Nucleic Acids

Nucleic acids are the information-carrying molecules of cells. DNA (deoxyribonucleic acid) stores the genetic instructions for making proteins, while RNA (ribonucleic acid) plays several roles in translating those instructions into functional proteins. Both are polymers of nucleotides.

By the end of this lesson you should be able to: describe the three components of a nucleotide and the differences between DNA and RNA nucleotides; explain phosphodiester-bond formation and strand directionality; account for the antiparallel double helix and complementary base pairing; use Chargaff's rules to calculate base composition; and describe semi-conservative replication and the Meselson–Stahl evidence.

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Nucleotide Structure

Key Definition: A nucleotide is the monomer of a nucleic acid. It consists of three components: a pentose sugar, a nitrogenous base, and a phosphate group.

DNA Nucleotides

• Pentose sugar: deoxyribose (C₅H₁₀O₄ — one fewer oxygen atom than ribose, specifically at the 2ʹ carbon position).
• Nitrogenous bases: adenine (A), thymine (T), cytosine (C), guanine (G).
• Phosphate group: PO₄³⁻, attached to the 5ʹ carbon of the sugar.

RNA Nucleotides

• Pentose sugar: ribose (C₅H₁₀O₅).
• Nitrogenous bases: adenine (A), uracil (U), cytosine (C), guanine (G).
• Phosphate group: same as DNA.

Purine and Pyrimidine Bases

The bases are classified into two groups:

• Purines (two-ring structure): adenine and guanine.
• Pyrimidines (single-ring structure): cytosine, thymine (DNA only), and uracil (RNA only).

Exam Tip: A useful mnemonic — Pyrimidines have a "y" in their name (cytosine, thymine, uracil — well, you can remember uracil as the odd one out). Alternatively, remember: the pure gold ring is a double ring — purines have two rings.

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Phosphodiester Bonds and Polynucleotide Chains

Nucleotides join together by condensation reactions between the phosphate group of one nucleotide and the hydroxyl group on the 3ʹ carbon of the sugar of the next nucleotide, forming a phosphodiester bond (a covalent bond with a phosphate group linking two sugar molecules).

This creates a sugar–phosphate backbone with alternating sugar and phosphate groups. The nitrogenous bases project out from the backbone.

A polynucleotide chain has directionality:

• The 5ʹ end has a free phosphate group on the 5ʹ carbon of the terminal sugar.
• The 3ʹ end has a free hydroxyl group on the 3ʹ carbon of the terminal sugar.

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DNA Structure — The Double Helix

James Watson and Francis Crick proposed the double helix model of DNA in 1953, building on X-ray crystallography data from Rosalind Franklin and Maurice Wilkins, and Chargaff's rules about base composition.

Key Features

1. Two polynucleotide strands wind around each other to form a right-handed double helix.
2. The strands are antiparallel: one strand runs 5ʹ→3ʹ while the other runs 3ʹ→5ʹ.
3. The sugar–phosphate backbones are on the outside of the helix.
4. The nitrogenous bases point inward and are joined by hydrogen bonds following complementary base pairing rules:
   • Adenine (A) pairs with thymine (T) — held by two hydrogen bonds.
   • Guanine (G) pairs with cytosine (C) — held by three hydrogen bonds.
5. A purine always pairs with a pyrimidine, ensuring a constant width of the helix (approximately 2 nm).

Chargaff's Rules

Erwin Chargaff (1950) found that in any DNA molecule:

• The amount of adenine equals the amount of thymine ([A] = [T]).
• The amount of guanine equals the amount of cytosine ([G] = [C]).
• Therefore, [purines] = [pyrimidines].

These observations were crucial evidence supporting Watson and Crick's complementary base pairing model.

Stability of the Double Helix

• The large number of hydrogen bonds between base pairs provides cumulative stability.
• Hydrophobic stacking interactions between adjacent base pairs (perpendicular to the helix axis) also contribute to stability.
• The sugar–phosphate backbone (with its covalent phosphodiester bonds) provides structural strength.

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RNA Structure

Feature	DNA	RNA
Sugar	Deoxyribose	Ribose
Bases	A, T, C, G	A, U, C, G
Strands	Double-stranded	Usually single-stranded
Helix	Double helix	No helix (though local folding can occur, e.g., in tRNA)
Function	Stores genetic information	Involved in protein synthesis (mRNA, tRNA, rRNA)
Location	Nucleus (mainly), mitochondria, chloroplasts	Nucleus, cytoplasm, ribosomes
Stability	Very stable (double helix, H bonds, deoxyribose)	Less stable (single strand, ribose has extra –OH making it more reactive)

Types of RNA

• mRNA (messenger RNA): carries the genetic code from DNA in the nucleus to ribosomes in the cytoplasm; the sequence of codons (triplets of bases) specifies the amino acid sequence of the polypeptide.
• tRNA (transfer RNA): small, clover-leaf shaped molecules; each carries a specific amino acid to the ribosome; has an anticodon that base-pairs with the complementary codon on mRNA.
• rRNA (ribosomal RNA): a structural and catalytic component of ribosomes; makes up about 60% of ribosome mass.

