Email Questions & Answers
Hello Class - Several students seem to be confused about the "committed step" in glycolysis. An ideal point of regulation of a certain pathway is the so-called "committed step", which is defined as the first pathway-specific irreversible reaction (the emphasis is on pathway-specific). In the case of glycolysis, this is the reaction catalyzed by phosphofructokinase; NOT the hexokinase-catalyzed reaction, because glucose-6-P has several metabolic fates OTHER than glycolysis (have a look at p. 21 of the booklet).
I was working on the last lecture. I have question about the enzyme involved in the phosphorylsis. Does Glycogen phosphorylase enzme work only at the ends and cleave the gylcogen producing Glc-1-P? and debranching enzyme take care of the branching structure? Also,I am little bit confuse about Glycolysis. in the reaction #6 when 1,3-bisPGA is produced from the oxidation of the G3P the energy that is produced is stored in the thioester bonds? and is used to phosphorylate the G3P?
A: Yes, glycogen phosphorylase works only from the end of a glycogen chain and produces Glc-1-P; whereas the debranching enzyme takes care of the branch point structure by cleaving the (1-6) bond. As for rxn #6, yes, the thioester intermediate stores the energy released during the oxidation of the aldehyde to the carboxylate, which is then used to form 1,3-bisPGA (by replacing the sulfhydryl group with a phosphate).
I had a question about lactic acid formation. In the Garrett and Grisham textbook, it says that lactate is produced in muscles when they are under anerobic conditions, mainly in strenous exercise. It also says that it is also produced in areas of the body that have limited blood flow, like the cornea of the eye. Since the book gives this as an example for lactic acid formation, does this suggest that the burning of the eyes(tired eyes)is caused by the lactate formation? Is it analogous to the "burning" of muscles caused by lactate formation?
A: That is an interesting thought to which I don't know the correct answer. I don't think that the "burning" of tired eyes (a sensation) can be compared to the "burning" (the biological oxidation) of lactate in the muscles. I believe I will talk about "eyes" at the end of class in the context of "reactive oxygen species" because the eyes are particularly exposed to the atmosphere and oxygen, which is quite reactive and thus not entirely harmless. Also, think of ozone (O3), which is formed at a higher rate during really hot days and is likely responsible for those symptoms.
I have question from glycoloysis of fructose. why both the end of fructose are not phosphorylated before the molecule is being cleaved into two diff molecues that is DHAP and G3P? is there any specific reason?
A: I think there are two reasons: First, we have to "kick-start" glycolysis, and that is accomplished to coupling reaction 1 and 3 to the hydrolysis of ATP. Second, the products of the aldolase reaction need to be phosphorylated in order to keep them within the cell for the reasons discussed in class.
Hello Dr. Abel, one question is dazzling me, but I appreciate your help. Will our bodies break down d-aminoacids and use it as a source of Energy?
A: That's an interesting question. The "aminotransferases" I introduced in class when we talked about the malate-OAA shuttle are strictly stereospecific for L-amino acids; so this route (taking off the amino group and leaving behind a keto group) is not possible, unless there are D-amino acid-specific aminotransferases; may be some organisms have those. An alternative would be the racemerization of a D-amino acid to the L-stereoisomer, which would be a PLP-dependent reaction (see p49 of booklet or textbooks). On p. 49, the electrophilic "sink" can destabilize the bond between C-H, C-R1, and C-COOH. Depending on the exact enzyme, that can lead to amino group transfer, decarboxylation, transfer of R1 or removal and reattachment of the "H" (which would result in such a racemerization). Another possibility is oxidative degradation in peroxisomes by D-amino acid oxidases (may be true only for certain organisms). I am not sure how much energy that would yield. But in any case, D-amino acids are very rare.
I was just wondering if there is a "committed step" in the TCA cycle similar to that of the third reaction of Glycolysis.
A: That is indeed a tricky question as three reactions of the TCA cycle that are essentially irreversible (large negative delta G), in addition to the PDH complex that provides the acetyl-CoA substrate for the first TCA cycle reaction. As shown on p. 46 of the booklet, all four reactions (PDH, CS, IDH, a-KGA-DH) are tightly regulated (in concert) by allosteric effectors (ATP, NADH, etc.). In a strict sense, we can consider the citrate synthase (CS) reaction as the committed step of the TCA cycle, which is the first irreversible reaction of the TCA pathway.
