SAJC 2012 H2-Bio-TYS-ANS

2012 UCLES ‘A’ Level H2 Biology Mark Scheme 1 2012 ‘A’ Level H2 Biology Mark Scheme PAPER 1 (MCQ) 1 D 21 B 2 A 22 A 3 D 23 A 4 D 24 B 5 A 25 C 6 B 26 D 7 B 27 C 8 C 28 A 9 B 29 A 10 D 30 D 11 D 31 A 12 D 32 B 13 A 33 C 14 D 34 B 15 B 35 B 16 A 36 B 17 C 37 C 18 C 38 A 19 D 39 C 20 C 40 C

2012 UCLES ‘A’ Level H2 Biology Mark Scheme 2 PAPER 2 (CORE) QUESTION 1 (a) 1 Metaphase. (b) 1 Centromeres divide during anaphase ; 2 Sister chromatids separate to form independent chromosomes ; 3 Chromosomes pulled to opposite poles of the cell by the shortening of kinetochore microtubules. (c) 1 Centrioles are found in pairs at each pole and their position is important in determining the polarity of the cells ; 2 Centrioles play a role in organising the assembly of the spindle fibres, which aid in the separation of sister chromatids during nuclear division. Examiner’s comment: Most candidates correctly outlined the role of the centrioles in organising the synthesis of the spindle fibres, which then led to the separation of the chromatids. Fewer candidates mentioned that there was a pair of centrioles at each pole and that their position was important in determining the polarity of the cells. (d) 1 During prophase, the nuclear envelope disintegrates and disappears ; 2 This allows free movement of chromosomes to align at the metaphase plate ; 3 During telophase, the nuclear envelope reappears and encloses the separated chromosomes at opposite poles of the cell ; 4 This prevents entanglement of the de-condensing chromosomes and prepares the cell for cytokinesis. Examiner’s comment: While the majority of the candidates correctly described what happens to the nuclear envelope during mitosis, very few provided a full response to the question by explaining the changes that they had described. (e) Structures in region A: chromosomes (consisting of replicated sister chromatids) ; Structures in region B: spindle microtubules.

2012 UCLES ‘A’ Level H2 Biology Mark Scheme 3 QUESTION 2 (a) 1 Chloride ions are charged ; REJECT polar / hydrophilic 2 Unable to diffuse through the hydrophobic core of the phospholipid bilayer of the membrane ; 3 Transmembrane CFTR protein has a water-filled pore lined with hydrophilic amino acids which allows the diffusion of the chloride ions down their concentration gradient ; Examiner’s comments: Some candidates incorrectly referred to chloride ions as polar. (b) 1 Hydrophobic interactions between the non-polar tails of the phospholipids and the hydrophobic R-groups of the non-polar amino acids in contact with the phospholipids ; 2 Hydrophilic interactions between the polar phosphate heads and the hydrophilic R-groups of the polar amino acids on the exterior of the protein. (c) 1 ATP binds to the ATP binding domain of the CFTR protein ; 2 This changes the 3D conformation of the protein (and hence opens the channel). Examiner’s comments: Many candidates did not make use of the information provided in the introduction to the question and the diagram, giving accounts that incorrectly referred to active transport and pumping of ions. (d) 1 Deletion mutation. (e) 1 Loss of amino acid results in a change in the primary structure of the protein ; 2 Since site A is near the ATP binding domain, the change in primary structure results in changes in the 3D conformation in this domain such that it is no longer complementary to ATP and is unable to bind to it ; 3 The pore remains closed and prevents the facilitated diffusion of chloride ions out of the epithelial cells. (f) 1 When one copy of the normal dominant allele is present, functional CFTR protein will still be expressed (cystic fibrosis is a recessive disease where two copies of the alleles must be mutated for the disease to manifest) ; 2 Facilitated diffusion of chloride ions out of the epithelial cells can still occur, hence no accumulation of chloride and sodium ions in the cell that will cause water to be drawn in and result in thick mucus formation on the surface of the epithelial cell.

2012 UCLES ‘A’ Level H2 Biology Mark Scheme 4 QUESTION 3 (a) A capsid (icosahedral) head ; B tail sheath. C tail fiber. D base plate (b) 1 (Inside the bacterial cell), viral DNA is used as a template for replication of more copies of the viral DNA genome ; 2 Host RNA polymerase binds to the promoter of the viral DNA template and synthesize messenger RNA (mRNA) of viral genes ; 3 Viral mRNA is translated by host ribosomes to produce viral proteins and enzymes (to direct the assembly of new phages). (c) 1 (viral DNA incorporated into the bacterium’s DNA to become pro-phage), shield virus from unfavourable environmental conditions (potential survival strategy to maintain phage population when host abundance is too low for the population to be maintained by lytic infection); 2 (integrated viral DNA replicates along with host chromosome), increase number of copies of viral DNA / more pro-phage in bacteria population, increases the chance of propagation and hence survival of viruses.

2012 UCLES ‘A’ Level H2 Biology Mark Scheme 5 QUESTION 4 (a) 1 Transcription factors contain a DNA-binding domain which recognize and bind to specific DNA sequence in the promoter / TATA box ; 2 Nucleotide sequence of TATA box has complementary shape to DNA-binding domain ; Examiner’s comments: A number of candidates referred to complementary base pairing between transcription factors and DNA indicating uncertainty about the difference between complementary base pairing (only between nucleic acids) and the complementary shape of the binding site of specific proteins and the molecules that they bind to. (b) 1 unwinding of the double helix of DNA exposes the promoter sequence 2 for general transcription factors and RNA polymerase II to bind to, forming the transcription initiation complex for the subsequent initiation of transcription Examiner’s comments: Candidates engaging effectively with this question recognized the significance of the process in exposing bases for RNA polymerase attachment and subsequent transcription. (c) 1 eukaryotic genes contain both coding and non-coding sequences / exons and introns, which are transcribed ; 2 Introns are excised and exons ligated together ; 3 to make the coding sequences continuous in the mature mRNA for translation ; Examiner’s comments: Candidates were familiar with the processing of pre-mRNA molecules and its significance.

2012 UCLES ‘A’ Level H2 Biology Mark Scheme 6 QUESTION 5 (a) 1 Epistasis refers to gene interaction where the expression of one gene masks the expression of another gene. 2 (Given the 9:7 ratio observed in the F2 generation) expression of a pair of recessive alleles of the first gene locus (i.e. aa) masks the expression of the alleles of the second gene locus, B. 3 The expression of a pair of recessive alleles at the second gene locus (i.e. bb) also masks the expression of the alleles at the first gene locus, A. (b) F1 phenotype: purple x purple F1 genotype: AaBb x AaBb ; Parental gametes: AB Ab aB ab x AB Ab aB ab ; F2 genotypes in Punnett square ; F2 genotypes correspond to phenotype ; AB Ab aB ab AB AABB Purple AABb Purple AaBB Purple AaBb Purple Ab AABb Purple AAbb White AaBb Purple Aabb White aB AaBB Purple AaBb Purple aaBB White aaBb White ab AaBb Purple Aabb White aaBb White aabb White F2 phenotypes: purple flowers white flowers F2 phenotypic ratio: 9 : 7 (c) 1 The critical χ2 value at p=0.05 and degrees of freedom of 1 is 3.841 ; 2 As the calculated χ2 value of 1.60 is smaller than 3.841, p value is larger than 0.05 / More than 5% probability that any differ