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

Molecular Basis of Inheritance Exercise 109

Solution 1

Nitrogenous bases are nitrogen-containing molecules which have chemical properties similar to those of bases.

Nitrogenous bases: Adenine, Thymine, Uracil, Cytosine

Nucleosides are formed by a nitrogenous base and pentose sugar covalently bonded together by a glycosidic bond.

Nucleosides: Cytidine, Guanosine

Solution 2

begin mathsize 12px style table attributes columnalign left end attributes row cell According space straight to space straight Chargaff apostrophe straight s straight rules colon end cell row cell text Molar amount of adenine is always equal to the molar amount of thymine.  end text end cell row cell text Similarly, molar concentration of guanine is equal to the molar concentration of cytosine. end text end cell row cell therefore text  A = T and G = C end text end cell row cell because text  C = 20 ;  end text therefore text  G = 20  end text end cell row cell because text  C + G = 40;  end text end cell row cell therefore text  A + T = 100 - 40 = 60 end text end cell row cell because text  A = T;  end text therefore text  A = T = 30 end text end cell row cell text The percentage of adenine in ds DNA is 30%. end text end cell end table end style

Solution 3

The sequence of the complementary strand in the 3′→5′ direction:


Solution 4

The synthesis of RNA occurs on the template or non-coding strand of DNA. Therefore, to find out the sequence of mRNA, we need to find the sequence of the template strand.

DNA coding strand


Template strand


mRNA strand




Solution 5

Watson and Crick observed that the nitrogenous bases form a complementary pair between the two polynucleotide chains of DNA. Based on the X-ray diffraction data, they proposed that DNA consisted of a double helix of two chains with sugar phosphate on the outside and nitrogen bases on the inside.

Further, they proposed that the two chains are antiparallel with 5'-3' orientation of the other. The two chains are twisted helically just as a rope ladder with rigid steps twisted into a spiral.

This property of a double helix model of DNA led them to hypothesise the semi-conservative mode of DNA replication, where the two strands separate and act as a template for the synthesis of a new complementary strand.

Solution 6

DNA template



DNA-dependent DNA polymerase

Catalyses the polymerisation deoxynucleotides

DNA-dependent RNA polymerase

Catalyses the transcription of all types of RNA (in bacteria)

DNA-dependent RNA polymerase-I

Transcribes rRNA

DNA-dependent RNA polymerase-II

Transcribes the precursor of mRNA (hnRNA)

DNA-dependent RNA polymerase-III

Transcribes tRNA

RNA template



RNA-dependent RNA polymerase

Synthesises of RNA in some viruses

Reverse transcriptase

Synthesises cDNA

Solution 7

To confirm the constituents of genetic material, Hershey and Chase conducted the bacteriophage experiment.

Normally, T2 bacteriophage joins the walls of E. coli by means of its tail fibre which releases lysozyme to rupture the bacterial cell wall.

Some bacteriophages were grown in a medium which contained radioactive phosphorus (32P) and others were grown in a medium which contained radioactive sulphur (35S).

The bacteriophages grown in the presence of radioactive phosphorus (32P) contained radioactive DNA, because DNA contains phosphorus and not the protein. Similarly, the bacteriophages grown on radioactive sulphur (35S) contained radioactive protein but not radioactive DNA, because sulphur is a constituent of amino acids.

These two types of phages were made to infect normal bacterial cells in two samples. It was found that phages grown in radioactive phosphorus passed their radioactivity to daughter phages, while the phages containing radioactive sulphur did not transfer their radioactivity to daughter phages.

This indicates that only DNA and not the protein coat entered the bacterial cells, thereby proving that DNA is the genetic material which is passed on from one generation to another.

Solution 8 (a)

Differences between repetitive DNA and satellite DNA:

Repetitive DNA

Satellite DNA

  • DNA in which certain base sequences are repeated several times is called repetitive DNA.
  • DNA in which large portions or stretches of DNA are tandemly repeated is called satellite DNA.
  • Repetitive DNA sequences are transcribed.
  • Satellite DNA sequences are not transcribed.
  • Repetitive DNA cannot be classified further.
  • Satellite DNA can be classified further into micro-satellites and mini-satellites depending on the base composition, length of segment and the number of repetitive units.

Solution 8 (b)

Differences between mRNA and tRNA:



 It constitutes about 5% of the total RNA in the cell.

 It constitutes about 15% of the total RNA in the cell.

 It consists of about 75-6000 bases.

 It consists of about 73-93 bases.

