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1. A full-term female infant failed to gain weight and showed metabolic acidosis in the neonatal period. A physical examination at 6 months showed failure to thrive, hypotonia, small muscle mass, severe head lag, and a persistent acidosis (pH 7.0 to 7.2). Blood lactate, pyruvate, and alanine were greatly elevated. Treatment with thiamine did not alleviate the lactic acidosis. Which of the following enzymes is most likely deficient in this patient?

a) Alanine amino transferase

b) Phosphoenolpyruvate carboxy kinase

c) Pyruvate carboxylase

d) Pyruvate dehydrogenase

e) Pyruvate kinase

 The right answer is- d)

The symptoms are suggestive of Pyruvate dehydrogenase complex deficiency. Pyruvate dehydrogenase complex (PDC) deficiency (PDCD) is one of the most common neurodegenerative disorders associated with abnormal mitochondrial metabolism.

Pyruvate dehydrogenase, a multienzyme complex, catalyzes the conversion of pyruvate to acetyl co A. The major fate of Acetyl co A is oxidation in citric acid cycle, which is a chief metabolic cycle to derive energy from carbohydrates. Malfunction of this cycle deprives the body of energy. The impaired conversion of pyruvate to Acetyl co A leads to elevation of blood pyruvate, lactate and alanine. The persistent metabolic acidosis is due to accumulation of lactates in blood that results in nonspecific symptoms (e.g., severe lethargy, poor feeding, tachypnea), especially during times of illness, stress, or high carbohydrate intake. Similar symptoms though mild in nature can also be observed in Thiamine deficiency, as TPP is requires as a coenzyme for PDH complex. Since in the given case the patient is resistant to supplementation with thiamine, the deficiency of thiamine is ruled out.

Alanine transferase catalyzes the conversion of pyruvate to alanine (transamination). It cannot be the defect as alanine levels are also elevated.

 Phospho enol pyruvate carboxy kinase catalyzes the conversion of Oxaloacetate to phosphoenol pyruvate. The deficiency is rare.

Pyruvate carboxylase catalyzes the conversion of pyruvate to oxaloacetate.

Pyruvate kinase catalyzes the conversion of phosphoenol pyruvate to pyruvate.

The clinical manifestations in the given case are only suggestive of PDH complex deficiency.

2. A 3- year-old female child whose growth rate has been in the lower 10th percentile over the last year presents with chronic, nonproductive cough and diarrhea with foul-smelling stools. She is diagnosed as having cystic fibrosis. For which of the following vitamins is this child most likely to be at risk of deficiency?

a)  Vitamin C

b) Vitamin B6

c) Folic acid

d) Retinol (vitamin A)

e) Riboflavin (vitamin B2)

The right answer is- d) – Cystic fibrosis, is an autosomal recessive disorder affecting approximately 1 in 2500 white individuals. It is caused due to defective chloride ion channels of exocrine glands and epithelial tissues involving pancreas, sweat glands, and mucous glands in the respiratory, digestive, and reproductive tracts. Affected patients usually have abnormal mucus secretion resulting in recurrent respiratory infections, gastrointestinal obstruction and pancreatic enzyme dysfunction.

The protein cystic fibrosis transmembrane conductance regulator (CFTR) is defective, leading to abnormal chloride transport. Because cystic fibrosis leads to pancreatic damage and diminution of the ability to secrete HCO3− and pancreatic digestive enzymes, as a result fat and protein are absorbed poorly. Retinol is a fat soluble vitamin that must be absorbed along with lipid micelles; hence the absorption of retinol is grossly affected.  The absorption of other fat-soluble vitamins E, D, and K is also decreased.

Vitamins C, B6, folic acid and riboflavin are water-soluble vitamins and their absorption is not significantly affected.

3. A 54-year-old male who was diagnosed with HIV (human immuno deficiency virus) infection 2 years back, is currently in the terminal stage. He is now cachectic and having a difficult time obtaining any caloric intake yet he refuses to take a naso gastric or gastric feeding tube. Since his muscle and organs are metabolically active, which of the following amino acids will produce both glucose and ketone bodies as an energy source?

a) Alanine

b) Tyrosine

c) Proline

d) Glycine

e) Leucine

The right answer is b) Tyrosine.

Tyrosine is the only amino acid in the given options that is both glucogenic as well as ketogenic. Upon metabolism tyrosine produce one molecule of fumarate and one molecule of acetoacetate. Fumarate is converted to Oxaloacetate (through intermediate formation of malate) which can be converted to glucose. Acetoacetate is cleaved to form 2 molecules of acetyl co A under the effect of Thiolase enzyme.  Since Acetyl co A is a precursor of ketone bodies, thus it represents the ketogenic component of Tyrosine.

