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BIO00004C - Part II

Examination Candidate Number: _____________ Desk Number: _____________

University of York Department of Biology B. Sc Stage 1 Degree Examinations 2017-18

Molecular Biology and Biochemistry - Part II Time allowed: 1 hour and 30 minutes Total marks available for this paper: 50

● Answer all questions in the spaces provided on the examination paper ● The marks available for each question are indicated on the paper ● A calculator will be provided

For marker use only: 1-13 14-24 25

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BIO00004C - Part II

Answer all questions in the spaces provided Multiple Choice Questions Only circle one answer per question. +1 mark for every correct answer, -1/3 mark for every incorrect answer. If a candidate does not choose any answer, there will be no penalty. 1. ATP carries energy because: a) breaking the bond between its last two phosphate groups generates a low energy state. b) it can easily phosphorylate phosphoenolpyruvate during glcuconeogenesis. c) the ribose moiety in ATP is generated by the pentose phosphate pathway. d) it is one of the main energy stores in the human body. 2. Fatty acids are converted into glucose in humans: a) to allow quicker ATP production during exercise. b) fatty acids cannot be converted into glucose in humans. c) when blood glucose levels are low. d) to prevent muscle loss during starvation. 3. Why is NADH rather than NADPH produced as an electron carrier during glycolysis? a) Because NADH can be phosphorylated by 1,3-bisphosphoglycerate kinase in the process to give NADPH. b) Because glycolysis is an ATP producing catabolic process, and hence needs NADH as electron carrier. c) The question is not correct, since NADH and NADPH can both be produced during glycolysis. d) Because NADH has a significantly larger redox potential that allows it to generate more ATP. 4. When glucose conversion to pyruvate is routed through the pentose phosphate pathway, the amount of ATP produced per glucose: a) is more than the amount produced during glycolysis b) is the same as produced during glycolysis c) is less than the amount produced during glycolysis page 2 of 12

BIO00004C - Part II

d) can be more or less than that produced in glycolysis dependent on the presecence of oxygen. 5. During starvation the brain: a) is fed by excess glucose generated from keton bodies. b) does not produce any CO2 because it cannot obtain glucose. c) still uses a significant proportion of the body’s energy expenditure. d) starts to use fatty acids delivered to it through the blood stream. 6. FAD is: a) a permanent cofactor of complex II in the electron transport chain. b) does not have enough redox-power to oxidise a -OH group, but can oxidise a carbon-carbon single bond. c) is a major energy carrier between beta-oxidation and the electron transport chain. d) all of the above are true. 7. A short sprint produces the same amount of CO2 per second as a long sprint because: a) sprinting preferentially uses fatty acids as a fuel source. b) during sprinting the muscles use fermentation to produce ethanol. c) energy generation during sprinting is done anaerobically. d) the pentose phosphate pathway cannot generate NADH from NAD+. 8. The reversible reactions in glycolysis are: a) also used in gluconeogenesis in the liver. b) used to make sure glycolysis is tightly regulated. c) not really reversible, this is just an artifact of considering the standard Gibbs free energy (ΔG 0) rather than the actual ΔG. d) ensuring that fermentation can produce CO2. 9. Fermentation in yeast is essential when no oxygen is present, otherwise:

a) NADH would accumulate to high levels and inhibit glycolysis. b) the food industry would have to use bacterial systems for producing beer. c) pyruvate kinase would be inhibited. page 3 of 12

BIO00004C - Part II

d) all of the above are correct. 10. Glucose-6-phosphate is an important metabolite in humans because: a) it is the main product of glycogen degradation. b) it traps glucose in the cell and can initiate several different glucose utilizing pathways. c) it is a key intermediate in glycolysis whose production can tightly regulate this pathway. d) it can be generated from ribose-5-phosphate through the pentose phosphate pathway. 11. The term beta-oxidation refers to: a) the second carbon after the carbonyl group in a fatty acid ester being oxidised. b) a process in which a double bond is formed to then generate an -OH group that can be oxidised, reminiscent of the first half of the TCA cycle. c) the production acyl-CoA from fatty acids. d) the process in which dogs metabolise fat to excrete its products in their urine. 12. Ketone bodies are: a) generated from glucose during starvation to allow more sustained feeding of the brain with energy. b) used to generate energy via catabolism through glycolysis and the TCA cycle. c) the reason why a low-carb diet allows weight loss, as they allow the brain to catabolise fat. d) not generated from pyruvate because the reaction catalyzed by pyruvate dehydrogenase is irreversible. 13. The following are statements comparing ⍺-ketoglutarate dehydrogenase (KDH) and pyruvate dehydrogenase (PDH). Which one is correct? a) KDH and PDH are both multienzyme complexes that combine two different enzymatic activities to convert their substrate to its product. b) KDH and PDH both oxidize their substrate and thereby generate FADH2 that is sent to the electron transport chain. c) The enzymatic reaction catalysed by these two enzymes releases four of the page 4 of 12

