Frequently Asked Questions #1

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Last entry: Oct. 24, 1996

 

1. Can I use Curtis instead of Purves? instead of Becker? Answer

2. In your handout about the Flow of Glucose in E. Coli, glucose is shown on the bottom of the sketch undergoing biosynthetic pathways and intermediates to become monomers. Is glucose a monomer of polysaccharides or is it only the "initial material" required to make monomers? Answer

3. In reference to hydrocarbons and their inability to dissolve in water is this because they are non-polar and therefore not likely to have hydrogen bonding to water molecules as occurs with NaCl in aqueous solutions? Answer

4. Is an "induced dipole" called such because partial charges are created when the atoms are forced to come together? Answer

5. Are the cyclo glucose structures called alpha because the hydroxyl group on carbon 1 axial while the same hydroxyl group on carbon 1 equatorial in beta? Answer

6. Ethanol is hydrophilic -- is this because of the hydroxyl group and more importantly the partial negative charge of the oxygen in the functional group? Answer

7. You mentioned that hydrophobic forces would get stronger at higher temperatures -- Why? Answer

8. I'm speaking in behalf of myself and a few other kids who have had the same confusion. Some of us have been buying both Becker and Purves, while others have only bought one of the two, saying only one of the two books is enough. It seems that the handout says either book is fine, but the handout also says that any college bio book is fine. It just seems wierd -- as if we're not basing the class on a standard text at all. If that's the case it's fine with me, but it just seems so opposite from what a traditional science class should be like. P.S. Just in case I don't get to talk to you in person tomorrow I have this question also: I have a biology textbook called: Biological Science, 5th Edition, by Keeton and Gould (copyright 1993). It's the book I used in Advanced Biology in highschool, but it's used in colleges, at least Yale's Intro to Bio course. When you mention in the handbook that "Keeton and Gould is okay" is this the book (note edition and year) that you mean? Answer

9. Isn't there a mistake in the Exponential Growth handout, on line 3 of the second paragraph, it says " If we started with 100 cells, after 1 gen. we would have 200 and after 4 gens. 400, etc." ? Answer

10. By what means does dialysis separate a protein? I thought it was used as a semi-permeable sac to allow wanted species into the protein and unwanted species out of a protein. Answer

11. How do you determine the sequence of a polypeptide when given different subpeptide sequences? i.e. 2-9D What is the order of amino acids in the polypeptide? Answer

12. Regarding problem 2-18, I was confused with this question because I didn't understand if "molecule" was used for "total protein" in the answer. It was a very in-depth problem. Is there any way to go over it in class? Answer

13. There's precious little information in Becker and Purves about protein/polypeptide separation and purification techniques. Can you suggest any other reading in these areas? Answer

14. I don't understand the real meaning of the Km value and how it relates to an enzyme or rate of the reaction for that matter. It could be defined as [s} when Vmax = 1/2 Vmax but that doesn't really help me to understand what it is used for. It can also be defined as a "kinetic paramater" or as compared to [s] when it is much greater than or much less than the Km. I am having some trouble understanding the whole picture. Is there a definition of the actual Km in terms of how it affects an enzyme or the rate of the reaction? Answer

15. What are Km and Vmax and what do they represent? Answer

16. Is there an equation that relates the aformentioned to turnover rate? Answer

17. Why isn't Km affected by [E] when Km = Vmax/2? I see why Vmax is affected by [E] as Vmax = k3 [E]. The more enzyme there is, the more quickly product can be formed. Answer

18. Also, I have a concern about the lectures. It seems to me that you are rushing through the material. Is it that there is a lot of material to cover and not enough time to cover it? If not, I was wondering if you could slow down a bit? Your teaching is otherwise enjoyable. Answer

19. My question concerns 3-5 c., in which the different Km values and the substrate value are given. In terms of answering this question, what are the parameters of a saturating value of substrate (how do you draw the conclusion that this concentration of substrate is saturating in both cases relative to the Km values)? Answer

20. Once that assumption is made, does it logically follow that Vmax is reached in both cases due to this saturation? Answer

21. Also, does that mean that the sick person's enzyme takes longer to reach the same Vmax, or merely requires more substrate, and in fact the functioning of the "sick" enzyme occurs essentially at the same rate, and therefore doesn't cause disease? Answer

22. Additionally, is the slower Vmax value for a non-competitive inhibitor an indication that fewer functioning enzyme molecules are available, as opposed to a measure of the impaired enzyme's inhibited functioning? Answer

23. And, finally, the Km values of a non-competitively inhibited reaction must be different from the Km value of its non-inhibited counterpart, due to the differing Vmax values? In recitation and lecture, I have written that Km doesn't change in non-competitively inhibited reactions. Answer

24. Please recall problem 2-9E, in which we take given subpeptides and arrange them in a different order to get a new peptide. Clearly the two peptides will have the same charge and molecular weight. But, by re-arranging the order of the AA, is it not possible that the solubility (hydrophobicity) of the peptide will be effected a bit, for the change in ordering will cause a change in folding? So perhaps certain non-polar groups will be brought into closer proximity with other non-polar groups. Maybe the change will not be enough for separation by paper chromatography, but am I correct in thinking there MAY be SOME change in solubility as a result of the change in folding? Answer

25. Non-competitive inhibitors bind to the catalytic site of an enzyme, I believe. Is there anything else of note which occurs at the catalytic site-- is this, for example, where co-factors may bind? Answer

26. Could you explain to me exactly what is the difference between a noncompetative and an allosteric inibitor? They both seem to do the same thing, ie. bind somewhere apart from the active site on an enzyme and cause a conformational change in its shape. Answer

27. After the three lectures on the krebs cycle and related subjects I am as confused as the image presented on the last handout you gave us that showed everything. I have read becker and scanned purvis. Is there a way to synthesize the main points and organize them in a "reasonable" fashion? Answer

28. I am a bit confused on one of the problems on problem set #4. In problem #4-7, part A, the answer says that "given 10 hours and enzyme the reaction should reach equilibrium; at equilibrium, K/Q = 2 regardless of starting concentration." I understand that K/Q = 2 at equilibrium regardless of starting concentration, but it is NOT the case that equilibrium will be reached regardless of starting concentration, correct? Am I correct in thinking that in the current problem, the reaction will proceed toward equilibrium because, even though the standard free energy is positive going from left to right, the free energy (delta-G) will be negative because the concentration of Q is sufficiently greater than the concentration of K? If you asked the same problem, only the starting concentration of K was 6 mmoles and the starting concentration of Q was 5 mmoles, then even with enzyme, the reaction would not reach equilibrium, would it? In that case the free energy will be positive going from left to right-- the concentration of K with respect to Q will not be great enough to compensate for the positive standard free energy. Answer

29. I'm a little puzzled by the test we took on Tuesday in your 2005 course. In class, we spent a good deal of time on protein separation, and yet it failed to appear in even a single part of any of the three questions. Similarly, inhibition and pathway problems were emphasized in the problem sets, and neglected on the exam. What confused me most was the first question, dealing with disaccarides. The time spent on this subject was small in comparison with the topics I mentioned above. While I understand that every topic cannot find its way to the exam, I felt that the three questions you gave did not adequately measure our understanding of the material given during these first weeks. If there had been more problems covering a wider range of material, perhaps the test would have felt more balanced. As it was, I left feeling as if all the studying I put into adequately rounding my understanding of all the topics simply did not pay off when it came to the test. In the future, I was hoping you might try to clue us in to the topics you were going to emphasize on the exam, either through the amount of time you devoted to them during your lectures, or with a few words before the actual test. Answer

