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# Question of the day

A 56-year-old man with longstanding poorly controlled diabetes mellitus visits his primary care physician for a follow-up after a recent hospitalization. The patient experienced an episode of acute renal failure while in the hospital, and his creatinine level rose to 3.4 mg/dl (normal, 0.7-1.5 mg/dl).

Creatinine a marker of kidney function, is produced from which of the following precursors?

A. Glutamine, Aspartic acid and CO2

B. Glutamine, Cysteine and Glycine

C. Serine and Palmitoyl Co A

D. Glycine and Succinyl Co A

E. Glycine, Arginine and S-Adenosyl Methionine.

The correct answer is E- Glycine, Arginine and S-Adenosyl Methionine

Creatinine is an anhydrous product of creatine.

Creatine also called, “Methyl guanidoacetic acid”, is produced from Glycine, Arginine and Methionine in the human body primarily in the kidney and liver (figure-1).



Figure-1- In the first step Glycine and Arginine react to form Guanido acetic acid; this reaction takes place in the kidney. In the second step, (in the liver), methylation of guanido acetic acid takes place to form Methyl guanido acetic acid (creatine). Creatine is reversibly phosphorylated to creatine-P by creatine kinase. Creatine phosphate is a high energy compound. Creatinine is produced by the loss of water from creatine.


Creatine is transported in the blood for use by muscles. Approximately 95% of the total creatine is located in skeletal muscle. It is also present in liver, testes and brain. It can occur in free form or phosphorylated form. The phosphorylated form is called ‘Creatine-Phosphate’, ‘Phosphocreatine’ or ‘phosphagen’.

In the skeletal muscle creatine is reversibly phosphorylated, The reaction is catalyzed by Creatine kinase, also called Creatine phospho kinase (CPK). The physiological role of creatine phosphate is to form a rapidly available reserve of energy-rich phosphate groups.

There are 3 main isoenzymes of CPK, which are present in brain (CPK-BB), myocardium (CPK-MB) and skeletal muscle (CPK-MM). High CPK-MB in plasma is an early marker of Acute myocardial infarction, whereas high CPK-MM is a marker of muscle injury or muscular disorder.

Creatine is excreted in the form of Creatinine, which is formed by removal of one molecule of water from creatine.

Creatine————–> Creatinine + H20

The reaction is non- enzymatic and irreversible. About 2 % of the total Creatine is converted daily to creatinine so that the amount of creatinine produced is related to the total muscle mass and remains approximately the same in plasma and urine from day-to-day unless the muscle mass changes.

Creatinine is mostly eliminated through kidney by glomerular filtration. Normal excretion ranges between 1-2 g/day.  Estimation of serum and urinary creatinine is undertaken to assess the renal functions. Creatinine clearance estimation is undertaken to determine Glomerular filtration rate (GFR). The normal serum creatinine ranges as follows :

1) < 12 Years-0.25-0.85 mg/dL

2) Adult male 0.7-1.5 mg/dL

3) Adult female 0.4 -1.2 mg/dL

It is higher in males because of the muscle mass.

As regards other options

A. Glutamine, Aspartic acid and CO2- these three along with ATP are needed for the synthesis of pyrimidine nucleotides.

B. Glutamine, Cysteine and Glycine- there is no peptide or compound formed from these amino acids. Glutamic acid, cysteine and glycine are required for the formation of Glutathione.

C. Serine and Palmitoyl Co A condense to form Sphingol, which is required for the synthesis of sphingolipids.

D. Glycine and Succinyl Co A are required for the heme biosynthesis.







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Laboratory results for a patient with uncontrolled Type I diabetes mellitus reveal hyperglycemia (456 mg/dL) and hypertriglyceridemia (395 mg/dL). The most likely cause of the hypertriglyceridemia in this patient is which of the following?

A. Deficiency in apoprotein C-II

B. Increased hepatic triglyceride synthesis

C. Decreased lipoprotein lipase activity

D. Deficiency in LDL receptors

E. Absence of hormone-sensitive lipase.


The correct answer is-C- Decreased Lipoprotein lipase activity.

Circulating lipoproteins are just as dependent on insulin as is the plasma glucose, it is because, the lipoprotein lipase that catalyzes the degradation of circulating lipoproteins is activated by insulin.

Lipoprotein lipase


Lipoprotein lipase is located on the walls of blood capillaries, anchored to the endothelium by negatively charged proteoglycan chains of Heparan sulfate. It has been found in heart, adipose tissue, spleen, lung, renal medulla, aorta, diaphragm, and lactating mammary gland, although it is not active in adult liver.

Lp L

Figure-1- Action of lipoprotein lipase (LpL).

It is not normally found in blood; however, following injection of heparin, lipoprotein lipase is released from its heparan sulfate binding into the circulation.  Due to this reason ‘heparin’ is called a “clearing factor”, as it causes release of lipoprotein lipase that promotes utilization and clearance of lipoproteins from the plasma.

Activators and inhibitors of lipoprotein lipase

Both phospholipids and apo C-II are required as cofactors for lipoprotein lipase activity (figure-1), while apo A-II and apo C-III act as inhibitors. Hydrolysis takes place while the lipoproteins are attached to the enzyme on the endothelium.

Hypertriglyceridemia in Type 1 Diabetes Mellitus

Dyslipidemia is a common metabolic abnormality in uncontrolled diabetes mellitus. Dyslipidemia includes; hypercholesterolemia, hypertriglyceridemia, raised LDLc and LDLc, but low levels of HDLc.

Biochemical basis of dyslipidemia in uncontrolled DM

One major role of insulin is to stimulate the storage of food energy following the consumption of a meal.This energy storage is in the form of glycogen in hepatocytes and skeletal muscle. Additionally, insulin stimulates synthesis of triglycerides in hepatocytes and promotes their storage in adipose tissue. In opposition to increased adipocyte storage of triglycerides is insulin-mediated inhibition of lipolysis (insulin inhibits adipolysis because the enzyme hormone- sensitive lipase, that catalyzes adipolysis is stimulated by Glucagon and inhibited by Insulin).

Lipolysis and its implications in Type 1 DM

In uncontrolled IDDM since the lipolysis is promoted, thus  there is a rapid mobilization of triglycerides leading to increased levels of plasma free fatty acids. The free fatty acids are taken up by numerous tissues (however, not the brain) and metabolized to provide energy. Free fatty acids are also taken up by the liver.

Increased fatty acid oxidation

Normally, the levels of malonyl-CoA are high in the presence of insulin. These high levels of malonyl-CoA inhibits carnitine acyl Transferase I, the enzyme required for the transport of fatty acyl-CoA’s into the mitochondria where they are subject to oxidation for energy production. Thus, in the absence of insulin, malonyl-CoA levels fall and transport of fatty acyl-CoA’s into the mitochondria increases. Mitochondrial oxidation of fatty acids generates acetyl-CoA which can be further oxidized in the TCA cycle.

Suppressed TCA cycle and its implications

a) Ketosis

In hepatocytes the majority of the acetyl-CoA is not oxidized by the TCA cycle but is metabolized into the ketone bodies, Acetoacetate and β-hydroxybutyrate.These ketone bodies leave the liver and are used for energy production by the brain, heart and skeletal muscle. In IDDM, the increased availability of free fatty acids and ketone bodies exacerbates the reduced utilization of glucose furthering the ensuing hyperglycemia. Production of ketone bodies, in excess of the body’s ability to utilize them leads to ketoacidosis. In diabetics, this can be easily diagnosed by smelling the breath. A spontaneous breakdown product of acetoacetate is acetone which is volatilized by the lungs producing a distinctive odor.

b) Hypercholesterolemia

The unutilized Acetyl Co A, due to suppressed TCA cycle, is also channeled towards the pathway of cholesterol biosynthesis resulting in hypercholesterolemia.

Basis of hypertriglyceridemia

Normally, plasma triglycerides are acted upon by lipoprotein lipase (LPL). In particular, LPL activity allows released fatty acids to be taken from circulating triglycerides for storage in adipocytes. The activity of LPL requires insulin and in its absence a hypertriglyceridemia results (figure-2).

Clinical pearls

In type 1 diabetes, moderately deficient control of hyperglycemia is associated with only a slight elevation of LDL cholesterol and serum triglycerides and little if any change in HDL cholesterol. Once the hyperglycemia is corrected, lipoprotein levels are generally normal. However, in obese patients with type 2 diabetes, a distinct “diabetic dyslipidemia” is characteristic of the insulin resistance syndrome. Its features are a high serum triglyceride level (300–400 mg/dL), a low HDL cholesterol (less than 30 mg/dL), and a qualitative change in LDL particles, producing a smaller dense particle whose membrane carries supranormal amounts of free cholesterol. These smaller dense LDL particles are more susceptible to oxidation, which renders them more atherogenic. Since low HDL cholesterol is a major feature predisposing to macro vascular disease, the term “dyslipidemia” has preempted the term “hyperlipidemia,” which mainly denoted the elevated triglycerides. Measures designed to correct the obesity and hyperglycemia, such as exercise, diet, and hypoglycemic therapy, are the treatment of choice for diabetic dyslipidemia, and in occasional patients in whom normal weight was achieved, all features of the lipoprotein abnormalities cleared.