BIO00004C - Part II

six carbons of glucose as CO2. d) Both enzymes are part of the TCA cycle, which is why this cycle can produce so much energy. Most of the SAQs were answered well. Students had the main problem with Q13. While it could be argued that the fact that KDH and PDH are composed of 3 rather than 2 enzymes is a less significant detail (many wrongly ‘tipped’ A to be correct), the correct answer - where the CO2 is released from glucose, is important conceptual knowledge. 14. Electrons donated to the mitochondrial respiratory chain by succinate are transferred to oxygen via: a) PSII and PSI plus intermediate redox components b) PSII and Plastocyanin c) Complexes II, III and IV a) Complexes I, II, III and IV -well answered by most 15. Aldolases in the Calvin cycle catalyse: a) transfer of a five-carbon to rubisco b) the reduction of phosphoglycerate c) a two-carbon unit from a ketose to an aldose d) a condensation reaction between two compounds that each contain three carbons -well answered by most 16. The electron acceptor of Complex III in the mitochondrial electron transport chain is: (a) Cytochrome C (b) Plastocyanin (c) Succinate (d) Ubiquinone -well answered by most

17. The difference between light harvesting chlorophylls and reaction centre chlorophylls is: a) the former transfer light energy in the form of redox energy, the latter page 5 of 12

BIO00004C - Part II

in the form of vibrational energy b) the former transfer light energy in the form of resonance energy, the latter in the form of vibrational energy c) the former transfer light energy in the form of resonance energy, the latter in the form of redox energy d) the former transfer light energy in the form of redox energy, the latter in the form of resonance energy -well answered by most 18. If reaction (1), Succinate → Fumarate (E’o = 0.03 V) and reaction (2), FADH2 →FAD + + 2H + (E’o = -0.22 V), are coupled, then: a) reaction (1) would proceed from left to right and is an oxidation b) reaction (2) would proceed from left to right and is an oxidation c) reaction (1) would proceed from right to left and is a reduction d) reactions (1) and (2) would proceed from left to right and are an oxidation and a reduction respectively -This was not answered well: with these delta G values FADH2 would be oxidised to FAD while fumarate is reduced to succinate (though in a real cell it is usually the other way round because the actual, non-standard, delta G values are quite different). Thus, reaction 2 goes from left to right and is an oxidation.

19. The colour(s) light primarily absorbed by chlorophyll are: a) Blue and Red b) Green and Red c) Green and Blue d) Yellow and Red -well answered by most 20. Plants can harvest energy from light with wavelengths other than the above because: a) They have multiple chlorophyll pigments with varying absorption spectra b) Many chlorophylls are bound to proteins that absorb light at different wavelengths c) Plants have other pigments that absorb light that is passed on by chlorophylls. page 6 of 12

BIO00004C - Part II

d) Plants have other pigments that absorb light at other wavelengths. -well answered by most; note that though proteins can impact, it is the pigments that absorb light, not the proteins 21. Cyclic electron transport: a) involves cycling of electrons between Photosystem II and the Cytochrome b6f complex and produces NADPH but not ATP b) involves cycling of electrons between Photosystem I and Plastocyanin and produces NADPH but not ATP c) involves cycling of electrons between Photosystem II and the Cytochrome b6f complex and produces ATP but not NADPH d) involves cycling of electrons between Photosystem I and Plastocyanin and produces NADPH but not ATP - answer is the same as b) -well answered by most 22. Mitochondrial and chloroplast complexes contain the following transition element(s): a) Cu, Fe, Mn b) Fe, Cu c) Cu, Mn, Zn d) Cu, Fe, Mg -less well answered (note, neither Mg nor Zn is a transition metal) 23. Transition elements in these complexes are directly involved in: a) the transport of protons across the membrane b) providing structure in the form of cytochromes c) the transport of electrons d) oxidation/reduction of NADH -well answered by most 24. What are the two constituents of the proton motive force (PMF) and where would you find a PMF in a plant cell? a) PMF is made up of an electrical component (membrane potential or delta psi) and a chemical component (the proton gradient or delta pH) and is found across the mitochondrial inner membrane and thylakoid membrane. b) PMF is made up of an electron gradient (membrane potential or delta psi) and a chemical component (the proton gradient or delta pH) and is found across the mitochondrial inner membrane and plasma membrane. c) PMF is made up of an electrical component (membrane potential or delta psi) page 7 of 12

