Category: Quiz

Hyperbilirubenimia

Hyperbilirubinemia

Q. 1

Unconjugate hyperbilirubinemia is seen in

 A

Physiological jaundice

 B

Breast milk jaundice

 C

Gilbert syndrome

 D

All

Q. 1

Unconjugate hyperbilirubinemia is seen in

 A

Physiological jaundice

 B

Breast milk jaundice

 C

Gilbert syndrome

 D

All

Ans. D

Explanation:

Ans. is ‘a’ i.e., Physiological jaundice; ‘b’ i.e., Breast milk jaundice; ‘c’ i.e., Gilbert syndrome


Q. 2

Unconjugated hyperbilirubinemia is seen in

 A

Rotor syndrome

 B

Dubin-Johnson syndrome

 C

Biliary atresia

 D

Crigler-Najjar syndrome

Q. 2

Unconjugated hyperbilirubinemia is seen in

 A

Rotor syndrome

 B

Dubin-Johnson syndrome

 C

Biliary atresia

 D

Crigler-Najjar syndrome

Ans. D

Explanation:

Ans. is ‘d’ i.e., Crigler-Najjar syndrome

Predominantly Unconjugated fliperbilirubinenrict

Excess production of bilirubin Hemolytic anemias

Resorption of blood from internal hemorrhage (e.g. alimentary

tract bleeding, hematomas)

Ineffective erythropoiesis syndromes (e.g. pernicious anemia, thalassemia)

Reduced hepatic uptake

Drug interference with membrance carrier systems

Some cases of Gilbert syndrome Impaired bilirubin conjugation

Physiologic jaundice of the newborn (decreased UGT I A I activity,

decreased excretion)

Breast milk jaundice (b-glucrurondases in milk)

Genetic deficiency of UGT 1 AI activity (Crigler-Najjar syndrome types I and II)

Gilbert syndrome (mixed etiologies)

Diffuse hepatocellular disease (e.g. viral or drug-induced hepatitis, cirrhosis)

Predominantly conjugated hyperbilirubinemia

Deficiency of canalicular membrane transporters (Dubin-Johnson syndrome, Rotor syndrome) Impaired bile flow.


Q. 3

Causes of conjugated hyperbilirubinemia is ‑

 A

Rotor syndrome

 B

Breast milk jaundice

 C

Crigler najjar

 D

Gilbert syndrome

Q. 3

Causes of conjugated hyperbilirubinemia is ‑

 A

Rotor syndrome

 B

Breast milk jaundice

 C

Crigler najjar

 D

Gilbert syndrome

Ans. A

Explanation:

Ans. is ‘a’ i.e., Rotor syndrome

Quiz In Between


Q. 4

Indirect hyperbilirubinemia are seen in

 A

Dubin-Johnson syndrome

 B

Rotor syndrome

 C

Gilbert syndrome

 D

Gallstone

Q. 4

Indirect hyperbilirubinemia are seen in

 A

Dubin-Johnson syndrome

 B

Rotor syndrome

 C

Gilbert syndrome

 D

Gallstone

Ans. C

Explanation:

Ans is ‘c’ i.e., Gilbert syndrome


Q. 5

Unconjugated hyperbilirubinemia in neonate is seen in all of the following except –

 A

Physiological jaundice

 B

Dubin johnson syndrome

 C

Hypothyroidism

 D

Hemolytic anemia

Q. 5

Unconjugated hyperbilirubinemia in neonate is seen in all of the following except –

 A

Physiological jaundice

 B

Dubin johnson syndrome

 C

Hypothyroidism

 D

Hemolytic anemia

Ans. B

Explanation:

Ans. is ‘b’ i.e., Dubin Johnson Syndrome


Q. 6

Autosomal dominant familial nonhemolytic hyperbilirubinemia occurs in all except –

 A

Crigler-Najjar syndrome

 B

Dubin – Johnson syndrome

 C

Gilbert syndrome

 D

Cryoglobulinemia

Q. 6

Autosomal dominant familial nonhemolytic hyperbilirubinemia occurs in all except –

 A

Crigler-Najjar syndrome

 B

Dubin – Johnson syndrome

 C

Gilbert syndrome

 D

Cryoglobulinemia

Ans. B

Explanation:

Ans. is ‘b’ i.e., Dubin-Johnson syndrome

Dubin-Johnson syndrome is an autosomal recessive disorder.

o Note sure about option d.

Quiz In Between


Q. 7

Hyperbilirubinemia in a child can be due to

 A

Breast milk jaundice

 B

Cystic fibrosis

 C

Fanconi’s syndrome

 D

All

Q. 7

Hyperbilirubinemia in a child can be due to

 A

Breast milk jaundice

 B

Cystic fibrosis

 C

Fanconi’s syndrome

 D

All

Ans. A

Explanation:

Ans. is ‘a’ i.e., Breast milk jaundice


Q. 8

Following are causes of unconjugated hyperbilirubinemia, except:

 A

Hemolytic anemia

 B

Large hematoma

 C

Rotor syndrome

 D

Megaloblastic anemia

Q. 8

Following are causes of unconjugated hyperbilirubinemia, except:

 A

Hemolytic anemia

 B

Large hematoma

 C

Rotor syndrome

 D

Megaloblastic anemia

Ans. C

Explanation:

Answer is C (Rotor syndrome)

Rotor’s syndrome is an Autosomal recessive inherited disorder characterized by a deject in biliary excretion leading to conjugated hyperbilirubinemia:

Indirect hyperbilirubinemia                                                              

Direct hyperbilirubinemia

A.   Hemolytic disorders

A.   Inherited conditions

1.    Inherited

1.     Dubin-Johnson syndrome

a.   Sperocyteosis, elliptocytosis

2.     Rotor’s syndrome

Glucose-6-phosphate dehydrogenase and pyruvate kinase deficiencies

b. Sickle cell anemia

 

2.    Acquired                         _

a. Microangiopathic hemolytic anemias

b. Paraoxysmal nocturnal hemoglobinuria

c. Immune hemolysis

 

B.    Ineffective erythropoesis

 

1.    Cobalamin, folate, thalassemia, and severe iron deficiencies

 

C. Drugs

 

1.    Rifampicin, probenbecid, ribavirin

 

D.   Inherited conditions

 

1.    Crigler-Najjar types I and II

 

2.    Glibert’s syndrome

 


Q. 9

A patient presents with unconjugated hyperbilirubinemia and presence of urobilinogen in urine. Which amongst the following is the least likely diagnosis:

 A

Hemolytic jaundice

 B

Crigler Najjar syndrome

 C

Gilbert’s syndrome

 D

Dubin Johnson syndrome

Q. 9

A patient presents with unconjugated hyperbilirubinemia and presence of urobilinogen in urine. Which amongst the following is the least likely diagnosis:

 A

Hemolytic jaundice

 B

Crigler Najjar syndrome

 C

Gilbert’s syndrome

 D

Dubin Johnson syndrome

Ans. D

Explanation:

Answer is D (Dubin Johnson Syndrome)

Dubin Johnson syndrome is associated with conjugated hyperbilirubinemia & not unconjugated hjperbilirubinemia. Dubin Johnson Syndrome results from a hereditary defect in excretion of conjugated bilirubin across the canalicular membrane and leads to conjugated hyperbilirubinemia.

Dubin Johnson syndrome is an inherited disorder charachterized by defective excretion of conjugated bilirubin from hepatocytes into biliary canaliculi. It thus presents with a clinical picture similar to obstructive jaundice with conjugated hyperbilirubinemia and absence of urobilinogen in urine.

Hemolytic Anemia typically presents with unconjugated hyperbilirubinemia and elevated urinary urobilinogens. Gilberts syndrome and Cri2ler Najjar syndrome also present with unconjugated hyperbilirubinemia. Urinary Urobilinogens are however not elevated in these conditions. Urobilinogen may never the less be present in urine (N or in these conditions

Quiz In Between


Q. 10

Conjugated hyperbilirubinemia is seen in all EX­CEPT:

March 2013

 A

Dubin Johnson syndrome

 B

Rotor syndrome

 C

Gilbert syndrome

 D

None of the above

Q. 10

Conjugated hyperbilirubinemia is seen in all EX­CEPT:

March 2013

 A

Dubin Johnson syndrome

 B

Rotor syndrome

 C

Gilbert syndrome

 D

None of the above

Ans. C

Explanation:

Ans. C i.e. Gilbert syndrome

Gilbert syndrome presents with unconjugated hyperbilirubinemia


Q. 11

Unconjugated hyperbilirubinemia is seen in all of the following except:   

March 2010

 A

Crigler Najjar Syndrome

 B

Physiological jaundice

 C

Dubin-Johnson syndrome

 D

Gilbert syndrome

Q. 11

Unconjugated hyperbilirubinemia is seen in all of the following except:   

March 2010

 A

Crigler Najjar Syndrome

 B

Physiological jaundice

 C

Dubin-Johnson syndrome

 D

Gilbert syndrome

Ans. C

Explanation:

Ans. C: Dubin-Johnson Syndrome

Dubin-Johnson syndrome is an autosomal recessive disorder that causes an increase of conjugated bilirubin without elevation of liver enzymes (ALT, AST).

This condition is associated with a defect in the ability of hepatocytes to secrete conjugated bilirubin into the bile.

The conjugated hyperbilirubinemia is a result of defective endogenous and exogenous transfer of anionic conjugates from hepatocytes into the bile.

Pigment deposition in lysosomes causes the liver to turn black.

Other causes of conjugated/direct hyperbilirubinemia:

  • Hepatocellular diseases:

– Hepatitis:

  • Neonatal idiopathic hepatitis
  • Viral (Hepatitis B, C, TORCH infections)
  • Bacterial (E. colt, urinary tract infections)

–        Total parenteral nutrition

–        Hepatic ischemia (post-ischemic damage)

–        Erythroblastosis fetalis (late, “Inspissated Bile Syndrome”)

Metabolic disorders:

  • Alpha-1 antitrypsin deficiency
  • Galactosemia, tyrosinemia, fructosemia
  • Glycogen storage disorders
  • Cystic fibrosis

Biliary tree abnormalities:

–         Extrahepatic biliary atresia: In first 2 weeks, unconjugated bilirubin predominates; elevated conjugated bilirubin is late.

–        Paucity of bile ducts

–        Choledochal cyst

–        Bile plug syndrome

Causes of unconjugated/indirect hyperbilirubinemia:

  • Increased lysis of RBCs (i.e., increased hemoglobin release)

–        Isoimmunization (blood group incompatibility: Rh, ABO and minor blood groups)

–        RBC enzyme defects (e.g., G6PD deficiency, pyruvate kinase deficiency)

–        RBC structural abnormalities (hereditary spherocytosis, elliptocytosis)

–        Infection (sepsis, urinary tract infections)

–        Sequestered blood (e.g., cephalohematoma, bruising, intracranial hemorrhage)

–        Neonatal Jaundice

–        Polycythemia

–        Shortened life span of fetal RBCs

Decreased hepatic uptake and conjugation of bilirubin

–        Immature glucuronyl transferase activity in all newborns: term infants have 1% of adult activity, preterm infants have 0.1%.

–        Gilbert Syndrome

–        Crigler Najjar Syndrome (Non-hemolytic Unconjugated Hyperbilirubinemia): inherited conjugation defect (very rare)

–        Breastmilk Jaundice (pregnanediol inhibits glucuronyl transferase activity)

Increased enterohepatic reabsorption

–        Breastfeeding jaundice (due to dehydration from inadequate milk supply)


Q. 12

Conjugated hyperbilirubinemia

 A

Dubin johnson syndrome

 B

Criggler naj jar syndrome

 C

Breast milk jandice

 D

Gilbert syndrome

Q. 12

Conjugated hyperbilirubinemia

 A

Dubin johnson syndrome

 B

Criggler naj jar syndrome

 C

Breast milk jandice

 D

Gilbert syndrome

Ans. A

Explanation:

Ans. is ‘a’ i.e., Dubin johnson syndrome

Breast milk jaundice –

  • Decrease bilirubin uptake across hepathocyte membrane.
  • Entero-hepatic recirculation.
  • Leads to indirect hyperbilirubinemia.

