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This lesson covers the structure and function of DNA and RNA — the two types of nucleic acid — as required by the Edexcel A-Level Biology B specification (9BI0), Topic 1: Biological Molecules. You need to understand the structure of nucleotides, the formation of polynucleotides, the double helix structure of DNA, complementary base pairing, and the differences between DNA and RNA.
Nucleic acids are large biological molecules that store and transmit genetic information. They are polymers of nucleotides.
There are two types:
| Nucleic Acid | Full Name | Location | Function |
|---|---|---|---|
| DNA | Deoxyribonucleic acid | Nucleus (and mitochondria/chloroplasts) | Stores the genetic code — the instructions for making proteins |
| RNA | Ribonucleic acid | Nucleus and cytoplasm | Involved in protein synthesis — mRNA, tRNA and rRNA |
A nucleotide is the monomer of nucleic acids. Each nucleotide consists of three components:
The bases are divided into two groups based on their chemical structure:
| Category | Bases | Number of Rings | Found In |
|---|---|---|---|
| Purines | Adenine (A) and Guanine (G) | Two rings (larger) | Both DNA and RNA |
| Pyrimidines | Cytosine (C), Thymine (T) and Uracil (U) | One ring (smaller) | C in both; T in DNA only; U in RNA only |
Key Definition: A nucleotide consists of a pentose sugar, a phosphate group and a nitrogenous base. A nucleoside is a nucleotide without the phosphate group (just sugar + base).
| Feature | DNA Nucleotide | RNA Nucleotide |
|---|---|---|
| Pentose sugar | Deoxyribose (–H at carbon 2) | Ribose (–OH at carbon 2) |
| Bases | A, T, C, G | A, U, C, G |
| Unique base | Thymine (T) | Uracil (U) |
The key chemical difference between deoxyribose and ribose is at carbon 2: deoxyribose has an –H group (one fewer oxygen atom, hence "deoxy-"), while ribose has an –OH group.
Nucleotides are joined together by condensation reactions to form polynucleotides (long chains of nucleotides). The bond that forms is called a phosphodiester bond.
A phosphodiester bond forms between:
This creates a sugar–phosphate backbone that runs along the length of the polynucleotide. The nitrogenous bases project from the backbone.
Each polynucleotide strand has two distinct ends:
By convention, the sequence of bases in a nucleic acid is written in the 5' → 3' direction.
Key Definition: A phosphodiester bond is a covalent bond between the phosphate group of one nucleotide and the sugar of the next, linking them together in a polynucleotide chain. It involves two ester bonds (one to each sugar), with the phosphate group forming a bridge.
DNA consists of two polynucleotide strands wound around each other to form a double helix. The structure was determined by James Watson and Francis Crick in 1953, based on:
| Feature | Detail |
|---|---|
| Two strands | Two polynucleotide chains wound around each other |
| Antiparallel | One strand runs 5' → 3', the other runs 3' → 5' |
| Complementary base pairing | A pairs with T (two hydrogen bonds); C pairs with G (three hydrogen bonds) |
| Hydrogen bonds | Hold the two strands together between complementary bases |
| Sugar–phosphate backbone | On the outside of the helix — provides structural support |
| Bases | On the inside of the helix — stacked on top of each other |
| Width | Approximately 2 nm |
| One complete turn | Every 10 base pairs (approximately 3.4 nm) |
| Major and minor grooves | Grooves run along the helix where proteins (e.g. transcription factors) can bind |
The bases on opposite strands are held together by hydrogen bonds according to strict pairing rules:
| Base Pair | Hydrogen Bonds | Mnemonic |
|---|---|---|
| Adenine (A) — Thymine (T) | 2 hydrogen bonds | A–T (2 letters, 2 bonds) |
| Cytosine (C) — Guanine (G) | 3 hydrogen bonds | C–G (3 bonds — stronger) |
A purine always pairs with a pyrimidine (one large + one small = constant width of 2 nm across the helix). This is why A pairs with T (not with G) and C pairs with G (not with T).
Chargaff's Rules follow from complementary base pairing:
Exam Tip: If told that 32% of bases in a DNA molecule are adenine, you can deduce: %T = 32%, %C + %G = 100% − 64% = 36%, so %C = 18% and %G = 18%. Practice these calculations — they appear regularly in exam papers.