BIO00004C - Part II

and a chemical component (the proton gradient or delta pH) and is found across the mitochondrial inner membrane, thylakoid membrane and plasma membrane. d) PMF is made up of a chemical proton gradient (or delta pH) and a chemical ATP gradient and found across the mitochondrial inner membrane and thylakoid membrane. -less well answered by most; note, that most plant, fungal and bacterial membranes will have a PMF (examples were givien in the lectures) 25. a) Which two reactions can be catalysed by Rubisco? (1 mark) The carboxylation or oxygenation of ribulose-1,5-BP -well answered by most, though a lot of people just mentioned ‘photorespiration’ (not a reaction) or ‘oxidation’ (not the same as ‘oxygenation’) b) Which one of the two is not deemed beneficial? Explain why. (2 marks) The oxygenation of ribulose (photorespiration) is damaging because it leads to loss of carbon in the form of CO 2 (1 mark) and to minimise this loss, plants use a pathway that consumes ATP (1 mark). -well answered by most

26. Explain in one or two sentences why, and in which direction, the pH of thylakoid lumen changes when plants are moved from the dark to the light. (2 marks) Light-driven proton pumping by the photosynthetic electron transport chain energises proton transport from the chloroplast stroma to the thylakoid lumen (1 mark) which acidifies the lumen, i.e. the pH goes down (1 mark). -well answered by most but some got into a muddle here with wrong signs and interpretations 27. Some plants have facultative CAM metabolism. In 2-3 lines, explain in what conditions such plants are likely to change from C3 to CAM metabolism and what advantage that would bring. (3 marks) Plants would change after onset of drought/salinity or increase in ambient temperature (1 mark). CAM leads to CO2 enrichment during the night (1 mark) which allows plants to keep stomata closed during the day and as such save water (1 mark). -reasonably well answered by most but some confusion with C4 metabolism

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BIO00004C - Part II

28. A) Rank the fuels glucose, fatty acids and glycogen for efficiency and preference. (3 marks) Generating most (i) to least (iii) amount of ATP per gram of fuel

Most (i) to least (iii) preferred by the human body as an energy source

i)

fatty acids

glucose

ii)

glycogen

glycogen

iii)

glucose

fatty acids

Answered generally well, although many students swapped glucose with glycogen - especially in the first column. This is not correct, as glycogen degradation results in glucose-phosphate, which saves energy expenditure during activation. b) The two rank orders do not agree. Explain why. (6 marks) Fatty acids generate more ATP per gram than either glucose or glycogen because they represent a less oxidised form of carbon (1 mark), but their degradation is slower (1 mark), and strictly dependent on oxygen (1 mark). Glucose also has a preferred status due to the brain’s dependence on blood glucose levels (1 mark). Glycogen, which generates more ATP than glucose because its main monomeric degradation product is glucose-1-phosphate (1 mark), will only be used once free glucose is not available (1 mark). A lot of marks were lost by just stating that fats have more energy than glucose/glycogen - something that was the point of the table - rather than saying why. Similarly, marks were lost by giving the same reason twice, just in a slightly altered form - such as “fats are degraded more slowly”, then later stating “glucose is degraded more quickly”. Describing details of fatty acid degradation and glycolysis did not provide marks, unless this was used to create an argument for the speed of energy generation, or the amount of energy generated. The argument that fatty acid degradation is less efficient because of the need for activation does not hold, as glucose also needs to be activated (by hexokinase and PFK).

29. Here is the open-chain form of D-Glucose:

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BIO00004C - Part II

(i)

Draw the open-chain form of L-Glucose. (1 mark)

(i)

Most students answered this correctly- L-Glucose and D-Glucose are enantiomers- a pair of molecules that are mirror images of each other. This was covered in Lecture 16. (ii)

D-Allose is a C3 epimer of D-Glucose. Draw the structure of D-Allose. (1 mark)

(ii)

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The majority of students answered this question correctly. Epimers were covered in Lecture 16 and in the Supported learning workshop. 30. Compare and contrast the structures and functions of cellulose and chitin. (5 marks)

A2. Similarities Both are linear polysaccharides (0.5 mark) Both are homopolysaccharides (0.5 mark) Monosaccharides linked by beta 1,4 glycosidic linkages (1 mark) Both have structural roles- e.g. cell walls (1 mark) Differences Chitin consists of N-acetyl-D-glucosamine units (0.5 mark) while Cellulose consists of beta D-glucose units (0.5 mark) Chitin is found in cell walls of fungi (and exoskeletons of insects and crustaceans) (0.5 mark) while Cellulose is found in plant cell walls (0.5 mark) Some students may wish to include a drawing/Figure as part of their answer. Most students answered this question very well but may have missed one or more of the key points in the suggested answer above. Some students compared cellulose to peptidoglycan rather than to chitin. Polysaccharides were covered in Lecture 17. 31. For the following phospholipid:

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BIO00004C - Part II

(i) Name the head group. (1 mark) (i) Choline (or Phosphocholine) Some students answered this question correctly. The different types of head group found in phospholipids were covered on slide 9 of Lecture 18

(ii) Using the alphanumeric system, name the unsaturated fatty acid. (1 mark) (ii) C18:1 Δ9 Most students recognised the unsaturated fatty acid and answered this question correctly. This naming system was covered in Lecture 18 and in the Supported learning workshop.

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