DHS 2022 Y5 H2 Promo Answer Scheme (Paper 1 and 2)

1 DUNMAN HIGH SCHOOL Promotional Examination Year 5 H2 Biology Promo Exam 2022 Mark Scheme Paper 1: Multiple Choice Question 1 B 6 C 11 C 2 C 7 D 12 A 3 A 8 D 13 D 4 A 9 D 14 B 5 B 10 B 15 A Q Explanation 1 Region P refers to the nucleoid region and region Q the nucleolus. Option A is incorrect because DNA in the bacterial nucleoid region (region P) is coiled around histone-like proteins, not histone proteins. Option C is incorrect because ATP synthesis does not occur in the bacterial nucleoid region (region P). The nucleolus (region Q) is also not the site of protein synthesis. Option D is incorrect because DNA is present in nucleolus (region Q). 2 The larger the organelle, the lower the centrifugal speed required for the organelle to sediment. Diagram shows the size of organelle. Sediment 1 will be the largest organelle – nucleus Sediment 2 – mitochondria Sediment 3 – endoplasmic reticulum Sediment 4 – ribosome 3 Statement 1 is correct: The reducing group is not bound in the first glucose. Statement 2 is correct: The first glucose is beta glucose. Statement 3 and 4 are false: The bond is a 1,2 glycosidic bond and this is not the bond in cellulose formation (1, 4 glycosidic bond). 4 The q uestion is indirectly asking for the change in membrane structure that increases the membrane fluidity. More cholesterol and more C=C double bonds in the hydrocarbon chain (aka higher unsaturation level) will increase membrane fluidity. 5 Option A addit ion of lysine will increase inhibition of the enzyme that converts aspartate to aspartylphosphate. This will lead to decrease in all downstream process/product, including methionine. Option B addition of aspartate semialdehyde will lead to increase in al l downstream process/product, including methionine.

2 Option C addition of threonine will increase isoleucine production. But will also lead to inhibition of enzyme that converts aspartate to aspartylphosphate. This will lead to decrease in all downstream process/product, including methionine. Option D addition of isoleucine inhibits the enzyme that converts threonine to α -oxobutyrate, which does not interfere with methionine synthesis pathway. 6 Change in quantity over time can be simplified as the rate of reaction. Especially change in product or substrate quantity over time. As quantity of product should increase over time, the graph should be 1, where the rate of reaction increases with incre asing temperature, until optimum temperature of the enzyme. Beyond optimum temperature, the rate decreases. Graph 3 is the reflection of graph 1. It represents the quantity of substrate because substrate decreases over time, hence the change is negative. But the change in rate of reaction with changing temperature is the same as product (graph 1). Graph 2 shows how enzyme quantity decreases with increasing temperature. 7 The strand on the left is running 5’ to 3’ from top to bottom. The bases with 2 rings are either A or G, and those with one ring are either T or C. A forms 2 hydrogen bonds (dotted lines) with T. G forms 3 hydrogen bonds with C. 8 The question stated that when all the molecules were mixed, the student successfully obtained many DNA molecules. This suggests that the set up is working. Set up Result 1 A few DNA strands, with some short double stranded regions of DNA-RNA hybrid. 2. A few DNA molecules. 3. Many DNA strands with long stretches of double stranded DNA regions. When something is removed in set up 1, DNA are in strands, instead of molecules. This suggest that the double stranded DNA is separated (by helicase). There are short double stranded regions of DNA-RNA hybrids suggest that primase had synthesized RNA primers. But there is no synthesis of daughter DNA strand, implying that DNA polymerase is missing. In set up 2, a few DNA molecules results. The DNA remains as double stranded DNA molecules, suggesting that helicase is absent to unwind the existing parental DNA molecules. In set up 3, long stretches of double stranded DNA implies that DNA polymerases has elongated the daughter strands. But them being in stretches along DNA strands implies that ligase is absent to join up the Okazaki fragments.

3 9 A child gets half of their DNA from their father and half from their mother. So every band in son R and daughter S must correspond to a band of the same length in their father’s or mother’s electrophoresis result. Band 4 matches son R to husband V. Band 8 matches daughter S to male X. 10 This is a factual question. Marker’s comments: Most students chose option A, thinking that the semi-conservative replication of the F plasmid occurs before the transfer of the single-stranded copy into the recipient cell. However, this is not possible as the F plasmid is naturally double- stranded and it is not possible for replication to occur for a double-stranded molecule before the transfer. Thus, the transfer of one copy from donor to recipient has to occur first before the other copy in the donor acts as a template for semi-conservative replication of the F plasmid. 11 The arg operon is similar to that of the trp operon. Statement 1 is correct as the operon is switched off since arginine acts as a co-repressor which activates the repressor. Statement 2 is correct as the use of a repressor is an example of negative regulation. Statement 4 is correct as the arg operon ultimately leads to the biosynthesis of arginine which is an example of an anabolic reaction. 12 A gamete has half the amount of DNA than that of a diploid cell. However, after the S phase, a diploid cell will have four times the amount of DNA due to DNA replication. Thus, among the options given, the diploid cell will only contain twice the amount of DNA before the S phase, which is the entire G1 phase. 13 Statement 1 is correct because the cell cycle would not progress past the G2 checkpoint where proteins check for correctly replicated DNA. This is before prophase where the nucleus is still intact. Statements 2, 3 and 4 are incorrect because these processes would have been checked by proteins at the M checkpoint where the nuclear envelope has already disintegrated and the nucleolus has disappeared, leaving a nucleus that is no longer intact. 14 Template DNA strand: 3’ ---------- GTA ACC GCA TCT CAG ATT ---------- 5’ Mutated DNA strand: 3’ ---------- GTA AUU GUA TUT UAG ATT ---------- 5’ Complementary mRNA strand: 5’ ---------- CAU UAA CAU AAA AUC UAA ---------- 3’ As 5’-UAA’-3’ is a stop codon, translation would be terminated prematurely and a truncated polypeptide would be synthesised. 15 The combination XXY can be formed by these gametic combinations: XY + X, where the abnormal sperm contains X and Y chromosomes → however, the only way that this can happen is through non- disjunction at meiosis I which is not provided in any of the options given. XX + Y, where the abnormal egg contains two X chromosomes → this is a possible product of non-disjunction at meiosis I. Options B and D are factually incorrect and do not happen in reality.

4 Paper 2 Section A: Structured Questions Question 1 (a) (i) glycine; R!: proline [1] (ii) hydrogen bond; Marker’s comments: Many students did not understand that the procollagen chain is the same as the polypeptide chain that makes up a tropocollagen (termed procollagen trimer in Fig 1). Hence they did not identify that the procollagen chain has a Gly-Pro-X repeat. Some of those who understood this, identified ‘proline’ as the most common amino acid, this is rejected because glycine is needed every 3rd position to enable to tight coiling of the triple helix. Also, refer to TYS 2015 paper 1 Q4, where a similar understanding that glycine appears more frequently than other amino acids.