which explanation best accounts for the distribution in the treated culture?

    The M checkpoint holds cells at metaphase until every kinetochore is attached to microtubules from both poles, so an unattached kinetochore keeps sending a wait signal and anaphase never begins The drug blocks DNA replication, so cells accumulate at the G₁ checkpoint before entering S phase The drug shortens prophase, so cells pass through it quickly and pile up in the following stage The drug prevents cyclin B from being degraded, so cells cannot exit mitosis once they have entered it

    Correct. Metaphase rises from 3% to 56% while anaphase and telophase stay at 1%, so cells are entering mitosis and stopping at exactly the transition the checkpoint guards. A single unattached kinetochore is enough to keep the wait signal on, which is why the cells sit at metaphase rather than proceeding with a chromosome missing.

    A block before S phase would raise the interphase percentage, and here it fell from 88% to 34%. The data show cells leaving interphase normally and being stopped later, which rules out an arrest at G₁.

    Prophase barely changed, 6% to 8%, and shortening a stage cannot by itself produce an eighteenfold accumulation in the next one. Cells pile up where they are held, not where they pass through quickly.

    Indestructible cyclin B does arrest cells in M phase, so the pattern is superficially similar, but the drug described acts on microtubule attachment, and the checkpoint it trips is what stops anaphase here. Match the mechanism to the stated action of the treatment, not just to the shape of the data.

Free response

1. Data: luminescence of Vibrio fischeri cultures held at a constant density of 5×10⁶ cells/mL. Each culture received a different concentration of synthetic autoinducer; light output was measured after 60 minutes.

   [Autoinducer] (nM)	0	1	5	20	50	100
   Light output (relative units)	0.5	0.6	1.2	22	180	240

   An investigator tested whether the autoinducer alone, rather than crowding, switches on luminescence. Using the investigation and the data, answer (a) through (d).

   1. Identify the independent and dependent variables, and describe the purpose of the 0 nM treatment.
   2. Explain the relationship between autoinducer concentration and light output shown by the data.
   3. Construct a graph of these data: place autoinducer concentration on the x-axis scaled from 0 to 100 nM and light output on the y-axis scaled from 0 to 250 relative units, label both axes with units, plot the six points, and connect them.
   4. Justify the claim that light output is controlled by autoinducer concentration rather than by cell density, using a feature of the investigation's design.
   Your response
   Scoring notes

   • (a) Accept: independent variable is autoinducer concentration in nM; dependent variable is light output in relative units. The 0 nM culture is the control: it establishes the baseline light produced by cells at this density with no autoinducer added (0.5 units), the value every treatment is compared against. Do not accept: the two variables reversed; "0 nM is the control" with no statement of what it establishes; naming cell density as the independent variable.
   • (b) Accept: output is essentially flat from 0 to 5 nM (0.5 to 1.2 units), rises steeply between 5 and 50 nM, and begins to level off by 100 nM: a threshold relationship rather than a proportional one, in which the luminescence genes stay off below a critical autoinducer concentration and switch on above it. Do not accept: "as autoinducer rises, light rises," which misses the threshold that is the feature of these data; an answer that cites no values from the table.
   • (c) Accept: x-axis = autoinducer concentration in nM from 0 to 100, evenly scaled; y-axis = light output in relative units from 0 to 250; both axes labeled with units; the six points (0, 0.5), (1, 0.6), (5, 1.2), (20, 22), (50, 180), (100, 240) plotted and connected. Do not accept: axes reversed; units omitted from the axis labels; a bar graph of unlabeled categories.
   • (d) Accept: cell density was held constant at 5×10⁶ cells/mL in every culture, making it a controlled variable that cannot account for the roughly 480-fold difference in output; the only variable that differed across the tubes was autoinducer concentration. Do not accept: repeating the trend described in (b); "the autoinducer caused it" with no reference to the constant density.
   Show a 4/4 response

   a The independent variable is the concentration of synthetic autoinducer in nM, and the dependent variable is light output in relative units. The 0 nM culture is the control: it shows how much light these cells make at this density with no autoinducer added; 0.5 units, so I have a baseline to compare every other tube against.

   b The relationship is a threshold, not a straight line. From 0 to 5 nM the light barely moves, 0.5 up to 1.2 units. Between 5 and 50 nM it jumps from 1.2 to 180, and by 100 nM it is 240 and flattening. Below a critical autoinducer concentration the luminescence genes stay off, and once the cells sense enough they switch on almost at once.

   c I would put autoinducer concentration on the x-axis from 0 to 100 nM and light output on the y-axis from 0 to 250 relative units, label both with those units, plot (0, 0.5), (1, 0.6), (5, 1.2), (20, 22), (50, 180) and (100, 240), and connect them into an S-shaped curve.

   d Every culture was held at 5×10⁶ cells/mL, so density was a controlled variable rather than something that changed with the light. Since tubes with identical density differed about 480-fold in output, crowding cannot be what turned the genes on: autoinducer concentration is the only thing that varied.

2. Blood glucose in a healthy adult is held near a set point of about 90 mg/dL. Answer (a) through (d).

   1. Describe the difference between negative feedback and positive feedback in terms of the effect the response has on the original stimulus.
   2. Explain how insulin and glucagon together return blood glucose to the set point both after it rises and after it falls.
   3. Predict what happens to the height of the glucose peak and to the time required to return to the set point in a person whose β cells release normal amounts of insulin but whose liver and muscle cells carry only about 30% of the normal number of insulin receptors.
   4. Justify your prediction in (c) by identifying which component of the feedback loop is impaired and stating what that does to the loop.
   Your response
   Scoring notes

   • (a) Accept: in negative feedback the response opposes the stimulus and drives the variable back toward the set point; in positive feedback the response reinforces the stimulus, so the variable moves further from its starting value until an event outside the loop ends the process. Do not accept: "negative is harmful and positive is helpful"; "negative decreases and positive increases" with no reference to the stimulus.
   • (b) Accept: rising glucose is detected by pancreatic β cells, which release insulin; insulin causes liver and muscle cells to take glucose up and store it as glycogen, so glucose falls and the stimulus for insulin is removed. Falling glucose triggers α cells to release glucagon, which drives glycogen breakdown and glucose release from the liver, so glucose rises. Do not accept: naming the two hormones with no effector action; reversing insulin and glucagon.
   • (c) Accept: the peak is higher than normal and the return to the set point takes longer. Both directional claims are required. Do not accept: "blood glucose is affected"; only one of the two claims; a prediction of no change.
   • (d) Accept: the impaired component is the effector arm, the target cells' response to insulin. The sensor and the signal are intact and insulin is released normally, but with about 30% of the receptors the same insulin concentration produces far less glucose uptake, so the loop's gain is low and the correction is slow and incomplete. Do not accept: "there is not enough insulin," which contradicts the stem; naming the receptor with no statement of the consequence for the loop.
   Show a 4/4 response

   a In negative feedback the response pushes back against whatever started it, so the variable heads back toward the set point. In positive feedback the response makes the original stimulus stronger, so the system runs further and further from where it started until something outside the loop stops it.

   b When glucose rises after a meal, β cells in the pancreas detect it and release insulin. Insulin makes liver and muscle cells pull glucose out of the blood and store it as glycogen, which brings the level down, and lowering glucose removes the signal for more insulin. When glucose drops too low, α cells release glucagon instead, the liver breaks glycogen back down and releases glucose, and the level climbs back toward 90 mg/dL.

   c The peak will be higher than a normal person's, and it will take longer than normal to get back down to about 90 mg/dL.

   d The sensor and the hormone are fine: this person releases normal insulin, so the problem is not detection or signaling. What is impaired is the effector arm: the target cells have only about 30% of the usual receptors, so the same amount of insulin triggers far less glucose uptake. A loop with a weak effector still runs, but with low gain, and low gain is exactly what a high peak and a slow correction look like.

3. Model: a signal-transduction pathway drawn as a vertical sequence of labeled boxes and arrows for a liver cell responding to epinephrine.

   At the top, epinephrine outside the cell binds a β-adrenergic receptor spanning the plasma membrane. An arrow leads to a G_s protein on the inner face of the membrane, labeled "1 activated receptor → 50 active G proteins." Each active G protein switches on one adenylyl cyclase, labeled "each cyclase → 300 cAMP." An arrow from cAMP leads to protein kinase A, labeled "4 cAMP bind to activate 1 PKA." From PKA an arrow leads to phosphorylase kinase, labeled "each PKA phosphorylates 120 phosphorylase kinase." A final arrow runs from phosphorylase kinase to glycogen phosphorylase and then to glucose leaving the cell. A side arrow shows phosphodiesterase converting cAMP to AMP.

   Using the model, answer (a) through (d).

   1. Identify the ligand and the second messenger in the model, and describe the role of the G_s protein.
   2. Explain how the model accounts for a liver cell responding to epinephrine present in the blood at about 10⁻¹⁰ M.
   3. Calculate the number of phosphorylase kinase molecules activated per activated receptor. Show your setup.
   4. A drug inhibits the phosphodiesterase shown in the model. Justify the claim that the response will last longer without its peak becoming larger.
   Your response
   Scoring notes

   • (a) Accept: the ligand is epinephrine and the second messenger is cAMP. The G_s protein is the relay between the activated receptor and adenylyl cyclase: the receptor switches it on, and it in turn switches on the cyclase. Do not accept: naming cAMP as the ligand or epinephrine as the second messenger; "the G protein passes the signal along" with no named partner on either side.
   • (b) Accept: each step activates many molecules at the next step, so the step factors multiply rather than add and one bound hormone molecule produces hundreds of thousands of downstream events; a concentration far too low to act directly is therefore sufficient. Do not accept: "the signal gets bigger" with no reference to multiplying step factors; a number quoted with no explanation.
   • (c) Accept: cAMP per receptor \(= 50 \times 300 = 15{,}000\); PKA activated \(= 15{,}000/4 = 3{,}750\); phosphorylase kinase \(= 3{,}750 \times 120 = 450{,}000\), i.e. \(4.5 \times 10^{5}\) per activated receptor. The setup must show the division at the cAMP-to-PKA step. Do not accept: \(50 \times 300 \times 4 \times 120 = 7.2 \times 10^{6}\), which multiplies where the model divides; a final number with no setup.
   • (d) Accept: phosphodiesterase is the enzyme that removes cAMP, so inhibiting it disables the off switch rather than the production step; cAMP already made persists and PKA stays active longer, but nothing in the drug's action increases the rate at which adenylyl cyclase produces cAMP, so the peak is still set by reception and transduction. Do not accept: "the drug makes more cAMP"; restating the claim; a justification that never states what phosphodiesterase does.
   Show a 4/4 response

   a The ligand is epinephrine and the second messenger is cAMP. The G_s protein sits just inside the membrane and works as a relay: the activated receptor switches it on, and it then switches on adenylyl cyclase.

   b Every arrow in the model multiplies instead of adding. One receptor turns on 50 G proteins, each cyclase makes 300 cAMP, and each PKA goes on to phosphorylate 120 more enzymes. Because those factors multiply, a single hormone molecule ends up moving hundreds of thousands of downstream molecules, which is how a hormone at 10⁻¹⁰ M can produce a whole-cell response.

   c cAMP made per receptor \(= 50 \times 300 = 15{,}000\). It takes 4 cAMP to activate one PKA, so PKA \(= 15{,}000/4 = 3{,}750\). Each PKA phosphorylates 120, so \(3{,}750 \times 120 = 450{,}000\), or \(4.5 \times 10^{5}\) phosphorylase kinase molecules per activated receptor.

   d Phosphodiesterase is the only thing in the model that gets rid of cAMP. Blocking it removes the off switch, so cAMP that has already been made hangs around and PKA keeps working after the hormone is gone: the response lasts longer. But the drug does nothing to the receptor, the G protein, or the cyclase, and those are what set how fast cAMP is made, so the highest level the response reaches is unchanged.

4. Data: 250 cells from an onion root tip scored by stage. For these cells the complete cell cycle takes 18 hours.

   Stage	Interphase	Prophase	Metaphase	Anaphase	Telophase
   Cells counted (of 250)	185	35	12	6	12

   Using the data, answer (a) through (d).

