Block 3 Lipid Synth & Storage
Questions and Answers
- 1.
What is the most positive activator of the process shown below? 8 acetylCoA + n ATP + 14NADPH -7 palmitate + 8 CoASH+ nADP + nPi + 14NADP
- A.
Acetyl CoA
- B.
Citrate
- C.
Malonyl CoA
- D.
Malate
- E.
Oxaloacetate
Correct Answer
B. CitrateExplanation
Citrate is a potent activator of acetylCoA carboxylase for fatty acid synthesis.Rate this question:
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- 2.
When adipose tissue stores triglyceride arriving from the liver or intestine, glycolysis must also occur in the adipocyte. Which of the following products or intermediates of glycolysis is required for fat storage?
- A.
Glycerol
- B.
Glucose 6-phosphate
- C.
Pyruvate
- D.
Acetyl CoA
- E.
Dihydroxyacetone phosphate
Correct Answer
E. Dihydroxyacetone phosphateExplanation
To reform triglycerides trom the incoming fatty acids, glycerol 3-P must be
available.The adipose can produce this only from DHAP in glycolysis.Rate this question:
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- 3.
Abetalipoproteinemia is a genetic disorder characterized by malabsorption of dietary lipid, steatorrhea (fatty stools), accumulation of intestinal triglyceride, and hypolipoproteinemia. A deficiency in the production of which apoprotein would most likely account for this clinical presentation?
- A.
ApoB-lOO
- B.
ApoB-48
- C.
ApoC-II
- D.
ApoA-I
- E.
ApoE
Correct Answer
B. ApoB-48Explanation
Apo B-48is required for intestinal absorption of dietary fat in the form of chylomicrons.
Apo B-lOOformation is also impaired in these patients, but this would not
explain the clinical symptoms described.Rate this question:
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- 4.
Abetalipoproteinemia is a genetic disorder characterized by malabsorption of dietary lipid, steatorrhea (fatty stools), accumulation of intestinal triglyceride, and hypolipoproteinemia Patients with abetalipoproteinemia exhibit membrane abnormalities in their erythrocytes with production of acanthocytes (thorny-appearing cells). This unusual red cell morphology would most likely result from malabsorption of
- A.
Palmitic acid
- B.
Ascorbic acid
- C.
Arachidonic acid
- D.
Folic acid
- E.
Linoleic acid
Correct Answer
E. Linoleic acidExplanation
The genetic defect would result in malabsorption of the 3 fatty acids listed, but
only linoleate is strictly essential in the diet. Absorption of water-soluble ascorbate and
folate would not be significantly affected.Rate this question:
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- 5.
A patient with a history of recurring attacks of pancreatitis, eruptive xanthomas, and increased plasma triglyceride levels (2000 mg/dL) associated with chylomicrons, most likely has a deficiency in
- A.
Lipoprotein lipase
- B.
LDLreceptors
- C.
HMG-CoA reductase
- D.
ApoB-48
- E.
ApoB-lOOreceptor
Correct Answer
A. Lipoprotein lipaseExplanation
These are the clinical features of lipoprotein lipase deficiency (TypeI lipoproteinemia).
LDLreceptor defects would result in elevated LDLs.HMG-CoA reductase and
ApoB-lOOhave no direct relationship to chylomicrons. ApoB-48 deficiencywould result
in decreased production of chylomicrons.Rate this question:
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- 6.
Uncontrolled phagocytosis of oxidized LDL particles is a major stimulus for the development of foam cells and fatty streaks in the vascular sub endothelium. This process may be inhibited by increased dietary intake of
- A.
Vitamin E
- B.
Vitamin B6
- C.
Vitamin D
- D.
Vitamin B12
- E.
Vitamin K
Correct Answer
A. Vitamin EExplanation
Only vitamin E is an antioxidantRate this question:
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- 7.
A 42-year-old man presents with a chief complaint of intermittent claudication during exercise. His family history is significant for the presence of cardiovascular disease on his father's side, but not on his mother's side. Physical exam reveals xanthelasmas and bilateral tendon xanthomas. A plasma lipid profile reveals a cholesterol level of 340 mg/dL, with a high LDL/HDL ratio. He is given instructions for dietary modifications and a prescription for Zocor (simvastatin). The clinical findings noted in this patient are most likely caused by deficient production of
- A.
Lethicin cholesterol acyltransferase
- B.
ApoB-lOO receptors
- C.
Fatty acyl-CoA synthetase
- D.
VLDL from LDL
- E.
Cholesterol ester transfer protein
Correct Answer
B. ApoB-lOO receptorsExplanation
The findings are indicative of heterozygous Type lla familial hypercholesterolemia,
an autosomal dominant disease. Deficient CETP,LCATor fatty acid CoA synthetase
would not elevate LDLcholesterol.VLDLare not produced from LDL.Rate this question:
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- 8.
A 42-year-old man presents with a chief complaint of intermittent claudication during exercise. His family history is significant for the presence of cardiovascular disease on his father's side, but not on his mother's side. Physical exam reveals xanthelasmas and bilateral tendon xanthomas. A plasma lipid profile reveals a cholesterol level of 340 mg/dL, with a high LDL/HDL ratio. He is given instructions for dietary modifications and a prescription for Zocor (simvastatin). The anticholesterolemic action of simvastatin is based on its effectiveness as a competitive inhibitor of the rate-limiting enzyme in cholesterol biosynthesis. The reaction product normally produced by this enzyme is
- A.
Squalene
- B.
Methylmalonate
- C.
Lanosterol
- D.
Mevalonate
- E.
Acetoacetate
Correct Answer
D. MevalonateExplanation
Must know that mevalonate precedes squalene and lanosterol in the pathway,
and that methylmalonate and acetoacetate are not associated with cholesterolgenesis.Rate this question:
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- 9.
A 42-year-old man presents with a chief complaint of intermittent claudication during exercise. His family history is significant for the presence of cardiovascular disease on his father's side, but not on his mother's side. Physical exam reveals xanthelasmas and bilateral tendon xanthomas. A plasma lipid profile reveals a cholesterol level of 340 mg/dL, with a high LDL/HDL ratio. He is given instructions for dietary modifications and a prescription for Zocor (simvastatin). From a Lineweaver-Burk plot, the Km and Vmaxof this rate-limiting enzyme were calculated to be 4 X 10-3 M and 8 X 102 mmol/h, respectively. If the above experiment is repeated in the presence of simvastatin, which of the following values would be obtained? Km (M) V max (mmolJh)
- A.
4 X 10(-3) 3 X 10(2) ( ) = to the power of
- B.
2 X 10(-3) 1 X 10(2) ( ) = to the power of
- C.
4 X 10(-3) 9 X 10(2) ( ) = to the power of
- D.
8 X 10(-3 ) 8 X 10(2) ( ) = to the power of
- E.
8 X 10(-3) 9 X 10(2) ( ) = to the power of
Correct Answer
D. 8 X 10(-3 ) 8 X 10(2) ( ) = to the power ofExplanation
With a competitive inhibitor, there will be an increase in Km with no change in
Vmax'Option A would be for a noncompetitive inhibitor (V max decreased, Km unaltered).Rate this question:
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Mar 21, 2023Quiz Edited by
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Nov 07, 2011Quiz Created by
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