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PRINCIPLES OF INHERITANCE AND VARIATIONS
Genetics term was given by W. Bateson.
Genetics = Collective study of heredity & Variations.
Heredity = Transmission of genetic characters from parent to offsprings.
Variation = individuals of same species have some differences, these are called variation.
History of reserches in genetics.
G.J. Mendel - Father of Genetics.
W. Bateson - Father of Modern Genetics.
Morgan - Father of Experimental genetics
He performed experiment on Drosophila & proposed various concepts, like linkage, Sex
linkage, Crossing over, Criss-cross inheritance, Linkage map on Drosophila.
A. Garrod = Father of human genetics & Biochemical genetics. Garrod discovered first human Metabolic genetic disorder which is called alkaptonuria(black urine disease). In this disease enzyme homogentisic acid oxidase is deficient. He gave the concept 'One mutant gene - one metabolic block'
SOME GENETICAL TERMS
1. Factors : Unit of heredity which is responsible for inheritance and appearance of characters.
These factors were referred as genes by Johannsen(1909). Mendel used term "element" or "factor".
Morgan first used symbol to represent the factor. Dominant factors are represented by capital letter
while recessive factor by small letter.
2. Allele : Alternative forms of a gene which are located on same position [loci] on the homologous
chromosome is called Allele. Term allele was coined by Bateson.
T
T
T
t
t
t
3. Homozygous : A zygote is formed by fusion of two gametes having identicle factors is called homozygote and organism developed from this zygote is called homozygous. Ex. TT, RR, tt
4. Heterozygous : A zygote is formed by fusion of two different types of gamete carrying different factors
is called heterozygote (Tt, Rr) and individual developed from such zygote is called heterozygous.
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5.
6.
7.
8.
9.
The term homozygous and heterozygous are coined by Bateson.
Hemizygous : If individual contains only one gene of a pair then individual is said to be Hemizygous.
Male individual is always Hemizygous for sex linked gene.
Phenotype : It is the external and morphological appearance of an organism for a particular character.
Genotype : The genetic constitution or genetic make-up of an organism for a particular character.
Genotype & phenotype terms were coined by Johannsen.
Phenocopy : If different genotypes are placed in different environmental conditions then they produce
same phenotype. Then these genotypes are said to be Phenocopy of each other.
Hybrid vigour/Heterosis : Superiority of offsprings over it's parents is called as Hybrid vigour & it
develops due to Heterozygosity.
Hybrid vigour can be maintained for long time in vegetaively propagated crops.
Hybrid vigour can be lost by inbreeding (selfing) because inbreeding induces the Homozygosity in
offsprings. Loss of Hybrid vigour due to inbreeding, is called as inbreeding depression.
MENDELISM
Experiments performed by Mendel on genetics and description of mechanisms of hereditory processes and formulation of principles are known as Mendelism.
Mendel postulated various experimental laws in relation of genetics.
Gregor Johann Mendel (1822 -1884) Mendel was born on July 22,1822 at Heinzendorf in Austria at
Silesia village. Mendel worked in Augustinian Monastery as monk at Brunn city, Austria.
In 1856-57, he started his historical experiments of heredity on pea(Pisum sativum) plant. His experimental work continued on pea plant till 1863 (19th century).
The results of his experiments were published in the science journal, "Nature For schender varein" in
1865.
This journal was in German language. Title was "verschue uber Pflangen Hybridan".
This journal was published by 'Natural History society of Bruno'.
A paper of Mendel by the name of "Experiment in plant Hybridization" was published in this journal.
Mendel was unable to get any popularity. No one understood of him. He died in 1884 without getting
any credit of his work (due to kidney disease (Bright disease)
After 16 years of Mendel's death in 1900, Mendel's postulates were rediscovered.
Rediscovery by three scientists independently.
1. Carl Correns (Germany) - (Experiment on Maize)
2. Hugo deVries (Holland) - (Experiment on Evening Primerose)
He republished the Mendel's results in 1901 in Flora magazine
3. Erich von Tschermak Seysenegg (Austria) - (Experiment on different flowering plants)
The credit of rediscovery of Mendelism goes to three scientists.
Correns gave two laws of Mendelism.
Law of Heredity/Inheritance/Mendelism Ist Law - Law of segregation.
IInd Law - Law of independent assortment.
Mendel experiments remain hidden for 34 years.
Mendel published his work on inheritance of characters in 1865 but for several reasons, it remained unrecognised till 1900. Firstly, communication was not easy (as it is now) in those days and
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his work could not be widely publicised. Secondly, his concept of genes (or factors, in Mendel's
words) as stable and discrete units that controlled the expression of traits and, of the pair of alleles
which did not 'blend' with each other, was not accepted by his contemporaries as an explanation for the
apparently continuous variation seen in nature. Thirdly, Mendel's approach of using mathematics to
explain biological phenomenon was totally new and unacceptable to many of the biologists of his
time. Finally, though Mendel's work suggested that factors (genes) were discrete units, he could not
provide any physical proof for the existence of factors or say what they were made of.
Reasons for Mendel's success :
1. Mendel studied the inheretance of one or two characters at a time unlike his predecessors who had
considered many characters at a time. (Kolreuter-Tobacco plant, John Goss & Knight -Pea plant).
2. Selection of Material Selection of garden Pea plant is suitable for studies ; which have the following advantages :
i) Pea plant is annual plant with short life cycle of 2-3 months so large no. of offsprings can be analysed
within a short period of time.
ii) It has many contrasting traits.
iii) Natural self pollination is present in pea plant.
iv) Cross pollination can be performed in it artificially so hybridization can be made possible.
v) Pea plant easy to cultivate.
vi) Pea seeds are large. In addition to pea, Mendel worked on rajama and honey bee.
3. Mendel quantitatively analyse the inheritance of qualitative characters.
4. He maintained the statistical records of all the experiments.
Mendel's work : Mendel studied 7 characters or 7 pairs of contrasting traits.
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In Wrinkled seed free sugar is more in place of starch.
Two of the genes are on chromosome 1st and three are on chromosome 4th. These genes are located far
apart on the chromosome except genes controlling plant height and pod shape.
Technique of Mendel









He developed a technique Emasculation and Bagging for hybridization in plants.
Flowers of pea plant are bisexual. In this method one considered as male and another as female.
Removal of anther from a bisexual flower in immature stage is called Emasculation.
Emasculation is done to prevent self pollination.
Emasculated flowers covered by bags, this is called bagging.
Bagging is only used to prevent undesirable cross pollination.
Mature pollen grains are collected from male plants and spread over emasculated flower.
Seeds are formed in the female flower after pollination.
The plants that are obtained from these seeds are called First Filial generation or F 1 generation according to Mendel. The plants of F1 generation are self pollinated & F2 generation is produced.
Petal
Stigma
Anther
Stamen
Carpel
Removal of anthers
(Emasculation)
Parent
Parent
Transfer of Pollen
(Pollination)
Steps in making a cross in Pea
MONOHYBRID CROSS
When we consider the inheritance of one character at a time in a cross this is called monohybrid cross.
First of all, Mendel selected tall and dwarf plants.
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Tall
(pure)
Parent
Dwarf
(pure)
All tall (impure)
F1 generation
Self pollination
Dwarf
Tall
:
F2 generation 3
2 tall(impure)
1tall (pure)
All tall
(phenotypic ratio
or basic ratio or
Mendelian ratio)
1 Dwarf (pure)
(Selfing)
(Selfing)
(Selfing)
1
3 tall : 1dwarf
All tall
Checker Board Method :
First time, it was used by
Draft
tt
Tall
TT
Reginald. C. Punnett (1875 - 1967)
The representation of generations to analyse in
the form of symbols of squares. Male gamets
t
T
Gametes
Gametes
lie horizontally and female gametes lie vertically.
T
Phenotypic ratio : tall : draft
t
Tt
3 : 1
Genotypic ratio : TT : Tt : tt
1 : 2:1
Conclusions (results) of Monohybrid Cross
Ist Conclusion (Postulate of paired factors) :
According to Mendel each genetic character is
controlled by a pair of unit factor. It is known as
conclusion of paired factor or unit factor.
IInd Conclusion (Postulate of Dominance):
This conclusion is based on F1 - generation.
When two different unit factors are present in
single individual, only one unit factor is able to
express itself and known as dominant unit factor.
Another unit factor fails to express is the
recessive factor. In the presence of dominant unit
factor recessive unit factor can not express and it
GENETICS
Tt
F1 generationTt
7
Tt
Tall
Tt
Selfing
Tall
Tt
t
T
Gametes
T
Gametes
t
Tt
Tt
Tt
F2 generation
tt
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is known as conclusion of dominance.
Dwarf
tt
Tall
TT
F1 generation
Tt
All tall
There are two exceptions of law of dominance. [A] Incomplete dominance, [B] Co-dominance,
IIIrd Conclusion (Law of segregation):
During gamete formation ; the unit factors of a pair segregate randomly and transfer inside different
gamete. Each gamete receives only one factor of a pair; so gametes are pure for a particular trait. It is
known as conclusion of purity of gametes or segregation.
Tt
gametogenesis
Tt gamete
gamete Tt
There is no exception of Law of segregation. The segregation is essential during the meiotic division
in all sexually reproducing organisms. (Nondisjunction may be exception of this law).
FORK LINE METHOD FOR GAMETES FORMATION
To find out the composition of factors inside the gamete, we use fork line method.
AaBb = 4 types of gamete
B
AB = 1/4 = 25%
A
- Ab = 1/4 = 25%
b






B
=
1/4 = 25%
ab =
b
Type of gamete / phenotypic category = 2n
Type of genotype = 3n
No. of zygote produced by selfing of a genotype = 4n
n = No of hybrid character or heterozygous pair.
1/4 = 25%
a
-
aB
DIHYBRID CROSS
A cross in which study of inheritance of two pairs of contrasting traits.
Mendel wanted to observe the effect of one pair of heterozygous on other pair.
Mendel selected traits for dihybrid cross for his experiment as follows
[1] Colour of cotyledons  Yellow (Y) & Green (y) [2] Seed form  Round (R) and Wrinkled (r)
yellow and round characters are dominant and green and wrinkled are recessive characters.
Mendel crossed, yellow and round seeded plants with green and wrinkled seeded plants.
All the plants in Fgeneration had yellow and round seeds.
When F1 plants were self pollinated to produce four kinds of plants in F 2 generation such as yellow
round, yellow-wrinkled, green round and green wrinkled, there were in the ratio of 9:3:3: 1. This ratio
is known as dihybrid ratio.
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Gametes
Selfing
Gametes
Gametes
Phenotypic ratio : yellow round : green round : yellow wrinkled : green wrinkled
9
:
3
:
3
:
1
Expression of yellow round (9) and green wrinkled (1) traits shows as their parental combination.
Green Round and yellow wrinkled type of plants are produced by the results of new combination.
Genotype :
YYRR
YYRr
YyRR
YyRr
YYrr
Yyrr
yyRR
yyRr
yyrr
1
2
2
4
1
2
1
2
1
Thus genotype : 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1
Conclusion (Law of Independent Assortment):
The F2 generation plant produce two new phenotypes, so inheritance of seed colour is independent
from the inheritance of shape of seed. Otherwise it can not possible to obtain yellow wrinkled and
green round type of seeds.
This observation leads to the Mendel's conclusion that different type of characters present in plants
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assorted independently during inheritance.
This is known as Conclusion of Independent Assortment. It is based on F 2 - generation of dihybrid
cross. The nonhomologous chromosome show random distribution during anaphase-I of meiosis.
Explanation :
A pure yellow and round seeded plant crossed with green and wrinkled seeded plant which are having
genotype YYRR and yyrr to produced F1 generation having YyRr genotype.
Both the characters recombine independently from each other during gamete formation in F 1 generation . Factor (R) of pair factor (Rr) is having equal chance to (Y) factor or (y) factor of gametes during
recombination to form two type of gametes (YR) and (yr).
Similarly (r) factor also having equal chance with (Y) factor or (y) factor of gametes to form a two type
gametes - (Yr) and (yr).
Thus, total four types of gametes - (YR), (yR), (Yr), and (yr) are formed.
Therefore, during the gametes formation in Fx generation , independent recombination is possible.
The law of independent assortment is most criticised. Linkage is the exception of this.
BACK CROSS
A back cross is a cross in which F1 individuals are crossed with any of their parents.
1) Out Cross : When F1 individual is crossed with dominant parent then it is termed out cross. The
generations obtained from this cross, all possess dominant character, so the any analysis can not possible in F1 generation.
2) Test Cross : When Fx progeny is crossed with recessive parent then it is called test cross.
[a] Monohybrid Test Cross : The progeny obtained from the monohybrid test cross are in equal
proportion , means 50% is dominant phenotypes and 50% is recessive phenotypes.
It can be represented in symbolic forms as follows.
Recessive parent
F1 progeny hybrid
tt
Tt
T
t
Tt Monohybrid test
tt cross ratio = 1:1
t
[b] Dihybrid Test Cross : The progeny is obtained from dihybrid test cross are four types and each of
them is 25%.
Recessive
parent
ttrr
F1 dihybrid
TtRr
RT
Rt
tr
Tr
tr TtRr Ttrr ttRr ttrr
Dihybrid test cross ratio = 1:1:11
The ratio of Dihybrid test cross = 1:1:1:1
Conclusion: In test cross phenotypes and genotypes ratio are same.
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Test cross helps to find out the genotype of dominant individual.
Homozygous
recessive
Homozygous
recessive
ww
ww
w
w
w
W
W
Ww
WW
W
w
Dominant phenotype
(Genotype unknown)
Result All flowers are violet
Interpretation
w
Half of the flowers are violet
and half flowers are white
Unknown flower is
homozygous dominant
Unknown flower is heterozygous
Diagrammatic representation of a test cross
RECIPROCAL CROSS
When two parents are used in two experiments in such a way that in one experiment "A" is used as the
female parent and "B" is used as the male parent, in the other experiment "A" will be used as the male
parent and "B" as the female parent, such type of a set of two experiments is called Reciprocal cross.
Characters which are controlled by karyogene present on autosomes are not affected by Reciprocal
cross. In case of cytoplasmic inheritance and sex linkage result will change by Reciprocal cross.
a) TT
x
tt
b) TT
x
tt
(Female)
(Male)
(Male)
(Female)

All Tall

All Tall
1. If the cell of an organism heterozygous for three pairs of genes represented by AaBbCc, undergoes meiosis then possible type of gametes will be
1) 4
2) 2
3) 8
4) 12
2. Which postulate of Mendel is still universal in nature?
1) Postulate I
2) Postulate II
3) Postulate III
4) Postulate IV
3. When aaBBcc is crossed with AaBbCc then the ratio of hybrid for all the three genes is
1) 1/8
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2) 1/4
3) 1/16
11
4) 1/32
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4. According to Mendelism which pair of character is showing dominance ?
1) Terminal position of flower and green colour of seed coat
2) Wrinkled seeds and green colour of seed coat
3) Yellow pod and round seeds
4) Green pod and axial position of flower
5. A plant of F1 generation with genotype AABbCc. On selfing of this plant what is phenotypic
ratio in F2 generation?
1) 3 : 1
2) 9 : 3 : 3 : 1
3) 1 : 1
4) 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1
6. A character which is expressed in a hybrid is called
1) Dominant
2) Recessive
3) Co-dominant
4) Epistatic
7. Alleles are
1) Alternate forms of a gene
2) homologous chromosome
3) Pair of sex chromosome
4) none of these
8. When F1 generation hybrid tall Tt is crossed with dwarf tt parent, it is a case of
1) Dihybrid cross
2) test cross
3) Crossing over
4) Reciprocal cross
9. The ultimate biological unit which controls heredity, is called :
1) Genome
2) Chromosome
3) Genotype
4) Gene
10. In case of inheritance of one gene, 3 : 1 phenotypic ratio can be explained on the basis of 1) Incomplete dominance
2) Codominance
3) Dominance
4) Linkage
1) 3
2) 3
3) 1
4) 4
5) 2
6) 1
7) 1
8) 2
9) 4
10) 3
GENE INTERACTION
Gene interaction is two types :
i) Allelic interaction/intragenic interaction
ii) Non allelic interaction/Intergenic interaction
i) Allelic interaction/intragenic interaction: Allelic interaction takes place between allele of same
gene which are present at same locus. Example of Allelic interaction are as follows
1) Incomplete dominance : According to Mendel's law of dominance, dominant character must be
present in Ft generation. But in some organisms, F1 generation is different from the both parents.
Both factors such as dominant and recessive are present in incomplete dominance but dominant factors is unable to express its character completely, resulting Intermediate type of generation is formed
which is different from the both parents. Some examples are 12
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a) Flower colour in Mirabilis jalapa :
Incomplete dominance was first discovered by
P generation
Correns in Mirabilis jalapa. This plant is called
as '4 O' clock plant 'or' Gul-e-Bans' Three different types
White (rr)
Red (RR)
of plant are found in Mirabilis on the basis of flower
colour, such as red , white and pink.When plants with
r
red flowers is crossed with white flower, plants with
R
pink flower obtained in F1 generation. The reason of
this is that the genes of red colour is incompletely
dominant over the genes of white colour. When, F1
generation of pink flower is self pollinated then the
F1 generation
phenotypic ratio of F2 generation is red, pink, white is
1:2:1 ratio in place of normal monohybrid cross ratio
All pink (Rr)
3:1. The ratio of phenotype and genotype of F2
generation in incomplete dominance is always same.
Gametes
Gametes
b) Flower colour in Antirrhinum majus :
R
R
Incomplete dominance is also seen in flower colour
of this plant.This plant is also known as 'Snapdragon'
r
r
or 'Dog flower'. Incomplete dominance is found in
this plant which is the same as Mirabilis.
c) Feather colour in Andalusian Fowls :
F2 generation
Incomplete dominance is present for their feather
colour. When a black colour fowl is crossed with a
white colour fowl, the colour of F1 generation is blue.
Phenotypic : red : pink : white
2) Co-dominance : In this phenomenon, both the
1 : 2 : 1
gene expressed for a particular character in F1
Genotypic : Rr : Rr : rr
hybrid progeny. There is no blending of characters,
1 : 2 :1
where as both the characters expressed equally.
Examples Co-dominance is seen in animals for coat colour.
when a black parent is crossed with white parent, a roan colour F1 progeny is produced.
When we obtain F2 generation from the F1 generation, the ratio of black ; black-white (Roan); white
animals is 1:2:1
Note : F2 generation is obtained in animals by sib-mating cross.
WHITE
BLACK
R2R2
R1R1
F1 generation
R1R2 (Roan)
Sib-mating cross
R1
R2
R1 R1R1 R1R2
R2 R1R2 R2R2
R1 R1=Black-1 R1R2=Roan-2 R2R2=White-1
It is obvious by above analysis that the ratio of phenotype as well as genotype is 1:2:1 in co-dominance.
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Sp. Note : In incomplete dominance, characters are blended phenotypically, while in co-dominace,
both the genes of a pair exhibit both the characters side by side and effect of both the character is
independent from each other.
Other Examples of Co-dominance :
ii) AB blood group inheritance (IAIB)
iii) Carrier of Sickle cell anaemia (HbA Hbs)
3) Multiple allele : More than 2 alternative forms of same gene called as multiple allele. Multiple
allele is formed due to mutation. Multile allele located on same locus of homologous chromosome.
A diploid individual contains two alleles and gamete contains one allele for a character.
n( n  1)
If n is the number of allele of a gene then number of different possible genotype =
2
Example of multiple allele :
1. ABO blood group  ABO blood groups are determined by three alleles - IA, IB, and i
IA = dominant
IB = dominant
i = recessive
Possible phenotypes - A, B, AB, O
Blood group
Genotype
Antigen or agglutinogen
Antibody or agglutinin
A
IAIA, IAi
A
b
B
IBIB,IBi
B
a
AB
IAIB
A&B
none
O
ii
none
a&b
3(3  1)
 6 genotype
2
2. Coat colour in rabbits  4 alleles
3. Eye colour in Drosophila  15 alleles
4. Self incompatibility genes in plants (Tobacco)  4 alleles
4) Lethal gene : Gene which causes death of individual when it comes in homozygous condition
called lethal gene. Lethal gene may be dominant or recessive both, but mostly recessive for lethality.
Many of these genes which do not cause definite lethality are called semilethals.
1. Lethal gene was discovered by L. Cuenot in coat colour of mice.
Yellow body colour(Y) was dominant over normal brown colour(y).
Gene of yellow body colour is lethal.
So homozygous yellow mice are never obtained in population. It dies in embryonal stage.
When yellow mice were crossed among themselves segregation for yellow and brown body colour
was obtained in 2 : 1 ratio.
Yy
Yy
Possible genotype number 
Y
y
Y
YY Yy
y
Yy
yy
YY - death in embryonal stage modified ratio = 2:1
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2. In plant lethal gene was first discovered by E. Baur in Snapdragon (Antirrhinum majus)
Golden leaves (G)
Snapdragon
Green leaves (g)
Golden
Gg
Golden
Gg
G
g
G GG Gg
g Gg gg
Modified ratio : 2:1
Homozygous golden leaves are never obtained.
5) Pleiotropic gene : Gene which controls more than one character is called pleiotropic gene. This
gene shows multiple phenotypic effect.
For example :
1) In pea plant :
Seed coat colour
Single gene influences
Red spot in the axil of leaf
Flower colour
2) Sickle cell anaemia - Gene Hbs provide a classical example of pleiotrophy. It not only causes
haemo- lytic anaemia but also results increased resistance to one type of malaria that caused by the
parasite Plasmodium falciparum. The sickle cell Hbs , allele also has pleiotropic effect on the development of many tissues and organs such as bone, lungs, kidney, spleen, heart.
Micrograph of the red blood cells and the amino acid composition of the relevant portion of pchain of haemoglobin :
a) From a normal individual; (b) From an individual with sickle-cell anaemia
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(ii) Non allelic interaction/Intergenic interaction
When interaction takes place between non allele is called non allelic gene interaction. It changes or
modifies other non allelic gene.
Examples of nonallelic interaction.
1. Complementary Gene : Two pair of non allelic genes are essential in dominant form to produce a
particular character.
Such genes that act together to produce an effect that neither can produce, it's effect separately are
called complementary genes.
Both types of gene must be present in dominant form.
Example Colour of flowers in Lathyrus odoratus
C- P
Purple coloured
C - pp
colour less
cc - P
colour less
cc - pp
colour less
Raw material
Gene C
Chromagen
CCPP
(Coloured)
Gene P
Anthocyanin (purple)
ccpp
(Colourless)
F1 generation CcPp (coloured)
Thus phenotypic ratio of complementary genes = Coloured : Colourless 9 : 7
2. Epistasis : When, a gene prevents the expression of another non-allelic gene, then it is known as
epistatic gene and this phenomenon is known as Epistasis.
Gene which inhibit the expression of another non alleleic gene is called epistatic gene and expression
of gene which is suppressed by epistatic gene called hypostatic gene.
Epistasis is of two types
1) Dominant epistasis (2) Recessive epistasis
1) Dominant epistasis : Example - Fruit colour in summar squash (Cucurbita pepo)
Y = Dominant allele for yellow colour of fruit
y = Recessive allele for green colour of fruit
W = Epistatic gene over Y and y gene and forms white colour of fruit.
Following types of offsprings will be obtained in a Mendelian pattern of cross-
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It is obviously clear by above analysis, the phenotypic ratio of F2- generation in epistasis is - 12:3:1
3. Supplementary gene and Recessive Epistasis A pair of gene change the effect of another non
allelic gene, is called supplementary gene.
Example Coat colour in Mice.
If alleles, C = Black coat colour
c = Albino (Colourless coat) or (It has no effect)
A = Supplementary gene
When black coat mice crossed with albino mice, the F1 generation is Agouti.
It means, here the effect of non allelic gene is changed.
Black
Albino
CC aa
cc AA
F1 generation
Cc Aa [Agouti]
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Polygenic inheritance first described by Nilsson - Ehle in kernal colour of wheat.
Nilsson - Ehle said that kernal colour of wheat is regulated by two pairs of gene.
RRBB
x
rbb
Red
White

F1-gen.
RrBb (intermediate)

4 :
6 :
4
:
1
light
intermediate very
white
red
red
light
3
2
1
0
1 Red : 14 intermediate : 1 white
Example - 2 : Colour of the skin in Human.
The inheritance of colour of skin in human studied by Devenport.
Human skin colour is regulated by 3 gene pairs.
When a Negro Black (AA BB CC) phenotype is crossed with white (aa bb cc) phenotype, intermediate
phenotype produced in F1 generation.
Negro Black
x
White
AA BB CC
aa bb cc
F2-gen.
1 :
Full
red
Number of dominant gene - 4

Aa Bb Cc  F1 - generation
[Intermediate brown/mullato]
 (Inbreeding)
F2- generation

Phenotypic ratio will be :
Negro
: Very Dark : Dark
Black
Brown
Brown
1
:
6
: 15
: Mullato
:
20
: Light
Brown
:
15
:
:
Very light
Brown
6
: White
:
1
Frequencies
20/64
15/64
6/64
1/64




CYTOPLASMIC INHERITANCE
Inheritance of characters which are controlled by cytogene or cytoplasm is called cytoplosmic inheritance. Genes which are present in cytoplasm called 'cytogene' or 'plasmagene' or extra nuclear gene.
Total cytogene present in cytoplasm is called 'Plasmon'.
A gene which is located in the nucleus is called ' karyogene .
Inheritance of cytogene in organisms occurs only through the female. Because female gamete has
karyoplasm, simultaneously it has cytogene because of more cytoplasm.
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 The male gamete of higher plant is called male nucleus. It has very minute [equivalent to nil] cytoplasm, so male gamete only inherited karyogene.
 Thus, inheritance of cytogene occurs only through female, (also called maternal inheritance)
If there is a reciprocal cross in this condition, then results may be affected.
 Cytoplasmic inheritance involving essential organelles like, Chloroplast and mitochondria called as
organellar inheritance.
Example of Organellar Inheritance : (True examles of cytoplasmic inheritance)
a) Plastid inheritance in Mirabilis jalapa - cytoplasmic inheritance first discovered by Correns in
Mirabilis jalapa. In Mirabilis jalapa branch (leaf) colour is decided by type of plastid present in leaf
cells. So it is an example of cytoplasmic inheritance.
Branch colour
a) Pale
x
Green
b) Green
x
Pale

F1- Pale

F1- Green
(c) Variegated
x
Pale or Green or Variegate

F1 - Pale, Green and Variegated
b) Male sterility in maize plant : Gene of male sterelity present in mitochondria. If a normal male
plant crossed with a female plant which has genes of male sterility then all the generation of male
become sterile because a particular gene was present with female which inherited by female.
c) Albinism in plant: Gene of albinism found in chloroplast. Gene of albinism in Maize is lethal.
d) Petite form in yeast (mitochondrial gene) : Petite is mutant form of yeast. This mutant form is
slow growing on culture medium.
e) Iojap inheritance in Maize : Iojap is characterized by constrasting strip of green and white colour
of leaves.
f) Poky Neurospora (mitochondrial gene) : Poky is mutant form of Neurospora. It is slow growing
on culture medium.
1. In case of co-dominance the monohybrid ratio of phenotypes in F 2 generation is
1) 3 : 1
2) 1 : 2 : 1
3) 1 : 1 : 1 : 1
4) 2 : 2
2. Which cross yields red, white and pink flower variety of Snapdragon flower.
l) RRxRr
2) Rr xRR
3) RrxRr
4) Rrxrr
3. In polygenic inheritance, a trait is controlled by two pairs of genes. Two individuals which are heterozygous for both genes, crossed each other, such type of cross produces what phenotypic ratio ?
1) 1 : 2 : 1
2) 9 : 3 : 3 : 1
3) 1 : 4 : 6 : 4 : 1
4) 1:6:15:20:15:6:1
4. If dominant C and P genes are essential for the development of purple colour in Lathyrus odoratus
flowers. What would be the ratio of purple and white colour in a cross between CcPp xcc Pp.
1) 3 : 5
2) 9 : 7
3) 2 : 6
4) 4 : 4
5. In mother has blood group AB, father has A group then Which of the following blood group will
not found in the offspring.
1) A
2) B
3) AB
4) 0
1) 2
GENETICS
2) 3
3) 3
4) 1
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CHROMOSOMAL THEORY OF INHERITANCE
This theory was proposed by Walter Sutton and Theodor Boveri (1902).
In 1900, three Scientists (de Vries, Correns and von Tschermak) independently rediscovered Mendel's
results on the inheritance of characters. Also, by this time due to advancements in microscopy that
were taking place, scientists were able to carefully observe cell division. This led to the discovery of
structures in the nucleus that appeared to double and divide just before each cell division. These were
called chromosomes (colouredbodies, as they were visualised by staining). By 1902, the chromosome
movement during meiosis had been worked out. Walter Sutton and Theodore Boveri noted that the
behaviour of chromosomes was parallel to the behaviour of genes and used chromosome movement to
explain Mendel's laws. Recall that you have studied the behaviour of chromosomes during mitosis
(equational division) and during meiosis (reduction division). The important things to remember are
that chromosomes as well as genes occur in pairs. The two alleles of a gene pair are located on homologous sites on homologous chromosomes.
Meiosis I Meiosis II Germ cells
G2
G1
anaphase anaphase
Bivalent
Meiosis and germ cell formation in a cell with four chromosomes.
You can see how chromosomes segregate when germ cells are formed
During metaphase of meiosis I, the two
Possibility I
Possibility II
One
long orange and
One long orange and
chromosome pairs can align at the metaphase
short green chromosome
short red chromosome
and long yellow and
and long yellow and
short red chromosome
short green chromosome
at the same pole
at the same pole
Meiosis I - anaphase
Meiosis I - anaphase
plate independently of each other. To
understand this, compare the chromosomes
Meiosis II - anaphase
Meiosis II - anaphase
Germ cells
Germ cells
of four different colour in the left and right
columns. In the left column (Possibility I)
orange and green is segregating together.
But in the right hand column (Possibility II)
the orange chromosome is segregating with
the red chromosomes.
Independent assortment of chromosomes
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Sutton and Boveri argued that the pairing and separation of a pair of chromosomes would lead to the
segregation of a pair of factors they carried. Sutton united the knowledge of chromosomal segregation
with Mendelian principles and called it the chromosomal theory of inheritance.
Following this synthesis of ideas, experimental verification of the chromosomal theory of inheritance
by Thomas Hunt Morgan and his colleagues, led to discovering the basis for the variation that sexual
reproduction produced. Morgan worked with the tiny fruit files, Drosophila melanogaster, which were
found very suitable for such studies. They could be grown on simple synthetic medium in the laboratory. They complete their life cycle in about two weeks, and a single mating could produce a large
number of progeny flies. Also, there was a clear differentiation of the sexes - the male and female flies
are easily distinguisable. Also, it has many types of hereditary variations that can be seen with low
power microscopes.
Male
Female
Drosophila melanogaster
LINKAGE
Collective inheritance of character is called linkage. Linkage first time seen by Bateson and Punnett
in Lathyrus odoratus and gave coupling and repulsion phenomenon. But they did not explain the
phenomenon of linkage. Sex linkage was first discoverd by Morgan in Drosophila & coined the term
linkage. He proposed the theory of linkage.
Linkage and independent assortment can be represented in dihybrid plant, as In case of linkage in dihybrid (AaBb)
In case of independent assortment in dihybrid (AaBb)
A
B
a
b
A
a
B
b
It produces two types of gamete
It produces four types of gamete
AB : ab
AB : ab : aB : Ab
Theory of linkage
1. Linked genes are linearly located on same chromosome. They get separated if exchange (crossing
over), takes place between them.
2. Strength of linkage  1/ distance between the genes. It means, if the distance between two genes is
increased then strength of linkage is reduced and it proves that greater is the distance between genes,
the greater is the probability of their crossing over.
Crossing over obviously disturbs or degenerates linkage. Linked genes can be separated by crossing
over.
3. Linkage group All the genes which are located on one pair of homologous chromosome form one
linkage group. Genes which are located on homologous chromosomes inherit together so we consider
one linkage group.
No. of Linkage group = haploid no. of homologous chromosomes.
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2n
n
Pair Linkage group
Pea
14
7
7
7
Maize
20 10
10
10
Drosophila
8
4
4
4
Barley
14
7
7
7
Mouse
42
21
21
21
Factors affecting crossing over (C.O) :
1) Distance  = C.O.  2) Temperature  = C.O. 
3) X-Ray  = C.O. 
4) Age t  = C.O. 
5) Sex - Male C.O.  (Crossing over totally absent in male Drosophila.)
Arrangement of linked Genes on Chromosomes
The arrangement of linked genes in any dihybrid plant is two types.
[a] Cis - Arrangement : When, two dominant genes located on one chromosome and both recessive genes
located on another chromosome, such type of arrangement is termed as cis- arrangement. Cis-arrangement
is an original arrangement.
Two types of gamete can be produced in cis-arrangement  (AB) and (ab).
A
a
B
b
[b] Trans-arrangement : When a chromosome bears one dominant and one recessive gene, and another
chromosome also possess one dominant and one recessive gene, such type of arrangement is called
trans-arrangement.Trans-arrangement is not an original form. It is due to crossing over. Two types of
gamete also formed in trans-arrangement but it is different from cis-arrangement (Ab) and (aB).
A
a
b
B
Types of Linkage There are two types of linkage 1. COMPLETE LINKAGE : Linkage in which genes always show parental combination. It never
forms new combination.
Crossing over is absent in it. Such genes are located very close on the chromosomes. Such type of
linkage very rare in nature, e.g. male Drosophila, female silk moth.
2. INCOMPLETE LINKAGE : When new combinations also appear along with parental combination in offsprings, this type of linkage is called incomplete linkage, the new combinations form due to
crossing over.
Application of Linkage :
Distance can be identified by the incomplete linkage. It's unit is centi Morgan.
1
1

