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NCERT Solutions · Class 12 Biology · Chapter 5

Chapter 5: Molecular Basis of Inheritance (Genetics and Evolution)

Exam-ready answers to all 14 exercises of Chapter 5, Molecular Basis of Inheritance (NCERT Class 12 Biology, 2026-27): bases and nucleosides, Chargaff's rule calculation, complementary strand and mRNA sequences, semi-conservative replication, nucleic acid polymerases, the Hershey–Chase experiment, repetitive DNA, template and coding strands, ribosomes, the lac operon, promoter, exons, the Human Genome Project, DNA fingerprinting and key terms. All 14 questions are answered, with the key answer highlighted.

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Exercises

1
Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Solution
  • Nitrogenous bases: adenine, thymine, uracil, cytosine
  • Nucleosides (base + sugar): cytidine, guanosine

Bases: adenine, thymine, uracil, cytosine. Nucleosides: cytidine, guanosine.

2
If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
Solution

By Chargaff's rule, C = G and A = T.

  • C = 20%, so G = 20%; together 40%.
  • A + T = 100 − 40 = 60%, and A = T, so A = 30%.

Adenine = 30%.

3
If the sequence of one strand of DNA is written as follows: 5′-ATGCATGCATGCATGCATGCATGCATGC-3′. Write down the sequence of complementary strand in 5′→3′ direction.
Solution

The complementary strand pairs A with T and G with C and runs antiparallel. Written 5′→3′:

5′-GCATGCATGCATGCATGCATGCATGCAT-3′

5′-GCATGCATGCATGCATGCATGCATGCAT-3′

4
If the sequence of the coding strand in a transcription unit is written as follows: 5′-ATGCATGCATGCATGCATGCATGCATGC-3′. Write down the sequence of mRNA.
Solution

The mRNA has the same sequence as the coding strand, with U in place of T:

5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′

5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′

5
Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Solution

The complementary base pairing between the two strands (A with T, G with C). Since the sequence of each strand fixes the sequence of the other, Watson and Crick proposed that the two strands separate and each acts as a template for a new complementary strand. Each daughter DNA then has one old (parental) and one new strand: semi-conservative replication.

Complementary base pairing: each separated strand can act as a template, so each new DNA keeps one old strand.

6
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.
Solution
  • DNA-dependent DNA polymerase: DNA template → DNA (replication)
  • DNA-dependent RNA polymerase: DNA template → RNA (transcription)
  • RNA-dependent DNA polymerase (reverse transcriptase): RNA template → DNA (in retroviruses)
  • RNA-dependent RNA polymerase: RNA template → RNA (in some RNA viruses)

DNA-dependent DNA polymerase, DNA-dependent RNA polymerase, RNA-dependent DNA polymerase (reverse transcriptase) and RNA-dependent RNA polymerase.

7
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Solution

They grew bacteriophages in two different media:

  • one with radioactive phosphorus (³²P), which is incorporated into DNA (DNA has phosphorus, proteins do not);
  • one with radioactive sulphur (³⁵S), which is incorporated into protein (proteins have sulphur, DNA does not).

These labelled phages infected E. coli; then the cultures were blended and centrifuged to separate the phage coats from the bacteria. Bacteria infected with ³²P-phages were radioactive, while with ³⁵S-phages the radioactivity stayed outside, in the supernatant. So DNA, not protein, entered the bacteria and is the genetic material.

They labelled DNA with ³²P and protein with ³⁵S; only ³²P (DNA) entered the bacteria.

8
Differentiate between the followings: (a) Repetitive DNA and Satellite DNA (b) mRNA and tRNA (c) Template strand and Coding strand.
Solution

(a) Repetitive DNA: DNA sequences repeated many times in the genome. Satellite DNA: a type of repetitive DNA, made of short sequences repeated in tandem, that forms a separate small (satellite) peak on density-gradient centrifugation; it does not code for proteins and is used in DNA fingerprinting.

(b) mRNA: carries the genetic code (codons) from DNA to the ribosome; linear. tRNA: an adapter molecule that reads the codon with its anticodon and brings the specific amino acid; clover-leaf (inverted L) shape.

