Genetics : Molecular basis of Inheritance

Biology Class - 1248
Anticodon : A sequence of three nitrogenous bases on tRNA which is com-
plementary to the codon on mRNA.
Genome : Sum total of genes in haploid set of chromosomes.
DNA Polymorphism : The variations at genetic level, where an inheritable
mutation is observed, in a population at high frequency.
Satellite DNA: The repetitive DNA sequences which form a large portion of
genome and have high degree of polymorphism but do not code for any proteins.
Operon : A group of genes which control a metabolic pathway.
Exons : The regions of a gene which become part of mRNA and code for
different regions of proteins.
Introns : The regions of a gene which are removed during the processing of
mRNA.
Euchromatin : The region of chromatin which is loosely packed and tran-
scriptionally active, it stains lighter.
Heterochromatin : The chromatin that is more densely packed, stains dark
and is transcriptionally inactive.
Splicing : The process in eukaryotic genes in which introns are removed and
the exons are joined together to form mRNA.
Bioinformatics : Science of use of techniques including statistics, storing
as data bases, analysing, modelling and providing access to various aspects of
biological information usually on the molecular level.
Central Dogma :
Replication form : The Y shaped structure formed when double stranded
DNA is unwound upto a point during its replication.
VNTR : Variable Number of Tandem Repeats
www.tiwariacademy.com

49Molecular basis of Inheritance
YAC : Yeast Arti®cial Chromosome
BAC : Bacterial Arti®cial Chromosome
SNPs : Single Nucleotide polymorphism
HGP : Human Genome Project
hnRNA : Heterogenous nuclear RNA. It is precursor of mRNA.
Friedrich 1869 First identi®ed and isolated a acidic substance from
Meischer pus cell and named it ‘Nuclien’.
Altman 1889 Separated protein from nuclear substance and named
it nucleic acid
Kossel 1893 Discover nitrogen bases (Adenine, guanine, cytocine,
Thymine, uracil)
T.H. Morgan 1910
Frederick 1928 Provide ®rst clear-cut evidance that DNA in the
Grif®th hereditary material while working on streptococus
pneumoniae Biochemical nature of genetic material
was not de®ned
Avery, 1944 Discover that transforming principle is DNA, not a
Macleod and protein or RNA. First identi®cation that DNA is the
McCarty hereditary material
Erwin 1950 Purine and pyrimidine components occur in equal
Chargaff amount in a DNA molecule.
A + G = T + C
Harshey and 1952 Performed experiment with Escherichia Coli and
chase bacteriophage and show that it is the viral DNA and
not protein that passed from virus to bacteria and
therefore DNA serves as the genetic material.
Wilkins and 1952 Produce X-ray diffraction data of DNA.
Frankline
Watson and 1953 Double helical structure of DNA.
Crick
Messelson 1958 Experimentaly proved the semiconservative nature
and Stahl of DNA replication.
Jacob and 1961 Proposed operon model - genetic material has a
Monod number of functional unit called operon.
Alec Jaffery 1985 Discovered the technique of DNA ®nger printing.

Chemical Structure of Polynucleotide Chain (DNA/RNA) : A nucleotide
has three components.
1. Nitrogen base
(i) Purines : Adenine and Guanine
(ii) Pyrimidines : Cytosine, Thymine and Uracil Thymine in DNA and
Uracil in RNA.
2. Pentose Sugar : Ribose (in RNA) or Deoxyribose (in DNA).
3. Phosphate Group
 Nitrogen base is linked to pentose sugar through N-Glycosidic linkage.
 Nitrogen base + Sugar = Nucleoside
 Phosphate group is linked to 5¢-OH of a nucleoside through phosphoester
linkage.
 Nucleoside + Phosphate group = Nucleotide
 Two nucleotides are linked through 3’-5 phosphodiester linkage to form a
dinucleotide
 A polynucleotide chain has free phosphate group at 5’ end of ribose sugar
and a free 3¢-OH group at other end.
RNA is highly reactive than DNA : In RNA nucleotide has an additional
OH group at 2¢. positions in the ribose; RNA is also catalytic.
Double-helix Structure of DNA : Proposed by Watson and Crick in 1953.
(i) DNA is made up of two polynucleotide chains.
(ii) The backbone is made up of sugar and phosphate and the bases project
inside.
(iii) Both polynucleotide chains are antiparallel i.e. on chain has polarity
5¢-3¢ and other chain has 3¢-5¢.
(iv) These two strands of chains are held together by hydrogen bonds i.e.
A = T, C º G.
(v) Both chains are coiled in right handed fashion. The pitch of helix is
3.4 nm with 10 base pairs in each turn.

