b) Glycerol kinase
d) Hormone sensitive lipase
e) Acetyl co A carboxylase
Answer- b) The right answer is Glycerol kinase. It is absent in adipose tissue, that is the reason glycerol released by hydrolysis of triglycerides cannot be reutilized there . It is a waste product. It is transported through blood to liver where it is first phosphorylated to form Glycerol-3-P, by glycerol kinase and then converted to Dihydroxy acetone phosphate by the action of glycerol-3-P dehydrogenase. DHAP can be either used for glucose production or oxidized through glycolysis as per the need. In adipose tissue glycerol-3-p is obtained through glycolysis from dihydroxy acetone phosphate for esterification to form triglyceride.
Hexokinase and Phosphofructokinase are enzymes of glycolytic pathway, they are present in muscle. Hormone sensitive lipase is the enzyme for breakdown of triglycerides; it is present in adipose tissue. Acetyl co A carboxylase is the rate limiting enzyme of fatty acid synthesis, it catalyzes the first step of conversion of acetyl co A to malonyl co A, and it is present in adipose tissue.
2. A postpartum woman from a rural community recently gave birth to a baby boy with the aid of a midwife at home. She now brings the baby to the hospital because of continued bleeding and oozing from the umbilical stump. It is likely that this bleeding diathesis is secondary to a deficiency of which vitamin?
a) Vitamin A
b) Vitamin K
c) Vitamin E
d) Vitamin D
c) Folic acid
The right answer is-b) Vitamin K. Newborn infants are at risk of developing vitamin K deficiency, and this coagulation abnormality leads to serious bleeding. Transplacental transfer of vitamin K is very limited during pregnancy, and the storage of vitamin K in neonatal liver is also limited. This makes the newborn infant uniquely vulnerable to hemorrhagic disorders unless exogenous vitamin K is given for prevention of bleeding immediately after birth.
Vitamin A has a role in vision, repair, reproduction and maintenance of epithelial structures. Deficiency is known to cause ophthalmological defects leading eventually to blindness.
Vitamin E is primarily an antioxidant and has also a role in reproduction.
Vitamin D – Its main function is in the regulation of calcium absorption and homeostasis. It has no role in coagulation of blood. Deficiency causes rickets in children and osteomalacia in adults.
Folic acid has a role in nucleotide synthesis and acts as a carrier of one carbon fragments. Deficiency causes megaloblastic anemia.
3. A 40- year-old male presents with excruciating pain in his left flank. He is diagnosed to have a kidney stone and is prescribed citrate to help prevent future stone formation. In the TCA cycle, Citrate is isomerized to Isocitrate and that is subsequently oxidized to α- Keto glutarate by Isocitrate dehydrogenase, a regulatory enzyme of TCA cycle. Which of the following would be most likely to inhibit Isocitrate dehydrogenase?
b) Acetyl co A
e) Co ASH
Answer- The right answer is- c)
Isocitrate dehydrogenase is inhibited by excessive concentration of NADH. The three NAD+ dependent enzymes of TCA cycle; Isocitrate dehydrogenase, α-keto glutarate dehydrogenase and Malate dehydrogenase are inhibited by accumulation of NADH. Since three reactions of the TCA cycle as well as PDH utilize NAD+ as co-factor it is not difficult to understand why the cellular ratio of NAD+/NADH has a major impact on the flux of carbon through the TCA cycle. Thus, activity of TCA cycle is immediately dependent on the supply of NAD+, which in turn, is dependent upon the functioning of electron transport chain and oxidative phosphorylation.
It cannot be ADP, because ADP excess is a sign of low energy state of cell that in itself is a stimulus for the TCA activity to restore the energy balance. On the contrary excess ATP concentration inhibits the TCA cycle (respiratory control of TCA cycle). Acetyl co A, in excess inhibits PDH complex, but it has no effect on activity of Isocitrate dehydrogenase. Since isocitrate dehydrogenase is NAD+ dependent enzyme, excess FADH2 will not affect the rate of action of enzyme. Similarly, Co ASH concentration can affect PDH complex, but it has no effect on isocitrate dehydrogenase activity.