As regards other options

Deficiency in apoprotein C-II

Apo CII is an activator of lipoprotein lipase, but its concentration is not decreased in diabetes mellitus.

Increased hepatic triglyceride synthesis

The Acetyl Co A carboxylase, the key regulatory enzyme of fatty acid biosynthetic pathway, is activated by insulin, thus, de novo fatty acid synthesis is decreased in insulin deficiency. However, the fatty acids mobilized from adipose tissue are used for esterification to form triglycerides which are transported from liver as VLDL. The excess flux of fatty acids that cannot be transported out as VLDL results in fatty liver. The hypertriglyceridemia in uncontrolled DM can be because of excess hepatic triglyceride synthesis, but mainly it is due to non degradation of circulating chylomicrons (carriers of dietary lipids) and VLDL (carriers of endogenous triglycerides) by the inactive lipoprotein lipase (figure-2).

Deficiency in LDL receptors

LDL receptors internalize LDL, their deficiency cannot cause hypertriglyceridemia, and otherwise also, they are not deficient in diabetes mellitus.

Absence of hormone-sensitive lipase

Hormone- sensitive lipase catalyzes the breakdown of triglycerides in adipose cells. It is activated by glucagon and catecholamines; inhibited by insulin. It is not absent; instead it is overactive in uncontrolled type 1 diabetes mellitus.

Hyper tgs 


Figure-2- Diabetic dyslipidemia. In normal health, VLDL released from liver, carrying endogenous triglycerides and Chylomicrons released from intestinal cells carrying dietary lipids, are acted upon by lipoprotein lipase (LPL), and the resultant fatty acids are taken up by peripheral cells, whereas the lipoprotein remnants are taken up by liver. VLDL is converted to LDL, through intermediate formation of IDL (intermediate density lipoprotein). In diabetes mellitus, in the absence of active LPL, the lipoprotein metabolism is disturbed resulting in hypertriglyceridemia, hypercholesterolemia, small dense LDL, low HDL and  a fatty liver.










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What is the basis of this statement “fats burn in the flame of carbohydrates”?

A)   Fats are hydrolyzed in the presence of carbohydrates

B)   Fatty acids and glucose are simultaneously oxidized

C)   Acetyl co A is the common product of fatty acid and glucose oxidation

D)   Acetyl co A is oxidized completely in the presence of oxaloacetate in TCA cycle


The correct answer is D) – Acetyl co A is oxidized completely in the presence of oxaloacetate in TCA cycle.

Fats burn in the flame of carbohydrates means fats can only be oxidized in the presence of carbohydrates.

Oxidation of fats

Fats includes triglycerides, cholesterol (free or esterified), free fatty acids and their derivatives such as phospholipids and other complex lipids. The end product of metabolism of almost all these fats is Acetyl co A. Hence, Acetyl co A can be considered a fat derivative. There are many other sources (figure-1), but fatty acid oxidation is the major source of Acetyl co A.

Sources of Acetyl Co A

Figure-1- Sources of Acetyl co A:  The major contribution of Acetyl Co A is from fatty acid oxidation, the pathways of fatty acid oxidation (minor and major), end up forming Acetyl Co A. Pyruvate, ketogenic amino acids and acetylcholine also contribute to Acetyl co A pool. The ketone bodies are synthesized from and are metabolized to Acetyl Co A.

Fate of Acetyl Co A

Acetyl Co A can be metabolized in many ways (figure-2); It is a precursor of fatty acids, cholesterol, steroids, ketone bodies, and Acetylcholine. It is also required for the detoxification of xenobiotics, but the major fate involves complete oxidation in the TCA cycle to provide energy.

Fate of Acetyl Co A

Figure-2- Central role of Acetyl co A: Acetyl Co A can be utilized in multiple ways depending upon the cell type and under different cellular conditions. Under low energy states, Acetyl co A is completely oxidized in the TCA cycle to provide energy.

Oxidation of Acetyl Co A in TCA cycle

The citric acid cycle is the central metabolic hub of the cell. It is the final common pathway for the oxidation of fuel molecule such as amino acids, fatty acids, and carbohydrates. It is the gateway to the aerobic metabolism of any molecule that can be transformed into an acetyl group or dicarboxylic acid. The citric acid cycle (Krebs cycle, tricarboxylic acid cycle) includes a series of oxidation-reduction reactions in mitochondria that result in the oxidation of an acetyl group to two molecules of carbon dioxide and reduce the coenzymes that are reoxidized through the electron transport chain, linked to the formation of ATP.

A four- carbon compound (oxaloacetate) condenses with a two-carbon acetyl unit to yield a six-carbon tricarboxylic acid (citrate). An isomer of citrate is then oxidatively decarboxylated. The resulting five-carbon compound (α-ketoglutarate) also is oxidatively decarboxylated to yield a four carbon compound (succinate) (Figure-3).

Overview of TCA

Figure-3- An overview of TCA cycle

Oxaloacetate is then regenerated from succinate. Two carbon atoms enter the cycle as an acetyl unit and two carbon atoms leave the cycle in the form of two molecules of carbon dioxide. This is considered as the complete oxidation of Acetyl co A having two carbon atoms in its structure.1 acetate unit generates approximately 12 molecules of ATP.

The four-carbon molecule, oxaloacetate that initiates the first step in the citric acid cycle is regenerated at the end of one passage through the cycle. The oxaloacetate acts catalytically: it participates in the oxidation of the acetyl group but is itself regenerated. Thus, one molecule of oxaloacetate is capable of participating in the oxidation of many acetyl molecules.

Sources and Fate of Oxaloacetate

Oxaloacetate can be synthesized from or metabolized to Aspartate (figure-4) in reversible transamination reaction. Aspartate can be used for the synthesis of purines and pyrimidines; it is also used in urea cycle. The major source of Oxaloacetate is pyruvate in a reaction catalyzed by Pyruvate carboxylase.

Pyruvate is mainly used up for Anaplerotic reactions to compensate for oxaloacetate concentration.  Thus without carbohydrates (Pyruvate), there would be no Anaplerotic reactions to replenish the TCA-cycle components. With a diet of fats only, the acetyl CoA from fatty acid degradation would not get oxidized and build up due to non- functioning of TCA cycle. Thus fats can burn only in the flame of carbohydrates -Figure-5

 Role of oxaloacetate

Figure-4- Role of Oxaloacetate in TCA cycle. Oxaloacetate acts as a catalyst of TCA cycle; it starts the cycle and is regenerated at the end of the cycle.


It can now be well concluded that the complete oxidation of Acetyl Co A, a derivative of fats cannot take place without the availability the derivative of carbohydrates, that is oxaloacetate,. Hence the statement “fats burn in the flame of carbohydrates”, is absolutely correct.

 Fats and carbs

Figure-5- Significance of oxaloacetate in TCA cycle operation.

As regards other options

A)   Fats are hydrolyzed in the presence of carbohydrates -is incorrect. Fats are not hydrolyzed in the presence of carbohydrates.

B)   Fatty acids and glucose are simultaneously oxidized- It is also not true. The location, enzymes, the nutritional states, the regulatory hormones are different for these oxidative processes. Fatty acid oxidation mainly takes place during period of starvation, in the presence of glucagon or catecholamines whereas glycolysis takes place in the well fed state in the presence of insulin as the regulatory hormone.

C)   Acetyl co A is the common product of fatty acid and glucose oxidation- The end product undoubtedly is Acetyl Co A but without oxaloacetate, Acetyl co A cannot be oxidized.

Therefore, D)   Acetyl co A is oxidized completely in the presence of oxaloacetate in TCA cycle, is the correct option.









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Case detail

A 3-year-old girl was brought into the Emergency Room. She was cold and clammy and was breathing rapidly. She was obviously confused and lethargic. Her mother indicated that she had accidentally ingested automobile antifreeze (ethylene glycol) while playing in the garage. Following gastrointestinal lavage and activated charcoal administration, a nasogastric tube for ethanol was administered. How will ethanol help in relieving the symptoms?

A. Conjugate with ethylene glycol to form a soluble compound

B. Induce the alcohol dehydrogenase enzyme

C. Competitively inhibit the metabolism of ethylene glycol

D. Promote the excretion of metabolite of ethylene glycol

E. All of the above.

The correct answer is- C- Competitively inhibits the metabolism of ethylene glycol.

