Nucleic Acids: Study Guide (Biology 1.6)

Information written in a chemical alphabet

Every cell has to store instructions and pass them on. It does this with nucleic acids: DNA (deoxyribonucleic acid) and RNA (ribonucleic acid). Like the other macromolecules in this unit, they are polymers: long chains of smaller monomers joined by covalent bonds. For nucleic acids, the monomers are nucleotides.

The key idea of this topic is simple. The backbone of a nucleic acid is the same all along its length. What changes from one nucleotide to the next is the base. So the information is carried in the order of the bases, much as the meaning of a sentence is carried in the order of its letters.

The nucleotide: three parts

Every nucleotide has three parts:

1. A five-carbon sugar (a pentose). DNA uses deoxyribose; RNA uses ribose.
2. A phosphate group, attached to the sugar’s carbon 5′.
3. A nitrogenous base, attached to the sugar’s carbon 1′. There are five bases in this topic: adenine (A), guanine (G), cytosine (C), thymine (T) and uracil (U). DNA uses A, G, C and T. RNA uses A, G, C and U.

The carbons of the sugar are numbered 1′ to 5′ (“one prime” to “five prime”). Two of these numbers matter most. Carbon 5′ carries the phosphate. Carbon 3′ carries an –OH (hydroxyl) group. Carbon 2′ is where the two sugars differ: ribose has an –OH there, while deoxyribose has only an –H. “Deoxy” means “missing an oxygen”.

Building a strand: the 5′ and 3′ ends

Nucleotides join into a strand when the phosphate of one nucleotide forms a covalent bond with the 3′ –OH of the sugar in the previous nucleotide. Repeating this gives a sugar–phosphate backbone, with the bases sticking out to one side. This is a dehydration synthesis reaction of the kind you met in Topic 1.3: overall, the equivalent of a water molecule is lost as each link forms. Hydrolysis (adding water) breaks these links again.

Because every link joins a 5′ phosphate to a 3′ –OH, a strand has direction, like a one-way street:

• At one end the last nucleotide has a free phosphate on carbon 5′. This is the 5′ end.
• At the other end the last sugar has a free –OH on carbon 3′. This is the 3′ end.
• Sequences are written from 5′ to 3′ unless the ends are labelled otherwise. So 5′-ATGC-3′ and 5′-CGTA-3′ are different molecules.

Growth happens at the 3′ end only. When a cell makes a new strand, each new nucleotide is joined by a covalent bond to the free 3′ –OH of the growing strand. The strand therefore grows in the 5′ → 3′ direction. A nucleotide cannot be added to the 5′ end, because that end has no free 3′ –OH to bond to.

Background (not needed for this topic): in cells, each incoming nucleotide arrives carrying three phosphate groups. Two are released as the link forms, and this release supplies the energy. You will see the enzymes involved in Unit 6.

DNA: an antiparallel double helix

A DNA molecule is usually made of two strands wound around each other in a double helix. The sugar–phosphate backbones are on the outside. The bases point inward and pair up across the middle.

Two rules describe the structure:

1. Complementary base pairing. A always pairs with T, and C always pairs with G. The pairs are held together by hydrogen bonds, the same type of weak attraction you met with water in Topic 1.1. An A–T pair has two hydrogen bonds; a C–G pair has three.
2. Antiparallel strands. The two strands run in opposite directions. If one strand runs 5′ → 3′ from top to bottom, its partner runs 5′ → 3′ from bottom to top.

Reading Figure 2. Read the left strand from its 5′ end: 5′-AGCT-3′. Read the right strand from its 5′ end, which is at the bottom: 5′-AGCT-3′. The two strands have the same sequence in this short example only by chance; usually they differ. Notice three features you should be able to describe in any such diagram:

• The backbones (alternating sugar and phosphate) are on the outside and are joined by covalent bonds.
• The bases meet in the middle and are joined by hydrogen bonds, which are much weaker. This is why the two strands can be separated (for example, by heating) without breaking either strand.
• Each 5′ end sits opposite a 3′ end. That is what antiparallel means.

Background: A and G are larger bases, and C, T and U are smaller ones. Each pair has one large and one small base, so every rung of the ladder is the same width. You will not be asked to draw the bases.

DNA and RNA compared

Feature	DNA	RNA
Sugar	deoxyribose (–H on carbon 2′)	ribose (–OH on carbon 2′)
Bases	A, G, C, T	A, G, C, U
Pairing	A–T, C–G	A–U, C–G (where RNA pairs)
Strands	typically double-stranded	typically single-stranded
Main job	long-term store of genetic information	copies and uses that information (Unit 6)

Notice the word typically. Some viruses carry single-stranded DNA or double-stranded RNA, and a single RNA strand can fold back and pair with itself in places. The rules still apply wherever pairing happens.

Worked example 1: writing a complementary strand

Question. One strand of a DNA molecule is 5′-ATGCTTAG-3′. (a) Write the complementary DNA strand, from 5′ to 3′. (b) Write the sequence of an RNA strand that could pair with the original strand, from 5′ to 3′.