Packaging of DNA Helix
 The average distance between the two adjacent base pairs is 0.34 nm
(0.34 × 10–9m or 3.4°A)

 The number of base pairs in Escherichia coli is 4.6 × 106.
 DNA Packaging in Prokaryotes : DNA is not scattered throughout the
cell. DNA (negatively Charged) is held by some proteins (has positive
charges) in a region termed as nucleoid. The DNAin nucleoid is organised
in large loops held by proteins.
 DNApackaging in Eukaryotes : There is a set of positively charged basic
proteins called histones. Eight histone molecules combines together to
form histone octamer.
 The negatively charged DNA is wrapped around positively charged
histone octamer to form as structure called nucleosome.
 Histone H1 is situated outside of nucleosomal DNA in linker region.
 Nucleosomes constitute the repeating unit of a structure in nucleus called
chromatin.
 The beads-on-string structure in chromatin is packaged to form chromatin
®bres that are further coiled and condensed at metaphase stage of cell
division to form chromosomes.
 Thepackagingofchromatinathigherlevelrequiresadditionalsetofprotein
that collectively are reffered to as Non-histone chromosomal (NHC)
proteins. At places chromatin is density packed to form darkly staining
heterochromatin. At other places chromatin is loosely packed to form
euchromatin.
 Euchromatin is said to be transcriptionally active chromatin, whereas
heterochromatin is inactive.

Transforming Principle :
Frederick Grif®th (1928) performed experiments with Streptococcus
phenumoniae and mice. This bacterium has two strains.
1. S-strain (Virulent)-which possess a mucilage coat and has ability to cause
pneumonia.
2. R-strain (Nonvirulent) which do not possess mucilage coat and is unable
to cause pneumonia.
 Griffth injected R-strain bacteria into mice.
No disease noticed and mice remain live.
 On injecting S-strain bacteria into mice.
Mice died due to pneumonia.
 When heat-killed S-strain bacteria were injected into mice No
pneumonia symptoms noticed and nice remain live.
 He than injected a mixture of R-strain bacteria (Non virulent) and heat
killed S-strain bacteria (virulent) into mice mice died due to
pnuemonia.
 Moreover Grif®th recovered living S-strain (virulent) bacteria from the
dead mice.

Conclusion : He concluded that presence of heat-killed S-strain bacteria
caused transformation of some R-strain bacteria into virulent by a chemical
substance, called ‘transforming principle’. But biochemical nature of the genetic
material was not de®ned by him.
Chemical Nature of Transforming Principle
In 1944, Avery MacLeod and McCarty worked to determine the chemical
nature of ‘transforming principle’.
The puri®ed biochemicals from heat killed S-cells :
 Proteins Transformation takes place So, protein is not a
‘transforming principle’.
 RNA
RNases Transformation takes place So, RNA is not a
‘transforming Principle’.
 DNA
DNases Transformation inhibited. Therefore, DNA is the
‘Transforming Principle’.
Hershey and Chase Experiment : In 1952, Hershey and Chase performed
an experiment on bacteriophages (Virsues that infect bacteria) and proved that
DNA is the genetic material.
Bacteriophage Bacteriophage
Radioactive (S35) Radioactive (P32)
Labelled protein coat labelled DNA
¯ ¯
Infection : E. coli E. coli
Blending : Viral coats removed from the bacteria.
¯ ¯
Centrifugation : Viral particles separated from the bacerial cell.
¯ ¯
No radioactive (S35) Radioactive (P35)
Detected in bacterial cells detected in bacterial
but detected in cells but not in
supernatant supernatant
Conclusion : DNA is the genetic material.