Crigler naj jar & Gilbert syndrome (deficiency of glucuronyl transferase)

  • Decrease conjugation leads to Indirect hyperbilirubinemia.
  • Defect in hepatocyte secretion of conjugated bilirubin.
  • Leads to direct hyperbilirubinemia

Quiz In Between



Metabolism of Amino Acids

Metabolism of Amino Acids

Q. 1

A vitamin B6 deficiency reduces the effectiveness of transaminase enzymes. Which amino acid is formed from transamination of a-ketoglutarate?

 A

Glycine

 B

Glutamine

 C

Asparagine

 D

Glutamate

Q. 1

A vitamin B6 deficiency reduces the effectiveness of transaminase enzymes. Which amino acid is formed from transamination of a-ketoglutarate?

 A

Glycine

 B

Glutamine

 C

Asparagine

 D

Glutamate

Ans. D

Explanation:

Glutamate is formed from a-ketoglutarate by the enzymes aspartate aminotransferase and alanine aminotransferase.


Q. 2

Transamination reaction is :

 A

Net deamination with splitting of NI-b

 B

c and d both

 C

Transaminase enzyme & pyridoxial P01 binding is covalent

 D

Glutamate is formed

Q. 2

Transamination reaction is :

 A

Net deamination with splitting of NI-b

 B

c and d both

 C

Transaminase enzyme & pyridoxial P01 binding is covalent

 D

Glutamate is formed

Ans. D

Explanation:

C i.e. Transaminase enzyme & pyridoxial PO4 binding is covalent; D i.e. Glutamate is formed


Q. 3

Co-enzyme used in transamination

 A

NAD

 B

Biotin

 C

Pyridoxal phosphate

 D

Riboflavin

Q. 3

Co-enzyme used in transamination

 A

NAD

 B

Biotin

 C

Pyridoxal phosphate

 D

Riboflavin

Ans. C

Explanation:

C i.e. Pyridoxal phosphate

–                        Pyridoxal phosphate (active B6) dependent conditions (in which it is used in treatment) are Homocystinuria, Oxaluria, Cystathioninuria and Xanthurenic acid uriaQ, Mn – “HOCX or Homo Ox Siton Zen”

– Homocystinuria is vitamin B6, B12 and folate. dependentQ

Maple syrup urine disease is d/t defective branched chain a-ketoacid dehydrogenase enzyme; and it may be a/ w thiamin (vitamin B1) deficiencyQ

Methylmalonyl acidttria is seen in vitamin B12 deficiencyQ.

Figlu (N- formimimino-glutamate) uria following a dose of histidine occurs in folate (folic acid) deficiency (Histidine load test). AICAR (amino imidazole carboxamide ribosyl 51-P) uria also occurs in folate deficiency.

Quiz In Between


Q. 4

Transamination of pyruvate and glutamic acid leads to the formation of

 A

Oxaloacetate

 B

a-ketoglutarate

 C

Aspartate

 D

Malate

Q. 4

Transamination of pyruvate and glutamic acid leads to the formation of

 A

Oxaloacetate

 B

a-ketoglutarate

 C

Aspartate

 D

Malate

Ans. B

Explanation:

 

Transamination of pyruvate and glutamic acid (or glutamate) 1/t formation of alanine and a-keto (oxo) glutarate.


Q. 5

Co-enzyme used in transamination ‑

 A

NAD

 B

Biotin

 C

Pyridoxal phosphate

 D

Riboflavin

Q. 5

Co-enzyme used in transamination ‑

 A

NAD

 B

Biotin

 C

Pyridoxal phosphate

 D

Riboflavin

Ans. C

Explanation:

Ans. is ‘c’ i.e., Pyridoxal phosphate

Pyridoxal phosphate (active form of vitamin B6) is the coenzyme for transamination reactions.


Q. 6

Transamination of alanine results in the formation of- 

 A

 Pyruvate

 B

Oxaloacetate

 C

 Aspartate

 D

 Arginine

Q. 6

Transamination of alanine results in the formation of- 

 A

 Pyruvate

 B

Oxaloacetate

 C

 Aspartate

 D

 Arginine

Ans. A

Explanation:

Quiz In Between


Q. 7

Coenzyme not required in formation of glutamate-

 A

 Thiamine pyrophosphate

 B

Pyridoxial phosphate

 C

Niacin

 D

None of the above

Q. 7

Coenzyme not required in formation of glutamate-

 A

 Thiamine pyrophosphate

 B

Pyridoxial phosphate

 C

Niacin

 D

None of the above

Ans. A

Explanation:

During transamination reaction glutamate is formed. Pyridoxial Phosphate acts as coenzyme.


Q. 8

True about transamination reaction are all except-

 A

Transfer of alpha amino group from alpha amino acid to keto acid

 B

Alpha ketoglutarate is the most common receptor

 C

Threonine does not undergo transamination

 D

Biotin is required as a coenzyme.

Q. 8

True about transamination reaction are all except-

 A

Transfer of alpha amino group from alpha amino acid to keto acid

 B

Alpha ketoglutarate is the most common receptor

 C

Threonine does not undergo transamination

 D

Biotin is required as a coenzyme.

Ans. D

Explanation:

Pyridoxial phosphate is a coenzyme in transamination reaction.

 


Q. 9

Glutamine in blood acts as-

 A

NH3 transporter

 B

 Toxic element

 C

Store energy

 D

Abnormal metabolite

Q. 9

Glutamine in blood acts as-

 A

NH3 transporter

 B

 Toxic element

 C

Store energy

 D

Abnormal metabolite

Ans. A

Explanation:

Glutamine is the major form of transport of ammonia

Quiz In Between


Q. 10

Ammonia is detoxified in brain to-

 A

Uric acid

 B

GABA

 C

Urea

 D

 Glutamine

Q. 10

Ammonia is detoxified in brain to-

 A

Uric acid

 B

GABA

 C

Urea

 D

 Glutamine

Ans. D

Explanation:

The brain is a rich source of glutamine synthase and predominantly detoxifies ammonia by synthesis of glutamate.


Q. 11

Which amino acid binds with NH4+ covalently and makes it non-toxic for transportation-

 A

Serine

 B

Aspartate

 C

Glutamate

 D

Histidine

Q. 11

Which amino acid binds with NH4+ covalently and makes it non-toxic for transportation-

 A

Serine

 B

Aspartate

 C

Glutamate

 D

Histidine

Ans. C

Explanation:

Q. 12

Co factors for glutamate dehydrogenase-

 A

NAD

 B

FADH2

 C

FMN

 D

 FAD

Q. 12

Co factors for glutamate dehydrogenase-

 A

NAD

 B

FADH2

 C

FMN

 D

 FAD

Ans. A

Explanation:

Hepatic L- glutamate dehydrogenase can use either NAD+ or NADP+.

Quiz In Between


Q. 13

Increased alanine during prolonged fasting represents-

 A

Increased breakdown of muscle proteins

 B

Impaired renal function

 C

Decreased utilization of amino acid from Glucogenesis

 D

Leakage of amino acids from cells due to plasma membrane leakage

Q. 13

Increased alanine during prolonged fasting represents-

 A

Increased breakdown of muscle proteins

 B

Impaired renal function

 C

Decreased utilization of amino acid from Glucogenesis

 D

Leakage of amino acids from cells due to plasma membrane leakage

Ans. A

Explanation:

During prolonged fasting there is increased gluconeogenesis. Alanine is provided by the muscle is one of the substrates for gluconeogenesis  and is called Glucose Alanine cycle.

So plasma level of alanine increases in prolonged starvation.


Q. 14

Amino acid absorption is by-

 A

Facilitated transport

 B

Passive transport

 C

Pinocytosis

 D

Active transport

Q. 14

Amino acid absorption is by-

 A

Facilitated transport

 B

Passive transport

 C

Pinocytosis

 D

Active transport

Ans. D

Explanation:

 Free amino acids are absorbed across the intestinal mucosa by sodium-dependent active transport. There are several different amino acid transporters, with specificity for the nature of the amino acid side-chain. Transporters of Amino Acids.

  • For Neutral Amino Acids
  • For Basic Amino acids and Cysteine
  • For Imino Acids and Glycine
  • For Acidic Amino Acids
  • For Beta Amino Acids (Beta Alanine)

Meisters Cycle

  • For absorption of Neutral Amino acids from Intestines, Kidney tubules and brain.
  • The main role is played by Glutathione. (GSH)
  • For transport of 1 amino acid and regeneration of GSH 3 ATPs are required.

Quiz In Between



Heme Catabolism

Heme Catabolism

Q. 1

False statement about bilirubin is

 A

Bilirubin circulating in plasma by covalently binding with albumin

 B

Bilirubin is taken up across the sinusoidal (basolateral) membrane of hepatocytes by a carrier-mediated mechanism

 C

Conjugated bilirubin in plasma undergoes stool elimination

 D

Conjugated bilirubin is then directed primarily toward the canalicular (apical) membrane

Q. 1

False statement about bilirubin is

 A

Bilirubin circulating in plasma by covalently binding with albumin

 B

Bilirubin is taken up across the sinusoidal (basolateral) membrane of hepatocytes by a carrier-mediated mechanism

 C

Conjugated bilirubin in plasma undergoes stool elimination

 D

Conjugated bilirubin is then directed primarily toward the canalicular (apical) membrane

Ans. C

Explanation:

Bilirubin transport:

  • Bilirubin circulates in plasma noncovalently, bound to albumin. 
  • It is taken up across the sinusoidal membrane of hepatocytes by a carrier-mediated mechanism. 
  • Bilirubin uptake is mediated by a liver-specific sinusoidal organic anion transport protein, (OATP1B1, SLC21A6)
  • Then bilirubin is directed by cytosolic binding proteins (e.g., glutathione S-transferase B, fatty acid binding protein) to the ER.
  • It is conjugated with uridine diphosphate (UDP)–glucuronic acid by the enzyme bilirubin UDP–glucuronyl transferase (B-UGT).
  • Conjugated bilirubin is directed toward the canalicular membrane, and it is transported into the bile canaliculus by an adenosine triphosphate (ATP)-dependent pump. 
  • The responsible protein is multidrug resistance–associated protein-2 (MRP2, ABCC2)
  • Small amounts of bilirubin glucuronides are secreted across the sinusoidal membrane via MRP3 (ABCC3)
  • Conjugated bilirubin in plasma undergoes renal elimination 
Ref: Sleisenger and Fordtran’s, E-9, P-324

Q. 2

Heme is converted to bilirubin mainly in:

 A

Kidney.

 B

Liver.

 C

Spleen

 D

Bone marrow.

Q. 2

Heme is converted to bilirubin mainly in:

 A

Kidney.

 B

Liver.

 C

Spleen

 D

Bone marrow.

Ans. C

Explanation:

C i.e. Spleen

Breakdown of heme to bilirubin occurs in macrophages of the reticuloendothelial system mainly in the spleenQ also in the liver and bone marrow.


Q. 3

Bilirubin is secreted by:

 A

Bile Salts

 B

Bile pigments

 C

Secretin

 D

CCK.

Q. 3

Bilirubin is secreted by:

 A

Bile Salts

 B

Bile pigments

 C

Secretin

 D

CCK.

