DNA Replication & Gene Expression

How to Use This Guide

• DNA Structure — A2.1 nucleotide components, double helix, base pairing, antiparallel strands
• DNA Replication — A2.2 semi-conservative replication, enzymes, leading and lagging strand synthesis
• Transcription & Translation — A2.3 gene expression from DNA to protein, codon table, ribosome function
• Gene Expression & Epigenetics — A2.4 regulation of gene expression, lac operon model, DNA methylation, histone modification
• HL / AHL Only — deeper enzyme detail, Okazaki fragments, lac operon, miRNA regulation
• MCQ Practice — styled like real IB Paper 1 questions
• Exam Alerts — the traps and mistakes that cost marks

Aligned to IB Biology 2025 syllabus — A2.1 DNA Structure, A2.2 DNA Replication, A2.3 Transcription & Translation, A2.4 Gene Expression

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Before diving in, here is the big picture: DNA is a master blueprint stored in the nucleus. When a cell needs to copy itself, it duplicates the blueprint precisely (replication). When a cell needs to build something, it reads a section of the blueprint and makes a working copy (transcription), then uses that working copy as instructions to build a protein (translation). Gene expression is the control system that decides which sections of the blueprint get read, when, and in which cells.

Jump to section: DNA Structure (A2.1) · DNA Replication (A2.2) · Transcription & Translation (A2.3) · Gene Expression & Epigenetics (A2.4) · MCQ Practice · Quick Reference

Videos on this page: DNA Replication · Transcription & Translation

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Section 1: DNA Structure (A2.1)

DNA contains all the heritable information of a cell, but this only works because its structure is extraordinarily stable and remarkably precise. Understanding the structure explains every other topic in this section — replication depends on base pairing, transcription depends on the antiparallel strands, and gene regulation depends on the accessibility of the double helix.

1.1 The Double Helix

DNA is a double-stranded helix — two polynucleotide chains wound around a common axis, first modelled by Watson and Crick in 1953 using X-ray diffraction data from Franklin and Wilkins.

Key structural features:

• Two antiparallel strands (one runs $5' \rightarrow 3'$; the other runs $3' \rightarrow 5'$)
• The sugar-phosphate backbone forms the outer rails of the helix
• Nitrogenous bases project inward from each strand and form pairs across the helix
• The helix makes one full turn every 10 base pairs
• The major groove and minor groove alternate along the outside — transcription factors bind here

1.2 Nucleotides and Base Pairing

Each nucleotide in DNA consists of three covalently bonded components:

1. A deoxyribose sugar (5-carbon; lacks the OH group at $2'$ carbon, unlike ribose in RNA)
2. A phosphate group (attached to the $5'$ carbon of the sugar)
3. A nitrogenous base — one of four types:
   • Purines (double ring): Adenine (A), Guanine (G)
   • Pyrimidines (single ring): Thymine (T), Cytosine (C)

Chargaff’s Rules — from measurements of base composition in many organisms:

• $[\text{A}] = [\text{T}]$ and $[\text{G}] = [\text{C}]$ in any DNA molecule
• Therefore: $[\text{A}] + [\text{G}] = [\text{T}] + [\text{C}]$ (purines = pyrimidines)
• The ratio $\frac{[\text{A}]+[\text{T}]}{[\text{G}]+[\text{C}]}$ varies between species — this is species-specific

Why 3 H-bonds for G—C? Guanine and cytosine have three complementary groups that form hydrogen bonds, whereas adenine and thymine form only two. This makes G—C pairs stronger. DNA with a higher G—C content has a higher melting temperature ($T_m$).

Quick Recall — Section 1

Try to answer without scrolling up:

1. Name the four bases in DNA and state the base pairing rules.
2. What type of bond holds the two strands of the double helix together?
3. DNA strands run antiparallel. What does this mean?

