3rd Biochemistry Lecture Exam (Batch 2025)

Question 1

1. Where and in what form in the mammalian cell is glycogen
   stored?
   A. mitochondrial matrix, globules
   B. golgi complex, droplets
   C. cytoplasm, granules
   D. lysosomes, micelles

Answer 1

C. cytoplasm, granules

Glycogen is a branched-chain polysaccharide made exclusively from α-D-glucose. The primary glycosidic bond is an α(1→4) linkage. After an average of eight to ten glucosyl residues, there is a branch containing an α(1→6) linkage . A single molecule of glycogen can have a molecular mass of up to 108 daltons. These molecules exist in discrete CYTOPLASMIC GRANULES that also contain most of the enzymes necessary for glycogen synthesis and degradation.
(Lippincott - p. 126)

Question 2

2. Which is the most important function of stored liver glycogen?
   A. maintain blood glucose to normal levels
   B. supply dependent tissues with glucose during starvation
   C. fuel reserve for synthesis of ATP when liver glycogen
   stores are depleted
   D. reserve buffer for energy needed for muscle activity

Answer 2

A. maintain blood glucose to normal levels

2. Glycogen stores, although small, are extremely important.
   a. LIVER GLYCOGEN - used to maintain blood glucose levels during the early stages of fasting.
   b. Muscle glycogen - oxidized for muscle contraction. It does not contribute to the maintenance of blood glucose levels under any conditions.

(BRS Biochemistry - p. 2)

Question 3

3. Which statement is involved in synthesis of glycogen?
   A. addition of glucose to the non-reducing end of a small
   molecule of glycogen
   B. synthesis of the protein primer
   C. elongation of chains forming amylopectin chains
   D. transfer of “limit dextrin” to an existing branch

Answer 3

C. elongation of chains forming amylopectin chains

Question 4

4. Which serves as a primer for glycogenesis possessing an
   auto-glycosylating activity?
   A. small molecule of glycogen
   B. glycogenin
   C. existing glycogen molecule
   D. UDP-glucose

Answer 4

B. glycogenin

B. Primer requirement and synthesis
Glycogen synthase catalyzes the α(1→4) linkages in glycogen. This enzyme cannot initiate chain synthesis using free glucose as an acceptor of a molecule of glucose from UDPglucose.
Instead, it can only elongate already existing chains of glucose and, therefore, requires a primer. A fragment of glycogen can serve as a primer. In the absence of a fragment, the homodimeric protein glycogenin can serve as an acceptor of glucose from
UDP-glucose (see Fig. 11.5). The side-chain hydroxyl group of tyrosine-194 in the protein is the site at which the initial glucosyl unit is attached. Because the reaction is catalyzed by
GLYCOGENIN itself via AUTOGLYCOSYLATION, glycogenin is an enzyme. Glycogenin then catalyzes the transfer of at least four molecules of glucose from UDP-glucose, producing a short,
α(1→4)-linked glucosyl chain. This short chain serves as a primer that is able to be elongated by glycogen synthase, which is recruited by glycogenin, as described in C.
below. (Note: Glycogenin stays associated with and forms the core of a glycogen granule.)
(Lippincott Illustrated Reviews 8th Edition, p. 273)

Question 5

5. Which statement is true of the primer used for glycogenesis?
   A. Has a single tyrosine residue to which glucose is added
   B. A homodimer with two active sites to where glucose is
   attached
   C. Involves a nucleophilic attack on the outer PO4 of UDP-
   glucose
   D. the released PPi ensures the completion of the reaction

Answer 5

B. A homodimer with two active sites to where glucose is attached

B. Primer requirement and synthesis
Glycogen synthase catalyzes the α(1→4) linkages in glycogen. This enzyme cannot initiate chain synthesis using free glucose as an acceptor of a molecule of glucose from UDPglucose.
Instead, it can only elongate already existing chains of glucose and, therefore, requires a primer. A fragment of glycogen can serve as a primer. In the absence of a fragment, the HOMODIMERIC protein glycogenin can serve as an acceptor of glucose from UDP-glucose (see Fig. 11.5). The side-chain hydroxyl group of tyrosine-194 in the protein is
the site at which the initial glucosyl unit is attached. Because the reaction is catalyzed by glycogenin itself via autoglucosylation, glycogenin is an enzyme. Glycogenin then catalyzes the transfer of at least four molecules of glucose from UDP-glucose, producing a short, α(1→4)-linked glucosyl chain. This short chain serves as a primer that is able to be elongated by glycogen synthase, which is recruited by glycogenin, as described in C. below. (Note: Glycogenin stays associated with and forms the core of a glycogen granule.)
(Lippincott Illustrated Reviews 8th Edition, p. 273)

Question 6

6. Which reactions are needed to form UDP-glucose utilized for glycogenesis?
   A. Isomerization of F6PO4 to G6PO4, hydrolysis of PPi
   B. Conversion of G6PO4 to G1PO4, nucleophilic attack of
   UTP’s phosphoryl O2 by G1PO4’s alpha PO4
   C. Change G6PO4 to G1PO4, subsequent hydrolysis of
   released PPi to 2 Pi
   D. Hydrolysis of a phosphoanhydride bond & release of PPi

Answer 6

C. Change G6PO4 to G1PO4, subsequent hydrolysis of
released PPi to 2 Pi

A. Uridine diphosphate glucose synthesis
α-D-Glucose attached to uridine diphosphate (UDP) is the source of all the glucosyl residues that are added to the growing glycogen molecule. UDP-glucose (Fig. 11.4) is synthesized from glucose 1-phosphate and UTP by UDP–glucose pyrophosphorylase (Fig. 11.5). Pyrophosphate (PPi), the second product of the reaction, is hydrolyzed to two inorganic phosphates (Pi) by pyrophosphatase. The hydrolysis is exergonic, which ensures that the UDP–glucose pyrophosphorylase reaction proceeds in the direction of UDP-glucose production. (Note: Glucose 1-phosphate is generated from glucose 6-phosphate by
phosphoglucomutase. Glucose 1,6-bisphosphate is an obligatory intermediate in this reversible reaction [Fig. 11.6].)
(Lippincott Illustrated Reviews 8th Edition, p. 272)

Question 7

7. What kind of bond is formed between first glucose unit and the tyrosine residue of glycogenin?
   A. glycosidic bond
   B. N-linked glycosidic bond
   C. O-linked glycosidic bond
   D. ester-linked bond

