Author: Renu Maurya

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



Glycolysis Cycle

Glycolysis Cycle


GLYCOLYSIS (EMBDEN- MEYERHOF PATHWAY)

  • Glycolysis is conversion of glucose into pyruvate (aerobic) or lactate (anaerobic)
  • Glycolysis takes place in all cells of the body.
  • The only metabolic fuel for mature erythrocytes in fed & starving state is glucose.
  • Glycolysis occurs in cytosol as all the enzymes are present here.
  • Steps involved in the cycle are-

1. Step 1- Glucose to glucose-6- phosphate (first rate limiting step)

  • Hexokinase– found in all tissues
  • Irreversible reaction.
  • No effect by feeding or insulin or starvation nor specific for glucose metabolism.
  • Glucokinase-
  • Found in liver and pancreatic β cells.
  • High Km for glucose & specific glucose.
  • Not inhibited by glucose-6-phosphate.
  • Induced by Insulin following a meal and important role in regulation of blood glucose
  • Glucose-6-phosphate has many fates as- Glycolysis, glycogenesis, gluconeogenesis, pentose phosphate pathway.

2. Step 2- Glucose-6-phosphate is isomerised to fructose-6-phosphate by phosphohexose isomerase

3. Step 3- Fructose-6-phosphate is further phosphorylated to fructose 1, 6- biphosphate by phosphofructokinase

  • This is an irreversible.
  • It is a rate limiting step & committed step and 1 ATP utilized.
  • Phosphofructokinase- major regulatory enzyme
  • Phosphofructokinase– allosterically inhibited by citrate

             The 1st, 2nd & 3rd steps are the energy investment phase.

4. Step 4- Fructose 1, 6 biphosphate splits into glyceraldehyde 3-phophate & dihydroxyacetone phosphate

  • Catalyzed by enzyme Aldolase.
  • Aldolase is Lyase.
  • Step 4 is called as Splitting phase.

5. Step 5-Glyceraldehydes 3- phosphate is oxidised to 1, 3- bisphospho glycerate  

  • Enzyme glyceraldehyde – 3- phosphate dehydrogenase.
  • An inorganic phosphate is added. (Iodoacetate & arsenate inhibits enzyme)

6. Step 6-  1, 3- biphosphoglycerate forms 3 –phosphoglycerate

  • Enzyme phosphoglycerate kinase 
  • Synthesis ATP.
  • This is an example of substrate level phosphorylation

7. Step 7- 3- phosphoglycerate converted to 2- phosphoglycerate 

  • Enzyme phosphogluco mutase (irreversible reaction)

8. Step 8- 2- phosphoglycerate is converted to phosphoenol pyruvate 

  • Enzyme enolase which requires Mg2+ or Mn2+inhibited by fluoride.

9. Step 9- Phosphophenol Pyruvate (PEP) is dephosphorylated to pyruvate 

  • Enzyme pyruvate kinase (ATP formed)
  • This is an irreversible step.

10. Step 10- In anaerobic condition, pyruvate is reduced to lactate

  • Enzyme lactate dehydrogenase.
  • 6th, 7th, 8th, 9th, 10th step is called energy generation phase.

Significance of Gycolysis-

  • In strenuous exercise, anaerobic glycolysis forms the major source of energy for muscles.
  • Tissues depending mainly on glucose as metabolic fuel- White fibres of skeletal muscle, erythrocytes, brain, GIT, renal medulla, skin.

Production of ATP in glycolysis-

  • Under anaerobic conditions, 2 ATP are synthesized.
  • Under aerobic conditions, 8 or 6 (7) ATP are synthesized.
  • Glycolysis under glycogen, 1 ATP is generated.
  • In anaerobic glycolysis, 3 ATP from glycogen.
  • The inhibition of glycolysis by oxygen is known as Pasteur effects.

Exam Important

  • Principal route for Carbohydrate metabolism is glycolysis.
  • Only pathway which can operate ananerobically and aerobically.
  • Mature Erythrocytes which lack mitochondria are completely dependent on glucose for metabolic fuel.
  • Glucokinase is induced by Insulin following a meal.
  • Fructose 6 phosphate to Fructose 1,6 bisphophate is the major regulatory step in glycolysis and is a committed step also.
  • Irreversible steps of glycolysis are- Hexokinase, Phosphofructokinase, Pyruvate kinase.
Don’t Forget to Solve all the previous Year Question asked on Glycolysis Cycle

Module Below Start Quiz

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



Isomerism

Isomerism


ISOMERISM

  • Isomerism in CarbohydratesDifferent compounds or structures having same molecular formula is called isomers and the phenomenon existing is called isomerism. 
  • Carbon to which four different substituent groups are attached called as Chiral or asymmetric carbon atom. 
  • Lebervon’t Hoff RuleRelation between asymmetric carbon atom and number of stereoisomers possible. (No. Of isomers= 2n) 
  • Types of Isomers In Carbohydrates- 1) Stereoisomerism        2) Optical isomerism

1)  Stereoisomers- compounds which have same structural formula but differ in spatial configuration of H and OH group around asymmetric carbon atoms.

  • Enantiomers (D and L isomerism) – *D sugars are naturally occurring. Examples are
    • D Glucose and L Glucose
    • D Fructose and L Fructose
    • D Mannose and L Mannose
    • D Glyceraldehyde and L Glyceraldehyde.    
  • Optical Activity- The presence of asymmetric carbon atom causes optical activity.
  •  When a beam of plane-polarized light is passed through a solution of carbohydrates, it will rotate the light either to right or to left. 
  • Depending on rotation it is called as

 a. Levorotatory (l or -) anticlockwise                                b.  Dexarotatory (d or +) clockwise

      (Rotate plane polarised light to left)                                    (Rotate plane polarised light to right)

     D Fructose                                                                                D glucose ( Dextrose) 

  • Racemic Mixture Equimolar mixture of optical isomers which has no net reaction of plane polarized light. 
  • Epimerism (Diastereoisomerism)When sugars are different from one another, only in configuration with regard to a single carbon atom, other than the Glucose reference carbon atom, they are called epimers. 

           Epimers of glucose

  1. 2nd Epimer of Glucose – Mannose
  2. 3rd Epimer of Glucose – Allose
  3. 4th Epimer of Glucose – Galactose. 
  • AnomerismFormation of ring structure in monosaccharides results in creation of an additional asymmetric carbon called Anomeric Carbon atom.
    • Difference in orientation of H and OH group around the anomeric carbon atom results in Anomerism and the resulting isomers are called α and β anomers.
    • Examples of Anomerism:
  1. αDGlucose and  β DGlucose
  2. α D Fructose and β D Fructose. 
  • Mutarotationis the change in the specific optical rotation of plane polarised light with time.
    • The optical rotation of α glucose is +112.20
    • The optical rotation of β glucose is 18.70  ( 190)
    • The optical rotation of Fructose is -920
    • An equilibrium with a constant value is +52.70 
  • Benedict’s Reaction- Standard lab test to diagnose Diabetes mellitus. 
  • Shapes of Osazones-
  • Needle-shaped/Broomstick/sheaves of corn – Glucose, Fructose, Mannose
  • Pincushion with pins/Hedgehog/ Flower of Touch –me-not – Lactose
  • Sunflower Petal-shaped –  Maltose 
  • Sucrose is also called as invert sugar.