Reveal answers

1. Adenine (A) pairs with Thymine (T); Cytosine (C) pairs with Guanine (G). A-T has 2 hydrogen bonds; C-G has 3.
2. Hydrogen bonds between complementary bases.
3. One strand runs 5’ to 3’ and the other runs 3’ to 5’ — they run in opposite directions.

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Section 2: DNA Replication (A2.2)

Before any cell can divide, it must copy its entire genome so that each daughter cell receives a complete set of instructions. Replication is extraordinarily accurate — the error rate is approximately $10^{-9}$ per base pair — and this accuracy relies on a team of coordinated enzymes acting at the replication fork.

2.1 Semi-Conservative Replication

DNA replication is semi-conservative: each new double helix consists of one original (parental) strand and one newly synthesised strand. This was proven by the Meselson-Stahl experiment (1958), which used heavy ($^{15}\text{N}$) and light ($^{14}\text{N}$) isotopes of nitrogen to track DNA strands across generations.

flowchart TD
    A["Original DNA\n(Both strands parental)"] --> B["Helicase unwinds &\nbreaks H-bonds"]
    B --> C["Two template strands\nexposed at replication fork"]
    C --> D["Primase adds\nRNA primer (5'→3')"]
    D --> E["DNA Pol III extends\nnew strand (5'→3')"]
    E --> F["Leading strand —\ncontinuous synthesis"]
    E --> G["Lagging strand —\ndiscontinuous synthesis\n(Okazaki fragments)"]
    F --> H["DNA Pol I replaces\nRNA primers with DNA"]
    G --> H
    H --> I["DNA Ligase seals\nnicks between fragments"]
    I --> J["Two identical\ndouble-stranded DNA molecules"]

Each origin of replication creates a replication bubble: two replication forks moving in opposite directions away from the origin. Eukaryotes have multiple origins of replication per chromosome (essential given chromosome size); prokaryotes typically have one origin (oriC).

2.2 Enzymes of Replication

2.3 Leading and Lagging Strand Synthesis HL

Because DNA polymerase can only synthesise new DNA in the $5' \rightarrow 3'$ direction, the two template strands are replicated differently:

Leading strand (template runs $3' \rightarrow 5'$):

• DNA Pol III synthesises continuously in the $5' \rightarrow 3'$ direction
• Only one primer needed per origin
• Synthesis moves toward the replication fork (same direction as fork movement)

Lagging strand (template runs $5' \rightarrow 3'$):

• Must be replicated in short fragments, each starting with a new RNA primer
• These fragments are called Okazaki fragments (~100—200 nt in eukaryotes; ~1000—2000 nt in prokaryotes)
• Each Okazaki fragment is synthesised $5' \rightarrow 3'$, but moves away from the replication fork
• After synthesis: Pol I removes primers, fills gaps; Ligase joins fragments

Quick Recall — Section 2

Try to answer without scrolling up:

1. What does “semi-conservative” mean in the context of DNA replication?
2. Name the enzyme that unwinds the double helix.
3. Why is one strand called the “lagging strand”?

Reveal answers

1. Each new DNA molecule consists of one original (parental) strand and one newly synthesised strand.
2. Helicase.
3. Because DNA polymerase can only synthesise in the 5’ to 3’ direction, the lagging strand must be copied in short fragments (Okazaki fragments) away from the replication fork, which is slower than continuous synthesis on the leading strand.

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Watch: DNA Replication — Leading & Lagging Strands

Khan Academy · 10 min · Leading and lagging strand synthesis, Okazaki fragments, and key replication enzymes — helicase, DNA polymerase, and ligase

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Section 3: Transcription & Translation (A2.3)

Gene expression converts stored genetic information into functional proteins. This happens in two stages: transcription (DNA → RNA, in the nucleus) and translation (RNA → protein, at the ribosome). Together, these form the central dogma of molecular biology: DNA → RNA → Protein.

