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Osazone test

This test is used for the identification of sugars. It involves the reaction of monosaccharide with phenyl hydrazine, a crystalline compound. All reducing sugars form osazones with excess of phenyl hydrazine when kept at boiling temperature. Each sugar has a characteristic crystal form of osazones.

Which “one” of the following shapes corresponds with the shape of sucrosazone crystals?

A. Needle shaped

B. Petal shaped

C. Rhombic plates

D. Powder puff shaped

E. None of the above.

Check the details at,

http://usmle.biochemistryformedics.com/osazone-test/

 

 

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Q.1- Which out of the following is an important storage site of glycogen?

A) Adipose tissue

B) Cardiac muscle

C) Spleen

D) Kidney

E) Liver.

Q.2- Which of the following is an active form of glucose to initiate the process of glycogenesis?

A) Adenosyl Glucose

B) UDP glucose

C) AMP Glucose

D) Glucose-6-Phosphate

E) Glucose-1-P

Q.3- Choose the correct statement describing the role of branching in the structure of Glycogen-

A) Branching provides compaction

B) Branching increases solubility

C) Branching increases the rate of glycogen synthesis and degradation

D) Branching takes place after every 4-6 glucose residues

E) Branching is carried out by Glycogen synthase enzyme.

Q.4- Glycogen synthase is highly active under the conditions of :

A) Excess glycogen stores

B) Starvation

C) High fat feeding

D) High carbohydrate feeding

E) Uncontrolled diabetes mellitus

Q.5- Which out of the following is not an enzyme involved in glycogen degradation?

A) Glucose-6-phosphatase

B) Phosphorylase

C) Phosphoglucomutase

D) Amylo (1-4) to (1-6) Glucan transferase

E) Amylo (1-4) to (1-4) Glucan transferase

Q.6- What is the net energy output when glucose obtained through the action of phosphorylase on glycogen is oxidized anaerobically through glycolysis ?

A) 4 ATP

B) 2 ATP

C) 3 ATP

D) 8 ATP

E) 38 ATP

Q.7- Which of the following enzymes of glycogen degradation uses phosphate for cleavage of the glycosidic bond?

A) Phosphorylase

B) Glucose-6-phosphatase

C) Debranching enzyme

D) Phosphoglucomutase

E) All of the above.

Q.8- Von Gierke’s disease is characterized by the deficiency of which of the following enzymes?

A) Phosphorylase

B) Glucose-6-phosphatase

C) Debranching enzyme

D) Phosphoglucomutase

E) Branching enzyme

Q.9- Mc Ardle’s syndrome causes muscle cramps and muscle fatigue with increased muscle glycogen. Which of the following enzymes is deficient?

A) Hepatic hexokinase                                                  

B) Muscle Phosphorylase

C) Muscle Debranching enzyme                               

D) Muscle Hexokinase  

E) Muscle phospho fructokinase                                              

Q.10- Which of the following enzymes generates free glucose during the breakdown of glycogen in skeletal muscle?

A) Phosphorylase                                           

B) α-1-6-amyloglucosidase         

C) Debranching enzyme               

D) Glucose-6-phosphatase         

E) Phosphoglucomutase

Q.11- Which of the following enzymes of glycogen metabolism is absent in the skeletal muscle?

A) Phosphorylase                                           

B) α-1-6-amyloglucosidase         

C) Debranching enzyme               

D) Glucose-6-phosphatase         

E) Phosphoglucomutase                              

Q.12- An 8-year-old child has been brought to emergency with dehydration, acidotic breathing and increased appetite from the past few weeks. The child has been diagnosed with diabetic keto acidosis. Which of the following biomolecules is the most likely precursor of ketone bodies?

A) Acetyl co A

B) Propionyl co A

C) Succinyl co A

D) Citrate

E) Oxalo acetate.

Q.13-   Which of the following is the best predictor of glycemic status of an individual with long-standing diabetes mellitus?

A) Fasting blood glucose

B) Plasma insulin level

C) C-Peptide level

D) HbA1c level

E) Glucose tolerance test

Q.14- The major cause of hyperglycemia in type 2 Diabetes mellitus is-

A) Auto immune destruction of beta cells of pancreas

B) Insulin resistance

C) Over activity of alpha cells of pancreas

D) Insulin resistance followed by insulin secretory defect

E) All of the above.

Q.15- Which of the following is not a laboratory finding in Diabetic keto acidosis?

A) Hyperglycemia

B) Glycosuria

C) Ketonuria

D) Low pH

E) High bicarbonate

Q.16- Which of the following complications is not observed in type 1 DM?

A) Lactic acidosis

B) Hyperosmolar non ketotic coma

C) Hypoglycemia

D) Microalbuminuria

E) Retinopathy

Q.17- Which of the following is an absolute contraindication of Glucose tolerance test ?

A) Family H/O DM

B) Blood glucose high but no symptoms

C) Long standing DM

D) Renal glycosuria

E) Gestational DM

Q.18- All are true of clinical manifestations of Von Gierke’s disease except:

A) Short stature

B) Xanthomas

C) Gouty arthritis

D) Visual impairment

E) Protuberant abdomen

Q.19- Which of the following is the most important cause of complications in DM?

A) Sorbitol formation

B) Glycation of tissue proteins

C) Oxidative stress

D)  All of the above

E) None of the above.

Q.20- Which of the following is not a cause of hyperglycemia?

A) Hypothyroidism

B) Pancreatitis

C) Hypercorticism

D) Hyperadrenalism

E) Viral infections.

Q.21- Which of the following is not a diabetes prone state ?

A) Impaired fasting glycemia

B) Impaired glucose tolerance

C) Gestational diabetes mellitus

D) All of the above.

E) None of the above.

Key to answers

1)- E, 2)- B, 3)- C, 4)- D, 5)- D, 6)- C, 7)- A, 8)- B, 9)- B, 10)-B, 11)- D, 12)- A, 13)- D, 14)-D,15)-E, 16)-B, 17)- C, 18)- D, 19)- D, 20)-D.

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Q.1. When fatty acid oxidation predominates in the liver, mitochondrial pyruvate is likely to be-

a)-Carboxylated to phosphoenol pyruvate for entry in to gluconeogenesis

b)- Oxidatively decarboxylated to acetyl co A for entry in to ketogenesis

c)- Reduced to lactate for entry in to gluconeogenesis

d)- Oxidatively decarboxylated to acetyl co A for oxidation in TCA cycle

e)- Carboxylated to oxaloacetate for entry in to gluconeogenesis

Answer- The right answer is- e) – Mitochondrial pyruvate is carboxylated to oxaloacetate for entry in to gluconeogenesis. This is the first step of gluconeogenesis. Gluconeogenesis is the process of converting noncarbohydrate precursors to glucose or glycogen. The major substrates are the glucogenic amino acids, and lactate, glycerol, and propionate.

Fatty acid oxidation takes place during conditions of prolonged fasting /starvation or in diabetes mellitus. Under these conditions Insulin to glucagon ratio is reversed with more circulating concentration of Glucagon than insulin. As a result, a state of catabolism is there with more lipolysis, glycogenolysis and protein catabolism. After 12–18 hours of fasting, liver glycogen is almost totally depleted. Gluconeogenesis meets the needs of the body for glucose when sufficient carbohydrate is not available from the diet or glycogen reserves. A supply of glucose is necessary especially for the nervous system and erythrocytes.

The option a) – Carboxylation of pyruvate to phosphoenol pyruvate for entry to gluconeogenesis is not correct. Pyruvate is first carboxylated to form oxaloacetate and then it is decarboxylated and phosphorylated to form phosphoenol pyruvate that gets converted to 2-phosphoglycetare to gain entry in to the pathway of gluconeogenesis.

Pyruvate can be oxidatively decarboxylated to form acetyl co A but under conditions of starvation or diabetes mellitus, the equilibrium is more towards formation of glucose. PDH complex remains inhibited

in insulin deficiency and by acetyl co A, that is the product of fatty acid oxidation. TCA is also in a state of suppression due to utilization of oxaloacetate for glucose production.

Pyruvate can be reduced to lactate, but for entry in to the pathway of gluconeogenesis, lactate is not the direct substrate, it has to be converted to pyruvate that is channeled to the pathway of gluconeogenesis.

Thus out of all the most suited option is e) – Mitochondrial pyruvate is carboxylated to oxaloacetate for entry in to gluconeogenesis.

Q.2- An 8-year old boy, a known diabetic (type 1 DM), has been brought to emergency room in a state of coma. His breathing is rapid and deep, and his breath has a fruity odor. His blood glucose is 480 mg/dL. The attending physician has administered IV fluids, insulin, and potassium chloride. A rapid effect of insulin in this situation is to stimulate

a) gluconeogenesis in liver

b) fatty acid release from adipose

c) glucose transport in muscle

d) ketone utilization in the brain

e) glycogenolysis in the liver

 Answer- The right answer is- c) Glucose transport in muscle.

The child is suffering from diabetic ketoacidosis, the commonest complication of Type 1 diabetes mellitus. Fruity odor of breath is due to the presence of acetone, one of the ketone bodies (the other two are acetoacetate and beta hydroxy butyrate). Acetone  is excreted through lungs. High blood glucose is due to non utilization or extra synthesis of glucose in the presence of reversed insulin to glucagon ratio.

In the conditions of non utilization of glucose, fats are alternatively oxidized to provide energy. The extra Acetyl co A produced by fatty acid oxidation is diverted to the pathway of ketogenesis.

As regards other options-

Insulin does not promote gluconeogenesis, rather it inhibits it.

Similarly fatty acid release from adipose tissue (adipolysis) is an action of glucagon and catecholamines, insulin inhibits this action also.

Ketone utilization in brain is also not the correct option. By promoting glucose utilization, insulin inhibits ketosis; in fact ketosis occurs only when glucose is not available for utilization as in starvation, low carbohydrate/high fat diet, or diabetes mellitus.

Glycogenolysis is also not the correct answer. Insulin promotes glycogenesis, it is an anabolic hormone, and it prevents all the catabolic processes including glycogenolysis.

In diabetic ketoacidosis, Insulin promotes glucose uptake through GLUT4 transporters in skeletal, cardiac muscle and adipose tissue. It also promotes glucose utilization by stimulating the enzymes of pathways of glucose utilization.

IV fluids are given to treat dehydration as DKA is mostly associated with polyuria. Potassium chloride is given to maintain potassium balance.

For details of DKA follow the link

http://www.namrata.co/case-study-diabetic-ketoacidosis/

Q.3-A 10-month-old child is being evaluated for the underlying cause of a hemolytic anemia. The oxygen dissociation curve for hemoglobin in his erythrocytes is compared with the curve obtained with normal cells. The results are shown in the image. A deficiency of which of the following enzymes is most likely to account for the hemolytic anemia in this patient?

a) Glucokinase

b) Glucose-6-P dehydrogenase

c) Pyruvate carboxylase

d) Glutathione reductase

e) Pyruvate Kinase

Oxygen dissociation curve 

Answer- The right answer is -e) – Pyruvate kinase.

There is right ward shift of oxy Hb dissociation curve, showing more unloading of oxygen that might be due to more concentration of 2 3 BPG. Out of the given options, it is pyruvate kinase deficiency that can cause more formation of 2, 3 BPG.

Glucokinase deficiency can cause hemolytic anemia, but there is decreased formation of 2, 3 BPG also, there cannot be right ward shit of oxy Hb dissociation curve. Glucokinase deficiency or impaired activity is found in type 2 DM.

Glucose-6-P dehydrogenase deficiency causes hemolytic anemia but that is due to met Hb formation as a result of impaired activity of glutathione reductase caused because of reduced availability of NADPH.

Pyruvate carboxylase deficiency causes impaired formation of oxaloacetate, it can cause hypoglycemia. This enzyme is mitochondrial, hence is absent in red blood cells. Thus this is not the right option.

Impaired activity of glutathione reductase is associated with hemolytic anemia, but 2, 3 BPG concentration is not affected.

Hence out of all the options, Pyruvate kinase deficiency is the most suited option. Pyruvate kinase lies at the end of the glycolytic pathway in RBCs followed only by lactate dehydrogenase. Pyruvate kinase activity is critical for the pathway and therefore critical for energy production. If ATP is not produced in amounts sufficient to meet the energy demand, then those functions are compromised. Energy is required to maintain the Na+/K+ balance within the RBC and to maintain the flexible discoid shape of the cell. In the absence of sufficient pyruvate kinase activity and therefore ATP, the ionic balance fails, and the membrane becomes misshapen. Cells reflecting pyruvate kinase insufficiency rather than a change in membrane composition are removed from the circulation by the macrophages of the spleen. This results in an increased number of circulating reticulocytes and possibly bone marrow hyperplasia, which is a biological response to lowered RBC count as a result of hemolysis of erythrocytes. Important intermediates proximal to the PK defect influence erythrocyte function. Two- to 3-fold increases of 2, 3-bisphosphoglycerate levels result in a significant rightward shift in the hemoglobin-oxygen dissociation curve. Physiologically, the hemoglobin of affected individuals has an increased capacity to release oxygen into the tissues, thereby enhancing oxygen delivery.

