Biomolecules

JEE Chemistry · 116 questions · Page 12 of 12 · Click an option or "Show Solution" to reveal answer

Q111

Which of the following tests cannot be used for identifying amino acids?

A Ninhydrin test

B Barfoed test

C Xanthoproteic test

D Biuret test

Correct Answer

Option B

Solution

Barfoed test is used to detect the presence of monosaccharide (reducing sugars) in solutions not amino acids.

Q112

Fructose and glucose can be distinguished by :

A Barfoed's test

B Fehling's test

C Seliwanoff's test

D Benedict's test

Correct Answer

Option C

Solution

Seliwanoff's test is used to distinguish aldose and ketose.

When Seliwanoff's reagent added to a solution containing ketoses, red colour is formed rapidly indicating a positive test.

While in case of aldose, a slower forming light pink colour observed.

Q113

The correct statement (s) among I to III with respect to potassium ions that are abundant within the cell fluid is /are I. They activate many enzymes II. They participate in the oxidation of glucose to produce ATP III. Along with sodium ions they are responsible for the transmission of nerve signals

A III only

B I and II only

C I and III only

D I, II and III only

Correct Answer

Option D

Solution

All the statements are correct.

K+ being metallic unipositive ions work as enzyme activators.

These also participate in many reactions of glycolysis and Kreb’s cycle to produce ATP from glucose.

Being unipositive these are also equally responsible for nerve signal transmission along with Na+.

Q114

Number of cyclic tripeptides formed with 2 amino acids A and B is :

A 2

B 3

C 5

D 4

Correct Answer

Option D

Solution

The cyclic tripeptides possible with amino acids $A$ and $B$ will be AAA, BBB, ABA, BAB There are are 4 possible cyclic structures.

Q115

The increasing order of pka of the following amino acids in aqueous solution is : Gly, Asp, Lys, Arg

A Asp < Gly < Arg < Lys

B Gly < Asp < Arg < Lys

C Asp < Gly < Lys < Arg

D Arg < Lys < Gly < Asp

Correct Answer

Option C

Solution

pKa = $-$ logKa $\therefore$ Acidic strength (Ka) $\propto$

{1 \over {pKa}}

$\therefore$ Order of acidic strength : Asp > Gly > LYs > Arg $\therefore$ Order of pKa is : Asp < Gly < Lys < Arg.

Q116

Among the following compounds most basic amino acid is -

A Lysine

B Asparagine

C Histidine

D Serine

Correct Answer

Option A

Solution

Compound Pl value Histidine 7.6 Serine 5.7 Lysine 9.8 Asparagine 5.4

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