definition of gene (repaired)
DESCRIPTION
agriTRANSCRIPT
Definition of gene, allele, homozygous, heterozygous, genome, phenotype,
genotype, monohybrid, dihybrid, polyhybrid, backcross, test cross, mendels’ laws
-contributions.
Genetics:
The branch of biology concerned with the study of heredity and variation.
Gene:
A segment of DNA which specifies a single polypeptide, also a segment of
DNA which codes for an RNA molecule (eg. R RNA, tRNA genes) or may serve as a
binding site for regulatory protein.
Allele:
An alternative form of a gene; allele’s occupy the same position or locus on the
homologous chromosomes; alleles do not show complementation, and yield typical
monohybrid ratio in F2.
Homozygous :
An individual having two copies of the same allele for one or more genes under
consideration.
(Adj: homozygous / homozygosity ) – Noun.
Heterozygous :
An individuals having two different alleles for one or more genes.
( Adj: heterozygous, heterozygosity) – noun.
Genome:
A complete set of chromosomes of diploid species : all the members of a
genome are distinct from each other in gene content and often in morphology;
members of a genome do not pair.
Phenotype :
The observable characteristic (of appearance, anatomy etc) of an organism.
Genotype:
The genetic constitution of an organism.
Monohybrid:
The progeny derived by mating or crossing two individuals / strains which
differ for one gene .
Dihybrid:
The progery from a cross between two homozygous parents differing for two
genes ( or characters):- an individual heterozygous for two genes. ( Aa Aa)
Poly hybrid:
The progeny from across between two homozygous parents differing for more
than 2 genes (or characters) an individual heterozygous for three genes ( Aa Aa Aa).
Back cross:
The cross of a F1 hybrid to one of its parents.
Test cross:
The cross of an F1 hybrid with an individual / strain having the recessive
phenotype for the concerned trait.
The Laws of Mendel
A number of scientists had worked plant hybridization during the 18 th and 19th
centuries prior to mendel. Some of the more notable scientists from among them
koelreuter ,John Goss, sargaret ,Gartner ,Darwin,Hebert,Lecog,Vichura and Naudin.
Koelreuter conducted extensive studies on hybridization in tobacco between 1760 and
1766; he noted the uniformity and heterosis in F1 ( first filial generation; filial –
progeny) and the appearance of increased variation in F2. Gartner (1722-1850) ,
Naudin (1815 – 1909), Darwin (1809 – 1882) and others confirmed the observation
and conclusion of koelreuter. Gartner used a back –cross programme to transform /
convert one species into another. Essentially, he transferred the nucleus of one species
into the cytoplasm of another species. He made such a transfer in 30 species belonging
to atleast 8 genera including Nicotiana. The important conclusions are available from
these studies.
i. In hybrids obtained by mating two different varieties of a species / two
distant species, the characters expressed will be dominant. The essence
of the concept of dominance is present.
ii. Characters of F1 and F2 progeny produced from reciprocal crosses are
identical. The observations from the reciprocal crosses clearly
demonstrates that the contribution of male and female parent to the
characters of a progeny are identical.
iii. F1 progeny are uniform in their characters, i.e., all the plants in F1 from
a cross are similar to each other .But F2 generation shows large variation
for different characteristics which is due to consequence of segregation
and recombination.
iv. In F2 generation, some plants have characters similar to one
parent ,while some others are similar to the other parent in their
appearance. The appearance of the parental forms in F2 was termed as
reversion.
v. Some plants in F2 have entirely new character forms, differ from those
of the two parents.
vi. These contributions by earlier workers, was not a plausible explanation
for data, which was so brilliantly accomplished by Mendel.
The reasons for the failure of the experimental approach of Mendel
predecessors, was successfully explained by Mendel as follows;
i. These scientists studied the plant as a whole i.e its total
appearance of a large number of characters.
ii. As a result of the above, the plants could not be classified into
few clear-cut classes.
iii. More concerned about the description of various forms
appearing in the progeny. Determination of frequencies of the
data was not made.
iv. Not maintained accurate and separate record.
v. Complete control on pollination in the F1 was lacking.
vi. The F1 was an inter-specific hybrid exhibiting partial
considerable sterilizing, would disturb the ratio of various
characters in F2.
vii. Number of plants studied in F2 was relatively small.
viii. Most of the characters studied – quantitative not qualitative.
Lethal genes, pleiotropy, Phenocopy, penetrance, expreosivity, allelic
interacticor, types, complete dominance, incomplete dominance, co dominance
and over dominance with examples.
A lethal gene causes the death of all the individuals carrying this gene in the
appropriate genotype before these individuals reach adulthood. The appropriate
genotype for an allele would depend on its dominance relationship with its other allele
for an allele producing a recessive effect on survival ( sec, recessive lethal).
i. The appropriate genotype would be the homozygous state.
ii. While for an allele having a dominance effect on survival both
homozygous and heterozygous states would be the appropriate
genotype.
The ‘Y’ gene in mice has a dominant phenotypic effect on coat colour, but is a
recessive lethal. The two characteristics feature of a recessive lethal are given below,
i. They are always present in the heterozygotes for a recessive lethal gene
yields a 2:1 ratio ( instead of the typical 3:1 ratio)
Example – 1
Parents – Yy x Yy
Yellow Yellow
O Y y
Y YY
Dies
Yy
Yellow
y Yy yy
Y y Y yGametes ----
Progeny
Phenotypic ratio: 2Y: 1grey
Genotypic ratio:2Yy:1yy
yellow Grey
Thus, the heritance of yellow coat colour in mice may be explained as follows.
The dominant allele Y is a recessive lethal and that YY embryos do not survive. As a
result, all the yellow mice are heterozygotes ( Yy) for this gene. Therefore mating of
yellow females with yellow males would produce the following four zygotes.
i. One fourth YY, 2 one half Yy and one forth yy. Homozygous YY
embryos will die at an early stage. Heterozygous Yy embryos will
develop normally and give rise to yellow mice while yy embryo will
develop in to grey individuals. Thus the phenotypic ratio obtained in the
yellow x yellow matings will be 2 : 1 in place of the typical
monohybrid ratio of 3:1.
Example – II
Inheritance of a gene producing albino seedlings in plants.
Albino seedlings almost white, are unable to produce their own food and dies as soon
as seed reserve.
Parents – Aa x Aa
Green Green
A a
A AA
Green
Aa
Green
A a A aGametes ----
Phenotypic ratio: 3:1 Green : albino Genotypic ratio : 1:2:1 AA: Aa : aa
a Aa
Green
aa
albino
(dies)
Dominant lethal
Some lethal genes reduce viability in the heterozygous state, as well such genes are
known as dominant lethals.
An example of a dominant lethal is the epiloia gene in human being. This gene
causes abnormal skin growths, senere mental defects and multiple tumors in the
heterozygote’s. So that they die before reaching adulthood. Dominant lethal, therefore
cannot be maintained in the population while recessive lethal are maintained in the
heterosing state. Thus the dominant lethal have so be produced in every generation
through mutation.
Conditional lethals:
Lethal genes that require specific condition for their legal action are term as
conditional lethal. Many mutants Dropsophila, Neurosperr, barey, maize and many
other organism.
Balanced lethal:
Lethal genes linted in repulsion phase; they are maintained in this phase due to
light lonkge or crossing over suppression only their heterozygots …………….
Plecotropy:
A single gene affecting more than one character. Such genes are known as
pliotropis genes and the conditions is termed as pleiotropy. An example of a
pleiotropic gene in human being to the gene and ( some forms denoted as ….. which
produces. Sickle cell anemis in the homozygotes. More than 50% of the individual
homozygous for this gene (ss) die before the age of 20 years.
Penetrance:
In general, genes express themselves in all the individuals present in the
appropriate genotype; This is coorplete penetrance. But many genes do not produce
the concerned phenotype in all the individuals which carry them in the appropriate
genotype. Such a Situation in known as incomplete preference When a gene is present
in the appropriate genotype, the percent individuals in which it is able to express itself
is a measure of its peretrance. Example – chlorophyee deficeeing gene. In lina beans
has a penetrance of 10%.
Expressivity:
The ability of a gene to produce identical genotype in all the individuals
carrying it in the appropriate genotype is known as complete expressivity.
Incomplete expressivity:
They produce variable pyenotype in all the individuals that have this gene in
the appropriate genotype.
Example – Partial chlorophyle deficiency in the cotyledenory leaves of lima
beem (Varying degree of chlorophyll deficiency.
Gene interaction:
The phenomenon of two or more gene affecting the expression of each other in
various ways in the development of a single character of an organism is known as
gene interation.
Complete dominance:
The phenotype produced by heterozygote’s is identical with that produced by
homozygotes for the concerned dominant allele.
aa AA
Aa Aa Aa
Incomplete Complete Over
Dominance dominance dominance
Location of the heterozygote Aa in relation to the two hornozygotes ( aa and
AA) complete, incomplete and over dominance relation.
Incomplete dominance:
In many cases, the intensity of phenotype produced by the heterozygote is less
than that produced by the homozygotes for the concerned dominant allele. Therefore,
the phenotype of below zygotes falls between those of the homozygous for the
concerned alleles. Such a situation is known as imcomplete / partial dominance.
For example
Parents RR X rr
Red white
O R r
R RR Rr
r Rr rr
Phenotypic ratio - 1 Red : 2 Pink : 1 white
Genotypic ratio - 1 RR : 2 Rr : 1 rr
Incomplete dominance for flower colour in four O clock plant ( Mirablis
jalaba)
Codominance:-
In codominance, both the alleles of gene express them levels in the
heterozygotes.
R r
Rrpink
R r
Gametes
Gametes
F1
Red
Pink
White
Example: Blood group antigeas of man present excellent examples of co dominance.
Parents - IAIB X IAIB
Blood group AB Blood group AB
O IA IB
IA IA IA
BG-A
IA IB
BG-AB
IB IA IB
BG-AB
IB IB
BG-B
Phenotypic ratio: IA : 2AB : 1BB
Genotypic ratio : 1 I A I A : 2 IAI B : 1IB IB
One of the most widely known and the earlier revegnised, human bloced
groups is the ABO blood group. These blood group arise due to the presence / absence
of an antigen on the surface of red blood cells, these antigens are produced by the
gene 1.One dominant allele of this, gene, IA produces antigen A which give rise to
blood group ‘A’.
Another dominant allele of the gene 1, IB produces antigen B which is
responsible for blood group B. the heterozygote IA I B both alleles IA and IB produce
their respective antigens, as a result, the heterozygote’s are classified in the AB blood
group.
IA IB IA IB
Gametes ----
Over dominance:
In some casen, the intensity of character governed by them is greater in
heterozygous then in the two concerned hernozygots. This situation is known as over
dominance.
The white eye (ee) gene of drosophila extubits over dominance for some of
the eye pigments.
Example- Sepiapteridine who himmerblans.
Allele us produces white eye in the homozygous state curue while its
completely dominant allele W give rise the normal dull red colour both in the
homozygous as well as heterozygous states (WW and Ww). Eyepigments
sepiapteridine and himmel blaus are present in low concenteatwins of these pigments
in crore homozygotes., While WW homozygotes have relatively higher concentration
of these pigments. However, …..heterozygotes for the gene (Ww) have an appreciably
higher concentration of these two pigments than the two hornzygotes ( WW and ww).
Non-allelic interaction types of epistasin – non allelic interaction without
modification in mendelian ratio- Bateson and Punnethe’s experiment of foul comb
shape, epistasis with modifications of mendelian rations – 12:3:1, 9:3:4, 9:6:1, 15:1,
9:7, 13:3.
The phenomenon of two or more characters affecting the expression of each
other in valious ways in the development of character of an organism is known as
gene interaction.
Types of gene interaction
i. Typical dihybrid ratio ( for a single trait) 9:3:3:1
ii. Duplicate gene action (15:1)
iii. Complementary gene action (9:7)
iv. Supplementary gene action (9:3:4)
v. Inhibitory gene action (13:3)
vi. Masking gene action (12:3:1)
vii. Polymeric gene action ( 9:6:1)
The type of gene interaction produces the typical dihybrid ratio of 9:3:3:1 in F2
for a single character.
Obviously, the concerned characters is contributed by genes exhibiting full
dominance.
The dominant alleles of each of the two genes produce separate forms of the
character ( Phenotype) when they are alone ie. When the dominant allele of one gene
is present with the homozygous recessive rose allele of the other locus.
But when the dominant alleles of both the genes are present together, they
produce a distinct phenotype, the humozugous recessive state at both the locus
giving rife to yet another phenotype.
In chickens, comb shape is governed by two genes, p and r. The dominant
allele of gene p (P) alone (with rr) produes pea comb, while that of gene, r (R) alone
(with pp) produces rose comb. But when both the dominant genes P and R are
present together e.g. PPRR, they give rise to walnut comb. Recessive condition at
both lou e.g. pprr given rise to distinct comb shape called single. When a breed of
poultry homozygous for pea comb (PPrr) is crossed with another breed homozygous
for rose comb ( ppRR) I the F1 ( PpRr) has walnut comb as it has the dominant
alleles of both the genes Pand R. segregate for the two genes produce 16 possible
zygotic combinations in F2. Nine of these combinations have
Parents PPrr X PPRR
pea Rose
Pr PR
PpRrWalnut
Gametes
F1
O PR Pr pR Pr
PR PPRR
Walnut
PPRr
Walnut
PpRR
Walnut
PpRr
Walnut
Pr PPPr
Walnut
PPrr
Pea
PpRr
Walnut
Pprr
Pea
pR PpRR
Walnut
PpRr
Walnut
ppRR
Rose
PP Rr
Rose
pr PpRr
Walnut
Pprr
Pea
ppRr
Rose
Pp rr
Single
Phenotypic ratio: 9 walnut : 3pea : 3 Rose :1 single
Typical dihybrict ratio (9:3:3:1) in case of a single character e.g. comb shape
poultry.
Duplicate gene action (15:1)
Duplicate dominant epitasis
Characters showing duplicate gene action are determined by two completely
dominant genes. These dominant genes produce the same phenotype whether they are
alone (i.e. with the recessive allele of the other gene) or together, the contracting
phenotype is produced only when both the genes are in the homozygous recessive.
Parents DW1DW1 DW2Dw2 X dw1dw1 dw2dw2
Non floating floating
DW,DW2 dw1 dw2
Dw1 dw1 DW2 dw2 Non floating
Gametes
F1
O DW1 DW2 DW1 dw2 dw1 DW2 dw1 dw2
DW1 DW2 DW1 DW1
DW2 DW2
DW1 DW1
DW2 dw2
DW1 dw1
DW2 DW2
DW1 dw1
DW1 dw2
DW1 dw2 DW1 DW1
DW2 dw2
DW1 DW1
dw2 dw2
DW1 dw1
DW2 dw2
DW1 dw1
DW2 dw2
dw1 DW2 DW1 Dw1
DW2 DW2
DW1 dw1
DW2 dw2
DW1 Dw1
DW2 DW2
dW1 dw1
DW2 dw2
dw1 dw2 DW1 DW1
DW2 dw2
DW1 DW1
dw2 dw2
dw1 dw1
DW2 dw2
dw1 dw1
dw2 dw2
Phenotypic ratio : 15: 1
Non floating : floating
Fig: Duplicate gene interaction in the development of floating habit in rice.
