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

Rajasthan Board RBSE Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance Textbook Exercise Questions and Answers.

RBSE Class 12 Biology Solutions Chapter 6 Molecular Basis of Inheritance

RBSE Class 12 Biology Molecular Basis of Inheritance Textbook Questions and Answers

Question 1. 
Group the following as nitrogenous bases and nucleosides : Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer:
Following is the grouping:
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 percent of adenine in the DNA.
Answer:
According to Chargaffs rule, the DNA molecule should have an equal ratio of purine (adenine and guanine) and pyrimidine (cytosine and thymine). It means that the number of adenine molecules is equal to thymine molecule and the number of guanine molecules is equal to cytosine molecules.
%A = %T and %G = %C
If double stranded DNA has 20% of cytosine, than according to law, it would have 20% of guanine.
Thus, percentage of G + C content = 40%.
The remaining 60% represent both A + T contents. Since adenine and thymine are always present in equal numbers, the percentage of adenine molecule is 30%. 

Question 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.
Answer:
Complementary sequence in 3' → 5 is :
3'-TACGTACGTACGTACGTACGTACGTACG-5' 
Then in 5' → 3' direction, complementary strand will be: 
5'-GCATGCATGCATGCATGCATGCATGCAT-3' 

Question 4. 
If the sequence of the coding strand in a transcription is written as follows: 
5'-ATGCATGCATGCATGCATGCATGCATGC-3' 
Write down the sequence of mRNA 
Answer:
In - mRNA thymine is replaced by Uracil (U). Therefore the coding sequence will be: 5'-GCAUGCAUGCAUGCAUGCAUGCAUGCAUGCAU-3' 

Question 5. 
Which property of DNA double helix led Watson and Crick to hypothesise semi - conservative mode of DNA replication? Explain.
Answer:
The two strands of DNA show complementary base pairing. This property of DNA led Watson and Grick to suggest semi - conservative mode of DNA replication in which one strand of parent DNA remains conserved while the other complementary strand is formed new. 

Question 6. 
Depending upon 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.
Answer:
DNA template:

• DNA polymerase for DNA replication,
• RNA polymerase for RNA synthesis or transcription.

RNA template:

• RNA - dependent RNA polymerase for synthesis of RNA in some RNA viruses.
• Reverse transcriptase to synthesis cDNA (complementary DNA) over RNA template.

Question 7. 
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material? Answer:
Hershey and Chase conducted experiment on T₂ - phage, a parasite on bacterium Escherichia coli and prepared two types of T₂ - phages.
(a) One containing radioactive sulphur and
(b) Other radioactive phosphorus.

Both phages were made to infect normal bacteria in two separate samples. After, infection the bacterial cells ruptured liberating daughter phages. By the test for radioactivity it was found that radioactive phosphorus was passed on to the daughter phages while radioactive sulphur was found in the protein coat left outside the bacterial cell.

Question 8. 
Differentiate between the following:
(a) Repetitive DNA and Satellite DNA.
(b) mRNA and tRNA.
(c) Template strand and Coding strand.
Answer:
(a) Difference between Repetitive DNA and Satellite DNA.

(b) Difference between mRNA and tRNA

(c) Difference Template Strand and Coding Strand

Question 9. 
List two essential roles of ribosomes during translation.
Answer:
Two essential roles of ribosome during translation are:

• One of the rRNA (23S in prokaryotes) acts as a peptidyl transferase ribozyme for formation of peptide bonds.
• Ribosome provides sites for attachment of mRNA and charged tRNAs for polypeptide synthesis.

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?
Answer:
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 lack operon system is switched on, ß - galactosidase is formed, which convertes lactose into glucose and galactose. As soon as all the lactose is consumed, repressor again becomes active and causes the lac operon to switch off (shut down).

 
 Question 11. 
 Explain (in one or two lines) the function of the following:
(a) Promotor
(b) tRNA
(c) Exons.
Answer:
(a) Promotor: It is the segment of DNA which lies adjacent to the operator and functions as the binding site for RNA polymerase to carry transcription if allowed by operator.

(b) tRNA: It acts as a adaptor molecule that picks up a particular amino acid from cellular pool and takes the same over to a site of mRNA for incorporation into polypeptide chain.

(c) Exons: These are the coding segments present in primary transcription which after splicing are joined to form functional mRNA.

Question 12. 
Why is the Human Genome project called a mega project?
Answer:
Human genome project is called a mega project, because of following reasons:

• Sequencing of more than 3 x 10⁹ bp.
• High expenditure of more than 9 billion dollors.
• Identification of all the genes present in human genome.
• Identification of all the alleles of genes and their functions.
• Storage of data for sequencing would require space equal to 3300 books of 1000 pages each if each page contains 1000 letters.

Question 13. 
What is DNA fingerprinting? Mention its application.
Answer:
DNA fingerprinting is a technique of determining nucleotide sequence of certain areas (VNTRs) of DNA which are unique to each individual. Each person has a unique DNA fingerprint. Unlike a conventional fingerprint that occurs only on the fingerprints and can be altered by surgery a DNA fingerprint is the same for every cell, tissue and organ of a person, it can not be changed any known treatment.

Applications of DNA fingerprinting are as follows:

• Paternity disputes can be solved by DNA fingerprinting.
• DNA fingerprinting technique is being used to identify genes connected with hereditary diseases.
• It is useful in detection of crime and legal Pursuit.
• It can identify racial groups, their origin, historical migrations and invasions.

Question 14. 
Briefly describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics.
Answer:
(a) Transcription: It is a DNA directed synthesis of RNA in which the RNA transcribed on 3' → 5' template strand of DNA in 5' → 3' direction.
(b) Polymorphism: Variation of genetic level occurred due to mutation, is called polymorphism. Such variations are unique at particular site of DNA, forming satellite DNA. The polymorphism in DNA sequences is the basis of genetic mapping and DNA finger printing.
(c) Translation: Protein synthesis from mRNA, tRNA, rRNA.
(d) Bioinformatics: Computational method of handling and analyzing biological databases.