I have a question about one of the multiple choice questions on today's midterm. Letter "G" asks which of the cofactors is inactivated by arsenite poisoning and the correct answer is marked Lipoic Acid. Wouldn't Coenzyme A also be inactivated by arsenite reacting with CoA's terminal sulfhydryl group? Thanks!
A: There is a little catch: The question states 'neighboring' sulfhydryl groups. That's important for arsenite poisoning; it will not work on single sulfhydryl groups because a stable complex cannot be formed. You can find this example in one of the textbooks.
not see how it reacts to perform as a reductant.
Hi professor, what makes lactose a reducin sugar? I understand that it has to do with an aldehyde group exposed, however, looking at the howorth projection, I do not see it as being any more reactive than sucrose. I do
A: If you look at the structure of lactose on p29 in the booklet, you will see that carbon 1 of the glucose residue is "free", that means it is not part of the glycosidic bond connecting both monosaccharides. However, carbon 1 of glucose is the "semiacetal" of the aldehyde (the hydroxyl group of C5 attacked the C1 carbon (aldehyde), which resulted in ring closure. Since this ring can be opened (without breaking a glycosidic bond such as in sucrose), the aldehyde group can be formed, which can function as a reductant (or reducing agent); thus, we call lactose a "reducing sugar".
I was going through the steps of glycolysis and got slightly confused about the 4th and 5th steps. I understand that aldolase is used to cleave the carbon skeleton of F-1,6-bisP to form DHAP and G3P (GA3P) and then the more favored DHAP gets converted to G3P. What I don't understand is how the aldol condensation mechanism (pg. 25 of class booklet) fits into these steps of glycolysis. I guess DHAP is substrate 1 and attacks G3P, which acts as substrate 2, and the result is the formation of a "joined product" - F-1,6-bisP. But in glycolysis, the reaction happens in the opposite direction, so why are we studying the reverse reaction? Is the main idea that without aldolase, the reverse reaction is favored? Also, in my notes I have something written about "PFK-1" forming an intermediate from F-1,6-bisP, that then reacts with aldolase to produce the products. What is this PFK-1, I can't find it in the booklet. And, do we have to know the mechanism for the DHAP --> G3P transformation?
A: The aldolase-catalyzed reaction, see number 4 on page 23 of the booklet, is reversible and plays a role in glycolysis (cleavage of F1,6BP) as well as in gluconeogenesis (formation of F1,6BP from DHAP and GA3P), which we will discuss later in class. Thus the enzyme "aldolase" and its mechanism is important for both pathways (glycolysis and gluconeogenesis). That is the reason why the reaction mechanism (forward and reverse) is depicted on p. 25.
The products of the aldolase-catalyzed "forward" reaction in glycolysis are DHAP and GA3P (formed initially at a ratio of 1:1). However, DHAP is the more stable molecule. That means GA3P isomerizes spontaneously (without the need for an enzyme) to DHAP. At equilibrium, we find 96% DHAP and only 4% GA3P in the reaction mix. It is important to realize that only GA3P is the substrate for reaction #6 (not DHAP!). To make sure, that there is always enough GA3P for reaction #6 around, we have an additional enzyme, called triose phosphate isomerase. Although this enzyme does not effect the equilibrium of reaction #5 (DHAP still dominates with 96%), the triose phosphate isomerase increases the RATE of the isomerization reaction, providing GA3P "fast enough" for reaction #5, so the spontaneous isomerization does not become rate-limiting.
PFK-1 (phosphofructokinase-1) is given in the table on page 22. By the way, this is an excellent example when it makes sense to consult a textbook.
I don't think that I talked about the reaction mechanism of the DAHP-to-GA3P conversion, you don't need to know that.
I have a quick question about today's lecture. In the example about step 6 of glycolysis, we had the example of acetaldehyde being oxidized by NAD+ to form Acetate. I understand that Delta E is E(oxidant) - E(reductant), and the E values of the half reactions can be found in Table 3 of the class guide (written as reductions). What I don't understand was why we did -0.32V-(-0.58V) when that -0.58V value would be for the reduction of acetate to acetaldehyde and the reverse reaction would be +0.58V. Maybe I'm missing some key concept in doing these problems and I'm hoping that you can help me clarify. Thanks in advance.