   1. Identify the stage in which these cells spend the most time, and describe the assumption that allows a count of cells in a stage to be converted into the time spent in that stage.
   2. Explain why so few cells are observed in anaphase.
   3. Calculate the number of minutes a cell spends in metaphase and the mitotic index for this root tip. Show your setup.
   4. A second root tip has a mitotic index of 12%. Justify the claim that the first tip is dividing more actively, using both mitotic indices.
   Your response
   Scoring notes

   • (a) Accept: interphase, with 185 of the 250 cells. The assumption is that the cells were fixed at random moments in their cycles, so the proportion of cells caught in a stage equals the proportion of the cycle that stage occupies. Do not accept: "most cells are in interphase" as the answer to the second task, which restates the data instead of stating the assumption; naming a stage with no assumption given.
   • (b) Accept: anaphase is the briefest stage, once cohesin is cleaved the separation of sister chromatids to the poles takes only a few minutes, so at any instant only a small fraction of the population is in it. Do not accept: "anaphase is rare"; "the cells were not dividing"; an answer that never connects a short duration to a small count.
   • (c) Accept: cycle \(= 18 \times 60 = 1080\) min; metaphase \(= (12/250) \times 1080 = 51.84\) min (accept 51.8 or about 52 min with correct setup); mitotic index \(= (250 - 185)/250 = 65/250 = 26.0\%\). Units required for the time. Do not accept: reporting 12/250 as the time; a mitotic index computed from metaphase cells alone; an answer with no setup.
   • (d) Accept: 26.0% against 12% means that at any instant more than twice as large a fraction of the first tip's cells are in M phase; because the mitotic index stands for the fraction of the cycle spent dividing, the first tip's cells are completing cycles more often. Do not accept: "26 is bigger than 12" with no reference to what the index measures; a comparison that treats the index as a count rather than a proportion.
   Show a 4/4 response

   a Cells spend the most time in interphase, 185 of the 250 scored were in it. The assumption is that the root tip was fixed at one random moment, so the fraction of cells caught in a stage is the same as the fraction of the cycle that stage takes up.

   b Anaphase is the shortest part of the cycle. Once the cohesin holding the sister chromatids together is cleaved, the spindle pulls them to opposite poles in only a few minutes, so hardly any cells are caught in it at a given instant: 6 out of 250 here.

   c The cycle is \(18 \times 60 = 1080\) min. Metaphase \(= (12/250) \times 1080 = 51.84\) min, about 52 minutes. Mitotic index \(= (250 - 185)/250 = 65/250 = 0.260\), or 26.0%.

   d The mitotic index is the fraction of cells caught in M phase, which stands for the fraction of the cycle spent dividing. At 26.0% the first tip has more than twice the fraction of the second tip's 12%, so its cells spend more of their time in division and finish cycles more often. That is what dividing more actively means, and both numbers come from the same kind of count.

5. Data: percentage of cells in each phase of the cell cycle 12 hours after a DNA-damaging dose of UV, for a human cell line with normal p53 and a line in which both p53 genes have been knocked out.

   Phase	p53 normal, no UV	p53 normal, + UV	p53 null, no UV	p53 null, + UV
   G1 (%)	52	88	50	51
   S (%)	28	5	30	29
   G2 + M (%)	20	7	20	20

   Using the data, answer (a) through (d).

   1. Identify the checkpoint at which p53 acts, and describe the change in the distribution of the p53-normal cells after UV exposure.
   2. Explain how p53 produces that change at the molecular level.
   3. Calculate the change in the percentage of cells in G₁ caused by UV in each cell line, and state which line arrests.
   4. Justify the claim that a cell that loses p53 function is more likely to accumulate mutations, using the data.
   Your response
   Scoring notes

   • (a) Accept: the G₁ checkpoint (the restriction point). After UV the p53-normal cells shift into G₁, from 52% to 88%, while S falls from 28% to 5% and G₂ + M from 20% to 7%: cells are being held before S phase. Do not accept: naming the G₂ or M (spindle) checkpoint; "the cells stopped dividing" with no statement of where they accumulated.
   • (b) Accept: DNA damage activates p53, which acts as a transcription factor and induces the CDK inhibitor p21; p21 blocks the G₁ cyclin–CDK complex so the cell cannot pass the restriction point, and it arrests in G₁ while repair enzymes work or is directed to apoptosis if the damage is too severe. Do not accept: "p53 stops the cell cycle" with no named intermediate or target; describing p53 as an enzyme that repairs DNA itself.
   • (c) Accept: p53 normal, \(88 - 52 = +36\) percentage points; p53 null, \(51 - 50 = +1\) percentage point; only the p53-normal line arrests in G₁. Do not accept: a ratio such as 88/52 offered as the change; omitting one of the two lines; failing to state which line arrests.
   • (d) Accept: the p53-null line barely changes after UV, 29% of its cells are still in S phase, so damaged cells enter S phase and replicate the damage instead of pausing to repair it, and both daughter cells inherit the resulting mutation; the 36-point shift into G₁ in the normal line is the pause that prevents this. Do not accept: "p53 is a tumor suppressor" as the whole justification; a claim that cites no number or comparison from the table.
   Show a 4/4 response

   a p53 acts at the G₁ checkpoint. After UV the p53-normal cells pile up in G₁ (52% before, 88% after), while S phase drops from 28% to 5% and G₂ plus M from 20% to 7%. They are being held before S phase instead of moving through it.

   b UV damage activates p53, and p53 is a transcription factor. It switches on the gene for p21, a CDK inhibitor, and p21 blocks the G₁ cyclin–CDK complex the cell needs in order to pass the restriction point. With that complex inhibited the cell cannot enter S phase, so it waits in G₁ while repair enzymes work, or is sent to apoptosis if the damage is too severe to fix.

   c p53 normal: \(88 - 52 = +36\) percentage points in G₁. p53 null: \(51 - 50 = +1\) percentage point. Only the p53-normal line arrests.

   d In the knockout line UV barely changed anything, 29% of the cells were still in S phase 12 hours after the damage, against 5% in the normal line. Those cells are replicating damaged DNA instead of pausing, so the damage gets copied into both daughter cells and becomes a permanent mutation. The 36-point pile-up in G₁ in the normal line is the pause that prevents exactly that, and the knockout has lost it.

Lesson 5.1 · Unit 5 · CED topic 5.1

Meiosis: one replication, two divisions

Sexual reproduction has an arithmetic problem. If a human egg and sperm each carried 46 chromosomes, the zygote would carry 92 and the genome would double every generation. Meiosis is the fix: one round of replication followed by two divisions, so the chromosome number halves exactly once.

And it is not halved by throwing chromosomes overboard: each gamete gets one complete set. That precision comes from one event mitosis never performs, in meiosis I, homologous chromosomes pair and then separate from each other, while sister chromatids stay attached.

Lesson 5.2 · Unit 5 · CED topic 5.2

Where the variation comes from

Halving the chromosome number is only half of what meiosis is for. The other half is shuffling. A population with no variation has nothing for selection to act on, and meiosis plus fertilization generates that variation three ways (crossing over, independent assortment, and random fertilization) before a single mutation is needed.

The same machinery can fail, and when it does the result is a gamete with the wrong chromosome number. That failure, nondisjunction, is worth understanding as a slip in exactly one of the two divisions.

Lesson 5.3 · Unit 5 · CED topic 5.3

Mendel's laws, probability, and the chi-square test

Before Mendel, inheritance was assumed to blend, like paint. His pea crosses killed that idea: a trait that vanished in the F₁ came back intact in the F₂, so the hereditary factors had to be discrete particles that stay whole. We now know why: the particles are alleles, and meiosis separates them.

Mendel's laws are really statements about meiosis, published in 1866 and not tied to chromosomes until the Sutton–Boveri theory of 1902. That is why crosses can be worked with probability rules instead of grids, and why counted offspring can be tested against a predicted ratio.

Lesson 5.4 · Unit 5 · CED topics 5.4–5.5

Beyond simple dominance, and beyond the genotype

Mendel chose seven traits that each behaved as one gene with two alleles and clean dominance. Most traits do not. One allele can be only partly dominant, two can both show, a gene can have dozens of alleles or dozens of effects, and whole genes can override each other.

And even a fully specified genotype does not fix the phenotype. Temperature, soil chemistry, nutrition, and population density all feed into what an organism actually looks like: the same DNA, read in a different place, gives a different answer.

Lesson 5.5 · Unit 5 · CED topic 5.6

Chromosomal and extranuclear inheritance

Not every gene sits on a chromosome you have two of. In mammals, females are XX and males XY, so a male has only one copy of every gene on the X: he is hemizygous, and a single recessive allele there is expressed with nothing to mask it. That asymmetry produces inheritance patterns Mendel never saw.

Two more departures follow. Chromosomes can be miscounted during meiosis, giving a zygote an extra or missing copy. And mitochondria and chloroplasts carry their own small DNA, which the egg supplies and the sperm does not.

Unit 5 quiz · 15 multiple-choice · 5 free-response

Unit 5 quiz: Heredity

Fifteen multiple-choice questions and five short free-response questions covering meiosis, the three sources of genetic diversity, Mendelian crosses and chi-square, non-Mendelian patterns and environmental effects, and sex-linked, chromosomal, and extranuclear inheritance: click an option to see why each answer works or fails, and write each FRQ on paper before you open the model response.

Multiple choice

1. Figure: a dividing cell from an organism with 2n = 6, drawn from a light micrograph.

   Six chromosomes, each clearly made of two sister chromatids, lie across the middle of the cell. They are arranged as three side-by-side pairs, so the row is three units long rather than six. Spindle fibers reach each chromosome from one pole only, and the two members of each pair are attached to opposite poles. No nuclear envelope is visible.

   The cell shown is in which stage of division?

   Metaphase I of meiosis Metaphase II of meiosis Anaphase I of meiosis Metaphase of mitosis

   Correct. Three side-by-side pairs of replicated chromosomes are three tetrads, which is what a 2n = 6 cell makes, and the one-pole attachment per chromosome is the arrangement that lets whole homologs, not chromatids, be pulled apart in anaphase I.

   Metaphase II follows the reduction division, so each cell would hold only three replicated chromosomes, not six, and they would line up singly with fibers from both poles. Counting the chromatids in the figure gives twelve, twice what a haploid cell in meiosis II contains.

   In anaphase I the homologs have already separated and are moving toward opposite poles, so the chromosomes would be in two groups away from the equator. Here everything is still aligned in a single row at the center, which defines a metaphase, not an anaphase.

   Mitotic metaphase in a 2n = 6 cell would show six replicated chromosomes lined up singly, six units long, with spindle fibers from both poles attached to each centromere. The figure shows paired chromosomes with one-pole attachment, which mitosis never produces because it does not pair homologs at all.

2. Data: chromosome and DNA content of one grasshopper cell (2n = 10) as it proceeds through meiosis, measured at four points.

   Per cell	End of G1	End of G2	End of meiosis I	End of meiosis II
   Chromosomes	10	10	5	?
   DNA (pg)	6.0	12.0	6.0	?

   Which pair of values correctly completes the final column?

   10 chromosomes and 3.0 pg 5 chromosomes and 6.0 pg 5 chromosomes and 3.0 pg 2.5 chromosomes and 1.5 pg

   The DNA value is right but the chromosome count is not: the diploid number is restored only at fertilization, never by a meiotic division. Meiosis II separates sister chromatids within an already haploid cell, so the count stays at 5 and cannot climb back to 10.

   The chromosome count is right and the DNA value is not. Anaphase II pulls the two chromatids of each chromosome apart into different cells, so the 6.0 pg held by a cell at the end of meiosis I is divided between two products, giving 3.0 pg each.

   Correct. Meiosis II is an equational division: chromosome number stays at 5 because each chromosome keeps its single centromere, while the DNA halves from 6.0 to 3.0 pg as the chromatids separate. Check the bookkeeping: four products at 3.0 pg account for the 12.0 pg present at the end of G₂.

   Chromosomes are counted as whole physical structures, so a fractional chromosome number is not possible. This choice comes from halving both rows again, as though a third division followed; meiosis is exactly two divisions after one round of replication.

3. Which event occurs during meiosis I but never during mitosis?

   The nuclear envelope breaks down before chromosomes attach to the spindle Sister chromatids are pulled toward opposite poles of the cell DNA is replicated during the S phase that precedes the division Homologous chromosomes synapse and exchange segments at chiasmata

   Nuclear envelope breakdown, spindle assembly, and reassembly at telophase occur in mitosis and in both meiotic divisions. Shared machinery like this is common in distractors because it is true: it just does not answer the question being asked.

   Sister chromatids separate in mitotic anaphase and again in anaphase II, so this is the one event the two processes share most obviously. It is also precisely what does not happen in anaphase I, where the chromatids stay joined while the homologs are pulled apart.

   Both processes are preceded by a single S phase that doubles the DNA; that is why a meiotic cell needs two divisions to reach the haploid amount. Replication is the shared setup, not the distinguishing step.

   Correct. Synapsis pairs the two homologs into a tetrad and crossing over swaps segments between non-sister chromatids, and neither happens in mitosis, where homologs behave as unrelated chromosomes. That pairing is also what makes the reduction to haploid possible in the first place.

4. Data: offspring of a test cross in tomato. An F₁ plant heterozygous for stem height and fruit skin was crossed to a dwarf, peach-skinned plant, and 1,400 offspring were scored.

   Offspring phenotype	Tall, smooth	Dwarf, peach	Tall, peach	Dwarf, smooth
   Number	621	604	98	77

   What is the map distance between the two loci?

   5.5 map units 12.5 map units 25.0 map units 50 map units, because the genes assort independently

   This is \(77/1400\), only one of the two recombinant classes. A single crossover between the loci produces both reciprocal products in roughly equal numbers, so 98 and 77 must be added before dividing by the total.

   Correct. The two frequent classes are parental, so the recombinants are \(98 + 77 = 175\) out of 1,400, and \(175/1400 \times 100\% = 12.5\%\), which by definition is 12.5 map units. The near-equality of 98 and 77 is the sign that both came from the same crossover event.

   This doubles the recombination frequency, a step that has no justification: one map unit is defined as 1% recombinant offspring, with no factor of two. The doubling error usually comes from thinking each crossover involves two chromatids and therefore needs correcting.

   Unlinked genes give a 1:1:1:1 test-cross ratio, which here would be four classes of 350. Instead two classes hold 1,225 of the 1,400 offspring, so the alleles are travelling together far more often than chance allows and the genes must be linked.

5. In a human primary oocyte, the sister chromatids of chromosome 21 fail to separate during meiosis II. Which statement best describes the four products of that meiosis?

