2010 a Level H2 Biology P2 Ans

2010 A Level Examination H2 Biology Paper 2 Solutions PAPER 1

PAPER 2 Question 1 (a) A – Mitochondrion;; 1. encolosed by double membrane;; 2. inner membrane is folded to form cristae;; B – Rough endoplasmic reticulum;; 1. network of folded membranes forming sheets/ flattened membranous sacs (called cisternae);; 2. presence of ribosomes on surface of ER;; Examiners’ comments: Candidates must ensure that when asked for ‘visible features’ that they only state 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 - site of ATP synthesis;; R: energy is produced B – site of protein synthesis/ protein transport/ vesicle formation;; (ii) A provides ATP for protein synthesis at B;; Mitochondrial proteins are synthesised at B before transported to A;; (iii) Compartmentalisation;;

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2010 A Level Examination H2 Biology Paper 2 Solutions different conditions are required to maintain optimum conditions for enzyme reaction/ to increase surface area;; AVP: maintain high concentrations of reactants at specific sites;; provide different local environments that facilitate biochemical pathways;; prevent intermediates of one pathway from mixing with those of another;; ensure that harmful substances are enclosed within structures and hence, kept away from the rest of the cell;; Question 2 (a) DNA polymerase only works in one direction from 5’ to 3’, adding new nucleotides to the free 3’ end;; the two parental DNA strands are antiparallel;; (b) both strands of parental DNA separate and act as templates for the synthesis of daughter DNA strands;; each daughter DNA double helix comprises one parental DNA strand and one daughter DNA strand;; (c) deoxyribonucleoside triphosphate/ ATP/ helicase/ DNA ligase/ RNA primer/ primase [any three] (d) DNA polymerases may insert the wrong nucleotides, or too many or too few into the sequence;; Repair enzymes failed to recognised these improperly paired nucleotides;; (e) Most of the descriptions given centred on either sickle-cell anaemia or oncogenes and were well known. 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. Sickle-cell anaemia caused by a single missense mutation/ base pair substitution of the -globin gene;; due to this mutation, the glutamic acid (hydrophilic) in the normal polypeptide is changed to a valine (hydrophobic) in the mutant polypeptide;; this alteration generates a sticky patch on the surface of the -chain , the deoxygenated form of the mutant Hb (HbS) is insoluble in erythrocytes and forms crystalline arrays;; the erythrocytes of affected individuals become rigid and their transit through capillaries is blocked, causing severe pain and tissue damage;; Question 3 (a) tail sheath;; base plate;; (b) 1. 2. 3. 4. 5. 6.

T4 phage tail fibres bind to specific receptor sites; on host bacterium cell wall; The tail sheath contracts; driving a hollow tube; through the bacterial cell wall; The T4 phage dsDNA genome is injected into the bacterial cytoplasm; Page 2 of 8

2010 A Level Examination H2 Biology Paper 2 Solutions (c) Feature Integration of viral genome Lysis of host cell Number of host cell infected Initiation of lytic pathway Synthesis of viral protein

Lytic cycle Viral genome is not incorporated into the bacterial chromosome/ the host cell’s genome is hydrolysed; Replication of virus culminates in lysis and death of host cell; Only 1 host cell is infected;

Lysogenic cycle Viral genome is incorporated into the bacterial chromosome;

The phage genome replicates without killing the host bacterial cells; The host cell divides, viral DNA also replicates, results in many daughter cells infected with prophage; Viral DNA can immediately initiate the Prophage enters the lytic pathway lytic pathway after entering the host cell; upon stimulation; Synthesis of phage proteins occurs Transcription of prophage genes is immediately after entry; blocked by lambda repressor;

(d) (i) Presence of capsid enclosing the viral genome;; Capsomeres are arranged in a precise and symmetrical pattern around the nucleic acid;; (ii) Absence of tail/sheath/collar;; Presence of RNA instead of DNA;; Question 4 (a) A- Nucleosome;; Reject: nucleosome core, solenoid B - Deoxyribonucleic Acid or linker Deoxyribonucleic acid;; Reject: abbreviation (b) Packaging of eukaryotic genomic DNA into nucleosomes 1. makes it compact/ condensed; thus DNA can be packed / contained within the cell’s nucleus; 2. shields the nitrogenous bases from the enzymes in the nucleus; thus preventing DNA mutation; reject: prevent degradation by enzymes 3. prevents the long DNA molecules (of different chromosomes); from getting knotted or tangled; 4. prevent the access of transcription factor if promoter is part of the nucleosome, regulating gene expression/ transcription;; [any three] (c) methylated histones are bound to proteins; OR which prevents histones from being acetylated by histone acetyltransferase ; positive charge on histone tail is maintained; thus histone tail continues to interact with other nucleosomes; to result in a condensed chromatin structure; implication: transcription factors/RNA polymerase are not able to access the gene;; (d) only a particular sub-set of proteins are involved in cell differentiation; only certain genes need to be expressed; large numbers of genes need not be expressed/turned off; i.e. chromatin needs to be of the condensed structure; a high proportion of genome is non-coding/ large size of genome;;

