NEET Biology Molecular Basis of Inheritance MCQs Set 07

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Chapter-wise Objective Questions: Molecular Basis of Inheritance

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Multiple Choice Questions

Question 1: In eukaryotes as well as prokaryotes those DNA sequences that appear in mature or processed RNA are known as
(a) Introts
(b) Exons
(c) Recons
(d) Mutons
Show Answer & Explanation

Answer: (b) Exons

Explanation:
1. Exons are the sequences that appear in mature processed RNA.

Teacher's Note:
1. Hershey and Chase (32P in DNA, 35S in protein) proved DNA is the genetic material, and Meselson and Stahl (15N, CsCl gradient) proved semi-conservative replication.
2. The transcription unit has promoter, structural gene and terminator; template strand is 3'->5', and eukaryotic hnRNA is processed by capping, tailing (poly A) and splicing.

Question 2: Regarding to role of RNA in protein synthesis find out the odd one
(a) m-RNA - provides the template
(b) t-RNA - brings aminoacids
(c) r-RNA - read genetic code
(d) sn-RNA splicing
Show Answer & Explanation

Answer: (c) r-RNA - read genetic code

Explanation:
1. The tRNA, not the rRNA, reads the genetic code; rRNA has structural and catalytic roles in the ribosome.

Teacher's Note:
1. The genetic code is triplet, degenerate, nearly universal and non-overlapping; tRNA is the adapter and mRNA has start AUG and stop codons UAA, UAG, UGA.
2. Lac operon is a negative regulation system in which lactose (allolactose) is the inducer; HGP found about 30,000 genes and over 3 billion bp.

Question 3: DNA dependent RNA polymerases mediate synthesis of RNA over DNA, called transcription. Which of the following statements about it is wrong?
(a) In bacteria mRNA does not require any processing to become active
(b) In eukaryotes there is a clear cut 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 snRNAs
Show Answer & Explanation

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

Explanation:
1. Presence, not absence, of introns in eukaryotic genes is thought to be a relic of antiquity (introns early hypothesis); the other statements are true.

Teacher's Note:
1. Hershey and Chase (32P in DNA, 35S in protein) proved DNA is the genetic material, and Meselson and Stahl (15N, CsCl gradient) proved semi-conservative replication.
2. The transcription unit has promoter, structural gene and terminator; template strand is 3'->5', and eukaryotic hnRNA is processed by capping, tailing (poly A) and splicing.

Question 4: Which of the following was not involved in deciphering of genetic code
(a) Physi cist geor ge Gamow 's permut ation combination of 43 bases
(b) H.G. Khorana's based synthesis of RNA molecules with defined combination of bases
(c) Severo ochoa enzyme for polymerising DNA with defined sequences
(d) Marshall Nirenberg's cell free system for protein synthesis
Show Answer & Explanation

Answer: (c) Severo ochoa enzyme for polymerising DNA with defined sequences

Explanation:
1. Severo Ochoa's enzyme (polynucleotide phosphorylase) polymerised RNA, not DNA, with defined sequences.

Teacher's Note:
1. The genetic code is triplet, degenerate, nearly universal and non-overlapping; tRNA is the adapter and mRNA has start AUG and stop codons UAA, UAG, UGA.
2. Lac operon is a negative regulation system in which lactose (allolactose) is the inducer; HGP found about 30,000 genes and over 3 billion bp.

Question 5: Which of the following mutation forms the genetic basis of proof that codon is a triplet and it 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 insertions and deletions proved that the genetic code is a triplet read continuously.

Teacher's Note:
1. DNA is a polymer of deoxyribonucleotides joined by 3'-5' phosphodiester bonds; nitrogenous base joins sugar by an N-glycosidic bond, and phosphate joins sugar by a phosphoester bond.
2. In the Watson-Crick model the two strands are antiparallel, base pairs A=T (two hydrogen bonds) and G≡C (three), pitch 3.4 nm with 10 bp per turn.

