SAJC 2009 H2-Bio-TYS-ANS

2009 UCLES ‘A’ Level H2 Biology Mark Scheme 1 2009 ‘A’ Level H2 Biology Mark Scheme PAPER 1 (MCQ) 1 A 21 B 2 C 22 C 3 A 23 D 4 D 24 A 5 D 25 C 6 A 26 B 7 B 27 B 8 D 28 B 9 B 29 B 10 D 30 C 11 C 31 B 12 D 32 B 13 D 33 C 14 D 34 A 15 C 35 B 16 B 36 B 17 B 37 B 18 C 38 C 19 C 39 B 20 A 40 C

2009 UCLES ‘A’ Level H2 Biology Mark Scheme 2 PAPER 2 (CORE) QUESTION 1 (a) A – mRNA B – polypeptide C – cisterna/lumen of rough endoplasmic reticulum (reject RER/cristae) (b) Location Ribosome present Ribosome absent Nucleus √ Chloroplast √ Vacuole √ Mitochondria √ 4 c o r r e c t = 2 3 c o r r e c t = 1 (c) 1 allows the movement of synthesized polypeptide across the membrane; 2 functions as a receptor to bind the signal recognition particle/signal peptide of the polypeptide; 3 functions to hold the ribosome so that the polypeptide can be inserted into the ER lumen. (d) 1 transport vesicle carrying the protein buds off the rough endoplasmic reticulum and transported to and fuse at the cis face of the Golgi body; 2 secretory vesicle buds off the trans face of the Golgi body and transports the protein to the cell surface membrane; 3 Membrane of the vesicle fuses with the plasma membrane and releases the protein via exocytosis; 4 microtubules aid in the transport of the vesicles from the ER to the Golgi body and to the plasma membrane Examiner’s comments: Many candidates unnecessarily wrote about the roles of the RER and Golgi in relation to the proteins. (e) 1 through protein pore/protein channels on cell surface membrane

2009 UCLES ‘A’ Level H2 Biology Mark Scheme 3 QUESTION 2 (a) Bacterial chromosome Plasmid 1 larger, with more genes smaller, with fewer genes 2 has genes coding for production of enzymes for cell metabolism has genes coding for genetic markers such antibiotic resistance Examiner’s comments: Need to mention exactly what the genes are coding for. (b) (i) 1 Sex pilus of the F+ donor cell makes contact with the F- recipient cell 2 Replication and transfer of F plasmid from F+ into F- cell (ii) 1 Foreign DNA taken up by bacteria through its cell wall and plasma membrane 2 and is incorporated into bacterium’s DNA via genetic recombination (homologous recombination or site-specific insertion) (c) 1 Acquire antibiotic resistant genes 2 Acquire genes that confer ability to use a new metabolite 3 Acquire genes for other xenobiotic resistance ( A xenobiotic is a chemical which is found in an organism but which is not normally produced or expected to be present in it) Examiner’s comments: Avoid making vague statements about helping bacteria to survive, increase variation or having same characteristics as donor. (d) 1 phage attaches to specific receptors on bacterial cell wall and injects viral DNA into bacterial cell 2 expression of viral genes produces viral proteins like enzymes that breakdown host bacteria DNA into smaller pieces 3 Packaging of host bacteria DNA, together with viral DNA into the capsid of newly assembled phages 4 Upon lysis, resultant phages infect other bacteria which acquire the original bacterial DNA Examiner’s comments: - Be specific in terms used: refer to bacterial DNA and viral DNA (not chromosomes or genomes) during description. - There is no need to describe generalized and specialized transduction separately as the potential marking points in both are the same for the overall process of transduction.

