bibc 102 final

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Bibc102Feng, Final exam, FA10 page 10-1 BIBC102, METABOLIC BIOCHEMISTRY Gen-Sheng Feng Fall, 2010 Mid-term accounts for 40% Final exam accounts for 60% However, if the final scored 10 points better than the mid-term, 70% for final and 30% for mid-term. Overall grades: > 90 A+ 84 – 89 A 77 – 83 A- 70 – 76 B+ 63 – 69 B 56 – 62 B- 49 – 55 C+ 42 – 48 C 35 – 41 C- 28 – 34 D+ 21 – 27 D < 20 F Average: 56 Standard deviation: 20

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Page 1: BIBC 102 final

Bibc102Feng, Final exam, FA10 page 10-1

BIBC102, METABOLIC BIOCHEMISTRY Gen-Sheng Feng

Fall, 2010   Mid-term accounts for 40% Final exam accounts for 60% However, if the final scored 10 points better than the mid-term, 70% for final

and 30% for mid-term. Overall grades: > 90 A+ 84 – 89 A 77 – 83 A- 70 – 76 B+ 63 – 69 B 56 – 62 B- 49 – 55 C+ 42 – 48 C 35 – 41 C- 28 – 34 D+ 21 – 27 D < 20 F Average: 56 Standard deviation: 20

Page 2: BIBC 102 final

Bibc102Feng, Final exam, FA10 page 10-2

Bibc102, Metabolic Biochemistry Gen-Sheng Feng Fall, 2010

FINAL EXAM 8:00 – 11:00 am, Dec. 9, 2010

    Student  ID  #                       First  name:            Last  Name:                    Questions  and  scores:           1.  (4)                   2.  (4)                   3.  (14)                   4.  (16)                   5.  (12)                   6.  (16)                   7.  (12)                   8.  (6)                   9.  (4)                   10.  (12)                 11.  (4)                   12.  (6)                   13.  (12)                 14.  (14)                 15.  (14)                   Total:              

Page 3: BIBC 102 final

Bibc102Feng, Final exam, FA10 page 10-3

1. What is the common chemical component shared by Coenzyme A, NAD+ and FAD

molecules? Note: it is unnecessary to draw out the detailed chemical structure of the component, just write down the name (4 points).

These three enzyme cofactors contain Adenosine.

2. Amino acids can be used as building materials for biosynthesis of purines but not for pyrimidines in nucleotides. Is this statement correct? and why? (4 points).

This is not correct. Amino acids can be used for biosynthesis of both purines and pyrimidines.

Page 4: BIBC 102 final

Bibc102Feng, Final exam, FA10 page 10-4

3. Protein phosphatase 1 (PP1) coordinately regulates hepatic glycogen synthesis and breakdown, in order to maintain constant blood glucose levels. One important mechanism underlying the central role of PP1 in glucose homeostasis is its reciprocal regulation of glycogen phosphorylase and glycogen synthase activities. Please illustrate how PP1 does so, no need to draw out the detailed chemical structures of enzyme substrates or products. (14 points).

PP1 dephosphorylates and inactivates glycogen phosphorylase, which catalyzes glycogen breakdown. Meanwhile, PP1 also dephosphorylates but activates glycogen synthase, facilitating glycogen synthesis. Therefore, the combined effect of PP1 activation leads to increase in glycogen synthesis accompanied by decrease in glycogen breakdown. When blood glucose level is high after a big meal, insulin secretion is increased, stimulating PP1 activity and a net increase in conversion of glucose to glycogen, resulting in decrease of blood glucose levels.

Page 5: BIBC 102 final

Bibc102Feng, Final exam, FA10 page 10-5

4. Explain why glucose can be eventually converted to fatty acid, but fatty acid can NOT be directly converted to glucose in mammals. However, elevated levels of Acetyl-CoA derived from fatty acid breakdown can facilitate gluconeogenesis, why? (16 points).

Glucose can be degraded to Acetyl-CoA, a starting material for fatty acid synthesis. However, there is no enzyme in mammals that can convert Acetyl-CoA, the product of fatty acid β-oxidation, to any of the molecules needed for gluconeogenesis. Elevated levels of Acetyl-CoA can stimulates pyruvate carboxylase activity that catalyzes conversion of pyruvate to oxaloacetate, thereby promoting gluconeogenesis.

Page 6: BIBC 102 final

Bibc102Feng, Final exam, FA10 page 10-6

5. Most fatty acids have even numbers of carbon atom, this is because that a 2-carbon unit is added to extend the chain in each cycle of fatty acid biosynthesis. However, the most critical building material is a 3-carbon molecule. Draw out its molecular structure (no detail is needed for the CoA part) and explain why this molecule instead of acetyl-CoA is used directly for FA biosynthesis. (12 points).

The use of activated malonyl groups rather than acetyl groups is what makes the condensation reactions thermodynamically favorable. The methylene carbon (C-2) of the malonyl group, sandwiched between carbonyl and carboxyl carbons, is chemically situated to act as a good nucleophile. In the condensation step, decarboxylation of the malonyl group facilitates the nucleophilic attack of the methylene carbon on the thioester linking the acetyl group to β-ketoacyl-ACP synthase, displacing the enzyme’s —SH group. Coupling the condensation to the decarboxylation of the malonyl group renders the overall process highly exergonic.

