JEE Main Biomolecule Previous Year Questions With Solutions

Biomolecule previous year solved questions are given in this article. The activities performed by living beings include responses to specific natural mixtures. Such mixtures are called biomolecules. These are fundamental natural atoms, which are engaged with the upkeep and metabolic cycles of living beings. These non-living atoms are the real infantrymen of the skirmish of food of life. Instances of some normal biomolecules are proteins, nutrients, carbs and lipids. The significant subjects of biomolecules incorporate, carbohydrates, proteins, lipids and nucleic acids.

The inquiries given here to give you a thought regarding what kind of questions to anticipate from the subject biomolecule. Infinity Learn gives solutions made by our master team.

1. Observation of “Rhumann’s purple” is a corroborative test for the presence of:

(1) Starch

(2) Reducing sugar

(3) Cupric particle

(4) Protein

Solution:

Rhumann’s purple is ninhydrin. It is utilized to recognize alpha-amino corrosive or carboxylic corrosive. While contact with the free amines (proteins), a dark blue or purple tone known as Ruhemann’s purple is created. So observation of “Rhumann’s purple” is a corroborative test for the presence of protein.

Consequently choice (4) is the answer.

2. Amalgamation of every atom of glucose in photosynthesis includes:-

(1) 18 atoms of ATP

(2) 10 atoms of ATP

(3) 8 atoms of ATP

(4) 6 atoms of ATP

Solution:

12H₂O + 12 NADP + 18 ADP → 6O₂ + 18 ATP + 12 NADPH ( light response)

6CO₂ + 12 NADPH + 18 ATP → C6H12O6 + 12 NADP + 18 ADP + 6H₂O (dull response)

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ (Net response)

18 atoms of ATP is engaged with the combination of every particle of glucose in photosynthesis.

Henceforth choice (1) is the answer.

3. Collection of which of the accompanying atoms in the muscles happens because of enthusiastic activity:-

(1) Pyruvic corrosive

(2) L-lactic corrosive

(3) Glycogen

(4) Glucose

Solution:

Because of overwhelming activity, the collection of L-lactic corrosive in the muscles occurs. This is a consequence of anaerobic breath.

Subsequently choice (2) is the answer.

4. Which of the nutrients given beneath is water-solvent?

(1) Vitamin E

(2) Vitamin K

(3) Vitamin C

(4) Vitamin D

Solution:

L-ascorbic acid is a water-solvent nutrient. It is likewise called ascorbic corrosive.

Consequently choice (3) is the answer.

5. Biuret test isn’t given by:-

(1) proteins

(2) starches

(3) polypeptides

(4) urea

Solution:

The Biuret test is given by an amide linkage. It is available in proteins, polypeptides and urea. It is absent in carbs.

Consequently choice (2) is the answer.

6. RNA is not the same as DNA since RNA contains

(1) ribose sugar and thymine

(2) ribose sugar and uracil

(3) deoxyribose sugar and thymine

(4) deoxyribose sugar and uracil

Solution:

RNA is not quite the same as DNA since RNA contains ribose sugar and uracil base.

Henceforth choice (2) is the answer.

7. Which base is present in RNA and yet not in DNA?

(1) Uracil

(2) Thymine

(3) Guanine

(4) Cytosine

Solution:

Uracil is available in RNA, yet not present in DNA.

Consequently choice (1) is the answer.

8. The term anomers of glucose implies to

(1) isomers of glucose that vary in configu­rations at carbons one and four(C-land C-4)

(2) a combination of (D)- glucose and (L)- glucose

(3) enantiomers of glucose

(4) isomers of glucose that vary in the arrangement at carbon one (C-l)

Solution:

An anomer is an epimer at the hemiacetal or hemiketal carbon in a cyclic saccharide, a molecule called the anomeric carbon. Anomer is an illustration of a stereoisomer. The carbon molecule at position 1 is anomeric.

Henceforth choice (4) is the answer.