Answer 7

C. O-linked glycosidic bond

N- and O-glycosides:
If the group on the non-carbohydrate molecule to which the sugar is attached is an –NH2 group, the structure is an N-glycoside and the bond is called an N-glycosidic link. If the group is an –OH, the structure is an O-glycoside, and the bond is an O-glycosidic link. [Note: All sugar–sugar glycosidic bonds are O-type linkages.]
(Lippincott Illustrated Reviews 5th Edition, p. 86)

Question 8

8. Which serves as substrates for UDP-glucose synthesis?
   A. UTP & glucose-1-PO4
   B. GTP & fructose 1-6bisPO4
   C. UTP & glucose-6-PO4
   D. ATP & fructose 2-6bisPO4

Answer 8

A. UTP & glucose-1-PO4

A. Uridine diphosphate glucose synthesis
α-D-Glucose attached to uridine diphosphate (UDP) is the source of all the glucosyl residues that are added to the growing glycogen molecule. UDP-glucose (Fig. 11.4) is synthesized from GLUCOSE 1-PHOSPHATE AND UTP by UDP–glucose pyrophosphorylase (Fig. 11.5). Pyrophosphate (PPi), the second product of the reaction, is hydrolyzed to two inorganic phosphates (Pi) by pyrophosphatase. The hydrolysis is exergonic, which ensures that the UDP–glucose pyrophosphorylase reaction proceeds in the direction of UDP-glucose production. (Note: Glucose 1-phosphate is generated from glucose 6-phosphate by
phosphoglucomutase. Glucose 1,6-bisphosphate is an obligatory intermediate in this reversible reaction [Fig. 11.6].)
(Lippincott Illustrated Reviews 8th Edition, p. 272)

Question 9

9. Which enzyme synthesizes UDP-glucose?
   A. phosphoglucomutase
   B. glycogen synthase
   C. nucleoside diphosphokinase
   D. UDP-Glucose-pyrophosphorylase

Answer 9

D. UDP-Glucose-pyrophosphorylase

A. Uridine diphosphate glucose synthesis
α-D-Glucose attached to uridine diphosphate (UDP) is the source of all the glucosyl residues that are added to the growing glycogen molecule. UDP-glucose (Fig. 11.4) is synthesized from glucose 1-phosphate and UTP by ““UDP-glucose-phosphorylase”” (Fig. 11.5). Pyrophosphate (PPi), the second product of the reaction, is hydrolyzed to two inorganic phosphates (Pi) by pyrophosphatase. The hydrolysis is exergonic, which ensures that the UDP–glucose pyrophosphorylase reaction proceeds in the direction of UDP-glucose production. (Note: Glucose 1-phosphate is generated from glucose 6-phosphate by
phosphoglucomutase. Glucose 1,6-bisphosphate is an obligatory intermediate in this reversible reaction [Fig. 11.6].)
(Lippincott Illustrated Reviews 8th Edition, p. 272)

Question 10

10. At least how many glucose units will be removed from a
    straight chain to create a branch thru creation of an αlpha 1,6
    link?
    A. six
    B. seven
    C. eight
    D. eleven

Answer 10

A. six

1. Branch synthesis: Branches are made by the action of the branching enzyme, amylo-α(1→4)→α(1→6)-transglycosylase. This enzyme removes a set of ““six to eight”” glucosyl residues from the nonreducing end of the glycogen chain, breaking an α(1→4) bond to another residue on the chain, and attaches it to a nonterminal glucosyl residue by an
   α(1→6) linkage, thus functioning as a 4:6 transferase. The resulting new, nonreducing end (see “i” in Fig. 11.5), as well as the old nonreducing end from which the six to eight residues were removed (see “o” in Fig. 11.5), can now be further elongated by glycogen synthase.
   (Lippincott Illustrated Reviews 8th Edition, p. 276)

Question 11

11. Which is the activity of the branching enzyme of the
    glycogenesis?
    A. amylo α1,4 –> α1,4 glucan transferase
    B. amylo α1,4 –> α1,6 transglucosidase
    C. amylo α1,6 glucosidase
    D. α1,4 –> α1,4 transglycosylase

Answer 11

B. amylo α1,4 –> α1,6 transglucosidase

IV. Glycogenolysis
B. Branch removal
Branches are removed by the two enzymic activities of a single bifunctional protein, the debranching enzyme (see Fig. 11.8). First, oligo-α(1→4)→α(1→4)-glucantransferase activity removes the outer three of the four glucosyl residues remaining at a branch. It next transfers them to the nonreducing end of another chain, lengthening it accordingly. Thus, an ““α(1→4)”” bond is broken and an α(1→4) bond is made, and the enzyme functions as a 4:4 transferase. Next, the remaining glucose residue attached in an α(1→6) linkage is removed hydrolytically by ““amylo-α(1→6)-glucosidase activity””, releasing free (nonphosphorylated) glucose. The glucosyl chain is now available again for degradation by glycogen phosphorylase until four glucosyl units in the next branch are reached.

Question 12

12.Glycogen synthase action will commence after how many glucose units have been incorporated into the tyrosine residue of each of the active sites of glycogenin?
A. six
B. eight
C. nine
D. ten

Answer 12

B. eight

Question 13

13. Which is the specific reaction in the breaking down
    glycogen to glucose 1-PO4?
    A. nucleolysis
    B. electrolysis
    C. phosphorolysis
    D. thiolysis

Answer 13

C. phosphorolysis

Glycogen phosphorylase sequentially cleaves the α(1,4) glycosidic bonds between the glucosyl residues at the nonreducing ends of the ““glycogen”” chains by simple ““phosphorolysis”” (producing ““glucose 1-phosphate””) until four glucosyl units remain on each chain at a branch point. (Lippincott Illustrated Reviews 8th Edition, p. 396)

Question 14

14. Which statement refers to removal of glucose as free
    glucose carried out through glycogenolysis?
    A. Initial removal of glucose from the reducing end of glycogen
    B. Remodeling of the glycogen molecule by the use of
    debranching enzymes
    C. Removal of glucose units as free glucose using glycogen
    phosphorylase
    D. Cleavage of alpha 1,4 glycosidic links up to the branch with
    alpha 1,6 bond

Answer 14

B. Remodeling of the glycogen molecule by the use of
debranching enzymes ???

Question 15

15. Glycogen phosphorylase action will stop at how many glucose units away from a branch?
    A. two
    B. three
    C. four
    D. five

Answer 15

C. four

IV. DEGRADATION (GLYCOGENOLYSIS)
A. Chain shortening
Glycogen phosphorylase sequentially cleaves the α(1→4) glycosidic bonds between the glucosyl residues at the nonreducing ends of the glycogen chains by simple phosphorolysis (producing glucose 1-phosphate) until ““four glucosyl units”” remain on each chain at a branch point (Fig. 11.7). The resulting structure is called a limit dextrin, and phosphorylase cannot degrade it any further (Fig. 11.8). (Note: Phosphorylase requires pyridoxal phosphate [a derivative of vitamin B6] as a coenzyme.)
(Lippincott Illustrated Reviews 8th Edition, p. 276)