3.1 Transcription

Before the mechanics, here is the key idea: only one strand of DNA is used as a template for any given gene, and the RNA copy produced is a complementary, antiparallel copy of that template strand.

Stages of transcription:

1. Initiation — RNA polymerase binds to the promoter region upstream of the gene. The promoter contains consensus sequences (e.g. TATA box) that are recognised by transcription factors. HL
2. Elongation — RNA polymerase unwinds the DNA locally and moves $3' \rightarrow 5'$ along the template strand, synthesising a complementary mRNA strand in the $5' \rightarrow 3'$ direction. Uracil (U) pairs with adenine (A) in the template — RNA contains no thymine.
3. Termination — RNA polymerase reaches a terminator sequence and dissociates, releasing the pre-mRNA transcript.

Post-transcriptional processing (eukaryotes only): HL

• 5’ cap (methylated guanosine) is added — protects mRNA from degradation and aids ribosome attachment
• Poly-A tail (100—250 adenine nucleotides) added at $3'$ end — increases stability and aids export
• Splicing — introns (non-coding sequences) are removed by the spliceosome; exons (coding sequences) are joined together
• The mature mRNA is exported from the nucleus through nuclear pores to the cytoplasm

3.2 Translation

Translation takes place at the ribosome in the cytoplasm (or on the rough endoplasmic reticulum for secreted proteins). It is the process of decoding the mRNA sequence into an amino acid sequence.

Key players:

Ribosome subunits:

• Eukaryotes: 80S ribosome = 60S large subunit + 40S small subunit
• Prokaryotes: 70S ribosome = 50S large subunit + 30S small subunit

Stages of translation:

1. Initiation — the small ribosomal subunit binds to the $5'$ cap of mRNA and scans for the start codon (AUG). Initiator tRNA (carrying methionine) binds at the P site. Large subunit joins.
2. Elongation — a charged tRNA enters the A site; its anticodon base-pairs with the mRNA codon. A peptide bond forms between the growing polypeptide (in P site) and the new amino acid (in A site). The ribosome translocates one codon in the $5' \rightarrow 3'$ direction; the now-empty tRNA moves to the E site and exits.
3. Termination — a stop codon (UAA, UAG, or UGA) enters the A site; no tRNA matches. Release factors cause the ribosome to dissociate and the polypeptide is released.

Polyribosomes (polysomes): Multiple ribosomes can translate the same mRNA simultaneously, increasing the rate of protein production.

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Watch: Transcription & Translation

Khan Academy · 9 min · Transcription — mRNA synthesis, 5’ cap, poly-A tail, and splicing of introns in eukaryotes

Khan Academy · 11 min · Translation — codons, anticodons, tRNA, ribosome, and how the polypeptide chain is assembled

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Section 4: Gene Expression & Epigenetics (A2.4)

Not all genes are active in every cell at every time. A liver cell and a neuron contain identical DNA, yet they are structurally and functionally completely different — because different genes are switched on. Gene expression regulation allows a single genome to produce hundreds of specialised cell types.

4.1 Regulation of Gene Expression

Gene expression can be regulated at multiple levels:

• Transcriptional regulation — controlling whether RNA polymerase transcribes a gene (most common and energetically efficient)
• Post-transcriptional regulation — controlling mRNA processing, stability, or translation (e.g. by miRNA) HL
• Post-translational regulation — modifying the protein after synthesis (e.g. phosphorylation, cleavage)

The lac Operon as a Model HL

The lac operon in Escherichia coli is the classic model for transcriptional regulation. It controls the expression of genes needed to metabolise lactose, and only switches them on when lactose is present AND glucose is absent.

