NCERT Solutions For Class 12 Biology chapter 6-Molecular Basis Of Inheritance

NCERT Solutions for class-12 Biology Chapter 6 Molecular Basis of Inheritance is prepared by Physics Wallah senior teacher team, primary focus while solving these questions of class-12 in NCERT textbook, Do read theory of this Chapter while going before solving the NCERT questions. You can download and share NCERT Solutions of Class 12 Biology from Physics Wallah.

Chapter 6 Molecular Basis of Inheritance

 

Answer the following Questions.

Question1. Group the following as nitrogenous bases and nucleosides:

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

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

Nucleosides present in the list are cytidine and guanosine.

Question2. If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.

Solution :
The percent of cytosine = 20 (given) % of C = % of G

Therefore, the percent of guanine = 20
The percent of thymine + adenine will be 100 - (20 + 20) = 60
Therefore, the percent of adenine will be 60/2 = 30

Question3. 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 DNA strands are complementary to each other with respect to base sequence. Hence, if the sequence of one strand of DNA is

5'- ATGCATGCATGCATGCATGCATGCATGC − 3’

Then, the sequence of complementary strand will be

3'- TACGTACGTACGTACGTACGTACGTACG − 5’

Therefore, the sequence of nucleotides on DNA polypeptide

5'- GCATGCATGCATGCATGCATGCATGCAT− 3’

Question4. 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 coding strand in a transcription unit is

5’− ATGCATGCATGCATGCATGCATGCATGC-3’

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

3’ − TACGTACGTACGTACGTACGTACGTACG-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.

Hence, the sequence of mRNA will be

5’ − AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3’

Question5. Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.

Solution :
The property of DNA double helix led Watson and Crick are:

  • Two strands running opposite to each other, wherein bases will always pair with their counterpart-A with T and G with C (specific pairing).
  • If H bonds break and bases of one strand lie exposed, unpaired, they will easily pair up with free nucleotides as well.

This type of arrangement in DNA molecule led to the hypothesis that DNA replication is semi-conservative. where the two strands separate and act as a template for the synthesis of a new complementary strand.

Question6. 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 uses a DNA template to catalyse the polymerisation deoxynucleotides.
  • DNA-dependent RNA polymerase catalyses the transcription of all types of RNA (in bacteria).
  • DNA-dependent RNA polymerase-I transcribes rRNA.
  • DNA-dependent RNA polymerase-II transcribes the precursor of mRNA (hnRNA).
  • DNA-dependent RNA polymerase-III transcribes tRNA.

Question7. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?

Solution :

chapter 6-Molecular Basis Of Inheritance

According to Hershey and Chase experiment:

  • They grew some bacteriophages on a medium that contained radioactive phosphorus and some in another medium that contained radioactive sulphur.
  • Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein as phosphorus is present only in DNA.
  • Viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.
  • It was found that bacteria which were infected with bacteriophages 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 indicated that proteins did not enter the bacteria from the viruses.
  • This was a clear proof that DNA is the genetic material that is passed from virus to bacteria.

Question8. Differentiate between the followings:

(a) Repetitive DNA and Satellite DNA

(b) mRNA and tRNA

(c) Template strand and Coding strand

Solution :
(a) Repetitive DNA and satellite DNA

Repetitive DNA

Satellite DNA 

1.

Repetitive DNA are DNA sequences that contain small segments, which are repeated many times.

Satellite DNA are DNA sequences that contain highly repetitive DNA.

(b) mRNA and tRNA

mRNA

tRNA 

1.

mRNA or messenger RNA acts as a template for the process of transcription.

tRNA or transfer RNA acts as an adaptor molecule that carries a specific amino acid to mRNA for the synthesis of polypeptide.

2.

It is a linear molecule.

It has clover leaf shape.

(c) Template strand and coding strand

Template strand

Coding strand 

1.

Template strand of DNA acts as a template for the synthesis of mRNA during transcription.

Coding strand is a sequence of DNA that has the same base sequence as that of mRNA (except thymine that is replaced by uracil in DNA).

2.

It runs from 3’ to 5’.

It runs from 5’to 3’.

Question9. List two essential roles of ribosome during translation.

Solution :
The important functions of ribosome during translation are as follows.

(a) Ribosome acts as the site where protein synthesis takes place from individual amino acids. It is made up of two subunits.

The smaller subunit comes in contact with mRNA and forms a protein synthesizing complex whereas the larger subunit acts as an amino acid binding site.

(b) Ribosome acts as a catalyst for forming peptide bond. For example, 23s r-RNA in bacteria acts as a ribozyme.

Question10. 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 :
Lac operon is a segment of DNA that is made up of three adjacent structural genes, namely, an operator gene, a promoter gene, and a regulator gene. It works in a coordinated manner to metabolize lactose into glucose and galactose.

