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IB Biology D1.3: Mutation and gene editing worked solutions

Worked solutions to every practice question in D1.3: Mutation and gene editing of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

[1 mark]

Which process can produce a new allele of a gene? (A) Crossing over in meiosis (B) Mutation of the gene (C) Independent assortment (D) Random fertilisation

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B. A new allele is a changed base sequence, and only mutation changes it. Crossing over, independent assortment and random fertilisation reshuffle alleles that already exist.

Final answer

Question A2

[1 mark]

Which statement about gene mutation is correct? (A) Mutations are aimed at the bases an organism needs to change. (B) Mutations can occur at any base, but some bases mutate more often. (C) Mutations occur only after exposure to a mutagen. (D) Every base of a gene has the same chance of mutating.

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B. A mutation can fall on any base, though some bases change more often than others. A and D are wrong because mutation is neither directed nor equal at every base, and C is wrong because errors in replication also cause mutations.

Final answer

Question A3

[1 mark]

A section of a gene reads CTAGGCAT. After a mutation it reads CTAGGCCAT. Which mutation has occurred? (A) A substitution (B) An insertion (C) A deletion (D) A deletion followed by a substitution

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B. The mutant has nine bases and the original eight, and the extra base is a C, so this is an insertion. A substitution keeps the length and a deletion shortens it.

Final answer

Question A4

[1 mark]

A base substitution occurs in the middle of the coding sequence of a gene. Which result is possible? (A) Every amino acid after the change is different. (B) Only one amino acid changes, or none changes. (C) The gene becomes one base shorter. (D) The reading frame shifts by one base.

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B. A substitution changes one codon, so either one amino acid differs or, because the code is degenerate, none does. A and D describe a frameshift, and C describes a deletion, not a substitution.

Final answer

Question A5

[1 mark]

A mutation occurs in a skin cell exposed to ultraviolet light. Which statement is correct? (A) It will be inherited by the person's children. (B) It can be inherited only by a daughter. (C) It is not inherited and it may contribute to skin cancer. (D) It will be present in every cell of the person's body.

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C. A skin cell is a somatic cell, so its mutation is not in the gametes and is not inherited, but it can contribute to skin cancer. The mutation is not in every cell of the body and children cannot receive it.

Final answer

Question A6

[1 mark]

Which of these can act as a mutagen and as a carcinogen? (A) Visible light (B) X-rays (C) Radio waves (D) Infrared at low intensity

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B. X-rays and other ionising radiation damage DNA, and damage that leads to cancer makes the agent a carcinogen. Visible light, radio waves and low-intensity infrared do not damage DNA bases.

Final answer

Question A7

[1 mark]

The mRNA AUGUCAGCAGUU is read in codons from the start. The fifth base, C, is deleted. What is the result? (A) One amino acid is lost and the rest is unchanged. (B) A stop codon appears at the second codon. (C) One amino acid is replaced by another. (D) The polypeptide is unchanged.

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B. The new sequence is AUGUAGCAGUU, which reads AUG UAG, and UAG is a stop codon at the second position. One deletion is not a multiple of three, so every codon after it is regrouped.

Final answer

Question A8

[1 mark]

The coding strand of a gene contains the triplet GAA. A substitution changes it to TAA. Use the table of mRNA codons to predict the effect on the polypeptide. mRNA codons: GAA Glu; GAG Glu; UAA Stop; UAG Stop. (A) No change in the polypeptide (B) Glu is replaced by another amino acid (C) The polypeptide ends at this position (D) The polypeptide is longer

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C. GAA is transcribed to the mRNA codon GAA, which codes for Glu. TAA becomes UAA, a stop codon, so translation ends at that position and the polypeptide is shorter.

Final answer

Question A9

HL[1 mark]

What is gene knockout technology used for? (A) To increase the expression of a gene (B) To study the function of a gene (C) To add a new gene to an organism (D) To repair a mutated gene

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B. A knockout makes a gene inoperative so that the changes in the organism show what the gene normally does. It does not raise expression, add a gene or repair a mutation.

Final answer

Question A10

HL[1 mark]

What is the role of the guide RNA in CRISPR-Cas9 gene editing? (A) It cuts both strands of the DNA. (B) It pairs with the target sequence and directs Cas9 to it. (C) It repairs the cut in the DNA. (D) It supplies the correct template.

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B. The guide RNA is complementary to the target and brings Cas9 to the matching DNA. Cas9 does the cutting, and the cell's own enzymes repair the cut.

Final answer

Tutorial practice

Question B1

[3 marks]

Outline how a base substitution in a gene could lead to a shorter polypeptide.

