2009 a Level H2 Biology P2 Ans

July 24, 2017 | Author: joannetzy | Category: Chemical Synapse, Active Site, Action Potential, Neurotransmitter, Depolarization
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2009 A Level H2 Biology

Paper 2 Section A Question 1 (a) A - mRNA B - polypeptide or protein being synthesised C - cisterna or lumen of the RER Reject: RER/ crista (b) Nucleus

– ribosome absent


– ribosome present


2009 A Level H2 Biology


– ribosome absent

Mitochondria – ribosome present (c) channel protein allows transport of large molecules, i.e. polypeptide chain across the membrane channel protein functions as a receptor, i.e. it binds the signal recognition particle channel protein holds / positions the ribosome for protein synthesis Note: do not describe the protein channels in general terms (d) a transport vesicle buds off from the RER carrying the protein to the Golgi apparatus for modification from where a secretory vesicle buds off from the Golgi apparatus and takes it to the cell surface membrane Protein is released out of the cell by exocytosis – fusion of membrane of vesicle & cell surface membrane ref to the role of microtubules in the transport of the vesicles (e) protein channels are present in the cell surface membrane of prokaryotic cells, and that these might be involved in protein secretion. Reject: description about protein synthesis.


2009 A Level H2 Biology

Question 2 (a) plasmid is smaller than bacterial chromosome, i.e. it has fewer genes plasmid has antibiotic resistance genes, which are not found on the bacterial chromosome Accept: any other named example or genes code for enzyme responsible for cell metabolism Reject: mere mention of genes “for survival” (b) (i) Conjugation refers to the direct transfer of genetic material of one bacterial cell to another, through the temporary joining of the two cells.  The sex pilus is produced by the donor cell and allows it to attach to a recipient cell. A temporary cytoplasmic mating bridge is formed between donor and recipient cells.  The F plasmid DNA separates into 2 single strands and one of the strands is transferred across the mating bridge to the recipient cell which acts as a template for the synthesis of a complementary daughter strand in each respective cell. Key words: contact between cells, sex pili , transfer of the plasmid (ii) Transformation refers to the modification of a bacterium’s genotype by the uptake, incorporation and expression of naked, foreign DNA from the surrounding environment.  The foreign DNA of closely related species is recognised by cell-surface proteins of the bacterium and transported into the cell.  The foreign DNA crosses over with a homologous region on the bacterial chromosome. Key words: A bacterium takes up foreign DNA which is then incorporated into its own DNA. (c) 1. antibiotic resistance/other xenobiotic resistance 2. the ability to use a new metabolite Tutor’s comments: Question asked you to suggest two potential benefits – you need to be specific. Mere mention of “helping bacteria to survive or to increase variation” would not suffice (d) 1. In the phage lytic reproductive cycle, phage enzymes hydrolyse the bacterial chromosome into small pieces of DNA. 2. a small random fragment of bacterial DNA may be packaged within the capsid by mistake. 3. The defective phage is released. It attaches to another bacterium and injects the piece of bacterial DNA acquired from the first cell (donor). 4. Some of the donor bacterial DNA can replace the homologous region of the recipient cell’s chromosome by DNA recombination and will subsequently be expressed by the cell. From examiner’s report: Bacterial transduction was well known, but sometimes credit could not be given because of imprecise statements e.g. bacterial DNA and viral DNA were often referred to in terms of chromosomes or genomes. Most knew that a phage was involved and that it inserted its DNA into a bacterium. That the resultant phages infected other bacteria was also well known as was the fact that the original bacterial DNA was acquired by the newly infected cell. Many candidates described both generalised and specialised transduction when only one was required since, in effect, the potential mark points in both are the same.


