Biomolecules Test

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Biomolecules Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Which of the vitamins given below is water soluble?

A
Vitamin $$C$$

B
Vitamin $$D$$

C
Vitamin $$E$$

D
Vitamin $$K$$

Solution

Vitamins $$B$$ and $$C$$ are water soluble whereas vitamins $$A, D, E, K$$ and $$P$$ are fat soluble.

Option C is correct.
Question 2
1 / -0

Amylopectin is composed of:

A
$$\alpha-D-$$glucose, $$C_1-C_4$$ and $$C_1-C_6$$ linkages

B
$$\alpha-D-$$glucose, $$C_1-C_4$$ and $$C_2-C_6$$ linkages

C
$$\beta-D-$$glucose, $$C_1-C_4$$ and $$C_2-C_6$$ linkages

D
$$\beta-D-$$glucose, $$C_1-C_4$$ and $$C_1-C_6$$ linkages

Solution

Solution:- (A) $$\alpha-D-$$glucose, $$C_1-C_4$$ and $$C_1-C_6$$ linkages
Amylopectin is a homoplymer of $$\alpha -D-$$glucose where $$C_1-C_4$$ linkage and $$C_1 - C_6$$ linkage are present.
Question 3
1 / -0

Among the following, the incorrect statement is:

A
Cellulose and amylose has 1,4-glycosidic linkage

B
Maltose and lactose has 1,4-glycosidic linkage

C
Lactose contains $$\beta$$- D-galactose and $$\beta$$-D-glocose

D
Sucrose and amylose has 1,2-glycosidic linkages

Solution

Option (D) represents incorrect statement.
Amylose contains $$1 \rightarrow 4 \alpha$$ glycoside bonds.
Sucrose has 1,2-glycosidic linkages.
Question 4
1 / -0

The two functional groups present in a typical carbohydrate are:

A
$$-OH$$ and $$-COOH$$

B
$$-CHO$$ and $$-COOH$$

C
$$>C = O$$ and $$-OH$$

D
$$-OH$$ and $$-CHO$$

Solution

Carbohydrates are primarily hydroxyl carbonyl compounds. Hence, $$ C = O$$ and $$OH$$ bonds are present.
Question 5
1 / -0

Complete hydrolysis of starch gives:

A
glucose and galactose in equimolar amounts

B
galactose and fructose in equimolar amounts

C
glucose only

D
glucose and fructose in equimolar amount

Solution

Starch is a polymer of $$\alpha-D-$$glucopyranose.

Complete hydrolysis of starch gives glucose only.

$$\underset {starch} {(C_6H_{10}O_5)_n} + nH_2O \xrightarrow [393K,2-3atm] {dil H_2SO_4} \underset {glucose} {nC_6H_{12}O_6}$$
Question 6
1 / -0

Synthesis of each molecule of glucose in photosynthesis involves:

A
$$10$$ molecules of ATP

B
$$8$$ molecules of ATP

C
$$6$$ molecules of ATP

D
$$18$$ molecules of ATP

Solution

$$12H_{2}O+12NADP+18ADP\xrightarrow[]{Light reaction}6O_{2}+18ATP+12 NADPH $$

$$
6CO_{2}+12NADPH+18\ ATP \xrightarrow[]{Dark reaction}C_{6}H_{12}O_{6}+12NADP+18ADP+6H_{2}O
$$

Net reaction: $$6CO_{2}+6H_{2}O\rightarrow C_{6}H_{12}O_{6}+6O_{2}$$
Question 7
1 / -0

A tetrapeptide has $$-COOH$$ group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary structures) with $$NH_{2}$$ group attached to a chiral centre is:

A
4

B
3

C
6

D
2

Solution

Since $$-COOH$$ group is attached to alanine, one terminal of the tetrapeptide contains alanine. Thus, its position is fixed.

Glycine is achiral whereas valine and phenylalanine are chiral. Thus, $$-NH_2$$ group is attached to either valine or phenylalanine.

Thus following are the four possible combinations of the tetrapeptide.

$$H_2N-Val-Gly-Phe-Ala-COOH$$

$$H_2N-Val-Phe-Gly-Ala-COOH$$

$$H_2N-Phe-Gly-Val-Ala-COOH$$

$$H_2N-Phe-Val-Gly-Ala-COOH$$

Therefore, the correct option is A.
Question 8
1 / -0

Transition state structure of the substrate formed during an enzymatic reation is

A
Temporary but stable.

