Nucleic Acids & Protein Synthesis
From the molecular architecture of DNA to the precise mechanism of polypeptide synthesis. Watson and Crick's double helix, semi-conservative replication, the universal triplet code, transcription and translation in eukaryotes, and how gene mutations propagate from base sequence to phenotype.
Structure of nucleic acids and replication of DNA
Nucleic acids store genetic information in the sequence of their bases and serve as the template for protein synthesis. DNA is the long-term hereditary molecule; RNA is the working messenger that transfers the message to the cytoplasm. Both are polymers of nucleotides, but key chemical differences allow them to play complementary roles.
The 9700 (2025–2027) syllabus does not require structural formulae for nucleotides or bases. Students must know what each component is and how they connect, but not the detailed organic chemistry. Don't waste time memorising ring structures.
Structure of a nucleotide
A nucleotide is the basic monomer of nucleic acids. Each nucleotide has three components joined by covalent bonds:
A 5-carbon sugar at the centre of each nucleotide. DNA nucleotides contain deoxyribose; RNA nucleotides contain ribose. The two sugars differ by one oxygen atom — ribose has an −OH at the 2′ carbon; deoxyribose has just an −H. This single difference makes RNA more reactive (and shorter-lived) than DNA.
A negatively charged phosphate group attached to the 5′ carbon of the pentose sugar. The phosphate provides the negative charge of nucleic acids and forms the backbone link between adjacent nucleotides.
A nitrogen-containing ring molecule attached to the 1′ carbon of the pentose sugar. The base carries the genetic information.
Five bases occur in nucleic acids: A (adenine), G (guanine), C (cytosine), T (thymine), U (uracil). DNA uses A, G, C, T; RNA uses A, G, C, U.
ATP — a phosphorylated nucleotide
ATP (adenosine triphosphate) is a special phosphorylated nucleotide. Its structure is:
- The pentose sugar is ribose
- The base is adenine
- Instead of one phosphate, ATP has three phosphate groups linked in a chain
ATP is the universal energy currency of cells. Hydrolysis of ATP to ADP + Pi (inorganic phosphate) releases energy that powers active transport, muscle contraction, anabolic reactions including DNA replication and protein synthesis, etc. ATP is covered in detail in Topic 12.
Purines and pyrimidines
The five bases divide into two structural classes:
| Class | Ring structure | Bases | Found in |
|---|---|---|---|
| Purines | Double ring (one 6-membered + one 5-membered, fused) | Adenine (A), Guanine (G) | Both DNA and RNA |
| Pyrimidines | Single ring (one 6-membered) | Cytosine (C), Thymine (T), Uracil (U) | C in both; T in DNA only; U in RNA only |
The width of the DNA double helix is uniform (~2 nm). For this to be possible, every base pair must consist of one big base (purine, 2 rings) plus one small base (pyrimidine, 1 ring). A purine–purine pair would be too wide; a pyrimidine–pyrimidine pair would be too narrow. The geometry rules out anything except A=T and G≡C as accurate (Watson-Crick) pairs.
The DNA double helix
A DNA molecule consists of two polynucleotide strands wound around each other to form a double helix. The structure was deduced by Watson and Crick in 1953, drawing on Rosalind Franklin's X-ray diffraction data and Erwin Chargaff's base-pairing rules.
Within each strand, nucleotides are joined by phosphodiester bonds — covalent bonds between the phosphate group of one nucleotide and the 3′ carbon of the next sugar. The repeating sugar-phosphate-sugar-phosphate sequence forms the strong outer backbone of the strand. The bases project inwards.
The two strands run in opposite directions. One strand runs 5′ to 3′; the other runs 3′ to 5′. This antiparallel arrangement is essential for accurate base pairing because the sugar geometry requires it. It also has consequences during replication: the lagging strand is made in short fragments, often called Okazaki fragments. The term is useful, but the key 9700 requirement is understanding discontinuous lagging-strand synthesis and DNA ligase joining the fragments — covered in the replication section below.