2009 A Level H2 Biology

Question 3 From examiner’s report: In this question it was possible to gain credit in different sections for the same relevant idea, for example, that of complementary base pairing. (a) Requirement: to outline the role of telomeres 1. To prevent the loss of genetic information, i.e. refer to inability of DNA polymerase to be able to replicate to the end of the chromosome; disposable buffer effect of telomere in DNA replication; 2. To prevent chromosomes from sticking together, i.e. after many cell division cycles, telomeres become extremely short  cannot bind telomeric capping proteins, exposing bare ends of dsDNA that are unstable and tend to stick together 3. Role of telomeres in apoptosis, i.e. telomere shortening can induce replicative senescence which blocks cell division. Senescence involves p53 and pRb pathways and leads to the arrest of cell proliferation. (b) (i) Requirement: to describe the function of telomerase reverse transcriptase 1. Complementary base pairing of RNA template to the ends of the telomere 2. Binding of telomerase reverse transcriptase to the DNA 3. Telomerase reverse transcriptase active site binds (i) RNA template; (ii) DNA and (iii) free nucleotides 4. Elongation of DNA, i.e. formation of phosphoester bonds between DNA and free nucleotides 5. Repeat sequencing There were fewer references to the binding of the enzyme to DNA and the role of its active site. (ii) Requirement: to describe the part played by the telomerase RNA 1. forms complementary base pairs with the end of the DNA and 2. provides a template for the complementary base pairing of new deoxyribonucleotides and 3. repeats the sequences. (c) (i) Refer to Fig. 3.1 Many candidates failed to deduce the relevant base sequence on the telomerase RNA. (ii) Requirement: to suggest why parts of the molecules are double stranded Reference to double-stranded nature being achieved by complementary base pairing double stranded nature of parts of telomerase RNA would stabilise the molecule double stranded nature would lead to an overall shape that allows the telomerase DNA to fit into the active site of the enzyme.


2009 A Level H2 Biology

(iii) 1. Reference to the DNA carrying the gene that encodes telomerase RNA 2. Reference to DNA as a template for transcription, process is catalysed by RNA polymerase 3. dsDNA template unwinds and separate 4. Complementary base pairing of free ribonucleotides to the DNA template 5. RNA polymerase catalyse phosphoester bond formation Question 4 (a) That the locus is the position of a gene on a chromosome (b) The expected number of phenotypes with no linkage would give a 1:1:1:1 ratio and therefore divided the total by four to give 660 for each phenotype. Reject: calculation based on 9:3:3:1 ratio (c) the genes are on the same chromosome Reject: use of the word “allele” (d) There were many clearly labelled and fully correct genetic diagrams that showed the parental genotypes, the gamete genotypes including recombinants, the subsequent offspring genotypes and phenotypes. Many candidates did not use a convention to indicate linkage e.g. sticks or loops. Some candidates annotated the recombinants. (e) 1. Reference to recombinant frequency and formula 2. The recombinant frequency would be low if the genes were close together 3. Reference to map units 4. If the expected phenotype ratio is obtained then there is no linkage.


2009 A Level H2 Biology

Question 5 Candidates found this to be the most difficult question on the paper. (a) Cytochrome b gene was present in all of the species and so it was a good basis for comparison. The gene changes very slowly Mutations are silent There is a restriction on change and it is passed down the maternal line. (b) Mutation, specifically base-pair substitution, was the cause of these changes in the DNA sequence. Warning: do not give a laundry list of all mutation Tutor’s comments: A cause of such mutations is also creditworthy (c) In the neutral theory of molecular evolution no selective advantage or disadvantage is involved and there is no effect on the phenotype. The plot of the line is straight  the rate of mutation is steady Mutations are silent and that there are a small number of changes were not often seen. (d) From the examiner’s report: Most candidates cited the involvement of selective advantage or disadvantage but failed to relate their answers to the context in the question and simply stated Darwin’s theory as those with a selective advantage survived, reproduced and passed on their genes and vice versa. Other relevant points were in relation to the different selection pressures on the different islands as each island was colonised and that this was an example of adaptive radiation. Far too many candidates still describe the passing on of traits rather than genes.