Complementary gene action (9:7)
Or
Duplicate recessive epitasis
In this type of gene interaction, the production of one of the two phenotypes of
a trait requires the presence of dominant nlleles of both the genes controlling the
N N N N
N N N N
N N N N
N N N N
concerned trait, when any one of the two / both the genes are present in the
homozygous recessive state, the contrasting phenotype to produced.
In sweet pea, the development of purple (or coloured) flower requires the
presence of two dominant genes, C and R, e.g., CCRR. When either C(eg., cc RR) or
eg. CC rr) or both the genes ( eg., cc RR) or R (eg. CC rr) or both the genes (eg. cc rr)
are present in homozygous recessive condition, purple flower colour as a result of
which white flowers are obtained.
Therefore complementary gene action modifies the typical 9:3:3:1 ratio in to a
9:7 ratio in F2
Parents CCRR X ccrr
Purple White
CR cr
CcRrPurple
Gametes
F1
O DW1 DW2 DW1 dw2 dw1 DW2 dw1 dw2
CR CCRR CCRr CcRR CcRr
Cr CCRr CCrr CcRr Ccrr
cR CcRR CcRr ccRR ccRr
cr CcRr Ccrr ecRr ccrr
P- Purple W- white
Phenotypic ratio : 9:7
Purple : White
Fig : Complementary gene interaction in the development of of lower colour in sweet
pea giving rise to the 9:7 ratio in F2.
Supplementary gene action ( 9:3:4)
In this gene interaction, the dominant alleles of one of the two genes governing
a character produces a phonotypic effect. However, the dominant allele of the other
gene does not produce a phenotypic effect of its own, but when it is present with the
dominant allele of the first gene it modifies the phenotypic effect produced by that
gene.
When a maize in bred with puple grains ( RR Pr Pr) is crossed with an inbred
having while grains and the genotype rr pr Pr, the F1 ( Rr Pr Pr) plants produce purple
grains. In the F2, 9 …. Of the 16 zygote combination will have dominant allele of
both the genes R and Pr, they will develop into puple grains. There (3) out of the 16
zygotes will have dominant allele of the gene R but will be homozygous for Pr (e.g.
RR Pr Pr) ; these grains will develop into red colour since the recessive allele pr has
no effect on colour production. Three other zygotes will be homozygous or but will
have the dominant allele of pr ( e.g. rr Pr Pr) ; these seeds will be white since rr is
P P P P
P W P W
P P W W
P W W W
unable to produce any colour and Pr does not produce any colour. The remaining one
zygote will be homozygous recessive for both the genes ( rr pr Pr0 and will produce
white seed. As a resoult, the 9:3:3:1 ratio is modified into 9 puple : 3 red : 4 white
parents - (purple) RR Pr Pr x rr Pr Pr ( white)
O RPr RPr rPr RprRPr RRPrPr RRPrPr RrPrPr RrPrPr
RPr RRPrPr RRPrPr RrPrpr Rrprpr
rPr RrPrPr RrPrPr rrPrPr rrPrPr
rpr RrPrpr Rrprpr rrPrpr rrprpr
P – Purple, R – Red , W-White
Phenotypic ratio : 9 purple : 3 red: 4 while
Supplementary gene action in the development of grain (ale ………….) colour in
maize.
Inhibitory gene action (13:3)
Dominant inhibitory epitasis
Inhibitory gene action, me of the two completed dominant genes produces the
concerned phenotype of the character, while its recessive allele in homozygous state
P P P P
P R P R
P P W W
P R W W
RPr rpr
Rr Pr pr
Gametes
F1 Purple
produces the contrasting phenol…… The second dominant gene has 20 effect of its
own on the character in question. However it has the ability to stop the expression of
the dominant allele of the first gene. As a result when the two dominant genes are
present together they produce the same phenotype as that produced by the recessive
homozygote of the first gene. The recessive allele of the second gene does not affect
the development of the character in any way. Then in inhibitory gene, one dominant
gene is capable of producing a character only if its to expression is not preventeel by
another dominant gene known as inhibitory gene and denoted by I. The 9:3:3:1 ratio is
modified to 13:3 ratio in to care.
An example of inhibitory gene interaction occurs in the development of
aleuronic colour is maize. A dominant gene R produces red colour, while its recessive
allele ‘r’ produces no colour. Another dominant gene I does not produce any colour by
itself, it only percents the colour production by R, when both I and R are present
together. The recessive allele I does not affect in any way the colour production in
maize aleurone. As a result, red colour (or any other colour) in alewone is produced
only when R is present with the homozygous recessive state of the inhibitory locus
(e.g. RRii)
Parents RR ii x rr II
Gametes Ri rI
Rr Ii
WhiteF1
Red White
O RI Ri rI Ri
RI RRII RRIi RrII RrIi
Ri RRIi RRiI RrIi Rrii
rI RrII RrIi RrIi rrIi
Ri RrIi Rrii rrIi rrii
W= White
R = Red
Fig: inhibitory gene action for seed colour in maize modifying to dihybrid ratio of
(9:3:3:1) into 13:3.
Masking gene action ( 13:3:1) Dominant epitasis
In this interaction , the two genes affecting the same character produce district
phenotypes when they are alone.
But when both the genes are present together, the expression of one gene
masko the expression of the other. When both the genes are present in the recessive
state, a different phenotype is produced.
This type of gene infraction us district from that of inhibitory gene actions.
In this case, one gene does not inhibit the expression of the other gene, as this
case of inhibitory gene action.
In fact, both the genes express themselves when they are present together, but
the expression of one gene is so intense or strong that the expression of the other gene
can not be observed. That is why such a gene interaction has been formed masking
gene action.
In barley, seed coat, colour is govern by two dominant gene y B and Y Gene B
leads to the development of Black colour. The other gene Y produces yellow and
coat, while its recessive allele (yy) gives rise to white seed coat colour.
W R W W
W W W W
W
W
R
W
W
W
R
W
When the dominna talleles of B and Y are present together, both the genes
express thselves. Howerver, the black colour produced by gene B is so intense that it
does not permit the detection of yellow colour produced by to Y gene.
Parents BB yy x bb YY
O BY By bY by
BY BBYY BBYy BbYY BbYy
By BBYY BByy BbYy Bbyy
bY BbYY BbYy bbYy bbYy
by BbYy Bbyy bbYy bbyy
12: bend : 3 Yellow : 1 White
B B B B
Gametes By
by
BbYy
Black
F1
Black yellow
B B B B
B B Y Y
B Y Y B
Fig: Masking gene action in the control of seed coat in barley.
Polymeric gene action (9:6:1)
In polymeric gene action, the two genes controlling a character produce
identical phenotype when they are alone ( e.e. with the homozygous recessive
condition of the other gene).
But when both the genes are present together, their phenotypic effect is
enhanced as if the effects of the two genes were cumulative or additive.
In this case both the genes show complete dominance.
If the two genes showing polymeric gene action also show an absence of
dominance an addition gene effect will be the consequence.
In barley, two completely dominant genes A and b affect the length of auns, the
this needle like extensions of lemma. Gene A or B alone ( AA bb and BB)
respectively, genes rise to medium length awn. The effect of A is the same as that
of B.
But when both the genes A and B present together, they produce long ….
Indicating that the effects of A and B on own length are added together.
Individuals homozygous recessive for both these genes are own less
. Parents AA BB x aa bb
Gametes AB
ab
AaBb
Long own
F1
Long own Own less
O BY By bY by
BY BBYY BBYy BbYY BbYy
By BBYy BByy BbYy Bbyy
bY BbYY BbYy bbYy bbYy
by BbYy Bbyy bbYy bbyy
9 long awn : 6 Medium awn : 1 awn less
Fig: Polymeric gene action determining awn development in Barley.
Gregor Johnmendel
The choice of pea for hybridization by mendel was the consequence of the
understanding of the problems of such studies .
Pea plant varieties available commercially
Several characters had two contrasting forms and easily distingshable.
Pea plant has ensured self pollination
Flowers are very large, emasculation and pollination of flower is quite easy.
Single season crop.
Seeds are large, no problem ingemination.
Reasons for mendel’s success
Success depend on kis ability for accurate and incisive analysis of the reasons
for failure of earlier workers
L M M M
L L L L
L
L
M
L
L
M
M
M
Diagnosed the ……. Of their experiential material techniques, and approaches
and avoided them.
Studied the inheritance of only one pair of contrasting characters at a time.
Selected pea varieties that had clearly different forms of one / more
characters. ( round / wrinkled seeds, yellow and green, cotyledons, green and
yellow puds).
Classification is based on the contrasting characters.
Carried out the experiments care and elaborateness.
Formulated appropriate hypotheses on the basis of the explanation.
On the basis of the findings, mendel proposed the first of the two fundamental
principles of genetics, the law of segregation. The law of segregation can be
explained more clearing the by making the following supposition.
1. A character is produced by a specific gene.
2. Each gene has two alternative forms these form are known as alleles.
3. The two alleles of gene govern the development of contrasting forms of the
characters governed by the gene.
4. Each somatic cell of an organism has two coping of each gene.
Therefore
Parents WW x ww
Gametes
W w
WasRound
F1
Round Wrinkled
W wGametes
O W w
W WWO WwO
W O
Ww
O
ww
III. The low of independent assortment
Studies on the inheritance of only one characters at a time enable mendel to
stimulate the concept of gene. The essential of the concept are as follows,
i. The development of each character is controlled by a gene.
ii. Each gene exists in two alternative forms called alleles, which govern the
contrasting forms of a character.
iii. The genes are particulate so that the two alleles a gene do not modify teach
other when they exist together in the same cell.
iv. Each somatic cell has two copies of a gene ( identical or distilment alkles)
while gamets have only one copy.
v. The alleles of a gene separate and pass into different gamets of the hybrid.
vi. Genes are the units of inheritance passed from one generation to the next.
Genetic explanation for independent assortment, depicted diagrammatically as
follows,
Parents
Phenotypic ratio:
3 Round : 1 Wrinkled
Genotypic ratio :
1WW: 2Ww : 1ww
WWGGRound yellow
Xwwgg
wrinkled, green
Gametes WG wg
WwGgRound yellow
F1
WG
WG WGWG
O WG Wg wG Wg
WG WWGG(RY)
WWGg(RY)
WwGG(Ry)
WwGg)
(RY)
Wg WWGg(RY)
WWgg(RG)
Wwgg(RY)
Wwgg(RG)
wG WwGG(Ry)
WwGg(RY)
wwGG(WY)
wwGg(wy)
wg WwGg(Ry)
Wwgg(RG)
Wwgg(Wy)
Wwgg(WG)
Characters
Multiple alleles – characteristic features, study of blood group, coat colour in rabbits
and self incompatibility in plants.
Generally assumed that a gene has two alternate forms called alleles.
One of the two alleles of a gene is dominant over the other, which is recessive.
Many cases, several alleles of a single gene are known, each governing a
distinct form of the concerned trait.
This situation is called multiple allelism.
The many alleles of a single gene are called multiple alleles.
Examples of multiple alleles
ABO blood group in man
Fur colour in rabbit
Self incompatibility in plants
Wing type in drosophila
Eye colour in drosophila etc.
1. Fur colour in Rabbit.
In rabbit, there are >2 alternate forms of genes, which controls coat colour. C
causes wild type and its alleles.
CC, Ccch, Cch, Cca Agouti ( wild type)
cchcch, cchch cchc Chinchilla ( salivary gray hair)
chch chc Himalayan ( white except black feet nose ear tail)
cc Albino ( complete white)
Agouti This has full colour and is also known as wild type. This colour is
dominant over all the remaining colour and produces agouti colour in F1 and 3:1 ratio
in F2 when crossed with any of the other three colored rabbits. C represents this
colour.
Chinchilla This lighter than agouti. This colour is dominant over Himalayan
and albino and produces chinchilla in F1 and 3:1 ratio in F2 when crossed either
Himalayan or albino . This is represented by cch
hh allele – produces a temperature – sensitive form of tyrosinase – involved in
production of malaria ( shin and hair pigment).
Dominance relationship ( > ch > ch > c.
Agouti Chinchilla Himalayan Albino
(C) (cch) (ch) (c)
Thus the variation in fur colour in rabbits is due to multiple alleles of a single gene.
2. ABo Blood group in man
Antibodies are a class of proteins, referred as immunoglobulins. It is usually
found in the serum or plasma. Each antibody molecule has antigen binding sites.
Antibodies are produced by B lymphocytes. All the antibody molecules produced by
a single lymphocyte have the same antigen binding specificity. Every individual has a
highly heterogenous population of lymphocytes.
The presence of antibody can be demonstrated by its specific reaction with an
antigen.
Antigen An antigen refers to a substance or agent, which, when introduced
into the system of a vertebrate animal like cow, goat, man etc induces the production
of specific antibody, which binds specifically to this substance. Antigens are located
in the red blood corpuscles ( RBC). If a person has a particular antigen in his RBCs;
his serum has usually antibodies against the other antigen. In human RBC two types
of antigens viz A and B are present. Depending upon the presence or absence of
antigen A and B, the blood group in man is of four types vi A,B,AB and O. A person
with blood group A has antigen A on the surface of RBCs; persons with blood group
B will have antigen B; those with blood group AB have antigens A and B; and those
with blood group O have no antigen on the surface of their RBCs.
Blood
GroupGenotype
Antigen
found
Antibody
present
Compatible blood
group
A IAIA, IAi A B A and O
B IBIB, Ibi B A B and O
AB IAIB AB None A,B,AB,O
O Ii None AB O
Recent studies show that antigen is galactosamine and B is galactose.
Antibodies A B AB and None are naturally present in the serum of individuals having
A,B,AB, and O blood group respectively. The agglutination or coagulation of RBCs
lead to clotting of blood due to interaction between antigen and antibody.
The blood group B cannot be retransferred to an individual having blood group
A, because the recipient has antibody against antigen B, which is present on the RBCs
of blood group B. Similarly the reverse transfusion is not possible. The blood group
AB does not have antibody against antigen A and B. hence individuals with AB blood
group can accept all types of blood, viz, A,B, AB and O. Such individuals are known
as universal acceptors or recipients. The O blood group does not have any antigen and
has antibody against antigen A and B. It can not accept blood group other than O
Individuals with blood group O are known as universal donors, because transfusion of
blood group O is possible with all the four blood types.
