NCERT Solutions | Class 12 Biology Chapter 6

NCERT Solutions | Class 12 Biology Chapter 6 | Molecular Basis of Inheritance 

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

CBSE Solutions | Biology Class 12

Check the below NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Pdf free download. NCERT Solutions Class 12 Biology  were prepared based on the latest exam pattern. We have Provided Molecular Basis of Inheritance Class 12 Biology NCERT Solutions to help students understand the concept very well.

NCERT | Class 12 Biology

NCERT Solutions Class 12 Biology
Book: National Council of Educational Research and Training (NCERT)
Board: Central Board of Secondary Education (CBSE)
Class: 12th
Subject: Biology
Chapter: 6
Chapters Name: Molecular Basis of Inheritance
Medium: English

Molecular Basis of Inheritance | Class 12 Biology | NCERT Books Solutions

You can refer to MCQ Questions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.


NCERT Solutions for Class 12 Biology Chapter 6, Question 1.

Group the following as nitrogenous bases and nucleosides:
Adenine, cytidine, thymine, guanosine, uracil, and cytosine.

Solution:

Adenine, Guanosine, Thymine, Uracil, and Cytosine are nitrogenous bases. (Adenine and Guanosine ? Purine, Thymine, Uracil and Cytosine ? Pyrimidine) Cytidine is a nucleoside.

NCERT Solutions for Class 12 Biology Chapter 6, Question 2.

If a double-stranded DNA has 20 percent of cytosine, calculate the percent of adenine in the DNA.

Solution:

According to Chargaff’s rule, in a double-stranded DNA, the total number of cytosine molecules will be equal to the number of guanine molecules and the number of adenine molecules will be equal to the number of thymine molecules. Therefore, if a double-stranded DNA has 20 percent of cytosine then the guanine will also be 20 per cent. The remaining 60% will consist of adenine and thymine in equal amount. Thus adenine will be 30%.

NCERT Solutions for Class 12 Biology Chapter 6, Question 3.

If the sequence of one strand of DNA is written as follows:
5′-ATGCATGCATGCATGCATGCA
TGCATGC-3′
Write down the sequence of complementary strand in 5′ -> 3′ direction.

Solution:

5′-GCATGCATGCATGCATGCAT G C ATG CAT-3′.

NCERT Solutions for Class 12 Biology Chapter 6, Question 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:

If the sequence of coding strand is :
5′ – ATGCATGCATGCATGCATGCATGCATGC – 3′
Then template strand is :
3′ – TACGTACGTACGTACGTACGTACGTACG – 5′
The mRNA will be formed on the template strand in 5′ —> 3’ direction. Thus mRNA sequence will be:
5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′
Thymine in DNA is substituted by uracil in RNA.

NCERT Solutions for Class 12 Biology Chapter 6, Question 5.

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

Solution:

The two strands of DNA show complementary base pairing. This property of DNA led Watson and Crick to suggest a semi-conservative mechanism of DNA replication in which one strand of a parent is conserved while the other complementary strand formed is new.

NCERT Solutions for Class 12 Biology Chapter 6, Question 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 polymerases and DNA dependent RNA polymerases.

NCERT Solutions for Class 12 Biology Chapter 6, Question 7.

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

Solution:

They raised 2 types of bacteriophages
  • On radioactive phosphorous (32P)
  • On radioactive sulphur (35S).

35S gets into protein and 32P into DNA When both bacteriophages infected bacteria differently and by shaking them, the viral protein coat was separated

After raising these bacteria it was found that those infected with 32P bacteriophage ? radioactivity were found. But with 35S ? no radioactivity was found.

NCERT Solutions for Class 12 Biology Chapter 6, Question 8.

Differentiate between the following:
  1. Repetitive DNA and Satellite DNA
  2. Template strand and Coding strand
  3. mRNA and tRNA

Solution:

  1. Differences between repetitive DNA and satellite DNA are as follows:
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.1
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.2
  2. Differences between template strand and coding strand are as follows:
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.3
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.4
  3. Differences between mRNA and tRNA are as follows:
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.5
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.6
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.7
    NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Q8.8

NCERT Solutions for Class 12 Biology Chapter 6, Question 9.

