2013 HCI H2 Biology Prelims Paper 2 Answers

HWA CHONG INSTITUTION / 2013 JC2 H2 BIOLOGY / PRELIMINARY EXAMINATION / MARK SCHEME 1 HWA CHONG INSTITUTION 2013 JC2 H2 BIOLOGY PRELIMINARY EXAMINATION MARK SCHEME QUESTION 1 (a)(i) Identify the class of carbohydrate molecule of which agarose is an example. [1] polysaccharide (ii) Plants contain a carbohydrate called amylose. Amylose does not contain galactose. Describe one similarity and another difference in structure between agar ose and amylose. [2] similarity: 1. monosaccharides of both agarose and amylose are held together by glycosidic bonds 2. both agarose and amylose are unbranched difference: 1a. coiled/helical structures in amylose 1b. but linear/straight structures in agarose 2a. alternate glucose is not inverted/ same orientation in amylose 2b. but alternate galactose is inverted 180 o in agarose 3a. α(1,4) glycosidic bond is present in amylose 3b. different types of glycosidic bonds are present in agarose (iii) Suggest why bacteria are unable to metabolise agarose. [2] 1. lack of bacterial enzyme 2. to hydrolyse the type of glycosidic bonds present in agarose 3. substrate not complementary to active site of enzyme (b) (i) Describe how the structure of domain A of the cholera toxin is maintained. [3] 1. primary structure folds into secondary structures 2. maintained by hydrogen bonds formed between C=O and NH groups 3. further folding 4. into a unique 3D conformation 5. maintained by hydrogen bonds, ionic bonds, disu lfide bonds and hydrophobic interactions between R- groups of amino acid residues (ii) Explain how a globular protein like cholera toxin differs from a fibrous protein, such as collagen. [2] 1a. cholera toxin being a globular prot ein is soluble due to the presence of hydrophilic amino acids on the surface of the protein 1b. collagen being a fibrous protein is insoluble due to t he presence of mainly hydrophobic amino acids in the protein 2a. cholera toxin being a globular protein is more compact/ spherical in shape 2b. collagen being a fibrous protein is elongated in shape /forms multimolecular parallel filament to strands/ collagen fibrils and collagen fibres 3a. cholera toxin being a globular protein is made up of fi xed, non-repetitive specific sequence of amino acids 3b. collagen being a fibrous protein is ma de up of repetitive sequence of amino acids 4a. cholera toxin being a globular protein has a relati vely unstable structure ref. to weak non-covalent bonds 4b. collagen being a fibrous protein has a stable st ructure ref. to extensive intra- and inter-molecular hydrogen bonds/ covalent crosslinks 5a. cholera toxin being a globular protein perfor ms metabolic functions since they are soluble 5b. collagen being a fibrous protein performs structur al functions since they have high tensile strength PAPER 2

HW (c) QUE (a) ( ( Fig. (b) ( ( WA CHONG INS Suggest h 1. rec e 2. bin d 3. resu 4. ch o 5. inv a 6. for m ESTION 2 i) Name stru X: Heteroc Y: Euchro ii) Account f 1. Re g 2. Th u 3. He t 4. He t 2.2 shows p i) Complete ii) State the 1. Hy d 5’ ; STITUTION / 20 how the chole eptor binding ds/attach to r ulting in a co olera toxin en agination/ inf mation of a v uctures X an chromatin omatin for the differe gion X is mor us heterochro terochromati terochromati part of a DNA Fig. 2.2 by i importance o drogen bond 3’ 13 JC2 H2 BIO era toxin ent g domain B o receptors on onformationa nters by rece folding of epi vesicle enclos d Y. ence in the s re electron d omatin is mo n is transcrip n is wound a A molecule. ndicating the of hydrogen s allow for th LOGY / PRELIM ers epithelia of cholera tox n cell membra l change in r ptor mediate ithelial cell m sing cholera structures X a dense / darkly ore condense ptionally inac around deace e polarity of t bonds in DN he formation MINARY EXAM l cells. xin is comple ane receptor ed endocytos membrane toxin and Y. y stained tha ed / tightly pa ctive as comp etylated histo Fig. 2 the DNA mo NA structure. of double st MINATION / MAR ementary to r sis an region Y acked / coiled pared to euc ones 2b lecule in the tranded DNA RK SCHEME receptors on d than euchr hromatin boxes provid A / a double h 3’ ; 5’ n intestinal e romatin ded. helix pithelial cells [Total: 12 m 2 [2] s arks] [2] [2] [1] [2]

HWA CHONG INSTITUTION / 2013 JC2 H2 BIOLOGY / PRELIMINARY EXAMINATION / MARK SCHEME 3 2. Hydrogen bonds hold t he two polynucleotide strands together 3. Hydrogen bonds hold the complementary nucleotides / bases together 4. Many hydrogen bonds give stability to DNA molecule 5. Hydrogen bonds can be broken for tr anscription / DNA replication to occur (c) Explain how the data in Table 2.1 helps to confirm the arrangement of bases in DNA. [3] 1. In all the organisms, percentage of adenine to thymine and percentage of guanine to cytosine are approximately 1:1 ratio / equal / similar 2. For example, in yeast, percentage of adenine is 31.3% which is very similar to thymine with percentage at 32.9% and percentage of guanine is 18.7% which is very similar to cytosine with percentage at 17.1%. Accept any one example 3 . This shows that adenine and guanine base pair with thymine and cytosine respectively (d) (i) State how the result for the virus differs from those for all the organisms given in table 2.1. [1] 1. The percentage of adenine to thymine and gua nine to cytosine are not similar / not 1:1 (ii) Give a reason for your answer to d(i). [1] 1. Virus is made up of single-stranded DNA / DNA that is not a double helix [Total: 12 marks] QUESTION 3 (a) Describe the role of lactose in the regulation of lac operon. [3] 1. Lactose / allolactose acts as an in ducer that binds the lac repressor 2. Lac repressor is inactivated and dissociates from the operator 3. RNA polymerase can bind to the promoter, re sulting in transcription of structural genes (b) Identify the location of the point mutation in the operon of strain A. Give a reason for your answer. [ 2] 1. The mutation occurred in the promoter / operator 2. RNA polymerase cannot recognise / bind to mutant promoter and hence, transcription initiation cannot occur or mutant operator is permanently bound to lac repressor and hence RNA polymerase cannot recognise and bind promoter 3. Plasmids used in rescue experiments bear functional promoter and operator and hence, RNA polymerase can bind to this functional promoter and initiate transcription (c) Suggest why strain B cannot be “rescued” at all. [2] 1. Mutation occurred in the lacI gene 2. Resulting in expression of a constitutively active / hyperactive repressor / constitutively active repressor cannot be inactivated by allolactose 3. Hence, even if the genes are insert ed under the control of the functional lac promoter and operator, the constitutively active repressor will always bind to lac operator and prevent RNA polymerase from binding to the promoter / continues to inhibit the transcription of the new genes (d) Describe how the formation of separate mature protei ns from these polycistronic mRNAs in HIV maturation and in bacteria differs. [3] Bacteria HIV maturation 1. The polycist