Genotypes of progenies obtained due to crosses between various self – sterility
types of Nicotiana.
Seed parents Pollen parent
S1S2 S2S3 S3S4 S4S5
S1S2 S3S2
S3S1
S3S1
S3S2
S4S1
S4S2
S4S1
S4S2
S5S1
S5S2
S2S3 S1S2
S1S3
S4S2
S4S3
S4S2
S4S3
S5S2
S5S3
S3S4 S1S3
S1S4
S2S3
S2S4
S2S3
S2S
S5S2
S5S4
S4S5 S1S4
S1S5
S2S4
S2S5
S2S4
S2S5
S3S4
S3S5
S3S4
S3S5
Features of multiple alleles
Multiple alleles map at the same locus
There is no crossing over between the members of multiple allelic series
( exception exist)
The wild type ( normal) allele of a multiple allelic series is almost
always dominant over the alleles producing mutant phenotypes. The
mutant alleles may show partial or complete dominance.
Multiple alleles always control the same triat of an individual.
A homozygous mutant allele of a multiple allelic series is crossed with
another tain homozygous for another mutant allele belonging to the
same series, the F1 individuals show the mutant phenotype only: they do
not show the wild type phenotype. In other words, multiple alleles do
not show complementation.
Further, F2 generations from such crosses show typical monohybrid
ratio for the concerned trait.
The consideration of Rh (rhesus) type is important in blood tansfusion.
Each blood group has generally two types of Rh group, viz positive and
negative. The same type of Rh is compatible for blood transfusion.
Opposite type leads to reaction resulting in death of the recipient.
Self sterility in nicotiana
Multiple alleles have been associated with self – sterility or self incompatibility
in several groups of plants. Self – sterility is the phenomenon in which the pollen
grains from a plant fail to bring about fertilization I the ovules of the same plant. As
early as 1764 Koelreuter described self-sterility in tobacco, Nicotiana. In 1925 E.M.
East discovered self – sterility in nicotiana is governed by alleles of multiple allelic
series of gene S. Different alleles of this multiple allelic series were designated as
S1,S2,S3,S4,S5,etc., None of the cross – fdertilizing tobacco plants were homozygous,
(i.e., S1S1 or S2S2) but all plants were heterozygous (e.e., S1S2,S3S4,S5S6, etc). When
crosses were attempted between different S1S2 plants, it was observed that pollen tubes
did not develop normally, but pollen from S1S2 were effective on stigmas of plants
with other alleles, for example S3S4.
When crosses were made between seed parents with S1S2 and pellen parents
with S2S3, two kinds of pollen tubes were distinguished. Pollen grains carrying S2
were not effective, but the pollen grains carrying S3 were capable of fertilization.
Thus, from the cross S1S2 x S2S3, two kinds of progeny, S1S3, were produced. From a
cross S1S2 x S3S4, all the pollens were effective and four kinds of progeny resulted:
S1S3, S1S4, S2S3 and S2S4. Some combinations are summarized.
Multiple factor hypotheses – Nilson – Ehle – Wheat Kernel experiment –
Polygenes – Transgressive segregation – Quantitative Vs Qualitative charactors and
modifiers.
Linkage – Coupling and repulsion – experiment of Bateson and Purnet –
Chromosomal theory of linkage of Morgan – complete and incomplete linkage.
The multiple factor hypothesis was originally postulate by Yale in 1906. But
experimentally probed by the existence of multiple factors was provided by Nilsson –
Ehle in 1908; therefore he is credited with the concept of multiple factor inheritance.
In studies on the inheritance of seed colour in wheats and Oats, Nilsson – Ehle
obtained 3:1, 15:1 and 63:1 ratio between coloured and white seeds from different
crosses. It is dear from these ratio that in these crosses, seed colour was governed by
one (3:1 ratio in F2), two (15:1 ratio in F2) or three (63:1 ratio in F2) genes.
Thus red seeds from crosses showing 15:1 ratio could be classified into four
distinct classes on the basis of colour intensity, these classes were dark red, medium –
dark red medium red and light red, and they were present in the ratio 1:4:6:4:1. Thus
what was apparently 15:1 ratio ( indicating). Duplicate gene action tarned out to be a
1:4:6:4:1 ratio on a closer examination. This ratio ( 1:4:6:4:1) was the first clear cut
demonstration for the existence of multiple factors, that is genes with small and
cumulative effects and lacking dominance.
In order to explain the 1:4:6:4:1 ratio, Nilsson – Ehle made the following
assumptions.
i. In crosser, showing 15:1 ratio int eh F2, seed colour is governed by two genes (
This in crosses showing 63:1 ratio, three genes would be involved.
ii. One of the alleles of each colour gene produced seed colour and is called
positive allele; it is generally denoted by a capital letter e.g., R1, R2 etc.,
The other allele of each colour gene does not produce any colour and is
known as negative allele; it is ordinarily denoted by the correspond small
letter eg. r1, r2 etc.,
iii. These genes do not show dominance; so that the heterozygote for a colour gnee
e.g. R1r1 is intermediary in colour between the two homozygotes (R1R2 and
r1r2). This may be stated little differently two positive alleles of a colour gene
( R1R1) produce tince the intensity of red colour of that produced by a single
positive allele ( R1r1)
iv. Each of these genes ( positive alleles) has a small, equal almost equal, effect
on seed colour.
v. The effects of positive alleles of different colour genes are additive in the small
manner as these of the two positive alleles of a single colour gene. Thus, the
intersity of the seed colour depends on the number of positive alleles of all the
genes affecting seed colour present in a seed and not on which of these genes is
present in the positive or the negative form.
In other words, a character is governed by several genes; the positive alleles of
all the genes governing the train are similar to each other in their action ( effect) and
that their effects are additive or cumulative in ratare. This is the essence of multiple
factor hypothesis.
Parents R1R1R2R2 X R1r1r2r2
Gametes R1R2 r1r2
R1r1R2r2
Medium red F1
R1R2
R1r2 R1R2
r1r2
Red White
O R1R2 R1r2 R1R2 r1r2
R1R2
R1R1R2R2
Dark red
R1R1R2r2
Medium red
R1r1R2R2
Medium dark red
R1r1R2r2
Medium red
R1r2 R1R1R2r2
Medium dark red
R1R1r2r2
Medium red
R1r1R2r2
Medium red
R1r1r2r2
Light red
R1R2
R1r1R2R2
Medium Dark red
R1r1R2r2
Medium Dark red
r1r1R2R2
Medium red
r1r1R2r2
Light red
R1r2 R1r1R2r2
Medium Dark red
R1r1r2r2
Light red
r1r1R2r2
Light red
r1r1r2r2
white
Inheritance of seed colour in wheat
Phenotype Dark red Medium dark
red
Medium red Light red White
Genotype
and
frequency
(i) R1R1R2R2
(ii) (4)
(i) R1r1R2R2(2)
(ii) R1R1R2r2(2)
(i) R1r1R2r2(4)
(ii) R1R1r2r2(2)
(i) R1r1r2r2(2)
(ii) r1r1R2r2(2)
(i) r1r1r2r2
(1)
Posit….. (i) 4 (i) 3
(ii) 3
r1r1R2R2 (1)
(i) 2
(ii) 2
(iii) 2
(i) 1
(ii) 1
(i) O
Phenotype
frequency
1
(DR)
4
(MDR)
6
(MR)
4
LR
1
W
Poly genes :-
Genes having individually small but cumulative effect on a character they
govern quantitative character.
Transgressive segregation :-
The appearance of individuals in the F2 or a subsequent generation which
exceed the parental limits with respect to one or male characters.
Modifying gene :-
A gene that increase the phenotypic expression of a major gene; usually,
several modifying genes act in an additive manner and generate a continous variation
in an otherwise qualitative trait.
Quantitative characters Vs qualitative characters
The two characteristic feather of quantitative character are
i. Continuous variation
ii. Marked influence of the environment on their expression.
iii. Quantitative characters are heritable and their inheritance follows the same
medallion principle. Which was developed for qualitative characters
showing discontinuous variation.
Chromosomal theory of linkage of morgan – complete and incomplete
Linkage – The tendancy of two or more genes to stay together during
inheritance is known as linkage.
Consequence of the concerned genes being located in the same chromosome.
Do not show independent segregation.
So, the ratio’s obtained in F2 and test cross generations are significantly
different from the expected rations of 9:3:3:1 and 1:1:1:1 respectively in case
of two linked genes.
The frequencies of parental combinations are markedly more than expected,
white these of the … character combinations are considerately lower.
Synteny
Coupling:
In linkage, the dominant alleles of two or more genes present in the same
chromosome and hence, linked together; contributed by the same parent.
Linkage :
A the tendency of genes to stay together during inheritance, due to the genes
being located relatively close to each other in the same chromosome. Produces typical
distortion of test cross ratio ( of F1 ratio as well )
In maize, a dominant gene ‘C’ produces colourred seeds, while its recessive
allele ‘C’ determines colourless seeds. Another dominant gene ‘Sh’ gonerns full
seeds, whereas its recessive. Allele ‘sh’ gives rise to shrunken seeds. When plant
shaving coloured full seeds.
(CCSH Sh) were crossed with those having colourless shrunken seeds ( cc sh &
h), F1 seeds were coloured full (Ca Sh & h). Out of 8, 368 seeds obtained from the
test cross ( Ca Sh sh), 4032 ( 48.2%) were coloured full, 4.035 (48.3%) colourless
shrunken, 149 ( 1.7%) were coloured shrunken and 152 (1.8%) were colourless full.
Clearly, the four phenotypic classer coloured full and colourless shrunken had much
higher frequencies than the expected 25%. These two character combinations.
(Coloured full and colourless shrunken) are referred to as parental combinations,
parental phenol types or parental types since they are the character combinations
present in the two parents (of the F1 used for the test cross). The remaining two
phenotypic classes, coloured shrunken and colourless full, are far less than frequent
than expected (25%) these two phenotypes are called recombinant phenotypes or
recombinant types since they are generated by a reshuffling of the characters of the
two parents (of the F1 used in testcross).
In the above example, it appears as if the two dominant genes, C and Sh have a
strong affinity for each other so that the frequencies of coloured full and colourless
shrunted phenotypes are greater than expected. This situation is referred to as coupling
phase, and due to the presence of the dominant given C and Sh in the same
chromosome.
Coupling phase linkage between genes C and sh in maize.
C
Sh Sh Sh Sh
C e e
C sh e shGametes
Colourless shrunken (Test cross parent )
X
C sh
e sh
C sh
C sh
X
C sh
C sh
c sh
c sh
Parents
Coloured full Colour less shrunken
F1
C sh C ShC Sh
C Sh
C Sh
C sh
C sh
C sh
C sh
C sh
C sh
C sh
C sh
Coloured Shrunken 4,032(48.2%)Parental
type
Coloured full
149(1.7%)
Coloured Shrunken 152(1.8%)
Recombinant types
Coloured full 4,035(48.3%)Parental
type
Repulsion:
A linkage between the dominant alleles (s) of one ( on more) gene (s) and the
recessive allele (s) of another (several the) gene (s). In such a case, one parent
involved in a cross combination contributes the dominant allele of one or (more) gene,
while the second paran provide, the dominant allele (s) of the other gene (s).
Repulsion phase
Test cross C
Sh Sh Sh Sh
C e e
C sh E shGametes
Coloured full Colourless shrunken (Test cross parent )
X
C sh
e sh
C sh
C sh
X
C sh
C sh
c sh
e sh
Parents
Coloured shrunken
Colour less full
Types of linkage
The linkage may be classified as (1) complete or (2) incomplete depending
upon the absence (compiete linkage) or the presence ( incomplete linkage) of
recombinance phenotypes in a test cross progeny
Complete linkage (lack of crossing over) is known is male drosophila.
When males heterozygous for two linked aulosomal genes are mated with
douple recessive (having recessive alleles of both the gfenes in homozygous state)
femals, only the two parental character combinations (only those present into the two
parents of the helerozygous male) are recovered in the progeny; and there is a
complete absence of the recombinant types. This absence of recombinants is due to
the absence of crossing over in male.
Drosophila:
In most other cases, however, linkage as rule is incomplete. But some genes
may be so closely linked that they may show a very low frequency of recombination.
Such genes are called fighting linked.
Crossing over – significance of crossing over – cytological proof for crossing
over- sterns experiment.
Crossing over:
C sh C ShC Sh
C Sh
C Sh
Gametes
Testers progeny
C sh
C sh
C sh
C sh
C sh
C sh
C sh
C sh
Coloured Shrunken 21,379(47.9%)Parental
type
Coloured full
639(1.4%)
Coloured Shrunken 672(1.5%)
Recombinant types
Coloured full 21,906(49.1%)Parental
type
Exchange of strictly homologous segments between non- sister chromatics of
the homologous chromosomes.
Significance of crossing over:-
After the second dission of meiosin two of the four resulting cells will contain
chromosoms with the recombination of genes brought about by the crossing over of
the chromosoms. In this way, new combination of linked genes occur.
Cytological proof of crossing over:-
Since linked genes are located in the same chromosome, they would show
recombination only when there is exchange of the concerned segments between the
homologous chromosomes i.e., crossing over.
Morgan had arrived at the conclusion in 1911 based on genetic studies and
theoretical considerations.
Experimental evidence corroborating this conclusion was presented in 1931
independency by curt sterm in drosophila and by creightom and medintock in maize.
The expected results from the experiment of in drosophila designed to
establish the etymological basis of crossing over are as follows.
Car B
+
+
Y segment
XCar B
Red BarCarnation
normal
Stern used a drosophila femalw in which one X chromosome was shorler then
normal; this chromosome had the recessive gene car ( Carnation eye colour) and the
dominant gene B ( bar eye shape). The other x chromosome of this female was of
normal lengths, but a segment of the Y chromosome was translocated into its short
arm; this chromosome had the dominant gene car+ ( wild type allele of car, producing
dull red eye colour) and the recessive gene b+ ( wild type allele of B, producing
normal ovate eye shape). Storm test crossed the female to a car B+ male. As expected
the following four types of flies were recovered in the test cross progeny. Red,
normal ( Car+ B+); red, bar ( Car+B) carnation, normal c car B+) and carnation, bar
(Ccar B; alleles contributed by the test cross parent are not shown.
Two of the four phenotypes viz red normal and carnation bar, are non- cross
over or non recombinant types. Therefore, the carnation bar individuals are expected
Crossing Over
Car
B
Gametes
Car B +
B Car +
++
Car + +
to carry one short X, while the red normal one would have one long x with a Y –
segment. In contrast the remaining two phenotypes viz., red bar and carnation normal
are cross over or recombinant types.
