Questions on mutation, gene expression and cancer

Question 1

Explain how a mutation in the CFTR gene impairs the functioning of the gaseous exchange system and of the digestive system in a person with cystic fibrosis.

Accepted Answer

B1 the CFTR protein is a chloride ion channel in the cell-surface membrane of epithelial cells; when it is missing or faulty, too few chloride ions are secreted from the cells B1 sodium ions are absorbed into the cells and water follows by osmosis, rather than leaving the cells, so the mucus is dehydrated and becomes thick and sticky B1 cilia cannot move the thick mucus, so it collects in and blocks bronchioles, and the alveoli beyond are not ventilated B1 the surface area available for gas exchange is reduced, so less oxygen diffuses into the blood; bacteria trapped in the mucus cause repeated infections that scar the lungs B1 thick mucus blocks the pancreatic duct, so pancreatic enzymes such as lipase, amylase and proteases do not reach the duodenum B1 food is incompletely digested and the products are poorly absorbed, made worse by thick mucus lining the small intestine, so the person struggles to gain weight

Question 2

Explain why the deletion of a single base near the start of a gene usually destroys the protein, while the deletion of three bases at the same point usually does not.

Accepted Answer

B1 the code is non-overlapping and read in triplets from a fixed starting point B1 losing one base moves every base after it one place forward, so the reading frame is shifted and every triplet downstream is read from a new position B1 all the amino acids after the deletion are therefore potentially wrong, and a stop codon usually appears early by chance, so the polypeptide is shortened as well as scrambled and will not fold into anything B1 three is a multiple of three, so the frame is not shifted: one amino acid is removed and every triplet after it is read exactly as before

Question 3

Compare a proto-oncogene with a tumour suppressor gene, referring to what the product of each normally does and to how many alleles must be affected before division is lost from control.

Accepted Answer

B1 a proto-oncogene codes for a protein that stimulates cell division when a growth signal arrives, whereas a tumour suppressor gene codes for a protein that halts the cell cycle at a checkpoint or triggers apoptosis B1 a proto-oncogene does harm when a mutation makes it permanently active or present in too many copies, whereas a tumour suppressor gene does harm when its product is lost B1 one faulty allele is enough for an oncogene, because a permanently active protein acts whatever the other allele makes, whereas both alleles of a tumour suppressor gene must go, because one working copy still makes the protein B1 in both cases the outcome is the same: cells pass checkpoints they should not and divide repeatedly, so a mass of cells accumulates

Question 4

A research group wishes to obtain stem cells from embryos left over from IVF treatment. Evaluate the case for allowing this work.

Accepted Answer

B1 for: embryonic stem cells are pluripotent and can give rise to any cell type of the body, so they offer a range of treatment that multipotent adult stem cells cannot B1 for: the embryos are surplus from IVF and would otherwise be discarded, and many people hold that there is a moral difference between using such an embryo and creating one in order to destroy it B1 against: those who hold that a human embryo has the full moral status of a person from fertilisation onwards regard its destruction as the deliberate ending of a human life, which the good done with the cells does not license B1 judgement: British law takes a middle position, licensing such research through the Human Fertilisation and Embryology Authority up to fourteen days; induced pluripotent cells narrow the disagreement, since they are pluripotent and need no embryo, though they can carry mutations acquired in the adult cell and a tendency to form tumours

Question 5

A person inherits about 70 new mutations that neither parent carried. A human body cell contains 6.2 × 10⁹ base pairs of DNA in total, and about 1.5% of the genome codes for protein. Calculate the rate of new mutation per base pair per generation, and calculate how many of the 70 new mutations would be expected to fall within protein-coding DNA.

Accepted Answer

M1 the rate is the number of new mutations divided by the number of base pairs in which one could occur, so 70 ÷ (6.2 × 10⁹) A1 1.1 × 10⁻⁸ per base pair per generation M1 the number expected in coding DNA is 1.5% of 70, so 0.015 × 70 A1 about 1 mutation, with the other 69 falling in non-coding DNA, where most change no polypeptide at all

Question 6

The table shows part of the coding strand of a gene, and the same region of four mutant alleles of it, each carrying a single mutation. Identify the type of mutation in each mutant allele, and identify in each case whether the reading frame of the triplets that follow it is shifted.

Accepted Answer

B1 mutant 1 is a substitution: the eighth base has been replaced and the length is unchanged, so the frame is not shifted and only that one triplet is altered B1 mutant 2 is an insertion of a single base, an extra A after the ninth base, so the sequence is one base longer and every triplet after the insertion is read from a new position, which is a frame shift B1 mutant 3 is an inversion: the six bases from the fourth to the ninth read as the original six reversed, so the length is unchanged and the frame is not shifted, but the reversed region codes for a different sequence of amino acids B1 mutant 4 is a duplication of the triplet TAC, which may also be described as an insertion of three bases; three is a multiple of three, so the frame is not shifted and one extra amino acid is added

Question 7

The table gives five-year survival for patients with a tumour of the same tissue, grouped by how far the tumour had spread when it was found; the figures are constructed for this question rather than measured. Suggest why survival falls so steeply from the first group to the third, referring to what distinguishes a benign tumour from a malignant one.

