Lesson 1.6 · Unit 1 · CED topic 1.6
Nucleic acids
A cell has to store the instructions for building every protein it will ever need, copy them accurately each time it divides, and read them when a protein is needed. Nucleic acids do all of that. They store information in a very simple way: as a sequence. The order of four kinds of subunit along a strand is the message, much as the order of letters makes the words on this page. Every other topic in genetics depends on understanding the structure in this lesson.
Describe the structure and function of DNA and RNA. The CED states that the molecular structure of specific nucleotides is beyond the scope of the exam. You should know the parts of a nucleotide and how they are arranged, not draw each base.
The nucleotide: three parts
The monomer of a nucleic acid is the nucleotide. Each nucleotide has a five-carbon sugar, a phosphate group, and a nitrogenous base. (EK 1.6.A.1)
- Adenine (A) and guanine (G) are larger bases with two rings (purines).
- Cytosine (C), thymine (T), and uracil (U) are smaller bases with one ring (pyrimidines).
- DNA uses A, T, G, and C. RNA uses A, U, G, and C: uracil takes the place of thymine.
The sugar is deoxyribose in DNA and ribose in RNA. Ribose has an extra −OH on carbon 2′; deoxyribose has only a hydrogen there, which is why it is “deoxy.” The ′ symbol (read “prime”) separates the numbering of the sugar's carbons from the numbering of the atoms in the base.
A strand has direction
Nucleotides join in a linear sequence with two different ends, defined by the sugar: the 5′ end has a free phosphate on the 5′ carbon, and the 3′ end has a free hydroxyl on the 3′ carbon. Covalent bonds join the 3′ hydroxyl of one nucleotide to the phosphate on the 5′ carbon of the next. This is a phosphodiester bond. The repeating sugar and phosphate form the sugar-phosphate backbone, and the bases extend from it. During nucleic acid synthesis, new nucleotides are always added to the 3′ end of the growing strand.
The direction matters because everything the cell does with a nucleic acid (copying it, reading it) proceeds in a particular direction along the strand. By convention a sequence is written from the 5′ end to the 3′ end.
The double helix
DNA is an antiparallel double helix: two strands of nucleotides run in opposite directions, one 5′ to 3′ and the other 3′ to 5′. The bases point inward and pair through hydrogen bonds: adenine pairs with thymine (A–T, two hydrogen bonds) and cytosine pairs with guanine (C–G, three hydrogen bonds). In RNA, adenine pairs with uracil (A–U). Each pair joins one large purine to one small pyrimidine, so the helix has a constant width. Strands that fit together this way are complementary.
Notice what holds each part together. The sugar-phosphate backbone of a strand is held by strong covalent bonds, so a strand does not fall apart. The two strands are held to each other by many individually weak hydrogen bonds (lesson 1.1), so they can be pulled apart when the cell needs to copy or read the sequence, and then come back together. A stretch rich in G–C pairs has three hydrogen bonds per pair instead of two, so it takes more energy (a higher temperature) to separate.
| DNA | RNA | |
|---|---|---|
| Sugar | Deoxyribose | Ribose |
| Bases | A, T, G, C (contains thymine) | A, U, G, C (contains uracil) |
| Strands | Typically double stranded | Typically single stranded |
| Main role | Stores genetic information | Carries and uses the information (Unit 6) |
One strand of DNA is 5′-GATTAC-3′. Write the complementary strand, give it in the conventional 5′ to 3′ direction, and count the hydrogen bonds between the strands.
Pair each base. G pairs with C, A with T, T with A, T with A, A with T, and C with G. Written directly under the first strand, the partner reads 3′-CTAATG-5′ (it runs the opposite way).
Write it 5′ to 3′. Reverse the order: 5′-GTAATC-3′. Graders accept either form only if the 5′ and 3′ ends are labeled correctly.
Count hydrogen bonds. The pairs are G–C (3), A–T (2), T–A (2), T–A (2), A–T (2), C–G (3): \(3 + 2 + 2 + 2 + 2 + 3 = 14\) hydrogen bonds.