30. I have a question about the isomerizations which occur in the glycolytic pathway and the Kreb's cycle: Do these isomerizations require even a minimal energy input, or rather do the atoms rearrange themselves spontaneously because the new structure is more stable? Answer

31. I have a question about the isomerization of citrate to isocitrate in the Krebs cycle. I imagine this was addressed in lecture but I must not have understood its importance at the time. The goal of this step is to convert citrate, a tertiary alcohol, into isocitrate, a secondary alcohol which is more readily oxidizeable. However, since the two ends of the citrate molecule are identical, the molecule is symmetrical, and either end could accept the OH group to complete the isomerization. Thus, the only way we could keep track of which carbons come from where is if the enzyme controlling the isomerization differentiates between the CH2COO- that comes from acetyl and the one that was already there from oxaloacetate. Can it really do this, or is there something else that I'm missing? Answer

32. I am a little confused, perhaps I wrote something down wrong... When S>>Km, K3 = Vmax/Eo = turnover # Is this equality true?? If not, what is the equality for the turnover #? Answer

33. I have a question about problem 5-3-c-iii. If yeast are only getting their energy from substrate phosphorylation of lactate and not the NADH2+, shouldn't they die because they only get 1 GTP per Lactate, and this is consumed transporting the NADH2+ to the mitochondria to oxidize it? Answer

34. Regarding prob. 5-5 C: I thought electron transport directly affected the proton gradient, such that if ATP production increases and the processes of e-transport and oxidative phosphorylation are coupled, then NADH2 reoxidation would also increase. Answer

35. Also regarding problem 5-5C, what other factors influence the lowering of the pH in a cell and are these constant factors that should always be considered, or is it just in this problem that the test tube environment allows for outside pH manipulation? Answer

36. Problm 5-7: More ATP and not less? I am missing this point. Answer

37. Problem 5-9 D: Is this answer implying that the only way energy can be derived from pyruvate is via glycolysis, and not through fermentation? Or glycolysis and then Krebs? Answer

38. Problem 5-11 C: this is related to 5-5C, in that the manipulation of the proton gradient, in the case of oxidative phophorylation being uncoupled from e-transport, must then be achieved by another source. Is this the case? Is the proton gradient being increased or relieved by another oxidizing agent? Answer

39. In Becker, they use an analogy for this proton pump, by saying it's like a water turbine pump that has driven water so far up that no more can be pumped up due to back pressure, and furthermore, water is poured onto the top of the standpipe, increasing this back pressure. What is actually increasing this back pressure? Is it only the reverse action of the ATPase? Answer

40. In glycolysis, what determines if pyruvate progresses to lactate or ethanol and CO2? Is is just the presence of the appropriate enzyme? Answer

41. In Krebs Cycle, your handout lists the carboxylic groups as COOH and in Becker page 335, the carboxylic groups are listed as ionized (COO-). At intracellular pH, what form are the carboxylic groups? Answer

42. To prepare for the Krebs Cycle, when pyruvate converts to acetyl-CoA, where do the protons come from to make NAD+ to NADH and H+? I see that the electron from the COO- is transfered to NAD+ to reduce it to NAD and the H from the CoA-SH makes NADH, and the carboxylic group splits off into CO2. But where does the extra proton come from to make the H+ for NADH and H+? What am I missing? Answer

43. #4-9c: Why isn't the .001 mole of ATP the limiting factor in the > glycolysis equation (instead of the .01 mole of NAD) since, for every > mole of glucose, 2 moles of ATP are consumed before 4 are generated? Answer

44. #4-10b: Are the answers to subparts b & c dependent on the exact same analysis of symmetric compounds, namely that: in b: at any point in time you have an equally likely chance of finding the radioactive C in the top or bottom carboxyl group because succinic acid looks the same upside-down as right-side up (ie there really is no "right"-side up); and in c. there's an equal chance of the radioactive C being in either carboxyl group of OAA because back in a previous step, when fumaric acid was oxidized into malic acid, the oxidation occurred closer to one or other of the carboxyl groups of fumerate because of the 50/50 chance of finding fumerate in one or another position? Answer

45. #4-15c: Is an alternative analysis possible -- instead of presuming that rxn 1 cannot proceed forward to produce G6P unless ATP is first present as a reactant, can you assume that rxn 1 represents a coupled rxn where the breakdown of ATP drives the phosphorization of glucose (which would otherwise be uphill) Answer

46. and then consider whether the reverse rxn for the breakdown of G6P would be spontaneous if it were uncoupled from the ATP rxn (calculating the delta G of the reverse G6P rxn to then be -4kcal/mole)? Are you prevented from doing this b/c you'd need a new enz (not hexokinase) to drive the uncoupled rxn? Answer

47. #5-3B(i): Should the pathway include as an initial step the oxidation of lactate into pyruvic acid (using NAD as the oxidizing agent)? This would mean, if calculating energy, that more energy produced as a result of greater amt of NADH2 generated. (or does the K.C. encompass this step? Answer

48. #5-4B: I understand that the higher pH outside the vesicle indicates that protons are being pumped into the vesicle instead of outside and thus that the inner membrane must be inside out. I also understand that oxphos (ATP synthase complex) is not functioning (as indicated by fact that ADP concentration not changing and proton concentration not in steady state). BUT those 2 facts are distinct from each other, right? I.e., couldn't oxphos continue even with an inside-out inner membrane? Answer

49. #5-9: With respect to the givens, if bacteria uses less glucose per unit time with aerobic respiration, wouldn't its growth rate be faster in the presence of oxygen (given the same amount of glucose)? Answer

50. #5-9E: I don't understand the Answer Key's statement, "assuming regulation works to prevent abortive entry into Krebs etc.)." What is an "abortive entry" and how would it bypass a backup due to a bad/missing enzyme down the line? Answer

51. #5-10C: I understand the rationale of the problem, but in terms of the details of the Krebs Cycle, where exactly would malonic acid inhibit the Cycle? Is malonic acid related to malic acid? Answer

52. If cyanide interferes with only 1 cytochrome, why does ATP production stop completely? Can't the rest of the cytochromes do the job, only slower? Won't H+ still be pumped out, and consequently let back in by the FoF1 complex? Answer

53. In the conversion of one mole of glucose to ethanol and carbon dioxide, which 2 carbon atoms of the glucose molecule come off as carbon dioxide? I know it is not 1 or 6. Answer

54. Since we get 10 NADH2 and 2 FADH2 produced per glucose molecule, it seems we should have 12 H2O to account for rather than What is the story with the 6 H2O not accounted for in the equation for glucose breakdown. Are they simply understood? Answer

55. For problem 5-10B, how are we getting 36 ATP from three acetate? Answer

 

 

 

 

 

1. Can I use Curtis instead of Purves? instead of Becker?

For Purves, yes; for Becker, no. See Syllabus.

2. In your handout about the Flow of Glucose in E. Coli, glucose is shown on the bottom of the sketch undergoing biosynthetic pathways and intermediates to become monomers. Is glucose a monomer of polysaccharides or is it only the "initial material" required to make monomers?

Both. Glucose serves as both a monomer (in the formation in E. coli of disaccharides such as sucrose and maltose) and polysaccharides such as glycogen) and as the starting material for all else organic in the E. Coli cell.

3. In reference to hydrocarbons and their inability to dissolve in water is this because they are non-polar and therefore not likely to have hydrogen bonding to water molecules as occurs with NaCl in aqueous solutions?

Yes.