| Function | Explanation |
|---|---|
| Stores genetic information | The sequence of bases along the DNA strand encodes the instructions for making proteins (genes) |
| Replication | DNA can be copied precisely before cell division, ensuring each daughter cell receives an identical copy of the genetic information |
| Gene expression | Specific sections of DNA (genes) are transcribed into mRNA, which is then translated into proteins |
| Heredity | DNA is passed from parent to offspring, ensuring the transfer of genetic information between generations |
| Structural Feature | Functional Advantage |
|---|---|
| Double-stranded | Each strand is a template for the other during replication — ensures accuracy |
| Complementary base pairing | Allows precise copying (replication) and transcription |
| Two hydrogen bonds (A–T) and three (C–G) | Hold strands together firmly but can be separated by helicase for replication and transcription |
| Sugar–phosphate backbone | Provides structural stability (covalent phosphodiester bonds are strong) |
| Very long molecules | Can carry vast amounts of genetic information (human genome ≈ 3.2 billion base pairs) |
| Helical structure | Allows compact packaging within the nucleus (wound around histones to form chromatin) |
| Base sequence is stable | DNA is chemically stable (deoxyribose and the double-stranded structure make it resistant to hydrolysis) |
RNA (ribonucleic acid) is a polynucleotide that differs from DNA in several important ways:
| Feature | DNA | RNA |
|---|---|---|
| Sugar | Deoxyribose | Ribose |
| Bases | A, T, C, G | A, U, C, G |
| Strands | Double-stranded | Usually single-stranded |
| Length | Very long (millions of nucleotides) | Shorter (hundreds to thousands of nucleotides) |
| Stability | Chemically stable (long-term storage) | Less stable (short-lived — easily hydrolysed) |
| Location | Nucleus (mainly) | Nucleus and cytoplasm |
| Type | Abbreviation | Function |
|---|---|---|
| Messenger RNA | mRNA | Carries the genetic code from DNA in the nucleus to the ribosome in the cytoplasm for translation |
| Transfer RNA | tRNA | Carries specific amino acids to the ribosome during translation; has an anticodon that pairs with the mRNA codon |
| Ribosomal RNA | rRNA | Structural and catalytic component of ribosomes — the site of protein synthesis |
Exam Tip: When comparing DNA and RNA, always mention at least three differences: sugar (deoxyribose vs ribose), bases (T vs U), and number of strands (double vs single). For full marks, also mention relative stability and function.
Although DNA replication is covered in more detail in later topics, the basic principle is essential here because it relates directly to the structure of DNA:
This is called semi-conservative replication because each new DNA molecule conserves (retains) one of the original strands.
Exam Tip: The Meselson and Stahl experiment (1958) provided evidence for semi-conservative replication using ¹⁵N (heavy nitrogen) and ¹⁴N (light nitrogen) isotopes and density gradient centrifugation. Be prepared to describe and interpret this classic experiment.
This lesson sits in Edexcel 9BI0 Topic 1 — Biological Molecules, on the structure of nucleic acids: the nucleotide (pentose + phosphate + nitrogenous base); the phosphodiester backbone (condensation between the 3'-OH of one nucleotide and the 5'-phosphate of the next); DNA as a right-handed antiparallel double helix with complementary base pairing (A=T, two H-bonds; G≡C, three H-bonds); RNA as a typically single-stranded polynucleotide with ribose and uracil; and the three principal RNA species (mRNA, tRNA, rRNA) and their roles (refer to the official Pearson Edexcel 9BI0 specification for exact wording). Content statements paraphrase to: distinguish purines (A, G — two-ring) from pyrimidines (T, C, U — single-ring); explain why purine–pyrimidine pairing maintains a constant helix diameter (~2 nm); state Chargaff's rules for dsDNA (%A = %T, %G = %C); contrast deoxyribose (2'-H) with ribose (2'-OH); identify the 5'-end (free phosphate on C5) and 3'-end (free OH on C3) so the strands run antiparallel (5'→3' opposite 3'→5'); and link DNA structure to its functions in storage and replication. The lesson reactivates synoptically through Topic 3 (replication, transcription, translation), Topic 5 (mtDNA and cpDNA in endosymbiosis), Topic 6 (PCR and DNA profiling), Topic 8 (gene therapy and CRISPR-Cas9), and lesson 10 (ATP — a close structural relative of nucleotides).
Question (8 marks): A sample of double-stranded DNA is analysed and found to contain 30% adenine bases.
(a) State the percentage of each of the other three bases in the sample, showing your reasoning. (3)
(b) A second nucleic-acid sample contains 30% adenine, 18% uracil, 22% guanine and 30% cytosine. Identify the molecule and justify your answer using the data. (3)
(c) Explain, using the geometry of the double helix, why DNA must be antiparallel rather than parallel for complementary base pairing to work. (2)
Solution with mark scheme:
(a) M1 (AO1) — apply Chargaff's rules for dsDNA: %A = %T and %G = %C; total = 100%.
A1 (AO2) — %T = 30% (pairs with A); remaining (100 − 30 − 30) = 40% split equally between G and C → %G = 20%, %C = 20%.
A1 (AO3) — Chargaff equivalence holds because every A pairs with a T across strands, and every G with a C — a geometric consequence of antiparallel complementary pairing.
(b) M1 (AO2) — the molecule contains uracil, not thymine, so it is RNA.
A1 (AO2) — Chargaff's rules do not hold (%A ≠ %U; %G ≠ %C), confirming the molecule is single-stranded — there is no obligatory cross-strand pairing to enforce equimolarity.
A1 (AO3) — ssRNA can fold intramolecularly (e.g. tRNA cloverleaf) for local pairing, but global composition is not constrained by Chargaff's rules.
(c) M1 (AO1) — the two strands run 5'→3' and 3'→5' in opposite directions; this places the bases on each strand in the correct orientation to form planar H-bonded base pairs.
A1 (AO3) — a parallel arrangement would point the bases the same way on both strands, misaligning the H-bond donors and acceptors of A/T and G/C; base pairs could not form correctly and the helix could not close to a constant ~2 nm diameter.
Total: 8 marks (a: M1 A1 A1; b: M1 A1 A1; c: M1 A1). A* candidates name Chargaff's rules, identify uracil as the diagnostic feature of RNA, and give the geometric reason for antiparallel orientation rather than just stating it.
Question (6 marks): Compare and contrast the structure of DNA and RNA, and explain how their structural differences relate to their biological functions of long-term information storage and short-term information transfer respectively.
Mark scheme decomposition by AO:
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