Around 1950 Erwin Chargaff measured the percentage of each base in DNA from different organisms. The values below are approximate historical data. Describe the pattern in the data and explain what it suggests about DNA structure.
| Source | % A | % T | % G | % C | A/T | G/C |
|---|---|---|---|---|---|---|
| Human | 30.9 | 29.4 | 19.9 | 19.8 | 1.05 | 1.01 |
| E. coli | 26.0 | 23.9 | 24.9 | 25.2 | 1.09 | 0.99 |
| Yeast | 31.3 | 32.9 | 18.7 | 17.1 | 0.95 | 1.09 |
Describe. In every organism the percentage of A is close to the percentage of T, and the percentage of G is close to the percentage of C (both ratios are near 1.0). The overall proportion of A+T versus G+C differs from species to species: human DNA is 60.3 percent A+T, E. coli 49.9 percent.
Explain. If A always pairs with T and G always with C, then each A on one strand has a T opposite it on the other strand, so the totals match: %A = %T and %G = %C. The data are therefore evidence for base pairing, and since each pair has one purine and one pyrimidine, the total purines (A + G) should equal the total pyrimidines (T + C). That also fits (the ratio is 1.03 for human DNA).
The small departures from 1.0 come from measurement error. The amounts of A+T versus G+C vary between species because organisms have different sequences, but the pairing rule is the same.
- Always label the ends. A complementary strand written without 5′ and 3′ labels, or with them in the wrong places, can lose the point.
- Two kinds of bond. Covalent bonds (phosphodiester) link nucleotides along one strand; hydrogen bonds link the bases of the two strands. A question about “what holds the two strands together” wants hydrogen bonds.
- New nucleotides go on the 3′ end, so a strand is made in the 5′ to 3′ direction.
- RNA has U where DNA has T, ribose where DNA has deoxyribose, and is usually a single strand.
- For a percentage problem: %A = %T, %G = %C, and all four add to 100. If \(\%G = 28\), then \(\%C = 28\) and \(\%A = \%T = (100 - 56)/2 = 22\).
- A DNA strand has the sequence 5′-TACGGATTC-3′. Write the complementary strand in the 5′ to 3′ direction.
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Pairing under the strand gives 3′-ATGCCTAAG-5′. Reversed so that it reads 5′ to 3′, it is 5′-GAATCCGTA-3′.
- List three structural differences between DNA and RNA.
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(1) DNA contains deoxyribose; RNA contains ribose. (2) DNA contains thymine; RNA contains uracil in its place. (3) DNA is typically double stranded; RNA is typically single stranded.
- In a sample of double-stranded DNA, 28% of the bases are guanine. Find the percentage of each of the other three bases.
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Cytosine pairs with guanine, so C = 28%. The remaining \(100 - 56 = 44\)% is A plus T, and since A pairs with T they are equal: A = 22% and T = 22%.
- A 12-base-pair DNA fragment contains 7 G–C pairs and 5 A–T pairs. How many hydrogen bonds hold the strands together, and how many phosphodiester bonds are in the whole fragment?
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Hydrogen bonds: \(7 \times 3 + 5 \times 2 = 21 + 10 = 31\). Each strand has 12 nucleotides and so \(12 - 1 = 11\) phosphodiester bonds; two strands have \(2 \times 11 = 22\).
- Two DNA molecules have the same length. One is 70% G–C and the other is 30% G–C. Predict which needs the higher temperature to separate its strands, and justify your prediction.
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Prediction: the 70% G–C molecule needs the higher temperature. Justification: the strands are held together by hydrogen bonds, and each G–C pair has three hydrogen bonds while each A–T pair has two. The molecule with more G–C pairs has more hydrogen bonds in total, so more energy is needed to break them and separate the strands.
- A growing DNA strand ends with a nucleotide whose 3′ carbon carries a free hydroxyl group. At which end of the strand will the next nucleotide be added, and which part of the incoming nucleotide bonds to it?
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It is added at the 3′ end. The 5′ phosphate of the incoming nucleotide forms a covalent (phosphodiester) bond with the free 3′ hydroxyl of the last nucleotide in the chain, so the strand grows in the 5′ to 3′ direction.