Stern observed a very also correspondence between the expectations described
above and the results actuating obtained. He concluded as follows
1. During m…….. there is a exchange of precisely homologous chromatic
segments between homologous chromosomes.
2. Crossing over is responsible for the recombining between linked genes.
3. Strength of linkage and recombination – two point and three point test cross –
double cross over, interference and coincidence genetic map.
Linkage studies in made Drosophila
One of the gene produces a black body rather then the more common gray one,
and the other produces ting, vestigial wings rather than the normal long wings.
Suppose we cross a male homozygous for both these recessive genes (cleak and
vestigial) with a wild type female which is homozygous for the dominant genes for
gray body and long wings. The members of the F1 generation are all of the wild type,
for both mutant genes are recessive. Now let us obtain virgin females from these
offspring’s and male them with males which are homozygous for the double
recessive. You may remember that this is what we call a test cross. If these two genes
are on different chromosomes we will expect a ratio of approximately graph long; 1
black, vestigial ; 1 black long; 1 gray vestigial on the basis of independent assortment.
The actual results however are as follows.
Wild type (gray, long) - 965
Black vestigial - 944 Parented
Black long - 206
Gray, vestigial - 185 recombinations
Total - 2300
Total corss overs, 391. % of cross over, 17
Strength of linkage and recombinations determining distance between genes
As we look at these figures, it is immediately evident that we do not have a
1:1:1:1 ratio we can therefore conclude that these two mutant genes are one the same
chromosome; that is, they are linked. The ratio is about 1:1 for the parental types, with
a lower recombinations which have resulted form crossing over.
The percentage of crossing over which is obtained between different linked
genes various according to the distance between the genes on the chromosomes. It is
evident that crossing over will not produce new combinations of genes unless it occurs
between the linked genes. Hence those genes which are closest together on the
chromosome will have the least amount of crossing over between them and
correspondingly greator amount of crossing over will be found between genes that are
farther apart. The % of crossing over therefore indicates the relation distance between
the linked genes being studied.
Double cross over:-
In linkage, the recombinant type produced by simultaneous crossing over on
both sides of a gene; it is the least frequent class in the test cross progeny of a three
point cross.
The linkage studies on the genes for black body and vestigial lings in
Drosophils showed about 17% crossing over. Let us make other crosses to accept this,
when we cross a black cinnabar fly with one of the wild type of the opposite sex and
test cross the offspring, it is to find 9%crossing over and 9.5% crossing over when
cinnabar vestigial fly with a wild type fly and followed by test cross. This
discrepancy is due to the occurrence of double crossing over, that is, two cross over
occurring simultaneously in the same cell between these two loci.
If the genes studied are aloes together on the chromosome, however double
crossing over cannot occur between them. If a crossing over occurs at one point, a
second cross over will not occur within a certain distance of it. This phenomenon to
known as interference.
Coeffiueint of Interference
The observed frequencies of double cross overs are lower then that expected
values.
This is interpreted as follows:- the occurrence of crossing over in one region of
a chromosome interferes with its occurrence in the neighbouring segments: this is
called interference.
Coefficient of coincidence:-
The estimate of coefficient of coimidence indicating the degree of agreement
between the observed and the expected frequencies of double cross overs.
Two point test cross:
The percentage of crossing over between two linked genes is calculated by test
cross in which a F1 dihybrid is crossed with a double recessive parent. Such crosses
because involve crossing over at two points, so called two point test cross.
three point test cross.
As three point test cross or tre hybrid test cross ( blousing three genes) gives
us information regarding relative distance between these genes and also shows us the
linear order in which these genes showed be present on chromosomes. Such a three
point test cross may be carried out if three points or gene lou on chromosome pair can
be identified by marker gene.
Chromosome mapping / genetic map
In any organism, when a sufficient number of genes have been located, it is
possible to construct chromosome maps which show the number of chromosomers,
the linkage group associated with each pair of chromosomes, the sequence of genes on
each of the chromosome pairs, and the relative distance between the genes on each of
the chromosome pairs.
Sex determination – chromosomal mechanism of sex determination and its
types. Geric balance theory of sex determination of Bridges.
In most cases, male and female individuals differ for many characteristic
features which may be grouped into two categories;
1. Primary and
2. Secondary sex characterization
Primary sex character gonads :-
Gamete producing organs og male and female individuals.
Secondary sex characters :-
The development of secondary sex character is indeed by hormones produced
by the somatic elements of gonads.
Chromosomal mechanism of sex determination and its types
i. Ina vast majority of animals, male and female individuals ordinary differ from
each other in respect of either the number or the morphocogy of the
homologues of one chromosome pair, the chromosome is regered to as sex
chromosome or allosome.
ii. On the other hand, those chromosomes whose number and morphology do not
differ between males and femals of a species are called autosomes.
iii. The two types of sex chromosomes, X and Y.
The X chromosomes is found in both males anafemaless although on sex has
only one, while the other has two X chromosomes. Eg., of Y chromosome occurring
species, O ……., Drosophila , humans etc.,
The different mechanisms of chromosomal sex determination may be grouped
into five classes.
i. XX female : XO male
ii. XO female : XX male
iii. XX female : XY male
iv. XX male : XY female
v. Diploid (21n) female and haploid (n) male.
But some animal species have more than two x or more then one Y
chromosomes. However most animal species have only one pair of sex chromosomes.
XX female, XO male
In grasshopper, Protenor and many other insects, especially those belonging to
orthoptera, femals have two X chromosomes . While male have only me X
chromosome.
Consequently, the somatic cells of females have one chromosomes more then
those of males.
During cogenesis, the two X chromosomes in female pair regularly and in AI,
the two X chromosome separate and pass to the opposite poles. Thus each of the four
eggs produced by one meiotic ever, receives one X chromosomes. Clearly at the
eggs produced by females are alike with respect to the X chromosome.
Hence females are known as homogametic sex, i.e., the sex producing only one
type of gametes.
In the males, the single X chromosome remain unpaired during prophale I and
MI, generally it passes to one of the two poles at AI. As a result, two of four sperms
produced by one meixotic event receive one X chromosome, white the remaining two
sperms do not receive any X chromosome. Thus half of the sperms in makes have one
chromosome, while the other half have non; therefore males are known as
heterogametic sex.
The union of a sperm having aa X chromosome with an egg produce a zygote
having two x chromosomes (xx); such zygotes develop into female individuals. But
when a sperm without an X chromosome fertilizer an egg, an Xo zygote is abtained.
Such zygotes develop into males. Thus, one half of the progeny from each mating are
females, while the other half are males.
Parents X X x X O
Female Male
Gametes X X O
Progeny XX X
OFemale Male
Grass hipper Heterogametic sex
Fumea ( heterogametic sex)
The above figure demonstrates the chromosomal basis of sex determination in
grasshopper and fum…..
XX Female XY Male
Parents X X x X Y
Female Male
Gametes XX
Prop…. XX X
OMale Female
Parents XO XXX
Female Male
Gametes
X X Y
Progeny
XX X
YFemale Male
Human
XX
XXX XY
Male Female
XY XXX
Female Male
Y
Bird
XY female, XX male
This system of sex determination operates in birds, reptiles, some insects e.g.,
silk works etc., This scheme is essentially. The opposite of that found in mammals
etc. Here, the femals have XY chromosome constitutionl therefore it is the
heterogametic sex as half the eggs have an X, while the rest hve a Y chromosome.
The male of these species have two X chromosomes ( XX); as a result, male is the
homogametic sex since all the sperms produced by males, have one X chromosome.
Fertilization of an X containing egg with a sperm gives rise to an XX zygote, . Which
develops into a male. An XY zygote is produced when a Y containing egg is fertilized
by a sperm, such a zygote develops into a female.
Diploid (2n) female, Haploid (n) male
This system of sex determination of found mainly in Hymenoptera : honey bee,
anto , termites. In these species the somatic chromosome number of female is
deiploid, while that of males in only haploid.
Genic Balance thory of sex determination of Bridges
In 1916, Bridges discovered XXY females and Xo males in Drosophila
studying the inheritance of vermillion eye gene located in the X chromosome.
This clearly showed that XX and XY chromosome constitution were not
essential for femaleness and maleness, respectively and that Y chromosome did not
play a role in sex determination. A little later, bridges obtained triploid females; these
females when mated with normal diploid males produced a number of aneuploid
situations. By correlating the sex of an individual with its chromosome constitution,
bridges developed the genic balance theory of sex determination . This theory states
that the sex of an individual to determined by a balance between the genes for
maleness and those for femaleness present in the individual. In Drosophila, genes for
maleness are present in autosomes, while those for femaleness, are located in the X
chromosome, In essence, the sex of an individual in determined by the ratio of the
number of its X chromosome and that of its auto somal sets, this ratio is termed as
sex index and is expressed is follows;
Sex indeed = Number of X chromosomes (= X)
Number of auto somel sets (- A)
= X/A
Individuals with the sex index of 1.0 are normal females irrespective of
whether they are XX, XXX or XXXX. Similarly, flies having the ax index of 0.5 are
normal males ( xo and xxoo, xxyy and xxy tetraploid flies). Those flies that have a
sex index between 1.0 and 0.5 develop into inter sexes. An excess of sex index of 1.0
called super females.
Sex expression in drosophila as a function of the ratio of the number of X
chromosomes and the number of autosomal sets present in an individual.
(Sex expression in drosophila as a function of the ratio of X chromosomes and
the number of auto somal sets present in an individual).
(sex ratio = X/A)
Ploidy Number of X
chromosomes
(=X)
Number of
auto somal sets
( =4)
Sex indux
( X/A)
Sex expression
2n 3 2 3/2=1.5 Super female
3n 4 3 4/3=1.33 Super female
3n 3 3 3/3=1.0 Female
2n 2 2 2/2=1.0 Female
4n 3 4 ¾=0.75 Intersex
3n 2 3 2/3=0.67 Intersex
2n 1 2 ½=0.5 Male
4n 2 4 2/4=0.5 Male
3n 1 3 1/3=0.33 Super female
Sex linked inheritance – Criss cross in heritance – reciprocal deference –
holandric genes – sex influenced and sex limited inheritance – sex determination in
plants – melandrium, papaya and maize.
Sex linkage is the consequence of a gene being located in X or sex
chromosome which occur in different numbers in the two sexes ( x), one sex and only
x in the other
Or the association between a character and the sex during inheritance, the
concerned genes is located in the X chromosome; in human beings, several genetic
diseases are sex – linked.
Characteristics of sex linked inheritance
The characteristic features of inheritance of a sex linked. Trait may be
summarized as follows.
1. The frequency of individuals showing a recessive sex linked trait is markedly
higher in the heterogametic sex than that in the homogametic sex. Dr. Vs, red
green colour blind human males is 8% as compared to only 0.5% females
showing this trait.
2. Ordinary, genes governing sex linked traits are not transmitted from male
parents directly to their male progeny. For eg. white eye gene (w) is not
transmitted from male drosophila to its male progeny. This is because male
individuals receive their x chromosome from their mothers; their male parents
contribute only the Y chromosome.
3. A male transmits its sex linked genes to all its daughters, since females receive
one of their two x chromosomes from their fathers. The daughters in farm
transmits this gene to half of their male progeny.
Thus, a sex linked recessive gene is transmitted from a male, to its female
progeny and then to half the male progeny of send females. In other words sex linked
genes passes from male to female then back to male; such an inheritance pattern is
known as criss – cross inheritance.
4. Sex linked genes are not located in the Y chromosome. Consequently, the
heterogametic sex (male humans, mice, Drosophila etc., and female birds)
hemizygous for such genes, i.e. it has only one allele of sex linked gene.
Therefore, recessive allele of sex linked genes express themselves in a
hemizygous condition.
5. Genes located in the region of the X chromosome is homologous to a segment
of the Y chromosome do not show sex linked inheritance if their allele is
located in the Y Chromosome.
Sex limited and sex influenced genes
Sex – limited genes are those which produce characteristics that are expressed
in only one of the sexes. They are often confused with sex inked genes, but are
entirely different in mode of inheritance.
Whereas sex linked genes are those which are located on the X or Z
chromosomes, sex limited genes may be located on any of the chromosomes.
Sex linked genes may be expressed in both sexes, although they usually show
more abundantly in one sex, white sex limited genes show in only one sex.
Sex limited genes are responsible for what we commonly call secondary sexual
characteristic as well as primary sexual characters.
Eg. -Beard in man ( sex limited)
- Milk production is cattle (sex limited)
- Birds (Brillant plumage of the peacock (sex limited)
- Baldness in man ( sex influenced trait)
Genic sex determination
Sex determination in some plant species . e.g. ( Papaya (Carica Papaya)
spinach, Vitis cinerea, Asparagm etc., is postulated to be governed by a single gene.
Postulated monogenic sex determination in papaya; the gene ‘M’ has three
alleles, viz., m, m1 and m2.
Genotype Suicidal Sex- expression
Mm Vital Female
M1m Vital Male
M2m Vital Hermaphrodite
M1m1, M1M2, & M2M2 Lethal ( all dies) 1
Maize plants are generally monoecism; both male and female flowers are
produced on the same plant.
Conversion of ordinarily monoecious plants into male and female plants
(dioecious) by two recessive genes ba and ts.
Genotype Female flowers Male flowers Sex Expression
Baba Ts Ts Normal Normal Monoecious
ba ba Ts Ts Radimentary Normal Male
BaBa ….. (Table
seed)
Normal Develop tato
female flowers
Female
bs bs ts ts
( Barren silk)
Rudimentary Develop into
female flowers
Female
Westergard and warmke studied sex determination in Melandrium and seed
plant in the pink family. In melandrium album the diploid female plants have twenty
four – chromosomes – twenty two ( 22) autosomes and two x chromosomes.
Diploid males have the same chromosome number, but show a cytological
distinction between one of the pasis into an X – and Y – chromosome.
These consist of three types –
(i). Females - 44 autosomes
4 x chromosomes
(ii). Males - 44 antosomes
2 x chromosomes
2 Y chromosomes
(iii). Male - 44 autosomes
3 x chromosome
1 y chromosome.
Hence, Y chromosome is the sex determining element and carious the genes to
produce maleness, that one Y chromosome, even in the presence of three x
chromosomes, is sufficient to produce a male.
Holandric genes: gene located on a Y chromosome.
Y – linked gene.
In cytoplasmic inheritance, generally the character of only one of the two
parents (usually the female parent) is transmitted to the progeny. As a result,
reciprocal crosses exhibit consistent differences for such characters, and there is lack
of segregation in the F2 and the subsequent generations. Such inheritance in also
referred to as extra nuclear inheritance extra chromosomal in heritance and material
inheritance.
Genes showing cytoplasmic inheritance for particular character which located
out side the nucleus and in the cytoplasm & they are referred to as plasma genes /
cytoplasmic genes, cytogenes, extra nuclear genes or extra chromosomal genes. Sum
total of genes present in the cytoplasm of a cell is known as Plasmon, while all the
genes present in a plastid constitute a plastid.
