Question

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:

Answer 1

Answer:

101

Explanation:

Provided that

$S_1 = S_2 = same\ V_{max}$

And,

$S_1\ k_M = 2.0mM\\S_2\ k_M = 20mM$

Now we expect the same

{S} (0.1mM)

This determines that $S_1$ generates a  higher rate of product formation as compared to the $S_2$

So we can easily calculate the $V_{max}$ for either of $S_1$ or $S_2$ as we know that Tube 1 is $S_2$ and tube 2 is $S_1$

As we know that

$V_0 = V_{max}\ {S} / (K_M + {S})$

As the rates do not include any kind of units so we do not consider the units for $V_{max}$

Now the calculation is

$0.5 = V_{max} (0.1\ mM) / (20\ mM + 0.1\ mM)$

$V_{max} = 0.5 (20.1\ mM) / 0.1\ mM$

= 100.5

≈ 101