4. Is an "induced dipole" called such because partial charges are created when the atoms are forced to come together?

Not exactly. The molecules needn't be "forced" to come together. As 2 molecules approach each other, say from just random motion, each will have a particular distribution of charge. A molecule that happens to have an excess of negative charge (because it is a fluctuating dipole) will repell the electrons on a molecule it approaches very closely. This will leave a positive charge on the second molecule, so that the 2 molecules can now form a bond by electrical attraction. Thus the bond is "induced." For the puposes of this course we need only realize that as two molecules, any two molecules, approach each other very closeley, their is a natural tendency to form a weak bond. The distance is critical (bond energy falls off as a function of the 6th power of the distance, as I recall), so this van der Waals bonding takes place only within a very very narrow range of distance.

5. Are the cyclo glucose structures called alpha because the hydroxyl group on carbon 1 axial while the same hydroxyl group on carbon 1 equatorial in beta?

Alpha and beta are used to indicate the different positions the hydroxyl could take. If we are drawing the flat (inaccurate) form of the ring, the depiction is Up or Down. But if we are drawing the chair form, the depiction, and the difference, is equatorial (out) or axial (UP or DOWN, parallel to a north-south AXIS). If you study the two types of depictionside by side, in hand-out or in the texts, you can satisfy yourself that at each position the relationship between Up and Axial or equatorial can be different. Up and Down will always be true relaative to each other at a given carbon. But Up at one carbon could be axial while up at another carbon could be equatorial. Diffficult to show this without pointing to the pictures. Come to office hours if you are still confused.

6. Ethanol is hydrophilic -- is this because of the hydroxyl group and more importantly the partial negative charge of the oxygen in the functional group?

Yes, as well as the concomitant partial positive charge on the hydrogen.

7. You mentioned that hydrophobic forces would get stronger at higher temperatures -- Why?

Because the water molecules will be more disordered at higher temperature, their organization into a cage around the hydrophobic molecules will result in a larger difference between the (highly) disordered and ordered states.

8. I'm speaking in behalf of myself and a few other kids who have had the same confusion. Some of us have been buying both Becker and Purves, while others have only bought one of the two, saying only one of the two books is enough. It seems that the handout says either book is fine, but the handout also says that any college bio book is fine. It just seems wierd -- as if we're not basing the class on a standard text at all. If that's the case it's fine with me, but it just seems so opposite from what a traditional science class should be like. P.S. Just in case I don't get to talk to you in person tomorrow I have this question also: I have a biology textbook called: Biological Science, 5th Edition, by Keeton and Gould (copyright 1993). It's the book I used in Advanced Biology in highschool, but it's used in colleges, at least Yale's Intro to Bio course. When you mention in the handbook that "Keeton and Gould is okay" is this the book (note edition and year) that you mean?

It is true that C2005 is not based on a standard textbook, or any particular textbook, so it may be non-traditional. The Becker text is required and will suffice for all but the genetics part of the course. The genetics part of the course is substantial, so you do need a text that covers geentics at an introductory level. We like Purves et al., but any general biology inroductory college level text may also be used, incuding 1993 edition of Keeton, which is a fine book. You could avoid buying a general intro text and do your genetics reading at the library, but these texts will also be used in the second semester, when the non-Becker text (Purves or whatever you have) will be used for physiology and developmental biology.

9. Isn't there a mistake in the Exponential Growth handout, on line 3 of the second paragraph, it says " If we started with 100 cells, after 1 gen. we would have 200 and after 4 gens. 400, etc." ?

Yes, that's a mistake. It should read: If we started with 100 cells, after 1 gen. we would have 200 and after 2 gens. 400, etc. {Or alternatively: If we started with 100 cells, after 1 gen. we would have 200 and after 4 gens. 1600, etc.} Sorry.

10. By what means does dialysis separate a protein? I thought it was used as a semi-permeable sac to allow wanted species into the protein and unwanted species out of a protein.

Indeed, you are correct, dialyis is used usually to remove small molecules away from large molecule such as proteins, and not as a means to separate different proteins from each other.

11. How do you determine the sequence of a polypeptide when given different subpeptide sequences? i.e. 2-9D What is the order of amino acids in the polypeptide?

Once you have the sequence of the subpeptides, they themselves must be arranged in the right order to give the sequence of amino acids in the entire polypeptide. We hardly covered this in lecture, although it is among the problems.

To determine the order, you need TWO sets of sub-peptides, each produced by a different proteolytic enzyme, that is, two enzymes that cut the polypeptide at different places (e.g., one cuts after arg and lys whereas the other cuts after phe and tyr). Knowing the sequences of every subpeptide in the two sets, you can then align them via their overlapping sequences, a tedious but logical operation. Actually kind of a challenging puzzle, as presented in the problem set.

12. Regarding problem 2-18, I was confused with this question because I didn't understand if "molecule" was used for "total protein" in the answer. It was a very in-depth problem. Is there any way to go over it in class?

"Molecule" is a flexible term. The entire native protein can be considered one molecule, as in "a molecule of hemoglobin has a molecular weight of 64,000." Similarly, a molecule of hemoglobin can be denatured to produce 4 molecules of subunits, 2 alpha chain molecules and 2 beta chain molecules.

13. There's precious little information in Becker and Purves about protein/polypeptide separation and purification techniques. Can you suggest any other reading in these areas?

Most biochemistry textbooks treat most of the protein purification methods mentioned in lecture. However, the treament is often more detailed than is necessary or desirable for this introductory course. We prepared a rather lengthy handout to address this deficiency of treatment in the recommended texts.

I have recently read the section on protein separation techniques in Lodish et al.'s Molecular Cell Biology (3rd Ed., Freeman) and find it close to the level of the lecture. Check the index for electrophoresis to find the subject.

14. I don't understand the real meaning of the Km value and how it relates to an enzyme or rate of the reaction for that matter. It could be defined as [s} when Vmax = 1/2 Vmax but that doesn't really help me to understand what it is used for. It can also be defined as a "kinetic paramater" or as compared to [s] when it is much greater than or much less than the Km. I am having some trouble understanding the whole picture. Is there a definition of the actual Km in terms of how it affects an enzyme or the rate of the reaction?

The rate of an enzyme catalyzed reaction is highly dependent on the concentration of substrate that is presented to the enzyme. This dependence is described inthe Michaelis-Menton equation:

Vo = (Vmax x [S])/(Km + [S]) where Km = (k2 + k3)/k1 by definition

From this equation, one can see that Km influences Vo (the initial rate of the reaction) as it relates to [S]. The influence is not very obvious from looking at the equation, as S is both in the numerator and the denominator. The resulting curve of Vo vs. S, using this equation, starts out as a straight line at low S but then slows down and eventually flattens out at high S. The flattening out is when the enzyme is saturated with S, so that adding more S no longer can increase the rate by "driving" the reaction. Enzymes that have a low Km will be saturated with S at lower values. These enzymes can capture S more efficiently, even at low S concentrations; they have a high affinity for S. In contrast, enzymes with high Km's don't saturate unless high concentrations of S are present; they have trouble binding S, they have a lower affinity for S. The relationship between an enzyme's Km and its affinity for S can also be rationalized by seeing that Km ~= k2/k1 if we assume that k3 << k2; and that k2/k1 equals the dissociation constant for the reaction ES -> E + S, the DISsociation constant of the enzyme-substrate complex. (The dissociation constant is simply the equilibrium constant for the dissociation reaction.) So a high Km means better dissociation, or less affinity of the enzyme for its substrate. And vice-versa.