Mt – DNA – Mitochondrial DNA
Cp – DNA – Chloroplast DNA
Mt – DNA + Cp – DNA – organelle DNA
Cytoplasm – mitochondria – cytosol.
Cyt………… inheritance is due to the plasma genes located in cell organelles
(plstid and mitochondria ) that are integral constituents of normal cells.
All other cases of non – mendelian inheritance are not cases of cytoplamic
inheritance.
Characteristics of cytoplasmic inheritance
1. Reciprocal difference: - Reciprocal crosses showed marked, differences for the
characters governed by plasme gene. If plasma genes transmitted from only
one parent, liniparental inheritance.
2. Lack of segregation: In genral F2, F3 and the subsequent generation do not
show segregation for a cytoplasmically inherited trait, as the F1 receive
plasma genes from me parent only.
3. Irregular segregation in Biparantal inheritance.
4. Somatic segregation: Plasma genes generally show somatic segregation
during mitosin which may be of rare occurrence.
5. Association with organelle DNA
6. Nuclear transplantation : Nuclear of a cell is removed and replaced by a
nuclear of another genotype from a different cell, it is governed by the
genotype of cytoplasm and not by that of nuclear.
7. Transfer of nuclear genome through backcrosses.
The nuclear of a variety or specious may be transferred into the
cytoplasm of another species / variety through repeated backerossing with the
farmer, which is used as the recurrent O parent. Lines produced in this way are
known as alloplassure lines. Since they have nuclear and gytoplasm from for
different species.
8. Mutagenesis
Some mutagens, e.g. ethidium bromide are highly specific mutagen
for plasma genes.
9. Heck of chromosomal location:
Plasma genes cannot be located in linkage maps.
10. Lack of association with a parasite, symbionts or virus.
The known cases of true cytoplasm inheritance are concerned with either
chloroplast or mito chondrial traits and are usually associated with their DNA.
Such eases are, therefore, often referred to as organelle inheritance, plastid
inheritance and mitochondrial inheritance.
Chloroplast inheritance
Chloroplast characteristics are controlled by both cp DNA ( Chloroplast DNA)
as well as nuclear DNA. The machinery for replication and transcription of cp DNA
and the soluble enzymes of stroma are provided by nuclear genes, while plastid r RNA
is produced by cp – DNA alone. Most other components of plas fid structure and
function are controlled jointly by cp- DNA and nuclear DNA. It is ….. wonder, then
that chloroplast differentiation and function are affected by several nuclear genes.
More than 75 such genes are known in barley.
Mitochondrial inheritance
Mitochondria originate from pre existing mitochondria only, and they contain
DNA. Many plasma genes are beliened to be located in mitochondrial DNA ( mt-
DNA). Available evidence shows that generally mitochondria from only one of the
parents are transmitted to the progeny. Therefore, characters governed by the genes
located in mt – DNA show cytoplasmic inheritance. Cyplasmic male sterility.
Cytoplasmic male sterility ( Cms) is produced by plasma genes located in the
cytoplasm, which may be a part of mt- DNA, cp – DNA and plasmid like elements of
mitochondria or may be produced by RNA various.
Cons show typical cytoplasmic inheritance progeny from cms and a normal
male fertile strain are all male sterile. As a result, a cms stain has to be pollinated by a
male. Fertile strain in every generation for its maintenance (ms in extensively used in
hybrid seed production in crops like maize, jowar, bajra etc.,
Parents X
Progeny X
rr
s
rr
F
Male sterile ( A line)
Male fertile (maintainer line)
B line
rr
s
rr
F
Male sterile
Male fertile
rr
s
rr
F
Male sterile
Male fertile
Progeny
The same as before.
II. Restoration of male fertility in the presence of crns cytoplasm ( denoted by S)
by a dominant restorer gene R. In the presence of the recessive allele, r, cms produces
male sterility.
Plasmids
Many species of bacteria passen, in addition to their chromosome proper,
other genetic elements which lead an independent existence within the cell, These are
called plasmids. Some plamids can unite with the chromosome by crossing over and
are known as episomes. Plasmids carrying antibiotic resistance genes called R-
plasmids . The R’ Plasmids are also called episomes and are genetic elements which
can exist in two alternative stages, independently in the cytoplasm and integrated into
the bacterial chromosomes.
The act of uniting with the chromosome is known as integration and the
recerse is excision. Bacterial plasmids take many from and are found in nature as sex
factors, colicin factors, phages and as R factors responsible for the transmission of
drug resistance Many of the properties of plasmids known to be stared with the extra
chromosomal DNA of eukaryotic cells.
Male sterile
Male sterile
Male fertile
rr
F
rr
S
Male fertile
RR
s
S
Rr
DNA – the genetic materials – Griffith experiments – Expt of A very, meleod and Mc
carthy, confirmation by Hershey and Chase, RNA as genetic material frankel, conrat
and singer experiment.
In 1903, Sutton and Boveri postulated genes were located in chromosomes,
this is known as the chromosomal theory of inheritance.
Properties of the genetic material :
The chemical of which genes are composed must possess the following
properties.
i. Genetic material must be replicated in enormous copies which must be
distributed to the daughter cells with a very high preciston, so that genes
passed to next generation without alternation.
ii. Genetic material must be able to express itself by exercising a control on the
development of the character it governess.
iii. Able to stage the highly variable information necessary for gene function.
iv. Causes new genetic variation through mutation.
Griffith experiment:
The experiments of Griffith are briefly described here. When live cells of the
virulent strain III R were injected into mice, they did not suffer from pneumonia and
all the mice died due to precumonia. But mice injected with heat killed cells of the
virulent strain III s. indicating that all the cells were killed by the heat treatment.
However, when mice were injected with a mixture of heat killed IIIs cells and live II
R cells, some of them died due to pneumonia. Diplococcus cells isolated from dead
mice were of the type III s. Since all the cells of the heat killed IIIs culture were dead,
it was postulated that some of the cells of II R had changed into the IIIs type due to
the influence of dead IIIs cells present in the onixture. This phenomenon was called
transformation, and that component IIIs cells which induce the conversion of IIR cells
into IIIs was named the transforming principle.
A.
B.
C.
D.
Strain III s (Live)
Injected into mice
All mice dead due to pneumonia
III s ( heat killed)
Injected into mice
All mice alive (no pneumonia )
Strain II R ( Live)
Injected into mice
All mice alive (no pneumonia)
Live II R
Live II R
Injected into mice
Some mice dead From dead
mice
Isolated
Live III S
Experiments of Avery, Macleod and Me Carty ( 1944)
A.
B.
C.
D.
F.
G.
Culture
In Petri plates
Rough II R colonies
Culture
Strain III S
No Colony
Culture
DNA is olated from IIIs cells
No Colony
Culture
+ anti body II R
IIIS colonies
+
Live II RHeat Killed III S
Live II R
+ + Protease Culture
+ anti body II R IIIS colonies
Live II R
+ + R Nase
Culture
+ anti body II R IIIS colonies
H.
The findings of Avery and coworkers presented in controvertible evident for
DNA to be the transforming principle. But in spite of this, DNA could not be
universally accepted as the genetic material, which may be due to lack of information
an the molecular mechanism of transformation.
Experments of Hershey and chase ( 1952)
----------------------------------------------- picture - not dra
Thus, the studies of Hershey and chase clearly showed that DNA of T2
transmitted protecis are not transmitted. There fore atleast in T2 phase, DNA is the
genetic material. As a result, the finding of Hershey and chase led to the universal
acceptance of DNA as the genetic material.
RNA as a genetic material ( 1957)
Frantkel – conrat and singer
Live II R
+ + D Nase Culture
+ anti body II R
No colony
A.
B.
C.
It is evident from these findings that only RNA ( and not the protein) of TMV
has the capacity to produce the disease, and that the type of protein present in the virus
particles is defermered by the RNA. Clearly RNA is the genetic material in TMV also
function as the genetic material.
Strain ATMV
Separation
RNA
+
Reassociation n Crofection
TMVParticles
Symptom of strain A
Strain BTMV particles Prokin
RNA
+ Infection
TMVParticles
Symptom of strain B
+
RNA
Infection
Hybrid TMV
Symptom of
strain B
Bobn….
Strain BTMV particles
Strain B
RNA
+ Infection
Hybrid TMV
Symptom of
strain B
Bobn….
Strain BTMV particles
Structure of DNA – Watson and crick Model
- Machanisms of DNA replication
- Proot for semi conservative of DNA rep.
Nuckic acids are polymers made up of hundreds, thousands or million of
nucleotides linked together by phosphodiester bond. In both DNA or RNA the
intermediate linkage invoine 3’ and 5’ hydroxyl of sugar i.e., it is 3’ 5’ – phosphor
diester bond that joins adjacent nucleotides and thus form a polynucleotide structure.
Nucleosides and nucleotides
The linkage of a Purina or pyramiding base with a sugar molecule,
ribose or ceoxribose results in the formation a nucleoside.
The meclesides found in RNA are called rib nucleosides where as in
DNA they are called as deoxyribonucleosides.
Pentose sugar Nitrogenous Base
= Nucleoside
Pentose sugar
Nitrogenous Base
= Nucleoside
Phosphate
Chargaff’s rule of base equivalence
E. chargaff established that when DNA from particular species is
subjected to chemical hyolroly in order to release its component purines and
pyrimidney the total amount of adenine released is equal to the total amount of
thymine and similarly in ease of cytosine are equal to guanine.
e.g. A-T
G-T
Erwin chargaff’s conclusion
1. DNA specimen studies and subsequent investigation led to the establishment
of base composition.
2. The base composition of DNA varies from one species to another.
3. The base composition of given species doesnot change at any time.
4. The difference in DNA and RNA nucleotides is the presence of hydroxyl
group at pentose sugar in RNA. The remarks of ‘oxy’ group give it a name as
‘beoxy’ for DNA.
5. Presence of OH group in RNA makes sucephsble for chemicals and
enzymatic degradation.
DNA
DNA is a polynucleotide comprising of fair types .
Nitrogenous base, adenine (A) and guanine are double ring componunds and
form purine where as thymine (T) and cytosine (C) are single ring compared
forming pyrimidines.
DNA sugar present in DNA is a pentose specially named B-D-2 deoxyribo
turanose which linked to nihogenow base on end, C1-1 and phosphanic ….. at
the other end of 31 or 51.
Phosphate molecule along with deoxyribose form the main back bone of the
DNA molecule.
The correct structure of DNA was first f……. by J.D. Watson and F.H.C. crisk
in 1953. Their double helix stnicture of DNA was based on two major kinds of
evidences.
1. It was observed that the concentration of thymine was always equal to the con.
Of adenine. Similarly in case of cytosine and quinine. It indicates that thymine
and adinine as well as cytosine and quinine were present in DNA some fixed
relationships. It was ascertained that the total amount of pyrine concentration
are equal to the total amount of pyramiding concentration.
2. X-ray diffraction studies indicated that DNA was a highly ordered,
neultistranded structure with respecting substructures spaced every 3.4Ao along
the axis of the molecule.
Postulates of weson and cricks model
DNA exists as double helix in which two polynucleotide chains are coiled
about one another in a spiral way.
Each polynucleotide chain consists of a sequence of nucleotides liked together
by phosphodiester lo… joining adjacent deoxy ribose sugar moieties.
Two polypeptide chains are held together in their helical conformation by
hydrogen bonding between the two chains perpendicular to the axis of the
molecul like the steps of spiral stairease.
The base paring is specific, adenine always pairs with thomine and guanine
always with cytosine. These specific base pairing results from the hydrogen
bonding capacities of the bases in their normal configuration.
Adenine linked by thyamine in two H-bond Guanine linked by cytosine
for three H-bond.
The two DNA strands are said to be complemented i.e., if the sequence of
micleotides in one strand is known, the sequence of nucleotides other shand
can be accurately predicted due to this property of complementary.
The base pairs in DNA all stanted 3.4Ao apart with 10 base paise in turn of
double helix.
The sugar phosphate (3) backbone of two complementary stands are
antiparalld.
One stand 3’ OH 5-P
Another stand 5’ P 31OH
31
51
A - - - TG - - - CA - - - T
3.4Ao[- - - - - - - - -
34Ao
10Ao
20Ao
51
B1
Types of DNA
Based on, number of residues per turn(n), the spacing of residues along
with the helices abrs(h), rotation of base pair, dinection of rotation, qrouke depth etc.,
Four type: A,B,C, and Z.
S.No Characters A B C Z
1. Helix sense Right hand ( Rh) Right hand
(Rh)
Rh Lh
2. Residue per
turn
Right hand ( Rh) 10 9 12
3. Helical dia
meter
23oA 20Ao 19Ao 19oA
4. Shape Broadest Intermediate Inter Most
elongate
5. Major Nerrow and very
deep
Wide and
deep
Wide &
deep
Flat
6. Minor
grooves
Very broad and
shallow
Narrow and
quite deep
Narrow and
deep
Vary narrow
and deep
7. Conditions 70% Rh, High
salt, condition
92% RH
low ions
62% low
ions
High salt
DNA replication
Thymine dimer is not removed by the nucleotide excision repair system, this
postreplication repair must be repeated after each round of DNA replication.
Error prone repair system
The DNA repair systems described so far are quire accurate. However, when the
DNA of E. coli cells is heavily damaged by mutagenic agents such as UV light, the
cells take some drastic steps in their attempt to survive. They go through the so called,
SOS response, during which a whole battery of DNA repair, recombination and
replication proteins are synthesized. Two of these proteins, encoded by the umu C and
umu D ( uv mutable genes), encode proteins that allow DNA replication to proceed
across damaged segments of template strands, even though the nucleotide sequences
in the damaged region cannot be accurately replicated. This error – prone repair
system eliminates gaps in the damaged nucleotides in the template strands but, in so
doing, sharply increases the frequency of replication errors. The SOS response
appears to be a somewhat desperate and risky attempt to escape lethal effects of
heavily damaged DNA. When error – phone repair system is operative mutation rates
increase sharply. Distinction can be made based on the pattern of methylation in
newly replicated DNA.
In E. coli, the A in GATC sequences is methlated subsequent to its synthesis.
Thus an interval occurs during which the template strand is methy lated and the newly
synthesized strand is unmethylated and the newly synthesized strand is unmethylated.
This mismatch repair system uses this difference in methylation state to excise the
mismatched nucleotide in the nascent strand and replace it with the correct nucleotide
by using the methylated parental strand of DNA as template.
