6 molecular basis of inheritance extra

MOLECULAR BASIS
OF INHERITANCE
Part (I)
Class XII- (CH 6) BIOLOGY Sushama Mudgal 1

1. What is DNA polymerase?
2. What is central dogma?
3.What is Nucleosome model?
4.What is the aim of Human Genome Project?
5. How DNA fingerprinting is done?
2

Deoxyribonucleic acid (DNA) and ribonucleic acid
(RNA) are the two types of nucleic acids found in
living systems.
Transcription
Translation
Reverse transcription
3

Anticodon : A sequence of three nitrogenous bases on
tRNA which is complementary 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.
4

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 transcriptionally active, it stains lighter.
 Heterochromatin : The chromatin that is more
densely packed, stains dark and is transcriptionally
inactive.
5

 Cistron : A segment of DNA that is equivalent to a
gene and that specifies a single functional unit (such
as a protein or enzyme)
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.
6

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
DNA M-RNA Protein synthesis
Transcription Translation
7

YAC : Yeast Artiﬁcial Chromosome
BAC : Bacterial Artiﬁcial Chromosome
 SNPs : Single Nucleotide polymorphism
HGP : Human Genome Project
 hnRNA : Heterogenous nuclear RNA. It is
precursor of mRNA.
8

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
9

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.
10

 (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 ” 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.
15

The length of DNA is usually defined as number of
nucleotides (or a pair of nucleotide referred to as base
pairs) present in it.
 This also is the characteristic of an organism.
For example, a bacteriophage known as φ 174 has 5386
nucleotides,
 Bacteriophage lambda has 48502 base pairs (bp),
 Escherichia coli has 4.6 × 106 bp, and
haploid content of human DNA is 3.3 × 109 bp.
17

1. 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 DNA in nucleoid is organised in large loops held
by proteins.
18

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.
20

The beads-on-string structure in chromatin is
packaged to form chromatin ﬁbres that are further
coiled and condensed at metaphase stage of cell
division to form chromosomes.
 The packaging of chromatin at higher level requires
additional set of protein that collectively are
referred 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.
21

The linker DNA is a
variable. The length of
linker DNA varies from
organism to organism.
Typically, it is ranging
from 20 to 60bp. The
length of the
nucleosome DNA is
147bp.
22

Physical Bases of inheritance was understood by
Mendel and Sutton and Bovrei (1926). But What
molecule is responsible for heredity was not clear.
1. Transformation : In 1928, Frederick Griffith, in
Streptococcus pneumoniae (bacterium responsible
for pneumonia), witnessed a miraculous
transformation in the bacteria.
23

Oswald Avery, Colin MacLeod and Maclyn McCarty
(1933-44), the genetic material was thought to be a
protein.
They purified biochemical (proteins, DNA, RNA, etc.)
from the heat-killed S cells to see which ones could
transform live R cells into S cells.
 They discovered that DNA alone from S bacteria
caused R bacteria to become transformed.
25

1. Proteins Transformation takes place
So, protein is not a transforming principle.
 RNA Transformation takes place So, RNA
is not a transforming Principle.
 Digestion with DNase did inhibit transformation,
suggesting that the DNA caused the transformation.
 They concluded that DNA is the hereditary material,
but not all biologists were convinced.
Proteases
RNase
26

MOLECULAR BASIS
OF INHERITANCE
Part(II)
Class XII- (CH 6) BIOLOGY Sushama Mudgal 27

Can you think of any
difference between DNAs and
DNase?
28

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.
29

 Experiment I Experiment II
 Bacteriophage Bacteriophage
Radioactive (S35) Radioactive (P32)
 Infection : E. coli E. coli
 Blending : Viral coats removed from the bacteria.
 Centrifugation
Conclusion : DNA is
the genetic material.
30

(i) It should be able to generate its replica
(Replication).
 (ii) It should chemically and structurally be stable.
 (iii) It should provide the scope for slow changes
(mutation) that are required for evolution.
(iv) It should be able to express itself in the form of
'Mendelian Characters’.
32

Watson-Crick model for
semiconservative DNA
replication
35

 It was shown first in Escherichia coli and
subsequently in higher organisms, such as plants and
human cells.
 Matthew Meselson and Franklin Stahl performed the
experiment in 1958.
36

 In eukaryotes, the replication of DNA takes
place at S-phase of the cell-cycle.
 The replication of DNA and cell division cycle
should be highly coordinated.
 A failure in cell division after DNA replication
results into polyploidy(a chromosomal
anomaly).
39

 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 caesium 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.
40

1.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
2.DNA polymerase attaches to the replication fork
and add nucleotides complementary to the parental
DNA strand. The direction of polymerisation is 5’-3’.
3. DNA polymerase cannot initiate the
polymerisation itself, so a small segment of RNA
called primer is attached at replication start point
42

4. 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
5. 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.
6. Accuracy of polymerisation is maintained by Proof
reading and any wrong base added is removed by
DNA polymerase
44