2009 A Level H2 Biology

Question 6 Candidates found this to be an easy question. (a) (i) A – chromatids B – centromere C – spindle fibres / kinetochore microtubules (ii) In interphase DNA replicates To form two identical chromatids attached at centromere (b) (i) Anaphase was named almost without exception. (ii) In anaphase, centromeres divide (R: splits) so that chromatids are now daughter chromosomes, which are pulled to the poles (R: ends) of the cell by spindle fibres which are attached to the centromeres

Question 7 (a) At a higher light intensity, the carbon dioxide assimilation rate levels off, i.e. photosynthesis was at maximum rate The systems were saturated with light, i.e. light was no longer a limiting factor Something else was the limiting factor, e.g. temperature. (b) At the lowest light intensity there would be a low rate or no photosynthesis (occurring below the compensation point) However, respiration would still occur Carbon dioxide would be given off Reject: (i) respiration occurs instead of photosynthesis and (ii) although there was no light reaction, there is a dark reaction which uses up carbon dioxide to give a negative value (c) At point A, the rate of respiration equals the rate of photosynthesis and that this is the compensation point. references to there being no net gain in dry mass and no growth as any products of photosynthesis are used up in respiration


2009 A Level H2 Biology Section B Both essay questions generally resulted in a high mark.

Question 8 (a) Examiners’ Comments: The mode of action of enzymes was generally well known by many candidates. Points commonly made referred to the enzyme lowering the activation energy of the reaction and then being available for reuse, their specificity and lock and key or induced fit mechanisms leading to an enzyme-substrate complex. Other points made were in relation to the temporary bonds between the substrate and the active site with references to ionic and hydrogen bonds and stress in the substrate. Less commonly seen points which were creditworthy were for the idea of collisions between enzymes and substrate, the fact that the fit between substrate and active site can be referred to in terms of a chemical fit and that at the end of the reaction the products leave the active site. In relation to the latter point it was not uncommon to see the term substrate instead of product leaving the active site. A significant minority of candidates wrote a basic minimum about the mode of action of enzymes but wrote a lot about factors affecting enzyme action which were not relevant to answering the question asked and consequently gained very little credit.

1. 2. 3. 4. 5. 6. 7.

enzymes lower the activation energy; thus increasing rate of reaction collision between substrate and enzyme causes the substrate molecule to bind the enzyme molecule at its active site to form an enzyme-substrate complex; the substrates are held in the active site by hydrogen or ionic bonds; the enzyme molecule holds the different substrate molecules in close proximity or in the correct orientation relative to one anotehr; thus increasing the probability of a reaction occurring; once the reaction has occurred, the products leave the active site; the unchanged enzyme molecule is then available for reuse;

Lowering of activation energy: 8. certain bonds in the substrate molecule may be placed under physical stress; 9. the R-group of amino acid residues at the active site can change the charge/electron distribution on the substrate which will increase the reactivity of the substrate; 10. Enzymes are highly specific in the reactions they catalys; 11. act on molecules that have specific functional groups; 12. act on particular type of chemical bonds; 13. act on a particular stereo or optical isomer; 14. shape of substrates must be complementary to that of the enzyme active site; Specificity: 15. lock and key hypothesis; 16. substrate is the key; 17. the enzyme's active site -the "lock"; 18. once products are formed, they no longer fit into the active site; OR 19. induced fit hypothesis; 20. shape of the active site of enzyme may not be exactly complementary to that of substrate; 21. substrates enter the active site, induces a conformational change in the enzyme, substrates fit more snugly into the active site; (b) Examiners’ Comments: The effect of pH on an enzyme catalysed reaction was very well known and often the answers usefully incorporated a graph which was well annotated. The majority of descriptions included all of the relevant points in that enzymes have an optimum pH at which there is maximum rate of reaction and that above and below that pH the rate of reaction decreases. The causes of rate decrease included references to hydrogen and ionic bonds being affected, particularly in relation to R groups in the active site with the consequent failure to form an enzyme substrate complex and the eventual denaturation of the enzyme.