A: Your way and my way of calculation will lead to the same result. The directionality of a process is very important in thermodynamics/bioenergetics. That is the reason why we have certain "conventions". For example, +deltaG means the consumption of energy and -deltaG means the release of energy. Another example is the sign (-/+) of the reduction potentials of reduction "half reactions" (always written as reductions!) relative to the test electrode. Based on this convention, deltaE is calculated as the difference between E of the oxidant and E of the reductant (deltaE = Eox - Ered). This is the approach of "biochemists" (see textbooks). I know that "chemists" use your way, but the outcome is the same. If this is not clear enough we can talk about this in the review sessions.
i went to the review session last night and we talked about the specific enzymes correlating to each step of glycolysis. however, im not clear about WHY for steps 1,3, and 7 (and 10) are TRANSFERASE when for steps 7 and 10, it's cleaving a phospahate group, so shouldn't that be HYDROLASE (KINASE?) and i don't rememeber what you said if we needed to know the common name or systematic class name? or both?? thank you!!
A: If you look closer at reactions 1, 3, 7 and 10, the phosphate is NOT cleaved using water, BUT it is rather TRANSFERRED from ATP to the other substrate (Glc for rxn 1; F69 for rxn 2; 3PGA for rxn.7; and Pyr for rxn. 10). Note that for the enzymes of reactions 7 and 10 are named in the opposite direction, in accordance with the guidelines given on p. 24 of the booklet. You need to know only the common names of the enzymes.
I was working on problem set 1, and I don't understand question #5b and 5c. Why do we use the delta G = delta G' + RTLn[P]/[R] instead of just delta G = -RTLn[P]/[R]? Only the concentrations have changed, not the standard temperature.
A: The equation to calculate delta G (dG) of a reaction is dG = dGo' + RTln [initial P]/[initial R]. You could use just dG = RTln [initial P]/[initial R] if (and that's a big IF) dGo' is zero, meaning that the reaction is at equilibrium when we have equal amounts of [P] and [R] (which is rarely the case). What really matters is the ratio of Q ([initial P]/[initial R]) to Keq ([P @ equilibrium]/[R @ equilibrium]), meaning how far off are we from equilibrium.
If Q=Keq, we are already at equilibrium (EQ).
If Q<Keq, we are not yet at EQ, having relatively more [R], or less [P] than under EQ conditions. This means the reaction will move forward to produce more P until EQ is achieved (dG is therefore NEGATIVE).
If Q>Keq, we are also off EQ, but we have relatively more [P], or less [R] than under EQ conditions. This means the reaction will move backwards to produce more R until EQ conditions are achieved (dG is therefore POSITIVE).
Try to understand these equations below (they say what I tried to describe in words)
dGo' = -RTlnKeq (under "standard conditions", i.e. we try to figure out how a reaction "behaves" if we start out with the same molar concentrations of R and P)
dG = dGo' + RTlnQ Q=[initial P]/[initial R] or
dG = -RTlnKeq + RTlnQ or
dG = RTlnQ - RTlnKeq or
dG = RTln Q/Keq
I'm a little confused about the wording on your final essentials. In the concept portion, you state you'd like us to review "strategies" for amino acid degradation, nitrogen excretion, and N and S assimilation. What exactly are you looking for?
A: For amino acid degradation you should know that the amino group is separated from the carbon backbone. The amino group is excreted as ammonia, urea or uric acid, depending on the species. The remaining carbon backbone can be either glucogenic, ketogenic or both. Animals/humans can assimilate N and S only in its fully reduced form (amino group, sulfhydryl group), therefore animals and humans are dependent on plants/bacteria for food, which can reduce nitrate and sulfate.
In the finals essentials list under the enzymes/reactions section, I was wondering what differences we needed to know between Cys/Met biosynthesis in animals and plants.
A: You need to know that humans use methionine to make cystein, whereas bacteria and plants do it the other way around.
In the essentials under the concepts section, I was wondering what exactly you wanted us to know for the hormonal regulation and integration of metabolism question. Is it only regarding the information on page 110, or does in encompass other things?