Ethylene glycol is the major ingredient of almost all radiator fluid products. It is used to increase the boiling point and decrease the freezing point of radiator fluid, which circulates through the automotive radiator. Ethylene glycol is added to prevent the radiator from overheating or freezing, depending on the season.

As with ethyl alcohol and methanol, ethylene glycol is metabolized by alcohol dehydrogenase to form glycoaldehyde. Through interaction with aldehyde dehydrogenase, ethylene glycol is then metabolized to glycolic acid (GA). A profound acidosis often ensues and is attributable to the glycolic acid in circulation. The patient may develop hyperventilation that results from acidemia. This glycolate is then transformed into glyoxylic acid. This glycolate is then transformed into glyoxylic acid. At this point, the molecule may be transformed into the highly toxic oxalate.


 Ethylele glycol

Figure- Metabolism of ethylene glycol

With the formation of oxalate crystals in the urine, calcium oxalate crystals form and accumulate in blood and other tissues. The precipitation of calcium oxalate in the renal cortex results in decreased glomerular filtration and renal insufficiency. Calcium is consumed in circulation, and hypocalcemia may occur.

The rate-dependent step of ethylene glycol metabolism is the alcohol dehydrogenase–catalyzed step. Ethyl alcohol binds much more easily to alcohol dehydrogenase than ethylene glycol or methanol does. Because ethanol is the preferential substrate for alcohol dehydrogenase, the presence of ethanol may essentially block metabolism of ethylene glycol.

Treatment of ethylene glycol poisoning

Emergency and Supportive Measures

For patients presenting within 30–60 minutes after ingestion, the stomach is emptied by gastric lavage. Charcoal is not very effective but should be administered if other poisons or drugs have also been ingested.

Specific Treatment

Patients with significant toxicity (manifested by severe metabolic acidosis, altered mental status, and markedly elevated osmolar gap) should undergo hemodialysis as soon as possible to remove the parent compound and the toxic metabolites. Treatment with folic acid, thiamine, and pyridoxine may enhance the breakdown of toxic metabolites.

Ethanol blocks metabolism of the parent compounds by competing for the enzyme alcohol dehydrogenase. The desired serum ethanol concentration is 100 mg/d. Fomepizole (4-methylpyrazole; Antizol), blocks alcohol dehydrogenase and can be used instead of ethanol.

As regards other options

Alcohol does not form any complex with ethylene glycol, it does not induce alcohol dehydrogenase enzyme and even it does not promote the excretion of metabolites of ethylene glycol.






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Which of the following occurs in non shivering thermogenesis?

A) Glucose is oxidized to lactate

B) Fatty acids uncouple oxidative phosphorylation

C) ATP is spent for heat production

D) Glycogen is excessively degraded

E) Fatty acids are excessively oxidized

The correct answer is C- Fatty acids uncouple oxidative phosphorylation

Basic Concept

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 (figure-1).This process, which takes place in mitochondria, is the major source of ATP in aerobic organisms.

Electron transport chain

Figure-1-Flow of electrons in the electron transport chain and ATP synthase complex. 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. The resulting uneven distribution of protons generates a pH gradient and a transmembrane electrical potential that creates a proton-motive force. ATP is synthesized when protons flow back to the mitochondrial matrix through an enzyme complex. 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.

Thus, the oxidative phosphorylation or in other words, oxidation of fuels and the phosphorylation of ADP are coupled processes.

Uncouplers of oxidative phosphorylation

This tight coupling of electron transport and phosphorylation in mitochondria can be disrupted (uncoupled) by

  1. 2,4- Dinitrophenol ,
  2. 2,4  dinitrocresol
  3. CCCP( chloro carbonyl cyanide phenyl hydrazone)
  4. FCCP
  5. Valinomycin
  6. High dose of Aspirin
  7. The antibiotic Oligomycin completely blocks oxidation and phosphorylation by blocking the flow of protons through ATP synthase complex.

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.

Physiological uncouplers

1) Long chain fatty acids

2) Thyroxin

3) 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

4) 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. This UCP-1 channel is activated by fatty acids (as in the given case) – Figure-2.

Brown adipose tissue

Figure-2- Process of uncoupling of oxidative phosphorylation in the Brown adipose tissue.

Thus the most appropriate answer in the given situation is – uncoupling of oxidative phosphorylation by fatty acids.

As regards other options

A) Glucose is oxidized to lactate- under anaerobic conditions or in the cells lacking mitochondria. The released energy is captured as 2 mols of ATP; the surplus energy is released as heat to maintain body temperature.

C) ATP is never spent for heat production

D) Glycogen is not excessively degraded in non shivering thermogenesis

E) Fatty acids are not excessively oxidized to yield extra energy as heat.

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.






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During acute myocardial infarction, the oxygen supply to an area of heart is reduced forcing the cardiac muscle cells to switch to anaerobic oxidation. Under this condition the activity of which of the following enzymes is increased by the increasing concentration of AMP?

A.  Phosphofructokinase-1

B.  Pyruvate kinase

C.  Citrate synthase

D.  Lactate dehydrogenase

E.  Succinate dehydrogenase

The correct answer is- A) Phosphofructokinase-1.

Lactate dehydrogenase is not the right option, Although the level of Lactate dehydrogenase rises in circulation after myocardial injury but it is not the enzyme that is affected by the increasing concentration of AMP.

Basic concept

AMP, a marker of low energy state, is a positive allosteric modifier for Phosphofuctokinase-1,  which is a rate limiting enzyme of glycolysis. Phospho fructokinase is the “valve” controlling the rate of glycolysis.  

ATP is an allosteric inhibitor of this enzyme.

In the presence of high ATP concentrations, the Km for fructose-6-phosphate is increased, glycolysis thus “turns off.” ATP elicits this effect by binding to a specific regulatory site that is distinct from the catalytic site. AMP reverses the inhibitory action of ATP, and so the activity of the enzyme increases when the ATP/AMP ratio is lowered. In other words, glycolysis is stimulated as the energy charge falls, or the AMP concentration rises.

Phosphofructokinase-1 is also regulated by D-fructose-2,6-bisphosphate, a potent allosteric activator that increases the affinity of phosphofructokinase-1 for the substrate fructose-6-phosphate. Stimulation of phosphofructokinase is also achieved by decreasing the inhibitory effects of ATP. Fructose-2,6-bisphosphate increases the net flow of glucose through glycolysis by stimulating phosphofructokinase-1 and, by inhibiting fructose-1,6-bisphosphatase, the enzyme that catalyzes this reaction in the opposite direction.

AMP and Fructose 2,6 Bisphosphate are thus positive allosteric modifiers of Phosphofructokinase-1 (PFK-I), whereas ATP and Citrate are negative modifiers (Figure-1)

Citrate inhibits phosphofructokinase by enhancing the inhibitory effect of ATP.

Inhibition of glycolysis by citrate ensures that glucose will not be committed to these activities if the citric acid cycle is already saturated.

 Regulation of PFK-1

Figure-1- Phosphofructokinase-1 is regulated primarily by the energy needs of the cells. During the low energy conditions, glycolysis is stimulated to compensate for the energy. Reverse occurs in the presence of excess ATP (high energy state), the glycolysis is turned off.

Acute Myocardial infarction

Myocardial infarction (MI) is the irreversible necrosis of heart muscle secondary to prolonged ischemia. This usually results from an imbalance of oxygen supply and demand. This usually results from plaque rupture with thrombus formation in a coronary vessel, resulting in an acute reduction of blood supply to a portion of the myocardium.
The reduction of blood supply causes the change of aerobic mode of glycolysis to anaerobic, as a result lactate is the end product of glycolysis.

Lactate dehydrogenase(LDH) catalyzes the reversible conversion of pyruvate and lactate.


Figure-2- Reaction showing the interconversion of pyruvate and lactate.

The appearance of LDH in the circulation generally indicates myocardial necrosis. It begins to rise in 12 to 24 hours following MI, and peaks in 2 to 3 days, gradually dissipating in 5 to 14 days.

Measurement of LDH isoenzymes is necessary for greater specificity for cardiac injury. There are 5 isoenzymes (1 through 5). Ordinarily, isoenzyme 2 is greater than 1, but with myocardial injury, this pattern is “flipped” and 1 is higher than 2.

LDH-5 from liver may be increased with centrilobular necrosis from passive congestion with congestive heart failure following ischemic myocardial injury.

 As regards other options

B.  Pyruvate kinase- It catalyzes the conversion of phosphoenolpyruvate to pyruvate, the last step of glycolysis.  AMP affect the activity of this enzyme also, but this is not a rate limiting enzyme of Glycolysis.