Mismatch repair of DNA in E. coli requires the products of four genes, mutH,
mutL, mutS and mutU. The MutS protein recognizes mismatches and binds to them to
initiate the repair process. MutH and MutL proteins then join the complex. MutH
contains a GATC-specific endonuclease activity that cleaves the unmethylated strand
at hemimethylated GATC sites either 5’ or 3’ to the mismatch. The incision sites may
be 1000 nucleotide pairs or more from the mismatch. The subsequent excision process
requires MutS, MutL, DNA helicase II (MutU) and an appropriate exonuclease. If the
incision occurs at GATC sequence 5’ to the mismatch, a 5` - 3` exonuclease like E.
coli exonuclease VI is requied. If the incision occurs 3` to the mismatch, a3-5`
nuclease activity like that of E. coli exonuclease I required. After the excision process
has removed the mismatched nucleotide from the unmethylated strand, DNA
polymerase fills the gap and DNA ligase seals the nick.
Post – replication repair
During the DNA replication, the presence of a thymine dimer in a template
strand block the operation of DNA polymerase III, thus preventing the addition of
nucleotides in the new strand. DNA polymerase restarts DNA synthesis at some
position past the dimer, leaving a gap in the nascent strand opposite the dimer in the
template strand.
At this point, the original nucleotide sequence has been lost from both strands
of this progeny double helix. The damaged DNA molecule is repaired by a
recombination – dependent repair process mediated by the E. coli rec A gene product.
The Rec A protein, which is required for homologous recombination, stimulates the
exchange of single strands between homologous double helices.
During postreplication repair, the Rec A protein binds to the single strand of
DNA at the gap and mediates pairing with the homologous segment of the sister
double helix. The gap opposite the dimer is filled with the homologous DNA strand
from the sister DNA molecule. The resulting gap in the sister double helix is filled in
by DNA polymerase, and the nick is sealed by the DNA ligase. The thymine dimer
remains in the template strand of the original progeny DNA molecule, but the
complementary strand is now intact. If the excision repair pathways remove larger
defects like thymine dimmers.
Base excision repair
Base excision repair can be initiated by any group of enzymes called DNA
glycolyases that recognize abnormal bases in DNA. Each glycolyase recognizes a
specific type of altered base, such as deaminated bases, oxidized bases etc. The
glycolyases cleave the glycosidic bond between the abnormal base and 2-
dexoxyribose, creating apurinic or apyrimidnic sites ( AP sites) with missing bases.
These AP sites are recognized by AP endonuclease, which act together with
phosphodiesterases to excise the sugar – phosphate groups at sites where no base is
present. DNA polymerase then replaces the missing nucleotide according to the
specifications of the complementary strand, and DNA ligase seals the nick.
Nucleotide excision repair
Nucleotide excision repair removes larger lesions like thymine dimmers and
bases with bulky side groups from DNA. In nucleotide excision repair, a unique
excision nuclease activity produces cuts on either side of the damaged nucleotide(s)
and excises an oligonucleotide containing the damaged base9s). This nuclease called
an exinuclease to distinguish it from the endonucleases and exonucleases.
The nucleotide excision repair in E. coli needs the products of three genes,
uvrA, uvrB and uvr(fig). A trimeric protein containing two UvrA polypeptides and
one UvrB polypeptide recognizes the defect in DNA, binds to it and uses energy from
ATP to bend the DNA at the damaged site. The UvrA dimer is then released and the
UvrC protein binds to the UvrB-DNA complex, The UvrB protein cleaves the
phosphodiester bond from the damaged nucleotide(s) on the 3’ side, and the UvrC
protein hydrolyzes the phosphodiester linkage from the damage on the 5’ side. The
uvrD gene product, DNA helicase II, releases the excised di-decamer. In the last two
steps of the pathyway, DNA polymerase I fill in the gap, and DNA ligase seals the
remaining nick in the DNA molecule.
Mismatch repair
The mismatch repair is carried out by the 3’ -5’ exonuclease activity built into
DNA polymerase. DNA polymerase proofreads DNA strands during their synthesis,
removing any mismatched nucleotides at the 3’ termini of growing strands.
Mismatches often involve the normal four bases in DNA. For example, a T may be
mispaired with a G. Because both T and G are normal components of DNA, mismatch
repair systems need some way to determine whether the T or G is the correct base at a
given site. The repair system makes this distinction by identifying the template strand,
which contains the original nucleotide sequence, and the newly synthesized strand,
which contains the misincorporated base (the error). This occasional mistakes that
occur during replication. DNA repair / damage can occur as the result of exposure to
environmental stimuli such as alkylating chemicals or ultraviolet or radioactive
irradiation and free radicals generated spontaneously in the oxidizing environment of
the cell. These phenomena can, and do, lead to the introduction of mutations in the
coding capacity of the DNA. Mutations in DNA can also, but rarely, arise from the
spontaneous tautomerization of the bases ( the rare imino form of adenine can form a
stable hydrogen bond with cytosine and the enol form of thymine can pair with
guanine). Modification of the DNA bases by alkylation (predominately the
incorporation of –CH3 groups) occurs on purine residues. Methyulation of G residues
allows them to base pair with T instead of C. A unique activity of O6 – alkylguanine
transferase removes the alky1 group from G residues.
Mutations in DNA are of two types. Transition mutations result from the
exchange of one purine, or pyrimidine, for another purine, or pyrimidine.
Transversion mutations result from the exchange of one purine, or pyrimidine, for
another purine, or pyrimidine. Transversion mutations result from the exchange of a
purine for a pyrimidine or visa versa. These mutations are supposed to be kept at a
tolerable level. E. coli cells possess at least five distinct mechanisms for the repair of
defects in DNA.
1. Light – dependent repair or photoreactivation
2. Excision repair
3. Mismatch repair
4. Post – replication repair and
5. Error – phone repair Mammals seem to possess all of the repair mechanisms
found in E. coli except photo reactivation.
Light dependent repair
Light dependent repair or photoreactivation of DNA in bacteria is carried out
by a light activated enzyme called DNA photolyase. When DNA is exposed to UV
light, thymine dimmers are produced by covalent cross-linkages between adjacent
thymine residues. DNA photolyase binds to thymine dimmers in DNA and uses light
energy to cleave the covalent cross – links (Fig). Photolyase will bind to the thymine
dimmers in DNA in the dark, but it cannot catalyze cleavage of the bonds joining the
thymine moieties without energy derived from visible light specifically light within
blue region of the spectrum.
Excision repair
Excision repair of demaged DNA involves at least three steps. In step1, a DNA
repair endonuclease or endonuclease – containing enzyme complex recognizes, binds
to, and excises the damaged base or bases in DNA. In step 2, a DNA polymerase fills
in the gap by using the undamaged complementary strand of DNA as template. In
stept 3, the enzyme DNA ligase seals the break left by DNA polymerase to complete
the repair process. There are two major types of excision repair base excision repair
systems remove abnormal or chemically modified bases from DNA, whereas
nucleotide nucleotide is added it supplies another free 3’ OH group.
The primer for both leading and lagging strand synthesis is a short RNA
oligonucleotide that consists of 1 to 60 bases; the exact number depends on the
particular organism. This RNA primer is synthesized by copying a particular base
sequence from one DNA strand and differs from a typical RNA molecule, in that after
its synthesis the primer remains hydrogen bonded to the DNA template. In bacteria
two different enzymes are known that synthesis primer RNA molecules – RNA
polymerase and Primase. The DnaB protein complex moves along the other pare ntal
strand, prepriming it so that primase will synthesis a primer RNA. Pol III holoenzyme
adds nucleotides to the primer, thereby synthesizing a precursor fragment. This
synthesis continues up to the primer of the preceding precursor fragment.
Apart from this function, DNA polymerases also has 3-5 exonuclease, 5-3
exonuclease and endonuclease activity and so they can perform nick translation and
strand displacement. By nick translation the RNA is removed and replaced by DNA is
removed and replaced by DNA. Once the RNA is gone, DNA ligase seals the nick,
thereby joining the precursor fragment to the lagging strand. Pol II moves back along
the DNA ( in the direction of advancement of the fork) until it encounters the next
primer and the process continue again and again.
Since each strand has 5’ –P terminus and 3’- OH terminus, strand growth is
said to proceed in the 5’-3’ direction (Fig). The advance of the replication fork
continues until replication is completed. An unsolved question is how the reates of
growth of the leading and lagging strands are coordinated.
Termination
In a unidirectionally replicating molecule, replication terminates at the origin.
In bidirectionally replicating molecule, it may be of two types: 1 there might be
definite termination sequence 2. Two growing points collide and termination occurs
where ever the collision point happens to be.
Replication in Eukaryotes
The complete mechanism of initiation, elongation and termination of linear
DNA molecule and chromatin replication has not yet been elucidated. However, it is
believed that there might be multiple replication forks exist during replication.
Similarly different isoforms of DNA polymerases have been identified in eukaryotes
with specific function.
Fidelity of DNA replication
There is no single molecule whose integrity is as vital to the cell as DNA.
Thus, in the course of hundreds of millions of years there have evolved efficient
systems for correcting the..
DNA Replication
Genetic information is trnaferred from parent to progeny organisms by a
faithful replication of the parental DNA molecules. At the biochemical level,
replication is defined as a template – directed nucleic acid synthesis reaction where
the template and nascent (growing) strand are the same type of nucleic acid.
Replication is a polymerization reaction and can be divided into stages of initiation,
elongation and termination.
Replication of do DNA is a complicate process that is not completely
understood due to the following facts;
1. A supply of energy is required to unwind the helix.
2. The single strands resulting from the unwinding tend to form intranstrand base
pairs.
3. A single enzyme can catalyze only a limited number of physical and chemical
reactions and many reactions are needed in replication.
4. Several safeguards have evolved that are designed both to prevent replication
errors and to eliminate the rare errors that do occur.
5. Both circularity and the enormous size of the DNA molecules impose
geometric constraints on the replicative system and how this fit into the system
has to be understood.
Models for the replication of DNA
The replication of cellular DNA was originally conceived as two models:
conservative and dispersive. In conservative model, the parental DNA remains
unchanged and gets passes to one daughter cell, whereas newly synthesized DNA gets
passed to the other. But in dispersive replication, new DNA synthesis is interstitial
(small openings), and each daughter cell receives a mixture of parental and newly
synthesized DNA.
In a third, semi conservative model, proposed by Watson and Crick, the
parental strands remain unchanged, but the duplex is sperated into two halves. Each
parental strand acts as a template for replication and the daughter duplexes have one
parental strand and one daughter strand each. The semi conservative model holds for
cellular DNA, but the single stranded genomes of viruses and some plasmids replicate
conservatively – the structure of the single parental strand is conserved following
replication.
Meselson- Stahl Experiment: Proof for Semi conservative DNA replication
In 1958, Mathew Meselson and Franklin Stahl showed that the replication of
bacterial chromosomal DNA was semi conservative. E. coli were grown for many
generations in a medium containing N so that, their DNA became universally labeled
with the isotope (heavy DNA). The cells were then shifted to a medium containing.
One gene one polypeptide hypothecs
May be regarded as applicable to only those genes which species polypeptide
sequences, that is structural and regulator gene.
One gene – one enzgme hypothesis
Simply states that lach gene controls a single Regalation of gene expression –
Operon model of Jacob and monad – structural genes and regulator genes.
The control on gene activity , i.e. protein production permits the function of
only those genes given five is terned gene regulation.
Regulation of transcription
The production of in RNA from different genes (transcription ) is regulated
through two basic control mechanism)
The first mechanism produces negative control through specific repressor
proteins which present transcription of specific genes.
The second mechanism is termed positive control as it earplugs specific factors
for the initiation and termination of transcription and specific promoters of
transcription for specific groups of genes.
The regulation of transcription is the universal control mechanism for turning
on and off specific genes or groups of genes as per the needs of cells and issues.
Ordinarily, the term ‘regulation of gene action’ is exclusively applied to the regulation
of transcription.
Regulation of translation
The modes of translation regulation may be grouped in the following
eategories based on the nature of repressor and the net results produced.
1. A protein or proteins coded by in RNA binds to a site on the mRNA and
prevents its translation, thereby blocking its own synthesis.
e.g. Proteins of ribosome’s(auto genous repression )
The ribosomal protein quickly associate with r RNA and together become
assembled into ribosomes. But when r proteins accumulate due to the non- availability
of r RNA for ribosome assembly.
Ribosomal protein operon
or RNA gene
RNA
Translation
Normally
RNA
Assembled into ribosomes
Ribosomal proteins of or proteins accumulate Cr RNA not available
Ribosomal proteins RNA for ribosomal proteins
No translation
2. A protein binds to its own mRNA or to some other component of the protein
synthesis machinery this leads to the rapid degradation of mRNA, b- tabulins
utilized for the assembly of microtubules.
3. A short stretch of RNA pairs with the mRNA by complementary base pairing
and prevent its transcription.
The r proteins bind to the upstream region of their own m RNA; the interaction
prevents the translation of mRNA.
The operon concept:
An operon is a group of structural genes whose transcription is regulated by the
co-ordinated action of a (i) regulator (2) a promoter (p) and (3) an operator (o) gene.
The regulator gene produces a protein known as repressor which binds to the operator
gene.
The regulator gene produces a protein known as repressor which binds to the
operator gene. The operator gene is located at the beginning of the operon from which
the transcription begins an dis continous iwht the structural genes transcribed first i.e.
it is just upstream of the first structural gene. The promoter gene is located fast
upstream of the operator and provides the binding site for RNA poly merase which
carries out transcription.
Only the DNA strand oriented in the 3’ to 5’ direction the initiation point
( promoter) is transcribed by RNA polymerase; This strand is known is the antisense
strand. The base sequence of the antisense strand is complementary to that of the
RNA produced by the gene. The DNA strand oriented in the 5’ to 3’ direction is not
transcribed and is designated as the sense strains. The base sequence of sense strand is
the same as that of the RNA produced by the gene.
The perons may be termed as compact or divided base, on the linkage
relationship among the gnes of the operon. In a compact operion all the genes of the
operon (regulator, promoter, operator and structural) are located in a single region of
the chromosome next to each other. E.g. the lac operon.
Split gene, Introm, exons, modern concept of gene – cistron, muton, recon –
complementation test.
The members of multiple alkali series are located in a single gene ( multiple
alkali series) or in two or more separate but closely linked genes producing a single
character (pseudo alkali series0 is generally based on
(i) Recombination (ii) us-term test.
According to the classical concept a gene is qot sub divisible , that is by
crossing over does not occur within a gene it always occur between two separate
genes. Also, gene is the unit of function, recombination and mutation.
Thus gene is considered to control the inheritance of one character (unit of
function ) to be indivisible by recombination ( unit of recombination) and to be the
smallest unit capable of mutation (unit of mutation).
In 1940’s it become evident that a gene controlled a single biochemical
reaction by directing the production of a single enzyme.
It was soon discovered that one gene produces a single polypeptide and not
one enzyme.