 Its molecular weight is about 25000-20,00,000 daltons.

 Its molecular weight is about 25000 daltons.

 It carries coded information for several amino acids.

 It carries coded information for association with only one amino acid and an anticodon for its incorporation in a polypeptide.

 It is linear and never coiled.

 It is cloverleaf-shaped and folded.

 It is synthesised by RNA polymerase II in the cell.

 It is synthesised by RNA polymerase III in the cell.

 Nitrogen bases are not modified.

 Nitrogen bases may be modified.

 It undergoes additional processing, i.e. capping and tailing.

 It does not undergo any additional processing.

 It is short lived and degrades after protein syntheses.

 It is quite stable and degrades very slowly.

 It is called template or nuclear or messenger RNA as it carries genetic information provided by DNA.

 It is called soluble or adaptor RNA as it carries amino acids to mRNA during protein synthesis.



Solution 8 (c)

Differences between template strand and coding strand:

Template strand

Coding strand

 It is the strand of DNA on which transcription occurs.

 It is the strand of DNA which does not participate in transcription.

 The polarity of the strand is 3' 5'.

 The polarity of the strand is 5' 3'.

 The nucleotide sequence of the strand is complementary to the base sequence present on mRNA.

 The nucleotide sequence of the strand is the same as the base sequence present on mRNA except for the presence of thymine instead of uracil.


Solution 9

Essential role of ribosome during translation:

  • It provides a surface for the binding of mRNA in the groove of the smaller subunit of ribosome.
  • It also provides sites for the attachment of charged tRNA for polypeptide synthesis.
  • The larger subunit of ribosome has a P site which contains peptidyl transferase ribozyme (23S rRNA in prokaryotes) which acts as a catalyst for the formation of peptide bonds.

Solution 10

When lactose is added in the medium containing E. coli, the lac operon is switched on. The lactose acts as an inducer and makes the repressor inactive. On adding lactose, the enzyme beta-galactosidase is synthesised in E. coli. This enzyme hydrolyses lactose to galactose and glucose. The bacteria use these sugars as a source of energy. In the absence of lactose, the regulatory gene produces a repressor protein which binds on the operator gene and makes it inactive. This prevents RNA polymerase from transcribing the operon. So, the lac operon shuts down sometime after the addition of lactose in the medium.

Solution 11

(a) Promoter: It is one of the three components of a transcription unit which takes part in transcription. It is located upstream at the 5¢ end adjacent to the operator. It acts as an initiation signal which functions as a recognition centre for RNA-polymerase and transcription factors (TATA box) provided the operator gene is switched on.

(b) tRNA: It acts as an adaptor molecule which reads the code and  transfers amino acids from the cellular pool to the A site of mRNA embedded in the ribosomes for synthesis of polypeptides.

(c) Exons: These are the coding or expressed regions in eukaryotic genes present in the primary transcript. They undergo splicing and join to become a part of mRNA during transcription. They code for different regions of the amino acid.

Solution 12

The Human Genome Project is called a mega project because

  • It aimed to sequence more than 3 × 109 base pairs in the entire human genome.
  • It targeted the identification of all the genes and their alleles present in the human genome.
  • It incurred a high expenditure of more than 9 billion dollars.
  • It required bioinformatics data basing and other high speed computational devices for analysis, storage and retrieval of information.
  • It developed ways to map the human genome with high levels of precision.
  • The magnitude and the requirements for this project opened new areas and avenues for people.

Solution 13

DNA fingerprinting is a method of identifying DNA by locating the differences in the arrangement of nucleotides in those specific regions of the DNA sequence which are repeated several times.

Applications of DNA fingerprinting:

  • To identify criminals or potential suspects in forensic laboratories and relieving people wrongly accused of crimes.
  • To determine the real parents of a given offspring and resolve paternity disputes.
  • To determine the actual parents of a lost child.
  • To infer blood relationships in the members of the same family.
  • To determine the sex of badly damaged bodies of accident victims or of archaeological specimens.

Solution 14 (a)

  • Transcription is the process of copying the genetic information coded in the DNA into an mRNA molecule.
  • The transcribed RNA moves out of the nucleus to the ribosomes in the cytoplasm to direct protein synthesis.
  • Transcription occurs in the nucleus during the G1 and G2 phases of the cell cycle during interphase.

