C2005/F2401 '09 -- Answers to Recitation Problems #4  

1. Hint: Is V vs S linear? 

Answer: (> 1/2 the Vmax but < the Vmax). 
In words:  
As [S] increases, rate of increase of V increases more slowly than value of [S].  
Value of V starts to plateau, so adding more S doesn’t increase V very much at large values of S. 
Doubling [S] won’t double V.  
Graphical solution: 
Show a line rising from 2X Km to the V curve, and then left to the y-axis to intersect it between 1/2 Vmax and Vmax.
 

2A. 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?

2A.  Answer: 33.3 per second or 2000 per min.

How to get the Vmax: Since the substrate concentration of lactose used was equal to the Km of the enzyme, the velocity in that experiment must have been one-half the Vmax. So the Vmax 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 Km, and the V at [S] = Km, you can calculate Vmax (as in this question); but you can’t say how much [S] is needed to get to Vmax (see question 1). That’s because V approaches Vmax as [S] increases, but V only reaches Vmax at infinite [S].


2B.  Hint: Do you have enough information to calculate the Vmax?

2B.  Answer: Can’t predict. Vmax = k3Eo, but we do not know the k3 for this substrate. Like the Km, it need not be the same as for galactose-glucose (lactose) calculated in part A. It is important to realize that the Km and Vmax are independent variables, determined by different aspects of the enzyme/substrate interaction. If the enzyme acts on a different substrate, and the Km changes, the k3 (&  therefore the Vmax) may or may not change.

 
2C. Hint: What has changed, the Km, the Vmax, or both?

2C.   Vmax = X. Vmax w/ cellobiose = the Vmax in the absence of the inhibitor. In competitive inhibition., the Vmax can always be reached by adding excess substrate.  In other words, the Km has changed, but the Vmax is the same.  With the inhibitor cellobiose present, the (apparent) Km is increased -- the amount of lactose needed to reach ½ Vmax is greater. So you can’t use the V at the old Km value (measured w/o cellobiose) and assume that it = 1/2 Vmax. 


3. A. Hint: Remember that ΔGo's are additive.

    Answer: The ΔGo for reaction (A)  is ___-61.5_____ .
   
How is this calculated? First add up (A), (B*) = reverse of (B), & (C) = same reactants and products as reaction 6.

    (A) Gald-3-P + 1/2 O2 <--> 3-PGA    + H2O                         ΔGo = x kcal/mole

    (B*) NAD + H2O <--> NADH2 + 1/2 O2                               ΔGo = + 53 kcal/mole

    (C) 3-PGA + Pi <--> 1,3-diPGA                                             ΔGo = + 10 kcal/mole

Sum = reaction 6: Gald-3-P + NAD + Pi <--> 1,3-diPGA + NADH2         ΔGo = +1.5 kcal/mole (value given for rxn 6)

Now solve for x: 
    53 + 10 + x = +1.5
    x = -53 -10 + 1.5 = -61.5 kcal/mole = ΔGo for reaction (A)


3B1. Hint: How are ΔG and ΔGo related?

        Answer: -1.3 kcal/mole. Here is the calculation: 

     ΔG = ΔGo + RT lnQ

        Q = (10-6)(10-5) / (10-3)(10-3)(10-3) = 10-11 / 10-9 = 10-2; RT = 0.6 kcal/deg-mole

     ΔG = +1.5 + 0.6 X 2.3 log (10-2) = 1.5 + (1.38 X [-2]) = 1.5 - 2.76 = -1.26 = ~ -1.3 kcal/mole

Or, writing it all out in one step:

    ΔG = +1.5 + 0.6 X 2.3 log (10-6)(10-5) / (10-3)(10-3)(10-3) = +1.5 + (0.6 X 2.3 log 10-2), or

    ΔG = +1.5 + 0.6 X ln (10-6)(10-5) / (10-3)(10-3)(10-3) = +1.5 + 0.6 X ln 10-2


3B2. Hint: What changes with concentration (of reactants and/or products)? ΔG ? ΔGo ?

    Answer: +1.5 kcal/mole. ΔGo is a constant, given here as +1.5.


3B3. Hint: How is Keq related to ΔG and/or ΔGo ?

    Answer: <1. Since ΔGo = -RT ln Keq, then since ΔGo here is positive, and RT is positive, then ln Keq must be negative, and so Keq must be <1.


3B4. Hint: What determines direction? Keq ? ΔG ? ΔGo ? How can a reaction "go" in one direction, if the Keq favors the "other" direction?

Answer: To the right, since ΔG is negative -- even though ΔGo is positive (and so equilibrium lies to the left). 
Remember it is  ΔG that determines direction, not ΔGo . ΔG = ΔGo + RT lnQ. In this example, the size of the negative RT lnQ term (-2.76) overrides the relatively small size of the positive ΔGo term (+1.5), so that the overall  ΔG is negative, and reaction goes to the right. In other words, the relatively high concentrations of the starting materials, and the relatively low concentrations of the products, push the reaction to the right. This reaction is an example of how a reaction with a positive ΔGo (corresponding to an unfavorable equilibrium constant) can occur in the "uphill" direction as part of a pathway. The reaction is pushed "uphill"  because the previous steps in the pathway provide substrate and the following steps remove the products, keeping the ratio of [products]/[substrates] at a low value.