15. What are Km and Vmax and what do they represent?

Km has several defintions, that is several ways of looking at the same thing:

1- Km = (k2 + k3) /k1, from the derivation of the Michaelis Menton equation, this is by definition. The small k's are rate constants defined by the laws of mass action: e.g., for th ereaction A -> B, th erate of A disappearance is dA/dT = k1A - k2B, the forward reaction being driven by the concentration of A and the back reaction being driven by the concentration of B.

2- Km is the concentration of substrate (S) at which the initital velocity (Vo) of an enzymatic reaction achieves half the maximal velocity (Vmax). Vmax is the maximum initial velocity at which a given amount of enzyme (Eo) can catalyze the reaction when substarte is not limiting, that is, when the substrate concentration is saturation, when increasing the substrate concentration produces no further increase in initial velocity.

3- For many enzymatic reaction, k3 is much less than k2, and so Km approximates k2/k1 (from (1) above). k2/k1 represents the equilibrium constant for the dissociation of the enzyme-substrate complex (ES)into free enzyme (E) and substrate (S). Km in these cases then is an inverse indicator of the affinity of the enzyme for its substrate, since a relatively higher Km means the ES complex will tend to DISSOCIATE more.

16. Is there an equation that relates the aformentioned to turnover rate?

Turnover number is the number of moles of substarte that the a mole of enzyme can convert to product per second when the enzyme is satutated with substrate, that is, at the Vmax odf the reaction. Fromthe Michaelis-Menton equation, at infinite (S), Vo = Vmax = k3Eo, and rearranging, k3= Vmax/Eo. k3 is then the turnover number, which can also be visualized as the maximum number of molecules of substrate that one molecule of enzyme can convert to product in one second.

17. Why isn't Km affected by [E] when Km = Vmax/2? I see why Vmax is affected by [E] as Vmax = k3 [E]. The more enzyme there is, the more quickly product can be formed.

Km is defined in the derivation of the Michaelis-Menton equation as = (k2 + k3)/k1. The small k's, in turn, are defined as rate CONSTANTS; that is, they are the constants, which, when multiplied by the reactants' concentrations, determine the rate of the reaction. Consider the association of substrate with enzyme (ignoring the conversion to product to simplify this argument): E + S = ES. The rate of ES formation can be DEFINED as d(ES)/dt= k1(E)(S)-k2(ES), by the law of mass action. The equilibrium CONSTANT for this reaction is Keq=(ES)eq/(E)eq(S)eq. At equilibrium, d(ES)/dt must equal 0. So k1(E)eq(S)eq=k2(ES)eq and k1/k2=(ES)eq/(E)eq(S)eq = Keq. So here again we can see that k1 and k2 are constants.. Similarly for k3 and k4. So Km is a constant for a given enzyme and its given substrate. It does not vary with (S). It does not vary with E, with the amount of E. It modifies the amount of E, that is, E is multiplied by its components (the little k's) to determine the rate of the reaction (via k3) and the affinity of binding (via k1 and k2)

18. Also, I have a concern about the lectures. It seems to me that you are rushing through the material. Is it that there is a lot of material to cover and not enough time to cover it? If not, I was wondering if you could slow down a bit? Your teaching is otherwise enjoyable.

I am talking as slow as I dare.

19. My question concerns 3-5 c., in which the different Km values and the substrate value are given. In terms of answering this question, what are the parameters of a saturating value of substrate (how do you draw the conclusion that this concentration of substrate is saturating in both cases relative to the Km values)?

Looking at the shape of the Vo vs. S curves that we have drawn, that result from plotting the Michaelis-Menton equation, one can take note of the Km on the abscissa (X-axis), which is 10-5M for the normal person, and then look at how Vo is behaving at 10 times this value, which is the 10-4M given as the substrate concentration. One can see in this way that at 10-4M, you are at 10 times the Km, you are out near the plateau value, which is the Vmax. For the sick person, with the even lower Km of 10-6M, you are even a bit closer to the Vmax, at 100 times the Km at 10-4M substrate. you ar every near the Vmax.

20. Once that assumption is made, does it logically follow that Vmax is reached in both cases due to this saturation?

Yes. Each person's Vmax is reached at saturation, by definition, since the max in Vmax refers to the fact that a maximum responsiveness to increasing S has been achieved. That is, Vmax is none other than the maximum Vo than can be achieved with respect to [S], for a given amount of enzyme.

21. Also, does that mean that the sick person's enzyme takes longer to reach the same Vmax, or merely requires more substrate, and in fact the functioning of the "sick" enzyme occurs essentially at the same rate, and therefore doesn't cause disease?

Your use of the word "longer" confuses me. In a Vo vs. S plot, it is not time that is the independent variable on the abscissa, but rather S concentration. Time figures in only in the consideration of the Vo points themselves that make up the curve, each ordinate (y-axis) value representing a rate, or product appearance in the reaction per minute. If by "longer' you were speaking only in analogy, meaning you need more S to achieve a certain Vo, then you have it backwards: the sick person's enzyme doesn't require MORE [S] to achieve a given Vo for a given amount of enzyme, it takes LESS [S], because its Km is, contrary to expectations, LOWER than the normal person's enzyme. That is, the sick person's enzyme is even better than the normal person's inasmuch as it binds the substrate with more affintity. Thus one is barking up the wrong tree, probably, in looking at this enzyme as a cause of the person's illness.

22. Additionally, is the slower Vmax value for a non-competitive inhibitor an indication that fewer functioning enzyme molecules are available, as opposed to a measure of the impaired enzyme's inhibited functioning?

I do not see the difference in these 2 possibilities: they are one and the same. It is because the enzyme is impaired that it is unavailable for catalyzing the reaction.

23. And, finally, the Km values of a non-competitively inhibited reaction must be different from the Km value of its non-inhibited counterpart, due to the differing Vmax values? In recitation and lecture, I have written that Km doesn't change in non-competitively inhibited reactions.

A non-competitive inhibitor, by definition, does not compete with the substrate for binding to the enzyme. Since Km is a measure of substrate binding to the enzyme, a non-competitive inhibitor does not affect Km. The non-competitively inhibited enzyme molecule still bind subsrate, but those molecules that also have bound the inhibitor are unable to catalyze the reaction. In a sense, catalyically speaking, it is as if they were not there. So the Vmax is lowered, since with less enzyme you cannot get as much catalysis, as high a Vo; you cannot achieve the old Vmax (i.e., the uninhibited Vmax). Mechanistically, a non-competitive inhibitor could act by binding to those parts of the active site that help catalyze the reaction, but not those that help bind the substrate.

24. Please recall problem 2-9E, in which we take given subpeptides and arrange them in a different order to get a new peptide. Clearly the two peptides will have the same charge and molecular weight. But, by re-arranging the order of the AA, is it not possible that the solubility (hydrophobicity) of the peptide will be effected a bit, for the change in ordering will cause a change in folding? So perhaps certain non-polar groups will be brought into closer proximity with other non-polar groups. Maybe the change will not be enough for separation by paper chromatography, but am I correct in thinking there MAY be SOME change in solubility as a result of the change in folding?

A good point. One can certainly expect that two proteins with the same amino acid composition but a different order of those amino acids, would fold differently, and thus could well expose to the solvent, in the native state, different amounts of hydrophobic groups, and thus affect the solubility of the folded proptein. The difference in solubility might be manifested by a lower solubility in water, or a different sensitivity to insolubility in high salt solutions. But the solubilities exploited in in paper chromatography are exhibited by denatured polypeptides, since the organic solvents used represent conditon too harsh for most proteins to retain their native folded state (large polypepides would probably precipitate and so cannot be analyzed by paper chromatography). In the denatuired condition, a peptide would be likely to have a solubility that reflects the overall composition of hydrophobic amino acids.