Exam Important

  1. Monosaccharide with no Asymmetric Carbon atom- Dihydroxyacetone.
  2. Aldose sugar is converted to ketose and vice versa by enzyme isomerase.
  3. Glucose and mannose are C2 epimers and glucose & galactose are C4 epimers.
  4. Maltose is a reducing disacharide.
  5. Lactose is a reducing dissacharide.
  6. Sucrose is a non-reducing sugar.
  7. Benedict’s quantitative reagent detects any reducing sugar.
  8. Fehling solution contains CuSO4 and Ruchelle salt .
  9. Glucose oxidase method is used for glucose.
  10. Perioxidase and oxidase enzymes are used for estimation of glucose. 
Don’t Forget to Solve all the previous Year Question asked on Isomerism

Module Below Start Quiz

Derived Sugars

Derived Sugars

Q. 1

Amino sugar are formed from-

 A

a)      Glucose- 1- phosphate

 B

b)      Glucose -6- phosphate

 C

c)      Fructose- 1- phosphate

 D

d) Fructose- 6- phosphate

Q. 1

Amino sugar are formed from-

 A

a)      Glucose- 1- phosphate

 B

b)      Glucose -6- phosphate

 C

c)      Fructose- 1- phosphate

 D

d) Fructose- 6- phosphate

Ans. D

Explanation:

They are synthesized from fructose -6- phosphate


Q. 2

Gluconic acid formation requires which enzyme?

 A

a)      Glucose oxidase

 B

b)      Glucose dehydrogenase

 C

c)      Glucokinase

 D

d) Glucose dehydratase

Q. 2

Gluconic acid formation requires which enzyme?

 A

a)      Glucose oxidase

 B

b)      Glucose dehydrogenase

 C

c)      Glucokinase

 D

d) Glucose dehydratase

Ans. A

Explanation:

Q. 3

Glucose is converted to glucoronate by

 A

a)      Oxidation of aldehyde group

 B

b)      Oxidation of terminal group

 C

c)      Oxidation of both

 D

d)      None

Q. 3

Glucose is converted to glucoronate by

 A

a)      Oxidation of aldehyde group

 B

b)      Oxidation of terminal group

 C

c)      Oxidation of both

 D

d)      None

Ans. B

Explanation:

glucose is converted to glucuronic acid by oxidation of only terminal alcohol group

Quiz In Between



Derived Sugars

Derived Sugars


  • Oxidation of Sugar- Under mild oxidation, aldehyde group is oxidised to Aldonic Acid.
  • *Glucose to Gluconic Acid
  • Mannose to Mannonic Acid
  • Galactose to Galactonic Acid.
  • Oxidation of aldehyde group from glucose in blood glucose estimation results in Gluconic Acid.
  • Oxidation of terminal alcohol group leads to the production of Glucuronic acid.
  • Under strong oxidation-
  • Glucose to glucosaccharic acid
  • Mannose to mannaric acid
  • Galactose to mucic acid

 Sugar Alcohols-

Glucose Sorbitol
Mannose Mannitol Reduces intracranial pressure by forced diuresis
Galactose Dulcitol/Galactitol Osmotic effect causes cataract in Galactosemia &Diabetes
Fructose Sorbitol and Mannitol  

 

  • Deoxy Sugar- Oxygen of the hydroxyl group may be removed to form deoxy sugars.
  • Deoxyribose– is an important part of nucleic acid.
  • Oxygen is removed from the 2nd position
  • Feulgen Staining is specific for 2-deoxy sugar(DNA) in tissues.(*Schiff’s agent)
  • L-fuctose is present in the blood antigen.  
  • Amino Sugars– Amino group substituted for hydroxyl group in the second carbon atom of monosacharides to form Amino Sugars. 
Glucosamine Seen in hyluronic acid, heparin & blood group substance.
Galactosamine (Chondrosamine) Present in chondroitin of cartilage, tendons & bone.
Mannosamine Constituent of glycoproteins.
  • An unusual amino sugar with 9 carbon atom is Sialic Acid.
  • The principal Sialic Acid found in human body is N Acetyl Neuraminic Acid (NANA) and is derivative f N-acetylmannose and pyruvic acid.
  • Erythromycin is an antibiotic which contains amino sugar.
  • GLYCOSIDES- When the monosaccharide is condensed with an alcohol, phenol or sterol by O- Glycosidic Linkage to form glycoside.
  • Digitonin is a cardiac stimulant.
  • Quabain in cardiac effect
  • Phlorhizin produces renal damage.
  • Antibiotic- Streptomycin, Puromycin.
  • Benedict’s Reaction- Standard lab test to diagnose Diabetes mellitus.
  • Shapes of Osazones-
  • Needle-shaped/Broomstick/sheaves of corn – Glucose, Fructose, Mannose
  • Pincushion with pins/Hedgehog/ Flower of Touch –me-not – Lactose
  • Sunflower Petal-shaped –  Maltose

METHOD OF ESTIMATION OF GLUCOSE-

 Reductometric Methods
  1. Nelson Somogyi Method
  2. Folin-Wu Method
  3. O- Toluidine Method
Enzymatic Method
  1. Hexokinase Method
  2. Glucose-Oxidase Perioxidase Method (GOD-POD)
  3. Highly specific method 
  4. Used in dry nanlysis technique                                                       

 

TESTS FOR CARBOHYDRATES

General test for all carbohydrates Molisch test
Tests for reducing substances Benedict’s Test
Test to differentiate monosaccharides and Diasaccharides Barfoed’s Test

Moore’s Test

Fehling’s Test

Test to differentiate Aldoses and Ketoses Seliwanoff’s Test

Rapid Furfural Test

Fougler’s Test

Test to detect Deoxy Sugar Feulgen Staining
Test for Pentoses Bial’s Test
Test for Galactose Mucic Acid Test
  • Reaction In Glucose oxidase and Perioxidase Method is- (GOD-POD)
  • Glucose + H2O2 + O2  GOD→   Gluconic Acid + H2O2
  • 2H2O2 + 4 aminoantipyrine + PHB  POD→ __ Quinoneminine dye + H2O (RED COLOR)

Exam Important

  1. Derived Sugars include- Acid sugars, Sugar alcohols, Deoxy sugars, Amino sugars, Glycosides, Furfural derivatives.
  2. The immediate precursor of Glucosamine is Fructose 6 phosphate.
Don’t Forget to Solve all the previous Year Question asked on Derived Sugars

Module Below Start Quiz

Chemistry of Carbohydrates

Chemistry of Carbohydrates

Q. 1

Which of the following are enantiomers:

 A

D-Galactose & L-Glucose

 B

d-Galactose & 1-Glucose

 C

D-Mannose & L-Mannose

 D

All Correct

Q. 1

Which of the following are enantiomers:

 A

D-Galactose & L-Glucose

 B

d-Galactose & 1-Glucose

 C

D-Mannose & L-Mannose

 D

All Correct

Ans. C

Explanation:

C i.e. D-Mannose


Q. 2

Epimer combination (s) is/are :

 A

D-glucose & D-fructose

 B

D-mannose & D-talose

 C

D-glucose & D-mannose

 D

None of these

Q. 2

Epimer combination (s) is/are :

 A

D-glucose & D-fructose

 B

D-mannose & D-talose

 C

D-glucose & D-mannose

 D

None of these

Ans. C

Explanation:

C, i.e. D-glucose

–  Carbohydrate isomers that differ in configuration around only one carbon atom are called epimers. D-glucose and D­mannose are C2 epimers whereas D- glucose and D- galactose differ only at C4. However, D-mannose & D- galactose are not epimersQ as they differ at 2 carbons (2 & 4).

Enantiomers (optical isomers or stereoisomers) are pairs of structures that are mirror images of each other but not identical , similar to left and right hand which are the same but opposite. These are non super imposable and are designated as D and L-sugar. Each enantiomer often shows different chemical reactions with other enantiomers. Because of presence of many enantiomers in living beings, there is usually a marked difference in effects of two stereoisomers. For example only one of drug stereoisomer produces desired effect while the other does not.


Q. 3

Hexose sugar is not present in:    

 A

Ribose

 B

Glucose

 C

Fructose

 D

Galactose

Q. 3

Hexose sugar is not present in:    

 A

Ribose

 B

Glucose

 C

Fructose

 D

Galactose

Ans. A

Explanation:

 

Hexose sugar is present in:

  • Glucose, “blood sugar”, the immediate source of energy for cellular respiration
  • Galactose, a sugar in milk (and yogurt), and
  • Fructose, a sugar found in honey
  • Ribose is a pentose sugar

Quiz In Between


Q. 4

Ketone bodies are formed in:        

 A

Liver

 B

Pancreas

 C

Kidneys

 D

Lungs

Q. 4

Ketone bodies are formed in:        

 A

Liver

 B

Pancreas

 C

Kidneys

 D

Lungs

Ans. A

Explanation:

 

Ketone bodies

  • They are three different water-soluble biochemicals that are produced as by-products when fatty acids are broken down for energy in the liver.
  • Two of the three are used as a source of energy in the heart and brain while the third (acetone) is a waste product excreted from the body.
  • In the brain, they are a vital source of energy during fasting.
  • The three endogenous ketone bodies are

– Acetone,

Acetoacetic acid, and

–  Beta-hydroxybutyric acid


Q. 5

The L or D form of a sugar is determined by its relation to: 

 A

Fructose

 B

Glycogen

 C

Glyceraldehyde

 D

Glucose

Q. 5

The L or D form of a sugar is determined by its relation to: 

 A

Fructose

 B

Glycogen

 C

Glyceraldehyde

 D

Glucose

Ans. C

Explanation:

An optical isomer can be named by the spatial configuration of its atoms.

The D/L system does this by relating the molecule to glyceraldehyde.

Glyceraldehyde is chiral itself, and its two isomers are labeled D and L


Q. 6

Alpha D Glucose and Beta D Glucose are examples of ?

 A

Epimers

 B

Anomers 

 C

Opitcal Isomers 

 D

Stereoisomers

Q. 6

Alpha D Glucose and Beta D Glucose are examples of ?

 A

Epimers

 B

Anomers 

 C

Opitcal Isomers 

 D

Stereoisomers

Ans. B

Explanation:

Q. 7

Which of the following is an aldose sugar ‑

 A

Ribulose

 B

Fructose

 C

Glyceraldehyde

 D

All of the above

Q. 7

Which of the following is an aldose sugar ‑

 A

Ribulose

 B

Fructose

 C

Glyceraldehyde

 D

All of the above

Ans. C

Explanation:

Quiz In Between



Chemistry of Carbohydrates

Chemistry of Carbohydrates


CARBOHYDRATES

(Hydrates of Carbon)

  • Main source of energy in the body as Glycogen (45-65% of total energy).
  • Produces energy from carbohydrates is 4 k calories/g (16 k joules/g).
  • Functions- Energy storage in the form of starch and glycogen structural basis.
  • Molecular formula for carbohydrate is Cn(H2O)n.
  • Carbohydrates are polyhydroxy aldehydes or ketones.
  • Carbohydrate broadly classified in 4 parts
  1. Monosaccharides
  2. Disaccharides
  3. Oligosaccharides
  4. Polysaccharides
  1. Homopolysaccharides
  2. Heteropolysaccharides
  •  Monosaccharides Cn(H2O)n– classified into –

According to the number of carbons-

  1. Trioses
  2. Tetroses
  3. Pentoses
  4. Hexoses
  5. Heptoses
  6. Nanoses

According to the fucntional group-

  1. Aldoses
  2. Ketoses
  •   Glucose is an aldohexose.
  •   Frusctose is a ketohexose. Present in seminal fluid. Present in honey, fruit juice.
    •  Glyceraldehyde (triose) is the simplest monosaccharide.
    •  Glucose is the main source of energy present in blood often called as Dextrose.
    • Galactose epimerised to glucose in liver.
    • Glucose is the main source of energy for organs- Brain, Renal medulla, Cornea, Retina, RBC, testis.
    • The only metabolic fuel for mature erythrocytes in fed state and starving state is Glucose.
    • Hexose is the reducing sugars.
  • Disaccharides (Cn(H2O)n-1)    two monosaccharides units are linked by glycosidic bond. Divided into 2- 

A.) Reducing sugar(free functional group present)

Disaccharides Sugar Units Linkage
Maltose αD Glucose+ α D Glucose α 1+ α4 linkage( α1 —— 4 linkage)
Isomaltose α  DGlucose+  α DGlucose α1+ α6Linkage (α1 —– 6 linkage)
Lactose(milk sugar) DGalactose+β D Glucose β1 —— β4 linkage
Lactulose α D Galactose+ β D Fructose α1——- β4 linkage

B) Non reducing sugar- No free functional group present.

Disaccharides Sugar Units Linkage
Trehalose (Sugar of insect  hemolymph, yeast and fungi) αDGlucose + αDGlucose   α1 —– α1 linkage
Sucrose (Cane Sugar) αDGlucose

+β DFructose

   α1 —– β2 linkage
  • Polysaccharides-

a)Homopolysaccharides

  • Glycogen- Storage form of Glucose in animals, so called as animal starch.
  • Glycogen present in muscle and liver.
  • Starch (glucan or glucosan)

Homopolysaccharides are made of glucose only and is found in plants.

Ø Chitin- Made up of N Acetyl D Glucosamine joined by β1 —- 4 linkage.

