NEET Biology Molecular Basis of Inheritance MCQs Set 13

Biology Objective Questions and Answers: Molecular Basis of Inheritance

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Multiple Choice Questions

Question 1: If the sequence of one strand of DNA is 5'-ATGCATCG-3', the sequence of the complementary strand in the 5' to 3' direction is
(a) TACGTAGC
(b) CGATGCAT
(c) ATGCATCG
(d) ATCGTACG
Show Answer & Explanation

Answer: (b) CGATGCAT

Explanation:
1. The complementary strand runs antiparallel. Complement of 5'-ATGCATCG-3' is 3'-TACGTAGC-5', which written 5' to 3' is CGATGCAT.

Teacher's Note:
1. Write the complement and then reverse it.

Question 2: NHC structural proteins are
(a) Basic proteins rich in lysine, arginine
(b) Regulatory proteins
(c) Catalytic proteins rich in tryptophan and arginine
(d) Required for packaging of chromatin at higher levels
Show Answer & Explanation

Answer: (d) Required for packaging of chromatin at higher levels

Explanation:
1. Non-histone chromosomal (NHC) proteins are needed for packaging of chromatin at higher levels. Histones are the basic proteins rich in lysine and arginine.

Teacher's Note:
1. Histones make nucleosomes; NHC proteins fold higher.

Question 3: Which of the following bond is not present in DNA?
(a) 1'-9' N-glycosidic bond
(b) 3'-5' phosphodiester bond
(c) 1'-1' N-glycosidic bond
(d) 1'-2' N-glycosidic bond
Show Answer & Explanation

Answer: (d) 1'-2' N-glycosidic bond

Explanation:
1. DNA contains N-glycosidic bonds from C1' of sugar to N9 of purines or N1 of pyrimidines, and 3'-5' phosphodiester bonds. A 1'-2' N-glycosidic bond does not exist.

Teacher's Note:
1. Purines use N9, pyrimidines use N1.

Question 4: How many types of DNA polymerases are present in bacteria?
(a) Five
(b) Three
(c) Two
(d) One
Show Answer & Explanation

Answer: (b) Three

Explanation:
1. Bacteria have three DNA polymerases: Pol I, Pol II and Pol III.

Teacher's Note:
1. Pol III is the main replicating enzyme.

Question 5: Synthesis of leading and lagging strand require
(a) Single primer
(b) Single and many primers respectively
(c) Many and single primers respectively
(d) Many primers
Show Answer & Explanation

Answer: (b) Single and many primers respectively

Explanation:
1. The leading strand is made continuously after a single primer, whereas the lagging strand is made in pieces needing many primers.

Teacher's Note:
1. Okazaki fragments need a primer each.

Question 6: For the strand separation and stabilisation during DNA replication which of the following set of enzymes and proteins are required?
(a) SSBP, gyrase, primase
(b) Topoisomerase, helicase, ligase
(c) Gyrase, ligase, primase
(d) Topoisomerase, helicase, SSBP
Show Answer & Explanation

Answer: (d) Topoisomerase, helicase, SSBP

Explanation:
1. Topoisomerase relieves twisting stress, helicase unwinds the helix and SSBP stabilises single strands.

Teacher's Note:
1. Primase makes primers, ligase seals nicks.

Question 7: In eukaryotes, the RNA polymerase that synthesises tRNA is RNA polymerase ______ and is also responsible for formation of ______ rRNA.
(a) II, 5.8 S
(b) I, 5 S
(c) III, 5 S
(d) II, 18 S
Show Answer & Explanation

Answer: (c) III, 5 S

Explanation:
1. RNA polymerase III synthesises tRNA, 5S rRNA and snRNAs; polymerase I makes 28S, 18S and 5.8S rRNA.

Teacher's Note:
1. Pol I rRNA, Pol II mRNA, Pol III tRNA.

Question 8: What is correct for bacterial transcription?
(a) mRNA requires processing to become active
(b) Translation can begin when mRNA is fully transcribed
(c) Transcription and translation takes place in the same compartment
(d) Rho factor initiates the process
Show Answer & Explanation

Answer: (c) Transcription and translation takes place in the same compartment

Explanation:
1. In bacteria there is no nucleus, so transcription and translation occur in the same compartment (the cytoplasm), often simultaneously.

Teacher's Note:
1. Prokaryotic mRNA needs no processing.

Question 9: Which of the following is not required during post transcriptional processing in eukaryotes?
(a) Methyl guanosine triphosphate
(b) Ligase
(c) ScRNA
(d) SnRNA
Show Answer & Explanation

Answer: (c) ScRNA

Explanation:
1. Processing needs capping (methyl guanosine triphosphate), splicing (snRNA) and tailing; scRNA is not needed.

Teacher's Note:
1. Splicing uses snRNPs.

Question 10: Which of the following feature is correct for bacteria?
(a) Presence of intervening sequences in DNA
(b) DNA does not show coiling
(c) Linear ss-DNA representing single chromosome
(d) DNA can be chromosomal as well as extrachromosomal
Show Answer & Explanation

Answer: (d) DNA can be chromosomal as well as extrachromosomal

Explanation:
1. Bacterial DNA is chromosomal (nucleoid) as well as extrachromosomal (plasmid).

Teacher's Note:
1. Bacteria lack introns.

Question 11: In-vitro template independent RNA synthesis is a feature of
(a) RNA polymerase
(b) Reverse transcriptase
(c) Ochoa enzyme
(d) DNA polymerase
Show Answer & Explanation

Answer: (c) Ochoa enzyme

Explanation:
1. Polynucleotide phosphorylase (Ochoa's enzyme) can make RNA in vitro without a template.

Teacher's Note:
1. Used by Nirenberg.

