UNIVERSITY OF YORK BSc Stage 2 Degree Examinations Department: BIOLOGY. Title of Exam: Cell Biology

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1 Examination Candidate Number: Desk Number: UNIVERSITY OF YORK BSc Stage 2 Degree Examinations Department: BIOLOGY Title of Exam: Cell Biology Time allowed: 1 hour and 30 minutes Total marks available for this paper: 80 This paper has two parts: Section A: Short answer questions (40 marks) Answer all questions in the spaces provided on the examination paper Section B: Problem questions (40 marks) Answer all questions in the spaces provided on the examination paper Instructions: The marks available for each question are indicated on the paper A calculator will be provided For marker use only: Total as % DO NOT WRITE ON THIS BOOKLET BEFORE THE EXAM BEGINS DO NOT TURN OVER THIS PAGE UNTIL INSTRUCTED TO DO SO BY AN INVIGILATOR SECTION A: Short answer questions Answer all questions in the spaces provided page 1 of 14

2 Mark total for this section: 40 Q1. a) Compare and contrast paracellular and transcellular transport (2 marks) In paracellular transport substances cross a cell layer by passing through the intercellular space between the cells (1 mark) whereas in transcellular transport substances cross by passing through cells, crossing both apical and basolateral membranes (1 mark). This was generally answered well. b) Give an example of where in the body each is used (2 marks) Examples given in the lecture for paracellular transport = magnesium and calcium ions crossing kidney epithelia (1 mark) for transcellular = absorption of glucose across intestinal epithelium (1 mark) Most common reason for dropping marks was to mention which substances are transported by para/transcellular transport, but fail to mention where in the body these processes are used. Some answers mentioned synapses as using paracellular transport - this is incorrect. c) Discuss how key features of tight junctions regulate paracellular and transcellular transport across epithelial tissue. (4 marks). Central to paracellular transport is the fact that tight in junctions do not form an impermeable barrier between epithelial cells, but rather act as molecular sieves (1 mark). Specificity (e.g. size, charge of molecules that can pass) conferred by different claudins present in the tight junction (1 mark). Central to transcellular transport is the role tight junctions play in maintaining the separate identity of the apical and basolateral membranes (membrane proteins can t cross the barrier that they create), allowing the two to function as separate entities (2 marks). This was generally answered well, although some answers confused paracellular transport across an epithelial layer being used to transport electrical current with synchronisation of cardiac myocytes; a function of gap junctions. LO: Describe the structure and function of different types of cell adhesion page 2 of 14

3 molecules and complexes. Q2. The figure below represents a flow cytometry analysis of a mixed population of cells: a) Annotate the figure to indicate which phases of the cell cycle are represented by the labels A, B and C. (3 Marks) A= G1 Phase (1 Mark), B= S Phase (1 Mark), C=G2 (and/or M) Phase (1 Mark) Most people got this correct. b) If the population of cells above were normal human cells, what ploidy would the cells in phase A and Phase C be? (2 Marks) page 3 of 14

4 A are 2N (diploid) (1 Mark). C are 4N (1Mark). Most people got this correct. c) The cells above are a mixed population, but it is often useful to synchronise a population of cells so that they are all in the same phase of the cell cycle. Name two ways of achieving natural synchrony of the cell cycle in cultured mammalian cells. (2 Marks) Growing cells until they experience quiescence due to contact inhibition (1 Mark), or withdrawing serum from the cell culture medium (1 Mark). Most people got this question correct. Some struggled to name contact inhibition or serum withdrawal specifically. LO: Explain the mechanisms involved in the control of the cell cycle. Q3. a) What would the effects of N-ethylmaleimide be on membrane traffic? (1 mark) Inhibition/block (by inhibiting NSF) Most people gained full marks here b) Give an example of a protein tethering complex (1 mark) HOPS, CORVET, GARP, COG etc Most people gained full marks here c) If a synaptic vesicle membrane contains Qa, Qb and R SNAREs, what SNARE classification type would be required for a functional SNARE complex and in what membrane would it have to be for neurotransmitter release (2 marks). Qc SNARE (1 mark) in the plasma membrane/synaptic cleft membrane (1 mark) Most people gained full marks here LO: Describe secretory and endocytic pathways within in a cell. Q4. a) Which amino acid side chain typically provides a ubiquitin acceptor site within a polypeptide? (1 mark) page 4 of 14

