Class 12 Biology CH 6 Molecular Basis of Inheritance
1. Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Ans: Nitrogenous Bases – Adenine, Uracil and Cytosine, Thymine; Nucleosides – Cytidine, guanosine.
2. If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
Ans: In a DNA molecule, the number of cytosine molecule is equal to guanine molecules & the number of adenine molecules are equal to thymine molecules. As a result, if a double stranded DNA has 20% of cytosine, it has 20% of guanine. The remaining 60% includes both adenine & thymine which are in equal amounts. So, the percentage of adenine is 30%.
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.
Ans: If the sequence of one strand of DNA is written as follows:
5′ – ATGCATGCATGCATGCATGCATGCATGC – 3′
The sequence of the complementary strand in 5′ —> 3′ direction will be:
5′ – GCATGCATGCATGCATGCATGCATGCAT – 3′
4. If the sequence of the coding strand in a transcription unit is written as follows: 5′-ATGCATGCATGCATGCATGCA TGCATGC-3′ Write down the sequence of mRNA.
Ans: mRNA: 5′ -A U G CAU G CAU G C AU G CA UGCAUGCAUGC-3′.
5. Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain
Ans- Watson and Crick noted that the two DNA strands are anti-parallel, complementary to each other with regards to their base sequences which facilitates each strand to serve as a template to synthesise a new strand. This organisation of the DNA molecule furthered the hypothesis that the replication of DNA is semi-conservative. In other words, the double-stranded DNA molecule splits, and in turn, each of the separated strands serves as a template to synthesise a new complementary strand. Subsequently, each DNA molecule will have one parental strand and a newly synthesised daughter strand. As only one parental strand is conserved in each of the daughter molecules, the mode of replication is termed as semi-conservative.
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.
Ans: (i) DNA dependent DNA polymerase – synthesis.
(ii) DNA dependent RNA polymerase – synthesis.
(iii) RNA dependent DNA polymerase – Retroviral nucleic acid.
(iv) RNA dependent RNA polymerase – cDNA synthesis.
7. How did Hershey and Chase differentiate between DNA and protein in their experiment white proving that DNA is the genetic material?
Ans: Alfred Hershey and Martha Chase (1952) worked with viruses that infect bacteria called bacteriophages. In 1952, they chose a bacteriophage known as T2 for their experimental material.
They grew some viruses on a medium that contained radioactive phosphorus (p32) and some others on medium that contained radioactive sulphur (s35). Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protem does not. Similarly, viruses grown on radioactive sulphur contained radioactive protein but not radio’active DNA because DNA does not contain sulphur.
Radioactive phages were allowed to attach to E. coli bacteria. Then, as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge. ,
Bacteria which was infected with viruses 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 indicates that proteins did not enter the bacteria from the viruses. DNA is therefore the genetic material that is passed from virus to bacteria.
8. Differentiate between the followings: (a) Repetitive DNA and Satellite DNA (b) mRNAand tRNA (c) Template strand and Coding strand
(a) Repetitive DNA and Satellite DNA
|Repetitive DNA||Satellite DNA|
|These are DNA sequences containing small segments repeated several times||These are repetitive DNA sequences containing highly repetitive DNA|
|Vary in length from several base pairs to hundreds and thousands||Shorter in length and are close to a hundred base pairs long|
|Can be segregated from bulk DNA by density gradient centrifugation, because of which they appear as light bands||Can be segregated from bulk DNA through density gradient centrifugation, because of which they appear as dark bands and small peaks.|
(b) mRNA and tRNA
|The messenger RNA serves as a template for the transcription process||The transfer RNA serves as an adaptor molecule carrying a particular amino acid to the mRNA to synthesise polypeptide|
|mRNA is a linear molecule||Resembles a clover-shaped leaf|
|Gets attached to the ribosome only||Gets attached at one end to the ribosome and the other end to an amino acid|
(c) Template Strand and Coding Strand
|Template Strand||Coding Strand|
|Serves as a template for the mRNA synthesis during transcription||Serves as a complementary strand of the template strand|
|Consists of a sequence that is complementary to the mRNA||Consists of a sequence that is identical to the mRNA, except that thymine in DNA is replaced by uracil in mRNA|
|Template strand runs from 3′ to 5′||The coding strand runs from 5′ to 3′|
9. List two essential roles of ribosome during translation.
Ans: Two essential roles of ribosomes during translation are-
(i) they provide surface for binding of mRNA in the groove of smaller subunit of ribosome.
(ii) As larger subunit of ribosome has peptide transferase on its ‘P’ site, therefore, it helps in joining amino acids by forming peptide bonds. .
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?
Ans: 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 lac operon system is switched on, β-galactosidase is formed, which converts lactose into glucose and galactose. As soon as all the lactose is consumed, repressor again becomes active and causes the system to switch off (shut down).
11. Explain (in one or two lines) the function of the followings: (a) Promoter (b) tRNA (c) Exons
Ans: Promoter: It is one of the three components of a transcription unit that takes part in transcription. It is located at the start 5′ end and provides site for attachment of transcription factors (TATA Box) and RNA.
polymerase: tRNA: It takes part in the transfer of activated amino acids from cellular pool to ribosome for their taking part in protein formation.
Exons: In eukarytoes, DNA is mosaic of exons and introns. Exons are coding sequences of DNA which are transcribed and translated both.
12. Why is the Human genome project called a mega project?
Ans: Human genome project is called a mega project because
(i) it required bioinformatics data basing and other high speed computational devices for analysis, storage and retrieval of information.
(ii) it generated lot of information in the form of sequence annotation.
(iii) it was carried out in number of labs and coordinated on extensive scale.
13. What is DNA fingerprinting? Mention its application.
The technique of DNA fingerprinting is helpful in identifying and analysing variations in different individuals at the DNA level. It is based on the principle of variability and polymorphism in DNA sequences.
Its applications are as follows:
- To identify potential crime suspects in forensic science
- Helpful in establishing family and paternity relationships
- Useful in identifying and preserving the commercial varieties of livestock and crops
- Useful to discover and know more about the evolutionary history of an entity, thus tracing the linkages between different entities.
14. Briefly describe the following: (a) Transcription (b) Polymorphism (c) Translation (d) Bioinformatics
Ans: Transcription: It is the process of synthesising RNA from a DNA template, in which the RNA is transcribed on 3*—>5’ template strand of DNA in 5’—>3’ direction.
Polymorphism: Variation at genetic level arisen 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.
Translation: It is the process wherein amino acids are polymerized for the formation of a polypeptide chain, a ribosome, by reading an mRNA molecule.
Bioinformatics: It is a Computational method of handling and analysing biological databases. It explains practical issues that arise from the analysis and management of biological data.