In lac operon, lactose acts as an inducer. It binds to the repressor and inactivates it. Once the lactose binds to the repressor, RNA polymerase binds to the promoter region. Hence, three structural genes express their product and respective enzymes are produced. These enzymes act on lactose so that lactose is metabolized into glucose and galactose.

After sometime, when the level of inducer decreases as it is completely metabolized by enzymes, it causes synthesis of the repressor from regulator gene. The repressor binds to the operator gene and prevents RNA polymerase from transcribing the operon. Hence, the transcription is stopped. This type of regulation is known as negative regulation.

chapter 6-Molecular Basis Of Inheritance

Question11. Explain (in one or two lines) the function of the followings:

(a) Promoter

(b) tRNA

(c) Exons

Solution :

  • Promoter: Promoter is a region of DNA that helps in initiating the process of transcription. It serves as the binding site for RNA polymerase.
  • tRNA: tRNA or transfer RNA is a small RNA that reads the genetic code present on mRNA. It carries specific amino acid to mRNA on ribosome during translation of proteins.
  • Exons: Exons are coding sequences of DNA in eukaryotes that transcribe for proteins.

Question12. Why is the Human Genome project called a mega project?

Solution :
Human genome project was considered to be a mega project because it had a specific goal to sequence every base pair present in the human genome. It took around 13 years for its completion and got accomplished in year 2006. It was a large scale project, which aimed at developing new technology and generating new information in the field of genomic studies. As a result of it, several new areas and avenues have opened up in the field of genetics, biotechnology, and medical sciences. It provided clues regarding the understanding of human biology.

Question13. What is DNA fingerprinting? Mention its application.

Solution :
DNA fingerprinting is a technique used to identify and analyze the variations in various individuals at the level of DNA. It is based on variability and polymorphism in DNA sequences.

Application

  • It is used in forensic science to identify potential crime suspects.
  • It is used to establish paternity and family relationships.
  • It is used to identify and protect the commercial varieties of crops and livestock.
  • It is used to find out the evolutionary history of an organism and trace out the linkages between groups of various organisms.

Question14. Briefly describe the following:

(a) Transcription

(b) Polymorphism

(c) Translation 

(d) Bioinformatics

Solution :

  • Transcription: The process of copying genetic information from one strand of the DNA into RNA is known as transcription. RNA is assembled simply based on complementarity of the DNA strand, only uracil is substituted in place of thymine. Only a small segment of DNA that codes for a polypeptide is copied.
  • Polymorphism: The variation in DNA arising through mutation at non-coding sequences is known as Polymorphism. Such variations are unique to specific sites of DNA and can occur due to deletion, insertion or substitution of bases. It can be observed by making fragments of DNA sample and separating them through electrophoresis.The polymorphism in a DNA sequence is the basis of genetic mapping of the human genome as well as DNA fingerprinting. 
  • Translation: It refers to the process of polymerisation of amino acids to form a polypeptide. The order and sequence of amino acids are defined by the sequence of bases in the mRNA. It occurs in cytoplasm in both prokaryotes and eukaryotes.
  • Bioinformatics: It is the application of computer science and information technology which deals with handling, storing of huge information of genomics, processing information, analyzing data and creating new knowledge.
  • Bioinformatics: Bioinformatics is the application of computational and statistical techniques to the field of molecular biology. It solves the practical problems arising from the management and analysis of biological data. The field of bioinformatics developed after the completion of human genome project (HGP). This is because enormous amount of data has been generated during the process of HGP that has to be managed and stored for easy access and interpretation for future use by various scientists. Hence, bioinformatics involves the creation of biological databases that store the vast information of biology.

It develops certain tools for easy and efficient access to the information and its utilization. Bioinformatics has developed new algorithms and statistical methods to find out the relationship between the data, to predict protein structure and their functions, and to cluster the protein sequences into their related families.

NCERT Solutions For Class 12 Biology Chapter Wise.

  1. Chapter 1 Reproduction in Organisms

  2. Chapter 2 Sexual Reproduction In Flowering Plants

  3. Chapter 3 Human Reproduction

  4. Chapter 4 Reproductive Health

  5. Chapter 5 Principles of Inheritance And Variation

  6. Chapter 6 Molecular Basis of Inheritance

  7. Chapter 7 Evolution

  8. Chapter 8 Human Health And Disease

  9. Chapter 9 Strategies For Enhancement In Food Production

  10. Chapter 10 Microbes In Human Welfare

  11. Chapter 11 Biotechnology And Processes

  12. Chapter 12 Biotechnology And Its Applications

  13. Chapter 13 Organisms And Populations

  14. Chapter 14 Ecosystem

  15. Chapter 15 Biodiversity And Conservation

  16. Chapter 16 Environmental Issues

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