Q.4- A 54- year-old man with Type 1 diabetes is referred to an ophthalmologist for evaluation of developing cataract. Blood Biochemistry results are shown below-

 Fasting blood glucose   198 mg/dl

Hemoglobin A                  15 gm/dl

Hemoglobin A 1c             10% of total Hb

Urine ketones                   Positive

Urine glucose                    Positive

Which of the following enzymes is most strongly associated with cataract formation on this patient ?

a) Galactokinase

b) Aldose reductase

c) Glucokinase

d) Galactose-1-P uridyl transferase

e) Aldolase

 Answer- The right answer is -b) Aldolase reductase. The patient is suffering from uncontrolled diabetes mellitus. Cataract, ketosis and glycosuria, the complications of diabetes mellitus are evident from the history and blood biochemistry results.

Aldose reductase is responsible for reduction of glucose to sorbitol. Sorbitol accumulation is responsible for most of the diabetic complications including cataract,

Diabetic cataract is due to sorbitol accumulation as well as due to non enzymatic glycation of lens proteins.

Galactokinase is not the right option. Galactokinase is the enzyme for phosphorylation of galactose. Galactokinase deficiency in galactosemia can lead to accumulation of excess galactose that can also get converted to galacitol by aldolase reductase. Galacitol accumulation is responsible for premature cataract in galactosemia.

Glucokinase is the enzyme for phosphorylation of glucose (1st step of glycolysis); it is not associated with cataract in diabetes mellitus.

Galactose-1-P uridyl transferase is the enzyme for conversion of galactose-1-P to UDP galactose. Deficiency is responsible for classical galactosemia. It has no role in diabetic cataract.

Aldolase enzyme has two isoenzymes. Aldolase A is responsible for cleavage of Fr-1, 6 Bisphosphate to Glyceraldehyde-3-P and Dihydroxy acetone- phosphate, whereas Aldolase B is responsible for cleavage of Fructose-1-P to form glyceraldehyde and dihydroxy acetone phosphate in fructose metabolism.

Both of these two enzymes are not responsible for Diabetic cataract.

Q.5- A physician is examining a patient who exhibits fasting hypoglycemia and needs to decide between a carnitine deficiency and a carnitine acyl transferase 2 deficiency as the possible cause. A blood test for confirmation of diagnosis has been ordered. The levels of which one of the following would help in confirmation of diagnosis?

a) Glucose

b) Ketone bodies

c) Insulin

d) Acyl-carnitine

e) Carnitine

 Answer- The right answer is d) – Acyl carnitine.

Activated long-chain fatty acids are transported across the mitochondrial membrane by conjugating them to carnitine, a zwitterionic alcohol, ß-hydroxy-Υ-trimethyl ammonium butyrate-  (CH3)3N+—CH2—CH(OH)—CH2—COO–,

The acyl group is transferred to the hydroxyl group of carnitine to form acyl carnitine. This reaction is catalyzed by carnitine acyl transferase I. Acyl carnitine is then shuttled across the inner mitochondrial membrane by a translocase. The acyl group is transferred back to CoA on the matrix side of the membrane. This reaction, is catalyzed by carnitine acyl transferase II. Finally, the translocase returns carnitine to the cytosolic side in exchange for an incoming acyl carnitine

With a carnitine deficiency, fatty acids cannot be added to carnitine, and acyl-carnitine would not be synthesized.  Hence the Acyl carnitine level would be low.

With a carnitine acyl-transferase II deficiency, the fatty acids are added to carnitine, but the acyl-carnitine cannot release the acyl group within the mitochondria. This will lead to an accumulation of acyl carnitine, which will lead to an accumulation in the circulation.

Hence acyl carnitine level would help to differentiate between carnitine deficiency and Carnitine acyl transferase II deficiency.

The end result of either deficiency is a lack of fatty acid oxidation, such that ketone body levels would be minimal under both conditions, and blood glucose levels would also be similar in either condition.

Hypo glycemia and low ketone body level are characteristic of carnitine as well as associated enzyme deficiencies but their levels cannot discriminate between either of these two conditions.

Insulin release is not affected by either deficiency, and carnitine levels, normally low, would not be significantly modified in either deficiency.

 

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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

http://www.namrata.co/minor-pathways-of-fatty-acids-oxidation-a-quick-review/

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

http://www.namrata.co/inhibition-of-oxidative-phosphorylation/

 

 

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1. Which of the following enzymes is not present in adipose tissue?

a) Hexokinase

b) Glycerol kinase

c) Phosphofructokinase

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?

a) ADP

b) Acetyl co A

c) NADH

d) FADH2

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

c) β-1,4

d) α 1,2

e) β-1,6

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?

a) Histamine

b) GABA

c) Creatinine

d) Epinephrine

e) Serotonin

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?
a) Niacin

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?

a) IDL

b) Cholesterol

c) Cholesteryl esters

d) HDL

e) Chylomicrons

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 ?

a) 38

b) 129

c) 92

d) 36

e) 131

 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

(2 ATP used for activation of fatty acids)

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1. A 24-year-old female presents with complaints of intestinal bloating, gas, cramps, and diarrhea following a meal including dairy products. A lactose-tolerance test confirms that she has a deficiency of lactase. Which of the following dairy products could you recommend that would be least likely to cause her difficulties in the future?

a) Condensed milk

b) Cottage cheese

c) Ice cream

d) Skim milk

e) Yogurt

2. A 25 year-old female meets with her family physician. She is interested in starting a family soon and is looking for advice on what nutritional supplements would be beneficial during pregnancy. Which of the two supplements should be considered as the most important?

a) Vitamin K and copper

b) Selenium and thiamine

c) Iron and folate

d) Vitamin C and D

e) Vitamin A and E

3. A 16 -year-old girl presents with severe menstrual camping, caused by increased prostaglandin production. Prostaglandins are synthesized from which of the following?

a) Arachidonic acid

b) Glucose

c) Acetyl co A

d) Palmitic acid

e) Propionyl co A

4. A 45 -year-old man presents with multiple gunshot wounds to the abdomen. He undergoes emergent surgery during which a part of his intestine is resected. After surgery he is placed on total parenteral nutrition (TPN), an intravenous form of nutrition.TPN supplies essential fatty acids also in the diet. Which of the following is an essential fatty acid?

a) Palmitic acid

b) Linoleic acid

c) Elaidic acid

d) Oleic acid

e) Erucic acid

5. A 35 year-old woman presents with severe upper abdominal pain, nausea and vomiting. She is diagnosed with pancreatitis (an inflammation of pancreas). Her serum triglyceride level is 1500 mg/dl (normal 150-199 mg/dl). Triglycerides are primarily synthesized in which of the following tissues?

a) Skeletal muscle

b) Heart muscle

c) Liver

d) Spleen

e) Red blood cells

6. An 8 -year-old boy has been brought to emergency room with generalized tonic/clonic seizure. The child has a history of epilepsy and ataxia. Blood biochemistry reveals lactic acidosis. Upon questioning the parents state that their child was born with the disease, “Pyruvate carboxylase deficiency”. Which of the following products is the fate of pyruvate when the reaction is catalyzed by pyruvate carboxylase?

a) Lactate

b) Alanine

c) Acetyl co A

d) Oxaloacetate

e) Acetaldehyde

7. The catabolism of   1 mol of glucose to lactate in the glycolytic pathway is accompanied by the reduction of how many moles of O2?

a) 0

b) 2

c) 4

d) 8

e) 10

8. After digestion of a piece of cake that contains flour, milk and sucrose as its primary ingredients, the major carbohydrate products entering the blood are-

a) Glucose

b) Galactose and fructose

c) Glucose and galactose

d) Glucose and fructose

e) Fructose, glucose and galactose

9. Which of the following enzymes is not present in muscle?

a) Glycogen synthase

b) Lactate dehydrogenase

c) Glucose-6-Phosphatase

d) Hexokinase        

e) Phosphorylase

10. A 43-year-old man with a long history of poorly controlled hypertension presents to the emergency with a severe headache. His blood pressure is found to be dramatically elevated at 220/150 mm Hg. Which of the following products derived from amino acids, might bring down his highly elevated blood pressure?

a) Melanin

b) Nitric oxide

c) GABA

d) Dopamine

e) Serotonin

 

Answers

 1 – e) – Lactose intolerance is the inability or insufficient ability to digest lactose, a sugar found in milk and milk products. Lactase breaks down lactose into two simpler forms of sugar called glucose and galactose, which are then absorbed into the bloodstream. Lactose intolerance is caused by a deficiency of the enzyme lactase, which is produced by the cells lining the small intestine. Enzyme levels are highest shortly after birth and decline with aging, despite a continued intake of lactose.

Although the body’s ability to produce lactase cannot be changed, the symptoms of lactose intolerance can be managed with dietary changes. Most people with lactose intolerance can tolerate some amount of lactose in their diet. Gradually introducing small amounts of milk or milk products may help some people adapt to them with fewer symptoms. Lactose-free, lactose-reduced milk, Soy milk and other products may be recommended.

The microorganisms that convert milk to yogurt (Streptococcus salivarius thermophilus and Lactobacillus delbrueckii  bulgaricus) metabolize most of the lactose in the milk. Yogurt is also a good source of dietary calcium. Thus yogurt would cause least difficulties and can be safely recommended to the patient.

2- c)- Pregnancy is a time of increased metabolic demand, and two of the most important supplements are Iron and folic acid ;Iron to prevent anemia and folic acid to prevent neural tube defects in the developing fetus. Copper and selenium are the trace elements that are rarely deficient. Thiamine is present in grain products. Vitamin C and D are sufficiently obtained from the diet. Vitamin A derivative are teratogenic and therefore should be avoided during pregnancy. Vitamin K deficiency is common in newborns and often the supplementation is required after birth.

3- a)- Prostaglandins are synthesized from Arachidonic acid; they cannot be synthesized from glucose, acetyl co A, Palmitic acid or propionyl co A.

Arachidonate, which may be obtained from the diet, but is usually derived from the 2nd position of phospholipids in the plasma membrane by the action of phospholipase A2, is the substrate for the synthesis of the prostanoids – prostaglandins (PG2) and thromboxane (TX2) series by the cyclooxygenase pathway, or the leukotrienes- LT4 and Lipoxins -LX4 series by the lipoxygenase pathway, with the two pathways competing for the arachidonate substrate.

4- b)- Linoleic acid is the right answer.  It has 18 carbon atoms and two double bonds (ω6), octadecadienoic acid. Linoleic (ω6), α-Linolenic (ω3) and Arachidonic acid (ω6) are the essential fatty acids. Arachidonic acid is relatively non-essential since it can be synthesized from Linoleic acid in the body.

Palmitic acid is a saturated fatty acid with 18 carbon atoms. Elaidic acid, Oleic acid and Erucic acid are monounsaturated fatty acids (ω9). Elaidic and oleic are Trans and cis isomers respectively with 18 carbon atoms. Elaidic acid is the major Trans fat found in hydrogenated vegetable oils. Erucic acid has 22 carbon atoms and one double bond.

5- c) Liver is the right answer. Triglycerides are mainly synthesized in the liver but they can also be synthesized in the adipose tissue and intestinal cells. In the liver they are packaged in VLDL and are secreted in to the blood. The triglycerides synthesized in the intestine are packaged in chylomicrons and are transported to peripheral cells for utilization. Triglycerides cannot be synthesized in skeletal, heart muscle, spleen and red blood cells. They are stored in adipose tissue and liver.

6- d) The right answer is (d) – Oxalo acetate. Pyruvate carboxylase catalyzes the conversion of pyruvate to oxaloacetate. It is the first step of gluconeogenesis and is also considered an “anaplerotic” or “filling up” or “CO2 fixation reaction” to replenish the oxaloacetate concentration in conditions of sudden influx of Acetyl co A, for the optimum functioning of TCA cycle. Biotin is required as a cofactor in this energy requiring reaction.

Lactate is the product of lactate dehydrogenase catalyzed reaction. Alanine is produced by transamination reaction catalyzed by Alanine transferase (ALT) also called SGPT (Serum glutamate pyruvate transaminase).

Acetyl co A is the product of pyruvate dehydrogenase reaction. Acetaldehyde is the product of pyruvate decarboxylase catalyzed reaction that does not occurs in human beings. This reaction is common in micro organisms and is commercially used for fermentation of glucose for wine production.

