Study Notes on Molecular Basis of Inheritance

Study Notes on Molecular Basis of Inheritance


  • DNA is a long polymer of deoxyribonucleotides. The length of DNA usually defined as a number of nucleotides (or a pair of nucleotide referred to as base pairs) present in it.
  • This is also the characteristics of organisms. For example, a bacteriophage known as xl74 has 5386 nucleotides.
  • The Bacteriophage lambda has 48502 base pairs (bp). Escherichia coli has 4.6 x 10″ bp and haploid content of human DNA 1s 3.3 x 10″ bp.


  • Let us recapitulate the chemical structure of a polynucleotide chain (DNA or RNA).
  • A nucleotide has three components -a nitrogenous base, 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)
  • The 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.
  • The 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.
  • More nucleotides can be joined in such a manner to form a polynucleotide chain.
  • A polymer thus formed has at one end a free phosphate moiety at 5-end of a ribose sugar, which is referred to as 5′-end of a polynucleotide chain.
  • Similarly, at the other end of the polymer, the ribose has a free 3′-OH group which is referred to as 3-end of the polynucleotide chain.
  • The backbone in a polynucleotide chain is formed due to sugar and phosphates.
  • The nitrogenous bases linked to the sugar moiety project from the backbone.

The salient features of the Double-helix structure of DNA are as follows:

  • It is made of two polynucleotide chains, where the backbone is
  • (ul) The two chains have antiparallel polarity. It means, if one chain has the polarity 53′, the other has 3 +5′.
  • () The bases in two strands are paired through hydrogen bond Because of this, the genetic would be constituted by sugar-phosphate, and the bases project inside.
  • (H-bonds) forming base pairs (bp). Adenine forms two hydrogen bonds with Thymine from the opposite strand and vice versa.
  • Similarly, Guanine is bonded with Cytosine with three H-bonds.
  • As a result, always a purine comes opposite to a pyrimidine. This generates approximately uniform distance between the two strands of the helix.


  • The two chains are coiled in a right-handed fashion. The pitch of the helix is 3.4 nm (a nanometre is one-billionth of a metre, that is 10° m).
  • There is roughly 10 bp in each turn. Consequently, the distance between a bp in a helix is approximately equal to 0.34 nm.
  • The plane of one base pair stacks over the other in a double helix.
  • In addition to H-bonds, confers stability of the helical structure Compare the structure of purines and pyrimidines.
  • The proposition of a double helix structure for DNA and its simplicity in explaining the genetic implication became revolutionary.
  • Very soon, Francis Crick proposed the Central dogma in molecular biology, which states that the genetic information flows from DNA to RNA to Protein.
  • Packaging of DNA Helix Taken the distance between two consecutive base pairs as 0.34 nm (0.34×10 m).
  • The length of DNA double helix in a typical mammalian cell is calculated (simply between two consecutive bp, that is, 6.6 x 10 bp x by multiplying the total number of bp with distance 22 metres.
  • A length that is far greater than the 0.34 x 10 m/bp). It comes out to be approximately dimension of a typical nucleus (approximately 109m).
  • In prokaryotes, such as. E Coli, though they do not have a defined nucleus, the DNA is not scattered throughout the cell.
  • The DNA (being negatively charged) is held with some proteins (that have positive charges) in a region termed as ‘nucleoid”.
  • The DNA in nucleoid is organised in large loops held by DNA Core proteins.
  • In eukaryotes, this organisation is much more complex. There is a set of positively charged.
  • basic proteins called histones. A protein acquires charge depending upon the abundance of amino acids residues with charged side chains.
  • The Histones are rich in the basic amino acid residues lysines and arginines.
  • Both the amino acid residues carry positive charges in their side chains.
  • Histones are Organised to form a unit of eight molecules called histone.
  • The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome A Optical nucleosome contains 200 bp of the helix.
  • A Nucleosomes constitute the repeating unit of a structure in helix thread-like stained (coloured) bodies seen is called chromatin structure when viewed under an electron microscope
  • The beads-on-string structure in chromatin is packaged to form chromatin fibres that are further coiled and condensed at the metaphase stage of cell division to form chromosomes.
  • The packaging of chromatin at a higher level requires an additional set of proteins that collectively are referred to as-Non-histone Chromosomal (NHC) proteins.
  • In a typical nucleus, some region of chromatin are loosely packed (and stains light) and are referred to as euchromatin.
  • The chromatin that is more densely packed and stains dark are called as Heterochromatin.
  • Euchromatin is said to be transcriptionally active chromatin, whereas heterochromatin is inactive.