Ø Cellulose- chief constutuent of plant cell wall.

Human cannot digest cellulose as they lack the hydrolose β 1, 4 glycosidic bonds.

Ø Iulin(Fructosan)- Homopolysaccharide of Fructose in β 2 —–1 linkage.

Used in clearance test to determine GFR.

  • Ø Dextran- used as a plasma volume expander. ( Eg. Human albumin, Dextran, Hydroxyethyl starch (Hetastarch), Degraded gelatin polymer.
  • Ø Glycogen

b) Heteropolysaccharides-(Heteroglycans)

  • Glycoaminoglycans (Mucopolysaccharides) – unbranched heteropolysaccharide
  • Pectins
  • Agarose
  • Agar

Important Glycoaminoglycans present in the body-

1)Hyluronic acid (Hyluronan) Skin, Synovial fluid, bone, cartilage, vitreous humor, loose connective tissue, umbilical cord.
2)Chondroitin Sulphate Cartilage, bone, CNS
3)Keratan Sulphate I & II Cornea, Cartilage, loose CT
4)Heparin (anticoagulant) Mast cells, liver, lung, skin
5) Heparan sulphate Skin, kidney
  • Glycoaminoglycan with no uronic acid is Keratan sulphate.
  •  Mucin Clot Test (Rope test)—the test detects hyluronate in the synovial fluid.
  • Normal synovial fluid forms a tight ropy clot in the presence of acetic acid.
  • Mucopolysaccharidoses (MPS)- Herediatry progressive disease caused by mutation of genes coding for Lysosomal Enzymes needed to degrade glcoaminoglycan results in Intralysosomal accumulation of gycoaminoglycans.
  • Diseases Caused by Mucopolysaccharidosis (MPS)

1. Hurler’s Disease (MPS-I H)-

  • Inheritence- autosomal recessive
  • Defective enzyme- L-Iduronidase
  • Clinical features-
    • Premature death
    • Hepatosplenomegaly
    • coarse facial features
    • corneal clouding
    •  large tongue
    • prominent forehead
    •  joint stiffness
    •  short stature
    • skeletal dysplasia
    • obstructive airways
    • Valvular heart disease

2. Scheie Disease (MPS-IS)

  • Inheritence- autosomal recessive
  • Defective enzyme- L- Iduronidase
  • Clinical features-
    • joint stiffness
    • aortic valve disease,
    • corneal clouding
    •  mild dysostosis multiplex

3. Hunter Disease (MPS II)-

  • Inheritence- X-linked Recessive
  • Defective enzyme- Iduronate Sulphatase
  • Clinical Features-
    • slower progression of somatic and central nervous system (CNS) deterioration
    • large tongue
    • prominent forehead
    •  joint stiffness
    •  short stature
    • skeletal dysplasia
    • obstructive airways
    • Valvular heart disease
    • Mental retardation

Maroteaux-Lamy Disease (MPS VI)

  • Inheritence- autosomal recessive
  • Defective enzyme- N Acetyl Galactoseamine 4 sulphatase (Aryl Sulphatase B)
  • Clinical features-
    • Premature death
    • Hepatosplenomegaly
    • coarse facial features
    • corneal clouding
    •  large tongue
    • prominent forehead
    •  joint stiffness
    •  short stature
    • skeletal dysplasia
    • obstructive airways
    • Valvular heart disease
  • Isomerism in Carbohydrates- different compounds having same molecular formula.
  • Carbon to which four different substituent groups are attached called as Chiral or asymmetric carbon atom.
  • Lebervon’t Hoff Rule- Relation between asymmetric carbon atom and number of stereoisomers possible. (No. Of isomers= 2n)
  • Types of Isomers In Carbohydrates- 1) Stereoisomerism        2) Opticalisomerism

Stereoisomers- compounds which have same structural formula but differ in spatial configuration of H and OH group around asymmetric carbon atoms.

Enantiomers (D and L isomerism) – *D sugars are naturally occurring. Examples are

  • D Glucose and L Glucose
  • D Fructose and L Fructose
  • D Mannose and L Mannose
  • D Glyceraldehyde and L Glyceraldehyde

Optical Activity-

    The presence of asymmetric carbon atom causes optical activity. When a beam of plane-polarized light is passed through a solution of carbohydrates, it will rotate the light either to right or to left.
  • Depending on rotation it is called as

Levorotatory (l or -) anticlockwise                                    – Dexarotatory (d or +) clockwise

(Rotate plane polarised light to left)                                        (Rotate plane polarised light to right)

D Fructose                                                                                   D glucose ( Dextrose)

  • Racemic Mixture – Equimolar mixture of optical isomers which has no net reaction of plane polarized light.
  • Epimerism (Diastereoisomerism) – When sugars are different from one another, only in configuration with regard to a single carbon atom, other than the reference carbon atom, they are called epimers.

Epimers of glucose

  1. 2nd Epimer of Glucose – Mannose
  2. 3rd Epimer of Glucose – Allose
  3. 4th Epimer of Glucose – Galactose.
  • Anomerism- Formation of ring structure in monosaccharides results in creation of an additional asymmetric carbon called Anomeric Carbon atom.
    • Difference in orientation of H and OH group around the anomeric carbon atom results in Anomerism and the resulting isomers are called α and β anomers.
    • Examples of Anomerism:

a)  αDGlucose and  β DGlucose

b) α D Fructose and β D Fructose.

  • Mutarotation- is the change in the specific optical rotation of plane polarised light with time.
    • The optical rotation of α glucose is +112.20
    • The optical rotation of β glucose is 18.70  ( 190)
    • The optical rotation of Fructose is -920
    • An equilibrium with a constant value is +52.70
  • Benedict’s Reaction- Standard lab test to diagnose Diabetes mellitus.
  • Shapes of Osazones-
  • Needle-shaped/Broomstick/sheaves of corn – Glucose, Fructose, Mannose
  • Pincushion with pins/Hedgehog/ Flower of Touch –me-not – Lactose
  • Sunflower Petal-shaped –  Maltose
  •  Oxidation of Sugar- Under mild oxidation, aldehyde group is oxidised to Aldonic Acid.
  • Glucose to Gluconic Acid
  • Mannose to Mannonic Acid
  • Galactose to Galactonic Acid.
  • Oxidation of aldehyde group from glucose in blood glucose estimation results in Gluconic Acid.
  • Oxidation of terminal alcohol group leads to the production of Glucuronic acid.
  • Under strong oxidation-
  • Glucose to glucosaccharic acid
  • Mannose to mannaric acid
  • Galactose to mucic acid
  • Sugar Alcohols-
Glucose Sorbitol
Mannose Mannitol Reduces intracranial pressure by forced diuresis
Galactose Dulcitol/Galactitol Osmotic effect causes cataract in Galactosemia &Diabetes
Fructose Sorbitol and Mannitol  
  • Deoxy Sugar- Oxygen of the hydroxyl group may be removed to form deoxy sugars.
  • Deoxyribose– is an important part of nucleic acid.
  • Oxygen is removed from the 2nd position
  • Feulgen Staining is specific for 2-deoxy sugar(DNA) in tissues.(*Schiff’s agent)
  • L-fuctose is present in the blood antigen.  
  • Amino Sugars– Amino group substituted for hydroxyl group in the second carbon atom of monosacharides to form Amino Sugars.
Glucosamine Seen in hyluronic acid, heparin & blood group substance.
Galactosamine (Chondrosamine) Present in chondroitin of cartilage, tendons & bone.
Mannosamine Constituent of glycoproteins.
  •  An unusual amino sugar with 9 carbon atom is Sialic Acid.
  • The principal Sialic Acid found in human body is N Acetyl Neuraminic Acid (NANA) and is derivative f N-acetylmannose and pyruvic acid.
  • Erythromycin is an antibiotic which contains amino sugar.
  • GLYCOSIDES- When the monosaccharide is condensed with an alcohol, phenol or sterol by O- Glycosidic Linkage to form glycoside.
  • Digitonin is a cardiac stimulant.
  • Quabain in cardiac effect
  • Phlorhizin produces renal damage.
  • Antibiotic- Streptomycin, Puromycin.