Meselson and Stahl’s Experiment :
 Meselson and Stahl performed the experiment in 1958 on E. coli to prove
that DNA replication is semiconservative.
 E. coli was grown in 15NH4CI for many generations.
 N15 was incorporated into newly synthesised DNA.
 This heavy DNA could be differentiated from normal DNA by
centrifugation in cesium chloride (CsCl) density gradient.
 Then they transferred these E.coli into medium with normal 14NH4Cl.
 After 20 minutes, it was found that all the DNA molecules of daughter
cells were hybrid-First generation.
 After 40 minutes, it was found that 50% DNA molecules were hybrid and
50% were normal-second generation.
DNA replication
DNA strands start separating from ori (origin of replication). This unwinding
is catalysed by many enzymes. Y-shaped structure is formed at ori called
replication fork
¯
DNA polymerase attaches to the replication fork and add nucleotides
complementary to the parental DNA strand. The direction of polymerisation
is 5¢-3¢.
¯
DNA polymerase cannot initiate the polymerisation itself, so a small segment
of RNA called primer is attached at replication start point
¯
DNA polymerase adds nucleotides on one of the template strand called as
leading strand (the template with polarity 3¢-5¢). In this strand nucleotides
are added continuously therefore called as continuous replication
¯
On the other strand the replication is discontinuous, small fragment of DNA are
formed called okazaki fragments which are later joined by DNA ligase. This
strand is called as lagging strand.
¯
Accuracy of polymerisation is maintained by Proof reading and any wrong
base added is removed by DNA polymerase

Transcription in Prokaryotes : In prokaryotes the process of transcription
is completed in three steps :
1. Initiation : RNApolymerase binds with initation factor (sigma factor) and
then binds to promotor site.
2. Elogation : RNA polymerase separates from sigma factor and adds
nucleoside triphosphate as substrate. RNA is formed during the process
following the rule of complementary and remains bound to enzyme RNA
polymerase.
3. Termination : On reaching terminator region RNApolymerase binds with
who factor (terminator factor) As a result nascent RNA separates.

Transcription in Eukaryotes :
 In eukaryotes three types of RNA polymerases found in the nucleus.
(In addition to the RNA polymerase found in the organelles) are involved
in transcription.
RNA Polymerase I : Transribes rRNAs.
RNAPolymerase II : Transcribes hnRNA(which is precursor of mRNA).
 RNA Polymerase III : Transcribes tRNA, 5 srRNA and sn RNA.
 The primary transcription has both exon and intron regions.
 Introns which are non-coding regions removed by a process called
splicing.
 hnRNA undergoes who additional process :
(a) Capping : An unusual nucleotide (methylguanosine triphosphate) is
added to 5’–end of hnRNA.
(b) Tailling : Adenylate residues (200-300) are added at 3–end. It is
fully processed hnRNA. Now called mRNA is transported out of the
nucleus.

Genetic Code
(i) The codon is triplet 61 codons code for amino acids and 3 codons
function as stop codons (UAG, UGA, UAA)
(ii) Onecodoncodesforonlyoneaminoacid,hencethecodonisunambiguous
and speci®c.
(iii) Some amino acids are coded by more than one codon, hence called as
degenerate.
(iv) The codon is read in mRNA in a contiguous fashion. There are no
punctuations.
(v) The code is nearly universal.
(vi) AUG has dual functions. It codes for Methionine (met) and it also acts
as initiator codon.

tRNA, the Adapter Molecule
 tRNA has an anticoden loop that has bases complementary to the code,
and also has an amino acid aceptor and through which it bind to amino
acids.
Translation
 Translation refers to the process of polymerization of amino acids to
form a polypeptide. The order and sequence of amino acids are de®ned
by the sequence of bases in the mRNA. 20 amino acids participate in
naturally occuring protein synthesis.

 First step is—charging of tRNA or aminoacylation of rRNA-here amino
acids are activated in the presence of ATP and linked to speci®c tRNA.
 Initiation : Ribosome binds to mRNA at the start codon (AUG) that is
recognized by the initiator tRNA.
 Elongation phase : Here complexes composed of an amino acid linked
to tRNA. Sequentially bind to the appropriate codon in mRNAby forming
complementary base pairs with rRNA codon. The ribosomes move from
codon to codon along with mRNa. Amino acids are added one by one,
translated into polypeptide sequences.
 Termination : Release facotors binds to the stop codon (UAA, UAG,
UGA) translation and releasing the complete polypeptide from the
ribosome.
Lac Operon
 The concept of operon was proposed by Jacob and Monod. Operon is a
unit of prokaryotic gene expression.