(a)

1. Write the given strand with its ends: 5′-ATGCTTAG-3′.
2. Under each base write its partner (A↔T, G↔C). Because the strands are antiparallel, the new strand runs the other way, so label it 3′ on the left: 3′-TACGAATC-5′.
3. Rewrite it from its 5′ end by reading right to left: 5′-CTAAGCAT-3′.

Check. Read the answer from its 3′ end (right to left): T-A-C-G-A-A-T-C. Each base pairs with the original read from its 5′ end: A–T, T–A, G–C, C–G, T–A, T–A, A–T, G–C. All eight pairs are correct.

(b) RNA uses U instead of T. Pairing with the original strand: under A write U, under T write A, under G write C, under C write G. That gives 3′-UACGAAUC-5′, which is 5′-CUAAGCAU-3′.

Interpretation. A sequence without labelled ends is ambiguous. The most common mistake is to write the complement (TACGAATC) and label it 5′ to 3′ from left to right. That describes a different molecule.

Worked example 2: using base pairing to calculate percentages

Question. (a) A sample of double-stranded DNA contains 27% thymine. Calculate the percentage of each other base. (b) A second DNA sample contains A 31%, T 19%, G 22% and C 28%. What can you conclude about its structure?

(a)

1. In double-stranded DNA, every T is paired with an A. So %A = %T = 27%.
2. A and T together make 27 + 27 = 54%.
3. The rest, 100 − 54 = 46%, is G and C.
4. Every G is paired with a C, so they are equal: G = C = 46 ÷ 2 = 23% each.

Check. 27 + 27 + 23 + 23 = 100%.

(b)

1. If the sample were double-stranded, A would equal T and G would equal C.
2. Here A (31%) is not equal to T (19%), and G (22%) is not equal to C (28%).
3. Thymine is present, so it is DNA, not RNA.
4. Conclusion: the sample is most likely single-stranded DNA, for example the genome of a virus. In a single strand the bases are not paired, so nothing forces A = T or G = C.

Interpretation. The pairing rules apply only to bases that are paired. They let you predict the composition of double-stranded molecules, and they let you use composition data as evidence about structure.

Worked example 3: comparing two DNA segments

Question. Two double-stranded DNA segments are each 10 base pairs long. One strand of each is shown.

• Segment 1: 5′-GCATGCGTAA-3′
• Segment 2: 5′-ATTATGCATA-3′

(a) For each segment, count the hydrogen bonds holding the two strands together. (b) Count the covalent links between nucleotides in one strand of a 10-nucleotide segment. (c) Predict which segment separates into single strands more easily when heated, and explain.

(a)

1. Segment 1 has 3 G and 2 C, so 5 C–G pairs and 5 A–T pairs. Hydrogen bonds = (5 × 3) + (5 × 2) = 15 + 10 = 25.
2. Segment 2 has 1 G and 1 C, so 2 C–G pairs and 8 A–T pairs. Hydrogen bonds = (2 × 3) + (8 × 2) = 6 + 16 = 22.

(b) Ten nucleotides in a row need one link between each neighbouring pair: 10 − 1 = 9 covalent links per strand (18 for both strands).

(c) Segment 2 should separate more easily. It has fewer C–G pairs, so fewer hydrogen bonds hold its strands together (22 against 25). Heating breaks hydrogen bonds between the strands, not the covalent backbone, so each strand stays intact.

Interpretation. This is a prediction from a model. Real separation temperatures also depend on other factors, such as salt concentration and how neighbouring base pairs stack on one another, so test such predictions with data.

Common misconceptions

• “The information is in the sugars and phosphates.” The backbone repeats the same pattern all along. The information is in the order of the bases.
• “Hydrogen bonds hold the nucleotides of one strand together.” No. Nucleotides in a strand are linked by covalent bonds. Hydrogen bonds join bases on opposite strands.
• “Both strands run 5′ to 3′ in the same direction.” They are antiparallel: each 5′ end faces a 3′ end.
• “A new nucleotide can be added at either end.” Only the 3′ end has the free –OH that the new nucleotide bonds to, so strands grow 5′ → 3′.
• “The complement of 5′-ATGC-3′ is 5′-TACG-3′.” The complement must be reversed to read 5′ to 3′: 5′-GCAT-3′.
• “A = G and T = C in DNA.” The equal pairs are A = T and G = C, and only in double-stranded DNA.
• “RNA has thymine” or “DNA has uracil.” In this course, DNA uses T and RNA uses U.
• “RNA is just DNA with one strand missing.” RNA also has a different sugar (ribose) and uracil in place of thymine.

Where this leads

Nucleic acids return in Unit 6, where you will see how DNA is copied and how the information in its base sequence is used to make RNA and then proteins. Proteins are the next macromolecule: continue with Topic 1.7, Proteins. To check this topic first, try the practice questions, then use the revision notes and the checklist to consolidate. The previous topic is Topic 1.5, Lipids.