An unknown concentration of enzyme A catalyzes the reaction X => Y in one test t
ID: 536098 • Letter: A
Question
An unknown concentration of enzyme A catalyzes the reaction X => Y in one test tube. An unknown concentration of enzyme B catalyzes the reaction X => Z in another test tube.
Enzyme A
Enzyme B
Vmax (nM/s)
200
200
KM (mM)
10
20
kcat (1/s)
2000
1000
Would one molecule of enzyme A consume substrate X at a rate less than, greater than, or equal to the rate at which one molecule of enzyme B would consume X? Explain your answer.
Enzyme A
Enzyme B
Vmax (nM/s)
200
200
KM (mM)
10
20
kcat (1/s)
2000
1000
Explanation / Answer
The rate constant for the interaction of substrate (S) and Enzyme (E) is given as kcat/KM and it can be used as a measure of catalytic efficiency.
i.e; catalytic efficiency (CE) = kcat/KM.
With this, we can compare the preference of an enzyme for different substrates.
Here, for enzyme A , CE = 2000/10 = 200
and for enzyme B, CE = 1000/20 = 50
Enzyme A clearly has a preference for the substrate X.
So one molecule of enzyme A would consume substrate X at a rate greater than the rate at which one molecule of enzyme B would concume substrate X.
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