Ans. A

Explanation:

A i.e. Bile salts

Substances that increase the secretion of bile are called as cholerecticsQ. Bile salts are amongst the most important physiological cholerectionQ

Quiz In Between


Q. 4

Bilirubin is the degradation product of –

 A

Albumin

 B

Globulin

 C

Heme

 D

Transferrin

Q. 4

Bilirubin is the degradation product of –

 A

Albumin

 B

Globulin

 C

Heme

 D

Transferrin

Ans. C

Explanation:

Ans. is ‘c’ i.e., Heme

Bilirubin metabolism

o Bilirubin is the end product of heme degradation.

o The heme is derived from –

(i)       Senescent erythrocytes by mononuclear phagocytic system in the spleen, liver and bone marrow (major source).

(ii)     Turnover of hemoproteins (e.g. cytochrome p.450).

o Heme is oxidized to biliverdin by heme oxygenase.

o Biliverdin is then reduced to bilirubin by biliverdin reductase.

o Bilirubin is transported to liver in bound form with albumin.

o There is carrier mediated uptake of bilirubin in the liver.

o This bilirubin is conjugated with glucuronic acid by UDP glucuronyl transferase (UGT1A1) to from conjugated bilirubin (bilirubin glucronides).

o Conjugated bilirubin is excreted into bile.

o Most of the conjugated bilirubin is deconjugated and degraded to urobilinogen.

o The most of the urobilinogen is excreted in the feces.

o Approximately 20% of the urobilinogen is reabsorbed in the ileum and colon and is returned to the liver, and promptly rexcreted into bile —> Enterohepatic circulation.

o The small amount that escapes this enterohepatic circulation is excreted in urine.


Q. 5

One gm of Hb liberates mg of bilirubin

 A

40

 B

34

 C

15

 D

55

Q. 5

One gm of Hb liberates mg of bilirubin

 A

40

 B

34

 C

15

 D

55

Ans. B

Explanation:

Ans. is ‘b’ i.e., 34

o Bilirubin is the end product of catabolism of hemoglobin.

o 1 gm of hemoglobin yields 35mg of bilirubin.


Q. 6

Bilirubin bound inside hepatocyte to ‑

 A

Albumin

 B

Ubiquinone

 C

Ligandin

 D

Globulin

Q. 6

Bilirubin bound inside hepatocyte to ‑

 A

Albumin

 B

Ubiquinone

 C

Ligandin

 D

Globulin

Ans. C

Explanation:

Ans. is ‘c’ i.e., Ligandin

Bilirubin metabolism

Bilirubin is the end product of heme degradation.

The heme is derived from –

i) Senescent erythrocytes by mononuclear phagocytic system in the spleen, liver and bone marrow (major source).

ii) Turnover of hemoproteins (e.g. cytochrome p.450).

Heme is oxidized to biliverdin by heme oxygenase.

Biliverdin is then reduced to bilirubin by biliverdin reductase.

Bilirubin is transported to liver in bound form with albumin.

Bilirubin is transferred to hepatocytes where it is bound to ligandin.

There is carrier mediated uptake of bilirubin in the liver.

This bilirubin is conjugated with glucuronic acid by UDP glucuronyl transferase (UGT1A1) to from conjugated bilirubin (bilirubin glucronides).

Conjugated bilirubin is excreted into bile.

Most of the conjugated bilirubin is deconjugated and degraded to urobilinogen.

The most of the urobilinogen is excreted in the feces.

Quiz In Between



Citric Acid Cycle

citric acid cycle

Q. 1

All of the following vitamins are required in citric acid cycle, EXCEPT:

 A

Riboflavin

 B

Niacin

 C

Thiamin

 D

Ascorbic acid

Q. 1

All of the following vitamins are required in citric acid cycle, EXCEPT:

 A

Riboflavin

 B

Niacin

 C

Thiamin

 D

Ascorbic acid

Ans. D

Explanation:

Four of the B vitamins are essential in the citric acid cycle and hence energy-yielding metabolism: (1) riboflavin, in the form of flavin adenine dinucleotide (FAD), a cofactor for succinate dehydrogenase; (2) niacin, in the form of nicotinamide adenine dinucleotide (NAD), the electron acceptor for isocitrate dehydrogenase, α-ketoglutarate dehydrogenase, and malate dehydrogenase; (3) thiamin (vitamin B1), as thiamin diphosphate, the coenzyme for decarboxylation in the α-ketoglutarate dehydrogenase reaction; and (4) pantothenic acid, as part of coenzyme A, the cofactor attached to “active” carboxylic acid residues such as acetyl-CoA and succinyl-CoA
 
Ref: Harper 28th edition, chapter 17

Q. 2

The citric acid cycle is inhibited by which of the following?

 A

Fluoroacetate

 B

Fluorouracil

 C

Arsenic

 D

Aerobic conditions

Q. 2

The citric acid cycle is inhibited by which of the following?

 A

Fluoroacetate

 B

Fluorouracil

 C

Arsenic

 D

Aerobic conditions

Ans. A

Explanation:

Fluoroacetate can be converted to fluorocitrate, which is an inhibitor of aconitase. Arsenic is not a direct inhibitor, but arsenite is an inhibitor of lipoic acid–containing enzymes such as α-ketoglutarate dehydrogenase. Malonate, not malic acid, is an inhibitor of succinate dehydrogenase. The citric acid cycle requires oxygen and would be inhibited by anaerobic, not aerobic, conditions. Fluorouracil is a suicide inhibitor of thymidylate synthase and blocks deoxythymidylate synthesis. 

Ref: Bender D.A., Mayes P.A. (2011). Chapter 17. The Citric Acid Cycle: The Catabolism of Acetyl-CoA. In D.A. Bender, K.M. Botham, P.A. Weil, P.J. Kennelly, R.K. Murray, V.W. Rodwell (Eds), Harper’s Illustrated Biochemistry, 29e.

Q. 3

In vivo control of citric acid cycle is effected by:

 A

Acetyl CoA

 B

Coenzyme A

 C

ATP

 D

Citrate

Q. 3

In vivo control of citric acid cycle is effected by:

 A

Acetyl CoA

 B

Coenzyme A

 C

ATP

 D

Citrate

Ans. C

Explanation:

C i.e. ATP, E i.e. NADH

Quiz In Between


Q. 4

Which end product of citric acid cycle is used in detoxification of ammonia in brain?

 A

Oxaloacetate

 B

Alphaketoglutarate

 C

Succinate

 D

Citrate

Q. 4

Which end product of citric acid cycle is used in detoxification of ammonia in brain?

 A

Oxaloacetate

 B

Alphaketoglutarate

 C

Succinate

 D

Citrate

Ans. B

Explanation:

Q. 5

All are components of Citric acid cycle EXCEPT:

 A

Fumarase

 B

Malonate

 C

Succinate dehydrogenase

 D

Alpha-ketoglutarate dehydrogenase

Q. 5

All are components of Citric acid cycle EXCEPT:

 A

Fumarase

 B

Malonate

 C

Succinate dehydrogenase

 D

Alpha-ketoglutarate dehydrogenase

Ans. B

Explanation:

 

Malonate competitively inhibits succinate dehydrogenase, which converts succinate to fumarate


Q. 6

Which of the following glucogenic amino acid enters the citric acid cycle at succinyl CoA

 A

Phenylalanine

 B

Tyrosine

 C

Isoleucine

 D

Tryptophan

Q. 6

Which of the following glucogenic amino acid enters the citric acid cycle at succinyl CoA

 A

Phenylalanine

 B

Tyrosine

 C

Isoleucine

 D

Tryptophan

Ans. C

Explanation:

Ans. is ‘c’ i.e., isoleucine

Quiz In Between


Q. 7

Not an intermediate product of citric acid cycle is:

 A

Acyl Co-A

 B

Succinyl Co-A

 C

Citrate

 D

a-ketoglutarate

Q. 7

Not an intermediate product of citric acid cycle is:

 A

Acyl Co-A

 B

Succinyl Co-A

 C

Citrate

 D

a-ketoglutarate

Ans. A

Explanation:

Ans. a. Acetyl Co-A


Q. 8

Substrate level phosphorylation is seen in reaction catalyzed by which enzyme of citric acid cycle‑

 A

Pyruvate kinase

 B

Succinate thiokinase

 C

Phosphoglycerate kinase

 D

All of the above

Q. 8

Substrate level phosphorylation is seen in reaction catalyzed by which enzyme of citric acid cycle‑

 A

Pyruvate kinase

 B

Succinate thiokinase

 C

Phosphoglycerate kinase

 D

All of the above

Ans. B

Explanation:

Ans. is ‘b’ i.e., Succinate thiokinase 

Quiz In Between



Myoglobin

Myoglobin

Q. 1

Binding of oxygen to myoglobin:

 A

Is characterized by the sigmoidal saturation curve

 B

Occurs at multiple sites

 C

Is characterized by the sigmoid and multiple hyperbolic saturation curve

 D

Is characterized by the hyperbolic saturation curve

Q. 1

Binding of oxygen to myoglobin:

 A

Is characterized by the sigmoidal saturation curve

 B

Occurs at multiple sites

 C

Is characterized by the sigmoid and multiple hyperbolic saturation curve

 D

Is characterized by the hyperbolic saturation curve

Ans. D

Explanation:

Myoglobin molecule contains a single O2 binding site and thus the saturation process is characterized by a simple hyperbolic curve, in contrast to hemoglobin, sigmoid saturation curve of which shows positive cooperativity of multiple oxygen binding sites.

Ref: Lippincott’s Biochemistry, 5th Ed  page 26


Q. 2

Hemoglobin unlike myoglobin shows:

 A

Sigmoid curve of oxygen dissociation

 B

Positive cooperativity

 C

Hills coefficient of one

 D

A & B

Q. 2

Hemoglobin unlike myoglobin shows:

 A

Sigmoid curve of oxygen dissociation

 B

Positive cooperativity

 C

Hills coefficient of one

 D

A & B

Ans. D

Explanation:

A i.e. Sigmoid curve of oxygen dissociation & B i.e. Positive co-operativity

Unlike Myoglobin, hemoglobin shows:

–  Tetrameric structure

–  Can bind four 02 moleculeQ

–  Sigmoid saturation kineticsQ

–  Co-operativity or co-operative binding kinetics: property that permits it to bind a maximal quantity of 02 at respiratory organs & to deliver a maximal quantity of 02 at

peripheral tissue. Here binding of Hb with 02 facilitates binding of other 02 molecules thus positive co-operativity.


Q. 3

The oxygen dissociation curve of myoglobin & hemoglobin is different due to‑

 A

Hb can bind to 2 oxygen molecules

 B

Cooperative binding in Hb

 C

Myogloobin has little oxygen affinity

 D

Hemoglobin follows a hyperbolic curve

Q. 3

The oxygen dissociation curve of myoglobin & hemoglobin is different due to‑

 A

Hb can bind to 2 oxygen molecules

 B

Cooperative binding in Hb

 C

Myogloobin has little oxygen affinity

 D

Hemoglobin follows a hyperbolic curve

Ans. B

Explanation:

Ans. is `b’ i.e., Cooperative binding in Hb

  • Cooperative binding is responsible for sigmoid shape of the oxygen-hemoglobin dissociation curve.
  • As myoglobin is monomeric (consists of one polypeptide chain only), it can bind only one molecule of oxygen and for the same reason myoglobin cannot show the phenomenon of cooperative binding. Hence, the oxygen‑myoglobin dissociation curve is hyperbola as compared to sigmoid shape of Hb-O2 curve.