5 lysine Most people gained full marks here b) If a mutation in a cell caused the rate of deubiquitination to be higher than ubiquitination, what would be the consequences on: i) mitochondria (2 marks) Damaged mitochondria would not be cleared by mitophagy (1mark) and this would lead to an increase in ROS and which would lead to apoptosis (1 mark) Mitophagy wasn t mentioned by a lot of people and the outcome of having damaged mitochondria. ii) Epidermal growth factor receptor (2 marks) Receptor would not be endocytosed and the receptor would not be downregulated (1 mark) potentially leading to uncontrolled growth (1 mark). Most people mentioned that the receptor would not be degraded but the majority failed to say what the biological consequence would be. LO: Describe secretory and endocytic pathways within in a cell. Q5. Outline the molecular mechanisms involved in Fas-receptor mediated destruction of influenza-infected cells. (5 marks) Following docking of cytotoxic T-cells onto infected cells (achieved through recognition by the T cell receptor of viral peptides presented in complex with the infected cells histocompatibility complex). i) FasL on the cytotoxic T-cell recognises the Fas-receptor on the target cells (1) ii) Fas/FasL recognition results in death receptor signalling procaspase 8 activation (DISC) (1) iii) leading to other caspase (3/7) activation (1). iv) Also Bid is cleaved to tbid which releases cyt-c from mitochondria and cas-9 activation (1). v) All leading to apoptosis of viral infected cell (1). Most people gained full marks here LO: Provide an overview of apoptosis and its role in health and disease. Q6. Draw and label the basic structure of a type I collagen molecule. (3 marks) Sketch needs to show: Triple helical structure of alpha chains (1) Amino acid sequence Gly-X-Y repeats (1) X and Y often Proline, Hydroxyproline or Hydroxylysine (1) page 5 of 14

6 Examples LO: Explain the composition and function of the extracellular matrix. Feedback: There were quite a few different and interesting interpretations offered in the answers, marks were given as above. Marks also award for other correct annotations, e.g. glycosylation. Q7. An experimental protocol is being designed to differentiate induced pluripotent stem cells (ipscs) into corneal tissue to help treat impaired eyesight. a) Provide one advantage and one disadvantage of using ipscs compared to adult limbal stem cells. (2 marks) Two from: Advantages: High proliferative/telomerase activity - plentiful cell supply. Autologous source, earlier developmental stage. Disadvantages: Teratoma risk (pluripotent), technically more difficult. Possibly viral transduction risk, low generation efficiency. Feedback: Quite a number of answers gave pluripotency as an advantage. In much more general biological terms, pluripotency could be considered an advantage if differentiation into multiple cell types was required, but not in this case. Disadvantages were mostly correct. b) Part of the protocol involves growing the ipscs on different extracellular matrix (ECM) substrates, namely type IV collagen and heparan sulfate proteoglycans. Give reasons why these particular ECM substrates were chosen. (3 marks) Example marks: Type IV collagen (and heparan sulfate proteoglycans) is a component of the basement membrane (1) and forms a flat, planar structure, page 6 of 14

7 suitable for corneal development (1). Heparan sulfate proteoglycans are able to modify growth factor function (e.g. FGFs, Wnt signalling) to enhance corneal differentiation (1) Feedback: Answered well. Needed to explain why these particular substrates were used, not ECM generally to the cells could adhere. c) How would you determine whether the ipscs had differentiated into corneal cells in vitro? (3 marks) Example marks: Expression of corneal markers (1), which was not specifically covered in the lectures, but could suggest keratins (1) by for example immunocytochemistry/western blotting (1). Microscopy/morphology examination would be less satisfactory, but mechanical and translucency testing are valid (1). This question covers material from different lectures and requires ideas and concepts to be integrated. LOs: Describe the properties and regulation of embryonic and adult stem cells. Provide examples of how stem cells may be used in therapy. Feedback: Some good answers. The main reasons for losing marks were not testing differentiation, missing techniques or describing in vivo rather than in vitro assays. The teratoma assay would not be appropriate but some marks were given for in vivo testing corneal function. page 7 of 14