C.  Citrate synthase-catalyzes the condensation of Acetyl co A with oxaloacetate to form Citrate, the first step of TCA cycle.

D.  Lactate dehydrogenase- catalyzes the interconversion of pyruvate and lactate.

E.  Succinate dehydrogenase  is an enzyme of TCA cycle, it catalyzes the conversion of succinate to fumarate.

Thus the most appropriate answer is Phosphofructokinase-1, which is affected by the rising concentration of AMP to stimulate the overall pathway of Glycolysis.

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Case details

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.

Basic concept

2,3-BPG, is  the most concentrated organophosphate in the erythrocytes.  It is synthesized from Glucose by Luebering-Rapoport pathway which is a diversion from the main glycolytic pathway.

Luebering-Rapoport pathway

In erythrocytes, the reaction catalyzed by phosphoglycerate kinase may be bypassed, to some extent by the reaction of bisphosphoglycerate mutase, which catalyzes the conversion of 1,3-bisphosphoglycerate to 2,3-bisphosphoglycerate, followed by hydrolysis to 3-phosphoglycerate and Pi, catalyzed by 2,3-bisphosphoglycerate phosphatase (Figure-1). This alternative pathway involves no net yield of ATP from glycolysis. However, it does serve to provide 2,3-bisphosphoglycerate, which binds to hemoglobin, decreasing its affinity for oxygen, and so making oxygen more readily available to tissues .

 RL Shunt

Figure-1- RL Shunt- Formation and breakdown  of 2,3-bisphosphoglycerate. 

Significance of 2,3 BPG

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 (figure-2).

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.

States of Hb 

Figure-2- Oxygen binding and unloading. 2,3 BPG being negatively charged binds to positively charged Histidine and Lysine residues of hemoblobin, stabilizing the tight binding state (T) or low affinity state that promotes oxygen unloading.

2,3-BPG can help to  prevent tissue hypoxia in conditions where it is most likely to occur. Conditions of low tissue oxygen concentration such as high altitude (2,3-BPG levels are higher in those acclimated to high altitudes), airway obstruction, anemia or congestive heart failure will tend to cause RBCs to generate more 2,3-BPG in their effort to generate energy by allowing more oxygen to be released in tissues deprived of oxygen.

This release is potentiated by the Bohr effect in tissues with high energetic demands.

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 (figure-3).

 Adult and fetal Hb

Figure-3- Adult Hb and fetal Hb.

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 (figure-4).

 Flow of oxygen

Figure-4-Flow of oxygen from maternal Hemoglobin to fetal hemoglobin. Maternal Hb has low affinity for oxygen (due to bound 2,3 BPG) whereas the affinity of fetal Hb for oxygen is more due to unavailable binding sites for 2,3 BPG.








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Case details

A 55-year-old man suffers from cirrhosis of liver. Toxins such as ammonia are not properly metabolized by the liver and can damage brain. Which of the following compounds should be in highest concentration in brain as a result of detoxification of ammonia?

A. Alpha Ketoglutarate



D.GABA (Gamma amino butyric acid)

E. Asparagine 

The correct answer is C-Glutamine

Basic concept

Ammonia is produced as a result of various metabolic activities. It has to be detoxified immediately else can prove toxic to the brain cells and other tissues.

Mechanism of Ammonia detoxification

1) The first line of defense- Glutamate condenses with ammonia to produce Glutamine. The reaction catalyzed can be represented as follows (figure-1)


 Reaction catalyzed by Glutamine synthetase

Figure-1- Glutamate to Glutamine conversion is catalyzed by Glutamine synthetase. It is an energy requiring process, ATP acts as a source of energy.

Glutamine is transported to liver.The nitrogen of glutamine can be converted to urea in the liver (Figure -2).

 Reaction by Glutaminase

Figure-2- Hydrolytic release of the amide nitrogen of glutamine as ammonia,is catalyzed by glutaminase in liver. Ammonia thus released is detoxified producing urea.

2) Second line of defense

In conditions of excess ammonia release the second line of defense involves the formation of Glutamate from Alpha ketoglutarate (intermediate of TCA cycle) that can be subsequently used for Glutamine synthesis as explained above . The reaction can be represented as follows :


 role of glutamate

Figure-3- Ammonia detoxification. In the first step the reaction is catalyzed by Glutamate dehydrogenase, a unique enzyme that can use either of NAD+ or NADP+ as a coenzyme. The reaction is reversible, but in the liver the reaction is directed towards  Alpha ketoglutarate formation and the released ammonia is used for urea synthesis. Glutamate to Glutamine is an energy requiring irreversible reaction catalyzed by Glutamine synthetase.

Implication of Ammonia Intoxication

In the process of detoxification of Ammonia, some of the biologically important compounds are depleted whereas some are produced in highly excess amounts to cause toxicity.

Glutamine is the final product of detoxification which is transported out of the brain cells in exchange with tryptophan. Tryptophan is a precursor of Serotonin, the excess of which causes a state of hyper excitation.

Glutamate and Alpha ketoglutarate are depleted in this process of ammonia detoxification.

Decreased Glutamate, produces less GABA(figure-4), which is an inhibitory transmitter, thus again the result is a state of hyper excitation.

 GABA synthesis

Figure-4- Glutamate is decarboxylated to produce Gamma AminoButyric acid (GABA). The reaction is catalyzed by Glutamate decarboxylase, that requires the presence of Vitamin B6.

The second compound that is decreased is Alpha ketoglutarate, an intermediate of TCA cycle and the depletion of which causes overall suppression of TCA cycle resulting in a state of energy depletion.

The symptoms of ammonia intoxication include- Slurring of speech, blurring of vision, tremors, convulsions,coma and death. The biochemical basis of such symptoms is energy depletion and hyperexcitation due to excess serotonin formation and decreased GABA synthesis. In Cirrhosis of liver, the conversion of ammonia to urea is impaired, the resulting hyperammonemia), proves toxic to brain. In severe liver disorders blood and CSF Glutamine levels are increased (a diagnostic feature of hepatic encephalopathy).

As regards other options

A. Alpha Ketoglutarate- its concentration is decreased, as explained above.

B. Glutamate- Its concentration is also decreased

C. Glutamine concentration is increased.

D.GABA (Gamma amino butyric acid)- levels are decreased.

E. Asparagine- levels are not much affected in ammonia intoxication.




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Case Details

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 correct answer is B- Mevalonate.

Basic concept

Biosynthesis of cholesterol

The biosynthesis of cholesterol may be divided into five steps:

(1) Synthesis of Mevalonate from acetyl-CoA.

(2) Formation of Isoprenoid units from Mevalonate by loss of CO2.

(3) Condensation of six isoprenoid units forms Squalene.

(4) Cyclization of Squalene gives rise to the parent steroid, Lanosterol.

(5)Formation of cholesterol from lanosterol.

Always remember the formula

2+2= 4

4+2= 6






Details- (See the figure-1 and 2)

i) 2+2= 4

  • Initially, two molecules of acetyl-CoA (2+2) condense to form Acetoacetyl-CoA (4) catalyzed by cytosolic thiolase.

ii) 4+2=6

  • Acetoacetyl-CoA condenses with a further molecule of acetyl-CoA (4+2) catalyzed by HMG-CoA synthase to form HMG-CoA that is reduced to Mevalonate (6) by NADPH catalyzed by HMG-CoA reductase (figure-1).

Reaction catalyzed by HMG Co A reductase

Figure-1- Reaction catalyzed by HMG Co A reductase. HMG Co A reductase is inhibited by Statins by the mechanism of competitive inhibition and by bile acid, cholesterol and Mevalonate by feedback inhibition


  • The synthesis of Mevalonate is the committed step in cholesterol formation.
  • The enzyme catalyzing this irreversible step, 3-hydroxy-3-methylglutaryl CoA reductase (HMG-CoA reductase), is an important control site in cholesterol biosynthesis,
  • It is the site of action of the most effective class of cholesterol-lowering drugs, the HMG-CoA reductase inhibitors (statins).

iii) 6-1=5

  • Decarboxylation (6-1 =5) yields Isopentenyl pyrophosphate (5), an activated isoprene unit that is a key building block for many important biomolecules (figure-2).

iv) 5+5=10

  • Isopentenyl pyrophosphate is isomerized by a shift of the double bond to form dimethylallyl pyrophosphate that condenses with another molecule of Isopentenyl pyrophosphate (5+5 =10) to form the ten-carbon intermediate Geranyl pyrophosphate (10).

v) 10+5=15

A further condensation with Isopentenyl pyrophosphate forms Farnesyl pyrophosphate (15). 

v) 2×15=30

Two molecules of Farnesyl pyrophosphate (15+15) condense at the pyrophosphate end to form Squalene (30)

vi) 30-3=27

  • Squalene can fold into a structure that closely resembles the steroid nucleus
  • Before ring closure occurs, Squalene is converted to Squalene 2, 3-epoxide by a mixed-function oxidase in the endoplasmic reticulum, Squalene peroxidase.
  • The methyl group on C14 is transferred to C13 and that on C8 to C14 as Cyclization occurs, catalyzed by oxidosqualene: lanosterol cyclase (figure-2).