A: You need to know the important concepts and the bigger picture: role of the three hormones discussed in metabolism, function of cAMP and phosphorylation of regulatory enzymes, what pathways are how regulated by what hormone.
A: You need to have a general understanding how trigylerides are made (glycerol plus acyl-CoA) and know about the general functions of glutathione (redox buffer, detoxification). The material on p. 93/94 will not be covered in detail, but you should know the function of THF and that Met is derived from Cys in plants/bacteria, whereas Cys is made from Met (essential amino acid) in humans.
I was wondering if we need to know anything about membrane lipid synthesis or cholestrol for the final?
A: I guess you should know that cholesterol is made from acetyl-CoA units, and that membrane lipids are made from glycerol-P and acyl-CoA, that's all.
Based on the sample problem #3 on the sample midterm, how do we know that oleic acid needs 8 rounds of beta-oxidation and that these then yield 9 acetyl CoA molecules? I'm having a hard time finding it in my biochemistry book.
A: You don't need a textbook for your question, just remember that each round of beta-oxidation cleaves off two carbons as an acetyl-CoA from the carboxylate end of a fatty acid. So if you have a fatty acid of 18 carbons, like oleic acid (it has also one double bond), then you need 8 rounds of beta-oxidation (8 x 2 = 16) to hack oelic acid into nine C2 pieces. Note that the last (8.) round will work on butyryl-CoA (4 carbons), which will be cleaved into TWO acetylc-CoA.
Could you please explain the overall stoichiometry for the PPP on page 68?
A: Yeah, I admit this is not very clear and I will have to change that next year. This stoichiometry applies to the special situation when only NADPH and no ribose is required. In that case, 6x ribulose-5-P (derived from 6x Glc-6-P) are converted into 4x Fru-6-P and 2x glyceraldehyde-3-P (first line on p. 68). Then, 4x Fru-6-P and 2x glyceraldehyde-3-P are converted by gluconeogenesis into 5x glucose-6-P. The 2x glyceraldehyde-3-P (or better 1 GA3P and 1 DHAP) react (aldolase reaction) to F1,6BP, which is hydrolyzed (requires 1 extra water) by F1,6BPase to F6P and Pi and further isomerized to G6P. The second line on p. 68 reflects these extra reactions. The third line, makes the point that, if we go this route, glucose can be completely oxidized into CO2, but the hydrogens are conserved as NADPH for biosynthetis purposes.
I just had a question regarding the Midterm Essentials handout. I noticed that in the TCA cycle you required us to know that structures of the reactants, products, etc, but in other pathways you do not. For example, in the Pentose Phosphate Pathway, you write that you want us to know the reactions, but not specificall the structures. Thus, would it be suffice to just learn the names of the compounds and the enzymes and intermediates used, and not the specific structures?
A: For the PPP you need only to know the structures of the substrate (G6P), initial intermediates (R5P, Ru5P), and final products (F6P, glyceraldehyde-3P). For the rest of the PPP it is sufficient if you know the names/abbreviations of the intermediates and types of reactions (e.g., isomerization, epimerization, C2-transfer, C3-transfer).
I have a couple of questions that arose in todays review session. What is the importance of the pentose phosphate pathway with respect to other cycles/pathways? Also, how do you get 12 ATP from 1 Acetyl-CoA??
A: The PPP provides NADPH for biosynthetic pathways and interconvertes sugars from the triose stage (C3) all the way up to the heptose stage (C7), so it is central to carbohydrate metabolism. One round of Acetyl-CoA degradation via the TCA cycle and the ETC yields 1GTP or ATP, 3 NADH (=9ATP) and 1 FADH (=2 ATP).
Besides being an electron carrier (1e or 2e?), does CoQ have any other function? Also, I was wondering if we need to know how to identify the differences between the cytochromes (a,b,c). Lastly, how is the lysine residue of the biotin coenzyme function in the ATC cycle?
A: The primary function of CoQ is that of an electron carrier. You don't have to know the differences between each cytochrome, but you need to be aware of the fact that each cytochrome has a unique reduction potential. I don't know what you mean with ATC cycle. If you mean TCA cycle, there is no biotin-dependent enzyme, unless you consider the anaplerotic reactions; pyruvate carboxylase is biotin-dependent, and the lysin residue attaches biotin covalently to the enzyme.
