Molecular Basis Of Inheritance Class 12th Biology CBSE Solution

Class 12th Biology CBSE Solution

Exercises
Question 1.

Group the following as nitrogenous bases and nucleosides:

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


Answer:

Nitrogenous bases are Adenine, Thymine, Uracil and Cytosine.

Nucleoside are Cytidine and Guanosine.


Explanation: There are two types of nitrogenous bases: Purines (Adenine and Guanine) and Pyrimidines (Cytosine, Uracil and Thymine).


A nitrogenous base when linked to a pentose sugar through a N-glycosidic linkage forms a nucleoside like adenosine, guanosine, cytidine and uridine.



Question 2.

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


Answer:

Its given that Cytosine = 20%. Hence according to Erwin Chargaff, Guanine = 20%. So, Cytosine + Guanine = 40% which makes Adenine + Thymine = 60%. Hence Adenine = 30%.


Explanation: Adenine is paired to Thymine by 2 hydrogen bonds. Cytosine is paired to Guanine by 3 hydrogen bonds. According to Erwin Chargaff, the ratios between Adenine & Thymine and Cytosine & Guanine are constant and equals one.


%A = %T and %C = %G



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:

The sequence of complementary strand in 5' 3' direction will be 5'-GCATGCATGCATGCATGCATGCATGCAT-3'.


Explanation: The two strands of the DNA have opposite polarity. 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 complementary strand in 5' 3' direction will be 5'-GCATGCATGCATGCATGCATGCATGCAT-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.


Answer:

The sequence of mRNA will be 5'-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3'.


Explanation: In a RNA strand thymine is replaced with uracil. The coding strand of mRNA is the same as that of DNA with the only difference being uracil instead of thymine. If the coding strand is 5'-ATGCATGCATGCATGCATGCATGCATGC-3' then the mRNA sequence will be 5'-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3'.



Question 5.

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


Answer:

The property of DNA double helix that led Watson and Crick to hypothesise semi-conservative of DNA replication was that the two strands in the DNA are complementary and anti-parallel based on their base sequences. They suggested that the two strands would separate and act as a template for the synthesis of new complementary strands. After the replication, each DNA molecule would have one parent strand and one newly synthesized daughter strand.



Question 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.


Answer:

There are two types of nucleic acid polymerases:

a. DNA-dependent DNA polymerase


b. DNA-dependent RNA polymerase


DNA-dependent DNA polymerase use a DNA template to synthesize a new DNA strand and DNA-dependent RNA polymerase use a DNA template to synthesize a new RNA strand.



Question 7.

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


Answer:

Alfred Hershey and Martha Chase grew some viruses that infect bacteria called bacteriophages on two mediums, one that contained radioactive phosphorous and one that contained radioactive sulfur. Viruses grown in radioactive phosphorous had radioactive DNA and those grown in radioactive sulfur had radioactive protein. When these bacteriophages were allowed to attach to E.coli bacteria, the bacteria that radioactive DNA were radioactive but those that were infected with radioactive protein were not radioactive. This indicated that DNA was passed from the virus to the bacteria. Therefore it was proved by Chase and Hershey that DNA is the genetic material.



Question 8.

Differentiate between the followings:

Repetitive DNA and Satellite DNA


Answer:




Question 9.

Differentiate between the followings:

mRNA and tRNA


Answer:




Question 10.

Differentiate between the followings:

Template strand and Coding strand


Answer:




Question 11.

List two essential roles of ribosome during translation.


Answer:

Ribosome have the following roles during translation:

(i) Ribosomes are responsible for synthesizing proteins.


(ii) Ribosomes act as catalyst in the formation of peptide bonds.



Question 12.

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:

The lac operon consists of a regulatory gene and 3 structural genes. The lactose acts as an inducer. In the presence of an inducer, such as lactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase access to the promoter and transcription proceeds. The repressor of the operon is synthesised. The repressor protein binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon. Regulation of lac operon by repressor is referred to as negative regulation.



Question 13.

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

(a) Promoter

(b) tRNA

(c) Exons


Answer:

(a) Promoter:


Promoter defines the start process of the transcription and is located at the start of the strand. It also provides binding site to RNA polymerase.


(b) tRNA:


tRNA or transfer RNA is a type of RNA that brings amino acids and reads the genetic code.


(c) Exons:


Exons are the coding sequences or expressed sequences in eukaryotes.



Question 14.

Why is the Human Genome project called a mega project?


Answer:

Human genome project was a mega project that aimed to sequence every base in human genome. This project has provided much new information about genomes. It was a 13-year long project which was launched in the year 1990 and completed in 2003. Many new areas and avenues have opened up because of the project. It helped understand the human biology better.



Question 15.

What is DNA fingerprinting? Mention its application.


Answer:

DNA Fingerprinting is a technique to find out variations in individuals of a population at DNA level.

Its applications are as follows:


(i) Used in forensic science to identify suspects


(ii) Used to find out history of an organism


(iii) Used to find out paternity and family relations.



Question 16.

Briefly describe the following:

Transcription


Answer:

Transcription:


The process of copying genetic information from one strand of DNA into RNA is called transcription. Only a segment of DNA and only one strand is copied into RNA. A transcription unit in DNA contains three regions, a Promoter, the Structural gene, a Terminator. There is single DNA-dependent RNA polymerase that catalyses transcription in bacteria. RNA polymerase binds to promoter and initiates transcription. It also facilitates opening of the helix and continues elongation. Only a short stretch of RNA remains bound to the enzyme. Once the polymerases reaches the terminator region, the RNA falls off and the RNA polymerase. This results in termination of transcription.




Question 17.

Briefly describe the following:

Polymorphism


Answer:

Polymorphism:


Polymorphism is variation at genetic level that arises due to mutations. New mutations may arise in an individual either in somatic cells or in the germ cells. The germ cell mutation can be transferred from parents to offsprings. Polymorphism in DNA sequence is the basis of genetic mapping of human genome as well as of DNA fingerprinting. If an inheritable mutation is observed in a population at high frequency, it is referred to as DNA polymorphism. There is a variety of different types of polymorphisms ranging from single nucleotide change to very large scale changes.



Question 18.

Briefly describe the following:

Translation


Answer:

Translation:


Translation refers to the process of polymerization of amino acids to form a polypeptide. In the first phase amino acids are activated in the presence of ATP and linked to their cognate tRNA. The cellular factory responsible for synthesising proteins is the ribosome. When the small subunit of ribosome encounters an mRNA, the process of translation of the mRNA to protein begins. For initiation, the ribosome binds to the mRNA at the start codon (AUG) that is recognized only by the initiator tRNA. During the elongation phase, complexes composed of an amino acid linked to tRNA, sequentially bind to the appropriate codon in mRNA by forming complementary base pairs with the tRNA anticodon. The ribosome moves from codon to codon and amino acids are added one by one, translated into polypeptide sequences dictated by DNA and represented by mRNA. At the end, a release factor binds to the stop codon, terminating translation and releasing the complete polypeptide from the ribosome.




Question 19.

Briefly describe the following:

Bioinformatics


Answer:

Bioinformatics:


Human Genome Project was closely associated with the rapid development of a new area in biology called Bioinformatics. The enormous amount of data that had been generated also required the use of high speed computational devices for data storage and retrieval, and analysis. Hence, bioinformatics was developed to create biological databases that stored information. It solved problems of management of data. Bioinformatics has developed new tools and algorithms for efficient use and methods to find relations between the data.