7- a) Production of lactate through glycolysis takes place under anaerobic conditions. No oxygen is used; hence the right answer is zero (0). Glycolysis can provide energy in the absence of oxygen; it is the unique pathway that can operate both aerobically as well as anaerobically. One mol of glucose yields 2 moles of lactate, with a yield of 2ATP by substrate level phosphorylation. In aerobic respiration via Electron transport chain O2 is used as a final acceptor and is reduced to water.

8) -e) – As per the ingredients the products would be- Glucose, galactose and fructose.

Flour- Starch-Starch is a polymer of glucose. The digestion of starch would produce glucose.

Milk- Lactose (milk sugar). Lactose is a disaccharide of galactose and glucose. Thus the digestion products of milk would be galactose and glucose.Sucrose is a disaccharide, made up of glucose and fructose. Digestion by sucrase in intestine would produce glucose and fructose.At the end, the net products of digestion entering blood would be – Glucose, galactose and fructose.

9- c) – Glucose-6-phosphatase is not present in muscle and does not contribute to blood glucose. Glucose-6-p produced from gluconeogenesis or from degradation of glycogen is channeled towards glycolytic pathway in muscle. Lactate dehydrogenase is not absent in muscle. In fact excessive lactate accumulation is the reason for muscle fatigue and the lactate produced from pyruvate by the action of lactate dehydrogenase is transported through blood to liver for glucose production (cori’s cycle).Hexokinase is the first enzyme of glycolytic pathway, glycogen synthase is the rate controlling enzyme of glycogenesis synthesis and phosphorylase is the rate limiting enzyme of glycogen degradation, all these three are present in muscle.

10- b) Hypertensive emergencies require immediate lowering of blood pressure. This may be accomplished with IV nitroprusside, a short-acting vasodilator with a rapid onset of action that allows for minute-to-minute control of blood pressure. Nitroprusside acts by releasing Nitric oxide (NO). Nitric oxide was previously referred as endothelial derived relaxing factor. It is a smooth muscle relaxant and a potent vasodilator. Nitroglycerin, amyl nitrite, (isobutyl nitrite or similar) and other nitrite derivatives used in the treatment of heart disease are also  converted to nitric oxide, which in turn dilates the coronary arteries, thereby increasing  the blood supply.

Melanin is derived from Tyrosine and is the major pigment of skin and hair.

GABA (gamma amino butyric acid), a decarboxylation product of glutamic acid is an inhibitory neurotransmitter.

Dopamine is derived from decarboxylation of DOPA, which is a product of tyrosine. Dopamine is used for raising the blood pressure in patients of shock. It is a stimulatory neurotransmitter.

Serotonin is derived from tryptophan; it is also a vasoconstrictor and a stimulatory neuro transmitter. Serotonin is involved in mood elevation and depression.

 

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Vitamin E

  • A fat soluble vitamin
  • Also called Anti infertility vitamin
  • Vitamin E acts as a chain-breaking antioxidant and is an efficient free radical scavenger.
  • It protects low-density lipoproteins (LDLs) and polyunsaturated fats in membranes from oxidation.
  • A network of other antioxidants (e.g., vitamin C, glutathione) and enzymes maintains vitamin E in its reduced state.

Structure of vitamin E

  • Vitamin E derivatives have a chromane ring (tocol) system.
  • An isoprenoid side chain is attached to the chromane ring (Figure-1)
  • There are eight naturally occurring tocopherols
  • Vitamin E is a collective name for all stereoisomers of tocopherols and tocotrienols.
  • The most biologically active is α-tocopherols, but β-, γ-, δ-tocopherols, 4 tocotrienols, and several stereoisomers may also have important biological activity.

 Structure of vitamin E

 

 

 

 

 

 

 

 

 

Figure-1- Alpha tocopherol- 5,7,8-trimethyl tocol

Absorption and Metabolism

  • Absorption requires the presence of bile salts
  • After absorption, vitamin E is taken up by chylomicrons to the liver
  • A hepatic α -tocopherol transport protein mediates intracellular vitamin E transport and incorporation into very low-density lipoprotein (VLDL).
  • The transport protein has particular affinity for α -tocopherol; thus this natural isomer has the most biologic activity.

Requirement

  • Vitamin E is widely distributed in the food supply and is particularly high in sunflower oil, safflower oil, and wheat germ oil;
  • γ tocotrienols are notably present in soybean and corn oils.
  • Vitamin E is also found in meats, nuts, and cereal grains, and small amounts are present in fruits and vegetables.
  • The RDA for vitamin E is 15 mg/d (34.9 μmol or 22.5 IU) for all adults.
  • Diets high in polyunsaturated fats may necessitate a slightly higher requirement for vitamin E.

Functions of vitamin E

1) It acts as a lipid-soluble antioxidant in cell membranes, and is important in maintaining the fluidity of cell membranes.

Antioxidant role of vitamin E

  • Reactive oxygen species (ROS) are molecular oxygen metabolites that are highly reactive with lipids, proteins, and DNA, causing oxidative damage to these cellular macromolecules.
  • This damage, termed oxidative stress, accumulates over time and is thought to contribute to both disease pathology and the aging process
  • Cellular mechanisms that exist to counteract ROS include stabilization by enzymes such as superoxide dismutase and Catalase,
  • Direct scavenging by antioxidant molecules such as glutathione (GSH), a major intracellular antioxidant; cysteine, a precursor of GSH and a major extracellular antioxidant in plasma
  • Vitamin E, is a major lipid soluble antioxidant; and
  • Ascorbate, is an intracellular and extracellular antioxidant.

The main function of vitamin E is as a chain-breaking, free-radical trapping antioxidant in cell membranes and plasma lipoproteins.

  • By reacting with the lipid peroxide radicals formed by peroxidation of polyunsaturated fatty acids, it gets converted to tocopheroxyl radical.
  • The resultant radical (oxidized form) is relatively unreactive, and ultimately forms nonradical compounds.
  • Commonly, the tocopheroxyl radical is reduced back to tocopherol by reaction with vitamin C from plasma. (See Figure-2)

Synergism between vitamin E, C, Selenium and Glutathione

  • Ascorbate (Vitamin C) is essential for maintaining vitamin E in its reduced, active form.
  • Ascorbate is oxidized to dehydroascorbate in plasma and that is recycled back to ascorbate by GSH as well as by several enzyme systems in erythrocytes, neutrophils, endothelial cells and hepatocytes (See figure-2).
  • GSH itself gets oxidized during this process and is converted back to its reduced form by Glutathione reductase utilizing NADPH as the reductant.
  • GSH is also required by Selenium containing Glutathione Peroxidase enzyme for decomposing H2O2.
  • A synergism is observed between selenium and vitamin E .
  • The synergism is  related to the process of antioxidation, wherein tocopherols tend to prevent oxidative damage to polyunsaturated fats in cell membranes, selenium, as part of seleno-enzyme glutathione peroxidase,catalyzed the destruction of lipid hydro peroxides.
  • This explains how these two nutrients play separate but interrelated role sin the cellular defense system against oxidative damage.
  • In high concentration, the tocopheroxyl free radical can penetrate further into cells and, potentially, propagate a chain reaction.
  • Therefore, vitamin E may also have pro-oxidant actions, especially at high concentrations. This explains the bleeding observed in vitamin E toxicity

 

 Synergism between vitamin C, E and Selenium

 

 

 

 

 

 

 

 

 

 

Figure- 2- A synergism is observed between Vitamin E, C and G-SH dependent Glutathione peroxidase. Vitamin E during the process of breaking the lipid peroxidation chain gets converted to oxidized form (tocopheroxyl ) that is reconverted back to reduced form by ascorbic acid (vitamin C), which in turn  itself gets converted to oxidized or dehydroascorbate form. The regeneration of reduced form takes place by reduced glutathione(G-SH), and  that gets oxidized  during this process and converted back to reduced form by glutathione reductase using NADPH as the hydrogen donor. Free radicals can also be quenched by Glutathione peroxidase, a Se containing Metalloenzymes, that requires the presence of reduced G-SH. Adequate activity of glutathione peroxidase reduces the amount of Vitamin E in diet and similarly adequate concentration of vitamin E in diet reduces the requirement of selenium.

2) Other functions of vitamin E

  • It also has a (relatively poorly defined) role in cell signaling.  
  • Vitamin E also inhibits prostaglandin synthesis
  • As an antioxidant, vitamin E plays a protective role in many organs and systems.
  • Vitamin E is necessary for maintaining a healthy immune system, and it protects the thymus and circulating white blood cells from oxidative damage.
  • Also, it may work synergistically with vitamin C in enhancing immune function.
  • Recent research evidence indicates that the combined use of high doses of vitamin C and vitamin E helps prevent Alzheimer’s disease
  • In eyes, vitamin E is needed for the development of the retina and protects against cataracts and macular degeneration.
  • In experimental animals vitamin E deficiency causes infertility

Vitamin E Deficiency

  • Dietary vitamin E deficiency is common in developing countries;
  • deficiency among adults in developed countries is uncommon and is usually due to fat malabsorption.
  • Absorption of vitamin E depends on normal pancreatic biliary function, biliary secretion, micelle formation, and penetration across intestinal membranes. Interference with any of these processes could result in a deficiency state.
  • The main symptoms are hemolytic anemia and neurologic deficits.
  • Vitamin E deficiency causes fragility of RBCs and degeneration of neurons, particularly peripheral axons and posterior column neurons.

Diagnosis

  • Low α-tocopherol level or low ratio of plasma α-tocopherol to plasma lipids
  • Measuring the plasma α-tocopherol level is the most direct method of diagnosis.
  • In adults, vitamin E deficiency is suggested if the α-tocopherol level is < 5 μg/mL (< 11.6 µmol/L).
  • Because abnormal plasma lipid levels can affect vitamin E status, a low ratio of plasma α-tocopherol to plasma lipids (< 0.8 mg/g total lipid) is the most accurate indicator in adults with hyperlipidemia.

Treatment

Supplemental α-tocopherol

  • If malabsorption causes clinically evident deficiency, α-tocopherol 15 to 25 mg/kg orally once/day should be given.
  • However, larger doses given by injection are required to treat neuropathy during its early stages or to overcome the defect of absorption

Prevention

Although premature neonates may require supplementation, human milk and commercial formulas have enough vitamin E for full-term neonates.

Vitamin E Toxicity

  • Large amounts of vitamin E for months to years may not cause apparent harm.
  • Occasionally, muscle weakness, fatigue, nausea, and diarrhea occur.
  • The most significant risk is bleeding.
  • However, bleeding is uncommon unless the dose is > 1000 mg/day.

 

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Although vitamin E has no precise metabolic role but still it is considered an important vitamin, what is its significance in the body?

a) Mineralization of bones

b) Antioxidant

c) Coagulation of blood

d) Vision

e) Coenzyme for metabolic reactions

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ATP SYNTHASE COMPLEX

 ATP synthase is embedded in the inner membrane, together with the respiratory chain complexes .

  • Several subunits of the protein form a ball-like shape arranged around an axis known as F1, which projects into the matrix and contains the phosphorylation mechanism .
  • F1 is attached to a membrane protein complex known as F0, which also consists of several protein subunits (Figure-1).
  • F0 spans the membrane and forms a proton channel.
  • The flow of protons through F0 causes it to rotate, driving the production of ATP in the F1 complex.

 ATP Synthase complex

Figure-1-The enzyme complex consists of an F0 sub complex which is a disk of “C” protein sub units. Attached is a Υ subunit in the form of a “bent axle.” Protons passing through the disk of “C” units cause it and the attached Υ subunit to rotate. The Υ subunit fits inside the F1 sub complex of three α and three β subunits, which are fixed to the membrane and do not rotate.

OXIDATIVE PHOSPHORYLATION

Chemiosmosis

  • As the electrons are transferred, some electron energy is lost with each transfer.
  • This energy is used to pump protons (H+) across the membrane from the matrix to the inner membrane space. A proton gradient is established (Figure-2)
  • The higher negative charge in the matrix attracts the protons (H+) back from the intermembrane space to the matrix.
  • The accumulation of protons in the intermembrane space drives protons into the matrix via diffusion.
  • Most protons move back to the matrix through ATP synthase.
  • ATP synthase uses the energy of the proton gradient to synthesize ATP from ADP + Pi

 

 Proton motive force

Figure-2- Showing electrochemical gradient across the inner mitochondrial membrane. The chemiosmotic theory, proposed by Peter Mitchell in 1961, postulates that the two processes are coupled by a proton gradient across the inner mitochondrial membrane so that the proton motive force caused by the electrochemical potential difference (negative on the matrix side) drives the mechanism of ATP synthesis.

 P: O Ratio

  •  Defined as the number of inorganic phosphate molecules incorporated in to ATP for every atom of oxygen consumed.
  • Oxidation of NADH yields 3 ATP molecules(P: O ratio 3, Latest concept 2.5)
  • Oxidation of FADH2 yields 2 ATP molecules (P: O ratio 2, Latest concept 1.5)

Inhibition of Oxidative phosphorylation

Oxidative phosphorylation is susceptible to inhibition at all stages of the process.  