Transcription in Prokaryotes :
 In prokaryotes the process of transcription is
completed in three steps :
 1. Initiation : RNA polymerase binds with initiation
factor (sigma factor) and then binds to promotor
site.
46

 2. Elongation : 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 RNA polymerase binds with rho factor
(terminator factor) As a result nascent RNA
separates.
47

 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.
RNA Polymerase II : Transcribes hnRNA (which is
precursor of mRNA).
 RNA Polymerase III : Transcribes tRNA, 5 srRNA and
sn RNA.
49

The primary transcription has both exon and intron regions.
 Introns which are non-coding regions removed by a process
called splicing.
50

 hnRNA undergoes Two additional process :
(a) Capping : An unusual nucleotide
(methylguanosine triphosphate) is added to 5end
of hnRNA.
(b) Tailling : Adenylated residues (200-300) are
added at 3’end. It is fully processed hnRNA. Now
called mRNA is transported out of the nucleus.
51

MOLECULAR BASIS OF
INHERITANCE
Part (3)
XII Biology /Ch 6 Sushama Mudgal 54

The process of translation requires transfer of genetic
information from a polymer of nucleotides to a polymer of
amino acids.
Determining the biochemical nature of genetic material
and the structure of DNA was very exciting, the proposition
and deciphering of genetic code were most challenging
it required involvement of scientists from several disciplines
– physicists, organic chemists, biochemists and geneticists.
 It was George Gamow, a physicist, who argued that since
there are only 4 bases and if they have to code for 20 amino
acids, the code should constitute a combination of bases
55

(i) The codon is triplet. 61 codons code for amino
acids and 3 codons do not code for any amino acids,
hence they function as stop codons. (so total 64)
(ii) One codon codes for only one amino acid, hence,
it is unambiguous and specific.
 (iii) Some amino acids are coded by more than one
codon, hence the code is degenerate.
57

 (iv) The codon is read in mRNA in a contiguous
fashion. There are no punctuations.
(v) 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.
(vi) AUG has dual functions. It codes for Methionine
(met) , and it also act as initiator codon.
58

A) If following is the sequence of nucleotides in
mRNA, predict the sequence of amino acid coded by
it (take help of the checkerboard):
-AUG UUU UUC UUC UUU UUU UUC
B) Now try the opposite. Following is the sequence
of amino acids coded by mRNA. Predict the
sequence of nucleotide in the RNA:
Met-Phe-Phe-Phe-Phe-Phe-Phe
59

C) Can you now correlate which two properties of
genetic code you have learnt?
60

Point mutation is a change in a single base pair of
DNA by substitution, deletion, or insertion of a
single nitrogenous base. An example of point
mutation is sickle cell anaemia. It involves mutation
in a single base pair in the beta-globin chain of
haemoglobin pigment of the blood. Glutamic acid in
short arm of chromosome II gets replaced with
valine at the sixth position.
61

 Francis Crick knew that there has to be a
mechanism to read the code and also to link it to
the amino acids, because amino acids have no
structural specialities to read the code uniquely.
 He postulated the presence of an adapter molecule
that would on one hand read the code and on other
hand would bind to specific amino acids.
 The tRNA, then called sRNA (soluble RNA), was
known before the genetic code was postulated.
64

CLOVERLEAF MODEL OF tRNA
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.
65

Translation refers to the process of polymerization
of amino acids to form a polypeptide.
 The order and sequence of amino acids are deﬁned
by the sequence of bases in the mRNA.
20 amino acids participate in naturally occurring
protein synthesis.
66

First step : charging of tRNA or aminoacylation
of rRNA-here amino acids are activated in the
presence of ATP and linked to speciﬁc tRNA.
(1) Initiation : Ribosome binds to mRNA at the
start codon (AUG) that is recognized by the
initiator tRNA.
67

 (2) Elongation phase : Here complexes composed of
an amino acid linked to tRNA. Sequentially bind to
the appropriate codon in mRNA by 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.
 (3) Termination : Release factors binds to the stop
codon (UAA, UAG, UGA) translation and releasing
the complete polypeptide from the ribosome.
68

TRANSLATION OCCURS ON THE
SEAT OF RIBOSOME
69

A translational unit in mRNA is the sequence of RNA
that is flanked by the start codon (AUG) and the
stop codon and codes for a polypeptide.
 An mRNA also has some additional sequences that
are not translated and are referred as untranslated
regions (UTR).
The UTRs are present at both 5'-end (before start
codon) and at 3'-end (after stop codon).
71

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.
72

 Operatoracts as switch for operon.
 Lactose is an inducer.
 Operator : Act as switch for operon.
 Gene zCodes for b-galactosidase Gene yCodes for
permease Gene aCodes for transacetylase.
73

 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
75

Human Genome Project was a 13-year project
coordinated by the U.S. Department of Energy and the
National Institute of Health.
 During the early years of the HGP, the Wellcome Trust
(U.K.) became a major partner; additional contributions
came from Japan, France, Germany, China and others.
The project was completed in 2003.
Human genome Project was launched in the year 1990.
76