TESTS FOR CARBOHYDRATES

General test for all carbohydrates Molisch test
Tests for reducing substances Benedict’s Test
Test to differentiate monosaccharides and Diasaccharides Barfoed’s Test

Moore’s Test

Fehling’s Test

Test to differentiate Aldoses and Ketoses Seliwanoff’s Test

Rapid Furfural Test

Fougler’s Test

Test to detect Deoxy Sugar Feulgen Staining
Test for Pentoses Bial’s Test
Test for Galactose Mucic Acid Test

METHODS OF ESTIMATION OF GLUCOSE-

Reductometric Methods
  1. Nelson Somogyi Method
  2. Folin-Wu Method
  3. O- Toluidine Method
Enzymatic Method 1. Hexokinase Method

2. Glucose-Oxidase Perioxidase Method (GOD-POD)

3. Highly specific Method

4. Used in dry ananlysis technique like Glucometer.

  • Reaction In Glucose oxidase and Perioxidase Method is-

Glucose + H2O2 + O2     GOD          Gluconic Acid + H2O2

2H2O2 + 4 aminoantipyrine + PHB    POD       Quinoneminine dye + H2O (RED COLOR)

Exam Important

1. General Formula for Carbohydrates-

  1. Cn(H2O)n-1
  2. Cn(H2O)n
  3. Both of above
  4. None of the above

Answer- Cn(H2O)n

Explaination- The general formula for carbohydrates is Cn(H2O)n  where n= no of carbon atoms.

2. Glycogen is the main source of energy for human beings. What is the percentage of that-

  1. 70-90% of total energy
  2. 25-35% of total energy
  3. 10-20% of total energy
  4. 45-65% of total enegy

Answer- 45-65% of the total energy

Explanation- Primary source of energy for human beings is Carbohydrates as Glycogen which is around 45-65% of total energy.

3. In a starving state in which form does body have energy?

  1. Glucose
  2. Fructose
  3. Galactose
  4. Mannose

Answer- Glucose

Explaination- Glucose is the main metabolic fuel for mature erythrocytes in a starving state.

4. Reducing sugar except-

  1. Glucose
  2. Sucrose
  3. Fructose
  4. Mannose

Answer- Sucrose

Explanation- Sucrose is a non reducing sugar.

5. Which of the following is not an aldose?

  1. Glucose
  2. Mannose
  3. Fructose
  4. Galactose

Answer- Fructose

Explanation-

Triose Glyceraldehyde
Tetrose Erythrose
Pentose Ribose, Xylose, Arabinose
Hexose Glucose, Galactose, Mannose
Heptose

6. Lactose is a constituent of-

  1. Maltose
  2. Mannose
  3. Fructose
  4. Galactose

Answer- Galactose

Explanation- Galactose is a constituent of Lactose(milk sugar) synthesized in the mammary gland for synthesis of lactose.

7. Glucose detection can be done by all except-

  1. Glucose oxidase
  2. Ferric Chloride test
  3. Dextrostix
  4. Folin-Wu method

Answer- Ferric Chloride test

Explanation- Ferric Chloride test is the test done in Alkaptonuria and

Phenyl Ketonuria.

8. Which of the following are enantiomers-

  1. D-Galactose and L-Glucose
  2. D-Galactose and I-Glucose
  3. D-Mannose and L-Mannose
  4. D-mannose and I-Mannose

Answer- D-Mannose and L-Mannose

Explanation- examples of enantiomers-

  • D Glucose and L Glucose
  • D Fructose and L Fructose
  • D Mannose and L Mannose
  • D Glyceraldehyde and L Glyceraldehyde

9. Cellulose is-

  1. Complex Lipoprotein
  2. Starch Polysaccharide
  3. Nonstarch Polysaccharide
  4. Complex Lipoprotein

Answer- Nonstarch polysaccharide

Explanation- Foods contain a wide variety of other polysacdnarides that are collectively known as nonstarch polysaccharides. They are not digested by human enzymes/ and are the major components of dietary fiber.

Examples are cellulose from plant cell walls (a glucose polymer) and inulin, the storage carbohydrate in some plants (a fructose Polymer)’

10. Mucopolysaccharide that doesn’t contains Uronic acid residue is-

  1. Heparan Sulfate
  2. Heparin
  3. Chondritin sulphate
  4. Keratan sulphate

Answer- keratin sulphate

Explanation-

GAG with no Uronic Acid- Keratan Sulfate

GAG with no Sulfate group- Hyaluronic Acid

GAG not covalently linked to protein- Hyaluronic Acid.

Don’t Forget to Solve all the previous Year Question asked on Chemistry of Carbohydrates

Module Below Start Quiz

Antioxidant

Antioxidant

Q. 1

Allopurinol is used in organ preservation as:

 A

Antioxidant

 B

Preservative

 C

Free radical scavenger

 D

Precursor for energy metabolism

Q. 1

Allopurinol is used in organ preservation as:

 A

Antioxidant

 B

Preservative

 C

Free radical scavenger

 D

Precursor for energy metabolism

Ans. C

Explanation:

Free radical scavenger [Ref www.essentialpharma.com/Eghtesad5-18-07.ppt]

Organ preservation is the supply line for organ transplantation. Currently, the liver, pancreas, and kidney can be successfully preserved for up to two days by flushing the organs with the University of Wisconsin (UW) solution and storing them at hypothermia (0-5° C). The UW solution is effective because it uses a number of cell impermeant agents (lactobionic acid, raffinose, hydroxyethyl starch) that prevent the cells from swelling during cold ischemic storage. Additionally, the UW solution contains glutathione, allopurinol and adenosine, agents that may stimulate recovery of normal metabolism upon reperfusion by augmenting the antioxidant capacity of the organs (glutathione) or by stimulating high-energy phosphate generation (adenosine) upon reperfusion and allopurinol as free-radical scavenger.