2009 A Level H2 Biology

1. enzymes work within narrow ph range; 2. at optimum ph, the rate of an enzyme catalysed reaction is at its maximum; 3. tthe intramolecular bonds that maintain the secondary and tertiary structures of the enzyme molecule are intact; 4. the frequency of effective collisions between enzyme and substrate molecules is the highest; 5. the number of enzyme-substrate complexes formed per unit time is at its highest; 6. at ph above or below the optimum, the changes in ph alter the charges in the acidic and basic r groups of the enzyme molecule; 7. this alteration of charges in the r groups disrupts bonds that helps to maintain the specific shape of the active site of the enzyme; 8. ionic bonding; and 9. hydrogen bonds; 10. the conformation of the enzyme molecule would be affected;and 11. denaturation of the enzyme occurs; 12. the active site no longer fits the substrate; 13. hence, less / no enzyme-substrate complexes are formed per unit time;and 14. the rate of reaction decreases; 15. annotated graph;;

(c) Examiners’ Comments: The effect of non-competitive inhibitors on enzyme activity was very well known and again many answers usefully included an annotated graph. The points most commonly seen were that the inhibitor binds to an allosteric site on the enzyme altering the tertiary structure of the active site so that the substrate cannot bind to the active site and therefore the rate of the reaction decreases. Many candidates also knew that increasing the amount of substrate will not reduce the effect. Other valid points which were rarely seen were that allosteric inhibitors can be reversible or irreversible and that some non-competitive inhibitors bind to the active site irreversibly.

1. a non-competitive inhibitor has no structural similarity to the substrate; 2. it binds the enzyme at a point other than the active site; 3. interaction between the non-competitive inhibitor and the enzyme causes the enzyme molecule to change its shape; 4. the active site unreceptive to the substrate; or 5. leaving the enzymes less effective in catalysing the conversion of the substrate to the product; 6. a proportion of the enzyme molecules out of action, resulting in a drop in the effective enzyme concentration; 7. the rate of reaction will not reach maximum value; 8. the effect of non-competitive inhibition cannot be overcome by increasing the substrate concentration; 9. allosteric inhibitors binds to allosteric site stabilizes the inactive form of the enzyme & vice versa; 10. the binding can be reversible or irreversible; 11. some non-competitive inhibitors bind to the active site irreversibly; Question 9 (a) The transmission of an action potential along a myelinated neurone was very well known by the vast majority of candidates. Descriptions usually started with a reference to the resting potential of –60 to –70 mV inside the membrane and the involvement of the sodium-potassium pump in the movement of sodium and potassium ions. This was then followed by the events involved in depolarisation when the potential difference changes to +40 mV inside the membrane once the threshold potential has been reached because of the opening of sodium ion channels allowing sodium ions to enter the axon. Finally the events involved in repolarisation which restore the resting potential because sodium channels close


2009 A Level H2 Biology but potassium channels open allowing potassium ions to leave the axon. All of the above was usually couched in the context of the myelinated neurone, in terms of the Nodes of Ranvier leading to saltatory conduction. Other creditworthy points which were less commonly seen were in relation to hyperpolarisation occurring before the resting potential is restored, the fact that local circuits are involved and that there is a refractory period before another action potential can occur. Sometimes candidates began their description at the events involved in depolarisation and therefore missed the marks allocated to the setting up of a resting potential. Depolarisation cannot occur unless there is a resting potential.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24.