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Semi-Conservative Replication of DNA

DNA replication ensures that each daughter cell receives an identical copy of the genetic material. The process is described as semi-conservative because each new DNA molecule contains one original (parental) strand and one newly synthesised strand.

Steps of Replication

1. Helicase unwinds and separates the two strands of the double helix by breaking the hydrogen bonds between complementary base pairs, creating a replication fork.
2. Each exposed strand acts as a template. Free DNA nucleotides align opposite their complementary bases on the template strand (A with T, G with C).
3. DNA polymerase catalyses the formation of phosphodiester bonds between adjacent nucleotides, synthesising the new complementary strand in the 5ʹ→3ʹ direction only.
4. Because the two template strands are antiparallel, one new strand (the leading strand) is synthesised continuously toward the replication fork, while the other (the lagging strand) is synthesised in short fragments (Okazaki fragments) away from the fork.
5. DNA ligase joins the Okazaki fragments together by forming phosphodiester bonds between them, completing the lagging strand.
6. The result is two identical DNA molecules, each consisting of one parental strand and one new strand.

Meselson and Stahl Experiment (1958)

This classic experiment provided direct evidence for semi-conservative replication:

1. Escherichia coli bacteria were grown for many generations in a medium containing heavy nitrogen (¹⁵N), so that all their DNA contained ¹⁵N (heavy DNA).
2. The bacteria were then transferred to a medium containing light nitrogen (¹⁴N) and allowed to replicate.
3. DNA was extracted after each generation and separated by density-gradient centrifugation in caesium chloride (CsCl).

Results:

• Generation 0 (before transfer): all DNA was heavy (¹⁵N–¹⁵N) — one band at the bottom of the gradient.
• Generation 1: all DNA was intermediate density (¹⁵N–¹⁴N) — one band in the middle of the gradient. This ruled out conservative replication (which would have produced one heavy and one light band).
• Generation 2: DNA was 50% intermediate (¹⁵N–¹⁴N) and 50% light (¹⁴N–¹⁴N) — two bands, one in the middle and one near the top. This ruled out dispersive replication (which would have produced a single band of lighter-than-intermediate density).

These results are consistent only with the semi-conservative model of replication.

Exam Tip: You may be asked to predict or explain the banding pattern for subsequent generations. In generation 3, 25% of molecules would be intermediate and 75% light. In general, after n generations, 2 out of 2ⁿ molecules contain a ¹⁵N strand (intermediate), and the rest are entirely ¹⁴N (light).

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Worked Example: Chargaff and Meselson–Stahl Calculations

Base composition (Chargaff's rules). Because A pairs with T and G with C, the percentages of A and T are equal and the percentages of G and C are equal, and all four must sum to 100%.

Worked question: A double-stranded DNA sample is found to contain 20% cytosine. Calculate the percentage of each of the other three bases, and state the ratio of hydrogen bonds to base pairs implied by this composition.

Solution:

• $C = 20\%$C=20%, so by complementary pairing $G = 20\%$G=20% as well.
• The purines and pyrimidines account for the remainder: $A + T = 100 - (20 + 20) = 60\%$A+T=100−(20+20)=60%. Since $A = T$A=T, each is $30\%$30%. So $A = 30\%$A=30%, $T = 30\%$T=30%, $G = 20\%$G=20%, $C = 20\%$C=20%.
• Hydrogen bonds: each A–T pair has 2 bonds, each G–C pair has 3. Per 100 base pairs there are $(30 \times 2) + (20 \times 3) = 60 + 60 = 120$(30×2)+(20×3)=60+60=120 hydrogen bonds, i.e. a mean of 1.2 bonds per base — a useful proxy for thermal stability. A DNA molecule richer in G–C therefore denatures at a higher temperature, which is why "melting temperature" rises with G–C content.

Exam Tip: The single most common Chargaff slip is to treat a single-stranded value as double-stranded. Chargaff's rules ($A = T$A=T, $G = C$G=C) apply to a double-stranded molecule as a whole; they need not hold for one strand alone. Check whether the question specifies double-stranded before equating A with T.

Replication banding (Meselson–Stahl). After $n$n rounds of replication of fully-heavy DNA in light medium, the number of molecules is $2^n$2n, of which exactly 2 still contain one original heavy strand (and are therefore intermediate-density); the remaining $2^n - 2$2n−2 are fully light.

