NEET Biology Molecular Basis of Inheritance MCQs Set 06

Biology Objective Questions and Answers: Molecular Basis of Inheritance

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Multiple Choice Questions

Question 1: Which of the following is not the feature of human genome ?
(a) Less than 2 percent of the genome code for protein
(b) Chromosome 1 has fewest gene (231)
(c) Repetitive sequences make up very large portion of human genome
(d) The functions are unknown for over 50% of the discovered genes
Show Answer & Explanation

Answer: (b) Chromosome 1 has fewest gene (231)

Explanation:
1. Chromosome 1 has the most genes (2968), not the fewest, so statement 2 is not a feature of the human genome.

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 2: The sequence of which chromosome number was completed in May 2006 ?
(a) Chromosome number 1
(b) Chromosome number 2
(c) Chromosome number 5
(d) Chromosome number 10
Show Answer & Explanation

Answer: (a) Chromosome number 1

Explanation:
1. The sequence of chromosome 1 was the last to be completed, in May 2006.

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 3: Which of the following pyrimidine base is common in both DNA and RNA
(a) Adenine
(b) Guanine
(c) Cytosine
(d) Thymine
Show Answer & Explanation

Answer: (c) Cytosine

Explanation:
1. Cytosine is the pyrimidine present in both DNA and RNA; thymine is only in DNA and adenine and guanine are purines.

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: By which of the following bond phosphoric acid remain linked with 5' carbon of sugar in one nucleotide
(a) Phosphoester bond
(b) Phosphodiester bond
(c) N-Glycosidic bodn
(d) Hydrogen bond
Show Answer & Explanation

Answer: (a) Phosphoester bond

Explanation:
1. The phosphate is linked to the 5' carbon of the sugar by a phosphoester bond.

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 5: The backbone in a polynucleotide chain is formed due to
(a) Sugars and nitrogenous bases
(b) Phosphates and nitrogenous base
(c) Nitrogenous bases and histones
(d) Sugar and phosphates
Show Answer & Explanation

Answer: (d) Sugar and phosphates

Explanation:
1. The backbone of a polynucleotide chain is formed by sugars and phosphates.

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 6: In RNA, every nucleotide residue has an additional – OH group at which of the following position
(a) 2' position of deoxyribose
(b) 1' possition of ribose sugar
(c) 3' position of ribose sugar
(d) 2' position of ribose sugar
Show Answer & Explanation

Answer: (d) 2' position of ribose sugar

Explanation:
1. RNA has an extra -OH group at the 2' position of the ribose sugar.

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 7: In addition to hydrogen bonding which of the following feature confers stability to helical structure
(a) Phosphodiester bond
(b) Pairing between one purine and one pyrimidine
(c) Glycosidic linkage bet ween sug ar and nitrogenous base
(d) The plane of one base pair stacks over the other
Show Answer & Explanation

Answer: (d) The plane of one base pair stacks over the other

Explanation:
1. Stacking of one base pair over the next adds stability to the helix in addition to hydrogen bonds.

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 8: Which of the following is responsible for constant distance between two polynuclestide chains in DNA
(a) Antiparallel polarity of two polynucleotide strands
(b) Hydrogen bonding
(c) Pairing between one purine and one pyrimidine
(d) All the above
Show Answer & Explanation

Answer: (c) Pairing between one purine and one pyrimidine

Explanation:
1. Pairing of a purine with a pyrimidine keeps a constant distance between the two strands.

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 9: Phosphoric acid remain associated with which of the following carbon of sugar in a nucleotide :-
(a) Ist
(b) 3rd
(c) 4th
(d) 5th
Show Answer & Explanation

Answer: (d) 5th

Explanation:
1. The phosphate group is attached to the 5' carbon of the sugar in a nucleotide.

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 10: Positive charge and basic nature of histone is due to abundance of
(a) Lysines and tryptophanes
(b) Arginines & threonines
(c) Lysines and arginines
(d) Tryptophanes and threonines
Show Answer & Explanation

Answer: (c) Lysines and arginines

Explanation:
1. Histones are rich in the basic amino acids lysine and arginine, which give them positive charge.

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: Negative charge of DNA is due to which of the following constituent
(a) Sugar
(b) Nitrogenous base
(c) Phosphoric acid
(d) Hydroxyl group (–OH) present on sugar
Show Answer & Explanation

Answer: (c) Phosphoric acid

Explanation:
1. The phosphate groups give DNA its negative charge.

