MBB 222 Problem Set 2 answers.pdf

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Molec Biol & Biochem
MBB 222
Frederic Pio

Problem Set 3 MBB 222 – Molecular Biology and Biochemistry Fall 2013 First Name: _____________________________________________________ Last Name: _____________________________________________________ Student Number: __________________________________ Tutorial Day and Time: ___________________________________________ Total marks for Problem Set =44 marks Your score = _______ INSTRUCTIONS st  Problem set is due on Friday, November 1 , 2013 in class.  Hand in your problem set to your Teaching Assistant for marking in class.  Write your answers in pen. Answers written in pencil will not be accepted.  Write CLEARLY! Zero marks will be given to questions with answers that are not legible!!  This problem set is worth 2% towards your final mark in the course. QUESTIONS A segment of a protein is provided below in which the R-groups specify each amino acid. Answer the following questions: a) Indicate with an arrow where chymotrypsin would cleave this segment of the polypeptide. (1 mark) b) Circle the atom that would form a transient covalent bond with chymotrypsin. (1 mark) c) Indicate which part of the polypeptide will be released first (P1), and which will be released second (P2) following chymotrypsin activity? (1 mark) 1 | P a g e ANSWER:  The arrow is at the peptide bond between phenylalanine and serine. (1 mark)  The carbon atom of the carbonyl group would form a transient covalent bond with chymotrypsin. (1 mark)  P1 would involve the polypeptide segment with serine (0.5 mark)  P2 would involve polypeptide segment with alanine, aspartic acid and phenylalanine. (0.5 mark) Which of the following statements is false? (2 marks) A. Hypocapnia leads to a left Bohr shift of the oxygen-dissociation curve which means that hemoglobin would have a high affinity for oxygen. B. Metabolic acidosis leads to a right Bohr shift of the oxygen-dissociation curve which means that hemoglobin would readily release oxygen. C. At peripheral tissues, fetal hemoglobin has a higher affinity for oxygen compared to adult hemoglobin. D. A drop in body temperature leads to left Bohr shift of the oxygen-dissociation curve. E. An decrease in 2,3-biphosphoglyercate leads to a greater release of oxygen from hemoglobin at the peripheral tissues. ANSWER: E (2 marks)… An increase in 2,3-biphosphoglyercate leads to a greater release of oxygen from hemoglobin at the peripheral tissues. Which of the following statements is true? (2 marks) A. An enzyme with a low k catnd high K ms more efficient in metabolizing substrate into product than an enzyme with a high k catnd low K m. B. A single molecule of 2,3-biphosphoglyercate binds preferentially to the “relaxed” state of hemoglobin compared to the “tense” state. C. Oxygen binds reversibly to the heme group of the multimeric protein myoglobin. D. During the catalytic mechanism of chymotrypsin, a stronger electrophile in the form of a reactive alkoxide ion is generated after His57, a base catalyst, abstracts a proton from the hydroxyl group of Ser195. E. In chymotrypsin, the role of the oxyanion hole is to bind the substrate carbonyl group in a position that helps form the transition state. ANSWER: E. (2 marks) 2 | P a g e If 20 µg of pure enzyme (Mr30,000) working at Vmaxcatalyzes the conversion of 100 µmol of substrate into product in 2 minutes, what is the turnover number (k ) of this enzyme (in units of min ). (2 marks for the work cat + 1 mark for answer = 3 marks in total) ANSWER: Amount of enzyme is 20 x 10 g, which is (20 x 10 g)/(3 x 10 g/mol) = 6.67 x 10 -10mol of enzyme. The rate of product formation is 100 x 10 mol/ 2 min = 50 x 10 mol of product per minute. The -6 -10 4 -1 turnover number is therefore (50 x 10 mol/min)/(6.67 x 10 mol of enzyme) = 7.5 x 10 min . (1 mark for answer + 2 marks for work) The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: V 0 Substrate added (µmol/min) (mmol/L) ————————————— 217 0.8 325 2 433 4 488 6 647 1,000 ————————————— The K mor this enzyme is approximately (2 marks): A. 1 mM B. 1000 mM C. 2 mM D. 4 mM E. 6 mM ANSWER: C (2 marks) The deoxyribonucleotide polymer (5')GTGATCAAGC(3') could only form a double-stranded structure with (2 marks): A. 5'-CACTAGTTCG-3' B. 5'-CACUAGUUCG-3' C. 5'-CACUTTCGCCC-3' D. 5'-GCTTGATCAC-3' E. 5'-GCCTAGTTUG-3' ANSWER: D (2 marks) The composition (mole fraction) of one of the strands of a double-helical DNA is [A] = 0.30, and [G] = 0.24. Calculate the following, if possible. If impossible, write “I.” (0.5 mark each x 8 = 4 marks) For the same strand:  [T] = _______  [C] = _______  [T] + [C] = _______ 3 | P a g e For the other complementary strand:  [A] = _______  [T] = _______  [A] + [T] = _______  [G] = _______  [C] = _______ ANSWER: (a) I; (b) I; (c) 0.46; (d) I; (e) 0.30; (f) I; (g) I; (h) 0.24 (0.5 mark each) A closed circular DNA molecule has 3675 base pairs and a topological linking number (Lk) of 345. Use 10.5 base pairs per turn in answering the following questions: a. What would the writhe number (Wr) be for this molecule in order for it to optimize its base pairing? Show your work too. (1 mark) b. Is the writhe right- or left-handed? (1 mark) c. Is the molecule positively or negatively supercoiled? (1 mark) d. How could you increase the Lk to 346? (1 mark) e. How would the Lk change if the DNA molecule was denatured by heat? (1 mark) ANSWER:  The molecule would have 3675 bp/10.5 base pairs per turn or 350 turns to optimize its base pairing. Since Lk = Tw + Wr, 345 = 350 +Wr, and Wr = -5. (1 mark)  Since the writhe number is negative it is right-handed. (1 mark)  Since the writhe number is right-handed, the molecule is negatively supercoiled. (1 mark)  Treat the DNA to one cycle of topoisomerase I, which will increase Lk by 1 and partially relax the supercoiling. (1 mark)  The linking number would not change since heat only disrupts non-covalent interactions so the strands would remain intact. (1 mark) If DNA replication were conservative (i.e., daughter DNA was comprised of two newly-synthesized strands of DNA), what would sort of bands would you expect from CsCl density gradient centrifugation in the Meselson- 14 Stahl experiment after one generation in N media? When addressing the bands think about the number of bands that will be seen, the density of each band and what the bands represent. (3 marks) ANSWER: There would be 2 bands (1 mark), one more dense band (0.5 mark) corresponding to the parent 15 DNA with all N (0.5 mark), and one les
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