(c) Template strand: the DNA strand with 3′→5′ polarity that is actually copied during transcription. Coding strand: the other strand (5′→3′); it is not copied, and its sequence matches the mRNA (with T in place of U).

(a) Repetitive DNA in general vs tandem repeats forming a satellite band. (b) Carries codons vs carries amino acids (anticodon). (c) Strand copied (3′→5′) vs strand matching mRNA (5′→3′).

9
List two essential roles of ribosome during translation.
Solution
  • It provides the site where the mRNA binds and where tRNAs with amino acids come together to read the codons.
  • The large subunit's rRNA acts as a ribozyme (peptidyl transferase) that catalyses the formation of peptide bonds between amino acids.

It binds mRNA and tRNAs, and catalyses peptide bond formation (peptidyl transferase).

10
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?
Solution

Lactose (allolactose) acts as the inducer: it binds the repressor and inactivates it, so the structural genes are transcribed and β-galactosidase is made. This enzyme breaks lactose down into glucose and galactose. Once the lactose is used up, there is no inducer left; the free repressor binds the operator again and switches off transcription. So the operon shuts down when lactose is exhausted.

The enzymes break down the lactose; without the inducer, the repressor binds the operator again.

11
Explain (in one or two lines) the function of the followings: (a) Promoter (b) tRNA (c) Exons.
Solution
  • (a) Promoter: a DNA sequence at the 5′ end (upstream) of the structural gene where RNA polymerase binds to start transcription.
  • (b) tRNA: reads the mRNA codon through its anticodon and brings the matching amino acid to the ribosome during translation.
  • (c) Exons: the coding sequences of a split gene that remain in the mature mRNA (after introns are removed) and are translated into protein.

(a) Binding site for RNA polymerase to start transcription. (b) Brings amino acids to the ribosome. (c) Coding parts kept in mature mRNA.

12
Why is the Human Genome project called a mega project?
Solution
  • The human genome has about 3 × 10⁹ base pairs; sequencing it was a huge task needing great resources.
  • It cost about 3 billion US dollars (about 9 billion at the rate of 3 dollars per base pair) and took 13 years (1990–2003).
  • It needed high-speed computers to store and analyse enormous data, and many countries and scientists worked together.

The genome is huge (3 × 10⁹ bp), so it needed enormous money, time, computing power and international cooperation.

13
What is DNA fingerprinting? Mention its application.
Solution

DNA fingerprinting is a technique to identify differences in specific regions of DNA (repetitive DNA, VNTRs) between individuals. Since these patterns are unique to each person (except identical twins), it can identify a person.

Applications:

  • Forensic science: identifying criminals from blood, hair or semen.
  • Settling paternity and maternity disputes.
  • Identifying bodies and studying genetic diversity, evolution and populations.

Identifying individuals from differences in repetitive DNA (VNTRs); used in forensics, paternity disputes and population studies.

14
Briefly describe the following: (a) Transcription (b) Polymorphism (c) Translation (d) Bioinformatics.
Solution
  • (a) Transcription: copying of genetic information from one strand of DNA (template) into RNA by RNA polymerase. It has three steps: initiation (at the promoter), elongation and termination (at the terminator). In eukaryotes, the primary transcript (hnRNA) undergoes splicing, capping and tailing.
  • (b) Polymorphism: the existence of more than one variant (allele) of a DNA sequence at a locus in a population, arising from inheritable mutations; it is the basis of DNA fingerprinting.
  • (c) Translation: the synthesis of a polypeptide using the sequence of mRNA codons; tRNAs bring amino acids, and ribosomes join them by peptide bonds (initiation at AUG, elongation, termination at a stop codon).
  • (d) Bioinformatics: the use of computers and databases to store, retrieve and analyse biological data, such as DNA and protein sequences (e.g. data from the Human Genome Project).

(a) DNA → RNA copying. (b) Many variants of a DNA sequence in a population. (c) mRNA → protein. (d) Computer analysis of biological data.

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