 The lac operon consists of one regulatory gene (the i-gene) and three
structural genes (z, y and a).
 The i-gene codes for repressor of lac operon.
 Promoter - It is the site where RNA-polymerase binds for transcription.
 Operator—acts as switch for operon.
 Lactose is an inducer.
 Operator : Act as switch for operon.
 Gene z—Codes for b-galactosidase
Gene y—Codes for permease
Gene a—Codes for transacetylase.
In the absence of Inducer (lactose)
Repressor (i-gene) binds with operator (o)
¯
Operator (O) turns off
¯
RNA polymerase stops the transcription
¯
structural genes (z, y and a ) do not produce lac mRNA and enzymes
In the presence inducer (lactose)
Repressor binds to inducer (lactose)
¯
Operator (O) turns ON
¯
RNA polymerase starts the transcription

Structural genes (z, y and a) produce mRNA and enzymes
( -galactosidase, permease and transacetylase respectively)
Human Genome Project was a 13 year project coordinated by the U.S.
Department of energy and National institute of Health, it was completed in 2003.
Important goals of HGP
(i) Identify all the approzimately 20,000-25,000 genes in human DNA.
(ii) Determinate the sequence of the 3 millon chemical base pairs that make
up human DNA.
(iii) Store this information in database.
(iv) Transfer the related technologies to other sectors, such as industries.
(v) Address the ethical, legal and social issues (ELSI) that may arise from
the project.
Methodologies-2 major approaches
Expressed Sequences Tags (ESTs) Sequence Annotation
(identifying all genes that are (Blind approach of simply
expressed as RNA sequencing the whole set of genome)

Steps for Sequencing :
 DNA isolated from cell and converted into fragments.
 DNA is cloned for ampli®cation is suitable host using specialssed vectors.
 Commonly used hosts—Bacteria, Yeast
 Commonly used Vectors—BAC (Bacterial Arti®cal chromosomes)
YAC (Yeast Arti®cial Chromosomes)
International Rice Genome Sequencing Project (IRGSP)
 Rice bene®ts from having the smallest genome of the major cereals, dense
genetic maps.
 The IRGSP, formally established in 1998, pooled the resources of
sequencing groups in 10 nations (Japan, Korea, UK, Taiwan, China,
Thailand, India, United States, Canada and France)
 Estimated Cost— $ 200 million.
 India joined in June 2000 and chose to sequence a part of chromosome 11.
 Tools used in sequencing were :
BAC (Bacterial Arti®cial chromosomes)
PAC (P1-Phase derived arti®cial chromosomes)
 How Sequenced
Shotgun sequencing involved—generation of short DNA fragments that
are then sequenced and linearly arranged.
It enables full coverage of the genome in a fraction of time required for
the atternative BAC sequence approach.
 Salient Features of Rice Genome
Rice is monocarpic annual plant, wind pollinated. It is with only 389 base
pairs.

The world’s ®rst genome of a crop plant that was completely sequenced.
2,859 genes seem to be unique to rice & other cereals.
Repetitive DNA is estimated to constitute at least 505 of rice genome. The
transposon content of rice genome is at least 35%.
 Applications
To improve ef®ciency of Rice breeding.
To improve nutritional value of rice, enhance crop yield by improving
seed quality, resistance to pests and diseases and plant hardiness.
DNA Fingerprinting :
It is a technique of determine nucleotide sequence of certain areas of DNA
which are unique to each individual.
Principle of DNA Fingerprinting : Short nucleotide repeats in the DNA
are very speci®c in each individual and vary in number from person to person
but are inherited. These are Variable Number Tandem Repeats. (VNTRs.) Each
individual inherits these repeats from his/her parents which is used as genetic
markers. One half of VNTR alleles of the child resembles that of mother and
other half the father.
Steps/Procedure in DNA Fingerprinting
 Extraction of DNA—using high speed refrigerated centriguge.
 Ampli®cation—many copies are made using PCR
 Restriction Digestion—using restriction enzymes DNA is cut into
fragments.
 Separation of DNA fragments—using electrophoresis agarose polymer
gel
 Southern Blotting : Separated DNA sequences are transferred on to
nitrocellulose or nylon membranes.

 Hybridization : The nylon membranes exposed to radio active probes.
 Autoradiography : The dark bands develop at the probe site.
Applications of DNA Fingerprinting
(i) identifycriminalsiftheirDNAfromblood,hairfollicle,skin,bone,saliva,
Sperm etc is available in forensic labs.
(ii) determine paternity
(iii) Verify whether a hopeful immigrant is really close relative of an already
established resident.
(iv) identity racial groups to rewrite biological evolution.

