NCERT Exemplar: Molecular Basis of Inheritance - 1 | Biology Class 12 - NEET PDF Download

MULTIPLE CHOICE QUESTIONS 

Q.1. In a DNA strand the nucleotides are linked together by:
(a) Glycosidic bonds
(b) Phosphodiester bonds
(c) Peptide bonds
(d) Hydrogen bonds
Ans. (b)

Q.2. A nucleoside differs from a nucleotide. It lacks the:
(a) Base
(b) Sugar
(c) Phosphate group
(d) Hydroxyl group
Ans. (c)
Solution.
Nucleoside = Base + sugar
Nucleotide = Base + sugar + phosphate group

Q.3. Both deoxyribose and ribose belong to a class of sugars called:
(a) Trioses
(b) Hexoses
(c) Pentoses
(d) Polysaccharides
Ans. (c)
Solution.
Both deoxyribose and ribose belong to a pentoses class of sugars.

Q.4. The fact that a purine base always pairs through hydrogen bonds with a pyrimidine base in the DNA double helix leads to:
(a) The antiparallel nature
(b) The semiconservative nature
(c) Uniform width throughout DNA
(d) Uniform length in all DNA
Ans. (c)
Solution.
A purine base always paired through hydrogen bonds with a pyrimidine base leads to, in the DNA double helix, uniform width throughout DNA. 

Q.5. The net electric charge on DNA and histones is:
(a) Both positive
(b) Both negative
(c) Negative and positive, respectively
(d) Zero
Ans. (c)

Q.6. The promoter site and the terminator site for transcription are located at:
(a) 3' (downstream) end and 5' (upstream) end, respectively of the transcription unit
(b) 5' (upstream) end and 3' (downstream) end, respectively of the transcription unit
(c) The 5' (upstream) end
(d) The 3' (downstream) end
Ans. (b)

Q.7. Which of the following statements is the most appropriate for sickle cell anaemia?
(a) It cannot be treated with iron supplements
(b) It is a molecular disease
(c) It confers resistance to acquiring malaria
(d) All of the above
Ans. (d)
Solution.
Sick/e cell anaemia: It is a molecular disease, autosomal recessive disorder and an example of pleiotropy. It cannot be treated with iron supplements and it confers resistance to acquiring malaria.

Q.8. Which of the following is true with respect to AUG?
(a) It codes for methionine only
(b) It is an initiation codon
(c) It codes for methionine in both prokaryotes and eukaryotes
(d) All of the above
Ans. (d)
Solution.
AUG—It codes for methionine only in both prokaryotes and eukaryotes, it is also an initiation codon.

Q.9. The first genetic material could be:
(a) Protein
(b) Carbohydrates
(c) DNA
(d) RNA
Ans. (d)
Solution.
The first genetic material could be RNA but ideal genetic material is DNA.

Q.10. With regard to mature mRNA in eukaryotes:
(a) Exons and introns do not appear in the mature RNA
(b) Exons appear but introns do not appear in the mature RNA
(c) Introns appear but exons do not appear in the mature RNA
(d) Both exons and introns appear in the mature RNA
Ans. (b)
Solution.
In eukaryotes exons appear but introns do not appear in the mature RNA.

Q.11. The human chromosome with the highest and least number of genes in them are respectively:
(a) Chromosome 21 and Y
(b) Chromosome 1 and X
(c) Chromosome 1 and Y
(d) Chromosome X and Y
Ans. (c)
Solution.
• Chromosome one have 2968 genes (highest)
• Chromosome Y have 231 genes (lowest)

Q.12. Who amongst the following scientists had no contribution in the development of the double helix model for the structure of DNA?
(a) Rosalind Franklin
(b) Maurice Wilkins
(c) Erwin Chargaff
(d) Meselson and Stahl
Ans. (d)
Solution.
Meselson and Stahl give the experimental proof of semiconservative DNA replication. He had no contribution to the development of the double helix model for the. structure of DNA.

Q.13. DNA is a polymer of nucleotides which are linked to each other by 3’-5’ phosphodiester bond. To prevent polymerisation of nucleotides, which of the following modifications would you choose?
(a) Replace purine with pyrimidines
(b) Remove/Replace 3' OH group in deoxy ribose
(c) Remove/Replace 2' OH group with some other group in deoxy ribose
(d) Both ‘b’ and ‘c’
Ans. (b)
Solution.
DNA is a polymer of nucleotides which are linked to each other by 3′-5′ phosphodiester bond. Remove/replace 3’OH group in deoxyribose, to prevent polymerisation of nucleotides.

