DNA Structure and Base Pairing Revision Notes | A Level AQA Chemistry

1. Why DNA Structure Matters in A-Level Chemistry

DNA carries the genetic code for every living organism, but from a chemistry perspective it is a fascinating macromolecule held together by the same types of covalent and hydrogen bonds you have already studied in organic chemistry. Understanding its structure at the molecular level — the components, the bonding, and the geometry — is exactly what AQA expects you to be able to explain and evaluate.

This lesson builds directly on your knowledge of hydrogen bonding, functional groups, and carbonyl/amino chemistry. You will see how those principles scale up to explain one of the most elegant molecular structures in nature. The Data Booklet provides the structures of all four bases, 2-deoxyribose, and the phosphate ion — so your job is not to memorise every atom, but to understand how they connect and why the pairing rules are what they are.

2. The Three Components of a Nucleotide

Think of a nucleotide as a three-part assembly. Each part has a distinct chemical identity, and the order in which they connect is fixed. Getting this sequence right is the foundation for everything else in this topic.

The phosphate ion ($\text{PO}_4^{3-}$) is the outermost component. The phosphate groups are acidic — they can donate protons — which is why DNA is called an acid (the name deoxyribonucleic acid reflects this). At physiological pH, these groups are deprotonated and carry a negative charge, making the overall molecule negatively charged. The 2-deoxyribose sugar sits in the middle, acting as the scaffold that links the phosphate to the base. The base is the variable part: it is the base sequence along the strand that encodes genetic information.

The prefix '2-deoxy' tells you something chemically precise: compared with ribose (the sugar in RNA), 2-deoxyribose is missing the hydroxyl ($-\text{OH}$) group at the 2' carbon, replaced by a hydrogen atom. This single structural difference makes DNA more chemically stable than RNA, which is one reason DNA is the long-term storage molecule for genetic information.

The four bases divide into two structural families. Adenine and guanine are purines — they have a fused double-ring structure. Cytosine and thymine are pyrimidines — they have a single-ring structure. This size difference is not incidental: a purine always pairs with a pyrimidine, which keeps the width of the double helix constant along its entire length.

3. Hydrogen Bonding Between Base Pairs

Now that you know what a nucleotide is, the next question is: how do two separate strands of nucleotides hold together? The answer lies in hydrogen bonding between the bases — and the geometry of those hydrogen bonds is what makes the pairing rules so specific.

Recall from your earlier study of hydrogen bonding that an H-bond forms between a hydrogen atom covalently bonded to a highly electronegative atom (N or O) and a lone pair on another electronegative atom. The bases in DNA carry $-\text{NH}$ and $\text{C=O}$ groups positioned precisely so that adenine's donor and acceptor groups are geometrically complementary to thymine's, and guanine's to cytosine's. No other combination produces the correct geometry for stable hydrogen bonds across the helix.

Adenine and thymine form two hydrogen bonds: one between an $-\text{NH}$ on adenine and a $\text{C=O}$ on thymine, and one between a lone pair on N of adenine and an $-\text{NH}$ on thymine. Guanine and cytosine form three hydrogen bonds, making the G–C pair more thermally stable. A DNA sequence with a higher proportion of G–C pairs requires more energy to separate the two strands — a fact that has real consequences in molecular biology and is a favourite context for AQA evaluation questions.

It is worth pausing to appreciate why this specificity matters chemically. The hydrogen bonds themselves are individually weak (around 20 kJ mol$^{-1}$ each), but DNA contains millions of base pairs, so the cumulative effect of all those hydrogen bonds — plus additional stabilisation from hydrophobic stacking interactions between adjacent base pairs — makes the double helix remarkably stable under normal physiological conditions.

4. From Nucleotides to a Polynucleotide Strand

Individual nucleotides link together to form a polynucleotide strand through covalent bonds between the phosphate group of one nucleotide and the sugar of the next. This creates a repeating sugar–phosphate backbone, with the bases projecting inward. The backbone is the structural scaffold; the bases are the information-carrying component.

The linkage between nucleotides is a phosphodiester bond — a covalent bond in which the phosphate group bridges two sugar molecules. This bond forms between the 3' carbon of one deoxyribose and the 5' carbon of the next via the phosphate, giving the strand a defined directionality: a 5' end (where the phosphate is free) and a 3' end (where the hydroxyl group is free). This directionality is chemically significant and is why the two strands in a double helix run antiparallel to each other.

Worked example: Tracing the Connectivity of a Dinucleotide

Describe, in order, the covalent bonds that connect a phosphate ion, 2-deoxyribose, and adenine in a single nucleotide, and then explain how a second nucleotide (carrying thymine) would extend the chain.

