VJC 2023 H2 9744 P3 Ans

1 Victoria Junior College Biology Department 2023 H2 Preliminary Exams Paper 3 Proposed answers Marking abbreviations: A: Accept, R: Reject, BOD: benefit of doubt, AW: alternative wording, AVP: Any valid point, NAQ: not answering question, ECF: error carried forward Question 1 Cells can respond to their environment through regulating the expression of certain genes. Unicellular eukaryotes such as yeasts respond to changes in the glucose concentration in the environment. Glucose is the preferred carbon source for Saccharomyces cerevisiae (Baker’s yeasts) which metabolizes glucose by a purely glycolytic process (fermentation), producing ethanol even under aerobic conditions. Presence of glucose represses the uptake and metabolism of other carbon sources eg. galactose, maltose. When the glucose has been consumed, the cell switches to aerobic metabolism of the ethanol. (a) (i) State the difference in the number of ATP produced from one molecule of glucose between fermentation and aerobic respiration. [1] • 30 ATP; (ii) Suggest why yeast cells undergo fermentation even in the presence of oxygen. [2] • (idea of) Take advantage of the high glucose concentration quickly taking up glucose from the environment due to competition from other yeast cells; • (idea of ethanol as a reserve) when glucose is depleted, yeast can then fully oxidise ethanol to obtain ATP; (R: directly lift from text without further explanation) • So that repression on other carbon sources eg. galactose (monosaccharide) and maltose (disaccharide) is removed; • ATP produced via fermentation is faster as it involves fewer steps compared to complete oxidation; • AVP (b) Several components of the signalling pathway involved in glucose repression have been identified. These are shown in Fig.1.2. Snf1 is a protein kinase and binding of Mig1 to the promoter interferes with the binding of other proteins.

2 Fig 1.2 (i) With reference to the information given in Fig 1.2, explain how the presence of glucose in the environment is able to bring about the repression of GAL1 gene. [4] 1. Glucose taken up by facilitated diffusion via transport/channel proteins into cytoplasm; 2. Presence of glucose/ High glucose inactivates the Snf1 protein kinase; resulting in Mig1 not being phosphorylated (A: cytoplasmic Mig1 dephosphorylate) and; 3. Mig1 enters the nucleus, and binds to the promoter of GAL1; 4. Preventing the attachment of other transcription factors and RNA polymerase, prevents the assembly of transcription initiation complex and no transcription of GAL1gene; Mig1 binds to promoter sites with these features: ● 17 base pairs long ● Includes a region of five repeating adenine-thymine pairs ● Includes a region of six repeating cytosine-guanine pairs Promoter sites to which Mig1 binds are known as Mig1-binding promoter sites. (ii) Explain how Mig1 recognizes and binds to these sites. [2] • Mig1 has a DNA binding domain; • that has a specific 3D conformation which is complementary in shape (and charge) to the base sequence of the binding sites; R: complementary base pairing Scientists analysed the yeast genome to look for DNA that matches the features shown by of Mig1-binding promoter site. Analysis of four chromosomes revealed the presence of 26 Mig1- binding promoter sites. yeast chromosome number of Mig1-binding promoter sites A 1 B 9 C 2 D 14

3 Table 1.1 Since five different enzymes coded by five different genes are required for galactose metabolism, the expected number for of Mig1-binding promoter sites for an individual diploid yeast is 10. (iii) Explain why the expected number of Mig1-binding sites is 10. [2] 1. Since yeast is a eukaryote, every gene would have its own promoter => 5 genes – 5 promoters/ 5 Mig1-promoter binding sites; R: 5 genes 5 Mig1-promoter binding sites without explanation (in bold) 2. A diploid cell will have chromosomes occurring in pairs – homologous chromosomes/ 2 copies of each gene on homologous chromosomes and hence the number is 2x5 = 10; (iv) Suggest reasons for the difference in the number of Mig1-binding promoter sites seen in Table 1.1 and the expected number. [2] Note: Difference: Total number of Mig1-binding promoter sites is 26 for the 4 chromosomes, much larger than the expected 10/ idea of more than the number of genes involved in galactose metabolism and only 4 chromosomes are shown; Possible reasons: [Any 2] • Mig1 regulates genes involved in other pathways; • There could be several copies of the five genes; • Some galactose genes are found on the same chromosome – because only 4 chromosomes analysed; (c) Changes in gene regulation was also observed in Aedes mosquitoes infected with dengue virus. Scientists carried out experiments to study how the mosquitoes respond to the presence of dengue virus (DENV) after a blood meal. Fig 1.2 Experimental setup Source: https://pubmed.ncbi.nlm.nih.gov/35814655/ (i) Explain the importance of the control group in this experiment.[2] 1. Show that any changes in gene expression/ fecundity between the control and experimental group; 2. Is due to the dengue virus present in the experimental group; 3. control sets the baseline for the level of gene expression in the ovary for the uninfected group, for comparison with the experimental;

4 Fig 1.3 shows the fecundity (number of eggs laid per mosquito) for the two groups of female mosquitoes. Fig 1.3 (**) – indicates that the p < 0.05 (ii) Suggest two reasons for the variation seen in the fecundity of the individual mosquitoes within each group. [2] Any two of the following: For both groups 1. Differences in the genetic makeup of the mosquitoes resulting in them having different combination of alleles that determine the number eggs laid; A: Additive gene effects 2. Since female mosquitoes need to take a blood meal before they can lay eggs, differences in the amount of blood taken up by each mosquito can result in different amounts of nutrients available for egg laying; 3. The mosquitoes are of different ages 7-10 days. There might be a differences in the fertility of the mosquitoes based on age; (Note: the positive relationship is seen more for body size than age) For experimental group: 4. different amount of blood taken up will mean different amount of the viruses (taken up), and this would affect the extent of infection in the mosquitoes; different ability of the immune systems of the mosquitoes to overcome the infection as more nutrients and energy will be directed to fighting the infection/production of antibodies and less for egg laying; 5. AVP; A study of the ovaries obtained from the infected mosquitoes showed the cells infected with the dengue virus exhibit reduced ● cytochrome c oxidase activity ● synthesis of ribosomal proteins ● RNA binding to proteins (iii) Based on the information provided, explain the difference in the fecundity between the two groups of mosquitoes in Fig. 1.3. [4] 1. Difference: Overall reduction in fecundity of the mosquitoes infected with Denv2- average of 75 eggs/mosquito compared to 100 eggs/mosquito in control; 2. Reduced cytochrome c oxidase activity leads to decrease ATP production for the synthesis of protein/ reduce energy required for egg production; 3. Reduced synthesis of ribosomal proteins leads to a decrease in ribosomes synthesis, resulting in less protein involved in egg production will be translated;

5 4. Reduced RNA binding to proteins leads to less rRNA associating with ribosomal proteins, less ribosomes will be formed and less protein synthesis; (d) Two genes were found to be upregulated in the infected mosquitoes. To investigate the role of these two genes ( Oatp, amd), scientists carried out further studies using mosquito cell culture. Three cell cultures we