Operon components:

• Promoter — site where RNA polymerase binds
• Operator — DNA sequence where the repressor binds (between promoter and structural genes)
• Structural genes — lacZ ($\beta$-galactosidase), lacY (permease), lacA (transacetylase)
• Repressor gene (lacI) — always expressed; produces the lac repressor protein

Regulation logic:

Condition	Repressor state	Transcription
No lactose, glucose present	Repressor binds operator	OFF (genes not needed)
Lactose present, no glucose	Allolactose (inducer) binds repressor → repressor changes shape → releases operator	ON (lactose needs to be metabolised)
Both lactose and glucose present	Repressor off operator, but low cAMP → weak transcription	Weak (cell prefers glucose)

cAMP and catabolite repression (when glucose is absent, cAMP is high, activating CAP protein that enhances transcription) ensures the lac genes are fully active only when lactose is the only carbon source available. This is positive regulation by CAP alongside negative regulation by the repressor.

Post-Transcriptional Regulation by miRNA HL

MicroRNA (miRNA) are small non-coding RNA molecules (~22 nucleotides) that regulate gene expression after transcription.

Mechanism:

1. miRNA genes are transcribed to form a long primary miRNA (pri-miRNA)
2. The pri-miRNA is cleaved in the nucleus by the enzyme Drosha into a ~70 nt hairpin structure called the pre-miRNA, which is exported to the cytoplasm
3. The pre-miRNA is processed by the enzyme Dicer into a short double-stranded RNA (~22 nt)
4. One strand is loaded into the RISC (RNA-induced silencing complex)
5. RISC uses the miRNA as a guide to find complementary sequences in the 3’ UTR of target mRNA
6. RISC either cleaves the mRNA (if match is perfect) or represses translation (if match is partial)

Result: the target gene’s protein is NOT produced (or is produced at reduced levels), even though the gene was transcribed. This is called RNA interference (RNAi).

miRNAs are crucial for development and cell differentiation — a single miRNA can regulate hundreds of target genes simultaneously.

4.2 Epigenetics

Epigenetics is the study of heritable changes in gene expression that do NOT involve changes to the DNA sequence itself. The DNA code is unchanged — what changes is how accessible or readable that code is.

The two major epigenetic mechanisms are:

1. DNA Methylation

• Addition of a methyl group ($-\text{CH}_3$) to cytosine bases, typically at CpG dinucleotides
• Methylation of a promoter region generally silences gene expression by preventing transcription factor binding
• Methylation patterns are heritable through cell division (maintained by methyltransferase enzymes)
• Example: X-chromosome inactivation (Barr body formation in mammalian females) is maintained by methylation

2. Histone Modification

• DNA is wrapped around histone proteins to form nucleosomes (fundamental unit of chromatin)
• Chemical groups can be added to the histone tails:
  • Acetylation (addition of acetyl groups) → loosens chromatin (euchromatin) → gene expression ON
  • Deacetylation → tightens chromatin (heterochromatin) → gene expression OFF
  • Methylation of histones can either activate or repress, depending on which histone residue is modified
• The combination of histone modifications across a region is called the histone code

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MCQ Practice

Question 1

A DNA molecule contains 34% guanine. What is the percentage of adenine?

A. 34%

B. 32%

C. 16% ← CORRECT

D. 68%

Why: G = 34%, so by Chargaff’s rules C = 34%. Therefore A + T = 100 − 68 = 32%, and since A = T, adenine = 16%. Option A confuses G with A. Option B gives A + T combined. Option D gives G + C combined.

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Question 2

During DNA replication, what is the function of DNA ligase?

A. Unwind the double helix at the replication fork

B. Synthesise new nucleotides in the $5' \rightarrow 3'$ direction

C. Seal the nicks between Okazaki fragments on the lagging strand ← CORRECT

D. Remove RNA primers from the template strand

Why: Ligase joins the $3'$ end of one fragment to the $5'$ end of the next by forming a phosphodiester bond — it seals the nick. Option A is helicase. Option B describes DNA Pol III (although ligase does form a bond, it is not synthesising new nucleotides). Option D is DNA Pol I.

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Question 3

Which of the following is a difference between transcription and DNA replication?

