1. Measurement of the rate constants for simple enzymatic reaction obeying Michaelis- Menten kinetics gave the following results: =3x10-5 = 30μM

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1 1. Measurement of the rate constants for simple enzymatic reaction obeying Michaelis- Menten kinetics gave the following results: k 1 = 2 x 10 8 M -1 s -1, k 2 = 1 x 10 3 s -1, k 3 = 5 x 10 3 s -1 a) What is (the dissociation constant for the enzyme-substrate complex? b) What is K m (the Michaelis constant)? = k 2 /k 1 = 5x10-6 M = 5μM K m = k 2 +k 3 k 1 =3x10-5 = 30μM c) What is k cat (the turnover number)? k cat = k 3 = 5x10 3 s -1 d) What is k cat /K m? 1.67x10 8 M -1 s -1 e) Does this enzyme approach catalytic perfection? yes..almost diffusion ~10 9 f) What is Vmax if 2 nmol/ml and saturating substrate are used? V max = [E] total *k cat =2x10-9 mol ml -1 *5x10 3 s -1 = 1x10-5 mol ml -1 s -1 g) What are Vmax and K m using 4 nmol/ml enzyme and saturating substrate. V max = 2x10-5 mol ml -1 s The turnover number of an enzyme is 40,000,000 per second. The Km is 0.11 M. a) What is Vmax using 3 nmol/l? V max = [E] total *k cat = 3x10-9 mol/10 3 ml*4x10 7 s -1 =1.2x10-4 mol ml -1 s -1 = 12 mmol ml -1 s -1 V max =12 M s -1 b) What is v when [S] = 0.75 M? v = V max [S ] K m +[S ] =12Ms -1 *0.75M/(0.11M+0.75M)= 9M2s -1 /0.86M = 10.5 Ms -1 c) What is the catalytic efficiency? efficiency = kcat/km = 4x10 7 s -1 /0.11M =3.6x10 8 M - 1s -1 1 of 5

2 3. Fetal hemoglobin (HbF) contains serine in place of the cationic histidine at residue 143 in HbA, between the β chains in the central cavity of the enzyme. a) The P50 for HbF is 18 torr whereas the P50 for HbA is 26 torr. How do these values explain the efficient transfer of oxygen from maternal blood to the fetus? The affinity of fetal hemoglobin for oxygen is ~1.5 times that of matrernal hemoglobin, thus it will be able to effectively compete for oxygen (effect transfer). b) What is the likely explanation for the difference in the P50 values? Negatively charged BPG (bisphosphoglycerate) is known to bind in the central cavity. Replacement of a positive residue with a polar/neutral residue will likely reduce binding, destabilizing the T state, allowing T to R transitions with an accompanying increase in O2 binding. c) Sketch a plot of fraction binding versus O 2 for each variant. 1 Fraction Bound PO 2 torr 2 of 5

3 4. An antibody binds and antigen with a = 5 x 10-8 M. Calculate the antigen concentration when the ratio bound is: a) 0.2 b) 0.5 c) 0.6 d) 0.8 fraction bound = [AX ] [A ]+[AX ] =0.5 when X = 5x10-8 M = [A ][X ] [AX ] such that [AX] = [A ][X ] such that [ A][ X ] fraction bound = = [ A][ X ] = [ X ] = [ X ] = [X ] [A ]+ [ A][ X ] [A ](1+ [ X ] K ) (1+ [ X ] D K ) ( D K + [ X ] D K ) +[X ] D and [X] = fraction bound*( )/(1-fraction bound) so that a) [X] = 0.2*5x10-8 M/(1-0.2)=1.25x10-8 M, b) 5x10-8 M, c) 7.5x10-8 M, d) 2x10-7 M 5. You are studying a novel enzyme with site-directed mutagenesis and make the following kinetic measurements: Variant k cat (s -1 ) K m (M) Wild-type 4 x x 10-6 E99Q 4 x x 10-3 A12K 4 x x 10-3 H135A 2 x x 10-6 a) Which residues are involved in substrate binding? The residues that affect Km are E99 and A12. b) Which residues are involved in catalysis? The residues that affect kcat are A12 and H135. c) Which variant is the least efficient? efficiency is kcat/km. A12K affects both, thus it is least efficient. (wt 2x10 12 M s -1, E99Q 2x10 9 M s -1, A12K 2x10 5 M s -1, H135A 1x10 7 M s -1 ) 6. Catalytic triad groupings of amino acid residues increase nucleophilic character of active site serine, threonine or cysteine residues in enzymes that catalyze cleavage of amide and ester bonds. Using chymotrypsin as a model, draw the expected arrangement of the catalytic triads in the following enzymes. a) Human cytomegaloviral protease: His, His, Ser 3 of 5

4 b) β-lactamase: Glu, Lys, Ser, c) Asparaginase: Asp, Lys, Thr d) Hepatitis A protease: Asp, His, Cys 7. Estimate Vmax and Km using the following data: [S] (M) v (μm/min) 2.5 x x x x x x x x Vmax 140 μm/min, Km 1x10-5 M (at Vmax/2 or 70 µm/min) 8. Prostaglandins are a class of eicosanoids, fatty acid derivatives responsible for producing fever and inflammation. The first enzyme in the conversion of arachidonic acid to prostaglandin is inhibited by ibuprofen with the following kinetics: Arachidonic acid (mm) Rate( mm/min) Rate with 10 mg/ml ibuprofen (mm/min) a) What are K m and Vmax for the enzyme? Vmax = 1/ = 51.5 mm/min, Km = -( )/(0.0116)=1.67 mm b) What type of inhibitor is ibuprofen? Vmax no change, Km change = competitive c) What is the K I of ibuprofen? Km = -( )/(0.0203) = mm, Km = (1+[I]/K I ) such that K I = (1+10mg/ml(mmol/206.30mg)/1.67mM = 0.59 mm 4 of 5

5 y = x y = x wt +I Linear (+I) Linear (wt) /[S] 9. The enzyme urease enhances hydrolysis of urea at ph 8.0 by a factor of If a quantity of urea can be hydrolyzed in 5 minutes by the enzyme, how long would it take to hydrolyze this same quantity in the absence of the enzyme in a sterile solution? If it takes the enzyme 5 minutes, and the enzyme increases the rate by 10 14, the normal rate would be 5x10 14 min or 5x10 14 min (hr/60min) (day/24hr) (year/365day)=9.5x10 8 yrs or 950 million years 10. A solution of hexokinase loses 50% of its activity in 12 minutes when heated at 45 C but only 3% when in the presence of large concentrations of its substrate glucose. Describe possible modes of interaction between the enzyme and substrate and explain how the substrate delays loss of activity. Glucose must be stabilizing the structure, most likely through hydrogen bonding with the hydroxyl groups on the glucose with hydroxyl and/or amide groups in the protein binding site. 5 of 5

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