Question 16

16. Which stimulates the inactive glycogen phosphorylase in
    the muscles?
    A. AMP
    B. ATP
    C. ADP
    D. GMP

Answer 16

A. AMP

IV. DEGRADATION (GLYCOGENOLYSIS)
A. Allosteric regulation of glycogenesis and glycogenolysis
2. Glycogenolysis activation by AMP: Muscle glycogen phosphorylase (myophosphorylase), but not the liver isozyme, is active in the presence of the high AMP concentrations that occur under extreme conditions of anoxia and ATP depletion. AMP binds to glycogen phosphorylase b, causing its activation without phosphorylation (see Fig. 11.9). Recall that AMP also activates phosphofructokinase-1 of glycolysis, allowing glucose from glycogenolysis to be oxidized.
(Lippincott Illustrated Reviews 8th Edition, p. 284)

Question 17

17. In the muscles, which will convert inactive glycogen phosphorylase to the T state?
    A. AMP
    B. glucose 6-PO4
    C. glucose
    D. GTP

Answer 17

B. glucose 6-PO4

IV. DEGRADATION (GLYCOGENOLYSIS)
3. Glycogenolysis activation by calcium
a. Muscle phosphorylase kinase activation: During muscle contraction, there is a rapid and urgent need for ATP. It is supplied by the degradation of muscle glycogen to ““glucose 6-phosphate’”, which enters glycolysis. Nerve impulses cause membrane depolarization, which promotes Ca2+ release from the sarcoplasmic reticulum into the sarcoplasm of myocytes. The Ca2+ binds the CaM subunit, and the complex activates muscle phosphorylase kinase b (see Fig. 11.9).
(Lippincott Illustrated Reviews 8th Edition, p. 285)

Question 18

18. In the liver, which will convert active glycogen phosphorylase to the T state?
    A. AMP
    B. glucose 6-PO4
    C. glucose
    D. GTP

Answer 18

C. glucose

IV. DEGRADATION (GLYCOGENOLYSIS)
3. Glycogenolysis activation by calcium
b. Liver phosphorylase kinase activation: During physiologic stress, epinephrine is released from the adrenal medulla and signals the need for blood glucose. This ““glucose”” initially comes from hepatic glycogenolysis. Binding of epinephrine to hepatocyte α1-adrenergic GPCR activates a phospholipid-dependent cascade that results in movement of Ca2+ from the ER into the cytoplasm. A Ca2+–CaM complex forms and activates hepatic phosphorylase kinase b. Note that the released Ca2+ also helps to activate protein kinase C that can phosphorylate and inactivate
glycogen synthase a. (Lippincott Illustrated Reviews 8th Edition, p. 285-286)

Question 19

19. Which debranching enzyme’s activity will expose the
    glucose unit linked by an α1,6 glycosidic bond?
    A. amylo α1,4 –> α1,4 glucan transferase
    B. amylo α1,4 –> α1,6 transglucosidase
    C. amylo α1,6 glucosidase
    D. α1,4 –> α1,4 transglycosylase

Answer 19

A. amylo α1,4 –> α1,4 glucan transferase

IV. DEGRADATION (GLYCOGENOLYSIS)
B. Branch removal
Branches are removed by the two enzymic activities of a single bifunctional protein, the debranching enzyme (see Fig. 11.8). First, oligo-α(1→4)→α(1→4)-glucantransferase
activity removes the outer three of the four glucosyl residues remaining at a branch. It next transfers them to the nonreducing end of another chain, lengthening it accordingly. Thus, an
α(1→4) bond is broken and an α(1→4) bond is made, and the enzyme functions as a 4:4 transferase. Next, the remaining glucose residue attached in an α(1→6) linkage is removed
hydrolytically by amylo-α(1→6)-glucosidase activity, releasing free (nonphosphorylated) glucose. The glucosyl chain is now available again for degradation by glycogen phosphorylase until four glucosyl units in the next branch are reached.

Question 20

20. Which is the sequence of enzymes used in the process of glycogenolysis?
    A. alpha 1,6 glucosidase, glycogen phosphorylase, amylo 1,4 to 1,4 glucan transferase
    B. glycogen phosphorylase, alpha 1,6 glucosidase, amylo 1,4 to 1,4 glucan transferase
    C. amylo 1,4 to 1,4 glucan transferase, glycogen phosphorylase, alpha 1,6 glucosidase
    D. glycogen phosphorylase, amylo 1,4 to 1,4 glucan transferase, alpha 1,6 glucosidase

Answer 20

D. glycogen phosphorylase, amylo 1,4 to 1,4 glucan
transferase, alpha 1,6 glucosidase

IV. DEGRADATION (GLYCOGENOLYSIS)
A. Chain shortening
Glycogen phosphorylase sequentially cleaves the α(1→4) glycosidic bonds between the glucosyl residues at the nonreducing ends of the glycogen chains by simple phosphorolysis (producing glucose 1-phosphate) until four glucosyl units remain on each chain at a branch
point (Fig. 11.7). The resulting structure is called a limit dextrin, and phosphorylase cannot degrade it any further (Fig. 11.8). (Note: Phosphorylase requires pyridoxal phosphate [a derivative of vitamin B6] as a coenzyme.)
B. Branch removal
Branches are removed by the two enzymic activities of a single bifunctional protein, the
debranching enzyme (see Fig. 11.8). First, oligo-α(1→4)→α(1→4)-glucantransferase
activity removes the outer three of the four glucosyl residues remaining at a branch. It next transfers them to the nonreducing end of another chain, lengthening it accordingly. Thus, an α(1→4) bond is broken and an α(1→4) bond is made, and the enzyme functions as a 4:4 transferase. Next, the remaining glucose residue attached in an α(1→6) linkage is removed hydrolytically by amylo-α(1→6)-glucosidase activity, releasing free (nonphosphorylated) glucose. The glucosyl chain is now available again for degradation by glycogen
phosphorylase until four glucosyl units in the next branch are reached.
(Lippincott Illustrated Reviews, p. 278-279)

Question 21

21. Which enzyme when dephosphorylated helps maintain glycogen phosphorylase inactive in the skeletal muscle?
    A. protein kinase A
    B. phosphorylase kinase
    C. calcium calmoldulin
    D. protein kinase G

Answer 21

B. phosphorylase kinase

???