25. Non-competitive inhibitors bind to the catalytic site of an enzyme, I believe. Is there anything else of note which occurs at the catalytic site-- is this, for example, where co-factors may bind?

Co-factor may be an ambiguous term in this respect. If the co-factor is bound to the enzyme so tightly that the enzyme always has it bound, i.e., it is a part of the enzyme, a prosthetic group, then the "co-factor" (i.e., the prosthetic group) can be considered to be part of the active site and part of the catalytic site. More usually, the co-factor is a recyclable SUBSTRATE for th enzyme, as in the oxidative co-factor NAD. There is a substrate binding site within the active site for the co-factor, and it behaves in all respects like a second substrate. In this case, although the co-factor is "at" or "in" the catalytic site, it is not part of it as part of the enzyme, but as an outside temporary occupant.

26. Could you explain to me exactly what is the difference between a noncompetative and an allosteric inibitor? They both seem to do the same thing, ie. bind somewhere apart from the active site on an enzyme and cause a conformational change in its shape.

My explanation of the distinction was: The active site can include side chains that carry out the catalysis but do not contribute to the specificity of binding. That is, some active sites can comprise a substrate binding site and a distinct catalytic site. For active sites sites such as these, an inhibitor could interfere with the cataltyic site without having any effect on the ability of the substrate to bind to the substrate-binding site. Catalysis will be inihibited, but the addition of an excess of substrate will not relieve this inhibition, since the substrate is already binding without interference. This type of inhibition must be distinguished from competitive inhibition, which DOES affect substrate binding. So we can call this inhibition NON-competitive, in distinction.

Allosteric inhibitors bind to sites that are remote from the active site, remote from the catalytic site and the substrate binding site, even if they are separable. Allosteric inhibitors bind to specially designed allosteric sites, designed to fit the inhibitor, as these are physiological inhibitors. Upon binding its allosteric inhibitor, an enzyme will alter its shape such that the active site is distorted at its remote location. In its distorted state, it cannot catalyze the reaction, usually because it cannot bind substrate. The remote site is often on a separate polypeptide within the quaternary structure of the enzyme.

27. After the three lectures on the krebs cycle and related subjects I am as confused as the image presented on the last handout you gave us that showed everything. I have read becker and scanned purvis. Is there a way to synthesize the main points and organize them in a "reasonable" fashion?

I need to know the source of your confusion in order to provide you with a useful answer. The "outline of energy metabolism" summarizes where the ATPs will be coming from in about as simple a manner as possible, although you should gain a more detailed understanding of the pathways than this summary of course. Perhaps you should read Purvis completely to get an only slightly simpler picture, and then reread your notes or Becker. To make this task easier, you could use your notes as a guide to know what to skip in these chapters, as both text include subjects that I omit.

The final handout was extremely complex and difficult to read, but it is far from confused; just the opposite. In any case, it is to be viewed from about 8 feet away, to see the forest in this case, not the trees.

28. I am a bit confused on one of the problems on problem set #4. In problem #4-7, part A, the answer says that "given 10 hours and enzyme the reaction should reach equilibrium; at equilibrium, K/Q = 2 regardless of starting concentration." I understand that K/Q = 2 at equilibrium regardless of starting concentration, but it is NOT the case that equilibrium will be reached regardless of starting concentration, correct? Am I correct in thinking that in the current problem, the reaction will proceed toward equilibrium because, even though the standard free energy is positive going from left to right, the free energy (delta-G) will be negative because the concentration of Q is sufficiently greater than the concentration of K? If you asked the same problem, only the starting concentration of K was 6 mmoles and the starting concentration of Q was 5 mmoles, then even with enzyme, the reaction would not reach equilibrium, would it? In that case the free energy will be positive going from left to right-- the concentration of K with respect to Q will not be great enough to compensate for the positive standard free energy.

Your mistake is in thinking that equilibrium will be reached only if you have some certain ratio of reactants initially. The opposite is the case. Equilibrium will always be reached in a closed system, eventually, no matter how much of whatever you start with. All reaction proceed toward equilibrium. Indeed how could they do otherwise? They will follow the delta G until they reach equilibrium. And the "eventually" is actually quite quickly, when a good catalyst such as an enzyme is present. That the enzyme in this problem was sufficient to allow the reactants to reach equilibirum was indicated by the fact that there was no further change betwen 1 hour and 10 hours. That information tells you that equilibrium was reached (with no further net change in reactant concentrations). What about the "closed system" qualification? Well, equilibrium will not be reached if you are constantly putting more reactants into the system, or taking products away. If you mix simple soluble reactants in a test tube, that is not happening, so you do achieve equilibrium. In a living cell, however, you are almost always interfering with this achievement of equilibrium: eating more food, breathing out carbon dioxide (letting product escape), excreting, removing the product of one reaction by having it react in a subsequent reaction, etc. So chemical reactions in living systems are seldom at equilibrium. But equilibrium constants are nevertheless very useful in biochemistry, as they allow us to easily measure delta Go, which in turn allows us to see what reactions are or are not "spontaneous" and so where the problems lie in coupling particular reactions to make them energetically favorable.

We can use your examples and delta G to make this point mathematically: When the initial concentrations of K and G are 50 and 5 mM (as in 4-7A), then delta G for the reaction, as written, K -> Q, will initially be -1 kcal/mole. The negative value means that the reaction will indeed run left to right, which is the direction it must take if it is to achive a final Q/K ratio of 0.5 (Q/K being the equilibrium constant of the reaction written in this direction). Using your proposed initial concentration of 6 and 5 mM for K and Q, the delta G works out to be +0.3 kcal/mole. The positive number means that the reaction will run right to left, just the direction it needs to in order to proceed toward the equilibrium ratio of Q/K=0.5. I leave it to you to confirm my delta G calculations.

29 I'm a little puzzled by the test we took on Tuesday in your 2005 course. In class, we spent a good deal of time on protein separation, and yet it failed to appear in even a single part of any of the three questions. Similarly, inhibition and pathway problems were emphasized in the problem sets, and neglected on the exam. What confused me most was the first question, dealing with disaccarides. The time spent on this subject was small in comparison with the topics I mentioned above. While I understand that every topic cannot find its way to the exam, I felt that the three questions you gave did not adequately measure our understanding of the material given during these first weeks. If there had been more problems covering a wider range of material, perhaps the test would have felt more balanced. As it was, I left feeling as if all the studying I put into adequately rounding my understanding of all the topics simply did not pay off when it came to the test. In the future, I was hoping you might try to clue us in to the topics you were going to emphasize on the exam, either through the amount of time you devoted to them during your lectures, or with a few words before the actual test.

I agree that an element of chance is present when an exam does not include all of the important topics discussed in lecture, and I can understand your frustration in having mastered some of these important topics only to find that this mastery was not to be recognized. However, I purposely decided to limit the number of topics for 2 reasons: (1) I would like to minimize the time element in testing a student's understanding of the material. I don't much care if it takes one student twice as long as another to figure out the right answer, as long as it is the right answer, i.e., the basic understanding is there. Consequently I want to keep the exam reasonably short. (2) That said, then my choice is to cover all the important topics relativly lightly, or to cover less topics, but to ask more probing questions that require more time to answer. I opted for the latter.