Q.14. Discontinuous synthesis of DNA occurs in one strand, because:
(a) DNA molecule being synthesised is very long
(b) DNA dependent DNA polymerase catalyses polymerisation only in one direction (5’ → 3’)
(c) It is a more efficient process
(d) DNA ligase joins the short stretches of DNA
Ans. (b)

Q.15. Which of the following steps in transcription is catalysed by RNA polymerase?
(a) Initiation
(b) Elongation
(c) Termination
(d) All of the above
Ans. (b)

Q.16. Control of gene expression in prokaryotes take place at the level of:
(a) DNA-replication
(b) Transcription
(c) Translation
(d) None of the above
Ans. (b)

Q.17. Which of the following statements is correct about the role of regulatory proteins in transcription in prokaryotes?
(a) They only increase expression
(b) They only decrease expression
(c) They interact with RNA polymerase but do not affect the expression
(d) They can act both as activators and as repressors
Ans. (d)
Solution.
Regulatory proteins are the accessory proteins that interact with RNA polymerase and affect its role in transcription. They can act both as activators and as repressors.

Q.18. Which was the last human chromosome to be completely sequenced:
(a) Chromosome 1
(b) Chromosome 11
(c) Chromosome 21
(d) Chromosome X
Ans. (a)
Solution.
The Human genome project was completed in 2003. The sequence of chromosome 1 was completed only in May 2006.

Q.19. Which of the following are the functions of RNA?
(a) It is a carrier of genetic information from DNA to ribosomes synthesising polypeptides.
(b) It carries amino acids to ribosomes.
(c) It is a constituent component of ribosomes.
(d) All of the above.
Ans. (d)
Solution.
Functions of mRNA:
• Carrier of genetic information from DNA to ribosomes synthesising polypeptides.
• Carries amino acids to ribosomes.
• Constituent component of ribosomes.

Q.20. While analysing the DNA of an organism a total number of 5386 nucleotides were found out of which the proportion of different bases were:
Adenine = 29%, Guanine = 17%, Cytosine = 32%, Thymine = 17%. Considering the Chargaff’s rule it can be concluded that:
(a) It is a double stranded circular DNA
(b) It is single stranded DNA
(c) It is a double stranded linear DNA
(d) No conclusion can be drawn
Ans. (b)
Solution.
In the DNA of an organism a total number of 5386 nucleotides were found out of which the proportion of different bases were: Adenine = 29%, Guanine = 17%, Cytosine = 32%, Thymine = 17%. Considering the Chargaff’s rule it can be concluded that it is a single stranded DNA.

Q.21. In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called:
(a) A-DNA
(b) B-DNA
(c) CDNA
(d) RDNA
Ans. (c)
Solution.
In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called c-DNA (Complementary DNA).

Q.22. If Meselson and Stahl's experiment is continued for four generations in bacteria, the ratio of N¹⁵/N¹⁵: N¹⁵/N¹⁴: N¹⁴/N¹⁴ containing DNA in the fourth generation would be:
(a) 1:1:0
(b) 1:4:0
(c) 0:1:3
(d) 0:1:7
Ans. (d)
Solution.
• If Meselson and Stahl’s experiment is continued for four generations in bacteria, the ratio of 15_N/15_N : 15_N/14_N : 14_N/14_N containing DNA in the fourth generation would be 0 : 1 : 7.
• After third generation (60 min.) bacteria contains 25% hybrid (15_N14_N) in 1 :3 ratio.
• After 4th generation (80 min.) bacteria contains 12.5% hybrid and 87.5% light DNA in 1 : 7 ratio.

Q.23. If the sequence of nitrogen bases of the coding strand of DNA in a transcription unit is:
5' - A T G A A T G - 3',
the sequence of bases in its RNA transcript would be;
(a) 5' - A U G A A U G - 3'
(b) 5' - U A C U U A C - 3'
(c) 5' - C A U U C A U - 3'
(d) 5' - G U A A G U A - 3'
Ans. (a)
Solution.
Coding strand: 5′-AT G A AT G – 3′
Template strand: 3′ – TACTTAC-5″
RNA transcript: 5′-AUGAAUG-3′

Q.24. The RNA polymerase holoenzyme transcribes:
(a) The promoter, structural gene and the terminator region
(b) The promoter and the terminator region
(c) The structural gene and the terminator region
(d) The structural gene only
Ans. (d)

Q.25. If the base sequence of a codon in mRNA is 5'-AUG-3', the sequence of tRNA pairing with it must be:
(a) 5' - UAC - 3'
(b) 5' - CAU - 3'
(c) 5' - AUG - 3'
(d) 5' - GUA - 3'
Ans. (b)
Solution.
Base sequence of a codon in mRNA is 5′-AUG-3′, the sequence of tRNA pairing with it must be complementary, i.e., 3′ – UAC – 5′