Hemoglobin – O2  binding

  • Each molecule of hemoglobin can combine with upto four molecules of oxygen. Combination with the first molecule alters the conformation of the hemoglobin molecule in such a way as to facilitate combination with the next oxygen molecule. In light of this, if we look at the curve, as the PO2 starts rising from 0 mm Hg upwards, initially all hemoglobin molecules in blood starts combining with their first oxygen molecule. This is the most difficult molecule to combine with. Hence saturation rises only slowly with initial rise in PO2. As PO2 rises further, hemoglobin molecules combine with their second, third and fourth molecules, which are progressively easier to combine with. Hence saturation rises steeply between PO2 of 15 mm Hg and 40 mm Hg. When PO2 rises still further, oxygen finds most of the hemoglobin molecules carrying four molecules of oxygen each. Since no molecules of hemoglobin can carry more than four molecules of oxygen, there is not much scope for more O2 combining with hemoglobin. Hence the curve becomes almost flat again beyond the PO2 of 60 mm Hg.
  • Thus, the primary reason for the sigmoid shape of the oxygen-hemoglobin dissociation curve is that out of the four molecules of oxygen that can combine with a hemoglobin molecules, the first combines with the greatest difficulty and binding of an oxygen molecules increases affinity to next O2 molecule. This phenomenon is termed as cooperative binding or cooperativity, i.e., a molecule of O2 binds to a hemoglobin tetramer more readily if other O2 molecules are already bound.

Myoglobin O2  binding

  • Myoglobin is present in higher concentration in red (slow) muscle fibers. Myoglobin has greater affinity for oxygen than hemoglobin and its P50 is only 5 mm Hg (as compared to PO2 of hemoglobin which is about 26 mm Hg). Therefore, myoglobin-oxygen dissociation curve is shifted far to the left than Hb-O2 dissociation curve. It has shape of hyperbola as compared to sigmoid shape of Hb-O2 curve because it binds 1 molecule of O2 per mole (in comparison to Hb which binds 4 molecules of O2 per mole). The role of myoglobin is to bind O2 at very low PO2 and release them at even lower PO2, for example in exercising muscles where PO2 close to zero.

Q. 4

True about O2 binding to myoglobin

 A

Sigmoid shaped curve

 B

More affinity than hemoglobin

 C

Binds 4 molecule of O2 to myoglobin

 D

P50 is 26 mmHg

Q. 4

True about O2 binding to myoglobin

 A

Sigmoid shaped curve

 B

More affinity than hemoglobin

 C

Binds 4 molecule of O2 to myoglobin

 D

P50 is 26 mmHg

Ans. B

Explanation:

Ans. is ‘b’ i.e., More affinity than hemoglobin

Myoglobin is present in higher concentration in red (slow) muscle fibers.

Myoglobin has greater affinity for oxygen than hemoglobin and its P50 is only 5 mm Hg (as compared to P50 of hemoglobin which is about 26 mm Hg).

Therefore, myoglobin-oxygen dissociation curve is shifted far to the left than Hb-O2 dissociation curve.

It has shape of hyperbola as compared to sigmoid shape of Hb-O2 curve because it binds 1 molecule of 02 per mole (in comparison to Hb which binds 4 molecules of O2 per mole).

The role of myoglobin is to bind O2 at very low PO2 and release them at even lower PO2, for example in exercising muscles where PO2 close to zero.

Quiz In Between



Structure of Insulin

Structure of Insulin

Q. 1

Plasma half life of insulin is:

March 2013

 A

1 minute

 B

10 minutes

 C

1 hour

 D

2 hours

Q. 1

Plasma half life of insulin is:

March 2013

 A

1 minute

 B

10 minutes

 C

1 hour

 D

2 hours

Ans. B

Explanation:

Ans. B i.e. 10 minutes

Degradation of insulin

  • Once an insulin molecule has docked onto the receptor and effected its action, it may be released back into the extracellular environment, or it may be degraded by the cell.
  • The two primary sites for insulin clearance are the liver and the kidney.
  • The liver clears most insulin during first-pass transit, whereas the kidney clears most of the insulin in systemic circulation.
  • Degradation normally involves endocytosis of the insulin-receptor complex, followed by the action of insulin-degrading enzyme.
  • An insulin molecule produced endogenously by the pancreatic beta cells is estimated to be degraded within about one hour after its initial release into circulation (insulin half-life – 4-6 minutes)

Q. 2

Crystallization and storage of insulin requires ‑

 A

Mn++

 B

Zn++

 C

Cu++

 D

Ca++

Q. 2

Crystallization and storage of insulin requires ‑

 A

Mn++

 B

Zn++

 C

Cu++

 D

Ca++

Ans. B

Explanation:

Ans. is `b’ i.e., Zn++

Zinc ions are essential for the formation of hexameric insulin and hormone crystallization.

Following the synthesis of proinsulin, zinc promotes the formation of proinsulin hexamers and increases its solubility before conversion to insulin.

Proinsulin binds 30 zinc ions, of which 2 to 4 are coordinated within the molecule. These zinc ions appear to be important for the solubility of proinsulin hexamers.

With the removal of C-peptide from each proinsulin monomers, the resulting insulin increases its coordination of zinc to up to six ions per hexamer, which decreases its solubility and increases its crystallization within the secretory vesicle.

The insulin, stored as crystalline hexamer, is resistant from proteolytic attack within the vesicle.

Upon release of insulin into the bloodsteam, zinc is also released and the insulin becomes soluble in the blood.


Q. 3

How many polypepide chains does insulin compose of ?

 A

4 chains

 B

3 chains 

 C

2 chains 

 D

5 chains 

Q. 3

How many polypepide chains does insulin compose of ?

 A

4 chains

 B

3 chains 

 C

2 chains 

 D

5 chains 

Ans. C

Explanation:

  • Insulin is two-chain polypeptide hormone, with 51 amino acids.
  • Structure: 
    • 2 chains – A & B chain.
    • A – chain – 21 amino acids; B – chain – 30 amino acids.
  • Half-life-: About 5 mins

Quiz In Between



Disorders of urea cycle

Disorders of urea cycle

Q. 1

Inherited hyperammonemia is a result of deficiency of which enzyme of Krebs-Henseleit urea cycle?

 A

Malate dehydrogenase

 B

Isocitrate dehydrogenase

 C

N-acetyl glutamate synthetase

 D

Succinate dehydrogenase

Q. 1

Inherited hyperammonemia is a result of deficiency of which enzyme of Krebs-Henseleit urea cycle?

 A

Malate dehydrogenase

 B

Isocitrate dehydrogenase

 C

N-acetyl glutamate synthetase

 D

Succinate dehydrogenase

Ans. C

Explanation:

Inherited hyperammonemias are a group of six diseases caused by inborn deficiencies of the enzymes of the Krebs-Henseleit urea cycle.

The enzymes involved are:
1. N-acetyl glutamate synthetase
2. Arbamyl phosphate synthetase (CPS)
3. Ornithine transcarbamylase (OTC)
4. Argininosuccinic acid synthetase (citrullinemia)
5. Argininosuccinase deficiency
6. Arginase deficiency

Most Severe Cases: In the most severe forms of the hyperammonemic disorders, the infants are asymptomatic at birth and during the first day or two of life, after which they refuse their feedings, vomit, and rapidly become inactive and lethargic, soon lapsing into an irreversible coma. Profuse sweating, focal or generalized seizures, rigidity with opisthotonos, hypothermia, and hyperventilation have been observed in the course of the illness.

These symptoms constitute a medical emergency, but even with measures to reduce serum ammonia, the disease is usually fatal.

Ref: Ropper A.H., Samuels M.A. (2009). Chapter 37. Inherited Metabolic Diseases of the Nervous System. In A.H. Ropper, M.A. Samuels (Eds), Adams and Victor’s Principles of Neurology, 9e.


Q. 2

Citrullinemia is due to deficiency of ‑

 A

Argininosuccinate lyase

 B

Argininosuccinate synthase

 C

Arginase

 D

Ornithine transcarbamylase

Q. 2

Citrullinemia is due to deficiency of ‑

 A

Argininosuccinate lyase

 B

Argininosuccinate synthase

 C

Arginase

 D

Ornithine transcarbamylase

Ans. B

Explanation:

Q. 3

In urea cycle, hydrolysis of arginine forms ‑

 A

Citrulline

 B

Ornithine

 C

Carbomoyl phosphate 

 D

Arginosuccinase

Q. 3

In urea cycle, hydrolysis of arginine forms ‑

 A

Citrulline

 B

Ornithine

 C

Carbomoyl phosphate 

 D

Arginosuccinase

Ans. B

Explanation:

Ans. is ‘b’ i.e., Ornithine 


Q. 4

In urea cycle which defect is an X linkeddisease ‑

 A

Ornithine transcarbamylase

 B

Aspartate transcarbamylase

 C

Arginase

 D

Argininosuccinate synthase

Q. 4

In urea cycle which defect is an X linkeddisease ‑

 A

Ornithine transcarbamylase

 B

Aspartate transcarbamylase

 C

Arginase

 D

Argininosuccinate synthase

Ans. A

Explanation:

Ans. is ‘a’ i.e., Ornithine transcarbamylase 

Quiz In Between



Glycolysis Cycle

glycolysis cycle

Q. 1

Glycolysis occurs in

 A

Cytosol

 B

Mitochondria

 C

Nucleus

 D

Lysosome

Q. 1

Glycolysis occurs in

 A

Cytosol

 B

Mitochondria

 C

Nucleus

 D

Lysosome

Ans. A

Explanation:

Cytosol [Ref. Harper 26/e, p 136]

Reactions/Pathways                                Site                                         

Glycolysis

Kreb’s cycle

Electron transport chain

1-11V1P shunt

Fatty acid synthesis

Fatty acid oxidation

Oxidation of very long chain fatty acids

Glycogenesis

Glycogenolysis

Gluconeogenesis

Urea cycle

Cytosol

Mitochondria

Mitochondria

Cytosol

Cytosol

Mitochondria

Peroxisomes

Cytosol

Cytosol

Both cytosol & mitochindria

Both cytosol & mitochondria



Q. 2 Net ATP’s formed in glycolysis are:
 A 5
 B 8
 C 10
 D 15
Q. 2 Net ATP’s formed in glycolysis are:
 A 5
 B 8
 C 10
 D 15
Ans. B

Explanation:

8


Q. 3

What is the net amount of ATP’s formed in aerobic glycolysis?

 A

5

 B

8

 C

10

 D

15

Q. 3

What is the net amount of ATP’s formed in aerobic glycolysis?

 A

5

 B

8

 C

10

 D

15

Ans. B

Explanation:

During aerobic glycolysis the number of net ATPs formed are 8.
Steps involved in the formation of ATP during glycolysis are:
  • Conversion of 2 molecules of glyceraldehyde 3 phosphate to 1,3 bisphoglycerate release 2 molecules of NADH which yield 6 ATP.
  • Conversion of 2 molecules of 1,3 BPG to 3 phosphoglycerate yield 2 ATP.
  • Conversion of 2 molecules of phosphoenol pyruvate to pyruvate yield 2 ATP.
Steps involved in the consumption of ATP during glycolysis are:
  • Conversion of glucose to glucose 6 phosphate 
  • Conversion of fructose 6 phosphate to fructose 6 bis phosphate
  • Total ATP formed during glycolysis  : 10
  • ATP utilised  during glycolysis          :   2
  • Net ATP formed during glycolysis    :   8
Net ATP produced during anaerobic glycolysis is only 2.
 
Ref: Medical Biochemistry By N. Mallikarjuna Rao page 160.

Quiz In Between


Q. 4

Net ATP’s formed in glycolysis are:

 A

5

 B

8

 C

10

 D

15

Q. 4

Net ATP’s formed in glycolysis are:

 A

5

 B

8

 C

10

 D

15

Ans. B

Explanation:

Aerobic glycolysis is a process of splitting of glucose into two molecules of pyruvate with the synthesis of ATP. Net ATP formed in aerobic glycolysis is 8.
 