8 SECTION B: Problem questions Answer all questions in the spaces provided Mark total for this section: 40 Q8. You have developed an assay to follow formation of actin filaments from purified actin monomers. Your assay starts with purified actin monomers labeled with a fluorescent probe. Upon polymerization the fluorescence of the probe increases, allowing polymerization to be measured. Data from a typical time course is shown and plotted on the left hand graph. The right hand graph shows the equilibrium distribution of actin in free subunits (monomers) and in filaments, as a function of actin concentration. a) While optimizing conditions for the assay you establish a requirement for a physiological buffer (ph7.4) and the addition of another molecule in order for the actin monomers to form filaments. What is that molecule and why is it required here? (2 marks) ATP (1 mark) Nucleotide binding is required for structural integrity of G-actin (1 mark This was generally answered well, especially the first part (ATP), the most common reason for dropping marks in the second part was stating that ATP was required as actin is an ATP-ase, while it is true that actin is an ATPase, this activity is NOT required for actin polymerisation. b) Speculate as to why the intensity of fluorescence at time zero is not zero. (1 mark) page 8 of 14

9 Background fluorescence of the actin monomers (speculation that nucleation occurs instantaneously would gain half marks, not full because if that were the case then one would expect that lag phase to show a slight increase in signal over time). This was generally answered well, although some answers appeared to be based upon the polymerisaiton being followed inside a cell whereas the question states that the assay follows formation of actin filaments from purified actin monomers. c) What stages of actin filament assembly do the three phases of the polymerization curve on the left hand graph reflect? (3 marks) (i) Nucleation (1 mark) (ii) growth phase/elongation (1 mark) (iii) equilibrium where rate of addition of actin at filament ends = rate of release/treadmilling (1 mark) This was generally answered well. d) Estimate the critical concentration of actin polymerization for this experiment. (1 mark) 0.1 mg/ml (1 mark) This was generally answered well - most common incorrect answer was 0.2mg/ml. e) How would the curve on the left change if you doubled the initial concentration of actin monomers used in the assay? Would the concentration of free (monomeric) actin at equilibrium be higher or lower than in the original experiment? (4 marks) If the starting concentration of actin were doubled, the lag phase during which nucleation occurs would be shorter (1 mark), the growth phase would be more rapid (steeper) (1 mark), and the mass of polymer at equilibrium would be twice as great (1 mark)- N.B. students do not need to have correctly answered either part A) or part B) to get these marks as stating that (i) would be shorter, (ii) would be steeper and the plateau of (iii) would be twice as high would suffice. The concentration of free actin monomers at equilibrium the critical concentration ( C c) would be the same regardless of the initial actin concentration (1 mark). All parts of this question were answered well, although not many answers gave all the required information. LO: Explain the structure and function of the cytoskeleton. page 9 of 14

10 f. At 1.4 mg/ml pure tubulin, microtubules formed from 13 protofilaments grow at a rate of 2 μm/min. At this growth rate how many αβ-tubulin dimers (8 nm in length) are added to the ends of a microtubule each second? (4 marks). A growth rate of 2 μm/min (2000 nm/60 sec = 33 nm/sec) corresponds to the addition of 4.2 αβ-tubulin dimers [(33 nm/sec) X (αβ-tubulin/8 nm) = 4.17 dimers/sec] to each of the 13 protofilaments, or ~54 αβ-tubulin dimers/sec to the ends of a microtubule. This was generally answered well - the most common reason for dropping a mark was to give the answer of 4.17 (or 4) and not multiply this by 13 protofilaments. LO: Explain the structure and function of the cytoskeleton. Q9. A patient arrives in the clinic and examination of the patient s blood identifies high levels of the lysosomal hydrolase cathepsin-b. a) Suggest two possible explanations for this observation. (2marks) Lysosome exocytosis (1 mark) or mis-sorting of the hydrolase (1 mark) For those that attended the workshop most people recognised that mutations in the M6PR or tagging of the enzyme would result in the observed phenotype. b) The cathepsin-b protein from the patient and a normal individual were subjected to SDS-PAGE and transferred to two membranes. One membrane was western blotted for cathepsin-b (image on the left) and the other membrane was incubated with a mannose-binding lectin and the lectin visualised (image on the right). Explain the observations and how this causes elevated cathepsin-b in the patient s blood. (4 marks). Lack of lectin binding suggests that the protein does not contain a mannose (1 mark). The fact that the cathepsin B from the patient is smaller would suggest that it does not have a mannose group (1 mark). It is likely that the cathepsin B has not had a mannose-6 Phosphate (1 mark) added page 10 of 14

