NCERT Based & Simplified

1. Finding out the chemical responsible for Mendelian inheritance (factor), its structure,& How it works is 


2. DNA act as genetic material in all cellular organism either it is prokaryotic or eukaryotic.


o RNA also act as messenger ,adaptor & structural RNA.

o RNA also act as enzyme molecule(Ribozyme) @ RNA World Hypothesis


●Polymer of Nucleotide –>POLYNUCLEOTIDE

●The length of DNA is expressed in base pair (bp)

●Specific for particular organism.

●Bp is number of nucleotide pairs in DNA

●Bacteriophage -Φ174 5386 nucleotides.

●Bacteriophage –>48502 bp

●E.coli –> 4.6×106 bp

●Human –>3.3 x109 bp in haploid cell.


o A nucleotide has three component -a nitrogenous base (either purine or pyrimidine)

,deoxyribose pentose sugar& a phosphate group.

   o ADENINE & GUANINE are purine

   o CYTOSINE & THYMINE are pyrimidine.

  o N- base joins pentose sugar with N-glycosidic linkage  Nucleoside.

  o Nucleoside joins with phosphate with phosphodiester bond  Nucleotide

5. Two nucleotide join each other with 3’–> 5’ phosphodiester linkage

o One end of polynucleotide has PO4-3 at 5rd carbon of Pentose sugar

o Other free end has -OH group at 3rd carbon of Pentose sugar

o The backbone of nucleotide is made of sugar & phosphate

o Nitrogenous base lie perpendicular to this backbone.

6. Meischer discovered the DNA & named it Nuclein.

7. WATSON & CRICK proposed DNA double helical model. –>1953, Noble prize –>1962

o For this they took X-ray diffraction data of DNA given by Wilson, Stoke, Willikins & Morrins. Helical nature given by Franklin

o They also used information given by CHARGAFF that purine is always equal to pyrimidine in

DNA structure.

8. Specific base pairing (A with T & C with G) makes the two strands of DNA complementary to each other & also basis of DNA replication. DNA model also states :

9. Antiparallel property of two strand

i.e. if one strand has the polarity 5’to 3’ end then the other strand of DNA has 3’to 5’ end polarity.

10. Between purine & pyrimidine there is hydrogen bond in base pairing.

o (2 H bond between A&T

/ 3 H bond between G&C)

11. The two chain are coiled in Right handed direction.

12. The pitch of helix is 3.4nm & there are roughly 10bp in each turn.

13. Consequently ,the distance between a base pair (bp) is in a helix is approximately equal to 0.34nm.

14. The plane of one base pair stack over the other in double helix. This property gives stability to double helix . 15.Hydrogen bond between base pair also provide stability to double helix of DNA.

16. Central dogma in molecular biology states the genetic information flow from DNA –> RNA –> Protein. (Proposed by Francis Crick)

17. We can calculate the length of DNA helix by multiplying the number of base pairs of an organism by 0.34×10 -9 m (the distance between two consecutive base pairs ).

18. Length of typical mammalian cell DNA is (6.6×109) x (0.34×10-9). It is roughly 2.2meters.

19. If the length of DNA helix is given ,we can find the number of base pair by dividing the length by 0.34 nm. 20.DNA has negative charge  Because of phosphate group

21. In prokaryotic cell , negatively charged DNA is held with some positively charged protein in a region termed as NUCLEOID.

22. In eukaryotic cell, the packaging of DNA is more complex.

o There is a set of positively charged protein called HISTONES which are rich in basic amino acids residues LYSINE & ARGININE.

o They carry positive charge in their side chain.

23. Histones make octamer (8 units) around which negatively charged DNA is wrapped & form the structure called NUCLEOSOME.

 24. A typical nucleosome has 200 bp of DNA helix.

25. Chromatin is nothing but repeating units of nucleosomes .

o In nucleus it looks like stained threads.

o In chromatin ,nucleosome looks like “BEADS ON STRING”

26. To find out the total number of nucleosome in an organism you have to divide the total number of base pair by 200.

27. In eukaryotes more packing of chromatin is require during cell division which is done with the help of NON HISTONE CHROMOSOMAL PROTEIN (NHC).