A. Transcription uses deoxyribonucleotides; replication uses ribonucleotides

B. Transcription produces a single-stranded product; replication produces a double-stranded product ← CORRECT

C. Transcription requires a primer; replication does not

D. Transcription occurs in the cytoplasm; replication occurs in the nucleus

Why: mRNA is single-stranded; the replicated DNA is double-stranded. Option A has the types reversed — transcription uses ribonucleotides (to make RNA); replication uses deoxyribonucleotides. Option C has it backwards — replication requires a primer; transcription does not. Option D has it backwards for eukaryotes — both occur in the nucleus (transcription in nucleus; replication in nucleus).

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Question 4

A ribosome is at a codon that reads 5’-UAA-3’. What happens next?

A. A tRNA carrying tyrosine enters the A site

B. The ribosome translocates one codon in the $3' \rightarrow 5'$ direction

C. Translation terminates and the polypeptide is released ← CORRECT

D. The mRNA is degraded immediately

Why: UAA is one of the three stop codons (UAA, UAG, UGA). No tRNA has a matching anticodon, so a release factor binds, triggering release of the polypeptide and dissociation of ribosome subunits. Option A is wrong — UAA codes for no amino acid (tyrosine codons are UAU and UAC). Option B gets the direction wrong (ribosomes move $5' \rightarrow 3'$ along mRNA). Option D is not triggered directly by a stop codon.

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Question 5 HL

In the lac operon, what is the role of allolactose?

A. It directly activates RNA polymerase binding to the promoter

B. It acts as an inducer by binding the repressor and preventing it from binding the operator ← CORRECT

C. It methylates the operator region to allow transcription

D. It is the substrate for the enzyme encoded by the lacZ gene

Why: Allolactose (a metabolite of lactose) binds to the lac repressor protein, causing a conformational change that reduces the repressor’s affinity for the operator. When the repressor is released from the operator, RNA polymerase can transcribe the structural genes. Option A is the role of CAP (with cAMP), not allolactose. Option C describes an epigenetic mechanism, not operon regulation. Option D confuses the regulatory role of allolactose with the metabolic substrate of the lacZ enzyme.

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Question 6

Which statement about epigenetic changes is correct?

A. Epigenetic changes always involve a mutation in the DNA base sequence

B. Histone acetylation tightens chromatin and silences gene expression

C. DNA methylation of a promoter region typically reduces gene expression ← CORRECT

D. Epigenetic changes are never passed from parent cell to daughter cell

Why: Methylation of CpG sites in a promoter prevents transcription factor binding and typically silences the gene. Option A is the definition of a mutation — epigenetic changes do NOT alter the DNA sequence. Option B has the effect of acetylation reversed — acetylation loosens chromatin (euchromatin) and activates expression; it is deacetylation that silences. Option D is false — DNA methylation patterns are heritable and are maintained during cell division by methyltransferase.

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Question 7

Which row correctly describes the differences between mRNA, tRNA, and rRNA?

	mRNA	tRNA	rRNA
A	Carries anticodon	Carries amino acid	Found in nucleus
B	Carries genetic code	Carries anticodon	Structural component of ribosome
C	Produced in cytoplasm	Carries codons	Catalyses peptide bonds
D	Single-stranded only	Double-stranded	Structural only

B is CORRECT ← CORRECT

Why: mRNA carries the genetic code (codons); tRNA carries the anticodon and the amino acid; rRNA is both structural and catalytic (peptidyl transferase activity). Option A assigns the anticodon to mRNA (wrong — that is tRNA). Option C says mRNA is made in the cytoplasm (wrong — transcription occurs in the nucleus). Option D says tRNA is double-stranded (wrong — tRNA is single-stranded but folds into a cloverleaf shape).