Glycolysis: is the process by which glucose or other hexoses are converted into the three-carbon compound pyruvate. All the reactions takes place in cytoplasm.
  • Aerobic glycolysis: is a process of splitting of glucose into two molecules of pyruvate with the synthesis of ATP.
  • Anaerobic Glycolysis: glycolysis is the only process in which ATP is generated anaerobically. This is of importance because RBC which does not have a mitochondria is wholly dependant on the anaerobic energy production. The byproduct of anaerobic glycolysis is lactate.
Glycolytic pathway: occurs in cytoplasm consists of 10 steps. The first five steps result in one molecule of glucose is converted to 2 glyceraldehyde-3-phosphate molecules at the expense of two molecules of ATPs. The second five steps results in the production of 2 ATP molecules per one molecule of glucose.
  • Rate limiting step: Phosphofructokinase the commited step of glycolysis.
  • Irreversibles steps:
               Hexokinase or glucokinase
               Phosphofructokinase
               Pyruvate kinase
 
No of ATP generated is:
Aerobic glycolysis 8
Anaerobic glycolysis 2
 

Enzyme

Reducing Equivalents

ATP

Glucokinase

 

– 1 ATP

Phosphofructokinase

 

– 1 ATP

Glyceraldehyde 3 phosphate dehydrogenase

2 NADH

 6 ATP

Phosphoglycerate kinase

 

 2 ATP

Pyruvate kinase

 

 2 ATP

NET ATP

 

 8 ATP

 
Ref: Essentials of Medical Biochemistry: With Clinical Cases, By Chung-Eun Ha, N. V. Bhagavan, Page 115

Q. 5

What is the net ATP’s formed in glycolysis?

 A

5

 B

7

 C

10

 D

15

Q. 5

What is the net ATP’s formed in glycolysis?

 A

5

 B

7

 C

10

 D

15

Ans. B

Explanation:

ATP formation in glycolysis:

Reaction Catalyzed by

Method of ATP Formation

ATP per mol of Glucose

Glyceraldehyde 3-phosphate dehydrogenase

Respiratory chain oxidation of 2 NADH

5

Phosphoglycerate kinase

Substrate-level phosphorylation

2

Pyruvate kinase

Substrate-level phosphorylation

2

   

Total 9

 

Consumption of ATP for reactions of hexokinase and phosphofructokinase

-2

   

Net 7

Ref: Bender D.A., Mayes P.A. (2011). Chapter 18. Glycolysis & the Oxidation of Pyruvate. In D.A. Bender, K.M. Botham, P.A. Weil, P.J. Kennelly, R.K. Murray, V.W. Rodwell (Eds), Harper’s Illustrated Biochemistry, 29e.


Q. 6

Which of the following statements about anaerobic glycolysis is INCORRECT?

 A

Anaerobic glycolysis yields two molecules of lactate and two molecules of ATP

 B

There are two oxidation-reduction steps in the anaerobic glycolysis.

 C

The lactated formed in the muscles in the condition of an oxygen deficit can be recycled through the Cori cycle.

 D

The extraction of energy from the molecule of glucose in the anaerobic glycolysis happens due to the net change in the oxidation state of carbon

Q. 6

Which of the following statements about anaerobic glycolysis is INCORRECT?

 A

Anaerobic glycolysis yields two molecules of lactate and two molecules of ATP

 B

There are two oxidation-reduction steps in the anaerobic glycolysis.

 C

The lactated formed in the muscles in the condition of an oxygen deficit can be recycled through the Cori cycle.

 D

The extraction of energy from the molecule of glucose in the anaerobic glycolysis happens due to the net change in the oxidation state of carbon

Ans. D

Explanation:

Although there are two oxidation-reduction steps is the anaerobic glycolysis, and the energy for the synthesis of two ATP molecules is released, there is no net change in the oxidation state of carbon. The first oxidative reaction is catalyzed by glyceraldehyde-3-phosphate dehydrogenase to yield NADH. Later, NADH is spent for the reduction of pyruvate to the lactate which is catalyzed by lactate dehydrogenase. The ratio between C, H, and O atoms is the same for both glucose, C6H12O6 and lactic acid C3H6O3.

 
Ref: Bender D.A., Mayes P.A. (2011). Chapter 18. Glycolysis & the Oxidation of Pyruvate. In D.A. Bender, K.M. Botham, P.A. Weil, P.J. Kennelly, R.K. Murray, V.W. Rodwell (Eds), Harper’s Illustrated Biochemistry, 29e.

Quiz In Between


Q. 7

Which of the following occurs during glycolysis?

 A

Glucose is reduced to pyruvate in the cytosol of all cells.

 B

The rate-limiting step is the formation of fructose-6-phosphate.

 C

Glucose is phosphorylated by aldolase.

 D

Pyruvate, NADH, and ATP are produced.

Q. 7

Which of the following occurs during glycolysis?

 A

Glucose is reduced to pyruvate in the cytosol of all cells.

 B

The rate-limiting step is the formation of fructose-6-phosphate.

 C

Glucose is phosphorylated by aldolase.

 D

Pyruvate, NADH, and ATP are produced.

Ans. D

Explanation:

In virtually all cells of the body, glycolysis is the primary pathway for carbohydrate catabolism.
During glycolysis, glucose undergoes oxidation to form pyruvate, NADH, and ATP. The initial reaction of glycolysis involves phosphorylation of glucose to glucose-6-phosphate via the enzyme hexokinase (glucokinase in liver).
The first committed step involves the phosphorylation of fructose-6-phosphate to fructose-1,6-diphosphate by phosphofructokinase (PFK). The reactions catalyzed by hexokinase and PFK are two of three irreversible reactions occurring in glycolysis.
The other irreversible reaction involves pyruvate kinase. In terms of the energetics of the reactants and products, it is important to remember that every intermediate in this pathway between glucose and pyruvate contains phosphate.
 
Ref: Bender D.A., Mayes P.A. (2011). Chapter 18. Glycolysis & the Oxidation of Pyruvate. In D.A. Bender, K.M. Botham, P.A. Weil, P.J. Kennelly, R.K. Murray, V.W. Rodwell (Eds), Harper’s Illustrated Biochemistry, 29e.

Q. 8

Within the RBC, hypoxia stimulates glycolysis by which of the following regulating pathways?

 A

Hypoxia stimulates pyruvate dehydrogenase by increased 2,3 DPG

 B

Hypoxia inhibits hexokinase

 C

Hypoxia stimulates release of all glycolytic enzymes from band 3 on RBC membrane

 D

Activation of the regulatory enzymes by high PH

Q. 8

Within the RBC, hypoxia stimulates glycolysis by which of the following regulating pathways?

 A

Hypoxia stimulates pyruvate dehydrogenase by increased 2,3 DPG

 B

Hypoxia inhibits hexokinase

 C

Hypoxia stimulates release of all glycolytic enzymes from band 3 on RBC membrane

 D

Activation of the regulatory enzymes by high PH

Ans. C

Explanation:

During Hypoxia, the glycolytic enzymes that bind in the same region of band 3 of Hb are released from the membrane resulting in an increased rate of glycolysis. Increased glycolysis increases ATP production and the hypoxic release of ATP.
 
Ref: Oxygen Transport to Tissue, Xxxiii, edited by Martin Wolf, David K Harrison, 2012, Page 188.

Q. 9

Phosphofructokinase-1 occupies a key position in regulating glycolysis and is also subjected to feedback control. Which among the following is the allosteric activators of phosphofructokinase-1?

 A

Fructose 2, 3 bisphosphate

 B

Fructose 2, 6 bisphosphate

 C

Glucokinase

 D

PEP

Q. 9

Phosphofructokinase-1 occupies a key position in regulating glycolysis and is also subjected to feedback control. Which among the following is the allosteric activators of phosphofructokinase-1?

 A

Fructose 2, 3 bisphosphate

 B

Fructose 2, 6 bisphosphate

 C

Glucokinase

 D

PEP

Ans. B

Explanation:

The most potent positive allosteric activator of phosphofructokinase-1 and inhibitor of fructose 1,6-bisphosphatase in the liver is fructose 2,6-bisphosphate.

  • It relieves inhibition of phosphofructokinase-1 by ATP and increases the affinity for fructose 6-phosphate. 
  • It inhibits fructose 1,6-bisphosphatase by increasing the Km for fructose 1,6-bisphosphate. 
  • Its concentration is under both substrate (allosteric) and hormonal control (covalent modification).
Phosphofructokinase-1 is inhibited by citrate and by normal intracellular concentrations of ATP and is activated by 5′ AMP.
 
Ref: Bender D.A., Mayes P.A. (2011). Chapter 20. Gluconeogenesis & the Control of Blood Glucose. In D.A. Bender, K.M. Botham, P.A. Weil, P.J. Kennelly, R.K. Murray, V.W. Rodwell (Eds), Harper’s Illustrated Biochemistry, 29e.

Quiz In Between


Q. 10

Glycolysis is the metabolic pathway that involves 10 enzyme mediated steps. It occur in which of the following cell organelle?

 A

Cytosol

 B

Mitochondria

 C

Nucleus

 D

Lysosome

Q. 10

Glycolysis is the metabolic pathway that involves 10 enzyme mediated steps. It occur in which of the following cell organelle?

 A

Cytosol

 B

Mitochondria

 C

Nucleus

 D

Lysosome

Ans. A

Explanation:

Glycolysis, the major pathway for glucose metabolism, occurs in the cytosol of all cells. Glycolysis is both the principal route for glucose metabolism and also the main pathway for the metabolism of fructose, galactose, and other dietary carbohydrates. 

Ref: Bender D.A., Mayes P.A. (2011). Chapter 18. Glycolysis & the Oxidation of Pyruvate. In D.A. Bender, K.M. Botham, P.A. Weil, P.J. Kennelly, R.K. Murray, V.W. Rodwell (Eds), Harper’s Illustrated Biochemistry, 29e.

 


Q. 11

Enzymes of glycolysis are found in:

 A

Cytosol

 B

Cell membrane

 C

Mitochondria

 D

Ribososmes

Q. 11

Enzymes of glycolysis are found in:

 A

Cytosol

 B

Cell membrane

 C

Mitochondria

 D

Ribososmes

Ans. A

Explanation:

All of the enzymes of glycolysis are found in the cytosol.
           
Ref: Harper 28th edition, chapter 18.

Q. 12

Common to both glycolysis and pentose phosphate pathway is:

 A

Glucose 6 phosphate

 B

NAD

 C

ATP

 D

All of the above

Q. 12

Common to both glycolysis and pentose phosphate pathway is:

 A

Glucose 6 phosphate

 B

NAD

 C

ATP

 D

All of the above

Ans. A

Explanation:

Although glucose 6-phosphate is common to both pathways, the pentose phosphate pathway is markedly different from glycolysis. Oxidation utilizes NADP rather than NAD, and CO2, which is not produced in glycolysis, is a characteristic product. No ATP is generated in the pentose phosphate pathway, whereas it is a major product of glycolysis
Ref: Harper 28th edition, chapter 21.

Quiz In Between


Q. 13

Which of the following is an energy-requiring step of glycolysis?

 A

Pyruvate carboxylase

 B

Phosphoenolpyruvate carboxykinase

 C

Phosphoglycerate kinase

 D

Hexokinase

Q. 13

Which of the following is an energy-requiring step of glycolysis?

 A

Pyruvate carboxylase

 B

Phosphoenolpyruvate carboxykinase

 C

Phosphoglycerate kinase

 D

Hexokinase

Ans. D

Explanation:

Hexokinase catalyzes the conversion of glucose to glucose-6-phosphate in the energy-requiring first step of glycolysis. ATP is also required in the conversion of fructose-6-phosphate to fructose 1,6-bisphosphate by PFK. ATP is generated in the conversion of 1,3-bisphosphoglycerate to 3-phosphoglycerate by phosphoglycerate kinase and in the conversion of phosphoenolpyruvate to pyruvate by PK. Both phosphoenolpyruvate carboxykinase and pyruvate carboxylase are energy-requiring reactions except that these occur in the gluconeogenesis pathway.

