Download Biochemistry 440: Problem Set 8 - Citric Acid Cycle and Related Topics - Prof. Rachel Klev and more Assignments Biochemistry in PDF only on Docsity!
1. What are alternate names for the Citric Acid Cycle?^ Problem Set 8
TCA (tricarboxylic acid) or Krebs cycle (Hans Krebs)
2. The standard free energy change for converting malate to oxaloacetate is highly unfavorable (∆G’o (^) = 30 kJ/mol). This reaction is driven forward by maintaining low concentrations of oxaloacetate. What is the maximum concentration of oxaloacetate that allows the reaction to go forward? Assume [malate] = 1mM, [NADH] = 1μM and [NAD+] = 1μM, and all are in steady state. 0= - 30 = 5* log ([oxal]/1x1030+5log([oxaloacetate][1- (^3) ) μM]/1mM][1μM]) - - 6 = log[oxal]6 = log[oxal] – – log 1 x 10(-3) -^3 - 10 9 = log[oxal]- (^9) M = [oxal] 1 nM = [oxal] (just slightly less than this and the reaction will proceed forward). 3. The conver A. Which steps are the strongly exergonic steps?sion of pyruvate into CO 2 is highly regulated at the strongly exergonic steps.
Pyruvate to Acetyl ketoglutarate; alpha-CoA; Acetyl-ketoglutarate to succinyl-CoA+Oxaloacetate to Citrate; Isocitrate to alpha-CoA - B. If you added the following to crude cellular lysate and examined the first regulated step identified in part A, what changes would you expect to see in the reaction (if any) and what is the physical mechanism of that regulation? i. ii. NADH:ADP: NADH inhibNo affect its the PDH complex allosterically iii. iv. Ca2+:AMP: Calcium allosterically activites PDHAMP allosterically activates PDH v. Pyruvate: This is just for your information and will not be part of the exam, since we did not teach the covalent modification mechanisms for PDH this year: => PDH increased [pyruvate] inhibits this kinase and thus activates PDH.-E1 is inactivated by phosphorylation through a specific kinase;
4. What intermediates between glucose and the end of the citric acid cycle can be used to generate 12 of the 20 amino acids.
See Fig. 16 i. Alpha-ketoglutarate: Glutamine, Proline, Arginine, Glutamate-15 in our textbook: ii. iii. Oxaloacetate: Aspartate, AsparaginePhosphoenolpyruvate: Serine, Glycine, Cysteine, Phenylalanine, Tyrosine, and Tryptophane
5. A culture of respiring mammalian cells is briefly incubated with media containing glucose labeled on C6. You isolate the following citric acid cycle intermediates and look for 14 C- the position of the 14 C label: 1) acetyl CoA, 2) isocitrate, 3) succinate, 4) malate. A. Which carbon(s) in each of the above intermediates is/are radioactively labeled? Assume that the timing is such that the initial labeled compound has only been through one turn of the citric acid cycle. B. How would your answers change if you fed the cells with glucose labeled on C5 instead? acetyl CoA isocitrate SCoA │ SCoA│ COO│ -^14 │COO- C=O │ 14 │C =O (^14) │C H 2 CH│ 2 (^14) CH 3 CH 3 HC—COO- (^) HC—COO- C6-glu C5-glu HO—^ │CH^ HO—│CH │ COO (^) - │COO- C6-glu C5-glu succinate malate COO │ -^14 │C OO-^ (1/2) COO│ -^14 │COO-^ (1/2) (^14) │C H 2 (1/2) CH│ 2 HO— (^14) │C H (1/2) HO—CH│ (^14) │C H 2 (1/2) CH│ 2 14 │C H 2 (1/2) CH│ 2 COO-^14 COO-^ (1/2) COO-^14 COO-^ (1/2) C6-glu C5-glu C6-glu C5-glu
6. You are taking biochemistry because D to eradicating tuberculosis in developing world. When your biochemistry professor tells your. Paul Farmer has inspired you to dedicate your life the glyoxylate cycle has never been observed in vertebrates, you realize that you have found an ideal drug target. You are determined to use this knowledge to target Mycobacterium tuberculosis. A. Why do organisms such as Mycobacterium require the glyoxylate cycle? How does the glyoxylate cycle affect other metabolic pathways? The glyoxylate cycle allows organis synthesize glucose. The glyoxylate cycle also bypasses the steps of the citric acid cyclems such as Mycobacterium to use acetate to which involves decarboxylation so that net formation of citric acid cycle intermediates such as succinate can be made. If those steps were not bypassed, it would be two carbons entering, and two carbons exiting and no net intermediates would be produced. The glyoxylate cycle uses substrates from the degradation of fatty acids. The products and intermediates of the glyoxyla two cycles must be regulated (at isocitrate dehydrogenase).te cycle can be substrates in the citric acid cycle, thus the B. Which enzymes of the glyoxylate cycle would provide good drug targets? Why? Isocitrate lyase and malate synthase would be the be unique to the glyoxylate cycle. The other enzymes of the glyoxylate cycle are isozymesst drug targets because they are of enzymes that participate in the citric acid cycle, and inhibiting those enzymes in microorganisms would likely also inhibit the same enzymes in the host. The goal is to exploit differences between host and pathogen such that pathogen designed with the least effect on the host. -specific drugs can be C. You design a library of substrate and product analogues for use in various enzymatic biological assays. You find that the compounds 3-bromopyruvate and 3-nitropropionate and (shown below) reduce growth of Mycobacterium when used together. Which enzyme might they target? How do they inhibit this enzyme? Why are they most effective when used together? The inhibitors target the enzyme isocitrate lyase. 3 are analogues for the enzymatic products glyoxylate and succinate, respectively. They-bromopyruvate and 3-nitropropionate inhibit by binding tightly in the active site. Each inhibitor usually occupied by one product. fills the part of the active site
7. Compound 3852 is an inhibitor of 2,4 dienoyl Give the number of acetyl-CoA molecules obtained from oxid-CoA reductase (called reductase in lecture).ation of the following 20 carbon general names of the enzymes that will catalyze the steps of oxidation. fatty acids in the mitochondrion in the presence of Compound 3852. Give the A. cis 343 - , cis-344 in your textbook).- 9,12 icosadienoic acid (for a refresher on fatty acid nomenclature, see page The reductase is only needed to reduce the unsaturated bond starting at an even carbon, in this case 12. The first 8 carbons can be oxidized to 4 acetyl CoA. The first step to oxidize the 10 bond with the 4,5.th^ carbon forms a 2, This cannot be resolved because the reductase is inactive.3 trans bond which then forms a conjugated double So oxidation stops with 4 acetyl CoA. Enzymes will be: 3X (dehydrogenase, hydratase, dehydrogenase, thiolase); 1X (isomerase, hydratase, dehydrogenase, thiolase), one more dehydrogenase-step; B. cis-, cis- 9,18-icosadienoic acid. 7 acetyl CoA see logic above. 3X (dehydrogenase, hydratase, dehydrogenase, thiolase); 1X (isomerase, hydratase, dehydrogenase, thiolase); 3X (dehydrogenase, hydratase, dehydrogenase, thiolase); one more dehydrogenase-step; C. trans-,cis- 9,12- icosadienoic acid. 4 acetyl CoA because an isomerase can isomerize a trans double bond, see Figure 17 3X (dehydrogenase, hydratase, dehydrogenase, thiolase); 1X (isomerase, hydratase,-10. dehydrogenase, thiolase); one more dehydrogenase-step; D. Give the number of glucose molecules that would produce the same number of acetyl CoA molecules from parts A, B, and C. - 2, 3.5, 2