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A base in the gene is replaced by a different base (1). The mRNA codon becomes a stop codon, UAA, UAG or UGA (1). Translation ends at that codon, so the polypeptide is shorter than normal (1). The same change in the middle of the gene could also be silent or missense, so the answer must name the stop codon.

Final answer

A substitution changes one codon of the mRNA into a stop codon (UAA, UAG or UGA). Translation stops there, so the polypeptide is shorter and probably not functional.

Question B2

[4 marks]

Compare and contrast substitution, insertion and deletion mutations.

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Any four of: all three change the base sequence of the gene (1); all can occur at any position and are random (1); a substitution replaces one base, so the length of the gene is unchanged (1); an insertion adds bases (1); a deletion removes bases (1); an insertion or deletion whose size is not a multiple of three shifts the reading frame (1); a substitution changes at most one codon and cannot shift the frame (1). A comparison must give a similarity and a difference.

Final answer

All change the base sequence and can occur anywhere. A substitution replaces a base and keeps the length. An insertion adds bases and a deletion removes them. An insertion or deletion not in multiples of three causes a frameshift; a substitution never does.

Question B3

[3 marks]

The commonest cystic fibrosis allele deletes three bases from the CFTR gene, and the protein lacks one phenylalanine. (a) Explain why the other amino acids in the protein are unchanged. [1] (b) Explain how the loss of one amino acid can cause a disease. [2]

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(a) Three bases is a whole codon, so the reading frame is kept and only one amino acid is lost (1). (b) The sequence of amino acids is changed, so the folded shape of the protein is altered (1); the misfolded channel cannot do its normal function of carrying chloride ions across the membrane (1).

Final answer

(a) Three is a multiple of three, so the groups of three after the deletion are unchanged. (b) The missing amino acid changes the folding, so the protein is misfolded and does not reach the cell surface to work as a chloride channel.

Question B4

[7 marks]

A base substitution in an enzyme gene gives a different amino acid at the enzyme's active site and causes a recessive disorder. Explain how the substitution leads to the disorder in a person with two copies of the allele.

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Any seven of: one base in the gene is replaced by another (1); the base sequence of the transcribed mRNA changes at one codon (1); a different tRNA anticodon pairs with it (1); a different amino acid is added to the polypeptide (1); the change is in the active site, so the shape of the folded enzyme changes (1); the substrate can no longer bind, so the enzyme has little or no activity (1); the substrate accumulates or the product is not made, so the symptoms appear (1); a person with one working allele still makes enough enzyme, so the disorder shows only with two copies (1). The chain must run DNA, mRNA, amino acid, shape, function, phenotype.

Final answer

The changed base alters the mRNA codon, so a different amino acid is placed in the active site. The tertiary structure changes, the substrate no longer fits, and the enzyme is inactive. The substrate builds up or the product is missing, which gives the symptoms. Two copies are needed because one working allele makes enough enzyme.

Question B5

[5 marks]

Mutations along a gene of 600 bases were counted in six regions of 100 bases. Mutations per 100-base region: 1 to 100: 2; 101 to 200: 3; 201 to 300: 15; 301 to 400: 3; 401 to 500: 2; 501 to 600: 5. (a) State the region with the most mutations. [1] (b) Calculate the percentage of all mutations that fall in this region. [1] (c) Deduce whether mutation is equally likely at every position. [1] (d) Explain why the data are consistent with mutation being random. [2]

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(a) Bases 201 to 300 (1). (b) 15\div30\times100=50\% (1). (c) No, mutation is not equally likely at every position (1): an even spread would give 30\div6=5 mutations per region, and region 3 has three times that. (d) Mutations are found in every region, so none is excluded (1); no mechanism directs a change toward a trait the cell needs, and the rate differs between sites without the outcome being predictable (1).

Final answer

(a) Bases 201 to 300. (b) 15/30=50\%. (c) No: an even spread would give 5 per region. (d) Mutations occur in every region and nothing aims them at a need; some sites change more often, but which base mutates next cannot be predicted.

Question B6

[6 marks]

Bacteria were exposed to ultraviolet light for different times and the mutant colonies on each plate were counted (mean of five plates). Mean mutant colonies per plate for exposure of 0 s, 10 s, 20 s and 30 s: 12, 36, 84 and 132. (a) Describe the trend, quoting values. [2] (b) Calculate how many times greater the number of mutants is after 20 s than at 0 s. [1] (c) State the purpose of the plates exposed for 0 s. [1] (d) Evaluate the claim that the data show that ultraviolet light is a mutagen. [2]

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(a) Mutant colonies increase as the exposure time increases (1), from 12 at 0 s to 132 at 30 s, an increase of 120 (1). (b) 84\div12=7 times (1). (c) The 0 s plates are the control: they show the mutation rate without ultraviolet light, so the effect of the light can be separated (1). (d) The steady rise with exposure supports ultraviolet light as a mutagen (1); the limits are that only one species was tested, no spread is given for the means, and only four exposure times were used (1).

