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1) Yes. Although the mechanism by which a deficiency in the desaturase results in dis-inhibition (or activation) of the AMPK hasn't been determined. Yes. The result is a decrease in fatty acid synthesis and an increase in beta-oxidation. 2)YES, oleoyl (C:18, cis-delta nine) could be a substrate for the other desaturases, C#6, C#5, or C#4. 3) The correct term is polymorphism of G6PDH, since individuals with this common polymorphism do have active G6PDH, but the enzyme is less stable. The result of this polymorphism is that older red blood cells will have decreased levels of G6PDH activity. (Recall that rbcs cannot carry out new protein synthesis). When the malaria parasite infects a rbc there is an increased demand for ribose and NADPH, and this demand is met by G6PDH, in concert with the pentose phosphate pathway. However, when the malaria parasite infects rbcs of individuals that have the unstable G6PDH, the rbcs will tend to lyse. This is because [NADPH/NADP+] will fall, antioxidants such as glutathione (GSH), and vitamins C and E will not be maintained in their reduced state. As a result the rbc membrane lipids will be oxidized and the rbcs will lyse. When this happens the malarial parasite is released prematurely and killed. Fortunately, the patient responds to the resulting anemia by releasing immature reticulocytes, and so this is not lethal to the patient. 4) As we age, there is a tendency to become deficient in vitamin B12, as there is a general decline in our ability to absorb it. B12 deficiency leads to pernicious anemia, the symptoms are anemia and neurodegeneration. Treatment with folate solves the anemia, but not the neurodegeneration. It is therefore argued by some that the supplementation with high folate may actually mask B12 deficiencies in older individuals. 5) Yes, the final is 8-11 AM Friday December 16th, in CS24.
for the Spring 2000 153CH exam, problem #7, is it possible to attach arachidonic acid with ethanolamine to make anandamide? thanks
As shown in the answer key, ethanolamine reacts with arachidonyl-CoA to form anandamide. Fatty acids must be activated before they can be oxidized, elongated, or desaturated. By analogy, it is necessary to activate arachidonic acid. Thus, the key shows arachidonic acid + CoASH + ATP --> arachidonylCoA + AMP + PPi. Question 7 also states:"Reactants can include amino acids, lipids, intermediates of the central metabolic pathways and common coupling factor coenzymes." Since ethanolamine is not one of these, the complete answer should show the mechanism by which it is generated.
Hello, Will any of Dr. Edwards lecture be on the final? Thanks.
Yes. The slides of Dr. Edwards talk were provided in class and are also posted on the voh site. There is also reading in Voet and Voet that pertains to this lecture (pp. 954-957).
Yes.
i don't understand the lecture on desaturases where NADPH is oxidized and O2 are reduced when introducing the double bond. i have in my notes that its a 4 electron reduction--2 from the substrate and 2 from NADPH. the carbons are being oxidized rigth? so the 4 electrons is being used to reduce oxygen, which drives the rxn? thanks
You are exactly correct -- the desaturase reaction (the oxidation of the fatty acid) is coupled to the thermodynamically favorable reduction of dioxygen to 2 H2O. The reduction of dioxygen to two waters requires 4 e-. Two e- are donated by NADPH, and two e- come from the substrate (in this example stearoyl CoA). This is a good example of an "oxidation driven by reduction".
1) Bring a regular function scientific calculator. 2) The final will be cumulative, with slightly more emphasis on the recent material covered since the second midterm 3) The exam will be twice the length of the midterms -- however, I will provide more space for your answers. therefore roughly twice the number of questions, but maybe seven pages instead of six.
for methionine synthesis from homoserine, how come the cysteine doesen't directly attack the homoserine, but succinyl coa has to be used first?
Curious no? In the analogous reaction discussed in the metabolism of homocys of mammals: homocys + serine --> cystathionine + H2O; there was no need for succinylCoA. So why the big need for synthesis of methionine from cysteine + homoser? I think it is reasonable to suppose that plants and proks that need to synthesize methionine de novo need the extra input of a good leaving group to make this reaction go in the direction of net synthesis of methionine. In the case of the analogous reaction in mammals, the pathway is acting to salvage homocys, and may not need to be as strongly driven.
when do we use N5 methyl THF as opposed to N5,N10 methylene THF as the methyl donor?
