In the conversion of A into D in the biochemical pathway shown, enzymes EA, EB, and EC have the 𝐾M values 10−2 M, 10−4 M,and 10−6 M, respectively.

If all of the substrates and products are present at a concentration of 10−4 M and the enzymes have approximately the same 𝑉max, which step will be rate limiting?

Question

In the conversion of A into D in the biochemical pathway shown, enzymes EA, EB, and EC have the 𝐾M values 10−2 M, 10−4 M,and 10−6 M, respectively.

If all of the substrates and products are present at a concentration of 10−4 M and the enzymes have approximately the same 𝑉max, which step will be rate limiting?

BlackTom AI · Accepted Answer

We are analyzing a three-step pathway A -> B -> C -> D with each step catalyzed by a different enzyme having Km values: KmA = 10^-2 M, KmB = 10^-4 M, KmC = 10^-6 M. All substrates/products are at 10^-4 M and Vmax are approximately equal across enzymes. To compare rates, we use the Michaelis-Menten form for each step at a given [S]: v ≈ Vmax × [S] / (Km + [S]). Since Vmax is assumed similar for EA, EB, and EC, we can compare the fractional velocities f = [S] / (Km + [S]) for each step.

Option 1: the second step. Here [S] = 10^-4 M and Km for EB is 10^-4 M. The fractional velocity is f2 = 10^-4 / (10^-4 + 10^-4) = 10^-4 / 2×10^-4 = 0.5. This step proceeds at about half of its maximum rate, given the chosen concentrations.

Option 2: the first step. For EA, [S] = 10^-4 M and Km = 10^-2 M. The fractional velocity is f1 = 10^-4 / (10^-2 + 10^-4) ≈ 10^-4 / 0.0101 ≈ 9.9×10^-3, which is about 0.01 of Vmax. This step is very slow relative to Vmax because the substrate concentration is far below Km.

Option 3: the third step. For EC, [S] = 10^-4 M and Km = 10^-6 M. The fractional velocity is f3 = 10^-4 / (10^-6 + 10^-4) ≈ 10^-4 / 1.01×10^-4 ≈ 0.99. This step operates near its maximal rate given the substrate concentration, since [S] >> Km for this enzyme.

Option 4: They will all be rate limiting. Based on the calculations above, the fractional velocity for the first step is much smaller than for the second and third steps, and the third step is near Vmax. Therefore, it is not accurate to call all steps rate limiting under these conditions.

In summary, the second step has a moderate rate (about half of Vmax), the first step is severely constrained by a Km much larger than [S], and the third step is near its maximum; thus the first step is the slowest, i.e., rate-limiting under these conditions.

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