2009 UCLES ‘A’ Level H2 Biology Mark Scheme 4 QUESTION 3 (a) 1 DNA polymerase unable to replicate to the end of the chromosome without a free 3’OH end of a pre-existing nucleotide 2 Prevents loss of genetic information by acting as a disposable buffer blocking ends of chromosome 3 Prevents chromosomal DNA ends from being recognized as double-stranded breaks and initiating apoptosis/unintentional cell death 4 Prevents chromosomal end-to-end fusions with its binding to proteins to form a protective nucleoprotein cap (b) (i) 1 Binds to 3’ end of DNA template at the enzyme’s active site 2 Elongation of DNA via complementary base pairing using RNA as a template , catalyzing formation of phosphodiester bonds between nucleotides 3 The process is repeated as the telomerase moves in the 5’ to 3’ direction of the growing chain Examiner’s comments: Before describing the process of reverse transcription, need to mention about the binding of DNA template to enzyme’s active site. (ii) 1 Forms complementary base pairs with the end of the DNA / allows extension of DNA from its 3’ end 2 Serves as a template for the complementary base pairing of new deoxyribonucleotides 3 5’TTAGGG3’ sequence is repeated (c) (i) Circle RNA template sequence 3’-CAAUCCCAAUC-5’. This will code for the telomeric sequence TTAGGG. (ii) 1 Stablilises telomerase RNA molecule by complementary base pairing 2 Contributes to specific shape to fit stably into telomerase enzyme (iii) 1 DNA contains gene coding for transcription of telomerase RNA 2 RNA polymerase binds to and unzips double-stranded DNA 3 One of the two strands acts as a template for transcription of a complementary RNA strand

2009 UCLES ‘A’ Level H2 Biology Mark Scheme 5 QUESTION 4 (a) 1 refers to the position of a gene on a chromosome 2 different alleles of a gene occupy the same gene locus (b) 1 expected numbers if no linkage for all phenotypes : 660 2 total : 2640 (c) 1 genes found on the same chromosome (d) F1 phenotypes: Red eye, Normal wings x Purple eye, Ve stigial wing F1 genotypes: RN/rn rn/rn F1 gametes: RN rn Rn rN r n rn rn rn F2 genotypes shown in Punnett square F1 gametes RN rn Rn rN rn RN/rn rn/rn Rn/rn rN/rn F2 genotypes RN/rn rn/rn Rn/rn rN/rn F2 phenotypes Red eye, Purple eye, Red eye, P u r p l e e y e , N o r m a l w i n g V e s t i g i a l w i n g V e s t i g i a l w i n g Normal wing F2 phenotypic ratio 1139 : 1195 : 151 : 155 ( P a r e n t a l P h e n o t y p e s ) ( R e c o m b i n a nt phenotypes) Examiner’s comments: Phenotypic ratio is not 1:1:1:1 as there is linkage of genes. Many candidates did not use a convention to indicate linkage e.g. sticks or loops. Some candidates annotated the recombinants. (e) 1 calculation of recombination frequency (no. of recombinants / total no. of offsprings x 100%) where low recombinant frequency suggests close proximity of genes; 2 distance between genes represented as map units i.e. 1 map unit = 1% recombinant frequency; 3 if expected phenotypic ratio is obtained, it indicates no linkage between genes; Examiner’s comments: **Low recombinant frequency implies genes close proximity; BUT high recombinant frequency does not mean genes far apart.

2009 UCLES ‘A’ Level H2 Biology Mark Scheme 6 QUESTION 5 (a) 1 gene was present in all of the species, so making it a good basis for comparison 2 slow mutation rate in gene / gene is highly conserved / restriction on change / silent mutations 3 found in mitochondria, therefore it is passed down the maternal line (b) 1 Gene mutation base substitution 2 Due to errors in DNA replication (c) 1 Neutral mutations do not confer any selective advantage or disadvantage 2 Rate of mutation is steady / plot of the line is straight 3 Silent mutations, no effect on the phenotype (d) 1 Genetic variations exist among the Hawaiian honey creepers. 2 Different selection pressures on the different islands as each island was colonised by honey creepers, an example of adaptive radiation 3 Honey creepers with a selective advantage in the particular environment survived till reproductive age and pass on their genes to offsprings QUESTION 6