The two strands are held together by hydrogen bonds between bases on opposite strands. The pairing rules are absolute (Chargaff's rules):
- Adenine pairs with thymine via 2 hydrogen bonds
- Guanine pairs with cytosine via 3 hydrogen bonds
Because A always pairs with T and G with C, the sequence on one strand specifies the sequence on the other. This complementary base pairing is the structural basis for accurate DNA replication and for transcription — both processes use one strand as a template to build a complementary new strand.
G–C pairs have 3 hydrogen bonds; A–T pairs have 2. A DNA region with high GC content needs more energy to separate the strands (e.g. higher temperature). This is biologically and clinically important — thermophilic bacteria have GC-rich genomes; PCR primer design depends on melting temperatures determined largely by GC content.
DNA vs RNA — key differences
| Feature | DNA | RNA |
|---|---|---|
| Pentose sugar | Deoxyribose | Ribose |
| Bases | A, G, C, T | A, G, C, U (replaces T) |
| Strands | Double-stranded (helix) | Single-stranded (often folded) |
| Length | Very long (millions of base pairs per chromosome) | Short (typically hundreds to thousands of bases) |
| Stability | Very stable (long-term storage) | Less stable (degrades within hours-days) |
| Location (eukaryotes) | Almost all in the nucleus (also a small amount in mitochondria/chloroplasts) | Nucleus and cytoplasm |
| Function | Stores genetic information | Transfers information from DNA to ribosomes (mRNA, tRNA, rRNA) |
Genes and the storage of genetic information
The genetic information a cell needs to build all its proteins is stored in the sequence of bases along its DNA. A gene is a length of DNA that codes for a polypeptide. The bases are read in groups of three (triplets), and each triplet codes for one amino acid in the resulting polypeptide.
A typical human genome contains around 20,000 protein-coding genes spread across 23 pairs of chromosomes. The total genetic information amounts to roughly 3 billion base pairs of DNA per haploid set.
Semi-conservative DNA replication
Before a cell divides, it must duplicate its DNA so that each daughter cell receives a complete copy. The mechanism is semi-conservative replication: each new DNA molecule consists of one original (parental) strand and one newly synthesised strand.
- The double helix is unwound and the hydrogen bonds between complementary bases are broken, separating the two strands — this exposes single-stranded templates. (The enzyme that does this is helicase; knowledge of helicase by name is not required by 9700.)
- The two separated strands act as templates for new strand synthesis
- Free deoxyribonucleoside triphosphates align opposite their complementary bases on each template strand (A–T, G–C) by hydrogen bonding
- The enzyme DNA polymerase catalyses the formation of phosphodiester bonds between adjacent nucleotides, building new strands in the 5′→3′ direction only
- Because DNA polymerase can only work in the 5′→3′ direction, the two strands are synthesised differently (see below)
- The enzyme DNA ligase joins the short DNA fragments on the lagging strand into a continuous strand by forming phosphodiester bonds
- Result: two identical DNA double helices, each with one original (parental) strand and one new (daughter) strand — semi-conservative
Leading and lagging strands
Because DNA polymerase can only add nucleotides in the 5′→3′ direction, and the two template strands run antiparallel, the two new strands are synthesised by different mechanisms:
The leading strand template runs 3′→5′ in the direction of the replication fork. DNA polymerase can synthesise continuously in the 5′→3′ direction, following the fork. Only one primer is needed to start synthesis. The result is one long, unbroken new strand.
The lagging strand template runs 5′→3′ relative to the fork, which is opposite to the direction of polymerase movement. DNA polymerase must work away from the fork, synthesising short fragments called Okazaki fragments. Each fragment needs its own primer.
Once synthesis is complete, DNA ligase joins the Okazaki fragments together by sealing the gaps between them — forming a continuous lagging strand. This is why DNA ligase is essential.
DNA ligase catalyses the formation of a phosphodiester bond between the 3′-OH of one DNA fragment and the 5′-phosphate of the next, sealing the nick in the sugar-phosphate backbone. It is specifically required on the lagging strand to join the Okazaki fragments. Without DNA ligase, the lagging strand would remain as a series of short unconnected fragments.
