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

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NCERT Solutions for Class 12 Biology

Page No. 125


Q1. Group the following as nitrogenous bases and nucleosides:
Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.

Answer: Nitrogenous Bases – Adenine, Uracil and Cytosine, Thymine; Nucleosides – Cytidine, guanosine.

Q2. If a double-stranded DNA has 20 percent of cytosine, calculate the percentage of adenine in DNA.

Answer: As per Chargaff’s rule, DNA molecules are required to have an equal ratio of purine(adenine and guanine) and pyridine(cytosine and thymine). This is to say that the number of adenine molecules is equivalent to the cytosine molecule.

Percentage of adenosine = percentage of thymine,

Percentage of guanine = percentage of cytosine

Hence, according to the law, if the double stranded DNA has 20% of cytosine, it should have 20% of guanine. Therefore, the percentage of G + percentage of C = 40%

The other 60% indicates both A + T percentage molecule. As adenine and thymine is always found in equal numbers, the adenine content is 30%.

Q3. If the sequence of one strand of DNA is written as follows:

Write down the sequence of the complementary strand in the 3′ → 5′ direction.


Q4. If the sequence of a coding strand in a transcription unit is written as follows:

Write down the sequence of m-RNA.

Answer: If the sequence of coding strand is :
Then template strand is :
The mRNA will be formed on the template strand in 5′ —> 3’ direction. Thus mRNA sequence will be:
Thymine in DNA is substituted by uracil in RNA.

Q5. Which property of DNA double helix led Watson and Crick to hypothesise the semi-conservative mode of DNA replication? Explain.

Answer: Watson and Crick observed that the two strands of DNA are anti-parallel and complementary to each other with respect to their base sequences. This type of arrangement in DNA molecule led to the hypothesis that DNA replication is semi-conservative. It means that the double stranded DNA molecule separates and then, each of the separated strand acts as a template for the synthesis of a new complementary strand. As a result, each DNA molecule would have one parental strand and a newly synthesized daughter strand.

Since only one parental strand is conserved in each daughter molecule, it is known as semi-conservative mode of replication.

Q6. 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.

Answer: There are two different types of nucleic acid polymerases.

(1) DNA-dependent DNA polymerases
(2) DNA-dependent RNA polymerases

The DNA-dependent DNA polymerases use a DNA template for synthesizing a new strand of DNA, whereas DNA-dependent RNA polymerases use a DNA template strand for synthesizing RNA

Q7. 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 (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.

Q8. Differentiate between the followings:

(a) Repetitive DNA and Satellite DNA


Repetitive DNASatellite DNA
1) It is the non-coding DNA with multiple copies of identical sequences which may lie in tandem or interspersed.
2)These can be few base pairs to hundreds or thousands of base pairs.
3) It appears as light bands.
1) It refers to non-coding tandem repeat sequences.
2)These are generally short sequence repeats (up to 60 base pair long).
3)It appears as small dark bands.

(b) mRNA and tRNA


1) It is called messenger RNA and carries the codes for amino acid sequence.
2) It is a linear molecule.
3) It is synthesised by RNA polymerase II
1) It is called transfer RNA as it carries amino acids to the site of protein synthesis.
2) It has clover leaf shape.
3) It is synthesised by RNA polymerase III.

(c) Template strand and Coding strand


Template strandCoding strand
1) It is the strand which is transcribed into RNA.
2) It is called anti sense strand.
3) It has 3’→ 5′ polarity.
1) It has the same sequence as mRNA.
2) It is called sense or non-template strand.
3) It has 5’→ 3′ polarity.

Q9. List two essential roles of ribosome during translation.

Answer: Two essential roles of ribosome during translation are:

  • Ribosomes are sites where synthesis of proteins occurs from individual amino acids. It consists of two subunits – larger subunit serves as an amino acid binding site whereas smaller subunit attaches to the mRNA forming a protein synthesizing complex
  • Since large subunit of ribosome has two different sites to attach to tRNA, it facilitates amino acids to come closer for peptide bond formation. Also, ribosome behaves as a catalyst for the formation of peptide bond. Example – 23s r-RNA acts as a ribozyme in bacteria

Q10. In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then why does lac operon shut down sometime after the addition of lactose to the medium?

Answer: The lac operon is regulated by the amount of lactose in the medium where the bacteria are grown. When the amount of lactose is exhausted in the medium, the lac operon shuts down

Q11. Explain (in one or two lines) the function of the followings :

(a) Promoter

Answer: Promoter – Promoter is a region of DNA that helps in initiating the process of transcription. It serves as the binding site for RNA polymerase.

(b) tRNA

Answer: tRNA – tRNA or transfer RNA is a small RNA that reads the genetic code present on mRNA. It carries specific amino acid to mRNA on ribosome during translation of proteins.

(c) Exons

Answer: In eukaryotes, these are coding sequences of DNA which transcribe proteins. In between, exons comprise of long non-coding sections of DNA known as introns.

Q12. Why is the human genome project called a mega project?

Answer: Human genome project was considered to be a mega project because it had a specific goal to sequence every base pair present in the human genome. It took around 13 years for its completion and got accomplished in year 2006. It was a large scale project, which aimed at developing new technology and generating new information in the field of genomic studies. As a result of it, several new areas and avenues have opened up in the field of genetics, biotechnology, and medical sciences. It provided clues regarding the understanding of human biology.

Q13. What is DNA fingerprinting? Mention its application.

Answer: The technique which is used to identify and analyse the variation in DNA in every individual is known as DNA fingerprinting.

The various applications of DNA fingerprinting are as follow:

1) In forensic science, it is used for identifying potential crime suspects.
2) It is used for finding out paternity and family relationships.
3) It is used for the identification and protection of commercial crop varieties and livestock.
4) It is used to find out the evolutionary relationship and linkage between the various

Q14. Briefly describe the following:

(a) Transcription

Answer: Transcription: The process of copying genetic information from one strand of the DNA into RNA is known as transcription. RNA is assembled simply based on complementarity of the DNA strand, only uracil is substituted in place of thymine. Only a small segment of DNA that codes for a polypeptide is copied.

(b) Polymorphism

Answer: It is a form of genetic variation wherein different nucleotide sequences can be present at different sites in a molecule of DNA. There is a high frequency of this heritable mutation to be observed in a population which emerges as a result of mutation either in the germ cells or somatic cells. The germ cell mutation can be passed from parents to the offsprings which leads to the accumulation of different mutations in a population causing variation and polymorphism in the population. This has a crucial role to play in the evolution and speciation process.

(c) Translation

Answer: Translation: It refers to the process of polymerisation of amino acids to form a polypeptide. The order and sequence of amino acids are defined by the sequence of bases in the mRNA. It occurs in cytoplasm in both prokaryotes and eukaryotes.

(d) Bioinformatics

Answer: It is the application of statistical and computational methodologies to the molecular biology stream. It explains practical issues that arise from the analysis and management of biological data. The stream of bioinformatics developed after accomplishing human genome project as huge amount of data was produced during the process which has to be stored and managed for easy access and analysis for later use.

Thus, bioinformatics includes creating biological databases which stores huge data of biology. It comes up with a few tools for efficient and easy access to data which can be used. Also, bioinformatics brought in new algorithms and statistical techniques to figure out the dynamics between data, to predict structure of protein and their functionalities and to group protein sequences into their associated families.

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