Final answer

(a) The number of mutant colonies rises as exposure increases, from 12 to 132. (b) 84/12=7 times. (c) The 0 s plates show the number of mutants that arise without ultraviolet light. (d) The rise supports ultraviolet light as a mutagen. Limits: one species and no spread given for the means.

Question B7

HL[5 marks]

A patient has an inherited condition caused by a single-base substitution in one gene. Explain how CRISPR-Cas9 could be used to correct this mutation.

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A guide RNA is designed that is complementary to the DNA at the site of the mutation, next to a PAM (1). The guide RNA pairs with the target and directs Cas9 to it (1). Cas9 cuts both strands of the DNA at that site (1). A template with the correct sequence is supplied (1). The cell repairs the cut by copying the template, so the mutated base is replaced by the correct one (1). Without a template, plain repair usually adds or loses bases and knocks the gene out instead.

Final answer

A guide RNA is made complementary to the sequence at the mutation, beside a PAM. It leads Cas9 to the site and Cas9 cuts both strands. A DNA template with the correct base is supplied, and the cell repairs the cut by copying the template, so the base is corrected.

Question B8

HL[5 marks]

The gene fin1 was knocked out in zebrafish. Fin length is given as mean \pm standard deviation, n=50 fish per group. Fin length: two working copies 6.0 mm (SD 0.4); one working copy 5.8 mm (SD 0.4); no working copy 3.0 mm (SD 0.5); n = 50 per group. (a) Calculate the percentage by which the knockout fish are shorter than the wild type. [1] (b) Deduce the function of the gene. [1] (c) Suggest why fish with one working copy are almost normal. [1] (d) Evaluate the strength of this evidence. [2]

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(a) (6.0-3.0)\div6.0\times100=50\% (1). (b) The gene is needed for normal fin growth (1). (c) One working copy produces enough functional product, so the allele is recessive and the fish look nearly normal (1). (d) Evaluation, two of: n=50 per group and the means differ by much more than the standard deviations, so the difference is unlikely to be chance (1); the wild-type group is an appropriate control (1); other genes may contribute to fin growth, and other traits were not measured, so the function is only partly known (1).

Final answer

(a) (6.0-3.0)/6.0=50\% shorter. (b) fin1 is needed for normal fin growth. (c) One working copy makes enough product, so the heterozygote is near normal. (d) The sample is large and the means are well separated, but another gene could contribute and the gene's effect on other traits was not measured.

Advanced practice

Question C1

HL[6 marks]

The table compares one ribosomal protein gene in four species with the human gene. Identity is the percentage of positions that are the same as in the human sequence. Silent positions are positions where a change would not alter the amino acid. The values are illustrative. Million years since common ancestor, protein identity and silent-site identity to the human gene: chimpanzee 6, 100 percent, 98 percent; mouse 80, 99, 70; chicken 310, 98, 45; zebrafish 430, 97, 36.. (a) Describe the trends shown by the data. [2] (b) Explain these trends using the hypothesis that the gene product is essential. [2] (c) State what the data suggest about the hypothesis that the gene mutates slowly. [1] (d) Suggest why a ribosomal protein would be essential. [1]

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(a) Protein identity stays high, from 100 to 97 percent (1), while silent-site identity falls steeply, from 98 to 36 percent, as the time since the common ancestor increases (1). (b) The silent positions differ, so mutations are occurring in the gene (1); mutations that change the protein were not kept because individuals with a faulty ribosomal protein survive and breed less, so the sequence is conserved by function (1). (c) The mutation rate in the gene is not low, so the data do not support the slower-mutation hypothesis (1). (d) Ribosomes make all proteins, so a damaged ribosomal protein is likely to be lethal (1).

Final answer

(a) Protein identity falls only from 100 to 97 percent over 430 million years while silent-site identity falls from 98 to 36 percent. (b) The silent positions have changed, so mutation has occurred; changes that alter the protein were lost because organisms with a damaged ribosomal protein did not breed. (c) It does not support the slow-mutation hypothesis. (d) The protein is needed to make every other protein.

Question C2

HL[6 marks]

Discuss the ethical issues raised by using CRISPR-Cas9 to edit the genes of human embryos.