The experiments are given below prove that DNA is the genetic material:

  1. Transforming Principle: The transformation experiments, conducted by Frederick Griffith in 1928, are great evidence in establishing the nature of genetic material.
  2. He performed a series of experiments by selecting two strains of bacterium Streptococcus pneumonia (also called Pneumococcus).
  3. These are S4 and R-IL S-Ill strain/smooth or capsulated type have a mucous  (Polysaccharide) coat and produce so much shiny colonies in a culture plate.
  4. These are virulent and cause pneumonia.
  5. R-Il strain/rough or non-capsulated type have no mucous coat and produce rough colonies.

These are non-virulent and do not cause pneumonia, The experiment can be described in the following four steps:

(a) S strain Injected into mice  – Mice die

(b) R strain Injected into mice  -Mice live

(c) S strain (heat-killed) Injected into mice  live

(d) S-strain (heat-killed) + R-strain (live)

Injected into mice – Mice die.


  • The unequivocal proof that DNA is the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952).
  • They worked with viruses that infect bacteria called bacteriophages.
  • The bacteriophage attaches to the bacteria and its genetic material then enters the bacterial cell.
  • The bacterial cell treats the viral genetic material as if it was it’s own and subsequently manufactures more virus particles.
  • Hershey and Chase worked to discover whether it was protein or DNA from the viruses that entered the bacteria.
  • They grew some viruses on a medium that contained radioactive phosphorus and some others on a medium that contained radioactive sulfur.
  • Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. Similarly, viruses are grown on radioactive
  • The sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulfur.
  • Radioactive phages were allowed to attach to E. coli bacteria.
  • The infection proceeded, the viral coats were removed from the bacterias, 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.
  • RNA used to act as a genetic material as well. As a catalyst.
  • There are some important biochemical reactions in living systems that are catalysed by RNA catalyst (ribozyme) and not by protein enzymes e.g., Ribonuclease P (Cleavage), (Splicing), Peptidyl transferase (peptide bond formation).
  • RNA being a catalyst was reactive and hence unstable.
  • Therefore, DNA has evolved from RNA with chemical modifications that make it more stable. DNA is double-stranded and has complementary strand further resists changes by evolving a process of repair.
  • RNA is an adapter, structural molecule and in some cases catalytic.
  • From the above discussion it is very much clear both RNA and DNA can function as genetic material.
  • The DNA being stable is preferred for storage of genetic material For the transmission of genetic information RNA is better material.


  • Watson and Crick had immediately proposed a scheme for DNA replication while proposing the consistence structure of DNA.
  • The scheme suggested that the two strands would separate and act as a template for the synthesis of new complementary strands.
  • After the completion of replication, each DNA molecule would have one parental and one newly synthesised strand.
  • This scheme was termed as semiconservative DNA


  • The process of copying genetic information from one strand of the DNA into RNA is known as transcription DNA replication.
  • the principle of complementarity governs the process of transcription, except the adenosine which forms base pair with uracil instead of thymine.
  • Unlike DNA replication where total DNA of an organism gets duplicated, in transcription only a segment of DNA and only one of the strands is copied into RNA.
  • Here only one strand is template strand while in replication both strands are template.
  • There are two explanations for both the strands of DNA not being copied during transcription.
  • It both strands act as a template, they would code for RNA molecule with different sequences.
  • If they code for proteins, the sequence of amino acids in the proteins would be different. Hence, one segment of the DNA would be coding for two different proteins.
  • This would complicate the genetic information transfer machinery
  • The two RNA molecules if produced simultaneously would be complementary to each other, hence would form a double-stranded RNA.
  • This would prevent the translation of RNA into protein during DNA. Transcription Unit
  • The segment of DNA that takes part in transcription is called transcription unit it has three components:
  1. The promotor
  2. The structural gene
  3. The terminator.