   All four products carry 23 chromosomes, but one has two copies of chromosome 21 Two products carry 46 chromosomes and two carry 23 All four products are aneuploid: two carry 24 chromosomes and two carry 22 Two products carry the normal 23 chromosomes, one carries 24, and one carries 22

   A product with two copies of chromosome 21 has 24 chromosomes, not 23: adding a chromosome changes the count. This choice confuses an extra chromosome with an extra chromatid, which is the distinction anaphase II actually resolves.

   No meiotic product is diploid. Meiosis II separates chromatids within cells that are already haploid, so the products range from 22 to 24 chromosomes; a 46-chromosome cell would require both divisions to be skipped.

   This is the outcome of nondisjunction in meiosis I, where both homologs go to the same pole and every downstream cell inherits the mistake. A meiosis II error happens after the two cells have already separated, so only one of them can be affected.

   Correct. The cell that divides normally yields two products with 23 chromosomes; the cell whose chromatids fail to separate yields one product with both copies of chromosome 21 (24 chromosomes) and one missing it entirely (22). Half the products are normal, which is the signature of a meiosis II error.

6. Description of a karyotype prepared from cultured cells of a newborn and arranged by chromosome size.

   Forty-five chromosomes are present. Twenty-two matched autosomal pairs are arranged in order; no autosome appears in one copy or in three. There is a single X chromosome and no second sex chromosome.

   Which statement about this karyotype is best supported?

   A gamete carrying no sex chromosome fused with a normal gamete, and these cells contain no Barr body The cells contain one Barr body, because one X chromosome is always inactivated An extra X was gained, which is why the total is odd Nondisjunction of chromosome 21 during meiosis produced the gamete responsible

   Correct. Notation 45,X arises when a gamete that received no sex chromosome, from nondisjunction in either meiotic division, fertilizes or is fertilized by a normal gamete. Only X chromosomes beyond the first are condensed into Barr bodies, so a cell with a single X has none.

   X inactivation silences every X past the first, not the only X a cell has: a typical XX female cell shows one Barr body and a typical XY cell shows none. Silencing the lone X here would leave the cell with no expressed copy of hundreds of essential genes.

   Gaining a chromosome would raise the total above 46, not lower it to 45. The count is odd because one chromosome is absent, which is what makes 45,X a monosomy rather than a trisomy.

   Trisomy 21 would show three copies of chromosome 21 and a total of 47, and the description states explicitly that every autosome is present as a matched pair. The missing chromosome here is a sex chromosome, so the nondisjunction involved the X–Y pair or the two X chromatids.

7. Data: F₂ phenotypes from a monohybrid cross in peas, with the counts expected under a 3:1 ratio.

   Phenotype	Tall	Dwarf	Total
   Observed	792	208	1,000
   Expected (3:1)	750	250	1,000

   Using \(\chi^{2} = \sum (o-e)^{2}/e\) and the critical value 3.841, which conclusion is best supported?

   \(\chi^{2} = 9.41\); reject the null hypothesis, because the deviation is larger than chance sampling is expected to produce \(\chi^{2} = 9.41\); fail to reject the null hypothesis, because \(df = 3\) and the critical value is 7.815 \(\chi^{2} = 0.22\); fail to reject the null hypothesis \(\chi^{2} = 9.41\); the data prove that the alleles of this gene do not segregate

   Correct. The terms are \((792-750)^{2}/750 = 1764/750 = 2.352\) and \((208-250)^{2}/250 = 1764/250 = 7.056\), summing to 9.408. Two phenotype classes give \(df = 1\), and 9.41 exceeds 3.841, so the null hypothesis of a 3:1 ratio is rejected.

   The chi-square value is computed correctly but the degrees of freedom are not: \(df\) is the number of phenotype classes minus one, and there are two classes here, not four. Using 7.815 flips a rejection into a non-rejection, which is why counting classes first is worth the five seconds.

   This is what you get by dividing the raw deviation by the expected value instead of the squared deviation: \(42/750 + 42/250 = 0.224\). Squaring is what makes the statistic insensitive to the direction of the deviation and puts it on the scale the critical values were built for.

   A chi-square test never proves anything, and rejecting a 3:1 ratio does not implicate segregation specifically. Reduced viability of one genotype, a second gene affecting height, or a scoring bias would all produce this result, so the honest conclusion names the ratio, not the mechanism.

8. In the cross AaBbCc × AaBbcc, with all three genes unlinked, what is the probability that an offspring shows the dominant phenotype for gene A and the recessive phenotypes for genes B and C?

   \(9/64\) \(3/64\) \(3/32\) \(1/16\)

   This multiplies \(3/4 \times 3/4 \times 1/4\), which is the probability of dominant A, dominant B, and recessive C. Reading which gene needs the recessive phenotype is half the work in these items; the stem asks for recessive at both B and C.

   This uses \(1/4\) for the C gene, as if both parents were Cc. The second parent is cc and contributes only c, so a Cc × cc cross gives \(1/2\) recessive offspring, not \(1/4\).

   Correct. Treat each gene separately and multiply: \(Aa \times Aa\) gives \(3/4\) dominant, \(Bb \times Bb\) gives \(1/4\) bb, and \(Cc \times cc\) gives \(1/2\) cc, so \(3/4 \times 1/4 \times 1/2 = 3/32\), about 9.4%. The product rule gets this in seconds; a 32-box grid does not.

   This is \(1/4 \times 1/4\), the answer for recessive at A and B with the C gene ignored altogether. Every gene named in the stem has to appear as a factor, even when its probability is \(1/2\) rather than \(1/4\).

9. A breeder has a purple-flowered plant and needs to know whether it is PP or Pp, where purple is dominant to white. Which cross settles the question, and what result identifies a heterozygote?

   Cross it to a white (pp) plant; if it is Pp, about half of 40 offspring, roughly 20, will be white Cross it to a homozygous purple (PP) plant; if it is Pp, about one quarter of the offspring will be white Self-pollinate it; if all 40 offspring are purple, the plant must be Pp Cross it to another purple plant of unknown genotype; any white offspring shows the first plant is PP

   Correct. A test cross to the homozygous recessive contributes only p, so each offspring's phenotype reports the allele it received from the unknown parent. A Pp plant donates p half the time, giving about 20 white offspring in 40; a PP plant donates P every time and gives none.

   A PP partner donates P to every offspring, so every offspring has at least one dominant allele and all of them are purple whatever the unknown plant is. A cross that gives the same result for both hypotheses cannot distinguish them.

   Selfing does work in principle, but the interpretation here is backwards: a Pp plant selfed gives about \(1/4\) white offspring, so 40 purple offspring point toward PP, not Pp. The test cross is preferred because it produces a 1:1 split, which is easier to detect in a small litter than 3:1.

   White offspring are pp and need a p allele from each parent, so a white offspring shows the plant is Pp: the opposite of what this choice claims. An unknown second parent also leaves the result uninterpretable when no white offspring appear.

10. Data: coat colors among 160 F₂ Labrador retriever puppies from crosses between black F₁ dogs of genotype BbEe.

    Coat color	Black	Chocolate	Yellow
    Number of puppies	88	33	39

    Which statement best explains why the yellow class is larger than the chocolate class?

    The yellow allele is dominant to the chocolate allele, so more genotypes produce yellow The E gene is epistatic to the B gene, so both B_ee and bbee puppies are yellow, 4 of 16, while only bbE_ is chocolate at 3 of 16 The two genes are linked on the same chromosome, which inflates two of the three classes The difference is sampling error; with a larger litter the two classes would be equal

    Yellow and chocolate are not alleles of one gene, so no dominance relationship exists between them: yellow is an ee phenotype at one locus and chocolate is a bbE_ phenotype involving both. Treating the three coat colors as three alleles of a single gene is the usual first move here, and it cannot generate a 9:3:4 ratio.

    Correct. Any ee dog fails to deposit pigment in the coat and is yellow no matter what the B gene says, so the \(3/16\) B_ee and \(1/16\) bbee classes merge. Expected counts in 160 puppies are 90 black, 30 chocolate, and 40 yellow, and the observed 88, 33, 39 match closely.

    Linkage changes which allele combinations travel together and would distort the ratio in a test cross, but it cannot merge two phenotype classes into one. The clue that this is epistasis rather than linkage is that the yellow class is exactly the sum of two of the four dihybrid classes.

    Sampling error explains the gap between 39 and the expected 40, not the gap between the chocolate and yellow classes, which is predicted by the model before any puppies are counted. Expected values of 30 and 40 are not equal, so no sample size would make the observed classes converge.

11. A woman with type O blood has a child with type B blood. She names a man with type AB blood as the father. Which statement is best supported?

    He is excluded, because any child of a type AB father and a type O mother must be type AB The blood types confirm that he is the father He is excluded, because a man with type AB blood cannot father a child with type B blood He cannot be excluded: the mother is ii and donates i, and a man who is I^AI^B can donate I^B, giving an I^Bi child with type B blood

    I^A and I^B are codominant with each other but both are dominant to i, so a child who receives I^A from the father and i from the mother is type A, and one who receives I^B is type B. Only a child carrying both I^A and I^B is type AB, which is impossible with an ii mother.

    Blood typing can exclude a man but can never confirm one, because a large share of the population carries the same alleles. Any type AB or type B man in the population would be equally compatible with this child, so the evidence is consistent with him rather than specific to him.

    A type AB man is I^AI^B and donates one of those two alleles to each child, so half his children by a type O mother are expected to be type B. The exclusion rule works the other way round: an AB man cannot father a type O child, because he carries no i.

    Correct. Work backwards from the child: type B is I^BI^B or I^Bi, and since the mother can only supply i, the child is I^Bi and needs I^B from the father. A type AB man has one, so nothing in these data excludes him.

12. Data: twelve cuttings taken from a single hydrangea plant, rooted and then grown one season in soils adjusted to four pH values, three cuttings per soil.

    Soil pH	4.5	5.5	6.5	7.5
    Flower color	deep blue	blue-purple	pink-purple	pink
    Cuttings scored	3	3	3	3

    Which conclusion is best supported by these data?

    The cuttings must differ genetically, because plants with the same genotype cannot differ in flower color Low soil pH caused a mutation in the pigment gene, and the mutation was expressed as blue flowers Genetically identical plants produced different phenotypes because soil pH changes the availability of soil aluminum to the pigment Flower color in hydrangea shows incomplete dominance, with purple as the heterozygous phenotype

    Cuttings from one plant are clones, so this conclusion contradicts the design of the investigation rather than following from it. The whole point of using cuttings is to hold the genotype constant so that any variation in phenotype has to come from somewhere else.

    Mutations are random changes in base sequence, not directed responses that happen to produce a useful or matching phenotype in every one of three plants per treatment. A mutagen would also not produce the same color in all three cuttings at each pH, and reversing the color when the plant is moved would be impossible.

    Correct. All twelve plants carry identical DNA, and the only variable that differs among the groups is soil pH, which controls how much aluminum the roots can take up and therefore the color of the pigment complex in the petals. Phenotype is genotype plus environment, and this design isolates the environmental half.

    Incomplete dominance is a relationship between two alleles at one locus, which would require genetically different plants and would give a fixed ratio of colors within one environment. Here the color tracks a measured environmental gradient across clones, which no allele-based model can explain.

13. Description of a three-generation pedigree for a rare human trait. Squares are males, circles are females, and filled symbols are affected.

    Generation I: I-1, an affected male, is married to I-2, an unaffected female. Their children are II-1, an unaffected female, and II-3, an unaffected male. II-1 married II-2, an unaffected male from outside the family; II-3 married II-4, an unaffected female from outside the family. Generation III: II-1 and II-2 have III-1, an affected male, and III-2, an unaffected female. II-3 and II-4 have III-3, an unaffected male, and III-4, an unaffected female.

    Which mode of inheritance is best supported by this pedigree?

    Autosomal dominant Y-linked X-linked recessive Mitochondrial

    A dominant allele is expressed in every carrier, so an affected child must have an affected parent. III-1 is affected while both II-1 and II-2 are unaffected, which rules dominance out at once and tells you the allele is recessive.

    A Y-linked allele passes from father to every son and to no daughter, so II-3 would be affected and would pass the trait to III-3. Both are unaffected, and the trait instead reappears through I-1's daughter, which a Y-linked allele can never do.

    Correct. I-1 passes his only X to both daughters, so II-1 is an obligate carrier, and her son III-1 received her X^a with no second X to mask it. The same allele cannot reach III-3 through II-3, because a father gives his son a Y, which is exactly the skipping pattern the pedigree shows.

    Mitochondrial DNA is inherited from the egg alone, so an affected father transmits to none of his children and an affected mother to all of hers. Here the affected individual in generation I is male and his grandson is affected through a daughter, the opposite of maternal transmission.

14. An X-linked recessive allele has a frequency of \(q = 0.04\) in a large population that mates at random. Approximately what percentage of males and of females are affected?

    4% of males and 4% of females 4% of males and 0.16% of females, about 25 affected males for every affected female 0.16% of males and 4% of females 8% of males and 0.16% of females, because males have one X and females have two

    This treats the two sexes as equivalent, which is exactly what X-linkage rules out. A male is hemizygous, so one copy of the allele is enough; a female needs two, and those are very different probabilities when \(q\) is small.

