Exercise 10.21 A carboxypeptidase was found to have a Michaelis constant KM of 2

Question

Exercise 10.21 A carboxypeptidase was found to have a Michaelis constant KM of 2.00 M and a ,t of 150 s-1 for its substrate A. Part A What is the initial rate of reaction for a substrate concentration of 500 M and an enzyme concentration of 0.01 M? Express your answer using two significant figures and include the appropriate units. Uo= 1.1x10-6 M-s-1 Submit My Answers Give Up Correct Part B The presence of 5.00 mM of a competitive inhibitor decreased the initial rate above by a factor of 2. What is the dissociation constant for the enzyme-inhibitor complex. K , where K! = [E] [1] / [EI ? Express your answer using two significant figures and include the appropriate units. Ki= 1.4x10-3 M Submit My Answers Give Up Correct

Explanation / Answer

Ans. Given, For substrate B

            [S] = 5.0 uM

            Kcat = 100 s-1

            [E] = 0.01 uM

            Km = 10.0 uM

Now,

            Vmax for substrate B = Kcat x [E]

            Or, Vmax = 100 s-1 x 0.01 uM = 1.0 uM s-1

# Calculate Vo using MM equation-         Vo = Vmax [S] / (Km + [S])

Putting the values in above equation-

            Vo = (1.0 uM s-1 x 5.0 uM) / (10.0 uM + 5.0 uM) = 0.333 uM s-1

            Hence, Vo for substrate B under given conditions, VB = 0.333 uM s-1

# From #A      , Vo for substrate A, VA = 1.1 x 10-6 M s-1 = 1.1 uM s-1

Now,

            VB / VA = 0.333 uM s-1 / 1.1 M s-1 = 0.303

Therefore, VB / VA = 0.303

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