Molecular Biology Questions and Answers – Chemical Structures of Nucleic Acids

This set of Molecular Biology Multiple Choice Questions & Answers (MCQs) focuses on “Chemical Structures of Nucleic Acids”.

1. Which macromolecule is not abundantly found though being of critical importance for biological mechanisms?
a) Proteins
b) Lipids
c) Nucleic acids
d) Polysaccharides
View Answer

Answer: b
Explanation: Though lipids is of essential use in the cell for the formation of cell wall it is a micro molecule and is generally found in association with either phosphate or polysaccharide. In case of the other three they are all macromolecules and are of essential use for the cell. For example, proteins such as enzymes regulates biological functions, nucleic acids carry genetic information and polysaccharides function as either storage of energy or acts as structural polymers.

2. Which of the following is wrongly paired?
a) Proteins – peptide bond
b) Nucleic acid – hydrogen bond
c) Polysaccharide – glycosidic bond
d) Phospholipids –phosphate linkage
View Answer

Answer: b
Explanation: Nucleic acids, that is, DNA and RNA show phosphodiesterase linkage which is the major type of linkage. Without the phosphodiester bonds between the phosphate and adjacent 3’OH sugar molecule the backbone will not be formed. Therefore, the nucleotides would not be able to attach and bond to form a nucleic acid.

3. With respect to nucleosides which of the following is paired correctly?
a) Purine – Adenosine, Thymidine
b) Purine – Guanosine, Thymidine
c) Pyrimidine – Uridine, Cytidine
d) Pyrimidine – Uridine, Adenosine
View Answer

Answer: c
Explanation: As we know purines have a fused ring structure with 9 element backbones ring structure. They are of two types Adenosine and Guanine. Again, the other three residues cytidine, Thymidine and Uridine are pyrimidines having a ring structure of 6 elements in the core backbone. Thus the correct pair among the above options is pyrimidine – uridine, cytidine.

4. Which of the following is not a component of the nucleic acid backbone?
a) Pentose sugar
b) Phosphate group
c) Nucleotide
d) Phosphodiesterase bond
View Answer

Answer: c
Explanation: A nucleic acid backbone is mainly composed of a pentose-phosphate unit which acts as a monomer. The repeat of this monomer is attached together by the phosphodiesterase linkage, thus, giving rise to the backbone. The nucleotides attached to the sugar moieties forms the side chain which gives rise to the hydrogen linkage with its complementary strand.

5. According to Chargaff’s rule the two strands of DNA has ___________
a) Same molecular weight
b) Same amount of A and G
c) Different amount of A and G
d) Different molecular weight
View Answer

Answer: d
Explanation: According to Chargaff’s rule the two strands have equal number of A and T residues and equal number of G and C residues. Thus when in one strand A is more automatically in the other T is more. Thus, A being a pyrimidine has a higher molecular weight than T, which is a purine, and so the two strands have different molecular weight. The same happens in case of G and C also.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

6. In one strand of a double stranded DNA the rate of occurrence of A is 3 times C in consecutive 10 bases. So how many G will be there in 100 base pairs of a DNA duplex?[Consider G=T in one strand].
a) 30
b) 20
c) 40
d) 60
View Answer

Answer: c
Explanation: Let’s consider C = 1
Therefore, A = 3C = (3*1) = 3
Now, A+C = (3+1) = 4
Again G = T [given]
And A+C+G+T = 10 [given]
Now replacing T with G and putting the value of A+C
We get,
4 + 2G = 10
2G = 6
G = 3, T = 3, A = 3, C = 1
Thus in 100 bases in one strand there are (3*10 = 30) G residues
Now for the complementary strand G = C residues [Chargaff’s rule]
Therefore, C = G = 1 residue in every 10 bases
That is, G = 1*10 = 10 in 100 bases
Therefore total number of G residues = 30 + 10 = 40 in 100 base pairs of a DNA duplex.

7. In a diploid organism with 30,000 bases haploid genome contains 23% A residues. What is the number of G residues in the genome of this organism?
a) 16000
b) 16200
c) 16500
d) 14200
View Answer

Answer: b
Explanation: By Chargaff’s rule, A = T = 23%
Therefore, G + C = [100 – (A+T)]
G + C = [100 – 46]
G + C = 54
G = C = 27 [By Chargaff’s rule]
Therefore, G = 27%
Now, each cell is diploid thus it contains (2*30,000 = 60,000) bases
So, G = 27% of 60,000
G = 16,200 bases.

8. Which of the following is not a characteristic of nucleotide bases?
a) Planar
b) Heterocyclic
c) Aliphatic
d) Ubiquitous
View Answer

Answer: c
Explanation: The nucleotide bases have a 6 member ring structure as one component which is common for all five bases. Thus, they are considered as aromatic molecules due to the presence of a benzene ring structure which provides the molecules with an aromatic property.

9. Which of the following factors do not provide to the separation of DNA fragments during electrophoresis?
a) Chargaff’s rule
b) Matrix density
c) Ethidium bromide
d) Size
View Answer

Answer: c
Explanation: Ethidium bromide only helps in tagging the DNA molecules to make the visible under the UV radiation due to its fluorescence property, but does not take any part in the separation of DNA molecules during electrophoresis.

10. Which one of the following is not a function of a nucleotide?
a) Nucleic acid monomer
b) Ribozyme
c) Energy carrier molecules
d) Receptors
View Answer

Answer: d
Explanation: Nucleotides do not form receptor molecules. Receptor molecules are generally polysaccharide molecules or protein molecules that can have structural configurations. Energy carrying nucleotide molecules such as ATP helps binding and unbinding of signal molecules to these receptors thus facilitating different biological functions.

Sanfoundry Global Education & Learning Series – Molecular Biology.

To practice all areas of Molecular Biology, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.