Accepted Answer

B1 survival falls from 372 of 400, which is 93%, to 244 of 400, which is 61%, and then to 48 of 400, which is 12% B1 the first group's tumour behaves as a benign one: it has remained localised and has not metastasised, and the supporting features are that it is contained, often slow growing and made of cells much like those of the tissue it came from, so removing it removes the disease B1 the second group's tumour is malignant: its cells have invaded the surrounding tissue, they have often lost the appearance and the function of that tissue, and surgery may leave cells behind from which the tumour regrows B1 the third group's tumour has metastasised, shedding cells into the blood and lymph which have established secondary tumours in other organs, so there is no single site to remove and any treatment has to reach the whole body

Question 8

The base change responsible for lactase persistence lies inside an intron of a neighbouring gene, not inside the lactase gene itself. Explain how it can still change how much lactase is made, without changing the lactase protein's amino acid sequence.

Accepted Answer

B1 the base change lies within an intron of a neighbouring gene, in a sequence that acts as an enhancer for the lactase gene rather than inside the lactase gene's own coding sequence B1 because the change is not in the lactase gene's own coding sequence, it does not alter a single codon, so the amino acid sequence of lactase is identical whether or not a person carries it B1 what the enhancer change affects is how much lactase mRNA is transcribed, keeping the lactase gene switched on into adulthood rather than being silenced after weaning B1 this shows that a mutation in a control sequence changes how much protein is made rather than changing what the protein is

Question 9

In some cases of β-thalassaemia the mutation lies at the boundary between an intron and an exon of the β-globin gene rather than in its coding sequence. Suggest how this could reduce the amount of normal β-globin made, and suggest why sequencing only the exons of the gene might miss this mutation.

Accepted Answer

B1 a mutation at an intron-exon boundary can disrupt the sequence the spliceosome uses to recognise where to cut, so splicing occurs at the wrong position B1 the mature mRNA then retains intron sequence or loses exon sequence, most likely shifting the reading frame or introducing a premature stop codon, so translation gives a shortened or non-functional polypeptide, or none at all B1 because the mutation lies at the very edge of an intron rather than inside an exon, sequencing that reads only exon sequence could fail to cover the position where the change lies B1 the coding sequence read on its own would then appear completely normal, so the cause of the reduced β-globin would be missed unless the intron-exon boundary itself were sequenced

Question 10

Explain why a person born with one faulty BRCA1 allele does not have cancer at birth, but is at a greatly increased lifetime risk of developing it.

Accepted Answer

B1 BRCA1 is a tumour suppressor gene, and in most cells one working allele still makes enough functional protein to enforce the checkpoint or trigger the apoptosis it is responsible for B1 every cell in that person's body already carries one faulty allele, so only a single further mutation, in the one remaining working allele of any one cell, removes the protein from that cell completely B1 someone born with two working alleles would need that same single event to happen twice, independently, in the same cell, which is far less likely during a lifetime B1 the inherited faulty allele does not itself cause cancer, but it removes one of the two events needed and greatly raises the chance that some cell in the body will eventually lose both

Question 11

Compare an embryonic stem cell with an induced pluripotent stem cell, referring to potency, the need for an embryo, and the risk of tissue rejection.

Accepted Answer

B1 an embryonic stem cell is naturally pluripotent, taken from an embryo about four to five days old, whereas an induced pluripotent stem cell is made by treating an adult, differentiated cell with a small set of transcription factors that return it to a pluripotent state B1 obtaining embryonic stem cells destroys the embryo they came from, whereas producing induced pluripotent stem cells uses no embryo at all B1 tissue grown from a patient's own induced pluripotent stem cells carries a much lower risk of rejection, since it shares the patient's own antigens, whereas embryonic stem cells come from a different individual and can trigger an immune response B1 both are pluripotent in principle, but induced pluripotent cells can carry mutations acquired during the donor adult cell's life and have a tendency to form tumours; embryonic stem cells also form tumours if undifferentiated cells are transplanted, so both need the same safety controls

Question 12

A tumour has lost function of both copies of a tumour suppressor gene: one copy silenced by promoter hypermethylation and the other carrying a point mutation in its coding sequence. Evaluate whether treating this tumour with a drug that inhibits DNA methylation alone would restore normal control of cell division.

Accepted Answer

B1 for: removing methyl groups from the hypermethylated copy could restore transcription of that allele and let it make functional protein again, exactly as it does in a tumour where methylation silences both copies B1 against: the second copy carries a point mutation in its coding sequence, so even once it is transcribed and translated normally the protein produced could still have the wrong primary structure and remain non-functional, since demethylation cannot correct a changed base B1 against: with only one of the two alleles genuinely restored, the cell may still make too little functional protein to enforce its checkpoint fully if, as for a few suppressor genes, one copy is not enough B1 judgement: the drug would very likely restore control from the demethylated allele alone, since one working copy is usually enough, but the control would be less secure than in a tumour silenced by methylation in both copies, because one further mutation in that allele would remove it again

Question 13

A healthy woman has had several miscarriages. Her karyotype shows a reciprocal translocation between chromosome 4 and chromosome 11. Explain why she is healthy, and why her pregnancies often end in miscarriage.