Distance b/w linked gene
Crossing over
Genetic map/Linkage map/chromosome map - In genetic map different genes are linearly arranged
according to % of recombination (  Distance) between them.
With the help of genetic map we can find out the position of a particular gene on chromosome. Genetic
map is helpful in the study of genome.
Strength of linkage 
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Cross A
Cross B
Parental
F1 generation
Parental
type (98.7%)
Recombinant
types(1.3%)
Parental
type(62.8%)
Recombinant
types(37.2%)
F2 generation
Results of two dihybrid crosses conducted by Morgan.Cross A shows crossing between gene y
and w; Cross B shows crossing between genes w and m. Here dominant wild type ailetes are
represented with (+) sign in superscript. Note : The strength of linkage between y and w is
higher than w and m.
SEX LINKAGE
When the genes are present on sex-chromosome is termed as sex linked gene and such phenomenon is
known as sex-linkage.
Types of sex linkage
1. X-linkage : Genes of somatic characters are found on x-chromosome. The inheritance of x-linked
character may be through the males and females.
Example of X-linkage
[i] Eye colour in Drosophila : Eye colour in Drosophila is controlled by a X-linked gene.
If a red eyed colour gene is represented as '+' and white eyed colour represented as 'w', then on basis of
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this different type of genotypes are found in Drosophila.
Gene for red eye is dominant (+) and white colour of eye is recessive (w)
Homozygous red eyed female
= X+X+
Heterozygous red eyed female
= X+Xw
Homozygous white eyed female
= XWXW
Hemizygous red eyed male
= X+Y
Hemizygous white eyed male
= X WY
It is clear by above different types of genotype that female either homozygous or heterozygous for eye
colour. But, for the male eye colour, it is always hemizygous.
[ii] Haemophilia : Haemophilia is also called "bleeder's disease" and first discovered by John Otto
(1803). The gene of haemophilia is recessive.
On the basis of x-linked, following types of genotype are found.
Xh X = Carrier female
XhXh = Affected female
XhY = Affected male.
But, XhXh type of female dies during embryo stage because in homozygous condition, this gene becomes lethal and causes death.
Haemophilia -A  due to lack of factor -VIII (Antihaemophilic globulin AHG). It shows lethality in
homogygous condition.
Haemophilia B or Christmas disease - due to lack of factor - IX (Plasma thromboplastin component)
Haemophilia - C (Autosomal disorder)  due to lack of factor - XI (Plasma Thromboplastin antecedent)
[iii] Colour Blindness : The inheritance of colour-blindness is alike as haemophilia, but it is not a
lethal disease so it is found in male and female.(discovered by Homer)
Three types of colour blindness are [a] Protanopia It is for red colour.
[b] Deuteranopia It is for green colour
[c] Tritanopia For blue colour blindness. Colour blindness is checked by ishihara - chart.
Other examples of X-linkage
[iv] Diabetes insipidus (recessive).
[v] Duchenne muscular dystrophy (recessive).
[vi] Pesudoricketes (Dominant)
[vii] Defective enamel of teeth (Dominant)
2. Y- linkage : The genes of somatic characters are located on Y- chromosome. The inheritance of
such type of character is only through the males. Such type of character is called Holandric character.
These characters found only in male.
e.g. (1) Gene which forms TDF /sry-gene
2) Hypertrichosis (excessive hair on ear pinna.)
Gene which is located on differential region of Y - chromosome is known as Holandric gene.
Types of Inheritance of sex linked characters
1. Criss cross inheritance (Morgan) : In criss-cross inheritance male or female parent transfer a Xlinked character to grandson or grand daughter through the offspring of opposite sex.
a) Diagenic (Diagynic) : Inheritance in which characters are inherited from father to the daughter and
from daughter to grandson.
Father  daughter  grand son.
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b) Diandric : Inheritance in which characters are inherited from mother to the son and from son to
grand daughter.
Mother  Son  Grand-daughter..
2. Non criss-cross inheritance : In this inheritance male or female parent transfer sex linked character
to grand son or grand daughter through the offspring of same sex.
a) Hologenic(Hologynic) : Mother  Daughter  Grand-daughter (female to female)
b) Holandric : Father  Son  Grand-son (male to male)
Sex-Limited Character
These characters are present in one sex and absent in another sex. But their genes are present in both
the sexes and their expression is depend on sex hormone.
Example : Secondary sexual characters  these genes located on the autosomes and these genes are
present in both male and female, but effect of these are depend upon presence or absence of sexhormones.
For example - genes of beard-moustache express their effects only in the presence of male hormone testosterone.
Sex Influenced Characters
Genes of these characters are also present on autosomes but they are influenced differently in male and
female. In heterozygous condition their effect is different in both the sexes.
Example Baldness Gene of baldness is dominant (B).
Genotype
Male
Female
BB
Baldness present Baldness present
bb
Baldness absent Baldness absent
Bb
Baldness present Baldness absent
Gene Bb shows partiality in male and female, Baldness is found in male due to effect of this gene, but
baldness is absent in female with this genotype.
SEX DETERMINATION
Establishment of sex through differential development in an individual at an early stage of life, is
called sex determination. There are different methods for sex determination in organisms like allosomic
sex determination, haplodiploidy, genic balance etc.
Sex Determination on the basis of fertilization.
Three types 1. Progamic - Sex is determined before fertilization, eg. - drone in honey bee
2. Syngamic - Sex is determined during fertilization, eg. - most of plants & animals
3. Epigamic - Sex is determined after fertilization, eg. - Female in honey bee.
Mechanism of sex determination :
[1] Allosomic determination of sex Chromosomes are of two types a) Autosomes or somatic chromosomes - These regulate somatic characters.
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b) Allosomes or Heterosomes or Sex chromosomes These chromosomes are associated with sex determination. Term "Allosome" & "Heterosome" were
given by Montgomery.
Sex chromosomes first discovered by "Me Clung" in grass hopper X- Chromosome discovered by
"Henking" and called 'x-body'.
Wilson & Stevens proposed chromosomal theory for sex determination.
Figure : Determination of sex by chromosomal differences: (a,b) Both in humans and in
Drosophila, the female has a pair of XX chromosomes (homogametic) and the male XY (heterogametic) composition; (c) In many birds, female has a pair of dissimilar chromosomes ZW
and male two similar ZZ chromosomes
1) XX - XY type or Lygaeus type : This type of sex determination first observed by Wilson & Stevens
in Lygaeus insect. Two typesa) XX female and XY male : In this type of sex determination female is Homogametic i.e produces
only one type of gamete
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A+X
2A + XX (Female)
gametes
A+X
Male is heterogametic (male produces two types of gamete)
A+X
gametes
2A + XY(Male)
A+Y
In male X-chromosome containing gametes is called "Gynosperm" and Y- chromosome containing
gamete is called "Androsperm".
eg. Man and dioecious plants like Coccinea, Melandrium
b) XY female and XX male or ZW female and ZZ male : In this type of sex determination female is
Heterogametic i.e produces two types of gamete and male individual is homogametic i.e produces one
type of gamete.
It is found in some insects like butter flies, moths and vertebrates like birds, fishes and reptiles.
In plant kingdom this type of sex determination is found in Fragaria elatior.
2) XX female and XO male or "Protenor type" In this type of sex deternination deficiency of one
chromosome in male. In this type, female is homogametic and male is heterogametic.
A+X
2A + XX (Female)
A+O
homogametic
heterogametic
2A + XO(Male)
A+X
A+X
Example
- Grass hopper
- Squash bug Anasa
- Cockroach
- Ascaris and in plants like - Dioscorea sinuta & Vallisneria spiralis
Haploid - diploid mechanism (Sex determination in Honey Bee) In insects of order Hymenoptera which includes ants,honey bees, wasps etc.
Sex determination takes place by sets of chromosomes.
Haploid (One set)  Male
Diploid (two sets)  Female
In honey bee, male individual (Drone) develops from unfertilized eggs (Haploid). Male is always
parthenote. Queen and worker bees develop from diploid eggs i.e. fertilized egg.
Feeds on Royal jelly
Fertilized egg
diploid larva
(2n)
Bee bread
Queen
(Fertile female)
Worker
(Sterile female)
[2] Genic balance theory : C.B. Bridges proposed genic balance theory for sex determination in Drosophila.
According to Bridges in Drosophila Y-chromosome is heterochromatic so it is not active in sex deterGENETICS
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mination In Drosophila sex determination takes place by sex index ratio.
No. of x chromosomes X
=
No. of set of Autosomes A
In Drosophila gene of femaleness (Sxl- gene) (Sxl=Sex lethal gene) is located on x-chromosome and
gene of maleness is located on autosome
Gene of male fertility is located on y-chromosome and in Drosophila, y-chromosome plays additional
role in spermatogenesis and development of male reproductive organ, so y-chromosome is essential
for the production of fertile male.
Sex index ratio=
Sex index ratio
a)
(b)
X
= 1  female (2A + XX), (3A + XXX)
A
X
= 0.5  male
A
(2A + XY)=Fertile male
(2A + XO)=Sterile male
c)
X
= 1.5  Super female or meta female (sterile) (2A + XXX)
A
d)
X
= less than 0.5  Super male or meta male (Sterile) (3A + XY)
A
e)
X
= In between 0.5 and 1  Intersex (Sterile) (3A+XX)
A
Cytological basis of sex identification  Barr body technique or Lyon's hypothesis  Interphasic nucleus of human female contains two X- chromosomes. Out of two, one X- chromosome
becomes heterochromatin and other X- chromosome is euchromatin. By staining X- heterochromatin,
it appears as a dense body which is called Barr body. (Facultative hetrochromatin) v
 No. of Barr body  (No. of X chromosomes - 1)
 So in a Normal female (2A + XX)  One Barr body
 Normal male (2A + XY)  Barr body absent
 Turner syndrome (Sterile female) (2A + XO)  No. Barr body
 Klinefelter syndrom (Sterile male)(2A + XXY)  One Barr body
 Drum stick which occurs in blood of female of mammals, is also a type of barr body. Drum stick is
absent in neutrophils of Male.
1. An exception to the law of independent assortment is
1) Dominance
2) Incomplete dominance
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3) Segregation
4) Linkage
2. Experimental proof for sex-linked gene was given by
1) Morgan
2) Muller
3) Mendel
4) Johannsen
3. X-linked recessive gene easily express in
1) male
2) female
3) equal in male and female
4) not easily express
4. Maize has 10 pairs of chromosomes. How many linkage groups are present.
1) 05
2) 10
3) 20
4)40
5. The machanism of sex determination in birds shows
1) Male heterogamety
2) Both heterogametic
3) Female heterogamety 4) Both homogametic
1) 4
2) 1
3) 1
4) 2
5) 3
HUMAN GENETICS
1.
2.
3.
4.
5.
6.
The study (analysis) of genetic characters and aspects like genetic improvements among humans are
included in human genetics. This is also known as eugenics (=well born).
Eugenics is a term derived from Greek language 'Eugenes' meaning "Well born". First of all, Sir
Francis Galton, 1883 proposed the idea of improvement in human species through change in hereditary characters in a scientific manner and named it Eugenics. Because of this Sir Francis Galton is
known as "father of Eugenics".
To find out various facts, scientists have to perform a number of experiments, but it is not possible to
do so in humans. The following problems are faced in studying human genetics Cells of human body are relatively smaller and number of chromosomes present in them is more.
It is not possible to perform various experiments on humans in the laboratory.
Due to greater life span of humans, a lot of time is required to study their genetic characteristics.
Rate of reproduction is slow in humans.
Individuals of human species are generally heterozygous for various characters and it is very difficult
to find homozygous individuals.
Due to controlled hybridization and long lived life among humans, it is not possible to study many
generations in easy way.
Despite of above mentioned problems humans are considered suitable for genetical experiments due
to 1. Longer life span of humans, the abnormalities that require relatively long period to express themselves can be easily studied.
2. By pedigree study and analysis of families, many genetic characters of man can be traced out.
3.To study many diseases (like haemophilia, colour blindness etc.) and intelligence quotient (I.Q.)
humans are more preferable than any other organisms.
GENETICS
29
Devices Used in Human Genetical Studies
The study and analysis of human genetics is performed by many methods like pedigree analysis,
statistical analysis and human karyotyping. Of these the important ones, that is, pedigree analysis and
human karyotype is being described here.
1. Pedigree Analysis
Study of ancestoral history of man of transmission of genetic characters from one generation to next, is
pedigree analysis. Dwarfism, albinism, colour blindness, haemophilia etc. are genetically transmitted
characters. To study and analyse them a pedigree of genetic facts/data and following symbols are used.
Symbol used in Pedigree
1.
Normal Male
2.
Normal Female
3.
Mating (marriage)
The siblings are indicated in chronological order of birth
4.
5.
Sex unspecified
6. Twin
If monozygotic
If dizygotic
7.
,
Affected male and female individual
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,
8.
Heterozygous for autosomal recessive
Carrier female of sex linked recessive character or disease
9.
10.
Death of individual
11.
Abortion or still birth (sex unspecified)
12.
Consanguineous marriage
13.
Parent with male child affected with disease
14.
Five unaffected offsprings
5
Pedigree analysis provides valuable informations regarding genetical make up of human beings. If any
genetic disease is occuring in a family, then pedigree analysis provides guidance to forthcoming parents about their future progenies for example- polydactyly in humans.
A representative pedigree is shown in Figure for dominant and recessive traits, discuss with your teacher
and design pedigrees for characters linked to both autosomes and sex chromosome.
Examples :
(a)
(b)
Representative pedigree analysis of
a) Autosomal dominant trait (Mytotonic Dystrophy)
(b) Autosomal recessive trait (Sickle Cell Anaemia)
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1.
2.
3.
4.
POPULATION GENETICS
Study of gene frequency in a population is called population genetics.
Gene pool - A gene pool is the sum total of genes in reproductive gametes of a population.
Gene-flow - Migration of gene from one population to another population by cross fertilization.
Genetic load - The existance within the population of disadvantegeous allele in heterozygous genotype is known as genetic load
Gene frequency - Gene frequency is defined as proportion of different alleles of a gene in a population. Ex. In a population of 100 individuals of MN blood group 50 MM, 20 MN, 30 NN find out the
frequency of M&N.
MM - 50
MN - 20
NN - 30
Total M gene - 50 x 2 +20 =120
Total N gene - 30 x 2 +20 =80
Total genes = 200
Freq. of M gene P=
M
120
=
=0.6
M+N 200
N
80
=
=0.4
M+N 200
0.6 + 0.4= 1
p+q=1
Hardy Weinberg Law 1908 G.H.Hardy (English methematician) & German Physician, W.Weinbergh independently discovered that an equilibrium is established between frequencies of allele in random mating population
and these gene frequency remain constant from generation to generation.
This law applicable when factors like mutation, selection & migration, are absent.
Hardy Weinberg theorem or Hardy weinbergh law
A  P, a  q, p+q=1)
p2 + 2pq + q2 =1
AA Aa aa
In this equation frequency of A  P or the Frequency of Homozygous dominant will be AA-P 2
In this equation frequency a  q the frequency of homozygous recessive - q2
the frequency of Aa = 2pq
The frequency of different genotype produced due to random mating will depend upon the gene frequency and equilibrium is stablished after one single generation of random mating.
Q. In a random population frequency of recessive phenotype is 0.09. What is the frequency of
heterozygous genotype?
Sol. q2 = 0.09
q = 0.3
p=1-q
p = 1 - 0.3  0.7
Frequency of heterozygote = 2pq
= 2 x0.7 x0.3
= 0.42 = 42%
Freq. pf N gene
1. Which of the following symbols and its representation is correct
1)
= unaffacted female
2)
= male affacted
3)
= affacted female
4)
= affacted female
2. The presence of recessive trait in a large population is found to be 16%. The frequency of dominant trait in that population is
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1) 0.84
2) 0.42
3) 0.56
4) 0.96
3.
The pedigree shows
1) Dominant inheritance 2) Recessive inheritance
3) Sex linked recessive inheritance
4) Cytoplasmic inheritance
4. In a random mating population frequency of dominent allele is 0.7. What will be the frequency
of homozygous dominant phenotype.
1) 0.49
2) 0.09
3) 0.3
4) 0.21
5. If a couple has four girls, the probability of fifth child being male, is
1) 50%
2) 25%
3) 75%
4) 100%
1) 4
2) 1
3) 1
4) 1
5) 1
GOLDEN KEY POINTS
Monohybrid cross
Phenotypic ratio = 3:1
Genotypic ratio = 1:2:1
Test cross ratio = 1:1
Dihybrid cross
Phenotypic ratio = 9 : 3 : 3 : 1
Genotypic ratio =1:2:2:4:1:2:1:2:1
Test cross ratio = 1 : 1 : 1 : 1
Complementary gene = 9:7
Dominant epistasis = 12 : 3 : 1
Recessive epistasis = 9:3:4
Coupling = 7 : 1 : 1 : 7
Repulsion = 1 : 7 : 7 : 1
Incomplete dominance (Monohybrid cross)
Phenotypic ratio =1:2:1
Genotypic ratio = 1:2:1
Co-dominance (Monohybrid cross)
Phenotypic ratio =1:2:1
Genotypic ratio =1:2:1
Trihybrid phenotypic ratio = 27:9:9:9:3:3:3:1
Type of phenotype = 2n
Type of gametes = 2n
Type of genotype = 3n
Total possible zygotic combination = 4n
n(n+1)
2
Modified ratio of lethal gene in monohybrid cross = 2:1
Polygenic inheritance
1 : 4 : 6 : 4 : 1 (For two gene)
1:6:15:20:15:6:1(For three gene)
Type of phenotype in polygenic intheritance = (2n + 1)
Possible genotype number (In multiple alleles) =
Contribution of each dominant allele =
GENETICS
Maximum expression - Minimum expression
Total number of dominant allele
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MOLECULAR BASIS OF INHERITANCE
NUCLEIC AICDS
F. Meischer discovered nucleic acid in nucleus of pus cell and called it "nuclein". The term nucleic
acid was coined by "Altman".
Nucleic acids are polymer of nucleotides.
= Nitrogen base + pentose sugar + phosphate
On the basis of structure nitrogen bases are broadly of two types
1. Pyrimidines - Consist of one pyrimidine ring. Skeleton of ring composed of two nitrogen and four
Carbon atoms, e.g. Cytosine, Thymine and Uracil.
O
NH2
N3
2
O
4
1
O
CH3
HN
5
6
O
N
HN
O
N
H
N
H
CYTOSINE
URACIL
THYMINE
2. Purines - Consist of two rings i.e. one pyrimidine ring (2N + 4C) and one imidazole ring (2N + 3C)
e.g. Adenine and Guanine.
NH2
6
N1
5
O
N
8
2
4
3
N
9
H2N
N
H
ADENINE
Pentose Sugar :
CH2OH
H
H
OH
O
OH
H
H
OH
N
HN
7
N
N
H
GUANINE
CH2OH
H
O
O
OH
H
H
OH
H
H
HO
P
OH
OH
Ribose
Deoxy ribose
Phosphate
Nitrogen base forms bond with first carbon of pentose sugar to form a nucleoside. Nitrogen of first
place (N1) forms bond with sugar in case of pyrimidines while in purines nitrogen of ninth place (N g)
forms bond with sugar.
Phosphate forms ester bond (covalent bond) with fifth Carbon of sugar to form a complete nucleotide.
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H
H
N
N
H
O
O
P
O
8
7
5
9
4
N
O
5
H
3
N
1
N
2
H
O
CH2
4
6
H
3
H 1
2H
H
OH
Deoxyribose
Nucleoside
Nucleotide
Types of Nucleosides and Nucieotides
1. Adenine + Ribose = Adenosine
Adenosine + Phosphate = Adenylic acid
AMP)
2. Adenine + Deoxyribose = Deoxy adenosine
Deoxy adenosine + P = Deoxy adenylic acid
dAMP)
3. Guanine + Ribose = Guanosine
Guanosine + P = Guanylic acid
GMP)
4. Guanine + Deoxyribose = Deoxy guanosine
Deoxy guanosine + P = Deoxy guanylic acid
dGMP)
5. Cytosine + Ribose = Cytidine
Cytidine + P = Cytidylic acid
CMP)
6. Cytosine + Deoxyribose = Deoxycytidine
Deoxycytidine + P = Deoxycytidylic acid
dCMP)
7. Uracil + Ribose = Uridine
Uridine + P = Uridylic acid
UMP)
8. Thymine + Deoxyribose = Deoxy thymidine
Deoxythymidine + P = Deoxythymidylic acid
dTMP)
DNA
Discovered by - Meischer
DNA term was given by - Zacharis
In DNA pentose sugar is deoxyribose sugar and four types of nitrogen bases A,T,G,C
Wilkins and Franklin studied DNA molecule with the help of X-Ray crystallography.
With the help of this study, Watson and Crick (1953) proposed a double helix molel for DNA. For this
model Watson, Crick and Wilkins were awarded by Noble Prize in 1962.
One main hallmark (main point) of double helix model is complementary base pairing between purine
and pyrimidine.
According to this model, DNA is composed of two polynucleotide chains.
Both polynucleotide chains are complementary and antiparallel to each other.
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In both strand of DNA direction of phosphodiester bond is opposite, i.e. If direction of phosphodiester
bond in one strand is 3-5' then it is 5'-3' in another strand.
Both strand of DNA held together by hydrogen bonds. These hydrogen bonds are present between
nitrogen bases of both strand.
Adenine binds to thymine by two hydrogen bonds and cytosine binds to guanine by three hydrogen
bonds. In a DNA molecule one purine always pairs with a pyrimidine. This generates approximately
uniform distance between the two strands of DNA.
In DNA plane of one base pair stacks over the other in double helix. This, in addition to H-bonds,
confers stability of the helical structure of DNA.
34 A
10 A
3.4 A
Chargaff s equivalency rule - In a double stranded DNA amount of purine nucleotides is equals to
amount of pyrimidine nucleotides.
Purine = Pyrimidine
[A] + [G] = [T] + [C]
 A  + G  =1
 T  +  C
A+T
=1 = constant for a given species, i.e. species specific.
G+C
In a DNA A + T > G + C  A-T type DNA. Base ratio of A - T type of DNA is more than one. eg.
Eukaryotic DNA
In a DNA G + C>A + T  G-C type DNA. Base ratio of G - C type of DNA is less than one.
eg. Prokaryotic DNA
Melting point of DNA depends on G - C contents.
More G - C contents means more melting point.
Tm = Temperature of melting.
Tm of prokaryotic DNA > Tm of Eukaryotic DNA
DNA absorbs U.V. rays means 2600A wavelength.
Denaturation and renaturation of DNA - If a normal DNA molecule is placed at high temperature
(80- 900C) then both strands of DNA will separate from each other due to breaking of hydrogen bonds.
Base ratio =
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It is called DNA-denaturation.
When denatured DNA molecule is placed at normal temperature then both strand of DNA attached
and recoiled to each other. It is called renaturation of DNA.
Configuration of DNA Molecule
Two strands of DNA arc helically coiled like a revolving ladder. Back bone of this ladder (Reiling) is
composed of phosphates and sugars while steps (bars) are composed of pairs of nitrogen bases.
Two chains have anti-parallel polarity. It means, if one chain has the polarity 5'  3', the other has
3'  5'.
Distance between two successive steps is 3.4 A0. In one complete turn of DNA molecule there are such 10
steps (10 pairs of nitrogen bases). So the length of one complete turn is 34 A0. This is called helix length.
Diameter of DNA molecule i.e. distance between phosphates of two strands is 20A 0.
Each step of ascent is represented by a pair of bases. At each step of ascent, the strands turns 360.
In nucleus of eukaryotes the DNA is associated with histone protein to form nucleoprotein. Histone
occupies major groove of DNA at 300 angle.
Bond between DNA and Histone is salt linkage (Mg+2).
Types of DNA
On the basis of direction of twisting, there are two types of DNA.
1. Right Handed DNA Clockwise twisting e.g. The DNA for which Watson and Crick proposed model was 'B' DNA.
DNA Helix Length No. of base pairs Distance between two pairs Diameter
A
28 A0
11 pairs
2.56 A0
23 A0
B
34 A0
10 pairs
3.4 A0
20 A0
0
0
C
31 A
9.33 pairs
3.32 A
19 A0
D
24.24 A0
8 pairs
3.03 A0
19 A0
2. Left handed DNA
Anticlockwise twisting e.g. Z-DNA - discovered by Rich. Phosphate and sugar backbone is zig-zag.
Helix length
- 45.6 A0
Diameter
- 18.4 A0
No. of base pairs
- 12 (6 dimers)
Distance between 2 base - pairs- 3.75 A0
Palindromic DNA - Wilson and Thomas

CC GG TA CC GG
GG CC AT GG CC

Sequence of nucleotides same from both ends.
PACKAGING OF DNA HELIX
The average distance between the two adjacent base pairs of 0.34 nm (0.34 x 10 -9 m or 3.4 A). Length
of DNA for a human diploid cell is 6.6 x 109 bp x 0.34 x 10-9 m/bp = 2.2 m. The length is far greater
than the dimension of a typical nucleus (approximately 10 -6 m).
The number of base pairs in Escherichia coli is 4.6 x 10 6. The total length is 1.36 mm. The long sized
DNA accommodated in small area (about 1  m in E. coli) only through packing or compaction. DNA
is acidic due to presence of large number of phosphate group. Compaction occurs by folding acid
attachment of DNA with basic proteins, polyamine in prokaryotes and histone in eukaryotes.
DNA packaging in Prokaryotes : DNA is found in cytoplasm in supercoiled state. The coils are maintained by
non histone basic protein like polyamines. DNA may also be involved. This compact structure of DNA is
called nucleoid or genophore.
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DNA packaging in Eukaryotes : It is carried out with the help of lysine and ariginine rich basic proteins called
histone. The unit of compaction in nucleosome. There are five types of histone proteins H1, H2A, H2B, H3 and
H4. Four of them occur in pairs to produce histone octamer (2 copies of each - H2A, H2B, H3 and H4), called
nubody or core of nucleosome.Their positively charged ends are directed outside. They attracted negatively
charged strands of DNA. About 200 bp of DNA is wrapped over nu body to complete about 1 - turns.
This forms a nucleosome of size 110 x 60 A0 (11 x 6 nm). DNA present between two adjacent nucleosome
is called linker DNA. It is attached to H( histone protein. Length of linker DNA varies from species to
species. Nucleosome chain gives a beads on string appearance under electron microscope. The nucleosomes furthers coils to form solenoid. It has diameter of 30 nm as found in chromatin. The beads on string
structure in chromatin is packaged to form chromatin fibres that are further coiled and condensed at metaphase
stage of cell division to form chromosomes. The packaging at higher level requires additional set of proteins
(acidic) that collectively are referred to as non-histone chromosomal (NHC) proteins.
Non-Histone chromosomal proteins are of three types :
i) Scaffold or structural NHC protein
ii) Functional NHC protein e.g., DNA polymerase, RNA polymerase
iii) Regulatory NHC protein e.g., HMG (High mobility group proteins that controls gene expression)
DNA
H1 histone
Histone
octamer
Core of histone molecule
Basic unit of DNA compaction
EM picture - Beads-on-String
(Nucleosome)
In a typical nucleus, some region of chromatin are loosely packed (and stains light) and are referred to as
enchromatin. The chromatin that is more densely packed and stains dark is called as heterochromatin, specifically euchromatin is said to be transcriptionally active and heterochromatin is transcriptionally inactive.
Super solenoid
Solenoids
Chromatid
Chromosome
Various steps in the folding and superfilding of basic chromatin components to generate an
eukaryotic chromosome
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THE SEARCH FOR GENETIC MATERIAL
The experiments given below prove that DNA is the genetic material.
I) Evidence from bacterial transformation : The transformation experiments conducted by Frederick
Griffith in 1928, are of greater importance in establishing the nature of genetic material.
He used two strains of bacterium Diplococcus or Streptococcus pneumoniae or Pneumococcus i.e., SIII and R-II.
i) Smooth (S) or capsulated type which have a mucous coat and produce shiny colonies. These bacteria are virulent and cause pneumonia.
ii) Rough (R) or non-capsulated type in which mucous coat is absent and produce rough colonies.
These bacteria are nonvirulent and do not cause pneumonia.
The experiment can be described in following four steps :
a) Smooth type bacteria were injected into mice. The mice died as a result of pneumonia caused by
bacteria.
S strain  injected into mice  Mice died
b) Rough type bacteria were injected into mice. The mice lived and pneumonia was not produced
R strain  injected into mice  Mice lived
c) Smooth type bacteria which normally cause disease were heat killed and then injected into the mice. The
mice lived and pneumonia was not caused.
S strain (heat killed)  Injected into mice  Mice lived
d) Rough type bacteria (living) and smooth type heat-killed bacteria (both known not to cause disease)
were injected together into mice. The mice died due to pneumonia and virulent smooth type living
bacteria could also be recovered from their bodies.
S strain (heat killed) + R strain (living)  injected into mice  Mice died
1 Living
1 Living
encapsulated
bacteria injected
into mouse
2
Mouse died
3 Colonies of
encapsulated
bacteria were
isolated from
dead mouse
(a)
nonencapsulated
bacteria injected
into mouse
1 Heat-killed encapsulated 1 Living nonencapsulated
bacteria injected into
mouse
2 Mouse remained healthy 2 Mouse remained healthy
3 A few colonies of
nonencapsulated bacteria
were isolated from mouse,
phagocytes destroyed
nonencapsulated bacteria
3 Non colonies were
isolated from mouse
and heat-killed
encapsulated bacteria
injected into mouse
2 Mouse died
3 Colonies of encapsulated
bacteria were isolated
from dead mouse
(c)
(b)
Bacterial transfomation experiments conducted by Griffith
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He concluded from fourth step of the experiment that some rough bacteria (nonvirulent) were transformed into smooth type of bacteria (virulent). This occurred pherhaps due to absorption of some
transforming substance by rough type bacteria from heat killed smooth type bacteria. This transforming substance from smooth type bacteria caused the synthesis of capsule which resulted in production
of pneumonia and death of mice. Therefore, transforming principle appears to control genetic characters (for example, capsule as in this case). However, the biochemical nature of genetic material was
not defined from this experiments.
Biochemical characterisation of Transofrming Principle :
Later, Avery, Macleoid and McCarty (1944) repeated the experiment in vitro to identify the biochemical nature of transforming susbtance. They proved that this substance is DNA.
Pneumococcus bacteria cause disease when capsule is present. Capsule production is under genetic
control.
In the experiments, rough type bacteria (non-capsulated and non-virulent) were grown in a culture
medium to which DNA extract from smooth type bacteria (capsulated and virulent) was added. Later,
the culture showed the presence of smooth type bacteria also in addition to rough type. This is possible
only if DNA of smooth type was absorbed by rough type bacteria which developed capsule and became virulent. This process of transfer of characters of one bacterium to another by taking up DNA
from solution is caleld transformation. When DNA extract was treated with DNase (an enzyme which
destroys DNA), transformation did not occur. The transformation occurs when proteases and RNases
were used. This clearly shows that DNA is the genetic material.
II) Evidence from experiments with bacteriophage : T2 bacteriophage is a virus that infects bacterium Escherichia coli and multiplies inside it. T2 phage is made up of DNA and protein coat. Thus, it
is the most suitable material to determine whether DNA or protein contains information for the production of new virus (phage) particles. Hershey and Chase (1952) demonstrated that only DNA of the
phage enters the bacterial cell and , therefore, contains necessary genetic information for the asssembly
of new phage particle.
The functions of DNA and proteins could be found out by labelling them with radioactive tracers.
DNA contains phosphorus but not sulphur. Therefore, phage DNA was labelled with P32 by growing
bacteria infected with phages in culture medium containing 32P04. Similarly, protein of phage contains sulphur but no phosphorus. Thus, the phage protein coat was labelled with S35 by growing
bacteria infected with phages in another culture medium containing 35S04. After the formation of
labelled phages. Three steps were followed, i.e., infection, blending, centrifugation.
1. Infection : Both type of labelled phages were allowed to infect normally cultured bacteria in separate experiments.
2. Blending : These bacteria cells were agitated in a blender to break the contact between virus and
bacteria.
3. Centrifugation : The virus particles were separated from the bacteriam by spinning them in a
centrifuge.
After the centrifugation the bacterial cells showed the presence of radioactive DNA labelled with P32
while radioactive protein labelled with S35 appeared on the outside of bacteria cells (i.e., in the medium).
Labelled DNA was also found in the next generation or phage. This clearly showed that only DNA enters
the bacterial host andno the protein. DNA, therefore, is the infective part of virus and also carries all the
gnetic information. This provided the unequivocal proof that DNA is the genetic material.
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Bacteriophage
Radioactive (35S) labelled
protein capsule
Radioactive (32S)
labelled DNA
1. Infection
2. Blending
3. Centrifugation
1)
2)
3)
4)
No Radioactive (35S)
detected in cells
Radioactive (35P)
detected in cells
Radioactive (35S)
dected in supernatant
No Radioactivity
detected in supernatant
Properties of Genetic material :
Following are the properties and functions which should be fulfilled by a substance if it is to qualify as
genetic material.
It should chemically and structurally be stable.
The genetic material should be able to transmit faithfully to the next generation, as Mendelian characters.
The genetic material should also be capable of undergoing mutations.
The genetic material should be able to generate its own kind (replication)
This can be concluded after examining the above written qualities, that DNA being more stable is
preferred as genetic material, as
a) Free 2'OH of RNA makes it more labile and easily degradable. Therefore DNA in comparison is
more stable.
b) Presence of thymind at the place of uracil also confers additional stability of DNA.
c) RNA being unstable, mutates at a laster rate.
RNA WORLD
RNA was the first genetic material. There are evidences to suggest that essential life processes, such as
metabolism, translation, splicing etc. evolved around RNA. RNA used to act as a genetic material as
well as a catalyst, there are some important biochemical reactions in living systems that are catalysed
by RNA catalysts and not by protein enzymes (e.g., splicing) RNA being a catalyst was reactive and
hence unstable. Therefore, DNA has evolved from RNA with chemical modifications that make it
more stable. DNA being double stranded and having complementary strand further resists changes by
evolving a process of repair. RNA is adapter, structural molecule and in some cases catalytic. Thus
RNA is better material for transmission of information.
DNA REPLICATION
D.N.A. is the only molecule capable of self duplication so it is termed as a " Living molecule"
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All living beings have the capacity to reproduce because of this characteristic of D.N.A.
D.N.A replication takes place in "S - Phase" of the cell cycle. At the time of cell division, it divides in
equal parts in the daughter cells.
SEMI CONSERVATIVE MODE OF DNA REPLICATION
Semi conservative mode of D.N.A. replication was first proposed by Watson & Crick. Later on it was
experimentally proved by Meselson & Stahl (1958) on E- Coli and Taylor on Vicia faba (1958).To
prove this method, Taylor used Radiotracer Technique in which Radioisotopes (tritiated thymidine = 1H3)
were used. Meselson and Stahl used heavy isotope (N15) and Cairns (1963) used radioactive thymidine.
Due to the replication of active Thymidine containing D.N.A., two D.N.A. molecules were obtained in
which 50% radioactivity was found.
When these two D.N.A. molecules containing active Thymidine were made to replicate, the next time
four D.N.A. molecules were obtained. Out of these 4 D.N.A. , 2 D.N.A. molecules were radioactive
and remaining were not radioactive.
In the same sequence, the obtained D.N.A. molecules were further made to replicate then also the no
: of radioactive D.N.A. remains 2.
Generation II
Generation I
15
14
N-DNA
N-DNA
14
40 min
20 min
15
N-DNA
14
N-DNA
14
15
N 15N
14
N
15
N
14 14
N
N
14
N-DNA
N-DNA
N 15N
Separation of DNA by Centrifugation
MECHANISM OF DNA REPLICATION
The following steps are included in D.N.A. replication
1) Unzipping (Unwinding)
The separation of 2 chains of D.N.A. is termed as unzipping. And it takes place due to the breaking of
H-bonds. The process of unzipping starts at a certain specific point which is termed as initiation point
or origin of replication. In prokaryotes there occurs only one origin of replication but in eukaryotes
there occur many origin of replication i.e. unzipping starts at many points simultaneously. At the place
of origin the enzyme responsible for unzipping (breaking the hydrogen bonds) is Helicase (= Swivelase).
In the process of unzipping Mg+2 act as cofactor. Unzipping takes place in alkaline medium.
SSB (single stranded DNA binding protein) prevents the reformation of H-bonds.
Topoisomerase (in prokaryotes also called as DNA gyrase) release the tension arises due to supercoiling.
Note : The process of D.N.A. replication takes a few minutes in prokaryotes and a few hours in
Eukaryotes.
2) Formation of New Chain
To start the synthesis of new chain, special type of R.N.A. is required which is termed as R.N.A. Primer.
The formation of R.N.A. primer is catalysed by an enzyme - R.N.A. Polymerase (primase). Synthesis of
RNA- primer takes place in 5'  3' direction. After the formation of new chain, this R.N.A. is removed.
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For the formation of new chain Nucleotides are obtained from Nuceloplasm. In the nucleoplasm, Nucleotides are present in the form of triphosphates like dATP, dGTP, dCTP, dTTP etc.
5' 3'
Template DNA
(Parental strands)
Continuous
synthesis
3' 5'
Discontinous
sysnthesis
5'
Newly synthesised
3'
strands
Replicating Fork
During replication, the 2 phosphate groups of all nucleotides are separated. In this process energy is
yeilded which is consumed in D.N.A. replication. So, it is clear that D.N.A. does not depend on
mitochondria for it's energy requirements.
Energetically replication is a very expensive process.. Daoxyribonucleoside triphosphase serve duas
purposes in addition to acting as substrates they provide energy for polymerisation.
The formation of new chain always takes place in 5'  3' direction. As a result of this, one chain of
D.N.A. is continuously formed and it is termed as Leading strand. The formation of second chain
begins from the centre and not from the terminal points, so this chain is discontinuous and is made up
of small segments called Okazaki Fragments. This discontinuous chain is termed as Lagging strand.
Ultimately all these segments joined together and a complete new chain is formed.
The Okazaki segments are joined together by an enzyme DNA Ligase.(Khorana)
The formation of new chains is catalysed by an enzyme DNA Polymerase. In prokaryotes it is of 3
types:
1) DNA - Polymerase I : This was discovered by KORNBERG (1957). So it is also called as 'Kornberg's
enzyme'. Kornberg also synthesized DNA first of all, in the laboratory. This enzyme functions as
exonuclease. It separates RNA - primer from DNA and also fills the gap.It is also known as DNArepair enzyme.
2) DNA - Polymerase II : It is least reactive in replication process. It is also helpful in DNA-repairing
in absence of DNA-polymerase-I and DNA polymerase-III
3) DNA - Polymerase III : This is the main enzyme in DNA - Replication. It is most important. It was
discovered by Delucia and Cairns. The larger chains are formed by this enzyme. This is also known as
Replicase.
In the semi conservative mode of replication each daughter DNA molecule receives one chain of
polynucleotides from the mother DNA - molecule and the second chain is synthesized.
Special Point :
All DNA polymerase I, II and III enzymes have 5'-3" polymerisation activity and 3'-5" exonuclease
activity.
Any failure in cell division after DNA replication result into polyploidy.
Difference between DNAs and DNase is that DNAs menas many DNA and DNase means DNA digestive enzymes.
3'
5'
1. nu body of nucleosome consists of
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2.
3.
4.
5.
6.
7.
8.
9.
10.
1) H1 and H2A
2) H2A and H2B
3) H3 and H4
4) Both (2) and (3)
Radioactive element used to label DNA of bacteriophage in Wareing-blender experiment of
Hershey and Chase was
1) S35
2) P32
3) N15
4) C14
Bonding between deoxyribose and base in pyrimidine nucleoside molecule is
1) l'-l' glycosidic linkage2) 1-6' glycosidic linkage
3) l'-9' glycosidic linkage
4) l'-4' glycosidic linkage
Tm (melting temperature) value of DNA is high when it contains
1) A + T > G + C
2) G + C > A + T
3) A + T = G + C
4) A + G = T + C
Select an incorrect statement regarding RNA molecule
1) It has highly reactive 2'-OH group
2) It shows high rate of mutation than DNA
3) It is genetic material in some viruses
4) It follows Chargaff rule
In Meselson and Stahl's experiment, heavy isotope 15N was used in the form of
2) 15NH4C1
3) 15KNO3
4) 15NH4NO3
1) 15NaNO3
15
Assuming that 50 heavy (i.e. containing N ) DNA molecules replicated twice in a medium containing N14, we expect
1) 100 half heavy and half light and 150 light DNA molecules
2) 100 half heavy and half light and 100 light DNA molecules
3) 50 heavy and 150 light DNA molecules
4) 50 heavy and 100 light DNA molecules
The enzyme which has polymerising activity in 5'  3' direction but exonuclease activity in
3'  5' direction only is :
1) RNA polymerase III
2) DNA polymerase II
3) DNA polymerase I
4) All of these
DNA polymerase I is involved in
1) Removal of RNA primer
2) Filling of gap
3) Joining of okazaki fragments
4) Both (1) and (2)
DNA replication in lagging strand of most of the eukaryotic organism is
1) Conservative and continous
2) Semi-conservative but discontinous
3) Conservative and semi-discontinous
4) Semi-conservative but continous
1) 4
2) 2
3) 1
4) 2
5) 2
6) 2
7) 2
8) 2
9) 4
10) 2
RIBO NUCLEIC ACID (RNA)
Structure of RNA is fundamentally same as DNA, but there are some differences.
The differences are as follows :
1) In place of De-oxyribose sugar in DNA, there is present Ribose sugar in RNA.
2) In place of nitrogen base Thymine in DNA, there is present uracil in RNA.
The presence of thymine at the place of uracil also provide additional stability to DNA.
3) RNA is made up of only one polynucleotide chain i.e. R.N.A. is single stranded.
Exception :
RNA found in Reo - virus is double stranded, i.e. it has two polynucleotide chains.
 x 174 bacteriophage (single stranded DNA) has 5386 nucleotides,  - bacteriophage has 48502 base
pairs, Escherichia coli has 4.6 x 106 base pairs and 6.6 x 109 base pairs in human (2n).
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2-OH groups present at every nucleotide in RNA as reactive group and makes RNA labile and easily
degradable and RNA also has catalytic function so it is more reactive so DNA is chemically less
reactive and structurally more stable as compared to RNA.
DNA is more stable so preferred for storage of genetic information but for the transmission of genetic
information RNA is better.
RNA being a catalyst was reactive and hence unstable. Therefore DNA has evolved from RNA with
chemical modification that make it more stable. DNA being double stranded and having complementary strand further resists changes by evolving a process of repair.
Types of RNA :
1. Genetic RNA or Genomic RNA - In the absence of DNA, sometime RNA works as genetic material and it transfers informations from one generation to next generation.
eg. Reo virus, TMV, QB bacteriophage.
2. Non-genetic RNA - 3 types i) r - RNA
(ii) t - RNA
(iii) m - RNA
RNA functions as adapter, structural and in same cases as a catalytic (Ribozyme)
i) Ribosomal RNA (r - RNA)
This RNA is 80% of the cell's total RNA r RNA was discovered by Kuntze.
It is found in ribosomes and it is produced in nucleolus.
It is the most stable form of RNA.
There are present 80s type of ribosomes in Eukaryotic cells. Their subunits are 60s and 40s. In 60s sub
unit of ribosome three types of r-RNA are found - 5s, 5.8s, 28s
40s sub unit of ribosome has only one type of r-RNA i.e. 18s.
So 80s ribosome has total 4 types of r-RNA.
Prokaryotic cells have 70s type of ribosomes and its subunits are 50s and 30s.
50s sub unit of ribosome contains 2 types of r-RNA i.e. 5s and 23s.
30s sub unit of ribosome has 16s type of r-RNA.
So 70s RNA has total 3 types of r-RNA.
Function :
At the time of protein synthesis, r-RNA provides attachement site to t-RNA and m-RNA and attaches
them on the ribosome.
The bonds formed between them are known as Salt linkages. It attaches t-RNA to the larger subunit of
ribosome and m-RNA to smaller subunit of ribosome.
ii) Transfer - RNA (t-RNA) :
It is 10-15% of total RNA.
It is synthesized in the nucleus by DNA.
It is also known as soluble RNA (sRNA)
It is also known as Adapter RNA.
It is the smallest RNA (4s).
Function : At the time of protein synthesis it acts as a carrier of amino-acids.
Discovery : t-RNA was discovered by Hogland, Zemecknike and Stephenson.
Structure : The structure of t - RNA is most complicated.
A scientist named Holley presented Clover leaf model of its structure. In two dimensional structure the
t-RNA appears clover leaf like but in three dimensional structure (by Kim) it appears L-shaped.
The structure of tRNA is looks like a clover leaf but in actual structure, the tRNA is a compact molecule which looks like inverted 'L'.
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3'
5'G
A
C
C
Acceptor arm
DHU Loop
(8-12 bases)
T  c Loop
(7 bases)
Extra arm
Recognition Loop
(7 bases)
Anticodon/N odoc
The molecule of t - RNA is of single stranded.
There are present three nucleotides in a particular sequence at 3' end of t - RNA and that sequence is
CCA.
All the 5' ends i.e. last ends are having G (guanine).
3' end is known as Acceptor end.
t-RNA accepts amino acids at acceptor points. Amino acid binds to 3' end by its - COOH group.
The molecule of t - RNA is folded and due to folding some complementary nitrogenous bases come
across with each other and form hydrogen bonds.
There are some places where hydrogen bonds are not formed, these places are known as loop.
Loops :
There are some abnormal nitrogenous bases in the loops, that is why hydrogen bonds are not formed.
e.g. (i) Inosine (I)
ii) Pseudouracil (VF)
iii) Dihydrouridine (DHU)
A) T C Loop or Attachment loop :
This loop connects t - RNA to the larger subunit of ribosome.
B) Recognition Loop (Anticodon loop) :
This is the most specific loop of t-RNA and different types of t-RNA are different due to this loop.
There is a specific sequence of three nucleotides called Anticodon, is present at the end of this loop.
t-RNA recognizes its place on m - RNA with the help of Anticodon.
The anticodon of t-RNA recognises its complimentary sequence on m-RNA. This complimentary
sequence is known as codon.
C) DHU Loop :
It is also known as Amino - acyl synthetase recognition loop. Amino - acyl synthetase is a specific type
of enzyme. The function of this enzyme is to activate a specific type of amino acid, after activation this
enzyme attaches the aminoacid to the 3' end of t-RNA.
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There are 20 types of enzymes for 20 types of aminoacids.
The function of DHU loop is to recognize this specific Aminoacyl synthetase enzyme.
iii) Messenger RNA (m -RNA)
The m - RNA is 1 - 5% of the cell's total RNA.
Discovery : Messenger RNA was discovered by Huxley, Volkin and Astrachan. The name m-RNA
was ; given by Jacob and Monad.
The m - RNA is produced by genetic DNA in the nucleus. This process is known as Transcription.
It is least stable RNA.
Both DNA and RNA are able to mutate. In fact, RNA being unstable, mutate at faster rate so virus
having RNA genome and having shorter life span mutate and evolve faster.
RNA was the first genetic material.
TRANSCRIPTION
Formation of RNA over DNA template is called transcription. Out of two strand of DNA only one
strand participates in transcription and called Antisense strand".
If both strands act as a template during transcription they would code for RNA molecule with different
sequence and If they code for proteins the sequence of aminoacid in these protein would be different
and another reason that if the two RNA molecule produced they would be complementary to each
other and form a ds RNA which prevent translation of RNA.
A gene is defined as the functional unit of inheritance. It is difficult to literally define a gene in terms
of DNA sequence, because the DNA sequence coding for tRNA or rRNA molecule is also define a
gene (But information of protein is present on the DNA segment which code mRNA. So generally it is
reffered for it)
The segment of DNA which contains signal for the synthesis of one polypeptide is known as "Cistron".
RNA polymerase enzyme is involved in transcription.
In eukaryotes there are three types of RNA polymerases.
RNA polymerase-I for 28s rRNA, 18s rRNA, 5.8s rRNA synthesis.
RNA polymerase-II for m-RNA synthesis.
RNA polymerase-III for t-RNA, 5s rRNA, SnRNA synthesis.
In eukaryotes RNA polymerase enzyme is made up of 10-15 polypeptide chains.
Prokaryotes have only one type of RNA polymerase which synthesizes all types of RNAs.
RNA polymerase (Core enzyme) of E. Coli has five polypeptide chains  ,  ' ,  ,  and  .
 polypeptide chain is also known as a factor (sigma factor).
Core enzyne + Sigma factor  RNA Polymerase
(  ,  ' , , , )
( )
A transcription unit in DNA is defined primarily by the in three gigons in the DNA
i) A promoter,
ii) The structural gene
iii) A terminator
Transcription start site
Promoter
3'
Structural gene
Template strand
5'
3'
5'
Coding strand
Following steps are present in transcription GENETICS
Terminator
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1) INITIATION
DNA has a "Promoter site" where RNA polymerase binds and a "Terminator site" where transcription
stops.
Sigma factor (a) recognises the promoter site of DNA.
With the help of sigma factor RNA polymerase enzyme attached to a specific site of DNA called
"Promoter site"
In prokaryotes before the 10 N2 base from "Starting point" a sequence of 6 base pairs (TATAAT) is
present on DNA, which is called "Pribnow box".
In eukaryotes before the 20 N2 base from "Starting point" a sequence of 7 base pairs (TATAAAA) or
(TATATAT) is present on DNA which is called "TATA box or Hogness box"
At promoter site RNA polymerase enzyme breaks H-bonds between two DNA strands and separates
them.
One of them strand takes part in transcription. Transcription proceeds in 5'  3' direction.
Ribonucleoside triphosphate come to lie opposite complementary nitrogen bases of anti sense strand.
These Ribonucleotides present in the form of triphosphate ATP, GTP, UTP and CTP. When they are
used in transcription, pyrophosphatase hydrolyse two phosphates from each activated nucleotide. This
releases energy.
This energy is used in the process of transcription.
2) ELONGATION
RNA polymerase enzyme establishes phosphodiester bond between adjacent ribonucleotides.
Sigma factor separates and core enzyme moves along the anti sense strand till it reaches terminator site.
3) TERMINATION
When RNA polymerase enzyme reaches at terminator site, it separates from DNA templet.
At terminator site on DNA, N2 bases are present in palindromic sequence.
In most cases RNA polymerase enzyme can recognise the 'Terminator site' and stop the synthesis of
RNA chain, but in prokaryotes, it recognises the terminator site with the help of Rho factor (p factor).
Rho (p) factor is a specific protein which helps RNA polymerase enzyme to recognise the terminator
site.
3'
3'
5'
DNA helix
5'
Promoter
RNA polymerase
Initiation
Sigma factor
5'
3'
5'
RNA
Elongation
Termination
Terminator
3'
3'
5'
5'
3'
RNA polymerase
RNA
Rho factor
SPLIT GENE
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Discovered by sharp and Roberts. They awarded by Nobel Prize in 1993. Gene which contains non
functional part along with functional part is known as split gene. Non functional part is called intron
and functional part is called exon. By transcription split gene produces a RNA which contains coding
and non coding sequence and called hn RNA (Hetero genous nuclear RNA). This hn RNA is unstable.
Now 7 methyl guanonsine is added to its 5' end, and a cap like structure is formed. It is called capping
and 200 nucleotides of adenylic acid are added to its 3' end, which is called poly 'A' tail, Now it
becomes stable. By the process of RNA splicing hn-RNA produces functional m-RNA that is exonic
RNA. In RNA splicing non coding parts is removed with the help of ribonuclease enzyme and coding
part join together with the help of RNA ligase. Some specific proteins are also helpful in RNA splicing called 'Small nuclear ribonucleoprotein' or 'SnRNP' or 'Snurps'. These SnRNP proteins combine with some other proteins and SnRNA to form spliceosome complex. This spliceosome complex
uses energy of ATP to cut the RNA, releases the non-coding part and joins the coding-part to produce
functional RNA. Non coding part of hn RNA remained inside the nucleus and not translated in to
protein. Only coding part moves from nucleus to cytoplasm and translated into protein. Mostly Eukaryotic genes are example of split gene, but gene which forms histone and interferon protein are non
split gene. It contains only and only exonic part.
Mostly prokaryotic genes are example of non split gene.
In euckaryotes after transcription splicing process also occured.
The split gene represent an ancient (primitive) feature of gene.
Presence of intron is a primitive character.
The splicing process represent the dominance of RNA world.
3
functional part Non functional part Functional part
5
Exon
Exon
Intron
Antisense Strand of DNA
Transcription
Coding part
Non coding part
Coding part
3
HnRNA(unstable)
Stabilization
Coding part
7mG cap
Non coding part
Spliceosome
complex
ATP
Capping
Coding part
Splicing
Ribonuclease
RNA lygase
by Guanyl transferase
7mG cap
[AAA...]
Poly 'A' tail
5' end
m-RNA
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49
[AAA...]
3' end Poly 'A' tail
Tailing
(by Poly A Polymerase)
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Capping
5'
5'
3'
3'
3' mRNA
Intron
Cap
m
5' G ppp
Exon
RNA splicing
5'
m
Polyadenylation
3'
G ppp
Poly A tail
5'
m
G ppp
3'
m
G ppp
Messenger RNA
5'
Process of Transcription in Eukaryotes