1. Phosphate ion $\rightarrow$ covalent bond $\rightarrow$ 5' carbon of 2-deoxyribose — The phosphate attaches to the 5' position of the sugar. This is the outermost bond in the nucleotide.
2. 2-Deoxyribose $\rightarrow$ covalent bond (N-glycosidic bond) $\rightarrow$ adenine base — The base attaches to the 1' carbon of the sugar via a nitrogen atom on the base. This is the N-glycosidic bond.
3. 3'-OH of first deoxyribose $\rightarrow$ phosphate of second nucleotide $\rightarrow$ 5' carbon of second deoxyribose — The phosphate of the incoming nucleotide bridges the 3' carbon of the first sugar and the 5' carbon of the second sugar, forming a phosphodiester bond and extending the chain.
4. 5'-phosphate–deoxyribose(A)–3'–5'–phosphate–deoxyribose(T)–3'-OH — The chain now has a defined 5' to 3' direction. The bases (A and T) project away from the backbone and are available for hydrogen bonding with a complementary strand.

Answer: A nucleotide is assembled as: phosphate ion → 2-deoxyribose (via 5' carbon) → base (via N-glycosidic bond). Nucleotides are then linked 3' to 5' through phosphodiester bonds, creating a directional polynucleotide strand with a free 5'-phosphate at one end and a free 3'-hydroxyl at the other.

5. The Double Helix: Two Complementary Strands

With two polynucleotide strands formed, the final structural picture comes together. The two strands align so that their bases face inward and pair up according to the complementary base-pairing rules. The paired strands then twist around each other to form the double helix — the iconic structure first described by Watson and Crick in 1953, using X-ray diffraction data.

Three structural features define the double helix. First, the two strands are antiparallel: one strand runs 5' to 3' in one direction, and the complementary strand runs 5' to 3' in the opposite direction. Second, the strands are complementary: the base sequence of one strand completely determines the base sequence of the other, because of the specific A–T and G–C pairing rules. Third, the overall shape is a right-handed helix, with the sugar–phosphate backbones on the outside and the base pairs stacked on the inside like rungs on a twisted ladder.

The antiparallel arrangement is not arbitrary — it is a direct consequence of the geometry of the phosphodiester backbone and the hydrogen-bonding geometry of the base pairs. Only when the strands run in opposite directions do the donor and acceptor groups on complementary bases align correctly to form stable hydrogen bonds. This is a point that examiners frequently probe with 'explain' questions, so it is worth being precise: the antiparallel orientation is required for complementary base pairing to occur.

6. Applying Base-Pairing Rules: A Quantitative Perspective

The complementary base-pairing rules have a quantitative consequence known as Chargaff's rules: in any DNA sample, the molar proportion of adenine equals that of thymine, and the molar proportion of guanine equals that of cytosine. This follows directly from the fact that every A on one strand is paired with a T on the other, and every G is paired with a C. AQA questions sometimes give you base composition data and ask you to calculate or evaluate the proportion of a specific base — the step-by-step below shows you how to handle this.

Worked example: Calculating Base Composition from Chargaff's Rules

A sample of double-stranded DNA is analysed and found to contain 22% adenine by mole. Calculate the percentage of guanine in this DNA sample. Show your working.

1. If A = 22%, then T = 22% — Because every adenine on one strand pairs with a thymine on the complementary strand, the molar percentages of A and T are always equal in double-stranded DNA.
2. A + T = 22% + 22% = 44% — These two bases together account for 44% of all bases in the sample.
3. G + C = 100% − 44% = 56% — The four bases must sum to 100%, so G and C together make up the remaining 56%.
4. G = C = 56% ÷ 2 = 28% — Because every guanine pairs with a cytosine, G and C are present in equal proportions. Dividing 56% equally gives 28% each.
5. A + T + G + C = 22% + 22% + 28% + 28% = 100% ✓ — All four base percentages sum to 100%, confirming the answer is internally consistent.

Answer: Guanine = 28%

7. Exam Technique and Common Misconceptions

AQA questions on DNA structure tend to use the command words 'describe', 'explain', and 'evaluate'. Each demands a different type of response. A 'describe' answer states what the structure is or what happens — no reasons required. An 'explain' answer must give the chemical reasoning: name the bond type, identify the groups involved, and state the consequence. An 'evaluate' answer requires you to weigh evidence and reach a conclusion — for example, comparing the thermal stability of two DNA sequences with different G–C content.

When writing about hydrogen bonding in base pairs, precision in language earns marks. Do not write 'the bases attract each other' — write 'hydrogen bonds form between the N–H groups and C=O groups of complementary bases'. Similarly, when describing the double helix, use the term 'antiparallel' explicitly and state that the sugar–phosphate backbone is on the outside with the bases on the inside. These are the phrases that appear in mark schemes.

Base pairing questions become simpler if you separate the two rules: A pairs with T, and G pairs with C. Once one base percentage is known, the complementary base must match it and the remaining percentage is split equally between the other complementary pair.