4 . A 50-year-old female presents with severe abdominal pain. Her serum amylase and lipase levels are abnormally elevated and she is diagnosed with pancreatitis. Which linkage between glucose residues is cleaved by amylase ?
a) α- 1,4
b) α- 1,6
d) α 1,2
Answer- a) alpha – 1, 4 is the right answer. Amylase acts on starch and glycogen to cleave α- 1, 4 glycosidic linkages. Both starch and glycogen are polymers of glucose. The two main constituents of starch are amylose (13–20%), which has a nonbranching helical structure, and amylopectin (80–85%), which consists of branched chains composed of 24–30 glucose residues united by α-1,4 linkages in the chains and by α-1 .6 linkages at the branch points. Glycogen, a more highly branched structure than amylopectin contains chains of 12–14 α-D-glucopyranose residues (in α-1 ,4 glucosidic linkage) with branching by means of α-1 ,6 glucosidic bonds. Thus the branch point contains α- 1, 6 linkage that cannot be cleaved by amylase. Mammals lack any enzyme that hydrolyzes the β-1 ,4 bonds, and so cannot digest cellulose that consists of β-D-glucopyranose units linked by β-1 ,4 bonds to form long, straight chains strengthened by cross-linking hydrogen bonds. α 1,2 linkage, more precisely- O-α-D-glucopyranosyl-(1,2)-β -D-fructofuranoside linkage is found in sucrose that is hydrolyzed by sucraseβ-1,6 linkages are not common in nutrients.
5 . A 12-year-old -girl presents with polyuria, polydipsia and polyphagia. Blood Biochemistry reveals a glucose level of 320 mg/dl . She is diagnosed with Type 1 Diabetes mellitus, a disease characterized by a deficiency of Insulin. Which of the following would occur in this patient ?
a) Increased stores of triacylglycerol in adipose tissue
b) Deceased conversion of fatty acids to ketone bodies
c) Increased fatty acids synthesis from glucose in liver
d) Decreased synthesis of cholesterol in liver
e) Increased conversion of fatty acids to acetyl co A
Answer- The right answer is -e) Increased conversion of fatty acids to acetyl co A. In diabetes mellitus, Insulin to glucagon ratio is reversed as a result there is a state of catabolism. Triglycerides stores in adipose tissue are decreased due to adipolysis under the influence of glucagon. In diabetes mellitus, de novo fatty acid synthesis in liver is deceased because in insulin deficiency the catalytic activity of Acetyl co A carboxylase (rate limiting – insulin dependent enzyme) is decreased, hence the overall pathway is inhibited. Cholesterol synthesis in diabetes mellitus is increased. It is expected that cholesterol synthesis would decrease in insulin deficiency since insulin stimulates the activity of HMG Co A reductase (rate limiting enzyme of cholesterol pathway), but on the contrary de novo cholesterol synthesis is increased.
In diabetes mellitus, due to non utilization of glucose, fatty acid oxidation takes place excessively to compensate for energy, as a result there is accumulation of acetyl co A (product of fatty acid oxidation). Excess acetyl co A (since TCA cycle is in a state of suppression) is channeled either towards ketogenic pathway or it is utilized for cholesterol synthesis. Uncontrolled diabetes mellitus is a commonest cause of hypercholesterolemia.
6. Which of the following is a compound formed from both a hydroxylation with an enzyme requiring vitamin C and subsequent methylation?
The right answer is-d) Epinephrine.
Neural cells convert tyrosine to epinephrine and norepinephrine. Tyrosine is first converted to DOPA by tyrosine hydroxylase enzyme. Dopa decarboxylase, a pyridoxal phosphate-dependent enzyme, forms dopamine. Subsequent hydroxylation by dopamine β-oxidase then forms norepinephrine. In the adrenal medulla, phenyl ethanolamine-N-methyltransferase utilizes S-adenosylmethionine to methylate the primary amine of norepinephrine, forming epinephrine.
Gamma Amino butyrate (GABA) functions in brain tissue as an inhibitory neurotransmitter by altering transmembrane potential differences. It is formed by decarboxylation of glutamate, a reaction catalyzed by L-glutamate decarboxylase
Glycine, arginine, and methionine all participate in creatine biosynthesis. Synthesis of creatine is completed by methylation of guanidoacetate by S-adenosylmethionine. Creatinine is an anhydrous product of creatine.
Serotonin (5-hydroxytryptamine), a potent vasoconstrictor and stimulator of smooth muscle contraction, is formed by hydroxylation of tryptophan to 5-hydroxytryptophan followed by decarboxylation.
7. A deficiency of biotin in higher animal is likely to be accompanied by which one of the following:
a) Defective oxidation of fatty acids to Acetyl co A
b) Decreased conversion of pyruvate to Acetyl co A
c) Defective synthesis of fatty acids
d) Decreased utilization of Acetyl co A in TCA cycle
e) Decreased formation of lactic acid in muscles
Answer- The right answer is c) – Defective synthesis of fatty acids.
Biotin is mainly required as a coenzyme for carboxylation reactions and the main examples are carboxylation of-i) pyruvate to oxaloacetate (first step of gluconeogenesis); ii) Acetyl co A to Malonyl co A (first step of fatty acid synthesis) and iii) Propionyl co A to D-Methyl malonyl co A (in the conversion of propionyl co A to Succinyl co A to gain entry to TCA cycle). In biotin deficiency, out of the given options, defective fatty acid synthesis is the most suited option because of the impaired conversion of acetyl co A to malonyl co A.