A) Site specific inhibitors -Specific inhibitors of electron transport chain are (figure-3) for example-

1) Inhibitors of complex IRotenone and amobarbital block electron transfer in NADH-Q oxidoreductase and thereby prevent the utilization of NADH as a substrate. In contrast, electron flow resulting from the oxidation of succinate is unimpaired, because these electrons enter through QH2, beyond the block. The other inhibitors are  ChlorpromazinePiericidin A and Guanethidine

2) Inhibitors of Complex II

Malonate is a competitive inhibitor of Complex II. The other inhibitors are Carboxin and TTFA

 Site specific inhibitors of ETC

Figure-3- Site specific inhibitors block the flow of electron at specific sites

3) Inhibitors of Complex III

BAL (British Anti Lewisite), Antimycin A, Naphthoquinone  and hypoglycemic agents  interfere with electron flow from cytochrome b in Q-cytochrome c oxidoreductase.

4) Inhibitors of Complex IV

Furthermore, 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.

B) Inhibitors of ATP synthase complex 

Oligomycin and dicyclohexyl carbodiimide(DCCD) prevent the influx of protons through ATP synthase. If actively respiring mitochondria are exposed to an inhibitor of ATP synthase, the electron transport chain ceases to operate. Indeed, this observation clearly illustrates that electron transport and ATP synthesis are normally tightly coupled.

C) Inhibition of ATP-ADP translocase

ATP-ADP translocase is specifically inhibited by very low concentrations of Atractyloside (a plant glycoside) or Bongregate (an antibiotic from a mold).Unavailability of ADP also inhibits the process of ATP formation. This is because oxidation and phosphorylation are tightly coupled; ie, oxidation cannot proceed via the respiratory chain without concomitant phosphorylation of ADP.

 ADP-ATP Transport system

Figure-4- ADP moves in to the mitochondrial matrix and newly synthesized ATP is transported out in to the cytoplasm to be used for cellular processes

D) Uncouplers of Oxidative phosphorylation

The tight coupling of electron transport and phosphorylation in mitochondria can be disrupted (uncoupled) by agents called uncouplers of oxidative phosphorylation. These substances carry protons across the inner mitochondrial membrane. In the presence of these uncouplers, electron transport from NADH to O2 proceeds in a normal fashion, but ATP is not formed by mitochondrial ATP synthase because the proton-motive force across the inner mitochondrial membrane is dissipated. This loss of respiratory control leads to increased oxygen consumption and oxidation of NADH. Indeed, in the accidental ingestion of uncouplers, large amounts of metabolic fuels are consumed, but no energy is stored as ATP. Rather, energy is released as  heat.

Examples

Physiological Uncouplers

  • Long chain fatty acids
  • Thyroxin
  • Brown Adipose tissue-Thermogenin (or the uncoupling protein) is a physiological uncoupler found in brown adipose tissue that functions to generate body heat, particularly for the newborn and during hibernation in animals
  • Calcium ions.

The regulated uncoupling of oxidative phosphorylation is a biologically useful means of generating heat. The uncoupling of oxidative phosphorylation is a means of generating heat to maintain body temperature in hibernating animals, in some newborn animals (including human beings), and in mammals adapted to cold. Brown adipose tissue, which is very rich in mitochondria (often referred to as brown fat mitochondria), is specialized for this process of non shivering thermogenesis. The inner mitochondrial membrane of these mitochondria contains a large amount of uncoupling protein (UCP), here UCP-1, or thermogenin,a dimer of 33-kd subunits that resembles ATP-ADP translocase. UCP-1 forms a pathway for the flow of protons from the cytosol to the matrix. In essence, UCP-1 generates heat by short-circuiting the mitochondrial proton battery(figure-5).

Thus, the manner in which biologic oxidative processes allow the free energy resulting from the oxidation of foodstuffs to become available and to be captured is stepwise, efficient, and controlled—rather than explosive, inefficient, and uncontrolled, as in many non biologic processes. The remaining free energy that is not captured as high-energy phosphate is liberated as heat. This need not be considered “wasted,” since it ensures that the respiratory system as a whole is sufficiently exergonic, allowing continuous unidirectional flow and constant provision of ATP. It also contributes to maintenance of body temperature.

 

Uncoupling of oxidative phosphorylation 

Figure-5- UCP-1 forms a pathway for the flow of protons from the cytosol to the matrix. The proton gradient is dissipated, oxidation proceeds without phosphorylation.

Pathological uncouplers

These compounds are toxic in vivo, causing respiration to become uncontrolled, since the rate is no longer limited by the concentration of ADP or Pi.

  •  2,4-dinitrophenol
  • 2, 4- dinitrocresol
  • CCCP( chloro carbonyl cyanide phenyl hydrazone)
  • FCCP
  • Valinomycin
  • High dose of Aspirin
  • The antibiotic Oligomycin completely blocks oxidation and phosphorylation by blocking the flow of protons through ATP synthase.
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An 18-year-old college student is brought to the emergency room unconscious, with a very high serum alcohol level. Alcohol metabolism can result in high NADH levels. When NADH enters the electron transport chain, which of the following is the correct order in which electron transfer occurs ?

a) NADH, coenzyme Q, cytochrome c, FMN,O2

b) NADH, cytochrome c, coenzyme Q, FMN,O2

c) NADH, FMN, coenzyme Q, cytochrome c, O2

d) NADH, cytochrome c, FMN, cytochrome c, O2

e) NADH, FMN, cytochrome c, coenzyme Q, O2

 

The right answer is (c).The electron flow is from NADH to O2 passing through FMN, coenzyme Q (ubiquinone) and cytochrome c (Figure-1). See the details below-

 

Overview of ETC

 Figure-1  – An overview of Electron transport chain . FeS represents iron sulfur center, and cytochrome b,c1 are components of complex III.

Basic concept of electron transport chain

Introduction-Most of the energy liberated during the oxidation of carbohydrates, fatty acids, and amino acids is made available within mitochondria as reducing equivalents (—H or electrons) .The NADH and FADH2 formed in glycolysis, fatty acid oxidation, and the citric acid cycle are energy-rich molecules because each contains a pair of electrons having a high transfer potential. The enzymes of the citric acid cycle and β-oxidation are contained in mitochondria, together with the respiratory chain (electron transport chain), which collects and transports reducing equivalents, directing them to their final reaction with oxygen to form water (Figure-2)

Role of ETC in ATP formation

 Figure-2- The energy trapped in the food is passed on to electron transport chain after a series of oxidative processes, in the form of reducing equivalents, where finally energy is released as ATP

Components of Electron Transport Chain

Electrons flow through the respiratory chain through a redox span of 1.1V from NAD+/NADH to O2/2H2O, passing through three large protein complexes (Figure-3)

1) NADH-Q oxidoreductase (Complex I), where electrons are transferred from NADH to coenzyme Q (Q) (also called Ubiquinone);

2) Q-cytochrome c oxidoreductase (Complex III), which passes the electrons on to cytochrome c; and

3) Cytochrome c oxidase (Complex IV) ,which completes the chain, passing the electrons to Oand causing it to be reduced to H2O (Figure-1).

4) Succinate Q reductase (Complex II),Some substrates with more positive redox potentials than NAD+/NADH (eg, succinate) pass electrons to Q via a fourth complex, succinate Q reductase (Complex II), rather than Complex I.

The four complexes are embedded in the inner mitochondrial membrane, but Q and cytochrome c are mobile. Q diffuses rapidly within the membrane, while cytochrome c is a soluble protein.

 

Components of ETC

 Figure-3- Showing the components of electron transport chain. Arrows show the flow of protons.

Electron flow in ETC 

NADH-Q oxidoreductase or Complex I is a large L-shaped multi-subunit protein that catalyzes electron transfer from NADH to Q, coupled with the transfer of  four H+across the membrane: Electrons are transferred from NADH to FMN initially, then to a series of Fe-S centers, and finally to Q (Figure-4).

NADH dehydrogenase complex

 Figure-4NADH-Q oxidoreductase is a large L-shaped multi-subunit protein that catalyzes electron transfer from NADH to Q, coupled with the transfer of  four H+across the membrane

In Complex II (succinate -Q reductase), FADH2 is formed during the conversion of succinate to fumarate in the citric acid cycle and electrons are then passed via several Fe-S centres to Q (Figure-3 and 5).

Flow of electrons in ETC

Figure-5- Showing the flow of electrons through four complexes. Coenzyme Q and cytochrome c are mobile carriers. Complex V is ATP synthase complex, meant for ATP production.

Iron-sulfur proteins (non-heme iron proteins, Fe-S) are found in Complexes I, II, and III (Figure-5).These may contain one, two, or four Fe atoms linked to inorganic sulfur atoms and/or via cysteine-SH groups to the protein (Figure-6). The Fe-S takes part in single electron transfer reactions in which one Fe atom undergoes oxidoreduction between Fe2+ and Fe3+.

 

Iron -sulphur centers

Figure-6- Iron sulfur proteins (Fe-S). (A) The simplest Fe-S with one Fe bound by four cysteines. (B) 2Fe-2S center. (C) 4Fe-4S center.

Flavoproteins (are important components of Complexes I and II (Figure-4). The oxidized flavin nucleotide (FMN or FAD) can be reduced in reactions involving the transfer of two electrons (to form FMNH2 or FADH2),but they can also accept one electron to form the semiquinone.

Coenzyme Q or Ubiquinone (ubi). Ubi accepts electrons and transports them to Complex III. Ubi is capable of moving in the membrane between Complex I and III, picking up electrons at Complex I and dropping them off at Complex III (Figure-3 and 5).

As is the case with Complex I, Complex III contains a series of proteins that transport electrons. As the electrons are moved from carrier to carrier within Complex III, the energy that is released moves hydrogen ions out of the mitochondrion (Figure-3).Thus, Complex III also serves as a hydrogen ion pump. Once again, there must be a mechanism by which electrons are removed from Complex III. This done by a small peripheral membrane protein known as cytochrome c. Cytochrome c transports the electrons from Complex III to Complex IV.

Reduced cytochrome c is oxidized by Complex IV (cytochrome c oxidase), with the concomitant reduction of O2 to two molecules of water(Figure-3 and 4).

This transfer of four electrons from cytochrome c to O2 involves two heme groups, a and a3,and Cu. Electrons are passed initially to a Cu centre (CuA), which contains 2 Cu atoms linked to two protein cysteine-SH groups (resembling an Fe-S), then in sequence to heme a, heme a3, a second Cu centre, CuB, which is linked to heme a3, and finally to O2.

By the time the electrons reach the final electron acceptor in Complex IV, they have very little energy left . Complex IV finally gives its electrons to molecular oxygen (O2), the final or terminal electron acceptor. The reduction of oxygen results in the formation of a water molecule from the oxygen (Figure-3 and 5).

For every pair of electrons donated to the ETC by an NADH, ½ O2(or one O atom) is reduced to form one water molecule. Two NADH’s would result in the use of 1 (½ + ½) O2 molecule, and so on. If there was no oxygen to take the electrons from Complex IV, the electrons would accumulate up the electron transport chain very quickly and NADH would no longer be able to donate electrons to the ETC (because the ETC would be filled with electrons).This would cause an accumulation of NADH and a shortage of NAD+(remember, NAD+ is regenerated when NADH donates its electrons to Complex I). Without NAD+, there would be no electron acceptors for the Krebs Cycle and all reactions that require NAD+ would stop. This would cause the entire Krebs Cycle to stop. Thus the Krebs Cycle would not run if there were no oxygen present, thus it is considered aerobic!

Proton Motive Force -The flow of electrons from NADH or FADH2 to O2 through protein complexes located in the mitochondrial inner membrane leads to the pumping of protons out of the mitochondrial matrix.(Figure-3) .The resulting uneven distribution of protons generates a pH gradient and a transmembrane electrical potential that creates a proton-motive force (Figure -7). ATP is synthesized when protons flow back to the mitochondrial matrix through an enzyme complex (Figure -3 and 4). Thus, the oxidation of fuels and the phosphorylation of ADP are coupled by a proton gradient across the inner mitochondrial membrane.

Proton motive force

 Figure-7-showing the electrochemical gradient created by pumping of protons across the inner mitochondrial membrane

In other words, the electron-motive force is converted into a proton-motive force and, finally, the proton-motive force is converted into phosphoryl transfer potential.

Like NADH, FADH2 contains electrons (and energy) from oxidation reactions. FADH2, however, contains less energy than does NADH. Because of this, FADH2 is unable to donate its electrons to Complex I of the ETC. Instead, FADH2 donates its electrons to Complex II. Complex II eventually passes the electrons on to Complex III via Ubiquinone. From then on, the electrons follow the same pathway as do those from NADH. Complex II, it turns out, does not move protons as the electrons are transported. This means that every NADH that donates electrons to the ETC will cause more hydrogen ions to be transferred than does each FADH2.