Worked question: What fraction of DNA molecules is intermediate-density after four generations?

Solution: Total molecules $= 2^4 = 16$=24=16; intermediate $= 2$=2; so the intermediate fraction is $\tfrac{2}{16} = \tfrac{1}{8} = 12.5\%$162​=81​=12.5%, and the light fraction is $87.5\%$87.5%. This quantitative prediction — halving the intermediate fraction each generation — is precisely why the experiment was decisive: neither the conservative nor the dispersive model reproduces this pattern.

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Summary

• Nucleotides consist of a pentose sugar, a phosphate group, and a nitrogenous base.
• DNA uses deoxyribose and the bases A, T, C, G; RNA uses ribose and the bases A, U, C, G.
• Phosphodiester bonds link nucleotides into polynucleotide chains with 5ʹ→3ʹ directionality.
• DNA is a double helix with antiparallel strands linked by complementary base pairing (A=T, G≡C).
• mRNA, tRNA, and rRNA each play distinct roles in protein synthesis.
• Semi-conservative replication involves helicase, DNA polymerase, and DNA ligase; the Meselson–Stahl experiment confirmed this model using density-gradient centrifugation with ¹⁵N/¹⁴N labelling.

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A-Level Deep Dive

Spec mapping

This lesson is mapped to AQA 7402 Section 3.1.5 — Nucleic acids (refer to the official AQA specification document for exact wording). It covers nucleotide structure (pentose, base, phosphate), the differences between DNA and RNA, complementary base pairing, the antiparallel double helix, the three RNA classes (mRNA, tRNA, rRNA), and semi-conservative replication including the Meselson–Stahl evidence. Examined directly on Paper 2 and synoptically on Paper 1 and Paper 3.

Historical context: the double helix model is associated with James Watson and Francis Crick (1953), informed by X-ray crystallographic data from Rosalind Franklin and Maurice Wilkins at King's College London and by Erwin Chargaff's base-composition rules. AQA expects you to name these contributors but, per platform citation standards, all references are paraphrased — never put verbatim quoted words in their mouths. Meselson and Stahl (1958) provided the definitive experimental confirmation of semi-conservative replication using ¹⁵N density-gradient centrifugation.

Why complementary base pairing produces constant helix width

A purine (two-ring) always pairs with a pyrimidine (one-ring). A-T and G-C base pairs are both approximately 1.1 nm wide, giving the double helix a uniform diameter of approximately 2 nm. If two purines tried to pair, the helix would be ~1.4 nm at that point and would bulge; if two pyrimidines paired, it would constrict. The constancy of width is therefore a consequence of the geometry of base pairing — examiners reward this mechanistic explanation rather than "purines pair with pyrimidines".

graph LR
    A["DNA nucleotide<br/>= deoxyribose + base + PO₄³⁻"] --> B["Phosphodiester bond<br/>5' phosphate ↔ 3' hydroxyl"]
    B --> C["Polynucleotide strand<br/>sugar-phosphate backbone<br/>directional 5' → 3'"]
    C --> D["Double helix<br/>2 antiparallel strands<br/>complementary base pairing"]
    D --> E["A=T (2 H-bonds)<br/>G≡C (3 H-bonds)"]
    D --> F["Width ≈ 2 nm<br/>constant"]
    D --> G["Replication template"]

    style D fill:#3498db,color:#fff
    style G fill:#27ae60,color:#fff

Synoptic links

This lesson connects to:

1. AQA 7402 Section 3.2 — Cells (organelles): DNA in the nucleus, mitochondrial DNA, chloroplast DNA. Nuclear DNA is packaged with histones into chromatin (eukaryotes); prokaryotic DNA is naked circular DNA in the nucleoid plus plasmids. Mitochondrial / chloroplast DNA reflects the endosymbiotic origin of these organelles.
2. AQA 7402 Section 3.4 — DNA and protein synthesis: the central dogma DNA → mRNA → protein is built directly on this lesson. Transcription requires RNA polymerase reading the template strand 3'→5' while synthesising mRNA 5'→3'. Translation uses tRNA anticodons base-pairing with mRNA codons at the ribosome.
3. AQA 7402 Section 3.8 — Gene expression control: epigenetic modifications (methylation of cytosine, histone modification), gene mutation, and gene technology (PCR, recombinant DNA, gel electrophoresis) all depend on nucleic acid chemistry from this lesson.
4. AQA 7402 Section 3.7 — Inheritance and evolution: DNA sequence variation underlies all evolutionary change; mitochondrial DNA is used for matrilineal phylogeny; horizontal gene transfer in bacteria depends on DNA structural features (plasmids, transposons).

Specimen question modelled on the AQA paper format