I consider my primary purpose is to teach the material, to convey and promote and understanding of the material. The means I have toward this end are the lectures, guidance on the reading, providing instructional problems, and, also, giving examinations. So I consider the examinations are one of the teaching tools. The exams are designed to test a deep understanding of the material, and to de-emphasize rote memorization. Usually I try to present students with a new situation, one that we have not discussed before: a Martian protein, a new drug, an enzyme with 3 substrates, etc. If properly designed, such questions cannot be completely answered by simple repetition of facts. In designing such questions, I occasionally fail, and the question becomes abstruse. I take this risk because I believe that a student anticipating an exam of this type will force him/herself to do the right kind of studying, studying that will result in a deeper understanding of the material, learning the principles and the reasoning and not just the facts. Despite the fact that most of the problems are questions of this type, my impression is that for many students the necessity of studying the material in this way is not brought home until after the first exam. So the exams are a teaching tool designed to get students to study in what we consider to be the right way.

Of course, I realize that the grades on these exams are also an important differentiator, important in a very real way for the future of many students. So it is also important that they be fair. But I don't consider the omission of some topics to be inherently unfair. I expect students to study all of the material presented; to leave out any important part is a not a safe strategy, and I don't want it to be.

I do not agree that the subject of polysaccharides was minor. I tried to emphasize the importance of the subject as a perfect example of the principle of how function is related to chemical structure, a recurring theme in this part of the course (I even went as far as to borrow my daughter's tinker toys to make the point). I am confident that I spent more time on sugars and polysaccharides than on ultracentrifugation or feedback inhibition.

Over the course of 4 exams, discrepancies in studying emphasis such as you described are likely to even out. Finally, we do have something of a safety net for a student who does poorly in one particular subject, in that the lowest exam grade is dropped in calculating a final grade.

30. I have a question about the isomerizations which occur in the glycolytic pathway and the Kreb's cycle: Do these isomerizations require even a minimal energy input, or rather do the atoms rearrange themselves spontaneously because the new structure is more stable?

Isomerizations can have either positive or negative standard free energy changes. These standard free energy changes are usualy smaller than those associated with hydrolyses or oxidation-reductions. See for example the standard free energy changes for the reactions of glycolysis, which include several isomerizations or similar rearrangements, i.e., reactions 2, 5, 8, and 9 as listed on the back of the glycolysis handout. All are less than 2 kcal/mole standard free energy change.

31. I have a question about the isomerization of citrate to isocitrate in the Krebs cycle. I imagine this was addressed in lecture but I must not have understood its importance at the time. The goal of this step is to convert citrate, a tertiary alcohol, into isocitrate, a secondary alcohol which is more readily oxidizeable. However, since the two ends of the citrate molecule are identical, the molecule is symmetrical, and either end could accept the OH group to complete the isomerization. Thus, the only way we could keep track of which carbons come from where is if the enzyme controlling the isomerization differentiates between the CH2COO- that comes from acetyl and the one that was already there from oxaloacetate. Can it really do this, or is there something else that I'm missing?

The answer is that the molecule indeed appears symmetrical and yet the enzyme can in fact distinguish between the 2 apparently identical ends. When citrate is formed by the addition of acetate to oxaloacetate, the acetate is being added to a non-symmetric molecule, and always gets added to the same side of the C=O group. Thus the donated acetate group maintains its identity in 3-dimensional space, even if it appears indistinguishable at first glance. The enzyme aconitase distinguishes the two -CH2-COOH groups by making a 3-point contact with this substrate. Imagine a citrate drawn with the labeled acetate-derived -CH2-COOH group at the top. If the enzyme always grabs an hydroxyl with its left hand, a -COOH with its right hand, and a -CH2-COOH with its knees, then the labeled -CH2-COOH group will be sticking up and out. Rotating the molecule will not change things: since the enzyme always has to grab the -OH with its left hand, it will have to rotate also, and the labeled acetate will always be the one sticking out. This ability to distinguish between two seemingly identical groups by a 3-point contact is known as the Ogston Effect and is usually described (in less anthropomorphic terms) in any biochemistry textbook. We leave it out to save time at this introductory level.

32. I am a little confused, perhaps I wrote something down wrong... When S>>Km, K3 = Vmax/Eo = turnover # Is this equality true?? If not, what is the equality for the turnover #?

When S>>Km, V=Vmax. Vmax is always k3 x Eo, it is defined that way. The Michalis-Menton Equation is derived as Vo=(k3EoS)/(Km + S). At S>>Km, this reduces to Vo=k3Eo, and is the maximum value the equation can produce for Vo. Thus we can define a term Vmax whic is always = k3Eo; for a given amount of enzyme (Eo) it is a constant, not depending on S. It is also a useful concept, as we imagine the enzyme (that amount of enzyme present) so saturated with S that it is insensitive to any further increase in S. Turnover number is defined as k3. It is, as well, at the same time, the maximum number of substrate molecules that can be converted to product per second per enzyme molecule. Also: from the definition of k3: ES -> P + E, dP/dT=k3[ES]-k4[P][E]. But at saturating substrate, there is no free E, only ES, and dP/dT is at a maximum with respect to dependence on S. So dP/dT= max.dP/dT = k3[Eo], and k3=(max.dP/dT)/Eo, where Eo is the total enzyme concentration present.

33. I have a question about problem 5-3-c-iii. If yeast are only getting their energy from substrate phosphorylation of lactate and not the NADH2+, shouldn't they die because they only get 1 GTP per Lactate, and this is consumed transporting the NADH2+ to the mitochondria to oxidize it?

For problem 5-3-c-iii: Your reasoning is correct based on the simplified view given in lecture of the "cost" of transporting cytosolic NADH2 being "one ATP". In reality, the mechanism of transport of the 2 electrons present on the NADH2 formed in glycolysis is more complex, being indirect. In most tissues (I am not sure about yeast), the mechanism involves transfer of these electrons to glycerol-3-phosphate, which then gets into the mitochondria at no cost (freely diffuses); it then gives up its electrons to FAD, and the resulting FADH2 can get only 2 ATPs, instead of the expected 3 for the original NADH2. Such a mechanism is necessary because NADH2 is usually at a higher concentration in the mitochondria than in the cytoplasm, so it is the negative free energy change associated with the transfer of electrons from NADH2 to FADH2 (via this shuttling glycerol-3-P) that is driving this transport. See Becker p.366-368 and Fig. 12-26 for a more complete answer. So, since it is electron transport that is driving the transport, and not ATP directly, the 36 vs. 38 comes in only when calculating how much ATP one can get from the protons pumped (2 less per glucose because each FADH2 yields only enough for 2 ATP vs. the 3 for NADH2). For the problem, DNP is only short-circuiting the ATP's generated by oxidative phosphorylation, that is, from proton pumping. DNP does not affect electron transport, which is providing the free energy for transporting the cytosolic NADH2's electrons into the mitochondria. These get a free ride now. No ATP "cost" because the cost is moot, as no ATP is being generated by oxidative phosphorylation under these circumstances. The substrate-level GTP (ATP) generated in the Krebs cycle would still be real and a net of one per lactate.

34. Regarding prob. 5-5 C: I thought electron transport directly affected the proton gradient, such that if ATP production increases and the processes of e-transport and oxidative phosphorylation are coupled, then NADH2 reoxidation would also increase.