GENETIC CODE
Term Given by George Gamow.
Discovered by Nirenberg, Mathai and Khorana.
The relationship between the sequence of amino acids in a polypeptide chain and nucleotide sequence
of DNA or m-RNA is called genetic code.
There occur 20 types of amino acids which participate in protein synthesis. DNA contains information
for the synthesis of any types of polypeptide chain. In the process of transcription, information is
transfered from DNA to m-RNA in the form of complementary N 2-base sequences.
m-RNA contains code for each amino acid and it is called codon. A codon is the nucleotide sequence
on m-RNA which codes for particular amino acid ; wherease the genetic code is the sequence of
nucleotides on m-RNA molecule, which contains information for the synthesis of polypeptide chain.
Triplet Code
The main problem of genetic code was to determine the exact number of nucleotide in a codon which
codes for one amino acid.
There are four types of N2-bases in m-RNA (A, U, G, C) for 20 types of amino acids.
If genetic code is singlet i.e. codon is the combination of only one nitrogen base, then only four codons
are possible A, C, G and U. These are insufficient to code for 20 types of amino acids.
Singlet code = 41= 4x1=4 codons
A
C
G
Codons
U
Singlet Code : 4 x 1 = 4 codons
If genetic code is doublet (i.e. codon is the combination of two nitrogen bases) then 16 codons are
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



In this case there occurs 64 codons in dictionary of genetic code.
64 codons are sufficient to code 20 types of amino acids.
H.G. Khorana artificially synthesized an mRNA.
Severo ochoa enzyme (RNA polymerase enzyme) is also helpful in polymerising RNA with defined
sequences in a template independent manner.
Third position
First position
Second position
U
A
C
G
UUU Phe
U UUC Phe
UUA Leu
UUG Leu
UCU Ser
UCC Ser
UCA Ser
UCG Ser
UAU Tyr
UAC Tyr
UAA Stop
UAG Stop
UGU Cys
UGC Cys
UGA Stop
UGG Trp
U
C
A
G
CUU Leu
CUC Leu
CUA Leu
CUG Leu
CCU Pro
CCC Pro
CCA Pro
CCG Pro
CAU His
CAC His
CAA Gin
CAG Gin
CGU Arg
CGC Arg
CGA Arg
CGG Arg
U
C
A
G
AUU He
AUC He
A AUA lie
AUG Met
ACU Thr
ACC Thr
ACA Thr
ACG Thr
AAU Asn
AAC Asn
AAA Lys
AAG Lys
AGU Ser
AGC Ser
AGA Arg
AGG Arg
U
C
A
G
C
GUU Val GCU Ala GAU Asp GGU Gly U
GUC
Val GCC Ala GAC Asp GGC Gly C
G
GUA Val GCA Ala GAA Glu GGA Gly A
GUG Val GCG Ala GAG Glu GGG Gly G
Triplet codons for the various amino acids
Characteristics of Genetic Code
i) Triplet in Nature
 A codon is composed of three adjacent nitrogen bases which specifies the one amino acid in polypeptide chain.
For Ex. :
In m-RNA if there are total 90 N2 - bases.
Then this m-RNA determines 30 amino acids in polypeptide chain.
In above example, number of nitrogen bases are 90 so codons  30 and 30 codons decide 30 amino
acids in polypeptide chain.
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







ii) Universality
The genetic code is applicable universally. The same genetic code is present in all kinds of living
organism including viruses, bacteria, unicellular and multicellular organisms.
iii) Non - Ambiguous
Genetic code is non ambiguous i.e. one codon specifies only one amino acid and not any other.
In this case one codon never code two different amino acids. Exception GUG codon which codes both
valine and methionine amino acids.
iv) Non - Overlapping
A nitrogen base is a constituent of only one codon.
v) Comma less
There is no punctuation (comma) between the adjacent codon i.e. each codon is immediately followed
by the next codon.
If a nucleotide is deleted or added, the whole genetic code read differently.
A polypeptide chain having 50 amino acids shall be specialized by a linear sequence of 150 nucleotides. If a nucleotide is added in the middle of this sequence, the first 25 amino acids of polypeptide
will be same but next 25 amino acids will be different.
vi) Degeneracy of Genetic code
There are 64 codons for 20 types of amino acids, so most of the amino acids (except two) can be coded
by more than one codon. Single amino acid coded by more than one codon is called "Degeneracy of
genetic code". This incident was discovered by Baumfield and Nirenberg.
Only two amino acids Tryptophan and Methionine are specified by single codon.
UGG for Tryptophan
AUG for Methionine.
All the other amino acids are specified or coded by 2 to 6 codons.
Leucine, serine and arginine are coded or specified by 6-codons.
Leucine = CUU, CUC, CUA, CUG, UUA & UUG
Serine = UCU, UCC, UCA, UCG, AGU, AGC
Arginine = CGU, CGC, CGA, CGG, AGA, AGG
Degeneracy of genetic code is related to third position (3'- end of triplet codon) of codon. The third
base is described as "Wobbly base".
Chain Initiation and Chain Termination Codon
 Polypeptide chain synthesis is signalled by two initiation codons AUG or GUG.
 G codes methionine amino acid in eukaryotes and in prokaryotes AUG codes N-formyl methionine. *Some times GUG also functions as start codon it codes for valine amino acid normally but when it is
present at starting position it codes for methionine amino acid.
 Out of 64 codons 3-codons are stopping or nonsense or termination codon.
Nonsense codons do not specify any amino acid.
UAA (Ochre)
UAG (Amber)
UGA (Opal)
Non-Sense Codons or
Stop codons
So only 61 codons are sense codons which specify 20 amino acid.
WOBBLE HYPOTHESIS
 It was propounded by CRICK.
 Normally an anticodon recognises only one codon, but sometimes an anticodon recognises more than
one codon. This is known as Wobbling. Wobbling normally occurs for third nucleotide of codon.
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 For e.g. anticodon AAG can recognise two codons i.e. UUU and UUC, both stands for phenyl alanine.
Types of m-RNA - m-RNA is of 2 types 1) Monocistronic : The m - RNA in which genetic signal is present for the formation of only one
polypeptide chain eg. Eukaryotes.
2) Polycistronic : The m-RNA, in which genetic signal is present for the formation of more than one
polypeptide chains eg. Prokaryotes.
 Non sense codons are found in middle position in polycistronic m-RNA.
CENTRAL DOGMA
 Central dogma was given by Crick.
 The formation (production) of m - RNA from DNA and then synthesis of protein from it, is known as
Central Dogma.
Transcription
Translation
 DNA 

 RNA 
 Protein
 It means, it includes transcription and translation.
Transcription
Translation
DNA
Replication
RNA
Protein
Reverse transcription
Reverse Transcription
 The formation of DNA from RNA is known as Reverse - transcription. It was discovered by Temin
and Baltimore in Rous - sarcoma virus. So it is also called Teminism.
 ss-RNA of Rous-Sarcoma virus (Retro virus) produces ds-DNA in host's cell with the help of enzyme
reverse transcriptase (DNA-polymerase). This DNA is called c-DNA (Complimentary DNA). Some
times this DNA moves in host genome. Such mobile DNA is called "Retroposon" (Oncogene).
TRANSLATION (Protein Synthesis)
1) Activation of Amino acid
 20 types of amino acids participate in protein synthesis.
 Amino acid reacts with ATP to form "Amino acyl AMP enzyme complex" , which is also known as
'Activated Amino acid'.
Amino acyl AMP-enzyme complex + PP
Amino acid + ATP 
t-RNA synthetase
Amino acyl
 This reaction is catalyzed by a specific 'Amino acyl t-RNA synthetase' enzyme.
 There is a separate 'Amino acyl t-RNA synthetase' enzyme for each kind of amino acid.
2) Charging of t-RNA (Loading of t-RNA)
 Specific activated amino acid is recognised by its specific t-RNA.
 Now amino acid attaches to the 'Amino acid attachment site' of its specific t-RNA and AMP and
enzyme are separated from it.
Amino acyl AMP-enzyme complex + t-RNA  Amino acyl t-RNA complex + AMP + enzyme
 Amino acyl t-RNA complex is also called 'Charged t-RNA'.
 Now Amino acyl t-RNA moves to the ribosome for protein synthesis.
3) Translation 3 steps A) Initiation of polypeptide chain
 In this step 30 's' and 50 's' sub units of ribosome, GTP, Mg+2, charged t-RNA, m-RNA and some
initiation factors are required.
 In prokaryotes there are three initiation factors present - IF1, IF2, IF3.
 In Eukaryotes more than 3 initiation factors are present. Ten initiation factors have been identified in
red blood cells elFl, eIF2, eIF3, eIF4A, eIF4B, eIF4C, eIF4D, eIF4F, eIF5, eIF6
 Initiation factors are specific protein.
 GTP and initiation factors promote the initiation process.
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 Both sub units of ribosome are separated with the help of IF3 factor.
 In prokaryotes with the help of "S D sequence" (Shine-Delgamo sequence) m-RNA recognises the
smaller sub unit of ribosome. A sequence of 8 N 2 base is present before the 4-12 N2 base of initiation
codon on mRNA, called "SD sequence". In Smaller subunit of ribosome, a complementary sequence
of "SD sequence" is present on 16 'S' rRNA, which is called "Anti Shine-Delgamo sequence" (ASD
sequence)
 With the help of 'SD' and 'ASD sequence' mRNA recognises the smaller sub unit of ribosome.
 While in eukaryotes, smaller sub unit of ribosome is recognised by "7mG cap".
 In eukaryotes, 18 'S' rRNA of smaller sub unit has a complementary sequence of "7mG cap".
 30 'S' m-RNA - complex
30 'S' sub unit + m-RNA 
Mg 2
IF3
This "30 'S' m-RNA - complex" reacts with 'Formyl methionyl t-RNA - complex' and "30 'S' m-RNA formyl methionyl t-RNA - complex" is formed. This t-RNA attaches with codon part of m-RNA. A GTP
molecule is required.
30 'S' m-RNA - complex + Formyl methionyl t-RNA - complex
GTP
IF2, IF3Mg +2
30 'S' m-RNA formyl methionyl t-RNA - complex
 Now larger sub unit of ribosome (50 'S' sub unit) joins this complex. The initiation factor released and
complete 70 'S' ribosome is formed.
In larger sub unit of ribosome there are three sites for t-RNA 'P' site = Peptidyl site.
'A' site = Amino acyl site.
'E' site = Exit site
Starting codon of m-RNA is near to 'P' site of ribosome, so t-RNA with formyl methionine amino acid
first attaches to 'P' site of ribosome and next codon of m-RNA is near to 'A' site of ribosome. So next
new t-RNA with new amino acid always attach at 'A' site of ribosome but in initiation step 'A' site is
empty.
E
AA2
AA1
AA1
P
5
AU G
E
A
P
5
3
AU G
A
3
B) Chain - Elongation :
 New tRNA with new amino acid is attaches at 'A' site of ribosome.
 The link between amino acid of 'P' site of t-RNA is broken and t-RNA of P-site is discharged so COOH of P-site A.A. becomes free.
 Now peptide bond takes place between - COOH group of P site amino acid and - NH2 group of A-site
amino acid.
 Peptidyl transferase enzyme induces the formation of peptide bond. In peptide bond formation, 23 'S'
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






r-RNA is also helpful. This r-RNA acts as an enzyme so it is also called "Ribozyme".
After formation of peptide bond t-RNA of P site released from ribosome via E-site and dipeptide
attaches with A site.
Now t-RNA of A site is transferred to P site and  A site becomes empty..
Now ribosome slides over m-RNA strand in 5'  3' direction. Due to sliding of ribosome on m- RNA,
new codon of m-RNA continuously available at A site of ribosome and according to new codon of mRNA new amino acid attaches in polypeptide chain.
Translocase enzyme is helpful in movement of ribosome (translocation). GTP provides energy for
sliding of ribosome.
In elongation process some protein factors are also helpful, which are known as 'Elongation factors'.
In prokaryotes three 'Elongation factors' are present - EF-Tu, EF-Ts, EF-G.
In Eukaryotes two elongation factors are present - eEFl, eEF2.
Peptide bond
AA2
AA1
E
P
AA2
AA1
A
E
translocation
5
3
AU G
AA2
AA1
E
P
AA3
A
P
5
3
AU G
AA3
A
5
3
AU G
C) Chain - Termination
 Due to sliding of ribsome over m-RNA when any Nonsense codon (UAA, UAG, UGA) available at A
site of ribosome, then polypeptide chain terminates.
 The linkage between the last t-RNA and the polypeptide chain is broken by three release factor called
RF1, RF2, RF3 with the help of GTP.
 In eukaryotes only one Release Factor is known - eRFl.
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AA1
AA2
AA3
AA4
E
AA5
P
A
5
3
UA A
 An mRNA also have some additional sequences that are not translated and are referred as untranslated
regions (UTR). The UTRs are present at both 5'end (before start codon) and at 3'end (after stop codon).
 The UTR(untranslated regions) present on mRNA are required, for efficient translation process (by
recognising the smaller subunit of ribosome by mRNA)
SPECIAL POINTS
1) The chargaff's rule is not valid (true) for RNA. It is valid only for double helical DNA. i.e. for RNA
it is A  U and G  C.
2) The duplication of DNA was first of all proved in E. coli bacterium.
3) E. coli Bacterium is mostly used for the study of DNA duplication.
4) Hargovind singh Khurana first of all recognised the triplet codon for Cysteine and Valine amino
acids.
5) Cytoplasmic DNA is 1 - 5% of total cell DNA.
6) Three scientists named Avery, Me - Leod and Me Carty (by their transformation experiments on
bacteria) Proved that DNA is a genetic material.
7) Hershey and Chase first of all proved that DNA is genetic material in bacteriophages.
8) Frankel and Conret proved, RNA as a genetic material in viruses (g-RNA).
9) AUC, ACU & AUU - These anticodons do not exist.
10) The structure formed by the combination of m - RNA and Ribosomes is known as polyribosomes/
Polysomes/ Ergosomes
11) The formation of t - RNA takes place from the heterochromatin part of DNA.
12) The formation of m - RNA takes place from the Euchromatin part of DNA.
13) m - RNA is least stable. It is continuously formed and finished.
14) In cytoplasm, t - RNA is present in the form of soluble colloid.
15) Nucleases : Nucleases are the breaking enzymes of nucleic acids. These are of two types
i) Endo-Nucleases : These break down the nucleic acids from the inside.
ii) Exo-nucleases : These break down the nucleic acids from the ends(terminal ends).
These separate each nucleotide.
16) Some Inhibitors of Bacterial Protein Synthesis :
Antibiotic
Effect
Tetracycline
Inhibits binding of amino-acyl tRNA to ribosome
Streptomycin
Inhibits initiation of translation and causes misreading
Chloramphenicol
Inhibits peptidyl transferase and so formation of peptide bonds
Erythromycin
Inhibits translocation of ribosome along mRNA
Neomycin
Inhibits interaction between tRNA and mRNA
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1. Unidirectional flow of information called central dogma was given by
1) F.H.C. Crick
2)Temin
3) Baltimore
4) Dulbecco
2. In eukaryotes, RNAPIII catalyses the synthesis of
1) All rRNA and tRNA
2) mRNA, HnRNA and SnRNA
3) 5S rRNA, tRNA and SnRNA
4) 28S, 18S and 5S rRNA
3. The core enzyme requires a factor for termination of RNA synthesis at some sites. This is known as
1) Sigma factor
2) Rho factor
3) Gamma factor
4) Alpha particle
4. If one strand of DNA has the base sequence ATCCACGACTAG and the second strand undergoes transcription what would be the base sequence on mRNA ?
1) TACGT GCTG ATC
2) ATCCACGACTAG
3) AUCCACGACUAG
4) AUGCACGACTAG
5. During protein synthesis, amino acid gets attached to tRNA with the help of
1) mRNA
2) Aminoacyl synthetase 3) Ribosome
4) rRNA
6. The first amino acid in any polypeptide chain of prokaryote is always
1) Formylated methionine
2) Formylated arginine
3) Lysine
4) Methionine
7. Which site of a tRNA molecule forms hydrogen bonds with mRNA molecule ?
1) Codon
2) Anticodon
3) 5' end of the t-RNA molecule
4) 3' end of the t-RNA molecule
8. To code the 50 aminoacids in a polypeptide chain, what will be the minimum number of nucleotides in its cistron?
1) 50
2) 153
3) 306
4) 300
9. A single anticodon can recognize more than one codon of m-RNA. This phenomenon is termed as
1) Richmond and Lang effect
2) Gene flow hypothesis
3) Wobble hypothesis
4) Transposability
10. The genetic code is called a degenerate code because
1) One codon has many meanings
2) More than one codon has the same meaning
3) One codon has one meaning
4) There are 64 codons present
1) 1
2) 3
3) 2
4) 3
5) 2
6) 1
7) 2
8) 3
9) 3
10) 2
REGULATION OF GENE EXPRESSION
 Constitutive genes (House-keeping genes) - These genes are expressed constantly, because their
products are constant needed for cellular activity, e.g. genes for glycolysis, gene of ATPase enzyme.
 Non-constitutive genes (Smart gene or Luxary gene) - These genes remain silent and are expressed
only when the gene product is needed. They are switched 'on' or 'off' according to the requirement of
cellular activities. Non-constitutive genes are of two types; inducible and repressible. The inducible
genes are switched- on in presence of a chemical substance called inducer, required for the functioning
of gene activity. The repressible genes continue to express themselves till a chemical, often an end
product of the metabolism inhibits or represses their activity. Such type of inhibition is called feed
back inhibition or feed back repression.
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 The mechanism which stimulates the expression of certain genes and inhibits that of others is called
regulation of gene expression.
 It is possible only if the organism has a mechanism of regulating gene activity by allowing some to
function and others to restrain their activity through switching on and switching off system. This
means, the genes are turned 'on' or 'off' as per requirement.
 A set of genes is 'switched on' when enzymes are required to metabolise a new substrate. The enzymes
produced by these genes metabolise the substrate.
 The molecules of metabolite that come to switch on of the genes are termed as inducers and the
phenomenon is called induction.
 Similarly, certain genes which are in their 'switch on' state, continue to synthesise a metabolite till the
later is produced in amount more than required or else, it is supplied to the cell from outside. In other
words, certain genes continue to express themselves till the end product of inhibits or repress their
expression. Inhibition by an end product is known as 'feed back repression'.
 Regulation of gene expression refers to a very broad term that may occur at various levels. Considering that gene expression results in the formation of a polypeptide, it can be regulated at several levels.
In eukaryotes, the regulation could be exerted at
i) transcriptional level (formation of primary transcript),
ii) processing level (regulation of splicing),
iii) transport of mRNA from nucleus to the cytoplasm,
iv) translational level.
OPERON CONCEPT
 In 1961, two French microbiologist Francis Jocob and Jacques Monad at the Pasteur Institute in Paris,
proposed a mechanism called operon model for the regulation of gene action in E. coli.
 An operon is a part of genetic material or DNA, which acts as a single regulated unit having one or
more structural genes-an operator gene, a promoter gene, a regulator gene.
p
i
p o
Repressor mRNA
z
y a In absence of inducer
Repressor binds to the operator region (o)
and prevents RNA polymerase from
transcribing the operon
Repressor
p
i
p o
Repressor mRNA
z
y a In presence of inducer
Transcription
lac mRNA
Translation
-galactosidase permease transacetylase
Inducer
(Inactive repressor)
 Operons are of two types (i) inducible (ii) repressible.
1. Inducible System (Lac operon of E. coli)
 An inducible operon system normally remains in switched off condition and begins to work only when
the substance to be metabolised by it is present in the cell. Inducible operon system generally occurs in
catabolic pathways, e.g. Lac operon of E. coli.
Active repressor + inducer = inactive repressor
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An inducible operon system consists of four types of genes
i) Structural genes - These genes synthesise mRNAs, which in turn synthesise polypeptide or enzyme over
the ribosomes. An operon may have one or more structural genes. Each structural gene of an operon is called
cistron. The lac operon (lactose operon) of Escherichia coli contains three structural genes (Z, Y and A). These
genes occur adjacent to each other and thus are linked. They transcribe a polycistronic mRNA molecule (a
single stretch of mRNA covering all the three genes), that helps in the synthesis of three enzymes-  galactosidase (breaks lactose into glucose and galactose), lactose permease (helps in entry of lactose in cell from
outside) and transacetylase (transfers an acetyl group from acetyl Co A to p galactosidase).
ii) Operator gene - It lies adjacent to the structural genes and directly controls the synthesis of mRNA
over the structural genes. It is switched off by the presence of a repressor. An inducer can take away the
repressor and switch on the gene that directs the structural genes to transcribe.
iii) Promoter gene - This gene is the site for initial binding of RNA polymerase. When the operator
gene is turned on, the enzyme RNA polymerase moves over it and reaches the structural genes to
perform transcription.
iv) Regulator gene - It produces a repressor that binds to operator gene and stops the working of the
operator gene.
v) Repressor - It is a protein, produced by the regulator gene. It binds to the operator gene so that the
transcription of structural gene stops. Repressor has two allosteric site (1) operator gene (2) effective
molecule (inducer/corepressor)
vi) Inducer - It is a chemical (substrate, hormone or some other metabolite) which after coming in
contact with the repressor, forms an inducer repressor complex. This complex cannot bind with the
operator gene, which is thus switched on. The free operator gene allows the structural gene to transcribe mRNA to synthesise the enzymes.
The inducer for lac operon of Escherichia coli is lactose (in fact allolactose an isomer of lactose).
When the sugar lactose is added to the culture of E. coli, a few molecules of lactose gets into the
bacterial cells by the action of the enzyme permease, a small amount of this enzyme is present in the
cell even when the operon is not working. These few lactose molecules are then converted into an
active form which acts as an inducer and binds to the repressor protein. The inducer repressor complex
fails to join with the operator, which is turned on. The three genes are expressed as three enzymes to
metabolise lactose. Allolactose is real inducer of lac operon.
Gratuitous inducer  Some molecules resemble with natural inducer but are not metabolize by the
enzyme.
Example : Isopropyl thio galactoside (IPTG): This resembles lactose and thus has the property of
induction. Such inducer which induces enzyme synthesis, without getting metabolized are called gratuitous inducer.
2. Repressible System (Tryptophan operon of E. coli.)
A repressible operon system is normally in it's switch on state and continue to synthesise a metabolise till
the latter is produced in amount more than required, or else it becomes available to the cell from outside.
Repressible operon system is commonly found in anabolic pathway, e.g. Tryptophan operon of E. coli.
[Inactive repressor + co-repressor = acitve repressor]
Tryptophan operon of Escherichia coli is an example of repressible system. It consists of the following:
i) Structural genes : These genes are meant for transcription of mRNA, which in turn synthesise
enzyme. Tryptophan operon has five structural genes E, D, C, B and A. They lie in continuation and
synthesise enzymes for five steps of tryptophan synthesis.
ii) Operater gene (trp O) : It lies adjacent to the structural genes and controls the functioning of the
structural genes. Normally, it is kept switched on, because the apo-repressor produced by the regulator
gene does not bind to it. The operator gene is switched off when a co-repressor is available alongwith
apo-repressor.
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iii) Promoter gene (trp P) : It marks the site at which the RNA polymerase enzyme binds. When the
operator gene is switched on, it moves from promotor gene to structural genes for transcription.
iv) Regulator gene (trp R) : It produces a regulatory protein called apo-repressor for (Inactive repressor) possible blocking the activity of operator gene.
v) Apo-repressor : It is a regulatory protein synthesised by regulator gene. When a co-repressor substrate is available in the cell, the apo-repressor combines with the co-repressor to form a apo-repressor
corepressor complex. This complex binds with the operator gene and switches it off. Presence of aporepressor alone, the operator gene is kept switched on because, by itself the apo-repressor is unable to
block the working of operator gene.
vi) Co-repressor : It is an end product of reactions catalysed by enzymes produced by the structural
genes.
In the presence of tryptophan some molecules of tryptophan act as co-repressor, co-repressor-bind
with inactive repressor, co-repressor repressor complex bind with-operator region and prevent the
binding of RNA polymerase to the promoter, the trp-operon is off.
Structural gene
P.G. O.G. E D C
R.G
B A
Transcription
Aporepressor
Polycistronic m-RNA
Translation
(inactive repressor)
Polypeptides
[OPERON-ON]
E1
E2
E3
Enzymes
Tryptophan
Chorismic acid
Structural gene
P.G. O.G. E D C
R.G
Aporepressor
+
Co-repressor
(Tryptophan)
B A
No- Transcription
No-Translation
[OPERON-OFF]
The repressor molecule has key role in regulation of lac-operon. Repressor molecule active or inactive. Active repressor may be rendered inactive by addition of an inducer while the inactive repressor
can be made active by addition of a co-repressor.
Because the product of regulator gene the repressor act by shutling off the transcripition of structural
gene the operon model, as originally proposed by Jocob & Monad is referred as -negative control system.
MUTATION
 Sudden heritable change in genetic material of an organism is called as Mutation.
 Mutation are source of discontinuous variation.
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 Frequency of mutation at present is 1 x 10-6 (1 cell in : 1 million-cell). But it will increase in future due
to pollution and destruction of ozone layer.
 Only those mutation are heritable which occur in germinal cell of an organism. While somatic mutations are non heritable. Somatic mutations are also heritable in vegetative propagated plants.
 Mutation word was given by Hugo De Vries.
 De Vries studied mutations in the plant Oenothera lamarckiana (evening primrose).
 De Vries proposed mutation theory of evolution.
 This theory was given in support of Darwinism because Darwin was unable to explain the source of
sudden large variations. Darwin called such variation as sport.
 Mutation was first observed by Seth Wright. He observed some short legged sheep (Ancon) variety in
a population of long legged sheep. It was an example of dominant germinal mutation.
 Morgan observed some white eyed male Drosophila in a population of red eyed Drosophila. It is
produced due to mutation.
 Muller is the discoverer of Induced Mutations. He induced mutations in Drosophila by the help of X-rays.
 Beadle and Tatum induced mutations in Neurospora by the help of U. V. rays, or X-rays.
U.V.rays
Mutant Neurospora 
 Mutant Neurospora
PROTOTROPH)
(AUXOTROPH)
Normal-Neurospora can be grown in minimal medium because Neurospora can make all essential
nutrients required for it. This is known as Prototroph.
Mutant Neurospora doesn't has capability to grow in minimal medium because due to mutation it loses
those genes which prepare some special nutrients for it. Eg. Arginine. When Arginine was given to
mutant Neurospora then the growth of Neurospora was normal. This form is known as Auxotroph.
 M.S. Swaminathan induced mutations in wheat by the help of y-rays to obtain good varieties for eg.
Sharbati Sonora, Pusa Lerma. Swaminathan established y garden in IARI-New Delhi (Pusa Institute).
Types of mutation :
I. CHROMOSOMAL MUTATION
II. GENE MUTATIONS
I. Chromosomal Mutations :
 Change in number or structure of chromosome.
Types of chromosomal mutation
1) Heteroploidy/Genomatic mutation - change in chromosome number.
2) Chromosomal aberration - change in structure of chromosome.
1. Heteroploidy / Genomatic mutation
 Change in number of one or few chromosomes in a set or number of entire set of chromosome.
It is of two types :
i) Euploidy - Change in number of chromosome sets.
ii) Aneuploidy - Change in number of chromosome in a set.
i. Euploidy :
 Change in number of sets of chromosome i.e. either loss or addition of sets of chromosomes.
 Monoploidy (x) - Presence of one set of chromosomes.
 Diploidy (2x) - Presence of two sets of chromosomes.
 Polyploidy - Presence of more than two sets of chromosomes.
It may be
Triploidy (3x)
Tetraploidy (4x)
Pentaploidy (5x)
Hexaploidy (6x)
Heptaploidy (7x)
Octaploidy (8x)
 Polyploid plants with even number of sets are always fertile, reproduce sexually and form seeds.
 Polyploid plants with odd number of sets are always sterile don't reproduce by sexual reproduction,
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