Alanine can be transaminated to pyruvate (catalyzed by alanine transferase) which can be channeled towards the pathway of gluconeogenesis, thus it is purely glucogenic.

Proline upon metabolism produces Alpha ketoglutarate, a TCA cycle intermediate that can be converted to oxaloacetate and thus it is glucogenic.

Glycine is also purely glucogenic. Glycine is converted to serine that produces pyruvate by non oxidative deamination.

Leucine is purely ketogenic. Upon metabolism it produces acetoacetate and acetyl CoA, both are ketogenic fractions.

4. A 40 -year-old man presents with chest pain that radiates to his left jaw and shoulder. He is diagnosed with a myocardial infarct (heart attack) and is prescribed a statin medication. Statins are competitive inhibitors of HMG CoA reductase, which converts HMG Co A to which of the following?

a) Isopentenyl pyrophosphate

b) Mevalonate

c) Geranyl pyrophosphate

d) Farnesyl pyrophosphate

e) Cholesterol

The right answer is- b) Mevalonate.

The reaction catalyzed by HMG Co A reductase is the principal regulatory step in the pathway of cholesterol synthesis and is the site of action of the most effective class of cholesterol-lowering drugs, the HMG-CoA reductase inhibitors (statins). )

Isopentenyl pyrophosphate is formed from Mevalonate. Mevalonate is phosphorylated sequentially by ATP by three kinases, and after decarboxylation the active isoprenoid unit, Isopentenyl pyrophosphate (C5), is formed.

Isopentenyl diphosphate is isomerized by a shift of the double bond to form dimethylallyl pyrophosphate, then condensed with another molecule of Isopentenyl diphosphate to form the ten-carbon intermediate Geranyl pyrophosphate (C10).

A further condensation of Geranyl pyrophosphate (C10) with Isopentenyl pyrophosphate forms Farnesyl pyrophosphate (C15). Cholesterol is the final product of the pathway.

5. Which of the following compounds is the direct precursor for the heme nitrogen atoms?

a) Glucose

b) Glycine

c) Succinyl co A

d) Alanine

e) Methionine

The right answer is- b) Glycine.

Heme is synthesized in living cells by a pathway that requires Succinyl-CoA, derived from the citric acid cycle in mitochondria, and the amino acid glycine. The product of the condensation reaction between Succinyl-CoA and glycine is α-amino-β-ketoadipic acid, which is rapidly decarboxylated to form -δ-amino levulinate (ALA). Through a number of further steps Heme is synthesized.

Glucose does not have nitrogen in its structure to contribute towards heme formation, whereas alanine and methionine do possess amino groups but they do not contribute to Heme nitrogen.

6. A pregnant woman is able to transfer oxygen to her fetus because fetal hemoglobin has a greater affinity for oxygen than does adult hemoglobin. Why is the affinity of fetal hemoglobin for oxygen higher?

a) The tense form of hemoglobin is more prevalent in the circulation of the fetus.

b) There is less 2, 3-BPG in the fetal circulation as compared to maternal circulation.

c) Fetal hemoglobin binds 2, 3-BPG with fewer ionic bonds than the adult form.

d) The Bohr effect is enhanced in the fetus.

e) The oxygen-binding curve of fetal hemoglobin is shifted to the right.

The right answer is -c). The enhanced uptake of maternal oxygen by fetal Hb is due to less binding of 2, 3 BPG with fetal Hb.

It is not due to more prevalence of tense form of fetal hemoglobin in the circulation.

It is also not due to less 2, 3 BPG in the fetal circulation, the Bohr Effect is not enhanced in the fetus and the oxygen -binding curve of fetal Hb is also not shifted to the right.

In Hb A (adult Hb) when 2, 3-BPG binds to deoxyhemoglobin, it acts to stabilize the low oxygen affinity state (T state) of the oxygen carrier, exploiting the molecular symmetry and positive polarity by forming salt bridges with lysine and histidine residues in the four subunits of hemoglobin.

The R state, with oxygen bound to a heme group, has a different conformation and does not allow this interaction. By selectively binding to deoxyhemoglobin, 2, 3-BPG stabilizes the T state conformation, making it harder for oxygen to bind hemoglobin and more likely to be released to adjacent tissues.

Fetal hemoglobin (HbF) exhibits a low affinity for 2, 3-BPG, resulting in a higher binding affinity for oxygen. This increased oxygen-binding affinity relative to that of adult hemoglobin (HbA) is due to HbF’s having two α/γ dimers as opposed to the two α/β dimers of HbA. The positive histidine residues of HbA β-subunits that are essential for forming the 2, 3-BPG binding pocket are replaced by serine residues in HbF γ-subunits so 2, 3-BPG has difficulties in linking to the fetal hemoglobin, hence the affinity of fetal hemoglobin for O2 increases. That’s the way O2 flows from the mother to the fetus.

