Carbohydrates MCQ


21 Questions MCQ Test Mock Test Series of IIT JAM Biotechnology | Carbohydrates MCQ


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This mock test of Carbohydrates MCQ for IIT JAM helps you for every IIT JAM entrance exam. This contains 21 Multiple Choice Questions for IIT JAM Carbohydrates MCQ (mcq) to study with solutions a complete question bank. The solved questions answers in this Carbohydrates MCQ quiz give you a good mix of easy questions and tough questions. IIT JAM students definitely take this Carbohydrates MCQ exercise for a better result in the exam. You can find other Carbohydrates MCQ extra questions, long questions & short questions for IIT JAM on EduRev as well by searching above.
QUESTION: 1

Sucrose, also known as table sugar is one of the most commonly used carbohydrate as a sweetener. It is a disaccharide made up of glucose and fructose. The kind of linkage that is found in sucrose is

Solution:

The glucose and fructose in sucrose are linked by glucose fructose linkage.

QUESTION: 2

Which of the following is a disaccharide.

Solution:

Trehalose, is a disachharides that is made up of α -D-glucopyranosyl-(1 → 1)-α -D-glucopyranoside units. While inulin is polysaccharides, raffinose is trisaccharide and cellulose is polysaccharides.

QUESTION: 3

Which of the following is NOT a hexose sugar 

Solution:

Idose, Allose, Altrose, Gulose, Galactose, Mannose, Stachyose and Glucose are all hexose sugars, while xylose is an example of pentose sugar.

QUESTION: 4

Which of the following carbohydrates would be the most abundant in the diet of strict vegetarians?

Solution:

Cellulose, the most abundant compound known, is the structural fiber of plants and bacterial walls. It is a polysaccharide consisting of chains of glucose residues linked by β 1 → 4 bonds. Since humans do not have intestinal hydrolases that attack β 1 → 4 linkages, cellulose cannot be digested but forms an important source of “bulk" in the diet. Lactose is a disaccharide of glucose and galactose found in milk. Amylose is an unbranched polymer of glucose residues in α-1.4 linkages. Glycogen is a branched polymer of glucose with both α-1.4 and α-1, 6 linkages. Maltose is a disaccharide of glucose, which is usually the breakdown pro duct of amylose.

QUESTION: 5

Which of the following sugars is a non- reducing sugar

Solution:

A reducing sugar is capable of acting as a reducing agent, because it has free aldehyde or a free ketone group. Sucrose and trehalose both don’t have free aldehyde and ketone group, so they are non-reducing.

QUESTION: 6

Which of the following anomeric form of Glucose is present in Starch 

Solution:

Starch is made up of repeating units of alpha D glucose. In aqueous solution, glucose occurs predominantly as cyclic (ring) structure in which the carbonyl group of glucose forms a covalent bond with oxygen of hydroxyl group at C-5 position of the same glucose molecule, resulting in formation of hemiacetal winch contains an additional asymmetric carbon atom. Thus cyclic form of glucose can exist in two stereoisomeric forms called α and β. Therefore, anomers are isomeric forms of monosaccharides that differ only in their configuration about the hemiacetal or hemiketal carbon atom

QUESTION: 7

Which of the following is the correct combination of C-2 epimer and C-4 epimer of Glucose respectively.

Solution:

Epimers are the pair of carbohydrates that have difference in the rotation of atoms across single carbon atom. If they differ across 2nd carbon atoms, molecules are named as C-2 Epimers. If they differ across 4th carbon atoms, molecules are named as C-4 Epimers.

QUESTION: 8

Which of the following is an example of hemiacetal

Solution:

Hemiacetal is forimed due to reaction between aldehyde and internal hydroxyl groups of the carbohydrates. If hemiacetal reacts with one more sugar through alcohol group they lead to the formation of acetal hi linear structure carbohydrates are neither hemiacetal nor acetal. Glucose is a aldehyde sugar and aldehyde sugars on cyclization tonus hemiacetal. Fructose is a ketose sugar and ketose sugars on cyclization form Hemiketal. Hemiacetals / Hemiketal contain an additional asymmetric carbon atom and thus can exist in 2 stereo isomeric forms α and β.