Question 22

22. Which enzyme, when insulin to glucagon ratio is high, is
    activated?
    A. protein phosphatase
    B. phosphorylase
    C. phosphorylase kinase
    D. protein phosphatase inhibitor

Answer 22

A. protein phosphatase

V. GLYCOGENESIS AND GLYCOGENOLYSIS REGULATION
A. Covalent activation of glycogenolysis
5. Phosphorylated state maintenance: The phosphate groups added to phosphorylase kinase and phosphorylase in response to cAMP are maintained because the enzyme that hydrolytically removes the phosphate, protein phosphatase-1 (PP1), is inactivated by inhibitor proteins that are also phosphorylated and activated in response to cAMP (see Fig. 11.9). Because insulin also activates the phosphodiesterase that degrades
cAMP, it opposes the effects of glucagon and epinephrine.

Question 23

23. Which glycogen storage disease involves a deficiency of a
    unique enzyme of gluconeogenesis?
    A. Pompe’s
    B. von Gierke’s
    C. Anderson’s
    D. Her’s

Answer 23

B. von Gierke’s

Gluconeogenesis : Enzymes that by-pass irreversible steps
Pyruvate carboxylase

Phosphoenolpyruvate carboxykinase

Fructose-1,6-phosphatase

Glucose-6-phosphatase (Type Ia: von Gierke’s)

Question 24

24. Which glycogen storage disease is caused by a deficiency
    of an important enzyme of glycolysis:
    A. Pompe’s
    B. von Gierke’s
    C. Tarui’s
    D. Her’s

Answer 24

C. Tarui’s

Glycolysis (Irreversible Steps) : Enzymes
Hexokinase/Glucokinase

Phosphofructokinase (Rate-limiting step) : Type VII : Tauri’s

Pyruvate Kinase

Question 25

25. Which glycogen storage disease is caused by a deficiency of muscle phosphorylase? A. Her’s B. McArdle’s C. Cori’s D. Fanconi

Answer 25

B. McArdle’s

Question 26

26. The products of the Pentose Phosphate Pathway are: A. 2 NADPH, 1 carbon dioxide, 2 Ribose B. 2 NADPH, 1 carbon dioxide, Ribose C. 2 NADPH, 2 carbon dioxide, Ribose D. 1 NADPH, 2 carbon dioxide, 2 Ribose

Answer 26

B. 2 NADPH, 1 carbon dioxide, Ribose

Question 27

27. Biosynthesis of Fatty Acids occur in the: A. Mammary Glands B. Ovaries C. Adrenals D. Liver

Answer 27

D. Liver

Question 28

28. Uses of NADPH include: A. Degradation of Nitric Oxide B. Oxidation of Hydrogen Peroxide C. Phagocytosis of Red Blood Cells D. Reductive Biosynthesis

Answer 28

D. Reductive biosynthesis

Question 29

29. A breast fed infant began to vomit frequently and lose weight. Several days later she developed jaundice, hepatomegaly and bilateral cataract. What is the possible cause for these symptoms. A. G6PD Deficiency B. Galactosemia C. Von Gierke's Syndrome D. Hereditary Fructose Intolerance

Answer 29

B. Galactosemia

Question 30

30.The Uronic Acid Pathway: A. Catalyzes conversion of glucose to glucuronic, pentoses, sialic acid. B. Energy producing reaction C. Occurs in the cytosol D. Alternative pathway for the oxidation of glucose

Answer 30

D. Alternative pathway for the oxidation of glucose

Question 31

31.The rate limiting step in glycolysis is catalyzed by this enzyme and is bypassed in the metabolism of fructose A. Phosphofructokinase I B. Glucose-6-Phosphate C. Glucose-1-Phosphate D. Glucokinase

Answer 31

A. Phosphofructokinase Glycolysis (Irreversible Steps) : Enzymes Hexokinase/Glucokinase Phosphofructokinase (Rate-limiting step) : Type VII : Tauri’s Pyruvate Kinase

Question 32

32.Hexokinase has a low affinity for: A. Fructose-6-Phosphate B. Fructose-1-Phosphate C. Fructose D. Sucrose

Answer 32

C. Fructose

Question 33

33. Fructose-1 Phosphate reaction to Glyceraldehyde is catalyzed by: A. Aldolase A B. Aldolase B C. Aldolase C D. All of the above

Answer 33

B. Aldolase B

Question 34

34. Hereditary Fructose Intolerance is caused by a deficiency in this enzyme: A. Aldolase A B. Aldolase B C. Aldolase C D. All of the above

Answer 34

B. Aldolase B

Question 35

35. Classic Galactosemia is caused by this enzyme deficiency: A. GALT B. Galactokinase C. Galactomutase D. Phosphogalactogen

Answer 35

A. GALT

Question 36

36. Fructose -6-Phosphate can be produced from: A. Mannose-6-Phospate B. Sedoheptulase-6-Phosphate C.Xylulose-7-Phosphate D. All of the Above

Answer 36

A. Mannose-6-Phosphate

Question 37

37. Familial fructokinase deficiency causes no symptoms because: A. Hexokinase can phosphorylate fructose B. Liver aldolase can metabolize it C. Excess fructose is excreted through feces D. Excess fructose is converted to glucose

Answer 37

A. hexokinase can phosphorylate fructose

Question 38

38. A medical student developed hemolytic anemia after taking the oxidizing malarial drug primaquine. This severe reaction is most likely due to: A. Glucose-6-phosphate dehydrogenase deficiency B. Scurvy C. Diabetes D. Glycogen phosphorylase activity

Answer 38

A. Glucose-6-phosphate dehydrogenase deficiency

Question 39

39. G6PD is most severe in RBC's because: A.The Pentose Phosphate Pathway is the sole source of NADPH in RBC's B. RBC's have nucleus and can produce ribulose-5-phosphate C. More NADPH is needed in RBC's to reduce glutathione D. Glucose-6-Phosphate is consumed in RBC's

Answer 39

A. The Pentose Phosphate Pathway is the sole source of NADPH in RBC's

Question 40

40. True about G6PD: A. characterized by hemolytic anemia B. does not affect life expectancy C. Common in Middle America, Europe, and Africa D. symptomatic

Answer 40

A. Characterized by hemolytic anemia

Question 41

41. Substances that may make people with G6PD symptomatic: A. Fava beans B. Anti Tuberculosis drugs C. Anti-Inflammatories D. All of the above

Answer 41

A. Fava beans

Question 42

42. True of Fructose metabolism: A. sucrose yields 2 fructose and 1 glucose B. insulin dependent C. bypasses phosphofructokinase-1 D. all of the above