Ref: Bender D.A., Mayes P.A. (2011). Chapter 18. Glycolysis & the Oxidation of Pyruvate. In D.A. Bender, K.M. Botham, P.A. Weil, P.J. Kennelly, R.K. Murray, V.W. Rodwell (Eds), Harper’s Illustrated Biochemistry, 29e. 

 


Q. 14

In glycolysis, the following forms as the byproduct:

 A

Pyruvate

 B

H2O

 C

H+

 D

All of the above

Q. 14

In glycolysis, the following forms as the byproduct:

 A

Pyruvate

 B

H2O

 C

H+

 D

All of the above

Ans. D

Explanation:

Glycolysis is the metabolic pathway that breaks down (catabolism) hexose (six-carbon) monosaccharides such as glucose, fructose, and galactose into two molecules of pyruvate, two molecules of ATP, two molecules of NADH, two water (H2O) molecules, and two hydrogen ions (H+).
 
Ref: Janson L.W., Tischler M.E. (2012). Chapter 6. Carbohydrate Metabolism. In L.W. Janson, M.E. Tischler (Eds), The Big Picture: Medical Biochemistry.

Q. 15

Enzyme catalyzing reversible step in glycolysis is/are:

 A

Glyceraldehyde-3-phosphate-dehydrogenase

 B

Enolase

 C

Phospho-glyceromutase

 D

All Correct

Q. 15

Enzyme catalyzing reversible step in glycolysis is/are:

 A

Glyceraldehyde-3-phosphate-dehydrogenase

 B

Enolase

 C

Phospho-glyceromutase

 D

All Correct

Ans. D

Explanation:

A, B, C i.e. Enolase, Phospho-glyceromutase, Glyceraldehyde-3-phosphate-dehydrogenase

Quiz In Between


Q. 16

About glycolysis true is:

 A

Occurs in mitochondria

 B

Complete breakdown of glucose

 C

Conversion of glucose to 3C units

 D

3 ATP’s are used in anaerobic pathway

Q. 16

About glycolysis true is:

 A

Occurs in mitochondria

 B

Complete breakdown of glucose

 C

Conversion of glucose to 3C units

 D

3 ATP’s are used in anaerobic pathway

Ans. C

Explanation:

  1. C i.e. Conversion of glucose to 3 C units
  • Enzymes of glycolysis – a process which converts 6 carbon glucose to 3 carbon unit pyruvate & lactateQ – are present in cytoplasm. Whereas complete oxidation of glucose (to CO) & H20) requires mitochondrial enzymes (of TCA cycle).
  • Out of total 9 enzymes used in glycolysis

–   3 enzymes – hexokinase, phosphofructokinase (PFK-1) and Pyruvate kinase are used in irreversible steps.

 

–   6 enzymes – phosphohexose isomerase, aldolase, glyceraldehydes 3 phosphate dehydrogenase, 1, 3- biphospho glycerate kinase, phosphoglyceromutase and enolase are used in reversible stepsQ.

–   2 enzymes used in energy utilizing steps are – hexokinase (using 1 ATP) and phosphofructokinase PFK-1 (using 1 ATP). This energy consumption is for 1 molecule of glucose.

–  3 enzymes used in energy producing steps are – glyceraldehydes 3phosphate dehydrogenase (producing 2 NADH = 5 ATP), 1, 3 – biphospho glycerate kinase (producing 2 ATP), and pyruvate kinase (producing 2 ATP). This energy production is for 1 molecule of glucose or 2 molecules of glyceraldehyde 3-P.

 


Q. 17

Compound that joints glycolysis with glycogenesis & glycogenolysis :

 A

Glucose 1, 6 bi phosphate

 B

Glucose 1 PO4

 C

Glucose 6 PO4

 D

Fructose 1, 6 bi phosphate

Q. 17

Compound that joints glycolysis with glycogenesis & glycogenolysis :

 A

Glucose 1, 6 bi phosphate

 B

Glucose 1 PO4

 C

Glucose 6 PO4

 D

Fructose 1, 6 bi phosphate

Ans. C

Explanation:

C i.e. Glucose 6 PO4

Glucose 6 phosphate is an important compound that joins several metabolic pathways viz. glycolysis, glycogenolysis, glycogenesis, gluconeogenesis and pentose phosphate pathwayQ


Q. 18

Irreversible step (s) in glycolysis is/are:

 A

Hexokinase

 B

Phosphofructokinase

 C

Pyruvate kinase

 D

All of the above

Q. 18

Irreversible step (s) in glycolysis is/are:

 A

Hexokinase

 B

Phosphofructokinase

 C

Pyruvate kinase

 D

All of the above

Ans. D

Explanation:

All of the above

Quiz In Between


Q. 19

In glycolysis, the first commited step is catalysed by :

 A

2 ,3 DPG

 B

Glucokinase

 C

Hexokinase

 D

Phosphofructokinase.

Q. 19

In glycolysis, the first commited step is catalysed by :

 A

2 ,3 DPG

 B

Glucokinase

 C

Hexokinase

 D

Phosphofructokinase.

Ans. D

Explanation:

D i.e. Phosphofructokinase


Q. 20

The rate-limiting enzyme in Glycolysis is :

 A

Phosphofructokinase

 B

Glucose-6-dehydrogenase

 C

Glucokinase

 D

Pyruvate kinase

Q. 20

The rate-limiting enzyme in Glycolysis is :

 A

Phosphofructokinase

 B

Glucose-6-dehydrogenase

 C

Glucokinase

 D

Pyruvate kinase

Ans. A

Explanation:

Ans:A i.e. Phosphofructokinase.

Rate-Limiting Enzymes 

  • Rate-limiting enzyme of Glycolysis :Phosphofructokinase-1 (PFK-1)
  • Rate-limiting enzyme of Gluconeogenesis :Fructose-1,6,biphosphatase
  • Rate-limiting enzyme of TCA cycle :Isocitrate dehydrogenase
  • Rate-limiting enzyme of Glycogen Synthesis :Glycogen synthase
  • Rate-limiting enzyme of Glycogenolysis :Glycogen phophorylase (phophorylase breaks phosphate bond, which means activated glycogen releases a lot of energy)
  • Rate-limiting enzyme of HMP Shunt :Glucose-6-Phosphate dehydrogenase (bad to lose this in RBCs)
  • Rate-limiting enzyme of de novo pyrimidine synthesis :Carbamoyl phosphate synthase II (CPS I is involved in urea cycle)
  • Rate-limiting enzyme of de novo purine synthesis :Glutamine-PRPP amidotransferase
  • Rate-limiting enzyme of Urea cycle :Carbamoyl phosphate synthetase I (CPS II is involved in pyrimidine synthesis)
  • Rate-limiting enzyme of fatty acid synthesis :Acetyl-CoA carboxylase (ACC)
  • Rate-limiting enzyme of fatty acid oxidation :Carnitine acyltransferase I
  • Rate-limiting enzyme of Ketogenesis :HMG-CoA synthase
  • Rate-limiting enzyme of Cholesterol synthesis :HMG-CoA reductase

Q. 21

Enzyme to both common in gluconegenesis and glycolysis pathway is :

 A

Phosphofructokinase

 B

Fructose 2,6-biphosphatase

 C

Hexokinase

 D

Glucose 6 phosphatase

Q. 21

Enzyme to both common in gluconegenesis and glycolysis pathway is :

 A

Phosphofructokinase

 B

Fructose 2,6-biphosphatase

 C

Hexokinase

 D

Glucose 6 phosphatase

Ans. A

Explanation:

A i.e. Phosphofructokinase

Quiz In Between


Q. 22

Glycolysis occurs in

 A

Cytosol

 B

Mitochondria

 C

Nucleus

 D

Lysosome

Q. 22

Glycolysis occurs in

 A

Cytosol

 B

Mitochondria

 C

Nucleus

 D

Lysosome

Ans. A

Explanation:

A i.e. Cytosol


Q. 23

In glycolysis, insulin affects all of the following en­zymes except:       

 A

Phosphofructokinase

 B

Pyruvate kinase

 C

Glucokinase

 D

Hexokinase

Q. 23

In glycolysis, insulin affects all of the following en­zymes except:       

 A

Phosphofructokinase

 B

Pyruvate kinase

 C

Glucokinase

 D

Hexokinase

Ans. D

Explanation:

 

The activation as well as the quantities of certain key enzymes of glycolysis, namely glucokinase (NOT hexokinase), phosphofructokinase and pyruvate kinase are increased by insulin.


Q. 24

Key glycolytic enzymes in glycolysis are all except:

 A

Phosphofructokinase

 B

Hexokinase

 C

Pyruvate kinase

 D

Glucose-1, 6, diphosphatase

Q. 24

Key glycolytic enzymes in glycolysis are all except:

 A

Phosphofructokinase

 B

Hexokinase

 C

Pyruvate kinase

 D

Glucose-1, 6, diphosphatase

Ans. D

Explanation:

 

Glycolysis/Embden-Meyerhof pathway is the sequence of reactions that converts glucose into pyruvate with the concomitant production of a relatively small amount of adenosine triphosphate (ATP)

It is the initial process of most carbohydrate catabolism, and it serves three principal functions:

  • Generation of high-energy molecules (ATP and NADH) as cellular energy sources as part of aerobic respiration and anaerobic respiration.
  • Production of pyruvate for the citric acid cycle as part of aerobic respiration
  • Production of a variety of six- and three-carbon intermediate compounds, which may be removed at various steps in the process for other cellular purposes

In eukaryotes and prokaryotes, glycolysis takes place within the cytosol of the cell.

Quiz In Between


Q. 25

What is the end product of anearobic glycolysis?

 A

Pyruvate

 B

Lactate

 C

Fats

 D

Cholesterol

Q. 25

What is the end product of anearobic glycolysis?

 A

Pyruvate

 B

Lactate

 C

Fats

 D

Cholesterol

Ans. B

Explanation:

Q. 26

Which is not a common enzyme for glycolysis and gluconeogenesis?

 A

Aldolase

 B

Glucose-6-phosphatase

 C

Phosphoglycerate mutase

 D

Phosphoglycerate kinase

Q. 26

Which is not a common enzyme for glycolysis and gluconeogenesis?

 A

Aldolase

 B

Glucose-6-phosphatase

 C

Phosphoglycerate mutase

 D

Phosphoglycerate kinase

Ans. B

Explanation:

 

Seven of the reactions of glycolysis are reversible and are used in the synthesis of glucose by gluconeogenesis.

Thus, seven enzymes are common to both glycolysis and gluconeogenesis :

(i) Phosphohexose isomerase;

(ii) Aldolase;

(iii) Phosphotriose isomerase,

(iv) Glyceraldehyde 3-phosphate dehydrogenase;

(v) Phosphoglycerate kinase;

(vi) Phosphoglycerate mutase;

(vii) Enolase.

Three reactions of glycolysis are irreversible which are circumvented in gluconeogenesis by four reactions. So, enzymes at these steps are different in glycolysis and gluconeogenesis.


Q. 27

Number of ATP produced by RBC when Glycolysis occurs through Rapoport Leubering pathway

 A

1

 B

2

 C

3

 D

4

Q. 27

Number of ATP produced by RBC when Glycolysis occurs through Rapoport Leubering pathway

 A

1

 B

2

 C

3

 D

4

Ans. A

Explanation:

 

Usually 2 ATP molecules are formed in glycolysis by substrate level phosphorylation.

Quiz In Between


Q. 28

Which of the following enzyme does not catalyzes irreversible step in glycolysis ‑

 A

Hexokinase

 B

Phosphoglycerate kinase

 C

Pyruvate kinase

 D

Phosphofructokinase

Q. 28

Which of the following enzyme does not catalyzes irreversible step in glycolysis ‑

 A

Hexokinase

 B

Phosphoglycerate kinase

 C

Pyruvate kinase

 D

Phosphofructokinase

Ans. B

Explanation:

Ans. is ‘b’ i.e., Phosphoglycorate kinase

Glycolysis is regulated at 3 steps which are irreversible.

