SAJC 2010 H2-Bio-TYS-ANS

2010 UCLES ‘A’ Level H2 Biology Mark Scheme 1 2010 ‘A’ Level H2 Biology Mark Scheme PAPER 1 (MCQ) 1 C 21 C 2 A 22 C 3 C 23 B 4 C 24 D 5 D 25 C 6 D 26 D 7 C 27 C 8 D 28 B 9 C 29 C 10 B 30 C 11 B 31 A 12 A 32 D 13 A 33 B 14 B 34 B 15 C 35 D 16 D 36 A 17 D 37 A 18 B 38 B 19 D 39 B 20 B 40 D

2010 UCLES ‘A’ Level H2 Biology Mark Scheme 2 PAPER 2 (CORE) QUESTION 1 (a) A – Mitochondrion;; 1 Presence of cristae;; 2 Double membrane ;; Reject rod-shaped, reference to cisternae B – Rough endoplasmic reticulum;; 1 Presence of ribosomes;; 2 Parallel stacks of cisternae;; Reject references to sac-like or tube-like structures as not visible on the diagram Examiner’s comments: The features of mitochondria were well known, although some talked about rod-shapes or confused cristae with cisternae. Candidates must ensure that when asked for ‘visible features’ that they only sta te features that they can see on the electron micrograph. Some listed sacs or tubes as features of the rough endoplasmic reticulum. These were not clearly visible on the diagram and reflect what the candidate had learnt rather than what they could see. (b) (i) A – Synthesis of ATP during aerobic respiration;; B – Protein synthesis (by the ribosomes attached to it) / transport of protein / vesicle formation;; (ii) 1 ATP synthesized by A is hydrolyzed to release energy for protein synthesis during translation e.g. amino acid activation;; 2 Ribosomes on B synthesizes respiratory enzymes/proteins needed for aerobic respiration;; Examiner’s comments: Majority of candidates explained how the products of A were useful in B, rather than vice versa (c) 1 Compartmentalisation so that different conditions are established to maintain optimum conditions for enzyme reactions; ; 2 Increase surface area for metabolic reactions e.g. attachment of more electron carriers and ATP synthase for oxidative phosphorylation;;

2010 UCLES ‘A’ Level H2 Biology Mark Scheme 3 QUESTION 2 (a) 1 DNA polymerase works only in one direction, from 5’ to 3’, adding new nucleotides to the available 3’-OH end of a pre-existing polynucleotide chain;; 2 DNA strands are antiparallel /DNA strands run in the 5’ 3’ and 3’ 5’ directions ;; (b) 1 The parental DNA molecule is a double helix consisting of two strands, each strand serves as a template for the synthesis of a new complementary daughter strand;; 2 each of the two new DNA molecules contains one new and one original/parental strand of DNA;; Examiner’s comments: Care must be taken when using the terms like ‘strand’ when describing a DNA molecule. Weaker candidates referred to the newly synthesized DNA molecule as being ‘a strand’. (c) 1 Helicase;; 2 (DNA) Primase;; 3 RNA primer ;; OVP 4 ATP;; 5 Single-stranded binding proteins;; 6 Deoxyribonucleotides;; 7 DNA ligase;; Examiner’s comments: Candidates must be encouraged to tailor the length of their answers to the space provided. They must also be precise with terminology used. Imprecise answers, such as ‘DNA primer’ and ‘RNA primase’ were not allowed. (d) 1 During DNA replication, complementary base pairing may result in one or more incorrectly paired bases which is not corrected by proof reading of DNA polymerase;; 2 Resulting in a substitution where one nucleotide in the gene sequence is being replaced by another nucleotide;; Examiner’s comments: Candidates need to use the marks allocated as a guideline to the number of points or detail that they include. Most candidates confined their answers to describing one form of mutation, such as substitution, but did not give clear enough additional detail for full credit to be awarded.

2010 UCLES ‘A’ Level H2 Biology Mark Scheme 4 (e) 1 In sickle cell anemia, substitution of thymine for adenine at the 17th nucleotide of the gene coding for β-globin chain of hemoglobin results in a change in (mRNA) codon from GAA to GUA and subsequent change of the 6th amino acid from glutamic acid to valine ;; 2 Glutamic acid is hydrophilic whereas valine is hydrophobic, the tertiary structure of the molecule changes due to the change in R groups of the amino acid;; 3 Hb S stick to each other via their hydrophobic regions and polymerize into long fibres inside the red blood cells, deforming them into sickle shape;; 4 Due to the sickle shape, red blood cells clump and clog small capillaries, obstructing other cells from moving through the capillaries, leading to other symptoms such as physical weakness, pain, or organ damage / Sickle-shaped red blood cells have a shorter lifespan compared to normal cells and hemolyse readily resulting in anemia and also making them ineffective in transporting oxygen gas;; Examiner’s comments: Candidates must ensure that they describe the whole process and not miss the initial effects in an attempt to include all of the detail they have learnt about a given mutation. Many candidates began their responses beyond the immediate effect of a mutation, namely the alteration in the base sequence or codon and the subsequent change in amino acid.

2010 UCLES ‘A’ Level H2 Biology Mark Scheme 5 QUESTION 3 (a) A – Capsid head B – Contractile sheath / Tail sheath C – Base plate (b) 1 Adsorption of phage through binding of attachment sites of tail fibres to receptor sites on bacterium;; 2 Release of lysozyme which degrades host bacterial cell wall;; 3 Tail sheath contracts to drive a hollow tube into bacterium cell membrane; Phage DNA injected into cytoplasm of bacterium; Examiner’s comments: Candidates need to be clear of the difference between the bacterial cell wall and bacterial cell membrane (c) Lytic cycle Lysogenic cycle 1 Arrest of bacterium host gene expression when phage enters No arrest of bacterium host gene expression when phage enters 2 Bacterial chromosome hydrolysed to provide nucleotides for phage genome replication No bacterial chromosome hydrolysis 3 Viral genome does not integrate into host genome Viral genome integrates into host genome, known as prophage 4 No latency period Latency period where bacteria genes are not expressed in bacterium host 5 Bacterium host is lysed No lysis of bacterium host (before induction phase) (d) (i) 1 Genome enclosed by protein capsid;; 2 Rod-shaped TMV similar to rod-shaped tail sheath;; (ii) 1 RNA genome in TMV vs DNA genome in T4 phage;; 2 No tail fibres in TMV vs presence of tail fibres in T4 phage;; 3 No tail sheath / collar in TMV vs presence of tail sheath in T4 phage;; [Any 2]

2010 UCLES ‘A’ Level H2 Biology Mark Scheme 6 QUESTION 4 (a) A Nucleosome;; B DNA;; Reject: Linker DNA (b) 1 To make long DNA molecule more compact to fit in the nucleus;; 2 To prevent breaking or damage to DNA;; 3 Allows for regulation of gene expression / transcription (by interactions between DNA and histones);; (c) (i) 1 Histone methylation results in the binding of proteins which lead to tighter nucleosomes / more compact chromosomes;; 2 Prevents RNA polymerase and transcription factors from binding to promoter to initiate transcription;; Examiner’s comments: Candidates must study diagrams carefully to help with answering questions. (ii) 1 many genes are not required in that one type of differentiated cells thus their gene expression is switched off;; 2 large size of genome (thus the need for making it compact to fit into nucleus);; 3 eukaryotic genome contains many non-coding regions (thus, need to make these regions compact to be packed into the nucleus and not interfere with areas where genes are actively expressed);; Examiner’s comments: Candidates need to carefully target the depth of their answers to the number of marks allocated. Many restricted their potential by repeating ideas that mos