NCERT Class 12 Biology Chapter 6 Notes Molecular Basis of Inheritance

NCERT Class 12 Biology Chapter 6 Notes Molecular Basis of Inheritance

Edited By Irshad Anwar | Updated on Jan 25, 2024 06:13 PM IST

NCERT Class 12 Biology Chapter 6 Notes explain that over the years after Mendel, the nature of the genetic material was investigated, resulting in the realisation that DNA is the genetic material in the majority of organisms. Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are the two types of nucleic acids found in living systems. Nucleic acids are polymers of nucleotides. CBSE notes for Class 12 Biology are helpful for CBSE board exam preparation. let’s go through the CBSE Notes for Class 12 Molecular Basis of Inheritance.

  • CBSE Class 12 revision notes for Molecular Basis of Inheritance are a great resource for information.
  • Molecular Basis of Inheritance Class 12 Revision Notes will also help you prepare for competitive exams like NEET.
  • Molecular Basis of Inheritance CBSE Notes for Class 12 Biology will boost your knowledge and interest in biology.
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Molecular Basis of Inheritance Class 12 Notes

In class 12 Biology chapter 6 notes, you will study that DNA acts as genetic material in most organisms, whereas RNA acts as genetic material in some viruses. RNA mostly functions a messenger. You will also study in Molecular Basis of Inheritance Class 12th Notes, that RNA has other functions as an adapter, structural, or as catalytic molecule. Along with NCERT notes, students can use NCERT solutions to exercise problems and CBSE previous year papers for board exam preparation.

NCERT Class 12 Biology Chapter 6 Notes: What Is DNA?

DNA, or deoxyribonucleic acid, is a molecule that carries most of the genetic instructions used in the development, functioning, and reproduction of all known living organisms and many viruses. It is a long, double-stranded helical structure located in the cell nucleus of eukaryotic organisms and in the cell's cytoplasm in prokaryotes (bacteria and archaea).

As per the central dogma of molecular biology, genetic information flows from DNA to RNA to protein.

The structure of DNA is often described as a double helix, resembling a twisted ladder. Each "rung" of the ladder consists of two nucleotide units, and these nucleotides are the building blocks of DNA. Each nucleotide is composed of a sugar molecule (deoxyribose), a phosphate group, and one of four nitrogenous bases: adenine (A), thymine (T), cytosine (C), and guanine (G).

Structure of Polynucleotide Chain

(i) A nucleotide has three parts, i.e. a nitrogenous base, a pentose sugar (deoxyribose in DNA and ribose in RNA), and a phosphate group.

(ii) Nitrogenous bases are purines, i.e. adenine, guanine and pyrimidines, i.e. cytosine, uracil and thymine.

(iii) Cytosine is common for both DNA and RNA and thymine is present in DNA. Uracil is present in RNA at the place of thymine.

(iv) A nitrogenous base is linked to the pentose sugar through an N-glycosidic linkage to form a nucleoside, i.e. adenosine and guanosine, etc.

(v) When a phosphate group is linked to 5′ —OH of a nucleoside through phosphodiester linkage, a corresponding nucleotide is formed.

(vi) Two nucleotides are linked through 3′ -> 5′ phosphodiester linkage to form a dinucleotide.

(vii) Several nucleotides can be joined to form a polynucleotide chain.

Molecular Basis of Inheritance class 12 notes, says that in the case of RNA, every nucleotide residue has an additional—OH group present at 2-position in the ribose. Also, the uracil is found at the place of thymine (5-methyl uracil).

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Class 12 Biology Chapter 6 Notes: Discoveries Related to Structure of DNA

(i) Friedrich Meischer in 1869, first identified DNA as an acidic substance present in the nucleus and named it as ‘nuclein’.

(ii) James Watson and Francis Crick, proposed a very simple double helix model for the structure of DNA in 1953 based on X-ray diffraction data.

(iii) Erwin Chargaff proposed that for a double-stranded DNA, the ratios between adenine and thymine and guanine and cytosine are constant and equals to one.

Salient Features of Double-helix Structure of DNA

(i) DNA is a long polymer of deoxyribonucleotides. It is made up of two polynucleotide chains, where the backbone is constituted by sugar-phosphate and the bases project inside.

(ii) In Molecular Basis of Inheritance Class 12th Notes, it is mentioned that the two chains have anti-parallel polarity, i.e. 5′ > 3′ for one, 3′ > 5′ for another.

(iii) The bases in two strands are paired through hydrogen bonds (H—bonds) forming base pairs (bp). Adenine forms two hydrogen bonds with thymine from the opposite strand and vice-versa. Guanine bonds with cytosine by three H—bonds. Due to this, purine always comes opposite to pyrimidine. This forms a uniform distance between the two strands.

