Download SMU-DDE-Assignments-Scheme of Evaluation Q. No

SMU-DDE-Assignments-Scheme of Evaluation
PROGRAM
SEMESTER
SUBJECT CODE &
NAME
BK ID
DRIVE
MARKS
BSC BIOTECHNOLOGY
3
BO 0048
GENETICS
B0763
WINTER 2014
60
Q.
No
Criteria
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Marks
Answer the following
Explain supplementary and lethal interaction of genes with suitable example.
section 3.3 & 3.4;Page 51-52)
Explaining Supplementary interaction of genes:
 Supplementary genes are two independent pairs
of dominant genes, which interact in such a way
that one dominant gene will produce its effect
whether the other is present or not.
 The second dominant, when added, changes
the expression of the first one but only in the
presence of first one.
 Example: inheritance of coat colour in Rats:
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 In rats and guinea pigs, coat colour is
governed by two dominant genes A and C,
the agouti coloured guinea pigs have
genotype CCAA.
 The black rat possesses gene for black
colour (C) but not the gene (A) for agouti
color.
 If gene for black colour is absent agouti is
unable to express itself and rats with a
genotype ccAA are albino.
 Here the presence of gene C produced
black colour and addition of gene A
changes its expression to agouti colour.
 The result is 9 agouti : 3 black : 4 albino.
Lethal interaction of genes:
 Genes may affect viability as well as the
phenotypic traits of an organism.
 Experiments have shown that animals carrying
certain genes cause impaired biochemical as
well as physical functions.
 Some genes do not affect the appearance of an
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organism but do influence viability.
Some genes have such serious effects that the
organism is unable to live. These are called
lethal genes.
If the lethal effect is dominant, all individuals
carrying the genes will die and the gene will be
lost.
Some dominant lethals have delayed effect so
that the organism lives for a time.
Recessive lethals carried in the heterozygous
condition have no effect but may be expressed
when cross between two carriers occur.
Example: Creeper chickens
 The dominant gene ‘C’ in chickens is
responsible for developmental changes
that result in aberrant forms called
‘Creepers’ and the homozygous genotype
‘CC’ is lethal.
 These birds have short, crooked legs and
are of little value.
 When two creepers were crossed, a ratio
of 2 Creepers to 1 normal instead of 3:1
appeared. This is the characteristic ratio for
lethal interactions.
 The creepers are heterozygous Cc and the
missing ¼ individuals were CC.
 The creeper gene in homozygous condition
produces severe malformations that the
offsprings die during incubation.
Describe the mechanism and kinds of crossing over.
The mechanism of crossing over includes following
steps:
1. Synapsis
 During prophase – 1 of meiosis the maternal
and paternal chromosomes of a homologous
pair come close together and pair at zygotene
stage. This pairing is called the synapsis.
 The homologous chromosomes pair precisely by
mutual attraction between the allelic genes.
 The paired chromosomes are called bivalents.
2. Duplication
 During diplotene stage of meiosis, each of the
homologous chromosomes in a bivalent splits
longitudinally into two sister chromatids.
 Thus the bivalent now consists of four
chromatids and is known as tetrad.
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(Unit 4;Section 4.3;Page 68-76)
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SMU-DDE-Assignments-Scheme of Evaluation
3. Crossing over
 During diplotene stage, when the paired
chromosome start separating, the chromatids
remain in contact at one or more points. These
points of contacts are known as chiasmata.
 At each chiasma two non-sister chromatids of
the bivalent break at the corresponding points
and then rejoin with the exchange of segments.
4. Terminalization
 After crossing over, the non-sister chromatids
start repelling each other.
 The chromatids separate from the centromere
towards the tip.
 The chiasmata also start moving in zipper-like
fashion towards the ends. The movement of
chiasmata is known as terminalization.
Explaining the kinds of crossing over:
Depending upon the number of chiasmata crossing
over can be of the following types :
1. Single crossing over
 When only one chiasma is formed all along the
length of a chromosome pair, it is known as
single crossing over.
 The chromatids of homologous chromosomes
contact and break only at one point along their
entire length
2. Double crossing over
 In double crossing over, chromatids break and
rejoin at two points.
 In double crossing over two chiasmata are
formed along the entire length of the
chromosome. They can be further divided as:
a) Two strand double cross overs
 In this kind of crossing over, the same two
chromatids are involved at both the crossing
over points.
 The chiasmata thus formed are known as
reciprocal chiasmata.
 As a result of this, two chromatids are noncrossovers and preserve the parental
combination of genes.
b) Three strand double cross overs
 In these crossovers, three chromatids are
involved, i.e. the second chiasma contains one
of the same chromatid, which has crossed at
first chiasma but the other chromatid is
different.
