NEET Biology Molecular Basis of Inheritance MCQs Set 09

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Multiple Choice Questions

Question 1: How much time is required for the replication of 4.6 x 106 bp in E. coli?
(a) 83 minutes
(b) 20 minutes
(c) 2 minutes
(d) 3 hours
Show Answer & Explanation

Answer: (b) 20 minutes

Explanation:
1. E. coli replicates at about 2000 bp per second.
2. With bidirectional replication from one origin, the genome of 4.6 x 106 bp is copied in roughly 18-20 minutes.

Teacher's Note:
1. Cell cycle of E. coli is about 20 minutes.

Question 2: What is the rate of polymerisation in E. coli?
(a) 20,000 bp per second
(b) 2000 nucleotides per second
(c) 2000 bp per minute
(d) 2000 bp per second
Show Answer & Explanation

Answer: (d) 2000 bp per second

Explanation:
1. The polymerisation rate in E. coli is about 2000 bp per second.
2. The unit is bp (base pairs), not individual nucleotides.

Teacher's Note:
1. Eukaryotic rates are much slower.

Question 3: During replication a large amount of energy is expended. The source of this energy is:
(a) Deoxyribonucleotide triphosphate
(b) Deoxyribonucleoside monophosphate
(c) Deoxyribonucleoside triphosphate
(d) Both deoxyribonucleotide triphosphate and deoxyribonucleoside monophosphate
Show Answer & Explanation

Answer: (c) Deoxyribonucleoside triphosphate

Explanation:
1. Deoxyribonucleoside triphosphates (dATP, dGTP, dCTP, dTTP) serve as substrates.
2. They also provide the energy for polymerisation.

Teacher's Note:
1. dNTPs are both building blocks and energy source.

Question 4: During transcription only one of the DNA strands is transcribed. Which reason explains this?
(a) Otherwise one segment of DNA would code for two different proteins
(b) Otherwise double-stranded RNA would be formed
(c) Otherwise antisense RNA would arise that would not participate in translation
(d) All of the above
Show Answer & Explanation

Answer: (d) All of the above

Explanation:
1. Both strands would give complementary RNAs.
2. These would code for two different proteins, form double-stranded RNA and produce non-functional RNA, so all reasons are valid.

Teacher's Note:
1. Only the template strand is transcribed.

Question 5: A transcription unit in DNA is defined primarily by three regions. These regions are:
(a) Promoter, regulator and structural gene
(b) Promoter, structural gene and terminator
(c) Promoter, regulator and terminator
(d) Promoter, regulator and operator gene
Show Answer & Explanation

Answer: (b) Promoter, structural gene and terminator

Explanation:
1. A transcription unit has a promoter, the structural gene and a terminator.
2. Regulator and operator genes belong to the operon model.

Teacher's Note:
1. Promoter is 5' to the gene; terminator is 3'.

Question 6: In the transcription unit, the promoter and terminator are defined on the basis of the:
(a) Coding strand
(b) Template strand
(c) Non-coding strand
(d) Antisense strand
Show Answer & Explanation

Answer: (a) Coding strand

Explanation:
1. By convention the promoter and terminator are located with respect to the coding (sense) strand.
2. The promoter is at the 5' end and the terminator at the 3' end of the coding strand.

Teacher's Note:
1. Coding strand has the same sequence as RNA (T in place of U).

Question 7: Which is an exclusive property of RNA polymerase during transcription?
(a) Initiation
(b) Elongation
(c) Termination
(d) Processing
Show Answer & Explanation

Answer: (b) Elongation

Explanation:
1. Initiation needs the sigma factor and termination needs the rho factor, which are accessory proteins.
2. Elongation is the step carried out by the RNA polymerase core enzyme itself.

Teacher's Note:
1. Sigma starts, rho stops.

Question 8: What is the length and constituent base of the tail in a functional mRNA?
(a) Poly U - 200-300 bp
(b) Poly A - 200-300 bp
(c) Poly C - 200-300 nucleotides
(d) Poly A - 200-300 nucleotides
Show Answer & Explanation

Answer: (d) Poly A - 200-300 nucleotides

Explanation:
1. Polyadenylation adds about 200-300 adenylate residues at the 3' end.
2. The tail is single-stranded, so its length is in nucleotides, not base pairs.