Note: The 9700 syllabus requires knowledge of DNA polymerase and DNA ligase in replication. Knowledge of helicase, primase, or other replication enzymes is not required.
Of the three models proposed in the 1950s (semi-conservative, conservative, dispersive), Meselson and Stahl's experiment in 1958 confirmed the semi-conservative mechanism. After replication, each new molecule is half-old, half-new — not entirely new (conservative) and not a mix of fragments (dispersive). The 9700 syllabus does not require detailed knowledge of the Meselson-Stahl experiment but candidates should know the mechanism.
Accuracy of replication
DNA replication is extremely accurate — in human cells, only about one error per 10 billion bases survives after all proofreading is done. This precision matters because every cell in the body inherits the genome through repeated rounds of replication, and accumulated errors lead to mutations, dysfunction, and disease.
- Complementary base pairing — only the correct base fits geometrically and forms the right number of hydrogen bonds
- DNA polymerase proofreading — the polymerase detects mismatched bases and removes them before adding the next nucleotide
- Mismatch repair systems — specialised enzymes scan the new DNA after replication and repair any errors that escaped proofreading
Even with these systems, occasional errors slip through and become permanent mutations — covered in section 6.2.
Which of the following correctly describes the structure of a nucleotide?
- A. A pentose sugar joined to a phosphate by a peptide bond, with a hydrogen-bonded base
- B. A pentose sugar joined to a phosphate group and a nitrogenous base by covalent bonds
- C. Two pentose sugars joined to a single phosphate, with bases bonded by hydrogen bonds
- D. A phosphate, a sugar, and an amino acid joined by ester bonds
DNA replication is described as semi-conservative.
(a) Describe the role of DNA helicase and DNA polymerase in DNA replication. [4]
(b) Explain what is meant by “semi-conservative” replication. [2]
(c) Suggest why accurate DNA replication is essential for the production of genetically identical daughter cells. [2]
(a) Roles of helicase and polymerase [4 marks]
- DNA helicase: unwinds the double helix and breaks the hydrogen bonds between paired bases [1]
- This separates the two strands so each can act as a template [1]
- DNA polymerase: joins adjacent free nucleotides on the template strand by forming phosphodiester bonds [1]
- It also proofreads — identifies and removes any mismatched bases before adding the next nucleotide / synthesises new strand in 5′ to 3′ direction [1]
(b) Semi-conservative meaning [2 marks]
- Each new DNA molecule contains one original (parental) strand and one newly synthesised strand [1]
- So half of each daughter molecule is “conserved” from the parent — hence “semi” conservative [1]
(c) Why accuracy matters [2 marks]
- Errors in DNA sequence are mutations that change the genetic information / are passed to all subsequent daughter cells [1]
- The daughter cells must have the same base sequence as the parent cell to remain genetically identical / produce the same proteins / function correctly [1]
Protein synthesis
Protein synthesis is the process by which the genetic information stored in DNA is translated into the precise sequence of amino acids that defines a polypeptide. In eukaryotes the process has two stages: transcription in the nucleus produces mRNA, and translation at the ribosome produces the polypeptide.
The genetic code
The genetic code is the set of rules linking nucleotide sequence in DNA/mRNA to amino acid sequence in polypeptides. It has four defining properties:
The code is read in groups of three bases. Each three-base unit (a codon in mRNA, a triplet in DNA) specifies one amino acid. With 4 bases and 3 positions, there are 4³ = 64 possible codons.
Each base is part of only one codon. The reading frame is fixed: starting from the first base, the next three are codon 1, the next three are codon 2, and so on, with no shared bases between codons.
Most amino acids are coded by more than one codon. Only 20 amino acids are encoded, but there are 64 possible codons — so most amino acids have multiple codons (synonymous codons). Three codons are stop codons (signal end of polypeptide); one codon (AUG) acts as the start codon and also encodes methionine.
The same code is used in nearly all organisms — bacteria, plants, animals, fungi. This universality is strong evidence that all life shares a common ancestor. (A few minor variations exist, e.g. in some mitochondria and protozoa, but these are not in syllabus.)