The newly formed cyclized structure is Lanosterol.

  • The formation of cholesterol from lanosterol takes place in the membranes of the endoplasmic reticulum and involves changes in the steroid nucleus and side chain.
  • 3 carbon atoms are lost (30-3).
  • The double bond at C8–C9 is subsequently moved to C5–C6 in two steps, forming Desmosterol.
  • Finally, the double bond of the side chain is reduced, producing cholesterol.

 Steps of cholesterol synthesis

Figure-2- Steps of cholesterol biosynthesis

As regards other options

A. Isopentenyl pyrophosphate (IPP) – is formed from Mevalonate- PP

C. Geranyl pyrophosphate (C10) is formed by condensation of Isopentenyl pyrophosphate and Dimethyl allyl pyrophosphate

D. Farnesyl pyrophosphate (C15) is produced by condensation of Geranyl pyrophosphate and IPP

E. Cholesterol- is the end product of this pathway.


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Case details

A 30- year-old pregnant woman has a sugar craving and consumes a hot fudge sundae. Her serum glucose level increases, which causes release of Insulin. Insulin is known to increase the activity of acetyl co A carboxylase, the rate limiting enzyme of fatty acid biosynthesis. Which of the following best describes this regulatory enzyme?

A. It is activated by carboxylation

B. It catalyzes a reaction that condenses an acetyl group with malonyl group

C. It catalyzes a reaction that requires biotin and ATP

D. It converts Malonyl co A to Acetyl co A

E. It is activated by malonyl Co A.

The correct answer is- C, It catalyzes a reaction that requires Biotin and ATP.

Basic concept

Acetyl co A carboxylase is the first enzyme of fatty acid biosynthesis that catalyzes the carboxylation of Acetyl co A to Malonyl CoA . The reaction catalyzed can be represented as follows :

 Reaction catalyzed by Acetyl co A carboxylase

Figure- Reaction catalyzed by Acetyl co A carboxylase. It is a carbon-carbon condensation, an energy requiring process.  ATP is the source of energy, and biotin is the coenzyme for this carboxylation process.

Acetyl co A carboxylase is the rate controlling enzyme in the pathway of lipogenesis. It is regulated by-

1) Allosteric modification-Acetyl-CoA carboxylase is an allosteric enzyme and is activated by citrate, which increases in concentration in the well-fed state and is an indicator of a plentiful supply of acetyl-CoA. Citrate converts the enzyme from an inactive dimer to an active polymeric form, with a molecular mass of several million. Inactivation is promoted by long-chain acyl-CoA molecules.

2) Feedback inhibition -The enzyme is inhibited by malonyl co A and palmitoyl co A, an example of negative feedback inhibition by a product of a reaction. Thus, if acyl-CoA accumulates because it is not esterified quickly enough or because of increased lipolysis or an influx of free fatty acids into the tissue, it will automatically reduce the synthesis of new fatty acid

3) Covalent Modification- Acetyl-CoA carboxylase is also regulated by hormones such as glucagon, epinephrine, and insulin via changes in its phosphorylation state

4) Induction and Repression-Insulin is an important hormone causing gene expression and induction of enzyme biosynthesis, and glucagon (via cAMP) antagonizes this effect. These mechanisms for longer-term regulation of lipogenesis take several days to become fully manifested. Prolonged consumption of high calorie or high carbohydrate diets causes an increase in acetyl co A carboxylase concentration by increasing the gene expression (induction). Conversely, a low-calorie diet or fasting causes a reduction in fatty acid synthesis by decreasing the synthesis of acetyl co A carboxylase (repression) .

Thus Insulin stimulates the synthesis of fatty acids by covalent modification and induction . It also provides glycerol-3-Phosphate through glycolytic pathway that can be used for esterification. That is the reason for weight gain in individuals on insulin therapy.

Based on the similar concept, obesity related with high carbohydrate diet can also be explained.

Excessive carbohydrate ingestion promotes triglyceride synthesis through following mechanisms-

1) Glycolysis yields pyruvate and hence Acetyl coA which is a precursor for fatty acid biosynthesis.

2) Glycolysis provides glycerol-3-p through dihydroxyacetone phosphate, that is used for esterification

3) HMP pathway provides NADPH which can be used for reductive biosynthesis.

All the pathways of glucose utilization are stimulated by Insulin and by these mechanisms, fatty acids are synthesized and esterified with glycerol to produce triglycerides. The adipose mass increases and the person becomes obese. That is the reason, low carbohydrate diet is recommended to those who want to lose weight.

As regards other options

Acetyl co A Carboxylase, the enzyme for fatty acid biosynthesis-

  • It is not activated by carboxylation
  • It  does not catalyze a reaction that condenses an acetyl group with malonyl group.
  • It does not convert Malonyl co A to Acetyl co A, instead Acetyl co A is converted to Malonyl co A .
  • It is  not activated by malonyl Co A. Malonyl co A, the product of this reaction inhibits this enzyme by feedback inhibition.





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Case details

An infant presents with lethargy, sweating, and irritability. He is admitted to the Pediatrics unit, where the attending pediatrician notices that his symptoms are pronounced when the feeding is delayed. After a series of tests, the child has been diagnosed with an enzyme deficiency that catalyzes the first step in the β-oxidation spiral, which is,

A. Fatty acid synthase

B. Acyl co A dehydrogenase

C. Enoyl Co A hydratase

D. β- Hydroxyacyl co A dehydrogenase

E. Thiolase

The correct answer is- Acyl co A dehydrogenase.

Acyl co A dehydrogenase is the first enzyme of β-oxidation spiral.

Basic concept

A saturated acyl Co A is degraded by a recurring sequence of four reactions (Figure)

1) Oxidation by flavin adenine dinucleotide (FAD)

2) Hydration,

3) Oxidation by NAD+, and

4) Thiolysis by Co A

The reaction catalyzed by acyl-CoA dehydrogenase  involves the removal of two hydrogen atoms from the 2( α)- and 3( β)-carbon atoms. The enzyme requires FAD as a coenzyme. (Figure).

Fatty acid oxidation is an important source of energy. Impaired activity of Acyl co A dehydrogenase or deficiency of any of the enzymes of beta oxidation leads to an energy deficit.

In the given case study, the infants presents with symptoms of hypoglycemia, due to an imbalance between demand and supply of glucose. In the conditions of impaired beta oxidation of fatty acids, glucose becomes the major source of energy, since glycogen stores are not well-developed and the substrates of gluconeogenesis are also not adequately available, the diet remains the only source of glucose. Any delay in feeding results in precipitation of symptoms of hypoglycemia.

Carnitine deficiency also produces hypoglycemia, due to similar reason of non- availability of energy from fatty acid oxidation (Carnitine acts as a transporter of fatty acids from cytoplasm to mitochondrial matrix since fatty  acids due to long hydrophobic chain cannot pass through inner mitochondrial membrane).


 Beta oxidation spiral

Figure- Steps of beta oxidation of fatty acids. The fatty acyl chain is shortened by two carbon atoms as a result of these reactions, and FADH2, NADH, and acetyl Co A are generated. Because oxidation is on the β carbon and the chain is broken between the α (2) – and β (3)-carbon atoms—hence the name – β oxidation.

As regards other options

Fatty acid synthase is a multienzyme complex involved in fatty acid synthesis.

Enoyl Co A hydratase- is an enzyme of beta oxidation (Figure); it catalyzes the second step of beta oxidation (hydration of fatty acids)

β- Hydroxyacyl co A dehydrogenase and Thiolase are also enzymes of beta oxidation spiral, but they catalyze the third and fourth steps (oxidation by NAD+ and thiolytic cleavage) respectively.

Thus the enzyme that catalyzes the first step in the β-oxidation spiral is Acyl co A dehydrogenase.





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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. Increase Insulin release

E. Decrease muscle protein break down.


The correct answer is- B- Gluconeogenesis.

Significance of glucose

Glucose is a universal fuel molecule required for all cell types especially brain cells and the cells lacking mitochondria.

The blood glucose concentration is maintained in a range of 60-100 mg/dl to provide a constant supply of glucose to all cells especially brain cells and the cells lacking mitochondria, the latter are totally dependent upon glucose for their energy needs. Brain cells can utilize ketone bodies alternatively under conditions of prolonged fasting/starvation but glucose remains the preferred source of energy.