They don't produce seeds but they may produce seedless fruits by parthenocarpy.
eg. Banana and seedless grapes.
Polyploidy is of two types :
a. Autopolyploidy :
It is repetition of same set of chromosomes, eg. AAA.
Cyanodon and Rose are example of Autortirploid plants These are sterile plants.
Reproduce by vegetative propagation.
b. Allopolyploidy:
More than one type of sets are present in these plants eg. AA BB.
These plants are obtained by intergeneric cross.
e.g Raphanobrassica is obtained by cross between Raddish and cabbage and first time obtained by
Russian Scientist Karpechenko
Raphanus Sativus
X
Brassica oleracea
2x = 18
2x = 18
Raphanobrassica
2x = 18
Sterile plant
Colchicine
Raphanobrassica fertile
4x = 36
Triticale :
Triticum durum
4x = 28
X
Triticale
3x = 21
Rye (Secale Cereale)
2x = 14
Sterile
Colchicine
Triticale
6x = 42
Triticale :
Triticum aestivum
6x = 42
X
Rye (Secale Cereale)
2x = 14
Tritical4x = 28Sterile
Colchicine
Triticale
8x = 56 Fertile
ii. Aneuploidy :
Loss or addition of chromosomes in a set of chromosomes.
Types of Aneuploidy :
1. Hypoaneuploidy (loss)
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 2n - 1 = Monosomy (loss of one chromosome in one set).
 2n - 1 - 1 = Double monosomy (loss of one chromosome from each set, but these are non homologus.)
2n - 2 = Nullisomy (loss of two homologus chromosome)
2. Hyperaneuploidy (addition)
 2n + 1 = Trisomy : addition of one chromosome in one set.
 2n + 1 + 1 = Double Trisomy : addition of one chromosome in each set.
 2n + 2 = Tetrasomy : addition of two chromosome in one set.
Cause of aneuploidy is chromosomal nondisjunction means chromosomes fail to separate during meiosis.
 Chances of aneuploidy are more in higher age female due to less activity of oocyte, so chances of
syndrome increase in children who are born from higher age female.
2. Chromosomal Aberrations :
Change in structure of chromosome.
i) Deletion:
Loss of a part or segment of chromosome which leads to loss of some gene is called as deletion.
It is of 2 types
a) Terminal deletion - Loss of chromosomal segment from one or both ends.
1
2
3
4
3
4
5
6
7
8
5
6
7
8
eg. The cry -du-chat syndrome is an example of terminal deletion in short arm of 5 th chromosome.
b) Intercalary deletion - Loss of chromosomal part between the ends.
1
2
3
4
1
2
5
6
7
8
5
6
7
8
ii) Inversion:
Breakage of chromosomal segment but reunion on same chromosome in reverse orders. It leads to
change in distance between genes on chromosome or sequence of genes on chromosome so crossing
over is affected. It is of 2 types
a) Paracentric - If inversion occur only in one arm and inverted segment does not include centromere.
12 3 4
5 6 7 8
12 3 4
5 7 6 8
b) Pericentric - In this type of inversion inverted segment include centromere.
1 2 3 4
5 6 7 8
1 2 6 5
iii) Duplication :
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Occurence of a chromosomal segment twice on a chromosome.
Example : In drosophila "Bar eye character" is observed due to duplication in X-chromosome. Bar
eye is a character where eyes are narrower as compared to normal eye shape.
ABCDEF
ABCD
ABCDEFEF
ABCDEF
iv) Translocation:
In this, a part of the chromosome is broken and may be joined with non homologous chromosome.
This is also known as Illegitimate crossing over (illegeal crossing over)
Three types of translocation A) Simple Translocation - When a chromosomal segment breaks and attached to the terminal end of
a non- homologous chromosome.
123456
1234
ABCDEF56
ABCDEF
B) Interstitial or shift translocation - If a segment of chromosome breaks and gets inserted in interstitial position of a non homologous chromosome.
123456
1234
ABCD56EF
ABCDEF
C) Reciprocal Translocation - Exchange of segments between two non-homologous chromosome.
123456
1 2 3 4 E F
ABCDEF
ABCD56EF
eg. Chronic myloid leukemia [C M L] is a type of blood cancer. This disease is a result of reciprocal
transiocation between 22 and 9 chromosome.
Note : If exchange of segments takes place in between homologous chromosomes then it is called
crossing over.
II. Gene Mutation or point mutation
Two types
A. Substitution
B. Frame shift mutation.
A. Substitution :
Replacement of one nitrogenous base by another nitrogenous base is called as substitution.
 It causes change in one codon in genetic code which leads to change in one amino acid in structure of
protein.
eg. Sickle cell anaemia
Change may not occur some time because for one animo acid more than one type of codons are
present.
Substitution is of two types
1. Transition :
Replacement of one purine by another purine or replacement of pyrimidine by another pyrimidine.
2. Transversion:
Replacement of purine by pyrimidine or pyrimidine by purine is called transversion.
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B. Frame shift mutation/Gibberish mutation :
Loss or addition of one or rarely more than one nitrogenous bases in structure of DNA.
Frame shift mutation is of two types
1. Addition
Addition of one or rarely more than one nitrogenous bases in structure of DNA.
2. Deletion
Loss of one or rarely more than one nitrogenous bases in structure of DNA.
Due to frame shift mutation complete reading of genetic code is changed. It leads to change in all animo
acids in structure of protein so a new protein is formed which is completely different from previous protein.
 So frame shift mutations are more harmful as compared to substitution, eg : Thallesemia (lethal genetic disorder)
MUTAGENS :
 Mutagens are those substances which cause mutations.
They are of two types
1. Physical mutagens
2. Chemical mutagens
1. Physical mutagens : Mainly includes radiations. Radiations are two types
i) Ionising  , ,  , X-ray
Ionising radiation like X-ray causes chromosomal aberrations and also induces transition. These radiations convert nitrogenous bases in their ions and ions perform forbidden pairing, which results in
transition.
ii) Non ionising U. V. rays.
U. V. rays has less penetration power and skin of higher organisms absorb radiations. So they don't
cause any effect in higher animals, but U. V. rays and radiations are effective mutagens in microbes
and due to more effect leads to death of microbes. So U. V. rays are used to sterilize operation theatre.
Radiations mainly cause chromosomal aberrations which cause major change in organisms. So chromosomal mutations are more important in evolution. By U.V. rays two adjacent thymine bind together
and form thymine-dimer.
2. Chemical mutagens : Chemical mutagens are more harmful than radiations because body is not
protected against chemicals.
Source of chemical mutagens are food, air and water.
Effect of radiation is localised, while chemical mutagens spread in complete body through blood
circulation and when they reach in gonads they cause germinal mutation.
Chemicals also cause chromosomal mutations.
Some chemical mutagens are
i) Mustard gas (first identified Chemical Mutagens)
ii) Nitrous acid (HNO2) HNO2 changes normal structure of nitrogenous base and changed nitrogenous base is called as Tautomer.
HNO2 cause deamination of nitrogenous base means they remove amino group from nitrogenous base.
- A* = T - T* = A -
I
- A* - C - T* - G -
II
-G C- CG -
deamination
deamination
deamination
Asenin 
 Hypoxanthine Guanin 
 Xanthin Cytosine 
 Uracil
In first DNA replication, Tautomer of adenine pairs with a normal cytosine and Tautomer of thymine
pairs with normal guanine.
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It is unusual pairing which is called as forbidden pairing so a wrong type of DNA is formed in cell.
In second DNA replication normal cytosine pairs with normal guanine and normal guanine pairs with
normal cytosine.
It is usual pairing so transition completes in two DNA replication (Tautomers always perform forbidden pairing)
iii) Base Analogues :
Those chemicals which are same as nitrogenous base in function. They are called base analogues or
duplicates of nitrogenous base.
eg. Aminopurine is base analogue to Adenine (purine) 5-Bromo uracil is base analogue to thymine
-A=T-T=A-
I
- A* = T - T* = A -
II
- A* - C - T* - G -
III
-G C- CG -
In I DNA replication base analogues get establish in normal structure of DNA.
In II DNA replication they perform forbidden pairing.
In III DNA replication transition is completed.
iv) Alkylating agents :
EMS = Ethyl methane sulphonate
MMS = Methyl methane sulphonate
These chemicals causes depurination means they remove one purine from structure of DNA. So a gap
is formed.
If this gap is filled by another purine then it is called as transition.
But if this gap is filled by pyrimidine then it is called as transversion.
So EMS and MMS may cause both transition and transversion.
v) Acredine and proflavin dyes:
They causes loss or addition of one or rarely more than one nitrogenous bases in structure of DNA,
Thus results in frame shift mutation.
vi) Other chemical mutagens
Carbon tetra sulphide Organic peroxide
Ethyl urethane
Colchicine
Pesticides
DDT (Dichloro Diphenyl Trichloro Ethane)
LSD (Lysergic acid diethylamide)
Antibiotics like Neomycin, Kenamycin , Streptomycin.





GOLDEN KEY POINTS
Mostly mutations are harmful.
Sometimes they are lethal which leads to death of organisms.
But sometimes they are beneficial which are used to obtain good varieties of plants and animals. It is
called as Mutation Breeding.
Mostly mutations are recessive and they never eliminate from a population.
Dominant lethal mutation always eliminate from a populaton either it is large or small.
Forward and Backward Mutation :
 Wild gene
Forward
Backward
Mutant gene
Muton (unit of mutation) :
 Smallest part of DNA which undergoes mutation.
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 It is one nucleotide.
Mis-sense mutation
 When a nucleotide change in genetic code cause the change of one amino acid of a polypeptide chain
it is called mis-sense mutation.
Non-sense mutation
 When a nucleotide change in one codon causes termination of polypepetide synthesis by producing
non-sense codon.
Same sense codon
 A change in one nucleotide in a codon does not change amino acid in polypeptide chain, because both
codons code same amino acid.
1. (2n-1) condition of chromosomes is called
1) tetrasomy
2) trisomy
3) monosomy
4) nullisomy
2. Given below is the representation of a kind of chromosomal mutation. It is
D
F
E
C
A B
A
D
C
B
E
F
G
1) Deletion
2) Inversion
3) Duplication
4) Reciprocal translocation
3. Which of the following has normal sex chromosome complements.
1) Down's syndrome
2) Klinefelter's syndrome (3) Super female
4) Turner's syndrome
4. Addition or deletion of a nitrogenous base causes
1) Frameshift mutation 2) Inversion
3) Transformation
4) Translocation
5. Trisomy of which chromosome is involved in Down's syndrome.
1) 8th
2) 13th
3) 21st
4) 22nd
1) 3





2) 2
3) 1
4) 1
5) 3
DNA FINGER PRINTING / DNA TYPING / DNA PROFILING / DNA TEST
It is technique to identify a person on the basis of his/her DNA specificity.
This technique was invented by sir Alec. Jeffery (1984).
In India DNA Finger printing has been started by Dr. V.K. Kashyap & Dr. Lai Ji Singh.
DNA of human is almost the same for all individuals but very small amount that differs from person to
person that forensic scientists analyze to identify people.
These differences are called Polymorphism (many forms) and are the key of DNA typing. Polymorphisms are most useful to forensic scientist. It is consist of variation in the length of DNA at specific
loci is called Restricted fragment. It is most important segment for DNA test made up of short repeti-
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





1.
2.
3.