    Correct. A male is affected whenever his single X carries the allele, so the frequency is \(q = 0.04\), or 4%. A female needs the allele on both X chromosomes, so the frequency is \(q^{2} = 0.0016\), or 0.16%, and \(0.04/0.0016 = 25\).

    The two values are the right numbers assigned to the wrong sexes. Squaring a frequency below one always makes it smaller, so the rarer outcome, \(q^{2}\), has to belong to the sex that needs two copies, which is the female.

    Nothing in the model doubles the male frequency; having one X means the probability is \(q\) itself, not \(2q\). The factor of two shows up in \(2pq\), the frequency of heterozygous carrier females, which is \(2(0.96)(0.04) = 7.7\%\) here.

15. Description of a pedigree for a human myopathy that causes exercise intolerance. Squares are males, circles are females, and filled symbols are affected.

    Generation I: I-1, an affected female, married I-2, an unaffected male. All four of their children are affected: II-1 (male), II-2 (female), II-3 (male), and II-4 (female). Generation III: II-1 married an unaffected woman, and neither of their two children is affected. II-2 married an unaffected man, and both of their children are affected.

    Which statement best explains the pattern of transmission in this pedigree?

    The allele is X-linked dominant, so an affected mother transmits it to all of her children The mutation is in mitochondrial DNA, so it passes from an affected mother to all of her children and from an affected father to none The allele is autosomal dominant with incomplete penetrance in males The allele is X-linked recessive, so affected males transmit it only to their daughters

    An X-linked dominant allele in a heterozygous mother reaches about half her children, not all four, and an affected father would transmit it to every daughter. Here II-1 is affected and has no affected children at all, which no X-linked model allows.

    Correct. Essentially all of a zygote's mitochondria come from the egg's cytoplasm, so every child of an affected mother inherits the mutant organelles while a father contributes none. The pedigree shows exactly that asymmetry: II-2's children are affected and II-1's are not.

    Incomplete penetrance would produce unaffected carriers scattered through the pedigree in both sexes, not a clean split by the sex of the transmitting parent. It also cannot explain why all of an affected mother's children are affected while none of an affected father's are.

    Under X-linked recessive inheritance an affected female passes her allele to every son, so all of I-1's sons would be affected but her daughters would only be carriers, and II-2 is affected, not a carrier. The variable severity mitochondrial disorders show is a better fit for the same data.

Free response

1. Data: F₂ phenotypes from a dihybrid cross in tomato. True-breeding tall, purple-stemmed plants were crossed to true-breeding dwarf, green-stemmed plants; the F₁ were all tall and purple-stemmed and were allowed to self-pollinate. 640 F₂ plants were scored.

   F2 phenotype	Tall, purple	Tall, green	Dwarf, purple	Dwarf, green
   Observed number	371	118	126	25

   Using the data, answer (a) through (d).

   1. Identify the null hypothesis being tested by a chi-square analysis of these data, and describe the phenotypic ratio that hypothesis predicts.
   2. Explain why a chi-square test rather than an inspection of the counts is needed to decide whether these data are consistent with independent assortment.
   3. Calculate the chi-square value for these data. Show the expected count for each phenotype class and your setup.
   4. The critical value at \(p = 0.05\) with \(df = 3\) is 7.815. Justify a conclusion about independent assortment of these two genes, using your calculated value as evidence.
   Your response
   Scoring notes

   • (a) Accept: the null hypothesis is that the two genes assort independently and the F₂ fits a 9:3:3:1 phenotypic ratio, so any deviation from that ratio is due to chance sampling. Do not accept: "the genes are linked" (that is the alternative hypothesis); restating the stem; naming 9:3:3:1 with no statement of what the hypothesis claims.
   • (b) Accept: an explanation that every real sample deviates from its expected ratio by chance, so a decision needs a measure of how large the deviation is relative to what chance alone produces; the test converts the four deviations into one number that can be compared with a threshold. Do not accept: "the numbers look close" or "to be more accurate" with no reasoning; a description of how to compute the statistic instead of why it is needed.
   • (c) Accept: expected counts of \(640 \times 9/16 = 360\), \(640 \times 3/16 = 120\) twice, and \(640 \times 1/16 = 40\); terms 0.336, 0.033, 0.300, and 5.625; \(\chi^{2} = 6.29\) (accept 6.28–6.30). Setup must be shown. Do not accept: a bare number with no expected values; expected counts that do not sum to 640; use of percentages in place of counts.
   • (d) Accept: \(\chi^{2} = 6.29\) is less than 7.815, so we fail to reject the null hypothesis and the data are consistent with independent assortment; the reasoning must name the comparison. Do not accept: "the genes assort independently" stated as proven; "we accept the null hypothesis"; a conclusion with no comparison to the critical value.
   Show a 4/4 response

   a The null hypothesis is that the height gene and the stem-color gene assort independently, so the F₂ should fall into a 9:3:3:1 ratio of tall purple to tall green to dwarf purple to dwarf green, and any difference from that is just chance.

   b Even if the genes really do assort independently, I would never get exactly 360:120:120:40 in a real sample of 640 plants, because which gametes happen to fuse is random. The chi-square test tells me how big my deviation is compared with the deviations chance alone produces, so I am not just eyeballing whether 25 is "close enough" to 40.

   c Expected: \(640 \times 9/16 = 360\), \(640 \times 3/16 = 120\), \(640 \times 3/16 = 120\), \(640 \times 1/16 = 40\). Terms: \((371-360)^{2}/360 = 0.336\), \((118-120)^{2}/120 = 0.033\), \((126-120)^{2}/120 = 0.300\), \((25-40)^{2}/40 = 5.625\). Adding them, \(\chi^{2} = 6.29\).

   d Since 6.29 is less than the critical value of 7.815 at \(df = 3\), I fail to reject the null hypothesis. Almost all of the statistic comes from the dwarf green class being 15 plants short, and a deviation that size still falls inside the range chance produces, so these data are consistent with the two genes assorting independently.

2. Description of a pedigree for a rare human metabolic disorder. Squares are males, circles are females, and filled symbols are affected.

   Family 1: I-1, an unaffected male, and I-2, an unaffected female, have three children: II-1, an affected female; II-2, an unaffected male; and II-3, an unaffected female. Family 2: I-3, an unaffected male, and I-4, an unaffected female, have two children: II-4, an affected male, and II-5, an unaffected female. II-2 and II-5 have married; they have no children yet. No one else in either family is affected.

   Using the pedigree, answer (a) through (d).

   1. Identify the mode of inheritance of this disorder, and describe the feature of the pedigree that rules out a dominant allele.
   2. Explain why the pedigree is not consistent with X-linked recessive inheritance.
   3. Calculate the probability that the first child of II-2 and II-5 is affected. Show your setup.
   4. Justify the claim that II-2 and II-5 are at higher risk of having an affected child than a couple chosen at random from the population, using evidence from the pedigree.
   Your response
   Scoring notes

   • (a) Accept: autosomal recessive, supported by affected children born to two unaffected parents in both families (I-1 × I-2 → II-1; I-3 × I-4 → II-4), which means carriers exist and the allele is not expressed in them. Do not accept: naming "recessive" with no evidence; citing only one family; claiming the trait skipped a generation without saying what that implies.
   • (b) Accept: II-1 is an affected female whose father I-1 is unaffected; under X-linked recessive inheritance an affected female must be X^aX^a and therefore must have received an X^a from her father, who would then be affected himself. Do not accept: "affected individuals are both male and female" alone, which is suggestive but not decisive; any answer that does not use a specific individual.
   • (c) Accept: \(2/3 \times 2/3 \times 1/4 = 4/36 = 1/9 \approx 11\%\), with the \(2/3\) values explained as the probability that an unaffected child of two carriers is a heterozygote. Setup required. Do not accept: \(1/4\) (ignoring the parents' genotype uncertainty); \(1/2 \times 1/2 \times 1/4\) (using \(1/2\) for the carrier probabilities); a bare fraction with no setup.
   • (d) Accept: each of them has an affected full sibling, so both sets of grandparents are obligate carriers and each of II-2 and II-5 is a carrier with probability \(2/3\), far above the carrier frequency for a rare allele in the general population; the named evidence must be the affected siblings. Do not accept: "it runs in the family"; repeating the \(1/9\) figure with no comparison; a claim with no reasoning attached.
   Show a 4/4 response

   a It is autosomal recessive. In both families two unaffected parents have an affected child (I-1 and I-2 have II-1, and I-3 and I-4 have II-4), and a dominant allele has to show up in anyone who carries it, so neither set of parents could have been hiding a dominant allele.

   b It is not X-linked recessive because II-1 is an affected female. She would have to be X^aX^a, and one of those X chromosomes came from her father I-1, who would then be X^aY and affected himself. He is unaffected, so the gene is on an autosome.

   c II-2 is unaffected and his parents are both carriers, so he is \(2/3\) likely to be Aa; the same reasoning gives II-5 a \(2/3\) chance. If both are carriers, \(1/4\) of their children are aa. So \(P = 2/3 \times 2/3 \times 1/4 = 4/36 = 1/9\), about 11%.

   d Both of them have an affected full sibling, which proves all four of their parents carry the allele; that is what makes each of them a \(2/3\) carrier instead of the very low carrier frequency you would expect for a rare recessive allele in the general population. Two \(2/3\) carriers marrying is exactly why the risk climbs to \(1/9\).

3. Data: a test cross in Drosophila melanogaster. A female heterozygous for two autosomal genes, eye color (wild type dominant to scarlet) and bristle shape (straight dominant to forked), was crossed to a male homozygous recessive for both genes. 1,000 offspring were scored.

   Offspring phenotype	Wild eyes, straight	Scarlet eyes, forked	Wild eyes, forked	Scarlet eyes, straight
   Number	452	448	52	48

   Using the investigation and the data, answer (a) through (d).

   1. Identify the dependent variable in this investigation, and describe the purpose of crossing the heterozygous female to a male that is homozygous recessive for both genes.
   2. Explain how the distribution of the four offspring classes shows that these two genes are not assorting independently.
   3. Calculate the recombination frequency, and convert it to a map distance in map units. Show your setup.
   4. Justify a claim about the two recombinant classes being nearly equal in number, using the mechanism of crossing over as your reasoning.
   Your response
   Scoring notes

   • (a) Accept: the dependent variable is the number (or frequency) of offspring in each of the four phenotype classes; the homozygous recessive father contributes only recessive alleles, so every offspring's phenotype reports directly which allele combination the egg carried. Do not accept: naming the variable with no description of the cross's purpose; "the father is the control" with no mechanism; calling eye color alone the dependent variable.
   • (b) Accept: independent assortment predicts four equal classes of about 250 each, but two classes hold 900 of the 1,000 offspring while the other two hold 100, so the parental allele combinations stay together far more often than chance allows, which happens when the loci are on the same chromosome. Do not accept: "the numbers are uneven" with no expected ratio named; asserting linkage with no comparison to 1:1:1:1.
   • (c) Accept: recombinants \(= 52 + 48 = 100\); \(\mathrm{RF} = 100/1000 \times 100\% = 10.0\%\), so the genes are 10 map units apart. Do not accept: using one recombinant class only; doubling the frequency; omitting the conversion statement that 1% recombination equals 1 map unit.
   • (d) Accept: a single crossover between the two loci is a reciprocal exchange between non-sister chromatids, so it produces one chromosome of each recombinant type at the same time, and the two classes should therefore appear in about equal numbers, 52 and 48 differ by an amount chance sampling easily explains. Do not accept: "they are close so it is random"; restating the numbers with no mechanism; attributing the equality to independent assortment.
   Show a 4/4 response

   a The dependent variable is the number of offspring falling into each of the four phenotype classes. The father is homozygous recessive for both genes, so he can only donate recessive alleles and never masks anything; that means each offspring's appearance tells me exactly which pair of alleles came in the egg.

   b If these genes were on different chromosomes, independent assortment would give four classes of about 250 each. Instead the wild-straight and scarlet-forked classes together hold 900 flies and the other two hold only 100, so the alleles the mother inherited together are staying together. That happens when both loci sit on the same chromosome and only a crossover can separate them.

   c The recombinants are the two rare classes: \(52 + 48 = 100\) out of 1,000. So \(\mathrm{RF} = 100/1000 \times 100\% = 10.0\%\), and since 1% recombination is defined as one map unit, the genes are about 10 map units apart.

   d A crossover is a reciprocal swap between two non-sister chromatids, so the same event that makes a wild-forked chromosome also makes a scarlet-straight one. The two recombinant classes therefore come out of the same events in equal numbers, and the observed 52 versus 48 is the kind of small gap random sampling of 1,000 flies produces.

