Asparaginase catalzyes the following reaction: Asparagine + H 2 O ? aspartate +
ID: 830341 • Letter: A
Question
Asparaginase catalzyes the following reaction:
Asparagine + H2O ? aspartate + NH4+
The enzyme has been isolated from two different organisms. Asparaginase A has a KM for asparagine of 10 ?M and a Vmax of 50 nmol-sec-1. Asparaginase B has a KM for asparagine of 20 mM and a Vmax of 200 nmol-sec-1.
A. Which enzyme would convert more asparagine to aspartate in 10 min if the asparagine concentration were 20 ?M? Explain your answer.
B. Which enzyme do you think is the more efficient enzyme and why?
Explanation / Answer
Using the Michaelis Menten equation i.e.
Vo = Vmax.S / (Km + S)
For Asparaginase A;
Vo = 50 x 10^-9. 20 x 10^-6 / (10 x 10^-6 + 20 x 10^-6)
Vo = 10^-12 / (30 x 10^-6) = 0.033 x 10^-6 mol/s
Thus in 10 mins (600 secs)
Amount of substrate converted = 0.033 x 10^-6 x 600 = 20 umol
For Asparaginase B;
Vo = 200 x 10^-9. 20 x 10^-6 / (20 x 10^-3 + 20 x 10^-6)
Vo = 4 x 10^-12 / (20.02 x 10^-3) = 0.199 x 10^-9 mol/s
Thus in 10 mins (600 secs)
Amount of substrate converted = 0.199 x 10^-9 x 600 = 119.4 x 10^-9 mol = 0.119 umol
A) Thus, the Asparaginase A converts more of the substrate in the stipulated time of of mins when compared to Asparaginase B. The lower the Km value the greater is the affinity of the enzyme for the substrate.
The Km value of the enzyme A is much lower than B, and thus it has higher affinity for the substrate and thus converts more of the substrate in the given time.
B) As the Asparaginase A has higher affinity for the substrate and converts more of the substrate into product in a given period of time, thus it is the more efficient enzyme in this case.
NOTE: The question marks have been taken as the symbol u for microMolar (10^-6), as that seemed the most appropriate interpretation.
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