Do we also need to memorize the inhibitors of the complexes in the ETC cycle?
A: No, but you need to know what each complex is doing.
I was wondering if the NAPDH used in beta oxidation of a linoleic acid (C9 C12) would be used in the calculation of ATP generated in fatty acid egradation? Would NADPH generate 3 ATP just like NADH?
A: Since NADPH/NADP is derived from NADH/NAD by ATP-dependent phosphorylation you could argue that the consumption of NADPH in beta oxidation of linolenic acid will affect overall ATP yield (3 ATP less or even 4 ATP if you take the NAD kinase reaction into account).
How many protons move across the inner membrane in complex 3 of the ETC, I wrote down 2 in class but I think I should have written 4.
How did you determine the number of H20's produced from the FA degradation example you did at the end of class on Thursday?
A friend and I were going over the Midterm Essentials link, and we're a little confused as to what will and won't be on the midterm because we haven't gone over the TCA cycle yet in class. Will we need to know it? Also, will we need to know the 5 co-factors that you mentioned at the end of last lecture? Thanks a lot!
A: As I mentioned in class yesterday, the TCA cycle won't be on the 1. midterm but the PDH reaction and its 5 co-facors will be. The reason why it is listed on the "Essential's" is because last year the 1. midterm was given on a TUE, so students had the weekend to study the material of lecture 6. Since there are only two days between lecture 6 and the first midterm this spring, I decided to leave the TCA cycle for the 2. exam.
Do we need to know all of the intermediates and names of enzymes for the urea cycle and lipid biosynthesis. You mentioned that we needed a general knowledge of it in the review session, I just wanted to get it cleared up.
A:Yes, you need to know the urea cycle and lipid biosynthesis (fatty acids only!) in detail because that was covered after the 2. midterm. The "general understanding" (which requires also knowledge of important structures and enzymes) refers to the cumulative part of the final exam (glycolysis etc.).
I was wondering.. from lecture 12, Dr. Callis went over a diagram in which PPP was added in with glycolysis. She gave us 3 different examples to show what rxns would run.. one being if we only need NADPH we need to run all three phases of PPP.The one i was wondering about was if we need R5P.. she mentioned that we need phase 3 and 2 to run because we want two ribose and running only phase 3 would only give us 1 ribose. I understand that if we run pase 1 we would deplete our source of NADPH, but my question is, if why cant we just do glycolysis and taking GA3P and F1,6 BP or is that not possible?Also, what do you mean on the essential page by role of activated sugars... do you just mean like when biotin attaches CO2 and activates it?
A: 1. PPP: We can envision 3 different situations in metabolism:
A) We need much, much more R5P than NADPH, let's say no NADPH is required. In this case, Phase I would not be useful because it produces NADPH (not needed) and it loses carbon as CO2 (waste of carbon that is better be used for R5P). To produce only R5P, we go Phase III backwards all the way up to Xu5P/R5P (see p. 68). Phase II just interconverts pentoses. However, in order to start Phase III backwards with GA3P and F6P, we need to convert our initial G6P by the first glycolytic reactions.
B) We need much more NADPH rather than R5P, let's say no R5P at all. In this case, we go through phase I, which produces 2 NADPH from one G6P. The other product is R5P. Since we do not need R5P at all, it is completely converted back to G6P using the following stoichiometry: 6 X C5 (pentose) ==> 5 x C6 (hexose). This happens by using Phase II and Phase III, but the end products of theses phases are GA3P and F6P. So, we need some gluconeogenetic reactions to produce G6P to start over with phase I of the PPP.
C) If we need both NADPH and R5P, then, phase I is sufficient.
2. Activated sugars are sugar nucleotides (such as UDP-Glc) that are the precursors for glycogen synthesis or synthesis of disaccharides etc.
Do we need to know the enzymes involved in the pathway and all intermediates of the regeneration phase of the Calvin cycle,(page 72 of the booklet) or just the general principal of the regeneration phase with the net of 1 triose carbohydrate for every 3 co2.
A: You need to know the enzyme and reaction of the "Carboxylation Phase". The "Reduction phase" is very similar to reactions in gluconeogenesis (look this up), and the "regeneration phase" shares many reactions with the pentose phosphate pathway. You also need to know the source of ATP and NADPH.
