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

Glossary of biochemical terms | Enzyme Problems Set1. | Frequently asked questions | Typical Examination Questions | C2005/F2402 '06 -- Key to Recitation Problems #1 | C2005/F2401 '06 Answers to Recitation Problems #2 1. Hint: Consider which molecules (or parts thereof) are hydrophobic or hydrophilic. | C2005/F2401 ’06– Recitation Problems #3 -- Answers | 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? | C2005/F2401 '06 -- Recitation Problems #7 -- Answers | T A C T C A T C G A 3’ ......................… Promotor; transcription right to left. |


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

(ii) By considering relative ATP yields -- Another way to get the relative ΔGo 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 ΔGo for reduction of NADH2 must be more negative.

(iii) Adding up the ΔGo values:

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

FADH2 + 1/2 O2 <--> FAD + H2O ΔGo = 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 ΔGo = +52 kc/m or some large positive number

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

Net: NAD + FADH2 <--> FAD + NADH + H+ ΔGo = 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 + Pi as hydrogen ions flow back across the membrane from right to left through the ATP synthetase of the FoF1 complex in the membrane.

A-3. NADox 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 (NADred) will decrease on the left as it gives up its electrons to the electron transport chain and is converted to NADox 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 NADox or NADHred. 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 NADred. So compared to the situation above, there will be less NADox and more NADred 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 NADred to NADox) will continue normally using the electron transport chain components in the membrane, so NADox will increase on the left side as before. (You have uncoupled ox. phos. from electron transport.) Some of the NADox generated on the left will diffuse into the right compartment through the tear, reducing the difference in NADox between the two sides. However NAD oxidation will continue on the left side only.


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