  • Tailing is the addition of adenine residues about 200-300 at 3′ end template-independent manner on newly formed hnRNA with the help of Poly A polymerase.
  • Splicing is the process of removal of introns and joining of exons in a defined order.
  • Introns are removed by small nuclear RNA (snRNA) and protein complex called small nuclear ribonucleoproteins an snRNPs
  • The fully processed hnRNA is now called mRNA and it is transported out of the nucleus for translation.
  • The split-gene arrangements represent probably an ancient feature of the genome.
  • The presence of introns is reminiscent of antiquity, and the process of splicing represents the dominance of RNA-world.


  • DNA (or RNA) carries all genetic information. It is expressed in the form of proteins which are made up 20 different types of essential amino acids.
  • The information about the number and sequence of these and acids forming protein is present in DNA and is passed on to mRNA during transcription.
  • Thus, genetic code is inter-relationship between nucleotides sequence of DNA or mRNA and amino acids sequence in the polypeptide.
  • It is an mRNA sequence containing coded information for one amino acid and consists of 3 nucleotides.
  • The proposition and deciphering of genetic code were the most challenging.
  • It requires the involvement of scientists from several disciplines – physicists, organic chemists, biochemists and genetics.
  • It was George Gamow, a physicist, who coined the term genetic code and argued that since there are on 4 bases and if they have to code for 20 amino acid code should constitute a combination of the base.
  • He suggested that in order to code for all the 20 amino acids, the code should be made up of 3 nucleotides.


  • It refers to the polymerisation of amino acid to form a polypeptide.
  • the order and sequence of amino acids are defined by the sequence of bases in the mRNA small subunit.
  • Ribosomes have two sites for binding aminoacyl URNA, P-site (peptidyl site) and (aminoacyl).
  • When the small subunit encounters an mRNA, the process of translation of the mRNA to begins

The steps of the translation mechanism are:

  • Activation of amino acids: In the presence of enzyme aminoacyl – tRNA synthetase (E), specific amino acid (AA) bind with ATPAA, +ATP_E. Mg>AA, -AMP-E, complex +Ppi.
  • (b) Charging of tRNA: The AA, AMP-E, complex reacts with specific tRNA. Thus, the amino acid is transferred to tRNA. As a result, the enzyme and AMP are liberated. It is also called as aminoacylation of IRNA AA,+AMP-E, complex + tRNA, →AA, tRNA, +AMP+E (Charged RNA)
  • (c) Formation of polypeptide chain: It completes in three steps:

(1).Chain initiation

  • It requires 3 initiation factors in prokaryotes and 9 initiation factors in eukaryotes Binding of mRNA with a smaller subunit of ribosomes (30S/40S).
  • Binding of 30S-mRNA with tRNA non-formylated methionine is attached with 1RNA n eukaryotes and formylated in prokaryotes 30S-mRNA + IRNAeGTP30S-mRNA-IRNA, (Attachment of larger subunit of the ribosome.
  • It is 50S in prokaryotes & 60S in eukaryotes.

(2).Chain elongation.

  • After the formation of the complete ribosome – mRNA-RNA complex, an amine acceptor site (A-site) is established next to the P-site.
  • It exposes mRNA Codon next to the initiation codon. A new aminoacyl tRNA Complex reaches the A-site and forms codon-anticodon bonding.
  • This requires elongation factor and energy, 6TP A peptide bond is formed between COOH group of a first amino acid (methionine) and NH, a group of second amino acid.
  • If two charged RNAs a close enough the formation of a peptide bond between them would be favoured energetically.

(3).Chain termination.

  • The termination of the polypeptide is signalled by one of the three termination codons (UAA, UAG UGA).
  • A GTP-dependent factor known as release factor is associated with termination codon.
  • It is eRF, in eukaryotes and RF, and RF2 in prokaryotes.
  • that help in terminating translation and releasing the complete polypeptide from the ribosomes.