These reactions are catalyzed by following key enzymes :‑

1) Hexokinase and glucokinase

2) Phosphofructokinase – I

3) Pyruvate kinase.


Q. 29

Number of ATP molecules and NADH formed in each cycle of glycolysis ‑

 A

4 ATP, 2 NADH

 B

2 ATP, 2 NADH

 C

4 ATP, 4 NADH

 D

2 ATP, 4 NADH

Q. 29

Number of ATP molecules and NADH formed in each cycle of glycolysis ‑

 A

4 ATP, 2 NADH

 B

2 ATP, 2 NADH

 C

4 ATP, 4 NADH

 D

2 ATP, 4 NADH

Ans. A

Explanation:

Ans. is ‘a’ i.e., 4 ATP, 2 NADH

Enegetics of glvcolysis

During glycolysis 2 ATP are utilized and 4 ATP are produced at substrate level. 2 reducing equalents NADH’ are produced and reoxidized by electron transport chain, to generata 5 ATP molecules (2.5 ATP per NADH’ molecule). Thus total 9 ATP molecules are produced and 2 are utilized, i.e., There is net gain of 7 ATP molecules in aerobic glycolysis.

In anaerobic conditions, the reoxidation of NADH by electron transport chain is prevented and NADH gets reoxidized by conversion of pyruvate to lactate by lactate dehydrogenase. Thus, in anaerobic glycolysis only 4 ATP are produced at substrate level. Therefore, there is net gain of 2 ATP molecules in anaerobic glycolysis.

Note : – Previous calculations were made assuming that NADH produces 3 ATPs and FADH2 generates 2 ATPs. This will amount to a net generation of 8ATPs per glucose molecule during glycolysis. Recent experiments show that these old values are overestimates and NADH produces 2.5 ATPs and FADH2 produces 1.5 ATPs. Thus, net generation is only 7ATPs during glycolysis.


Q. 30

Number of ATP molecules formed in anaerobic glycolysis ‑

 A

1

 B

2

 C

4

 D

8

Q. 30

Number of ATP molecules formed in anaerobic glycolysis ‑

 A

1

 B

2

 C

4

 D

8

Ans. C

Explanation:

 

Two different questions can be framed :‑

  • Number of ATP molecules produced in anaerobic glycolysis → 4
  • Number of ATP molecules gained in aerobic glycolysis  2

Quiz In Between


Q. 31

Fructose 2-6 bisphosphate (F26BP) regulates glycolysis at the level of ‑

 A

Glucose -6- phosphate

 B

Fructose -6- phosphate

 C

Glyceraldehyde -3- phosphate

 D

Phosphoenol pyruvate

Q. 31

Fructose 2-6 bisphosphate (F26BP) regulates glycolysis at the level of ‑

 A

Glucose -6- phosphate

 B

Fructose -6- phosphate

 C

Glyceraldehyde -3- phosphate

 D

Phosphoenol pyruvate

Ans. B

Explanation:

 

Regulation of glycolysis

  • Glycolysis is regulated at 3 steps which are irreversible. These reactions are catalyzed by following key enzymes : (1) Hexokinase and glucokinase, (2) Phosphofructokinase I, and (3) Pyruvate kinase.

Hexokinase and glucokinase

  • These enzymes catalyze the first step of glycolysis, i.e., Glucose —> Glucose-6-phosphate. Glucokinase is found in liver, Whereas hexokinase is found in all tissues. Kinetic properties of these two are different.
  • Hexokinase has low Km, i.e., high affinity for glucose, low Vmax, and is subjected to feedback inhibition by the reaction product, glucose-6-phosphate. Hexokinase is found in most of the tissue except liver and comes into play when blood glucose is low. It is not affected by feeding or insulin. Hexokinase is not specific for glucose metabolism, it is also involved in metabolism of fructose and galactose.
  • Glucokinase, on the other hand, is specific for glucose. It has high Km (i.e., low affinity for glucose), high Vmax and unlike hexokinase, it is not inhibited by glucose-6-phosphate. As it has low affinity for glucose (high km), it comes into play only when intracellular glucose concentration is high. It is induced by feeding and insulin. Glucagon inhibits glucokinase.
  • Function of hexokinase is to provide glucose-6-phosphate at a constant rate, according the needs of cells, i.e., function of hexokinase is to provide constant glucose utilization by all tissues of body even when blood sugar is low. Function of glucokinase in the liver is to remove glucose from blood after a meal, providing glucose-6­phosphate in excess of requirement for glycolysis so that it can be used for glycogen synthesis and lipogenesis.

Phosphofructokinase I

  • Phosphofructokinase I is the major regulatory enzyme of glycolysis. It catalyzes the 3rd reaction of glycolysis, i.e., fructose-6-P Fructose 1,6 bis-P. This reaction is irreversible and is the “rate -limiting step” for glycolysis. It is also the “commeted step”, meaning that once fructose 1,6 bisphophate is formed it must go for the glycolytic pathway only. So, most important control point for glycolysis is through regulation of phosphofructokinase I.
  • Phosphofructokinase – I is allosterically activated by : Fructose-6-phosphate, fructose 2,6-bisphophate, AMP, ADP, K+ and phosphate. It is allosterically inhibited by : ATP, citrate, Ca+2, Mg+2, and low pH. Phosphofructokinase is an inducible enzyme that increases its synthesis in response to insulin and decreases in response to glucagon.
  • Fructose 2,6-bisphosphate (F-2,6-BP) is the most important allosteric modulator (activator) of phosphofructokinase-I. Fructose 2,6-bisphosphate is synthesized as a side product of glycolysis. A bifunctional enzyme named PFK-2/Fructose 2,6 bisphosphatase is responsible for regulating the level of fructose 2,6 bisphosphate in the liver. Phosphofuctokinase-2 (PFK-2) activity of this bifunctional enzyme is responsible for synthesis of F-2,6-BP from fructose-6-phosphate and fructose 2,6 bisphosphatase activity is responsible for hydrolysis of F-2,6-BP back to fructose-6-phosphate.

Q. 32

Which of the enzyme of glycolysis is used ingluconeogenesis ‑

 A

Glucokinase

 B

PFK

 C

Pyruvate kinase

 D

Phosphotriose isomerase

Q. 32

Which of the enzyme of glycolysis is used ingluconeogenesis ‑

 A

Glucokinase

 B

PFK

 C

Pyruvate kinase

 D

Phosphotriose isomerase

Ans. D

Explanation:

Ans. is ‘d’ i.e., Phosphotriose isomerase 

Enzyme in gluconeogenesis

  • Seven of the reactions of glycolysis are reversible and are used in the synthesis of glucose by gluconeogenesis. Thus, seven enzymes are common to both glycolysis and gluconeogenesis: (i) Phosphohexose isomerase; (ii) Aldolase; (iii) Phosphotriose isomerase; (iv) Glyceraldehyde 3-phosphate dehydrogenase; (v) Phosphoglycerate kinase; (vi) Phosphoglycerate mutase; (vii) Enolase.
  • Three of the reactions of glycolysis are irreversible and must be circumvented by four special reactions which are unique to gluconeogenesis and catalyzed by : (i) Pyruvate carboxylase, (ii) PEP carboxykinase, (iii) Fructose-1, 6- bisphosphatase, (iv) Glucose-6-phosphatase.

Q. 33

Which of the enzyme of glycolysis is a part ofgluconeogenesis ‑

 A

Pyruvate kinase

 B

PFK

 C

Hexokinase

 D

Phosphoglycerate kinase

Q. 33

Which of the enzyme of glycolysis is a part ofgluconeogenesis ‑

 A

Pyruvate kinase

 B

PFK

 C

Hexokinase

 D

Phosphoglycerate kinase

Ans. D

Explanation:

Ans. is ‘d’ i.e., Phosphoglycerate kinase 

  • Seven of the reactions of glycolysis are reversible and are used in the synthesis of glucose by gluconeogenesis. Thus, seven enzymes are common to both glycolysis and gluconeogenesis: (i) Phosphohexose isomerase; (ii) Aldolase; (iii) Phosphotriose isomerase, (iv) Glyceraldehyde 3-phosphate dehydrogenase; (v) Phosphoglycerate kinase; (vi) Phosphoglycerate mutase; (vii) Enolase.
  • Three reactions of glycolysis are irreversible which are circumvented in gluconeogenesis by four reactions. So, enzymes at these steps are different in glycolysis and gluconeogenesis.

Reactions                                       Enzyme in glycolysis        Enzyme in gluconeogenesis

Glucose – Glucose-6-P                    Hexokinase/glucokinase   Glucose-6-phosphatase

Fructose-6-P – Fructose- I ,6-BP     Phosphofructokinase         Fructose-1-6-bisphosphatase

Phosphoenolpyruvate – Pyruvate    Pyruvate kinase                Pyruvate carboxylase PEP carboxykinase

Quiz In Between


Q. 34

True about glycolysis are all except ‑

 A

Provide nutrition to cancer cells

 B

Substrate level phosphorylation at pyruvate kinase

 C

Two carbon end product is formed

 D

NADPH is formed by glyceraldhyde-3-phosphate dehydrogenase

Q. 34

True about glycolysis are all except ‑

 A

Provide nutrition to cancer cells

 B

Substrate level phosphorylation at pyruvate kinase

 C

Two carbon end product is formed

 D

NADPH is formed by glyceraldhyde-3-phosphate dehydrogenase

Ans. C

Explanation:

Ans. is ‘c’ i.e., Two carbon end product is formed 

Important facts about glycolysis

  • An important biochemical significance is the ability of glycolysis to provide ATP in the absence of oxygen (anerobic glycolysis) and allows tissues to survive anoxic episodes.
  • It occurs in cytosol
  • 3 Carbon atoms end product (pyruvate or lactate) is produced.
  • Irreversible steps are catalyzed by : – Glucokinase/Hexokinase, phosphofructohnase-I, and pyruvate kinase. 
  • Reversible steps are catalyzed by : – Phosphohexose isomerase, aldolase, phosphotriose isomerase, glyceraldehyde 3-phosphate dehydrogenase, Phosphoglycerate kinase, Phosphoglycerate mutase, Enolase.
  • Energy (ATP) using steps are catalyzee by : – Hexokinase/glucokinase, phosphofurctokinase.
  • Energy (ATP) production at substrate level are catalyzed by : Phosphoglycerate kinase, Pyruvate kinase. 
  • Reducing equivalent (NADH) production is catalyzed by : Glyceraldehyde 3-phosphate dehydrogenase.
  • Cancer cells derive nutrition from glycolysis as they have lack of 02 supply because of lack of capillary network. Glycolysis (anaerobic glycolysis) is the only metabolic pathway in the body which can provide energy by glucose metabolism in anerobic conditions.

Q. 35

Anaerobic glycolysis occurs in all places except

 A

Muscles

 B

RBCs

 C

Brain

 D

Kidney

Q. 35

Anaerobic glycolysis occurs in all places except

 A

Muscles

 B

RBCs

 C

Brain

 D

Kidney

Ans. C

Explanation:

Ans. is ‘c’ i.e., Brain 

There are two types of glycolysis : –

  1.  Aerobic glycolysis : – It occurs when oxygen is plentiful and the final product is pyruvate, i.e., final step is catalyzed by pyruvate kinase (see the cycle above). Which is later converted to acetyl CoA by oxidative decarboxylation. There is net gain of 7 ATPs. Acetyl CoA enters TCA cycle.
  2. Anaerobic glycolysis : – It occurs in the absence of oxygen. The pyruvate is fermented (reduced) to lactate in single stage. The reoxidation of NADH (formed in the glyceraldehyde-3-phosphate dehydrogenase step) by respiratory chain is prevented as same NADH is utilized at lactate dehydrogenase step. So, there is no net production of NADH. Thus, there is net gain of 2 ATP only. Unlike pyruvate which is converted to acetyl CoA to enter into krebs cycle, lactate cannot be further utilized by further metabolic pathways. Thus, lactate can be regareded as dead end in glycolysis. Anaerobic glycolysis occurs in exercising skeletal muscle, RBCs, lens, some region of retina, renal medulla, testis and leucocytes.