Accepted Answer

B1 in a reciprocal translocation, sections of two non-homologous chromosomes swap places, so no genetic material is lost or gained B1 she still has two copies of every gene, so she is a balanced carrier and her phenotype is normal, provided no break point falls inside a gene B1 at meiosis each gamete receives one version of chromosome 4 and one of chromosome 11, so some gametes receive one normal and one translocated chromosome B1 those gametes are unbalanced, carrying one section twice and lacking another, so an embryo formed from one has extra copies of some genes and only one copy of others, and usually does not survive

Question 14

A drug that inhibits the enzymes adding methyl groups to DNA is being trialled in patients whose tumours carry an intact but heavily methylated tumour suppressor gene. Suggest why the drug might restore control of division in those cells, and suggest one reason it may also harm healthy cells.

Accepted Answer

B1 with fewer methyl groups on the promoter, transcription factors and RNA polymerase can bind to it again, so the gene is transcribed and mRNA is made B1 the tumour suppressor protein is then synthesised and the checkpoint it enforces is applied again, so cells with damaged DNA are held for repair or sent into apoptosis instead of dividing; no base had changed, so nothing needed repairing B1 the drug reduces methylation across the whole genome and in every cell it reaches, so genes that were properly silenced in healthy cells are switched on as well

Question 15

Outline the difference between an established use of stem cells and a use that remains at the clinical-trial stage, giving one named example of each.

Accepted Answer

B1 bone marrow transplantation, replacing a patient's blood-forming stem cells destroyed by leukaemia or its treatment, is an established use that has been standard care for decades B1 sheets of limbal stem cells restoring the corneal surface after a chemical burn are also licensed and established B1 treatments for conditions such as macular degeneration, Parkinson's disease, type 1 diabetes or spinal cord injury remain in clinical trials, with some encouraging early results but none yet in routine use

Question 16

Tests show that a man with cystic fibrosis produces normal sperm in his testes but has no sperm in his semen. Suggest an explanation, and suggest how he could still become the biological father of a child.

Accepted Answer

B1 the vas deferens, which carries sperm from the testis towards the urethra, is blocked by thick mucus or is absent, having failed to develop B1 so sperm are made but cannot leave the testis and join the semen B1 sperm could be collected directly from the testis by a minor operation and used to fertilise an egg by IVF, for example by injecting a sperm into the egg

Question 17

Most people with cystic fibrosis take capsules containing pancreatic enzymes with every meal. Explain why.

Accepted Answer

B1 thick, sticky mucus blocks the pancreatic duct B1 so lipase, amylase and proteases from the pancreas cannot reach the small intestine, and fats, starch and proteins are not fully digested B1 the capsules supply the enzymes with the food, so it is digested and the products can be absorbed, allowing the person to gain weight normally

Question 18

In chronic myeloid leukaemia, a translocation between chromosomes 9 and 22 joins part of the BCR gene to part of the ABL gene, which codes for an enzyme involved in signalling cells to divide. No base in either gene is substituted. Suggest how the translocation causes uncontrolled cell division, and suggest why the leukaemia is not passed on to the patient's children.

Accepted Answer

B1 the fused gene codes for a hybrid enzyme that is permanently active, or is transcribed under the control of the BCR promoter rather than ABL's own B1 so the cell receives a constant signal to divide; the ABL proto-oncogene has become an oncogene by being moved rather than by a change to its base sequence B1 the translocation arose in a blood stem cell during the patient's life, a somatic cell, so it is not present in the cells that give rise to gametes

Question 19

Name the kind of substitution that produces a stop codon, and name the property of the genetic code that allows a different substitution to change no amino acid at all.

Accepted Answer

B1 a nonsense substitution, which ends translation early B1 the code is degenerate, so more than one triplet codes for the same amino acid

Question 20

State the defining difference between a benign tumour and a malignant tumour.

Accepted Answer

B1 a malignant tumour's cells invade the surrounding tissue B1 and may spread through the blood or the lymph to form secondary tumours elsewhere in the body, which a benign tumour does not do

Question 21

Compare a gene mutation with a chromosome mutation.

Accepted Answer

B1 a gene mutation changes the base sequence within one gene, while a chromosome mutation changes the structure or number of whole chromosomes B1 a gene mutation affects one triplet or a few, while a chromosome mutation can lose, duplicate or move many genes at once, possibly without changing any base sequence

Allele	Base sequence of part of the coding strand
original	GCATACGGTATCCAT
mutant 1	GCATACGATATCCAT
mutant 2	GCATACGGTAATCCAT
mutant 3	GCATGGCATATCCAT
mutant 4	GCATACTACGGTATCCAT

Extent of the tumour when it was found	Number of patients	Number alive after five years
contained within a capsule in the tissue where it arose	400	372
invading the surrounding tissue, no secondary tumours found	400	244
secondary tumours present in other organs	400	48