C2005/F2401 '06 -- Answers to Recitation Problems #4

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

2A. The ΔG^o 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 ΔG^o = x kcal/mole

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

(C) 3-PGA + P_i <--> 1,3-diPGA ΔG^o = + 10 kcal/mole

Sum = reaction 6: Gald-3-P + NAD + P_i <--> 1,3-diPGA + NADH₂ ΔG^o = +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 = ΔG^o for reaction (A)

2B1. -1.3 kcal/mole:

ΔG = ΔG^o + 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

2B2. +1.5 kcal/mole. ΔG^o is a constant, given here as +1.5.

2B3. <1. Since ΔG^o = -RT ln K_eq, then since ΔG^o here is positive, and RT is positive, then ln K_eq must be negative, and so K_eq must be <1.

2B4. To the right, since ΔG is negative -- even though ΔG^o is positive (and so equilibrium lies to the left). Remember it is ΔG that determines direction, not ΔG^o . ΔG = ΔG^o + RT lnQ. In this example, the size of the negative RT lnQ term (-2.76) overrides the relatively small size of the positive ΔG^o 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 ΔG^o (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.

3A. Having trouble with this problem? Remember you are starting with DHAP, not glucose!

DHAP: (5) -- you need to metabolize (ferment) one DHAP to get 2 ATP made.
ADP: (10) -- 2 used per DHAP metabolized (& per 2 ATP made).
Pi: (5) -- one used per DHAP.
ATP: (0) -- not needed since you start with DHAP. You make ATP, but you don't use up any.
NAD: (0.1) -- catalytic amounts needed -- 1 NAD reduced and 1 NADH oxidized per DHAP metabolized (fermented).
CoA: (0) -- not involved (needed only for Krebs)

Explanation: If you check out the stoichiometry of the pathway, you see that two ADP's are phosphorylated to ATP per DHAP broken down. No input of ATP is needed; you have bypassed the steps in glycolysis that use ATP by starting with DHAP instead of glucose. (When you start with glucose, you have to put in 2 ATP to "prime the pump." When you start with DHAP, you don't need to "spend" any ATP to get started.) For each DHAP, one inorganic phosphate is picked up from solution, to make a di-phosphorylated intermediate. The two phosphates are then transferred in two separate steps to phosphorylate two ADP's to ATP. No CoA is involved -- it is needed for entry into the Krebs cycle, which is not occurring here. One NAD is needed for oxidation of each DHAP for step 6 of glycolysis. However the NAD reduced to NADH₂ in step 6 is recycled -- it is oxidized back to NAD in the last step of fermentation. So you only need a catalytic amount of NAD.

3B. Ethanol. The #1 carbon of DHAP (top as written on handout 7-2) becomes #3 carbon of glyceraldehyde-3 phosphate (bottom as written on handout). The phosphate group doesn't move -- you flip the DHAP molecule over and the hydroxyl on top becomes a carbonyl and the carbonyl at position 2 becomes a hydroxyl. The labeled carbon (now #3) becomes the methyl group of pyruvate. The pyruvate is subsequently decarboxylated to yield acetaldehyde and then reduced to ethanol in the ethanolic fermentation found in yeast. (If you forgot to flip the DHAP, then you would think that the radioactivity would be lost as CO₂.) Next best answer is lactate, if you forgot that yeast do ethanolic and not lactate fermentation. Third best answer is pyruvate, if you mistakenly added more than the minimum NAD to the reaction, such that fermentation is not necessary (though it would presumably take place anyway). Acetyl CoA would not be formed, as only the glycolytic and fermentation enzymes are present. The other compounds have no connection to this carbon.

3C. Yes. There is a net gain of ATP, and no net oxidation or reduction of NAD -- what is used is regenerated. Only one NAD is necessary in the glycolytic pathway converting DHAP to pyruvate, and that one NAD can be regenerated from NADH₂ in the final step of fermentation, the reduction of pyruvate to lactate (the usual product in E. coli).

Note: E. coli can't really live on DHAP because phosphorylated sugars can't get transported into the cells.

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