  • Glutathione——— as antioxidant
  • Adenosine——– as precursor for energy metabolism
  • Allopurinol —– as free-radical scavenger

Q. 2

Which of the following elements is known to influence the body’s ability to handle oxidative stress?

 A

Calcium

 B

Iron

 C

Potassium

 D

Selenium

Q. 2

Which of the following elements is known to influence the body’s ability to handle oxidative stress?

 A

Calcium

 B

Iron

 C

Potassium

 D

Selenium

Ans. D

Explanation:

Selenium is essential to the production of glutathione peroxidase, a constituent of the antioxidant defence system. 

Ref: Harper’s Biochemistry, 26th Edition, Pages 88, 166, 496 ; Ghai Essential Pediatrics 6th, Edition, Page 131

 


Q. 3

The following are biological antioxidants which act against the free radicals formed inside the body, EXCEPT:

 A

Selenium

 B

Glycine

 C

Uric acid

 D

Ascorbic acid

Q. 3

The following are biological antioxidants which act against the free radicals formed inside the body, EXCEPT:

 A

Selenium

 B

Glycine

 C

Uric acid

 D

Ascorbic acid

Ans. B

Explanation:

The peroxides that are formed by radical damage to lipids in membranes and plasma lipoproteins are reduced to hydroxy fatty acids by glutathione peroxidase, a selenium-dependent enzyme (hence the importance of adequate selenium intake to maximize antioxidant activity). Ascorbate, uric acid and a variety of polyphenols derived from plant foods act as water-soluble radical trapping antioxidants, forming relatively stable radicals that persist long enough to undergo reaction to non-radical products. Ubiquinone and carotenes similarly act as lipid-soluble radical-trapping antioxidants in membranes and plasma lipoproteins. Lipid peroxides are also reduced to fatty acids by reaction with vitamin E.
 
Ref: Bender D.A. (2011). Chapter 45. Free Radicals and Antioxidant Nutrients. 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. 4

Which of the following elements is known to influence the body’s ability to handle oxidative stress?

 A

Calcium

 B

Iron

 C

Potassium

 D

Selenium

Q. 4

Which of the following elements is known to influence the body’s ability to handle oxidative stress?

 A

Calcium

 B

Iron

 C

Potassium

 D

Selenium

Ans. D

Explanation:

D i.e. Selenium 

Selenium in the form of ‘selenocysteine’ is a component of enzyme glutathione peroxidaseQ. ‘Glutathione peroxidase serve to protect proteins, cell membranes, lipid and nucleic acids from oxidant molecules’. Glutathione prevents oxidation of hemoglobin to methemoglobinQ.


Q. 5

Which of the following reduces oxidative stress except

 A

Superoxide dismutase

 B

Catalase

 C

Glutathione peroxidase

 D

Xanthine oxidase

Q. 5

Which of the following reduces oxidative stress except

 A

Superoxide dismutase

 B

Catalase

 C

Glutathione peroxidase

 D

Xanthine oxidase

Ans. D

Explanation:

D i.e. Xanthine oxidase


Q. 6

Which of the following act as antioxidants:

 A

Glutathione peroxidase

 B

Vit. C

 C

Selenium

 D

All

Q. 6

Which of the following act as antioxidants:

 A

Glutathione peroxidase

 B

Vit. C

 C

Selenium

 D

All

Ans. D

Explanation:

 All Correct – A,B & C

Quiz In Between


Q. 7

Allopurinol is used in organ preservation as:

 A

Antioxidant

 B

Preservative

 C

Free radical scavenger

 D

Precursor for energy metabolism

Q. 7

Allopurinol is used in organ preservation as:

 A

Antioxidant

 B

Preservative

 C

Free radical scavenger

 D

Precursor for energy metabolism

Ans. C

Explanation:

Ans is c i.e. Free radical scavenger 

Organ preservation is the supply line for organ transplantation. Currently, the liver, pancreas, and kidney can be successfully preserved for up to two days by flushing the organs with the University of Wisconsin (UW) solution and storing them at hypothermia (0-5° C). The UW solution is effective because it uses a number of cell impermeant agents (lactobionic acid, raffinose, hydroxyethyl starch) that prevent the cells from swelling during cold ischemic storage. Additionally, the UW solution contains glutathione, allopurinol and adenosine, agents that may stimulate recovery of normal metabolism upon reperfusion by augmenting the antioxidant capacity of the organs (glutathione) or by stimulating high-energy phosphate generation (adenosine) upon reperfusion and allopurinol as free-radical scavenger.

  • Glutathione            as antioxidant
  • Adenosine              as precursor for energy metabolism
  • Allopurinol              as free-radical scavenger

Q. 8

All of the following are antioxidant except:

September 2005

 A

Vitamin A

 B

Vitamin B

 C

Vitamin C

 D

Vitamin E

Q. 8

All of the following are antioxidant except:

September 2005

 A

Vitamin A

 B

Vitamin B

 C

Vitamin C

 D

Vitamin E

Ans. B

Explanation:

Ans. B: Vitamin B

Antioxidants are a group of compounds that help to protect the body from the formation and elimination of free radicals. Free-radicals are formed from exposure to sunlight and pollution and also as a byproduct of cell metabolism. Alcohol, cigarette smoke, stress and even diet also affect the level of free-radical development in the body. Excellent antioxidants include Vitamin A, Vitamin E, Vitamin C, zinc, selenium, ginkgo biloba, grape seed extract, and green tea extract.


Q. 9

Antioxidant vitamin ‑

 A

Vitamin A

 B

Ascorbic acid

 C

Vitamin E

 D

All of the above

Q. 9

Antioxidant vitamin ‑

 A

Vitamin A

 B

Ascorbic acid

 C

Vitamin E

 D

All of the above

Ans. D

Explanation:

Quiz In Between


Q. 10

Role of selenocystein is important in ‑

 A

Hydroxylation of dopamine

 B

Oxidation of drugs

 C

Antioxidant mechanism

 D

None of the above

Q. 10

Role of selenocystein is important in ‑

 A

Hydroxylation of dopamine

 B

Oxidation of drugs

 C

Antioxidant mechanism

 D

None of the above

Ans. C

Explanation:

Ans. is ‘c’ i.e., Antioxidant mechanism


Q. 11

All are antioxidant except ‑

 A

Vitamin A

 B

Catalase

 C

Cystein

 D

Glutamine

Q. 11

All are antioxidant except ‑

 A

Vitamin A

 B

Catalase

 C

Cystein

 D

Glutamine

Ans. D

Explanation:

Ans. is ‘d’ i.e., Glutamine


Q. 12

Drug with beta adrenergic blocking, with antioxidant, with calcium channel blocking, with alpha 1 antagonist activity is ‑

 A

Esmolol

 B

Carvedilol

 C

Nebivolol

 D

Levobunolol

Q. 12

Drug with beta adrenergic blocking, with antioxidant, with calcium channel blocking, with alpha 1 antagonist activity is ‑

 A

Esmolol

 B

Carvedilol

 C

Nebivolol

 D

Levobunolol

Ans. B

Explanation:

Ans. is ‘b’ i.e., Carvedilol

Carvedilol

  • Carvedilol is a β1 + β2 + α1 adrenoceptor blocker with α : β blocking property of 1 : 9.
  • It has antioxidant and antimitotic property.
  • It produces peripheral vasodilation due to α-1 blockade as well as calcium channel blockade (direct effect).
  • It is used in hypertension and angina.
  • It is used as cardioprotective in CHF.
  • t1/2 is 2-8 hrs.
  • It is eliminated through liver and kidney.
  • It causes orthostatic hypotension.