Questions
VSA (I Mark)
1. Name the factors for RNA polymerase enzymes which recognises the start
and termination signals on DNA for transcription process in Bacteria.
2. RNA viruses mutate and evolve faster than other viruses. Why ?
3. Name the parts ‘X’ and ‘Y’ of the transcription unit given below.
4. Name the two initiating codons
5. Write the segment of RNA transcribed from the given DNA
3¢ – A T G C A G T A C G T C G T A – 5¢ – Template Strand
5¢ – T A C G T C A T G C A G C A T – 3¢ – Coding Strand.
SA-I (2 Marks)
6. The process of termination during transcription in a prokaryotic cell is being
represented here. Name the label a, b, c and d.
7. Give two reasons who both the standard of DNA are not copied during
transcription.
8. State the 4 criteria which a molecule must ful®ll to act as a genetic material.

SA-II (3 Marks)
9. Give six points of difference between DNA and RNA in their structure
chemistry and function.
10. Explain how does the hnRNA becomes the mRNA.
OR
Explain the process of splicing, capping and tailing which occur during
transcription in Eukaryotes.
11. Name the three major types of RNAs, specifying the function of each in the
synthesis of Polypeptide.
12. A tRNA is charged with the aminoacids methionine.
(i) Give the anti-codon of this tRNA.
(ii) Write the Codon for methionine.
(iii) Name the enzyme responsible for binding of aminoacid to tRNA.
LA (5 Marks)
13. State salient features of genetic code.
14. Describe the process of transcription of mRNA is an eukaryotic cell.
15. Describe the various steps involved in the technique of DNA ®ngerprinting.
Answers
VSA (1 Mark)
1. Sigma ( ) factor and Rho( ) factor
2. OH group is present on RNA, which is a reactive group so it is unstable and
mutate faster.
3. X Template strand, Y – Terminator.
4. AUG and GUG
5. 5¢ – U A C G U C A U G C A G C A U – 3¢ (In RNA ‘T’ is replaced by ‘U’)
SA-1 (2 Marks)
6. (a) DNA molecule
(b) mRNA transcript
(c) RNA polymers
(d) Rho factor

7. (a) IfboththestrandsofDNAarecopied,twodifferentRNAs(complementary
to each other) and hence two different polyeptides; if a segment of DNA
produces two polypeptides, the genetic information machinery becomes
complicated.
(b) The two complementary RNA molecules (produced simultaneously)
would form a double-stranded RNA rather than getting translated into
polypeptides.
(c) RNApolymerase carries out polymerisation in 5¢ – 3¢ direction and hence
the DNAstrand with 3¢ – 5¢ polarity acts as the template strand. (Any two)
8. (i) It should be able to generate its replica.
(ii) Should be chemically and structurally stable.
(iii) Should be able to express itself in the form of Mendelian characters.
(iv) Should provide the scope for slow changes (mutations) that are necessary
for evolution.
SA-II (3 Marks)
9. DNA RNA
(i) Double stranded molecules Single stranded molecules
(ii) Thymine as pyrimidine base Uracil as pyrimidine base
(iii) Pentose sugar is Deozyribose Sugar is Ribose
(iv) Quite stable and not very reactive 2¢-OH makes it reactive
(v) Dictates the synthesis of Perform other function in protein
Polypeptides synthesis.
(vi) Found in the nucleus. They are transported into the
cytoplasm.
10. hnRNA is precursor of mRNA. It undergoes
(i) Splicing : Introns are removed and exons are joined together.
(ii) Capping : an unusual nucleotide (methyl guanosine triphosphate is
added to the 5¢ end of hnRNA.
(iii) Adenylate residues (200-300) are added at 3¢ end of hnRNA.
Or
Refer ®g. 6.11, page 110, NCERT book. Biology-XII

11. (i) mRNA-(Messenger RNA) : decides the sequence of amino acids.
(ii) tRNA-(Transfer RNA) : (a) Recognises the codon on mRNA(b) transport
the aminoacid to the site of protein synthesis.
(iii) rRNA (Ribosomal RNA) : Plays the structural and catalytic role during
translation.
12. (a) UAC (b) AUG
(c) Amino-acyl-tRNA synthetase.
LA (5 Marks)
13. Refer page 6.9.1., Page No. 120 NCERT Biology XII.
14. Refer notes 35 and ®gure 6.11, page 110, NCERT Biology XII.
15. Refer points to remember. Steps involved in DNA ®ngerprinting.


Genetics : Molecular basis of Inheritance

More Related Content

What's hot (20)

Similar to Genetics : Molecular basis of Inheritance (20)

More from Eneutron (20)

Recently uploaded (20)

Genetics : Molecular basis of Inheritance