Question 6: A m-RNA also has some additional sequences that are not translated called UTR. The function of UTR is
(a) Charging of t-RNA
(b) Formation of peptide bond
(c) Helps in efficient translation
(d) Helps in translocation
Show Answer & Explanation

Answer: (c) Helps in efficient translation

Explanation:
1. Untranslated regions (UTRs) at the 5' and 3' ends of mRNA help in efficient translation.

Teacher's Note:
1. DNA is a polymer of deoxyribonucleotides joined by 3'-5' phosphodiester bonds; nitrogenous base joins sugar by an N-glycosidic bond, and phosphate joins sugar by a phosphoester bond.
2. In the Watson-Crick model the two strands are antiparallel, base pairs A=T (two hydrogen bonds) and G≡C (three), pitch 3.4 nm with 10 bp per turn.

Question 7: The t-RNA move away from ribosomes after translocation of ribosome in relation to m-RNA, is known as
(a) Acylated t-RNA
(b) Peptidyle t-RNA
(c) Deacylated t-RNA
(d) Charged t-RNA
Show Answer & Explanation

Answer: (c) Deacylated t-RNA

Explanation:
1. After translocation the tRNA that has lost its amino acid is deacylated tRNA and leaves the ribosome.

Teacher's Note:
1. Hershey and Chase (32P in DNA, 35S in protein) proved DNA is the genetic material, and Meselson and Stahl (15N, CsCl gradient) proved semi-conservative replication.
2. The transcription unit has promoter, structural gene and terminator; template strand is 3'->5', and eukaryotic hnRNA is processed by capping, tailing (poly A) and splicing.

Question 8: At which of the following levels does regulation of gene expression NOT occur in eukaryotes?
(a) Transcription level
(b) Processing level
(c) Transport of ribosomal subunits from the nucleus to the cytoplasm
(d) Translation level
Show Answer & Explanation

Answer: (c) Transport of ribosomal subunits from the nucleus to the cytoplasm

Explanation:
1. Eukaryotic gene expression is controlled at transcription, processing, transport of mRNA and translation, not at the level of ribosomal subunit transport.

Teacher's Note:
1. The genetic code is triplet, degenerate, nearly universal and non-overlapping; tRNA is the adapter and mRNA has start AUG and stop codons UAA, UAG, UGA.
2. Lac operon is a negative regulation system in which lactose (allolactose) is the inducer; HGP found about 30,000 genes and over 3 billion bp.

Question 9: In prokaryotes predominant site for control of gene expression is the
(a) Control of rate of processing of primary transcript
(b) Control of rate of transcription initiation
(c) Control of transport of m-RNA from nucleus to cytoplasm
(d) Control of Translation
Show Answer & Explanation

Answer: (b) Control of rate of transcription initiation

Explanation:
1. In prokaryotes, control of the rate of transcription initiation is the predominant site of regulation.

Teacher's Note:
1. DNA is a polymer of deoxyribonucleotides joined by 3'-5' phosphodiester bonds; nitrogenous base joins sugar by an N-glycosidic bond, and phosphate joins sugar by a phosphoester bond.
2. In the Watson-Crick model the two strands are antiparallel, base pairs A=T (two hydrogen bonds) and G≡C (three), pitch 3.4 nm with 10 bp per turn.

Question 10: HGP was closely associated with the rapid development of a new area in biology called as
(a) Biofortification
(b) Bioinformatics
(c) Biomining
(d) Biotransformation
Show Answer & Explanation

Answer: (b) Bioinformatics

Explanation:
1. Bioinformatics developed rapidly with HGP because of the large amount of data.

Teacher's Note:
1. Hershey and Chase (32P in DNA, 35S in protein) proved DNA is the genetic material, and Meselson and Stahl (15N, CsCl gradient) proved semi-conservative replication.
2. The transcription unit has promoter, structural gene and terminator; template strand is 3'->5', and eukaryotic hnRNA is processed by capping, tailing (poly A) and splicing.