11 and therefore not sorted to the lysosome and is therefore secreted instead (1 mark). This was either answered well (certainly for those that obtained full marks in part a) or not at all. The biggest reason for losing marks was not reading the question and students talking about a lectin receptor. Lectin is used to probe the western blot and wasn t a receptor in the cell. LO: Describe secretory and endocytic pathways within in a cell. c) When cells from the patient were treated with rapamycin, a drug which can induce autophagy, no effects were seen on the availability of nutrients. Given also the data above, suggest what might be wrong with the patient. (3 marks). This question was not answered well. Primarily marks were lost since students didn t refer to part b, since the question was asking given the data above. Suggests that nutrients are not released by the lysosome (1 mark). Lack of glycosylation leads to lack of hydrolases targeting to the lysosome (1 mark), leading to non-functional lysosomes (1 mark). Student may mention glycosylation defect which leads to lysosome defect. Q10. You incubate adipose cells with insulin, which promotes exocytosis of the glucose transporter GLUT4. The extracellular concentration of glucose, before insulin addition, is 10 mm. The cells after insulin addition, remove 50% of the extracellular glucose per minute. a) Plot a graph of extracellular glucose concentration vs time for the 5 minutes after the addition of insulin. (4 marks) (INSERT 6X6 square grid, with aprox 2cm space around grid) page 11 of 14

12 1 mark for each labelled axis and 2 marks for correctly plotted graph (1 for decrease - one for shape) The majority of students gained full marks here. The binding of insulin to its receptor causes the receptor to be endocytosed and degraded by the lysosome. To increase insulin sensitivity and subsequently glucose uptake, cells were treated with various compounds with the aim to inhibit insulin-receptor endocytosis. Glucose uptake was measured as above. The following table shows the drugs used and the rate of glucose uptake. Compound t ½ glucose uptake (min) Dynasore (inhibits dynamin) 0.5 Nocodazole 1 Vanadate (phosphatase inhibitor) 2 SP Pitstop-1 (clathrin inhibitor) 1 b) If you assume that the glucose uptake is wholly dependent upon the levels of the active cell-surface insulin-receptor what can you conclude about how the insulin-receptor is endocytosed? (5 marks) Most students failed to gain marks in what was meant to be a difficult question. The previous graph was there to indicate that the rate of glucose uptake (t1/2) was 1 min. If the t1/2 was 0.5 then this means glucose uptake is quicker, not slower. Likewise a t1/2 of 2 meant that glucose uptake was slower, not quicker. A lot of students had this the wrong way around. A lot of marks were also lost since students suggested correctly that nocodazole, SP and Pitstop-1 didn t do anything but didn t say that this meant that microtubules, JNK kinase and clathrin are not involved. Internalization requires dynamin (1 mark) it is independent of microtubules (1 mark), internalization requires phosphorylation (1 mark) but not JNK (1 mark) and is independent of clathrin (1 mark). LO: Describe secretory and endocytic pathways within in a cell. Q11. In an experiment, you continuously label all proteins in a batch of sea urchin eggs with a radioactive label and take samples every 10 minutes after fertilisation. You then run these samples on a gel using electrophoresis. page 12 of 14

13 You see a band ( BAND X ) that increases in intensity up to the 40 minute time point, but then rapidly and significantly reduces in intensity (indicated by the square brackets and label). The result is in the figure below. a) Give two possible reasons why BAND X drops in intensity after 40 minutes (when the sea urchin eggs enter M phase). (2 Marks) The protein it represents is either rapidly turned over / destroyed (1 mark) or is no longer produced (1 mark) This question was mostly answered well. b) What is the most likely name of the protein represented by BAND X? (1 Mark) Cyclin (1 mark) Most answered this question well. MPF was not credited as an answer, as this is a cyclin-cdk complex and migrates as two bands. You repeat the experiment, but this time you label the cells every ten minutes. You take samples and process them as before. You see a band ( BAND Y ) which represents the same protein as BAND X. However, this time the band is present at the same intensity at all time points. The result is in the figure below: page 13 of 14

14 c) What is the name of this labelling technique? (1 Mark) Pulse labelling (1 mark) Most people answered correctly. d) Explain why the different labelling techniques produce different results. (2 Marks) Pulse labeling reveals the rate of cyclin production (0.5 Marks), which is constant (0.5 marks). Whereas continuous labelling reveals the total amount of cyclin at any time (0.5 Marks), which varies (0.5 Marks). Many people struggled with this question meaning relatively few got full marks. However, few people got 0 marks either. Most people appreciated the difference in the proportion of protein labelled between the two techniques, but relatively few people then linked this back specifically to what was happening in the experiments presented. e) Given the results of the second experiment, which of the reasons given in (a) for the drop in intensity in the first experiment is correct? (1 Mark) The protein is being periodically destroyed / turned over (1 mark) Most people got this correct. LO: Explain the mechanisms involved in the control of the cell cycle. page 14 of 14