SAJC 2011 H2-Bio-TYS-ANS

2011 UCLES ‘A’ Level H2 Biology Mark Scheme 1 2011 ‘A’ Level H2 Biology Mark Scheme PAPER 1 (MCQ) 1 D 21 B 2 B 22 D 3 B 23 B 4 C 24 B 5 C 25 A 6 B 26 C 7 A 27 D 8 D 28 B 9 A 29 D 10 A 30 C 11 A 31 A 12 B 32 D 13 A 33 C 14 B 34 C 15 A 35 C 16 D 36 D 17 D 37 A 18 C 38 A 19 B 39 D 20 C 40 B

2011 UCLES ‘A’ Level H2 Biology Mark Scheme 2 PAPER 2 (CORE) QUESTION 1 (a) 1 Rate of reaction increases rapidly from 0 to 4 a.u. at low substrate concentration of 0 to 4 a.u. 2 Rate of reaction slows down as substrate concentration increases and eventually level-off to a plateau of 9.6 a.u. at a high substrate concentration of 80 a.u. Examiner’s comment: this required students to mention both the rapid initial increase followed by a levelling off to a plateau as well as quoting figures from both low and high substrate concentrations. (b) (i) (ii) 1 a competitive inhibitor has a close structural resemblance to the substrate, and therefore competes with the substrate for the same active site of enzyme 2 inhibitor binds to the active site, rate of reaction decreases 3 binding of the inhibitor to the active site is temporary and not permanent. At high substrate concentration, effect of inhibition is negligible / increasing the substrate concentration can decrease the effect of inhibition. (c) 1 toxin permanently binds to the active site of the enzyme and prevents hydrolysis of acetylcholine 2 acetylcholine accumulates and remains bound to receptors on the post-synaptic membrane 3 ligand-gated sodium ion channels would stay open, leading to repeated action potentials

2011 UCLES ‘A’ Level H2 Biology Mark Scheme 3 QUESTION 2 (a) A – (large ribosomal subunit ) ribosome B – polypeptide C – transfer RNA (tRNA) D – messenger RNA (mRNA) (b) 1 Attachment of an amino acid to its corresponding tRNA with specific anticodon by aminoacyl tRNA synthetases 2 which have an active site with 3D conformation complementary to the 3D conformations of the amino acid and the corresponding tRNA 3 catalyzes covalent attachment of the specific amino acid to its tRNA, 4 resulting in the formation of an aminoacyl-tRNA with 3D conformation that is no longer complementary to the conformation of the active site (and is released from the enzyme). (c) 1 The rRNA in ribosomes holds the tRNA and mRNA together in close proximity, via complementary base pairing using hydrogen bonds 2 positions the new amino acid for addition to the carboxyl end of the growing polypeptide 3 rRNA peptidyl transferase activity catalyzes formation of a peptide bond between the new amino acid and the polypeptide chain QUESTION 3 (a) Haemagglutinin 1 Binds to the sialic acid on host cell membrane 2 aids in the binding and entry into host cell Neuraminidase 3 Removes sialic acids on the envelopes of newly assembled viruses 4 aids in the release of newly assembled viruses from host cells and from each other (b) 1 phospholipids 2 glycoproteins 3 cholesterol Examiner’s comments: Any two components of the virus’s enveloped derived from host cell membrane.

2011 UCLES ‘A’ Level H2 Biology Mark Scheme 4 (c) 1 Haemagglutinin on the envelope of influenza virus binds to the sialic acid receptor on the cell surface membrane of the target cell 2 host cell membrane invaginates, engulfing the virus to form an endocytotic vesicle within the host cell 3 endosome membrane fuse with the viral envelope for the release of viral nucleocapsid 4 viral capsid is enzymatically removed for the release of the viral genome into the cytoplasm (d) (i) 1 influenza viruses from 3 different species infect a single cell 2 during assembly of new virus, there is reassortment / random assembly of RNA segments from the 3 different influenza 3 giving rise to novel combinations, antigenic shift (ii) 1 Mutations in key viral genes in human H1N1 due to an error-prone RNA polymerase (which has no proof-reading capability) 2 leading to subsequent potential for change in number and sequence of amino acids in polypeptide chain thus affecting the tertiary structure / 3D comformation of the protein structure (eg. glycoprotein haemagglutinin) QUESTION 4 (a) 1 Base mutation where there is a substitution of a single nucleotide (G in proto- oncogene ras to T in ras oncogene) causes change in a codon 2 resulting in change of amino acid from glycine to valine (in ras protein) 3 R-groups of glycine and valine are different and each form different type of bonds with other amino acids 4 results in different tertiary structure / 3D conformation which affects protein interactions with other molecules and alters the protein’s activity (resulting in ras protein triggering the kinase cascade even in the absence of growth factor resulting in increased cell division). (b) 1 gain of function mutations were usually dominant and require only a single mutation in a gain of function mutation 2 loss of function mutations were usually recessive and both genes required mutant alleles in loss of function mutation to have an effect 3 gain of function mutations result in gene product / protein e.g. ras protein gaining a new and abnormal function 4 while loss of function mutations result in gene product / protein e.g. p53 having less or no function

2011 UCLES ‘A’ Level H2 Biology Mark Scheme 5 (c) Any 2 points: 1 Radiation / UV light / X-ray / ionizing radiation 2 Carcinogens 3 Viruses (d) 1 translocation of c-myc proto-oncogene from chromosome 8 to chromosome 14 does not result in a change in gene sequence (Protein encoded carry out normal functions) 2 mutation of H-ras proto-oncogene results in a change in gene sequence, resulting in a change in amino acid sequence (hence mutated protein gained a new / abnormal function) 3 different regulatory sequence involved in the transcription of translocated c-myc proto-oncogene sequence 4 same regulatory sequence involved in the transcription of H-ras proto-oncogene and the oncogene (e) 1 DNA replication has taken place during interphase before cell division. 2 Chromosomes only become visible as they condense and shorten prior to cell division 3 results in each chromosome consisting of two sister chromatids joined at the centromere. Examiner’s comments: Candidates mistakenly referred to the pair of homologous chromosomes rather than addressing why each homologous chromosome appeared as a double structure in Fig. 4.3. (f) 1 c-myc oncogene has less effect compared to H-ras oncogene in causing tumours in mice. (state general trend for individual oncogene) 2 100% of mice without tumours in the first 100 days and 65% of mice without tumours at the end of 200 days for c-myc oncogene vs 75% of mice without tumours in the first 50 days and 45% of mice without tumours at the end of 200 days for H-ras oncogene 3 The combined effects of both oncogenes in causing tumours in mice is much greater than the sum of the individual tumour-causing effect of each oncogene (state general trend for combined oncogenes) 4 Only 30% of mice without tumours in the first 50 days, and 0% mice without tumours at about 180 days Examiner’s comments: Candidates needed to study Fig. 4.4 carefully before answering the question. Candidates making full use of the data presented in Fig. 4.4 to quote figures invariably achieved higher marks.

2011 UCLES ‘A’ Level H2 Biology Mark Scheme 6 QUESTION 5 (a) (i) 1 The fixed position of an allele or gene 2 on a particular chromosome (ii) 1 Alternative form of a gene 2 Present on the same gene locus (on a pair of homologous chromosomes) 3 May be dominant, recessive or co-dominant (b) Symbols XH – X chromosome with allele for normal blood clotting (no haemophilia) Xh – X chromosome