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 Here only one chromatid is non-cross over,
while the other three exchange their parts.
c) Four strand double cross overs
 In such cross overs all the four chromatids of
the tetrad are involved, two of them exchange
parts at first chiasma and the other two are
involved in the second chiasma.
 Such chiasmata are known as complementary
chiasmata.
 These produce four single crossovers.
3. Multiple Crossovers
 When crossing over occurs at more than two
places in the same chromosome pair and more
than two chiasmata are formed, this type of
crossing over is known as multiple crossing over.
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Describe sex linked inheritance with an example.
 Sex chromosomes (XX – XY) are primarily
concerned with the determination of sex but
they also carry genes for other body characters.
 Such traits whose genes are located on the sexchromosomes and follow sex during inheritance
are known as sex linked characters.
 The genes responsible for these traits are
termed sex linked genes and their mode of
inheritance is described as sex linked
inheritance.
 The sex linked inheritance can be X-linked or Ylinked inheritance.
 The most popular examples of sex linkage in
man are (i) Haemophilia (ii) Color blindness (iii)
Myopia (iv) Mitral stenosis (v) Juvenile glaucoma
and (vi) Muscular dystrophy.
 X-linked inheritance: Red green color blindness:
 Color perception is controlled by the cones
in the retina.
 The genes for the cone proteins are located
in the X chromosome.
 The red blindness is called protanopia and
the green blindness deutoranopia.
 Colorblindness is caused by X-linked
recessive gene.
 When a normal woman marries a colorblind
man, all her sons and daughters have
normal color vision, but when the daughters
are married to man with normal color vision
some colorblind sons are produced.
 If a colorblind woman is married to a normal
(Unit 5;Section 5.2;Page 106-110)
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man, all her sons are colorblind whereas all
the daughters have normal color vision.
 When these daughters having normal color
vision are married to a colorblind man, the
colorblind grandsons and granddaughters
are produced.
 It is observed that a color blind woman has
sons all colorblind and daughters all with
normal vision and a colorblind woman
always has a colorblind father and her
mother is a carrier.
 Following conclusions were drawn from the
above results.
1) Color blindness is more common in males
than in females.
2) Two recessive genes are needed for the
expression of color blindness in female,
whereas only one gene gains expression in
male.
3) Males are never carriers
4) Colorblind women always have colorblind
fathers and always produce colorblind sons.
5) Colorblind women produce colorblind
daughters only when their husbands are
colorblind.
6) Women with normal color vision, whose
fathers are colorblind, produce normal and
colorblind sons in approximately equal
proportion.
 Y-linked inheritance:
 The genes in Y chromosome in man are
exclusively transmitted through the male
line.
 These traits are also called holandric traits,
the genes, holandric genes and the
inheritance as holandric inheritance.
 The genes for hairy pinna is example for Ylinked genes.
 The sex determining genes such as H – Y and
TDF are located in Y chromosomes.
 17 genes are found to be present in the Y
chromosome of man.
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Explain in detail the factors that alter the gene equilibrium.
 Mutation:
 Mutation introduces new genes leading genetic
differences in the population.
 These new genes introduced may or may not
(Unit 7;Section 7.4;Page 165-170)
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persist in the population upon their utility.
The gene frequencies will also depend upon this
factor.
Example: If a dominant gene ‘A’ mutates to ‘a’,
then frequency of ‘a’ will replace ‘A’.
Quantitatively, let po be the initial frequency of
‘A’ and ‘u’ be the mutation rate with which ‘A’
changes to ‘a’.
In such a case, ‘a’ will appear with a frequency
of u x po in the first generation.
The frequency of ‘A’ will, therefore be reduced
by a factor pou and become po – pou = po (1 – u).
In the next generation there will, therefore, be
further change due to the change of A to a, thus
further reducing the frequency of A by a factor
po (1 – u) – po (1 – u) x u = po (1 – u) (1 - u) = po
(1 – u) 2.
In this manner, in ‘n’ generations, the frequency
of A will be reduced to po (1 – u) n.
Eventually the term (1 – u)n will approach zero
so that A will disappear after several
generations, if no reverse mutation takes place
and the mutant allele experiences no selection
pressure against it.
Selection
The gene frequencies may change due to
selection in favour of one of the two alleles of a
gene.
For example, if individuals with allele ‘A’ are
more successful in reproduction than the
individuals with a, the frequency of the former
will be higher.
The selection can be artificial or natural.
The factors influencing selection may include
temperature, humidity, food, sexual attractions,
etc.
There are several aspects of selection, like
fitness, gametic selection and zygotic selection.