Teacher's Note:
1. Cap: 5' end; tail: 3' end.

Question 9: DNA-dependent RNA polymerases mediate synthesis of RNA on DNA, called transcription. Which of the following statements is wrong?
(a) In bacteria, mRNA does not require any processing to become active
(b) In eukaryotes there is a clear division of labour among RNA polymerases
(c) Absence of introns in the RNA of eukaryotes is reminiscent of antiquity
(d) RNA polymerase III is responsible for synthesis of snRNA
Show Answer & Explanation

Answer: (c) Absence of introns in the RNA of eukaryotes is reminiscent of antiquity

Explanation:
1. Bacterial mRNA is functional immediately, and eukaryotes use RNA polymerases I, II and III for different RNAs.
2. The presence (not absence) of introns in eukaryotic genes is reminiscent of antiquity, so (c) is wrong.

Teacher's Note:
1. RNA pol III makes tRNA, 5S rRNA and snRNAs.

Question 10: Which of the following was not involved in deciphering the genetic code?
(a) George Gamow's permutation combination of 43 bases
(b) H. G. Khorana's synthesis of RNA molecules with defined combinations of bases
(c) Severo Ochoa's enzyme for polymerising DNA with defined sequences
(d) Marshall Nirenberg's cell-free system for protein synthesis
Show Answer & Explanation

Answer: (c) Severo Ochoa's enzyme for polymerising DNA with defined sequences

Explanation:
1. Gamow predicted a triplet code, Khorana made defined RNA sequences and Nirenberg developed a cell-free system.
2. Ochoa's enzyme (polynucleotide phosphorylase) polymerised RNA, not DNA, so the statement about DNA is incorrect.

Teacher's Note:
1. Ochoa: polynucleotide phosphorylase for RNA.

Question 11: Which mutation forms the genetic basis of the proof that a codon is a triplet and is read in a continuous manner?
(a) Chromosomal structural mutations
(b) Chromosomal numerical mutations
(c) Substitutional mutation
(d) Frame-shift insertion or deletion mutation
Show Answer & Explanation

Answer: (d) Frame-shift insertion or deletion mutation

Explanation:
1. Frame-shift mutations (insertions or deletions) changed the reading frame.
2. Their effects showed that the code is read as non-overlapping triplets in a continuous manner.

Teacher's Note:
1. Three insertions or deletions restore the frame.

Question 12: An adapter molecule that on one hand reads the code and on the other hand binds to specific amino acids is:
(a) mRNA
(b) rRNA
(c) tRNA
(d) hnRNA
Show Answer & Explanation

Answer: (c) tRNA

Explanation:
1. tRNA has an anticodon that reads the codon on mRNA.
2. Its amino-acid acceptor end binds a specific amino acid.

Teacher's Note:
1. Crick called it the adapter.

Question 13: Which of the following rRNAs shows structural as well as functional (catalytic) role in bacteria?
(a) 16S rRNA
(b) 23S rRNA
(c) 5S rRNA
(d) 28S rRNA
Show Answer & Explanation

Answer: (b) 23S rRNA

Explanation:
1. In bacteria 23S rRNA acts as a ribozyme for peptide bond formation.
2. It also forms part of the ribosome structure; 28S rRNA is the eukaryotic equivalent.

Teacher's Note:
1. 23S rRNA = peptidyl transferase in prokaryotes.

Question 14: An mRNA also has some additional sequences that are not translated, called UTRs. The function of UTRs is:
(a) Charging of tRNA
(b) Formation of peptide bonds
(c) Helping in efficient translation
(d) Helping in translocation
Show Answer & Explanation

Answer: (c) Helping in efficient translation

Explanation:
1. UTRs lie at the 5' and 3' ends of mRNA and are not translated.
2. They are required for efficient translation.

Teacher's Note:
1. UTR = untranslated region.

Question 15: The tRNA that moves away from the ribosome after translocation is known as:
(a) Acylated tRNA
(b) Peptidyl tRNA
(c) Deacylated tRNA
(d) Charged tRNA
Show Answer & Explanation

Answer: (c) Deacylated tRNA

Explanation:
1. After peptide transfer the tRNA at the P site has lost its amino acid.
2. It is deacylated (uncharged) and leaves the ribosome.

