1. 1
   Researchers investigating a mouse gene of unknown function, gene K, obtained mice from an international library of knockout mice in which gene K had been made inoperative. Knockout mice homozygous for the inoperative gene developed normally and were fertile. When fed a standard diet, their mean body mass at 20 weeks was 48 g, compared with 30 g for wild-type littermates kept in the same cages. The knockout mice ate about 25% more food per day than the wild-type mice, but their activity levels were the same.

   (a)

   What is meant by a gene knockout?

   [1 mark]

   • ARemoving the whole chromosome that carries the gene from the nucleus
   • BIncreasing the expression of a gene so that more of its protein is made
   • CChanging a gene so that it is inoperative, in order to investigate the gene's function
   • DTransferring the gene into bacteria so that its protein can be produced

   (b)

   Why were wild-type littermates kept in the same cages used for comparison?

   [1 mark]

   • AThey also lacked a functional gene K
   • BThey have a similar genetic background and environment, so differences can be attributed to the loss of gene K
   • CKnockout mice cannot survive unless they are kept with wild-type mice
   • DWild-type mice eat less food, so they make the difference easier to see

   (c)

   Deduce a possible function of gene K from the results and state one limitation of this conclusion.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   Cystic fibrosis is caused by recessive alleles of the CFTR gene, which codes for a chloride channel protein of 1480 amino acids in the plasma membrane of epithelial cells. The most common allele worldwide, ΔF508, lacks three adjacent bases, so the protein lacks a single phenylalanine. The protein folds incorrectly and most of it is broken down before it reaches the plasma membrane. Another, rarer allele has a single base deleted within the fifth codon of the gene.

   (a)

   Why does the ΔF508 deletion not cause a frameshift?

   [1 mark]

   • AThree bases are one codon, so the reading frame of the codons after the deletion is unchanged
   • BThe genetic code is degenerate, so deleted bases have no effect
   • CDeletions never change the reading frame of a gene
   • DThe deleted bases are replaced during DNA repair

   (b)

   What is the most likely effect of the single-base deletion in the fifth codon?

   [1 mark]

   • AThe protein lacks one amino acid but is otherwise normal
   • BThere is no effect, because the deletion is near the start of the gene
   • CThe protein has one amino acid substituted for another
   • DEvery codon after the deletion is altered, probably producing an early stop codon and a non-functional protein

   (c)

   Explain why the loss of just one amino acid out of 1480 can prevent the CFTR protein from functioning.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   Histone H4 is a protein of 102 amino acids that forms part of the core of each nucleosome, around which DNA is wound. The amino acid sequences of histone H4 from pea plants and from cattle differ at only 2 of the 102 positions, although plants and animals last shared a common ancestor more than 1.5 billion years ago. The DNA sequences of the histone H4 genes of the two species differ at many more positions than the amino acid sequences do, and most of these differences are at the third base of codons. By contrast, fibrinopeptides — short fragments cut off a blood-clotting protein and then discarded — differ at many positions even between different species of mammal.

   (a)

   Distinguish between a conserved and a highly conserved sequence, and deduce which term best describes histone H4.

   [3 marks]

   (b)

   Evaluate the hypothesis that histone H4 is highly conserved because its gene has a slower rate of mutation, rather than because of the functional requirements of the protein.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   In 2018 a scientist announced the birth of twin girls whose embryos had been edited using CRISPR–Cas9. The aim was to make the CCR5 gene inoperative, because the CCR5 protein on white blood cells is used by HIV to enter the cells, and people who naturally lack a functional CCR5 are highly resistant to HIV. The twins' father was HIV-positive, but effective methods already existed to prevent HIV passing to the children. Later analysis suggested that the edits did not match the natural variant and that not every cell in the embryos may have been edited. The work had not been approved by regulators, was widely condemned by scientists internationally, and the scientist was later imprisoned.

   (a)

   Explain how CRISPR–Cas9 could be used to make the CCR5 gene inoperative in an embryo, and compare the consequences of editing an embryo with editing the blood stem cells of an adult.

   [6 marks]

   (b)

   Discuss the ethical issues raised by this experiment and explain why there is an international effort to harmonise the regulation of genome editing.

   [6 marks]

   Total for question 4: 12 marks

End of questions