After looking over the "Essentials for Midterm 2" it does not look like we will be tested on the structures of gluconeogenesis. Is this true? Should we only be concerned with how it is different from glycolysis, regulation, and futile cycling?
A: Gluconeogenesis is largely a reversal of glycolysis with the exception of the reactions taking pyruvate to OAA and F1,6BP to F6P. Since all of the intermediates have been discussed in class and their structures were required knowledge of the first midterm, yes, you should know them, or at least be able to reconstruct them from their names.
I was wondering if questions like the ones we skipped in the first practice midterm will be on the second midterm? Also, on the practice midterm for midterm 2, the TA's said that the answer key is wrong for question 2a) third one down ("Cell switches from using glycogen as its energy source to utilizing saturated fatty acids as its energy source"). They said that the correct answer is "NO CHANGE." The answer key says, "INCREASE." Which answer is the correct answer?
A: The order of lectures this year is a bit different than last year and therefore also the material that was covered in the midterms. I will only ask questions on topics covered in lecture.
Oh, yes, I made a mistake in answer key for midterm 2. There is "No Change". Sorry
Q to Dr. Callis: 1. When does GNG undergo Path A? Does it go through Path A when there is sufficient NADH in the cell so you don't have to interchange between OAA and malate to make NADH?
A (Dr. Callis): Path A does depend on sufficient NADH for reduction of 1, 3-bisPGA to GA-3-P. So, for PEP to be converted in the mitochondria and transported out, there must be a continual source of NADH. This can come from lactate to pyruvate conversion.
2. In the overall accounting of GNG, we have: 2 Pyr + 4ATP + 2GTP + 2NADH + 2H+ + 6H20. Where do you get 6H20? I was only able to count 4 H20 on page63, both used up in the fructose and glucose rxns.
Source of waters: There is the hydration of PEP to form 2-PGA. That would consume 2 of them for 2 pyr. Hydrolysis of GLC and Fructose would consume 2 more. The other two are used to hydrate CO2 to make bicarbonate for the pyruvate carboxylase reaction for 2 PYR to 2 OAA.
3. You gave us a delta G for glycolysis: -74 kJ/mol. Is this delta G, or delta G standard?
That is delta G standard (zero prime).
4. What does the "syl" of glycosyl mean?
Means a carbohydrate linked to something. Not a free carbohydrate. glucosyl would be a glucose molecule attached to something.
5. You had given us a diagram about regulating flux to get NADPH and R5P. There were 3 rxns, and you asked which ones the cell would use in particular situations. You said that if the cell only needs NADPH, it would undergo I, II, III. Why not just I? You did mention that you go through all three rns to keep returning C to Glc 6P. However, it seems like in the next two situations (when cell needs R5P: go through II and III; when cell needs both: go through I and II), you don't worry about regenerating the substrate. Why is that?
I, II and III are phases of the PPP and contain more than one reaction. If only phase I operates, then Ribulose-5-P accumulates. This is not good, so the cell wants to return this carbon back into something that can be used- such as glucose-6-P.
6. When you need R-5-P, then this compound is converted into something else. It is gone! Also, you said that if you need a lot of R5P, you deplete your carbohydrates. Do you also deplete a lot of your carbohydrates when you need a lot of NADPH?
Yes, but slower. The rest of the carbon returns the c to GLC-6-P. For each GLC-6-P that goes through PPP, one CO2 is produced, the rest of the carbon can be converted to glycolytic intermediate.
For the midterm, what level of detail should we know about the electron transport chain.?Should we know exactly which molecules donate and accept the electrons in each complex, like on page 58 of the booklet,or just know which comples carries the elctrons through the chain like on page 57?
A: You should have an understanding of WHY electrons flow from NADH to oxygen, and this idea is depicted on p. 58. You also should know the general role of iron-sulfur clusters and heme groups in this process. As for p. 57, you should know the stoichiometry of the reactions catalyzed by each of the complexes; they are essentially multisubunit enzymes catalyzing redox reactions between NADH-CoQ, FADH2-CoQ, CoQ-Cytc, and Cytc-oxygen. You also need to know at which steps how many protons are pumped out of the matrix. So, you need to know a bit more than what is depicted on p. 57. But I went over this in class, and it is a good idea to read about it in a textbook.
