A carboxypeptidase was found to have a Michaelis constant KM of 2.00?M and a kca

Question

A carboxypeptidase was found to have a Michaelis constant KM of 2.00?M and a kcat of 150s?1 for its substrate A.

Part A What is the initial rate of reaction for a substrate concentration of 5.00 ?M and an enzyme concentration of 0.01 ?M? Express your answer using two significant figures and include the appropriate units. v0 = Request Answer

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, K1, where K1=[E][I]/[EI] ? Express your answer using two significant figures and include the appropriate units. K1 =

Part C A competing substrate B is added to the solution in part A. Its KM is 10.00 ?M and its kcat is 100 s?1. Calculate the relative rates of substrate reaction for equal concentrations of substrates; that is, calculate vB/vA. Express your answer using two significant figures. vB/vA =

Explanation / Answer

Answer

1. V0= Kcat [E]o [S] / Km + [S]
     = 150 [0.01] [5] / 2 + 5
     = 1.07 uM/s

2. V0= Kcat [E]o [S] / a Km + [S]

akmV0 + SV0 = KcatE0S
a = (Kcat[E]0 - V0) [S] / KmV0
= (150 [0.01] - 0.535) [5] / 2 (0.535)
= 4.5

a = 1+[I]/KI
KI = [I]/a-1
   = 5 / 4.5-1
   = 1.42mM

3. VA= KcatA [E]o [S] / KmA + [S]
   VB= KcatB [E]o [S] / KmB + [S]
VB/VA = KcatB [E]o [S] / KmB + [S] / KcatA [E]o [S] / KmA + [S]
Assume S=10     
= KcatB/KmB + [S] / KcatA / KmA + [S]
       = 100/10+10 / 150 / 2 + 10
       = 100/20 / 150 / 12
       = 5/32
       = 0.15

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