Q. 36

Reducing equivalants produced in glycolysis are transported from cytosol to mitochondria by ‑

 A

Carnitine

 B

Creatine

 C

Malate shuttle

 D

Glutamate shuttle

Q. 36

Reducing equivalants produced in glycolysis are transported from cytosol to mitochondria by ‑

 A

Carnitine

 B

Creatine

 C

Malate shuttle

 D

Glutamate shuttle

Ans. C

Explanation:

Ans. is ‘c’ i.e., Malate shuttle 

  • Most of the NADH and FADH2, entering the mitochondrial electron transport chain arise from citric acid cycle and 13-oxidation of fatty acids, located in the mitochondria itself.
  • However, NADH is also produced in the cytosol during glycolysis.
  • To get oxidized, NADH has to be transported into the mitochondria as respiratory chain (ETC) is located inside the mitochondria.
  • Since, the inner mitochondrial membrane is not permeable to cytoplasmic NADH, there are special shuttle systems which carry reducing equivalents from cytosolic NADH (rather than NADH itself) into the mitochondria by an indirect route.
  • Two such shuttle systems that can lead to transport of reducing equivalent from the cytoplasm into mitochondria are : –
  1. Malate shuttle (malate-aspartate shuttle system).
  2. Glycerophosphate shuttle.

Quiz In Between


Q. 37

Inhibition of glycolysis by increased supply of 02 is called ‑

 A

Crabtree effect

 B

Pasteur effect

 C

Lewis effect

 D

None

Q. 37

Inhibition of glycolysis by increased supply of 02 is called ‑

 A

Crabtree effect

 B

Pasteur effect

 C

Lewis effect

 D

None

Ans. B

Explanation:

Ans. is ‘b’ i.e., Pasteur effect 

Pasteur effect

  • It has been observed that under anaerobic condition a tissue or microorganism utilizes more glucose than it does under aerobic conditions.
  • It reflects inhibition of glycolysis by oxygen and is called pasteure effect.
  • The Pasteur effect is due to inhibition of the enzyme phosphofructokinase because of inhibitory effect caused by citrate and ATP, the compounds produced in presence of oxygen due to operation of TCA cycle. Crabtree effect
  • This is opposite of Pasteur effect, which represents decreased respiration of cellular systems caused by high concentration of glucose.
  • When oxygen supply is kept constant and glucose concentration is increased, the oxygen consumption by cells falls, i.e., relative anaerobiosis is produced when glucose concentration is increased in constant supply of oxygen.
  • It is seen in cells that have a high rate of aerobic glycolysis.
  • In such cells the glycolytic sequence consumes much of the available Pi and NAD+, which limits their availability for oxidative phosphorylation.
  • As a result, rate of oxidative phosphorylation decreases, and oxygen consumption also shows a corresponding fall.

Q. 38

Which activate Kinase of Glycolysis?

 A

a)     ATP

 B

b)    cAMP

 C

c)     Insulin

 D

d)    Glucagon

Q. 38

Which activate Kinase of Glycolysis?

 A

a)     ATP

 B

b)    cAMP

 C

c)     Insulin

 D

d)    Glucagon

Ans. C

Explanation:

Regulation of Carbohydrate Metabolism-

Enzyme

Inducer

Repressor

Activator

Inhibitor

Glycogen synthase

Insulin

Glucagon

Insulin, glucose-6-phosphate

glucagon

Hexokinase

 

Glucagon

 

Glucose-6-phosphate

Glucokinase

Insulin

Glucagon

 

Citrate, ATP, Glucagon

Phosphofructokinase

Insulin

Glucagon

5 AMP, fructose 6-phosphate, fructose 2,6-biphosphate, Inorganic phosphate

 

 

Pyruvate kinase

 

Insulin

 

Glucagon

 

Fructose 1-6- biphosphate, Insulin

 

ATP alanine glucagon norepinephrine

Pyruvate dehydrogenase

Insulin

Glucagon

CoA, NAD+, Insulin, ADP, pyruvate

Acetyl CoA, NADH, ATP

 


Q. 39

ATP is consumed at which stage of Glycolysis?

 A

a)     Enolase

 B

b)    Hexokinase

 C

c)     Pyruvate kinase

 D

d)    Isomerase

Q. 39

ATP is consumed at which stage of Glycolysis?

 A

a)     Enolase

 B

b)    Hexokinase

 C

c)     Pyruvate kinase

 D

d)    Isomerase

Ans. B

Explanation:

ATP is consumed at reactions catalysed by –> à hexokinase, phosphofructokinase I.

Quiz In Between


Q. 40

In glycolysis, the first committed step is catalysed by-

 

 A

a)     2, 3 DGP

 B

b)    Glucokinase

 C

c)  Hexokinase

 D

 

d)    Phosphofructokinase

Q. 40

In glycolysis, the first committed step is catalysed by-

 

 A

a)     2, 3 DGP

 B

b)    Glucokinase

 C

c)  Hexokinase

 D

 

d)    Phosphofructokinase

Ans. D

Explanation:

Phosphofructokinase catalyzes the commited step of glycolysis, meaning that once fructose l, 6 – bisphosphate is formed it must go for the glycolytic pathway only


Q. 41

The rate- limiting enzyme in glycolysis is-

 A

a)     Phosphofructokinase

 B

b)    Glucose 6- dehydrogenase

 C

c)     Glucokinase

 D

d)    Pyruvate kinase

Q. 41

The rate- limiting enzyme in glycolysis is-

 A

a)     Phosphofructokinase

 B

b)    Glucose 6- dehydrogenase

 C

c)     Glucokinase

 D

d)    Pyruvate kinase

Ans. A

Explanation:

Regulatory steps of Glycolysis are-

–         Hexokinase/ Glucokinase

–         Phosphofructokinase

–         Pyruvate kinase


Q. 42

Number of ATP molecules and NADH formed in each cycle of glycolysis?

 A

a)     4 ATP 2NADH

 B

b)    2 ATP 2 NADH

 C

c)     4 ATP 4 NADH

 D

d)    2 ATP 4 NADH

Q. 42

Number of ATP molecules and NADH formed in each cycle of glycolysis?

 A

a)     4 ATP 2NADH

 B

b)    2 ATP 2 NADH

 C

c)     4 ATP 4 NADH

 D

d)    2 ATP 4 NADH

Ans. A

Explanation:

– In glycoslysis 4 ATPs are produced in which 2 ATPs are utilized so a net gain of 2 ATPs

2 NADH molecules are produced.


Q. 43

Final product in anaerobic glycolysis is :

 A

a)     Pyruvate

 B

b)    Acetyl CoA

 C

c)     Lactate

 D

d)    Oxaloacetate

Q. 43

Final product in anaerobic glycolysis is :

 A

a)     Pyruvate

 B

b)    Acetyl CoA

 C

c)     Lactate

 D

d)    Oxaloacetate

Ans. C

Explanation:

the end product for anaerobic glycolysis is lactate

Quiz In Between



Catabolism of carbon skeleton of amino acids

Catabolism of carbon skeleton of amino acids

Q. 1

Which of the following is not synthesised from tyrosine?

 A

Norepinephrine

 B

Melatonin

 C

Thyroxine

 D

Dopamine

Q. 1

Which of the following is not synthesised from tyrosine?

 A

Norepinephrine

 B

Melatonin

 C

Thyroxine

 D

Dopamine

Ans. B

Explanation:

Q. 2

Which of the following is derived from tyrosine ‑

 A

Melatonin

 B

Serotonin

 C

Melanin

 D

Niacin

Q. 2

Which of the following is derived from tyrosine ‑

 A

Melatonin

 B

Serotonin

 C

Melanin

 D

Niacin

Ans. C

Explanation:

Ans. is ‘c’ i.e., Melanin

Tyrosine is a precursor of many important compounds such as catecholamines (epinephrine, norepinephrine ), dopamine), thyroxine, triiodothryonine, melanin.


Q. 3

Enzyme deficient in tyrosinemia type 1 ‑

 A

Phenylalanine hydroxylase

 B

Tyrosinase

 C

Fumarylacetoacetate hydroxylase

 D

Tyrosine transaminase

Q. 3

Enzyme deficient in tyrosinemia type 1 ‑

 A

Phenylalanine hydroxylase

 B

Tyrosinase

 C

Fumarylacetoacetate hydroxylase

 D

Tyrosine transaminase

Ans. C

Explanation:

 

Tyrosinemia

It is a defect in metabolism of tyrosine. It may be of following types :-

  1. Tyrosinemia type-I (tyrosinosis/hepatorenal syndrome) :- It is due to defect in fumarylacetoacetate hydroxylase deficiency. Patients with chronic tyrosinosis are prone to develop cirrhosis and hepatic carcinoma. There is cabbage like odor in acute tyrosinosis.
  2. Tyrosinemia type – II (Richer-Hanhart syndrome) :- It is due to deficiency of tyrosine transaminase (tyrosine aminotrans-ferase).
  3. Neonatal tyrosinemia : – It is due to deficiency of hydroxyphenyl pyruvate hydroxylase.

Quiz In Between



Isomerism

Isomerism

Q. 1

Identify the Carbohydrate marked as A and B in the Diagram ?

 A

A is Fructose and B is Glucose

 B

A is glucose and B is Fructose

 C

Both A and B is Glucose but Optical Isomers

 D

Both A and B is Fructose but Opitcal Isomers

Q. 1

Identify the Carbohydrate marked as A and B in the Diagram ?

 A

A is Fructose and B is Glucose

 B

A is glucose and B is Fructose

 C

Both A and B is Glucose but Optical Isomers

 D

Both A and B is Fructose but Opitcal Isomers

Ans. B

Explanation:

[img id=9023]


Q. 2

Identify the isomerism shown in the diagram ?

 A

Structural Isomerism 

 B

Stereoisomerism

 C

Opical Isomerism

 D

Anomerism

Q. 2

Identify the isomerism shown in the diagram ?

 A

Structural Isomerism 

 B

Stereoisomerism

 C

Opical Isomerism

 D

Anomerism

Ans. B

Explanation:

stereoisomers are isomeric molecules that have the same molecular formula and sequence of bonded atoms (constitution), but differ in the three-dimensional orientations of their atoms in space.


Q. 3

Alpha D Glucose and Beta D Glucose are examples of ?

 A

Epimers

 B

Anomers 

 C

Opitcal Isomers 

 D

Stereoisomers

Q. 3

Alpha D Glucose and Beta D Glucose are examples of ?

 A

Epimers

 B

Anomers 

 C

Opitcal Isomers 

 D

Stereoisomers

Ans. B

Explanation:

Q. 4

The conversion of an optically pure isomer into a mixture of equal amounts of both dextro and levo forms is called as-

 A

a)      Polymerization

 B

b)      Stereoisomerism

 C

c)      Racemization

 D

d)      Fractionation

Q. 4

The conversion of an optically pure isomer into a mixture of equal amounts of both dextro and levo forms is called as-

 A

a)      Polymerization

 B

b)      Stereoisomerism

 C

c)      Racemization

 D

d)      Fractionation

Ans. C

Explanation:

  •  Racemic Mixture – Equimolar mixture of optical isomers which has no net reaction of plane polarized light.

 

Quiz In Between



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