Name: Student Number

UNIVERSITY OF GUELPH CHEM 454 ENZYMOLOGY Winter 2003 Quiz #1: February 13, 2003, 11:30 13:00 Instructor: Prof R. Merrill Instructions: Time allowed = 80 minutes. Total marks = 34. This quiz represents 15% of the final grade. Please write all your answers in ink (not red ink). No examination materials may be removed from the examination room. Answers to parts A and B are to be written directly on this question booklet. Answers to part C are to be written in the answer booklet provided. Part A. Multiple-choice questions; circle the letter corresponding to the best answer. Sixteen (16) questions x 1 mark per question = 16 marks total. No marks will be deducted for incorrect answers. 1. Enzymes are potent catalysts. They: A) drive reactions to completion while other catalysts drive reactions to equilibrium. B) are consumed in the reactions they catalyze. C) are very specific and can prevent the conversion of products back to substrates. D) increase the equilibrium constants for the reactions they catalyze. E) lower the activation energy for the reactions they catalyze. 2. Which of the following statements is true of enzyme catalysts? A) To be effective, they must be present at the same concentration as their substrate. B) They can increase the equilibrium constant for a given reaction by a thousand-fold or more. C) They lower the activation energy for conversion of substrate to product. D) Their catalytic activity is independent of ph. E) They are generally equally active on D and L isomers of a given substrate. 3. The role of an enzyme in an enzyme-catalyzed reaction is to: A) ensure that the product is more stable than the substrate. B) make the free-energy change for the reaction more favorable. C) increase the rate at which substrate is converted into product. D) ensure that all the substrate is converted to product. E) do none of the above. 4. Which of the following statements is false? A) A reaction may not occur at a detectable rate even though it has a favorable equilibrium B) At the end of an enzyme-catalyzed reaction, the functional enzyme becomes available to catalyze the reaction again C) Substrate binds to an enzyme s active site D)For S P, a catalyst shifts the reaction equilibrium to the right E) Lowering the temperature of a reaction will lower the reaction rate 1

5. The concept of "induced fit" refers to the fact that: A) when a substrate binds to an enzyme, the enzyme induces a loss of water (desolvation) from the substrate. B) substrate binding may induce a conformational change in the enzyme, which then brings catalytic groups into proper orientation. C) enzyme-substrate binding induces an increase in the reaction entropy, thereby catalyzing the reaction. D) enzyme specificity is induced by enzyme-substrate binding. E) enzyme-substrate binding induces movement along the reaction coordinate to the transition state. 6. The benefit of measuring the initial rate of a reaction, V 0, is that at the beginning of a reaction: A) changes in [S] are negligible, so [S] can be treated as a constant. B) [ES] can be measured accurately. C) V 0 = V max. D) changes in K m are negligible, so K m can be treated as a constant. E) varying [S] has no effect on V 0. 7. Which of the following statements about a plot of V 0 vs. [S] for an enzyme that follows Michaelis-Menten kinetics is false? A) K m is the [S] at which V 0 = 1/2 V max. B) The shape of the curve is a hyperbola. C) The y-axis is a rate term with units of µμ/min. D) As [S] increases, the initial velocity of reaction, V 0, also increases. E) At very high [S], the velocity curve becomes a horizontal line that intersects the y- axis at K m. 8. The steady state assumption, as applied to enzyme kinetics, implies: A) K m = K s B) The maximum velocity occurs when the enzyme is saturated C)The ES complex is formed and broken down at equivalent rates D) The K m is equivalent to the cellular substrate concentration E) The enzyme is regulated 2

9. 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 m for this enzyme is approximately: A) 1 mm. B) 2 mm. C) 4 mm. D) 6 mm. E) 1,000 mm. 10. The double-reciprocal transformation of the Michaelis-Menten equation, also called the Lineweaver-Burk plot, is given by 1/V 0 = K m /(V max [S]) + 1/V max. To determine K m from a double-reciprocal plot, you would: A) take the x-axis intercept where V 0 = 1/2 V max. B) take the reciprocal of the x-axis intercept. C) multiply the reciprocal of the x-axis intercept by 1. D) take the reciprocal of the y-axis intercept. E) multiply the reciprocal of the y-axis intercept by 1. 11. To calculate the turnover number of an enzyme you need to know the: A) initial velocity of the catalyzed reaction at low [S]. B) initial velocity of the catalyzed reaction at [S] >> K m. C) K m for the substrate. D) enzyme concentration. E) both B and D. 12. In competitive inhibition, an inhibitor: A) binds at several different sites on an enzyme. B) binds reversibly at the active site. C) binds only to the ES complex. D) binds covalently to the enzyme. E) lowers the characteristic V max of the enzyme. 3