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2010 A Level Examination H2 Biology Paper 2 Solutions (e) reference to operator site; located in the promoter region; reference to an active repressor protein; that binds to the operator site; Reject: regulatory protein RNA polymerase; prevented from binding the promoter; Question 5 (a) continuous variation;; (b) characteristic is controlled by many genes;; each gene would have little overall effect;; each individual allele has similar and additive effects on the same character;; phenotypes are influenced by the environment;; (c) parents & F1 - individuals all genetically identical: all homozygous for parents, all heterozygous for F1; parents & F1 – variation is due to only environment; genotypes of F2 more varied than parent and F1: formed as a result of meiosis in the formation of gametes (from F1); processes such as independent assortment of homologous chromosomes;; crossing over (between linked genes forming new linkage groups) leads to new combination of alleles;; random fusion of gametes during fertilization;; each gene has its own number of alleles, leading to a large number of possible combinations of alleles;; phenotypes are influenced by the environment;; Question 6 (a) (i) ref to reduced NAD+ (i.e. NADH/H+); or reduced FAD (i.e. FADH2) as immediate source of electrons; ref. to NAD+ and FAD removing a pair of hydrogen atoms; from respiratory substrates in glycolysis, link reaction and krebs cycle;; (ii) electron pairs combine with oxygen atoms;; which are very electronegative each oxygen atom simultaneously picks up a pair of hydrogen ions from the aqueous solution; forming water as the final product; (iii) allows a proton gradient to develop across the inner mitochondrial membrane;; ref. to proton gradient as a source of potential energy for the synthesis of ATP;; (b) 1. electron flow along the ETC from A to B releases free energy in a series of small steps; 2. free energy released is coupled to the pumping of protons; 3. across the inner mitochondrial membrane; from the mitochondrial matrix into intermembranal space , i.e. C; 4. ref to inner mitochondrial membrane being impermeable to protons;;

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2010 A Level Examination H2 Biology Paper 2 Solutions (c) Features Sources of energy Location Electron donors Electron acceptors Establishment of proton gradient

Photophosphorylation energy for making ATP comes from light Thylakoid membrane of chloroplast For non-cyclic reaction: water; For cyclic reaction: PS I Non-cyclic acceptor: NADP+ Cyclic acceptor: PS I H+ pumped inwards, from stroma to thylakoid space

Oxidative Phosphorylation energy for making ATP comes from oxidation processes Inner membrane of mitochondria NADH, FADH2 Oxygen H+ pumped outwards, from matrix to intermembranal space

[Any three] Question 7 (a) ref subjected to different selection pressures favouring different phenotypes;; Birds with favourable phenotype survived and reproduced, passed on the alleles to the next generation;; ref to geographical isolation, no interbreeding and no gene flow;; (b) (i) the fossils shown are incomplete or damaged;; limited structural similarities of the fossil skulls, it can be subjective;; common ancestor skull is missing;; (ii) Quantifiable – differences in base sequence can be counted/ every nucleotide will be a point of comparison, e.g. if homologous regions of DNA from two species that are 1,000 nucleotides long provide 1,000 points of comparison;; Objective - Molecular character states are unambiguous: A, C, G and T are easily recognizable and one cannot be confused with another;; (c) insufficient time for evolution of flightless birds;; flightless birds are unable to cross the sea separating the islands;; Section B Question 8 (a) 1. E. coli regulates the production of the enzymes involved in lactose metabolism through the lactose operon; In the presence of lactose: 2. lactose enters bacterial cell; 3. lactose is converted to allolactose which binds to repressor protein; 4. results in a change in shape of repressor protein/inactive; 5. repressor protein no longer attached (to DNA/operator); 6. exposes operator site; 7. allows RNA polymerase to bind to the promoter; 8. transcription (of genes coding for proteins required to utilize lactose); 9. leading to synthesis of enzymes – -galatosidase, permease, transacetylase; 10. with higher lactose concentration, more repressor proteins are inactivated  more transcription;

