IB DP Biology: DNA replication, mutation and gene editing (DP Biology) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover IB Diploma Programme Biology sections D1.1 (DNA replication) and D1.3 (Mutation and gene editing), following the IB Biology guide, first assessment 2025. Questions 6, 7, 8 and 12 are HL only; the rest are SL and HL. Marks are given as a guide to depth, not as a copy of any exam tariff.

Read the study guide and revision notes first. Course pages: IB DP Biology hub and printable checklist.

Questions

1. A DNA molecule has both strands labelled with heavy nitrogen. It is placed in a medium with only light nitrogen and replicates 5 times.

(a) State how many molecules contain a heavy strand. [1] (b) Calculate the fraction of all molecules that contain a heavy strand. [1] (c) Explain why no molecule contains two heavy strands after round 1. [1]

2. (a) Describe the role of helicase at a replication fork. [2] (b) Explain how complementary base pairing helps make the copy accurate. [2]

3. A PCR run uses template DNA, two primers, free nucleotides and Taq polymerase.

(a) Describe what happens at the annealing and extension steps. [2] (b) State why Taq polymerase is used. [1] (c) Calculate the ideal number of copies of one target molecule after 28 cycles. [1] (d) Suggest why the real number is lower. [1]

4. DNA fragments are run on a gel. The ladder bands are 800 bp at 15 mm, 400 bp at 27 mm and 200 bp at 39 mm from the well. A sample band is at 33 mm.

(a) Explain why the fragments separate by size. [3] (b) Estimate the size of the sample band. [3]

5. Band sizes (bp) for three markers in a paternity test:

Marker	Mother	Child	Alleged father
1	120, 150	120, 175	175, 200
2	80, 95	95, 110	110, 110
3	210, 230	230, 260	140, 260

(a) Use the table to evaluate whether the alleged father could be the father. [3] (b) The chances that an unrelated man matches at the markers are 1 in 6, 1 in 14 and 1 in 22. Calculate the combined chance. [1] (c) Explain the effect of adding a fourth marker. [1]

6. (HL only) (a) State which end of a growing strand DNA polymerase adds nucleotides to. [1] (b) Distinguish between leading and lagging strand synthesis. [3]

7. (HL only) A circular chromosome of 1.8 × 10⁶ base pairs is copied from a single origin by two forks moving apart, each at 750 nucleotides per second (data given). Lagging strand fragments are 1200 nucleotides long.

(a) Calculate the time needed to copy the chromosome, in minutes. [3] (b) Calculate the number of Okazaki fragments made at one fork. [2] (c) Estimate the total number of RNA primers used. [1]

8. (HL only) (a) State the functions of primase and DNA ligase. [2] (b) Describe proofreading by DNA polymerase III. [2] (c) A polymerase makes 1 error per 10⁵ bases without proofreading and 1 per 10⁷ with it (data given). Calculate the expected errors in copying a 2.4 × 10⁶ bp genome in each case. [2]

9. An mRNA reads 5’ AUG AAA GAU CUC UGG UAA 3’.

(a) Give the amino acid sequence it codes for (Met, Lys, Asp, Leu, Trp). [1] (b) Classify each substitution and state its effect: AAA to AAG; GAU to GAA; UGG to UGA. [3] (c) The first C of CUC is deleted. Give the new sequence (Ser, Gly also occur) and name the mutation type. [2]

10. (a) State two causes of gene mutation. [2] (b) A new-mutation rate is 1.2 × 10⁻⁸ per base pair per generation. A 1500 bp coding region is present in two copies in each of 50 000 children. Calculate the expected number of new mutations in this region. [2] (c) State what randomness means for mutation. [1]

11. (a) Contrast the consequences of a mutation in a germ cell and a somatic cell. [3] (b) Explain why mutation is essential for evolution even though most mutations are harmful or neutral. [3] (c) A commercial genetic test reports a SNP linked to raised disease risk. Suggest two problems of interpreting the result. [2]

12. (HL only) Mice with gene M knocked out take a mean of 15.1 s to cross a balance beam; normal mice take 8.4 s.

(a) Calculate the percentage increase in time. [2] (b) Suggest what this indicates about gene M and one limitation. [2] (c) Outline how CRISPR-Cas9 could produce a knockout. [2] (d) Gene M is highly conserved. Give two hypotheses. [2]

Answers

1. (a) 2 molecules [1]. (b) 2/32 = 1/16 [1]. (c) Each heavy strand templates a new light strand, so every molecule is a hybrid [1]. Examiner insight: give the reason in terms of templates; a bare number answers only the first parts.

2. (a) Helicase unwinds the double helix [1] by breaking hydrogen bonds between the strands [1]. (b) Each base pairs only with its complement (A with T, G with C) [1], so the template sequence determines the new strand [1]. Examiner insight: name both the bond and the pairing rule; “it unzips DNA” alone is vague.