Fuel molecules during starvation

During starvation the energy needs are fulfilled by three types of fuels, glucose, fatty acids and ketone bodies.

Fatty acids due to long hydrophobic chains can’t cross through the Blood brain barrier (BBB), thus cannot be utilized by the brain. Ketone bodies although being relatively polar can cross through the blood brain barrier, yet they are utilized only under conditions of glucose deprivation. After several weeks of starvation, ketone bodies become the major fuel of the brain.

The oxidation of fatty acids and ketone bodies takes place in the mitochondria; hence the cells lacking mitochondria are left with no alternative fuels options except glucose to utilize. Thus an optimum glucose concentration has to be maintained for the proper functioning of these cells.

The sources of glucose include

1)      Diet

2)      Glycogen degradation, and

3)      Gluconeogenesis

Under conditions of inadequate dietary supply, the blood glucose levels are maintained initially by glucose supply from the stores (degradation of glycogen), the glycogen stores get exhausted within 14-16 hours. Thereafter the glucose is synthesized from the non-carbohydrate precursors (gluconeogenesis) such as lactate (the waste from the muscle), glycerol (the waste from the adipose tissue), and carbon skeleton of glucogenic amino acids (mainly obtained from the breakdown of muscle protein). The intermediates of TCA cycle and propionyl co A also serve as substrates for gluconeogenesis. The constant supply of substrates of gluconeogenesis, sustain the life of an individual.

In the given situation, after 30 hours of fasting, the only source of glucose is gluconeogenesis. Glycogenolysis cannot be expected after 30 hours.

As regards other options

  • Triacylglycerol synthesis cannot take place during conditions of starvation. Starvation is a state of catabolism. Glucagon the predominant hormone of this state, promotes adipolysis. Triacylglycerol synthesis takes place in the well fed state, in the presence of Insulin.
  • Increased insulin release is also incorrect, maximum insulin release occurs in the well fed state to promote utilization of glucose and to build up the body stores from the surplus nutrients. Under conditions of prolonged fasting as stated above, the insulin release is minimal, there is maximum release of glucagon to maintain blood glucose homeostasis.
  • Based on the similar concept of maintaining blood glucose homeostasis, decrease muscle protein breakdown during fasting is also inappropriate; there is rather a need for the constant supply of substrates of gluconeogenesis. Muscle protein breakdown can provide the carbon skeleton of amino acids, which can be further utilized for glucose production. The decreased insulin to glucagon ratio, and the decreased availability of circulating substrates, make this period of nutritional deprivation a catabolic state, characterized by initial degradation of glycogen, followed by degradation of triacylglycerol and protein.

Blood Glucose homeostasis (Summary)

Nutritional Status Source of blood glucose Cells using glucose as a fuel Major fuel of the brain
Well fed state Dietary glucose All cells Glucose
Post absorptive state Hepatic glycogenolysis (mainly) and gluconeogenesis All  cells except liver and skeletal muscle.

Adipose cells use at a reduced rate.

Fasting Hepatic  and renal gluconeogenesis and some hepatic glycogenolysis Brain cells and  the cells lacking mitochondria, Glucose and ketone bodies
Prolonged fasting / Starvation Gluconeogenesis (mainly) The cells lacking mitochondria Only ketone bodies


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An 18 month-old- child is left unattended while in the kitchen and ingests a small portion of rat poison found in the cupboard found under the sink. The ingredient fluoroacetate reacts with Oxaloacetate to form fluorocitrate. Which pathway of the body is inhibited by this poison?

A. Glycolysis

B. TCA cycle

C. Fatty acid oxidation

D. Fatty acid synthesis

E. HMP pathway

The right answer is -B) TCA cycle.

Fluoroacetate is structurally similar to acetate, which has an important role in cellular metabolism.

Fluoroacetate disrupts the citric acid cycle (also known as the Krebs cycle or TCA cycle- Tricarboxylic acid cycle) by combining with coenzyme A to form Fluoroacetyl CoA, which reacts with oxaloacetate in the presence of citrate synthase (host enzyme) to produce fluorocitrate (Figure-1).

 Activation of Fluoroacetate

Figure-1- Activation of fluoroacetate is brought by Citrate synthase, an enzyme of TCA cycle

Fluorocitrate binds very tightly to Aconitase, to inhibit its action, (Figure-2) thereby inhibiting the citric acid cycle as a whole. This is an example of suicidal inhibition.

 TCA cycle

Figure-2- TCA cycle and the action of Aconitase

The direct effect of Fluoroacetate is on TCA cycle, but indirectly Glycolysis and other pathways are also inhibited.

Effect on Glycolysis

This inhibition results in an accumulation of citrate in the blood. Citrate and fluorocitrate are allosteric inhibitors of phosphofructokinase-1 (PFK-1), a key enzyme in the breakdown of Glucose. As PFK-1 is inhibited cells are no longer able to metabolize carbohydrates, depriving them of energy.

Effect on fatty acid oxidation

Inhibition of TCA cycle inhibits fatty acid oxidation also due to non availability of oxidized coenzymes, and incomplete utilization of Acetyl co A. But these are late implications. TCA cycle is vital to life. Death occurs immediately after its suppression. Directly Fluoroacetate has no effect on enzymes of fatty acid oxidation.

 Similarly fatty acid synthesis and HMP pathway are not directly affected by Fluoroacetate.


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As electrons are received and passed down the transport chain, the electron carriers are first reduced with the acceptance of the electron and then oxidized with loss of the electron. A patient poisoned by which of the following compounds has the most highly reduced state of most of the respiratory chain carriers?

A. Antimycin A

B. Rotenone

C. Carbon monoxide

D. Puromycin

E. Chloramphenicol

The right answer is (c) – Carbon monoxide.

Basic concept

The electron transport chain contains three proton pumps linked by two mobile electron carriers. At each of these three sites (NADH-Q reductase, cytochrome c reductase and cytochrome oxidase); the transfer of electrons down the chain powers the pumping of the protons across the inner mitochondrial membrane (figure). As electrons pass through complexes I, III, and IV, protons are pumped across the inner membrane from the matrix to the intermembrane space. This sets up an electrochemical gradient consisting of a proton gradient (chemical), and a membrane potential (because the proton carries a positive charge). The electrochemical gradient represents a form of stored energy derived from the oxidation of NADH (or succinate). A physically distinct complex (ATP synthase or complex V) at a separate location in the inner membrane can exploit the electrochemical gradient to carry out the endergonic ATP synthesis (by oxidative phosphorylation).

 Inhibition of electron transport chain

The blockage of electron transfers by specific point inhibitors leads to a buildup of highly reduced carriers behind the block because of the inability to transfer electrons across the block.

Antimycin A blocks the complex III, Rotenone blocks complex I and Carbon monoxide (As well as cyanide, azide, hydrogen sulfide) block complex IV. Therefore a carbon monoxide inhibition leads to a highly reduced state of all the carriers of the chain. Puromycin and Chloramphenicol are inhibitors of protein synthesis and have no direct effect upon the electron transport chain.


Figure-1- Flow of electrons in the electron transport chain and the oxidative phosphorylation


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An unskilled worker in a water garden was sent to sweep up a spill of a white powder in the storage shed. Later he was found with labored breathing and convulsions. On further examination, the white powder was identified as rotenone. Respiratory distress is induced on rotenone exposure because it inhibits the complex that catalyzes which of the following?

A. Electron transfer from NADH to coenzyme Q

B. Oxidation of coenzyme Q

C. Reduction of cytochrome c

D. Electron transfer from cytochrome c to cytochrome a1/a3

E. Electron transfer from cytochrome a1/a3 to oxygen


The right answer is A) – Electron transfer from NADH to coenzyme Q

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 four large protein complexes;

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

ii) Succinate Q reductase (Complex II)-Some substrates with more positive redox potentials than NAD+/NADH (e.g., succinate) pass electrons to Q via succinate Q reductase (Complex II), rather than Complex I.

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

iv) Cytochrome c oxidase (Complex IV), which completes the chain, passing the electrons to O2 and causing it to be reduced to H2O (Figure-1).

Mobile complexes

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.


Figure-1- Components of electron transport chain. The three complexes I, III and IV acts as proton pumps, ATP is synthesized when protons flow back to the mitochondrial matrix through ATP synthase complex.

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

Electron flow in ETC (figure-2)

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-2). 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.

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. As is the case with Complex I, Complex III contains a series of proteins that transport electrons. The electrons are removed from Complex III 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).

This transfer of four electrons from cytochrome c to O2 involves two heme groups, a and a3, and Cu. 

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.