Electron transport does directly affect the proton gradient, as the energy from the transfer of electrons is used to pump the protons out of the mitochondria. However, it does not follow that the processes of electron transport and oxidative phosphorylation (i.e., ATP production) are directly coupled. To the contrary, they are NOT directly coupled. Oxidative phosphorylation depends only on the flow of protons into the mitochondria, through the FoF1 protein complex.

35. Also regarding problem 5-5C, what other factors influence the lowering of the pH in a cell and are these constant factors that should always be considered, or is it just in this problem that the test tube environment allows for outside pH manipulation?

As far as I know, it is only the pumping of protons out of the mitochodria using the process of electron transport that is responsible for the pH gradient in the living cell. Experimenters can affect this gradient at will, however, by adding acid (lowering the pH, increasing the [H+} concentration) or base (raising the pH, decreasing the [H+} concentration) to various preparations of mitochindira or mitochondrial membranes.

36. Problm 5-7: More ATP and not less? I am missing this point.

Let's say you ate 1 gram of carbohydrate, which represents X moles. If it were glucose, you would get 36X moles of ATP out of this by respiration. Now consider ATP, which has a molecular weight 3 times higher than glucose, as stated in the problem. So you get 0.33X moles of ATP in one gram. ATP is made up of the base adenine, the sugar ribose and 3 phophates. You would get one ATP from the ATP itself, or 0.33X moles. Then you would have ADP + Pi. Let's say you could get another from the second high energy phosphate bond in ADP, so that's 0.66X ATP. Now you could break down the AMP to Pi + adenine + ribose. Say ribose, a 5-carbon sugar, is metabolized similarly to glucose, so yields 5/6 of 36 * 0.33X moles of ATP, or 10X moles of ATPs generated, for a running total of 10.66. Now the adenine can be broken down in a manner analogous to the catabolism of protein, resulting in Krebs Cycle and glycolytic intermediates that can be coupled to ATP generation. If the adenine (5-carbons) were as good as ribose then we would get another 10 ATPs, maximum. So the end result is 20.66 ATPs, less than the 36 we get from eating glucose. So, ironically, glucose is a better source of ATP than ATP itself! That's because getting ATP from glucose or anything else means getting that third phosphate tacked back onto ADP, to form an ATP that can now be hydrolyzed back to ADP and release energy that can be coupled to otherwise energetically unfavorable reactions. The main point is that ATP you would eat here can just be used once in its direct energy-coupling way, for only 0.33X moles, a small fraction of the ADP-> ATP conversion that you can get from the catabolism of carbon compounds all the way to CO2 and H2O.

37. Problem 5-9 D: Is this answer implying that the only way energy can be derived from pyruvate is via glycolysis, and not through fermentation? Or glycolysis and then Krebs?

Glycolysis and then Krebs. In this mutant of B. examinus, glucose metabolism seems to be exactly the same (growth rate, yield) whether or not O2 is present. So some step between pyruvate -> CO2 + H2O must have been knocked out. Fermentation = glycolysis + the regeneration step for NAD (e.g., Pyr -> lactate). Pyruvate cannot yield energy from fermentation, since the only step of fermentation left is pyruvate -> lactate, which requires NADH2 and produces no ATP. Pyruvate cannot produce energy via glycolysis, since it is at the very end of the glycolytic pathway: going backwards would consume energy, not produce it, and would have a very very unfavorable delta Go. So the only way to get energy from the catabolism of pyruvate is to take it down through the Krebs cycle, and if O2 is present, through the electron transport chain as well. The mutant cannot do this, as it is oblivious to the presence of O2, and so can carry out only glycolysis, and so can do nothing with pyruvate.

38. Problem 5-11 C: this is related to 5-5C, in that the manipulation of the proton gradient, in the case of oxidative phophorylation being uncoupled from e-transport, must then be achieved by another source. Is this the case? Is the proton gradient being increased or relieved by another oxidizing agent?

In 5-5C, you have a situation, which, contrary to expectations, is producing only a single mole of ATP for each moel of acetate metabolized. Oxidative phosphorylation should produce many more ATPs than this, from all of those NADH2's produced in the Krebs Cycle running their electrons down to O2 and protons being pumped a etc. That you do get one mole of ATP suggest tha teh Krebs Cycle is still running, producing that ATP as GTP via substrate level phosphorylation. The NADH2 produced in the Krebs Cycle must be being reoxidized to NAD to shuttle back the the cycle to react with more acetate. So electron transport must be running okay, as this is how the NAD will be regenerated. But oxphos is not working. The proton gradient is not being formed or is not being utilized, somethng is amiss. It will m not be achieved any other way. It need not be. These are simply defective mitochondria in your test tube preparation. Maybe they came from a mutant, maybe they were damaged during their isolation from the cell. For the sake of the problem (and often unfortunately in real life in the lab), this happens. So your mistake is in seeking some redressing of this defect in some way.

39. In Becker, they use an analogy for this proton pump, by saying it's like a water turbine pump that has driven water so far up that no more can be pumped up due to back pressure, and furthermore, water is poured onto the top of the standpipe, increasing this back pressure. What is actually increasing this back pressure? Is it only the reverse action of the ATPase?

It is the H+ ions that are increasing the back pressure,and so running the pump backwards. In this case, backwards is the only direction you want the pump to run in. You want the ATPase to run inrverse,making ATP instead of using it. So you get your water from a completely independent source and pour it into the standpipe, and keep pouring, continuously, day and night, or as long as you have food. This independent source is the electron transport chain, it is independent of the ATP pump, the reverse-pump. The E.T.C is giving you this water (bucketfuls of H+ ions) right there at the top of the standpipe, as much as you need to keep the pump running in reverse.

40. In glycolysis, what determines if pyruvate progresses to lactate or ethanol and CO2? Is is just the presence of the appropriate enzyme?

It is the presence of oxygen, I believe. The enzymes, or the capacity to make the enzymes, for both fermentation and respiration are both present in E. coli and yeast. There are a few organisms (microorganisms) that are obligatorily fermentors; they cannot use oxygen. In additon, both glycolysis and the Krebs Cycle are highly regualted by feedback inhibition and activation (allosterically), with NADH and ATP being among the effectors that control the activity of phosphofructokinse (for glycolysis) and isocitrate dehydrogenase (for Krebs).

41. In Krebs Cycle, your handout lists the carboxylic groups as COOH and in Becker page 335, the carboxylic groups are listed as ionized (COO-). At intracellular pH, what form are the carboxylic groups?

Ionized at pH7, just as in amino acids.

42. To prepare for the Krebs Cycle, when pyruvate converts to acetyl-CoA, where do the protons come from to make NAD+ to NADH and H+? I see that the electron from the COO- is transfered to NAD+ to reduce it to NAD and the H from the CoA-SH makes NADH, and the carboxylic group splits off into CO2. But where does the extra proton come from to make the H+ for NADH and H+? What am I missing?

You can think of 2 protons from the -COOH and from the -SH groups of pyruvate and CoA. If you want to consider the fact that pyuvate is really ionized and its carboxyl protin has dissociated into the general H+ pool, then it could simply be retrieved from that pool. At pH7, there will be 10-7 M H+ ions available, and protons can be taken or donated to that pool as necessary. Overall, the 12 H's of glucose (C6H12O6) will end up on 6 molecues of water (after combining with oxygen), so all donations and withdrawals from the H+ pool even out over the course or respiration.

43. #4-9c: Why isn't the .001 mole of ATP the limiting factor in the > glycolysis equation (instead of the .01 mole of NAD) since, for every mole of glucose, 2 moles of ATP are consumed before 4 are generated?