Page 7: BIBC 102 final

Bibc102Feng, Final exam, FA10 page 10-7

6. What is β-oxidation in fatty acid breakdown, explain the chemical steps and structural changes for the relevant C-C bonds directly involved in the β-oxidation reactions. (16 points).

β-oxidation is the first step in fatty acid catabolism, in which a long chain fatty acid is oxidized to produce acetyl residues in the form of acetyl-CoA.

Page 8: BIBC 102 final

Bibc102Feng, Final exam, FA10 page 10-8

7. Each cycle of urea synthesis consumes 3 molecules of ATP. However, the actual expense of energy is less than that because of the “Krebs bicycle”. Is this correct? If so, explain this coupling process, no need to draw out the detailed structures, just write down the names of molecules involved. (12 points).

This is correct. In each cycle of urea production, a fumarate molecule is generated. Fumarate can enter the TCA cycle and produce NADH. One molecule of NADH will produce 2.5 ATP through oxidative phosphorylation in mitochondria. Therefore, the net energy expense for each urea cycle is less than 3 ATP due to coupling of metabolic pathways.

Page 9: BIBC 102 final

Bibc102Feng, Final exam, FA10 page 10-9

8. The name of cholesterol is mostly known in the public for its association with cardiovascular diseases. However, it is only the abnormally high amounts of cholesterol in the blood that are associated with the diseases. In fact, cholesterol has quite important physiological functions. List three functions, be brief! (6 points).

Cholesterol is a component of plasma membrane. Cholesterol is a precursor for synthesis of bile acids. Cholesterol is also a precursor for synthesis of steroid hormones. \

9. Glutamine and glutamate can provide the nitrogen source for biosynthesis of other amino acids but not for nucleotides. Is this correct? And why? (4 points).

No, it is NOT correct. Glu and Gln can provide the nitrogen source for both amino acids and nucleotides.

Page 10: BIBC 102 final

Bibc102Feng, Final exam, FA10 page 10-10

10. It was very surprising that Dr. Richard Hanson and colleagues detected high expression levels of phosphoenolpyruvate carboxykinase (PEP carboxykinase, or PEPCK) in adipocytes. This enzyme plays a critical role in gluconeogenesis. However, enormous experimental data indicated no gluconeogenic activity in the adipose tissue and, therefore, their observation of PEPCK expression in adipocytes was ignored for a while. Interestingly, more recent progress in biochemistry leads to appreciation of the significance of Dr. Hanson’s work, due to elucidation of a new pathway for TAG synthesis. What is this? (12 points).

Dr. Hanson’s observations led to elucidation of a glyceroneogenesis pathway, a truncated version of gluconeogenesis, in adipocytes.

Page 11: BIBC 102 final

Bibc102Feng, Final exam, FA10 page 10-11

11. In plants, photosynthesis generates O2, which is derived from splitting of what molecule? Do photosynthetic bacterial cells also produce O2 in their photosynthetic reactions? (4 points).

O2 is derived from splitting of H2O. Photosynthetic bacterial cells do NOT produce O2.

12. What is the common precursor of both triacylglycerols and glycerophospholipids? draw out its molecular structure. (6 points).

Phosphatidic acid is the precursor of both triacylglycerols and glycerophospholipids.

Page 12: BIBC 102 final

Bibc102Feng, Final exam, FA10 page 10-12

13. What are the common biochemical mechanisms for ATP synthesis in chloroplasts and mitochondria. (12 points).

The common biochemical mechanisms are: A). Electron transfer creates a proton gradient across the inner membrane of chloroplasts and mitochondria. B). The proton gradient drives ATP synthesis. C). The structure and function of the ATP synthase are very similar. D). In both cases, ADP and inorganic phosphate are the substrates for ATP synthesis.

Page 13: BIBC 102 final

Bibc102Feng, Final exam, FA10 page 10-13

14. What are ketone bodies? What is the precursor molecule for production of ketone bodies, and give one reason why ketone bodies are accumulated in diabetic patients. (14 points).

Acetone, Acetoacetate and D-β-hydroxybutyrate are collectively called ketone bodies. Acetyl-CoA are the precursor molecule for synthesis of ketone bodies. In diabetes, increased β-oxidation of fatty acid results in production of large amounts of acetyl-CoA. However, increased gluconeogenesis leads to decrease of oxaloacetate, a critical molecule required for TCA cycling. Therefore, impaired TCA cycling leads to accumulation of acetyl-CoA that in turn makes excess amounts of ketone bodies in diabetic subjects.

Page 14: BIBC 102 final

Bibc102Feng, Final exam, FA10 page 10-14

15. Fatty acids undergo β-oxidation in mitochondria to produce acetyl-CoA, but the biosynthesis of fatty acids from acetyl-CoA occurs in the cytosol. Describe how fatty acids enter the mitochondria and what is the molecular mechanism for transfer of acetyl-CoA from the mitochondrial matrix to cytosol. (14 points).

Fatty acid enters mitochondria via the carnitine transporter.

Acetate is shuttled out of mitochondria as citrate.