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

Here we provide NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance pdf, free NCERT solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance book pdf download. Now you will get step by step solution to each question.

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

Question 1.
Group the following as nitrogenous bases and nucleosides : Adenine, cytidine, thymine, guanosine, uracil and cytosine.
Solution:
Nitrogenous bases : Adenine, Thymine, Uracil and Cytosine.
Nucleosides : Cytidine and Guanosine.

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

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′.

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.

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Solution:
Complementary base pairing property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication. Watson & Crick observed that the nitrogenous bases are in complementary pairing in two strands of double helix of DNA molecule. Such an arrangement of DNA molecule led them to hypothesize the semi conservative mode of replication of DNA.

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:
The types of nucleic acid polymerases required for synthesis of DNA and RNA are :

1. DNA polymerase I, II, and III – They help in replication of DNA.
2. RNA dependent DNA polymerase – It helps in synthesis of DNA from RNA (reverse transcription).
3. DNA dependent RNA polymerase – It helps in synthesis of RNA from DNA (transcription).
In eukaryotes, there are at least three RNA polymerases in addition to those found in cell organelles.
• RNA polymerase I transcribes rRNA (28S, 18S and 5.8 S).
• RNA polymerase II transcribes the precursor of mRNA called heterogenous nuclear RNA (hnRNA).
• RNA polymerase III transcribes tRNA, 5SrRNA and snRNAs.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic
material?
Solution:
A. D. Hershey and Martha Chase performed an experiment to prove that DNA is the genetic material not the protein. They used T2 bacteriophages which infect Escherichia coil. To differentiate between DNA and protein during the experiment, they used radioactive isotopes of phosphorus (32P) and sulphur (35S) in the culture medium of bacterial hosts. They could not use radioactive isotnpes of carbon (14C) etc., because they are present in both molecules and would not distinguish the two. They grew two cultures of Eschcrichia cou. One culture was supplied with radioactive sulphur, 35S. The other culture was provided with radioactive phosphorus, 32P. Radioactive sulphur gets incorporated into sulphur containing amino acids and therefore, becomes part of bacterial proteins. Radioactive phosphorus gets incorporated into nucleotides which form nucleic acids, mostly DNA. Therefore, bacteria of both the cultures became labelled.
They, then introduced bacteriophage T2 in both the bacterial cultures. The virus entered the bacteria where it multiplied. The viral progeny was tested in both the cases. It was labelled, one type with radioactive protein and other type with radioactive DNA. These labelled phages were introduced in new bacterial cells. It was found that phages with labelled protein did not make the bacterial host labelled, while those with labelled DNA made the host labelled. It indicated that viral DNA is the genetic material (not proteins) which is transferred to the infected host.

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:

2. Differences between template strand and coding strand are as follows:

3. Differences between mRNA and tRNA are as follows:

Question 9.
List two essential roles of ribosome during translation
Solution:
Two essential roles of ribosome during translation are as follows:

• They provide surface for binding of mRNA in the groove of smaller subunit of ribosome.
• As larger subunit of ribosome has peptidyl transferase on its ‘P’ site, therefore it helps in joining amino acids by forming peptide bonds.

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:
Lac operon is switched on adding lactose in the medium, as lactose acts as inducer and make repressor inactive. Due to this switch on of lac operon system, (3-galactosidase is formed which converts lactose into glucose and galactose. As soon as lactose is consumed, repressor again become active and cause switch off (shut down) of system.

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.

Question 12.
Why is the Human Genome Project called a mega project?
Solution:
HGP (Human Genome Project) is called mega project because:

• it involved many countries (USA, UK, Japan, France, Germany, China) for determining the nucleotide sequences of genes
• it involved sequencing 3 x 109 base pairs costing 9 billion US dollars
• it required bioinformatics databasing and other high speed computational devices for analysis, storage and retrieval of information.

Question 13.
What is DNA fingerprinting? Mention its application.
Solution:
DNA fingerprinting is identification of difference in specific region 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.

Question 14.
Briefly describe the following:

1. Transcription
2. Polymorphism
3. Translation
4. Bioinformatics

Solution:

• Transcription – It is the process of copying genetic information from 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
1. a promoter,
2. the structural gene and
3. a terminator.
• 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.
• 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:
1. Activation of amino acids
2. Charging or aminoacylation of tRNA
3. Initiation
4. Elongation (Polypeptide chain formation)
5. 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.
• 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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