It can be shown that muton to smallest unit capable of mutation) and recon
smallest unit capable of recombination) represent a single nucleotide / base pair of a
DNA molecule.
Thus, the gene may now be defined as a segment of DNA which codes
( contains the information) for a single polypeptide (the functional unit). At the
operational level, the alleles of a single gene do not show complementation in a as –
trans test while those of different genes show complementation.
In trons and Exons
The genes contains sequences call exons (expressed sequences) and nitrous
(intervening sequences).
Exons all the sequenced that form the mature RNA and , as a result are
represented in the proteins coded by the concerned genes.
In contrast the intones are deleted during mRNA processing and are not
represented in the concerned proteins.
Exon A Exon B
Intron
5’
G – OH
P
P 3’
PG – P 3’
5’ OH
The excision of intones and the joining together of all the exons of a
gene in a proper sequence to yield the mature RNA is called splicing.
The two boundaries between the two exons and the intron lying between
then are known as splicing junction.
Exon A5’ P 3’
Exon B Spliced exons
G-----P OH
G – P - + P
Circulated intron
Trans – heterozygote
By comparing the phenotypes produced in cis – and trans heterozygote for any
two mutant alleles it is located in the sance gene or in two different genes. They are
placed in the same gene if their cis heterozygote’s produces the wild type phynotype.
White their trans heterozygote’s have a mutant phenotype. But if both their cis and
trans – heterozygotes have the type phenotype, they are considered to be located in
two different genes.
The production of wild type phenotype in a trans heterozygote for two mutant
alleles is known as complementation and such a study called complementation test.
Complementation test:- Two mutant alleles located in the same gene
Two mutant alleles located in two different genes.
GeneA
GeneB
GeneA
GeneB
a1 a2a1 a2
+ +Normal character
+ +Normal character
Cis- heterozygote
a1 +a1 +
+ a2 + b1
The results from complementation tests are precise, highly reliable and they
permit an operational demarcation of genes as follows, mutant alleles located in the
same gene do not show complementation, while those located in different genes show
complementation. Benzer proposed the term citron since the delineation of a gene is
based on complementation test, which is the trans portion of the cis – trans test.
Mid – Semester Examination
Batch : 2009 – 10 Semester : IIYear : I B.Sc., ( Ag.,) Time : 1 hrMax. Marks : 20 Date: …………….
PBG 101. Agricultural Botany, Genetics and Cytogenetics ( 2+1)
PART – A
Choose the correct answer. ( Answer all the questions)
A1. Chromosome number of Oryza sativa
a.2n=24 b. 2n=26 c.2n=28 d.2n=22.
A2. Chromosome number of Hexploid wheat (Triticum aestivum)
a. An =4x=42 b. 2n=6x=42 c.2n = 8x =42 d. 2n=3x=42
A3. The grain of sorphum
a. Achene b. Caryopsis c. Nut d. Schizocarp
A4. The botanical name of finger millet
a. Pennisetum glaucam b. Elensine coracana
c. Setaria italica d. Triticum aestivum
A5. Groundnut belongs to the family
a. Poaceae b. Papilonaceae
c. Asteraceae d. Solanaceae
A6. An example for protogyny
a. Pennisetum glacum b. Zea mays
c. Triticum aestivum d. Oryza sativa
A7. Nature of corolla in Pulses
a. Gamopetalous b. papilonaceous
c. gamosepalons d. polysepalous.
A8. Pigeon pea is
a. Self pollinated b. Cross pollinated
c. Often cross pollinated d. Highly self pollinated
A9. Botanical name for Mung bean
a. Vigna mungo b. Vigna radiala
c. Vigna ungliculata d. Glycine max
A10. The monohybrid ratio
a. 3:1 b. 15:1 c. 63:1 d. 9:6:1
A11. Back cross
a. Crossing of F1 with any one of the parents
b. Crossing of F1 with recessive parent
c. Crossing of F1 with dominant parent
d. Crossing of F1 wit both parents
A12. Alternative form of a gene is called
a. Allele b. genome c. Allelomorph d. geno type
A13. 13:3 is the ratio of gene interaction resulted thro’
a. Masking gene action b. Inhibitory gene action
c. Complementary gene action d. Supplementary gene action
A14. Homozygous
a. Two copies of the same allele
b. One copy of the same allele
c. Two copies of different allele
d. More than two copies of the same allele
A15. Dihybrid linkage ratio will be
a. 9:3:3:1 b. 1:1:1:1 c. 15:1 d. 3:1
PART – B
State True or False (Answer all the questions) 15x0.5=7.5
B1. Stem of rice is known as culm.
B2. Ray floret and Disc floret are found in sunflower ( Helianthus annceus L)
B3. Maize is a protandrous croptamen.
B4. Epipetalous stamen found in Groundnut.
B5. Gossypium herbacum is a old world cotton.
B6. Rape and Mustard belongs to Brassicaceae.
B7. Dimorphic branching in gingelly pattern observed.
B8. The chromosome number of ground nut ( Arachis hyposace. L.) is 40.
B9. Rice is a cross pollinated crop.
B10. The botanical name for sweet potato is ipomea batatas.
B11. Tuber is the modification of root in potato ( Solanum tuberosum)
B12. Fixed oil is otherwise known as Non- volatite oil.
B13. Starches are rich in carbohydrate complex.
B14. Single gene influences many character is called multiple allele.
B15. Quantitative character shows discontinuous variation.
PART – C
Answer any five questions ( brief answer) 5x1=5
C1. Genetics
C2. Tassel
C3. Peg formation
C4. Nobilization in sugarcane
C5. Pleiotropy
C6. Transgressive segregation.
Mechanism of DNA replication
DNA replication taken place at a y shaped structure called a replication fork. A
self correcting DNA polymerase …….. catalyses nucleotide polymerization, in a 5’ –
3’ deirection copying a DNA template strand with remarkable fidelity. Sincere the
two stands of a DNA double helix are antiparalled, the 5’ to 3’ DNA synthesis, can
taken place continuously on only are of the strands at a replication fork C the
leading strand). On the lagging straw short DNA fragment must be made by a
“backstitching” process. Because, the self correcting DNA polymerase cannot start a
new chain, these slapping strand DNA fragments are prioned by short RNA primer
molecules that are subsequenting erased and replaced with DNA.
DNA replication requires the cooperation of many protein. These include (i)
DNA polymerase and DNA primate to catalyee the nucleoside for phosphate
polymerization (2) DNA helicases and single strand DNA binding (SSB) protein to
help in opening up the DNA helix so that be copied (3) DNA ligase and an that
degrades RNA prisoners to red to gether the discontinuously synthesed lagging stand
DNA fragments and (4) DNA topoisomerases t help to relieve helical windly and
DNA tangling problems. May of these proteins associate with each other at a
replication fork to form a highly efficient replication machine through which the
activity and the spatial moments of the individual components are coordinated.
DNA, the genetic materials – Griffith experiment, experiment of avery, mxlesid
and earth.
DNA
Deoxyribonucleic Acid is the genetic material found in most of living
organisms including plant, animals and even some plant viruses. The criteria of being
a genetic material are specific and important for its functional life and stable
transmission.
The various erecteria, requirements for any meterials being genetic materials
are as follows.
1. It must contain biologically useful informstion that is maintained in a stable
form.
2. It must be reproduced and transmitted from cell to cell (or) generation to
generation.
3. It must be able to express itself so that other biotogical molecules will be
produced and maintained.
4. It must be capable of variation i.e. it should accept some minor variation such
change helpful for evolution of cell, organisms to new form.
Experiment proof to show that DNA is genetic material
F. Griffith experiment ( 1928)
He conducted a transformation experiment which laid the foundation for
experimental studies airing at DNA as a genetic material.
Page - 122------------------ Picture
Two strains of pneornococcus (Diplodocus pneumonia ) are which with
smdoth capsule ( S- III) strain and othe with rough capsule ( R- II) strain were
selected for experiment. S III was virulent when as – II was avirulent strain. He
injected heat killed virulent strain S- III alone and found no infection, death in rat.
The mice injected with mixture of R – II (living) and S – III ( hart – killed) died and
virulent pneumococcal could be isolated from these mice showing that transformation
has occurred virulent to virulent strain.
O.T Avery, C.M. Macleod and M.Mc carth ( 1944)
The above scientists reported epriffith’s experiment in an inviter system to know
the transforming principle involved in Griffith experiment.
The incubated R cells in the presence of highly purified DNA fraction obtained
from type S- III bacteria, and some virulent type S- III bacteria were recorded from the test
tube.
If the type S- III DNA fraction was first treated with beoxyribonuclease, , an
enzyme that degrades DNA, no transformation of R cells occurred in subsequent
experiment.
Present interpretation of the experiment is that purified S – III type DNA is capable
of entering the R – II cells and recombining with R – II DNA to produces small number
of living smooth, virulent S- III type recombinant colonies. So DNA was attributed the
role of transforming principle.
Living R – II type
Heat killed S- III type
Frontal – Control and sanger experiment
In several virus, e.g., TMV ( tobacco mosaic virus), DNA B absent. These
virus are composed of RNA and proffer. Their RNA is coiled like a spring, white the
protein molecules are arranged on the out side of the coil. Frankel – contract and
sanger demonstrated in 1957 that RNA function as a genetic material.
Proteins and RNA of TMV can be separated chemically; when they are
remixed appropriate condition they are mixed together to produce active TMV
particles. In ons experiment sanger and contract used either RNA or the proteins
isolated form TMV for infection of tobacco lenes. Mosaic symptom developed only
when RNA was used for infection (not in pre…… were. Used. Clearly indicated only
RNA of TMV is capable of producing the diseases, and hence appears to be the
genetic material.
Mibed together
Only R – Ii type colonies
Both R – Ii type and S – III type colonies
No colony reconnected
Transformation occurred
Stain A TMV parkeles
SeparationRNA
+
Protein
Infection Symptom of strain
A
CHROMOSOME
History
Holfmiester (1884) – He was first observed dark, rod like bodies in nuclens.
E. strasburger ( 1875) – discovered thread – like structure which appeaed in
cell division.
W. Waldeyer (1888) – coired the term chromosome.
Occurrence
Chromosomes are found in all living i.e. plants, Animal fungi, bacteria and
viruses. In all higher plant sand animals ( Eukaryotes) chromosomes are condensed
material found in nucleus, whereas in lower torms ( prokaryotes) chromosomes are
represented by loosly coiled a packed DNA thread present in the cell.
Numerology
Chromosome number is fixed for a species , the lowest number is seen in
Haplopappas gracilis ( 2r=4) and maximum of 2n = 1656 in some ophiolosum species.
RNA
+
Protein
+
Protein
Stain B
Separation Infection Symptom of strain
B
RNA
+
Protein
Stain A
Separation Infection Symptom of strain
A
Stain B
Separation Infection Symptom of strain
B
Single set of chromosome is represented by ‘n’ and organisms with single set
of chromosome are called as haploids, while the organisms with two complete set of
chromosomes are called as diploid and represented by ‘2n’.
S.No Organisms Chromosome Number (2n =)
1. Human ( Homo sapiens) 46
2. Canis familaris (oog) 78
3. Drosophila Melenogaster
(fruit fly)
8
4. Pisum sativum ( Pea) 14
5. Zea mays ( Maize) 20
6. Sacctraum afficinarum
( Mustard)
80
Size of chromosome
Size Example
1 micron Fungi
3 micron Drosophila
2 micron Human beings
200 micron Chironomou (Polytene chromosome)
Karyotype
Symmetric Asymmetric
Primitive type Ex. Pinus species
Advanced type Ex: Horno sapiens
Karuptype
The diagrammatic representation and arrangement of all chromosomes
according to their size is called as karyotype.
Idiogram
Diagrammatic representation according to their decreasing size is called as
“Idiogram”.
Structure of chromosome
Chromosome consists of following structure are as follows.
1. Centromere
2. Secondary construction
3. Nucleolar organizers
4. Telomeres
5. Satellites DNA.
Centromere
The joining position of two arms of a chromosome, which is darkly stained
body important for chromosomal movement during cell division. As the spindle fibus
join a proleineous body sumounding centromere called kinetochore. The position of
centromere differ based on chromosome.
Based on position of centromere, chromosome may be of following type
1. Metacenbic ( Centromere at central position)
2. Sub metacenbic (Centromere slightly towards one of the arms).
3. Areocentric ( Centromere at sub terminal position)
4. Telemetric ( Centromere at terminsl site)
Secondary constriction
Each chromosome has got a centromere ( primary constriction), but in some
cases besides primary there is another constriction called secondary construction it
helps in identifying the particular chromosome. Ex. Chr. 1, 10,13 16 and Y
chromosome in human.
Nucleolar Organizer ( Secondary constriction I )
Two homologous chromosome in diploid cell may have additional
constrictions called as. Nucleolar organizer. As they play an important role in the
formation of nucleus known as nucleolar organizer.
Satellite DNA
The segment of the chromosome bey and nucleolar organizer look like a
know like structure called a satellite DNA.
It has found in 13, 14, 15, 21, 22 and Y chromosome show nucleolar organizer
and satellite structures.
Centromere
Centromere
Telomeres
The tips of the chromosome are called telomeres. Telomeres are different in
structure and composition from rest of the chromosome. Telomeres are unique in
their function that they present the two chromosome from sticking to each other.
Telomere specifically designed region of the chromosome for attachment to nuclear
envelop from telophase to prophase through interphase.
Chromosome
During early stages of cell division chromosomal material becomes visible as a
very thin filaments, which are called as chromoremata. Chramatial and chromoremat
are all two names of the same structure.
Nucleosome model of chromosome
The model deseribing the association of histon proteins with DNA, forming
chromosomes was given by R.D. korenberg and J.P. Thonson.
DNA is a thread like structure and interacts with basic prolein called histories
to form the structural unit, nucleosome.
Nucleosome made up of core particle of each H2A, H2B, H3 and H4 histon
protein called octamer, around the DNA takes appeoximately two turns.
Nucleosomes joined together by linker DNA.
Further packing of chromation ( DNA + histone) gives rise to a metaphase
chromosome through different intermediate structure.
DNA + Histone protein
= Nucleosome
Chromosome
Special type of chromosome
Lamp brush chromosome
In diplotene stage of meotic division the yolk rich oocytes of many vertebrates
such as fishes, amphibians, reptile and birds contain exceptions large size
chromosome known as lampbhrush chromosome. The lampbrush chromosome were
first discovered by ruckert in 1892.
Structure
During early prophase lampbrush chromosomes are in the from of a pair of
honoloyous chromosome having few points of contact ( or) chiasmata. Under light
microscop each chromosome is seen to consists of an axis along which is a row of
dense granules or chromosome. From each chromosome arise lateral loops.
