C2005/F2401 '06 -- Key to Recitation Problems #5

Note: NADH₂ or NAD_red is used as shorthand for reduced NAD (= NADH + H⁺⁾.

1. E. coli can grow anaerobically in a minimal medium containing a mixture of glycerol and pyruvate, but not on glycerol or pyruvate alone.

ATP yield: You can't get any ATP from metabolizing pyruvate anaerobically, so the bacteria can't grow on pyruvate. One net ATP can be produced from metabolism of glycerol. One ATP is used up to produce DHAP -- it's used to phosphorylate glycerol to glycerol phosphate. 2 ATPs are produced in the reactions from DHAP to pyruvate. Net ATP generated = 2-1 =1.

NAD/NADH₂ balance: Metabolizing one molecule of glycerol generates one molecule of pyruvate, and converts 2 molecules of NAD (oxidized) to 2 molecules of NADH₂ (reduced). Reducing the pyruvate to lactacte oxidizes only one of the two reduced NADH₂ molecules. If glycerol is the only carbon source, the 2nd NADH₂ cannot be oxidized, so the NAD/NADH₂ balance cannot be maintained -- the cell will run out of oxidized NAD. In the presence of excess pyruvate in the medium, the 2nd NADH₂ molecule can be oxidized back to NAD by reducing the added pyruvate to lactate. Thus NAD/NADH₂ balance will not be a problem in the presence of glycerol plus pyruvate.

2. A. Positive. There are several ways to see this:

(i) By looking at the order in the electron transport chain: Reduced FAD feeds into the chain after reduced NAD -- that is, below it, on an energy scale. (See handout 8-2.) So NADH₂ should hand off hydrogens and/or electrons to FAD and not the other way around, since electrons go down the chain, not up it. In terms of affinity for electrons, the compounds "higher up" on the chain have a lower affinity for electrons, and tend to give them away to compounds further down the chain. The compounds "lower down" on the chain have a higher affinity for electrons, and tend to grab electrons from the compounds above. (Oxygen, which is on the bottom, has the highest affinity for electrons -- it gets them in the end.)

In terms of ΔG^o: if NADH₂ is "higher up" on the chain, then the distance from it to oxygen is greater, and so the ΔG^o for reduction of NADH₂ must be more negative (than for FADH₂). If you want to formally add up the ΔG^o values for oxidation/reduction of NAD and FAD to get the overall ΔG^o for the reaction shown in the problem, see (iii).

(ii) By considering relative ATP yields -- Another way to get the relative ΔG^o values: Oxidation of FADH₂ drives production of only 2 ATP's while oxidation of NADH₂ drives production of 3 ATP's. So you must get more energy from oxidation of reduced NAD, so the ΔG^o for reduction of NADH₂ must be more negative.

(iii) Adding up the ΔG^o values:

NADH + H⁺ + 1/2 O₂ <--> NAD + H₂O ΔG^o = -52 kc/mole or some large negative number if you don't remember exactly.

FADH₂ + 1/2 O₂ <--> FAD + H₂O ΔG^o = a much smaller negative number for the reasons given in (i) & (ii) -- since FAD enters the electron transport chain at a lower level, and since one derives less ATPs/electron pair from FADH₂ than from NADH + H⁺.

Inverting the top reaction and changing the sign:

NAD <--> NADH + H⁺ + 1/2 O₂ ΔG^o = +52 kc/m or some large positive number

FADH₂ + 1/2 O₂ <--> FAD + H₂O ΔG^o = a much smaller negative number

Net: NAD + FADH₂ <--> FAD + NADH + H⁺ ΔG^o = a large positive number plus a much smaller negative number = highly positive.

B. Less. In the bacteria, because there is so much enzyme present, all the NADH₂ will give up its electrons to FAD and then the FADH₂ will enter the electron transport chain "lower down" bypassing the first site of ATP formation. In humans, the NADH₂ will give up its electrons directly to the electron transport chain -- it will enter "higher up," before the first site of ATP synthesis. Therefore, in the bacteria, the electrons from NADH₂ will only yield 2 ATP, compared to the 3 ATP/NADH₂ in humans. Since there are 10 NADH₂'s produced from the oxidation of one molecule of glucose, the difference in ATP yield is considerable. In bacteria, the yield is 10 X 2 = 20 ATP vs. 30 ATP in humans, a difference of 10 less in the bacteria.

3. A-1. pH will increase in the left. Hydrogen ions will be pumped across the membrane in the direction from what had been the inside or matrix side (left) to what had been the intramembrane space (right). The pH on the right will fall and the pH on the left will rise. (The hydrogen ion concentration will increase on the right and fall on the left.)

A-2. Increase in the left. ATP will be synthesized from ADP + P_i as hydrogen ions flow back across the membrane from right to left through the ATP synthetase of the F_oF₁ complex in the membrane.

A-3. NAD_ox will increase on the left. The reduced NAD gives up its electrons to the electron transport chain in the membrane, interacting with components on the inside or matrix side (left) of the membrane. (The electron chain is embedded asymmetrically in the membrane, so the protein complex that binds NAD faces the matrix side.)

A-4. Neither. The reduced NAD (NAD_red) will decrease on the left as it gives up its electrons to the electron transport chain and is converted to NAD_ox in the membrane.

A-5. Neither. Pyruvate can be metabolized two ways in a cell, but neither can occur here. Pyruvate can be metabolized by the entrance reaction to the Krebs Cycle, which requires CoA (Coenzyme A) to run, and CoA is not included (nor are the enzymes of the matrix). Pyruvate can also be metabolized by conversion to lactic acid in fermentation, but the enzymes to do this are found in the cytoplasm, and are not included here.

B. pH, ATP, and NAD_ox or NADH_red. If there is no O₂ present to accept the electrons for the electron transport chain, then the electron transport chain will not be able to accept the electrons from NADr_ed. So compared to the situation above, there will be less NAD_ox and more NAD_red on the left. If there is no electron transport, no hydrogen ions will be pumped across the membrane, and no hydrogen ion gradient will be established. Therefore there will be no net flow of hydrogen ions back through the ATP synthetase and no ATP production (on the left).

C. Phosphorylation of ADP to ATP & establishment of a pH difference between the two chambers will be affected.

The electron transport chain will continue to pump hydrogen ions from left to right across the membrane, but hydrogen ions will not accumulate in the right chamber since the pumped-out hydrogen ions will return to the left chamber through the tear in the membrane. Thus no pH gradient will be established. With no pH gradient, there will be no flow of hydrogen ions though the ATP synthetase, and no ATP will be produced.
NAD oxidation (oxidation of NAD_red to NAD_ox) will continue normally using the electron transport chain components in the membrane, so NAD_ox will increase on the left side as before. (You have uncoupled ox. phos. from electron transport.) Some of the NAD_ox generated on the left will diffuse into the right compartment through the tear, reducing the difference in NAD_ox between the two sides. However NAD oxidation will continue on the left side only.

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