3. (a) Annealing (about 55 to 60 °C): primers bind to complementary ends of the target [1]. Extension (about 72 °C): Taq polymerase adds nucleotides to the primers [1]. (b) It is heat stable so it is not denatured at the strand-separating step [1]. (c) 2²⁸ = 268,435,456 [1]. (d) Not every strand is fully copied in each cycle, or reagents run out [1]. Examiner insight: temperatures are approximate values, so tie each to what happens; link “heat stable” to the denaturing step.

4. (a) DNA is negatively charged (phosphate groups) so moves towards the positive electrode [1]; the gel is a mesh [1]; small fragments pass more easily so travel further [1]. (b) Size halves every 12 mm [1]. The sample is 6 mm past 400 bp, so 400 × 2^(−6/12) [1] = about 280 bp [1]. Examiner insight: state the method and give the answer as an estimate with units.

5. (a) Marker 1: child’s 175 is not maternal and is present in the father [1]. Marker 2: child’s 110 is not maternal and is present in the father [1]. Marker 3: child’s 260 is present in the father, so he could be the father [1]. (b) 6 × 14 × 22 = 1848, so 1 in 1848 [1]. (c) A fourth marker lowers the chance of a false match further, so the result is more reliable [1]. Examiner insight: name the paternal band at each marker; “bands match” is too general.

6. (a) The 3’ end [1]. (b) Leading strand: made continuously [1], one primer needed [1]. Lagging strand: made discontinuously as Okazaki fragments, each with its own primer [1]. Examiner insight: use the terms continuous, discontinuous and Okazaki fragments, and compare both strands.

7. (a) Each fork copies 9 × 10⁵ bp [1]; 900 000 ÷ 750 = 1200 s [1]; = 20 min [1]. (b) 900 000 ÷ 1200 [1] = 750 fragments [1]. (c) About 750 per lagging strand, 1500, plus 2 for the leading strands = about 1502 [1]. Examiner insight: show the per-fork length; using the whole chromosome gives a time that is double.

8. (a) Primase makes RNA primers [1]; ligase joins fragments by sealing the backbone [1]. (b) Polymerase III detects a mismatched nucleotide at the 3’ end [1], removes it and adds the correct one [1]. (c) Without: 2.4 × 10⁶ ÷ 10⁵ = 24 [1]. With: 2.4 × 10⁶ ÷ 10⁷ = 0.24 [1]. Examiner insight: name the enzyme and the end of the strand it acts on.

9. (a) Met-Lys-Asp-Leu-Trp-stop [1]. (b) AAA to AAG: Lys still, silent [1]. GAU to GAA: Asp becomes Glu, missense [1]. UGG to UGA: stop codon, nonsense, polypeptide shortened [1]. (c) AUG AAA GAU UCU GGU AA: Met-Lys-Asp-Ser-Gly, with the stop codon no longer read [1]; a frameshift deletion [1]. Examiner insight: compare amino acids before naming a mutation; regroup into triplets after an indel.

10. (a) Errors in replication or repair that are not corrected [1]; mutagens, either chemical or radiation such as ultraviolet [1]. (b) 50 000 × 2 × 1500 × 1.2 × 10⁻⁸ [1] = 1.8 [1]. (c) A mutation can happen at any base, with no natural mechanism aiming it to change a trait [1]. Examiner insight: give specific mutagens, not “pollution”.

11. (a) Germ cell: can be inherited [1], present in all cells of offspring [1]. Somatic: not passed on but can cause cancer [1]. (b) Mutation creates new alleles [1]; selection acts on variation [1]; a rare beneficial allele can spread over time [1]. (c) Risk is probabilistic not certain [1]; results may cause worry or be misread without expert advice [1]. Examiner insight: link “source of variation” to selection acting on it.

12. (a) (15.1 − 8.4) ÷ 8.4 × 100 [1] = 80% (79.8) [1]. (b) Gene M helps balance or coordination [1]; limitation: other effects or small sample, so cause is uncertain [1]. (c) A guide RNA directs Cas9 to cut the gene [1]; repair errors disable it [1]. (d) Its product has a function that tolerates little change, so selection removes variants [1]; the region mutates more slowly [1]. Examiner insight: give a limitation tied to this data, not a generic one.

Where marks are usually lost

• Giving fractions without the 2 ÷ 2ⁿ reasoning.
• Not naming the bond helicase breaks.
• Stating “heat” without the temperatures’ purpose in PCR.
• Saying large fragments travel furthest on a gel.
• Using the whole genome length for a single fork.
• Calling a mutation silent without comparing amino acids.
• Missing the triplet regrouping after a deletion.
• Writing that mutations happen because an organism needs them.

Next steps

• Review the revision notes and study guide.
• Tick off your progress on the printable checklist and visit the IB DP Biology hub.
• Take the free 10-minute diagnostics to find weak spots.
• Book a free trial class.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization: sections D1.1 (DNA replication) and D1.3 (Mutation and gene editing).