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Any six of: editing an embryo could remove a serious inherited disease from a family (1); the benefit is large when the disease is untreatable (1); Cas9 may cut at the wrong place and the edit may not reach every cell, so the effects cannot be fully predicted (1); the edit is inherited by future generations who cannot consent (1); the long-term effects on the gene pool are unknown (1); treatment may be costly and available only to some people (1); editing to enhance a trait is different from treating a disease, and the line is hard to draw (1); regulation differs between countries, so misuse in one place has global effects and international agreement is needed (1). Both sides must appear for the discussion marks.

Final answer

Benefit: an inherited disease could be removed from a family line. Risk: off-target cuts and edits in only some cells with unknown long-term effects. The change is inherited by people who cannot consent. Cost may limit access, and edits for enhancement differ from treatment. Rules differ between countries, so international agreement is needed.

Question C3

[8 marks]

A short gene has the template strand 3'-TAC CGG TTT CTT ACT-5'. Use the table of mRNA codons. mRNA codons: AUG Met; GCC Ala; AAA Lys; AAG Lys; GAA Glu; GAG Glu; AAU Asn; UGA Stop. (a) Write the mRNA sequence that is transcribed. [1] (b) State the polypeptide it codes for. [1] (c) Mutation 1 changes the template codon CTT to CTC. Predict the effect on the polypeptide and explain it. [2] (d) Mutation 2 deletes the first T of the template triplet TTT. Write the new mRNA and the amino acids it codes for. [2] (e) Explain why mutation 2 is likely to be more harmful than mutation 1. [2]

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(a) The mRNA is the complement of the template: AUG GCC AAA GAA UGA (1). (b) Met-Ala-Lys-Glu, with the stop codon ending translation (1). (c) The substitution changes the template CTT to CTC, so the codon GAA becomes GAG (1); both code for Glu because the genetic code is degenerate, so the polypeptide is unchanged (1). (d) Deleting the first T of TTT gives TAC CGG TTC TTA CT, so the mRNA is AUG GCC AAG AAU GA (1) and the polypeptide Met-Ala-Lys-Asn, with the stop codon UGA no longer in frame (1). (e) The deletion shifts the reading frame, so every codon after it changes and the stop codon is lost (1); the substitution changes only one codon and, here, not the amino acid (1).

Final answer

(a) AUG GCC AAA GAA UGA. (b) Met-Ala-Lys-Glu. (c) Mutation 1 is a substitution: GAA becomes GAG, and both code for Glu, so the code is degenerate and the protein is unchanged. (d) Mutation 2 is a deletion that shifts the frame: AUG GCC AAG AAU GA, Met-Ala-Lys-Asn, and the stop codon is lost. (e) The deletion changes every amino acid after it and removes the stop; the substitution changes no amino acid.

Question C4

[7 marks]

Lung cells were taken from two 40-year-old adults. The mean number of mutations per cell was counted (illustrative values). Mean mutations per lung cell at age 40: non-smoker 1000; smoker of 20 cigarettes a day 2500.. (a) Calculate the percentage by which the mutation count is greater in the smoker. [1] (b) Explain how the mutations caused by smoke could lead to cancer. [2] (c) Explain why the mutations in a lung tumour are not passed to the person's children. [1] (d) Evaluate the claim that these data prove that smoking causes lung cancer. [2] (e) State one other cause of gene mutation. [1]

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(a) (2500-1000)\div1000\times100=150\% (1). (b) Smoke contains chemicals that are mutagens and damage DNA (1); a mutation in a gene that controls the cell cycle lets the cell divide without control, and its descendants form a tumour (1). (c) The mutations are in somatic cells, so they are not in the gametes and are not inherited (1). (d) The data show more mutations in the smoker but not tumours (1); the samples may be one person per group, so chance, age and other exposures could account for the difference (1). (e) Any one: ultraviolet light, X-rays, gamma rays, a base analogue, mustard gas or reactive oxygen species (1).

Final answer

(a) (2500-1000)/1000=150\%. (b) Smoke contains mutagens that damage DNA; a mutation in a gene that controls cell division can make a cell divide without control, and its clone forms a tumour. (c) The mutations are in somatic cells, not in gametes. (d) The data show an association between smoking and mutations but not cancer itself; the sample may be one person per group and age or other exposure may differ. (e) Ultraviolet radiation, X-rays or a chemical such as mustard gas.