tive nucleotide sequences, these are called VNTRs (variable number of tandem repeat).
VNTR's also called minisatellites were discovered by Alec Jeffery. Restricted fragment consist of
hypervariable repeat region of DNA having a basic repeat sequence of 11-60 bp and flanked on both
sites by restriction site. The number and position of minisatellites or VNTR in restriction fragmnt is
different for each DNA and length of restricted fragment is depend on number of VNTR.
Therefore, when the genome of two people are cut using the same restriction enzyme the length of
fragments obtained is different for both the people.
These variations in length of restricted fragment is called RFLP or Restriction fragment length polymorphism.
Restriction Fragment Length Polymorphism distributed throughout human genomes are useful for
DNA Finger printing.
DNA Fingerprint can be prepared from extremely minute amount of blood, semen, hair bulb or any
other cell of the body.
DNA content of 1 - Microgram is sufficient.
Technique of DNA Finger printing involves the following major stpes.
Extraction :
DNA extracted from the cell by cell lysis. If the content of DNA is limited then DNA can be amplified
by Polymerase chain reaction (PCR). This process is amplification.
Restriction Enzyme Digestion :
Restriction enzyme cuts DNA at specific 4 or 6 base pair sequences called restriction site.
Hae III (Haemophilus aegyptius) is most commonly used enzyme. It cuts the DNA, every where the
bases are arranged in the sequence GGCC. These restricated fragment transferred to Agarose Polymer
gel.
Gel Electrophoresis :
Gel electrophoresis is a method that separates macromolecules-either nucleic acid or proteins-on the
basis of size, electric charge.
A gel is a colloid in a solid form. The term electrophoresis describes the migration of charged particles
under the influence of an electric field. Electro refers to the energy of electricity. Phoresis, from the
Greek verb phoros, means "to carry across." Thus, gel electrophoresis refers to the technique in which
molecules are forced across a span of gel, motivated by an electrical current. Activated electrodes at
either end of the gel provides the driving force. A molecule's properties determine, how rapidly an
electric field can move the molecule through a gelatinous medium.
Many important biological molecules such as amino acids, peptides, proteins, nucleotides, and nucleic
acids posses ionisable groups and, therefore, at any given pH, exist in solution as electrically charged
species either as cation (+) or anions (-). Depending on the nature of the net charge, the charged
particles will migrate either to the cathode or to the anode.
By the gel electrophoresis these restricted fragments move towards the positive electrode (anode)
because DNA has -ve electric charge ( PO-34 ).
 Smaller Fragment more move towards the positive pole due to less molecular weight. So after the gel
electrophoresis DNA fragment arranged according to molecular weight.
 These separated fragments can be visualized by staining them with a dye that fluoresces ultraviolet
radiation.
4. Southern transfer / Southern blotting :
The gel is fragile. It is necessary to remove the DNA from the gel and permanently attaches it to a solid
support. This is accomplished by the process of Southern blotting. The first step is to denature the
DNA in the gel which means that the double-stranded restriction fragments are chemically separated
into the single stranded form. The DNA then is transferred by the process of blotting to a sheet of
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nylon. The nylon acts like an ink blotter and "blots" up the separated DNA fragments, the restriction
fragments, invisible at this stage are irreversibly attached to the nylon membrane the "blot".
This process is called Southern blot by the name of Edward Southern (1970).
5. Hybridization : To detect VNTR locus on restricted fragment, we use single stranded Radioactive
(P32) DNA probe which have the base pair sequences complimentary to the DNA sequences at the
VNTR locus. Commonly we use a combination of at least 4 to 6 separate DNA probes.
Labelled Probes are attached with the VNTR loci of restricted DNA Fragments, this process is called
Hybridization.
RESTRICTION
ENZYME
DIGESTION
GEL ELECTROPHORESIS
(Agargose polymer gel)
DENATURATION
BY ALKALI SOLUTION
SOUTHERN BLOTTING
NYLON OR NITROCELLULOSE
SHEET
HYBRIDIZATION
DNA PROBES
STEPS OF DNA FINGERPRINTING
6.
Autoradiography : Nylon membrane containing radio active probe exposed to X-ray. Specific bands
appear on X-ray film. These bands are the areas where the radioactive probe bind with the VNTR.
 This appears the specific restricted fragment length pattern. This length pattern is different in different
individual. This is called Restricted Fragment length Polymorphism (RFLP).
 These allow analyzer to identify a particular person DNA, the occurance and frequency of a particular
genetic pattern contained in this X-ray film. These x-ray film called DNA signature of a person which
is specific for each individual.?
 The probability of two unrelated individual having same pattern of location and repeat number of
minisatellite (VNTR) is one in ten billion (world population 6.1 billion)
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In India the centre for DNA finger printing and diagnosis (CDFD - center for DNA finger printing &
diagnosis) located at Hyderabad.
Application of DNA Finger printing
1. Paternity tests : The major application of DNA finger printing is in determining family relationships.
For identifying the true (biological) father, DNA samples of Child, mother and possible fathers are
taken and their DNA finger prints are obtained. The prints of child DNA match to the prints of biological parents.
2. Identification of the criminal : DNA finger printing has now become useful technique in forensic
(crime detecting) science, specially when serious crimes such as murders and rapes are involved. For
identifying a criminal, the DNA fingerprints of the suspects from blood or hair or semen picked up
from the scene of crime are prepared and compared. The DNA fingerprint of the person matching the
one obtained from sample collected from scene of crime can give a clue to the actual criminal.
Schematic representation of DNA fingerprinting : Few representative chromosomes have been
shown to contain different copy number of VNTR. For the sake of understanding colour
schemes have been used to trace the origin of each band in the gel. The two alleles (paternal
and maternal) of chromosome also contain different copy numbers of VNTR. It is clear that
the banding pattern of DNA from crime sceme matches with individual B, and not with A.
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HUMAN GENOME PROJECT
Genetic make-up of an organism or an individual lies in the DNA sequences. If two individuals differ,
then their DNA sequences should also be different, at least at some places. These assumptions led to
the quest of finding out the complete DNA sequence of human genome. With the establishment of
genetic engineering techniques where it was possible to isolate and clone any piece of DNA and
availability of simple and fast techniques for determining DNA sequences, a very ambitious project of
sequencing human genome was launched in the year 1990.
Human Genome Project (HGP) was called a mega project. You can imagine the magnitude and the
requirements for the project if we simply define the aims of the project as follows :
Human genome is said to have approximately 3 x 109 bp, and if the cost of sequencing required is US $
3 per bp (the estimated cost in the beginning), the total estimated cost of the project would be aproximately
9 billion US dollars. Further, if the obtained sequences were to be stored in typed form in books, and if
each page of the book contained 1000 letters and each book contained 1000 pages, then 3300 such books
would be required to store the information of DNA sequence from a single human cell. HGP was closely
associated with the rapid development of a new area in biology called as Bioinformatics.
Goals of HGP
Some of the important goals of HGP are as follows :
i) Identify all the genes in human DNA.
ii) Determine the sequences of the 3 billion chemical base pairs that make up human DNA.
iii) Store this information in databases.
iv) Improve tools for data analysis.
v) Transfer related technologies to other sectors, such as industries.
vi) Address the ethical, legal, and social issues (ELSI) that may arise from the project.
The project was completed in 2003. Knowledge about the effects of DNA variations among individuals can lead to revolutionary new ways to diagnose, treat and someday prevent the thousands of disorders that affect human beings. Besides providing clues to understanding human biology, learning
about non-human organisms, DNA sequences can lead to an understanding of their natural capabilities
that can be applied toward solving challenges in health care, agriculture, energy production, environmental remediation. Many non-human model organisms, such as bacteria, yeast, Caenorhabditis elegans
(a freeliving non-pathogenic nematode), Drosophila (the fruit fly), plants (rice and Arabidopsis), etc.,
have also been sequenced.
Methodologies : The methods involved two major approaches. (1) Expressed Sequence Tags (ESTs)
- Identifying all the genes that expressed as RNA. (2) Sequence Annotation - The blind approach of
simply sequencing the whole set of genome that contained all the coding and non-coding sequence,
and later assigning different regions in the sequence with functions. For sequencing, the total DNA
from a cell is isolated and converted into random fragments of relatively smaller sizes (recall DNA is
a very long polymer, and there are technical limitations in sequencing very long pieces of DNA) and
cloned in suitable host using specialised vectors. The cloning resulted into amplification of each piece
of DNA fragment so, that is subsequently could be sequenced with ease. The commonly used hosts
were bacteria and yeast, and the vectors were called as BAC (bacterial artificial chromosomes), and
YAC (yeast artificial chromosomes).
The fragments were sequenced using automated DNA sequencers that worked on the principle of a
method developed by Frederick Sanger. (Remember, Sanger is also credited for developing method
for determination of amino acid sequences in proteins). These sequences were then arranged based on
some overlapping regions present in them. This required generation of overlapping fragments for
sequencing. Alignment of these sequences was humanly not possible. Therefore, specialised computer
based programmes were developed. These sequences were subsequently annotated and were assigned
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to each chromosome. The sequence of chromosome I was completed only in May 2006 (this was the
last of the 24 human chromosomes -22 autosomes and X and Y- to be sequenced). Another challenging
task was assigning the genetic and physical maps on the genome. This was generated using information on polymorphism of restriction endonuclease recognition sites, and some repetitive DNA sequences known as microsatellites.
Salient Features of Human Genome Some of the salient observations drawn from human genome project are as follows :
i) The human genome contains 3164.7 million nucleotide bases.
ii) The average gene consists of 3000 bases, but sizes vary greatly, with the largest known human gene
being dystrophin at 2.4 million bases.
iii) The total number of genes is estimated at 30,000-much lower than previous estimates of 80,000 to
1,40,000 genes. Almost all (99.9 per cent) nucleotide bases are exactly the same in all people.
iv) The functions are unknown for over 50 per cent of discovered genes.
v) Less than 2 per cent of the genome codes for proteins.
vi) Repeated sequences make up very large portion of the human genome.
vii) Repetitive sequences are stretches of DNA sequences that are repeated many times, sometimes
hundred to thousand times. They are thought to have no direct coding functions, but they shed light on
chromosome structure, dynamics and evolution.
viii) Chromosome 1 has most genes (2968). and the Y has the fewest (231).
ix) Scientists have identified about 1.4 million locations where single-base DNA differences (SNPssingle nucleotide polymorphism, pronounced as ‘snips’) occur in humans, This information promises
to revolutionise the processes of finding chromosomal locations for disease-associated sequences and
tracing human history.
Organisms
Base pair
Gene No.
Bateriophage
10,000
----\---Lily
106 Billion B.P.
-------E.coli
4.7 million B.P.
4,000
S. cerviceae
12 Million B.P.
6,000
D. melangaster
180 Million B.P.
13,000
Caenorhabditis elegans 97 Million B.P.
18,000
Human
3 Billion B.P.
30,000
a) First prokaryotes in which complete genome was sequenced is Haemophilus influenzae.
b) First Eukaryote in which complete genome was sequenced is Saccharomyces cerviceae (Yeast).
c) First plant in which complete genome was sequenced is Arabidopsis thaliana (Small mustard plant).
d) First animal in which complete genome was sequenced is Caenorhabditis elegans (Nematode).
  - globin and insulin gene are less than 10 kilo base pair T.D.F. gene is the smallest gene (14 base
pair) and Duchenne muscular Dystrophy gene is made up of 2400 kilo base pair.(Longest gene)
1. DNA finger printing involves identifying differences in some specific
1) Repetitive DNA
2) Non repetitive DNA 3) Selfish DNA
2. Which of the following is produced by E-Coli in the lactose operon.
1) B galactosidase
2) Transacetylase
3) Permease
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4) All of the above
4) All of the above
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3. Maximum number of gene present on which chromosome number in human.
1) 1st
2) X
3) Y
4) 10th
4. In lac operon RNA polymerase binds with
1) Promoter gene
2) Opertor gene
3) Structural gane
4) Regulator gene
5. Fill the gap in following statement
Human genome have approximately _________ and the cost of sequencing was _______ per
base pair.
1) 4 x 109 bp, 9 billion US dollars
2) 9 billion US dollars, 4 x 19 9bp
4) 4.7 million bp, 9 billion US dollars
3) 3 x 109bp, 3 US dollars
1) 3
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
2) 3
3) 1
4) 4
5) 2
MENDELISM
On which plant Mendel had carried out his investigations
1) Garden - pea
2) Wild pea
3) Cow-pea
4) Pigeon pea
During breeding the removal of anthers from a flower is called
l)Anthesis
2) Pollination
3) Emasculation
4) Vasectomy
When a heterozygous tall pea plant of Fj generation upon self fertilization produces tall and
dwarf phenotypes it proves the principle of
1) Dominance
2) Segregation
3) Independant assortement
4) Inheritance & purity of gametes
Mendel formulated the law of purity of gametes on the basis of
1) Dihybrid cross
2) Monohybrid cross
3) Back cross
4) Test cross
A cross between AaBB X aaBB yields a genotypic ratio of
1) 1 AaBB: 1 aaBB
2) 1 AaBB : 3 aaBB
3) 3AaBB : 1 aa BB
4) All AaBb
In monohybrid cross what is the ratio of homozygous dominant and homozygous recessive individuals in F2-generation
1) 1:2:1
2) 2:1/1:2
3) 3:1/1:3
4)1:1
The cross between recessive to it's hybrid or it's F1 plant is called
1) Back cross
2) Test cross
3) Monohybrid cross
4) Dihybrid cross
What is the genotypic and phenotypic ratio of monohybrid test cross
1) 1:1
2) 1:2
3) 3:1
4) 1:2:1
Dihybrid cross proves the law of
1) 1:1
2) 1:2
3) 3:1
4) 1:2:1
How many types & in what ratio the gametes are produced by a dihybrid heterozygous
1) 4 types in the ratio of 9:3:3:1
2) 2 types in the ratio of 3:1
3) 3 types in the ratio of 1:2:1
4) 4 types in the ratio of 1:1:1:1
How many gametes are produced in F1 generation of a trihybrid
1) 3
2) 4
3) 8
4) 16
Which genotype represents a true dihybrid condition
1) tt rr
2) Tt rr
3) Tt Rr
4) TT Rr
Mendelian ratio 9:3:3:1 is due to
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14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
1) Law of segregation
2) Law of purity of gametes
3) Law of independent assortment
4) Law of unit characters
In a cross between a pure tall plant with green pod & a pure short plant with yellow pod. How
many short plants are produced in F2 generation out of 16
1) 1
2) 3
3) 4
4) 9
In a dihybrid cross between AABB and aabb the ratio of AABB, AABb, aaBb, aabb in F 2 generation is
1) 9:3:3:1
2) 1:1:1:1
3) 1:2:2:1
4)1:1:2:2
AABbCc genotype forms how many types of gametes
1) 4
2) 8
3) 2
4) 6
Who rediscovered the results of Mendel's experiments
1) DeVries, Tschemark, Correns
2) DeVries, Tschemark, Morgan
3) Tschemark, Morgan, Correns
4) Tschemark, Bateson, Punnet
Crossing AABB & aabb, the probability of AaBb would be in F 2 generation
1) 1/16
2) 2/16
3) 8/16
4) 4/16
In Mendel's experiments, colour of seed coat, nature of flower, position of flower, colour of pod,
height of stem, are called
1) Alleles
2) Genotype
3) Phenotype
4) All of the above
If 120 Plants are produced on crossing pure red and pure white flowered pea plants, than the
ratio of offsprings will be
1) 90 Red : 30 White
2) 30 Red : 90 White
3) 60 Red : 60 White
4) All Red
Pea plants were more suitable than cattle for Mendel's experiment because
1) There were no breeding records of cattles
2) Pea plants can be self-fertilised
3) Cattle are not easy to mantain
4) All pea plants have 2n chromosomes and fewer genetic traits
An individual with two identical members of a pair of genetic factors is called
1) Heteromorphic
2) Heterozygote
3) Homomorphic
4) Homozygote
Two allelic genes are located on
1) The same chromosome
2) Two homologous chromosomes
3) Two-non-homologous chromosomes
4) Any two chromosomes
The percentage of ab gametes produced by AaBb parent will be
1) 12.5
2) 25
3) 50
4) 75
How many character of pea pod were chosen by Mendel
1) 7
2) 5
3) 4
4) 2
Mendel's law of segregation is based on separation of alleles during
1) Gemete formation
2) Seed formation
3) Pollination
4) Embryonic development
In a cross 45 tall & 14 dwarf plants were obtained, the genotype of parents was
1) TT X TT
2) TT X Tt
3) Tt X Tt
4) TT X tt
When two hybrids Ttrr & Rrtt are crossed, the phenotypic ratio of offspring shell be
1) 3:1
2) 1:1:1:1
3) 1:1
4) 9:3:3:1
The allele which is unable to express its effect in the presence of another is called
1) Co-dominant
2) Supplementary
3) Complementary
4) Recessive
Which technique is used by Mendel for hybridisation
1) Emasculation
2) Bagging
3) Protoplast fusion
4) 1 & 2 both
Dihybrid plants forms how many types of pollen grains
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1) One
2) Two
3) Four
4) Eight
32. When flowers are unisexual then emasculation is done in
1) Female
2) Male
3) 1 & 2 both
4) None of these
33. How many plants are dihybrid in F2 generation of dihybrid cross
1) One
2) Two
3) Four
4) Sixteen
34. Mendel's conclusion first published in
1) Journal of plant breeding
2) Journal of genetics & plant breeding
3) Nature forschender verein
4) None
35. When a plant have two alleles of contrasting characters it is called
1)Homozygous
2) Dioecious
3) Heterozygous
4) Monoecious
36. From a single ear of corn, a farmer planted 200 kernels which produced 140 tall & 40 short
plants. The genotypes of these offsprings are most likely
1) TT, Tt & tt
2) TT & tt
3) TT & Tt
4) Tt & tt
37. A useful process for determining whether an individual is homozygous or heterozygous is
1) Cross-breeding
2) Self fertilization
3) Back - crossing
4) Test cross
38. Heterozygous tall plants were crossed with dwarf plants, what will be the ratio of dwarf plants
in the progeny
1) 50%
2) 25%
3) 75%
4) 100%
39. A pure tall plant can be differentiated from a hybrid tall plant
1) By measuring length of plant
2) By spraying gibberalins
3) If all plants are tall after self-pollination
4) If all plants are dwarf after self-pollination
40. If the cell of an organism heterozygous for two pairs of genes represented by Aa, Bb, undergoes
meiosis, then the possible genotypic combination of gametes will be
1) AB, Ab, aB, ab
2) AB, ab
3) Aa, Bb
4) A, a, B, b
41. Allele is the
1) Alternative form of gene pair
2) Total number of genes for a trait
3) Total number of chromosomes of a haploid set
4) Total number of genes present on a chromosome
42. Genetic constitution of an individual is represented by
1) Genome
2) Genotype
3) Phenotype
4) Karyotype
43. Genes do not occur in pairs in
1) Zygote
2) Somatic cell
3) Embryo
4) Gametes
44. "Like begets like" an important and universal phenomenon of life, is due to
1) Eugenics
2) Inheritance
3) dominance
4) Crossing-over
45. How many types of gametes are expected from the organism with genotype AABBCC
1) One
2) Two
3) Four
4) Eight
46. One of the following did not constitute the seven contrasting pairs of characters noticed by
Mendel
1) Height of the plants 2) Shape of the leaves
3) Shape of pod
4) Colour of pod
47. According to Mendelism which character is showing dominance
1) Terminal position of flower
2) Green colour in seed coat
3) Wrinkled seeds
4) Green pod colour
48. Due to the cross between TTRr X ttrr the resultant progenies showed how many percent plants
would be, tall, red flowered
1) 50%
2) 75%
3) 25%
4) 100%
49. Mendel obtained wrinkled seeds in pea due to deposition of sugars instead of starch. It was due
to which enzyme
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50.
51.
52.
53.
54.
55.
56.
57.
58.
59.
60.
61.
62.
63.
64.
1) Amylase
2) Invertase
3) Diastase
4) Absence of starch branching enzyme
A gene said to be dominant if
1) It express it's effect only in homozygous stage.
2) It expressed only in heterozygous condition
3) It expressed both in homozygous and heterozygous condition.
4) It never expressed in any condition.
A plant of F1-generation with genotype "AABbCC". On selfing of this plant what is the phenotypic ratio in F2-generation
1) 3 : 1
2) 1 : 1
3) 9 : 3 : 3 : 1
4) 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1
Which one of the following traits of garden pea studied by Mendel, was a recessive feature
1) Axial flower position 2) Green seed colour
3) Green pod colour
4) Round seed shape
A trihybrid cross is made between two plants with genotypes A/a B/b C/c how many offsprings
of such cross will have a genotype a/a b/b c/c
1) 1/64
2) 1/4
3) 1/16
4) 1/32
How is the arrangement of Mendel's selected seven characters on four chromosomes
1) One in ch. no. 1, 4 in ch. no. 4, one in ch. no. 5, and one in ch. no. 7
2) 2 in ch. no. 1, 3 in ch. no. 4, one in ch. no. 5 and one in ch. no. 6
3) 3 in ch. no. 1, 1 in ch. no. 4, 2 in ch. no. 5 and one in ch. no. 7
4) 2 in ch. no. 1, 3 in ch. no. 4, 1 in ch. no.5and 1 in ch. no. 7
When two different genotypes produce the same phenotype due to environmental difference,
then each one is known as
1) Phenotype
2) Phenocopy
3) Progeny
4) Independent offspring
When a red flower homozygous pea plant is crossed with a white flower plant what colour is
produced
1) Red
2) White
3) Pink (4) Red + white
If a heterozygous tall plant is crossed with a homozygous dwarf plant then what shall be the
percentage of dwarf in offspring
1) 25%
2) 100%
3) 75%
4) 50%
If a homozygous tall plant is crossed with a dwarf plant, what shall be the ratio of plants in
offsprings
1) All heterozygous tall
2) Two tall & Two dwarf
3) 1:2:1
4) All homozygous dwarf
How many different types of gametes can be formed by Fj progeny, resulting from the following
cross : AA BB CC x aa bb cc
1) 3
2) 8
3) 27
4) 64
In order to find out the different types of gametes produced by a pea plant having the genotype
AaBb, it should be crossed to a plant with the genotype
1) AaBb
2) aabb
3) AABB
4) aaBB
Law of independent assortment of Mendel was proved by
1) Monohybrid cross
2) Reciprocal cross
3) Dihybrid cross
4) Back cross
Mendel does not select which character in his experiment
1) Plant height
2) Plant colour
3) Pod shape
4) Pod colour
Genes controlling seven traits in pea studied by Mendel were actually located on
1) Seven chromosomes 2) Six chromosomes
3) Four chromosomes
4) Five chromosomes
Two crosses between the same pair of genotypes or phenotypes in which the sources of the ga76
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66.
67.
68.
69.
70.
71.
72.
73.
74.
75.
76.
77.
78.
79.
metes are reversed in one cross, is known as
1) Test cross
2) Reciprocal cross
3) Dihybrid cross
4) Reverse cross
If selfing occurs in the plant having genotype RrYy, then ratio of given genotype will be RRYY,
RrYY, RRYy, RrYy
1) 1:2:2:4
2)1:2:2:1
3) 1:1:1:1
4)2:2:2:1
The process of mating between closely related individuals is
1) Out-breeding
2) Inbreeding
3) Hybridisation
4) Heterosis
Marriages between close relatives should be avoided because it includes more
1) Recessive alleles to come together
2) Mutations
3) Multiple births
4) Blood group abnormalities
A self-fertilizing trihybrid plant forms
1) 8 different gametes and 32 different zygotes
2) 8 different gametes and 64 different zygotes
3) 4 different gametes and 16 different zygotes
4) 8 different gametes and 16 different zygotes
Segregation of genes take place during
1) Metaphase
2) Anaphase
3) Prophase
4) Embryo formation
An inherited character and its detectable variant is termed as
1) unit factor
2) trait
3) genctic profile
4) genotypic character
A trihybrid cross involve three pair of characters which will give rise to the Fj hybrids which are
heterozygous for three genes. How many types of gametes will be produced in both male and
female1) 2
2) 4
3) 6
4) 8
When an F1 individual is crossed with its either of the two parent. Then it is known as
1) Test cross
2) Back cross
3)Reciprocal cross
4) Monohybrid cross
If a homozygous red flowered plant is crossed with white plant, the offspring will be
1) All red flowered
2) All white flowered
3) Half red flowered
4) Half white flowered
How many types of genotypes are formed in F2 progeny obtained from self polination of a
dihybrid F1
1) 9
2) 3
3) 6
4) 1
If a dwarf plant is treated with gibberellins it becomes tall and this plant now crosses with pure
tall plant then progeny of first generation (F1) is
1) All dwarf
2) All tall
3) 75% tall and 25% dwarf
4) 75% dwarf and 25% tall
A test cross is performed
1) by selfing of F2-generation plants
2) by selfing of F1 generation plants
3) to determine whether F1-plant is homozgous or heterozygous
4) between a homozygous dominant and homozygous recessive plant
If a cross is made between AA and aa, the nature of Fi progeny will be
1) genotypically AA, phenotypically a
2) genotypically Aa, phenotypically a
3) genotypically Aa, phenotypically A
4) genotypically aa, phenotypically A
When a tall plant with round seeds (TTRR) is crossed with a dwarf plant with wrinkled seeds
(ttrr), the F1 generation consists of tall plants with rounded seeds. How many types of gametes F 1
plant would porduce
1) One
2) Three
3) Four
4) Eight
A pure tall and a pure dwarf plant were crossed to produce offsprings. Offsprings were self
crossed, then find out the ratio between true breeding tall to true breding dwarf
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80.
81.
82.
83.
84.
85.
86.
87.
88.
89.
90.
91.
92.
93.
94.
95.
1) 1 : 1
2) 3 : 1
3) 2 : 1
4) 1 : 2 : 1
If hybrid red flowered plants of pea are crossed back to pure red flowered parent, the progeny
will show
1) All red flowered plants
2) White flowered plants
3) 50% red and 50% white flowered plants
4) 3 Red : 1 white flowered plants
Mendel's Law was establisted when
1) Two F1 hybids are crossed
2) One parent is crossed with F1 hybrid
3) Two pure breeding contrasting characers are crossed
4) None of the above
Mendel's laws of inheritance are applicable on the plants which
1) Reproduce asexually
2) Reproduce sexually
3) Reproduce vegetatively
4) All of the above plants
A cross used to ascertain whether a dominant is homozygous or heteroxygous
1) Reciprocal
2) Back cross
3) Test cross
4) Monohybrid
Dihybrid test cross ratio proposed by Mendel is
1) 9:3:3:1
2) 1:1:1:1
3) 1:2:2:4:1:2:1:2:1
4) 3:1
A woman with albinic father marries an albinic man. The proportion of her progeny is
1) 2 normal : 1 albinic 2) All normal
3) All albinic
4) 1 normal : 1 albinic
A cross between pure tall pea plant with green pods and dwarf pea with yellow pods will produce tall F2 plants, out of 16,
1) 15
2) 13
3) 12
4) 7
Mendel's Principle of segregation means that the germ cells always receive
1) one pair of alleles
2) one quarter of the genes
3) one of the paired alleles
4) any pair of alleles
How many types of genetically different gametes will be produced by a heterozygous plant having the genotype AABbCc?
1) Two
2) Four
3) Six
4) Nine
Two allelic genes are located on
1) the same chromosome
2) two homologous chromosome
3) two non-homologous chromosome
4) any two chromosomes
The phenotypic ratio in a back cross between a trihybrid and homozygous recessive parent
would be
1) 1:1
2) 1:1:1:1
3) 1:1:1:1 : 1 : 1
4) 1:1:1:1:1:1:1:1
Among the seven pairs of contrasting traits in pea plant, studied by Mendel, the number of
traits related to flower, pod and seed respectively was
1) 2, 2, 2
2) 2, 2, 1
3)1, 2, 2
4) 1, 1, 2
Some of the dominant traits studied by Mendel were
1) Round seed shape, constricted pod shape and axial flower position
2) Yellow seed colour, inflated pod shape and axial flower position
3) Yellow seed colour, violet flower colour and yellow pod colour
4) Axial flower position, green pod colour and green seed colour
The colour based contrasting traits in seven contrasting pairs studied by Mendel in pea were
1) 1
2) 2
3) 3
4) 4
Mendel observed that all the F1 progeny plants.
1) resembled either one of the parents
2) resembled neither of the parents
3) resembled both of the parents
4) shows 3 : 1 ratio
Accoding to Mendel, "factors" or "genes"
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1) are the units of inheritance
2) contain information that is required to express a particular trait
3) Both 1 and 2
4) None of the above
96. The segregration of alleles is a random process and so there is a chance of a gametes containing
either allele.
1) 25%
2) 50%
3) 75%
4) 100%
97. The phenotype of any character will not be affected if the modified allele produces1) Normal enzyme
2) Non-functional enzyme
3) No-enzyme at all
4) 2 and 3 both
98. Gene which code for a pair of contrasting traits are known as 1) Allele
2) Non allele
3) Pseudoallele
4) Isoallele
99. The modified allele is equivalent to the unmodified allele when it produces1) A non functional enzyme
2) No enzyme
3) The normal enzyme
4)All the above
100. The recessive characters are 1) Only expressed in heterozygous condition
2) Only expressed in homozygous condition
3) Blend in heterozygous condition
4) Always impure
ALLELIC & NONALLELIC GENE INTERACTIONS
101. In Mirabilis & Antirrhinum plant the appearence of the pink hybrid (Rr) between cross of a red
(RR) and white (rr) flower parent indicates
1) Incomplete dominance
2) Segregation
3) Dominance
4) Heterosis
102. RR(red) is crossed with rr (white). All Rr offsprings are pink. This indicates that R-gene is
1) Hybrid
2) Incompletely dominant
3) Recessive
4) Mutant
103. In case of incomplete dominance the monohybrid ratio of phenotypes in F 2 generation is
1) 1:2:1
2) 3:1:1
3) 9:3:3:1
4) 2:3:1
104. A white flowered mirabilis plant rr was crossed with red coloured RR, If 120 plants are produced in F2 generation. The result would be
1) 90 uniformly coloured & 30 white
2) 90 Non-uniformly coloured & 30 white
3) 60 Non-uniformly coloured & 60 white
4) All coloured & No white
105. When the phenotypic and genotypic ratios resemble in the F2 generation it is an example of
1) Independant assortment
2) Qualitative inheritance
3) Segregation of factors
4) Incomplete dominance
106. In Mirabilis jalapa when homozygous red flowered and white flowered plants are crossed, all Fx
plants have pink coloured flowers. In F2 produced by selfing of F1 plants, red, pink, white flowered plants would appear respectively in the ratio of
1) 1:1:2
2) 2:1:1
3) 1:0:1
4) 1:2:1
107. In case of incomplete dominance, F2 generation has
1) Genotypic ratio equal to phenotypic ratio
2) Genotypic ratio is 3:1
3) Phenotypic ratio is 3:1
4) None
108. Incomplete dominance occurs in .1) Mirabilis
2) Antirrhinum
3) Andulasion fowl
4) All of the above
109. Which cross yields red, white & pink flowers variety of dog flower
1) RR X Rr
2) Rr X RR
3) Rr X Rr
4) Rr X rr
110. What shall be ratio in offspring when a roan cow is crossed with a white bull
1) 1:2:1
2)3:1
3) 1:1
4) All roan
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111. Which of the following is exception to Mendel's laws
1) Linkage
2) Incomplete dominance
3) Co-dominance
4) All of the above
112. In a dihybrid cross, when one pair of alleles show incomplete dominance, genotypic ratio comes
to
1) 3:6:3:1:2:1
2) 1:2:2:4:1:2:1:2:1
3) 9:3:3:1
4) 1:2:1
113. Which of the following is the example of codominance
1) HbA HbA, IA IB
2) Hbs Hbs , IA IB
3) HbA Hbs, IA IB
4) Hbs Hbs, IAIA
114. Which of the following conditions represent a case of co-dominant genes.
1) A gene expresses itself, suppressing the phenotypic effect of its alleles
2) Genes that are similar in phenotypic effect when present separately, but when together interact to
produce a different trait
3) Allele, both of which interact to produce a trait, which may resemble either of the parental type.
4) Alleles, each of which produces an independent effect in heterozygous condition.
115. The phenomenon of incomplete dominance was observed by
1) Devries
2) Correns
3) Tschermak
4) None
116. Mendel did not propose
1) Dominance
2) Incomplete dominance
3) Segregation
4) Independent assortment
117. A gene which suppresses the effect of another gene not located on the similar locus of the homologous chromosomes
1) Duplicate gene
2) Complementary gene
3) Epistatic gene
4) Supplementary gene
118. The phenomenon in which an allele of one gene suppresses the expression of an allele of another
gene is known as
1) Dominance
2) Inactivation
3) Epistasis
4) Suppression
119. When two independentaly assorting dominant genes interact with each other to produce particular phenotype but when they present alone they did not produce phenotype they are called
1) Complementary gene
2) Supplementary gene
3) Duplicate gene
4) Inhibitory gene
120. AB - Blood group shows
1) Co-dominance
2) Complete dominance
3) Mixed inheritance
4) Composite inheritance
121. ABO blood group is an example of
1) Epistasis
2) Multiple allelism
3) Pleotropism
4) Complementary genes
122. A child is blood group is 'O'. His parents blood group can not be
1) B & O
2) A & O
3) AB
4) A & B
123. If one parent has blood group A and the other parent has blood group B. The offsprings have
which blood group
1) AB only
2) O only
3) B only
4) A, B, AB, O
124. Ratio 9:7 is due to
1) Supplementary genes 2) Lethal genes
3) complementary genes 4) Epistatic genes
125. A man of A blood group marries a woman of AB blood group, which type of progeny would
indicate that man is heterozygous A
1) AB
2) A
3) O
4) B
126. A child of O blood group, has B-blood group father, the genotype of father would be
2) IBIB
3) IAIB
4) IBI0
1) I0I0
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127. When a red flowered plant was cross pollinated by white flowered one and the offspring were
self pollinated to obtain a phenotypic ratio of 1:2:1, it has to be a case of
1) Incomplete dominance
2) Dominance
3) Recessive epistasis
4) Pleurotropic effect of genes
128. Andalucian fowl exhibits
1) Phonotypic blending
2) Mosaic inheritance
3) Epistasis
4) Co-dominance
129. Epistatic gene differs from dominant gene in
1) Epistatic gene is non-allelic
2) Epistatic gene never express itself independently
3) Epistatic and hypostatic genes are present at different loci
4) All the above
130. A gene that shows it's effect on more than one character is
1) Polygene
2) Pleotropic gene
3) Multifactor gene
4) Multiple gene
131. In multiple allele system a gamete possesses
1) Two alleles
2) Three alleles
3) One allele
4) Several alleles
132. Blood grouping in humans is controlled by
1) 4 alleles in which IA is dominant
2) 3 alleles in which IA and IB are dominant
3) 2 alleles in which none is dominant
4) 3 alleles in which IA is recessive
133. Multiple alleles are present
1) In different chromosomes
2) At different loci on chromosome
3) At the same locus on homologous chromosomes4) At the non homologous chromosome
134. Which of the following is the example of pleiotropic gene
1) Haemophilia
2) Thalaessemea
3) Sickle cell anaemia
4) Colour blindness
135. Epistasis differs from dominance because
1) In epistasis one gene pair mask the expression of another pair of genes
2) Epistasis is an allelic interaction
3) Many genes collectively controls a particular phenotype
4) One gene pair independently controls a particular phenotype
136. A) Pleiotropic genes have multiple phenotypic effect.
B) Muliple alleles exhibit same phenotypic expression.
C) Polygenes exhibit continuous variation.
1) Statement (A), (B) and (C) are correct
2) Statement (A), (C) correct and (B) is incorrect
3) Statement (A), (B) and (C) are incorrect
4) Statement (B) and (C) are correct and (A) is incorrect
137. In a genetic cross having recessive epistasis, F2 phenotypic ratio would be
1) 9:6:1
2)15:1
3) 9:3:4
4)12:3:1
138. Sickle cell anaemia induces due to
1) Change of Amino Acid in a - chain of Haemoglobin
2) Change of Amino Acid in P - chain of Haemoglobin
3) Change of Amino Acid in both a and P chain of Haemoglobin
4) Change of Amino acid either a or p chain of Haemoglobin
139. What would be the colour of flower in F1 progeny as a result of cross between homozygous red
and homozygous white flowered Snapdragon
1) Red
2) White
3) Red and White
4) Pink
140. Incomplete dominance is found in
1) Pisum sativum
2) Antrrhinum majus
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3) Both Pisum sativum and Antirrhinum majus
4) None of these
141. In Mirabilis red (RR) and white (rr) flower produces pink (Rr) flower. A plant with pink flower
is crossed with white flower the expected phenotypic ratio is
1) red : pink : white (1 : 2: 1)
2) pink : white (1 : 1)
3) red : pink (1:1)
4) red : white (3 : 1)
142. A child with mother of 'A' blood group and father of AB' blood group will be
1) O
2) A
3) A and O
4) O and B
143. Epistasis implies
1) One pair of genes can completely mask the expression of another pair of genes
2) One pair of genes independently controls a particular phenotype
3) One pair of genes enhances the phenotypic expression of another pair of genes
4) Many genes collectively control a particular phenotype
144. The possible blood groups of children born to parents having A and AB groups are
1) O, A
2) A, B, AB
3) O, A, B
4) O, A, B, AB
145. A man with blood group B marries a female with blood group A and their first child is having
blood group B. What is the genotype of child
1) IAIB
2) IAI0
3) IBI0
4) IBIB
146. A child with mother of blood group A and father of blood group AB, will not have which of the
following blood group
1) A
2) B
3) AB
4) O
147. If mother has blood group B, father has A group, the offspring will be of
1) A
2) O
3) AB
4) any of the above
148. Two nonallelic genes produces the new phenotype when present together but fail to do so independently then it is called
1) Epistatisis
2) Polygene
3) Non complimentary gene
4) Complimenatry gene
149. Sickel cell anemia is the result of mutation in the haemoglobin gene
1) frame shift
2) deletion
3) point
4) none of the above
150. When both alleles of a pair are fully expressed in a heterozygote, theye are called
1) Lethals
2) Co-dominants
3) Semi-dominants
4) Recessive allele
151. In the inheritance of flower colour in dog flower plant, the Fj had a phenotype that
1) resembles both of the parents
2) did not resembles either of the two parents
3) resembles with only one parent
4) 1 and 3 both
152. The three different alleles of human ABO blood types will produce how many genotypes &
phenotypes respectively
1) 4 & 6
2) 6 & 4
3)6 & 6
4) 4 & 4
153. Other than pea plants it was found that sometimes the F1 had a phenotype that did not resemble
either of the two parents and was in between the two. It is due to
1) Complete Dominance
2) Incomplete Dominance
3) Co-Dominance
4) Complementary gene interaction
154. Which of the following material is good to understand incomplete dominance
1) Sweet Pea
2) Cattle
3) Snapdragon
4) Kernel colour in wheat
155. Find out the correct match
1) F1 resembled either of the two parents-Dominance
2) F1 resembled in between -incomplete dominance
3) F1 resembled both parent - Co-dominance
4) All are correct
156. Which of the following condition is true for codominance
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1) Phenotype of F1 resembled either of the two parents
2) Phenotype of F1 did not resemble either of two parents
3) Phenotype of F1 resembles both parents
4) None of these
157. Which of the followig is a good example of multiple allele
1) ABO blood groups
2) Size of starch grain in pea
3) Shape of seed
4) Flower colour in pea
158. In sickle cell anaemia
1) The mutant haemoglobin molecule undergoes polymerisation under low oxygen tension causing
the change in the shape of RBC
2) Substitution of Glutamic acid by valine at the sixth position of the a-chain of haemoglobin
3) The mutant haemoglobin undergoes polymerization under high oxygen tension causing the change
in shape of RBC
4) a-globin chain is modified
159. In a cross between true red flowered (RR) and true breeding white flowered (rr), snapdragon
plant, the F1(Rr) was pink. When the F1 was self pollinated the F2 resulted in the following ratio
1(RR) red; 2(Rr) pink; l(rr) white. Above condition can be explained by
1) True dominance
2) Incomplete dominance
3) Lethal gene
4) Independent assortment
160. In ease of ABO blood group allele IA and P if present together then
1) Ony IA allele expresses
2) Only IB allele expresses
4) None of these
3) Both IA and IB alleles express
POLYGENIC AND CYTOPLASMIC INHERITANCE
161. A polygenic inheritance in human beings is
1) skin colour
2) sickle cell anaemia
3) colour blindness
4) phenylketonuria
162. Which one carries extra nuclear genetic material
1) Plastids
2) Ribosomes
3) Chromosomes
4) Golgi-complex
163. When certain character is inherited only through the female parent, it probably represents the
case of
1) Mendelian nuclear inheritance
2) Multiple plastid inheritance
3) Cytoplasmic inheritance
4) Incomplete dominance
164. Cytoplasmic male sterility is inherited
1) Maternally
2) Paternally
3) Both
4) Bacteriophage multiplication
165. In which type of inheritance the results are affected by reciprocal cross
1) Nuclear
2) Cytoplasmic
3) Blending
4) All the above
166. The scientist who first discovered cytoplasmic - inheritance was
1) Correns
2) Rhoades
3) Mendel
4) Morgan
167. Extranuclear inheritance is a consequence of presence of genes in
1) Lysosomes and ribosomes
2) Mitochondria and chloroplasts
3) Endoplasmic reticulum and mitochondria
4) Ribosomes and chloroplast
168. Inheritance of skin colour in human beings is an example of
1) Complementary gene
2) Monogenic inheritance
3) Polygenic inheritance
4) Mendelian inheritance
169. Albinism in corn show
1) Pathogenic effect
2) Deficiency of light
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3) Deficiency of minerals
4) lethal gene effect
170. Polygenic genes show
1) Identical phenotype
2) Identical biochemistry
3) Different phenotype
4) Identical genotype
171. A dihybrid ratio of 1:4:6:4:1 is obtained instead of 9:3:3:1. This is an example of
1) Complementary gene
2) Supplementary gene
3) Polygenic inheritance
4) Incomplete dominance
172. An example of the quantitative trait in man is
1) Hair colour
2) Colour of eye
3) Skin colour
4) Shape of nose
173. Polygenic inheritance was first noted by
1) Davenport
2) Galton
3) Mendel
4) Kolreuter
174. In polygenic inheritance trait which controlled by three pairs of genes. Two individuals which
are heterozygous for three alleles, crossed each other. Such type of cross produces what phenotypic ratio
1) 1 : 2 : 1
2) 9 : 3 : 3 : 1
3) 1 : 4 : 6 : 4 : 1
4) 1 : 6 : 15 : 20 : 15 : 6 : 1
175. In totmato, genotype aabbcc produces 100g tomatoes and AABBCC produces 160g tomatoes.
What is contribution of each polygene in the production of tomatoes
1) 10 g
2) 20 g
3) 30 g
4) 40 g
176. A polygenic trait is controlled by 3 genes A, B and C. In a cross AaBbCc x AaBbCc, the phenotypic ratio of the offsprings was observed as
1 : 6 : x : 20 : x : 6 : 1
what is the possible value of x?
1) 3
2) 9
3) 15
4) 25
177. Gene for cytoplasmic male sterility in plants are generally located in the
1) chloroplast genome
2) mitochondrial genome
3) nuclear genome
4) cytosol
LINKAGE, SEX LINKAGE
178. What is the inheritance of colour blindness of both parents having a normal vision but mother
has a recessive gene for colour blindness
1) 50% Nil
2) 100%
Nil
3) Nil
100%
4) Nil
Nil
179. What would be the nature of children if a colour blind woman marries a normal man
1) Colourblind daughter & normal sons
2) Colourblind sons and carrier daughters
3) Normal sons & carrier daughters
4) Normal sons & Normal daughters
180. A colourblind man marries a normal lady whose father was colour blind. If it produces two sons
& two daughters, how many of them would be suffer
1) Both sons
2) Both daughters
3) One son & one daughter
4) Both sons & both daughters
181. A colourblind daughter is born when :
1) Father is colourblind, mother is normal
2) Mother is colourblind, father is normal
3) Mother is carrier, father is normal
4) Mother is carrier, father is colourblind
182. Hypertrichosis is