13. V max for an enzyme-catalyzed reaction: A) generally increases when ph increases. B) increases in the presence of a competitive inhibitor. C) is unchanged in the presence of a uncompetitive inhibitor. D) is twice the rate observed when the concentration of substrate is equal to the K m. E) is limited only by the amount of substrate supplied. 14. Both water and glucose share an OH that can serve as a substrate for a reaction with the terminal phosphate of ATP catalyzed by hexokinase (my whippin' boy). Glucose, however, is about a million times more reactive as a substrate than water. The best explanation is that: A) glucose has more OH groups per molecule than does water. B) the larger glucose binds better to the enzyme; it induces a conformational change in hexokinase that brings active-site amino acids into position for catalysis. C) water normally will not reach the active site because it is hydrophobic. D) water and the second substrate, ATP, compete for the active site resulting in a competitive inhibition of the enzyme. E) the OH group of water is attached to an inhibitory H atom while the glucose OH 15. What is the name of the enzyme catalyzing the following reaction: Dolichyl phosphate D-mannose + protein Dolichylphosphate + D-mannosyl-protein A) Dolichyl-phosphate-D-mannose-protein ligase B) Dolichyl-phosphate-D-mannose-protein mannosyltransferase B) Dolichyl-phosphate-D-mannose-protein transferase C) Dolichyl-phosphate-D-mannose isomerase 16. The catalytic power of an enzyme is: A) is the ratio of k cat /K M B) the number of substrate molecules product per min per mole enzyme C) the specificity factor for a given substrate D) a measure of the ability of an enzyme to increase the rate of a reaction 4

Part B. Short answer/calculation questions. Answer the following (five) questions with a short answer consisting of a few sentences or less (your answer can be in point form). Each question is worth 2 marks. Four (4) questions x 2 marks each = 8 marks total. 17. An enzyme can catalyze a reaction with either of 2 substrates, S 1 or S 2. The K m for S 1 was found to be 2.0 mm, and the K m for S 2 was found to be 20 mm. A student determined that the V max was the same for the two substrates. Unfortunately he lost the page of his notebook and needed to know the value of V max. He carried out two reactions, one with 0.1 mm S 1, the other with 0.1 mm S 2. Unfortunately, he forgot to label which reaction tube contained which substrate. Determine the value of V max from the results he obtained: Tube number Rate of formation of product 1 0.5 2 4.8 ANS: Given that S 1 = S 2 = same V max and S 1 K M = 2.0 mm and S 2 K M = 20 mm then we would expect at the same [S] (0.1 mm) that S 1 would yield a higher rate of product formation than S 2. Therefore, we can simply calculate the V max for either S 1 or S 2 knowing that Tube 1 must be S 2 and Tube 2 must be S 1. v o = V max [S]/(K M + [S]) The rates have no units so we will not include the units for the V max. Calculate the V max using the data from Tube 1 (S 2 ): 0.5 = V max (0.1 mm)/(20 mm + 0.1 mm) V max = 0.5(20.1 mm)/0.1 mm = 100.5 101 (no units in this question). 18. When 10 µg of an enzyme of M r 50,000 is added to a solution containing its substrate at a concentration one hundred times the K m, it catalyzes the conversion of 75 µmol of substrate into product in 3 min. What is the enzyme's turnover number? Ans: Because the velocity measured occurs far above K M, it represents V max. Ten mg of the enzyme represents 10 10 6 g/(5 10 4 g/mol) = 2 10 10 mol of enzyme. In 3 minutes, this amount of enzyme produced 75 µmol of product, equivalent to 25 10 6 mol of product per minute. The turnover number is therefore equal to (25 10 6 mol/min)/(2 10 10 mol) = 12.5 10 4 min 1. 5

19. An enzyme follows Michaelis-Menten kinetics. Indicate (with an "x") which of the kinetic parameters at the left would be altered by the following factors. Give only one answer for each. K M V max Neither Both X X X X (a) a competitive inhibitor (b) a noncompetitive inhibitor (c) 6 M urea (d) doubling [S] 20. Answer each of the following points (a) and (b), worth one mark each. a. Draw the structure of the cofactor pyridoxal phosphate O - HC =O O P O HO CH 2 O - CH 3 N H + pyridoxal phosphate (PLP) b. Write the general reaction for the class of enzyme known as a ligase. X + Y + ATP X - Y + ADP + P i X:Y ligase (ADP-forming) 6

