everybody calculated this Q wrongly!! please try

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everybody calculated this Q wrongly!! please try

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nadiabarati

02/21/06 - 07:40 AM

Enzymes A and B catalyse the same reaction , but Km for A is 0.000006mol/l and for B is 0.003 mol/l.If both enzymes have the same Vmax and if the activity of both enzymes is measured for substrate concentration of 0.003 mol/l , the relation of V(of enzymeA):V(of enzymeB) is aprox
a. 1:1
b. 2:1
c. 20:1
d. 1000:1

robin082006
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02/21/06 - 08:33 AM

Simple explanation:

Km of B = 0.003 mol/l and S=0.003 mo;/l, therefore Vb=1/2 Vmax

Km of A =0.000006 mol/l and S = 0.003 mol/l, greater than Km 500 times, therefore Va is likely to reach closely Vmax.

Thus, Va/Vb=2/1

Help it clear

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robin082006
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02/21/06 - 08:37 AM

other explanation

Va= Vmaxa*Sa/(Kma+Sa)

Vb= Vmaxb*Sb/(Kmb+Sb)

Since Vmax and S of A and B are the same.

So Va/Vb= (Kmb+Sb)/(Kma+Sa)= (0.003+0.003)/(0.000006+0.003) = nearly 2/1.

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