Q.26. The amino acid attaches to the tRNA at its:
(a) 5' - end
(b) 3' - end
(c) Anti codon site
(d) DHU loop
Ans. (b)

Q.27. To initiate translation, the mRNA first binds to:
(a) The smaller ribosomal sub-unit,
(b) The larger ribosomal sub-unit
(c) The whole ribosome
(d) No such specificity exists
Ans. (a)

Q.28. In E.coli, the lac operon gets switched on when:
(a) Lactose is present and it binds to the repressor
(b) Repressor binds to operator
(c) RNA polymerase binds to the operator
(d) Lactose is present and it binds to RNA polymerase
Ans. (a)

VERY SHORT ANSWER TYPE QUESTIONS

Q.1. What is the function of histones in DNA packaging?
Ans. Histone are the basis protein that are rich in arginine and lysine. Histones form octamer on which DNA is wrapped and form nucleosome that means it helps in packaging of DNA in eukaryotes.

Q.2. Distinguish between heterochromatin and euchromatin. Which of the two is transcriptionally active?
Ans. 

Q.3. The enzyme DNA polymerase in E.coli is a DNA dependent polymerase and also has the ability to proof-read the DNA strand being synthesised. Explain. Discuss the dual polymerase.
Ans. DNA polymerase helps in the replication of DNA molecules. DNA polymerase also helps in the proofreading, if any error/mutation occurred during replication.

Q.4. What is the cause of discontinuous synthesis of DNA on one of the parental strands of DNA? What happens to these short stretches of synthesised DNA?
Ans. The DNA-dependent DNA polymerases catalyse polymerisation only in one direction, that is 5′ → 3′. This creates some additional complications at the replicating fork. Consequently, on leading strand (the template with polarity 3′ → 5′), the replication is continuous, while on the lagging strain (the template with polarity 5′ → 3′), it is discontinuous. The discontinuously synthesised fragments are later joined by the enzyme DNA ligase.

Q.5. Given below is the sequence of coding strand of DNA in a transcription unit
3 'A A T G C A G C T A T T A G G – 5’
write the sequence of
a) its complementary strand
b) the mRNA
Ans. (a)Complementary strand:
5′ – TTACGTCGATAA T C C – 3′
(b) The mRNA
5′ – AAU GCAGCUAUUAGG-3′

Q.6. What is DNA polymorphism? Why is it important to study it?
Ans. Polymorphism (variation at genetic level) arises due to mutations. Allelic sequence variation has traditionally been described as a DNA polymorphism if more than one variant (allele) at a locus occurs in human population with a frequency greater than 0.01. In simple terms, if an inheritable mutation is observed in a population at high frequency, it is referred to as DNA polymorphism.
Polymorphism become very useful identification tool in forensic applications. Further, as the polymorphisms are inheritable from parents to children, DNA fingerprinting is the basis of paternity testing, in case of disputes.

Q.7. Based on your understanding of genetic code, explain the formation of any abnormal hemoglobin molecule. What are the known consequences of such a change?
Ans. The defect is caused by the substitution (trans version) of Glutamic acid (Glu) by Valine (Val) at the sixth position of the betaglobin chain of the haemoglobin molecule. The substitution of amino acid in the glob in protein results due to the single base substitution at the sixth codon of the betaglobin gene from GAG to GUG. The mutant haemoglobin molecule undergoes polymerisation under low oxygen tension causing the change in the shape of the RBC from biconcave disc to elongated sickle like structure.
Sickle-shaped red blood cells that obstruct capillaries and restrict blood flow to an organ resulting in ischaemia, pain, necrosis, and often organ damage.

Q.8. Sometimes cattle or even human beings give birth to their young ones that are having extremely different sets of organs like limbs/position of eye(s) etc. Comment.
Ans. There is a disturbance in co-ordinated regulation of expression of sets of genes associated with organ development.

Q.9. In a nucleus, the number of ribonucleoside triphosphates is 10 times the number of deoxy x10 ribonucleoside triphosphates, but only deoxy ribonucleotides are added during the DNA replication. Suggest a mechanism.
Ans. DNA polymerase is highly specific to recognise only deoxy ribonucleoside triphosphates. Therefore it cannot hold RNA nucleotides.

Q.10. Name a few enzymes involved in DNA replication other than DNA polymerase and ligase. Name the key functions for each of them.
Ans. (i) Helicase—opens the helix
(ii) Topoisomerases—removes the super coiling of DNA
(iii) Primase—synthesises RNA primer
(iv) Telomerase—to synthesises the DNA of telomeric end of chromosomes

Q.11. Name any three viruses which have RNA as the genetic material.
Ans. (i) TMV (Tobacco Mosaic virus)
(ii) HIV (Human Immuno Deficiency virus)
(iii) QB bacteriophage