Question 11: Which of the following is not a goal of HGP
(a) Identify all the approximately 20,000–25,000 genes
(b) Store this information in database
(c) Restrict the related technologies so that other sector donot benefited with it
(d) Address the ethical, legal and social issues
Show Answer & Explanation

Answer: (c) Restrict the related technologies so that other sector donot benefited with it

Explanation:
1. HGP aimed to share data and technology widely, not to restrict it.

Teacher's Note:
1. The genetic code is triplet, degenerate, nearly universal and non-overlapping; tRNA is the adapter and mRNA has start AUG and stop codons UAA, UAG, UGA.
2. Lac operon is a negative regulation system in which lactose (allolactose) is the inducer; HGP found about 30,000 genes and over 3 billion bp.

Question 12: The human genome project was coordinated by
(a) U.S. department of energy
(b) National institute of health
(c) Sanger centre
(d) Both 1 and 2
Show Answer & Explanation

Answer: (d) Both 1 and 2

Explanation:
1. HGP was coordinated by the U.S. Department of Energy and the National Institutes of Health.

Teacher's Note:
1. DNA is a polymer of deoxyribonucleotides joined by 3'-5' phosphodiester bonds; nitrogenous base joins sugar by an N-glycosidic bond, and phosphate joins sugar by a phosphoester bond.
2. In the Watson-Crick model the two strands are antiparallel, base pairs A=T (two hydrogen bonds) and G≡C (three), pitch 3.4 nm with 10 bp per turn.

Question 13: Which of the following organism was not used as model organism is human genome project
(a) Arabidopsis
(b) Caenorhabdities elegans
(c) Rice
(d) Hyacinthus orientalis
Show Answer & Explanation

Answer: (d) Hyacinthus orientalis

Explanation:
1. Hyacinthus orientalis was not a model organism in HGP; bacteria, yeast, C. elegans, Drosophila, Arabidopsis and rice were.

Teacher's Note:
1. Hershey and Chase (32P in DNA, 35S in protein) proved DNA is the genetic material, and Meselson and Stahl (15N, CsCl gradient) proved semi-conservative replication.
2. The transcription unit has promoter, structural gene and terminator; template strand is 3'->5', and eukaryotic hnRNA is processed by capping, tailing (poly A) and splicing.

Question 14: Approach of HGP focused on identifying all the genes that 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) focus on identifying genes expressed as RNA.

Teacher's Note:
1. The genetic code is triplet, degenerate, nearly universal and non-overlapping; tRNA is the adapter and mRNA has start AUG and stop codons UAA, UAG, UGA.
2. Lac operon is a negative regulation system in which lactose (allolactose) is the inducer; HGP found about 30,000 genes and over 3 billion bp.

Question 15: Automated DNA sequencers are based on the method developed by:
(a) Watson
(b) Chargaff
(c) Frederick Sanger
(d) Singer and Nicolson
Show Answer & Explanation

Answer: (c) Frederick Sanger

Explanation:
1. Automated DNA sequencers work on the principle of the method developed by Frederick Sanger.

Teacher's Note:
1. DNA is a polymer of deoxyribonucleotides joined by 3'-5' phosphodiester bonds; nitrogenous base joins sugar by an N-glycosidic bond, and phosphate joins sugar by a phosphoester bond.
2. In the Watson-Crick model the two strands are antiparallel, base pairs A=T (two hydrogen bonds) and G≡C (three), pitch 3.4 nm with 10 bp per turn.

Question 16: Regarding to salient features of human genome select out the incorrect one
(a) Human genome contains 3164.7 million nucleotide bases
(b) Human genome contain 30,000 genes
(c) y-chromosome has largest number of genes
(d) 1.4 million locations are associated with SNPs
Show Answer & Explanation

Answer: (c) y-chromosome has largest number of genes

Explanation:
1. The Y chromosome has the fewest genes (231), whereas chromosome 1 has the most.

Teacher's Note:
1. Hershey and Chase (32P in DNA, 35S in protein) proved DNA is the genetic material, and Meselson and Stahl (15N, CsCl gradient) proved semi-conservative replication.
2. The transcription unit has promoter, structural gene and terminator; template strand is 3'->5', and eukaryotic hnRNA is processed by capping, tailing (poly A) and splicing.

