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Carbohydrates MCQ - Biotechnology Engineering (BT) MCQ


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21 Questions MCQ Test Mock Test Series of IIT JAM Biotechnology 2025 - Carbohydrates MCQ

Carbohydrates MCQ for Biotechnology Engineering (BT) 2024 is part of Mock Test Series of IIT JAM Biotechnology 2025 preparation. The Carbohydrates MCQ questions and answers have been prepared according to the Biotechnology Engineering (BT) exam syllabus.The Carbohydrates MCQ MCQs are made for Biotechnology Engineering (BT) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Carbohydrates MCQ below.
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Carbohydrates MCQ - 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

Detailed Solution for Carbohydrates MCQ - Question 1

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

Carbohydrates MCQ - Question 2

Which of the following is a disaccharide.

Detailed Solution for Carbohydrates MCQ - Question 2

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.

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Carbohydrates MCQ - Question 3

Which of the following is NOT a hexose sugar 

Detailed Solution for Carbohydrates MCQ - Question 3

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

Carbohydrates MCQ - Question 4

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

Detailed Solution for Carbohydrates MCQ - Question 4

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.

Carbohydrates MCQ - Question 5

Which of the following sugars is a non- reducing sugar

Detailed Solution for Carbohydrates MCQ - Question 5

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.

Carbohydrates MCQ - Question 6

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

Detailed Solution for Carbohydrates MCQ - Question 6

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

Carbohydrates MCQ - Question 7

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

Detailed Solution for Carbohydrates MCQ - Question 7

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.

Carbohydrates MCQ - Question 8

Which of the following is an example of hemiacetal

Detailed Solution for Carbohydrates MCQ - Question 8

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.

Carbohydrates MCQ - Question 9

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

Detailed Solution for Carbohydrates MCQ - Question 9

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.

Carbohydrates MCQ - Question 10

Amino sugar present in bacterial cell wall is

Detailed Solution for Carbohydrates MCQ - Question 10

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.

Carbohydrates MCQ - Question 11

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

Detailed Solution for Carbohydrates MCQ - Question 11

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

Carbohydrates MCQ - 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. 

Detailed Solution for Carbohydrates MCQ - Question 12

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
Carbohydrates MCQ - 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.

Detailed Solution for Carbohydrates MCQ - Question 13

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
Carbohydrates MCQ - Question 14

During the process of mutarotation

Detailed Solution for Carbohydrates MCQ - Question 14

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
Carbohydrates MCQ - Question 15

Which of the following is true about the isomers of glucose

Detailed Solution for Carbohydrates MCQ - Question 15

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
Carbohydrates MCQ - Question 16

Which of the following polysachharides are heteropolysachharides.

Detailed Solution for Carbohydrates MCQ - Question 16

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
Carbohydrates MCQ - Question 17

Which of the following statement s are true about carbohydrates

Detailed Solution for Carbohydrates MCQ - Question 17

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
Carbohydrates MCQ - 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_________________ 


Detailed Solution for Carbohydrates MCQ - Question 18

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
Carbohydrates MCQ - 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 ___________ 


Detailed Solution for Carbohydrates MCQ - Question 19

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
Carbohydrates MCQ - 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).


Detailed Solution for Carbohydrates MCQ - Question 20

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
Carbohydrates MCQ - 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 ______________


Detailed Solution for Carbohydrates MCQ - Question 21

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

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