Oxidative phosphorylation- The reduction of molecular oxygen to water yields a large amount of free energy that can be used to generate ATP. Oxidative phosphorylation is the process in which ATP is formed as a result of the transfer of electrons from NADH or FADH2 to O2 by a series of electron carriers. This process, which takes place in mitochondria, is the major source of ATP in aerobic organisms.

P: O ratio

There is a net direct capture of two high-energy phosphate groups in the glycolytic reactions. Two more high-energy phosphates per mole of glucose are captured in the citric acid cycle during the conversion of Succinyl CoA to succinate. All of these phosphorylation occur at the substrate level. When substrates are oxidized via Complexes I, III, and IV in the respiratory chain (i.e., via NADH), 3 mol of ATP are formed per half mol of O2 consumed; ie, the P:O ratio = 3. On the other hand, when a substrate (e.g. succinate) via FADH2 is oxidized through Complexes II, III, and IV, only 2 mol of ATP are formed; i.e., P:O =2. These reactions are known as oxidative phosphorylation at the respiratory chain level. Taking these values into account, it can be estimated that nearly 90% of the high energy phosphates produced from the complete oxidation of 1 mol glucose is obtained via oxidative phosphorylation coupled to the respiratory chain.

To be continued……

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An 18-year-old college student is brought to the emergency room unconscious, with a very high serum alcohol level. Alcohol metabolism can result in high NADH levels. When NADH enters the electron transport chain, which of the following is the correct order in which electron transfer occurs ?

a) NADH, coenzyme Q, cytochrome c, FMN, O2

b) NADH, cytochrome c, coenzyme Q, FMN,O2

c) NADH, FMN, coenzyme Q, cytochrome c, O2

d) NADH, cytochrome c, FMN, cytochrome c, O2

e) NADH, FMN, cytochrome c, coenzyme Q, O2

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Collagen is synthesized intracellularly in the fibroblasts (or in osteoblasts of bone and chondroblasts of cartilages) as a large precursor, called preprocollagen. Like most secreted proteins,collagen is synthesized on ribosomes. It is then secreted in to the extra cellular matrix. After enzymatic modification,the mature collagen monomers aggregate and become cross linked to form collagen fibrils. The synthesis of collagen (Figure-3) involves three stages-

A) Intracellular steps

B) Secretion

C) Extra-cellular steps

A) Intracellular steps- The processes involved inside the cell in the synthesis of collagen are as folllows-

a) Formation of Pro- α chains– The newly synthesized polypeptide precursors of α chains contain, a leader sequence(a special amino acid sequence at the amino terminal end).This sequence acts as a signal that the polypeptide being synthesized is destined to leave the cell. The signal sequence directs the passage of the polypeptide chain in to the cisternae of the rough endoplasmic reticulum. Signal peptidase cleaves the signal sequence in the endoplasmic reticulum to yield a precursor of collagen called pro-α chain.

b) Hydroxylation- The pro -α chains are processed by a number of enzymatic steps within the lumen of the RER. The proline and lysine residues are hydroxylated by  prolyl and lysyl hydroxylases to form hydroxy proline and hydroxy lysine respectively (Figure-1). The reactions can be represented as follows-

 Hydroxylation of proline and lysine in collagen synthesis

Figure-1- Prolyl and lysyl hydroxylases require molecular oxygen,  ascorbic acid(vitamin C) and α-keto (oxo) glutarate for their action.

Impaired prolyl and lysyl hydroxylation , as in ascorbic acid (vitamin C) deficiency, leads to improper cross linking of collagen fibers , thus a weak collagen results with decreased tensile strength of the assembled fibers. The resulting disease, scurvy is manifested by  bleeding gums, bruises, weak bones, all due to weak collagen.

c) Glycosylation- Some of the hydroxy lysines are modified by glycosylation with galactose or galactosyl-glucose through an O-glycosidic linkage, a glycosylation site that is unique to collagen (Figure-2).

 glycosylation of collagen

Figure-2- The galactose and glucose residues are added sequentially by galactosyl and glucosyl transferases. The extent of glycosylation is different in different tissues

d) Assembly – After hydroxylation and glycosylation the  pro-α chains form procollagen, a precursor of collagen. The procollagen molecule contains polypeptide extensions (extension peptides) at both its amino and carboxyl terminal ends, neither of which is present in mature collagen. Both extension peptides contain cysteine residues. While the amino terminal propeptides forms only intrachain disulfide bonds, the carboxyl terminal propeptides form both intrachain and interchain disulfide bonds. Formation of these disulfide bonds, assists in the assembly of the three collagen molecules to form the triple helix, winding from the carboxyl terminal end. After formation of the triple helix, no further hydroxylation of proline or lysine or glycosylation of hydroxylysines can take place. Self-assembly is a cardinal principle in the biosynthesis of collagen (Figure-3)

B) Secretion- The procollagen  molecules are translocated to the Golgi apparatus where they are packaged in secretory vesicles. The vesicles fuse with the cell membrane, causing the release of procollagen molecules in to the extracellular space (Figure-3)

C) Extracellular steps– The processes involved outside the cell to form mature collagen molecules are as follows (Figure-3)-

a) Cleavage of amino and carboxyl terminal propeptides- Following secretion from the cell by way of the Golgi apparatus, extracellular enzymes called procollagen aminoproteinase and procollagen carboxyproteinase remove the extension peptides at the amino and carboxyl terminal ends, respectively, releasing triple helical collagen molecules. Cleavage of these propeptides may occur within crypts or folds in the cell membrane.

b) Formation of collagen fibrils– Once the propeptides are removed, the triple helical collagen molecules, containing approximately 1000 amino acids per chain, spontaneously assemble into collagen fibers. Collagen types that form long rod-like fibers in tissues are assembled by lateral association of these triple helical units into a “quarter staggered” alignment such that each is displaced longitudinally from its neighbor by slightly less than one-quarter of its length . This arrangement is responsible for the banded appearance of these fibers in connective tissues.

c) Cross link formation– Collagen fibers are further stabilized by the formation of covalent cross-links, both within and between the triple helical units. These cross-links form through the action of lysyl oxidase, a copper-dependent enzyme that oxidatively deaminates the Ʃ-amino groups of certain lysine and hydroxylysine residues, yielding reactive aldehydes. Such aldehydes can form aldol condensation products with other lysine- or hydroxylysine-derived aldehydes or form Schiff bases with the Ʃ-amino groups of unoxidized lysines or hydroxylysines. These reactions, after further chemical rearrangements, result in the stable covalent cross-links that are important for the tensile strength of the fibers. Histidine may also be involved in certain cross-links (Figure-4).

Steps of collagen synthesis

 

Figure-3- Showing the steps of collagen synthesis and formation of mature collagen

The cross links are essential for the achieving the tensile  strength necessary for the proper functioning of the connective tissue. Therefore any mutation that interferes with the ability of collagen to form cross linked fibrils affects the stability of the collagen.

 Overview of structure of collagen

Figure-4- Overview of collagen structure

Several collagen types do not form fibrils in tissues.

The types and organization of collagens are dictated by the structural role played by specific collagen in specific tissues. In some tissues collagen may be found in a gel like structure to give support as in extracellular matrix or the vitreous humor of eye. In other tissues, collagen may be bundled in tight, parallel fibers that provide greater strength, as in tendons. In cornea of eye, collage is stacked so as to transmit the light with a minimum of scattering. Collagen of bone occurs as fibers arranged at an angle to each other so as to resist mechanical from any direction.

The collagen can be organized in to three groups based on the location and functions in the body-

1. Fiber forming

Type I, II and III are the fibrillar collagens, and have the rope like structure. In the electron microscope, these linear polymers of fibrils have characteristic banding patterns, reflecting the regular staggered packing of the individual collagen molecules in the fibril. Type I fibers are found in supporting elements of high tensile strength (Table -1), whereas the fibers formed from type II collagen are restricted cartilaginous structures. The fibers derived from Type III collagen are found in more distensible tissues such as blood vessels.

2. Net work forming collagens

Type IV and VII form a three dimensional mesh work, instead  of making fibrils. This meshwork forms a major part of basement membranes (Table-1).

3. Fibril associated collagens

Type IX and XII bind to the surface of collagen fibrils, linking these fibrils to one another and to other components in the extra cellular matrix (Table-1).

 Table-1- Types and Tissue distribution of collagens

Type Tissue distribution
1.Fibril forming  
Type I Skin, bone, tendon, blood vessels and cornea
Type II Cartilage, intervertebral disc, vitreous body
Type III Blood vessels, fetal skin
2. Net work forming  
Type IV Basement membrane
Type VII Beneath stratified squamous epithelium
3.Fibril Associated  
Type IX Cartilages
Type X Tendons, ligaments and some other tissues

 

Once formed, collagen is relatively metabolically stable. However, its breakdown is increased during starvation and various inflammatory states. Excessive production of collagen occurs in a number of conditions, eg, hepatic cirrhosis.

Composition of different type of collagen

Collagen,  is not one protein but a family of structurally related proteins. The different collagen proteins have very diverse functions. The different types of collagen are characterized by different polypeptide compositions. Each collagen is composed of three polypeptide chains, which may be all identical or may be of two different chains (Table-2)

Type I collagen, the most abundant, is composed of two identical α1(I) chains and a third α2(I) chain that is the product of a different gene and differs slightly in its amino acid sequence. Type II collagen, the fibrillar collagen of cartilage, is composed of three identical α(II) chains. Type III collagen is composed of three identical α1(III) chains. It is found in small amounts in many tissues that contain type I collagen and in large amounts in major blood vessels. Type IV collagen is composed of three α chains synthesized from any of six different genes. It is found in basement membranes where it self-assembles with laminins and other macromolecules into a complex three-dimensional network to provide a diffusion barrier in the renal glomerulus, pulmonary alveolus, and other tissues.

Table-2- Composition of collagen types

Type of Collagen Composition
Type I [alpha1(I)]2alpha2(I).
Type II [alpha1(II)]3.
Type III [alpha1(III)]3
Type IV [alpha (IV)]3

 

Degradation of collagen and diseases- to be continued in the next post.

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A 56-year-old man with long-standing  poorly controlled diabetes mellitus visits his primary care physician for a follow-up after recent hospitalization. The patient experienced an episode of acute renal failure while in the hospital and his creatinine level rose to 4 mg/dl. Creatinine, a marker of renal function is produced from which of the following precursors ?

a) Glutamine, Aspartic acid and alanine

b) Glutamine, Cysteine and Glycine

c) Serine , Glycine and Methionine

d) Glycine, Arginine and Methionine

e) Glutamic acid, Cysteine and Glycine

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A 55-year-old male presents with difficult breathing and swollen ankles. He is found to have a failing heart, resulting in blood backing up in to his lungs (pulmonary congestion) and making it difficult for him to breathe. He is administered a drug  that inhibits Angiotensin converting enzyme (ACE). By inhibiting this enzyme, which of the following will change about the reaction it catalyzes ?

a) Energy of activation

b) Net free energy change

c) Equilibrium concentration of substrate

d) Equilibrium concentration of product

e) Thermodynamics

Answer- The right answer is – a) Energy of activation. The enzymes decrease the energy of activation for a reaction and speed up the reaction by many folds. The thermodynamics of a reaction, such as the free energy change and the equilibrium concentrations of the substrate and product remain unchanged.

Case details– It is a case of Cardiac failure.

The renin-angiotensin-aldosterone system (RAAS) plays an important role in regulating blood volume and systemic vascular resistance, which together influence cardiac output and arterial pressure.

As the name implies, there are three important components to this system: 1) renin, 2) angiotensin, and 3) aldosterone. Renin, which is primarily released by the kidneys, stimulates the formation of angiotensin in blood and tissues, which in turn stimulates the release of aldosterone from the adrenal cortex. It is called a system because each part influences the other parts and all are necessary for the whole to function correctly.

Renin is a proteolytic enzyme that is released into the circulation primarily by the kidneys. Its release is stimulated by:

1) sympathetic nerve activation (acting via β1-adrenoceptors)

2) renal artery hypotension (caused by systemic hypotension or renal artery stenosis)

3) decreased sodium delivery to the distal tubules of the kidney.

When renin is released into the blood, it acts upon a circulating substrate, angiotensinogen, that undergoes proteolytic cleavage to form the decapeptide angiotensin I. Vascular endothelium, particularly in the lungs, has an enzyme, angiotensin converting enzyme (ACE), that cleaves off two amino acids to form the octapeptideangiotensin II (AII), although many other tissues in the body (heart, brain, vascular) also can form AII (Figure-1).