1) Holandric character
2) X-Linked character
3) Diagenic character
4) Sex-influened character
183. In which of the following the inheritance takes place only by male
1) Nuclear
2) Cytoplasmic
3) co-dominance
4) Holandric inheritance
184. Which of the following is not a sex linked characters
1) Haemophilia
2) Colour blindness
3) Hypertrichosis
4) Baldness
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185. A gene located on Y-chromosome and therefore, transmitted from father to son is known as
1) Supplementary gene 2) Complementary gene 3) Duplicate gene
4) Holandric gene
186. The condition in which only one allele of a pair is present in a diploid organism is known as
1) Homozygous
2) Heterozygous
3) Hemizygous
4) Incomplete dominance
187. Baldness in man is a
1) Autosomal character
2) Sex linked character
3) Sex influenced character
4) 1 and 3 both
188. A colourblind man marries a daughter of colourblind father, then in the offsprings
1) All sons are colourblind
2) All daughters are colourblind
3) Half sons are colourblind
4) No daughter is colourblind
189. A woman with normal vision marries a man with normal vision and gives birth to a colourblind
son. Her husband dies and she marries a colourblind man. what is the probability of her children having the abnormality
1) 50% colourblind sons + 50% colourblind daughters
2) All sons colourblind & daughter carrier
3) All daughter colourblind & sons normal
4) 50% sons colourblind and all daughters normal
190. A single recessive trait which can express its effect should occur on
1) Any autosome
2) Any-chromosome
3) X-chromosome of female
4) X- chromosome of male
191. Sex- linked disorders are generally
1) Lethal
2) Recessive
3) Dominant
4) Not inherited
192. In Drosophila crossing over occurs in female but not in male. Gene A and B are 10 map unit
apart on chromosome. A female Drosophila with genotype AB/ab. and male Drosophila with
genotype AB/ab. How many type of gametes are produced by female and male Drosophila respectively
1) 4 types : 2 types
2) 2 types : 2 types
3) 4 types : 4 types
4) 4 types : one types
193. In a cross between individuals homozygous for (a, b) and wild type (+ +). In this cross 700 out of
1000 individuals were of parental type. Then the distance between a and b is
1) 70 map unit
2) 35 map unit
3) 30 map unit
4) 15 map unit
194. In maize coloured endosperm (C) is dominant over colourless (c) and full endosperm (R) is
dominant over shrunken (r). When a dihybrid of F1-generation was test crossed it produced
four phenotypes in the following percentage
Coloured - Full
= 45%
Coloured - Shrunken = 5%
Colourless - Full
= 4%
Colourless - Shrunken = 46%
From these data what would be distance between the two non allelic genes
1) 48 unit
2) 9 unit
3) 4 unit
4) 12 unit
195. What ratio is expected in offsprings if father is colour blind and mother's father was colour
blind
1) 50% daughter - colour blind
2) All the sons are colour blind
3) All the daughters colour blind
4) All the sons are normal
196. There are three genes a, b, c percentage of crossing over between a and b is 20%, b and c is 28%
and a and c is 8%. What is the sequence of genes on chromosome
1) b, a, c
2) a, b, c
3) a, c, b
4) None
197. The linkage map of X-chromosome of fruitfly has 66 units, with yellow body gene (y) at one end
and bobbed hair (b) gene at the other end. The recombination frequency between these two
genes (y and b) should be
1) 60%
2) > 50%
3) < 50%
4) 100%
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198. Mammary glands in female, moustaches and beard in human males are examples of
1) Sex linked traits
2) Sex limited traits
3) Sex differentiating traits
4) Sex-determining traits
199. When a cluster of genes show linkage behaviour they
1) Do not show a chromosome map
2) Show recombination during meiosis
3) Do not show independent assortment
4) Induce cell division
200. Genetic Map is one that
1) Establishes sites of the genes on a chromosome
2) Establishes the various stages in gene evolution
3) Shows the stages during the cell division
4) Shows the distribution of various species in a region
201. One of the genes present exclusively on the X- chromosome in humans is concerned with
1) Baldness
2) Red green colour blindness
3) Facial hair/Moustaches in males
4) Night blindness
202. The recessive genes located on X-chromosome in humans are always
1) Expressed in females
2) Lethal
3) Sub-lethal
4) Expressed in males
203. Lack of independent assortment of two genes A and B in fruit fly is due to
1) Crossing over
2) Repulsion
3) Recombination
4) Linkage
204. A normal woman, whose father was colour-blind is married to a normal man. The sons would be
1) All colour-blind
2) 75% colour-blind
3) 50% colour-blind
4) All normal
205. If father shows normal genotype and mother shows a carrier trait for haemophelia
1) All the female children will be carrier
2) A male child has 50% chances of active disease
3) Female child has probability of 50% to active disease
4) All the female children will be colourblind
206. Which of the following show linkage group in coupling phase
a
A
B
b
B
A
A
b
1)
2)
3)
4)
a
a
a
a
B
b
b
b
207. The longer the chromosome of an organism, the more genetic variability it gets from
1) Independent assortment
2) Linkage
3) Crossing over
4) Mutation
208. Which statement is incorrect about linkage
1) It helps in maintaining the valuable traits of new varieties
2) It helps in forming new recombinants
3) Knowledge of linkage helps the breeder to combine all desirable traits in a single variety.
4) It helps in locating genes on chromosome
209. A woman with normal vision, but whose father was colour blind, marries a colour blind man.
Suppose that the fourth child of this couple was a boy. This boy 1) Must have normal colour vision
2) May be colour blind or may be normal vision
3) Will be partially colour blind since he is heterozygous for the colour blind mutant allele
4) Must be colour blind
210. Haemophilia is more commonly seen in human males than in human females because 1) This disease is due to a Y-linked recessive mutation
2) This disease is due to an X-linked recessive mutation
3) This disease is due to an X-linked dominant mutation
4) A greater proportion of girls die in infancy
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211. If Mendel has chosen to study traits determined by linked genes he would not have discovered
1) Law of segregation
2) Law of dominance
3) Law of independant assortment
4) Law of unit character
212. Which law would have been violated if Mendel had chosen eight characters in garden-pea
1) Law of dominance
2) Law of segregation
3) Law of independant assortment
4) Law of purity of gametes
213. If Mendel might have studies 7 pairs of characters in a plant with 12 chromosomes, instead of
14, then
1) He could not discover independant assortment 2) He might have not discovered linkage
3) He might have discovered crossing-over
4) He might have not observed dominance
214. The first attempt to show linkage in plants was done in
1) Pisum sativum
2) Lathyrus odoratus
3) Zea mays
4) Oenothera lamarckiana
215. With increasing age the linkage becomes
1) Strong
2) Weak
3) Terminates
4) Remains unchange
216. Coupling and Repulsion theory produced by
1) Morgan
2) Bateson
3) Muller
4) De vries
217. If there were only parental combinations in F2 of a dihybrid cross then Mendel might have
discovered
1) Independant assortment
2) Atavism
3) Linkage
4) Repulsion
218. Linkage discovered in Drosophila by
1) Bateson
2) Morgan
3)Muller
4) Correns
219. A dihybrid plant with incomplete linkage on test cross may produce how many types of plants
1) 2
2) 4
3) 8
4) 1
220. How many linkage group are there in bacteria E.coli
1) One
2) Two
3) Four
4) None
221. If distance between gene on chromosome is more, then gene shows
1) Weak linkage
2) Strong linkage
3) Less crossing
4) 1 & 3 both
222. Linked gene shows
1) Always parental combination
2) Sometimes new combinations
3) Always new combination
4) New combination more
223. The number of linkage groups in a cell having 10 pairs of chromosomes are
1) 5
2) 10
3) 15
4) 20
224. The association of parental characters combinations in the offsprings of a dihybrid is excess to
non- parental combinations is said to be due to
1) Co-dominance
2) Blending inheritance 3) Linkage
4) Duplicate genes
225. Complete linkage is found in
1) Birds
2) Snakes
3) Female- Drosophila
4) Male - Drosophila
226. A phenomenon which works opposite to the linkage is
1) Independent assortment
2) Crossing-over
3) Segregation
4) Mutation
227. Cross over value (COV) of gene A and B is 5% while COV of genes B and C is 15% the possible
sequence of these genes on chromosome is
1) A-B-C
2) C-A-B
3) B-C-A
4) Both (1) & (2)
228. In female Drosophila the linked gene exhibit recombination during meiosis of gamete formation, but such a recombination does not occur during the formation of sperm in male Drosophila
1) Male Drosophila is sterile
2) Male Drosophila is parthenogenetic male
3) No crossing over occur in male Drosophila
4) Male Drosophila is haploid
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229. TDF gene is a
1) A gene present on X-chromosome
2) A segment of RNA
3) A proteinaceous factor
4) A gene present on Y-chromosome
230. A diseased man marries a normal woman. They get three daughters and five sons. All the daughters were diseased and sons were normal. The gene of this disease is
1) Sex linked dominant
2) Sex linked recessive
3) Sex limited character
4) Autosomal dominant
231. Who postulated the 'Chromosome Theory of Inheritance'
1) De Vries
2) Mendel
3) Sutton and Boveri
4) Morgan
232. Drosophila melanogaster has
1) 2 pairs of autosomes and 1 pair of sex chromosomes
2) 3 pairs of autosomes and 3 pairs of sex chromosomes
3) 1 pairs of autosomes and 3 pairs of sex chromosomes
4) 3 pairs of autosomes and 1 pairs of sex chromosomes
233. Which one of the following is associated with sex- linked inheritance
1) Night-blindness
2) muscular dystrophy 3) astigmatism
4) Polydactyly
234. Haemophilic female marries normal male, the theoretical ratio of their offsprings regarding
haemophilia will be
1) All offsprings are haemophilic
2) All girls are haemophilic
3) All sons are haemophilic
4) half daughters and half sons are haemophilic
235. Linkage was first studied by
1) Darwin
2) Morgan
3) Bateson and Punnett
4) Mendel
236. In man sex linked characters are mainly transmitted through
1) X-chromosome
2) Autosomes
3) Y-chromosome
4) X-chromosome, Y-chromosome & Autosomes
237. If a colourblind woman marries with a normal man. The offspring will be
1) All colourblind
2) All daughters normal and all son will be colourblind
3) All normal
4) All daughters will be colourblind and all sons will be normal
238. If a colourblind man is married to a normal woman, then from the point of view of disease, their
offsprings will be
1) All sons will be colourblind
2) All daughters will be colourblind
3) all sons and daughters will be normal
4) All sons and daughters will be colour blind 96
239. Walter Sutton is famous for his contribution to
1) Gentic engineering
2) Totipotency
3) Qantitative genetics
4) Chromosomal theory of inheritance
240. If a colour blind man marries a girl who is normal (homozygous) for this charactr, then genotypically
1) sons and daughters will be normal
2) sons wil be colour blind, daughters will be normal
3) sons will be normal, daughters wil be carriers
4) both sons and daughters will be colour blind
241. Frequency of crossing over will be relatively more if
1) distance between the two genes is less
2) distance between the two genes is more
3) linked genes are more
4) both (2) & (3)
242. Presence of recombinants is due to
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1) crossing over
2) linkage
3) lack of independent assortment
4) all of the above
243. Morgan coined the term ____ to describe the physical association of genes on a chromosome &
the term ____ to describe the generation of non-parental gene combinations.
1) Recombination; Linkage
2) Recombination; Non-recombination
3) Linkage; Non-recombination
4) Linkage; Recombination
244. Experimental verification of the chromosomal theory of inheritance done by Thomas Hunt ,
Morgan and his colleagues they worked with
1) Pea plant
2) Sweet pea plant
3) Snapdragon
4) Drosophila
245. Which is incorrect for Drosophila melanogaster
1) They could be grown on simple synthetic medium
2) Single mating could produce a large number of progeny
3) They complete their life cycle in about 7 weeks
4) There was a clear differentiation of the sexes.
246. Morgan and his group found that when genes were grouped on the same chromosome, some
genes were very tightly linked and showed
1) Very low recombination
2) Higher recombination
3) No recombination
4) 100% parental combination
247. Which statement is not true for Drosophila melanogaster1) They complete their life cycle about two weeks
2) Single mating produce large number of progeny flies
3) It has few hereditary variation that can be seen with high power microscope
4) It has clear differentiation of the sex
248. The experimental verification of the chromosomal theory of inheritance by
1) Boveri
2) Sutton
3) T.H. Morgan
4) Bateson
SEX DETERMINATION
249. How the sex of offsprings determined in humans
1) Sex chromosome of mother
2) Size of ovum
3) Size of sperm
4) Sex chromosome of father
250. Which of the following possess homogametic male
1) Plants
2) Man
3) Insect
4) Birds
251. Which chromosome set is found in male grass hopper
1) 2A + XY
2) 2A + XO
3) 2A + YY
4) 2A + XX
252. Genic balance theory for sex determination in Drosophila was proposed by
1) Pro. R.P.Roy
2) H.E.Warmke
3) C.B. Bridges
4) Mc. clung
253. No. of Bar Body in XXXX female
1) 1
2) 2
3) 3
4) 4
254. In Drosophila, the sex is determined by
1) The ratio of number of X-chromosomes to the sets of autosomes
2) X and Y chromosomes
3) The ratio of pairs of X-chromosomes to the pairs of autosomes
4) Whether the egg is fertilized or develops parthenogenetically
255. In Drosophila male differentiation is controlled by
1) No. of Y-chromosome
2) No. of X-chromosomes
3) Ratio between number of X-chromosome and the set of autosome
4) Sets of autosome
256. Sex determination ratio in an organism is given by X/A=1.5, then organism will be
1) male
2) female
3) super female
4) intersex
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257. Barr body is associated with
1) sex chromosome of female
2) sex chromosome of male
3) autosome of female
4) autosome of male
258. In male grass hoppers and moths there are two pairs of autosomes and
1) X only
2) X and Y
3) Y only
4) none of these
259. Which of the following symbols are used for representing sex chromosome of birds
1) ZZ - ZW
2) XX - XY
3) XO - XX
4) ZZ - WW
260. If somatic cells of a human male contain single Barrbody, the genetic composition of the person
would be
1) XYY
2) XXY
3) XO
4) XXXY
261. The theory where ratio between the number of X-chromosomes and number of complete sets of
autosomes will determine the sex is known as
1) Chromosome theory of sex determination
2) Genic balance theory of sex determination
3) Harmonal balance theory of sex determination 4) environmental sex determination
262. Sex determination in humans takes place by
1) sex chromosomes of father
2) measurement of sperm
3) measurement of ovum
4) sex chromosomes of mother
263. In Drosophila sex index of super female is
1) 1
2) 0.5
3) 1.5
4) 0.67
264. If X/A Ratio of two Drosophila is 0.6 and 0.33 respectively what would be their sex
1) Female & male
2) Super female & super male
3) Inter sex & super male
4) Inter sex and super female
265. Which of the following genotype represent intersex Drosophila
1) 2A + XXX
2) 2A + XXY
3) 3A + XXY
4) 2A + XY
266. In which organism female in homogametic & also have one chromosomes more than male.
1) Birds
2) Drosophila
3) Chicks
4) Grasshopper
267. Grasshopper is an example of 1) XO type of sex determination
2) XY type of sex determination
3) Environmental sex determination
4) Genic balance theory
268. Which of the following is responsible for sex determination in chick
1) Sperm
2) Egg
3) Somatic cell
4) Every cell of body
269. In which of the following sex is determined by female individual 1) Human
2) Drosophila
3) Birds
4) Grasshopper
270. Male heterogamy found in case of
1) XO type male in Grasshopper
2) XY type male in human
3) ZW male in birds
4) 1 and 2 both
271. In which of the following monosomic male is found
1) Human
2) Birds
3) Honey bee
4) Grasshopper
HUMAN GENETICS, POPULATION GENETICS
272. There are two alleles (A1 & A2) out of which one (A1) has nil abundance in a population then the
abundance of second allele (A2) is
1) 0.25
2) 1.00
3) 0.40
4) 0.50
273. If a normal woman marries an albino man and their offsprings are half albino, half normal the
woman is
1) Homozygous normal
2) Heterozygous normal
3) Homozygous recessive
4) Homozygous dominant
274. Which is a dominant trait
1) Colour blindness
2) Albinism
3) Haemophilia
4) Rh factor
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275. Parents are carrier for albinism. What will be the first three children
1) Some normal, heterozygous & albino
2) All normal
3) All heterozygous albino
4) No normal
276. If a cross is made between two individuals each having genotype Bb, two offsprings are obtained. Out of these first has dominant trait. What is the probability that the second offspring
will exhibit recessive trait
1) 1/4
2) 100
3) Zero
4) 3/4
277. In case of taster and non-taster human beings T is for dominance & t is for recessive gene.
Which of the following would not be able to taste PTC
1) TT
2) Tt
3) tt
4) None
278. A family has five girls and no son, the probability of the occurance of son in 6 th child is
1) 1/2
2) 1/5
3) 1
4) No chance
279. A tobacco plant heterozygous for albinism is self pollinated and 1200 seeds are subsequently
germinated. How many seedlings would have the parental genotype :
1) 900
2) 600
3) 1200
4) 300
280. Which one of the following character in man is controlled by recessive gene
1) Colourblindness
2) Woolly hair
3) Brachy-dactyly
4) Curly hairs
281. The migration of gene into a population from other population by interbreeding is called
1) Gene pool
2) Gene flow
3) Genetic drift
4) Gene erosion
282. What is the probability of three daughters to a couple in three children
1) 1/4
2) 1/8
3) 1/16
4) 3/8
283. In human right handedness is dominant over left handedness. What offsprings would be expected from two left handed parents
1) Only left handed
2) Only right handed
3) Left handed & right handed both
4) Neither left handed nor right handed
284. Probability of four son to a couple is
1) 1/4
2) 1/8
3) 1/16
4) 3/8
285. Down's syndrome is caused by an extra copy of chromosome number 21. What percentage of
offsprings produced by an affected mother and a normal father would be affected by this disorder
1) 50%
2) 25%
3) 100%
4) 75%
286. A male human is heterozygous for autosomal genes A and B and is also hemizygous for hemophilic gene h. What proportion of his sperms will be abh
1) 1/4
2) 1/8
3) 1/32
4) 1/16
287. Given below is a pedigree chart of a family with five children. It shows the inheritance of attached ear-lobes as opposed to the free ones. The squares represent the male individuals and
circles the female individuals
ATTACHED EAR LOBE
FREE EAR LOBE
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Which one of the following conclusions drawn is correct
1) The parents are homozygous recessive
2) The trait is Y-linked
3) The parents are homozygous dominant
4) The parents are heterozygous
288. Equilibrium of gene frequencies is Pq
1) P2 x 2Pq x q2 = 1
2)  
3) Hardy weinbergh law 4) Mutation
N
289. In a Random mating population of 28,800 individuals percentage of dominant homozygous individuals is 49% find out the percentage of heterozygous individual 1) 21%
2) 42%
3) 32%
4) 9%
290. Predict from the following chart
1) Character is dominant and carried by X chromosome
2) Character is carried by Y chromosome
3) Character is sex linked recessive
4) Character is autosomal recessive
291. In pedigree analysis symbol
is used for
1) Heterozygous for autosomal recessive
2) Affected individuals
3) Death
4) Carrier for sex linked recessive
292. Study the given pedigree carefully, the trait indicated is
: Normal male
: Affected male
: Normal female
: Affected female
1) Autosomal recessive
2) X-linked recessive
3) Maternal inheritance
4) Paternal inheritance
293. In a population that is in Hardy weinberg equilibrium, the frequency of a recessive allele for a
certain hereditary trait is 0.20. What percentage of the individual in the next generation would
be expected to show the dominant trait
1) 16%
2) 32%
3) 64%
4) 96%
294. Given pedigree shows inheritance of autosomal recessive gene. What is the genotype of given parents
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1) AA, aa
2) Aa, Aa
3) aa, Aa
4) aa, aa
295. A pedigree is shown below for a disease that is autosomal recessive. The genetic make up of the
first generation
I
II
III
1) AA, aa
2) Aa, Aa
3) Aa, aa
4) aa, aa
296. In a random mating population frequency of disease causing recessive allele is 80%. What would
be the frequency of carrier individual in population
1) 64%
2) 32%
3) 16%
4) 100%
297. In a random mating population frequency of dominant allele is 0.7. What will be the frequency
of recessive phenotype
1) 0.49
2) 0.09
3) 0.3
4) 0.21
298. A plant is heterozygous and is designated Bb and produces two kinds of gametes B and b. The
probability of b gamete fertilising B or b is
1) 1/2
2) 1/1
3) 0/1
4) 1/4
299. At a particular locus, frequency of A' allele is 0.6 and that of 'a' is 0.4. What would be the
frequency of heterozygotes in a random mating population at equilibrium
1) 0.24
2) 0.16
3) 0.48
4) 0.36
300. A normal woman whose father was albino, marries an albino man, what proportion of normal
and albino are expected among their offsprings
1) All normal
2) 2 normal : 1 Albino 3) All albino
4) 1 normal : 1 Albino
301. Albinism is determined by a recessive gene in man. The presence of albinism in 50% children
born to a couple proves that
1) Both parents are heterozygous for albinism
2) Father is homozygous normal and mother is heterozygous
3) Father is homozygous for albinism but mother is heterozygous
4) Both are homozygous
302. Family has 9 girls, Probability of son at 10th birth is
1) 50%
2) 100%
3) 25%
4) 75%
303. Polydactyly in man is due to
1) autosomal dominant gene
2) autosomal recessive gene
3) sex - linked dominant gene
4) sex - linked recessive gene
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304. Blue eye colour in human is recessive to brown eye colour. The expected children of a marriage
between a blue eyed woman and a brown eyed man who had a blue eyed mother will be
l) All black eyed
2) All blue eyed
3) All brown eyed
4) One blue eyed and one brown eyed
305. If the first seven children born to a particular pair of parents are all males, what is the probability that the eighth child will also be a male?
1) 1/2
2) 1/4
3) 1/8
4) 1/16
306. The existence within a population of non-beneficial alleles in heterozygous genotype is
1) genetic load
2) genetic drift
3) genetic flow
4) selection
307. Study the pedigree given below and assign the type of inheritance of the trait.
Normal male
Affected male
Normal female
Affected female
1) X-linked recessive
2) Y-linked
3) autosomal recessive
4) autosomal dominant
308. Given below is the pedigree of sickle cell anaemia, in a family
In this the RBC of both parents will be
1) Normal
2) Sickle shaped
3) Both normal & sickle shaped
4) Cannot be determined
309. Which of the following symbol is used for mating between relatives (Consangeineous mating)
1)
5
2)
3)
GENETIC MATERIAL, DNA
310. Which of the following may be true for RNA
1) A = U G =C
3) A = U = G = C
94
4)
2) A  U G  C
4) Purines = Pyrimidines
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311. Who proved DNA as genetic material
1) Griffith
2) Bacteria
3) PPLO
4) Hershey and chase
312. Circular and double stranded DNA occurs in
1) Golgibody
2) Mitochondria
3) Nucleus
4) Cytoplasm
313. Double helix model of DNA which was proposed by watson and crick was of1) C-DNA
2) B-DNA
3) D-DNA
4) Z-DNA
314. If there are 10,000 nitrogenous base pairs in a DNA then how many nucleotides are there
1) 500
2) 10,000
3) 20,000
4) 40,000
315. Double helix model of DNA is proposed by
1) Watson and Crick
2) Schleiden schwann
3) Singer and Nicholson
4) Kornberg and Khurana
316. Back bone in structure of DNA molecule is made up of
1) Pentose Sugar and phosphate
2) Hexose sugar and phosphate
3) Purine and pyrimidine
4) Sugar and phosphate
317. Substance common in DNA and RNA
1) Hexose Sugar
2) Histamine
3) Thymine
4) Phosphate groups
318. Nucleotide is
1) N2 - base, pentose sugar, and phosphoric acid 2) Nitrogen, Hexose sugar and phosphoric acid
3) Nitrogen base, pentose sugar
4) Nitrogen base, trioses and phosphoric acid
319. DNA differs from RNA in
1) Only Sugar
2) Nitrogen base only
3) Nitrogen base and sugar
4) None
320. Unit of nucleic acids are
1) Phosphoric acid
2) Nitrogenous bases
3) Pentose Sugar
4) Nucleotides
321. Which element is not found in nitrogenous base
1) Nitrogen
2) Hydrogen
3) Carbon
4) Phosphorus
322. DNA was first discovered by
1) Meischer
2) Robert Brown
3) Flemming
4) Watson & Crick
323. Nucleic acid (DNA) is not found in
1) Nucleus & nucleolus
2) Peroxysome & ribosome
3) Mitochondria & plastid
4) Chloroplast & nucleosome
324. DNA is not present in 1) Mitochondria
2) Chloroplast
3) Bacteriophage
4) TMV
325. A nucleic acid contains thymine or methylated uracil then it should be
1) DNA
2) RNA
3) Either DNA or RNA 4) RNA of bacteria
326. Prokaryotic genetic system contains
1) DNA & histones
2) RNA & histones
3) Either DNA or histones
4) DNA but no histones
327. A N2- base together with pentose sugar and phosphte forms (or) building - block unit of nucleic
acid is
1) Nusleoside
2) Polypeptide
3) Nucleotide
4) Aminoacid
328. One of the characteristics of DNA is
1) Uracil
2) Deoxyribose sugar
3) Single strandedness
4) Ability of protein synthesis
329. Which of the following is not a pyrimidine N2 base
1) Thymine
2) Cytosine
3) Guanine
4) Uracil
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330.The purine & pyrimidine pairs of complementry strands of DNA are held together by
1) H - bonds
2) 0 - bonds
3) C - bonds
4) N - bonds
331. Number of H - bonds between guanine and cytosine are
1) One
2) Two
3) Three
4) Four
332. Which purine & pyrimidine bases are paired together by H - bonds in DNA
1) AC & GT
2) GC & AT
3) GA & TC
4) None of the above
333. A single stranded DNA is present in4) Bacteria
1) TMV
2) Salmonella
3) Coliphage  x l74
334. What is the nature of the 2 strands of a DNA duplex
1) Identical & Complimentary
2) Antiparallel & complimentary
3) Dissimilar & non complimentary
4) Antiparallel & non complimentary
335. On an average, how many purine N2 bases are present in single coil of DNA
1) Four
2) Five
3) Ten
4) Uncertain
336. Distance between two nucleotide pairs of DNA is
3) 3.4 
4) 34 nm
1) 0.34 nm
2) 34 A0
337. Histone occupies the major groove of a DNA at an angle of
1) 600
2) 900
3) 450 to helix axis
4) 300 to helix axis
338. Strongest evidence that DNA is genetic material comes from
1) Chromosomes contain DNA
2) Transformation of bacterial cells
3) DNA is present in nucleus
4) DNA is not present in cytoplasm
339. Chargaaf's rule is given as 1) Purines  Pyrimidines
2) A + G = T+ C
3) A+ U = G + C
4) A+ T/ G + C = Const.
340. In RNA , Nucleotides are bonded by
1) H - bonds
2) Phospo diester bonds
3) Ionic bonds
4) Salt linkage
341. A nucleoside differs from a nucleotide is not having
1) Phosphate
2) Sugar
3) Phosphate & sugar
4) Nitrogen base
342. Wilkins X - ray diffraction showed the diameter of the DNA helix is
1) 10 A0
2) 20 A0
3) 30 A0
4) 40 A0
343. In the DNA of an animal percentage of Adenine is 30 then percentage of Guanine will be
1) 40
2) 30
3) 20
4) 70
344. Similarity in DNA and RNA
1) Both are polymer of nucleotides
2) Both have similar pyrimidine
3) Both have similar sugar
4) Both are genetic material
345. Length of one loop of B- DNA
1)3.4 nm
2) 0.34 nm
3) 20 nm
4) 10 nm
346. Extranuclear DNA is found in
1) Lysosome and chloroplast
2) Chloroplast and mitochondria
3) Mitochondria and lysosome
4) Golgi and E.R.
347. The following ratio is generally constant for a given species
1) A + G/ C + T
2) T + C / G + A
3) G +C / A + T
4) A + C / T + G
348. If base order in one chain of DNA is "ATCGA" then how many no. of H-bond found in DNA
duplex
1) 20
2) 12
3) 10
4) 11
349. In DNA purine nitrogen bases are
1) Uracil and Guanine
2) Guanine and Adenine
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3) Adenine and cytosine 4) None
350. Bond between phosphate and sugar in a nucleotide is
1) H-bond
2) Covelent bond
3) Phosphodiester bond 4) Sulphide bond
351. Two free ribonucleotide units are interlinked with
1) Peptide bond
2) covalent bond
3) Hydrogen bond
4) Phosphodiester bond
352. Short DNA segment has 80 thymine and 90 guanine bases. The total number of nucleotides are
1) 160
2) 40
3) 80
4) 340
353. D.N.A. strands are anti-parallel because of
1) H-bonds
2) Phosphate diester-bonds
3) Di-sulphide-bonds
4) Phosphate-bonds
354. Prokaryotic DNA is
1) double stranded round
2) single stranded round
3) double stranded straight
4) double stranded RNA as nucleic acid
355. Nucleoside is
1) Polymer of nucleic acid
2) Phosphoric acid + base
3) Phosphoric acid + sugar + base
4) Sugar + base
356. The back bone of RNA is consists of which of the following sugar
1) Deoxyribose
2) Ribose
3) Sucrose
4) Maltose 104
357. Retrovirus have genetic material
1) DNA only
2) RNA only
3) DNA or RNA only
4) None
358. Prions consist mainly of
1) Protein
2) DNA
3) RNA
4) Both DNA and RNA
359. DNA is acidic due to
1) Sugar
2) Phosphoric acid
3) Purine
4) Pyrimidine
360. T.M.V. contains
1) D.N.A.
2) R.N.A + Protein
3) D.N.A. + R.N.A.
4) D.N.A + Protein
361. R.N.A contains which of the following base, in place of Thymine of D.N.A.
1) Thymine
2) Uracil
3) Adenine
4) None of these
362. Genetic information are transferred from nucleus to cytoplasm of cell through
1) DNA
2) RNA
3) Lysosomes
4) ACTH
363. Base ratio of a DNA is 0.03 and the amount of A-T N 2 base contents is 30000, then what is the
amount of G-C, N2 base contents in this DNA?
1) 1000
2) 10000
3) 100000
4) None
364. If one strand of double stranded DNA, consists of the sequence 3-ATTCGTAC-5', then the complementary sequence must be 1) 5-UAAGCAUG-3'
2) 3-TAAGCATG-5'
3) 5-TAAGCATG-3'
4) 5'-TAAGCATG-3' in the reverse direction
365. Which of the following is a false statements ?
1) DNA is chemically less reactive, as compared to RNA
2) RNA mutate at a faster rate, as compared to DNA
3) Guanyl transferase enzyme helps in capping process during splicing of hn-RNA
4) Sweetness index of saccharine is 10,000
366. A DNA molecule contains 10,000 base pairs, then the length of this DNA molecule is
2) 0.34 x 10-5 meter
3) 34 x 10-5 meter
4) None of above
1) 3.4 x 10-5 meter
367. DNA molecule has uniform diameter due to ?
1) Double stranded
2) Presence of phosphate
3) Specific base pairing between purine and pyrimidine
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4) Specific base pairing between purine and purine
368. Following structure is related to which compound?
1) Adenine
H N
2) Guanine
O
3) Uracil
O
4) Thymine
N
H
369. If the sequence of bases in one strand of DNA is known then the sequence in other strand can be
predicted on the basis of
1) Antiparallel
2) Complementary
3) Polarity
4) Coiling
370. Polymer of deoxyribonucleotides is
1) DNA
2) RNA
3) Both (1) & (2)
4) None
371. The unequivocal proof that DNA is the genetic material came from the experiments of
1) Hershey and chaese (1952)
2) Frederic Griffith (1928)
3) Watson and Crick
4) Meselson and Stal (1958)
DNA REPLICATION
372. In process of replication deoxyribonucleoside triphosphate
1) acting as substrate
2) providing energy for polymerisation reaction
3) acting as an enzyme
4) both (1) & (2)
373. DNA polymerase is needed for
1) Replication of DNA
2) Synthesis of DNA
3) Elongation of DNA
4) All of above
374. DNA duplication occurs at
1) Meiosis - II
2) Mitotic interphase
3) Mitosis only
4) Meiosis and mitosis both
375. DNA Replication occurs at
2) G2 - stage
3) S - Stage
4) Mitotic phase
1) G0 & Gl
376. A DNA molecule in which both strands have radioactive thymidine is allowed to duplicate in an
environment containing non- radioactive thymidine. What will be the exact number of DNA
molecules that contains the radio active thymidine after 3 duplications
1) One
2) Two
3) Four
4) Eight
377. A bacterium with completely radioactive DNA was allowed to replicate in a non- radioactive
medium for two generation what % of the bacteria should contain radioactive DNA
1) 100 %
2) 50 %
3) 25 %
4) 12.5 %
378. DNA polymerase enzyme is required for the synthesis of
1) DNA from DNA
2) DNA from RNA
3) Both the above
4) DNA from nucleosides
379. In the base sequence of one strand of DNA is GAT , TAG ,CAT , GAC what shall be the sequence
of its complementary strand
1) CAT, CTG, ATC, GTA
2) GTA, ATC, CTG, GTA
3) ATC, GTA, CTG, GTA
4) CTA, ATC, GTA, CTG
380. Method of DNA replication in which two strands of DNA separates and synthesize new strands
1) Dispersive
2) Conservative
3) Semiconservative
4) Non conservative
381. During replication of a bacterial chromosome DNA synthesis starts from a replication origin
site and
1) RNA primers are involved
2) is facilitated by telomerase
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3) moves in one direction of the site
4) moves in bi-directional way
382. Which one of the following hydrolyses internal phosphodiester bonds in a polynucleotide chain
1) Lipase
2) Protease
3) Exonuclease
4) Endonuclease
383. The nature of DNA replication is
1) Conservation
2) Non conservative
3) Semi-consurvative
4) Cyanobacteria
384. The direction of D.N.A. replication is
1) From 5' end towards 3' end
2) From 3' end towards 5' end
3) Amino terminus to carboxy terminus
4) Carboxy terminus to amino terminus
385. Semiconservation replication of DNA was given by
1) Watson and Crick
2) Bateson and Punnett
3) Messelson and Stahl
4) Avery, McCarty and Mactleod
386. Which of the following enzyme is used in DNA multiplication
1) RNA polymerase
2) DNA endonuclease
3) Exonuclease
4) DNA Polymerase
387. Mode of DNA replication in E. coli is
1) Conservative and unidirectional
2) Semi conservative and unidirectional
3) conservative and bidirectional
4) Semi conservative and bidrectional
388. Which of the following enzyme is used to join DNA fragments
1) Terminase
2) Endonuclease
3) Ligase
4) DNA polymerase
389. Okazaki fragments are synthesised on
1) Leading strands of DNA only
2) Lagging strands of DNA only
3) Both leading and lagging strands of DNA
4) Complementary DNA
390. DNA replication includes
1) DNA ligase
2) DNA polymerase and ligase
3) RNA polymerase and ligase
4) All of these
391. During DNA replication in prokaryotes DNA is attached to
1) Chromosome
2) Mesosome
3) Nucleolus
4) Ribosome
392. In DNA replication, the primer is
1) A small deoxyribonucleotide polymer
2) A small ribonucleotide polymer
3) Helix destabilizing protein
4) Enzyme taking part in joining nucleotides of new strand
393. The strand of DNA, which does not code for anything is referred to as
1) Template strand
2) Antisense strand
3) Coding strand
4) Noncoding strand
394. During DNA replication discontinuosly synthesized fragments are later joined by the enzyme 1) Ligase
2) DNA polymerage
3) RNA primer
4) Primase
395. Replication fork is 1) Large opening of the DNA helix
2) Small opening of the DNA helix
3) Tightly coiled part of DNA helix
4) Loosely coiled part of DNA helix
396. The DNA dependent DNA polymerase catalyse polymerisation in2) 5'  3' direction
1) 3'  5' direction
3) Depend on the nature of template strand
4) both (1) & (2)
397. Main enzyme of DNA replication is
1) DNA dependent RNA polymerase
2) DNA dependent DNA polymerase
3) RNA dependent RNA polymerase
4) RNA dependent DNA polymerase
RNA, TRANSCRIPTION
398. The Process of copying genetic information from one strand of DNA into Y is termed as Z .
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Y
Z
1) Transcription
RNA
2) RNA
Transcription
3) DNA
Replication
4) Replication
RNA
399. Code in RNA corresponding to AGCT in DNA
1) TACA
2) UCGA
3) TCGA
4) AGUC
400. Which of the following is called adaptor molecule
1) DNA
2) m-RNA
3) t-RNA
4) RNA
401. Which may be attached with Adenine base in RNA
1) Guanine
2) Cytosine
3) Uracil
4) Thymine
402. The base sequence of nucleic acid is AGG, GGA, CGA, CCA from this it can be inferred that it
is a segment of
1) DNA
2) m - RNA
3) t - RNA
4) Data insufficient
403. RNA synthesis is controlled by
1) Rho- factor
2) Endo nuclease
3) Sigma factor
4) RNA - polymerase
404. In the base sequence of one starand of DNA is CAT, TAG , CAT , CAT, GAC what would be the
base sequence of its complementary m-RNA
1) GUA, GUA, CUG, AUC, CUG
2) AUG, CUG, CUC, GUA, CUG
3) GUA, AUC, GUA, GUA, CUG
4) GUC, CUG, CUG, CUA, CUU
405. The process by which DNA of the nucleus passes genetic information to m-RNA is called
1) Transcription
2) Translocation
3) Translation
4) T ransportation
406. A sequence of three consecutive bases in a t- RNA molecule which specifically binds to a complementary codon sequence in m RNA is known as
1) Triplet
2) Non - sense codon
3) Anti codon
4) Termination codon
407. t - RNA attach to larger subunit of ribosomes with the help of which loop 1) DHU-loop
2) T C loop
3) Anticodon loop
4) Minor loop
408. In bacteria the codon AUG stands for
1) Glycine
2) Methionine
3) N- formyl methionine 4) Alanine
409. In three dimensional view the molecule of t-RNA is
1) L-shaped
2) S-shaped
3) Y- shaped
4) E-shaped
410. During transcription, the DNA site at which RNA polymerase binds is called
1) Promoter
2) Regulator
3) Receptor
4) Enhancer
411. During transcription, if the nucleotide sequence of the DNA strand that is being coded is ATACG,
then the nucleotide sequence in the mRNA would be
l)TATGC
2) TCTGG
3) UAUGC
4) UATGC
412. Which form of RNA has a structure resembling clover leaf ?
1) rRNA
2) hnRNA
3) mRNA
4) tRNA
413. Which one of the following makes use of RNA as a template to synthesize DNA
1) DNA dependant RNA polymerase
2) DNA polymerase
3) Reverse transcriptase
4) RNA polymerase
414. During transcription holoenzyme RNA polymerase binds to a DNA sequence and the DNA assumes a saddle like structure at that point. What is that sequence called
1) CAAT box
2) GGTT box
3) AAAT box
4) TATA box
415. Which of the following is formed in nucleolus
1) r RNA
2) t RNA
3) m-RNA
4) DNA
416. cDNA probes are copied from the messenger RNA molecules with the help of
1) Restriction enzymes
2) Reverse transcriptase
3) DNA polymerase
4) Adenosine deaminase
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417. The enzyme responsible for transcription is
1) D.N.A polymerase-I
2) R.N.A. polymerase
3) Reverse transcriptase
4) D.NA. polymerase-HI
418. Transcription of DNA is aided by
1) RNA polymerase
2) DNA polymerase
3) Exo nuclease
4) Recombinase
419. If the base sequence in DNA is 5' AAAA 3' then the bases sequence in m-RNA is
1) 5' UUUU 3'
2) 3' UUUU 5'
3) 5' AAAA 3'
4) 3' TTTT 5'
420. Correct order of molecular weight is
1) DNA < r-RNA < t-RNA
2) DNA < m-RNA < r-RNA
3) t-RNA < m-RNA < DNA
4) t-RNA < DNA < m-RNA
421. The genes are responsible for growth and differentiation in an organism through regulation of
1) Translocation
2) Transformation
3) Transduction and translation
4) Translation and transcription
422. The scientist who was awarded Nobel prize in 1959 for in vitro synthesis of polyribonucleotide
1) Mendel
2) Calvin
3) Khorana
4) Ochoa
423. Method by which information reaches from DNA to RNA is
1) Transcription
2) Translation
3) T ransformation
4) T ransduction
424. DNA acts as a template for synthesis of
1) RNA
2) DNA
3) Both 1' and '2'
4) Protein
425. Which is soluble RNA
1) hnRNA
2) rRNA
3) mRNA
4) tRNA
426. Portion of gene which is transcribed but not translated is
1) exon
2) intron
3) cistron
4) codon
427. The smallest RNA is
1) r-RNA
2) m-RNA
3) t-RNA
4) None of these
428. The most abundant RNA of cell is
1) r-RNA
2) t-RNA
3) m-RNA
4) None of these
429. One strand of DNA (non template) has base sequence CAG, TCG, GAT. What will be the sequence of bases in m-RNA
1) AGC, CTA, CTA
2) GTC, AGC, CTC (3) CAG. UCG. GAU
4) GAC. TAG. CTA
430. Inverse transcription was discovered by
1) Watson and Crick
2) Khorana
3) Temin an Baltimore 4) Meischer
431. Growth hormone affects growth by controlling the production of... at cellular level
1) r-RNA
2) m-RNA
3) t-RNA
4) None of these
432. In a transcription unit promoter is said to be located towards
1) 3' end of structural gene
2) 5' end of structural gene
3) 5' end of template strand
4) 3' end of template strand
433. Mature eucaryotic m-RNA is recognised by
1) Shine dalgarno sequence at 5' end
2) 7-methyl guanosine at 5'end and polyadenine bases at 3' end
3) Anti shine dalgarno sequence at 5' end
4) Presence of coding and noncoding sequence
434. The term informosome is applied to
1) t-RNA protein complex
2) m-RNA protein complex
3) m-RNA + t - RNA complex
4) r - RNA + t - RNA complex
435. Which types of nitrogen bases are present in loops of t-RNA ?
1) Only unusual nitrogen bases
2) Only usual nitrogen bases
3) Both usual and unusual nitrogen bases
4) None of the above
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436. Transcription unit in DNA is
1) Promoter
2) Structural gene
3) Terminator
4) All
437. In DNA promoter is the site for the initiation of
1) Replication
2) Translation
3) Transcription
4) Both (2) & (3)
438. Main enzyme of transcription
1) DNA dependent DNA polymerase
2) DNA dependent RNA polymerase
3) RNA dependent RNA polymerase
4) RNA dependent DNA polymerase
439. Removal of introns and joining of exons is called
1) Capping
2) Tailing
3) Splicing
4) All
CODE, TRANSLATION
440. If genetic code is tetraplet then what is the possible number of codons wich code 20 types of
amino acids
1) 261
2) 64
3) 256
4) 43
441. A codon in m-RNA has
1) 3-bases
2) 2-bases
3) 1-base
4) Number of bases vary
442. A DNA strand is directly involved in the synthesis of all the following except1) Another DNA
2) t-RNA & m-RNA
3) r-RNA
4) Protein
443. Genetic code was discovered by1) Nirenberg & Mathei 2) Komberg & Crick
3) Khorana & Komberg 4) Gamow
444. Genetic code was deciphered by chemically synthesizing the trinucleotides by
1) Watson & Crick
2) Beadle & Tatum
3) Briggs & King
4) M.W. Nirenberg
445. Nirenberg synthesized an m-RNA containing 34 poly-Adenine (A-A-A-A-A-A--- ) and found a
polypeptide formed of 11 poly-lysine this proved that genetic code for lysine was
1) one-adenine
2) A-A doublet
3) A-A-A triplet
4) Many adenines
446. 64 Codons constitute genetic code because
1) There was 64 types of amino acid
2) 64 types of t-RNA
3) Genetic code is triplet4) There are 64 enzymes
447. Degeneracy of genetic code was discovered by
1) Me Clintock
2) Khorana
3) Ochoa
4) Baurnfield & Nirenberg
448. Genetic code consists of
1) Adenine & Guanine
2) Guanine & Cytosine
3) Cytosine & uracil
4) All
449. Which codon gives signal for the start of polypeptide (protein) chain synthesis1) AUG
2) UGA
3) GUA
4) UAG
450. The function of non-sense codons is
1) To release polypeptide chain from t-RNA
2) To form an unspecified amino acid
3) To terminate the message of a gene controlled protein synthesis4) To convert a sense DNA into non sense DNA
451. Termination of chain growth in protein synthesis is brought about by1) UUG, UGC, UCA
2) UCG, GCG, ACC
3) UAA, UAG, UGA
4) UUG, UAG, UCG
452. Genetic code determines1) Structural pattern of an organism
2) Sequence of amino acid in protein chain
3) Variation in offsprings
4) constancy of morphological trait
453. m - RNA is attached with
1) E.R.
2)Ribosome
3) Nucleus
4) Lysosome
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454. Sometimes the starting codon is GUG in place of AUG, GUG normally stands for
1) Valine
2) Glycine
3) Methionine
4) Tyrosine
455. Which one of the following triplet codes, is correctly matched with its specificity for an amino
acid in protein synthesis or as 'start' or 'stop' codon
1) UCG - Start
2) UUU - Stop
3) UGU - Leusine
4) UAC - Tyrosine
456. During translation initiation in prokaryotes, a GTP molecule is needed in
1) Formation of formyl-met-tRNA
2) Binding of 30S subunit of ribosome with mRNA
3) Association of 30 S-mRNA with formyl-met tRNA
4) Association of 50 S subunit of ribosome with initiation complex
457. Degeneration of a genetic code is attributed to the
1) First member of a codon
2) Second member of a codon
3) Entire codon
4) Third member of a codon
458. What would happen if in a gene encoding a polypeptide of 50 amino acids, 25th codon (UAU) is
mutated to UAA
1) A polypeptide of 24 amino acids will be formed
2) Two polypeptides of 24 and 25 amino acids will be formed
3) A polypeptide of 49 amino acids will be formed
4) A polypeptide of 25 amino acids will be formed
459. A sequence of how many nucleotides in messenger RNA makes a codon for an amino acid ?
1) Three
2) Four
3) One
4) Two
460. Which one of the following pairs is correctly matched with regard to the codon and the amino
acid coded by it ?
1) UUU-Valine
2) AAA-Lysine
3) AUG-Cysteine
4) CCC-Alamino
461. A strand of DNA has following base sequence 3-AAAAGTGACTAGTGA-5'. On transcription,
it produces an m-RNA which of the following anticodon of t-RNA recognizes the third codon of
this mRNA
1) AAA
2) CUG
3) GAC
4) CTG
462. Given below is sequence of the processed mRNA ready for translation