7. A 35-year-old female presents with severe dehydration and decreased urine output. Her blood urea nitrogen level is abnormally elevated because her kidneys are not able to excrete urea in the urine. In the production of urea, which of the following is an important intermediate?

a) Serine

b) Glutamate

c) Proline

d) Ornithine

e) Leucine

The right answer is-d) Ornithine

Urea is the end product of nitrogen/amino acid metabolism. Urea formation requires the participation of 6 amino acids which are aspartic acid, ornithine, citrulline, argino succinic acid, arginine and N-Acetyl glutamate. Out of these N -Acetyl glutamate is the only amino acid that acts as an allosteric activator of Carbamoyl phosphate synthase-1 enzyme, rest all participate in urea formation. Ornithine acts as a catalyst in the process of urea formation. Ornithine is a non standard amino acid, it is not incorporated in to tissue proteins but it participates in urea formation and polyamine synthesis.

Serine does not participate in urea formation. It is glucogenic, incorporated in to tissue proteins, required for synthesis of glycine, cysteine and sphingosine. Also participates in one carbon metabolism, forming O- glycosidic linkages and is present in at the active site of many enzymes.

Glutamate is a precursor of GABA, glutamine, glucogenic, a neurotransmitter and is incorporated in to tissue proteins. It does not participate in urea formation.

Proline also does not participate in urea formation; it is incorporated in to tissue proteins and is present at those places where kinks or bends are needed in the folding of the proteins since it is an imino acid. Collagen is rich in hydroxy proline (its hydroxylated form).

Leucine is a branched chain amino acid; it is purely ketogenic and has no role in urea formation.

8. Which of the following is a common compound shared by the TCA cycle and the urea cycle?

a) α-Keto glutarate

b) Argino succinic acid

c) Arginine

d) Fumarate

e) Aspartate

The right answer is- d) Fumarate.

Alpha ketoglutarate is an intermediate of TCA cycle; it has no connection with urea cycle.

Argino succinic acid is an intermediate of urea cycle and is formed by condensation of aspartic acid and Citrulline. It is not connected to TCA cycle, but upon breakdown it produces Arginine and fumarate. Fumarate is channeled towards TCA cycle. Fumarate is converted to malate and then to oxaloacetate through TCA cycle enzymes. Oxalo acetate is transaminated to form Aspartic acid that can be reutilized to form Argino succinic acid for continuation of urea cycle.

Aspartic acid is needed in urea formation but it is not an intermediate of TCA cycle.

Thus, Fumarate is the right answer.

9. Which enzyme often mal functions in diseases associated with the symptoms of high blood triglyceride levels and Steatorrhea?

a) Phospholipase D

b) Lipoprotein lipase

c) Thiokinase

d) Acetyl co A carboxylase

e) Pancreatic lipase

The right answer is- e) pancreatic lipase.

Steatorrhea occurs due to impaired digestion and absorption of lipids and other chief nutrients. It is characterized by passage of bulky stools and is often associated with deficiencies of fat soluble vitamins. Hypertriglyceridemia is a characteristic finding in Pancreatitis and there are several causes of Acute and chronic pancreatitis. In pancreatic disorders, pancreatic lipase deficiency results in impaired digestion of lipids as well as other nutrients causing Steatorrhea.

Phospholipase D enzyme is required for removal of nitrogenous base from phospholipid. It cannot cause the symptoms of high blood triglyceride levels and Steatorrhea.

Lipoprotein lipase is an enzyme for the degradation of triglacylglycerols present in chylomicrons and VLDL. Impaired activity can cause increase in serum triacylglycerol levels but Steatorrhea is unlikely to happen.

Thiokinase is the first enzyme of beta oxidation of fatty acids. It is required for the conversion of free fatty acid to fatty acyl co A (activation of fatty acid); it is also called Acyl Co A synthetase. This enzyme has nothing to do with lipid digestion or absorption and is not involved in the synthesis or degradation of triacylglycerols.

Acetyl co A carboxylase is an enzyme of fatty acid synthesis. It catalyzes the first step of conversion of acetyl co A to malonyl co A. It is the rate limiting step of de no fatty acid synthesis, malfunctioning of this enzyme cannot cause increase in TGs or Steatorrhea.

Thus the right answer is pancreatic lipase.

10. An infant is born with a high forehead, abnormal eye folds, and deformed ear lobes. He shows little muscle tone and movement. After multiple tests, he is diagnosed with Zellweger syndrome, a disorder caused by malformation of peroxisomes.
Which of the following is expected to be high in concentration in brain tissue of the affected individual?

a) Ketone bodies

b) Lactate

c) Cholesterol

d) Very long chain fatty acids

e) Glucose

The right answer is d), “Very long chain fatty acids”. Very long chain fatty acids are first trimmed in the peroxisomes till the length of C16 or C18, then they are transported to mitochondria in the conventional way though carnitine shuttle to be oxidized completely by beta oxidation.