4. An animal has a diploid chromosome number of 2n = 10. Answer (a) through (d) about meiosis in this organism.

   1. Describe one difference between anaphase I and anaphase II in terms of what structures separate and what happens to ploidy.
   2. Explain how crossing over during prophase I increases the genetic variation among gametes beyond what independent assortment alone can produce.
   3. Calculate the number of chromosomally distinct gametes this organism can produce from independent assortment alone, and the number of distinct zygote combinations possible when two such individuals mate. Show your setup.
   4. A mutant of this species cannot assemble the synaptonemal complex, so homologs never synapse and each replicated chromosome attaches to the spindle independently. Predict the fraction of meiosis I products that receive a balanced haploid set, and justify your prediction.
   Your response
   Scoring notes

   • (a) Accept: in anaphase I homologous chromosomes separate while sister chromatids stay attached, and ploidy falls from 2n to n; in anaphase II sister chromatids separate and ploidy stays at n. Do not accept: "chromosomes separate in both" with no distinction; describing only one of the two anaphases; saying meiosis II halves the chromosome number.
   • (b) Accept: independent assortment shuffles whole chromosomes, so each gamete gets one intact maternal or paternal chromosome per pair; crossing over exchanges segments between non-sister chromatids, producing chromosomes that carry maternal alleles at some loci and paternal alleles at others, which are combinations independent assortment cannot make. Do not accept: "it makes more variation" with no mechanism; confusing crossing over with independent assortment; describing crossing over without connecting it to allele combinations.
   • (c) Accept: \(n = 5\), so \(2^{5} = 32\) chromosomally distinct gametes, and \(32 \times 32 = 1{,}024\) zygote combinations. Do not accept: \(2^{10}\) (using the diploid number); \(32 + 32\); an answer with no setup shown.
   • (d) Accept: about \(1/32\) or 3.125%, justified by noting that without synapsis each of the 5 pairs segregates at random, giving each pair a \(1/2\) chance of sending its two homologs to opposite poles, and \((1/2)^{5} = 1/32\) for all five together. Do not accept: "most gametes will be abnormal" with no number; a number with no link between pairing and segregation; the claim that no gametes at all would be balanced.
   Show a 4/4 response

   a In anaphase I whole homologous chromosomes are pulled apart while their sister chromatids stay stuck together at the centromere, and that is the step where the cell goes from 2n to n. In anaphase II the sister chromatids finally separate, and ploidy does not change: the products are still haploid, just with unreplicated chromosomes.

   b Independent assortment only decides which whole chromosome of each pair goes to a pole, so every chromosome in a gamete would be an intact copy from one grandparent. Crossing over swaps segments between non-sister chromatids, so a single chromosome ends up carrying my grandmother's alleles at some loci and my grandfather's at others. Those mixed chromosomes are new combinations that no amount of shuffling whole chromosomes could produce.

   c With 2n = 10, \(n = 5\), so independent assortment gives \(2^{5} = 32\) chromosomally different gametes. Two individuals mating gives \(32 \times 32 = 1{,}024\) zygote combinations.

   d I predict about \(1/32\), or 3.1%, are balanced. Without a synaptonemal complex the homologs never pair, so nothing makes the two members of a pair face opposite poles; each pair segregates correctly by chance with probability \(1/2\), and the five pairs are independent, so \((1/2)^{5} = 1/32\). Everything else is aneuploid, which shows that pairing is what guarantees an even split.

5. Data: Himalayan rabbits from one inbred, genetically uniform line were raised from weaning at five ambient temperatures, ten rabbits per group. At twelve weeks the percentage of body surface covered by black fur was measured.

   Ambient temperature (°C)	5	15	25	30	35
   Black fur (% of body surface)	96	72	41	12	0

   Using the data, answer (a) through (d).

   1. Describe the relationship between ambient temperature and the percentage of body surface covered by black fur.
   2. Explain, in terms of the pigment-producing enzyme, why rabbits with identical genotypes differ in coat color across the five groups.
   3. Construct a graph of these data: put ambient temperature in °C on the x-axis with a scale from 0 to 40, put percent black fur on the y-axis with a scale from 0 to 100, label both axes with units, plot the five points, and draw a best-fit line. Then calculate the mean rate of change in percent black fur per degree Celsius between 5 °C and 35 °C.
   4. A student claims that the rabbits raised at 35 °C must carry a different allele of the pigment gene from those raised at 5 °C. Justify a rejection of that claim, using evidence from the investigation.
   Your response
   Scoring notes

   • (a) Accept: as ambient temperature rises the percentage of black fur falls, from 96% at 5 °C to 0% at 35 °C, a negative relationship, with the sharpest drop between 25 °C and 30 °C. Accept a correct statement of the inverse relationship with at least one pair of values cited. Do not accept: "temperature affects fur color" with no direction; restating the table row by row with no trend statement.
   • (b) Accept: all the rabbits carry the same temperature-sensitive form of the pigment-producing enzyme, which folds and functions at cool surface temperatures but denatures above roughly 33 °C, so pigment is deposited only where the skin is cool; warm-reared rabbits keep almost no cool skin, so almost no fur is pigmented. Do not accept: "the cold makes it black" with no enzyme named; "the gene turns off in the heat" with no mechanism; any claim that the genotype changes.
   • (c) Accept: temperature on the x-axis (0–40 °C) and percent black fur on the y-axis (0–100%), both axes labeled with units, all five points plotted correctly, and a best-fit line drawn; plus \((0 - 96)/(35 - 5) = -96/30 = -3.2\) percentage points per °C. Accept 3.2 stated as a decrease. Do not accept: reversed axes; an unlabeled axis or missing units; a rate with no sign or no units; connecting the dots in place of a best-fit line is acceptable only if the line is described as such.
   • (d) Accept: the rabbits come from one inbred, genetically uniform line and were assigned to temperature groups, so genotype was held constant by design and cannot be the source of the difference; the only variable that differed was ambient temperature, so the phenotype difference is environmental. Do not accept: "phenotype equals genotype plus environment" with no reference to this investigation; rejecting the claim with no evidence named; arguing only that alleles cannot change.
   Show a 4/4 response

   a The warmer the rabbits were raised, the less black fur they had. Black coverage falls steadily from 96% of the body surface at 5 °C to 0% at 35 °C, and the steepest part of the drop is between 25 °C and 30 °C, where it falls from 41% to 12%.

   b Every rabbit has the same allele for the pigment enzyme, and in this breed that enzyme only works at cool temperatures: above about 33 °C it denatures and stops making pigment. In a cold room most of the skin surface is cool enough for the enzyme to fold properly, so most of the fur grows in black; in a warm room almost no skin is cool enough, so the fur grows in white.

   c I put ambient temperature (°C) on the x-axis from 0 to 40 and percent black fur on the y-axis from 0 to 100, labeled both with units, plotted (5, 96), (15, 72), (25, 41), (30, 12), and (35, 0), and drew a best-fit line through them. Mean rate of change: \((0 - 96)/(35 - 5) = -96/30 = -3.2\) percentage points of black fur per °C.

   d I reject that claim. All fifty rabbits came from one inbred, genetically uniform line, so their pigment-gene alleles are the same before the experiment starts: genotype was controlled by the design. The only thing that differed between the 5 °C and 35 °C groups was the temperature they were raised at, so the difference in coat color has to be environmental, not allelic.

Lesson 6.1 · Unit 6 · CED topic 6.1

DNA as the genetic material, and the structure that explains it

By 1940 everyone agreed that chromosomes carried the genes. Most biologists also assumed the genes themselves were protein: proteins come in twenty monomer flavors and DNA in only four, so DNA looked too monotonous to spell anything. Three experiments overturned that, and all three run on the same logic: take a mixture that works, remove one component at a time, and see which removal kills the effect. Learn that logic by name: elimination is the control design, and Section II asks you to name the independent variable, the dependent variable, and the control in exactly these setups.

Lesson 6.2 · Unit 6 · CED topic 6.2

DNA replication: the fork, the enzymes, and Meselson–Stahl

Watson and Crick noted that complementary pairing hands you a copying mechanism for free: unzip the helix and each old strand specifies its new partner. But three models fit that sentence. Conservative replication returns the two original strands to one duplex and builds a wholly new one. Semiconservative gives every daughter duplex one old strand and one new. Dispersive chops and splices, so every strand is a patchwork of old and new. Deciding between them took a density label and a centrifuge.

Lesson 6.3 · Unit 6 · CED topic 6.3

Transcription and RNA processing

A eukaryotic cell will not risk its chromosomes in the cytoplasm, and the ribosomes are not coming into the nucleus. So the cell makes a disposable working copy of one gene at a time. That copy is RNA, and everything odd about it (why it is single-stranded, why it is short-lived, why it gets a cap and a tail before it leaves) follows from its being a message rather than an archive.

Lesson 6.4 · Unit 6 · CED topic 6.4

Translation and the genetic code

The message has four letters and the product has twenty. Two RNA letters per amino acid would give only \(4^{2} = 16\) combinations, not enough. Three give \(4^{3} = 64\), comfortably more than enough, and that surplus is why the code is redundant rather than tight. Translation is the machinery that reads those triplets and welds amino acids together in the order they specify.

Lesson 6.5 · Unit 6 · CED topics 6.5–6.6

Regulation of gene expression and cell specialization

A bacterium in a glucose broth has the genes for digesting lactose and no reason to build the enzymes. Your neurons and your pancreatic β cells carry the identical genome and could not be more different. Both facts are the same fact: having a gene and expressing it are separate things, and almost all of the control happens at transcription, before a single amino acid is spent.

Lesson 6.6 · Unit 6 · CED topic 6.7

Mutations and the sources of genetic variation

Meiosis, fertilization, and horizontal gene transfer all shuffle alleles. Only mutation makes new ones. Most are neutral, a minority are harmful, and a few are useful, but "useful" is never a property of the mutation by itself, only of the mutation in a particular environment. Keep that straight and you will avoid the single most-penalized error on this exam: writing that an organism mutated because it needed to.

Lesson 6.7 · Unit 6 · CED topic 6.8

Biotechnology: cutting, sorting, copying, and editing DNA

Every tool in this lesson was stolen from a cell. Restriction enzymes are a bacterium's defense against phage DNA. Plasmids are how bacteria trade genes. Taq polymerase came from a hot-spring bacterium that needed a polymerase surviving 95 °C. CRISPR is a bacterial immune memory. Once you see what each one did for its owner, the lab techniques stop being a list to memorize.

Unit 6 quiz · 15 multiple-choice · 5 free-response

Unit 6 quiz: Gene Expression and Regulation

Fifteen multiple-choice questions and five short free-response questions covering the experiments that identified DNA, replication, transcription and RNA processing, translation and the genetic code, operons and epigenetic regulation, mutations and horizontal gene transfer, and the tools of biotechnology: click an option to see why each answer works or fails, and write each FRQ on paper before you open the model response.

Multiple choice

1. Data: a repetition of the Hershey–Chase experiment. T2 bacteriophage were grown with either ³⁵S or ³²P, allowed to infect E. coli for ten minutes, blended to shear off the spent coats, and centrifuged. Heavy bacteria form the pellet; light phage ghosts stay in the supernatant.

   Labeled phage	Radioactivity in pellet (cells)	Radioactivity in supernatant (ghosts)
   35S, labels protein only	18%	82%
   32P, labels DNA only	72%	28%

   Which conclusion is best supported by these data?

   DNA is the material the phage injects, because most of the ³²P sediments with the cells that go on to produce progeny phage while most of the ³⁵S is discarded with the coats DNA and protein are injected in equal amounts, because both isotopes appear in both fractions The phage coat is built from DNA, because ³²P appears in the pellet Protein enters the host cell, because 18% of the ³⁵S label was recovered in the pellet

   Correct. Sulfur labels only protein and phosphorus only DNA, so the two rows track the two candidate molecules independently. Four-fifths of the protein label stays outside with the ghosts while nearly three-quarters of the DNA label goes in with the cells that make new phage, which identifies DNA as the injected genetic material.

   Radioactivity appears in both fractions because separation is imperfect, not because both molecules enter. The design answers the question through the difference between the two rows, and 18% versus 72% is not an equal split by any reading.

   DNA contains phosphorus in its sugar–phosphate backbone and the protein coat contains essentially none, which is exactly why ³²P is a DNA label. The coat is protein, which is why it carries the ³⁵S and is left behind in the supernatant.

   An 18% carryover is what you expect when blending does not shear every coat free of every cell, and it is the small fraction rather than the large one. The comparison that matters is 18% against 72%: the isotope that ends up inside the infected cells is the one in DNA.

2. In the Avery–MacLeod–McCarty experiment, extract from heat-killed S cells was divided among tubes treated with protease, with RNase, or with DNase before being mixed with live R cells. Why was it essential to include the protease and RNase treatments rather than testing DNase alone?

   Protease and RNase were needed to purify the DNA so that DNase could reach it They are comparison treatments that destroy their own targets just as thoroughly yet leave transformation intact, so the loss seen with DNase can be attributed to removing DNA rather than to handling the extract They demonstrated that the extract from heat-killed S cells contains no protein or RNA at all They tested whether the live R cells could be killed by enzyme treatment

   DNase acts on DNA in a crude mixture perfectly well, and the three enzymes were used on separate portions of the same extract rather than in sequence. Reading them as purification steps loses the entire logic of the design, which is elimination one component at a time.

   Correct. Without them, a critic could argue that any enzyme treatment, or simply the handling, ruins the extract. Protease and RNase show that the extract survives an equally destructive treatment aimed at a different molecule, so only the removal of DNA abolishes the transforming activity.

   The extract certainly contained protein and RNA: that is why those enzymes had something to digest. What the experiment shows is that neither of those molecules is required for transformation, which is a different claim from their being absent.

   The dependent variable is whether R cells are converted into S cells, not whether R cells survive. All three treatments had to leave the R cells alive for the assay to work at all, so killing them would have destroyed the experiment rather than served as a control.

3. Data: base composition, in mol %, of nucleic acid isolated from three sources.

   Sample	A	T	G	C
   P	31.0	31.0	19.0	19.0
   Q	17.5	17.5	32.5	32.5
   R	24.0	32.0	26.0	18.0

   Which statement is best supported by the data?

