Molecular Basis of Inheritance Class 12 Biology NCERT Textbook PDF

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NCERT Class 12 Biology Textbook Chapter 6 With Answer PDF Free Download

Molecular Basis of Inheritance

Chapter 6: Molecular Basis of Inheritance

In the previous chapter, you have learned the inheritance patterns and the genetic basis of such patterns. At the time of Mendel, the nature of those ‘factors’ regulating the pattern of inheritance was not clear.

Over the next hundred years, the nature of the putative genetic material was investigated culminating in the realization that DNA – deoxyribonucleic acid – is the genetic material, at
least for the majority of organisms.

In class XI you have learned that nucleic acids are polymers of nucleotides.

Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are the two types of nucleic acids found in living systems.

DNA acts as the genetic material in most organisms. RNA though it also acts as genetic material in some viruses, mostly functions as a messenger.

RNA has additional roles as well. It functions as an adapter, structural, and in some cases as a catalytic molecule.

In Class XI, you have already learned the structures of nucleotides and the way these monomer units are linked to form nucleic acid polymers.

In this chapter, we are going to discuss the structure of DNA, its replication, the process of making RNA from DNA (transcription), the genetic code that determines the sequences of amino acids in proteins, the process of protein synthesis (translation) and elementary basis of their regulation.

The determination of the complete nucleotide sequence of the human genome during the last decade has set in a new era of genomics.

In the last section, the essentials of human genome sequencing and its consequences will also be discussed.

Let us begin our discussion by first understanding the structure o the most interesting molecule in the living system, that is, DNA. In subsequent sections, we will understand why it is the most abundant genetic material, and what its relationship is with RNA.

6.1 THE DNA

DNA is a long polymer of deoxyribonucleotides. The length of DNA is usually defined as the number of nucleotides (or a pair of nucleotides referred to as base pairs) present in it. This also is the characteristic of an organism.

6.1.1 Structure of Polynucleotide Chain

Let us recapitulate the chemical structure of a polynucleotide chain (DNA or RNA). A nucleotide has three components – a nitrogenous base, a pentose sugar (ribose in case of RNA, and deoxyribose for DNA), and a phosphate group.

There are two types of nitrogenous bases – Purines (Adenine and Guanine), and Pyrimidines (Cytosine, Uracil, and Thymine). Cytosine is common for both DNA and RNA and Thymine is present in DNA.

Uracil is present in RNA at the place of Thymine. A nitrogenous base is linked to the pentose sugar through an N-glycosidic linkage to form a nucleoside, such as adenosine or deoxyadenosine, guanosine, or deoxyguanosine, cytidine or deoxycytidine, and uridine or deoxythymidine.

When a phosphate group is linked to 5′ -OH of a nucleoside through phosphodiester linkage, a corresponding nucleotide (or deoxynucleotide depending upon the type of sugar present) is formed. Two nucleotides are linked through 3′-5′ phosphodiester linkage to form a dinucleotide.

AuthorNCERT
Language English
No. of Pages31
PDF Size2.1 MB
CategoryBiology
Source/Creditsncert.nic.in

NCERT Solutions Class 12 Biology Chapter 6 Molecular Basis of Inheritance

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

Solution:

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 molecules. As adenine and thymine are always found in equal numbers, the adenine content is 30%.

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.

Solution:

The list includes two different types of nucleic acid polymerases:

  • DNA-dependent DNA polymerases
  • DNA-dependent RNA polymerases
  • RNA-dependent DNA polymerases
  • RNA-dependent RNA polymerases

To synthesize a new strand, the DNA-dependent DNA polymerases use a DNA template. But, DNA-dependent RNA polymerases utilise a DNA template strand to synthesize RNA.

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

Solution:

To prove DNA is the genetic material, Hershey and Chase studied and worked on bacteriophage and E.Coli. To label the protein coat and DNA of the bacteriophage, they made use of different radioactive isotopes.

In a medium containing radioactive phosphorous (32P), they cultivated some bacteriophages to detect DNA and a few more on a medium comprising radioactive sulfur (35S) to detect protein.

The labeled radioactive phages were then made to infect the bacteria – E.coli. Once infected, the protein coat of the bacteriophage was segregated from the bacterial cell by mixing and then subjected to the centrifugation process.

It was observed that in the supernatant, as the protein coat was lighter, the infected bacteria got settled at the bottom of the centrifuge tube. In case I – supernatant was observed to be radioactive indicating that protein did not enter the bacterial cell when infected.

But in case II – the bacterial cells were radioactive as they possess radioactive DNA. Thus, it was proved that DNA is a genetic material as it was transferred from viruses to bacteria.

Molecular Basis of Inheritance NCERT Textbook With Solutions PDF Free Download

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