Three types of RNA
| Type | Structure | Function |
|---|---|---|
| Messenger RNA (mRNA) | Single-stranded; linear; length matches the gene being expressed | Carries the genetic message from DNA in the nucleus to the ribosome in the cytoplasm |
| Transfer RNA (tRNA) | Single-stranded but folded into a clover-leaf shape (~76–90 nucleotides); has an anticodon at one end and an amino-acid attachment site at the other | Brings specific amino acids to the ribosome and pairs its anticodon with the matching mRNA codon, ensuring the correct amino acid is added |
| Ribosomal RNA (rRNA) | Several different rRNA molecules combined with proteins, forming the two subunits of the ribosome | Forms the structural and catalytic core of the ribosome — the site of polypeptide synthesis |
Transcription
Transcription is the process of producing an RNA copy of a gene, occurring in the nucleus. The product is messenger RNA (mRNA).
- Initiation: the enzyme RNA polymerase binds to a specific region (promoter) at the start of the gene
- Strand separation: RNA polymerase unwinds the DNA double helix and breaks the hydrogen bonds, exposing the bases of the template strand (only one of the two DNA strands is copied per gene)
- Elongation: free RNA nucleotides align opposite the template strand by complementary base pairing — A pairs with U (not T, because the new strand is RNA), T pairs with A, G with C, C with G
- RNA polymerase joins the RNA nucleotides by phosphodiester bonds, building the mRNA in the 5′→3′ direction
- Termination: when RNA polymerase reaches a termination signal, the mRNA is released; DNA strands re-anneal
Template strand (also called antisense strand): the DNA strand that is read — the mRNA is complementary to this strand.
Coding strand (also called sense strand): the DNA strand not read — its sequence is the same as the mRNA except U replaces T. Examiners sometimes give the coding strand sequence and ask for the mRNA — in that case, just replace T with U; no further change needed.
Modifying the primary transcript — introns and exons
In eukaryotes, the immediate product of transcription is called the primary transcript (or pre-mRNA). It contains:
- Exons — protein-coding regions that will be expressed
- Introns — intervening, non-coding regions that must be removed
Before leaving the nucleus, the primary transcript is processed:
- Introns are cut out (excised) from the primary transcript
- The remaining exons are joined together in their original order
- The result is the mature mRNA, which contains only coding sequences
Mature mRNA exits the nucleus through a nuclear pore and travels to a ribosome in the cytoplasm for translation.
Prokaryotes (bacteria) do not have introns in most of their genes, and they have no nuclear membrane. So bacterial mRNA does not need splicing, and translation can begin while transcription is still in progress — the two processes happen in the same compartment. The 9700 syllabus focuses on eukaryotic protein synthesis.
Translation
Translation is the synthesis of a polypeptide using the mRNA sequence as a template. It takes place at ribosomes in the cytoplasm (or attached to the rough endoplasmic reticulum). Three RNA molecules (mRNA, tRNA, rRNA) cooperate, plus amino acids and ATP for energy.
- Initiation: mRNA binds to the small ribosomal subunit. The first codon is the start codon (AUG)
- A tRNA with the complementary anticodon (UAC) binds to the start codon, bringing in its amino acid (methionine)
- The large ribosomal subunit attaches; the ribosome is now complete and ready to extend the polypeptide
- Elongation: the ribosome moves along the mRNA one codon at a time. For each codon, a tRNA with the matching anticodon delivers the next amino acid
- A peptide bond forms between adjacent amino acids on the ribosome (a condensation reaction, catalysed by the ribosome itself), and the previous tRNA is released
- This continues, codon by codon, until the ribosome reaches a stop codon (UAA, UAG, or UGA)
- Termination: no tRNA matches a stop codon. The completed polypeptide is released from the ribosome; the ribosomal subunits separate from the mRNA
A single mRNA molecule is typically translated by several ribosomes simultaneously, forming a polysome (or polyribosome). This allows many copies of the same polypeptide to be made quickly from one mRNA — a major efficiency boost when a cell needs lots of protein in a short time (e.g. haemoglobin in immature red blood cells).