   Sample P is single-stranded, because it has the lowest percentage of G + C Sample Q would separate into single strands at a lower temperature than sample P All three samples are double-stranded DNA, because in each one the total purines equal the total pyrimidines Sample R is single-stranded, because %A does not equal %T and %G does not equal %C

   A low G + C content is normal for double-stranded DNA, the A + T to G + C ratio varies widely among species, and sample P satisfies both Chargaff equalities exactly. Strandedness is read from the equalities, not from the size of the GC fraction.

   This reverses the relationship. Sample Q is 65% G + C and sample P is 38%, and each G–C pair is held by three hydrogen bonds rather than two, so Q requires a higher temperature to melt.

   The premise is true and the conclusion does not follow: in sample R, A + G = 50.0 and T + C = 50.0, yet A ≠ T. Chargaff's rule for double-stranded DNA is the stricter pairwise equality, because every A is hydrogen-bonded to a specific T and every G to a specific C.

   Correct. In sample R, A is 24.0 while T is 32.0, and G is 26.0 while C is 18.0. With no complementary partner strand, nothing forces those equalities, so a molecule that breaks both of them is single-stranded.

4. Figure: one replication fork of a bacterial chromosome, drawn as the parental duplex opens toward the right.

   The upper parental strand runs 3' at the left end to 5' at the right end; the lower parental strand runs 5' at the left to 3' at the right. Against the upper parental strand a single new strand is drawn as one long arrow whose head points right, toward the fork. Against the lower parental strand three short new arrows are drawn, each with its head pointing left, away from the fork, and each begins with a short segment shaded differently from the rest of the arrow.

   Which statement about the three short new segments is correct?

   They form because helicase unwinds the lower parental strand more slowly than the upper one They are leading-strand segments, and the differently shaded pieces are the single-strand binding proteins that hold them open They are Okazaki fragments of the lagging strand; each begins at its own RNA primer and is extended 5'→3' away from the fork, because DNA polymerase can add nucleotides only to a free 3'-OH They are built in the 3'→5' direction, which is why they point away from the fork

   Helicase unwinds the double helix as a unit, so both templates become available at the same rate. The asymmetry comes from the antiparallel arrangement of the two parental strands, which is a matter of orientation rather than speed.

   The leading strand is the continuous arrow on the upper template, drawn as a single piece pointing toward the fork. The differently shaded starts are RNA primers laid down by primase, not binding proteins: single-strand binding proteins coat the separated parental DNA and are not part of the new strand.

   Correct. Because the lower template runs 5'→3' toward the fork, a new strand copied from it must grow away from the fork to keep its own 5'→3' direction. That forces discontinuous synthesis: the polymerase waits for more template, a new primer is laid, and another fragment is built, later joined by DNA polymerase I and ligase.

   No DNA polymerase synthesizes 3'→5'; the enzyme attaches each incoming nucleotide's phosphate to the free 3'-OH at the end of the growing chain, which makes 5'→3' the only possible direction. The fragments point away from the fork despite that rule, not because of an exception to it.

5. Figure: mean telomere length versus number of population doublings for two cultures of human fibroblasts.

   Both axes are linear: population doublings from 0 to 60 on the x-axis, mean telomere length from 0 to 12 kilobases on the y-axis. Curve A, untreated fibroblasts, starts at 10 kb and falls as a straight line to 5 kb at 50 doublings, where the cells stop dividing. Curve B, the same fibroblasts after an active telomerase gene was introduced, starts at 10 kb and stays within 0.3 kb of 10 kb all the way to 60 doublings, and those cells are still dividing.

   Which statement best explains the difference between the two curves?

   Telomerase proofreads and repairs mismatched bases, so curve B cells accumulate fewer replication errors Curve A loses about 0.1 kb per doubling because the outermost lagging-strand primer has no upstream DNA to replace it; telomerase in curve B rebuilds that lost terminal sequence each round The telomerase gene makes DNA polymerase III run faster, so curve B cells finish replication before any DNA is lost Curve B cells replicate their DNA conservatively, so the parental chromosome is never shortened

   Proofreading and mismatch repair are functions of DNA polymerase III and the repair system; telomerase is a reverse transcriptase that extends chromosome ends using its own RNA template. Error rate and end shortening are separate problems, and the graph measures length, not fidelity.

   Correct. Curve A falls 5 kb over 50 doublings, which is \(5/50 = 0.1\) kb, about 100 base pairs, lost per division, exactly the end-replication problem. Telomerase adds repeats back onto the 3' overhang, which is why curve B holds steady and those cells keep dividing past the point where curve A arrests.

   Speed is irrelevant to the problem. However fast the polymerase runs, it still cannot replace the terminal RNA primer on the lagging strand, because replacement requires a free 3'-OH upstream and there is no DNA there to provide one.

   Replication is semiconservative in all cells, as Meselson and Stahl showed, and a change in that mechanism is not something a single introduced gene could produce. Conservative replication would also not solve end shortening, since the new duplex would still be built by the same primer-dependent polymerase.

6. The template strand of a short gene reads 3'-TACCCGAAGGTTACT-5'. Which choice gives the mRNA transcript and the coding strand correctly?

   mRNA 5'-AUGGGCUUCCAAUGA-3'; coding strand 5'-ATGGGCTTCCAATGA-3' mRNA 5'-AGUAACCUUCGGGUA-3'; coding strand 5'-ATGGGCTTCCAATGA-3' mRNA 5'-AUGGGCUUCCAAUGA-3'; coding strand 5'-TACCCGAAGGTTACT-3' mRNA 5'-UACCCGAAGGUUACU-3'; coding strand 5'-TACCCGAAGGTTACT-3'

   Correct. Pairing A→U, T→A, C→G, G→C down the template gives 5'-AUGGGCUUCCAAUGA-3', which reads AUG GGC UUC CAA UGA: start, Gly, Phe, Gln, stop. The coding strand is that same sequence with T for U, which is the check worth running: the mRNA always matches the coding strand, never the template.

   The coding strand is right, but the mRNA has been written in the wrong direction; this is the correct transcript read 3' to 5' and then labeled as though it were 5' to 3'. Because the template is given 3'→5' from left to right, its complement comes out 5'→3' in the same left-to-right order.

   The mRNA is right, but the coding strand cannot be the template: those are the two different strands of the duplex. The coding strand is the one whose sequence the mRNA reproduces, so it reads 5'-ATGGGCTTCCAATGA-3'.

   This copies the template letter for letter, swapping U for T, instead of pairing each template base with its complement. Transcription is complementary base pairing, so a template T specifies an A in the RNA, not a U, and a transcript beginning UAC has no start codon.

7. Data: three human protein-coding genes compared. In each gene the first and last exon are always retained, and every internal exon can independently be included or skipped during splicing.

   Gene	P	Q	R
   Pre-mRNA length (nt)	12,000	4,000	2,500
   Mature mRNA length (nt)	1,500	3,200	1,000
   Number of exons	9	4	3

   Which statement is best supported by the data?

   Gene P has the largest fraction of its transcript removed as introns, 87.5%, and can generate the most distinct mature mRNAs, \(2^{7} = 128\) Gene Q generates the most distinct mature mRNAs, because its mature mRNA is the longest Gene R has the largest fraction removed, because it loses 1,500 nucleotides to splicing All three genes generate the same number of mature mRNAs, because splicing removes every intron from each transcript

   Correct. Gene P removes \((12{,}000-1{,}500)/12{,}000 = 87.5\%\) against 20.0% for Q and 60.0% for R, and with 9 exons it has 7 internal exons that can each be kept or skipped, giving \(2^{7} = 128\) combinations versus 4 for Q and 2 for R.

   Mature mRNA length says nothing about splice-variant number; what matters is how many internal exons are optional. Gene Q has only 2 internal exons, so it tops out at \(2^{2} = 4\) mature mRNAs even though its finished transcript is the longest of the three.

   Gene R loses 1,500 nucleotides, but gene P loses \(12{,}000 - 1{,}500 = 10{,}500\), so R does not lose the most in absolute terms either. As a fraction, R removes 60.0% and P removes 87.5%, so the claim fails on both readings.

   Alternative splicing means the same pre-mRNA can be cut in more than one way, some internal exons are removed along with the introns that flank them, which is precisely why one gene can specify several proteins. If splicing were fixed, there would be exactly one mature mRNA per gene.

8. An mRNA reads 5'-AUGCACGGAUCCUAG-3'. How many amino acids does the polypeptide contain, and what anticodon does the tRNA that delivers the third amino acid carry?

   5 amino acids; anticodon 3'-CCU-5' 4 amino acids; anticodon 3'-CCU-5' 4 amino acids; anticodon 5'-GGA-3' 3 amino acids; anticodon 3'-CCU-5'

   This counts the stop codon as a fifth amino acid. UAG is read by a release factor, not by a charged tRNA, and no residue is added when it enters the A site: the polypeptide is simply let go.

   Correct. The triplets are AUG CAC GGA UCC UAG, which is Met–His–Gly–Ser–stop, so four amino acids are joined. The third amino acid is glycine from the codon 5'-GGA-3', and its anticodon pairs antiparallel and complementary: 3'-CCU-5'.

   An anticodon is complementary to its codon, not identical to it. Writing 5'-GGA-3' simply repeats the codon, which would leave the tRNA unable to base-pair with the message at all.

   The anticodon is right, but the initiator methionine is a real residue in the polypeptide as it comes off the ribosome. It is often cleaved off afterward by a separate enzyme, which is why sickle-cell is called a change at the sixth residue of the mature β chain, but the ribosome still made it.

9. Data: growth of wild-type Neurospora crassa and three arginine-requiring mutants on minimal medium and on minimal medium supplemented with one intermediate of the arginine pathway, which runs precursor → ornithine → citrulline → arginine.

   Strain	Minimal	+ ornithine	+ citrulline	+ arginine
   Wild type	grows	grows	grows	grows
   arg-1	none	none	none	grows
   arg-2	none	none	grows	grows
   arg-3	none	grows	grows	grows

   Which strain carries a defect in the enzyme that converts ornithine to citrulline?

   arg-1 arg-2 arg-3 Wild type, because it grows on every medium

   arg-1 grows only when arginine itself is supplied, so it is blocked at the last step, converting citrulline to arginine. Supplying ornithine or citrulline does not help because neither can get past that block.

   Correct. A mutant grows on any intermediate downstream of its block. arg-2 is rescued by citrulline and arginine but not by ornithine, which places the broken enzyme exactly between ornithine and citrulline.

   arg-3 is rescued by all three supplements, including the earliest one, so its block lies before ornithine, in the conversion of the precursor to ornithine. The rule to remember is that the more supplements rescue a strain, the earlier its block.

   Wild type is the control, and it grows everywhere precisely because it has every enzyme in the pathway. Its row establishes that the media themselves support growth, so a "none" in any mutant row reflects a missing enzyme rather than a bad plate.

10. E. coli growing in medium containing both lactose and glucose makes only a small amount of β-galactosidase. Which statement best explains this result?

    Glucose binds the operator directly and physically blocks RNA polymerase The lacZ gene is deleted from the chromosome whenever glucose is available Glucose binds the lac repressor and activates it, so the repressor stays on the operator Allolactose has pulled the repressor off the operator, but glucose keeps cAMP low, so the CAP activator is inactive and RNA polymerase binds the promoter only weakly

    The operator is bound by the repressor protein, not by a sugar, and glucose never contacts the DNA. Mistaking a small-molecule signal for a DNA-binding protein loses the two-layer design that makes catabolite repression work.

    Regulation changes how much a gene is transcribed; it does not delete or restore genes. A cell that discarded lacZ in glucose could never induce it again when lactose returned, which is the opposite of what the wild type does.

    This describes the trp operon's corepressor logic, not the lac operon's. The lac repressor is switched off by allolactose and has no binding site for glucose; glucose acts one level up, through cAMP and CAP.

    Correct. Two switches control this operon. Lactose removes the block, and low glucose supplies the push: when glucose is abundant, cAMP falls, CAP cannot bind upstream of the promoter, and transcription is weak even with the operator free. The cell builds the enzymes only when lactose is present and the better fuel is not.

11. A hepatocyte and a neuron taken from the same person contain identical DNA sequences but very different sets of proteins. Which statement best explains this difference?

    Mutations accumulated during development changed the gene content of each lineage Each cell type discarded the genes it does not need as it differentiated The two cell types run different sets of transcription factors, which bind different enhancers and promoters, so different genes are transcribed The two cell types use different genetic codes, so the same mRNA specifies different proteins

    Somatic mutations do accumulate, but they are rare, random, and scattered, so they cannot produce the reproducible, cell-type-specific protein profiles that define a hepatocyte or a neuron in every person. A mechanism that explains a reliable outcome cannot itself be random.

    The stem states that the DNA sequences are identical, and that is the experimental fact: every somatic cell descends by mitosis from one zygote and inherits the whole genome. Differentiation changes which genes are read, not which genes are present.

    Correct. A cell's identity is essentially the set of transcription factors it is expressing; those proteins bind enhancers and promoters and recruit the transcription machinery to a particular subset of genes. Epigenetic marks such as promoter methylation and histone acetylation keep that subset stable through mitosis.

    The genetic code is nearly universal (the same codon table runs in bacteria, yeast, and every cell of your body), which is why a bacterium can be engineered to build human insulin. Two cells of one person certainly do not read different codes.