Part C. Problems-based questions. Answer the following three (3) questions in the examination booklet provided. Please write your answers in pen (not red ink). Each question is worth 4, 3, and 3 marks, respectively. Three (3) questions (4 + 3 + 3) = 10 marks total. 21. You are working for the summer in a renowned researcher's laboratory and you purchase a vial of lyophilized enzyme (a phosphatase) from Calbiochem Laboratories. It reads on the label that the vial content is 23% proteinaceous material. You accurately weigh the vial and find that there is 2.5 mg of solid material. You add 5 ml of 0.1 M Tris buffer, ph 7.5 to the vial. From this stock solution you remove 15 µl and dilute this sample with Tris buffer to 1 ml. You further dilute this secondary stock by transferring 10 µl to a third vial and you make this up to a total volume of 2.5 ml. From the tertiary stock solution you pipet 4 µl into your assay tubes (n=4) and find that this amount of enzyme yields 4.2, 4.9, 5.2, and 4.4 ng (1 ng = 10-9 g) of dephosphorylated product, p-nitrophenol (MW 139.11) in a final reaction volume of 5 ml. The time for each assay was 600 sec. Calculate the average specific activity for the phosphatase activity in the original stock solution in Units/mg protein and Katal/µL solution. Also, report the total number of Units of enzyme that were present in the original vial. Ans: There is 23% of 2.5 mg solid: 2.5 (0.23) = 0.575 mg protein Primary stock: 0.575 mg/5 ml = 0.115 mg/ml Secondary stock: 15 µl 1000 µl (66.7 times); (0.115 mg/ml)/66.7 = 1.724 x 10-3 mg/ml Tertiary stock: 10 µl 2.5 ml (250 times); (1.724 x 10-3 mg/ml)/250 = 6.9 x10-6 mg/ml Assay: 4 µl (4.2, 4.9, 5.2, 4.4 ng PNP) Average = 4.7 ng ± 0.5 ng. Activity in assay: (4.7 ng PNP 1 µg/1000 ng 1 µmole/139.11 µg)/(0.004 ml 6.9 x 10-6 mg/ml 600 s 1 min/60 s = 122.4 µmoles min -1 mg prot -1 = 122.4 U/mg protein The specific activity doesn t change upon dilution so in the original stock the specific activity is 122.4 U/mg protein. Convert to katal/µl: (122.4 µmole/mg min) 1 mole/10 6 µmole 0.115 mg/1 ml 1 min/60s 1 ml/10 3 µl = 2.35 x 10-10 katal/µl Total Units in original stock: 122.4 U/mg 0.575 mg = 70.38 Units 7

22. Calculate the extent of inhibition and the velocity of an alkaline phosphatase-catalyzed reaction in the presence of 2.1 x 10-7 M p-nitrophenolphosphate substrate (K M = 2 x 10-6 M) and 4.9 x 10-7 M noncompetitive inhibitor (K i = 3.3 x 10-7 M). The velocity observed at 2 x 10-2 M p-nitrophenolphosphate in the absence of inhibitor is 455 nmol/l-min. vo [ S] v Ans: = o 2.1 10 M = 6 V K + [ S] 455nmol (2 10 M + 2.1 10 M ) max M 5 9.55 10 nmol( M ) v o = = 43.2nmol 6 2.21 10 M ' Vmax [ S] Vmax For non-competitive inhibition: v o = where V max = K M + [ S] [ I] (1 + ) Recall from Problem Set #3: 3.3 10 M V i = (3.3 10 M + 4.9 10 7 v v i o K i = ( K + [ I]) i * 43.2nmol M ) K i 3.3 10 M V i = 8.2 10 M * 43.2nmol = 17.39nmol 17.39nmol Relative velocity = v i /v o = *100% = 40.25% 43.2nmol Degree of inhibition = 100% 40.25% = 59.75%. 8

Name Student number 23. You have isolated a tetrameric NAD + -dependent dehydrogenase. You incubate this enzyme with iodoacetamide in the absence or presence of NADH (at ten times the K M concentration) and you periodically remove aliquots of the enzyme for activity measurements and amino acid composition analysis. The results are shown in the table below: (No NAD + Present) Time (min) Activity (U/mg) His Cys (Residues/mole) (NADH Present) Activity (U/mg) His Cys (Residues/mole) 0 1000 20 12 1000 20 12 15 560 18.2 11.4 975 20 11.4 30 320 17.3 10.8 950 20 10.8 45 180 16.7 10.4 925 19.8 10.4 60 100 16.4 10.0 900 19.6 10.0 (a) What can you conclude about the reactivities of the cysteinyl and histidyl residues of the protein? Ans: In the absence of the NAD + substrate there are four His residues that react with iodoacetamide per enzyme molecule and two Cys residues that react. The other residues are less reactive and may be sterically hindered from reacting or are buried in the enzyme folded structure. In the presence of the NAD + substrate there are two Cys residues that react as in the absence of substrate. However, the four His residues per enzyme that reacted in the absence of substrate are now protected by bound NAD + indicating that these His residues may be within the active site (or very nearby). (b) Which residue could you implicate in the catalytic active site? On what do you base the choice? Ans: His residues, four in total (likely one per subunit) are likely within the active site and hence are protected from the modifying reagent when the NAD + substrate is bound (saturating conditions). Substrate protection experiments are a good simple first step towards the identification of active site residues. (c) After 1 hour you dilute the enzyme incubated with iodoacetamide but no NADH. Would you expect the enzyme activity to be restored? Explain. Ans: No, because the iodoacetamide covalently reacts with Histidine at the N-1 or N- 3 of the imidazole ring which blocks the ability of the residue to act in acid/base reactions. 9