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An enzyme can catalyze a reaction with either of two substrates, S1 or S2. The Km for S1 was found to be 2.0 mM, and the Km, for S2 was found to be 20 mM. A student determined that the Vmax was the same for the two substrates. Unfortunately, he lost the page of his notebook and needed to know the value of Vmax. He carried out two reactions: one with 0.1 mM S1, the other with 0.1 mM S2. Unfortunately, he forgot to label which reaction tube contained which substrate. Determine the value of Vmax from the results he obtained:
Tube number: 1
Rate of formation of product:0.5
Tube number: 2
Rate of formation of product: 4.8

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batolisis

Answer:

100.5 ≈ 101

Explanation:

Km for S1 = 2.0 mM

Km for S2 = 20 mM

Given that : S1 = S2 =  hence Vmax for either S1 or S2 can represent

The Vmax can be calculated using the data Given and equation below

Vo = Vmax [s] / ( Km + [s] )  ------ (1)

Vo = 0.5

[s] = 0.1 mM

km = 20 mM

making Vmax subject of  equation 1

Vmax = 0.5 ( 20.1 mM) / (0.1 mM )

         = 100.5 ≈ 101

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