Angiotensin I is able to alter the blood pressure to some degree, but it isn’t strong enough to cause large changes. Instead, most angiotensin I is converted to angiotensin II, a much more powerful hormone that does cause large changes in blood pressure. (This conversion is shut down by drugs called ACE Inhibitors, an important type of high blood pressure medication.)

Angiotensin II is a strong hormone, and can act directly on blood vessels to cause blood pressure increases. It also has another even more important function – stimulating the release of aldosterone. Aldosterone is a very powerful vasoconstrictor that causes large increases in blood pressure, but is more important because it can actually change the baseline filtering activity of the kidneys. Aldosterone causes the kidneys to retain both salt and water, which – over time – increases the amount of water in the body. This increase, in turn, raises blood pressure.

 
Renin Angiotensin system

 

Figure-1- Renin angiotensin system.


Activation of the renin-angiotensin-aldosterone (RAA) system rapidly kicks in with heart failure, due to decreased renal perfusion caused by both a reduction in cardiac output and redistribution of blood away from nonessential organs (kidney). The kidney retains sodium and water in response to the perception of ineffective blood volume. The perception of decreased blood volume and the increase in sympathetic nervous activity stimulates renin release from the juxtaglomerular cells in the kidneys.

The compensatory mechanisms in heart failure eventually initiate a vicious cycle which leads to continued worsening and downward spiraling of the heart failure state. The peripheral vasoconstriction mediated by increased sympathetic activity, angiotensin II, and other possible mechanisms causes an increase in systemic vascular resistance or afterload. Afterload resists myocardial fiber shortening and further decreases cardiac output, which leads to further increases in sodium and water retention and sympathetic nervous activity (Figure-2).

Cardiac failure
Figure- 2 – Vicious cycle of cardiac failure

Therapeutic manipulation of this pathway is very important in treating hypertension and heart failure. ACE inhibitors, Angiotensin II receptor blockers and aldosterone receptor blockers, for example, are used to decrease arterial pressure, ventricular afterload, blood volume and hence ventricular preload, as well as inhibit and reverse cardiac and vascular hypertrophy.

Mechanism of action of ACE inhibitors- Frequently prescribed ACE inhibitors include perindopril, captopril, enalapril, lisinopril, and ramipril. ACE inhibitors block the conversion of angiotensin I to angiotensin II. They act by competitive inhibition.

General Mechanism of Action of Enzymes

Enzymes are catalysts and increase the speed of a chemical reaction without themselves undergoing any permanent chemical change. They are neither used up in the reaction nor do they appear as reaction products.

The basic enzymatic reaction can be represented as follows

 Enzymes

where E represents the enzyme catalyzing the reaction, S the substrate, the substance being changed, and P the product of the reaction.

Enzymes employ multiple mechanisms to facilitate catalysis.  The mechanism of action of enzymes can be explained by two perspectives-

1) Thermodynamic changes

2) Processes at the active site

1) Thermodynamic changes – All enzymes accelerate reaction rates by providing transition states with a lowered G for formation of the transition states. However, they may differ in the way this is achieved.

A chemical reaction of substrate S to form product P goes through a transition state S‡ that has a higher free energy than does either S or P. (The double dagger denotes a thermodynamic property of the transition state).

The difference in free energy between the transition state and the substrate is called the Gibbs free energy of activation or simply the activation energy, symbolized by ∆G‡.

The activation energy barrier suggests how enzymes enhance reaction rate without altering ∆G of the reaction: enzymes function to lower the activation energy, or, in other words, enzymes facilitate the formation of the transition state. The combination of substrate and enzyme creates a new reaction pathway whose transition-state energy is lower than that of the reaction in the absence of enzyme (see Figure 3). The lower activation energy means that more molecules have the required energy to reach the transition state. Decreasing the activation barrier is analogous to lowering the height of a high-jump bar; more athletes will be able to clear the bar. The essence of catalysis is specific binding of the transition state.

Mechanism of action of enzymes 

Figure-3 Enzymes Decrease the Activation Energy. Enzymes accelerate reactions by decreasing ∆G, the free energy of activation.

2) Processes at the active site-Enzymes use various combinations of four general mechanisms to achieve dramatic catalytic enhancement of the rates of chemical reactions. These are as follows-

a) Catalysis by Bond Strain: In this form of catalysis, the induced structural rearrangements that take place with the binding of substrate and enzyme ultimately produce strained substrate bonds, which more easily attain the transition state. Enzymes that catalyze lytic reactions that involve breaking a covalent bond typically bind their substrates in a conformation slightly unfavorable for the bond that will undergo cleavage. The resulting strain stretches or distorts the targeted bond, weakening it and making it more vulnerable to cleavage.

b) Catalysis by Proximity and Orientation: For molecules to react, they must come within bond-forming distance of one another. The higher their concentration, the more frequently they will encounter one another and the greater will be the rate of their reaction. When an enzyme binds substrate molecules at its active site, it creates a region of high local substrate concentration. Enzyme-substrate interactions orient reactive groups and bring them into proximity with one another.

c) Catalysis Involving Proton Donors (Acids) and Acceptors (Bases): Other mechanisms also contribute significantly to the completion of catalytic events initiated by a strain mechanism, for example, the use of glutamate as a general acid catalyst (proton donor). The ionizable functional groups of aminoacyl side chains and (where present) of prosthetic groups contribute to catalysis by acting as acids or bases. Acid-base catalysis can be either specific or general. By “specific” we mean only protons (H3O+) or OH ions. In specific acid or specific base catalysis, the rate of reaction is sensitive to changes in the concentration of protons but independent of the concentrations of other acids (proton donors) or bases (proton acceptors) present in solution or at the active site. Reactions whose rates are responsive to all the acids or bases present are said to be subject to general acid or general base catalysis.

d) Covalent Catalysis: In catalysis that takes place by covalent mechanisms, the substrate is oriented to active sites on the enzymes in such a way that a covalent intermediate forms between the enzyme or coenzyme and the substrate. One of the best-known examples of this mechanism is that involving proteolysis by serine proteases, which include both digestive enzymes (trypsin, chymotrypsin, and elastase) and several enzymes of the blood clotting cascade. These proteases contain an active site serine whose R group hydroxyl forms a covalent bond with a carbonyl carbon of a peptide bond, thereby causing hydrolysis of the peptide bond. Covalent catalysis introduces a new reaction pathway whose activation energy is lower—and therefore is faster—than the reaction pathway in homogeneous solution. The chemical modification of the enzyme is, however, transient. On completion of the reaction, the enzyme returns to its original unmodified state. Its role thus remains catalytic. Covalent catalysis is particularly common among enzymes that catalyze group transfer reactions.

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A 56- year- old female presents with difficulty opening her eye lids, as well as inability to raise herself from sitting position. She is diagnosed with “myasthenia gravis”, a disease of extreme fatigue, due to decreased concentration of Acetyl choline in her muscles. She has been prescribed physostigmine, a drug that increases the amount of available Acetyl choline, by competitively inhibiting acetylcholinestrase.

myashenia gravis

Figure- Inability to open eyelid in myasthenia gravis

Which of the following statements is not true of competitive inhibitors ?

a) Vmax remains the same

b) Apparent  Km in increased

c) Inhibitor is a structural analogue of the substrate

d) Inhibitor binds covalently to the enzyme

e) Increasing concentration of substrate can reverse the changes

 3- A 55-year-old male presents with difficult breathing and swollen ankles. He is found to have a failing heart, resulting in blood backing up in to his lungs (pulmonary congestion) and making it difficult for him to breathe. He is administered a drug  that inhibits angiotensin converting enzyme (ACE). By inhibiting this enzyme, which of the following will change about the reaction it catalyzes ?

a) Energy of activation

b) Net free energy change

c) Equilibrium concentration of substrate

d) Equilibrium concentration of product

e) Thermodynamics

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Refer to the following figures and answer the question

 Enzyme inhibitio1

 

 

 

 

 

 

 

 

 

 

 

Figure-1- Effect of enzyme inhibitor. Km value has increased, Vmax is unchanged

      Enzyme inhibition-2

 

 

 

 

 

 

 

 

 

 

 

Figure-2- Double reciprocal curve showing the effect of inhibitor. 

A 20 year- old man was brought to the emergency with vomiting, sweating, drooling and a decreased heart rate. History reveals that he was in a corn field when it was sprayed by a crop duster. The pesticide causing his symptoms is an organo phosphate that covalently binds to acetyl cholinesterase and inactivates the enzyme. The enzyme kinetics affected by the inhibitor are shown in the figures. What is the mechanism of inhibition of this enzyme ?

Case discussion– It is a case of Organophosphate poisoning. Organophosphate poisoning results from exposure to organophosphates (OPs), which cause the inhibition of acetyl cholinesterase (AChE), leading to the accumulation of acetylcholine (ACh) in the body.

Acetylcholine is a chemical neurotransmitter found widely in the body. It triggers the stimulation of post-synaptic nerves, muscles, and exocrine glands.

Acetylcholinesterase (generally referred to as cholinesterase) is  an enzyme that rapidly breaks down the neurotransmitter, acetylcholine, so that it does not over-stimulate post-synaptic nerves, muscles, and exocrine glands (Figure-3).

 Action of acetyl cholinesterase

 

 

 

 

 

 

 

 

 

 

 

Figure-3- The normal mechanism of action of Acetyl cholinesterase. Serine is present at the active site of the enzyme. Acetylcholine, attaches to the enzyme, Acetyl cholinesterase and is broken down to Acetic acid and Choline.

Acetylcholinesterase inhibitor (generally referred to as cholinesterase inhibitor) is a chemical that binds to the enzyme, cholinesterase, and prevents it from breaking down the neurotransmitter, acetylcholine. With toxic doses, the result is that excessive levels of the acetylcholine build up in the synapses and neuromuscular junctions and glands.

Thus, the primary manifestations of acute cholinesterase inhibitor toxicity are those of cholinergic (neurotransmitter) hyperactivity. There are also other delayed and chronic pathological effects of inhibitors of the cholinesterase enzyme which are less well understood.

Organophosphate (OP) compounds are a diverse group of chemicals used in both domestic and industrial settings. Examples of organophosphates include insecticide, nerve gases, ophthalmic agents  and herbicides etc

OPs are one of the most common causes of poisoning worldwide .There are around 1 million OP poisonings per year with several hundred thousand resulting in fatalities annually.

Exposure to organophosphates (OPs) is also possible via intentional or unintentional contamination of food sources. Although no clinical effects of chronic, low-level organophosphates (OPs) exposure from a food source have been shown.

Case details

The mechanism of action of the inhibitor shown in the graphs and the mechanism stated in the case study do not correspond with each other.

As per the graphs- The Km value has increased , while Vmax is constant. This is possible only in competitive enzyme inhibition.

The effects of competitive inhibitors can be overcome by raising the concentration of the substrate. Most frequently, in competitive inhibition the inhibitor, (I), binds to the substrate-binding portion of the active site and blocks access by the substrate. The structures of most classic competitive inhibitors therefore tend to resemble the structures of a substrate, and thus are termed substrate analogs. In effect, a competitive inhibitor acts by decreasing the number of free enzyme molecules available to bind substrate, ie, to form ES, and thus eventually to form product.

In noncompetitive inhibition, binding of the inhibitor does not affect binding of substrate. Formation of both EI and EIS complexes is therefore possible. However, while the enzyme-inhibitor complex can still bind substrate, its efficiency at transforming substrate to product, reflected by Vmax, is decreased. Noncompetitive inhibitors bind enzymes at sites distinct from the substrate-binding site and generally bear little or no structural resemblance to the substrate.

As per the details provided in the case- The Organophosphate  covalently binds to enzyme to bring about its inactivation. Since these covalent changes are relatively stable, the enzyme gets “poisoned” by the irreversible inhibitor.

 This is not possible in competitive enzyme inhibition.

Organophosphates inhibit AChE, causing OP poisoning by phosphorylating the serine hydroxyl residue on AChE, which inactivates AChE. AChE is critical for nerve function, so the irreversible blockage of this enzyme, causes acetylcholine accumulation, resulting in muscle overstimulation. (Figure-4)

 

Inactivation of acetylcholinesterase

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure-4- Organophosphorus compounds bind to the active site causing its phosphorylation to form a phosphorylated  inactive enzyme, that fails to catalyze the degradation of Acetyl choline. The phosphorylation occurs by loss of an organophosphate leaving group and establishment of a covalent bond with AChE.

Thus it is not a competitive enzyme inhibition,  it is non competitive enzyme inhibition, though the inhibitor is bound to the active site but it has brought about the inactivation of the enzyme.

In non competitive enzyme inhibition, Km remains unchanged but Vmax is deceased, thus the inhibition of Acetyl cholinesterase by Organophosphates is non competitive enzyme inhibition. The comparison of the two types of inhibitions can be seen in the figures-

non competitive enzyme inhibition

 

 

 

 

 

 

 

 

 

Figure-5- In competitive enzyme inhibition, Vmax is unchanged, Km is increased, while reverse occurs in non competitive enzyme inhibition, Km remains constant, but Vmax is decreased.