12. Sequence: the first six triplets of the template strand of a gene, written 3' to 5'.

    3'-TAC AGA GGT CTC TTC ACC-5'. A single base substitution changes the sixth template triplet from ACC to ACT. No other change is made to the gene.

    Which classification of this mutation, with its predicted effect on the polypeptide, is correct?

    Silent; the polypeptide is unchanged, because the code is redundant at the third codon position Missense; tryptophan at position 6 is replaced by cysteine Nonsense; the sixth mRNA codon becomes UGA, so translation terminates there and only five amino acids are joined Frameshift; every codon from the sixth onward is read out of register

    Redundancy often does make third-position changes silent, but not always, and this is one of the exceptions: the original codon UGG is the only codon for tryptophan, so any change to it must change the message. Checking the actual codon beats relying on the general rule.

    This is the outcome of a different substitution. Template ACC → ACA would give the mRNA codon UGU, which is cysteine, and that change would be missense. The substitution given here is to ACT, which specifies UGA.

    Correct. The template reads 3'-TAC AGA GGT CTC TTC ACC-5', so the mRNA is 5'-AUG UCU CCA GAG AAG UGG-3': Met–Ser–Pro–Glu–Lys–Trp. Changing the template's ACC to ACT makes that sixth codon UGA, a stop, so a release factor ends translation after five residues.

    Frameshifts are caused by insertions or deletions of a number of bases not divisible by three, which changes how the message is parsed downstream. A substitution swaps one base for another and leaves the total length, and therefore the reading frame, untouched.

13. Colonies were grown on a non-selective master plate, stamped with sterile velvet onto several plates containing an antibiotic, and resistant colonies appeared at the same coordinates on every independent replica. Which conclusion is best supported?

    Resistance alleles were already present in particular colonies on the master plate, before any cell had encountered the antibiotic The surviving bacteria altered their own DNA in response to the antibiotic, as individuals Resistance spread by conjugation between neighboring colonies on the replica plates The antibiotic induced resistance mutations, and it happened to induce them at the same coordinates on each plate

    Correct. Because each replica preserves the master plate's colony map, identical positions across replicas trace the resistance back to particular master-plate colonies, and picking that colony from the never-exposed master plate yields an already-resistant culture. Selection edits existing variation; it does not create it.

    Individual organisms do not evolve, and a bacterium cannot direct a mutation to the gene it needs. The variation arises randomly during replication, and the environment then determines which pre-existing variants leave descendants.

    Conjugation would transfer resistance to whichever cells happened to be adjacent on each plate, which would scatter survivors differently on every replica. The perfectly reproducible pattern is exactly what horizontal transfer after stamping cannot produce.

    Independent plates could not agree on coordinates by chance if the antibiotic were creating the mutations after stamping; matching positions can only mean the difference was already there in the master plate's colonies. This is the reversed cause and effect the exam penalizes most heavily.

14. Data: a 9,000 bp circular plasmid digested to completion with EcoRI alone, with BamHI alone, and with both enzymes together, then separated by gel electrophoresis.

    Digest	EcoRI only	BamHI only	EcoRI + BamHI
    Fragment sizes (bp)	9,000	6,000 and 3,000	4,000, 3,000, and 2,000

    Which statement is best supported by the gel?

    BamHI cuts once and EcoRI cuts twice The EcoRI site lies inside the 3,000 bp BamHI fragment EcoRI cuts the plasmid twice, because a circle must be cut twice before any fragment can be released EcoRI cuts once, and its site lies inside the 6,000 bp BamHI fragment, 4,000 bp from one BamHI site and 2,000 bp from the other

    This swaps the two enzymes. BamHI produces two fragments and so has two sites; EcoRI produces one full-length linear molecule and so has one.

    The 3,000 bp band appears unchanged in both the BamHI lane and the double-digest lane, which is the evidence that EcoRI does not cut inside it. The fragment that disappears in the double digest is the one containing the extra site.

    One cut in a circle produces one linear molecule of the full length, which is the single 9,000 bp band in the EcoRI lane. Two cuts would give two fragments summing to 9,000, as the BamHI lane shows.

    Correct. The single full-length EcoRI band means one site; BamHI's two bands summing to 9,000 bp mean two sites. In the double digest the 3,000 bp band survives untouched while the 6,000 bp fragment is replaced by 4,000 and 2,000, which places the EcoRI site within it and fixes both distances.

15. A student adds 0.25 µg of a plasmid carrying an ampicillin-resistance gene to competent E. coli, heat-shocks them, and brings the recovery mixture to a total volume of 1,000 µL. She spreads 200 µL on an LB/ampicillin plate and counts 96 colonies. What is the transformation efficiency?

    384 colonies per microgram 480 colonies per microgram 1,920 colonies per microgram 9,600 colonies per microgram

    This is \(96/0.25\), which ignores the fact that only one-fifth of the mixture was plated. The colonies you counted came from the DNA in 200 µL, not from all 0.25 µg, so the dilution has to be applied before dividing.

    480 is the number of colonies the whole 1,000 µL would have produced: a useful intermediate, but not an efficiency, because efficiency is colonies per microgram of plasmid DNA. The last step, dividing by 0.25 µg, is still missing.

    Correct. The fraction plated is \(200/1000 = 0.20\), so the DNA represented on the plate is \(0.25 \times 0.20 = 0.050\) µg, and \(96/0.050 = 1{,}920\) colonies/µg. Cross-check the other way: 96 colonies from one-fifth implies 480 in the whole mixture, and \(480/0.25 = 1{,}920\).

    This divides by 0.010 µg, as though one twenty-fifth of the mixture had been plated. Multiplying by the dilution factor twice, once on the volume and again on the mass, is the usual route to a five-fold overestimate, and the units do not catch it.

Free response

1. Data: a density-gradient investigation. E. coli were grown for many generations in medium whose only nitrogen source was ¹⁵N, then transferred at time zero into medium containing only ¹⁴N. A sample was taken each generation, its DNA was spun to equilibrium in a cesium chloride gradient, and the percentage of DNA at each of three band positions was recorded.

   Generations after transfer	0	1	2	3
   % at heavy position	100	0	0	0
   % at hybrid position	0	100	50	25
   % at light position	0	0	50	75

   Using the investigation and the data, answer (a) through (d).

   1. Identify the independent variable in this investigation, and describe the purpose of the generation-0 sample.
   2. Explain how the generation-1 result eliminates the conservative model of replication.
   3. Predict the percentage of DNA at the hybrid position after four generations, and show the calculation that supports your prediction.
   4. Justify the claim that replication is semiconservative rather than dispersive, using evidence from more than one generation.
   Your response
   Scoring notes

   • (a) Accept: the independent variable is the number of generations after the transfer to ¹⁴N medium; the generation-0 sample marks the position of fully heavy DNA in the gradient so that later bands can be located relative to it. Accept "a reference/control that establishes the heavy band position". Do not accept: naming ¹⁵N as the independent variable; naming the variable with no description of the sample's purpose; "it is the control" with no statement of what it establishes.
   • (b) Accept: the conservative model predicts that the two original strands stay together, so generation 1 would show a heavy band and a light band and no DNA in between; the data show 100% at a single intermediate position, which conservative replication cannot produce. Do not accept: "the DNA was hybrid, so it is semiconservative" with no statement of what conservative predicted; describing the conservative model without applying it to generation 1.
   • (c) Accept: 12.5%, supported by \(2/2^{4} = 2/16 = 0.125\): only the two original ¹⁵N strands can ever be in hybrid molecules, and the total number of molecules doubles each generation. Accept the equivalent \(1/2^{\,n-1}\) form. Do not accept: a value with no calculation; 6.25% (an extra generation); any answer implying the hybrid band disappears.
   • (d) Accept: dispersive replication predicts one band that drifts steadily lighter and never splits, because every strand is a patchwork; the data show a single hybrid band at generation 1 that then splits into two discrete bands at generations 2 and 3, with a constant hybrid position. The evidence must be tied to at least two generations. Do not accept: citing generation 1 alone (both models predict one intermediate band); a conclusion with no reference to what dispersive replication predicts.
   Show a 4/4 response

   a The independent variable is the number of generations the cells have grown in ¹⁴N after the switch. The generation-0 sample is taken before any replication in light medium, so it shows where fully heavy DNA sits in the gradient and gives me a reference position to compare every later band against.

   b If replication were conservative, the two original ¹⁵N strands would stay paired with each other and the brand-new duplex would be all ¹⁴N, so generation 1 should show two bands, one heavy and one light, and nothing in between. Instead 100% of the DNA sits at one intermediate position, which means every molecule contains one old strand and one new one.

   c Only the two original heavy strands exist, and they can never be in more than two molecules. After 4 generations there are \(2^{4} = 16\) molecules per starting duplex, so hybrid \(= 2/16 = 0.125 = 12.5\%\), and the other 87.5% is light.

   d The dispersive model says every strand is a patchwork of old and new, so there should be one band that gets steadily lighter each generation and never splits. The data show the opposite: at generation 1 there is one hybrid band, and at generations 2 and 3 it splits into two separate bands, with the hybrid band staying at exactly the same position while it shrinks from 50% to 25%. A fixed hybrid position plus a second, distinct light band is what semiconservative replication predicts and dispersive replication cannot give.

2. Data: β-galactosidase activity, in arbitrary units, for wild-type E. coli and three mutant strains grown in three media. Glycerol supplies carbon without lactose or glucose.

   Strain	Glycerol	+ lactose	+ lactose and glucose
   Wild type	4	2,000	80
   Strain 1	1,600	1,700	70
   Strain 2	5	60	55
   Strain 3	3	5	4

   Using the data, answer (a) through (d).

   1. Identify the strain that expresses the operon constitutively, and describe the evidence in the table that supports your choice.
   2. Explain why wild-type cells produce far less β-galactosidase in lactose plus glucose than in lactose alone.
   3. Calculate the fold induction by lactose for the wild type and for strain 1. Show your setup.
   4. Justify a conclusion about which component of the lac system is defective in strain 2, using evidence from the table.
   Your response
   Scoring notes

   • (a) Accept: strain 1, because its activity in glycerol is 1,600 units (nearly its fully induced level of 1,700), so it makes the enzyme with no lactose present, which is what constitutive means. Do not accept: naming strain 1 with no number cited; "it is always on" with no reference to the glycerol column; selecting strain 2 or 3.
   • (b) Accept: glucose lowers cAMP, so the CAP activator cannot bind upstream of the lac promoter and RNA polymerase binds only weakly; the repressor is already off the operator because allolactose is present, so the loss of activity is due to the missing activator, not to repression. Do not accept: "glucose represses the operon" with no mechanism; claiming the repressor rebinds in the presence of glucose; any answer that never mentions cAMP or CAP.
   • (c) Accept: wild type \(2{,}000/4 = 500\)-fold; strain 1 \(1{,}700/1{,}600 = 1.06\)-fold (accept ≈1.1 or "about 1-fold, essentially no induction"). Setup must be shown. Do not accept: differences instead of ratios (1,996 and 100); a ratio taken against the lactose-plus-glucose column; bare numbers with no setup.
   • (d) Accept: the cAMP–CAP activation system is defective, justified by the pattern that strain 2 is inducible but only to 60 units, a 12-fold rise, and that adding glucose barely changes it (55 versus 60, about a 9% drop) whereas glucose cuts the wild type by 96%. The repressor must still work because activity in glycerol is at background. Do not accept: "the repressor is broken" (contradicted by the glycerol column); naming CAP with no supporting numbers; a claim with no comparison to the wild-type glucose effect.
   Show a 4/4 response

   a Strain 1 is constitutive. In glycerol, with no lactose anywhere, it already shows 1,600 units, and adding lactose only takes it to 1,700, so it is making β-galactosidase all the time instead of only when the substrate is there. Either its repressor protein or its operator is broken.

   b In lactose alone, allolactose knocks the repressor off the operator and low glucose means cAMP is high, so CAP binds upstream and helps RNA polymerase onto the promoter. Adding glucose drops cAMP, CAP goes inactive, and without that activator the polymerase binds the lac promoter weakly, so transcription, and enzyme activity, falls to 80 units even though the operator is free.

   c Wild type: \(2{,}000/4 = 500\)-fold induction. Strain 1: \(1{,}700/1{,}600 = 1.06\)-fold, so essentially none.

   d Strain 2's CAP activation is broken. Its glycerol value of 5 units is background, so the repressor still holds the operator shut, and lactose does induce it, but only to 60 units, 12-fold instead of 500-fold. The clincher is the glucose column: glucose cuts wild type from 2,000 to 80, a 96% drop, but only takes strain 2 from 60 to 55, about 9%. A strain that barely notices glucose is a strain that was never getting the CAP boost in the first place.

3. Model: a scale drawing of a eukaryotic protein-coding gene, read left to right along the coding strand from 5' to 3'.

   The labeled segments, in order from left to right, are: a promoter of 150 base pairs; exon 1, 210 base pairs; intron 1, 1,500 base pairs; exon 2, 402 base pairs; intron 2, 900 base pairs; exon 3, 351 base pairs; and a polyadenylation signal at the right-hand end. Transcription begins at the first base of exon 1 and ends downstream of the polyadenylation signal. Assume that every base of the three exons is translated, including the stop codon.

   Using the model, answer (a) through (d).