(i) Identify all the approximately 20,000-
25,000 genes in human DNA;
 (ii) Determine the sequences of the 3 billion
chemical base pairs that make up human DNA;
 (iii) Store this information in databases
 (iv) Improve tools for data analysis;
77

(v) Transfer related technologies to other
sectors, such as industries;
 (vi) Address the ethical, legal, and social
issues (ELSI) that may arise from the project.
DNA sequences developed an understanding of
their natural capabilities that can be applied
toward solving challenges in health care,
agriculture, energy production, environmental
remediation.
78

1.Expressed
Sequences Tags (ESTs)
(identifying all genes
that are expressed as
RNA)
2.Sequence
Annotation
(Blind approach of
simply sequencing
the whole set of
genome)
79

DNA isolated from cell and converted into
fragments.
 DNA is cloned for amplification is suitable host using
specialised vectors.
 Commonly used hosts Bacteria, Yeast
 Commonly used Vectors BAC (Bacterial Artifical
chromosomes) YAC (Yeast Artificial Chromosomes
80

MOLECULAR BASIS OF
INHERITANCE
Part
XII Biology /Ch 6 Sushama Mudgal 81

(i) The human genome contains 3164.7
million nucleotide bases.
(ii) The average gene consists of 3000
bases, but sizes vary greatly, with the
largest known human gene being
dystrophin at 2.4 million bases.
82

(iii) The total number of genes is estimated at
30,000–much lower than previous estimates of
80,000 to 1,40,000 genes. Almost all (99.9 per
cent) nucleotide bases are exactly the same in
all people.
83

(iv) The functions are unknown for over
50 per cent of discovered genes.
(v) Less than 2 per cent of the genome
codes for proteins.
(vi) Repeated sequences make up very
large portion of the human genome.
84

(vii) 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.
85

(viii) Chromosome 1 has most genes (2968), and the
Y has the fewest (231).
(ix) 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.
86

 In the past, researchers studied one or a few
genes at a time.
 With whole-genome sequences and new high-
throughput technologies, we can approach
questions systematically and on a much
broader scale.
87

They can study all the genes in a genome,
for example, all the transcripts in a
particular tissue or organ or tumour, or
how tens of thousands of genes and
proteins work together in interconnected
networks to orchestrate the chemistry of
life.
88

Rice beneﬁts 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)
89

 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 Artiﬁcial chromosomes) PAC
(P1-Phase derived artiﬁcial chromosomes)
90

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
alternative BAC sequence approach.
91

 Salient Features of Rice Genome Rice is
monocarpic annual plant, wind
pollinated. It is with only 389 base pairs.
The worlds ﬁrst genome of a crop plant
that was completely sequenced.
92

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%
93

To improve efﬁciency of Rice breeding.
 To improve nutritional value of rice,
enhance crop yield by improving seed
quality, resistance to pests and diseases
and plant hardiness.
94

It is a technique of determine nucleotide
sequence of certain areas of DNA which are
unique to each individual.
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
95

 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.
96

 These sequences normally do not
code for any proteins, but they form a
large portion of human genome.
97

Since DNA from every tissue (such as blood,
hair-follicle, skin, bone, saliva, sperm etc.),
from an individual show the same degree of
polymorphism, they become very useful
identification tool in forensic applications.
Polymorphism (variation at genetic level)
arises due to different kinds of mutations.
98

Short nucleotide repeats in the DNA are very speciﬁc 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.
99

(i) isolation of DNA,
(ii) digestion of DNA by restriction
endonucleases,
 (iii) separation of DNA fragments by
electrophoresis,
100

(iv) transferring (blotting) of separated DNA
fragments to synthetic membranes, such as
nitrocellulose or nylon,
(v) hybridisation using labelled VNTR probe,
and
(vi) detection of hybridised DNA fragments by
autoradiography.
101

DNA extraction from the cells in high speed
refrigerator centrifuge
Amplification of DNA content by PCR
(Polymerase Chain Reaction)
DNA fragmentation by restriction endonucleases
Gel electrophoresis
104

Double stranded DNA split into single stranded
DNA
Southern Blotting or( transferring separated
DNA to nylon/ nitrocellulose sheet )
Nylon is immerse in the bath having probes or
markers(Hybridization)
105

Nylon membrane is pressed on x-ray film
(Autoradiography)
Dark band developed at the probe site
Probes or markers are radioactive synthetic DNA
complimentary to VNTR
106

Few representative chromosomes have been shown to
contain different copy number of VNTR.
 For the sake of understanding different colour schemes
have been used to trace the origin of each band in the gel.
 The two alleles (paternal and maternal) of a chromosome
also contain different copy numbers of VNTR.
 It is clear that the banding pattern of DNA from crime
scene matches with individual B, and not with A.
107

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.
108

(i) Identify criminals if their DNA from blood, hair
follicle, 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.
109

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6 molecular basis of inheritance extra