This set of Life Sciences Multiple Choice Questions & Answers (MCQs) focuses on “Nucleic Acids”.
1. The monomeric unit of nucleic acid are called ___________
a) Nucleotides
b) Nucleosides
c) Pyrimidines
d) Purines
View Answer
Explanation: Nucleotides are the monomeric unit of nucleic acid; therefore, these are considered as polynucleotides. It is made up of three components i.e. nitrogenous base, sugar, and an ion of phosphoric acid. Nucleotides are phosphate easters of the nucleoside.
2. Who discovered nucleic acid?
a) Watson and Crick
b) Griffith
c) Friedrich Miescher
d) Walter Gilbert
View Answer
Explanation: Friedrich Miescher first discovered nucleic acid from nuclei of pus cells from discarded bandages and called it nuclein. It was later revealed as a mixture of basic protein and phosphorus-containing acid, now called nucleic acid.
3. Name the pyrimidine base which is found in RNA but not in DNA?
a) Thymine
b) Uracil
c) Adenine
d) Guanine
View Answer
Explanation: Uracil is considered like thymine, but it lacks methyl group at the C5 position. Uracil is a component of RNA whereas it is not found in DNA.
4. Name those purine bases which are commonly found in DNA and RNA?
a) Adenine and guanine
b) Cytosine and thymine
c) Adenine and thymine
d) Cytosine and guanine
View Answer
Explanation: There are two types of nitrogenous bases in both RNA and DNA, purines and pyrimidines. Adenine and guanine are purines while cytosine and thymine come under pyrimidines.
5. Deoxyribose sugar is found in DNA.
a) True
b) False
View Answer
Explanation: Ribose sugar is found in RNA, if the hydroxyl group(OH) at the 2nd position of ribose sugar is replaced by hydrogen (H) it will convert to deoxyribose which is DNA’s sugar.
6. Which of these statements are TRUE for a glycosidic bond?
a) The bond between sugar and the nitrogenous base
b) The bond between the two sugar residues
c) The bond between sugar and phosphate
d) The bond between base and phosphate
View Answer
Explanation: The glycosidic bond formed between sugar and nitrogenous base during the formation of a nucleoside. This bond is held to be in β-configuration with respect to the ribose sugar leaving it’s base free to rotate around the bond.
7. What is the energy currency of the living cell?
a) GTP
b) NADH
c) FADH
d) ATP
View Answer
Explanation: ATP is the energy currency of the cell as it acts as a donor of high energy phosphate. It also acts as a chemical link between catabolism and anabolism. In its active form ATP is complexed with Mg2+ or Mn2+.
8. The primary structure of DNA and RNA proceeds in which direction?
a) 3’ —–> 5҆’
b) 5’ —–> 3’
c) 4’ —–> 6’
d) 3’ —–> 6’
View Answer
Explanation: The primary structure is described by the order in which bases occur in a polypeptide. The convention is to arrange them from left to right from 5’ end to 3’ end. 5’ end of the polypeptide is attached to phosphate while 3’ end is usually a free hydroxyl group.
9. Which bond is used to stabilize the double helix of DNA?
a) Hydrophobic bond
b) Hydrogen bond
c) Covalent bond
d) Ionic bond
View Answer
Explanation: Polynucleotide runs on the backbone of the phosphate-sugar chain. The base pairing of these nucleotide gives secondary structure. The secondary structure of DNA is its double helix which held together by hydrogen bonding between the bases.
10. Which of these statements are INCORRECT according to chargaff’s rules?
a) The molar ratio of A to T or C to G = 1
b) The sum of purine = sum of pyrimidine
c) The % C+G ≠ % A+T
d) The composition of DNA remains the same in all the species
View Answer
Explanation: Erwin Chargaff proposed two rules which played a significant role in the discovery of the double helix structure of DNA. According to its second rule, the composition of DNA varies from one species to other. Statement composition of DNA remains the same in all the species is incorrect.
11. A purified DNA sample of leaf contains 20.0 mole of guanine. Calculate the approximate percentage of purine residues?
a) 30%
b) 40%
c) 60%
d) 20%
View Answer
Explanation: According to chargaff’s rules, C=G= 20%; A=T
(A+G) + (T+C) =100%
(A+20) + (T+20) =100%
(A+20) + (A+20) = 100%
2A= 100-40%
A = 30%.
Sanfoundry Global Education & Learning Series – Life Sciences.
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