Class 12 Biology Chapter 5 is “Molecular Basis of Inheritance”. This chapters includes sub-topics such as –
- 5.1 The DNA
- 5.2 The Search for Genetic Material
- 5.3 RNA World
- 5.4 Replication
- 5.5 Transcription
- 5.6 Genetic Code
- 5.7 Translation
- 5.8 Regulation of Gene Expression
- 5.9 Human Genome Project
- 5.10 DNA Fingerprinting
NCERT Solutions for Class 12 Biology Chapter 5 “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 molecules is equal to the number of guanine molecules and the number of adenine molecules is equal to the number of thymine molecules. So if a double-stranded DNA contains 20% cytosine, it contains 20% guanine. The remaining 60% consists of adenine and thymine, which are present in equal amounts. The percentage of adenine is therefore 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: 5ATGCATGCATGCATGCATGCA 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 semiconservative mode of DNA replication? Explain
Ans: The antiparallel, double-stranded nature of the DNA molecule led Watson and Crick to hypothesise that DNA replication is semiconservative. They proposed that the two strands of the DNA molecule unwind and separate, with each strand serving as a template for the synthesis of a new (complementary) strand next to it. The template and its complement then form a new DNA double strand that is identical to the original DNA molecule.
The sequence of bases that should be present in the new strands can be easily predicted as they are complementary to the bases in the old strands. A will pair with T, T with A, C with G and G with C. In this way, two daughter DNA molecules identical to the parent molecule are formed, and each daughter DNA molecule consists of an old (parent) strand and a new strand. Since only one parent strand is retained in each daughter molecule, this type of replication is called semiconservative. Meselson and Stahl and Joseph Taylor later proved this in experiments.
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.
- DNA dependent RNA polymerase – synthesis.
- 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?
Alfred Hershey and Martha Chase (1952) worked with viruses that infect bacteria, the so-called bacteriophages. In 1952, they chose a bacteriophage called T2 for their experimental material.
They grew some viruses on a medium containing radioactive phosphorus (p32) and some others on a medium containing radioactive sulphur (s35). Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but no radioactive protein, since DNA contains phosphorus but protein does not. Similarly, viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA, as DNA does not contain sulphur. Radioactive phages were allowed to attach to E. coli bacteria. In the course of the infection, the viral envelopes were then removed from the bacteria by shaking them in a blender.
The virus particles were separated from the bacteria by spinning them in a centrifuge. Bacteria infected with viruses containing radioactive DNA were radioactive, indicating that the DNA was the material that passed from the virus to the bacteria. Bacteria infected with viruses containing radioactive proteins were not radioactive. This indicates that the proteins did not pass from the viruses into the bacteria. The DNA is therefore the genetic material that is passed from the virus to the bacteria.
8. Differentiate between the followings:
- Repetitive DNA and Satellite DNA
- mRNA and tRNA
- Template strand and Coding strand
Ans: (a) The main differences between repetitive DNA and satellite DNA are as following:
| 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. |
- The main difference between mRNA and tRNA are as following:
| mRNA | 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 |
- The main difference between template strand and coding strand are as follows:
| 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 a surface for the binding of mRNA in the groove of the smaller subunit of the ribosome.
(ii) Since the larger subunit of the ribosome has peptidyl transferase on its ‘P’ side, it helps in the joining of amino acids by forming peptide bonds.
- 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: The lac operon is switched on by the addition of lactose to the medium, as lactose acts as an inducer and renders the repressor inactive by binding 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 has been used up, the repressor becomes active again and ensures that the system shuts down.
- Explain (in one or two lines) the function of the followings:
- Promoter
- tRNA
- Exons
Promoter: It is one of the three components of a transcription unit involved in transcription. It is located at the 5′-end of the start and is the site where the transcription factors (TATA box) and the RNA polymerase attach.
tRNA: It is involved in the transfer of activated amino acids from the cellular pool to the ribosome so that they can participate in protein formation.
Exons: In eukaryotes, DNA consists of a mosaic of exons and introns. Exons are coding sequences of DNA that are both transcribed and translated.
12. Why is the Human genome project called a mega project?
Ans: Human genome project is called a mega project because
- It required a bioinformatics database and other high-speed computing facilities to analyse, store and retrieve information.
- It generated a lot of information in the form of sequence annotations.
- It was carried out in a number of laboratories and coordinated on a large scale.
13. What is DNA fingerprinting? Mention its application.
Ans: DNA fingerprinting or DNA typing is a technique of determining nucleotide sequences 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 fingertips and can be altered by surgery, a DNA fingerprint is the same for every cell, tissue and organ of a person. It cannot be changed by any known treatment. Applications of DNA fingerprinting are as follows:
Paternity disputes can be resolved by DNA fingerprinting.
- The DNA fingerprinting technique is used to identify genes associated with hereditary diseases.
- It is useful in the detection of crime and in law enforcement.
- It can identify racial groups, their origins, historical migrations and invasions.
14. Briefly describe the following:
Transcription
Polymorphism
Translation
Bioinformatics
Ans: Transcription: This is a DNA directed synthesis of RNA in which the RNA is transcribed on 3*—>5’ template strand of DNA in 5’—>3’ direction. Polymorphism: Variations at the genetic level caused by mutation are referred to as polymorphism. Such variations are unique to a particular site in the DNA and form satellite DNA. Polymorphism in DNA sequences is the basis of genetic mapping and DNA fingerprinting.
Translation: Protein synthesis from mRNA, tRNA, rRNA.
Bioinformatics: Computational method of handling and analysing biological databases.