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Quick Reference

DNA vs RNA: Key Differences

Feature	DNA	RNA
Sugar	Deoxyribose	Ribose
Bases	A, T, G, C	A, U, G, C
Strands	Double-stranded (usually)	Single-stranded (usually)
Location	Mainly nucleus	Nucleus and cytoplasm
Function	Long-term genetic storage	Gene expression intermediary
Stability	Very stable	Relatively unstable

Replication vs Transcription vs Translation

Feature	Replication	Transcription	Translation
Template	Both DNA strands	One DNA strand	mRNA
Product	DNA	RNA (mRNA/tRNA/rRNA)	Polypeptide
Enzyme	DNA Pol III	RNA polymerase	Ribosome (rRNA)
Location	Nucleus	Nucleus	Cytoplasm/RER
Direction	$5' \rightarrow 3'$	$5' \rightarrow 3'$ (product)	$5' \rightarrow 3'$ along mRNA
Primer needed?	Yes (RNA primer)	No	No

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Mixed Practice — Exam Style

1. [DNA Replication] During DNA replication, DNA polymerase III can only add nucleotides in the $5' \rightarrow 3'$ direction. What is the consequence of this for the lagging strand?

   A. The lagging strand is synthesised continuously in one long piece

   B. The lagging strand is synthesised in short fragments (Okazaki fragments) in the $5' \rightarrow 3'$ direction, away from the replication fork

   C. The lagging strand is synthesised by RNA polymerase, not DNA polymerase

   D. The lagging strand cannot be replicated and is lost after each division

2. [Transcription] During transcription, which strand of the DNA double helix serves as the template?

   A. Both strands are transcribed simultaneously

   B. The coding (sense) strand, read $3' \rightarrow 5'$, is transcribed

   C. The template (antisense) strand, read $3' \rightarrow 5'$ by RNA polymerase, is used to produce mRNA with the same sequence as the coding strand (with U instead of T)

   D. The choice of strand depends on the organism (prokaryote vs eukaryote)

3. [Gene Regulation] The lac operon in E. coli is repressed in the absence of lactose. When lactose is present, it acts as an inducer. The inducer molecule:

   A. Directly activates RNA polymerase to transcribe the structural genes

   B. Binds to the repressor protein, changing its shape so it can no longer bind to the operator, allowing transcription to proceed

   C. Degrades the repressor protein permanently

   D. Methylates the promoter region, permanently activating the operon

4. [Translation — Codons] The mRNA sequence $5'\text{-AUG-CGU-UAA-}3'$ codes for:

   A. Two amino acids followed by a stop codon (methionine, arginine, stop)

   B. Three amino acids (methionine, arginine, glutamine)

   C. One amino acid only (methionine is the start and stop codon)

   D. This sequence cannot be translated because UAA is a start codon

5. [Mutations — Distractor] A single base insertion (frameshift mutation) at position 3 of a coding sequence is generally more harmful than a single base substitution (point mutation) at the same position. The best explanation is:

   A. Insertions change the molecular mass of the protein more than substitutions

   B. Insertions cause a shift in the reading frame, altering every codon downstream and likely producing a non-functional protein; substitutions change at most one amino acid, and the genetic code’s degeneracy may mean the amino acid is unchanged

   C. Substitutions are always silent mutations; insertions are always missense mutations

   D. Insertions are more harmful only if they occur in an exon, not an intron

6. [DNA Replication — Enzymes] Helicase is to unwinding the double helix as ____________ is to sealing nicks between Okazaki fragments.

   A. DNA polymerase I

   B. Primase

   C. DNA ligase

   D. Topoisomerase

7. [Post-Transcriptional Modification] In eukaryotes, pre-mRNA is processed before leaving the nucleus. Which of the following is NOT a step in this processing?

   A. Addition of a 5’ methyl-guanosine cap

   B. Addition of a poly-A tail to the 3’ end

   C. Splicing out of introns by the spliceosome

   D. Methylation of the DNA template to prevent re-transcription

8. [Translation — Ribosomes] During translation, the ribosome moves along the mRNA in the $5' \rightarrow 3'$ direction. The A site, P site, and E site have specific functions. In what order does a tRNA carrying an amino acid move through these sites?