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2010 A Level Examination H2 Biology Paper 2 Solutions In the presence of both lactose and glucose: 11. only glucose is used; 12. In the presence of glucose, [cAMP] drops; 13. CAP is inactive and disengages from CAP binding site; 14. Though RNA polymerase binds to the promoter; 15. transcription occurs at a low level  low levels of enzymes is produced; (b) 1. Low blood glucose concentration below normal level is detected by -cells ; 2. Glucagon is secreted by  -cells; of the islet of Langerhans; that will bind with receptors on their liver (target) cells; 3. hormone-receptor complexes interact with G proteins and activate adenyl cyclase 4. which converts adenosine triphosphate (ATP) to cyclic adenosine monophosphate 5. cAMP activates protein kinase A, which subsequently phosphorylate relay proteins 6. transduced signal triggers cellular responses which include: a. activation of enzymes involved in glycogenolysis; i. the stored glycogen is converted to glucose; which can be metabolised for energy or released into the bloodstream; ii. enzymes involved in glycogenesis is inhibited; b. activation of enzymes involved in gluconeogenesis; i. Conversion of amino acids, fatty acids and glycerol to glucose; 7. Raising the blood glucose level to the norm; (c) 1. Reception: signalling molecule from outside the cell binds to a receptor protein located at the cell’s surface (cell membrane receptor) or inside the cell (intracellular receptor). 2. binding site of receptor is specific for a single chemical messenger (ligand), i.e. the complementary fit 3. Transduction: ligand-receptor interaction changes the receptor protein, initiating the process of transduction. 4. a multistep pathway whereby proteins are activated by: addition or removal of phosphate groups; or 5. production / release of second messengers e.g. cAMP or ions (e.g. Ca2+) 6. Response: The transduced signal finally triggers a specific cellular response in the nucleus or, in the cytoplasm 7. Activities include: a. Protein synthesis: i. Activation of protein synthesis: synthesis of mRNA, which will be translated in the cytoplasm into a specific protein ii. Inhibition of protein synthesis b. Protein Activity i. opening / closure of ion channels e.g. opening of voltage-gated calcium ion channels in synaptic knob as a result of depolarisation ii. activation / inhibition of metabolic enzymes e.g. glycogen synthetase in carbohydrate metabolism, cyclin-dependent kinases in cell cycle control iii. regulation of overall cell shape. Question 9 (a) 1. reference to a central carbon atom (i.e. the a. an amino group (-NH2); b. a carboxyl group (-COOH); c. a hydrogen atom; and

-carbon atom); to which is bonded:

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2010 A Level Examination H2 Biology Paper 2 Solutions d. an R-group (or side chain); which is variable amongst the 20 different amino acids; occurring in nature. 2. a condensation reaction occurs between the carboxyl group of one amino acid; and the amino group of another; 3. a molecule of water is removed in the condensation reaction;; 4. condensation reaction is catalysed by peptidyltransferase, a component of the large ribosomal subunit;;

(b) 1. primary structure is the specific sequence of amino acids in a polypeptide chain; a. specificity due to : (i) identity of amino acid at each position; and (ii) order of amino acids; b. held by peptide bonds; c. between amino and carboxyl groups; 2. in secondary structures, segments of the polypeptide chain are repeatedly; coiled or folded; as a result of a. held by hydrogen bonds; b. at regular intervals along the polypeptide backbone; 3. tertiary structure is specific 3-dimensional conformation; in which polypeptide chain usually bends and folds extensively; Accept: precise, compact, ‘globular’ shape a. held by disulfide bonds, ionic bonds, hydrogen bonds and hydrophobic interactions; b. formed between the R-groups of amino acid residues; c. that are far apart in the linear sequence of the polypeptide chain; 4. quaternary structure comprises more than one; polypeptide chain / subunit; a. held together by R-group interactions; b. between subunits; (c) 1. named globular protein: haemoglobin (Hb); Accept: enzyme 2. specific function: transport of oxygen; Accept: catalysis 3. globular in shape; a. many Hb molecules can be packed into a single red blood cell; 4. consists of four subunits; a. each subunit is capable of binding one O2 molecule; b. subunits exhibit cooperative binding of O2 molecules; c. cooperative binding maximises both the amount of O2 loaded at the lungs and the amount of O2 released at the tissues; 5. bulk of the hydrophobic amino acids are buried in the interior; of the globular structure whilst the hydrophilic amino acid residues are on the exterior; a. haemoglobin molecule is soluble in the aqueous medium of blood; 6. haem-binding site is lined with hydrophobic amino acid residues; a. to provide a hydrophobic environment for the binding of haem, which is largely hydrophobic 7. haem group consists of a porphyrin ring and an iron (II) ion, Fe2+; Fe2+ can combine reversibly with O2; Page 7 of 8

2010 A Level Examination H2 Biology Paper 2 Solutions a. enhances the release of O2 in metabolically active tissues such as muscle;

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