BA, BSc, and MSc Degree Examinations

Transcription

1 Examination Candidate Number: Desk Number: BA, BSc, and MSc Degree Examinations Department : BIOLOGY Title of Exam: Molecular Biology and Biochemistry Part I Time Allowed: 1 hour and 30 minutes Marking Scheme: Total marks available for this paper: 50 Instructions: Answer all ques ons in the spaces provided on the examina on paper The marks available for each ques on are indicated on the paper Materials Supplied: CALCULATOR For marker use only: Office 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 Page 1 of 9

2 Answer all questions in the spaces provided 1. a) Calculate the molar concentration (in mm) and the number of molecules in a solution in which 138 mg of Hydrofluoric acid (atomic mass ~ 20.01) has been dissolved in 15 L of water. Show your working. 138 mg =0.138 g (0.5 mark) 0.138(g)/15 (L) (0.5 mark) = (g/l) / 20.01(g/mole) (0.5 mark) = M (0.5 mark) Molar concentration =0.46 mm Number of molecules= Mx15L x 6x10^23 = 4.15 x10^21 (2 mark) Many students answered the first part of this question correctly; however, quite a few ignored the second half. For many students conversion of units was a challenge. b) You have 45 g of acetic acid (pka=4.75). How many moles of sodium acetate do you need to add to produce a buffer with a ph of The atomic masses for acetic acid and sodium acetate are and g/mol respectively. ph = pka + log ([A - ]/[HA]) 45g/60.01 g/mol= 0.75 moles of acetic acid ph = pka + log ([acetate]/[acetic acid]) 5.75 = log ([moles of acetate]/0.75) 1= log ([moles of acetate]/0.75) 1= log [moles of acetate] - log [0.75] 0.88= log [moles of acetate] moles of acetate= 7.59 moles Response to this question was variable. Whilst many students answered this question correctly; unit conversions and/or use of the correct equation (Henderson-Hesselbalch) were problematic for others. Learning outcomes: Basic calculations related to acid-base chemistry To be able to apply quantitative approaches to perform basic calculations related to acid-base chemistry. 2. a) What is the direction of DNA replication and what are the implications for how the lagging strand is synthesized? What are these products called? (3 marks) Page 2 of 9

3 5 to 3. Synthesis is discontinuous and requires the production of Okazaki fragments (2 marks). Almost all students answered this questions correctly. b) Which carbon in the ribose of RNA distinguishes it from DNA? How? 2 C, which is attached to an OH instead of H (in DNA) Almost all students answered this questions correctly. c) Name the type of chemical bond between the bases of double stranded DNA. H bond Almost all students answered this questions correctly. d) What is the hydrated form of DNA helix known as? B-DNA Response to this question was variable e) Which enzyme catalyses the synthesis of DNA? DNA polymerase Almost all students answered this questions correctly. f) What are the condensed, repetitive DNA regions near the centromeres and telomeres called? Heterochromatin Several students struggled with this question. Learning outcomes: To be able to describe and explain how covalent and non-covalent interactions bring about the assembly of cellular components and macromolecules To be able to describe the main chemical components of cells, their structural properties, how these relate to their functions, and how they are altered during cellular processes 3. a) All naturally occurring amino acids, except glycine, have at least two possible configurations at the α-carbon. Explain what is meant by having two different configurations. Two different arrangements of bonds within the same compound that cannot be interconverted without breaking covalent bonds. Page 3 of 9

4 b) Given the structure of glycine is: explain why glycine is the only amino acid that has only one configuration. The central carbon atom of glycine has only three different substituents. c) Glycine also displays more conformational flexibility than other amino acids. Explain why this is the case. (2 marks) The R group is hydrogen which reduces steric clashes between glycine and nearby residues. To be able to describe the main chemical components of cells, their structural properties, how these relate to their functions, and how they are altered during cellular processes 4. a) Indicate on the diagram below the position of one of the peptide bonds: Any one of the four CO-NH bonds. Page 4 of 9

5 b) Why is the peptide bond planar? It has partial double bond character and double bonds cannot rotate. To be able to describe the main chemical components of cells, their structural properties, how these relate to their functions, and how they are altered during cellular processes. To be able to describe and explain how covalent and non-covalent interactions bring about the assembly of cellular components and macromolecules 5. a) How does the hydrogen bonding found within β sheets differ from that found within α helices? (2 marks) α helices have repeating hydrogen bonding between residues within the same helix whereas β sheets have repeating hydrogen bonding between adjacent strands. b) Sketch out a hydropathy plot for a protein having a single transmembrane α helix flanked by two very hydrophilic domains. 0.5 marks for correct pattern and 0.5 marks for correct y axis. c) If the α helices within a membrane spanning protein contained many residues with ionisable side chains what might you hypothesise about the function of this protein? Page 5 of 9