   1. Identify the segments of the primary transcript that are absent from the mature mRNA, and describe the structure that removes them.
   2. Explain why the promoter sequence is not represented in the mature mRNA even though the gene cannot be expressed without it.
   3. Calculate the length of the mature mRNA's coding sequence and the number of amino acids in the polypeptide it specifies. Show your setup.
   4. A student claims that a single-base substitution anywhere inside intron 1 cannot affect the protein. Justify a response to that claim, using the model as your evidence.
   Your response
   Scoring notes

   • (a) Accept: intron 1 (1,500 nt) and intron 2 (900 nt) are transcribed but excised; the spliceosome, built from snRNPs, cuts them out and joins the flanking exons. Do not accept: naming the introns with no structure named; "enzymes remove them"; identifying the promoter or the poly-A tail as removed sequence.
   • (b) Accept: the promoter is the DNA sequence upstream of the transcription start site where RNA polymerase and general transcription factors assemble, and transcription begins downstream of it, so it is never copied into RNA; it acts as a binding site on DNA rather than as information carried in the message. Do not accept: "it is spliced out" (it is never transcribed); "it is not a gene"; a restatement that it is upstream with no consequence drawn.
   • (c) Accept: \(210 + 402 + 351 = 963\) nucleotides; \(963/3 = 321\) codons; one of those is the stop codon, so the polypeptide is 320 amino acids. Do not accept: including intron lengths; 321 amino acids (counting the stop codon as a residue); an answer with no setup.
   • (d) Accept: reject the claim, the spliceosome recognizes conserved sequences at each intron's 5' and 3' boundaries, and those bases lie inside intron 1, so a substitution at a splice site can prevent the cut and leave all 1,500 nucleotides of the intron in the mature mRNA; the ribosome then translates intron sequence until it reaches an in-frame stop codon, truncating or altering the protein. The evidence must name the splice site. Do not accept: "mutations can always matter" with no mechanism; agreeing with the student; claiming that a substitution anywhere in the intron would change the protein.
   Show a 4/4 response

   a Intron 1 (1,500 nt) and intron 2 (900 nt) are in the primary transcript but not in the mature mRNA. The spliceosome, a complex of snRNPs and proteins, recognizes each intron's boundaries, cuts it out, and ligates exon 1 to exon 2 and exon 2 to exon 3.

   b The promoter is a stretch of DNA upstream of where transcription starts. RNA polymerase and general transcription factors assemble on it to position the polymerase, and then the polymerase begins copying at the first base of exon 1, so the promoter is behind the enzyme the whole time and never gets transcribed. It works as a landing pad on the DNA, not as information in the message.

   c Coding sequence \(= 210 + 402 + 351 = 963\) nucleotides. That is \(963/3 = 321\) codons, and one of them is the stop codon, which does not add a residue. So the polypeptide is 320 amino acids long.

   d I would reject the claim. The spliceosome finds an intron by its conserved sequences at the 5' and 3' ends, and those bases are inside intron 1. If a substitution destroys the 5' splice site, intron 1 is not cut out and all 1,500 of its nucleotides stay in the mature mRNA, so the ribosome translates intron sequence until it hits an in-frame stop, usually within a few dozen codons, and the protein comes out truncated or completely different. So a base deep in the middle of the intron probably does nothing, but a base at its boundary is as damaging as a mutation in an exon.

4. A liquid culture of \(2.0 \times 10^{9}\) E. coli cells was grown from a single streptomycin-sensitive cell in antibiotic-free medium. The culture is then spread on plates containing streptomycin, and a small number of colonies grow. Answer (a) through (d).

   1. Identify the process that produced the streptomycin-resistance allele in this culture, and describe where in the cell the change occurred.
   2. Explain why exposure to streptomycin cannot have been the cause of the resistance alleles in this investigation.
   3. The spontaneous mutation rate at this locus is \(1.0 \times 10^{-8}\) per cell per generation. Calculate the expected number of resistant cells in the culture before it was plated. Show your setup.
   4. Justify the claim that the resistant cells were present before plating, using the design of a replica-plating experiment as your reasoning.
   Your response
   Scoring notes

   • (a) Accept: spontaneous mutation, an uncorrected error during DNA replication (or damage by a mutagen), in the bacterium's chromosomal DNA, in the gene encoding the ribosomal target of streptomycin. Accept "a point mutation in a chromosomal gene". Do not accept: "the bacteria adapted"; naming mutation with no location; horizontal gene transfer, which is excluded because the culture descends from one cell in pure medium.
   • (b) Accept: the culture was grown in antibiotic-free medium from a single cell, so every cell's genome was copied billions of times before any cell met streptomycin; the antibiotic was added only at plating and therefore could not have directed a change that was already present. Mutations are random with respect to what the cell needs. Do not accept: "mutations are random" with no reference to this culture; any answer that allows the antibiotic to induce a specific useful change.
   • (c) Accept: \((1.0 \times 10^{-8})(2.0 \times 10^{9}) = 20\) resistant cells. Setup must be shown. Do not accept: a number with no multiplication shown; \(2 \times 10^{-17}\) or other order-of-magnitude errors; an answer given as a rate rather than a count.
   • (d) Accept: in replica plating, colonies are grown on a non-selective master plate and stamped onto antibiotic plates, and resistant colonies appear at the same coordinates on every independent replica; cells picked from that position on the never-exposed master plate are already resistant, which shows the allele existed before exposure. Selection acts on existing variation. Do not accept: restating the claim; describing replica plating with no statement of what the matching coordinates prove; using the plate count from (c) as the justification.
   Show a 4/4 response

   a Spontaneous mutation produced it: an error during DNA replication that was not caught by proofreading or mismatch repair. The change is in the cell's chromosomal DNA, in the gene coding for the ribosomal protein that streptomycin binds, so the altered ribosome no longer binds the drug.

   b The whole culture grew from one sensitive cell in medium with no streptomycin in it, so all \(2.0 \times 10^{9}\) genomes were copied long before a single cell ever encountered the drug. Streptomycin was only added at the moment of plating, so it cannot have caused a change that had to exist already for those colonies to survive the plating. Mutations happen at random and the environment sorts them afterwards.

   c Expected resistant cells \(= (1.0 \times 10^{-8}\ \text{per cell})(2.0 \times 10^{9}\ \text{cells}) = 20\) cells. That handful is enough to explain the small number of colonies on the plate.

   d Replica plating settles it. You grow colonies on a plate with no antibiotic, press velvet onto it, and stamp several antibiotic plates so each replica keeps the same colony map. The resistant colonies come up at identical coordinates on every replica, which independent plates could never match by chance if the drug were creating the mutations, and when you go back and pick that colony from the master plate, which never met streptomycin, it is already resistant. That is direct evidence the allele was there first.

5. Data: an agarose gel. Lane 1 holds a size ladder; lane 2 holds a circular plasmid of unknown size digested to completion with EcoRI. Migration distance is measured from the well.

   Ladder band	1	2	3	4
   Distance migrated (mm)	15	25	35	45
   Fragment size (bp)	8,000	4,000	2,000	1,000

   Lane 2 contains exactly two bands, one at 20 mm and one at 30 mm.

   Using the gel, answer (a) through (d).

   1. Identify the electrode toward which the fragments migrate, and describe the feature of DNA that makes them move that way.
   2. Explain why smaller fragments migrate farther through the agarose than larger ones.
   3. Construct a calibration graph: put distance migrated in mm on the x-axis with a scale from 0 to 50, put \(\log_{10}\) of fragment size on the y-axis with a scale from 3.0 to 4.0, label both axes, plot the four ladder bands, and draw a best-fit straight line. Then use the line to estimate the size of each band in lane 2, and calculate the total size of the plasmid.
   4. Justify the claim that this plasmid carries exactly two EcoRI recognition sites, using evidence from the gel.
   Your response
   Scoring notes

   • (a) Accept: toward the positive electrode (anode), because the sugar–phosphate backbone carries a negative charge at every pH used for electrophoresis, so the whole molecule is an anion regardless of its base sequence. Do not accept: naming the electrode with no reason; attributing the charge to the nitrogenous bases; "DNA is attracted to the positive end" with no structural feature named.
   • (b) Accept: agarose is a porous mesh, and small fragments pass through the pores with less obstruction while large fragments are slowed by having to thread through, so in a fixed run time the small ones travel farther. Do not accept: "small ones are lighter"; "small ones move faster" with no reference to the gel matrix; restating the trend as its own explanation.
   • (c) Accept: distance on the x-axis (0–50 mm) and \(\log_{10}\) size on the y-axis (3.0–4.0), both labeled with units where applicable, the four ladder points plotted at (15, 3.90), (25, 3.60), (35, 3.30), (45, 3.00), and a best-fit straight line; then about 5,700 bp at 20 mm and about 2,800 bp at 30 mm (accept 5,500–5,900 and 2,700–2,900), and a plasmid of about 8,500 bp (accept 8,200–8,800). Do not accept: reversed axes; plotting raw size rather than its logarithm; interpolating on raw size, which gives 6,000 and 3,000 bp; a total with no sizes shown.
   • (d) Accept: the digest of a circular molecule produced two bands, and a circle must be cut twice before it yields two separate fragments, one cut converts a circle into a single full-length linear molecule. Therefore two sites. Do not accept: "two bands means two sites" stated for a linear molecule; counting the sites with no reference to the plasmid being circular; a claim with no evidence from the gel.
   Show a 4/4 response

   a The fragments move toward the positive electrode. Every nucleotide in the sugar–phosphate backbone carries a negatively charged phosphate group at the pH of the running buffer, so a DNA fragment of any sequence is a large anion and is pulled toward the anode.

   b Agarose sets as a mesh full of pores. A short fragment slips through the pores with little obstruction, while a long fragment has to worm its way through and gets hung up constantly, so in the same run time the short fragment ends up farther from the well.

   c I put distance migrated (mm) on the x-axis from 0 to 50 and \(\log_{10}\)(size in bp) on the y-axis from 3.0 to 4.0, labeled both, and plotted (15, 3.90), (25, 3.60), (35, 3.30), and (45, 3.00), then drew a best-fit line through them; the line drops 0.301 log units per 10 mm. At 20 mm, \(\log_{10}S = 3.90 - 0.15 = 3.75\), so \(S \approx 5{,}700\) bp. At 30 mm, \(\log_{10}S = 3.60 - 0.15 = 3.45\), so \(S \approx 2{,}800\) bp. Plasmid \(= 5{,}700 + 2{,}800 \approx 8{,}500\) bp.

   d The plasmid is circular, and cutting a circle once just opens it into one linear molecule of the full length: you would see a single band. Lane 2 has two bands, so the circle had to be cut in two places to release two separate pieces. That is the evidence for exactly two EcoRI sites, and the two fragment sizes tell me how far apart they are around the circle.

Lesson 7.1 · Unit 7 · CED topics 7.1–7.2

Natural selection and the evidence from the field

Darwin and Wallace presented the idea jointly in 1858, and its power is how little it assumes. If individuals differ, if some of that difference is inherited, and if more offspring are produced than the environment supports, then whichever variants leave more offspring become more common.

The trap is language. Selection does not try to improve anything and cannot reach into the future: it sorts variation that already exists, in the environment that exists now.

Lesson 7.2 · Unit 7 · CED topic 7.3

Artificial selection, and the variation selection acts on

Cabbage, broccoli, kale, kohlrabi, cauliflower, and Brussels sprouts are all Brassica oleracea: one species, one wild mustard ancestor, six different organs enlarged by breeders who kept what they liked. Maize came the same way from teosinte, a Mexican grass with a few hard-cased kernels in two rows, over roughly 9,000 years.

Artificial selection is natural selection with the environment swapped for a human decision, and it happens whether we mean it or not: every antibiotic course and every pesticide spray is a selection experiment run by accident.

Lesson 7.3 · Unit 7 · CED topics 7.4–7.5

Population genetics and Hardy–Weinberg equilibrium

Evolution is a change in allele frequencies over time, so to detect it you need to know what "no change" looks like. That is all Hardy–Weinberg is: a null model of a population in which nothing is happening, written precisely enough to test real data against.

Because it is a null hypothesis, the interesting outcome is failure: when observed genotype counts depart from the prediction, one of five conditions is being violated, and finding out which one is the biology.

Lesson 7.4 · Unit 7 · CED topics 7.4, 7.8

Mechanisms that change allele frequencies

Reject Hardy–Weinberg and the next question is which condition failed. Selection is one answer, and in small populations often not the main one: chance alone moves allele frequencies, and chance does not care whether an allele is useful.

Besides mutation, four mechanisms shift frequencies: drift, gene flow, nonrandom mating, and selection. Keep them separate: the exam hands you a scenario and asks which one is operating.

Lesson 7.5 · Unit 7 · CED topics 7.6–7.7

Evidence of evolution and common ancestry

No single observation would carry the argument for common ancestry. What carries it is that fossils, geography, anatomy, embryos, and molecules, five independent kinds of evidence, keep producing the same nested pattern of relatedness.

On the exam this is a reasoning task: you are handed data and asked which hypothesis it supports. Say what the evidence is, then say what it rules out.

Lesson 7.6 · Unit 7 · CED topic 7.9

Phylogeny and cladistics

A cladogram is not a picture of what happened. It is a hypothesis about branching order, built from data and revised when new data conflict with it.

What carries information is the branching order and the characters placed on the branches. What carries none is the left-to-right order of the tips, since any branch can be swapped around its node, or, in a plain cladogram, the length of the branches.