NCERT Class 12 Biology Chapter 6 Molecular Basis of Inheritance

NCERT Class 12 Biology Chapter 6 Molecular Basis of Inheritance Solutions to each chapter is provided in the list so that you can easily browse through different chapters NCERT Class 12 Biology Chapter 6 Molecular Basis of Inheritance and select need one. NCERT Class 12 Biology Chapter 6 Molecular Basis of Inheritance Question Answers Download PDF. NCERT Biology Class 12 Solutions.

NCERT Class 12 Biology Chapter 6 Molecular Basis of Inheritance

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Also, you can read the NCERT book online in these sections Solutions by Expert Teachers as per Central Board of Secondary Education (CBSE) Book guidelines. CBSE Class 12 Biology Solutions are part of All Subject Solutions. Here we have given NCERT Class 12 Biology Chapter 6 Molecular Basis of Inheritance Notes, NCERT Class 12 Biology Textbook Solutions for All Chapters, You can practice these here.

Chapter: 6

BIOLOGY

UNIT – II GENETICS AND EVOLUTION

TEXT BOOK QUESTIONS ANSWERS

Q. 1. Group the following as nitrogenous bases and nucleosides:

Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.

Ans: Nitrogenous bases present in the list are adenine, thymine, uracil, and cytosine. 

Nucleosides present in the list are cytidine and guanosine.

Q. 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, % A = % T and % G = % C.

If dsDNA has 20% of cytosine, then it has 20% of guanine. Thus, percentage of G + C content = 40%

Therefore, both A + T molecule = 60%. 

Since adenine and thymine are present in equal numbers, percentage of adenine molecule is 30%.

Q.3. If the sequence of one strand of d 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

5’-ATGCATGCATGCATGCATGCATGCATGC → 3′ 

Then, the sequence of complementary strand in 5′ to 3′ direction will be 3′-TACGTACGTACGTACGTACGTACGTACG→ 5′

Therefore, the sequence of nucleotides on DNA polypeptide in 5′ to 3 direction is

5’-GCATGCATGCATGCATGCATGCATGCAT→ 3′

Q. 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.

Ans: If the coding strand in a transcription unit is

5’-ATGCATGCATGCATGCATGCATGCATGC-3′ 

Then, the template strand in 3′ to 5′ direction would be

3′-TACGTACGTACGTACGTACGTACGTACG-5′ 

Sequence of mRNA is same as the coding strand of DNA but RNA has uracil instead of thymine.

Hence, the sequence of mRNA will be 

5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′

Q. 5. Which property of DNA double helix led Watson and Crick to hypothesize semiconservative mode of DNA replication? Explain.

Ans: Watson and Crick proposed semiconservative DNA replication. Two strands of DNA separate and act as a template for synthesis of new complementary strands. After the completion of replication, each DNA has one parental and one newly synthesized strand. This is semi conservative DNA replication

Q. 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: DNA-dependent DNA polymerases and RNA polymerase I,II and III.

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

Ans: Hershey and Chase worked with bacteriophage and E.coli to prove that DNA is the genetic material. They grew some bacteriophages on a medium containing radioactive phosphorus to identify DNA and some on a medium containing radioactive sulfur to identify protein.

Q. 8. Differentiate between the followings:

(a) Repetitive DNA and Satellite DNA.

Ans: Repetitive DNA and satellite DNA: Repetitive DNA is specific region in DNA with same sequences of bases repeated many times.

Satellite DNA is highly repetitive DNA sequences.

(b) mRNA and tRNA.

Ans: mRNA and tRNA: m RNA acts as a template for the process of transcription. It is a linear molecule.

t RNA acts an adaptor molecule that carries specific amino acid to mRNA for peptide synthesis. It has clover leaf shape.

(c) Template strand and Coding strand.

Ans: Template strand and coding strand: Template strand is DNA strand with polarity 3-5 and acts as a template for m RNA synthesis. 

Coding strand is DNA strand with polarity 5′-3’and has same base sequence as that of mRNA [except thymine which is replaced by uracil in DNA].

Q. 9. List two essential roles of ribosome during translation.

Ans: Two essential roles of ribosome during translation are:

(i) Ribosome is site of protein synthesis. It has two subunits. The smaller subunit with mRNA forms a protein synthesizing complex and the larger subunit acts as an amino acid binding site.

(ii) Ribosome functions as a catalyst for formation of peptide bond. [23s r-RNA in bacteria is enzyme ribozyme].

Q. 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: Lactose present in growth medium enters bacterial cell through the action of permease. Repressor is inactivated by interaction with the inducer like lactose. This allows RNA polymerase access to the promoter and transcription proceeds. Enzymes act on lactose and lactose is metabolized into glucose and galactose. 

Inducer or lactose + repressor = operon switched on.

When the level of inducer decreases on complete metabolism by enzymes, it causes synthesis of the repressor from regulator gene. The repressor binds to the operator gene prevents RNA polymerase from transcribing è operon. Hence, lac operon is switched off.

Q. 11. Explain (in one or two lines) the function of the followings:

(a) Promoter.

Ans: Promoter initiates transcription and serves as the binding site for RNA polymerase.

(b) tRNA.

Ans: tRNA: t RNA or transfer RNA reads the genetic code present on mRNA. It carries specific amino acid to mRNA on ribosome during translation of proteins.

(c) Exons.

Ans: Exons: Exons are coding sequences of DNA in eukaryotes that transcribe for proteins.

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

Ans:  Human genome project is called mega project because HGP led to development of combination of biology, information technology and computer science called Bioinformatics. Human Genome Project was coordinated by the U.S. Department of Energy and the National Institute of Health. Human genome project aimed to sequence every base in human genome. 

Q. 13. What is DNA fingerprinting? Mention its application.

Ans: DNA fingerprinting is technique to compare the DNA sequences of any two individuals. It works on the principle of polymorphism in satellite DNA.

Uses of DNA fingerprinting

(i) Forensic science: DNA fingerprinting is the basis of paternity testing, in case of disputes.

(ii) It differs from individual to individual in a population except monozygotic (identical) twins.

(iii) It is used in determining population and genetic diversities to study evolution.

Q. 14. Briefly describe the following:

(a) Transcription.

Ans: Transcription is the process of synthesis of RNA from DNA template. It starts at the promoter region of the template DNA and ends at the terminator region. The transcription requires the enzyme DNA dependent RNA polymerase. Three important events of transcription are Initiation, Elongation and Termination.

(b) Polymorphism.

Ans: Polymorphism: If an inheritable mutation is observed in a population at high frequency, it is called DNA polymorphism. Polymorphism (variation at genetic level). occurs due to mutations. Polymorphism in DNA sequence is the basis of genetic mapping of human genome and DNA fingerprinting.

(c) Translation.

Ans: Translation is the process of protein synthesis from m RNA in the ribosome. Sequence of bases in mRNA defines the order and sequence of amino acids. Translation requires ribosomes, amino acids, m RNA, t RNA and aminoacyl t RNA synthetase.

(d) Bioinformatics.
Ans: Bioinformatics: Combination of biology, information technology and computer science is called Bioinformatics. It is used in human genome project.

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