 Flow of electrons

Figure-2- Flow of electrons in the electron transport chain

Site specific inhibitors -Specific inhibitors of electron transport for example, rotenone and amobarbital block electron transfer in NADH-Q oxidoreductase and thereby prevent the utilization of NADH as a substrate. The central portion of the rotenone structure resembles the isoalloxazine ring of the FMN molecule, and when it binds to complex I, rotenone prevents the transfer of electrons from NADH to coenzyme Q. In contrast, electron flow resulting from the oxidation of succinate is unimpaired, because these electrons enter through QH2, beyond the block. Malonate is a competitive inhibitor of Complex II

BAL (British Anti Lewisite), Antimycin A interferes with electron flow from cytochrome b in Q-cytochrome c oxidoreductase.

Furthermore, electron flow in cytochrome c oxidase can be blocked by 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.





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A 45-year-old morbidly obese woman has been attempting to lose weight using a low- carbohydrate diet. After 2 months of little success, she confides in her son that she does add glucose to her coffee in the morning and after dinner but feels only some of this will be absorbed and should not be the cause of her limited success. Her son, a medical student, states that glucose is almost completely absorbed from the gut. What type of transport does glucose utilize for gastro intestinal absorption?

A. Active- Carrier mediated, against the concentration gradient and energy dependent

B. Facilitated- Carrier mediated, down the concentration gradient

C. Passive- Down the concentration gradient

D. Active and facilitated

E. Passive and facilitated

The correct answer iso D- Active and facilitated.

There are two separate mechanisms for the absorption of monosaccharides in the small intestine- Active and facilitated transport. Active transport is an energy dependent carrier mediated transport that involves transport of glucose against the concentration gradient. The carrier protein in active transport has two binding sites, one for glucose and the other for sodium. Thus it is a sodium dependent transport. Sodium transport is down the concentration gradient whereas the glucose transport is against the concentration gradient. Glucose and galactose are absorbed by a sodium-dependent process. They are carried by the same transport protein (SGLT 1), and compete with each other for intestinal absorption.

The second mechanism, facilitated transport also requires a carrier protein to speed up the process, but the energy is not invested in this transport and the flow is down the concentration gradient. Other monosaccharides (mainly) including glucose (but to a lesser extent) are absorbed by carrier-mediated facilitated diffusion. Fructose is mainly transported by Facilitated transport using GLUT-5 transport. Because they are not actively transported, fructose and sugar alcohols are only absorbed down their concentration gradient, and after a moderately high intake, some may remain in the intestinal lumen, acting as a substrate for bacterial fermentation.

Glucose is a polar molecule; the passive diffusion across the intestinal membrane is very-very slow. Thus only a small amount of glucose is absorbed by a passive diffusion.

 Passive versus Active transport

Figure- Passive diffusion versus active transport

Summary of Glucose transport

Features Passive diffusion Facilitated diffusion Active transport
Concentration gradient Down the concentration gradient from high to low. Down the concentration gradient from high to low. Against a concentration gradient from low to high
Energy expenditure none none Energy expenditure is in the form of ATP
Carrier protein/ transporter Not required required required
Speed Slowest mode Fast Fastest mode

Clinical significance

  • In SGLT- 1 deficiency, glucose is left unabsorbed and gets excreted in feces. Galactose is also malabsorbed since it is transported through the same transporter.
  • SGLT- 2, a similar glucose transporter is present in the renal tubular cells; the filtered glucose is reabsorbed back by this transporter. In its deficiency, glucose is not reabsorbed back, and is lost in urine, causing glycosuria.


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Chylomicrons, intermediate- density- lipoproteins (IDLs), low- density- lipoproteins (LDLs), and very low-density lipoproteins (VLDLs) are all serum lipoproteins. What is the correct ordering of these particles from the lowest to the highest density?

A. LDLs, IDLs, VLDLs and chylomicrons

B. Chylomicrons, VLDLs, IDLs, LDLs

C.VLDLs, IDLs, LDLs and Chylomicrons

D.LDLs, VLDLs, IDLs, Chylomicrons

E. Chylomicrons, IDLS, VLDLs and LDLs.

The correct answer is B- Chylomicrons, VLDLs, IDLs, and LDLs

Based on density Lipoproteins can be classified as follows

Since fat is less dense than water, the density of a lipoprotein decreases as the proportion of lipid to protein increases.  Lipoproteins with high lipid content will have low density and so float on centrifugation. Those with high protein content sediment easily and have a high density. They are separated by Ultracentrifugation. Depending upon the floatation constant (Sf), Five major groups of lipoproteins have been identified that are important physiologically and in clinical diagnosis. These are

 (i) Chylomicrons, derived from intestinal absorption of triacylglycerol and other lipids; Density is generally less than 0.95 while the mean diameter lies between 100- 500 nm

 (ii) Very low density lipoproteins (VLDL), derived from the liver for the export of triacylglycerol; density lies between 0.95- 1.006 and the mean diameter lies between 30-80 nm.

 (iii) Intermediate density lipoproteins (IDL) are derived from the catabolism of VLDL, with a density ranging intermediate between Very low density and Low density lipoproteins i.e. ranging between 1.006-1.019 and the mean diameter ranges between 25-50 nm.

 (iv)Low-density lipoproteins (LDL), representing a final stage in the catabolism of VLDL; density lies between 1.019-1.063 and mean diameter lies between 18-28 nm

 (iv) High-density lipoproteins (HDL), involved in cholesterol transport and also in VLDL and chylomicron metabolism. Density ranges between 1.063-1.121 and the mean diameter varies between 5-15 nm.

 Classification of lipoproteins

Figure-  The relationship of density and mean diameter of lipoproteins


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An -8 year-old boy is seen by an ophthalmologist for difficulties in seeing in all visual fields as well as slow eye movements. The ophthalmologist finds pigmentary retinopathy and ophthalmoplegia. The child is suspected to have Kearns- Sayre syndrome, a disorder due to a mutation in complex II of ETC. The electron transport from which substance would be impaired?

A. Malate

B. Isocitrate

C. Succinate

D. Pyruvate

E. Alpha Keto glutarate

Answer- The correct answer is C- Succinate.

Electrons flow through the respiratory chain through a redox span of 1.1 V from NAD+/NADH to O2/2H2O passing through three large protein complexes;

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 O2 and causing it to be reduced to H2O .

Some substrates with more positive redox potentials than NAD+/NADH (e.g., succinate) pass electrons to Q via, 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.


Figure- Flow of electrons through the respiratory chain complexes.

Malate, Isocitrate and alpha ketoglutarate are intermediates of TCA cycle, the electrons flow from them to NAD+ forming NADH that is regenerated back to its oxidized form by passing electrons to complex I of ETC. Similarly electron flow from Pyruvate is also through complex I.

Electron from Succinate pass to FAD forming FADH2 that is regenerated back to its oxidized form after passing its electrons to complex II of ETC.

Thus the correct answer for the given problem is Succinate.


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A young infant, who was nourished with a synthetic formula, had a sugar in the blood and urine. This compound gave a positive reducing sugar test but was negative when measured with glucose oxidase (specific test for detection or estimation of Glucose). Treatment of blood and urine with acid (which cleaves glycosidic bonds) did not increase the amount of reducing sugar measured. Which of the following compounds is most likely to be present in this infant’s blood and urine?
A. Glucose
B. Fructose
C. Maltose
D. Sorbitol
E. Lactose

The right answer is fructose.

Reducing sugars are usually detected by Benedict’s reagent, which contains copper sulphate, sodium citrate and sodium carbonate. Sodium carbonate makes the medium alkaline. Copper sulphate furnishes Cu2+ ions and sodium citrate prevents the precipitation of cupric ions as cupric hydroxide by forming a loosely bound cupric- sodium –citrate complex which on dissociation gives a continuous supply of cupric ions.

Benedict’s test


Carbohydrates with free aldehyde or ketone groups have the ability to reduce solutions of various metallic ions. Reducing sugars under alkaline conditions tautomerise and form enediols. Enediols are powerful reducing agents. They reduce cupric ions to cuprous form and are themselves converted to sugar acids. The cuprous ions combine with OH- ions to form yellow cuprous hydroxide which upon heating is converted to red cuprous oxide.


Take 5 ml of Benedict’s reagent. Add 8 drops of carbohydrate solution. Boil over a flame or in a boiling water bath for 2 minutes. Let the solution cool down.


Benedict’s test is a semi quantitative test. The color of the precipitate gives a rough estimate of a reducing sugar present in the sample (figure-1)

Green color- Up to 0.5 g %(+)

Green precipitate -0.5-1.0 g %(++)

Yellow precipitate -1.0-1.5 g %(+++)

Brick red precipitate- >2.0 G% (++++)

Negative benedict's testPositive benedict's test

(-ve)                (+ve)

Figure– The positive test is given by reducing sugars. The color of the precipitate determines the rough estimate of the reducing sugar present in the given sample.

