DNA and RNA are two types of genetic material found in organisms. DNA is present in the majority of organisms while RNA is found in some viruses. RNA is known to function as a messenger mainly. Here is a collection of important MCQs related to replication of DNA, transcription and translation, and the process of protein synthesis and regulation.

1. Read the following statements and choose the appropriate option.

Statement 1: In Griffith’s experiment, when the mice were injected with heat-killed R-strain and live S-strain, the mice survived.

Statement 2: In the Hershey-Chase experiment, radioisotope of phosphorus – ³²P was incorporated into the DNA of the viruses.

Statement 3: In the experiment conducted by Avery, MacLeod and McCarty, they found out that in the presence of RNase enzyme, transformation of the bacteria took place.

(a) Statements 1 and 2 are correct, but statement 3 is wrong.

(b) Statements 1 and 3 are correct, but statement 2 is wrong.

(c) Statements 2 and 3 are correct, but statement 1 is wrong.

(d) Statements 1, 2 and 3 are correct.

Answer: (c)

Explanation: In Griffith’s experiment, when the mice are injected with heat-killed R-strain and live S-strain, the mice die because the live S-strain is virulent in nature. Hershey and Chase used two isotopes ³²P and ³⁵S; the sulphur radioisotope got incorporated into the protein capsid and the phosphorus radioisotope got incorporated into the DNA of the viruses. Avery, MacLeod and McCarty conducted experiments where they used protease, RNase and DNase. They observed that transformation took place in the case of protease and RNase but not in the case of DNase.

2. Which is not correct according to Chargaff’s rule?

(a) A+T = G+C

(b) A+G = C+T

(c) (A+G)/(T+C) = 1

(d) Total number of purines = Total number of pyrimidines

Answer: (a)

Explanation: According to Chargaff’s rule, in a double stranded DNA

• Total amount of adenine = Total amount of thymine
• Total amount of cytosine = Total amount of guanine
• Total amount of purines = Total amount of pyrimidines

Hence, A+G = C+T

Or,

3. Statement 1:In eukaryotes, both exons and introns are transcribed to form hnRNA.

Statement 2: Splicing is required in prokaryotes.

(a) Statement 1 is false and statement 2 is true.

(b) Statement 1 is true and statement 2 is false.

(c) Both statement 1 and statement 2 are false.

(d) Both statement 1 and statement 2 are false.

Answer: (b)

Explanation: In eukaryotes, translation results in the formation of hnRNA which is made of both coding (exon) and non-coding regions (introns). The hnRNA then undergoes splicing to form a functional mRNA that is made up of exons only. On the other hand, splicing is not required in prokaryotes because they do not have any introns.

4. Fill in the following blanks with appropriate options.

(A) Information present in the DNA is transcribed in the form of codons in ___(i)___ for protein synthesis.

(B) ___(ii)___ is a constituent of ribosomes.

(C) Around 80% of the total cellular RNA is ___(iii)____.

(D) ____(iv)___ carries amino acids to the ribosomes and mRNA to carry out protein synthesis.

(a) rRNA, mRNA, tRNA, rRNA

(b) mRNA, rRNA, rRNA, tRNA

(c) mRNA, rRNA, tRNA, mRNA

(d) tRNA, mRNA, tRNA, rRNA

Answer: (b)

Explanation: RNA is ribonucleic acid that is made up of ribose sugar, phosphate group and nitrogenous bases. There are three types of RNA –

• mRNA: also called messenger RNA, it constitutes about 3-5% of total RNA content. It carries information from DNA for protein synthesis. Three nucleotides in an mRNA constitute a codon which codes for a particular amino acid.
• tRNA: stands for transfer RNA and constitutes about 10-15% of the total RNA content. It brings amino acids from the cytoplasm for protein synthesis.
• rRNA: stands for ribosomal RNA, it constitutes about 80% of the total RNA and is a part of the ribosome.

5. If DNA has 45000 base pairs, how many complete turns will the DNA molecule take?

(a) 4500

(b) 450

(c) 45000

(d) 45

Answer: (a)

Explanation: Following are the characteristics of a DNA double helix –

• It is made up of two polynucleotide chains.
• It is coiled in a right-handed helical manner.
• Pitch length = 34Å
• Number of base pairs in one turn = 10

Therefore, in a DNA with 45000 base pairs, it will take 45000/10 = 4500 turns.