Fatty acid oxidation does not require biotin; FAD and NAD+ are the required coenzymes in beta oxidation of fatty acids.
Biotin is also not needed as a coenzyme during the conversion of pyruvate to Acetyl co A. The enzyme complex, PDH complex catalyzing this reaction requires TPP, CoASH, Lipoic acid, FAD and NAD+ as coenzymes.
In the utilization of Acetyl co A in TCA cycle. biotin is not required as a coenzyme; FAD and NAD+ are required as coenzymes.
Formation of lactic acid in muscle takes place from pyruvate in a reaction catalyzed by lactate dehydrogenase which requires NADH as a coenzyme.
8 . A 54 -year-old Native- American living on an Indian reservation in southwest Arizona presents to the clinic with impaired memory, diarrhea and a rash on the face, neck, and dorsum of hands. It is likely that the patient has a deficiency of which of the following vitamins?
b) Folic acid
c) Vitamin C
d) Vitamin E
e) Vitamin B6
Answer- The right answer is a)- Niacin.
The clinical manifestations in the present case are of ‘Pellagra’, that occurs due to niacin deficiency.
Pellagra is manifested by Dementia, diarrhea, dermatitis and ‘death’ in untreated cases (4Ds). Pellagra is a common finding in Maize eaters. Maize lacks tryptophan which is an endogenous source of niacin, besides that niacin is also present in the bound form in maize that is biologically unavailable. Hence maize eaters always carry a predisposition to niacin deficiency.
Folic acid deficiency causes megaloblastic anemia.
Vitamin C deficiency causes Scurvy,
There is no specific disorder associated with Vitamin E deficiency, but generalized neurological symptoms and infertility have been found to be associated with vitamin E deficiency.
Deficiency of vitamin B6 is manifested by skin changes (pellagra), neurological changes and Sideroblastic anemia. Vitamin B6 is required as a coenzyme in the pathway of niacin synthesis from tryptophan. Hence B6 deficiency also leads to niacin deficiency.
9 . A 40-year-old man presents with severe pain in his legs upon walking. He is diagnosed with atherosclerotic plaques in the arteries of his legs. High level of cholesterol and LDL contribute to the formation of atherosclerotic plaques. Which of the following is metabolized to form LDL?
c) Cholesteryl esters
The right answer is- a) IDL.
IDL is produced from the metabolism of VLDL and IDL is further processed to form LDL.
VLDL s are produced in the liver and are secreted directly in to the blood as nascent VLDLs containing apoprotein B-100. They obtain apo CII and apo E from HDL. Apo CII is required for the activation of lipoprotein lipase whereas apo E acts as a ligand for internalization through apo B100/apo E receptors.
As VLDLs pass through the circulation triacylglycerols are degraded by lipoprotein lipase, causing VLDLs to shrink in size and become denser. Initially apo CII is returned to HDL and VLDL is converted to IDL (intermediate density lipoprotein) or VLDL remnant. IDL can be either taken up by cells through receptor mediated endocytosis that uses apo E as a ligand or apo E is returned to HDL and IDL is converted to LDL (low density lipoprotein). The primary function of LDL is to provide cholesterol to the peripheral tissues.
Cholesterol cannot be the answer, since it is through lipoproteins only that cholesterol is transported to the peripheral tissues.
Cholesteryl esters are esterified forms of cholesterol. A fatty acid is esterified at 3-OH group of cholesterol. Plasma contains both free and esterified cholesterol.
HDL transports cholesterol from peripheral tissues back to liver for degradation, it is antiatherogenic; hence it is cardio protective also called “Good cholesterol”.
Chylomicrons are transporters of dietary lipids. Upon metabolism in the similar way as VLDL, they are converted to chylomicron remnants that are removed from the circulation by liver.
10 . A 5-year-old boy presents with altered mental status, heart failure and muscle weakness. His serum level of glucose and ketone are abnormally low. He is diagnosed with primary carnitine deficiency. Had there been no carnitine deficiency, how many ATP molecules would have been produced form the complete oxidation of a fatty acid with 14 carbon atoms ?
The right answer is- c)- 92 ATPs.
Fatty acid with 14 carbon atoms will undergo 6 cycles of beta oxidation and yield 7 Acetyl co A.
Number of ATPs per cycle of beta oxidation
1FADH2 = 1.5 ATPs (At the level of Acyl co A dehydrogenase)
1NADH +H+ = 2.5 ATPs (At the level of Beta hydroxy acyl co A dehydrogenase)
Total ATP yield = 4/cycle
In 6 cycles = 6×4 = 24 ATPs
Energy yield per Acetyl co A = 3X2.5 +1.5 = 9 +1(substrate level phosphorylation= 10
Total energy yield through complete oxidation of Acetyl co As=10×7 = 70
Total energy yield = 70 +24 = 94
Net energy yield = 94-2 = 92
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