Extra point: An enzyme can catalyze a reaction with ether of two substrates, si

Question

Extra point: An enzyme can catalyze a reaction with ether of two substrates, si orS. The K for Ss was found to be 2.0 mM, and the Kn for 5, was found to be 20 mM. A student determined that the vr was the same for the two substrates. Unfortunately. he lost the page ofhils notebook and needed to know the value of Vaa He carried out two reactions: one with 0.1 mM S, the other with 0.1 miM S2. unfortunately, he forgot to label which reaction tube contained which substrate. Tube number Rate of frmation of preduct ImM/s D.5 4.8 Determine the value of Ynx: from the results he obtained (Show your work and reasoning process)

Explanation / Answer

Rate, v = Vmax [S]/(Km + [S])

Km1 = 2 mM

Km2 = 20 mM

Km1 < Km2

For equal value of Vmax, v is inversely related to Km. Therefore, substrate 1 will show higher rate of formation of product.

For substrate, S1

Km = 2 mM

[S1] = 0.1 mM

v1 = 4.8 mM/s

Rate, v1 = Vmax [S1]/(Km1 + [S1])

4.8 = Vmax * 0.1 / (2 + 0.1)

Vmax = 100.8 mM/s

For substrate, S2

Km = 20 mM

[S2] = 0.1 mM

v2 = 0.5 mM/s

Rate, v2 = Vmax [S2]/(Km2 + [S2])

0.5 = Vmax * 0.1 / (20 + 0.1)

Vmax = 100.5 mM/s

Average Vmax = (100.8 + 100.5)/2

= 100.65 mM/s

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