I have a question regarding the ATP yield in beta oxidation of FA. Yesterday you said in lecture that the total yield in the oxidation of C16 FA yields 131ATP, but we need to subtract 2ATP because of the initial activation of FA to acyl-FA. So when I was rewriting my notes, I could only find that 1ATP was used in the initial activation. Can you tell me where the other ATP is being used?
A: The initial activation of fatty acids uses indeed 1 ATP. However, this ATP is hydrolyzed to AMP and pyrophosphate (PPi); the PPi is then further hydrolyzed to two phosphates (2Pi). In total, two phosphoanhydride bonds are hydrolyzed in this activation reaction. In other words, it takes 2(new) ATPs to "recharge" the AMP to become an ATP again.
Q to Dr. Callis: This is a very basic question that may help me to understand all these mechanisms or cycles as a whole. I know that glycolysis deals with breaking down glucose to get energy in the form of ATP. However, these cycles that we learned last thursday, the PPP and the calvin cycle, what are their main fcn to our body. Meaning like, you have glycolysis through the electron transport chain to get ATP. What is it that these new cycles are doing?? Just returning glucose? rebuilding it? I get the process by which it takes, i just seem to misunderstnad what the overall purpose of these cycles are doing.
A (Dr. Callis): PPP has several functions. First, PPP functions to make NADPH. Glycolysis and TCA cycle make NADH, not NADPH. This NADPH is NOT oxidized in the electron transport chain, but oxidized in different reactions- you will see some later this quarter. One reaction you did see was to reduce glutathione. The second very important function for the PPP is to produce 5 carbon sugars. Remember glycolysis does not contain any 5 carbon sugars! So, organisms have to have a way to biosynthesize them from glucose. And that is accomplished by the reactions of the PPP.
Gluconeogenesis is a way to make glucose. If our diet is low on glucose, then we do need to make some, initially the liver makes glucose from other compounds to keep blood glucose at normal levels for the brain, which prefers to use glucose, but does not have any stores of it. Also, glucose is very commonly synthesized from lactic acid made in the muscle, exported from the muscle into the blood and the liver takes up the lactic acid and converts it back into glucose. The liver then uses up energy to "salvage" the waste from the muscle.
The last cycle, the Calvin cycle, only occurs in photosynthetic organisms. It is the set of reactions by which CO2 in the atmosphere is reduce to carbohydrate. While humans have many of the same reactions, they DO NOT have RUBISCO and the phosphatases. So, we can not fix CO2 to carbohydrate and depend on plants to do this.
is the glycogen metabolism (degradation/ synthesis) you mentioned we need to know is that the phosphorolysis of glycogen and starch on page 46 right? or is there some section i am missing. thanks
A: You need to know that glycogen/starch can be degraded by hydrolysis (extracellular) and phosphorolysis (intracellular). You should know the enzymes and intermediates involved and that intracellular phosphorolysis provides an advantage (produce already a phosphorylated glucose; save 1 ATP for priming of glycolysis.)
No knowledge of glycogen/starch synthesis is yet required. I forgot to erase "synthesis" from the 2002 ESSENTIAL list. Sorry.
i have some question regarding the alpha, beta keto acid decarboxylation, can u briefly describe to me da differences between the two.
A: Beta-keto acids are chemically unstable and can be easily (even non-enzymatically) decarboxylated. Alpha-keto acids are stable, and their enzymatic decarboxylatiopn requires a co-factor (TPP). The alpha-keto acid and TPP form an intermediate which now somehow resembles a beta-keto acid, facilitating decarboxylation.
>1) Do we need to know the A.A. structures for this midterm?
A: Not yet, but knowing their names may help.
>2) What is the Km...what is it's significance as it applies to hexokinase and glucokinase
A: This is a type of question any biochemistry text will answer.
>3) What is the significance of intramolecular oxidoreduction?
A: This is only relevant for anaerobic glycolysis when the cell is forced to regenerate NAD+ from NADH by a reduction. Therefore, glucose undergoes an oxidation and reduction. Thus, there is no net redox change in lactic acid compared with glucose. However, the oxidation state of each of the six carbons of the original glucose has changed (thus the term of "intramolecular" redox reaction). This rearrangement is sufficient to liberate some energy to produce ATP by substrate-level phosphorylation.