QUESTION: 9

Which of the following is formed if glucose is oxidized at both C1 and C6 positions. 

Solution:

Gluconic acid is formed due to oxidation at C1 alone, while Glucuronic acid is formed when oxidation occurs at C6 only. Oxidation product of glucose which is oxidized at both C1 and C6 is formed in presence of strong oxidizing agent and is known as Glucaric acid.

QUESTION: 10

Amino sugar present in bacterial cell wall is

Solution:

A sugar molecule in which OH (hydroxyl group) is replaced by a amino group, is called an amino sugar. Bacterial cell walls are made up of peptidoglycan. Peptidoglycan is a polymer consisting of sugars and amino acid. The sugar component consists of alternating residues of β (1, 4) linked X acetyl glucosamine (NAG) and N acetyl muramic acid. In N acetyl muramic acid (NAM), lactic acid is ether linked to the oxygen at C-3 of N acetyl glucosamine.

QUESTION: 11

Osazone are formed by a chemical reaction between carbohydrates containing a free aldehyde or keto group and

Solution:

Carbohydrates reaction with phenyl hydrazine leads to the formation of Osazone.


All reducing sugars form osazones. Glucose is a reducing sugar hence it forms osazone. Sucrose does not form osazone crystals because it is a non-reducing sugar as it has no free carbonyl group. Sucrose is made
up of glucose and fructose. The linkage in sucrose is Glucose

QUESTION: 12

This sugar is extracted from the roots of Dahelia and many other tuberous plants and useful in the determination of Glomerular filteration rate of kidneys. 

Solution:

Inulin is extracted from the roots of Dahelia and used for the determination of Glomerular filtration rate. The monomeric unit of inulin is fructose and glucose.

*Multiple options can be correct
QUESTION: 13

On adding iodine to three different tubes labelled A, B and C, the colour of A turned blue, colour of B turned reddish violet and colour of C remained red (colour of iodine), Provided that all the tubes contained carbohydrates which of the following is true.

Solution:

Amylose being a linear polymer without branching forms a helical shape in which iodine fits well, while iodine cannot bind to the sheet link structure of cellulose. The binding of iodine is mild in case of amylopectin, as amylopectin is a branched chain molecule, so cannot trap the molecule of iodine, meaning the colour of iodine remains reddish violet. Hence Amylose gives blue colour, amylopectin gives violet colour and cellulose does not change the colour of iodine.

*Multiple options can be correct
QUESTION: 14

During the process of mutarotation

Solution:

Mutarotation is the change in the optical rotation because of the change in the equilibrium between two anomers. when the corresponding stereocenters interconvert. Cyclic sugars show mutarotation as α and β anomeric forms interconvert.
During mutarotation optical activity does not change from positive to negative but increases or decreases. For example-In α glucose specific rotation is +112.2 but in its anomeric form β -Glucose, the specific rotation is +18.7.

*Multiple options can be correct
QUESTION: 15

Which of the following is true about the isomers of glucose

Solution:

Mannose and Glucose are C2 Epimers, while Glucose and Galactose are C4 epimers. Glucose and fructose are not epimers but functional isomers. Alpha D Glucose and Beta D glucose are anomers.
(Anomers and epimers are both diastereomers, but an epimer is a stereoisomer that differs in configuration at any single stereogenic center, while an anomer is actually an epimer that differs in configuration at the acetal/hemiacetal carbon.)

*Multiple options can be correct
QUESTION: 16

Which of the following polysachharides are heteropolysachharides.