Answer 42

C. Sucrose yields 2 fructose and 1 glucose

Question 43

43. Ribulose 5 Phosphate is an important product of Pentose Phosphate Pathway because: A. it is a substrate for fatty acid synthesis B. It is a substrate for steroid synthesis C. It is a substrate for nucleotide synthesis D. all of the above

Answer 43

D. All of the above

Question 44

44.Coenzyme responsible for the dehydrogenation of Glucose 6 Phosphate: A. NADP+ B. NADPH C. NADP+ and NADPH D. NAD

Answer 44

A. NADP+

Question 45

45.True of the Pentose Phosphate Pathway A. Occurs in the cytoplasm B. Has three irreversible reactions C. Produces ATP D. Produces Ribose

Answer 45

D. Produces ribose

Question 46

46.Major storage form of energy in the body. A. Glycogen B. Glucose C. Cellulose D. Starch

Answer 46

A. Glycogen

Question 47

47.The monosaccharides glucose and fructose linked by glycosidic bonds form which sugar: A. Lactose B. Maltose C. Sucrose D. Fructose

Answer 47

C. Sucrose

Question 48

48.The only tissue with glycerol kinase to phosphorylate glycerol. A. Brain B. Liver C. Pancreas D. Small intestines

Answer 48

B. Liver

Question 49

49.This sugar derivative is vital in the synthesis of collagen. A. Glucuronic acid B. Hyaluronic acid C. Ascorbic acid D. Sorbitol

Answer 49

C. Ascorbic acid Vitamin C (ascorbic acid) functions as a coenzyme in the hydroxylation of proline and lysine in the synthesis of collagen, a fibrous protein of the extracellular matrix. (Lippincott Illustrated Reviews 8th Edition, p. 1526)

Question 50

50.Compounds having same structural formula but differing in configuration around one carbon atom are: A. epimers B. enantiomers C. stereoisomers D.optical isomers

Answer 50

A. Epimers Structures of Monosaccharides: Straight Chain Structure (*Fischer Projection ) D & L Isomerism (Enantiomer): Mirror Image Dextrorotatory = Right; Levorotatory = Left ; Racemic Mixture = Both Levo and Dextro Anomerism (Assymetric Carbon) : -OH Alpha = Below/Right; Beta = Above/Left Epimer :same structure different one chiral carbon C-4 (Gal-Glu); C-2 (Man-Glu) Ring (Cyclic) Structure (Haworth Projection) Pyranose Ring = 6 Sides; Furanose Ring = 5 Sides Boat (Chair) Form.

Question 51

51.The polysaccharide found in the cell wall of insects is: A. cellulose B. hyaluronic acid C. chitin D. starch

Answer 51

C. chitin Cellulose: · Glucose residues joined by B-1, 4 linkages. Major component of vegetable fiber. Chief constituent of plant cell walls. Source of bulk in stool. *Linear. Hyaluronic acid: · Lubricant, Shock Absorber, Promotes Wound Healing (*Treatment for Osteoarthritis, replaces synovial fluid) Chitin: · Gives rigidity to the exoskeletons of crabs, lobsters, shrimps, insects and other arthropods, Cell walls of fungi, Linear polymer of N-acetylglucosamine in B-1, 4 linkages. Starch: · Storage of carbohydrates in plants

Question 52

52.A 10 yrs old female was brought to the emergency room secondary to alterations in behaviour. On lab exam ,ketones are found in her urine. Which of the following is the most likely source of ketones? A. glycogenolysis B. glycolysis C. gluconeogenesis D. fatty acid breakdown

Answer 52

D. fatty acid breakdown Primary source (ATP) : Glucose Secondary source (ATP) : Ketone Bodies (Low BG, Fasting State) Metabolic Products: Glycogenolysis · Glycogen → Glucose Glycolysis · Glucose → 2 Pyruvate Gluconeogenesis · Non-carbohydrate (OAA, Lactic acid, Glycerol) → Glucose Fatty acid breakdown (β-oxidation) · Fatty Acid → Acetyl-CoA o Ketogenesis : Acetyl – CoA → Ketone Bodies

Question 53

53.Aldose reductase is an enzyme abundant in the following tissues except: A. liver B. nerve tissue C. seminal vesicles D. muscles

Answer 53

D. muscles Synthesis of sorbitol: Aldose reductase reduces glucose, producing sorbitol This enzyme is found in many tissues, including the lens, retina, Schwann cells of peripheral nerves, liver, kidney, placenta, red blood cells, and in cells of the ovaries and seminal vesicles. (P139 – Lippincott)

Question 54

54.Which of the following compounds is not produced from dietary starch by salivary a-amylase? A. fructose B. maltose C. isomaltose D. maltotriose

Answer 54

A. fructose Carbohydrate Digestion: Monosaccharides (Readily Absorbed in the Small Intestines): Glucose, Galactose, Fructose. Alpha 1- 4 of Amylose, Amylopectin, Glycogen→ (Enzyme: Amylase) → Amylase Products: Maltotriose, Alpha-Limit Dextrin (A1-6, A1-4), Maltose. *Amylase Cannot cleave Terminal A1-4 Linkage.

Question 55

55.Digestion in the mouth starts with this enzyme cleaving the bonds of dietary starch and glycogen in a random manner requiring an optimum pH. A. ptyalin B. maltotriose C. dextrin D. isomaltase

Answer 55

A. ptyalin Digestion of carbohydrates begins in the mouth The major dietary polysaccharides are of plant (starch, composed of amylose and amylopectin) and animal (glycogen) origin. During mastication, salivary α-amylase acts briefly on dietary starch and glycogen, hydrolyzing random α(1→4) bonds. (P86-Lippincott) Saliva contains two major types of protein secretion: (1) a serous secretion that contains ptyalin (an α-amylase), which is an enzyme for digesting starches, and (2) mucus secretion that contains mucin for lubricating and for surface protective purposes. (P809-Guyton) Alpha 1- 4 of Amylose, Amylopectin, Glycogen→ (Enzyme: Amylase) → Amylase Products: Maltotriose, Alpha-Limit Dextrin (A1-6, A1-4), Maltose. *Amylase Cannot cleave Terminal A1-4 Linkage.