B
Permanent but unstable.

C
Temporary and unstable.

D
Permanent and stable.

Solution

During an enzymatic reaction, a substrate goes through a transition state before it becomes a product. It is proposed that the transition state is bound more strongly than either the substrate or the product, however, the transition state is inherently unstable and are transient.
Question 9
1 / -0

The difference between amylose and amylopectin is _________.

A
Amlyopetin have $$1 \rightarrow 4 \alpha$$-linkage and $$1 \rightarrow 6\beta$$- linkage

B
Amlyopetin have $$1\rightarrow 4 \alpha$$-linkage and $$1\rightarrow 6 \alpha$$- linkage

C
Amylose is made up of glucose and galactose

D
Amylose have $$1\rightarrow 4 \alpha$$-linkage and $$1\rightarrow 6\beta$$- linkage

Solution

Amylose is a straight-chain polymer of D-glucose units.

The bonding involved is $$1 \rightarrow4\ \alpha-$$ linkage. Amylopectin is a branched chain polymer of D-glucose units.

The bonding involved is $$1 \rightarrow4\ \alpha$$ - linkage and $$1\rightarrow 6\alpha$$- linkage.

Hence, the correct answer is option B.
Question 10
1 / -0

The major role of minor elements inside living organisms is to act as

A
Cofactors of enzymes.

B
Building blocks of important amino acids.

C
Constituent of hormones.

D
Binder of cell structure.
Solution
Correct Option: A

Explanation:
- Microelements work as non-protein cofactors in enzymes. Zn, Cu, Fe, Mg, Mn, etc are some mineral elements that acts as a cofactor.
These ions attach to the enzyme and activate the enzyme.
- It brings a conformational change in the structure of an enzyme which eventually accelerates the reaction.
- For example, zinc is a cofactor for carbonic anhydrase.

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Biomolecules Test - 21

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Biomolecules Test - 21

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Question 1

Vitamin $$C$$

Vitamin $$D$$

Vitamin $$E$$

Vitamin $$K$$

Solution

Question 2

$$\alpha-D-$$glucose, $$C_1-C_4$$ and $$C_1-C_6$$ linkages

$$\alpha-D-$$glucose, $$C_1-C_4$$ and $$C_2-C_6$$ linkages

$$\beta-D-$$glucose, $$C_1-C_4$$ and $$C_2-C_6$$ linkages

$$\beta-D-$$glucose, $$C_1-C_4$$ and $$C_1-C_6$$ linkages

Solution

Question 3

Cellulose and amylose has 1,4-glycosidic linkage

Maltose and lactose has 1,4-glycosidic linkage

Lactose contains $$\beta$$- D-galactose and $$\beta$$-D-glocose

Sucrose and amylose has 1,2-glycosidic linkages

Solution

Question 4

$$-OH$$ and $$-COOH$$

$$-CHO$$ and $$-COOH$$

$$>C = O$$ and $$-OH$$

$$-OH$$ and $$-CHO$$

Solution

Question 5

glucose and galactose in equimolar amounts

galactose and fructose in equimolar amounts

glucose only

glucose and fructose in equimolar amount

Solution

Question 6

$$10$$ molecules of ATP

$$8$$ molecules of ATP

$$6$$ molecules of ATP

$$18$$ molecules of ATP

Solution

Question 7

4

3

6

2

Solution

Question 8

Temporary but stable.

Permanent but unstable.

Temporary and unstable.

Permanent and stable.

Solution

Question 9

Amlyopetin have $$1 \rightarrow 4 \alpha$$-linkage and $$1 \rightarrow 6\beta$$- linkage

Amlyopetin have $$1\rightarrow 4 \alpha$$-linkage and $$1\rightarrow 6 \alpha$$- linkage

Amylose is made up of glucose and galactose

Amylose have $$1\rightarrow 4 \alpha$$-linkage and $$1\rightarrow 6\beta$$- linkage

Solution

Question 10

Cofactors of enzymes.

Building blocks of important amino acids.

Constituent of hormones.

Binder of cell structure.

Solution

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