                                                                                                                                                                                                                                                                           

Figure-6- Double reciprocal curve (Line weaver burk plot) showing the comparison between competitive and non competitive enzyme inhibition.

Lineweaver_Burk2

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Refer to the following graphs and answer the question

 Enzyme inhibitio1

 

 

 

 

 

 

 

 

 

 

 

Enzyme inhibition-2

 

 

 

 

 

 

 

 

 

 

 

 

 

A 20 year- old man was brought to the emergency with vomiting, sweating, drooling and a decreased heart rate. History reveals that he was in a corn field when it was sprayed by a crop duster. The pesticide causing his symptoms is an organo phosphate that covalently binds to acetyl cholinesterase and inactivates the enzyme. The enzyme kinetics affected by the inhibitor are shown in the figures. What is the mechanism of inhibition of this enzyme ?

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Gene regulation is significantly more complex in eukaryotes than in prokaryotes for a number of reasons-

1)  Large Genome

First, the genome being regulated is significantly larger. The E. coli genome consists of a single, circular chromosome containing 4.6 Mb. This genome encodes approximately 2000 proteins. In comparison, the genome within a human cell contains 23 pairs of chromosomes ranging in size from 50 to 250 Mb. Approximately 40,000 genes are present within the 3000 Mb of human DNA. It would be very difficult for a DNA-binding protein to recognize a unique site in this vast array of DNA sequences. Consequently, more-elaborate mechanisms are required to achieve specificity.

2) Complex Genome

Another source of complexity in eukaryotic gene regulation is the many different cell types present in most eukaryotes. Liver and pancreatic cells, for example, differ dramatically in the genes that are highly expressed.

3) Widely spread genes- No well-defined operons

Moreover, eukaryotic genes are not generally organized into operons. Instead, genes that encode proteins for steps within a given pathway are often spread widely across the genome.

4) Compact Genome

The DNA in eukaryotic cells is extensively folded and packed into the protein-DNA complex called chromatin. Histones are an important part of this complex since they form the structures known as nucleosomes and also  contribute significantly into gene regulatory mechanisms.

5) Uncoupled transcription and Translation

Finally, transcription and translation are uncoupled in eukaryotes, eliminating some potential gene-regulatory mechanisms.

Mechanism of regulation of gene expression in Eukaryotes

1) Chromatin Remodeling

Chromatin structure provides an important level of control of gene transcription. Large regions of chromatin are transcriptionally inactive while others are either active or potentially active. With few exceptions, each cell contains the same complement of genes (antibody-producing cells are a notable exception). The development of specialized organs, tissues, and cells and their function in the intact organism depend upon the differential expression of genes. Some of this differential expression is achieved by having different regions of chromatin available for transcription in cells from various tissues. For example, the DNA containing the β-globin gene cluster is in “active” chromatin in the reticulocyte but in “inactive” chromatin in muscle cells.

Formation and disruption of nucleosome structure

The presence of nucleosomes and of complexes of histones and DNA certainly provides a barrier against the ready association of transcription factors with specific DNA regions. The dynamics of the formation and disruption of nucleosome structure are therefore an important part of eukaryotic gene regulation and the processes involved are as follows-

i) Histone acetylation and deacetylation is an important determinant of gene activity. Acetylation is known to occur on lysine residues in the amino terminal tails of histone molecules (Figure-1). This modification reduces the positive charge of these tails and decreases the binding affinity of histone for the negatively charged DNA. Accordingly, the acetylation of histones could result in disruption of nucleosomal structure and allow readier access of transcription factors to cognate regulatory DNA elements. Different proteins with specific acetylase and deacetylase activities are associated with various components of the transcription apparatus.

 

Histone acetylation

Figure-1- Showing the Acetylation of lysine residues  in the amino terminal ends of Histones. The positive charge is removed after acteylation.

Thus, histone acetylation can activate transcription through a combination of three mechanisms: by reducing the affinity of the histones for DNA, by recruiting other components of the transcriptional machinery, and by initiating the active remodeling of the chromatin structure (Figure-2).

Histone actylation and chromatin remodeling

Figure-2- Acetylation of histones leads to disruption of nucleosomal structure and access of transcription machinery for transcription of required genes

ii)  Modification of DNA-The modification of DNA provides another mechanism, in addition to packaging with histones, for inhibiting inappropriate gene expression in specific cell types. Methylation of deoxycytidine residues (Figure-3) in DNA may effect gross changes in chromatin so as to preclude its active transcription. Acute demethylation of deoxycytidine residues in a specific region of the tyrosine aminotransferase gene—in response to glucocorticoid hormones—has been associated with an increased rate of transcription of the gene. However, it is not possible to generalize that methylated DNA is transcriptionally inactive, that all inactive chromatin is methylated, or that active DNA is not methylated.

DNA methylation

Figure-3- Methylation of deoxycytidine residues  in DNA preclude its active transcription.

iii) DNA binding proteins- The interactions between DNA-binding proteins such as CAP and RNA polymerase can activate transcription in prokaryotic cells. Such protein-protein interactions play a dominant role in eukaryotic gene regulation. In contrast with those of prokaryotic transcription, few eukaryotic transcription factors have any effect on transcription on their own. Instead, each factor recruits other proteins to build up large complexes that interact with the transcriptional machinery to activate or repress transcription.

A major advantage of this mode of regulation is that a given regulatory protein can have different effects, depending on what other proteins are present in the same cell. This phenomenon, called combinatorial control, is crucial to multicellular organisms that have many different cell types.

The binding of specific transcription factors to certain DNA elements may result in disruption of nucleosomal structure. Many eukaryotic genes have multiple protein-binding DNA elements. The serial binding of transcription factors to these elements may either directly disrupt the structure of the nucleosome or prevent its re-formation. These reactions result in chromatin-level structural changes that in the end increase DNA accessibility to other factors and the transcription machinery.

2) Enhancers and Repressors- Enhancer elements are DNA sequences, although they have no promoter activity of their own but they greatly increase the activities of many promoters in eukaryotes. Enhancers function by serving as binding sites for specific regulatory proteins. An enhancer is effective only in the specific cell types in which appropriate regulatory proteins are expressed. In many cases, these DNA-binding proteins influence transcription initiation by perturbing the local chromatin structure to expose a gene or its regulatory sites rather than by direct interactions with RNA polymerase.

Enhancer elements can exert their positive influence on transcription even when separated by thousands of base pairs from a promoter; they work when oriented in either direction; and they can work upstream (5′) or downstream (3′) from the promoter. Enhancers are promiscuous; they can stimulate any promoter in the vicinity and may act on more than one promoter.

The elements that decrease or repress the expression of specific genes have also been identified. Silencers are control regions of DNA that, like enhancers, may be located thousands of base pairs away from the gene they control. However, when transcription factors bind to them, expression of the gene they control is repressed.

Tissue-specific gene expression is mediated by enhancers or enhancer-like elements. Many genes are now recognized to harbor enhancer or activator elements in various locations relative to their coding regions. In addition to being able to enhance gene transcription, some of these enhancer elements clearly possess the ability to do so in a tissue-specific manner. Thus, the enhancer element associated with the immunoglobulin genes between the J and C regions enhances the expression of those genes preferentially in lymphoid cells.

3) Locus control regions and Insulators- some regions are controlled by complex DNA elements called locus control regions (LCRs). An LCR—with associated bound proteins—controls the expression of a cluster of genes. The best-defined LCR regulates expression of the globin gene family over a large region of DNA.

Another mechanism is provided by insulators. These DNA elements, also in association with one or more proteins, prevent an enhancer from acting on a promoter .

4) Gene Amplification- One way to increase the rate at which gene product can be increased is to increase the number of genes available for transcription of specific molecules. Among the repetitive DNA sequences are hundreds of copies of ribosomal RNA genes and tRNA genes. These genes preexist repetitively in the genomic material of the gametes and thus are transmitted in high copy numbers from generation to generation.

During early development of metazoans, there is an abrupt increase in the need for specific molecules such as ribosomal RNA and messenger RNA molecules for proteins that make up such organs as the eggshell. Such requirements are fulfilled by amplification of specific genes. Subsequently, these amplified genes (Figure-4)  presumably generated by a process of repeated initiations during DNA synthesis, provide multiple sites for gene transcription.

Gene amplification

 Figure-4- gene amplification increases the copy number of genes and hence increase in the amount of gene product

In some cases, a several thousand-fold increase in the copy number of specific genes can be achieved over a period of time involving increasing doses of selective drugs. It has been demonstrated in patients receiving methotrexate for cancer that malignant cells can develop drug resistance by increasing the number of genes for dihydrofolate reductase, the target of Methotrexate.

5. Gene Rearrangement- Gene rearrangement is observed during immunoglobulins synthesis. Immunoglobulins are composed of two polypeptides, heavy (about 50 kDa) and light (about 25 kDa) chains. The mRNAs encoding these two protein subunits are encoded by gene sequences that are subjected to extensive DNA sequence-coding changes. These DNA coding changes are needed for generating the required recognition diversity central to appropriate immune function.

IgG heavy and light chain mRNAs are encoded by several different segments that are tandemly repeated in the germ line. Thus, for example, the IgG light chain is composed of variable (VL), joining (JL), and constant (CL) domains or segments. For particular subsets of IgG light chains, there are roughly 250-300 tandemly repeated VL gene coding segments, five tandemly arranged JL coding sequences, and roughly ten CL gene coding segments. All of these multiple, distinct coding regions are located in the same region of the same chromosome (Figure-4).By having multiple VL, JL, and CL segments to choose from, an immune cell has a greater repertoire of sequences to work with to develop both immunologic flexibility and specificity.

However, a given functional IgG light chain transcription unit contains only the coding sequences for a single protein. Thus, before a particular IgG light chain can be expressed, single VL, JL, and CL coding sequences must be recombined to generate a single, contiguous transcription unit excluding the multiple nonutilized segments (ie, the other approximately 300 unused VL segments, the other four unused JL segments, and the other nine unused CL segments). This deletion of unused genetic information is accomplished by selective DNA recombination that removes the unwanted coding DNA while retaining the required coding sequences: one VL, one JL, and one CL sequence. (VL sequences are subjected to additional point mutagenesis to generate even more variability—hence the name.) The newly recombined sequences thus form a single transcription unit that is competent for RNA polymerase II-mediated transcription.

Gene rearrangement

 Figure-5-  Showing Immunoglobulin m RNA for a light chain formed by transcription of rearranged genes.

6. Alternative RNA Processing

Eukaryotic cells also employ alternative RNA processing to control gene expression. This can result when alternative promoters, intron-exon splice sites, or polyadenylation sites are used. Occasionally, heterogeneity within a cell results, but more commonly the same primary transcript is processed differently in different tissues.

Alternative polyadenylation sites in the immunoglobulin  (Ig M) heavy chain primary transcript result in mRNAs that are either 2700 bases long (m) or 2400 bases long (s). This results in a different carboxyl terminal region of the encoded proteins such that the m protein remains attached to the membrane of the B lymphocyte and the s immunoglobulin is secreted.

Alternative splicing and processing, results in the formation of seven unique -tropomyosin mRNAs in seven different tissues (Figure-6).

alternative splicing

 

Figure-6- The presence or absence of extra exon can alter the structure and hence the functions of a protein.

7. Class switching- In this process one gene is switched off and a closely related gene takes up the function.

For example- During intrauterine life embryonic Hb is the first Hb to be formed. It is produced by having two “Zeta” and two “Epsilon” chains. By the sixth month of intrauterine life, embryonic Hb is replaced by HbF consisting of “α2 and y2 chains. After birth HbF is replaced by adult type of Hb A 1(97%) and HbA2(3%). Thus the genes for a particular class of Hb are switched off and for another class are switched on.

Gene switching is also observed in the formation of immunoglobulins. Ig M is the formed during primary immune response, while Ig G is formed during secondary immune response.

8. mRNA stability -Although most mRNAs in mammalian cells are very stable (half-lives measured in hours), some turn over very rapidly (half-lives of 10–30 minutes). In certain instances, mRNA stability is subject to regulation. This has important implications since there is usually a direct relationship between mRNA amount and the translation of that mRNA into its cognate protein. Changes in the stability of a specific mRNA can therefore have major effects on biologic processes.

The stability of the m RNA can be influenced by hormones and certain other effectors.

The ends of mRNA molecules are involved in mRNA stability. The 5′ cap structure in eukaryotic mRNA prevents attack by 5′ exonucleases, and the poly(A) tail prohibits the action of 3′ exonucleases.