Question 56

56.Glucose is maintained in the blood as the sole energy source for the ______ in the non-starving state and as an available energy source for all other tissues A. liver B. brain C. lungs D. kidneys

Answer 56

B. brain A constant source of blood glucose is an absolute requirement for human life. Glucose is the greatly preferred energy source for the brain, and the required energy source for cells with few or no mitochondria, such as mature erythrocytes. Glucose is also essential as an energy source for exercising muscle, where it is the substrate for anaerobic glycolysis. (P125 – Lippincott)

Question 57

57.Special type of isomerism found in structures that are mirror images of each other; vast majority in humans are D-sugars A. epimers B. enantiomers C. stereoisomers D. anomers

Answer 57

B. enantiomers D & L Isomerism (Enantiomer): D & L isomers are mirror Images of each other (*Different Properties) Dextro = Right, Levo = Left: Pertaining to Hydroxyl Groups.

Question 58

58.The carbonyl carbon, either aldehyde or keto group , undergo this process to produce a new alcohol group. Fructose goes through this process to form sorbitol and mannitol. A. Oxidation B. Reduction C. Mutarotation D. Conformation

Answer 58

A. Oxidation Sugars are Reducing Agents LEORA

Question 59

59.Which of the following is correct? A. The Fischer projection is the representation of the cyclic sugar where carbon 1 is the farthest to the right, the plane of the ring is flat. B. Glycogen is synthesized from Beta-D-glucopyranose. C. The noncarbohydrate portion of the complex carbohydrate molecule is called the aglycone. D. Lactose and sucrose are both reducing sugars.

Answer 59

C. The noncarbohydrate portion of the complex carbohydrate molecule is called the aglycone. The Haworth projection is the representation of the cyclic sugar where carbon 1 is the farthest to the right, the plane of the ring is flat. Glycogen is synthesized from glucose (α1, 4 (Branches: α1, 6) Cellulose consist of Beta-D-glucopyranose units linked by β1 → 4 bonds Lactose (Reducing) and sucrose (non-reducing) Structure of Monosaccharides: Straight Chain Structure (*Fischer Projection ) Ring (Cyclic) Structure (Haworth Projection) Boat (Chair) Form

Question 60

60.Which among these tissues depend on the hormone insulin for glucose uptake? A. muscles B. intestines C. hepatocytes D. erythrocytes

Answer 60

A. muscles Glucose Transporters: GLUT-1 is abundant in erythrocytes and brain GLUT-2 is found in liver(hepatocytes) and the β cells of the pancreas. GLUT-3 is the primary glucose transporter in neurons GLUT-4 (which is insulin-dependent) is found in muscle and adipose tissue GLUT-5 is the primary transporter for fructose (instead of glucose) in the small intestine and the testes. (P106 – Lippincott)

Question 61

61.An athlete has been training for an upcoming triathlon. His coach recommended a high carb diet after the work out to ensure a muscle glycogen storage. Activity of glycogen synthase in the resting muscles is increased by the action of: A. fasting B. glucagon C. epinephrine D. insulin

Answer 61

D. insulin The (pancreatic) β cells are the most important glucose-sensing cells in the body. Like the liver, β cells contain GLUT-2 transporters (see p. 97) and have glucokinase activity, and thus can phosphorylate glucose in amounts proportional to its actual concentration in the blood. Ingestion of glucose or a carbohydrate-rich meal leads to a rise in blood glucose, which is a signal for increased insulin secretion (as well as decreased glucagon synthesis and release). Glucose is the most important stimulus for insulin secretion. (P310 – Lippincott)

Question 62

62.Deficiency with this enzyme is associated with crampy abdominal pain and passage of loose, watery stools after significant intake of dairy product. A. sucrase B. maltase C. lactase D. galactose

Answer 62

C. Lactase Lactose Intolerance (Lactase Deficiency) – Lactose is not degraded (Not absorbed) & goes to the colon; It attracts water (Carbohydrate); Stool = Watery, Acidic, Foul-Smelling, Explosive.

Question 63

63.Pancreatic enzymes have an absolute requirement for these ions once carbohydrate digestion resumes in the lumen of the small intestines. A. sodium B. chloride C. bicarbonate D. calcium

Answer 63

C. bicarbonate Further digestion of carbohydrates by pancreatic enzymes occurs in the small intestine When the acidic stomach contents reach the small intestine, they are neutralized by bicarbonate secreted by the pancreas, and pancreatic α-amylase continues the process of starch digestion. (P86 – Lippincott)

Question 64

64.This carbohydrate is absorbed from intestine by facilitated diffusion and enters cells; converted to intermediates in glycolytic pathway; A. Sucrose B. Ribose C. Ribulose D. Fructose

Answer 64

A. Sucrose Fructose Metabolism The major source of fructose is the disaccharide sucrose, which, when cleaved in the intestine, releases equimolar amounts of fructose and glucose. (P137 – Lippincott)

Question 65

65.During the well-fed state, the liver and _________produce glycogen from excess glucose. A. Brain B. Kidney C. Muscle D. Pancreas

Answer 65

C. Muscle Fluctuation of glycogen stores Liver glycogen stores increase during the well-fed state, and are depleted during a fast. Muscle glycogen is not affected by short periods of fasting (a few days) and is only moderately decreased in prolonged fasting (weeks). Muscle glycogen is synthesized to replenish muscle stores after they have been depleted following strenuous exercise. [Note: Glycogen synthesis and degradation are cytosolic processes that go on continuously. The differences between the rates of these two processes determine the levels of stored glycogen during specific physiologic states.] (P126 – Lippincott)

Question 66

66.Which maternal blood type is the most commonly associated with ABO incompatibility? A. Blood Type A B. Blood Type B C. Blood Type O D. Blood Type AB

Answer 66

D. Blood Type O SGD (2025)

Question 67

67.Autoantibodies targeting myelin proteins initiate the injury of white and gray matter of the central nervous system, thereby leading to progressive muscle weakness, paresthesia, vision changes and cognitive decline. A. Guillian Barre B. Multiple Sclerosis C. Transverse Myelitis D. Hypokalemia

Answer 67

B. Multiple Sclerosis SGD (2025)

Question 68

68.Before birth, this antibody s transferred from the maternal circulation to the fetus crossing the placenta. A. IgA B. IgG C. IgM D. IgE

Answer 68

B. IgG SGD (2025)

Question 69

69.An antibody that plays a crucial role in the immune function of mucous membranes FOUND main immunoglobulin found in secretions, including tears, saliva, sweat, colostrum and secretions from the genitourinary tract, gastrointestinal tract, prostate and respiratory epithelium, A. IgA B. IgG C. IgM D. IgE

Answer 69

A .IgA SGD (2025)

Question 70

70.Most common clinical presentation of ABO incompatibility among neonates. A. Fever B. poor suck C. jaundice D. vomiting

Answer 70

C. Jaundice SGD (2025)

Question 71

71.Which of the following scenario warrants immediate investigation? A. A fullterm presenting with jaundice at 18th hour of life. B. A preterm presenting with hyperbilirubinemia at 3rd day of life. C. A fullterm presenting with icteric sclerae 2nd day of life. D. A preterm presenting with hyperbilirubinemia at 2nd week of life.

