RI 2023 Nov P3 answers

2023 Nov P3 answers Question 1 Mitosis is a key process in the development of a fertlised egg into an adult human. Each mitotic cell division is preceded by DNA replication. (a) (i) Explain the significance of mitosis in the development of a fertilised egg into an adult human. [3] 1. mitosis results in increase in number of cells which is essential for biological processes like tissue growth in a multicellular organism; 2. and replacement of cells for tissue repair; 3. mitosis produces 2 genetically identical diploid daughter nuclei, each of which has the same number and same type of chromosomes (2 sets/homologous chromosomes) thereby contributing to genetic stability in all cells of the adult human; (ii) Name one environmental factor and one lifestyle factor that are associated with increasing the risk of mutations that lead to cancer. [2] environmental factor: exposure to ultraviolet light / ionising radiation / radioactivity lifestyle factor: smoking and exposure to carcinogens such as tar in cigarette smoke (iii) Explain how random errors that occur during the process of DNA replication can result in different types of changes to DNA sequences. [4] 1. Mistakes can occur during the synthesis of new daughter strand catalysed by DNA polymerase*, during DNA replication; OR Mistakes made during DNA replication could have escaped proofreading by DNA polymerase during DNA replication, A: DNA repair enzymes also not correct the mistake. 2. Substitution* mutation where a nucleotide/base is replaced by a different nucleotide; 3. Inversion* – a segment of nucleotide sequences separates from the allele and rejoins at the original position but it is inverted; 4. The length of DNA is unchanged but the sequence will be different; 5. Insertion /addition* – occurs when one or several nucleotides are inserted in a sequence 6. Deletion* – occurs when one or several nucleotides are removed from a sequence of base(s) 7. Resulting in a change in both length and sequence of the DNA; Pt 4 with 2 or 3; Pt 7 with 5 or 6 Must have point 4 or 7 to get full marks.

(iv) Suggest why it is important for doctors to be able to distinguish passenger mutations from driver mutations. [2] 1. passenger mutations do not result any change in the function of the cell and hence are not linked to the development of cancer so doctors will not need to prescribe drugs; 2. driver mutations result in the benign tumour formation and subsequently invasive cancer resulting in uncontrolled cell division and cancer development; 3. Doctors need to distinguish the two in order to make an accurate diagnosis/ suggest appropriate treatment if driver mutations are discovered in patients; (v) State the name given to the type of mutation in cancer development that results in a dominant driver mutation. [1] Gain-in-function mutation* (vi) State the name given to the type of gene that, when mutated, contributes to cancer development as a dominant driver mutation. [1] proto-oncogene* (vii) Use Fig. 1.1 to calculate the annual mutation rate (number of mutations per year) in this cell lineage during two periods of time: Annual mutation rate from formation of the zygote at conception to the start of adulthood = 5 mutations / 21.75 years = 0.23 21.75 years to include the gestation period which is the period from formation of the zygote at conception to the start of adulthood Annual mutation rate from formation of the zygote at conception to the start of adulthood = ……0.23…. year-1 Annual mutation rate from first development of the benign tumour to chemotherapy-resistant recurrence = (18-8)/(70-67) = 3.3 Annual mutation rate from first development of the benign tumour to chemotherapy-resistant recurrence = ……3.3…. year-1 (viii) Within a struggle for existence with surrounding cells, discuss whether the cell lineage that has the mutator phenotype and the mutation shown as a black triangle has a selective advantage compared to cell lineages without these features. [5] 1. The cell lineage with both mutator phenotype and black triangle mutation will have a selective advantage; Mutator phenotype has an increased rate of mutation

2. The mutator phenotype has an increased rate of mutation which resulted in all three types of mutations, passenger, driver, and chemotherapy resistance mutations; 3. Mutator phenotype will also result in a higher rate of mutation giving rise to greater variation; 4. And higher chance of forming favourable alleles/phenotypes for the cell e.g. faster growth, ability to gain resources; Black triangle represents chemotherapy resistance mutation 5. When exposed to chemotherapy, the cell is able to survive and continue to divide while other cells without this mutation will die; (b) With reference to Table 1.1, explain how suitable each patient is as a candidate for treatment with 5-fluorouracil. [3] A: suitable. 5-fluorouracil kills cancer cells who are actively dividing. The absence of mutation in DPYD in normal cells allows for breakdown of 5-fluorouracil in normal cells so it will not accumulate to toxic levels to kill normal cells. B: not suitable. The mutation in DPYD mutation is present in both normal cells and tumour cells. While this allows for accumulation of 5-fluorouracil to kill cells, both normal cells and cancer cells will be killed. C: suitable. The mutation in DPYD in cancer cells allow for accumulation of 5- fluorouracil, allowing for more effective killing of cancer cells. At the same time, DPYD is not mutated in normal cells so 5-fluorouracil can be broken down and normal cells are not killed. (c) (i) Identify two types of biomolecule in food that a person may eat from 12 to 6 hours before the scan. [1] proteins and fats (i) Suggest why people should not eat carbohydrates for 12 hours before the scan. [2] 1. Carbohydrates would be broken down into glucose in the digestive system and taken up into the bloodstream; 2. The presence of glucose in the bloodstream could lead to cancer cells taking up glucose instead of FDG, preventing the building of FDG in cancer cells; (ii) Table 1.2 compares ATP production in mammalian muscle cells at different stages of the respiration of glucose under aerobic conditions and under anaerobic conditions. Table 1.2 stage of respiration ATP production aerobic conditions anaerobic conditions glucose → pyruvate ✔ ✔ pyruvate → lactate ✘ ✘ pyruvate → acetate (acetyl) ✘ ✘ acetate → CO2 + H2O ✔ ✘

(iii) Use Table 1.3 to explain why cancer cells use significantly more glucose than normal cells. [4] 1. Mutation in AKT increases number of glucose transporters, resulting in an increased uptake of glucose into cancer cells; 2. Mutation in AKT also stimulates the activity of enzyme that phosphorylates glucose, committing glucose to the glycolytic pathway; 3. Mutations in MYC and ras activates directly and indirectly the expression of genes coding for glycolysis enzymes, increasing the concentration of enzymes and the rate of glycolysis; 4. Mutation in ras inhibits the action of pyruvate dehydrogenase, preventing the conversion of pyruvate to acetyl-coA; 5. p53 prevents the expression of gene coding for final electron transport protein, preventing the transport of electrons down the electron transport chain and the generation of a proton gradient / proton motive force; 6. Link reaction, Krebs cycle and oxidative phosphorylation which produce the bulk of ATP in aerobic respiration is unable to happen, hence glycolysis occurs at a faster rate to meet the ATP demands of the cancer cell; [Total: 32] Question 2 (a) (i) Compare the biochemical composition of the outer structure of a bacteriophage and the outer structure of the host cell. [2] 1. The outer structure of a bacteriophage is the capsid which is made up of a protein coat that encloses the viral