Quiz In Between


Q. 13

Antioxidant property is seen in

 A

Chromium

 B

Selenium

 C

Magnesium

 D

Iron

Q. 13

Antioxidant property is seen in

 A

Chromium

 B

Selenium

 C

Magnesium

 D

Iron

Ans. B

Explanation:

Ans. is `b’ i.e., Selenium 


Q. 14

True about Glutathione except

 A

Tripeptide

 B

Formed from glutamic acid, glycine, cysteine

 C

Act as antioxidant in reduced state

 D

All of the above

Q. 14

True about Glutathione except

 A

Tripeptide

 B

Formed from glutamic acid, glycine, cysteine

 C

Act as antioxidant in reduced state

 D

All of the above

Ans. D

Explanation:

Ans. is `d’ i.e., All of the above 

Glutathione

  • Glutathione is a tripeptide of glutamie acid, cysteine, and glycine. The molecule has a sulfhydryl (-SH) or thiol group on the cysteine, which accounts for its strong electron-donating character.
  • It exists in two forms : reduced glutathione or GSH. In the reduced state, the thiol group of cysteine is able to donate a reducing equivalent (H+ e-) to other unstable molecules, such as reactive oxygen species. In donating an electron, glutathione itself becomes reactive, but readily reacts with another reactive glutathione to form glutathione disulfide (GSSG) or oxidized glutathione. GSH can be regenerated from GSSG by the enzyme glutathione reductase.
  • While all cells in the human body are capable of synthesizing glutathione, liver glutathione synthesis has been shown to be essential. The liver is the largest GSH reservoir.
  • Because of its reducing property, reduced glutathione has potent antioxidant action.

Functions :

  • GSH is an extremely important cell protectant. It directly reduces reactive hydroxyl free radicals, other oxygen centered free radicals, and radical centers on DNA and other biomolecules.
  • GSH is the essential cofactor for many enzymes which require thiol-reducing equivalents, and helps keep redox-sensitive active sites on enzymes in the necessary reduced state. GSH is used as a cofactor by ‑
  • Multiple peroxidase enzymes, to detoxify peroxides generated from oxygen radical attack on biological molecules;
  • Transhydrogenases, to reduce oxidized centers on DNA, proteins, and other biomolecules; and
  • Glutathione S-transferases (GST) to conjugate GSH with endogenous substances (e.g., estrogens) and to exogenous electrophiles (e.g., arene oxides, unsaturated carbonyls, organic halides), and diverse xenobiotics.
  • GSH is a primary protectant of skin, lens, cornea, and retina against radiation damage, and the biochemical foundation of P450 detoxication in the liver, kidneys, lungs, intestinal epithelia, and other organs.
  • GSH acts as a carrier in transport of certain amino acids across membranes in the kidney.
  • Glutathione (GSH) participates in leukotriene synthesis.

Q. 15

Which of the following Vitamin is an antioxidant‑

 A

Vitamin D

 B

Vitamin E

 C

Vitamin K

 D

Vitamin B1

Q. 15

Which of the following Vitamin is an antioxidant‑

 A

Vitamin D

 B

Vitamin E

 C

Vitamin K

 D

Vitamin B1

Ans. B

Explanation:

Ans. is ‘b’ i.e., Vitamin E 

Vitamin E (a-Tocopheroll

  • Vitamin E is sometimes described as anti-sterility vitamin.
  • However, its anti-sterility function is seen only in some animals and not in human beings.
  • Vitamin E activiy is present in several tocopherols, the most important being a-, (3-, y- and 6- tocopherol. a Tocopherol is the most abundant and is taken as the standard.
  • The most important function of vitamin E in human beings is to act as an anti-oxidant.
  • It prevents the oxidation of other antioxidants, .e.g, carotenes, vitamin A and vitamin C.
  • Vitamin E is the lipid-soluble antixodant in cell membranes and plasma lipoproteins.
  • The main function of vitamin E is as a chain-breaking, free radical-trapping antioxidant in cell membranes and plasma lipoproteins by reacting with lipid peroxide radicals formed by peroxidation of polyunsatured fatty acid. o Vitamin E prevents rancidity of fats due to its antioxidant action.

Antioxidant vitamins are :-

  1. Vitamin -A (carotenes) Vitamin – C
  2. Vitamin – E

Quiz In Between


Q. 16

The main function of Vitamin C in the body is 

 A

Coenzyme for energy metabolism

 B

Regulation of lipid synthesis

 C

Involvement as antioxidant

 D

Inhibition of cell growth

Q. 16

The main function of Vitamin C in the body is 

 A

Coenzyme for energy metabolism

 B

Regulation of lipid synthesis

 C

Involvement as antioxidant

 D

Inhibition of cell growth

Ans. C

Explanation:

Ans. is ‘c’ i.e., Involvement as antioxidant 

Vitamin C (Ascorbic acid)

  • Ascorbic acid (Vitamin C) is also called antiscorbutic factor. It is very heat labile, especially in basic medium.
  • Ascorbic acid itself is an active form. Maximum amount of vitamin C is found in adrenal cortex.
  • Ascorbic acid functions as a reducing agent and scavanger of free radicals (antioxidant). Its major functions are:-
  • In collagen synthesis : – Vitamin C is required for post-translational modification by hydroxylation of proline and lysine residues converting them into hydroxyproline and hydroxylysine. Thus vitamin C is essential for the conversion of procollagen to collagen, which is rich in hydroxyproline and hydroxylysine. Through collagen synthesis, it plays a role in formation of matrix of bone, cartilage, dentine and connective tissue.
  1. Synthesis of norepinephrine from dopamine by dopamine-(3-monoxygenase (dopamine-f3-hydroxylase) requires Vitamin C.
  2. Carnitine synthesis
  3. Bile acid synthesis :- 7-ct-hydroxylase requires vitamin C.
  4. Absorption of iron is stimulated by ascorbic acid by conversion of ferric to ferrous ions.
  5. During adrenal steroid synthesis, ascorbic acid is required during hydroxylation reactions.
  6. Tyrosine metabolism : – Oxidation of P-hydroxy-phenylpyruvate to homogentisate.
  7. Folate metabolism : – Folic acid is converted to its active form tetrahydrofolate by help of Vitamin C.

