2016 VJC H2 Biology Prelims Paper 3 Answers

1    VICTORIA JUNIOR COLLEGE BIOLOGY DEPARTMENT JC2 PRELIMINARY EXAMINATIONS 2017 HIGHER 2 9744/3 Answers Note: A: Accept; R: Reject 1 Huntington's disease (HD) is a rare neurodegenerative disease. Fig. 1.1 shows a pedigree of HD across three generations (I to III). Fig. 1.1 (a) With reference to Fig. 1.1, account for the mode of inheritance of the disease. [3] ● HD is inherited in an dominant manner; ● every generation has affected offspring as long as one parent is affected ( I1);; ● a single defective allele is sufficient for trait; ● Inherited in an autosomal manner; ● male and female offspring are similarly affected; ● an affected male parent ( II2) can produce an affected son (III1);; (b) (i) Explain the likely effect of the abnormal increase in CAG repeats on HTT protein structure and function. [3] ● Production of an abnormally long polypeptide / longer than the normal polypeptide (R! premature termination since length of CAG repeats is associated with disease); / ● Alters primary structure of polypeptide; ● disrupts the R group interactions such as hydrogen bonding, ionic, hydrophobic interactions and disulfide bridges;; ● essential for correct/ extensive folding into tertiary structure with specific 3D shape;;/ idea of 3D shape/conformation or tertiary structure is affected;; ● normal function of protein is lost/ abnormal protein is made; Max 3m (ii) Suggest possible reasons why individuals having number of repeats ranging from 21- 39 do not develop the disease. [2] ● Insertion mutation of multiples of 3 that code for chain of 20 glutamines (A! less than 40 glutamines) do not drastically affect 3D shape/ structure and thus function of the protein;; ● Slight effect on protein function but not drastic enough to develop disease;; ● A chain of more than 40 glutamines affect interactions between R groups that lead to folding into specific tertiary structure of the protein to affect normal function and cause HD;;

2    (c) (i) Explain why PCR can be used for the diagnosis of HD. [2] ● PCR makes use of specific primers that flank the region of the HTT gene/ exon 1 that contains the CAG repeats;; ● to amplify the different fragment lengths to allow for differentiating between normal and mutant allele;; (ii) Explain how gel electrophoresis was used to detect the band patterns of the offspring in Fig.1.1. [4] ● During electrophoresis, negatively charged DNA fragments migrate through a gel towards the positive electrode;; ● under an electric field; ● agarose gel acts as a molecular sieve; ● Larger fragments (i.e. has more CAG triplets) move slower compared to shorter fragments;; ● Gel is stained with methylene blue and observed under white light;; / ethidium bromide and observed under uv light;; (d) Based on this information, draw in the band patterns (in Fig. 1.2) for individuals #6, #10 and #11. [2] Fig. 1.2 (e) Individuals with 6-35 CAG repeats will be unaffected. Offspring of individuals with 36-39 repeats are at increased risk for HD. Suggest how this increased risk can occur. [2] ● As the altered HTT gene is passed from one generation to the next, the size of the CAG trinucleotide repeat may increase in size due to errors in DNA replication of CAG repeat region during formation of gametes;; ● Since trait is dominant, they are at risk of having children who will develop HD when affected gamete with >40 repeats fuses with a healthy gamete;; 2 (a) Explain how the loss of control in the cell cycle can lead to cancer. [3] ● Loss of control means that the checkpoints regulating the stop and go signal of the cell cycle is lost; ● Failure to halt cell cycle /Cells continue to divide even if they have not properly completed the previous stage;

3    ● Even when DNA is mutated (R: cells are damaged); ● Leads to an accumulation of mutations; ● Which includes loss of function mutation in several TS genes;; ● And gain of function mutation in at least one proto-oncogene;; ● causing cell to undergo uncontrolled cell division; ● cells cannot repair DNA damage; they evade apoptosis; grow in the absence of growth factor; loss of contact inhibition etc etc ● Ref to cancer development being a multistep process; [max 3m] (b) Outline how such a mechanism is activated to be effective in its function. [4] 1) (cancer-derived) peptides presented via MHC1 on surface of cancerous cells; 2) Naïve cytotoxic T cells with receptors specific to peptides recognize and bind; (idea of specific binding by cytotoxic T cells) 3) cancerous cells recognized by macrophages / dendritic cells (@ other examples of immune cells); 4) and engulfed via phagocytosis; 5) Presentation of peptides via MHCII to naïve T helper cells; 6) which activates of T helper cells (ref clonal selection); 7) T helper cells release cytokines; 8) cause activation of cytotoxic T cells (ref clonal selection); 9) Which target cancerous cells and perform direct killing; 10) Via release of granzyme and perforin; (c) (i) Use the data in Fig. 2 to compare the effectiveness of the two drugs used to treat the tumours. [4] ● (similarity) Both drugs are effective for treatment of tumor A and B; ● total volume decreased compared to control; ● Vinblastine and T138067 are both equally effective against tumor A / T138067 is slightly more effective than Vinblastine against tumor A; ● QV;; eg. After 25 days, size of tumor A decreased from 600mm 3 to 220mm3 when vinblastine was added; while size decreased from 600mm3 to 160mm3 when T138067 was added; ● T138067 is more effective for treatment against tumor B than Vin; ● QV;; eg. After 25 days, size of tumor B decreased from 620mm 3 to 360mm3 when vinblastine was added; while size decreased from 600mm3 to 120mm3 when T138067 was added; ● Idea that Vinblastine is more effective against tumor A than tumor B; ● Idea that T138067 is more effective against tumor B than tumor A; (ii) Both Vinblastine and T138067 were able to bind to tubulin. Explain the effects of Vinblastine and T138067 as anti-cancer drugs. [3] 1) idea of preventing the polymerisation of tubulin / idea of tubulin being important component of spindle fibre; 2) thus prevent formation of spindle fibre Alternative: prevent shortening of spindle fibres; 3) spindle fibres cannot attach properly to chromosome during metaphase; 4) sister chromatids cannot be separated equally during anaphase; (reject if students describe as separation of chromosomes) 5) idea of preventing mitosis (nuclear division) from occurring; 6) idea of reduced number of cancer progeny cells formed / new cancer cells cannot be formed / new cancer cells are not viable; (iii) Suggest why the same tumor cells may respond differently to these two drugs? [3] ● difference in uptake due to presence of different receptors to take in the drug;; ● difference in efflux of drug due to presence of different transporter proteins that can export the drug out of cell;;

4    ● difference in stability of drug as it may be degraded to different extent (idea of susceptibility of drug to degradative enzymes);; ● different affinity of the drug to binding with tubulin leading to different extent of responses;; ● AVP;; 3 (a) With reference to the curve for Barley, explain the meaning of limiting factor. [3] Definition of limiting factor: (Any one below)  As a factor that is closest to its minimum value and changing the concentration (or idea of) of this factor will change the rate of reaction/ rate of CO2 uptake;; (A: if students make reference to either an increase ot decrease) OR  as a factor that will directly affect the rate of reaction/ rate of CO2 uptake if its value is changed;;  From CO2 concentration of 0 to 350ppm;  increase [CO2], increases