Transcription unit:

  • The segment of the DNA template which transcribes RNA is called the transcription unit. It comprises three regions—promoter, structural gene and terminator.
  • Promoter region: The promoter sequence of the nitrogenous bases provides the binding site for RNA polymerase for the initiation of transcription. It also helps in the differentiation of the template and coding strands of DNA.
  • Terminator region: The termination sites have two or more stretches of G ≡ C pairs arranged as inverted repeats followed by a string of adenine (poly A) which signals the end of transcription.
  • Structural gene: It is a segment of DNA with a specific sequence of nucleotides which code for a protein or polypeptide required for a morphological or functional trait of the cell.

Mechanism of transcription:

Event of transcription


Activation of ribonucleotides

  • The ribonucleotides present in the nucleoplasm are converted to active triphosphates by phosphorylation in the presence of ATP energy.

Recognition of promoter region and binding of RNA polymerase to promoter region on DNA


  • On receiving a signal from the cytoplasm, the histones associated with the DNA double helix from the gene to be transcribed are removed, exposing the base sequence in this region of DNA.
  • The sigma factor of RNA polymerase enzyme recognises the consensus sequence in the promoter region of DNA.
  • RNA polymerase binds to the promoter in the template strand of the DNA double helix.

Exposure of DNA bases and initiation of transcription

  • RNA polymerase causes local unwinding and separation of two strands of DNA and exposes the bases for the initiation of transcription.
  • The separation begins from the middle of the Pribnow box and produces a transcription bubble.
  • The 3’→ 5′ strand of DNA functions as the template strand.

Base pairing


  • The ribonucleotide triphosphate molecules ATP, GTP, CTP and UTP start binding to the nucleotides of the template strand of DNA from the initiation point onwards according to the base pairing rule. This is catalysed by the enzyme RNA polymerase.
  • RNA polymerase initiates and causes polymerisation, i.e. elongation of the polynucleotide chain in the 5’→ 3′ direction.

Conversion of ribonucleotide triphosphates to ribonucleoside monophosphates


  • After linking to the DNA template, various ribonucleotide triphosphates break off their high energy bonds.
  • This converts them to ribonucleoside monophosphate with the release of pyrophosphate groups (P~P).
  • Pyrophosphate undergoes hydrolysis by the enzyme pyrophosphatase, releasing energy.

Formation and elongation of RNA polynucleotide chain


  • In the presence of energy so released, each ribonucleotide monophosphate joins the ribonucleotide arrived earlier, thereby elongating the RNA chain.
  • RNA polymerase catalyses the formation of phosphodiester bonds between the successive nucleotides in the presence of divalent ions, Mg++ and Mn++.

Termination of transcription and separation of RNA chain


  • As transcription proceeds, the hybrid DNA–RNA molecule dissociates partly, freeing the RNA molecule under synthesis.
  • When the polymerase reaches a terminator signal on DNA, it leaves DNA. The fully formed RNA chain is now released.

Solution 14 (b)

 Polymorphism means variation at the genetic level which arises due to mutations.

 The sequences which do not code for any protein constitute the bulk of the human genome and exhibit plenty of polymorphism.

 DNA from tissues such as skin, blood, bone, saliva and hair follicles represents the same type of polymorphism.

 The polymorphism in a DNA sequence is the basis of genetic mapping of the human genome and DNA fingerprinting.

 The allele sequence variation is called DNA polymorphism if more than one variant at a locus occurs in the human population.

 The probability of such variation to be observed in a non-coding DNA sequence is higher as mutations in these sequences may not have any immediate effect.

 These mutations keep on accumulating in each generation and form a basis for polymorphism. It has an important role in evolution and speciation.

Major types of genetic polymorphism:

  • Allelic polymorphism: It occurs because of multiple alleles of a gene. The alleles possess different mutations which alter the structure and function of a protein formed by them resulting in a change in phenotype.
  • Single nucleotide polymorphism (SNP): It occurs because of a difference in a single base in the nucleotide sequences of individuals. SNPs occur once in every 1000 base pairs in the genome. They are very useful in locating alleles, identifying disease-related sequences and tracing human history.
  • Restriction Fragment Length Polymorphism (RFLP):  The specificity of restriction endonucleases allows them to cut the double-stranded DNA only at their recognition sites. This produces variable lengths of homologous DNA molecules containing minor sequence variations or polymorphisms called restriction fragment length polymorphism. The cleaved fragments in the digested DNA can be separated by electrophoretic techniques. RFLP is an important tool in genetic fingerprinting, identification of genes for genetic disorders, genome mapping, determination of risk of disease and paternity testing.