Solution:

Those polysaccharides which consist of two or more different kind of monosaccharide are called as heteropolysachharides
Cellobiose is a disaccharide with the formula C12H22O11. Cellobiose. a reducing sugar, coznsists of two β-glucose molecules linked by a β (1 → 4) bond.
Heparin: The most common disaccharide unit is composed of a 2-O-sulfated iduronic acid and 6-O-sulfated. N-sulfated glucosamine. IdoA(2S)-GlcNS(6S).
Peptidoglycan: The sugar component consists of alternating residues of β -(1, 4) linked X-acetylglucosamine (NAG) and X-acetylmuramic acid (NAM) C'hondroitin sulfate: linear heteropolysaccharide consisting of repeatins disaccharide units of slucuronic acid and salactosamine

*Multiple options can be correct
QUESTION: 17

Which of the following statement s are true about carbohydrates

Solution:

Statement (a) is correct. Allose is a hexose. It has 4 chiral centres. Therefore, the number of isomers of allose are 2n = 24 = 16.
Statement (c) is correct. Trehalose is a disaccharide formed by 1-1 glycosidic bond between the two alpha glucose units. Therefore in trehalose, anomeric carbon of both glucose are involved in glycosidic linkage so it is a non-reducing sugar and also does not exhibit mutarotation.
Statement (d) is incorrect
α.D glucose and β D glucose are anomers. Isomeric forms of monosaccharides that differ only in their configuration about the hemiacetal or hemiketal carbon atoms are called anomers. D glucose and D galactose are C-4 epimers.
Two sugars that differ only in the configuration around one carbon atom are called epimers. D galactose differs from D glucose in the configuration around 4th carbon atom.

*Answer can only contain numeric values
QUESTION: 18

If the number of asymmetric carbon atoms in a sugar are 3, the total number of stereoisomers will be 8. Let us assume that one of the structure is named as pintose, then the number of Diasteromers pintose will have are_________________ 


Solution:

The number of stereoisomers will be 8 (number of stereoisomers = 2n where n is number of chiral centres). So each molecule will have one enantiomer while remaining will be the Diasteromers. So in this case pintose will have one enantiomer and remaining 6 molecules will be its Diasteromers.

*Answer can only contain numeric values
QUESTION: 19

During a mutarotation experiment, the net optical activity at any given time point was observed as +100o.  The absolute optical activity of alpha anomer is +112.2º and beta anomer is +18.7º. The percentage of alpha anomers at that time will be ___________ 


Solution:

Let us assume that the percentage of alpha anomer is x, so the percentage of beta anomer will be
100-x. The net optical activity will be the result of individual contribution of each anomer.
i.e. 112.2(x/100) + 18.7 (100-x/100) = 100
112x - 18.7 x = 10000 – 1870
X = 86.9

*Answer can only contain numeric values
QUESTION: 20

The net optical rotation of a solution containing 20% sucrose hydrolyzed into glucose and fructose during the process of sugar inversion is _____________
Given – Optical activity of Glucose = + 52.7, Fructose = –92.0 and Sucrose = + 66.7).


Solution:

In order to calculate the net optical activity add the contribution of each component
(20% sucrose was hydrolyzed means – 20% glucose, 20% fructose and 80 % sucrose)
0.2 x 52.7 – 92 x .2 + 0.8 x 66.7 = 45.5

*Answer can only contain numeric values
QUESTION: 21

The stems of bamboo, a tropical grass, can grow at the phenomenal rate of 0.3 m/day under optimal conditions. It is given that the stems are composed almost entirely of cellulose fibers oriented in the direction of growth. Each D-glucose unit contributes ~0.5 nm to the length of a cellulose molecule. The number of sugar residues per second that must be added enzymatically to growing cellulose chains to account for the growth rate is ______________


Solution:

Fust, calculate the growth per second:
0.3 m/day = 3 x 10-6 m/s
(24 h/day)(60 min/h)(60 s/min)
Given that each glucose residue increases the length of the cellulose chain by 0.5 nm(5 x 10-10 m), the number of residues added per second is 3 x 10-6 ms = 6,000 residues/s
5 x 10-10 m/residue