   A. E site → P site → A site

   B. A site → P site → E site

   C. P site → A site → E site

   D. A site → E site → P site

9. [Mutations — Types] A point mutation changes a codon from GAA (glutamic acid) to GUA (valine). This mutation is most accurately described as:

   A. A silent mutation — the amino acid is unchanged

   B. A nonsense mutation — the codon now codes for a stop

   C. A missense mutation — a different amino acid is incorporated, potentially altering protein structure and function

   D. A frameshift mutation — the reading frame is disrupted

10. [Gene Expression — Central Dogma] A researcher discovers a virus that can copy its RNA genome into DNA inside a host cell. This process requires an enzyme called reverse transcriptase. This finding challenges which aspect of the original central dogma?

    A. That DNA can be transcribed to RNA

    B. That RNA can be translated to protein

    C. That information flows in one direction only — originally stated as DNA → RNA → protein, with no provision for RNA → DNA

    D. That proteins can be converted back to nucleic acids

Show Answers

1. B — Okazaki fragments. Because DNA polymerase can only add to the $3'$ end of a growing strand, the lagging strand must be synthesised in short fragments moving away from the replication fork (in the $5' \rightarrow 3'$ direction). These fragments are later joined by DNA ligase. A describes the leading strand. D is incorrect — the lagging strand is fully replicated.

2. C — The template (antisense) strand is read $3' \rightarrow 5'$ by RNA polymerase, producing mRNA $5' \rightarrow 3'$ with the same sequence as the coding strand (substituting U for T). B is a common error — the coding strand is NOT the template; it is the strand whose sequence the mRNA matches.

3. B — The inducer (allolactose) binds to the repressor, causing a conformational change that prevents the repressor from binding the operator. RNA polymerase is then free to transcribe the structural genes (lacZ, lacY, lacA). A incorrectly states that the inducer directly activates RNA polymerase.

4. A — AUG = methionine (start), CGU = arginine, UAA = stop codon. The result is a dipeptide (two amino acids), then translation terminates. UAA is one of three stop codons (UAA, UAG, UGA). D is wrong — UAA is not a start codon.

5. B — Frameshift mutations displace the reading frame for all codons downstream of the insertion, usually producing a completely different and non-functional polypeptide. Point mutations change at most one codon. C is wrong — substitutions can be missense, nonsense, or silent. D is wrong — the key factor is frame-shifting, not intron/exon location (although non-coding region mutations may have less effect).

6. C — DNA ligase seals the phosphodiester bonds between adjacent Okazaki fragments (and between the RNA primer replacement and the existing DNA). Primase (B) synthesises RNA primers. Topoisomerase (D) relieves supercoiling ahead of the replication fork.

7. D — DNA methylation of the template is not a standard step in eukaryotic pre-mRNA processing. The three real steps are: 5’ cap (A), poly-A tail (B), and splicing of introns (C). These modifications protect the mRNA from degradation and enable nuclear export.

8. B — A site → P site → E site. An incoming aminoacyl-tRNA enters the A (aminoacyl) site, where a peptide bond forms. The growing peptide shifts to the P (peptidyl) site. The tRNA then moves to the E (exit) site and is released. A describes the reverse order.

9. C — Missense mutation. GAA (Glu) → GUA (Val) changes one amino acid for another. This is not silent (A) because the amino acid changes. It is not nonsense (B) because no stop codon is introduced. It is not a frameshift (D) because no bases are inserted or deleted.

10. C — Reverse transcriptase (found in retroviruses like HIV) copies RNA into DNA, demonstrating that information can flow RNA → DNA. This contradicted the “central dogma” as originally described by Crick in 1958, which implied unidirectional information flow with no provision for reverse transcription.

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May 2026 Prediction Questions

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