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: Which of the following is actual sequence of packaging of DNA in eukaryotic cells
(a) DNA ® Chromatin ® Nucleosome ® Chromosome
(b) DNA ® Nucleosome ® Chromosome ® Chromatin
(c) DNA ® Nucleosome ® Chromatin ® Chromosome
(d) DNA ® Chromosome ® Chromatin ® Nucleosome
Show Answer & Explanation

Answer: (c) DNA ® Nucleosome ® Chromatin ® Chromosome

Explanation:
1. The order of packaging is DNA, nucleosome, chromatin and then chromosome.

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 13: The packaging of chromatin at higher level requires additional set of proteins that is known as
(a) Histone proteins
(b) NHC proteins
(c) Homeotic proteins
(d) Domain proteins
Show Answer & Explanation

Answer: (b) NHC proteins

Explanation:
1. Higher order chromatin packaging needs non-histone chromosomal (NHC) proteins.

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 14: Radioactive (35S) was detected in
(a) Supernatant
(b) Sediment
(c) Both
(d) Either 1 or 2
Show Answer & Explanation

Answer: (a) Supernatant

Explanation:
1. Radioactive 35S labels the protein coat, which stays outside the bacteria and is found in the supernatant.

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: Which of the folloiwng is not a criteria for determination of genetic material
(a) Ability of replication
(b) Chemically and structurally stable
(c) It should be non mutable
(d) Ability to expres itself in from of Mendelian characters
Show Answer & Explanation

Answer: (c) It should be non mutable

Explanation:
1. A genetic material must be able to mutate (slowly) for evolution, so 'non-mutable' is not a criterion.

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: Which fo the folloiwng reason is suitable to expalin that RNA is best for expression of characters
(a) It shows catalytic properties
(b) Presence of 2'–OH group on ribose sugar
(c) It can directly code for the synthesis of protein
(d) Presence of uracil
Show Answer & Explanation

Answer: (c) It can directly code for the synthesis of protein

Explanation:
1. RNA is considered best for expression because it can directly code for proteins in addition to being catalytic.

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 17: Which of the following evidence suggests that essential life processes evolved around RNA
(a) RNA used to act as genetic material
(b) RNA can act as catalyst
(c) RNA is highly reactive
(d) Both 1 and 2
Show Answer & Explanation

Answer: (d) Both 1 and 2

Explanation:
1. RNA acted as genetic material and as a catalyst, suggesting life processes evolved around 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 18: If E.Coli is allow to grow for 80 minutes in 15NH4Cl medium then what would be the proportion of hybrid and heavy density DNA molecules
(a) 1 : 7
(b) 7 : 1
(c) 14 : 2
(d) 1 : 4
Show Answer & Explanation

Answer: (a) 1 : 7

Explanation:
1. After 80 minutes (4 generations) there are 16 DNA molecules, of which 2 are hybrid and 14 are fully heavy, a ratio of 1 : 7.

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 19: Semiconservative replication of DNA in chromosomes was proved by
(a) Meselson & Stahl by using 15NH4Cl
(b) Taylor by using 15NH4Cl
(c) Meselson & Stahl by using tritiated thymidine
(d) Taylor by using tritiated thymidine
Show Answer & Explanation

Answer: (d) Taylor by using tritiated thymidine

Explanation:
1. Taylor used radioactive thymidine in Vicia faba to show semi-conservative replication in chromosomes.

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 20: Regarding to direction of DNA replication select out the correct one
(a) 5' ® 3' Template – continuous synthesis
(b) 3' ® 5' Template – discontinuous synthesis
(c) 3' ® 5' Template – continuous synthesis
(d) 5' ® 3' Template – synthesis leadingstrand
Show Answer & Explanation

Answer: (c) 3' ® 5' Template – continuous synthesis

Explanation:
1. The template strand is read 3' to 5' and the new leading strand is made continuously in the 5' to 3' direction.

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: During transcription only one of the strand of DNA get transcribed. Which of the following reason explain it
(a) Otherwise one segment of DNA would be coding for two different proteins
(b) Otherwise dsRNA comes in existance
(c) Otherwise antisense RNA arise which donot participate in Translation
(d) All the above
Show Answer & Explanation

Answer: (d) All the above

Explanation:
1. Transcribing both strands would give two different RNAs, possibly dsRNA that blocks translation, so only one strand is transcribed.

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 22: In transcription unit promoter and terminator are determined on the basis of
(a) Coding strand
(b) Template strand
(c) Noncoding strand
(d) Antisense strand
Show Answer & Explanation

Answer: (a) Coding strand

Explanation:
1. Promoter and terminator are defined with respect to the coding strand, as they are located at the 5' and 3' ends of the unit.

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.

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