Question 12: In protein synthesis, which of the following are required for the synthesis of charged tRNA?
(a) Amino acid, GTP, initiation codon, ribosome
(b) Amino acid, ATP, Mg++, enzyme, tRNA
(c) Amino acid, ATP, K+, enzyme, mRNA
(d) Aminoacyl tRNA, ribosome, initiation codon, release factor
Show Answer & Explanation

Answer: (b) Amino acid, ATP, Mg++, enzyme, tRNA

Explanation:
1. Charging of tRNA by aminoacyl-tRNA synthetase needs amino acid, ATP, Mg2+, the enzyme and tRNA.

Teacher's Note:
1. Activation precedes translation.

Question 13: The accessibility of promoter regions of bacterial DNA in many cases regulated by the interaction of proteins with sequences termed
(a) Regulators
(b) Structural genes
(c) Inhibitor genes
(d) Operators
Show Answer & Explanation

Answer: (d) Operators

Explanation:
1. The operator is the sequence to which the repressor protein binds, controlling access of RNA polymerase to the promoter.

Teacher's Note:
1. Operator controls transcription.

Question 14: Select the correct one (w.r.t. Wobble hypothesis)
(a) Third base of a codon lacks vibrating capacity
(b) Third base can establish H-bonds even with the non complementary anticodon
(c) Specificity of a anticodon is particularly determined by first two codon
(d) Major cause of degeneracy is the first two N-bases of codon
Show Answer & Explanation

Answer: (b) Third base can establish H-bonds even with the non complementary anticodon

Explanation:
1. By the wobble hypothesis the third base of a codon can form hydrogen bonds with a non-complementary base of the anticodon, so one tRNA reads several codons.

Teacher's Note:
1. Wobble explains degeneracy.

Question 15: If there are 81 million bases in RNA of human cell, then calculate the total number of introns present in cDNA
(a) 27 millions
(b) Zero
(c) Equal to ribonucleotides
(d) Half the number of ribonucleotides
Show Answer & Explanation

Answer: (b) Zero

Explanation:
1. cDNA is made from mature mRNA, which has had its introns removed, so it contains no introns.

Teacher's Note:
1. Spliced mRNA has exons only.

Question 16: Splicing is necessary for preparing a mature transcript and its movement to cytoplasm. It requires
(a) scRNA and proteins
(b) snRNA and proteins
(c) scRNA and snRNA
(d) scRNA only
Show Answer & Explanation

Answer: (b) snRNA and proteins

Explanation:
1. Splicing needs snRNAs and proteins (forming the spliceosome).

Teacher's Note:
1. snRNPs.

Question 17: Majority of unusual bases are found in tRNA, TC loop is
(a) First loop from 5-end of tRNA
(b) AA - tRNA synthetase binding loop
(c) Ribosomal binding loop
(d) Nodoc site
Show Answer & Explanation

Answer: (c) Ribosomal binding loop

Explanation:
1. The TψC loop is the ribosome-binding loop of tRNA; the DHU loop binds the synthetase.

Teacher's Note:
1. tRNA has many unusual bases.

Question 18: How many amino acids will be coded by the mRNA sequence 5'-CCCUCAUAGUCAUAC-3' if an adenosine residue is inserted after the 12th nucleotide?
(a) Five amino acids
(b) Six amino acids
(c) Two amino acids
(d) Three amino acids
Show Answer & Explanation

Answer: (c) Two amino acids

Explanation:
1. Codons: CCC UCA UAG ... The third codon UAG is a stop codon so only 2 amino acids are formed, even after the insertion (after nucleotide 12).

Teacher's Note:
1. Stop codon ends the chain.

Question 19: Identification and binding of RNA polymerase to the promoter sequence is a function of
(a) Rho factor
(b) Sigma factor
(c) Beta factor
(d) Omega factor
Show Answer & Explanation

Answer: (b) Sigma factor

Explanation:
1. Sigma factor helps RNA polymerase recognise and bind the promoter in bacteria.

Teacher's Note:
1. Rho terminates transcription.

Question 20: The DNA strand showing replication using Okazaki fragments also shows
(a) Continuous growth in 5' to 3' direction
(b) Discontinuous growth on the 5' to 3' parental strand
(c) Discontinuous growth on the 3' to 5' parental strand
(d) Involvement of one primer only
Show Answer & Explanation

Answer: (b) Discontinuous growth on the 5' to 3' parental strand

Explanation:
1. The lagging strand is made discontinuously (Okazaki fragments) on the 5' to 3' parental template strand.

Teacher's Note:
1. Lagging template runs 5' to 3'.

Question 21: Prokaryotic transcription uses only one type of RNA polymerase and
(a) it occurs in the cytoplasm only
(b) it is often coupled with translation
(c) it does not require splicing but capping is essential
(a) All are correct
(b) Both (b) & (c) are incorrect
(c) Both (a) & (c) are correct
(d) Only (c) is incorrect
Show Answer & Explanation

Answer: (d) Only (c) is incorrect

Explanation:
1. Prokaryotic transcription takes place in the cytoplasm and is coupled to translation. There is no capping in prokaryotes, so statement (c) is incorrect.

Teacher's Note:
1. Capping and tailing happen in eukaryotes.

Question 22: Pribnow box is a consensus of ________ bases, forming a binding site for E. coli RNA polymerase at promotor.
(a) TATAAT
(b) AGGAGG
(c) CAAT
(d) GC
Show Answer & Explanation

Answer: (a) TATAAT

Explanation:
1. The Pribnow box is the TATAAT consensus about 10 bases upstream of the start site, the RNA polymerase binding site.

Teacher's Note:
1. Also called -10 sequence.

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Molecular Basis of Inheritance Objective Questions & Solutions for NEET Biology

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