Answer 71

A. A fullterm presenting with jaundice at 18th hour of life. SGD (2025)

Question 72

72.Most dreaded complication of neonatal hyperbilirubinemia. A. Sepsis B. Kernicterus C. Bronze Baby Syndrome D. Autism

Answer 72

B. Kernicterus SGD (2025)

Question 73

73.Which of the following is true? A. ABO blood group antibodies appear birth, B. ABO levels can change with changes in diet and with bacterial colonization in the intestinal tract C. ABO antibodies are catabolized by immune stimulation from gastrointestinal microbiota D. The promotion of ABO-antibody production by dedicated gut microbiota may decrease the levels of these natural antibodies sufficiently to enhance this natural mechanism of protection against SARS-CoV-2.

Answer 73

b. ABO levels can change with changes in diet and with bacterial colonization in the intestinal tract SGD (2025)

Question 74

74.Based on the journal, patients with low levels of ABO antibodies are at higher risk of being infected with SARS Cov-2. A. True B. False

Answer 74

A. True SGD (2025)

Question 75

75. Which of the following statements is/are true? A. Naturally occurring, Anti-B is found in the serum of people with blood groups O and B. B. Anti-A Is found in the serum of people with blood groups B and A. C. Anti-A and anti-B bind to RBCs and activate the complement cascade, which lyses the RBCs while they are still in the circulation (intravascular hemolysis). D. HDN caused by ABO antibodies occurs almost exclusively in infants of blood group O Or B who are born to group A mothers.

Answer 75

C. Anti-A and anti-B bind to RBCs and activate the complement cascade, which lyses the RBCs while they are still in the circulation (intravascular hemolysis). SGD (2025)

Question 76

76.Compartment of the mammalian cell wherein glycogen is stored: A. mitochondrial matrix B. golgi complex C. cytoplasm (granules) D. lysosomes

Answer 76

C. cytoplasm (granules) Glycogen is a branched-chain polysaccharide made exclusively from α-D-glucose. The primary glycosidic bond is an α(1→4) linkage. After an average of eight to ten glucosyl residues, there is a branch containing an α(1→6) linkage . A single molecule of glycogen can have a molecular mass of up to 108 daltons. These molecules exist in discrete cytoplasmic granules that also contain most of the enzymes necessary for glycogen synthesis and degradation. (P126 – Lippincott)

Question 77

77.Important function of stored muscle glycogen: A. maintain blood glucose normal levels B. supply dependent tissues with glucose during starvation C. fuel reserve for synthesis of ATP when liver glycogen stores are depleted D. reserve buffer for energy needed for muscle activity

Answer 77

D. reserve buffer for energy needed for muscle activity Fluctuation of glycogen stores Liver glycogen stores increase during the well-fed state, and are depleted during a fast. Muscle glycogen is not affected by short periods of fasting (a few days) and is only moderately decreased in prolonged fasting (weeks). Muscle glycogen is synthesized to replenish muscle stores after they have been depleted following strenuous exercise. [Note: Glycogen synthesis and degradation are cytosolic processes that go on continuously. The differences between the rates of these two processes determine the levels of stored glycogen during specific physiologic states.] (P126 – Lippincott)

Question 78

78.Which of the following is/are true? A. Monosaccharides, the simplest carbohydrates may be of the D or L series B. Monosaccharides form rings (pyranose or furanose) C. The hydroxyl group on the anomeric carbon (carbonyl carbon) may be of the α or β configuration D. All of the above

Answer 78

D. All of the above Structures of Monosaccharides: Straight Chain Structure (*Fischer Projection ) D & L Isomerism (Enantiomer): Mirror Image Dextrorotatory = Right; Levorotatory = Left ; Racemic Mixture = Both Levo and Dextro Anomerism (Assymetric Carbon) : -OH Alpha = Below/Right; Beta = Above/Left Ring (Cyclic) Structure (Haworth Projection) Pyranose Ring = 6 Sides; Furanose Ring = 5 Sides Boat (Chair) Form.

Question 79

79.Which of the following best describes a reducing sugar? A. When the anomeric carbon is oxidized, another compound is reduced B. Contains an anomeric carbon that is attached to another structure C. Only the state of the hydroxyl on the anomeric carbon determines if the sugar is reducing or non-reducing D. All of the above

Answer 79

A. When the anomeric carbon is oxidized, another compound is reduced LEORA = Reducing Sugar If the hydroxyl group on the anomeric carbon of a cyclized sugar is not linked to another compound by a glycosidic bond, the ring can open. The sugar can act as a reducing agent, and is termed a reducing sugar. Such sugars can react with chromogenic agents (for example, Benedict’s reagent or Fehling’s solution) causing the reagent to be reduced and colored, with aldehyde group (R-CHO) of the acyclic sugar becoming oxidized. [Note: Only the state of the oxygen in the aldehyde group determines if the sugar is reducing or nonreducing.] (P84-Lippincott)

Question 80

80.Which is an example of a triose? A. Xylulose B. Glucose C. Erythrose D. Glyceraldehyde

Answer 80

D. Glyceraldehyde

Question 81

81.Hydrolysis of lactose yields: A. Galactose and glucose B. Galactose and fructose C. Glucose and fructose D. Fructose and galactose

Answer 81

A. Galactose and glucose

Question 82

82.The sugar which forms a major component of nucleic acids is: A. Ribose B. Mannose C. Galactose D. Maltose

Answer 82

A. Ribose There are two chemically distinct types of nucleic acids: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). DNA, the repository of genetic information, is present not only in chromosomes in the nucleus of eukaryotic organisms, but also in mitochondria and the chloroplasts of plants (P395 – Lippincott)

Question 83

83.In glucose, the orientation of the –H and –OH groups around the carbon atom 5 adjacent to the terminal primary alcohol carbon determines: A. D or L series B. Dextro or levorotatory C. α & β anomers D. epimers

Answer 83

A. D or L series D & L or Optical Isomerism/ Enantiomer: Found in the pairs of structures that are mirror images of each other. Refers to the orientation of hydroxyl (-OH) at the penultimate carbon (5th carbon or second to the last carbon) whether it is at the left or right.

