A. Hint: Can you calculate the Vmax? (What value of [S] was used here? Does that help you get the Vmax?) How do you get from Vmax to the turnover number?

A. Answer: 33.3 per second or 2000 per min.

How to get the V_max: Since the substrate concentration of lactose used was equal to the K_m of the enzyme, the velocity in that experiment must have been one-half the V_max. So the V_max was 2 X 10^-5M/minute.

How to get the T. O. #: Turnover number is k₃ = Vmax/[E], or 2 X 10^-5 M/minute / 10^-8M = 2000 per minute, or = 2000/60 = 33.3 per second.

Note: This answer may appear to contradict the answer to Q1. It doesn’t. If you know the K_m, and the V at [S] = K_m, you can calculate V_max (as in this question); but you can’t say how much [S] is needed to get to V_max (see question 1). That’s because V approaches V_max as [S] increases, but V only reaches V_max at infinite [S].

B. Hint: Do you have enough information to calculate the Vmax?

3B. Answer: Can’t predict. V_max = k₃E_o, but we do not know the k₃ for this substrate. Like the K_m, it need not be the same as for galactose-glucose (lactose) calculated in part A. Also the K_m and V_max are independent variables, determined by different aspects of the enzyme/substrate interaction. If the enzyme acts on a different substrate, and the K_m changes, the E_o (& therefore the V_max) may or may not change.

C2005/F2401 '06 -- Hints for Recitation Problems #4

What has changed, the Km, the Vmax, or both?

A. Remember that ΔGo's are additive.

B1. How are ΔG and ΔGo related?

B2. What changes with concentration (of reactants and/or products)? ΔG? ΔGo?

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