Zellweger syndrome, also called cerebrohepatorenal syndrome is a rare, congenital disorder (present at birth), characterized by the reduction or absence of Peroxisomes in the cells of the liver, kidneys, and brain. In Zellweger syndrome the Peroxisomal trimming is impaired, thus VLFA accumulate in brain and blood of affected patients.

The option ketone body is ruled out because the above said symptoms are not characteristic of ketosis.

Lactate is also ruled out because excess of lactate (lactic acidosis) is also not a congenital disorder; lactate accumulates either due to non utilization (as in liver disorders) or excessive production (as in anaerobic conditions).

Cholesterol excess, hypercholesterolemia is also not the right option; the symptoms are suggestive of a congenital disorder that has affected the brain and physical growth. There is no such clinical state associated with hypercholesterolemia and impaired mental development.

Hyperglycemia is also not the right choice. Hyperglycemia is never associated with malformation of peroxisomes.

Hence the right option is very long chain fatty acids. For further details of Zellweger syndrome follow the link

11. After excessive drinking over an extended period of time while eating poorly, a middle-aged man is admitted to the hospital with “high output” heart failure. Which of the following enzymes is most likely inhibited?

a) Aconitase

b) Citrate synthase

c) Isocitrate dehydrogenase

d) α-Ketoglutarate dehydrogenase

e) Succinate thiokinase

The right answer is d) – Alpha ketoglutarate dehydrogenase deficiency.

The patient is most probably suffering from cardiac beriberi. The criteria for diagnosing cardiac beriberi is- 1)  Signs of heart failure 2) signs of neuropathies 3) history of alcoholism or poor nutritional history, 4)exclusion of other signs of heart failure, 5) low red cell Transketolase activity 6) Response by thiamine administration.

The above said patient is a known alcoholic, mal nourished and has heart failure .The probable diagnosis is Thiamine deficiency, which can be confirmed by Erythrocyte Transketolase activity. Thiamine is required as a coenzyme for Pyruvate dehydrogenase complex, alpha ketoglutarate dehydrogenase complex, transketolase and alpha keto acid dehydrogenase complex.

Out of all the given options only alpha ketoglutarate dehydrogenase is the only enzyme which is Thiamine dependent and that could possibly be inhibited in beri-beri.

Aconitase is iron dependent enzyme of TCA cycle. All the enzymes enlisted are though enzymes of TCA cycle, but the deficiencies of Aconitase, citrate synthase, isocitrate dehydrogenase and succinate thiokinase are unknown. Alpha ketoglutarate dehydrogenases deficiency is unknown but its impaired activity is observed in thiamine deficiency. TCA cycle enzyme deficiencies are incompatible with life proving that TCA cycle is vital to life.

12. A 16-month-old girl was found to have ingested approximately 30 mL of an acetonitrile-based cosmetic nail remover when she vomited 15 minutes post ingestion. The poison control center was contacted, but no treatment was recommended because it was confused with an acetone-based nail polish remover. The child was put to bed at her normal time, which was 2 hours post ingestion. Respiratory distress developed sometime after the child was put to bed, and she was found dead the next morning.

Inhibition of which of the following enzymes was the most likely cause of this child’s death?

a) Cytochrome c reductase

b) Cytochrome oxidase

c) Coenzyme Q reductase

d) NADH dehydrogenase

e) Succinate dehydrogenase

 The right answer is-b) – Cytochrome oxidase.

 Acetonitrile is used mainly as a solvent. It is the chemical compound with the formula CH3CN. Acetonitrile has only a modest toxicity in small doses. It can be metabolized to produce hydrogen cyanide, which is the source of the observed toxic effects. Generally the onset of toxic effects is delayed, due to the time required for the body to metabolize acetonitrile to cyanide (generally about 2–12 hours). It has been used in formulations for nail polish remover, despite its low but significant toxicity. Acetone and ethyl acetate are often preferred as safer for domestic use, and acetonitrile has been banned in cosmetic products in the European Economic Area since March 2000.

Cyanide released from the metabolism of acetonitrile, is an inhibitor of Cytochrome oxidase. Electron flow in cytochrome c oxidase can be blocked by hydrogen sulphide (H2S), cyanide (CN-), azide (N3 -), and carbon monoxide (CO). Cyanide and azide react with the ferric form of heme a 3, whereas carbon monoxide inhibits the ferrous form. Inhibition of the electron-transport chain also inhibits ATP synthesis because the proton-motive force can no longer be generated.

The options given above represent the complexes of electron transport chain. There are site specific inhibitors for each of the complexes of electron transport chain. It is only cytochrome oxidase that is inhibited by Cyanide.

For further details- follow the link given below



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