For every 0.001 moles of ATP consumed to produce F-1,6-di-P, 0.004 moles will be PRODUCED, via the downstream steps in glycolysis. That is, until you run out of the NAD needed for one of those downstream steps. So ATP is not a problem, until then, and the reactionis liited to 0.01 moles, the amount of NAD.

44. #4-10b: Are the answers to subparts b & c dependent on the exact same analysis of symmetric compounds, namely that: in b: at any point in time you have an equally likely chance of finding the radioactive C in the top or bottom carboxyl group because succinic acid looks the same upside-down as right-side up (ie there really is no "right"-side up); and in c. there's an equal chance of the radioactive C being in either carboxyl group of OAA because back in a previous step, when fumaric acid was oxidized into malic acid, the oxidation occurred closer to one or other of the carboxyl groups of fumerate because of the 50/50 chance of finding fumerate in one or another position?

You have it right.

45. #4-15c: Is an alternative analysis possible -- instead of presuming that rxn 1 cannot proceed forward to produce G6P unless ATP is first present as a reactant, can you assume that rxn 1 represents a coupled rxn where the breakdown of ATP drives the phosphorization of glucose (which would otherwise be uphill)

Reaction 1 IS a coupled reaction. The enzyme hexokinase needs ATP to phosphorylate glucose because that is the mechansim of its catalysis, it has not activity in su imply tacking a phosphate onto glucose, regardles of the energy consdierations. The enzyme is oblivious to the energy consideration: it will catalyze the reaction by binding substarte or product, but the direction will be determined by the delta G. So the lack of ATP here should be focused first on the enzyme, leaving the uphillor downhill question for second consideration (not gotten to in this case.)

46. and then consider whether the reverse rxn for the breakdown of G6P would be spontaneous if it were uncoupled from the ATP rxn (calculating the delta G of the reverse G6P rxn to then be -4kcal/mole)? Are you prevented from doing this b/c you'd need a new enz (not hexokinase) to drive the uncoupled rxn?

Right. If you had an enzyme that would catalyze the hydrolysis of G-6-P without coupling it to ATP phosphorylation, the reaction would proceed enzymatically (i.e., fast) anf the direction would be toward hydrolysis. Such an enzyme (phosphatase) exists and was described iin problem 4-5.

47. #5-3B(i): Should the pathway include as an initial step the oxidation of lactate into pyruvic acid (using NAD as the oxidizing agent)? This would mean, if calculating energy, that more energy produced as a result of greater amt of NADH2 generated. (or does the K.C. encompass this step?

You are right. This step is a fermenatation step, but run in reverse, and is a necessary step in this pathway of oxidative lactate metabolism..

48. #5-4B: I understand that the higher pH outside the vesicle indicates that protons are being pumped into the vesicle instead of outside and thus that the inner membrane must be inside out. I also understand that oxphos (ATP synthase complex) is not functioning (as indicated by fact that ADP concentration not changing and proton concentration not in steady state). BUT those 2 facts are distinct from each other, right? I.e., couldn't oxphos continue even with an inside-out inner membrane?

Yes, it certainly could. The problem was set up to reflect a damage to the ATP synthetase system in the formation of these inside out vesicles, so 2 things were changed reltvier to normal mitochondria, and for you to figure out. In fact, in plant chloroplats, organelles that produce ATP by photosynthesis, the pumping and gradient work just this way: protoin anre pumped into an inner membrane-bound compartment and then let run back outside, just the reverse of mitchondria.

49. #5-9: With respect to the givens, if bacteria uses less glucose per unit time with aerobic respiration, wouldn't its growth rate be faster in the presence of oxygen (given the same amount of glucose)?

This guess assumes that energy metabolism is limiting growth, which is reasonable but not necessarily true. Perhaps it is a low level of one enzyme catalyzing one step in one biosynthetic pathway, say for histidine formation, for example. In this case, the cells would have plenty of ATP, but there would be a bottleneck in the growth rate and they slowly cranked out histidine necessary for all protein synthesis. The problem implies some such circumstance, in stating that the growth rate is the same + or - air.

50. #5-9E: I don't understand the Answer Key's statement, "assuming regulation works to prevent abortive entry into Krebs etc.)." What is an "abortive entry" and how would it bypass a backup due to a bad/missing enzyme down the line?

If oxphos is knocked out, then if there were no regulation of the Krebs Cycle, we might expect pyruvate to keep on entering, which it could do as long as electron trasnport were functioning to regenerate NAD and FAD. Pyruvate would be mostly wasted by such an exercise, and if the pyruvate went that way, we might expect that the mutant would use up more glucose in air. Most examples one could come up with do no require such provisional explanations.

51. #5-10C: I understand the rationale of the problem, but in terms of the details of the Krebs Cycle, where exactly would malonic acid inhibit the Cycle? Is malonic acid related to malic acid?

The point of the problem is that interruption of the Krebs Cycle at ANY point will interfere with the whole cycle, because it is a cycle, and must run at all steps to run very much at all. Compare this to a linear pathway with a block, in which case you could accumulate lots of an intermediate that is upstream of a block. Not so for a cylic pathway. Malonate inhibits succinate dehydrogenase; I confirmed this for myself by looking up malonate in the index of Becker, then Purves, finding it in the latter.

52. If cyanide interferes with only 1 cytochrome, why does ATP production stop completely? Can't the rest of the cytochromes do the job, only slower? Won't H+ still be pumped out, and consequently let back in by the FoF1 complex?

The electron transport chain consists of an ordered series of steps, this order being obligatory. Electrons are passed from NADH2 to an Iron-sulfur protein, then to Coenzyme Q, etc. etc. The ability of each member of this bucket brigade to accept electrons from the immediately upstream donor is contingent upon its release of the electrons to the immediately downstream recipient. If it cannot do this, the entire flow of electrons must grind to a halt. All components upstream of the block will be in a reduced state (carrying electrons), while all downstream components will be in the oxidized state (divested of electrons).

53. In the conversion of one mole of glucose to ethanol and carbon dioxide, which 2 carbon atoms of the glucose molecule come off as carbon dioxide? I know it is not 1 or 6.

3+4 ---> CO2

If we look at the equation for glucose breakdown:

C6H12O6 + 6O2 <---> 6CO2 + 6H2O

We know the 6CO2 come off in the two rounds of the kreb's cycle. But as for the H2O, obviously it is produced by the equation

NADH2 + 1/2O2 <---> NAD + H2O

54. Since we get 10 NADH2 and 2 FADH2 produced per glucose molecule, it seems we should have 12 H2O to account for rather than What is the story with the 6 H2O not accounted for in the equation for glucose breakdown. Are they simply understood?

Although the electrons are coming form glucose, some protons are being provided by water. Consider the oxidations that are going on at aldehydes (e.g., glceraldehyde-3-P). Think of water first adding to the C=O double bond (not an oxidation):

             H
 O           O                                        O
 "           |                                        "
-C-H  --->  -C-H   and now extract 2 H's to NAD -->  -C  to give acid
             |                                        |
             O                                        O
             H                                        H
 
 

So one of the H's came from the added water.

Aldehydes give acids when oxidized.

Alcohols give aldehydes when oxidized.

55. For problem 5-10B, how are we getting 36 ATP from three acetate?

Acetate going through the Krebs Cycle produces 3 NADH2's and 1 FADH2. These are worth (3 x 3) + (1 X 2) ATPs = 11 ATPs/acetate. Then there is also a substrate-level phosphorylation yielding GTP, which = 1 ATP. So total per acetate is 12 ATPs. So 3 acetates yields 3 x 12 = 36.