Biomolecules Test

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Biomolecules Test - 77

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Weekly Quiz Competition

Question 1
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Which of the following statements is wrong regarding amino acids?

A
The living systems in plants can synthesise essential amino acids.

B
The living systems in animals can synthesise only essential amino acids.

C
The living systems in animals can synthesise only non essential amino acids.

D
The living systems in plants can synthesise non essential amino acids.
Solution
- Essential amino acids:- Essential amino acids$$(9)$$ cannot be made by the body. As a result, they must come from food.
- Non-essential amino acids:- An amino acid that can be made by the body and so is not essential to the diet.
The living system in plants can synthesize all the amino acids while the living system of animals can only synthesize non-essential amino acids.
Question 2
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The active site of an enzyme is situated in its amino acids _________ .

A
Amino groups

B
Carboxyl groups

C
R-groups

D
All the above
Question 3
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Inducer is not required in which of the following types of enzyme?

A
Allosteric

B
Inhibitory

C
Constitutive

D
All of the above

Solution

Allosteric enzymes are the enzymes which possess an allosteric site different from the active site for binding of allosteric modulators. These enzymes cannot be induced but their activity can be modulated. Constitutive enzymes are the enzymes which are constantly synthesized and are required for basic house keeping functions. Inhibitory enzymes slow down the rate of metabolism by inhibiting key reactions. None of these enzymes shows induction where induction is defined as increased synthesis in response to specific molecules called inducers.
Question 4
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..................... enzymes have heme as a prosthetic group.
(i) Catalase
(ii) Carboxypeptidase
(iii) Succinic dehydrogenase
(iv) Peroxidase

A
(i) only

B
(i) and (ii) only

C
(ii) and (iii) only

D
(iii) and (iv) only

E
(i) and (iv) only

Solution

Some enzymes, for catalytic function, in addition to the apoprotein part require a tightly covalently linked non protein part called as prosthetic group. Heme is a iron containing prosthetic group used by enzymes like catalase and peroxidase.
Question 5
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Which among the following are compounds structurally similar to substrate and inducing enzyme synthesis?

A
Constitutive enzymes

B
Gratuitous inducer

C
Coordinated inducers

D
None of the above

Solution

An inducer is a substance that induces, i.e., initiates or increases the synthesis of enzyme molecules. Gratuitous inducer is an analogue of a natural inducer that is capable of inducing an operon while not serving as a substrate for the enzyme being induced.
So, the correct answer is option B.
Question 6
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If a man is exposed to an environment of 100% oxygen, then the concentration of which of the following enzyme is increased in lungs?

A
Superoxide dismutase

B
Catalase

C
Carbonic anhydrase

D
All of the above

Solution

Superoxide dismutases are protective enzymes and protect body against oxidative damage. Superoxide dismutases alternately catalyze the dismutation (or partitioning) of the toxic superoxide (O$$_{2}$$) radical into either ordinary molecular oxygen (O$$_{2}$$) or hydrogen peroxide (H$$_{2}$$O$$_{2}$$). Superoxide is produced as a by-product of oxygen metabolism and causes many types of cell damage. Hydrogen peroxide is also damaging, but less, and is degraded by other enzymes, such as catalase. Thus, SOD is an important antioxidant defense in nearly all living cells exposed to oxygen.
So, the correct answer is option A.
Question 7
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Which vitamin is incorporated into FAD structure?

A
Vitamine B$$_{12}$$

B
Vitamine B$$_{6}$$

C
Vitamine B$$_{2}$$

D
Vitamine C

Solution

Some enzymes require no chemical groups other than their amino acid residues for activity. Others require an additional chemical component called as cofactor. The cofactor can be either one or more inorganic ions such as ferrous, divalent magnesium or divalent manganese and zinc ions or a complex organic or metallo-organic molecule called as coenzyme. For example - FAD or flavin adenine dinucleotide which transfers electrons in biochemical reactions and is derived from riboflavin or vitamin B$$_2$$.
So, the correct answer is option D.
Question 8
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$$K_m$$ value of enzyme is substrate concentration at

A
10 V$$_{max}$$

B
0.5/1 V$$_{max}$$

C
1/2 V$$_{max}$$

D
4/5 V$$_{max}$$

Solution

The catalytic efficiency of an enzyme is described by the K$$_m$$ value or Michaelis Menten constant. The Michaelis constant is the substrate concentration at which the reaction rate is one half the maximum. The K$$_m$$ describes the affinity of enzyme for a substrate molecule. Greater the affinity lower is the K$$_m$$ value and sooner the V$$_{max}$$ can be attained and vice versa.
According to Michaelis-Menten equation $$K_m$$ is equal to substrate concentration at which the velocity is half the maximum. Michaelis and Menten proposed a hypothesis for enzyme for action according to which the enzyme molecule combines with a substrate complex which further dissociates to form product and enzyme back.
$$\displaystyle E+S \xrightarrow [K_1]{K_2} E - S \xrightarrow [ K_3 ] E + P : K_m (Michaclis-Menten Consant) = \frac{K_3+K_2}{K_1}$$
They gave the following equation: $$\displaystyle V_0 = \frac{V_{max}(S)}{(S)+K_m}$$
It is the statement of the quantitative relationship between the initial velocity $$V_0$$, the maximum velocity $$V_{max}$$ the initial substrate concentrations; all related through Michaelis Menten constant $$K_m$$.
E = free enzyme
S = Substrate
ES = enzyme substrate complex
P = product
K$$_1$$ = the rate constant for ES formation
K$$_2$$ = the rate constant for dissociation of ES into E and S
K$$_3$$ = the rate constant for dissociation of EP complex into E and P.
Question 9
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Compound $$(D)$$ is:

A

B

C

D
none of the above

Solution

As seen in the above figure, the compound $$(D)$$ is represented by option $$(A)$$. The ester group of compound $$(C)$$ is hydrolyzed to the carboxylic group in acidic medium to give compound $$(D)$$.
Question 10
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Directions For Questions

$$\text{D-Glyceraldehyde} \xrightarrow[(ii)\, Ba(OH)_2+H_2SO_4]{(i)\, KCN} \text{Product (B)}$$
$$\downarrow \Delta$$ $$\text{(D+E)}\xleftarrow{Na/Hg \ , H_3O^{+}}\text{Product (C)}$$

$$\text{(D)}\xrightarrow[HNO_3]{[O]} \text{Dibasic acid (optically active)(F)}$$

$$\text{(E)}\xrightarrow[HNO_3]{[O]} \text{Dibasic acid (optically inactive)(G)}$$

...view full instructions

Two isomer products are obtained in $$(C)$$. They are:

A
both $$\gamma$$-lactone

B
both $$\delta$$-lactones

C
one is $$\gamma$$-lactone and another is $$\delta$$-lactone

D
none

Solution

Both isomer products are $$\gamma$$ lactose.