BSc and MSc Degree Examinations

Transcription

1 Examination Candidate Number: Desk Number: 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 The marks available for each question are indicated on the paper Instructions: Answer all questions in the spaces provided on the examination 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 11

2 1. a) A dilute solution contains a 3:1 molar ratio of base to acid with pka = 6.3. Calculate the ph. (3 marks) Use the Henderson-Hasselbalch equation, ph = pka + log ([A-]/[HA]) (1) ph= log ([base]/[ acid]), where [base]/[ acid] is (3:1) log (3/1)= 0.48; (1) ph = = 6.78 (1) Most students answered this question well. b) Convert 0.07 mol into μmol μmol = 10-6 mol, 70,000 μmol or 7 x 10 4 μmol Most students answered this question well. The figure below shows the titration curve for Glutamic acid, an amino acid which has a CH 2 CH 2 COOH side chain. Titration curve for Glutamic acid c) Determine the isoelectric point (pi) for Glutamic acid. pi= ( )/2 = 3.22 (1 mark for correct answer. 0.5 mark for correct method but incorrect answer.) d) Describe how the structure of Glutamic acid would be different at ph 1, at ph 3 and at ph 5. (3 marks) At ph 1, the amino group, terminal carboxyl group and side chain carboxyl group would all be protonated (NH + 3,COOH and COOH). At ph 3, the terminal Page 2 of 11

3 carboxyl group will deprotonate (NH + 3,COO - and COOH). At ph 5, the side chain carboxyl group will also deprotonate (NH + 3,COO - and COO - ). Some answers did not mention which carboxyl groups (whether terminal or side chain) were deprotonated at each ph. LO: 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) Protein X is a DNA binding protein. What are two possible types of molecular interactions between protein X and DNA? Like any two macromolecules, DNA can interact with proteins via several intermolecular interactions including H bonds and electrostatic interactions (2). Other answers are possible (e.g. hydrophobic interaction). Not acceptable as a correct answer are covalent bonds or ionic bonds (as they are not intermolecular interactions)... Some students mistook the question and named proteins that interact with DNA instead of the type of molecular interactions. b) The diagram below represents a replication fork. Label the 5 and 3 ends of the two diverging strands, the direction of replication fork movement, and the leading and lagging strands. Answer: Page 3 of 11

4 Several students labelled the strands in relationship to lagging and leading strands answered incorrectly. c) Circle the nucleoside. Answer: d) What does denaturation of dsdna involve? Breaking H bonds (possibly by an enzyme or high temperature) Most students answered this question correctly. Page 4 of 11

5 e) (i) In the dntp below, draw an arrow to the part of the molecule that you would modify to stop polymerisation from happening after incorporation of the dntp into DNA. Answer (ii) What would you would change this to? Change OH to H Many students answered this question well. Remember that the nucleotide is already incorporated in the chain and so it s the free 3 OH that needs to be modified and not the 5 PO 4. f) Name the enzyme that removes the lagging strand RNA primers in prokaryotes during DNA replication. DNA polymerase I Most students answered this question well. Partial credit was given to exonuclease as an answer (the question asks you to name the enzyme). LOs: 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 Page 5 of 11

6 3. Hydrogen bonding is an important intermolecular force found within proteins. a) Compare the pattern and the strength of hydrogen bonding in parallel and antiparallel beta sheets. The hydrogen bonds occur at an angle in parallel sheets but are directly aligned in antiparallel sheets. Students may alternatively draw a diagram to show this). The hydrogen bonds in parallel sheets are weaker than in antiparallel sheets b) State two other types of intermolecular forces found within proteins. 2 maximum from: Electrostatic interactions, Hydrophobic interactions (1 mark) van der Waals forces, Disulphide bonds LO: To be able to describe and explain how covalent and non-covalent interactions bring about the assembly of cellular components and macromolecules 4. The plot below shows an image of the Ramachandran plot. Ramachandran plot 1 a) Briefly explain what this plot shows. (3 marks) The plot is used to show energetically favourable conformational regions of protein structure. The axes show the polypeptide backbone dihedral or Page 6 of 11

7 torsion angles- psi and phi. Alpha helices and beta sheets are found in the allowed regions of the plot. The figures below show two further Ramachandran plots for synthetic proteins consisting of one repeated amino acid- poly(glycine) (plot 2) and poly(proline) (plot 3). Ramachandran plot 2 for poly(glycine) protein Ramachandran plot 3 for poly(proline) protein Page 7 of 11