(v) The plane of one base pair stacks over the other in the double helix. This confers stability to the helical structure in addition to H—bonds.

The length of a DNA double helix is about 2.2 meters, Therefore, it needs special packaging in a cell.


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Histones: They are organized to form a unit of eight molecules called histone octamer. The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called a nucleosome. A typical nucleosome contains 200 bp of DNA helix.

NCERT Class 12 Biology Chapter 6 Notes: Nucleosomes constitute the repeating unit of a structure in nucleus called chromatin (thread-like stained structure). Under electron microscope, the nucleosomes in chromatin can be seen as beads-on-string. This structure in chromatin is packaged to form chromatin fibres that further coils and condense to form chromosomes at metaphase stage.

The packaging of chromatin at higher level requires additional set of proteins which are collectively called Non-Histone Chromosomal (NHC) proteins.

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CBSE class 12 Biology chapter 6 notes, says that in a nucleus, some regions of chromatin are loosely packed (stains light) called euchromatin (transcriptionally active chromatin). In some regions, chromatin is densely packed (stains dark) called heterochromatin (inactive chromatin).

Molecular Basis of Inheritance Class 12th Notes: DNA is the Genetic Material

Alfred Hershey and Martha Chase (1952) gave unequivocal proof that DNA is the genetic material.

In this experiments, bacteriophages (viruses that infect bacteria) were used. They grew some viruses on a medium that contained radioactive phosphorus and some others on sulfur containing radioactive medium.

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NCERT Class 12 Biology Chapter 6 Notes: Radioactive phages were allowed to attach to E. coli bacteria. As the infection proceeded, 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 that were infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria and bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicated that the proteins did not enter the bacteria from viruses.

Hence, it proved that DNA is a genetic material that is passed from virus to bacteria.

Molecular Basis of Inheritancenotes Class 12: Properties of Genetic Material

Properties of Genetic Material is given below:

(i) It became established that DNA is the genetic material from the Hershey-Chase experiment.

(ii) class 12th Biology chapter 6 notes clarifies that in some viruses, RNA was also reported as genetic material, e.g. Tobacco mosaic viruses, QB bacteriophage, etc.

(iii) Characteristics of a Genetic Material

  • (a) It should be able to replicate.

  • (b) It should be chemically and structurally stable.

  • (c) It should provide scope for slow changes (mutation) that are required for evolution.

  • (d) It should be able to express itself in the form of ‘Mendelian characters’.

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(iv) According to the above-mentioned rules, both the nucleic acids (DNA and RNA) have the ability to direct duplications.

Stability can be explained in DNA as the two strands being complementary if separated by heating come together in appropriate conditions.

In Molecular Basis of Inheritance Class 12th Notes, Francis Crick proposed that the central dogma in molecular biology, which states that the genetic information flows from DNA to RNA to Protien.

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CBSE Class 12 Biology Molecular Basis of Inheritance Notes: Replication Of DNA

DNA replication was proposed by Watson and Crick (1953) and termed semiconservative. According to Them:

(i) The two strands would separate and act as a template for the synthesis of new

complementary strands.

(ii) After replication, each DNA molecule would have one parental and one newly synthesized strand.

In NCERT Class 12 Biology Chapter 6 Notes, Experimental proof says that DNA replicates semi conservatively, comes first from E. coli and later from higher organisms, such as plants and human cells.

Matthew Meselson and Franklin Stahl performed the following experiments to prove this in 1958.

E. coli was grown in a medium containing 15NH4C1 as the only nitrogen source for many generations. 15N got incorporated into newly synthesized DNA (and other nitrogen-containing compounds). This heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient.

In class 12 Molecular Basis of Inheritance notes, it is mentioned that the researchers then transferred the cells into a medium containing 14NH4Cl, took samples at various intervals as the cells multiplied, and extracted the DNA that remained as double-stranded helices. DNA samples were separated independently on CsCl gradients to measure DNA densities.

NCERT Class 12 Biology Chapter 6 Notes, says that the DNA that was extracted from the culture, one generation (after 20 min) after the transfer from 15 N to 14N medium had a hybrid or intermediate density. DNA extracted from the culture after another generation (after 40 min) was composed of equal amounts of this hybrid DNA and of light DNA.

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Replication in DNA strand occurs within a small opening of the DNA helix, known as the replication fork.

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Biology Notes for Class 12 Molecular Basis of Inheritance: Origin of Replication

DNA polymerases cannot initiate the process of replication on their own. Also, replication does not initiate randomly at any place in DNA. So, there is a definite region in E.coli DNA where the replication originates. The region is termed as the origin of replication.