Question 17: Match the items:
A. SNPs B. Genes of chromosome 1 C. Total number of human genes D. Total nucleotides of the human genome
(i) 3164.7 million (ii) 1.4 million (iii) 30000 (iv) 2968
(a) A-(ii), B-(iii), C-(iv), D-(i)
(b) A-(ii), B-(iv), C-(iii), D-(i)
(c) A-(ii), B-(iv), C-(i), D-(iii)
(d) A-(iv), B-(ii), C-(iii), D-(i)
Show Answer & Explanation

Answer: (b) A-(ii), B-(iv), C-(iii), D-(i)

Explanation:
1. About 1.4 million SNPs, 2968 genes on chromosome 1, about 30000 genes in total and 3164.7 million nucleotide bases.

Teacher's Note:
1. The genetic code is triplet, degenerate, nearly universal and non-overlapping; tRNA is the adapter and mRNA has start AUG and stop codons UAA, UAG, UGA.
2. Lac operon is a negative regulation system in which lactose (allolactose) is the inducer; HGP found about 30,000 genes and over 3 billion bp.

Question 18: Due to high degree of polymorphism, size of VNTR varies in size from
(a) 0.1 – 2 kb
(b) 0.1 – 2000 kb
(c) 0.1 – 20 kb
(d) 0.1 – 200 kb
Show Answer & Explanation

Answer: (c) 0.1 – 20 kb

Explanation:
1. VNTR size varies from about 0.1 to 20 kb because of high polymorphism.

Teacher's Note:
1. Hershey and Chase (32P in DNA, 35S in protein) proved DNA is the genetic material, and Meselson and Stahl (15N, CsCl gradient) proved semi-conservative replication.
2. The transcription unit has promoter, structural gene and terminator; template strand is 3'->5', and eukaryotic hnRNA is processed by capping, tailing (poly A) and splicing.

Question 19: What is the basis of heredity
(a) Variations
(b) Inheritance
(c) Genetics
(d) Recombination
Show Answer & Explanation

Answer: (b) Inheritance

Explanation:
1. Inheritance, the passing of characters from parents to progeny, is the basis of heredity.

Teacher's Note:
1. The genetic code is triplet, degenerate, nearly universal and non-overlapping; tRNA is the adapter and mRNA has start AUG and stop codons UAA, UAG, UGA.
2. Lac operon is a negative regulation system in which lactose (allolactose) is the inducer; HGP found about 30,000 genes and over 3 billion bp.

Question 20: Change in a single base pair of DNA can be termed as
(a) Chromosomal aberrations
(b) Point mutation
(c) Genomatric mutation
(d) Frame shift mutation
Show Answer & Explanation

Answer: (b) Point mutation

Explanation:
1. A change in a single base pair of DNA is a point mutation.

Teacher's Note:
1. DNA is a polymer of deoxyribonucleotides joined by 3'-5' phosphodiester bonds; nitrogenous base joins sugar by an N-glycosidic bond, and phosphate joins sugar by a phosphoester bond.
2. In the Watson-Crick model the two strands are antiparallel, base pairs A=T (two hydrogen bonds) and G≡C (three), pitch 3.4 nm with 10 bp per turn.

Question 21: Which of the following cell cycle event is responsible for aneuploidy based chromosomal disorders
(a) Failure of G1 phase
(b) Failure ofDNA replication in s-phase
(c) Failure of segregation/Disjunction
(d) Failure of movement of chromosomes
Show Answer & Explanation

Answer: (c) Failure of segregation/Disjunction

Explanation:
1. Failure of segregation (non-disjunction) of chromatids or homologues during cell division causes aneuploidy.

Teacher's Note:
1. Hershey and Chase (32P in DNA, 35S in protein) proved DNA is the genetic material, and Meselson and Stahl (15N, CsCl gradient) proved semi-conservative replication.
2. The transcription unit has promoter, structural gene and terminator; template strand is 3'->5', and eukaryotic hnRNA is processed by capping, tailing (poly A) and splicing.

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