When one genotype can produce more offspring
than the other in the same environment, it
means relative reproductive success, called
fitness or adaptive value or selective value
Migration
Through migration, new alleles can be
introduced into a population from nearby
population.
Let us consider a large continental population
donating individuals (genes) to a small island
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population.
The rate of migration is ‘m’, which is equal to
the fraction of genes on the island that are
replaced by genes from the continent each
generation.
If qi is the frequency of a particular allele ‘a’, on
the island, and qc is the corresponding
frequency on the continent, then after
migration, the frequency on the island will be
qi = (1 – m) qi + mqc.
The change in the frequency of ‘a’ in one
generation will therefore be qi = qi’ – qi = m
(qc – qi) ; qi will become zero either when
migration stops (m = 0) or when the frequency
of ‘a’ on the island equals the frequency of ‘a’
on the continent (qi = qc).
The deviations in gene frequencies will also be
due to sampling errors.
Describe inheritance of shell coiling trait in Limnaea peregra as an example for non-Mendelian
inheritance.
(Unit 9;Section 9.2;Page 195-197)
 The direction of coiling in shells of the snail
Limnaea peregra is one of the best examples of
a maternal effect (non-mendelian inheritance).
 Some strains of Limnaea peregra have dextral
shells, which coil to the right; others have
sinistral shells, which coil to the left.
 This characteristic is determined by the
genotype of the mother (not her phenotype)
rather than by the genes of the developing snail.
 Allele S+ for right handed coiling is dominant
over allele S for coiling to the left.
 When crosses were made between females
coiled to the right and males coiled left, the F1
snails were all coiled to the right.
 The usual 3: 1 ratio was not obtained in the F2
because the phenotype SS was not expressed.
 Instead, the pattern determined by the
mother’s (P) genes (S+S+) was expressed in the
F1, and the F1 mother’s genotype (S+S) was
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expressed in the F2.
 When individuals were inbred, only progeny
that coiled to the left were produced.
 When the S+S+ or S+S snails were inbred,
however, they produced offspring that all coiled
to the right.
 From the reciprocal crosses between left-coiling
females and right-coiling males, all F1 progeny
SMU-DDE-Assignments-Scheme of Evaluation
were coiled to the left.
 The F2 all coiled to the right; but, when each F2
snail was inbred, those with the genotype SS
produced progeny that coiled to the left.
 Further investigation of coiling in snails has
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shown the spindle formed in the metaphase
of the first cleavage division influences the
direction of coiling.
The spindle of potential ‘dextral’ snails is
tipped to the right, but that of ‘sinistral’
snails is tipped to the left.
This difference in the arrangement of the
spindle is controlled by the genes of the
mother.
They determine the orientation of the
spindle, which in turn influences further cell
division and results in the adult pattern of
coiling.
The actual phenotypic characteristic,
therefore, is influenced directly by the
mother, with no immediate relation to the
genes in the egg, sperm or progeny.
Explain genetic disorders related to blood group incompatibilities.
(Unit 6;Section 6.3;Page 142-153)
Blood group Incompatibilities and Related Diseases
 Blood type incompatibility occurs when a
person's
immune
system
manufactures
antibodies that attack the antigens not on his or
her cells.
 A person with blood type A, for example, has
antibodies against type B antigen. If he or she is
transfused with type B blood, the anti-B
antibodies cause the transfused red blood cells
to clump and block circulation.
 ABO blood type is often followed by a 'positive'
or 'negative', which refers to another blood
group antigen, the Rh antigen.
 If one has Rh antigen he or she is Rh positive
and if not he or she is Rh negative. This is
determined by alleles of three genes called C, D
and E.
 The Rh antigen was originally identified in
rhesus monkeys, hence named Rh.
SMU-DDE-Assignments-Scheme of Evaluation
 The Rh blood group incompatibilities is the
result when a woman who is Rh negative
marries a man who is Rh positive, the first born
child of such a marriage is usually normal.
 During the following pregnancies the fetus may
be lost by still birth or it may be born in such an
anemic or jaundiced state that it lives only a few
hours or days after birth.
 The infant dies from Erythroblastosis fetalis, a
condition of anemia due to the breakdown of
RBC (haemolysis) in the fetus and jaundice.
 The blood vessels in the liver become clogged
with the broken RBCs and is termed haemolytic
jaundice. This is in response to destruction of
baby's RBC by the maternal Rh antibodies.
 In this condition a newborn infant may be given
blood transfusion to replace the blood with
damaged RBCs as well as Rh immunoglobulin
(Rhogam) is given to the mother within 36 hours
of delivery of child. This will destroy the fetal
red cell and prevents Rh sensitization.
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*A-Answer
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