‘5-AUG CUA UAC UAA CUG CCA UGC UAG-3’
How many amino acids will present in polypeptide chain corresponding to this mRNA
1) 7
2) 8
3) 6
4) 3
463. Protein synthesis in an animal cell occurs
1) On ribosomes present in cytoplasm as well as in mitochondria
2) On ribosomes present in the nucleolus as well as in cytoplasm
3) Only on ribosomes attached to the nuclear envelope and endoplasmic reticulum
4) Only on the ribosomes present in cytosol
464. Which one of the following statement is true for protein synthesis (translation)
1) Amino acids are directly recognized by m-RNA
2) The third base of the codon is less specific
3) Only one codon codes for an amino acid
4) Every t-RNA molecule has more than one amino acid attachment site
465. The drug streptomycin inhibits the process of
1) Prokaryotic translation
2) Eukaryotic translation
3) Prokaryotic transcripion
4) Eukaryotic transcription
466. Translation is the process in which
1) D.N.A. is formed on D.N.A template
2) R.N.A. is formed on D.N.A. template
3) D.N.A. is formed on R.N.A. template
4) Protein is formed from R.N.A. message
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467. In a polypeptide chain of 125 amino acids, if the 25th amino acid is mutated to UAA, then
1) A polypeptide of 124 amino acid is formed
2) A polypeptide of 25 amino acid is formed
3) A polypeptide of 24 amino acid is formed
4) Any of the above can be possible
468. The first codon discovered by Nirenberg and Mathii was
1) CCC
2) GGG
3) UUU
4) AAA
469. Which of the following is not correct about translation
1) It starts with AUG
2) Stopped at termination codon
3) Based on operon model
4) Occurs in nucleus
470. t-RNA attaches, amino acid at its
1) 3' end
2) 5' end
3) Anticodon
4) Loop
471. Arrangement of three successive bases in the genetic code signifies
1) Protein
2) Nucleic acid
3) Plasmids
4) Amino acids
472. Out of 64 codons only 61 codes forthe 20 different amino acids.. This character of genetic code is
called
1) Degeneracy
2) Non ambiguous nature
3) Redundancy
4) Overlapping
473. Anticodons are found in
1) m RNA
2) t RNA
3) r RNA
4) In all
474. One-gene-one enzyme hypothesis was proposed by
1) Beadle and Tatum
2) Jacob and Monod
3) Lederberg
4) Watson and Crick
475. How many ATP and GTP molecules are required respectively for incorporation of 25 amino
acids in peptide chain ?
1) 20 ATP 20 GTP
2) 25 ATP, 25 GTP
3) 50 ATP. 50 GTP
4) 25 ATP 50 GTP
476. Which of the following RNA play structural and catalytic role during translation.
1) m-RNA
2) t-RNA
3) r-RNA
4) All
477. Transfer of genetic information from a polymer of nucleotides to a polymer of amino acid is 1) Replication
2) Transcription
3) Translation
4) Reverse transcription
478. Translation refers to the process of 1) Polymerisation of nitrogen bases
2) Polymerisation of nucleotides
3) Polymerisation of nucleosides
4) Polymerisation of amino acids
479. Khorana & his collegeous synthesized an RNA molecule with repeating sequences of U G N2bases. The RNA with "UGU GUG UGU GUG" produced a tetra peptide with alternating sequences of cystein and valine. This prove that codon for cystein & valine is
1) UGG, GUU
2) UUG, GGU
3) UGU & GUG
4) GUG & UGU
REGULATION
480. Gene and cistron words are sometimes used synonymously because
1) One cistron contains many genes
2) One gene contains many cistrons
3) One gene contains one cistron
4) One gene contains no cistron
481. A gene containing multiple exons and at least one intron is termed as
1) split gene
2) operator gene
3) synthetic gene
4) epistatic gene
482. Gene which is responsible for the synthesis of a polypeptide chain is called
1) Promotor gene
2) Structural gene
3) Regulator gene
4) Operator gene
483. Which is true for tryptophan operon
1) It is the example of inducible operon
2) It is example of repressible operon
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co-repressor
3) on 
4) (2) and (3) both are correct
 offf
484. The end product of a metabolic pathway may bind a repressor to make the latter active enough
to bind to operator. In this case the end-product is called
1) Inducer
2) Aporepressor
3) Co-repressor
4) Regulator
485. Which is true for repressible operon
Inducer
1)Off 
2) Inactive repressor + Co-repressor = active
 on
Inducer
3) Active repressor + Inducer = inactive repressor 4) On 
 offf
486. What does "lac" refer to, in what we call the lac operon
1) Lactose
2) Lactase
3) Lac insect
4) The number 1,00,000
487. Which of the following is not produced by E.Coli in the lactose operon 1) P galactosidase
2) Thiogalactoside transacetylase
3) Lactose dehydrogenase
4) Lactose permease
488. A functional complex comprising a cluster of genes including structural gene, a promoter gene,
an operator gene and a regulator gene was discovered by
1) Beadle and Tatum (1958)
2) Watson and crick (1953)
3) Jacob and Monad (1961)
4) Britten and Davidson (1961)
489. Functioning of structural genes is controlled by
1) Operator
2) Promoter
3) Ligase
4) Regulator gene
490. Who explained the operon model for the first time
1) Francois Jacob
2) Jacques Monod
3) Francois Jacob and Jacques Monod
4) Beadle & Tatum
491. The accessibility of promotor regions of prokaryotic DNA by RNA polymerase is in many cases
regulated by the interaction of some protein with sequences termed as
1) Promoter
2) Operator
3) Regulator
4) Cistron
492. Regulation of lac operon by repressor is referred to as
1) Positive regulation 2) Negative regulation 3) Both (1) and (2)
4) None
493. Which is incorrect
1) i-gene codes for the repressor of lac operon
2) z-gene codes for the beta-galactosidase
3) y-gene codes for transacetylase
4) three gene products are required for metabolism of lactose
494. Which is the primary step for regulation of gene expression.
1) Transport of m-RNA from nucleus to the cytoplasm
2) Translational level
3) Processing level
4) Transcriptional level
495. Find out the correct sequence of structural gene in lac operon
1) y, a, z
2) a, z, y
3) z, y, a
4) z, a y
MUTATION
496. The concept of sudden genetic change which breeds true in an organism is visualized as
1) Natural selection
2) Inheritance of acquired characters
3) Mutation
4) Independent assortment
497. Mutation is
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1) An abrupt or discontinuous change which is inherited
2) A factor for plant growth
3) A change which affects parents only and is never inherited
4) A change which affects the offspring of F2 generation
498. The change of chromosomal parts between non homologous pairs of chromosome
1) Crossing over/Transduction
2) Translocation
3) Inversion
4) Transition
499. Which of the following can be called a mutation
1) The halting of the chromosome number at meiosis
2) The doubling of the chromosome after syngamy
3) The possession of an additional chromosome
4) All the above
500. Mutations are generally
1) Dominant
2) Recessive
3) Codominant
4) Incompeletely dominant
501. The earliest record of point mutation is
1) Short legged sheep by Sneth Wright
2) White eyed male Drosophila by Morgan
3) Autotrophic mutants of Neurospora by Beadle and Tatum
4) Mutants of Escherichia coli discovered by Joshua Lederberg
502. Short-legged variety of sheep is an example of
1) Recessive germinal mutation
2) Dominant germinal mutation
3) Recessive somatic mutation
4) Dominant somatic mutation
503. Genetic mutations occur in
1) DNA
2) RNA
3) Protein
4) RNA & protein both
504. Which of the following undergoes change in mutation
1) Chromosome
2) Structure of gene
3) Sequence of gene
4) Any of the above
505. The locus of mutation is
1) Gene
2) Chromosome
3) Centromere
4) Nucleus
506. Gene mutation is caused
1) Due to reproduction
2) Due to linkage
3) Due to change in sequence of N2 base
4) Due to changes in sequence of genes in DNA
507. The natural rate of mutation is
2) 1 x 10-6
3) 1 x 105
4) 1 x 1010
1) 1 x 10-10
508. H.J. Muller received Nobel Prize for
1) Discovering linked genes
2) Discovering the mutations induced by X-rays
3) His studies on the genetics of Drosophila
4) Proving that DNA was a genetic material
509. X-rays generally cause
1) Polyploidy
2) Frame shift mutations
3) Chromosomal aberrations
4) Paramutations
510. Which of the following causes mutation (not polyploidy)
1) Crossing-over
2) Nacl
3) Colchicine
4) y-rays
511. Non-ionizing radiations commonly used for inducing mutations in organisms are
1) UV-rays
2) Beta-rays
3) X-rays
4) Gamma-rays
512. The smallest unit of genetic material which upon mutation produce a phenotypic effect is
1) Mutons
2) Inducer gene
3) Mutator gene
4) Regulator gene
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513. Ultimate source of genetic variation is (OR) the process which provides raw material for evolution is
1) Sexual reproduction
2) Meiosis
3) Mutation
4) Independent assortment
514. To be evolutionary successful the mutation must occur in (OR) important mutations occur in
1) Somatoplasm
2) Germplasm
3) Karyolymph/Zygote
4) Ergastoplasm
515. Chemical mutagens are far more hazardous than radiations because
1) The exposure to chemicals is more prevalent
2) The organism possess protection for radiation but no protection for chemicals
3) The chemically induced mutations are more deliterious
4) The chemicals are synthetics
516. X-rays cause mutations by
1) Breaking spindle fibers
2) Rupturing nuclear envelope
3) Changing chromosome morphology
4) Inhibiting cytokinesis
517. Mutagens which are effective on replicating DNA only are
4)  and  rays
1) Base analogues
2) Alkylating agents
3) HNO2
518. Which of the following agents cause mutation through deamination
1) 2-Aminopurine
2) 5-Bromo uracil and X-rays
4) Alkyl sulphonates and mustard gas
3) HNO2
519. Haploids are preferred over diploids for mutation studies because
2) Recessive mutation is expressed in F2
1) Recessive mutation is expressed in F1
3) Dominant phenotype is expressed
4) Dominant phenotype is suppressed
520. Type of gene mutation which involves replacement of purine with pyrimidine or vice versa (OR)
The substitution of one type of base with another type of base is
1) Transduction
2) Transversion
3) Translocation
4) Transcription
521. Mutations induced by 5-Bromo uracil are
1) Transversional mutations
2) Transitional mutations
3) Frame shift mutations
4) Backward mutations
522. The minimum requirement for mutation is
1) Change of triplet codon
2) Change in single nucleotide
3) Change in whole DNA
4) Change in single strand of DNA
523. Which of the mutagen causes frame shift mutation
1) 2 Aminopurine
2) Proflavine
3) 5 Bromouracil
4)Methane sulphonates
524. Mutations are
1)Always useful
2) Mostly useful
3) Never useful
4) Rarely useful
525. If a mutation is not visible in successive generations it is called as a
1) Deletion
2) Dominant mutation
3) Recessive mutation
4) Segregation
526. Which of the following is most difficult to detect
1) Auxotrophic mutation
2) Lethal mutation
3) Recessive non-lethal mutation
4) Dominant mutation
527. Deamination of adenine and guanine by HNO- produces
1) Cytosine and uracil
2) Xanthine and hypoxanthine
3) Hypoxanthine and xanthine
4) Xanthine and uracil
528. Sickle cell anaemia is an example of
1) Frame shift mutation
2) Point mutation
3) Segmental mutation
4) Gibberish mutation
529. The most striking example of frame shift mutation was found in a disease called
1) Sickle cell anaemia
2) Colour blindness
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3) Laesh-Nyhn Syndrome
4) Thallesemia
530. Hexaploid wheat developed through
1) Hybridomas
2) Chromosome doubling
3) Hybridisation
4) Hybridisation and chromosome doubling
531. Point mutation is induced by
1) Adenine
2) Guanine
3) 3-cytosine
4) Bromouracil
532. If a diploid cell is treated with colchicine then it becomes
1) Triploid
2) Tetraploid
3) Diploid
4) Monoploid
533. A nutritionally wild type organism, which does not require any additional growth supplement is
known as
1) Holotype
2) Auxotroph
3) Prototroph
4) Phenotype
534. Which of the following is generally used for induced mutagenesis in crop plants 1) Gamma rays (from cobalt 60)
2) Alpha particles
3) X rays
4) UV (260nm)
535. Given below is the representation of a kind of chromosomal mutation :
What is the kind of mutation represented
A B
C
D
E
F
G
A
H
E
F
G
H
B
C
D
1) deletion
2) duplication
3) inversion
4) reciprocal translocation
536. The "cri-du-chat" syndrome is caused by change in chromosome structure involving
1) Deletion
2) Duplication
3) Inversion
4) Translocation
537. A class of mutation induced by addition or deletion of a nucleotide is called
1) Missense
2) Non-sense
3) Substitution
4) frame shift
538.Mutations occur in
1) Dominant genes
2) Recessive genes
3) Lethal genes
4) Mendel’s genes
539. Chromosomes with genes abcdefg becoming abedcfg is :
1) duplication
2) deletion
3) translocation
4) inversion
540. Gene mutation is :
1) mutation in the genes of DNA
2) mutation in the phosphodiester linkage
3) mutation in the chromosomes
4) change in the sequence of nitrogenous bases
541. The frequency of mutant gene in a population is expected to increase if that gene is
1) dominant
2) recessive
3) sex linked
4) favourable selected
542. Chromosome number 2n-1 is an example of
1) trisomy
2) euploidy
3) polyploidy
4) monosomy
543. A mutant micro-organism unable to synthesize a compound required for its growth but able to
grow if the compound is provided, is known as
1) Auxotroph
2) Prototroph
3) Autotroph
4) None of these
544. Rate of mutation is affected by :
1) temperature
2) X-rays
3) gamma and beta radiations
4) all of these
545. After a mutation at a genetic locus the character of an organism changes due to the change in
1) protein structure
2) DNA replication
3) protein synthesis
4) RNA transcription pattern
DNA FINGER PRINTING, HUMAN GENOME
546. DNA finger printing was invented by
1) Kary Muliis
2) Alec Jeffery
3) Dr. Paul Berg
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4) Francis Collins
GENETICS
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547. Which one of the following pairs of terms/names mean one and the same thing
1) Gene pool - genome
2) Codon - gene
3) Cistron - triplet
4) DNA Fingerprinting - DNA profiling
548 What is the first step in the Southern Blot technique
1) Denaturation of DNA on the gel for hybridization with specific probe
2) Production of a group of genetically identical cells
3) Digestion of DNA by restriction enzyme
4) Isolation of DNA from a nucleated cell such as the one from the scene of crime
549. Which step does not involve in DNA finger printing
1) Southern blotting
2) Gel electrophoresis
3) Restriction enzyme digestion
4) Northern blotting
550. The technique of transferring DNA fragment separated on agarose gel to a synthetic membrane
such as nitrocellulose is known as
1) Northern blotting
2) Southern blotting
3) Western blotting
4) Dot blotting
551. Western blotting is used for the identification of
1) DNA
2) RNA
3) Protein
4) All the above
552. Which of the following techniques are used in analyzing restriction fragment length polymorphism (RFLP)
a) Electrophoresis
b) Electroporation
c) Methylation
d) Restriction digestion
1) 'a' and 'c'
2) 'c' and 'd'
3) 'a' and 'd'
4) 'b' and 'd'
553. The approximate number of genes contained in the genome of Kalpana Chawla was
(1) 40,000
2) 30,000
3) 80,000
4) 1,00,000
554 The total number of nitrogenous bases in human genome is estimated to be about
1) 3.5 million
2) 35 thousand
3) 35 million
4) 3.1 billion.
555. The transfer of protein from electrophoretic gel to nitrocellulose membrane is known as
1) transferase
2) northern blotting
3) western blotting
4) southern blotting
556. The method of DNA fingerprinting involves the use of
1) Restriction enzymes
2) Taq polymerase
3) Oligonucleotide primers
4) All the above
557. Which of the following is not associated with HGP
1) Bioinformatics
2) Cloning vectors BAC & YAC
3) Automated DNA sequencers
4) VNTR
558. In density gradient centrifugation , the bulk DNA forms while satellite DNA forms
.
1) Major peak; Minor peak
2) Minor peak; Major peak
3) Major peak; Major peak
4) Minor peak; Minor peak
559. Which step is not correct in DNA finger printing
1) Isolation of DNA
2) Digestion of DNA by DNA ligase enzyme
3) Separation of DNA by electophoresis
4) Hybridisation using labelled VNTR probe
560. DNA fingerprinting method is very useful for
1) DNA tests for identity & relation ships
2) Forensic studies
3) Polymorphism
4)All of the above
1) 1
11) 3
21) 2
31) 3
41) 1
51) 1
GENETICS
2) 3
12) 3
22) 4
32) 4
42) 2
52) 2
3) 2
13) 3
23) 2
33) 3
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109
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60) 2
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GENETICS
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1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
AIPMT 2006
Phenotype of an organism is the result of
1) Mutations and linkages
2) Cytoplasmic effects and nutrition
3) Environmental changes and sexual dimorphism 4) Genotype and environment interactions
How many different kinds of gametes will be produced by a plant having the genotype AABbCC?
(1) Three
2) Four
3) Nine
4) Two
Test cross involves 1) Crossing between two genotypes with recessive trait
2) Crossing between two F1 hybrids
3) Crossing the F1 hybrid with a double recessive genotype
4) Crossing between two genotypes with dominant trait
In Mendel's experiments with garden pea, round seed shape (RR) was dominant over wrinkled
seeds (rr), yellow cotyledon (YY) was dominant over green cotyledon (yy). What are the expected phenotypes in the F2 generation of the cross RRYY x rryy ?
1) Only round seeds with green cotyledons
2) Only wrinkled seeds with yellow cotyledons
3) Only wrinkled seeds with green cotyledons
4) Round seeds with yellow cotyledons, and wrinkled seeds with yellow cotyledons
Which one of the following is the most suitable medium for culture of Drosophila melanogaster?
1) Moist bread
2) Agar-agar
3) Ripe banana
4) cow dung
If a colourblind woman marries a normal visioned man, their sons will be
1) All normal visioned
2) One-half colourblind and one-half normal
3) Three-fourths colourblind and one- fourth normal
4) All colourblind
In which mode of inheritance do you expect more maternal influence among the offspring?
1) Autosomal
2) Cytoplasmic
3) Y-linked
4) X-linked
Which antibiotic inhibits interaction between tRNA and mRNA during bacterial protein synthesis ?
1) Erythromycin
2) Neomycin
3) Streptomycin
4) Tetracycline
Amino acid sequence, in protein synthesis is de-cided by the sequence of
1) tRNA
2) mRNA
3) cDNA
4) rRNA
One gene-one enzyme hypothesis was postulated by
1) R. Franklin
2) Hershey and Chase
3) A.Garrod
4) Beadle and Tatum
One turn of the helix in a B-form DNA is ap-proximately
1) 20 nm
2) 0.34nm
3) 3.4 nm
4) 2 nm
Antiparallel strands of a DNA molecule means that
1) one strand turns anti-clockwise
2) the phosphate groups of two DNA strands, at their ends, share the same position
3) the phosphate groups at the start of two DNA strands are in opposite position (pole)
4) one strand turns clockwise
GENETICS
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13. Cri-du-chat syndrome in humans is caused by the
1) Fertilization of an XX egg by a normal Y-bearing sperm
2) Loss of half of the short arm of chromosome 5
3) Loss of half of the long arm of chromosome 5
4) Trisomy of 21st chromosome
14. Triticale, the first man-made cereal crop, has been obtained by crossing wheat with 1) Rye
2) Pearl millet
3) Sugarcane
4) Barley
AIIMS 2006
15. During protein synthesis in an organism, at one point the process comes to a halt. Select the
group of the three codons from the following from which any one of the three could bring about
this halt
1) UUU, UCC, UAU
2) UUC, IIA, UAC
3) UAG, UGA, UAA
4) UUG, UCA, UCG
16. Thymine is 1) 5-Methyl uracil
2) 4-Methyl uracil
3) 3-Methyl uracil
4) 1-Methyl uracil
17. In which one of the following combinations (1-4) of the number of the chromosomes is the present
day hexaploid whaet correctly represented
Combination
1)
(2)
3)
(4)
Monosomic
21
7
21
41
Haploid
28
28
7
21
Nullisomic
42
40
42
40
Trisomic
43
42
43
43
18.
19.
20.
21.
22.
23.
24.
AIPMT 2007
A human male produces sperms with the genotypes AB, Ab, aB, and ab, in equal proportions.
What is the corresponding genotype of this person
1) AaBb
2) AaBB
3) AABb
4) AABB
In pea plants, yellow seeds are dominant to green. If a heterozygous yellow seeded plant is crossed
with a green seeded plant, what ratio of yellow and green seeded plants would you expect in
generation
1) 50 : 50
2) 9 : 1
3) 1 : 3
4) 3 : 1
Inheritance of skin colour in humans is an example of
1) chromosomal aberration
2) point mutation
3) polygenic inheritance 4) codominance
Two genes R and Y are located very close on the chromosomal linkage map of maize plant.
When RRYY and rryy genotypes are hybridized, the F2 segregaion will show
1) Higher number of the recombinant types.
2) Segregation in the expected 9:3:3:1 ratio.
3) Segregation in 3:1 ratio.
4) Higher number of the parental types
Differentiation of organs and tissues in a developing organism, is association with
1) Developmental mutations
2) Differential expression of genes
3) Lethal mutations
4) Deletion of genes
Molecular basis of organ differentiation depends on the modulation in transcription by :
1) RNA polymerase
2) Ribosome
3) Transcription factor
4) Anticodon
The Okazaki fragments in DNA chain growth :
1) Result in transcription
2) Polymerize in the 3'-to-5' direction and forms replication fork
3) Prove semi-conservative nature of DNA repli-cation
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4) Polymerize in the 5'-to-3' direction and explain 3'-to-5' DNA replication
25. The two polynucleotide chains in DNA are :
1) Parallel
2) Discontinuous
3) Antiparallel
4) Semiconservative
26. In the hexaploid wheat, the haploid (n) and basic (x) numbers of chromosomes are
1) n= 7 and x=21
2) n=21 and x=21
3) n=21 and x=14
4) n=21 and x=l
27.
28.
29.
30.
31.
32.
33.
AIPMT 2008
Which of the following nitrogen base is not found in DNA1) Thymine
2) Cytosine
3) Guanine
4) Uracil
Polysome is formed by
1) A ribosome with several subunits
2) Ribosomes attached to each other in a linear arrangement
3) Several ribosomes attached to a single mRNA
4) Many ribosomes attached to a strand of endoplasmic reticulum
Which one of the following pairs of nitrogenous bases of nucleic acids, is wrongly matched with
the category mentioned against it ?
1) Guanine, Adenine - Purines
2) Adenine, Thymine - Purines
3) Thymine, Uracil - Pyrimidines
4) Uracil, Cytosine - Pyrimidines
In the DNA molecule
1) the proportion of Adenine in relation to thymine varies with the organism
2) there are two strands which run antiparallel one in 5'  3' direction and other in 3'  5'
3) the total amount of purine nucleotides and pyrimidine nucleotides is not always equal
4) there are two strands which run parallel in the 5'  3' direction
Which one of the following pairs of codons is correctly matched with their function or the signal
for the particular amino acid ?
1) AUG, ACG - Start/Methionine
2) UUA, UCA -Leucine
3) GUU, GCU -Alanine 4) UAG, UGA - Stop
Which of the following bond is not related to nucleic acid :
1) H-bond
2) Ester bond
3) Glycosidic bond
4) Peptide bond
Haploids are more suitable for mutation studies than the diploids. This is because
1) haploids are more abundant in nature than diploids
2) All mutations, whether dominant or recessive are expressed in haploids
3) Haploids are reproductively more stable than diploids
4) Mutagens penetrate in haploids more effectively than in diploids
AIPMT 2009
34. Sickle cell anemia is
1) Characterized by elongated sickle like RBCs with a nucleus
2) An autosomal linked dominant trait
3) Caused by substitution of valine by glutamic acid in the beta globin chain of haemoglobin
4) Caused by a change in a single base pair of DNA
35. The most popularly known blood grouping is the ABO grouping. It is named ABO and not ABC,
because "O" in it refers to having
1) No antigens A and B on RBCs
2) Other antigens besides A and B on RBCs
3) Overdominance of this type on the genes for A and B types
4) One antibody only - either anti-A or anti-B on the RBCs
GENETICS
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36. Select the incorrect statement from the following
1) Baldness is a sex-limited trait
2) Linkage is an exception to the principle of independent assortment in heredity
3) Galactosemia is an inborn error of metabolism
4) Small population size results in random genetic drift in a population
37. Study the pedigree chart given below
`
38.
39.
40.
41.
42.
What does it show
1) Inheritance of a recessive sex-linked disease like haemophilia
2) Inheritance of a sex-lined inborn error of metabolism like phenylketonuria
3) Inheritance of a condition like phenylketonuria as an autosomal recessive trait
4) The pedigree chart is wrong as this is not possible
What is not true for genetic code
1) It is unambiguous
2) A codon in mRNA is read in a non-contiguous fashion
3) It is nearly universal
4) It is degenerate
Removal of introns and joining the exons in a defined order in a transcription unit is called
1) Capping
2) Splicing
3) Tailing
4) Transformation
Semiconservative replication of DNA was first demonstrated in
1) Salmonella typhimurium
2) Drosophila melanogaster
3) Escherichia coli
4) Streptococcus pneumoniae
Whose experiments cracked the DNA and discovered unequivocally that a genetic code is a
"triplet"
1) Beadle and tatum
2) Nirenberg and Mathaei
3) Hershey and Chase
4) Morgan and Sturtevant
Point mutation involves
1) Deletion
2) Insertion
3) Change in single base pair
4) Duplication
AIPMT 2010
43. The genotype of a plant showing the dominant phenotype and can be detemined by
1) Pedigree analysis
2) Back cross
3) Test cross
4) Dihybrid cross
44. Which one of the following cannot be explained on the basis of Mendel's Law of Dominance?
1) Alleles do not show any belending and both the characters recover as such in F 2 generation
2) Factors occur in pairs
3) The discrete unit controlling a particular character is called a factor
4) Out of one pair of factors one is dominant and the other recessive
45. ABO blood groups in humans are controlled by the gene I. It has three alleles - I A, IB and i. Since
there are three different alleles, six different genotypes are possible. How many phenotypes can
occur?
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1) Four
2) Two
3) Three
4) One
46. Select the corrrect statement from the ones gives below with respect to dihybrid cross
1) Genes loosely linked on the same chromosome show similar recombinatios as the tightly linked ones
2) Tightly linked genes on the same chromosome show very few recombinations
3) Tightly linked genes on the same chromosome show higher recombination
4) Genes far apart on the same chromosome show very few recombinations
47. Select the two correct statements out of the four (a-d) given below about lac operon
a) Glucose or galactose may bind with the repressor and inactivated
b) In the absence of lactose the repressor binds witht the operator region
c) The z-gene codes for permease
d) This was elucidated by Francois Jacob and Jacque Monod
The correct statements are
1) (b) and (d)
2) (a) and (b)
3) (b) and (c)
4) (a) and (c)
48. Which one of the following symbols and its repre-sentation, used in human pedigree analysis is
correct?
1)
3)
= unaffected female
= mating between relatives
2)
= male affected
4)
= unaffected male
49. Satellite DNA is useful tool in :
1) Forensic science
2) Genetic engineering
3) Organ transplantation
4) Sex detemination
50. The one aspect which is not a salient feature of genetic code, is its being :
1) Universal
2) Specific
3) Degenerate
4) Ambiguous
51. Which one of the following does not follow the central dogma of molecular bilogy?
1) Chlamydomonas
2) HIV
3) Pea
4) Mucor
52.
53.
54.
55.
AIPMT (Pre.) 2011
When two unrelated individuals or lines are crossed, the performance of F1 hybrid is often
superior to both its parents. This phenomenon is called
1) Heterosis
2) Transformation
3) Splicing
4) Metamorphosis
Which one of the following conditions correctly describes the manner of determining the sex in
the given example ?
1) Homozygous sex chromosomes (ZZ) determine female sex in Birds.
2) XO type of sex chromosomes determine male sex in grasshopper
3) XO condition in humans as found in Turner Syndrome, determines female sex.
4) Homozygous sex chromosomes (XX) produce male in Drosophila
Test cross in plants or in Drosophila involves crossing:
1) Between two genotypes with dominant trait
2) Between two genotypes with recessive trait
4) The F1 hybrid with a double recessive genotype
3) Between two F1 hybrids
Which one of the following conditions of the zygotic cell would lead to the birth of a normal
human female child ?
1) One X and one Y chromosome
2) Two X chromosome
3) Only one Y chromosome
4) Only one X chromosome
AIPMT (Pre.) 2012
56. F2 generation in a Mendelian cross showed that both genotypic and phenotypic ratios are same
GENETICS
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as 1 : 2 : 1. It represents a case of
1) Monohybrid cross with complete dominance 2) Monohybrid cross with incomplete dominance
3) Co-dominance
4) Dihybrid cross
57. A certain road accident patient with unknown blood group needs immediate blood transfusion.
His one doctor friend at once offers his blood. What was the blood group of the donor ?
1) Blood group O
2) Blood group A
3) Blood group B
4) Blood group AB
58. A normal visioned man whose father was colour-blind marries a woman whose father was also
colour blind. They have their first child as a daughter. What are the chances that this child
would be colour-blind?
1) 25%
2) 50%
3) 100%
4) Zero percent
59. PCR and Restriction Fragment Length Polymorphism are the methods for
1) DNA sequencing
2) Genetic fingerprinting
3) Study of enzymes
4) Genetic transformation
AIPMT (Mains) 2012
60. Genetic transformation A test cross is carried out to :
1) assess the number of alleles of a gene.
2) determine whether two species or varieties will breed successfully.
3) determine the genotype of a plant at F2
4) predict whether two traits are linked.
61. Represented below is the inheritance pattern of a certain type of traits in humans. Which one of
the following conditions could be an example of this pattern?
1) Haemophilia
3) Phenylketonuria
2) Thalassemia
Female
Male
Mother
Father
Daughter
Son
4) Sickle cell anaemia
62. What is it that forms the basis of DNA Fingerprinting?
1) The relative amount of DNA in the ridges and grooves of the fingerprints.
2) Satellite DNA occurring as highly repeated short DNA segments
3) The relative proportions of purines and pyrimidines in DNA
4) The relative difference in the DNA occurence in blood, skin and saliva
63. Read the following four statements (A-D):
A) In transcription, adenosine pairs with uracil.
B) Regulation of lac operon by repressor is referred to as positive regulation.
C) The human genome has approximately 50,000 genes.
D) Haemophilia is a sex-linked recessive disease.
How many of the above statements are right?
1) Four
2) One
3) Two
4) Three
64. Which one of the following is a wrong statement regarding mutations?
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1) UV and Gamma rays are mutagens
2) Change in a single base pair of DNA does not cause mutation
3) Deletion and insertion of base pairs cause frame- shift mutations.
4) Cancer cells commonly show chromosomal aberrations.
65.
66.
67.
68.
69.
70.
71.
NEET-UG 2013
If two persons with 'AB' blood group marry and have sufficiently large number of children,
these children could be classified as 'A' blood group : 'AB' blood group 'B' blood group in 1 : 2
: 1 ratio. Modern technique of protein electrophoresis reveals presence of both 'A' and 'B' type
proteins in 'AB' blood group individuals. This is an example of
1) Complete dominance
2) Codominance
3) Incomplete dominance
4) Partial dominance 124
Which Mendelism idea is depicted by a cross in which the F1 generation resembles both the parents?
1) co-dominance
2) incomplete dominance
3) law of dominance
4) inheritance of one gene
Which of the following statements is not true of two genets that show 50% recombination frequency ?
1) If the genes are present on the same chromosome, they undergo more than one crossovers in every meiosis
2) The genes may be on different chromosomes
3) The genes are tightly linked
4) The genes show independent assortment
The incorrcct statement with regard to Haemophilia is :
1) A single protein involved in the clotting of blood is affected
2) It is a sex-linked disease
3) It is a recessive disease
4) It is a dominant disease
If both parents are carriers for thalessemia, which is an autosomal recessive disorder, what are
the chances of pregnancy resulting in an affected child?
1) 100%
2) No chance
3) 50%
4) 25%
Which enzyme/s will be produced in a cell in which there is a nonsense mutation in the lac Y gene ?
1) Lactose permease and transacetylase
2)  -galactosidase
3) Lactose permease
4) Transacetylase
DNA fragments generated by the restriction endonucleases in a chemical reaction can be separated by :
1) Restriction mapping
2) Centrifugation
3) Polymerase chain reaction
4) Electrophoresis
AIIMS 2013
72. In a given pedigree there is a inheritance of certain character of two families, which of the
following option is correct for the inheritance of that character?
Family A
Family B
1) Family A has three daughters & two sons
3) Family A has homozygous parents
GENETICS
2) Family A has three sons & two daughters
4) Family A has Y-linked disorder
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73.
74.
75.
76.
77.
78.
79.
80.
81.
82.
83.
84.
85.
AIPMT 2014
Fruit colour in squash is an example of
1) Recessive epistasis
2) Dominant epistasis
3) Complementary genes
4) Inhibitory genes
A man whose father was colour blind marries a woman who had a colour blind mother and
normal father. What percentage of male children of this couple will be colour blind ?
1) 25%
2) 0%
3) 50%
4) 75%
In a population of 1000 individuals 360 belong to genotype AA, 480 to Aa and the remaining 160
to aa. Based on this data, the frequency of allele A in the population is
1) 0.4
2) 0.5
3) 0.6
4) 0.7
Commonly used vectors for human genome sequencing are
AIIMS 2014
What is the correct sequence of DNA finger printing?
a-seperation of desired DNA by gel electrophoresis
b-Digestion by restriction endonuclease
c- Isolation of DNA
d- Hybridisation using labelled VNTR probe
e- Southern blotting
2) b  d  e  a  c
1) a  b  c  d  e
4) c  b  a  e  d
3) c  b  a  d  e
AIPMT 2015
A man with blood group A' marries a woman with blood group 'B'. What are all the possible
blood groups of their offsprings ?
1) A,B and AB only
2) A,B,AB and O
3) O only
4) A and B only
How many pairs of contrasting characters in pea plants were studied by Mendel in his experiments?
1) Six
2) Eight
3) Seven
4) Five
In sea urchin DNA, which is double stranded, 17% of the bases were shown to be cytosine. The
percentages of the other three bases expected to be present in this DNA are
1) G 17%, A 16.5%, T 32.5%
2) G 17%, A 33%, T 33%
3) G 8.5%, A 50%, T 24.5%
4) G 34%, A 24.5%, T 24.5%
The movement of a gene from one linkage group to another is called
1) Duplication
2) Translocation
3) Crossing over
4) Inversion
Gene regulation governing lactose operon of £. coli that involves the lac 1 gene product is
1) Negative and inducible because repressor protein prevents transcription
2) Negative and repressible because repressor protein prevents transcription
3) Feedback inhibition because excess of P-galactosidase can switch off trascription
4) Positive and inducible because it can be induced by lactose
Multiple alleles are present :
1) At different loci on the same chromosome
2) At the same locus of the chromosome
3) On non-sister chromatids
4) On different chromosomes
Which is the most common mechanism of genetic variation in the population of sexually reproducing organism?
1) Chromosomal aberrations
2) Genetic drift
3) Recombination
4) Transduction
Alleles are :
1) true breeding homozygotes
2) different molecular forms of a gene
3) heterozygotes
4) different phenotype
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86. A population will not exist in Hardy - Weinberg equilibrium if :
1) There are no mutations
2) There is no migration
3) The population is large
4) Individuals mate selectively
87.
88.
89.
90.
91.
92.
93.
94.
95.
Re-AIPMT 2015
A colour blind man marries a woman with normal sight who has no history of colour blindness
in her family. What is the probability of their grandson (son's son) being colour blind ?
1) 0.25
2) 0.5
3) 1
4) Nil
The term "linkage" was coined by :
1) W.Sutton
2) T.H. Morgan
3) T. Boveri
4) G.Mendel
Which of the following biomolecules does have a phosphodiester bond ?
1) Nucleic acids in a nucleotide
2) Fatty acids in a diglyceride
3) Monosaccharides in a polysaccharide
4) Amino acids in a polypeptide
A pleiotropic gene :
1) controls multiple traits in an individual
2) is expressed only in primitive plants
3) is a gene evolved during Pliocene
4) controls a trait only in combination with another gene
In his classic experiments on pea plants, Mendel did not use :
1) Flower position
2) Seed colour
3) Pod length
4) Seed shape
A gene showing codominance has :
1) both alleles independently expressed in the heterozygote
2) one allele dominant on the other
3) alleles tightly linked on the same chromosome
4) alleles that are recessive to each other
Identify the correct order of organisation of genetic material from largest to smallest :
1) Chromosome, genome, nucleotide, gene
2) Chromosome, gene, genome, nucleotide
3) Genome, chromosome, nucleotide, gene
4) Genome, chromosome, gene, nucleotide
Which one of the following is not applicable to RNA?
1) Chargaff's rule
2) Complementary base pairing
3) 5' phosphoryl and 3' hydroxyl ends
4) Heterocyclic nitrogenous bases
In the following human pedigree, the filled symbols represent the affected individuals. Identify
the type of given pedigree.
1) X-linked dominant 2) Autosomal dominant 3) X-linked recessive
4) Autosomal recessive
96. Satellite DNA is important because it :
1) Codes for enzymes needed for DNA replication
2) Codes for proteins needed in cell cycle
3) Shows high degree of polymorphism in population and also the same degree of polymorphism in an
individual, which is heritable from parents to children
4) Does not code for proteins and is same in all members of the population
GENETICS
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AIIMS 2015
97. In cells of superfemale with 47 chromosomes (44 + x x x) visible barr bodies are
1) 1
2) 0
3) 2
4) 3
98. rRNA is synthesised in
1) Nucleus
2) Golgi body
3) Cytoplasm
4) Nucleoplasm
99. Which of the following is involved in translation
1) DNA
2) mRNA, tRNA, DNA
3) mRNA, tRNA
4) Only mRNA
100. Male cat is either black or orange because of
1) Hemizygous - x
2) Heterozygous - x
3) Heterozygous - y
4) Hemizygous - y
101. Which set of RNA are involved in protein synthesis.
1) tRNA, mRNA, rRNA 2) tRNA, mRNA, hnRNA
3) hnRNA, mRNA, rRNA
4) hnRNA, tRNA, rRNA
102. There will be no Barr body in female suffering from
1) Turner syndrome
2) Kleinfelter syndrome
3) Down syndrome
4) Haemophilia
103. Grey is dominant (G) over black (g). Which of the following will most probably give 50% black
and 50% grey offspring ?
1) GG x gg
2) Gg x gg
3) GG x Gg
4) gg x gg
104. Haemophilic gene does not transfer from
1) Haemophilic father to son
2) Haemophilic mother to son
3) Haemophilic father to daughter
4) Haemophilic mother to son & daughter
NEET - I 2016
105. Which of the following most appropriately describes haemophilia ?
1) Recessive gene disorder
2) X-linked recessive gene disorder
3) Chromosomal disorder
4) Dominant gene disorder
106. Which of the following is required as inducer(s) of the expression of Lac operon ?
1) Glucose
2) Galactose
3) Lactose
4) Lactose and Galactose
107. A tall true breeding garden pea plant is crossed with a dwarf true breeding garden pea plant.
When the F1 plants were selfed the resulting genotypes were in the ratio of :
1) 1 : 2 : 1 :: Tall homozygous : Tall heterozygous : Dwarf
2) 1 : 2 : 1 :: Tall heterozygous : Tall homozygous : Dwarf
3) 3 : 1 :: Tall : Dwarf
4) 3 : 1 :: Dwarf : Tall
108. Match the terms in Column-I with their description in Column-II and choose the correct option:
Column-I
Column-II
a) Dominance
i) Many gene s govern a single character
b) Codominance
ii)In a heterozygousorganism only oneallele expresses itself
c) Pleiotropy
iii)In a heterozygousorganism both allelesexpress themselvesfully
d) Polygenic inheritance (iv)A single geneinfluences manycharacters
(a)
b)
c)
d)
1) (ii)
i)
iv)
iii)
2) (ii)
iii)
iv)
i)
3) (iv)
i)
ii)
iii)
4) (iv)
iii)
i)
ii)
109. Joint Forest Management Concept was introduced in India during :
1) 1960 s
2) 1970 s
3) 1980 s
4) 1990 s
110. Pick out the correct statements :
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a) Haemophilia is a sex-linked recessive disease
b) Down's syndrome is due to aneuploidy
c) Phenylketonuria is an autosomal recessive gene disorder.
d) Sickle cell anaemia is a X-linked recessive gene disorder
1) (a) and (d) are correct 2) (b) and (d) are correct
3) (a), (c) and (d) are correct
4) (a), (b) and (c) are correct
111. Which of the following is not required for any of the techniques of DNA fingerprinting available
at present?
1) Polymerase chain reaction
2) Zinc finger analysis
3 ) Restriction enzymes
4) DNA-DNA hybridization
NEET - II 2016
112. If a colour-blind man marries a woman who is homozygous for normal colour vision, then probability of their son being colour-blind is
1) 0.5
2) 0.75
3) 1
4) 0
NEET 2017
113. Thalassemia and sickle cell anaemaia are caused due to a problem in globin molecule synthesis.
Select the correct statement
1) Both are due to a quantitatitive defect in globin chain synthesis
2) Thalassemia is due to less synthesis of globin molecules
3) Sickle cell anaemia is due to a quantitative problem of globin molecules
4) Both are due to a qualitative defect in globin chain synthesis
114. The genotypes of a husband and wife are IAIB and IAi.
Among the blood types of their children, how many different genotypes and phenotypes are
possible ?
1) 3 genotypes; 4 phenotypes
2) 4 genotypes; 3 phenotypes
3) 4 genotypes; 4 phenotypes
4) 3 genotypes; 3 phenotypes
115. A disease caused by an autosomal primary non disjunction is
1) Klinefelter’s syndrome
2) Turner’s syndrome
3) Sickle cell anaemia
4) Down’s syndrome
116. Among the following characters, which one was not considered by Mendel in his experiments on
pea ?
1) Trichomes-Glandular or non-glandular
2) Seed-Green or yellow
3) Pod-Inflated or constricted
4) Stem-Tall or dwarf
117. Which one from those given below is the period for Mendel’s hybridisation experiments ?
1) 1840-1850
2) 1857-1869
3) 1870-1877
4) 1856-1863
118. The final proof for DNA as the genetic material came from the experiments of
1) Hershey and Chase
2) Avery, MacLeod and McCarty
3) Hargobind Khorana
4) Griffith
119. If there are 999 bases in an RNA that code for a protein with 333 aminoacids, and the base at position
901 is deleted such that the length of the RNA becomes 998 bases, how many codons will be altered ?
1) 11
2) 33
3) 333
4) 1
120. During DNA replication, Okazaki fragments are used to elongate
1) the lagging strand towards replication fork
2) the leading strand away from replication fork
3) the lagging strand away from the replication fork
4) the leading strand towards replication fork
121. Which of the following RNA’s should be most abundant in animal cell ?
1) tRNA
2) mRNA
3) miRNA
4) rRNA
GENETICS
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122. Spliceosomes are not found in cells of
1) fungi
2) animals
3) bacteria
4) plants
123. The association of histone H 1 with a nucleosome indicates that
1) DNA replication is occuring
2) the DNA is condensed into a chromatin fibre
3) the DNA double helix is exposed
4) transcription is occuring
1) 4
11) 3
21) 4
31) 4
41) 2
51) 2
61) 1
71) 4
81) 2
91) 3
101) 1
111) 2
121) 4
2) 4
12) 3
22) 2
32) 4
42) 3
52) 1
62) 2
72) 1
82) 2
92) 1
102) 1
112) 4
122) 3
3) 3
13) 2
23) 3
33) 2
43) 3
53) 2
63) 3
73) 2
83) 2
93) 4
103) 2
113) 2
123) 2
4) 4
14) 1
24) 4
34) 4
44) 1
54) 4
64) 2
74) 3
84) 3
94) 1
104) 1
114) 2
5) 3
15) 3
25) 3
35) 1
45) 1
55) 2
65) 2
75) 3
85) 2
95) 4
105) 2
115) 4
6) 4
16) 1
26) 4
36) 1
46) 2
56) 2
66) 1
76) 2
86) 3
96) 3
106) 3
116) 1
7) 2
17) 4
27) 4
37) 3
47) 1
57) 1
67) 3
77) 4
87) 4
97) 3
107) 1
117) 4
8) 2
18) 1
28) 3
38) 2
48) 3
58) 4
68) 4
78) 2
88) 2
98) 1
108) 2
118) 1
9) 2
19) 1
29) 2
39) 2
49) 1
59) 2
69) 4
79) 3
89) 1
99) 3
109) 3
119) 2
10) 4
20) 3
30) 2
40) 3
50) 4
60) 3
70) 2
80) 2
90) 1
100) 1
110) 4
120) 3
1. On crossing red & white flowered plants the ratio of red and white flowered plants in F2-generation was 60 :20, then on selfing the heterozygous red flowered plants, the offsprings would be
1) 72:24
2) 40:60
3) 52:48
4) 84:16
2. What is the ratio of one pair of contrasting characters in F2 of a dihybrid cross
1) 5:3
2) 3:1
3) 9:3:3:1
4) 1:2:2:4:1:2:1:2:1
3. What is the probability of homozygous plants for both dominant characters in F 2 generation of
a dihybrid cross
1) 1/16
2) 3/16
3) 4/16
4) 9/16
4. Which of the following is significance of dominance
1) Organisms with dominant genes are more vital
2) Harmful mutations are not expressed due to dominant gene
3) Heterosis is due to dominant gene
4) All the above
5. An offspring of two homozygous parents differing from one another by alleles at only one gene
locus is known as
1) Back cross
2) Monohybrid
3) Dihybrid
4) Trihybrid