28. Euchromatin is transcriptionally active & stains light due to less histones.

29. Heterochomatin is inactive & stains dark due to more histones

30. In 1928, Griffith performed experiment known as TRANSFORMATION EXPERIMENT on pneumonia bacteria (Streptococcus pneumoniae) in search of nature of transforming material which changed the R strain into virulent S strain but he could not able to find the chemical nature of this transforming factor. He concluded that the R strain of pneumonia bacteria had somehow transformed by the heat killed S strain.


o They used proteases, RNAase to digest the protein & RNA. These enzymes did not inhibit the transformation of R Strain into S Strain.

o DNAase could stop transformation

32. The unequivocal proof that DNA is the genetic material

o Given by Hershey & Chase. @ Bacteriophage experiment

o Used radioactive P (in DNA) & radioactive S (in protein)

o Only DNA of bacteriophage enter into bacterial cell & protein capsid remained outside the host cell.

33. DNA & RNA act as genetic material as they can generate replica (due to complementary base pairing)

34. DNA is predominantly found as genetic material than RNA as it is more stable (double helical) ,allow mutation which is require for evolution & able to express itself as Mendelian character.

35. 2’OH group in nucleotide in RNA is a reactive group which make RNA more labile & easily degradable. Uracil also makes it more unstable because of these reasons DNA was the superior choice as heredity material for cellular organisms

o Meselson & Stahl 

Gave experimental evidence for semiconservative mode of replication for E.coli for prokaryotic organism.

o Eukaryotic DNA also follow semiconservative method for replication

o Evidence for this was given by Taylor in vicea faba plant.


o The main enzyme is DNA dependent DNA polymerase.

o It uses DNA as template

o It catalyse the polymerisation of deoxyribonucleotides.

o Replication fork is small opening in DNA helix where replication occur.

o DNA Polymerase enzyme polymerise nucleotides in : 5” 3” direction.

o Continuous synthesis of DNA occurs–>

●Only in leading strand

●Strand having 3”–> 5” polarity.

o The other strand of parent DNA which has 5”–>3” polarity, the replication process is Discontinuous

●Small pieces called OKAZAKI FRAGMENT formed which later joined by DNA ligase enzyme.


●Definite region in DNA where replication fork originate.


●Transcription is copying of genetic information from one strand of DNA into mRNA .

●Only a segment of DNA (GENE ) is transcribed into mRNA not whole DNA (as in case of replication.)

Transcription unit in DNA is defined primarily by the three regions in the DNA:

●The promoter (towards 5”end of structural gene also called upstream)

●Structural gene (between promoter & terminator gene).

●Terminator gene.(towards 3”end of structural gene, downstream)

43. DNA dependent RNA polymerase also polymerise ribonucleotides in only 5”–> 3” direction.

44. Therefore the DNA strand which has polarity of 3”–> 5” transcribed into mRNA. This DNA strand is also called template strand

45. The DNA strand which is not transcribed & having 5”–> 3” polarity is called CODING STRAND but actually it does not code for anything.

46. Promoter is a DNA sequence that provides binding site for RNA polymerase, as well as it also defines template strand & coding strand.

47. RNA Polymerase alone can’t initiate transcription

●Require INITIATION FACTOR (sigma factor) start initiation But alone it is responsible for elongation of mRNA.

48. In prokaryotic cells ,mRNA starts translation during elongation process as there is no need of processing (splicing) of mRNA in prokaryotic cell.

49. Consequently, the transcription and translation can be coupled in bacteria.

50. Termination of mRNA is done by RNA polymerase with the help of termination factor (rho factor)

51. Cistron is the segment of DNA responsible for coding of polypeptide.

52. Structural gene can be monocistronic ie code for only one polypeptide and it is present in eukaryotes.

53. In eukaryotes, the monocistronic structural genes have interrupted coding sequences – the genes in eukaryotes are split gene.

●The coding sequences or expressed sequences are defined as exons.

●Exons are said to be those sequence that appear in mature or processed RNA.

●The exons are interrupted by introns.

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


Location: Inside Nucleus in nucleoplasm

Function: Process pre-mRNA by splicing out intronic nucleic acids & Producing mRNA which is then translated to protein in ribosomes

Mechanisms of action: Recognition of the intron/exon boundaries, Via splice sequences & multiple other weak signals, Catalysis of cut-and-paste reactions that remove introns and join exons

Contents: Five small nuclear RNAs (snRNAs or U(uridine-rich)-RNAs), Each snRNA is associated with specific factors, Complex forms a small nuclear ribonucleoprotein particles (snRNPs); Also have Proteins more than 100 types

54. Polycistronic structural genes are found in prokaryotes. They code more than one polypeptide.

55. In eukaryotic cells there are at least three RNA polymerases in the nucleus

●There is a clear cut division of labour.

●RNA Polymerase I -> transcribed rRNA(28S, 18S, and 5.8S)

●RNA polymerase II -> precursor of mRNA

▪︎(hnRNA – the heterogeneous nuclear RNA)

●RNA polymerase III -> tRNA, 5srRNA, and snRNAs (small nuclear RNAs).