List two essential roles of ribosome during translation

Solution:

Two essential roles of the ribosome during translation are:
  1. One of the RNA acts as a peptidyl transferase ribozyme for the formation of peptide bonds.
  2. The ribosome provides sites for attachment of mRNA and charged tRNA for polypeptide synthesis.

NCERT Solutions for Class 12 Biology Chapter 6, Question 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:

The lac operon is regulated by the amount of lactose in the medium where the bacteria are grown. When the amount of lactose is exhausted in the medium, the lac operon shuts down.

NCERT Solutions for Class 12 Biology Chapter 6, Question 11.

Explain (in one or two lines) the function of the followings:
  1. Promoter
  2. tRNA
  3. Exons

Solution:

  1. Promoter: It is located at the 5′ end of the transcription unit and provides site for attachment of transcription factors (TATA Box) and RNA polymerase.
  2. tRNA: It takes part in the transfer of activated amino acids from cellular pool to ribosome so that they can take part in protein formation.
  3. Exons: In eukaryotes, DNA is mosaic of exons and introns. Exons are coding sequences of DNA which are both transcribed and translated.

NCERT Solutions for Class 12 Biology Chapter 6, Question 12.

Why is the Human Genome Project called a mega project?

Solution:

The human genome was a megaproject that aimed to sequence every base in the human genome. The estimated cost of the project would be a billion (1 billion = 100 crores) US dollars.

NCERT Solutions for Class 12 Biology Chapter 6, Question 13.

What is DNA fingerprinting? Mention its application.

Solution:

DNA fingerprinting is the identification of differences in specific regions of DNA sequences based on DNA polymorphism, repetitive DNA, and satellite DNA.
Application of DNA fingerprinting: Settling, paternity disputes and identity of criminal by different DNA profiles in forensic laboratories.

NCERT Solutions for Class 12 Biology Chapter 6, Question 14.

Briefly describe the following:
  1. Transcription
  2. Polymorphism
  3. Translation
  4. Bioinformatics

Solution:

1. Transcription – It is the process of copying genetic information from the anti-sense or template strand of the DNA into RNA. It is meant for taking the coded information from DNA in nucleus to the site where it is required for protein synthesis. Principle of complementarity is used even in transcription. The exception is that uracil is incorporated instead of thymine opposite adenine of template. The segment of DNA that takes part in transcription is called transcription unit. It has three components
    • a promoter,
    • the structural gene and
    • a terminator.

2. Polymorphism – It is the variation at genetic level, arisen due to mutations. Such variations are unique at particular site of
DNA. They occur approximately once in every 500 nucleotides or about 107 times per genome. These are due to deletions, insertions, and single-base substitutions. These alterations in healthy people, occur in non-coding regions of DNA and do not code for any protein but are heritable. The polymorphism in DNA sequences is the basis of genetic mapping of human genome as well as DNA fingerprinting.

3. Translation – It is the mechanism by which the triplet base sequence of mRNA guides the linking of a specific sequence of amino acids to form a polypeptide chain (protein) on ribosomes in the cell cytoplasm. All the protein that a cell needs are synthesised by the cell within itself.
The raw materials required in protein synthesis are ribosomes, amino acids, mRNA, tRNAs and amino acyl tRNA synthetase. Mechanism of protein synthesis involves following steps:

    • Activation of amino acids
    • Charging or aminoacylation of tRNA
    • Initiation
    • Elongation (Polypeptide chain formation)
    • Termination

The ribosomes move along the mRNA ‘reading’ each codon in turn. Molecules of transfer RNA (tRNA), each bearing a particular amino acid, are brought to their correct positions along the mRNA, molecule base pairing occurs between the bases of the codons and the complementary base triplets of tRNA. In this way, amino acids are assembled in the correct sequence to form the polypeptide chain.

4. Bioinformatics – Bioinformatics is the combination of biology, information technology and computer science. Basically, bioinformatics is a recently developed science which uses information technology to understand biological phenomenon. It broadly involves the computational tools and methods used to manage, analyse and manipulate volumes of biological data.

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