In NCERT class 12 Biology chapter 6 notes, There are the following three types of RNAs:

(i) mRNA (messenger RNA) provides the template for transcription.

(ii) tRNA (transfer RNA) brings amino acids and reads the genetic code.

(iii) rRNA (ribosomal RNA) plays a structural and catalytic role during translation.

All three RNAs are needed to synthesize a protein in a cell.

NCERT Class 12 Biology Chapter 6 Notes: Transcription:

In transcription, only a segment of DNA is duplicated and on Iv one of the strands is copied into RNA. Both the strands are not copied because of these reasons:

• If both the strands code for RNA, two different RNA me’ ^cules and two different proteins would be formed, hence complicating the genetic information transfer machinery.

• Since two RNA produced would be complementary to each other, they would form a double-stranded RNA without translation, making the process of transcription futile.

A transcription unit in DNA is defined by three regions in the DNA which are as follows:

  • A promoter

  • The structural gene

  • A terminator

Transcription Unit and the Gene

  • A gene can be defined as the functional unit of inheritance.

  • NCERT Class 12 Biology Chapter 6 Notes says that a cistron is a segment of DNA coding for a polypeptide.

  • The structural gene in a transcription unit could be said as monocistronic (mostly in eukaryotes) or polycistronic (mostly in bacteria or prokaryotes).

  • The coding sequences or expressed sequences are defined as exons. Exons appear in mature or processed RNA. The exons are interrupted by introns.

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  • Introns or intervening sequences do not appear in mature or processed RNA.

Transcription in Prokaryotes

Transcription in prokaryotes occur in the following steps:

  • According to that Molecular Basis of Inheritance Class 12th Notes, single DNA-dependent RNA polymerase catalyzes the transcription of all types of RNA in bacteria.

  • RNA polymerase binds to promoter and initiates transcription (initiation).

  • It uses nucleoside triphosphates as substrate and polymerizes in a template dependent fashion following the rule of complementarity.

  • It also facilitates the opening of the helix and continues elongation.

  • Once the polymerase reaches the terminator region, the nascent RNA falls off, so also the RNA polymerase. This results in the termination of transcription.

  • RNA polymerase is only capable of catalyzing the process of elongation.

  • It associates transiently with initiation-factor

    • (a) and terminator factor

    • (b), to initiate and terminate the transcription, respectively. Thus, catalyzing all three steps.

In notes for class 12 Biology chapter 6, it is mentioned that since, the mRNA does not require any processing to become active and also since transcription and translation take place in the same compartment, many times the translation can begin much before the mRNA is fully transcribed. As a result, transcription and translation can be coupled in bacteria.

Process of Transcription in Eukaryotes


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In NCERT Class 12 Biology Chapter 6, the Significance of these complexities are:

(i) The split gene arrangements represent an ancient feature of the genome.

(ii) The presence of introns is reminiscent of antiquity.

(iii) The process of splicing represents the dominance of the RNA world

You can also read this Molecular Basis of Inheritance class 12 notes pdf and download them for free.

Chapter-wise Biology Notes For Class 12

CBSE Class 12 Biology notes are provided for 16 chapters. Along with notes, students can use NCERT solutions to Solve problems and CBSE previous year papers for board exam preparation.

Below are the links to the Biology Class 12 notes PDF download for all chapters:

Subject-wise NCERT Exampler solutions

Subject wise NCERT solutions

Frequently Asked Questions (FAQs)

1. What is a DNA Helix?

A DNA helix is a twisted ladder-like structure composed of two long strands of nucleotides. These strands are held together by hydrogen bonds between complementary pairs of nitrogenous bases (adenine with thymine, and cytosine with guanine), forming the double-stranded structure of DNA.

2. What is regulation of gene expression or gene regulation?

Gene expression regulation refers to the control of the synthesis of proteins or functional RNA molecules from a gene. It involves a complex interplay of various molecular mechanisms that can enhance or suppress the transcription and translation processes, ultimately influencing the level of gene activity within a cell. This regulation is crucial for maintaining proper cellular functions and responding to environmental changes.

3. Why DNA replication is semi-conservative?

DNA replication is semi-conservative because each newly synthesized DNA molecule contains one strand from the parent template DNA and one newly synthesized complementary strand. 

During replication, the DNA double helix unwinds, and each separated strand serves as a template for the synthesis of a new strand. This process ensures that genetic information is accurately passed on to the next generation with each newly replicated DNA molecule conserving one "old" strand and one newly synthesized strand. This mechanism was first demonstrated by the classic Meselson-Stahl experiment in 1958.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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