56. Splicing & hnRNA -> mRNA Processing in Eukaryotes:

In eukaryotic, the primary transcripts contain both the exons and the introns and are non-functional.

●Hence, it is subjected to a process called splicing

●Where the introns are removed and exons are joined in a defined order.

●hnRNA undergoes additional processing called as capping and tailing.

●In capping an unusual nucleotide (methyl guanosine triphosphate) is added to the 5′-end of hnRNA.

●In tailing, adenylate residues (200-300) are added at 3′-end in a template independent manner.

●It is the fully processed hnRNA, now called mRNA, that is transported out of the nucleus for translation.

57. 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


●The process of translation requires transfer of genetic information from a polymer of nucleotides to a polymer of amino acids.

●Genetic codes are present on mRNA.

●There are 20 amino acids & only 4 types of nucleotides in DNA.

●Genetic code is consist of three nucleotides (Total 64 codon)

o The salient features of genetic code are as follows:

●The codon is triplet. (Indicated by GAMMOV)

●Total 64 codon

●61 codons code for amino acids

●3 codons do not code for any amino acids (stop codons)

●One codon codes one amino acid  unambiguous and specific.

●Some amino acids are coded by more than one codon degeneracy

●Codon is readed in mRNA in a continuous fashion  No punctuations

●The code is nearly universal for example,

• ▪︎from bacteria to human UUU would code for Phenylalanine (phe).

• ▪︎Some exceptions to this rule have been found in mitochondrial codons, and in some protozoans.

●AUG has dual functions. It codes for Methionine (met) , and it also act as initiator codon

59. Deciphering of codon by using only one type of nucleotide in cell free system was done by NIRENBERG.

60. HARGOBIND KHORANA used two nucleotides deoxyadenosine triphosphate and deoxycytidine triphosphate to synthesize mRNA. so the sequence on mRNA was ACACACACA which translated into polypeptide having therionine and histidine aminoacid in alternate sequence.


62. UAA,UAG & UGA  nonsense codon / stop signal / chain termination codon .

o They don’t have message for any amino acid but they have message to stop THE TRANSLATION.

63. WOBBLE HYPOTHYSIS was proposed by CRICK.

o 3rd nucleotide of genetic code is wobble that is not specific.

o For example, Genetic code of ALANINE amino acid is GCU/GCC/GCA.

o First two nucleotide of genetic code is same but 3rd nucleotide is different.

64. Severo ochoa enzyme (polynucleotide phosphorylase) polymerise RNA with defined sequence in template independent system.(There is no need of DNA for synthesis of RNA) 

65. Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion is called as frameshift insertion or deletion mutations. 

66. Insertion or deletion of three or its multiple bases insert or delete in one or multiple codon hence one or multiple amino acids, and reading frame remains unaltered from that point onwards.

67. t-RNA

o Read the genetic code in mRNA through anticodon & bring specific amino acid for translation.

o Also called sRNA (soluble RNA),

o Known before the genetic code was postulated.

o Has an anticodon loop that has bases complementary to the code

o Also has an amino acid acceptor end to which it binds to amino acids.

o tRNA are specific for each amino acid.

68. For initiation, there is another specific tRNA that is referred to as initiator tRNA.

69. There are no tRNAs for stop codons.

70. The secondary structure of tRNA –> looks like a clover-leaf.

71. In actual structure, the tRNA is a compact molecule  –>Looks like inverted L


o Process of polymerisation of amino acids to form a polypeptide.

o The order and sequence of amino acids are defined by the sequence of bases in the mRNA. The amino acids are joined by peptide bond.

o Formation of a peptide bond requires energy.

o Therefore, in the first phase itself amino acids are activated in the presence of ATP and linked to their cognate tRNA – a process commonly called as charging of tRNA / aminoacylation of tRNA

o If two such charged tRNAs are brought close enough, the formation of peptide bond between them take place.

o The presence of a catalyst enhance the rate of peptide bond formation.

o Translation takes place in ribosome.

73. The Ribosome in translation

o In its inactive state, exists as two subunits; a large subunit and a small subunit.

o When the small subunit encounters an mRNA, the process of translation of the mRNA to protein begins

o There are two sites in the large subunit, for subsequent amino acids to bind to and thus, be close enough to each other for the formation of a peptide bond.

o The ribosome also acts as a catalyst (23S rRNA in bacteria is the enzyme- ribozyme) for the formation of peptide bond.

74. An mRNA also has some additional sequences (other than initiation codon & stop codon) that are not translated and are referred as untranslated regions (UTR).

