CBSE · NCERT · Class 12 Biology · Chapter 5

NCERT Solutions: Class 12 Biology Chapter 5 - Molecular Basis of Inheritance

14 textbook Q&A14 verifiedFree Content

Chapter-wise NCERT intext questions and exercise answers for Molecular Basis of Inheritance, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercises 14
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1Exercises14 questions
Q.1Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.v
Solution

Nitrogenous bases are purines or pyrimidines. A nucleoside is a nitrogenous base linked to a pentose sugar. Adenine, thymine, uracil and cytosine are bases; cytidine and guanosine are nucleosides.

Answer:

Nitrogenous bases: Adenine, Thymine, Uracil and Cytosine. Nucleosides: Cytidine and Guanosine.

Q.2If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.v
Solution

In double-stranded DNA, C pairs with G and A pairs with T. If cytosine is 20%, guanine is also 20%, so C + G = 40%. The remaining 60% is A + T. Since A = T, adenine = 60% / 2 = 30%.

Answer:

The DNA has 30 per cent adenine.

Q.3If 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.v
Solution

The given strand is 5'-ATGCATGCATGCATGCATGCATGCATGC-3'. Its complementary strand is antiparallel: 3'-TACGTACGTACGTACGTACGTACGTACG-5'. Written in the 5' to 3' direction, this becomes 5'-GCATGCATGCATGCATGCATGCATGCAT-3'.

Answer:

The complementary strand in 5' to 3' direction is 5'-GCATGCATGCATGCATGCATGCATGCAT-3'.

Q.4If the sequence of the coding strand in a transcription unit is written as follows: 5' -ATGCATGCATGCATGCATGCATGCATGC-3' Write down the sequence of mRNA.v
Solution

The mRNA has the same sequence as the coding strand except that uracil (U) replaces thymine (T). Therefore, 5'-ATGCATGCATGCATGCATGCATGCATGC-3' gives 5'-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3'.

Answer:

The mRNA sequence is 5'-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3'.

Q.5Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.v
Solution

Because A pairs with T and G pairs with C, each strand contains enough information to direct synthesis of its complementary strand. If the two parental strands separate, each can act as a template. The two daughter DNA molecules therefore each contain one parental strand and one newly synthesised strand, which is semi-conservative replication.

Answer:

The complementary base pairing of the two antiparallel DNA strands led Watson and Crick to hypothesise semi-conservative replication.

Q.6Depending 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.v
Solution

A polymerase is named by the template it reads and the product it synthesises. DNA to DNA synthesis is done by DNA-dependent DNA polymerase. DNA to RNA is done by DNA-dependent RNA polymerase. RNA to RNA is done by RNA-dependent RNA polymerase. RNA to DNA is done by RNA-dependent DNA polymerase, also called reverse transcriptase.

Answer:

The types are DNA-dependent DNA polymerase, DNA-dependent RNA polymerase, RNA-dependent RNA polymerase and RNA-dependent DNA polymerase.

Q.7How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?v
Solution

DNA contains phosphorus but not sulfur, so phages grown with radioactive phosphorus had labelled DNA. Proteins contain sulfur but DNA does not, so phages grown with radioactive sulfur had labelled protein. After infection, radioactivity from labelled DNA entered bacteria, while labelled protein remained outside, proving DNA was the genetic material.

Answer:

They labelled DNA with radioactive phosphorus and protein with radioactive sulfur.

Q.8Differentiate between the followings: (a) Repetitive DNA and Satellite DNA (b) mRNA and tRNA (c) Template strand and Coding strandv
Solution

Repetitive DNA is a broad term for repeated DNA sequences. Satellite DNA is repetitive DNA separated from bulk genomic DNA as smaller peaks and includes micro-satellites and mini-satellites. mRNA is the message used for protein synthesis, while tRNA has an anticodon loop and an amino acid attachment site. During transcription, RNA polymerase reads the template strand in 3' to 5' direction; the coding strand is not transcribed and is similar to the RNA product in sequence.

Answer:

(a) Repetitive DNA is DNA with sequences repeated many times; satellite DNA is a class of repetitive DNA that forms separate peaks in density-gradient centrifugation. (b) mRNA carries codon information from DNA to ribosomes; tRNA acts as an adapter that brings amino acids and reads codons through anticodons. (c) The template strand is transcribed into RNA; the coding strand has the same sequence as mRNA except T in place of U.

Q.9List two essential roles of ribosome during translation.v
Solution

During translation, the ribosome binds mRNA and positions charged tRNAs so that codon-anticodon pairing can occur. The larger ribosomal subunit also contains peptidyl transferase activity, an rRNA-catalysed reaction that forms peptide bonds between amino acids.

Answer:

The ribosome provides the site for mRNA-tRNA interaction and catalyses peptide bond formation.

Q.10In 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?v
Solution

Lactose, through allolactose, binds the repressor and prevents it from blocking the operator. This allows transcription of lac genes. After lactose is metabolised, the inducer level falls, the repressor again binds the operator, and transcription of the lac operon stops.

Answer:

The lac operon shuts down after lactose is consumed because the inducer is no longer available to inactivate the repressor.

Q.11Explain (in one or two lines) the function of the followings: (a) Promoter (b) tRNA (c) Exonsv
Solution

The promoter controls initiation and strand selection in transcription. tRNA works as an adapter molecule during translation. Exons are expressed sequences; after introns are removed, exons are joined to form functional RNA.

Answer:

(a) A promoter is the DNA site where RNA polymerase binds to start transcription. (b) tRNA carries a specific amino acid and matches it to an mRNA codon through its anticodon. (c) Exons are coding sequences that remain in mature RNA after splicing.

Q.12Why is the Human Genome project called a mega project?v
Solution

NCERT notes that if sequencing cost was estimated at US $3 per base, the total cost would be about 9 billion dollars. The sequence information from one human cell would fill thousands of books if printed. The project also required high-speed computational tools, databases, sequencing technology and ethical, legal and social planning. It ran as a 13-year international project completed in 2003.

Answer:

The Human Genome Project is called a mega project because it aimed to sequence about 3 billion human DNA base pairs, identify thousands of genes, store and analyse enormous data, and required international collaboration, advanced technology and huge funding.

Q.13What is DNA fingerprinting? Mention its application.v
Solution

The technique detects polymorphism in repetitive DNA such as VNTRs. Because these patterns are highly variable among individuals except identical twins, they can match biological samples to persons, resolve parentage disputes and study genetic diversity.

Answer:

DNA fingerprinting is a technique for identifying variation in specific repetitive DNA regions among individuals. It is used in forensic identification, paternity testing, and studies of genetic and population diversity.

Q.14Briefly describe the following: (a) Transcription (b) Polymorphism (c) Translation (d) Bioinformaticsv
Solution

In transcription, RNA polymerase uses the template DNA strand to produce RNA complementary to it. Polymorphism arises due to mutations; when an inheritable variant occurs in a population at a frequency greater than 0.01, it is called DNA polymorphism. In translation, ribosomes, mRNA and tRNAs assemble amino acids according to the genetic code. Bioinformatics developed rapidly with genome projects because huge sequence datasets require computational storage and analysis.

Answer:

(a) Transcription is synthesis of RNA from a DNA template. (b) Polymorphism is heritable genetic variation present in a population at appreciable frequency. (c) Translation is synthesis of a polypeptide from the codon sequence of mRNA. (d) Bioinformatics is the use of computational tools to store, retrieve and analyse biological data.