Fehling test is an alternative to Benedict’s test. It differs from Benedict’s test in that it contains sodium potassium tartrate in place of Sodium citrate and potassium hydroxide as an alkali in place of sodium carbonate in Benedict’s reagent. It is not a preferred test over Benedict’s test since the strong alkali present causes caramelisation of the sugars; hence it is less sensitive than Benedict’s reagent.

Positive Benedict’s test for urine signifies Glycosuria.

Glycosuria is a non-specific term. Glucosuria, lactosuria, galactosuria, pentosuria and fructosuria denote the presence of specific sugars in urine.

Causes of Glycosuria are:

a. Renal glycosuria

b. Diabetes mellitus

c. Alimentary glucosuria

d. Hyperthyroidism, hyperpituitarism and hyperadrenalism

e. Stress, severe infections, increased intracranial pressure

Lactosuria– in lactose intolerance

Galactosuria– in galactosemia

Fructosuria– in hereditary fructose intolerance

Pentosuria – in essential pentosuria

Examples of non-carbohydrate substances which give a positive Benedict’s reaction are:

a) Creatinine

b) Ascorbic acid

c) Glucuronates

d) Drugs: Salicylates, PAS and Isoniazid.

Glucose oxidase test is a specific enzymatic method for the determination and estimation of glucose present in a given sample. True glucose can be estimated by this method.

As regards other options

Glucose cannot be present since specific test is negative.

Sorbitol is non reactive to reduction test.

Maltose and lactose would have caused increase in the amount of reducing sugar upon acid hydrolysis.

Hence it is fructose which is reducing in nature but non reactive to glucose oxidase.


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The general structure of chromatin has been found to be remarkably similar in the cells of all eukaryotes. The most abundant proteins associated with eukaryotic DNA (somewhat more than half its mass) are histones, a family of basic proteins rich in the positively charged amino acids that interact with the negatively charged phosphate groups in DNA.

Which of the following amino acids acts as a component of histones and a precursor for Nitric oxide (NO)?

A. Arginine

B. Histidine

C. Lysine

D. Asparagine

E. Leucine

The correct answer is- A- Arginine.

Histones are nucleoproteins, rich in basic amino acids such as Arginine and Lysine. The net positive charge (due to basic amino acids), helps in binding of histones to negatively charged groups of DNA. Histones help in DNA packaging. The nucleosomes, the first level of DNA organization, are composed of DNA wound around a collection of histone molecules. Nucleosomes contain four types of histones: H2A, H2B, H3, and H4. H1 histones are the ones least tightly bound to chromatin (figure-1)

Role of histones

Figure-1- Role of histones in nucleosome formation

Arginine, a semi essential amino acid, is also a precursor for Nitric oxide that acts as a vasodilator and a smooth muscle relaxant. Nitric oxide is synthesized by nitric oxide synthase (NOS). The reaction catalyzed can be expressed as follows (figure-2)

Synthesis of Nitric oxide

Figure-2-Synthesis of nitric oxide, the reaction is catalyzed by Nitric oxide

As regards other options

B. Histidine- Histidine carries a net positive charge at physiological pH but it is not a predominant component of histones or a precursor for Nitric oxide.

C. Lysine- Lysine is a predominant component of histones, but it is not a precursor for nitric oxide.

D. Asparagine is an amide group containing amino acid; it is neither a component of histones nor a precursor for nitric oxide.

E. Leucine is also not a suitable option due to similar reasons.


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A 7-month-old baby girl, the second child born to unrelated parents was brought to Pediatrics outdoor department. History revealed that she did not respond well to breast-feeding and was changed entirely to a formula based on cow’s milk at 4 weeks. Between 7 and 12 weeks of age, she was admitted to the hospital twice with a history of screaming after feeding, but was discharged after observation without a specific diagnosis. Elimination of cow’s milk from her diet did not relieve her symptoms; her mother reported that the screaming bouts were worse after the child drank juice and that she frequently had gas and a distended abdomen. The child was diagnosed having ‘Hereditary fructose intolerance’. The mother of the child was instructed to eliminate fructose containing foods from the child’s diet and was strictly instructed to feed milk without table sugar. The table sugar (sucrose), a disaccharide, contains glucose and fructose linked as:

A. O-α-D-glucopyranosyl-(1->6)-α -D- fructofuranoside

B. O-β-D-glucopyranosyl-(1->6)-α -D- fructofuranoside

C. O-α-D-glucopyranosyl-(1->2)-β -D-fructofuranoside

D. O-α-D-glucopyranosyl-(1->2)-α -D-fructofuranoside

E. None of the above.

The correct answer is C- O-α D-glucopyranosyl-(1->2)-β -D-fructofuranoside. The onset of symptoms after ingestion of juice (fructose or fructose containing diet) is a sign of hereditary fructose Intolerance’.

Hereditary fructose intolerance is caused by deficiency of Aldolase B, the enzyme required for the metabolism of fructose. These patients are healthy and asymptomatic until fructose or sucrose (table sugar) is ingested (usually from fruit, sweetened cereal, or sucrose-containing formula). Elimination of dietary fructose is both a compulsory and therapeutic step.

In patients who are ill, elimination may also serve as a diagnostic test because all symptoms should completely resolve. With this treatment, as the patient matures, symptoms become milder, even after fructose ingestion, and the long-term prognosis is good.

Table sugar (sucrose) is a source of fructose and in Sucrose, the anomeric carbon atoms of a glucose unit and a fructose unit are joined; the configuration of this glycosidic linkage is α-for glucose and β-for fructose (figure).


Structure of sucrose

Figure- Structure of sucrose

Sucrose can be cleaved into its component monosaccharides by the enzyme sucrase.

An overview of properties of sucrose

  • Sucrose has no free reactive group because the anomeric carbons of both monosaccharides units are involved in the glycosidic bond. Therefore, sucrose neither shows reducing nor mutarotation characters.
  • Sucrose is called invert sugar because the optical activity of sucrose (dextrorotatory) is inverted after hydrolysis (by an acid or an enzyme (invertase or sucrase) into an equimolar mixture of its two components glucose (+52.5) and fructose (-92.5) and the optical activity of the mixture becomes levorotatory.  




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A 16-year-old girl was abducted when she went to a local store. Her body was found the next morning and the autopsy revealed that she had been sexually assaulted and strangled. Crime scene investigators were able to collect a semen sample from vaginal fluid as well as tissue samples from underneath the victim’s fingernails. DNA samples were obtained from three suspects besides the victim. A variable number of tandem repeats (VNTR) analysis was performed on the DNA samples from the evidence collected, the victim, and the suspect and the results were compared.

Which of the following techniques is the most appropriately applied for this analysis?

A. Thin layer chromatography

B. Polyacrylamide gel electrophoresis

C. Northern blot

D. Southern blot

E. Western blot

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Case details- VNTR Analysis

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A clone is a population of identical organisms derived from a single parental organism. In Molecular biology- A clone is a collection of molecules or cells, all identical to an original molecule or cell. If the original cells harbored a recombinant DNA molecule in the form of a plasmid, the plasmids within the millions of cells in a bacterial colony represent a clone of original DNA molecules; these molecules can be isolated and studied. The mechanism of cloning involves , i) Cutting DNA at precise locations, ii) Joining two DNA fragments covalently, iii) Selection of a small molecule of DNA capable of self replication ,iv) Moving recombinant molecules from the test tube in to a host cell and v) Selecting or identifying those cells that contain Recombinant DNA .

Which of the following is required for joining two DNA fragments covalently?

A. DNA polymerase

B. Restriction endonucleases

C. Primase

D. Reverse transcriptase

E. DNA ligase

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The fidelity of replication is very high, with an overall error rate of 10-9 to 10-10. Insertion of an inappropriate nucleotide that occurs during replication can be corrected by editing during the replication process. This proof reading function is performed by a 3′-5′ exonuclease activity associated with the polymerase complex. The post replication repair process also increases the fidelity of replication.
Which “one” of the following is not a post replication DNA repair mechanism?
A. 5′-3′- Exonuclease repair
B. Base excision repair 
C. Mismatch repair
D. Nucleotide excision repair
E. Direct repair

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Duchenne muscular dystrophy (DMD) is an X-linked disorder due to a frameshift mutation in the muscle protein “Dystrophin”. The patients have proximal muscle weakness and muscle fibrosis, with death occurring as a result of cardiac or respiratory failure.

Which ‘one’ of the following statements correctly describes a frameshift mutation?

A. The frameshift mutations occur due to substitution of the base analogs

B. The reading frame beyond the mutation remains unaltered

C. A frameshift mutation may not produce effects due to degeneracy of the genetic code

D. Insertion or deletion of three nucleotides does not alter the reading frame beyond the site

E. There is always a premature termination of the growing peptide chain due to generation of stop codon.

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