HCI 2023 JC2 9744 H2 Bio Prelim P2 MS (Sharing)

HWA CHONG INSTITUTION / 2022-2023 H2 BIOLOGY / Preliminary Examination Paper 2 / MS 1 HWA CHONG INSTITUTION (COLLEGE SECTION) 2023 JC2 9744 H2 BIOLOGY PRELIMINARY EXAMINATIONS PAPER 2 MARK SCHEME QUESTION 1 (a) Explain why scientists keep the protoplasts in a solution that has the same water potential as the cell. [2] any two: 1 to prevent, lysis or to prevent, shrinking 2 ref. to no net movement, of water (occurs) 3 prevent movement of water, by osmosis (b) The cellulose microfibrils visible in Fig. 1.2 will form cellulose fibres. Each cellulose molecule is a polymer of β-glucose. (i) Define the term polymer. [1] a macromolecule, that is made up of many monomers (ii) Explain how the structure of a cellulose molecule allows for the formation of the cellulose microfibrils and fibres. [3] cellulose chain max 2 1 monomers joined by β(1,4) glycosidic bonds 2 adjacent, monomers, rotated through 180° 3 straight chain cellulose microfibril and fibres max 2 4 parallel molecules, of cellulose 5 hydrogen bonds form, between the OH groups of neighbouring chains 6 (as such) formation of cross links 7 microfibrils associate with other non-cellulose polysaccharides, and are arranged in larger bundles to form macrofibrils 8 many macrofibrils bundle together to form a cellulose fibre 9 idea that between adjacent cellulose molecules, beginnings and ends in different places (c)(i) Explain why the mAb ZAC-3 produced against the core polysaccharide and lipid A components will not act against the O-polysaccharide of the LPS molecules. [2] any two: 1 (mAbs) specific / different, antigen-binding sites 2 (each mAb has) specific / different, 3D conformation 3 ZAC-3 has complementary shape to core polysaccharide and lipid A Or 2D6 complementary shape to O-polysaccharide

HWA CHONG INSTITUTION / 2022-2023 H2 BIOLOGY / Preliminary Examination Paper 2 / MS 2 (ii) The results of the tests showed that both mAbs were effective in causing agglutination (clumping) of bacteria and in preventing their motility. This suggests they may be useful for preventing cholera and for treating the disease. Justify the claim that mAb 2D6 and mAb ZAC -3 are useful for preventing cholera and for treating the disease. [3] any three: general points 1 (agglutination / motility prevented, so) bacteria less able to, colonise intestine 2 less / no, choleragen released 3 bacteria passed out in faeces not able to cause disease in others 4 ref. to phagocytosis more effective prevention / treatment 5 (to prevent disease) needs to be given, at early stages 6 passive immunity 7 idea that in addition to immune response (so increased effect) 8 ref. to quicker recovery (if a person has cholera) 9 useful when, there is antibiotic resistance / antibiotics cannot be given specific mAb 10 mAb ZAC-3 may be more effective for cholera caused by, wider range of V. cholerae 11 mAb ZAC-3 may be useful if exact form of V. cholerae not known 12 AVP [Total: 11] QUESTION 2 (a) Name the structures labelled X and Y in Fig. 2.1 [2] X = 3’ CCA stem Y = anticodon (b) Compare the structure of Z with the structure of DNA. [3] Similarities: S1 double-stranded S2 hydrogen bonds between complementary bases S3 sugar-phosphate backbone S4 phosphodiester bonds Differences: D1 deoxyribose in DNA and ribose in RNA D2 thymine as a base in DNA and uracil as a base in RNA D3 Z is single-stranded but DNA is double-stranded (c) With reference to Fig. 2.1, explain how tRNA acts as an adaptor molecule for translation. [4] 1 translate base sequence into amino acid sequence 2 3’ CCA stem as attachment site for amino acid 3 anticodon determine specific amino acid attached 4 anticodon forms complementary base pairs with codon (d) Suggest why each amino acid can be carried by more than one type of tRNA. [1] 1 idea of induced fit model 2 active site of aminoacyl-tRNA synthetase can bind to slightly different anticodons [Total: 10]

HWA CHONG INSTITUTION / 2022-2023 H2 BIOLOGY / Preliminary Examination Paper 2 / MS 3 QUESTION 3 (a) Suggest why the structural genes in trp operon are transcribed together. [1] ref. to structural genes under control of one promoter (b)(ii) trpR is a regulatory gene located upstream of the trp operon. Describe the differences between the functions of trpR and trpE. [2] 1 trpR codes for regulatory protein, while trpE codes for structural proteins 2 repressor protein regulates express ion of structural genes, while enzyme is involved in synthesis of amino acid tryptophan (c)(i) On Fig. 3.2, draw the positions of RNA polymerase and the repressor molecule when tryptophan is present. [2] 1 repressor with bound tryptophan, attached to operator 2 RNA polymerase not bound to promoter / blocked by repressor (c)(ii) Explain the significance of your answer in (c)(i). [3] 1 tryptophan binds to inactive repressor protein, to activate repressor protein which will bind to operator 2 prevents binding of RNA polymerase to promoter, which stops expression of structural genes 3 trp operon is repressed when tryptophan level is high (d) Describe and explain the population growth curve shown in Fig. 3.3. [4] 1 E. coli population increases initially, then levels off, increases again, then levels off again 2 glucose was metabolised to provide ATP, and when glucose was depleted there is no available ATP 3 plateau because delay in lac operon structural gene expression before lactose was hydrolyzed 4 lactose used as respiratory substrate before lactose was depleted / run out hence second plateau [Total: 12] QUESTION 4 (a) State what is meant by recessive mutation in this context. [2] 1 both copies of TYR gene must be mutated 2 mutation resulted in non-functional tyrosinase / no tyrosinase 3 (normal) allele mask the effect of recessive (mutant) allele (b) Explain how an insertion mutation in the TYR gene can lead to a lack of melanin in a person with albinism. [4] 1 frameshift 2 change in primary structure 3 change in specific 3D conformation 4 may introduce stop codon 5 shortened polypeptide / no tyrosinase produced 6 tyrosinase becomes non-functional 7 tyrosine not converted to DOPA / DOPA not converted to dopaquinone 8 dopaquinone not formed to produce melanin

HWA CHONG INSTITUTION / 2022-2023 H2 BIOLOGY / Preliminary Examination Paper 2 / MS 4 (c) Describe the role of PCR primers in this context. [1] flanking exon 2, thus providing free 3’ OH group for extension (d) Discuss if PCR followed by gel electrophoresis can detect all types of mutations in exon 2 of the TYR gene. [3] 1 no, cannot detect base substitutions 2 mutant exon 2 remains 216 bp 3 (amplified) mutant exon 2 and normal exon would not appear as separate bands [Total: 10] QUESTION 5 (a)(i) Identify the phase of nuclear division that cell B is undergoing. [1] Metaphase (a)(ii) Describe two observable differences between cell B and the cell at prophase. [2] cell B cell at prophase 1 absence of nuclear envelope presence of nuclear envelope 2 chromosomes aligned at metaphase plate chromosomes not aligned at metaphase plate 3 chromosomes being fully condensed chromosomes not fully condensed yet (b)(i) State the cell cycle chec