N5-methyl THF is used to methylate homocysteine to make methionine. In mammalian cells that is apparently the only methylation reaction that uses N5-methyl THF. N5,N10-methylene THF carries carbon at the oxidation state of formaldehyde, and so donates a -CH2OH. It is however used as a methyl donor in the reaction dUMP --> dTMP. In this special case DHF is the product, not THF. The overall reaction is: dUMP + N5,N10-methylene THF --> dTMP + DHF.
Wednesday's lecture: GSH and GSSG see Figs 26-46 abd 26-45 Heme synthesis pp1013-1016 Polyamines -- Not covered in V&V! See power point figures posted to VOH today. Creatine and Phosphocreatine -- page 571 Amino acids and Neurotransmitters -- pp. 1024-1026 Today's lecture NOS -- pp.671-673 Hope this helps! Cathy
Decarboxylation proceeds by moving the electrons from the O- of COO- to produce CO2. You are then left with a carbanion on the alpha C of methionine. What stabilizes this? PLP! The carbanion is protonated, generating the decarboxylated product of AdoMet.
How many ATPs should we count in the conversion of NADPH to NADP+ ?
Dear H, I count NADPH to NADP+ as worth 4 ATP. The reason behind this is that our inner mitochondrial membrane has the enzyme "transhydrogenase" (TH). TH catalyzes the reaction NADH + NADP+ (2 H+ P-side) ---> NAD+ + NADPH + (2 H+ N-side). [Note: P=cytosolic side; N=matrix). Thus, TH conducts 2H+ from the cytosol to the mitochondrial matrix and uses NADH to reduce NADP+. This "costs" us one NADH (= 3 ATP equivalents) and 2 H+ came in (which "costs" about one ATP, if they had instead come in via the Fo/F1 ATP synthase).
In several reactions involving THF cycles, we use glycine to replenish N5N10-Methylene-THF and this even happens when Glycine itself is converted to pyruvate. Therefore, how would we be able to count ATP production after glycine is converted to Serine? One glycine goes to serine then pyruvate and is broken down to yield ATP, but *another* glycine is used to replenish the methylene-THF when Gly is converted to Ser. The energy of the THF replenishing glycine then cannot be broken down to make ATPs (except for the NADH made in the process), so Glycine yields *no* net energy except for one NADH? Is this right?
Dear H, I am impressed by your analysis. I think it is exactly correct. I agree that catabolism of glycine would yield one net NADH!
I was looking over the mechanism for beta oxidation of unsaturated fatty acids (Linoleoyl-CoA), and I was wondering if the FAD/FADH2 count at the end was incorrect. In the 4th cycle of beta oxidation, an FADH2 is not produced since it is not a full round. At the end, during the 5th round of oxidation, shouldn't that also not be a full round since there is already a double bond present? If so, the final count of FAD/FADH2 would be 6, not 7? I was just wondering. Also, I'm a little confused as to where the numbers come from for the oxidation of amino acids on the urea cycle handout. Could you explain where they are derived from? (or maybe I'm just missing something)Thanks!
There are 3 FADH2 produced by the first three rounds of beta oxidation. The subsequent beta oxidation cycle (4th cycle) does not produce FADH2. However the following step (conversion of the C10 delta4-cis acyl CoA to the dienoyl acyl CoA (C10 delta2-trans, delta4-cis) produces one FADH2. There are three more FADH2 produced in the last 4 cycles of beta oxidation to give a total of 7 FADH2.
As far as the material that we are responsible for, how much emphasis should we put on lecture, and how much on the book? For some topics, the book goes into much much more detail that was discussed in lecture. Are we responsible for this additional information?
Dear Anon, I recommend using the text to help you understand the material covered in lecture and in the study questions. This will help you focus your reading of the text.
Dear Inmo, There is a cytosolic glycerol 3-P dehase that does use NAD+, and could oxidize Glycerol 3-P to DHAP. In the event this enzyme is used, then the ATP equivalents of 23-1=22 is correct. However, in tissues that are relying on oxidation of fatty acids from stores of triglycerides, then the glycerol 3-P dehase that is used is the one located on the outside of the mitochondrial inner membrane. This enzyme oxidizes glycerol 3-P to DHAP, and reduces FAD to FADH2 (and subsequently coenzyme Q to coenzyme QH2). This pathway predominates in muscle, brain and insect flight muscle. The cytosolic enzyme that is NAD dependent can work in concert with the mitochondrial enzyme and serve as a shuttle that delivers cytosolic NADH to the mitochondrial matrix as FADH2 (QH2). A diagram of this shuttle can be found on the lecture handout from 10/05/05.