Connected questions

Question X1

Connected questions[6 marks]

D1.3 + D1.1 + D4.1. A human cell copies 6\times10^{9} base pairs of DNA before it divides. DNA polymerase pairs a wrong base about once in every 10^{5} bases it copies. Proofreading corrects 999 of every 1000 of these errors. (a) Calculate the number of wrong bases inserted in one copying of the genome before proofreading. [1] (b) Calculate how many errors remain after proofreading. [1] (c) State the kind of mutation caused by one remaining wrong base pair. [1] (d) Explain how such an error in a cell that forms gametes could change the frequency of an allele in a population. [3]

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(a) 6\times10^{9}\div10^{5}=6\times10^{4}=60\,000 errors (1). (b) Only 1 in 1000 remains: 60\,000\div1000=60 errors (1). (c) A wrong base pair becomes a substitution (1). (d) The error in a cell that forms gametes creates a new allele that is inherited (1); if it gives an advantage, individuals that carry it survive and reproduce more (1); the frequency of the allele rises over generations, which is evolution by natural selection (1).

Final answer

(a) 6\times10^{9}/10^{5}=60\,000. (b) 60\,000\times0.001=60. (c) A substitution. (d) A new allele is made; if it helps survival or reproduction, individuals with it leave more offspring, and the frequency of the allele rises over the generations.

Question X2

Connected questions[6 marks]

D1.3 + D1.2. A template strand triplet reads CGA and codes for alanine. A substitution changes it to CGG. Both codons code for alanine. (a) State the mRNA codon transcribed from the template triplet CGA. [1] (b) State the anticodon of the tRNA that pairs with this codon. [1] (c) State the name for a base substitution that does not change the amino acid. [1] (d) Explain why the substitution CGA to CGG does not change the polypeptide. [3]

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(a) The mRNA codon is the complement of the template CGA: GCU (1). (b) The tRNA anticodon pairs with GCU: CGA (1). (c) A silent mutation (1). (d) The changed template CGG gives the mRNA codon GCC (1); GCU and GCC are both codons for alanine, because the genetic code is degenerate (1), so the tRNA brings the same amino acid and the polypeptide is unchanged (1).

Final answer

(a) GCU. (b) The tRNA anticodon is CGA, complementary to GCU. (c) Silent mutation. (d) The mRNA codon becomes GCC, and GCU and GCC both code for Ala because the genetic code is degenerate, so the tRNA that arrives carries the same amino acid and the polypeptide is unchanged.

Question X3

Connected questions[6 marks]

D1.3 + D2.1. Sunlight damages the DNA of a skin cell, and a mutation arises in a gene that restrains cell division. The cell divides repeatedly and forms a tumour. (a) State why the mutation cannot be passed on to the person's children. [1] (b) Explain how one mutated cell can give rise to a tumour. [3] (c) Suggest why the risk of cancer rises with age. [2]

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(a) The mutation is in a body (somatic) cell, so it is not in the gametes and is not inherited (1). (b) Any three: the mutated cell divides by mitosis (1); every daughter cell inherits the mutation, forming a clone (1); the mutation removes the restraint, so the cells keep dividing without control (1); the growing mass of cells is a tumour (1). (c) Any two: mutations accumulate over time (1); DNA repair becomes less effective with age (1); the total exposure to mutagens such as sunlight is longer (1).

Final answer

(a) The mutation is in a somatic cell, so it is not in the gametes. (b) The mutated cell divides by mitosis, and each daughter carries the mutation, so a clone forms; without restraint the cells keep dividing faster than they are lost, and a mass of cells forms. (c) Mutations accumulate through the years and DNA repair may fail with age.

Question X4

Connected questions[8 marks]

D1.3 + D3.2. Cystic fibrosis is caused by a recessive allele with a deletion of three bases. Two healthy parents are both heterozygous carriers. (a) Calculate the probability that a child of these parents has cystic fibrosis. [2] (b) Calculate the probability that two children are both unaffected. [2] (c) Explain how a deletion of three bases can cause a disease. [2] (d) Suggest why the harmful allele persists in the population. [2]

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(a) Cross Cc\times Cc gives CC, Cc, Cc, cc, so the probability of cc is 1/4 (2). (b) The chance of one unaffected child is 3/4, so both are unaffected with probability (3/4)^2=9/16=0.56 (2). (c) The deletion of three bases removes one codon and keeps the reading frame (1); the missing amino acid changes the shape of the protein so that it cannot work (1). (d) Carriers are healthy, so selection acts only against cc (1); most copies of the allele are in carriers and are passed on (1).

Final answer

(a) 1/4. (b) (3/4)^2=9/16 or 0.56. (c) The deletion of three bases removes one amino acid and the frame is kept, but the protein is misfolded and does not work. (d) Carriers have a working allele and are healthy, so selection acts only against homozygous recessives and the allele stays in the population.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Mutation and gene editing from the core ideas to full exam questions, with question types, worked examples and graded practice.

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