6. Pick the incorrect statements among the following:

(i) A typical nucleosome contains 400bp of DNA.

(ii) The positively charged DNA is wrapped around negatively charged histone octamer to form a structure called nucleosome.

(iii) Histones are rich in basic amino acid residues like asparagine and valine.

(iv) The packing of chromatin at a higher level requires an additional set of proteins that collectively are referred to as non-histone chromosomal proteins.

(a) (i), (ii) and (iii)

(b) (ii), (iii) and (iv)

(c) (ii) and (iii)

(d) (i), (iii) and (iv)

Answer: (a)

Explanation: A section of DNA that is wrapped around histone proteins is known as nucleosome. There are around 200 base pairs of DNA wrapped around a nucleosome. DNA is a negatively charged molecule due to the presence of phosphate groups. The histone proteins wrap around the negatively charged DNA to neutralise the molecule. Histones are rich in basic amino acids such as lysine and arginine and hence they are considered as positively charged.

7. If Meselson and Stahl’s experiment is continued for 4 generations in E.coli, then the ratio of ¹⁵N/¹⁵N : ¹⁵N/¹⁴N : ¹⁴N/¹⁴N in the end would be:

(a) 0:1:3

(b) 0:1:7

(c) 0:1:15

(d) 0:1:31

Answer: (b)

Explanation: Meselson and Stahl conducted an experiment to show whether the DNA replicates conservatively, semi-conservatively or dispersively. For this, he mixed ¹⁵N in the replication medium so that all the nitrogen in the DNA is replaced by the radioisotope. Next he mixed ¹⁴N in the medium to incorporate ¹⁴N in new DNA strands.

At 0 minutes, the medium contained ¹⁵N heavy bands of DNA, after 1st replication (20 minutes in E.coli) a hybrid band of ¹⁵N¹⁴N was observed. The second replication cycle showed 50% 14N14N and 50% ¹⁴N¹⁵N. The third cycle showed 75% ¹⁴N¹⁴N and 25% ¹⁴N¹⁵N. Finally the fourth cycle of replication showed 87.5% ¹⁴N¹⁴N and 12.5% ¹⁴N¹⁵N.

8. The given diagram represents the transcription unit. Select the correct labellings regarding it.

(a) A-Terminator, B-Promoter, C-Template strand, D-Coding strand

(b) A-Promoter, B-Terminator, C-Template strand, D-Coding strand

(c) A-Promoter, B-Terminator, C-Coding strand, D-Template strand

(d) A-Promoter, B-Template strand, C-Terminator, D-Coding strand

Answer: (b)

Explanation: In a transcriptional unit, the strand in the direction of 5’ to 3’ is considered as the coding strand and the strand in the direction of 3’ to 5’ is the template strand. The promoter is located upstream of the coding strand. It functions to initiate the process of transcription with the help of RNA polymerase enzymes. The terminator is located downstream of the coding strand and becomes the site where transcription is terminated.

9. 3’-AAATGCGCCATA-5’ is the sequence of nucleotides on a gene. After transcription, the sequence of nucleotides in the mRNA formed from it and the sequence of nucleotides in the corresponding coding strand of DNA will be

(a) 5’-UAU-GUT-CCA-UUU-3’ and 5’-AUA-CAU-GGU-AAA-3’

(b) 5’-UUU-ACG-CGG-UAU-3’ and 5’-TTT-ACG-CGG-TAT-3’

(c) 5’-UAU-CGC-GCA-UUU-3’ and 5’-AUA-GCG-CGU-AAA-3’

(d) 5’-UUU-ACC-TUG-UAU-3’ and 5’-AAA-UGC-UAC-AUA-3’

Answer: (b)

Explanation: The mRNA and the coding strand are the same except that in the mRNA sequence, uracil is found in the place of thymine and in the coding strand thymine is intact.

10. The enzyme DNA-dependent RNA polymerase and DNA-dependent DNA polymerase catalyse the polymerisation reaction in ___(i)____ and ___(ii)___ direction, respectively.

(a) (i) 5’-3’ , (ii) 3’-5’

(b) (i) 5’-3’ , (ii) 5’-3’

(c) (i) 3’-5’ , (ii) 5’-3’

(d) (i) 3’-5’ , (ii) 3’-5’

Answer: (b)

Explanation: DNA-dependent RNA polymerase are enzymes that synthesise RNA from a DNA template, whereas DNA-dependent DNA polymerase are enzymes that add nucleotides in a growing DNA strand, i.e., help in DNA replication. Both the enzymes work in the same direction of 5’ to 3’.

