19.2. Genetic technology applied to medicine
A subsection of Biology, 9700, through 19. Genetic technology
Listing 10 of 37 questions
9700_w14_qp_41
THEORY
2014
Paper 4, Variant 1
9700_w14_qp_42
THEORY
2014
Paper 4, Variant 2
Haemophilia A and haemophilia B are common hereditary disorders of blood clotting. Haemophilia A is a sex-linked genetic disorder that affects approximately 1 in 20 000 males worldwide. It is caused by a recessive allele of a gene coding for a clotting factor and results in excessive bleeding. There is currently no cure, but symptoms of haemophilia can be treated with a transfusion of a clotting factor to slow down the bleeding. State how genetic screening could reduce the number of cases of haemophilia. Some genetic disorders can be treated with gene therapy. Outline the aims of gene therapy. Suggest why haemophilia A is a suitable disorder for treatment with gene therapy. Haemophilia A and haemophilia B are caused by mutations in different blood clotting genes, F8 and F9 respectively. Both disorders have been treated with gene therapy involving the use of a vector. Table 4.1 shows the lengths, in kilobases , of the F8 and F9 genes. Table 4.1 haemophilia gene gene length / kb A F8 >8 B F9 1.4 With reference to Table 4.1, suggest why gene therapy using the F9 gene has been more successful than using the F8 gene. Two frequently used vectors in gene therapy are compared in Table 4.2. Table 4.2 feature vector adenovirus retrovirus genetic material of virus double-stranded DNA single-stranded RNA expression of inserted gene high gene expression gene expression in dividing cells only host immune response to virus high low With reference to Table 4.2, explain the advantages and disadvantages of using adenovirus rather than retrovirus as a vector.
9700_w15_qp_43
THEORY
2015
Paper 4, Variant 3
Huntington’s disease is caused by a dominant allele of the gene that codes for the production of the huntingtin protein. This protein affects the development of many different tissues, including brain tissue. • The Huntington allele contains several repeats of the base sequence CAG, which codes for glutamine. • This results in a polyglutamine section in the synthesised protein. • A gene with more than 39 CAG repeats produces a protein that does not fold properly and does not function. • Symptoms of Huntington’s disease usually first appear between the ages of 30 and 45 years. • There is no treatment for the disease, which is progressive and always fatal. • Some people with between 27 and 35 CAG repeats do not develop the disease, but may still pass on the Huntington allele to their children, who may develop the disease as the number of repeats tends to increase when gametes are produced. The amino acid sequences on either side of the polyglutamine section of the huntingtin protein are not changed by the presence of the CAG repeats in the Huntington allele. Explain why this is so. With reference to the information given, explain why Huntington’s disease cannot be treated with gene therapy. Young people who have a parent with Huntington’s disease can choose to be screened for the presence of the Huntington allele. State the probability that a young person who has one parent with Huntington’s disease will inherit the Huntington allele. Suggest one advantage and one disadvantage of screening for Huntington’s disease before any symptoms occur. advantage disadvantage A couple, in which one partner has the Huntington allele, may choose to use IVF (in vitro fertilisation) to have a child. Any embryos obtained from the IVF procedure can be screened in the following way: • carry out an embryo biopsy • use PCR • test for the presence of the Huntington allele • only implant embryos that do not contain the Huntington allele. State what is meant by the term embryo biopsy. Explain why PCR is used in this procedure. Outline two social or ethical implications of screening embryos in this way.
9700_w17_qp_42
THEORY
2017
Paper 4, Variant 2
Leber’s congenital amaurosis (LCA) is an autosomal recessive eye disease. LCA results in eye disorders, including severe loss of vision, at birth. LCA has been successfully treated by gene therapy, using a virus instead of a plasmid as the vector. Adeno‑associated virus (AAvectors containing the therapeutic allele were injected directly into the retina, the layer at the back of the eye containing the photoreceptor cells. People who had been blind from a young age were able to see again. There is a risk associated with the injection method used to deliver the vectors, as it might cause the retina to detach, damaging vision. This method of delivery was first used for LCA before being trialled on other retinal diseases that gradually reduce the vision of people as they get older. Suggest the main steps involved in creating recombinant DNA for this example of gene therapy. Explain why the fact that LCA is an autosomal recessive genetic disease makes it suitable for treatment with gene therapy. Suggest why the retinal injection method of gene therapy was used for LCA before it was trialled on other retinal diseases that gradually reduce the vision of people as they get older. Scientists tried to create an improved virus vector for gene therapy. step 1 – The scientists used a special form of the polymerase chain reaction (PCR). This form of PCR causes mutations in the DNA sequence of AAV by base substitution. step 2 – The viruses containing different base substitutions were tested. This was done by using the different viruses to deliver a new gene, the gene for green fluorescent protein (GFP), into the photoreceptor cells of mice, using the retinal injection method. step 3 – The best virus, known as 7m8, caused the photoreceptor cells in the retina of the mouse to fluoresce brightly, even when the recombinant virus was injected into the fluid inside the eye instead of into the retina itself. step 4 – The 7m8 virus was used to cure a mouse with LCA by injecting this virus containing the therapeutic allele into the fluid inside the eye of the mouse. Suggest how errors occurring during PCR can cause base substitution mutations in the DNA sequence of AAV. Explain why the photoreceptor cells of the mouse fluoresced in step 3. Predict the impact of the 7m8 AAV on treatment for age‑related retinal diseases.
9700_w18_qp_41
THEORY
2018
Paper 4, Variant 1
Questions Discovered
37