From DNA to polypeptide — the full flow
| Step | What happens | Where (eukaryote) | Key molecules |
|---|---|---|---|
| 1. Transcription | DNA template → primary transcript (RNA) | Nucleus | RNA polymerase, free RNA nucleotides |
| 2. RNA processing | Introns removed; exons joined → mature mRNA | Nucleus | Spliceosome (not in syllabus); “splicing” |
| 3. mRNA export | Mature mRNA passes out through nuclear pore | Nucleus → cytoplasm | Nuclear pores |
| 4. Translation | mRNA codons read; tRNA delivers amino acids; peptide bonds form | Cytoplasm (free ribosomes or rough ER) | Ribosome (rRNA + proteins), tRNAs, amino acids, ATP |
| 5. Folding (post-translation) | Polypeptide folds into 3D shape; may be modified or assembled into multi-subunit protein | Cytoplasm or ER/Golgi | Chaperone proteins (not in syllabus) |
Gene mutations
A gene mutation is a change in the sequence of base pairs in a DNA molecule. Mutations are the source of new alleles and the raw material of evolution — but most are neutral or harmful, and a few cause genetic disease. Three classes of point mutation are required for the syllabus:
One base is replaced by a different base. Only one codon is affected; the reading frame is preserved. Effect on the polypeptide depends on the change:
- Silent: degenerate code — new codon still codes for the same amino acid
- Missense: new codon codes for a different amino acid — one amino acid changed in polypeptide
- Nonsense: new codon is a stop codon — polypeptide is truncated, often non-functional
One or more bases is added to the DNA sequence. If the number inserted is not a multiple of 3, the reading frame is shifted from that point onwards (a frameshift mutation) — every codon downstream is changed, often producing a completely non-functional polypeptide.
One or more bases is removed from the DNA sequence. Like insertion, deletion of a non-multiple of 3 causes a frameshift and dramatically alters the polypeptide. Deletion of a multiple of 3 simply removes whole codons (one or more amino acids missing) without shifting the frame.
Sickle-cell anaemia is caused by a single substitution in the gene for the β-globin chain of haemoglobin. The codon GAG (glutamic acid) is changed to GTG (valine) at the 6th amino acid position. This single missense mutation produces haemoglobin S, which polymerises in low-oxygen conditions, distorting red blood cells into sickle shapes. The shape change reduces O2 transport, blocks small blood vessels, and shortens cell lifespan — demonstrating how a single base change can affect the phenotype dramatically.
Not every mutation changes the polypeptide. The genetic code is degenerate, so many substitutions produce the same amino acid. Don't say “every mutation changes the protein” — say “a mutation may alter the polypeptide, depending on which codon is affected and how”.
A length of mRNA contains the codon sequence: AUG–GCC–UAA. How many amino acids will the resulting polypeptide contain?
- A. 0 amino acids (the sequence has a stop codon)
- B. 1 amino acid
- C. 2 amino acids
- D. 3 amino acids
A gene encoding a particular polypeptide undergoes a single base substitution mutation.
(a) Describe what happens during transcription. [4]
(b) The mutation changes the codon CCG to CCA in the mRNA. Suggest, with reasoning, what effect this mutation will have on the polypeptide. [3]
(c) Compare the effects of substitution and frameshift mutations on the resulting polypeptide. [2]
(a) Transcription [4 marks]
- RNA polymerase binds to the gene at the start (promoter region) and unwinds the DNA, breaking the hydrogen bonds [1]
- The exposed template strand acts as a template; free RNA nucleotides align opposite their complementary DNA bases (A–U, T–A, G–C, C–G) [1]
- RNA polymerase joins the RNA nucleotides via phosphodiester bonds, building mRNA in the 5′ to 3′ direction [1]
- The primary transcript is then processed (introns removed, exons joined together) to produce mature mRNA, which leaves the nucleus through a nuclear pore [1]
(b) Effect of CCG → CCA [3 marks]
- Both CCG and CCA are codons for the same amino acid (proline) — the genetic code is degenerate [1]
- This is a silent mutation; the amino acid sequence of the polypeptide is unchanged [1]
- Therefore the polypeptide structure and function are unaffected [1]
Note: Students don't need to memorise the codon table — in P2 questions, codon tables are provided. Show the reasoning regardless.