11. Consider the following statements regarding eukaryotes.

I. RNA polymerase I transcribes rRNAs.

II. RNA polymerase II transcribes snRNAs.

III. RNA polymerase III transcribes hnRNA.

IV. RNA polymerase III transcribes tRNA.

Which of the statements given above are correct?

(a) I and II

(b) I and III

(c) I and IV

(d) I, II and III

Answer: (c)

Explanation: There are three types of polymerases found in eukaryotes –

12. A frameshift mutation will have minimum effect when it leads to ________

(a) insertion of 2 bases

(b) deletion of 1 base

(c) deletion of 3 bases

(d) deletion of 2 bases

Answer: (c)

Explanation: Frameshift mutations happen when there is deletion or insertion of one or more nucleotides in a DNA sequence so that the reading frame of the sequence is changed. Deleting or adding 3 bases at once removes or adds one whole codon which minimises the effect of this type of mutation.

13. Which among the following are enzymes coded by the lac operon?

(a) β-Galactosidase, Permease, Glycogen synthase

(b) β-Galactosidase, Permease, Transacetylase

(c) Permease, Glycogen synthase, Transacetylase

(d) β-Galactosidase, Permease, Phosphoglucose isomerase

Answer: (b)

Explanation: The lac operon has three structural genes that code for the following enzymes –

lacZ: β-Galactosidase

lacY: Permease

lacA: Transacetylase

14. Arrange the following steps of DNA fingerprinting in their correct sequence.

I. Autoradiography

II. Isolation of DNA

III. Southern Blotting

IV. Cleavage by restriction endonucleases

V. Gel Electrophoresis

(a) I, II, IV, V, III

(b) II, V, III, IV, I

(c) II, IV, V, III, I

(d) V, II, III, IV, I

Answer: (c)

Explanation: DNA fingerprinting is a technique which uses DNA to determine the unique identity of an individual. The following are the steps of DNA fingerprinting:

• Isolation of DNA
• Polymerase Chain Reaction
• Restriction digestion
• Gel electrophoresis
• Southern blotting
• Hybridisation
• Autoradiography

15. Which of the following properties of DNA polymerase necessitates the requirement of RNA primer in DNA replication?

(a) It gets denatured at high temperatures

(b) It can add a nucleotide only to a 5’ end

(c) It can add a nucleotide only to a 3’ end

(d) It allows only continuous replication

Answer: (c)

Explanation: DNA polymerase is an enzyme that can add nucleotides only to the free 3’ end. RNA primer thus creates a free 3’ end for this addition which was not available at the start of replication.

16. Match column I with column II with respect to the findings of the Human Genome Project and choose the correct option.

(a) 1-A, 2-C, 3-D, 4-B

(b) 1-D, 2-C, 3-B, 4-A

(c) 1-D, 2-B, 3-C, 4-A

(d) 1-C, 2-D, C-B, 4-A

Answer: (b)

Explanation: The human genome project was aimed at sequencing the entire genome of an organism. The findings of the project are as follows:

• There are a total of 3146.7 million nucleotide bases in the human genome.
• The dystrophin gene is the largest gene, having 2.4 million base pairs. It is located on the X chromosome.
• Maximum number of genes are present on chromosome 1, a total of 2968 genes.
• The Y chromosome was found to have the least number of genes, i.e., 31.

17. What type of bonds link the two complementary strands of the DNA molecule?

(a) Ester bonds

(b) Glycosidic bonds

(c) Purine-pyrimidine hydrogen bonds

(d) All of the above

Answer: (c)

Explanation: The nitrogenous bases pair with their complementary strands by hydrogen bonds which link the two complementary strands of DNA. Adenine pairs with thymine by two hydrogen bonds, while guanine pairs with cytosine by three hydrogen bonds.

18. During protein synthesis, UAG functions as the stop codon in mRNA. What should be the corresponding anticodon in the tRNA molecule?

(a) UAC

(b) CAU

(c) TAC

(d) It has no anticodon

Answer: (d)

Explanation: UAA, UGA, UGA are nonsense or stop codons that do not have any anticodons, i.e., they do not code for any amino acids.

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