>4) What is the significance of creatine/ arginine phosphate?
A: These are considered to be compounds with a high phosphoryl group transfer potential (see Table 5). Phosphocreatine functions as a short-term storage of energy to replenish ATP in the muscles during the very early phases of contraction before glycolysis kicks in. Arginine phosphate has a similar function in invertebrates but was covered in class only as another example of compounds with a high phosphoryl group transfer potential.
For problem 11 (problem set 1), how do you get -30.5 kJ/mol for the hydrolysis of ATP --> AMP + PPi. I thought it would be double the delta
G of the hyprolysis of ATP ---> AMP + PPi. Thank you for your help. Also, I was unable to attend lecture on Thursday. How far must we go for
the first midterm?
A: You are right IF you consider the SUBSEQUENT hydrolysis of PPi. Without that you hydrolyze only one bond, that between AMP and PPi
1. On page 34 of the Handbook which of the reactions are the sequential reactions? Is it any reaction that is irreversible, like reactions 2, 4,
5, 7, 8, 9?
A: All reactions act in a sequence. Essentially irreversible reactions have large negative DG and are indicated by a single arrow (1, 3, 10). Reactions that are reversible under physiological conditions are indicated by a double arrow. The coupling of reaction 1 and 3 to ATP hydrolysis is responsible that they are essentially irreversible. The coupling of reaction 10 to PEP hydrolysis (although ATP is produced) also makes it essentially irreversible because the -DG of PEP hydrolysis is so much larger than that of ATP hydrolysis (see Table 5).
2. For the oxidation number system in the case of a carboxylic acid (-C00) would you multiply the carbonyl oxygen's value (+1) by two because of the double bond?
A: Yes, the rules on p. 39 are self-explanatory.
3. On page 42, how does UDP-Glc function as a co-factor? What does it do that makes it a co-factor?
A: In the galactose cycle on p. 42, UDP-Glc and UDP-Gal are only required in catalytic amounts. The conversion of Gal-1-P to Glc-1-P would not proceed without the presence of UDP-Glc.
4. On page 45, I'm confused about the arrow pointing vertically downward from the Glycogen molecule to the monosaccaride glucose. I thought that
the glycogen molecules must be hydrolyzed to disaccarides, maltoses, etc. before they can be hydrolyzed to a monosaccaride, but the vertical arrow
indicates that glycogen is immediately hydrolyzed to a monosaccaride without being hydrolyzed to a disaccaride/maltose first. Can you explain
the vertical reaction/arrow?
A: Yes, the arrow is somewhat misleading. It means though that if the action of all glucosidases is considered (amylases, maltases etc.) is considered the endproducts of glycogen/starch hydrolysis is glucose. I will draw this in a better way next year.
5. Is Glycogen Phosphorylase a hexo-enzyme? (page 46)
A: No. Don't confuse with hexokinase (which is a common enzyme name). Hexokinases phosphorylate a variety of hexoses using ATP as the Pi donor.
6. On page 46, what did you mean when you said, "1 ATP is saved up for later for the phosphorylation step to dietary glucose?" for the
A: Phosphorolysis produces already a monophosphorylated glucose. Thus only 1 ATP (instead of 2 ATP for glucose) is required to prime glycolysis.
7. When you lectured on the material on page 40 - 46, did that material apply to Aerobic conditions, Anaerobic conditions, or both?
A: These reactions can occur in both conditions.
8. In phosphorolysis, I don't understand what you meant by the energy of glycosidic bonds are used instead of ATP to phosphorylate the glycogen...
How does this work?
A: Glycogen/starch are polymers. The glycosidic bond in glycogen is a link between a hydroxyl group on C4 of one glucose unit and an aldehyde (cyclic hemiacetal) on the C1 of another glucose unit. The polymerization reaction requires 1 ATP per glucose unit or glycosidic bond (see later in class). Thus, phosphorolysis of glycogen/starch recaptures this energy to produce Glc-1-P (remember, phosphorylation of glucose requires ATP to make it happen in glycolysis).