(c) Substitution vs frameshift [2 marks]
- Substitution: only one codon is affected; the reading frame is preserved; effect ranges from silent (no change) to nonsense (truncated polypeptide), but typically no more than one amino acid changes [1]
- Frameshift (insertion/deletion of a non-multiple of 3): the reading frame is shifted from the point of mutation onwards; every codon downstream is changed; typically results in a completely non-functional polypeptide [1]
Topic 6 Practice — Comprehensive
Mixed practice covering both sub-sections in 9700 P1/P2 style. Try each before revealing the answer.
A region of DNA contains 30% adenine. What percentage of the same DNA region is guanine?
- A. 30%
- B. 20%
- C. 40%
- D. 60%
Which of the following is found in RNA but not in DNA?
- A. Adenine
- B. Phosphate group
- C. Ribose
- D. Cytosine
During translation, the order of amino acids in the polypeptide is determined by:
- A. The order of codons on the tRNA
- B. The order of codons on the mRNA
- C. The order of bases on the rRNA
- D. The number of ribosomes attached to the mRNA
During the cell cycle, DNA must be replicated accurately so that daughter cells inherit identical genetic information.
(a) State the phase of the cell cycle in which DNA replication occurs. [1]
(b) Explain how complementary base pairing contributes to the accuracy of DNA replication. [3]
(c) Even with high accuracy, occasional errors occur. Explain how a substitution mutation in a gene could result in (i) no change to the polypeptide, and (ii) a non-functional polypeptide. [5]
(a) Phase of cell cycle [1 mark]
S phase (synthesis phase) of interphase [1]
(b) Complementary base pairing accuracy [3 marks]
- The two DNA strands separate; each acts as a template [1]
- Free DNA nucleotides can only pair correctly with complementary template bases (A with T, G with C) due to the geometry and the number of hydrogen bonds — A=T (2 H-bonds), G≡C (3 H-bonds) [1]
- This complementary pairing means each new strand has a sequence that exactly matches what was originally on the partner strand — the daughter molecules are identical to the parent [1]
(c)(i) No change [2 marks]
- The genetic code is degenerate — many amino acids are encoded by more than one codon [1]
- If the substitution changes a codon to another codon for the same amino acid (e.g. CCC to CCG, both proline), the polypeptide is unchanged — a silent mutation [1]
(c)(ii) Non-functional [3 marks]
- The substitution may change the codon to one for a different amino acid (missense) or to a stop codon (nonsense) [1]
- If a different amino acid is incorporated at a critical position (e.g. in the active site of an enzyme), the protein's tertiary structure may change, altering the active site shape and reducing or abolishing function [1]
- If a stop codon is introduced early, translation ends prematurely — the polypeptide is truncated and almost certainly non-functional [1]
Synoptic note: This question links DNA replication (Topic 6.1), the cell cycle (Topic 5.1), the genetic code and mutations (Topic 6.2), and protein structure and function (Topic 2.3 — tertiary structure dependent on R-group interactions).
Describe what happens during translation, from the binding of mRNA at the ribosome to the release of the completed polypeptide. Include the roles of mRNA, tRNA, codon, anticodon, ribosome, and peptide bond. [6]
Six creditable points (any six from the full description):
- mRNA binds to the small subunit of the ribosome; the ribosome reads the mRNA in codons (groups of 3 bases) [1]
- The first codon (AUG) is the start codon; a tRNA with the complementary anticodon (UAC) brings the first amino acid (methionine) [1]
- The large ribosomal subunit attaches; the ribosome moves along the mRNA one codon at a time [1]
- For each codon, a tRNA with the matching anticodon delivers the corresponding amino acid to the ribosome [1]
- A peptide bond forms (by condensation) between the amino acid carried by the new tRNA and the polypeptide chain — catalysed by the ribosome [1]
- The previous tRNA (now without amino acid) is released; the ribosome moves to the next codon [1]
- This continues until a stop codon (UAA, UAG, or UGA) is reached — no tRNA matches a stop codon; the polypeptide is released and the ribosome separates from the mRNA [1]
Topic 6 — Common Mistakes
- 🧬Saying tRNA carries codonstRNA carries an anticodon (and an amino acid). mRNA carries codons. The codon-anticodon distinction is one of the most-tested terminology pairs in Topic 6 — getting it wrong loses easy marks.