6 The protein might form a channel through the membrane. To be able to describe the main chemical components of cells, their structural properties, how these relate to their functions, and how they are altered during cellular processes. To be able to describe and explain how covalent and non-covalent interactions bring about the assembly of cellular components and macromolecules 6. a) What are coenzymes and why are they important for enzyme function? (2 marks) They are small organic compounds that act as transient carriers of specific functional groups. b) Explain the structural role of Zn 2+ ions in zinc finger DNA binding proteins and how this facilitates binding to DNA. (3 marks) Zn 2+ ions are coordinated by four amino acids within a 30 residue region (1 mark), stabilising this polypeptide loop that can then bind within the major groove of DNA. c) Ni 2+ ions can be exploited to help purify proteins. How? (2 marks) The gene of a target protein can be engineered to encode a histidine tag (1 mark) and these multiple histidines can then specifically bind Ni 2+ ions immobilised on a column matrix. To be able to describe the main chemical components of cells, their structural properties, how these relate to their functions, and how they are altered during cellular processes. To be able to describe and explain how covalent and non-covalent interactions bring about the assembly of cellular components and macromolecules. To be able to relate knowledge of biological molecules to health and disease and to their application in biotechnology Page 6 of 9

7 7. a) A series of assays using 2.87 μg ml -1 of an enzyme was carried out in a volume of 10 μl with varying substrate concentrations to measure the reaction rate (U = μmol.min -1 ). The measurements were used to generate the graph below. Use these data to determine the V max of the enzyme in units of specific activity. Ans: 2.87 μg/ml= μg/μl * 10 μl= μg= mg y = x y intercept = V max = 1/10.902= V max = 0.092(U -1 )/ (mg) V max = μmol.min -1.mg -1 Specific activity (V max ) = μmol.min -1.mg -1 Response to this question was variable. Many students did not report Vmax in units of specific activity. b) Circle three assumptions made concerning Michaelis-Menten kinetics. (3 marks) NB. Each correct answer scores +1 mark, each incorrect answer -1 mark, and unanswered questions score 0. i) substrate is available in very large amounts compared to enzyme Page 7 of 9

8 ii) enzyme is available in very large excess compared to substrate iii) the enzyme-substrate complex is not at equilibrium iv) product formation is irreversible v) product-substrate complex formation is a requirement vi) the enzyme substrate complex is at steady state vii) the induced-fit model is required to derive the reaction by stabilizing the transition state Correct answers: i, iv, vi (1 mark each) Many students answered this question correctly. Learning outcomes: To be able to explain theoretical frameworks (such as Michaelis Menten kinetics, the laws of thermodynamics and the chemiosmotic theory) that allow us to understand function of biological molecules and cells 8. a) What is the difference between competitive and non-competitive inhibition? Draw diagrams to support your answer. Ans: Non-competitive inhibition (2 marks each diagram). vs. Competitive inhibition green triangles = inhibitor Almost all students answered this question correctly. Page 8 of 9

9 b) How is aspartate transcarbamylase inhibited by cytidine triphosphate (CTP)? Is this competitive or non-competitive inhibition? CTP binds to regulatory subunit (r) dimers. This alters the conformation at the interface between catalytic c subunits. The active site is between monomers and so this long distance conformational change prevents substrate binding. Non-competitive Many students answered this question correctly, but a sizable minority did not give the right level of detail. Some students answered feedback inhibition. With the correct level of mechanistic detail, some marks were awarded, but the question specifically asks whether or not this inhibition is competitive or non-competitive. c) Suggest two advantages of using enzymes rather than chemicals as catalysts in biotechnological processes. (2 marks) Any 2 of: Easy to manipulate, source of wide variety of enzyme activities, reproducibility, relatively low cost Almost all students answered this question correctly. Learning outcomes: To be able to describe the main chemical components of cells, their structural properties, how these relate to their functions, and how they are altered during cellular processes To be able to relate knowledge of biological molecules to health and disease and to their application in biotechnology Page 9 of 9