Teacher's Note:
1. Charged tRNA carries an amino acid.

Question 16: The Human Genome Project was closely associated with the rapid development of a new area in biology called:
(a) Biofortification
(b) Bioinformatics
(c) Biomining
(d) Biotransformation
Show Answer & Explanation

Answer: (b) Bioinformatics

Explanation:
1. The huge amount of sequence data required computational storage and analysis.
2. This led to the rapid growth of bioinformatics.

Teacher's Note:
1. Bioinformatics = data analysis of biological information.

Question 17: The Human Genome Project was coordinated by:
(a) US Department of Energy
(b) National Institutes of Health
(c) Sanger Centre
(d) Both US Department of Energy and National Institutes of Health
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Answer: (d) Both US Department of Energy and National Institutes of Health

Explanation:
1. HGP was launched in 1990 and was coordinated by the US Department of Energy and the National Institutes of Health.
2. The Sanger Centre was one of the partners from the UK.

Teacher's Note:
1. Wellcome Trust (UK) joined later.

Question 18: The approach of HGP that focused on identifying all the genes that are expressed as RNA is known as:
(a) Expressed sequence tags
(b) Sequence annotation
(c) Polymerase chain reaction
(d) Dermatoglyphics
Show Answer & Explanation

Answer: (a) Expressed sequence tags

Explanation:
1. Expressed sequence tags (ESTs) identify the genes expressed as RNA.
2. Sequence annotation is the other approach, which sequences the entire genome.

Teacher's Note:
1. Two methodologies: ESTs and sequence annotation.

Question 19: Automated DNA sequencers worked on the principle of a method developed by:
(a) Watson
(b) Chargaff
(c) Frederick Sanger
(d) Singer and Nicolson
Show Answer & Explanation

Answer: (c) Frederick Sanger

Explanation:
1. Automated sequencers use the chain-termination method of Frederick Sanger.
2. Singer and Nicolson proposed the fluid mosaic model.

Teacher's Note:
1. Sanger method: dideoxy chain termination.

Question 20: Regarding the salient features of the human genome, select the incorrect one:
(a) The human genome contains 3164.7 million nucleotide bases
(b) The human genome contains about 30,000 genes
(c) The Y chromosome has the largest number of genes
(d) 1.4 million locations are associated with SNPs
Show Answer & Explanation

Answer: (c) The Y chromosome has the largest number of genes

Explanation:
1. The genome has 3164.7 million bases, about 30,000 genes (later 20,000-25,000) and 1.4 million SNPs.
2. Chromosome 1 has the most genes (2968) and the Y chromosome the fewest (231).

Teacher's Note:
1. Y chromosome has the fewest genes.

Question 21: If an inheritable mutation is observed in a population at a high frequency, it is referred to as:
(a) DNA polyploidy
(b) DNA polymorphism
(c) DNA redundancy
(d) Sequence annotation
Show Answer & Explanation

Answer: (b) DNA polymorphism

Explanation:
1. A heritable mutation seen in more than 1 percent of the population is a polymorphism.
2. Polymorphism is the basis of DNA fingerprinting.

Teacher's Note:
1. SNPs are a type of DNA polymorphism.

Question 22: A change in a single base pair of DNA can be termed:
(a) Chromosomal aberration
(b) Point mutation
(c) Genomic mutation
(d) Frame-shift mutation
Show Answer & Explanation

Answer: (b) Point mutation

Explanation:
1. A change in one base pair is a point mutation.
2. Frame-shift mutations are insertions or deletions of bases and chromosomal aberrations involve large segments.

Teacher's Note:
1. Sickle cell anaemia is a point mutation.

Question 23: Which cell cycle event is responsible for aneuploidy-based chromosomal disorders?
(a) Failure of G1 phase
(b) Failure of DNA replication in S phase
(c) Failure of segregation (non-disjunction)
(d) Failure of movement of chromosomes
Show Answer & Explanation

Answer: (c) Failure of segregation (non-disjunction)

Explanation:
1. If sister chromatids or homologues do not separate (non-disjunction), gametes get an extra or missing chromosome.
2. This produces aneuploidy such as Down's syndrome.

Teacher's Note:
1. Aneuploidy = gain or loss of individual chromosomes.

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