9. Which pyrimidine bases are available in DNA?

(1) cytosine and adenine

(2) cytosine and guanine

(3) cytosine and thymine

(4) cytosine and uracil

Solution:

In DNA cytosine and thymine, two pyrimidine bases present

Consequently choice (3) is the answer.

10. The optional construction of a protein alludes to

(1) a – helical spine

(2) hydrophobic collaborations

(3) grouping of a-amino acids

(4) fixed arrangement of the polypeptide back­bone

Solution:

The optional construction of protein incorporates a-helical back security and R-sheet structures.

Consequently choice (1) is the answer.

11. Which one of the accompanying possibilities is right?

(1) All amino acids with the exception of lysine are optically dynamic

(2) All amino acids are optically dynamic

(3) All amino acids with the exception of glycine are optically dynamic

(4) All amino acids with the exception of glutamic acids are optically dynamic

Solution:

Glycine contains two hydrogen particles at the alpha carbon.

All amino acids with the exception of glycine are optically dynamic.

Consequently choice (3) is the answer.

12. Which of the accompanying mixtures can be recognized by Molisch’s test?

(1) Nitro compounds

(2) Sugars

(3) Amines

(4) Primary alcohols

Solution:

Molisch’s test is a test for the presence of carbs. The development of a purple or a purplish-red ring at the resource between the H2SO4 and the analyte + Molisch’s reagent blend affirms the presence of carbs in the analyte. The two types of carbs are sugars and starches. So sugars can be distinguished by Molisch’s test.

Henceforth choice (2) is the answer.

13. The availability or non-availability of hydroxy gathering on which carbon molecule of sugar separates RNA and DNA?

(1) third

(2) fourth

(3) first

(4) second

Solution:

In RNA, ribose is available on the second carbon particle. In DNA, deoxyribose is available on the second carbon atom. Thus, the presence or nonappearance of hydroxy gathering on the second carbon particle of sugar separates RNA and DNA.

Thus choice (4) is the answer.

14. Which α-D (+)- glucose and β-D(+)- glucose are present?

(1) conformers

(2) epimers

(3) anomers

(4) enantiomers

Solution:

α-D (+)- glucose and β-D(+)- glucose is two anomers present.

Thus choice (3) is the answer.

15. Complete hydrolysis of cellulose gives

(1) D-ribose

(2) D-glucose

(3) L-glucose

(4) D-fructose

Solution:

Complete hydrolysis of cellulose gives D-glucose.

(C₆H₁₀O₅ )_n + nH₂O → nC₆H₁₂O₆

Thus choice (2) is the answer.

Also read: JEE Main Exam Mode

FAQs

Q. What are Carbohydrates?

Ans: Carbohydrates are biomolecules containing carbon, hydrogen and oxygen iotas. They are the wellspring of energy for living creatures. We can experience sugars every step of the way of our life. For instance, the paper that we utilize every day, the wood, and so forth contain cellulose. As far as biomolecules, the equivalent word of carbs is saccharides which incorporate a gathering of sugar, starch and cellulose. Sugars are named:

• Monosaccharides – They are the least complex sugar and can’t be hydrolyzed to more modest carbs. Models: glucose, fructose and galactose.
• Disaccharides – Two monosaccharides consolidated are known as disaccharides. They are the most straightforward polysaccharides Examples: Sucrose, lactose and maltose.
• Oligosaccharides – They include three to ten monosaccharide particles and are for the most part connected by an o – glycosidic bond.
• Polysaccharides – They are a long chain of monosaccharides that are limited by glycosidic linkage.

What are Amino acids?

Amino acids are biomolecules that contain two functional groups specifically: carboxylic corrosive and amine. They are the subsidiaries of carboxylic acids wherein one hydrogen particle of a carbon molecule is subbed by an amino gathering. The critical components of an amino corrosive are carbon, hydrogen, oxygen and nitrogen through different components may likewise be connected. Amino acids are characterized by the area of the primary useful gatherings like alpha, beta, gamma, or delta. Amino acids are the structure squares of proteins, lipids. Proteins are the second biggest part of muscles, cells or tissues in the human body.