Question 84

84.Which of the following is not a disaccharide? A. Hyaluronic acid B. Maltose C. Lactose D. Sucrose

Answer 84

A. Hyaluronic acid Hyaluronic acid : Heteropolysaccharide Maltose (Glu + Glu) Lactose (Glu + Gal) Sucrose (Glu + Fru)

Question 85

85.Which of the following is not a homopolysaccharide? A. Starch B. Heparin C. Glycogen D. Cellulose

Answer 85

B. Heparin Starch – Storage of carbohydrates in plants. Glycogen – Highly branched (Lesser degree than Dextrin). Consists of glucose molecules by A-1, 4 linkages in the chain with branching points every 8-12 residues (A-1, 6 linkages) Cellulose – Glucose residues joined by B-1, 4 linkages. Major component of vegetable fiber. Chief constituent of plant cell walls. Source of bulk in stool. *Linear.

Question 86

86.Which of the following will most probably be affected by ABO incompatibility A. A mother bearing her first child B. A patient who has had one abortion C. A fetus with erythroblastosis fetalis D. All of the above

Answer 86

C. A fetus with erythroblastosis fetalis SGD (2025)

Question 87

87.Citrate is needed during blood transfusion because it is A. An important component of blood B. Is an anticoagulant C. is an immunosuppressant D. D. is a steroid

Answer 87

B. Is an anticoagulant SGD (2025)

Question 88

88.What is the hormone that should be in high levels in the system so that glycogenesis will be stimulated? A. epinephrine B. glucagon C. insulin D. adrenaline

Answer 88

C. insulin Glycogen synthesis and degradation are reciprocally regulated to meet whole-body needs by the same hormonal signals, namely, an elevated insulin level results in overall increased glycogenesis and decreased glycogenolysis, whereas an elevated glucagon (or epinephrine) level causes increased glycogenolysis and decreased glycogenesis.(P135- Lippincott)

Question 89

89.In the muscles, glycogen phosphorylase in transformed into the active enzyme directly by: A. glycogen synthase kinase B. protein kinase A C. phosphorylase kinase D. adenylate kinase

Answer 89

C. phosphorylase kinase Activation of glycogen phosphorylase: Glycogen phosphorylase also exists in two forms: the dephosphorylated, inactive “b” form and the phosphorylated, active “a” form. Active phosphorylase kinase phosphorylates glycogen phosphorylase b to its active “a” form, which then begins glycogenolysis. Phosphorylase a is reconverted to phosphorylase b by the hydrolysis of its phosphate by protein phosphatase-1. (P132 – Lippincott)

Question 90

90.The reaction catalyzed by glycogen phosphorylase is carried out with the help of what ion? A. calcium B. hydroxyl ions C. hydrogen ions D. inorganic PO4

Answer 90

C. hydrogen ions

Question 91

91.Presence of carbon monoxide poisoning results to which of the following? A. Shifting of the oxygen dissociation curve to the left B. Shifting of the oxygen dissociation curve to the right C. Causes an increased delivery of oxygen to tissues D. Decreases the oxygen affinity of the remaining 3 oxygen-binding sites

Answer 91

A. Shifting of the oxygen dissociation curve to the left Binding of CO: Carbon monoxide (CO) binds tightly (but reversibly) to the hemoglobin iron, forming carbon monoxyhemoglobin (or carboxyhemoglobin). When CO binds to one or more of the four heme sites, hemoglobin shifts to the relaxed conformation, causing the remaining heme sites to bind oxygen with high affinity. This shifts the oxygen dissociation curve to the left, and changes the normal sigmoidal shape toward a hyperbola. As a result, the affected hemoglobin is unable to release oxygen to the tissues [Note: The affinity of hemoglobin for CO is 220 times greater than for oxygen.] (P32-Lippincott)

Question 92

92.Which can be omitted in managing patients with Thalassemia and other hemoglobinopathies? A. Iron supplements B. Folic Acid supplements C. Blood transfusions D. D. Chelation

Answer 92

A. Iron supplements

Question 93

93.This accounts for the largest fraction of total hemoglobin (90%): A. HbA B. HbF C. HbA2 D. HbA1c

Answer 93

A. HbA

Question 94

94.What happens to the oxygen dissociation curve of an anemic patient? A. There is a “shift to the left” with associated reduction of the affinity of hemoglobin to oxygen B. There is a “shift to the left” with associated increase of the affinity of hemoglobin to oxygen C. There is a “shift to the right” with associated reduction of the affinity of hemoglobin to oxygen D. There is a “shift to the right” with associated increase of the affinity of hemoglobin to oxygen

Answer 94

C. There is a “shift to the right” with associated reduction of the affinity of hemoglobin to oxygen Response of 2,3-BPG levels to chronic hypoxia or anemia: The concentration of 2,3-BPG in the RBC increases in response to chronic hypoxia, such as that observed in chronic obstructive pulmonary disease (COPD) like emphysema, or at high altitudes, where circulating hemoglobin may have difficulty receiving sufficient oxygen. Intracellular levels of 2,3-BPG are also elevated in chronic anemia, in which fewer than normal RBCs are available to supply the body’s oxygen needs. Elevated 2,3-BPG levels lower the oxygen affinity of hemoglobin, permitting greater unloading of oxygen in the capillaries of the tissues.

Question 95

95.Which of the following Red Cell Indices show the differences in the sizes of Red Blood Cells and is being postulated to be a predictor of risk for mortality in Carbon Monoxide poisoning? A. Mean Corpuscular Volume B. Red Cell Distribution Width C. Mean Corpuscular Hemoglobin D. Mean Corpuscular Hemoglobin Concentration

Answer 95

A. Mean Corpuscular Volume

Question 96

96.Currently, the only cure available to patients with sickle cell disease A. Crizanlizumab B. Stem Cell Transplantation C. Hydroxyurea D. Gene therapy

Answer 96

B. Stem Cell Transplantation

Question 97

97.The following are associated with smoking A. decrease in hematocrit, platelet activation and aggregation B. vasoconstriction C. decrease in circulating fibrinogen & thrombin production D. exogenous fibrinolytic damage

Answer 97

B. vasoconstriction

Question 98

98.The mode of inheritance of Beta thalassemia major is: A. Autosomal dominant B. Autosomal recessive C. X-linked recessive D. De novo mutation

Answer 98

B. Autosomal recessive

Question 99

99.Which B-Thalassemia syndrome is characterized by frequent transfusions, marked hepatosplenomegaly, craniofacial changes, and stunting of growth: A. Thalassemia intermedia B. Thalassemia major C. Silent carrier D. Thalassemia trait

Answer 99

B. Thalassemia major

Question 100

100. This hemoglobin is markedly elevated in Beta thalassemia major: A. HbA1 B. HbA2 C. HbF D. HbH

Answer 100

C. HbF