75. The UTRs are present at both 5′-end (before start codon) and at 3′-end (after stop codon). They are required for efficient translation process.

76. During translation the ribosome moves from codon to codon along the mRNA.

77. Amino acids are added one by one, translated into Polypeptide sequences dictated by DNA and represented by mRNA. At the end, a release factor binds to the stop codon, terminating translation and releasing the complete polypeptide from the ribosome.


Gene regulation is just activation or inactivation of gene according to cells need. The gene expression results in the formation of polypeptide.

●In eukaryotes ,the regulation of gene expression could be exerted at

●transcriptional level (formation of primary transcript)

●processing level (regulation of splicing)

●transport of mRNA from nucleus to the cytoplasm

●translational level

●In prokaryotes, control of the rate of transcriptional initiation is the predominant site for control of gene expression.

●In a transcription unit, the activity of RNA polymerase at a given promoter is in turn regulated by interaction with accessory proteins, which affect its ability to recognise start sites.

●These regulatory proteins can act both positively (activators) and negatively (repressors). The accessibility of promoter regions of prokaryotic DNA is in many cases regulated by the interaction of proteins with sequences termed operators.

●The operator region is adjacent to the promoter elements in most operons and in most cases the sequences of the operator bind a repressor protein. 

●Each operon has its specific operator and specific repressor.

●For example, lac operator is present only in the lac operon and it interacts specifically with lac repressor only.


●Proposed by Jacob &Monad

●To elucidate a transcriptionally regulated system in E.coli.

●In lac operon (here lac refers to lactose),

●A polycistronic structural gene is regulated by a common promoter and regulatory genes.

●Such arrangement is very common in bacteria (referred to as operon)

●Consists of one regulatory gene and three structural genes (z, y, and a)

●The i gene codes for the repressor of the lac operon.

●The gene codes for beta-galactosidase ( β -gal), responsible for the hydrolysis lactose into galactose and glucose.

●The y gene codes for permease, which increases permeability of the cell to β -galactosides. 

●The gene encodes a transacetylase.

●All the three gene products are required for metabolism of lactose.

●Lactose is the substrate for the enzyme beta-galactosidase and it regulates the lac operon by switching on and off of the operon. (Inducer) 

●In absence of inducer, the regulator gene code for repressor protein which bind the operator gene & prevent the RNA polymerase to transcribe the structural genes of operon (z,y,a genes).

●In presence of inducer (lactose in lac operon),this inducer bind the repressor protein which was coded by regulator gene so repressor protein become inactive & can not bind the operator gene so RNA polymerase transcribed the structural genes (z,y,&a).

●Regulation of lac operon by repressor is referred to as negative regulation.

●In normal condition there is no expression of operon ( switch off state)

●Term & condition of expression of Lac operon –>Glucose is absent & Lactose is present in medium then lac operon make enzymes.

●The repressor of the operon is synthesised (all-the-time – constitutively) from the i gene as lactose is generally absent

●a very low level of expression of lac operon has to be present in the cell all the time, otherwise lactose cannot enter the cells as permease is absent. 

●The expression of lac operon will continue till there is lactose in medium.


o Uses Genetic engineering techniques & DNA sequencers

o Launched in the year 1990. (mega project) Cost  9 billion US dollars; Completed in 2003

o Technology –>Use of high speed computational devices for data storage and retrieval, and analysis.

o HGP was closely associated with the rapid development of a new area in biology called Bioinformatics.

o Goals of HGP

o Identify all the approximately 20,000 – 25,000 genes in human DNA;

o Determine the sequences of the 3billion chemical base pairs that make up human DNA;

o Store this information in databases;

o Improve tools for data analysis;

o Transfer related technologies to other sectors, such as industries;

o Address the ethical, legal, and social issues (ELSI) that may arise from the project.

o The Human Genome Project was a 13-year project

o Coordinated by the U.S. Department of Energy and the National Institute of Health.

o During the early years of the HGP, the Wellcome Trust (U.K.) became a major partner;

o Additional contributions came from Japan, France, Germany, China and others.

o Knowledge about the effects of DNA variations among individuals can lead to revolutionary new ways to diagnose, treat and someday prevent the thousands of disorders that affect human beings.

o Non-human organisms used to know and compare DNA sequences  bacteria, yeast, Caenorhabditis elegans (a free living non-pathogenic nematode), Drosophila (the fruit fly), plants (rice and Arabidopsis)


●The methods involved two major approaches.