9. Specific motifs of regulatory proteins- Certain DNA binding proteins having specific motifs bind certain region of DNA to influence the rate of transcription. The specificity involved in the control of transcription requires that regulatory proteins bind with high affinity to the correct region of DNA. Three unique motifs—the helix-turn-helix, the zinc finger, and the leucine zipper—account for many of these specific protein-DNA interactions. The motifs found in these proteins are unique; their presence in a protein of unknown function suggests that the protein may bind to DNA. The protein-DNA interactions are maintained by hydrogen bonds and van der Waals forces.

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1) A 30- year- old female presents with headache and blurry vision. Her blood pressure is 200/90 mm Hg. Imaging reveals that he she has a tumor that is overproducing the hormone most responsible for regulating salt and water balance  for blood pressure control. That hormone is-

a) Growth hormone

b) Glucocorticoid

c) Aldosterone

d) Epinephrine

e) Glucagon

Answer- The right answer is-c)- Aldosterone.

Aldosterone increases Na+ absorption in the kidney, and water is also reabsorbed resulting in an increase in blood volume and blood pressure. Growth hormone stimulates release of insulin like growth factors  and is responsible for growth and development. Glucocorticoids are important in response to stress. Epinephrine is the “fight or flight” hormone that stimulated glycogenolysis and lipolysis. Glucagon is the hormone to increase blood glucose levels during periods of fasting and starvation.

2) A 60-year-old female presents with severe back pain for the past one week. Imaging reveals a compression fracture of one of the vertebrae and diffuse osteoporosis, which is a common condition resulting from calcium depletion in bones. As a treatment she should be prescribed-

a) Oxytocin

b) Parathyroid hormone

c) Estradiol

d) Calcitonin

e) Prolactin

Answer- The right answer is-d)- Calcitonin. Calcitonin inhibits release of calcium from bone and also decreases blood calcium levels. Calcium, vitamin D and Calcitonin may be prescribed to help treat osteoporosis. Parathyroid hormone promotes calcium and phosphate mobilization from bone, increasing calcium levels. Oxytocin, Prolactin and Estradiol are not significant for calcium level regulation

3) A 16-year-old female presents with a fever, productive cough and rust colored sputum. She is diagnosed with bacterial pneumonia. She is a known case of type 1 Diabetes mellitus. She injects herself subcutaneously every day with exogenous insulin. As insulin is absorbed into her blood it binds to insulin receptors that activate-

a) Tyrosine kinase

b) Adenylate cyclase

c) c AMP

d) Protein kinase C

e) Phospholipase C

Answer- The right answer is -a)- Tyrosine Kinase. Insulin binds to a cell surface receptor that acts as a Tyrosine kinase. Hormones such as epinephrine and Glucagon activate Adenylate cyclase which converts ATP to cAMP. Hormones such as Thyrotropin- releasing hormone(TRH) and Oxytocin activate protein kinase C. Phospholipase C acts on Phosphatidyl Inositol (membrane Phospholipid) to cleave it to Inositol triphosphate and diacylglycerol .

4) An intern is scrubbing into a complicated surgery that is anticipated to last for 15 hours. In preparation, the intern has not eaten from the past 15 hours. After 30 hours  of fasting which of the following is most important for maintenance of normal blood glucose ?

a) Glycogenolysis

b) Gluconeogenesis

c) Triacylglycerol synthesis

d) Increased insulin release

e) Decreased muscle protein break down

Answer- The right answer is b)- Gluconeogenesis. Approximately 2-3 hours after a meal, the liver maintains normal blood glucose level by glycogenolysis. Within 30 hours liver glycogen stores are depleted, leaving gluconeogenesis as the primary source for maintaining blood glucose levels. ketone bodies are generated, triacylglycerols are broken down and muscle protein breakdown increases.

 In the fed state, insulin is the main hormone while in the fasting state, glucagon level increases.

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1) A 30- year- old female presents with headache and blurry vision. Her blood pressure is 200/90 mm Hg. Imaging reveals that he she has a tumor that is overproducing the hormone most responsible for regulating salt and water balance  for blood pressure control. That hormone is-

a) Growth hormone

b) Glucocorticoid

c) Aldosterone

d) Epinephrine

e) Glucagon

2) A 60-year-old female presents with severe back pain for the past one week. Imaging reveals a compression fracture of one of the vertebrae and diffuse osteoporosis, which is a common condition resulting from calcium depletion in bones. As a treatment she should be prescribed-

a) Oxytocin

b) Parathyroid hormone

c) Estradiol

d) Calcitonin

e) Prolactin

3) A 16-year-old female presents with a fever, productive cough and rust colored sputum. She is diagnosed with bacterial pneumonia. She is a known case of type 1 Diabetes mellitus. She injects herself subcutaneously every day with exogenous insulin. As insulin is absorbed into her blood it binds to insulin receptors that activate-

a) Tyrosine kinase

b) Adenylate cyclase

c) c AMP

d) Protein kinase C

e) Phospholipase C

4) An intern is scrubbing into a complicated surgery that is anticipated to last for 15 hours. In preparation, the intern has not eaten from the past 15 hours. After 30 hours  of fasting which of the following is most important for maintenance of normal blood glucose ?

a) Glycogenolysis

b) Gluconeogenesis

c) Triacylglycerol synthesis

d) Increased insulin release

e) Decreased muscle protein break down

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A 20 year-old- male presents with intermittent ataxia (abnormal gait), paralysis of eye muscles, and confusion. After an extensive work up, he is diagnosed with Carnitine acyl transferase I(CAT-I) deficiency. The reaction catalyzed by CAT-I forms-

A) Fatty acyl co A

B) Fatty acyl Carnitine

C) Malonyl co A

D) Palmitoyl co A

E) Carnitine

 

Answer- The right option is B- Fatty acyl carnitine.

Fatty acids are activated on the outer mitochondrial membrane, whereas they are oxidized in the mitochondrial matrix. A special transport mechanism is needed to carry long-chain acyl CoA molecules across the inner mitochondrial membrane. Activated long-chain fatty acids are transported across the membrane by conjugating them to carnitine, a zwitterionic alcohol.

Carnitine (ß-hydroxy-Υ-trimethyl ammonium butyrate), (CH3)3N+—CH2—CH(OH)—CH2—COO, is widely distributed and is particularly abundant in muscle. Carnitine is obtained from foods, particularly animal-based foods, and via endogenous synthesis.

The transportation across the inner mitochondrial membrane through carnitine shuttle involves three steps-

1) The acyl group is transferred from the sulfur atom of CoA to the hydroxyl group of carnitine to form acyl carnitine. This reaction is catalyzed by carnitine acyl transferase I (also called carnitine palmitoyl transferaseI), which is bound to the outer mitochondrial membrane.

2) Acyl carnitine is then shuttled across the inner mitochondrial membrane by a translocase.

3) The acyl group is transferred back to CoA on the matrix side of the membrane. This reaction, which is catalyzed by carnitine acyl transferase II (carnitine palmitoyl transferase II), is simply the reverse of the reaction that takes place in the cytosol.

Finally, the translocase returns carnitine to the cytosolic side in exchange for an incoming acyl carnitine (Figure)

Figure- showing the transportation of acyl co A in to to the mitochondrial matrix through carnitine shuttle

This counter-transport system provides regulation of the uptake of fatty acids into the mitochondrion for oxidation. As long as there is free CoA available in the mitochondrial matrix, fatty acids can be taken up and the carnitine returned to the outer membrane for uptake of more fatty acids. However, if most of the CoA in the mitochondrion is acylated, then the fatty acid uptake is inhibited.

This carnitine shuttle also serves to prevent uptake into the mitochondrion (and hence oxidation) of fatty acids synthesized in the cytosol in the fed state; malonyl CoA (the precursor for fatty acid synthesis) is a potent inhibitor of carnitine palmitoyl transferase I in the outer mitochondrial membrane.

Short and medium chain fatty acids do not require carnitine for their transportation across the inner mitochondrial membrane.

Inherited CAT-I deficiency affects only the liver, resulting in reduced fatty acid oxidation and ketogenesis, with hypoglycemia. CAT-II deficiency affects primarily skeletal muscle and, when severe, the liver.

For further details read my post at-

 

http://www.namrata.co/fatty-acid-oxidation-subjective-questions-set-2/

 

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A 20 year-old- male has presented with intermittent ataxia (abnormal gait), paralysis of eye muscles, and confusion.  Blood glucose  has been found to be 52 mg/dl  but there are no ketone bodies in urine. 

After an extensive work up, he has been diagnosed with Carnitine acyl transferase I (CAT-I) deficiency. The reaction catalyzed by CAT-I forms-

A) Fatty acyl co A

B) Fatty acyl Carnitine

C) Malonyl co A

D) Palmitoyl co A

E) Carnitine

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The right answer is – D and E) Porphobilinogen synthase also called ALA dehydratase.

There are 2 enzymes in the pathway of Heme biosynthesis which are inhibited by Lead and these are ALA dehydratase/ Porphobilinogen synthase  and Ferrochelatase (Figure). As a result of inhibition, there is impaired heme biosynthesis.

One of the first manifestations of lead toxicity is anemia. Lead-induced anemia manifests as a microcytic, hypochromic anemia.

As regards other options- 

A) Cytochrome oxidase is a complex of electron transport chain.

B) Protoporphyrinogen oxidase is an enzyme for the conversion of Protoporphyrinogen IX  to Protoporphyrin IX (See figure)

C) ALA synthase is the first  enzyme of heme bio synthetic pathway, that catalyzes the condensation of Glycine and Succinyl co A (See figure)

lead

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure- showing steps of heme biosynthesis and the inhibition of ALA dehydratase and ferrochelatase  by lead.

For further details follow the link-

 

http://www.namrata.co/case-study-porphyria/

 

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A 3-year- old boy is brought to the emergency room with abdominal pain, mental status changes and fatigue. On taking history the physician finds that the child belongs to a very poor family, lives in an old house and has the habit of licking the paint chips that have crumbled in the window sills. The physician suspects lead poisoning. Lead typically interferes with which of the following enzymes ?

A) Cytochrome oxidase

B) Protoporphyrinogen oxidase

C) ALA synthase

D) Porphobilinogen synthase

E) ALA dehydratase

lead paint

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A 24- year-old boy presents with diarrhea, dysphagia, jaundice, and white transverse lines on the fingernails. The patient is diagnosed with arsenic poisoning, which inhibits which of the following enzymes ?

A) Citrate synthase

B)  Isocitrate dehydrogenase

C) Pyruvate Kinase

D) Alpha keto glutarate dehydrogenase

E) Succinate dehydrogenase

Answer-  The right answer is -D , Alpha ketoglutarate dehydrogenase complex. Arsenite (the trivalent form of Arsenic) forms a stable complex with the – thiol group of lipoic acid (Figure-1) making that compound unavailable to serve as a coenzyme. Arsenic poisoning is however due to inhibition of the enzymes that require lipoic acid as a coenzyme. The enzymes requiring lipoic acid  are- Pyruvate dehydrogenase complex, Alpha keto glutarate dehydrogenase complex and alpha keto acid dehydrogenase complex. The later enzymes is involved in the metabolism of branched chain amino acids. In the presence of Arsenite, induced lipoic acid deficiency causes decreased activity of said enzymes with the resultant accumulation of pyruvate, alpha keto glutarate and alpha keto acids of branched chain amino acids. There is inhibition of TCA cycle affecting brain, causing neurological manifestations and death.

1

Figure-1- showing structure of lipoic acid

Not only lipoic acid but all enzymes containing -SH groups are affected by Arsenic poisoning. Arsenate (Pentevalent form of Arsenic) can interfere with glycolysis at the step of Glyceraldeyde-3-P dehydrogenase thereby causing decreased ATP and NADH production by glycolysis, without inhibiting the pathway itself.The poison does so by competing with inorganic phosphate as a substrate for Glyceraldehyde-3-Phosphate dehydrogenase, forming a complex that spontaneously hydrolyzes to form 3-phosphoglycerate (Figure-2).Thus by bypassing the synthesis and dephosphorylation of 1,3 BPG, the cell is deprived of energy usually obtained from Glycolysis.

2

Figure-2- Arsenate competes with inorganic phosphate forming 1-Arseno-3- phosphoglycerate that spontaneously hydrolyzes forming 3-phosphoglycerate, thus no ATP is formed by substrate level phosphorylation in Glycolysis.

Arsenic  also competes with phosphates for adenosine triphosphate, forming adenosine diphosphate mono arsine, causing the loss of high-energy bonds.

In the given problem, only Alpha keto glutarate dehydrogenase complex is the enzyme requiring lipoic acid, the other enzymes like citrate synthase, Isocitrate dehydrogenase and Succinate dehydrogenase enzymes although are enzymes of TCA cycle but are unaffected in the arsenic poisoning. Similarly Pyruvate kinase (the enzyme catalyzing the last step of glycolysis ) is also not affected.

Evidences are there that it can cause inhibition of Hexokinase as well as Succinate dehydrogenase but are not widely proved.

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