Solution 14 (c)

  • Translation is the process of polymerisation of amino acids to form a polypeptide chain.
  • DNA acts as the repository of genetic information. The mRNA serves as a link between the DNA and polypeptide chain. It is called the one-way flow of information or central dogma.

Raw materials for protein synthesis: 

  • Ribosomes, amino acids, mRNA, tRNA, enzymes, ATP, GTP, soluble protein initiation and transfer factors, and various inorganic cations.

Mechanism of translation:

Activation of amino acids

 The enzyme aminoacyl-tRNA synthetase catalyses the union of amino acid and ATP to form a complex aminoacyl adenosine monophosphate or aminoacyl adenylate (AMP ~ AA) enzyme complex.

Charging of tRNA

 The enzyme-bound activated amino acids attach to the 3' end of their specific tRNA molecules and form aminoacyl tRNA in the presence of the enzyme aminoacyl-tRNA synthetase.

 The aminoacyl tRNA complexes are taken to the ribosomes for protein synthesis.

Formation of polypeptide chain

 The initiation factors IF1, IF2 and IF3, GTP, mRNA and 30S subunit collectively constitute the 30S preinitiation complex.

 The anticodon of tRNAf-Met base pairs with the initiation codon AUG, the first codon on the 5' end of the mRNA. This completes the formation of the 30S preinitiation complex.

 The 30S preinitiation complex joins to the 50S ribosomal subunit, producing the 70S complex. The energy required for this process is provided by the hydrolysis of GTP. Mg2+ ions are needed during this process.

 In the 70S initiation complex, the f-Met-tRNAfMet occupies the P site on the 50S ribosomal subunit and the anticodon on f-Met-tRNA pairs with the initiation codon on AUG in RNA.

 When the initiation codon is occupied by tRNA carrying formylated methionine, the second codon of mRNA lies close to the A site.

 The second aminoacyl tRNA with anticodon corresponding to the second codon binds with it and occupies the A site of the ribosome.

Elongation of polypeptide chain

 After the formation of the 70S mRNA-f-Met-tRNAf-Met complex, the elongation of the polypeptide chain begins by the regular addition of amino acids in the following steps:

  • Binding of AA-tRNA at the A site of the larger subunit of the ribosome: Incoming aminoacyl-tRNA complex (AA-tRNA) attaches to the acceptor, A site and base pairs with the mRNA codon present in the A site.
  • Formation of the peptide bond: The enzyme peptidyl synthetase or peptidyl transferase catalyses the formation of the peptide bond between the carboxyl group (-COOH) of the first amino acid and the amino group (-NH2) of the second amino acid present on the P- and A-sites of the ribosome, respectively.
  • Translocation: The enzyme translocase shifts the ribosome on mRNA by one codon so that the tRNA at the A site carrying the dipeptide moves to the P site and the next codon on mRNA is available at the A site. 

Termination of translation

 Elongation terminates when the A site of the ribosome reaches either of the termination codons-UGA, UAG and UAA. There are no tRNAs to recognise these termination codons.

 A releasing factor (RF) is required for the separation of the polypeptide chain from the terminal tRNA.

 The free ribosome dissociates into two subunits with the help of the dissociation factor (DF) or RF3.

Protein folding

 The unfolded primary structure of the polypeptide assumes a secondary or tertiary three-dimensional structure.

 Special proteins called molecular chaperones assist in correct folding.


Solution 14 (d)

 Bioinformatics is a combination of molecular biology, information technology and computer science.

 It stores huge information of genomics as databases, analyses, models and provides various aspects of biological information, especially related to genomics and proteomics.

Aims of bioinformatics:

 To spread scientifically investigated knowledge for the benefit of the research community.

 To transform biological sequences into sequences in the form of digital symbols and store them as databases.

 To develop a variety of methods and software tools for data analysis and interpretation.

Components of bioinformatics:

 Computational biology: It is concerned with the development of algorithmic tools to develop DNA sequences of various organisms, structure and function of RNA and protein sequences and clustering of protein sequences. It aims to align similar proteins and generate phylogenetic trees to examine evolutionary relationship.

 Bioinformation infrastructure: It includes development of the entire array of information management systems, analytical tools and communication networks which support biology.

Applications of bioinformatics:

 It enables grouping organisms and establishing phylogeny by comparing gene, nucleotide, protein and amino acid sequences.

 It provides information about the chemical combination, effects and side-effects of pharmaceutical drugs.

 It helps in the development of computer-aided drug design through the analysis of drug-ligand complexes and assessment of binding interactions.

• It is useful in diagnostics, healthcare and drug research.

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