8 b) Compare the data in plots 2 and 3 to the data shown in plot 1. Explain the differences seen. The allowable area is greater for Glycine (Figure 3) (0.5 mark) but more restricted for Proline (plot 3) (0.5 mark). Glycine has a H side chain which is flexible (0.5 mark). Proline has a ring side chain which is conformationally rigid and can cause more steric clashes (0.5 mark) LO: To be able to describe and explain how covalent and non-covalent interactions bring about the assembly of cellular components and macromolecules 5. DCG is an important tripeptide (Asp-Cys-Gly) found in mouse cells. Cys Gly Asp a) Using the amino acid structures above, Draw the structure of DCG below: 1 mark for the correct structure (below): In some answers, the peptide bonds between the amino acids were correctly drawn but the amino acids were not put in the correct order (Asp-Cys-Gly). Remember to begin your polypeptide chain with the N-terminus and end with the C-terminus. b) On your structure, label the chiral centres. Page 8 of 11

9 0.5 mark for each chiral centre (marked with asterisks above). 0 marks if C in glycine is marked. c) Explain the overall charge of DCG at physiological ph. Overall charge of -1 due to side chain carboxylic acid group of Asp d) DCG regularly forms homodimers. Draw the structure of a DCG homodimer below: 1 mark for the correct structure (below): This question caused some confusion. Those answers which showed the -S-Sdisulphide bond were awarded the mark. e) Suggest a technique to separate DCG from DCG homodimers in mouse cell extracts Size exclusion chromatography or Liquid Chromatography f) Recently, a DCG analogue called DSG (Asp-Ser-Gly) has been purified from human cells. It is believed that DSG is post-translationally modified. Suggest a possible post-translational modification. 1 mark for Phosphorylation or Glycosylation LO: 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 Page 9 of 11

10 6. a) You have calculated the extinction coefficient (ε 400 ) for p -nitrophenol as M -1 cm -1 from the data you have collected in an experiment. Calculate the rate (in μmoles.min -1 ) of a reaction with an absorbance reading of 0.45 over 3 minutes when 5 µl of 9 mg/ml enzyme has been used in an assay volume of 1.5 ml. (5 marks) Δ A/ Δ t= 0.45/3 = 0.15 (absorbance.min -1 ) correct equation (1) correct unit (1) 0.15/15= 0.01 mm.min -1 (1) 0.01 mm.min -1 x 1.5 ml= (μmoles.min -1 ) correct equation (1) correct unit (1) Few students answered this question completely. Remember the concentration of the enzyme does not play a role in the rate calculation in this question. b) In an enzyme-catalyzed reaction experiment, Vmax = 0.33 mol/sec and Km = 8 mm. Assuming the enzyme shows standard Michaelis-Menten kinetics, what is the rate of the reaction when [S] = 25 mm? v = Vmax[S]/(Km + [S]) (1) v = 0.33 x 25/(8 + 25) = 8.25/33= 0.25 mol/sec (1) Most students answered this question well. A mark was taken off if the answer did not include units and 0.5 mark was taken off for incorrect units. c) Briefly define Km and Vmax, then describe their dependence on enzyme concentration. (4 marks) V max can depend on enzyme concentration (1). V max reveals the turnover number of enzyme, which is the number of substrate molecules converted into product by an enzyme in a unit of time when the enzyme is fully saturated with substrate (1). V max is not an intrinsic property of the enzyme because it changes if you change the enzyme concentration. Km is not dependent on concentration of the enzyme and is an intrinsic property of the enzyme (1). Km defines the substrate concentration where V is half Vmax (1) Most students answered this question well. Some students defined Km as ½ Vmax, which is a measure of rate. But again Km is the concentration of substrate not the rate itself. LO: To be able to explain theoretical frameworks (such a Michaelis Menten kinetics, the laws of thermodynamics and the chemiosmotic theory) that allow us to understand function of biological molecules and cells 7. Page 10 of 11

11 a) What class of enzymes is used in biological washing powders to remove blood stains? Biological washing powders use proteases to remove stains. (1) Almost all students answered this question correctly. b) Describe what is meant by allosteric inhibition. Binding of an inhibitor at one site on a protein/enzyme which brings about an effect at a distance is allosteric inhibition. Most students answered this question well. c) What aspects of the mechanism of the catalytic triad of chymotrypsin is considered acid-base catalysis? Explain your answer. (4 marks) Serine is not normally chemically reactive (1), but is made reactive by another residue of the catalytic triad Histidine (1). When in close proximity (i.e. when the substrate is bound) His draws a proton from Ser and acts as a proton acceptor (a base) (1). Ser donates a proton; therefore, it is an acid (1). The final residue of the catalytic triad, Asp s carboxylic acid side chain can stabilise the newly positive charge of His. Many students did not answer this question completely. Mechanistic details, context, and which a.a. acts as a base and which acts as an acid are important when answering this questions. LOs: To be able to relate knowledge of biological molecules to health and disease and to their application in biotechnology 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 the space above this line should be sufficient for your answer Page 11 of 11