Biomolecules

NEET Biology · 137 questions · Page 6 of 14 · Click an option or "Show Solution" to reveal answer

Q51
Which one of the following statements is incorrect?
A In competitive inhibition, the inhibitor molecule is not chemically changed by the enzyme.
B The competitive inhibitor does not affect the rate of breakdown of the enzyme-substrate complex.
C The presence of the competitive inhibitor decreases the Km of the enzyme for the substrate.
D A competitive inhibitor reacts reversibly with the enzyme to form an enzyme-inhibitor complex.
Correct Answer
Option C
Solution

(c) Km increases but Vmax remains the same in competitive inhibition.

Q52
Which one of the following statements is correct, with reference to enzymes?
A Holoenzyme = Apoenzyme + Coenzyme
B Coenzyme = Apoenzyme + Holoenzyme
C Holoenzyme = Coenzyme + Co-factor
D Apoenzyme = Holoenzyme + Coenzyme
Correct Answer
Option A
Solution

(a) Holoenzyme consists of a protein part called apoenzyme and a non-protein called cofactor/coenzyme.

Q53
Prosthetic groups differ from co-enzymes in that
A They can serve as co-factors in a number of enzyme-catalysed reactions.
B They require metal ions for their activity.
C They (prosthetic groups) are tightly bound to apoenzymes.
D Their association with apoenzymes is transient.
Correct Answer
Option C
Solution

(c) Prosthetic groups are tightly bound to the apoenzyme but coenzymes are associated transiently.

Q54
The correct statement with regard to the secondary structure of DNA/RNA is
A RNA possesses a single strand helix structure and contains thymine as one of the four bases
B DNA possesses a double strand helix structure and contains thymine as one of the four bases
C RNA possesses a double strand helix structure and contains uracil as one of the four bases
D DNA possesses a single strand helix structure and contains uracil as one of the four bases
Correct Answer
Option B
Solution

RNA is typically single standard but it contains uracil, not thymine.

DNA is secondary structure has a double strand helix consisting of two polynucleotide chains.

Its four nitrogenous bases are adenine (A), guanine (G), cytosine (C) and thymine (T).

Q55
Sugar ' XX ' A. is found in honey B. is a keto sugar C. exists in α\alpha and β\beta-anomeric forms. D. Is laevorotatory. ' X ' is :
A Maltose
B Sucrose
C D-Glucose
D D-Fructose
Correct Answer
Option D
Solution

D-Fructose is found in honey and is a keto sugar.

 D-(-) - Fructose \text{ D-(-) - Fructose }
Q56
Fehling's solution 'A' is
A aqueous copper sulphate
B alkaline copper sulphate
C alkaline solution of sodium potassium tartrate (Rochelle's salt)
D aqueous sodium citrate
Correct Answer
Option A
Solution

Fehling's solution is a chemical reagent used to differentiate between water-soluble carbohydrates and ketone-functional groups, and as a test for monosaccharides.

It is used specially in the Fehling's test for reducing sugars.

The test involves two solutions, generally known as Fehling's A and Fehling's B, which are mixed together and added to perform the test.

Fehling's solution 'A' is an aqueous solution of copper(II) sulfate.

Thus, the correct option is: Option A: aqueous copper sulphate The function of Fehling's A is to provide copper(II) ions,

Cu2+Cu^{2+}

, which act as the oxidizing agent in the reaction with the reducing sugar.

When Fehling's Solution A and B are mixed and heated with a reducing sugar, the copper(II) ions are reduced to copper(I) oxide, which precipitates as a red solid, indicating a positive result.

Fehling's Solution B, on the other hand, contains alkaline sodium potassium tartrate, which helps to maintain the solution in an alkaline condition and keeps the copper(II) ions in solution.

Thus, in summary, Fehling's A consists of aqueous copper sulfate, which directly matches with Option A.

Q57
The incorrect statement about denaturation of proteins is :
A Uncoiling of the helical structure takes place
B It results due to change of temperature and/or pH
C It results in loss of biological activity of proteins
D A protein is formed from amino acids linked by peptide bonds
Correct Answer
Option D
Solution

\bullet Proteins are polymers of α\alpha-amino acids and they are connected to each other by peptide bond, but this is not denaturation process.

\bullet Due to denaturation globules unfold and helix get uncoiled and protein loses its biological activity.

Denaturation can be caused if protein in its native form, is subjected to physical change like change in temperature or chemical change like change in pH

Q58
The incorrect statement regarding enzymes is
A Enzymes are biocatalysts.
B Like chemical catalysts enzymes reduce the activation energy of bio processes.
C Enzymes are polysaccharides.
D Enzymes are very specific for a particular reaction and substrate.
Correct Answer
Option C
Solution

Enzymes are complex nitrogenous organic compounds which are produced by living plants and animals.

They are protein molecules of high molecular mass.

They are not polysaccharides.

Q59
Which of the following is a basic amino acid?
A Alanine
B Tyrosine
C Lysine
D Serine
Correct Answer
Option C
Solution

Lysine is a basic amino acid it contains more number of - NH 2 groups as compared to - COOH groups due to their symmetrical.

Q60
Sucrose on hydrolysis gives :
A α\alpha -D-Glucose + β\beta -D-Glucose
B α\alpha -D-Glucose + β\beta -D-Fructose
C α\alpha -D-Fructose + β\beta -D-Fructose
D α\alpha -D-Glucose + α\alpha -D-Fructose
Correct Answer
Option B
Solution

Sucrose is a disaccharide composed of two monosaccharide units, glucose and fructose, joined together via a glycosidic bond.

Upon hydrolysis, this bond is broken, and the two constituent monosaccharides are released.

Hydrolysis of sucrose results in the formation of one molecule of glucose and one molecule of fructose.

However, it is important to specify the configurations of these monosaccharides as they exist in specific forms.

Sucrose

Hydrolysis\overset{{Hydrolysis}}\longrightarrow

α\alpha-D-Glucose + β\beta-D-Fructose Glucose in sucrose is in the α\alpha-D form while fructose is in the β\beta-D configuration.

Thus, the correct answer reflecting the products of sucrose hydrolysis is: Option B: α\alpha-D-Glucose + β\beta-D-Fructose This reflects the correct stereochemistry and identity of the monosaccharides produced from the hydrolysis of sucrose.

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