- ♻Confusing transcription and translationTranscription = DNA → mRNA, in the nucleus. Translation = mRNA → polypeptide, at the ribosome. They use different enzymes and different molecular machinery. Mixing them up loses entire question parts.
- 🧸Saying both DNA strands are read during transcriptionOnly the template (antisense) strand is read for any given gene. The other strand is the coding (sense) strand and is not transcribed. Different genes may use different strands as their template, but for any one gene, only one strand is the template.
- 🤐Saying every mutation changes the polypeptideThe genetic code is degenerate — many substitutions are silent (no amino acid change). Use careful wording: a mutation may change the polypeptide, depending on which codon and what change.
- ❓Saying introns are removed during translationIntrons are removed during RNA processing in the nucleus, before the mature mRNA leaves for the ribosome. Translation only ever sees the mature mRNA (exons only). Saying introns are removed at the ribosome is a serious error.
- 🦴Forgetting that AUG is also the start codonAUG codes for methionine but also serves as the start codon for translation. Every polypeptide initially has methionine as its first amino acid (though it's often removed later). Stop codons UAA, UAG, UGA do not code for any amino acid — they signal termination.
- 🎯A–T has 3 hydrogen bonds, G–C has 2Reversed. A–T pairs have 2 hydrogen bonds; G–C pairs have 3. This is why GC-rich DNA needs more energy to denature than AT-rich DNA. The numbers (2 vs 3) and which pair has which are both testable.
- 📂RNA is the same as a single strand of DNAWrong on three counts: (1) RNA contains ribose, not deoxyribose; (2) RNA uses uracil instead of thymine; (3) RNA is typically much shorter than a DNA strand and is single-stranded by design (not just a separated DNA strand). Do not say “RNA is half of a DNA molecule”.
- ⚙Conservative replication is what happensNo — DNA replication is semi-conservative: each daughter molecule has one original strand and one new strand. Conservative replication (entirely new molecule alongside intact original) was disproved by the Meselson-Stahl experiment.
- 🏭Calling phosphodiester bonds “hydrogen bonds”Phosphodiester bonds are covalent bonds linking adjacent nucleotides within a strand (sugar-phosphate-sugar-phosphate backbone). Hydrogen bonds hold the two strands together via base pairing. Bonds within one strand vs between two strands are different things.
- 🤯Saying a frameshift “changes one codon”A frameshift mutation changes every codon downstream of the insertion or deletion point. That's why insertions/deletions are usually far more damaging than substitutions: dozens of amino acids may change in a single frameshift, plus stop codons can appear in the new frame.
Topic 6 is the foundation of molecular biology and connects forward to almost every later topic: protein synthesis underlies all enzymes (Topic 3) and signalling proteins (Topic 4); transcription regulation is the basis of differentiation in stem cells (Topic 5); mutations link to evolution (Topic 17) and inheritance (Topic 16); replication is the molecular event of S phase (Topic 5). Highest-yield items: nucleotide structure with all 3 components, complementary base pairing with hydrogen-bond counts, semi-conservative replication mechanism with helicase and polymerase, the four properties of the genetic code (triplet, non-overlapping, degenerate, universal), the difference between codon and anticodon, the steps of transcription and translation, intron-exon splicing, and the three mutation types and their effects. Many Topic 6 questions are highly synoptic — expect them to test understanding alongside Topics 2, 3, 5, and later A2 topics.