Q. 17

All have antioxidant property except

 A

Catalase

 B

Glutathione peroxidase

 C

Phosphorylase

 D

Superoxide dismutase

Q. 17

All have antioxidant property except

 A

Catalase

 B

Glutathione peroxidase

 C

Phosphorylase

 D

Superoxide dismutase

Ans. C

Explanation:

Ans. is ‘c’ i.e., Phosphorylase 

  • There are two types of antioxidant systems :‑

1. Enzymatic antioxidant system

  1. This include
  2. Catalase
  3. Superoxide dismutase (SOD)
  4. Glutathione peroxidase

2. Non-enzymatic antioxidant system

  1. This is further subdivided into
  2. Vitamins : Vitamin E, Vitamin A & beta carotene, Vitamin C.
  3. Minerals : Selenium, Copper, Zinc, Manganese
  4. Tissue proteins : Transferrin, ferritin, lactoferrin, ceruloplasmin
  5. Amino acids : Glutathione, Cysteine

Q. 18

Which of the following has antioxidant property‑

 A

Selenium

 B

Copper

 C

Zinc

 D

All

Q. 18

Which of the following has antioxidant property‑

 A

Selenium

 B

Copper

 C

Zinc

 D

All

Ans. D

Explanation:

Ans. is ‘d i.e., All             

  • The activity of the antioxidant enzymes depends on supply of minerals :‑
  1. Manganese
  2. Copper
  3. Zinc
  4. Selenium
  • Manganese, copper and zinc are required for the activity of superoxide dismutase. 
  • Selenium is required for the activity of glutathione peroxidase.

Quiz In Between


Q. 19

Glutathione requires which vitamin to act as antioxidant ‑

 A

Vitamin E

 B

Niacin

 C

Vitamin C

 D

Vitamin A

Q. 19

Glutathione requires which vitamin to act as antioxidant ‑

 A

Vitamin E

 B

Niacin

 C

Vitamin C

 D

Vitamin A

Ans. B

Explanation:

Ans. is ‘b’ i.e., Niacin 


Q. 20

All are true about vitamin E except ‑

 A

Act as antioxidant

 B

Prevent lipid peroxidation of cell membrane

 C

Water soluble vitamin

 D

Chemically tocopheral

Q. 20

All are true about vitamin E except ‑

 A

Act as antioxidant

 B

Prevent lipid peroxidation of cell membrane

 C

Water soluble vitamin

 D

Chemically tocopheral

Ans. C

Explanation:

Ans. is ‘c’ i.e., Water soluble vitamin 

  • Vitamin E is a fat soluble vitamin (not water soluble).
  • All other options are correct.

Quiz In Between



Antioxidant

Antioxidant


ANTIOXIDANT

  • Molecule that prevent oxidation of other molecule.
  • Oxidative stress produces free radical,contribute to diseases.
  • It is measured by  BAP Test and d-ROMs test.

NATURAL AND SYNTHETIC ANTIOXIDANT

Antioxidant Source
Vitamin C Citrus Fruits
Vitamin E Vegetable Oil
Polyphenolic (resveratol and Flavoniod) Fruits, Red wine, Tea,Choclate, Orgeno, Coffe, Olive oil,Soy
Careteniods (lycopene carotene, lutien) Fruits Vegetable egg
Synthetic Propylgallate, Butylated Hydroxyanisole, Butylated Hydroxytoulene, Tertiarbutylhydroquinone.

CLASSIFICATION OF ANTIOXIDANTS

  1. ANTIOXIDANT ENZYMES
    1. Catalase
    2. Glutathione peroxidase
    3. Glutathione reductase
    4. Super oxide dismutase(both Cu -Zn and Mn)
  2. METAL BINDING PROTIENS
    1. Ceruloplasmin
    2. Ferritin
    3. Lactoferrin
    4. Metallotheinein
    5. Transferrin
    6. Hemoglobib
    7. Myoglobin
  3. Bilirubin
  4. Caroteniod
    1. Beta Carotene
    2. Alpha carotene
    3. Lutien
    4. Lycopene
    5. Zeaxanthin
  5. Vitamins A,C,E,D
  6. Uric acids
  7. Thiols
  8. Other-Copper, Selenium,Zinc,Alpha lipoic acid,Glutathione

MECHANISM OF ACTION

  • Vit. E prevents peroxidation of membrane phosphor lipids
  • Vit.C reactivate oxygen and scavenges free radicals
  • Super oxide dismutase reduces superoxide.
  • Catalase and glutathione convert hydrogen peroxide.

 What are antioxidants used for?

  •  Antioxidants may play a role in the management or prevention –
  1. Some cancers
  2. Macular degeneration
  3. Alzheimer’s diseases
  4. Some arthritis-related conditions.
  • Destroys free radicals.
  • Promote growth of healthy cell.
  • Support immune system

ANTIOXIDATIVE STRESS

  • Strong reducing acid binds to deitary minerals(iron,zinc)
  • Prevent their absorbtion
  • Examples are oxalic acid, tannins and phytic acid, which are high in plant-based diets.
  • Nonpolar antioxidants such as eugenol have toxicity limits.
  • examples are oxalic acid, tannins and phytic acid, which are high in plant-based diets

Exam Important

ANTIOXIDANT

  • Molecule that prevent oxidation of other molecule.

NATURAL AND SYNTHETIC ANTIOXIDANT

Antioxidant Source
Vitamin C Citrus Fruits
Vitamin E Vegetable Oil
Polyphenolic (resveratol and Flavoniod) Fruits, Red wine, Tea,Choclate, Orgeno, Coffe, Olive oil,Soy
Careteniods (lycopene carotene, lutien) Fruits Vegetable egg
Synthetic Propylgallate, Butylated Hydroxyanisole, Butylated Hydroxytoulene, Tertiarbutylhydroquinone.

CLASSIFICATION OF ANTIOXIDANTS

  1. ANTIOXIDANT ENZYMES
    1. Catalase
    2. Glutathione peroxidase(selenium imp.component)
    3. Glutathione reductase
    4. Super oxide dismutase(both Cu -Zn and Mn)
  2. METAL BINDING PROTIENS
    1. Ceruloplasmin
    2. Ferritin
    3. Lactoferrin
    4. Metallotheinein
    5. Transferrin
    6. Hemoglobib
    7. Myoglobin
  3. Bilirubin
  4. Caroteniod
    1. Beta Carotene
    2. Alpha carotene
    3. Lutien
    4. Lycopene
    5. Zeaxanthin
  5. Vitamins A,C,E,D
  6. Uric acids
  7. Thiols
  8. Other-Copper, Selenium,Zinc,Alpha lipoic acid,Glutathione

MECHANISM OF ACTION

  • Vit. E prevents peroxidation of membrane phosphor lipids
  • Vit.C reactivate oxygen and scavenges free radicals
  • Super oxide dismutase reduces superoxide.
  • Catalase and glutathione convert hydrogen peroxide.
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