●Expressed Sequence Tags (ESTs) –> identifying all the genes that are expressed as RNA

●Sequence Annotation -> Simply sequencing the whole set of genome that contained all the coding and non-coding sequence, and later assigning different regions in the sequence with functions

●For sequencing, the total DNA from a cell is isolated and converted into random fragments of relatively smaller sizes and cloned in suitable host using specialised vectors.

●The cloning resulted into amplification of each piece of DNA fragment so that it subsequently could be sequenced with ease.

●The commonly used hosts were bacteria and yeast,


The fragments were sequenced using automated DNA sequencers that worked on the principle of a method developed by Frederick Sanger.

Specialised computer based programs were developed to arrange sequences. These sequences were subsequently annotated and were assigned to each chromosome.

The sequence of chromosome 1 was completed only in May 2006 (this was the last of the 24 human chromosomes – 22 autosomes and X and Y – to be sequenced).

To assign the genetic and physical maps on the genome –> By using information on polymorphism of restriction endonuclease recognition sites, and some repetitive DNA sequences known as microsatellites


●The human genome contains 3164.7 million bp.

●The average gene consists of 3000 bases, but sizes vary greatly,

●Largest known human gene being dystrophin at 2.4 million bases.

●The total number of genes is estimated at 30,000

●Almost all (99.9 per cent) nucleotide bases are exactly the same in all people.

●The functions are unknown for over 50 per cent of the discovered genes.

●Less than 2 per cent of the genome codes for proteins.

●Repeated sequences make up very large portion of the human genome.

●Repetitive sequences are stretches of DNA sequences that are repeated many times, sometimes hundred to thousand times. They are thought to have no direct coding functions, but they shed light on chromosome structure, dynamics and evolution.

●Chromosome1has most genes (2968)

●The Y chromosome has the fewest (231) genes

Scientists have identified about 1.4 million locations where single base DNA differences (SNPs

– single nucleotide polymorphism, pronounced as ‘snips’) occur in humans.

This information promises to revolutionise the processes of finding chromosomal locations for disease-associated sequences and tracing human history.

o Applications in Genetic disease & Cancer treatment


●DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called as repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times.

●These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation.

●The bulk DNA forms a major peak and the other small peaks are referred to as satellite DNA.

●Depending on base composition (A : T rich or G:C rich), length of segment, and number of repetitive units, the satellite DNA is classified into many categories, such as micro-satellites, mini-satellites etc.

●These sequences normally do not code for any proteins, but they form a large portion of human genome.

●These sequence show high degree of polymorphism and form the basis of DNA fingerprinting (RFLP – Restriction Fragment Length Polymorphism) .

●Polymorphism in DNA sequence is the basis of genetic mapping of human genome as well as of DNA fingerprinting,

●Polymorphism (variation at genetic level) arises due to mutations.

●Allelic sequence variation has traditionally been described as a DNA polymorphism if more than one variant (allele) at a locus occurs in human population with a frequency greater than 0.01.

If an inheritable mutation is observed in a population at high frequency, it is referred to as DNA polymorphism.

●The probability of such variation to be observed in non- coding DNA sequence would be higher as mutations in these sequences may not have any immediate effect/impact in an individual’s reproductive ability. These mutations keep on accumulating generation after generation, and form one of the basis of variability/polymorphism.

●The technique of DNA Fingerprinting was initially developed by Alec Jeffreys.

He used a satellite DNA as probe that shows very high degree of polymorphism. It was called as Variable Number of Tandem Repeats (VNTR). The technique involved Southern blot hybridisation using radiolabelled VNTR as a probe.

o It included

●isolation of DNA,

●digestion of DNA by restriction endonucleases,

●separation of DNA fragments by electrophoresis,

●transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon,

●hybridisation using labelled VNTR probe, and

●detection of hybridised DNA fragments by autoradiography.

o The VNTR belongs to a class of satellite DNA referred to as mini-satellite.

o A small DNA sequence is arranged tandemly in many copy numbers. The copy number varies from chromosome to chromosome in an individual. The numbers of repeat show very high degree of polymorphism.

o As a result the size of VNTR varies in size from 0.1 to 20 kb.

o Consequently, after hybridisation with VNTR probe, the autoradiogram gives many bands of differing sizes. These bands give a characteristic pattern for an individual DNA

o It differs from individual to individual in a population except in the case of monozygotic (identical) twins.

o The sensitivity of the technique has been increased by use of polymerase chain reaction

o Consequently, DNA from a single cell is enough to perform DNA fingerprinting analysis.

o In addition to application in forensic science, it has much wider application, such as in determining population and genetic diversities.

o Currently, many different probes are used to generate DNA fingerprints.

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