NCERT Solutions for Class 12 Biology Chapter 6: Molecular Basis of Inheritance

NCERT Solutions for Class 12 Biology Chapter 6 Free PDF Download

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1. Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.

Ans. Nitrogenous Bases – Adenine, Uracil and Cytosine, Thymine; Nucleosides – Cytidine, guanosine.

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

Ans. According to Chargaff’s rule:
In DNA – A = T and G = C
A + T + G + C = 100%
Given, Cytosine = 20%
So, G = C = 20%
Thus,
G + C = 20% + 20% = 40%
A + T + G + C = 100%
A + T = 100 – 40 = 60%
A = T
So, A will be equal to 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.

Ans. If the sequence of one strand of DNA is written as follows:
5′ – ATGCATGCATGCATGCATGCATGCATGC – 3′
The sequence of the complementary strand in
5′ → 3′ direction will be:
5′ – GCATGCATGCATGCATGCATGCATGCAT – 3′

4. If the sequence of the coding strand in a transcription unit is written as follows: 5–ATGCATGCATGCATGCATGCA TGCATGC–3′ Write down the sequence of mRNA.

Ans. If the coding stand in a transcription unit is
5′ – ATGCATGCATGCATGCATGCATGC ATCG – 3′
Then the template stand in 3′ to 5′ direction would be
3′ – TACGTACGTACGTACHTACGTACGTACG – 5′
It is known that the sequence of mRNA is same as the coding strand of DNA. However, in RNA, thymine is replaced by uracil(0).
Hence, the sequence of mRNA will be
5′ – AUGC AUGC AUGC AUGC AUGC AUGC AUGC – 3′

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

Ans. The antiparallel, double-stranded nature of the DNA molecule led Watson and Crick to hypothesise semi-conservative mode of DNA replication. They suggested that the two strands of DNA molecule uncoil and separate, and each strand serves as a template for the synthesis of a new (complementary) strand alongside it. The template and its complement, then form a new DNA double strand, identical to the original DNA molecule. The sequence of bases which should be present in the new strands can be easily predicted because these would be complementary to the bases present in the old strands. A will pair with T, T with A, C with G, and G with C. Thus, two daughter DNA molecules identical to the parent molecule are formed and each daughter DNA molecule consists of one old (parent) strand and one new strand. Since only one parent strand is conserved in each daughter molecule, this mode of replication is said to be semi-conservative. Meselson and Stahl and Joseph Taylor, later proved it by experiments.

6.Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesized from it (DNA or RNA), list the types of nucleic acid polymerases.

Ans. There are two different types of nucleic acid polymerases.
(i) DNA dependent DNA polymerases.
(ii) DNA dependent RNA polymerases.
The DNA: dependent DNA polymerases use a DNA template for synthesizing a new strand of DNA, whereas DNA dependent RNA polymerases use a DNA template for synthesizing RNA.

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

Ans. Alfred Hershey and Martha Chase (1952) worked with viruses that infect bacteria called bacteriophages. In 1952, they chose a bacteriophage known as T2 for their experimental material.
They grew some viruses on a medium that contained radioactive phosphorus (p32) and some others on medium that contained radioactive sulphur (s35). Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. Similarly, viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.
Radioactive phages were allowed to attach to E. coli bacteria. Then, as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge.
Bacteria which was infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria. Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicates that proteins did not enter the bacteria from the viruses. DNA is therefore the genetic material that is passed from virus to bacteria.

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

Ans. (a) Repetitive DNA and Satellite DNA

Repetitive DNA	Satellite DNA
It includes both highly repetitive and middle repetitive DNA.	It represents the highly repetitive DNA.
The middle repetitive DNA present in their RNA genes codes for the ribosomal RNAs.	It does not code for any protein or RNA.

(b) mRNA and tRNA

mRNA	tRNA
It acts a template for the process of translation i.e., amino acid synthesis.	It acts as an adaptor molecule that carries a specific amino acid to mRNA for the synthesis of polypeptide.
It is a linear molecule.	It has clover leaf shape.

9. List two essential roles of ribosome during translation.

Ans. Two essential roles of ribosomes during translation are as follows:
(i) They are responsible for synthesizing proteins.
(ii) They acts as catalyst in the formation of peptide bond.

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?

Ans. Lac operon is switched on, on adding lactose in medium, as lactose acts as inducer and makes repressor inactive by binding with it. When the lac operon system is switched on, b-galactosidase is formed, which converts lactose into glucose and galactose. As soon as all the lactose is consumed, repressor again becomes active and causes the system to switch off (shut down).

Template strand	Coding strand
It acts as a template for the synthesis of mRNA during transcription.	It has the sequence of the DNA that has the same base sequence as that of mRNA (except thymine that is replaced by uracil in DNA)
It runs from 3′ to 5′	It runs from 5′ to 3′

11. Explain (in one or two lines) the function of the followings:
(a) Promoter
(b) tRNA
(c) Exons

Ans. (a) Promotor: It is a region of DNA that helps in the inititations of transcription. It serves as the binding sets for the RNA polymerase.
(b) tRNA: It reads the genetic code on mRNA and carries specific amino acid to mRNA on ribosome during translation of proteins.
(c) Exons: Exons are the coding sequences of DNA in eukaryotes that transcribe for proteins.

12. Why is the Human Genome project called a mega project?

Ans. Human Genome project is called a mega project because:
(i) It required bioinformatics data basing and other high speed computational devices for analysis, storage and retrieval of information.
(ii) It generated lot of information in the form of sequence annotation. A total of 3 × 109 BP were to be sequenced.
(iii) It was carried out in number of labs and coordinated on extensive scale.

13. What is DNA fingerprinting? Mention its application.

Ans. DNA fingerprinting: It is a technique used to identify and analyse the variations in various individuals at the level of DNA. It is based on variability and polymorplism in DNA sequences.
Applications of DNA fingerprinting:
(i) It is used in forensic science to identify potential crime suspects.
(ii) It is used to establish paternity and family relationships.
(ii) It is used to find out evolutionary history of an organism.

14. Briefly describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics

Ans. (a) Transcription: It is process of synthesis of RNA for DNA. It completes in 3 steps.

Initiation: RNA polymerase binds to the promotor and providing the single stand template for transcription.
Elongation: RNA polymerase reads the template stand and add bases leading to the elongation of the chain.
Termination: It is the last step of transcription, where the terminator sequences mark the end of transcription.

(b) Polymorphism: It is a form of genetic variation resulting in the formation of several different types of individuals among the number of one particular species. The varying sequences of nucleotides are located at different positions in DNA results in variation and hence polymorphism in the population. It is responsible for speciation and evolution.

(c) Translation: The process of polymerization of amino acid to form a polypeptide is known as translation. It completes in 3 steps.

Initiation: Ribosome assemble around the mRNA that has to read and first tRNA carrying the amino acid methionine.
Elongation: It this step, amino acid chain is elongated.
Termination: The synthesized polypeptide chain has released.

(d) Bioinformatics: Bioinformatics is the use of information technology for collecting and analyzing complex biological data such as DNA sequences, protein codes etc. There has been an advancement in the field of molecular biology, there is the invention of computers, scientific technique and instrumentation.

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