In a resting neurone, the resting potential is about –60 to –70 mV inside the membrane; The sodium-potassium pump removes 3 Na+ and takes up 2 K+; There is a high concentration of Na+ outside and a high concentration of K+ inside the neurone; The plasma membrane of neurones is more permeable to K+ than to Na+; Therefore loss of K+ exceeds gain of Na+; When the neurone is stimulated, voltage-gated Na+ channels in a particular part of the neurone open; Na+ rush into the axon along a concentration/electrochemical gradient; This causes depolarisation of the membrane; If the Na+ influx achieves threshold potential, then additional voltage-gated Na+ channels open, triggering an action potential; Some of the local current generated by the action potential will then flow passively down the axon; depolarises the membrane in the adjacent region of the axon; via the opening the Na+ voltaged-gated channels in the neighboring membrane; The local depolarization triggers an action potential in this region; which then spreads again in a continuing cycle until the end of the axon is reached; As depolarisation is restricted only to the nodes of Ranvier in myelinated axons; Thus action potential “jumps” from one node of Ranvier to another along the axon -saltatory conduction; Behind the impulse, Na+ voltaged-gated channels close; voltage-gated K+ channels open; K+ begin to leave the axon along a concentration/electrochemical gradient; the outflux of K+ causes the neurone to become repolarised behind the impulse; The slow-moving gates of voltage-gated K+ channels remain open for a while longer after the repolarisation phase; This causes the membrane potential to become more negative than its normal resting potential – hyperpolarisation; Presence of a refractory period before another action potential can occur; Na+ is once again actively expelled in order to increase the external concentration and resting state is restored so allow the passage of another impulse;

(b) Most candidates gave very good explanations of how a nerve impulse is transmitted across a synapse. Although the fact that there are few calcium ions inside the synaptic knob before the presynaptic membrane is depolarised was not often seen, the rest of the events were well described. Upon depolarisation the calcium channels open and calcium ions enter which causes vesicles containing neurotransmitter to fuse with the presynaptic membrane. Once the neurotransmitter is released into the synaptic cleft it diffuses across and binds to receptors on the postsynaptic membrane. The subsequent opening of sodium channels allows sodium ions to enter and depolarise the membrane leading to an action potential.

1. 2. 3.

Synaptic transmission occurs as a nervous impulse arrives at the synaptic knob; causes gated calcium channels, in addition to gated sodium channels, to open; sodium ions (depolarisation) and calcium ions to rush into the cytoplasm of the presynaptic neurone;


2009 A Level H2 Biology 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.


The influx of Ca causes vesicles of acetylcholine to move to the presynaptic membrane and fuse with it; emptying their contents into the synaptic cleft via exocytosis; 2+ Ca move out of the synaptic knob by active transport; acetylcholine diffuses across the synaptic cleft; usually in less than 0.5 ms; acetylcholine bind temporarily to complementary shape receptor protein on the postsynaptic membrane; + + the shape of the protein changes, opening chemically-gated Na channels, Na rush into the cytoplasm; depolarising the postsynaptic membrane/an excitatory post-synaptic potential (EPSP) is formed; If the depolarisation reaches the threshold level, an action potential is generated; Acetylcholine will be hydrolysed (degraded) by the enzyme acetylcholinesterase to acetyl and choline; which will diffuse back across the synaptic cleft to the presynaptic membrane The acetyl and choline will then be recombined to form acetylcholine using ATP provided by the mitochondria;

(c) Although the question asks candidates to suggest reasons why nerve impulses only travel in one direction across the synapse, many candidates included in their answers the reasons which are relevant to why nerve impulses travel in one direction along a neurone and so accounts about refractory periods were common. However points relevant to the synapse were commonly seen in terms of there only being calcium ion channels and vesicles containing neurotransmitter on the presynaptic side and therefore the neurotransmitter can only move one way to bind with receptors which are only on the postsynaptic side.

1. 2. 3. 4.

Calcium ion channels;; and vesicles containing neurotransmitter are present in the presynaptic knob;; Receptors for neurotransmitters are found on the postsynaptic membrane;; The neurotransmitter diffuse down a concentration gradient from high to low, crossing the cleft by diffusion;; Therefore the neurotransmitter can only move one way to bind with receptors which are only on the postsynaptic side;;


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