Lamphrush chromosome consists of following sub structure
(i). Chromosomal axis
(ii). Chromere
(iii). Loops
Nucleosome
Linker DNA
Polytene Chromosome
Polytene chromosome occure in salivary glands, gut, trachea, malpigean
tubules of many insects of dipteral. This chromosome was first discoved by “kollar”
The polytene chromosome by drosophila the total length of about zoom.
Structure of polytene chromosome
(i). Bands and interbands
It bears light and douk bands along their length. The dark bands takes deep
colour by basic chromosomal studies. The disc shaped structures and occupy the
whole chromosome.
This inversion
Lateral loop
Chromo mere
Thick inversion
Chiasmata
The light band take lisp chromosomal stains. These are febrile composed
heterochromatin.
(ii). Putts and balbiani ring of poly tene chromosome
Putts and Balbianims ring of chromosome normally have occurred during
development stage. The swelling of dark bands and light bands leads to formation of
putts.
The process of putt formation controlled by certain specific gene and is a cyclic
process. The process of putt formation involves several process such as accumulatin
of acidic protein, despiralization of DNA, synthesies of m RNA and storge of newly
synthesized m-RNA in chromosome.
Chromonemata
Dark band
Light band
Chromosoms puff
Chromoretiata
Balbiani ring
Dark band
Linter band
Functions
The putts are related with synthesis of RNA and protein because, besides DNA
and proteins these contains large amount of RNA. The putts are chiefly mutable
activity of chromosome because of high con. RNA.
Variation in chromosome structure – deletion and duplication genetic and
cytological implications. Inversion and translocation – genetic and cytological
implication.
Homologous chromosomes contain an identical number and kind of genes in
the same sequence. Occassionally, spontaneous (without any known causal factor)
variation in chromosome number or structure do arise in nature; these variations are
called chromosomal aberrations.
Classes (i) structural and numerical
There are four types of structural aberrations
(i) Deletion or deficiency
(ii) Duplication or repeat
(iii) Inversion and
(iv) Translocation
Deletion ( a deciease)
Duplication ( an increase)
Inversion – ( change the gene sequence)
Translocation - (affect the kind of genes located in the affected chromosomes )
1. Deletion - Loss of a segment of a chromosome
Terminal - Loss of a segment includes telomere
Interstitial - Loss of a segment between telomere and centromere lost.
2. Duplication - A chromosome segment present in more than two copies
in the same nuclens.
1. Tandem duplication - additional chromosome segment located just ofter
the norma,l segment, gene sequence being the
same.
2. Reverse duplication - Same as above gene sequence inverted.
3. Displaced duplication - additional segment in the same chromosome but
away from normal segment.
4. Translocation
duplication - additional segment located in a non- homologous
chromosome.
Invession - A chromosome segment contains genes in a
Sequence which is reverse of the normal.
(i) . Paracentric - The inverted segment does not contain centromere.
(ii). Pericentric - The inverted segment contains centromere.
Translocation :- - Chromosome segment (s) integrated into
nonhonedlogous chromosomes (s).
1. Simple translocation - A segment of a chromosome integrated into a non
– homologous chromosome.
2. Reciprocal translocation - A segment of chromosome integrated into a non –
homolomogsus chromosome from which a
segment is integrated into the first one.
Deletion
Genetic effects:
(i) Generally produce some stricking genetic and morphological
consequences.
(a) Loss of segment has a deleterious effect on diploid organisms and most
deficieacies may produce dominant lethal effect.
(b) Deletion – (Bridges in 1917)
(c) Recessive alleles of the genes located in the deleted segment, present in
the normal chromosome of the heterozygote will express the mselves as
if they were dominant. This Psendodominance.
(d) Crossing over is absent
(e) Many deletions produce a observable morphological effect.
Cytological defection:-
Cytological detection of defectives for relatively lorger chromosome segments
present little difficultly.
(i). It produces characteristic loop.
(ii). Observed in giant chromosomes or Drosophila.
(iii). Described in maize by Barbara and Mclintock.
Duplication: -
Cytology
(i). Relatively larger duplication are detectable due to the presence of loops formed by
the duplicated segment during pachyletic stage.
(ii). Crossing over in the duplicated regions produces a dicentric chromosome, which
are likely to give rise to a bridge – breakage – function cycle.
Chromosomes of Diptera
Genetic effects:-
Some duplication may have district phenotypic effects and may be regarded as
a mutant allele. E.g. Bar duplication in Drosophila.
Mutations in a duplicated locus are toleracted by the organism since there still
remains one unaltered copy of the concerned gene in the genome to perform its
normal function. Thus all new low which have developed during evolution may well
owe their origin to gene duplication.
Inversion:-
Genetic and cytological effects :
(i) The presence of inversions may be detected either phenotypically
genetically or cytologically ( i) inversion heterozygotes ( individuals
having one inverted and one normal homologue) generally exhibit parted
male sterility.
(ii) Linkage map of the strain with that of the normal strain provides conclusive
evidence for the existence of inversion.
(iii) Absence of crossing over between two genes could be indicative of
inversion.
(iv) Cytological observation of inversion heterozygote’s level the presence of
inversion loops during pachytene stage and salivary glands chromosomes
of Dipleran insects.
(v) Inoverted band patterns of salivary gland chromosomes make it possible to
defect even small inversions.
Translocation: -
Genetic effects:-
The phenomenon of transolocation was first discovered in 1923 by pridges in
Drosaphils. Through genetic analysis of a shift of a segment from chromosome Ii into
chromosome II it produced phenotypic effects as well as lethality ( Recersive).
Cytology:-
In transolocation homozygote, normal bivalents formed and no detectable
cytogenetic peculiarity ( except for a possible change in chromosome morphology)
due to translocation).
In translation heterozygote, one member from each of the two chromosome
pairs involved in a reciprocal translocation. E.g. Drosophila showed cross
configuration intheir salivary gland chromosomes.
Variation in chromosome number – Euploid, aneuploid – types of euploids.
In somatic cells of a diploid organism, two copies of the same genome are
present ( i.e. 2n-2x) while their gametes contain a single genome ( that is n-x0 A
deviation from the diploid ( 2n – 2x) states represents a numerical chromosome
aberration which often referred to as heteroploidy: individuals possessing the variant
chromosome numbers are known as heteroploids.
Heteroploid may be classified as
(i). aneaploidy and (2) enploidy
A summary of the various common types of changes in chromosome number
Term Type of change Symbol
Heteroploid Change from the 2x state
A. Anenploid
Nullisomic
Monatomic
Double monatomic
Trisomic
One or a few chromosomes extra or
missing from 2n
One chromosome pair missing.
One chromosome missing
Two non – homologous (each from
adifferent pair) chromosome missing
One chromosome extra
2n + few
2n-2
2n-1
2n-1-1
2n+1
Double transonic
Tetrasomic
Euploid
Monoplood
Haploid
Two non- homologues (each from a
different pair) chromosomes extra.
One chromosome pair extra.
Number of geneores different from two
Only one genome present
Gametic chromosome number of the
concerned species present.
2n+1+1
2n+2
x
x
Term Type of change Symbol
1. Auto poly ploid More than two copies of the same
genome present
Auto triploid Three copies of the same genome 3x
Auto tetraploid Four copies of the same genome 4x
Auto pentaploid Five copies of the same genome 5x
Auto hexa ploid Six copies of the same genome 6x
Auto octa ploid Eight copies of the same genome Sx
2. Allo polyp lord Two or more distinct
Genomes generally each
Genome has two copies.
Allo tetraploid
Two distinct genomes
Each has two copies
(2x, +2x2)**
Allo hexaplolid
Three district genomes each has three
copies
(2x, +2x2 +2x3)**
Alloocutaplod Four distinct genomes; each has two
copies.
(2x, +2x2
+2x3+2x4)**
* 2n – somatic chromosome number of a specis whether diploid / polyploid.
n – gametic chromosome number
** x, the basic chromosome number / genomic number
** In general, this situation occur, other situation may also occur.
Homeologous chromosome :- Those chromosome which compersates for the deficex
of each other in nullesomic tetrasomic ( 2n-2+2) condition are termed as homeologous
chromosome.
e.g. chromosome IA, IB and ID of wheat are homologous with each other.
(i). They are similar in gene content reduced pairing.
(In wheat, this reduced homeologous pairing in due to the gene ph located in
chromosome 5B)
Production of aneuploids:-
Aneuploids individuals may be obtained in several ways (1)
(i) meiotic irregularities,
(ii) Non- disjunction or lagging of one chromosome, occur spontaneously, in
low frequencies and produce n +1 and n-1 gametes.
(iii) When such gametes unite with normal ( n) gametes, 2n+1 and 2 n-1
individuals are obtained.
2. Triploid plants are the best source of aneuploidy gametes.
3. Many univalents are regularly, present at MI of synaptic and desymaptic plants.
Consequently, most of their gametes are aneuploid and they produce many cheuploid
progeny.
4. The ring of four in translocation heterozygote’s may disjoin 3:1 ancuploid
gametes thus produced would gemmates tertiary trisomics and monosomics.
5. Progeny from a cross betweentetrasomic (2n = 2) and destine plants show a high
frequency of transmits.
(i) Poly ploid – auto and allopolyploids
(ii) Role of poly ploidy in evolution of crops – wheat, cotton, tobacco and brassica.
(iii) Types of aneuploids and their origin
Presence of more than two genomes in an individual is known as
polyploidy.
If all the geames present in an individual are identical, it is called
autopolyploid. But in an allopolyploid, two or more district genomes are present.
Segmental allopolyploidy:-
Genomes present in an individual may be ony parpally differenciated
( partially homologers)
Amphidiploids:
Naturally occurring allopolyploids ordinarily contain two copies of the each of
the genomes present and they show normal bivalent formation.
Auto polyploids produced through
Chromosome doubliny through
(i) Unreduced (2n) gametes produced spontaneously which ( yield tetraploid
zygotes).
(ii) Somatic cells going rising to teltra ploid buds.
(iii) Adventitious shorts produced followed by decapitation may be polyploidy
( solanaceal, - 36% of such shoots reperfed tctraploid.
(iv) Heat / cold treatments
(v) Through tissue culture, variability in Nicotiana, Detura, rice.
(vi) Some chemicals acenaphthena 8 hydrory quinolone, miltrows oxide.
(vii) Treatment of seeds / seedlings or shoot tips with colchicines.
Genetic effects:
(i) In many species, auto polyploids show an increase in general vigour and six,
the phenomenon is known as ‘gigatism”.
(ii) Some species relatively weaker and smaller in size.
(iii) Leaves of autopdyploids are larger and thicker, fruits flowers, seeds are
larger, but their number is so……
(iv) Compared to normal disomic plants.
(v) Cells, pollen grains and stomata of autopolyploids are relatively larger than
those of diploids.
(vi) All antopolyploids show variable sterility
(vii) Increase in fresh weight over the concerned normal discomic strains.
Cytology
In triploids, the three homdogues for each chromosome from either a trivalent
or a bivalent and a univalent.
During AI, ( Anaphase – I) the disjunction of trivalents and the distribution of
univalents is irregular producing a range of anenploid gametes.
As a result, triploids produce a range of aneuploid progeny. E.g. trisomic,
double trisomic etc., Highly sterite in bahana, watermelon, fertile in spinach 9
Spinacea devalla).
Role in Evolution:-
Sutopolyploidy has contributed to a limited extent in evolution of plant
species. Potato ( 4x) peacit (4x) coffec ( 4x) Lucerne (4x) banana ( 3x), sweet potato
(6x).
Allo Polyploidy:-
Natural allopolybloids most likely originated through chromosome doubling,
of F1 hybrids produced through chance natural hybridization between two district
species of the same genes or from different genera. And it may be generally a sterile
one.
Synthetic production of allopolyploids in achieved through chromosome
doubling with the help of colchicines. (or some other agent) of distant hybrids
( hybrid between two district species); such allopolyplids are often known as
synthetic allopolyplids.
Experimental production of an allopolyploid
Triticum turgidum x Secale cereal
AA BB RR
Two distinct species are hybridized
AB R
ABR Colchicine (Chromosome doubling)
AA BB RR
The F1 hybrid would be highly sterile, chromosome number of the hybrid is double by very colchicines.
Fertile allopolyplod (amphidiploids) species distinct from the two parental species)
Triticale Hexapodies
(Allohexaploid)
Amphieliploid
Parental species
Genomes
Gametes
Role of evolution :-
Allopoly ploids have been more successful as crop species than autopolyploid.
Origin of wheat:-
Parents
Origin of cotton ;
The upland cotton, Gossipier hissutum is an auto tetraploid . It has B large and
13 small chromosomes in its haploid complement ( η = 26)
G. Lirsutum is the hybrid between
G. herbaceum var. africannum ( old would cotton)
Genome – Triticum Monococum X Aegilops speltoides
AA BB
Gametes - AB
AB
AABB
I. Diccocum
Doubling
Aegilops Squarrosa X DD
AB D
ABD
Doubling
AA BB DD
Triticum aestivum
(Bread wheat)
(η = 13 small chromosome)
X
G. Raimondi ( Amerikan diploid cotton)
(η = 13 large abromosome)
Origin of Tabacco
Similarly nicotiama tabacum and Nicotiana rustica are also allotetraploids, each
having 24 chromosomes in their haploid complements Qn-48). Available evidence
indicates that N. tobacum is an amphidiploids derived from the cross N. Sylvestris x
N. tomentosa, each of the parental species having 12 chromosomes their haploid
complaments.
Origin of Brassica species
Several of brassica e.g. B. Junea, b. rapus and B. carinata are tetraploids.
The relationship between some diploid and certain naturally occurring
amphidiploids species of Bressica.
Applications in crop improvement
1. Triticale is the synthetic allopolyploid that has succeeded as a crop species, in
which have been released for commercial cultivation.
2. Raphavo brassia as a fodder crop.
B.nigraη =8(BB)
B.carinate
η = 17(BBCC)
B.junceaη = 18
(AA BB)
B.deracea
η = 9(CC)
B.compestrics η = 10(AA)
B.napusη = 19
(AACC)
3. Festaca lolium hybrids and the triploid (AAC) obtained by crossing B. napus
(AA CC) with B. compestics (as a fodder crop).
Other Application
1. The genetic base of B. napus widened by synthesizing from the parental
species (B. oleralea X B. compestris).
2. Bridging species between two diploid species which do not cross with each
other & with Fx hybrid is sterile.
N. Sylvestris X N. tabacum
N.digluta X N. tabacum
In this way, resistance to tobacco mosai virus ( TMV) was transferred from N.
syluestris to N. tabacum by using N. digluta ( amphidiploids) as a bridging species.
Refer for RNA types , protein synther characteristies of actual deegma origin
of aneuploid.
Synapsis.
********************************
N. digluta (chromosome dowbling fertile).
F1 ( Partially fertile)
But backcrops with N. tabacum is easy.