6. Cross AABb X aaBb yields AaBB : AaBb : Aabb:aabb offspring in the ratio of
1) 0:3:1:1
2) 1:2:1:0
3) 1:1:1:1
4) 1:2:1:1
7. Mrs. verma has a autosomal gene pair 'Bb' and she contain x-linked gene 'd' both on of her xchromosome. What is the percentage of gamete which contain 'bd' genes
1) 1/2 or 50%
2) 1/4 or 25%
3) 3/4 or 75%
4) 1 or 100%
8. Independent assortment of genes does not takes place when
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9.
10.
11.
12.
13.
14.
15.
16.
1) Genes are located on homologous chromosomes
2) Genes are linked and located on same chromosome
3) Genes are located on non-homologous chromosome
4) All the above
A dihybrid plant on self pollination, produced 400 phenotypes with 9 types of genotype. How
many seeds will have genotype TtRr
1) 200
2) 100
3) 50
4) 150
If two pea plants baving red (dominant) coloured flowers with unknown genotypes are crossed,
75% of the flowers are red and 25% are white. The genotypic constitution of the parents having
red coloured flowers will be
1) Both homozygous
2) One homozygous and other heterozygous
3) Both heterozygous
4) Both hemizygous
In a dihybrid cross where two parents differ in two pairs of contrasting traits like seed colour
yellow (YY) and seed colour green (yy) with seed shape round (RR) and seed shape wrinkled
(rr). The number of green coloured seeds (yy) among sixteen products of F2 generation will be
1) 2
2) 4
3) 6
4) 8
Select the incorrect statement for Gregor Mendel
1) He conducted hybridization experiments on garden pea for seven years.
2) He applied statistical analysis and mathematical logic for the first time to the problems in biology.
3) His experiments had a small sampling size.
4) He conducted artificial cross-polination experiments using several true-breeding pea lines.
"When two pairs of traits are combined in a hybrid, segregation of one pair of characters is
independent of the other pair of characters". This explains
1) Law of dominance
2) Law of segregation
3) Law of independent assortment
4) Postulate of paired factors
Which of the following is not observed in a monohybrid cross
1) Recessive parental trait is expressed without any blending in the F2-generation
2) Recessive parental trait is expressed without any blending in the F1- generation
3) Dominance also explains the proportion of 3: 1 obtained at the F 2
4) Genotype ratio is 1 : 2 : 1
Which statement is true
1) Characters segregate during formation of gametes
2) All characters show true dominance
3) The characters always blend in heterozygous condition
4) Mendelian disorder are determined by or absence or excess of one or more chromosome
If 3n is the theoretically possible number of different genotypes (when n = the number of chromosome pairs with each carrying one pairs of heterozygous alleles), the different genotypes
produced by pea plant is
1) 310
2) 312
3) 314
4) 37
AB
Female
17. gametes
GENETICS
?
Ab
aB
ab
AaBB AaBb aaBB aaBb
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18.
19.
20.
21.
22.
23.
24.
25.
26.
In the Punnet square given above, the genotype of the female parent and male parents respectively
1) AABB, AaBb
2) AaBB, AaBb
3) aaBB, AaBb
4) AAbb, AaBb
When AABBcc is crossed with AaBbCc then the ratio of hybrid for all the three genes is
1) 1/8
2) 1/4
3) 1/16
4) 1/32
In a plant, red fruit (R) is dominant over yellow fruit (r) and tallness (T) is dominant over
shortness (t). If a plant with RRTt genotype is crossed with a plant that is rrtt
1) All the offsprings will be tall with red fruit
2) 25% will be tall with red fruit
3) 50% will be tall with red fruit
4) 75% will be tall with red fruit
Out of three characters on chromosome no. 4, two characters indicate linkage and not mentioned by Mendel. These characters were
1) Pod form - stem length
2) Pod form - pod position
3) Pod form - pod colour
4) Pod position - stem length
The Punnett square shown below represents the pattern of inheritance in dihybrid cross when
yellow (Y) is dominant over white (y) and round (R) is dominant over wrinkled (r) seeds
YR Yr
yR
yr
YR
F
J
N
R
Yr
G
K
O
S
yR
H
L
P
T
yr
I
M
Q
U
A plant of type ‘H’ will produce seeds with the genotype identical to seeds produced by the
plants of
1) Type M
2) Type J
3) Type P
4) Type N
In a plant gene A' is responsible for tallness and its recessive allele 'a' for dwarfness and 'B' is
responsible for red flower colour and it's recessive allele 'b' for white flower colour. A tall and
red flowered plant with genotype AaBb crossed with dwarf and red flowered (aaBb). What is
the percentage of dwarf-white flowered offspring of above cross
1) 50%
2) 6.25%
3) 12.5%
4) 50%
In rabbit black skin (B) is dominant over brown skin (b) and short hair (S) is dominant over
long hair (s). If homozygous black-short haired male is crossed with a homozygous brown-long
haired female. All F1-offspring are heterozygous black-short haired. F1 male crossed with F1female. In F2 generation what is the percentage of homozygous black-short haired offspring
1) 50%
2) 12.5%
3) 6.25%
4) 18.75%
Which of the following points further strengthened Mendelism
1) law of independent assortment which was based on monohybrid cross
2) law of independent assortment which could be stated on the basis of segregation of gametes
3) incomplete dominance gave a new way to mendelism
4) a character controlled by a pair of unit factors
If Aabb xaaBb, then genotypic ratio of its progeny will be
1) 9 : 3 : 3 : 1
2) 1 : 2 : 1
3) 1 : 1 : 1 : 1
4) 4 : 1
Red and tall dominant character hybrid plant when crossed with recessive white dwarf plant
(RrTt x rrtt). What will be the ratio of respective four combinations red tall, red dwarf, white
tall and white dwarf plants in the next generation
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27.
28.
29.
30.
31.
32.
33.
34.
35.
1) 9 : 3 : 3 : 1
2) 15 : 1 : 0 : 0
3) 9 : 3 : 4 : 0
4) 4 : 4 : 4 : 4
A plant with genotype AABbCcDD is self pollinated. Provided that the four genes are independently assorting, what proportion of the progeny will show the genotype AAbbccDD?
1) 1/4
2) 1/16
3) 1/64
4) 1/256
A rooster with gray feathers was mated with a hen of same phenotype. Among their offspring 15
were gray, 6 black and 8 white. What phenotypes would you expect among the offspring resulting from mating of gray rooster and black hen?
1) all black
2) all gray
3) Equal proportion of black and gray
4) 1/4 gray and 3/4 black
In duroc jersey hog, the coat colour is dependant on two pairs of alleles, R and r and S and s.
Any genotype containing at least one R-gene and at least one S-gene results in red coat colour.The
double recessive genotype results in white coat colour. All other genotypes results in sandy coat
colour. If one hog with genotype "RrSs" mated with another hog with genotype "rrss" what
kind of offsprings will be produced by above cross
1) 9 red : 6-sandy : 1 white
2) 9 red : 3-sandy : 4 white
3) 12 red : 3-sandy : 1 white
4) 1 red : 2-sandy : 1 white
In a plant three dominant independently assorting gene A, B and C are essential for production
of purple pigment. If any of the genes or all three genes are present in recessive condition then
flower is colourless
A
B
C
Rawmaterial 
X 
 Y 
Z pigment
A purple plant with genotype AABBCC crossed with a colourless plant with genotype aabbcc
gives purple F1 hybrid. On selfing of F1- what proportion of coloured offspring in F2
1) 27/64
2) 1/64
3) 9/64
4) 37/64
A person with unknow blood group under ABO system, has suffered much blood bss in an
accident and needs immediate blood transfusion. His one friend who has a valid certificate of his
own blood type, offers for blood donation without delay. What would have been the type of
blood group of the donor friend ?
1) Type B
2) Type AB
3) Type O
4) Type A
A roan bull is bred to three cows. Cow A has the same genotype as the roan bull, cow B is red and
cow C is white, what proportions of roan cows are expected in the offsprings of each group of
cows
1) 2,1,1
2) 1,2,1
3) 1,1,2
4)3,1
If dominant C and P genes are essential for the development of purple colour in sweet pea
flowers, what would be the ratio of white and purple colour in a cross between CcPpx Ccpp
1) 5 : 3
2) 9 : 7
3) 2 : 6
4) 6 : 2
In a family, father has a blood group A' and mother has a blood group 'B'. Their children show
50% probability for a blood group AB' idicating that 1) Father is heterozygous
2) Mother is heterozygous
3) Either of parent is heterozygous
4) Mother is homozygous
In man, gene producing the disease phenyl ketonuria also produces a number of abnormal phenotypic traits, which are collectively syndrome. This gene results mental retardation, widely spaced
incisors, pigmented patches on the skin and excessive sweating such types of genes are called
(1) Polygene
2) Pleiotropic gene
3) Lethal gene
4) Supplimentary gene
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36. Which one of the following conditions though harmful in itself, is also a potential sviour from a
mosquito borne infectious disease:
1) Thalaessaemia
2) Sickle cell anaemia
3) Pernicius anaemia
4) Leukemia
37. Which of the four couples claiming the baby with O+ blood type are possibly the biological
parents of it?
1) AB- and A+
2) A+ and O3) O+ and AB+
4) B- and O38. When a red grain variety of wheat is crossed with another white grain variety a F ^hybrid is
produced. On selfing of this F1- hybrid, how many offsprings of F2 - generation resemble phenotypically to it's parents (let grain colour of wheat controlled by three gene pairs)
1) 2/16
2) 20/64
3) 15/64
4) 2/64
39. In a plant flower colour is the example of quantitative trait and controlled by one gene pair. How
many plants show parental phenotype in F2 generation
1) 2/16
2) 2/4
3) 2/64
4) 2/256
40. Grain colour in wheat is determined by three pairs of polygenes. Following cross AABBCC
(dark colour) x aabbcc (light colour), in F2 generation what proportion of the progeny is likely to
resemble either parent ?
1) None
2) Less than 5 per cent 3) One third
4) Half
41. The weight of fruit in a plant is determined by the number of dominant alleles of a certain
number of genes. If seven weight categories are noticed, how many gene sites would be involved?
1) two
2) three
3) four
4) five
42. A scientist performed the gene mapping experiments in maize. He mapped the genes on chromosomes on the basis of % crossing over between different genes. One map unit corresponds to one
% crossing over or recombination. The genes showing more than 50% recombination were not
supposed to be linked on same chromosome. In crossing over studies on maize, scientist observed
the following % crossing over between genes A, B, C, D -between. A and D 10%, between A and C
3%, between genes C and D 7%, between genes A and B 5%, and between genes C and B 8%.
On the basis of above observation find out the correct sequence of genes A, B, C and D on
chromosomes
1) BCDA
2) ABCD
3) BACD
4) DACB
43. A man and a woman,who do not show any apparent signs of a certain inherited disease, have seven
children (2 daughters and 5 sons). Three of the sons suffer from the given disease but none of the
daughters are affected. Which of the following mode of inheritance do you suggest for this disease
1) Sex-limited recessive(2) Autosomal dominant 3) Sex-linked recessive 4) Sex-linked dominant
44. A test cross of F1 flies +a/+b produced the following offspring
++/ab = 9
ab/ab = 9
+b/ab = 41
a+/ab = 41
What will be distance between linked gene
1) 82 cM
2) 18 cM (cis)
3) 20 cM
4) 18 cM (trans)
45. A cross between white eyed female and red eyed male Drosophila gives rise red eyed females and
white eyed male. Rarely the cross gives rise to white eyed females and red eyed males. This is due to
1) Loss of one X-chromosome
2) Segregation of X- chromosome in female
3) Non disjunction of X- chromosome in female 4) Non disjunction of X- chromosome in male
46. Human embryo have all the genetic instructions, it needed to become a male or female. The male
foetus have a master gene which acts as a biological switch, turning other genes on and off. Loss
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47.
48.
49.
50.
51.
52.
53.
54.
of this gene results in female which remain sexually immature. This master gene is located on
1) Homologous part of X-chromosome
2) Non homologous part of X-chromosome
3) All the autosome
4) Y-chromosome
A man and woman are both affected by vitamin D resistance rickets, which is a dominant sexlinked allele. All of the female offsprings of this couple are affected with rickets but some of the
male offsprings are not. What are the genotypes of the parents?
1) Both are homozygous for the trait.
2) The woman has two dominant alleles and man has one dominant allele.
3) Both parents have only recessive alleles.
4) Each parent has only one dominant allele.
Depending upon the distance between any two genes which is inversely proportional to the
strength of linkage, cross overs will vary from
1) 50-100%
2) 0-50%
3) 75-100%
4) 100-150%
What shall be the ratio of heterozygous, homozygous and colourblind hemizygous in offsprings
of a colour blind husband & a carrier wife
1) 1:1:2
2) 1:1:1
3) 2:1:1
4) 1:2:1
A and B genes are linked. What shall be genotype of progeny in a cross between AB/ab and ab/
ab :
1) AAbb and aabb
2) AaBb and aabb
3) AABB and aabb
4) None
Mendelian dihybrid and dihybrid with linkage are respectively realated with how many chromosomes
1) 1 pair & 2 pair
2) 2 pair & 1 pair
3) 2 pair & 2 pair
4) 1 pair & 1 pair
In Drosophila the XXY condition leads to femaleness whereas in human beings the same condition leads to Klienfelter's syndrome in male. It proves
1) In human beings Y chromosome is active in sex determination
2) Y chromosome is active in sex determination in both human beings and Drosophila
3) In Drosophila Y - chromosome decides femaleness
4) Y chromosome of man has genes for syndrome
Based on observation on monohybrid crosses Mendel draw some conclusion. Which of the following is not correct
1) Characters are controlled by discrete units called factors
2) Factors occur in pairs
3) In a similar pair of factors one member of the pair dominates the other
4) The postulate of dominance also explains the proportion of 3 : 1 obtained at the F 2
Which is incorrect
i) ABO blood groups are controlled by the gene I
ii) Gene I has four alleles
iii) IA and IB produce same type of sugar
iv) i or 1O produce different type of sugar
v) IA and IB are incomplete dominant
1) i, ii
2) v, ii
3) ii, iii, iv
4) ii, iii, iv, v
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55. Theoretically a normal phenotype is expressed when a particular substrate trasnform in to product but in which of following condition phenotype may be affected
1) When the modified allele produce normal enzyme
2) When the modified allele produce a non functional enzyme
3) When the unmodified allele produce no enzyme
4) All the above
56. The pedigree shows the occurence of albinism which is a recessive trait. If person 4 is homozygous, the carrier for the trait is
1
2
female
male
4
3
albinism
6
5
1) 1, 4, 5 and 6
2) 5 and 6
3) 1, 2 and 3
4) 1, 2, 5 and 6
57. This is paedigree for autosomal recessive disease albinism (aa) what is probability of H-I is
homozygous Normal
1
2
1
1
2
2
3
3
4
6
5
4
5
6
7
8
1) 1/3
2) 1/2
3) 2/3
4) 1/4
58. A human male is heterozygous for autosomal gene A, B and G. He is also haemizygous for hemophilic gene h. What proportion of his sperm will be abgh
1) 1/4
2) 1/8
3) 1/16
4) 1/32
59. Haemolytic jaundice is due to dominant gene but only 20% of the people develop this disease. A
heterozygous man marries a homozygous normal woman. What proportion of the children in
population would be expected to have this disorder
1) 1/5
2) 1/20
3) 1/10
4) 1/2
60. A pedigree is shown below for a disease that is autosomal dominant. The genetic make up of the
first generation is
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Generation - I
Generation - II
Generation - III
1) AA, Aa
2) Aa, aa
3) Aa, AA
4) Aa, Aa
61. An organism is able to live on a culture medium containing nutrient A. by the enzyme catalysed
reactions X
Y
A 
 B 
 C A mutant failed to survive on this medium bu grew when nutrient B was
added to it. Which gene of this mutant was defective
(1) Only X
2 Only Y
3) X and Y both
4) Neither X or Y
62. Given below is the pedigree of an autosomal dominant disorder-Myotonic dystrophy.
63.
64.
65.
66.
In this pedigree the genotype of all affected children will be
1) AA
2) Aa
3) AA or Aa
4) aa
A point mutation which involves change of A  G, C  T, C  G and T  A in DNA aree
1) Transition, Transition, Transversion, Transversion
2) Transition, Transversion, Transition, Transversion
3) Transversion, Transition, Transversion, Transition
4) None of the above
A segment of DNA has a base sequence : AAG, GAG, GAC, CAA, CCA-, Which one of the
following sequence represents a frame shift mutation
1) AAG, GAG, GAC, CAA, CCA2) AAG, AGG, ACC, AAC, CAA 3) ACG, GAG, GAC, CAG, CCA 4) AAG, GCG, GAC, CAG, CCA If the DNA-codons are ATGATGATG and a cytosine base is inserted at the begining which of
the following would be the result
1) A non sense mutation
2) CA, TGA, TGA, TG
3) CAT, GAT, GAT, G
4) C, ATG, ATG, ATG
What is an auxotroph
1) A plant that responds by bending towards the sun
2) A mutant (organism) which has lost its ability to synthesize one or more essential compounds
3) An organism that depends on another organism for meeting its nutritional requirements
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67.
68.
69.
70.
71.
72.
73.
4) A plant that is able to synthesize its own carbohydrates
The most likely reason for the development of resistance against pesticides in insects damaging
a crop is
1) Genetic recombination
2) Directed mutations
3) Acquired heritable changes
4) Random mutations
Male XX and female XY sometime occur due to :
1) Deletion
2) Transfer of segments in X and Y chromosomes
3) Anneuploidy
4) Hormonal imbalance
In recent years, DNA sequences (nudeotide sequence) of mt-DNA and Y chromosomes were
considered for the study of human evolution, because
1) They are small, and therefore, easy to study
2) They are uniparental in origin and do not take part in recombination
3) Their structure is known in great detail
4) They can be studied from the samples of fossil remains
A completely radioactive double stranded DNA molecule undergoes two round of raplication in
a non radioactive medium. What will be the radioactive status of the four daughter molecules
1) All four still contain radioactivity
2) Radioactivity is lost from all four
3) Out of four, three contain radioacitivity
4) Half of the number contain no radioacitivity
Consider the following sequence on m-RNA AUGGCAGUGCCA. Assuming that genetic code is
overlap then how many number of codon may be present on this genetic code
1) 9
2) 10
3) 8
4) 11
A normal DNA molecule is continuously replicate in N15 medium than what is the % of lighter
DNA in 4th generation.
1) 12-5%
2) 25%
3) 0%
4) 6-25%
Find out the sequence of binding of the following amino acyl - t-RNA complexes during translation to an m-RNA transcribed by a DNA segment having the base sequence. 3' ATACCCATGGGG
5'. Choose the answer showing the correct order of alphabets
Aminoacid
Aminoacid
a)
b)
Anticodon
C
C
Anticodon
C
A
U
G
Aminoacid
Aminoacid
c)
d)
Anticodon
G
G
Anticodon
G
A
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1) a, b, c, d
2) d, a, b, c
3) a, b, d, c
4) b, a, c, d
74. KHORANA synthesized two RNAs (a) with repeat sequence of AB and (b) with repeat sequence
of ABC the polypeptides coded by (a) & (b) are respectively
1) Homopolypeptides in both (a) and (b)
2) Heteropolypeptides in both
3) Homopolypeptide in (a) & peptide heteropoly in (b)
4) Heteropolypeptide in (a) & peptide homopoly in (b)
75. Which of the following m-RNA is translated completely
A) 5' AUG UGA UUA AAG AAA 3'
B) 5' AUG AUA UUG CCC UGA 3'
C) 5' AGU UCC AGA CUC UAA 3'
D) 5' AUG UAC AGU AAC UAG 3'
1) (A) and (B)
3) (C) and (D)
2) (B) and (D)
4) (A) and (D)
76. In a m-RNA sequence of N2-base is 5' AUG GUG CUC AAA' 3'. What is the correct sequence of
anticodons which recognizes codons of m-RNA
a)
b)
c)
d)
UUU
GAG
UAC
CAC
1) a, b, c, d
2) d, a, b, c
3) c, d, b, a
4) d, c, b, a
77. Suppose evolution on earth has occurred in such a way that there are 96 amino acids instead of
20. DNA has 12 different types of bases and DNA synthesis occur in the same way as today. The
minimum number of bases per DNA condon would be
1) 12
2) 8
3) 2
4) 3
78. Assume that their are 6 types of nitrogen bases available and 40 types of amino acid are available for
protein synthesis, then in genetic code each codon made up by minimum how many nitrogen bases ?
1) 3
2) 4
3) 5
4) 2
79. In a segment of DNA 3.2 kelobases are present. If DNA segment has 820 adenine molecules, then
what will be number of cytosine ?
1) 1560
2) 1480
3) 780
4) 740
80. Which statement is correct ?
a) Degeneracy of code is related to, third member of codon
b) Single codon, codes for more than one aminoacid
c) In codon first two bases are more specific
d) In codons, third base is wobble
e) Code is universal
1) a, b, c, d, e
2) a, b, d
3) a, c, d
4) a, c, d, e
81. Both the strand of DNA are not copied during transcription because
1) If both strands act as a template, they would code for RNA with different sequence
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2) The two RNA molecules, if produced simultaneously would be complementary to each other, hence
would form a double stranded RNA
3) They would code, for RNA molecules with same sequences
4) Both (1) and (2) are correct
82. The salient feature of DNA are
i) It is made of two polynucleotide chain
ii) Back bone is constituted by sugar and nitrogen base
iii) Two chain have parallel polarity
iv) Bases in two strands are paired through H- bonds
v) The two chain are coiled in a left handed fashion
1) i, iv, v
2) i, iv
3) i, ii, v
4) i, ii, iii, iv, v
83. Which is incorrect for genetic codei) The codon is triplet
ii) 64 codons code for amino acids
iii) Genetic code is unambiguous
iv) Genetic code is nearly universal
v) AUG has dual functions
1) only ii
2) ii & iii
3) iii, iv & v
4) All are correct
84. Which is correct i) t-RNA has an anticodon loop that has bases complementary to the code
ii) t-RNA has an amino acid acceptor end
iii) t-RNA are specific for each amino acid
iv) For initiation, there is specific t-RNA that is reffered to as initiator t-RNA
v) For termination there is specific t-RNA that is reffered to as terminator t-RNA
1) i, ii
2) i, ii, iii
3) i, ii, iii, iv
4) i, ii, iii, iv, v
85. E. coli cells with a mutated z gene of the lac operon cannot grow in medium containing only
lactose as the source of energy because
1) They cannot synthesize functional betagalactosidase
2) They cannot transport lactose from the medium into the cell
3) The lac operon is constitutively active in these cells
4) In the presence of glucose, E. coli cells do not utilize lactose
86. DNA fingerprinting refers to
1) Techniques used for identification of fingerprints of individuals
2) Molecular analysis of profiles of DNA samples
3) Analysis of DNA samples using imprinting devices
4) Techniques used for molecular analysis of different specimens of DNA
87. Select the incorrect statement.
1) DNA from single cell is enough to perform DNA fingerprinting analysis
2) DNA fingerprinting has much wider applications in determining population & genetic diversities.
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3) The VNTR belongs to a class of satellite DNA referred as microsatellite.
4) DNA fingerprint differs from individual to individual in a population except in th case of monozygotic twins.
88. DNA fragments separated by gel electrophoresis are shown. Mark the correct statement
-
1
2
3
+
1) Band 3 contains more positively charged DNA molecule than 1
2) Band '3' indicates more charge density than 1 and 2
3) Band 1 has longer DNA fragment than 2 and 3
4) All the bands have equal length and charges but differ in base composition
1) 1
11) 2
21) 4
31) 3
41) 2
51) 2
61) 1
71) 2
2) 2
12) 3
22) 3
32) 1
42) 3
52) 1
62) 2
72) 3
3) 1
13) 3
23) 3
33) 1
43) 3
53) 3
63) 1
73) 2
4) 2
14) 2
24) 4
34) 3
44) 4
54) 4
64) 2
74) 4
5) 2
15) 1
25) 3
35) 2
45) 3
55) 2
65) 3
75) 2
6) 2
16) 4
26) 4
36) 2
46) 4
56) 4
66) 2
76) 3
7) 1
17) 3
27) 2
37) 2
47) 4
57) 1
67) 4
77) 3
8) 2
18) 1
28) 3
38) 4
48) 2
58) 3
68) 2
78) 1
9) 2
19) 3
29) 4
39) 2
49) 2
59) 3
69) 2
79) 3
10) 3
20) 1
30) 1
40) 2
50) 2
60) 2
70) 4
80) 4
81) 4
82) 2
83) 1
84) 3
85) 1
86) 2
87) 3
88) 3
89) 1
90) 1
Directions for Assertion & Reason questions
These questions consist of two statements each, printed as Assertion and Reason. While answering
these Questions you are required to choose any one of the following four responses.
A) If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.
B) If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.
C) If Assertion is True but the Reason is False.
D) If both Assertion & Reason are false.
1. Assertion : Inheritance is the basis of heredity.
Reason : Inheritance is the process by which characters are passed on from parent to progeny.
1) A
2) B
3) C
4) D
2. Assertion : Human exploited the variations that were present naturally in the wild populations of
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3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
plants & animals.
Reason : Variation is the degree by which progeny resembles from their parents.
1) A
2) B
3) C
4) D
Assertion : During Mandel's investigations into inheritance patterns, it was for the first time that
statistical analysis & mathematical logics were applied to the problems in biology.
Reason : Mendel conducted hybridization experiments on garden pea for seven years & proposed
laws of inheritance.
1) A
2) B
3) C
4) D
Assertion : A true breeding line shows the stable trait inheritance and expression for several generations.
Reason : A true breeding line is one that have undergone continuous cross pollination.
1) A
2) B
3) C
4) D
Assertion : Mendel conducted artificial pollination experiments using several true-breeding pea lines.
Reason : Mendel selected 14 true-breeding pea plant varieties.
1) A
2) B
3) C
4) D
Assertion : Mendel's experiment had a small sampling size.
Reason : It gave greater credibility to the data that he collected.
1) A
2) B
3) C
4) D
Assertion : Mendel never supported blending inheritance.
Reason : He found that the F1 always resembled either one of the parents.
1) A
2) B
3) C
4) D
Assertion : According to Mendelism. contrasting traits did not show any blending.
Reason : Only one of the parental traits was expressed in the F1 while at the F2 stage both the traits
were expressed in the proportion of 1 : 1.
1) A
2) B
3) C
4) D
Assertion : Genes are the units of inheritance.
Reason : They contain the information that is required to express a particular trait in an organism.
1) A
2) B
3) C
4) D
Assertion : Alleles are slightly different form of the same gene.
Reason : They are the genes which code for a pair of contrasting traits of a character.
1) A
2) B
3) C
4) D
Assertion : Mendel proposed that in a true breeding, tall or dwarf pea variety the allelic pair of genes
for height are identical i.e. TT and tt respectively.
Reason : Alleles can be similar as in the case of homozygotes or can be dissimilar as in the case of
heterozygotes.
1) A
2) B
3) C
4) D
Assertion : The segregation of alleles is a random process and so there is a 50% chance of a gamete
containing either allele.
Reason : By the process of mitosis, the alleles of the parental pair separates from each other & only
one allele in transmitted to a gamete.
1) A
2) B
3) C
4) D
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13. Assertion : Punnett square is a graphical representation to calculate the probability of all possible
genotypes of offsprings in a genetic cross.
Reason : It was developed by a British geneticist, Reginald C. Punnett.
1) A
2) B
3) C
4) D
14. Assertion : Simply looking at the phenotype of a dominant trait, it is not possible to know the genotypic composition.
Reason : Mendel performed reciprocal crosses to know the genotype of dominant organism.
1) A
2) B
3) C
4) D
15. Assertion :Factors occurs in pairs in an organism.
Reason : In a dissimilar pair of factors one member of the pair dominates the other.
1) A
2) B
3) C
4) D
16. Assertion : Law of dominance is based on the fact that both the characters are recovered as such in F 2
generation.
Reason : In a pair of factors one member always dominates the other.
1) A
2) B
3) C
4) D
17. Assertion : Incomplete dominance made it possible to distinguish heterogyzous from homozygous.
Reason : In incomplete dominance F1 had a phenotype that did not resemble either of the two parents
and was in between the two.
1) A
2) B
3) C
4) D
18. Assertion : The unmodified allele, which represents the original phenotype is the dominant allele and
the modified allele is generally the recessive allele.
Reason : The modified allele could be responsible for production of a non functional enzyme or no
enzyme at all.
1) A
2) B
3) C
4) D
19. Assertion : There are six different combinations or genotypes of the human ABO blood types.
Reason : The gene (I) has three alleles IA, IB and
1) A
2) B
3) C
4) D
20. Assertion : In co-dominance and incomplete dominance, the genotypic & phenotypic ratios are same.
Reason : In case of co-dominance the F1 generation resembles both parents.
1) A
2) B
3) C
4) D
21. Assertion : Multiple alleles can be observed or studies in an organism, when its complete genome is
known.
Reason : Multiple alleles is a rare phenomenon.
1) A
2) B
3) C
4) D
22. Assertion : Multiple alleles can be found only when population studies are made.
Reason : Occasionally, a single gene product may produce more than one effect.
1) A
2) B
3) C
4) D
23. Assertion : Round seeds in pea plant may have large sized or intermediate sized starch grains.
Reason : It is an example of pleiotropy, multiple alleles and incomplete dominance.
1) A
2) B
3) C
4) D
24. Assertion : Dominance is not an autonomous feature of a gene or its product.
Reason : It depends on gene product or on the particular phenotype that we choose to examine, when
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25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
more than one phenotype is influenced by the same gene.
1) A
2) B
3) C
4) D
Assertion : According to Mendel, a dihybrid cross is the multiple of two monohybrid crosses.
Reason : When two pairs of traits are combined in a hybrid, segregation of one pair of characters is
independent of the other pair of characters.
1) A
2) B
3) C
4) D
Assertion : On the basis of Punnett square, the F2 generation of dihybrid plant will produce 16 different types of genotypes.
Reason : In Fx generation plant, there are four genotypes of gametes with a frequency of 25% of total
gametes produced.
1) A
2) B
3) C
4) D
Assertion : According to Mendel, factors are stable & discrete units that control the expression of
traits.
Reason : He could not provide any physical proof for the existance of factors.
1) A
2) B
3) C
4) D
Assertion : Each pair of chromosome segregate independently of another pair.
Reason : Each gene pair also segregates independently of another pair.
1) A
2) B
3) C
4) D
Assertion : The pairing & separation of a pair of chromosomes would lead to the segregation of a pair
of factors they carried.
Reason : All the factors of a chromosome shows independent separation.
1) A
2) B
3) C
4) D
Assertion : Drosophila were found very suitable for genetical studies.
Reason : They complete their life cycle in about two weeks and produces fix number of offsprings.
1) A
2) B
3) C
4) D
Assertion : In
grasshopper, some of the sperms bear X-chromosome whereas some do not.
Reason : Grasshopper is an example of XX-XY type of sex determination.
1) A
2) B
3) C
4) D
Assertion : XO type, XY type & ZW type are the example of male heterogamety.
Reason : Male produces two types of sperms.
1) A
2) B
3) C
4) D
Assertion : In birds, the females have one Z and one W chromosome, whereas male have a pair of Zchromosomes besides autosomes.
Reason : In birds, sex of the offsprings is decided by the temperature of surroundings when they are
released.
1) A
2) B
3) C
4) D
Assertion : Egg is responsible for sex of the chicks & not the sperm.
Reason : Birds shows female heterogamety.
1) A
2) B
3) C
4) D
Assertion : In each pregnancy there is always 50 percent probability of either a male or a female child
in human.
Reason : There is an equal probability of fertilization of the ovum with the sperm carrying either X or
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36.
37.
38.
39.
40.
41.
42.
43.
44.
45.
Y chromosome.
1) A
2) B
3) C
4) D
Assertion : Mutation results in changes in the genotype and the phenotype of an organism.
Reason : Mutation is a phenomenon, which results in alternation of DNA sequences.
1) A
2) B
3) C
4) D
Assertion :In human beings, the inheritance of a particular trait is done by the study of family history.
Reason : Such an analysis of traits in several generations of family is called the pedigree analysis.
1) A
2) B
3) C
4) D
Assertion : In human genetics, pedigree study provides a strong tool, which is utilized to trace the
inheritance of a specific trait, abnormality or disease.
Reason : The controlled crosses that can be performed in pea plant or some other organisms, are not
possible in case of human beings.
1) A
2) B
3) C
4) D
Assertion : Consanguineous mating results in inbreeding depression.
Reason : Consanguineous mating is represented by 0=0 symbols in pedigree.
1) A
2) B
3) C
4) D
Assertion :
symbol represents parents with affected male child.
Reason : It is a autosomal dominant disorder.
1) A
2) B
3) C
4) D
Assertion : Mendelian disorders are mainly determined by alteration or mutation in the single gene
Reason : These disorders are transmitted to the offsprings on the same lines as studied in the principle
of inheritance.
1) A
2) B
3) C
4) D
Assertion :An X-linked recessive trait shows transmission from carrier male to female progeny.
Reason : X-linked recessive trait shows non-crisscross inheritance.
1) A
2) B
3) C
4) D
Assertion :The family pedigree of Queen Victoria shows a number of haemophilic descendent as she
was a carrier of the disease.
Reason : Haemophilia is a sex linked dominant disease which shows its transmission from unaffected
carrier female to some of the male progeny
1) A
2) B
3) C
4) D
Assertion :Sickle-cell anaemia is caused by the substitution of glutamic acid by valine at the sixth
position of the beta globin chain of the haemoglobin molecule.
Reason : It is due to the single base substitution at the sixth codon of the beta globin gene from GUG
to GAG.
1) A
2) B
3) C
4) D
Assertion : In sickle cell anaemia, the shape of RBC becomes elongated sickle like from biconcave
disc.
Reason : The mutant haemoglobin molecule undergoes polymerization under low oxygen tension
causing the change in shape.
1) A
2) B
3) C
4) D
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46. Assertion : Gynacomastia is observed in Turner's syndrome.
Reason : Such a disorder is caused due to the presence of extra copy of X-chromosome.
1) A
2) B
3) C
4) D
47. Assertion : Trisomy or monosomy situation leads to a very serious consequences in the individual.
Reason : Down's syndrome is the trisomy of chromosome-21
1) A
2) B
3) C
4) D
48. Assertion : Male is haploid and female is diploid in honey bee.
Reason : Sex depends on paired and unpaired chromosomes in every organism.
1) A
2) B
3) C
4) D
49. Assertion : In multiple alleles many alleles are present in a population for a character.
Reason : All These alleles are present in an individual.
1) A
2) B
3) C
4) D
50. Assertion : In a person with AB-blood group, the erythrocytes carry both A and B antigen on their
surface.
Reason : The alleles IA and IB that produce AB blood group are codominant and both are expressed.
1) A
2) B
3) C
4) D
51. Assertion : A gamete contains a single allele for any trait
Reason : During gametogenesis, the two alleles of a trait segregate, one passing into each gamete at
random
1) A
2) B
3) C
4) D
52. Assertion : In Neurospora each gene either dominant or recessive express it's effect.
Reason : Neurospora contain many alleles of a gene.
1) A
2) B
3) C
4) D
53. Assertion : Male and female ratio is almost equal in this world.
Reason : Male and female genotype is XY and XX.
1) A
2) B
3) C
4) D
54. Assertion : Recessive characters are always pure
Reason : Recessive genes always express itself in homozygous condition
1) A
2) B
3) C
4) D
55. Assertion : Recessive mutation easily detectable in Neurospora
Reason : Neurospora is diploid
1) A
2) B
3) C
4) D
56. Assertion : Most of the X-linked disorders can not be eliminated easily from nature and our gene
pool.
Reason : X-linked disorders are due to recessive genes.
1) A
2) B
3) C
4) D
57. Assertion : Albinism is X-linked inheritance.
Reason : Females suffer more with disorder.
1) A
2) B
3) C
4) D
58. Assertion :Mutation play key role in the process of evolution.
Reason : Mutation can change the genotypic constituent.
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59.
60.
61.
62.
63.
64.
65.
66.
67.
68.
69.
70.
71.
1) A
2) B
3) C
4) D
Assertion : Somatic mutations are sometime inheritable.
Reason : Some organism show vegetative propagation.
1) A
2) B
3) C
4) D
Assertion : Colchicine induces polyploidy
Reason : Colchicine causes disjunction of chro-mosomes.
1) A
2) B
3) C
4) D
Assertion : Translocation is a illigal crossing over
Reason : Translocation is exchange of chromosomal segments between nonhomologous chromosome
1) A
2) B
3) C
4) D
Assertion : Frame shift mutation is more harmful then substitution.
Reason : Acredine and proflavin induces frame shift mutation
1) A
2) B
3) C
4) D
Assertion : Chemical mutagens are more harmful as compared to radiations.
Reason : CC14 is firstly discovered chemical mutagen.
1) A
2) B
3) C
4) D
Assertion : Mutations are main source of evolution.
Reason : Mutation causes variations.
1) A
2) B
3) C
4) D
Assertion : A nucleotide is an assemblage of three distinct components.
Reason : Nucleotide is made up of a heterocyclic compounc. me nosaccharide and nitrogenous base.
1) A
2) B
3) C
4) D
Assertion :Adenine and Guanine are substituted purines.
Reason : In Adenine and Guanine purine heterocyclic ring has either amino or amino and oxy groups.
1) A
2) B
3) C
4) D
Assertion : Normally the genetic code is degenerate.
Reason : One amino acid is coded by more than one codon.
1) A
2) B
3) C
4) D
Assertion : UGG is a chain termination codon.
Reason : It does not code any amino acid.
1) A
2) B
3) C
4) D
Assertion : TAC is anticodon of AUG codon.
Reason : Codon and anticodon are non- complementary to each other.
1) A
2) B
3) C
4) D
Assertion : There occur 64 codons in dictionary of genetic code
Reason : Genetic code is quadraplete
1) A
2) B
3) C
4) D
Assertion : Codon for methionine & tryptophan said to be degenerate
Reason : Methionine and tryptophan amino acids coded by more than one codons
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72.
73.
74.
75.
76.
77.
78.
79.
80.
81.
1) A
2) B
3) C
4) D
Assertion : Constant diameter of DNA double helix is 20 A
Reason : Purines always pair with pyrimidines and vice-versa
1) A
2) B
3) C
4) D
Assertion : In a DNA molecule, A-T rich parts melt before G-C rich parts.
Reason : In between A and T there are three FI- bond, whereas in between G and C there are two Hbonds.
1) A
2) B
3) C
4) D
Assertion : Prokaryotic DNA has high Tm as compare to eucaryotic DNA.
Reason : Prokaryotic DNA contains more G-C pairs as compare to eucaryotic DNA
1) A
2) B
3) C
4) D
Assertion : Wobbling process established the economy of t-RNA.
Reason : In wobbling process single t-RNA recognizes more than one codon of m-RNA.
1) A
2) B
3) C
4) D
Assertion : In prokaryotes, there occurs coupled transcription & translation.
Reason : In prokaryotes, translation occurs in cytoplasm and transcription in nucleoplasm.
1) A
2) B
3) C
4) D
Assertion : Replication and transcription occur in the nucleus but translation occurs in the cytoplasm.
Reason : Transcription and translation are unidirectional.
1) A
2) B
3) C
4) D
Assertion : At the time of Mendel, the real basis of inheritance was obscured.
Reason : At the time of Mendel, the nature of those factors regulating the pattern of inheritance was
not clear.
1) A
2) B
3) C
4) D
Assertion :Complementary base pairing confers a very unique property to the DNA chains.
Reason :If sequence of one strand is known then sequence in other strand can be predicted.
1) A
2) B
3) C
4) D
Assertion : Antiparallel polarity helps in stability of DNA.
Reason : It allows complementary pairing between base pairs.
1) A
2) B
3) C
4) D
Assertion : Base pairing between purine and pyrimidines allows uniform distance between two strands
of the helix.
Reason : Number of hydrogen bonds between pairing bases are constant.
1) A
2) B
3) C
4) D
82. Assertion : Hydrogen bonds between base pairs is the only mean to stabilise helical structure.
Reason : Plane of one base pair stacks over other have no role in stability of DNA helics.
1) A
2) B
3) C
4) D
83. Assertion : Positively charged histone proteins are essential for packaging negatively charged DNA.
Reason : Without histone protein DNA can not fold due to negative charge.
1) A
2) B
3) C
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84. Assertion : Packaging of chromatin at higher level requires additional set of proteins.
Reason : Non histone chromosomal proteins provide scaffold for packaging of chromatin.
85.
86.
87.
88.
89.
90.
91.
92.
93.
94.
1) A
2) B
3) C
4) D
Assertion : Euchromatin is said to be transcriptionally active chromatin.
Reason : Due to loose coiling of enchromatin RNA polymerase can easily move up on it.
1) A
2) B
3) C
4) D
Assertion : Unequivocal proof that DNA is the genetic material came from Griffith's transformation
experiment.
Reason : The biochemical nature of genetic material was defined from transformation experiment.
1) A
2) B
3) C
4) D
Assertion :DNA is predominant genetic material.
Reason : It is able to generate its replica.
1) A
2) B
3) C
4) D
Assertion : DNA is chemically and structurally more stable.
Reason : DNA have no catalytic properties.
1) A
2) B
3) C
4) D
Assertion : DNA is chemically and structurally more stable.
Reason : In DNA thymine is present instead of uracil.
1) A
2) B
3) C
4) D
Assertion : RNA is less stable then DNA.
Reason : In RNA 2' - OH group present on every nucleotide.
1) A
2) B
3) C
4) D
Assertion : RNA is not a predominant genetic material.
Reason : RNA being unstable, mutate at faster rate.
1) A
2) B
3) C
4) D
Assertion :RNA is best material for transmission of genetic informations in the same generation.
Reason : RNA can directly code for synthesis of proteins.
1) A
2) B
3) C
4) D
Assertion :RNA viruses having shorter life span and evolve faster.
Reason : In RNA rate of mutations is slower than DNA.
1) A
2) B
3) C
4) D
Assertion : In the same generation for transmission of genetic informations RNA is better than DNA.
Reason : The protein synthesising machinery has evolved around RNA.
1) A
2) B
3) C
4) D
95. Assertion : Essential life processes evolved around RNA.
Reason : Beside genetic material RNA also act as catalyst.
1) A
2) B
3) C
4) D
96. Assertion :In Meselson and Stahl experiment 15NH4C1 can be separated from 14N by radiography.
Reason :In 15NH4C1, 15N is a radioactive isotope.
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1) A
2) B
3) C
4) D
97. Assertion : Taylor and colleages in 1958 used CsCl density gradient centrifugation technique to
prove semiconservative DNA replication.
Reason : They use 15N heavy nitrogen which is not a radioactive isotope.
1) A
2) B
3) C
4) D
98. Assertion : Beside high rate of polymerisation of nucleotides DNA polymerase have to catalyse the
reaction with high degree of accuracy.
Reason : Any mistake during replication wold results into mutations.
1) A
2) B
3) C
4) D
99. Assertion : DNA replication is energetically a very expensive process.
Reason : Unwinding of DNA strands is an active process, while pairing of bases is a passive process.
1) A
2) B
3) C
4) D
100. Assertion : Deoxyribonucleotide triphosphates serve dual purposes.
Reason : In addition to acting as substrate, they provide energy for polymerisation.
1) A
2) B
3) C
4) D
101. Assertion : For long DNA molecule, the two strands of DNA can not be separated in it's entire length.
Reason : It needs very high energy investment.
1) A
2) B
3) C
4) D
102. Assertion : On template 5'  3' DNA replicate in discontinuous manner..
Reason : DNA polymerase catalyse polymerisation only in one direction that is 5'  3'.
1) A
2) B
3) C
4) D
103. Assertion : The replication of DNA and cell division cycle should be highly coordinated.
Reason : A failure in cell division after DNA replication results into chromosomal anomaly.
1) A
2) B
3) C
4) D
104. Assertion : Both the strands of DNA are not copied during transcription.
Reason : Both strands of DNA show antiparallel polarity.
1) A
2) B
3) C
4) D
105. Assertion : The DNA strand which does not code for anything is referred to as coding strand.
Reason : Coding strand sequence are same as RNA except thymine at the place of uracil.
1) A
2) B
3) C
4) D
106. Assertion : The split-gene arrangement complicates the definition of a gene in terms of a DNA
segment.
Reason : In eukaryotes the monocistronic structural genes have interrupted coding sequence.
1) A
2) B
3) C
4) D
107. Assertion : RNA polymerase associates with  -factor and  -factor to terminate & initiate the
transcription respectively.
Reason : Association with these factors will not alter the specificality of the RNA polymerase.
1) A
2) B
3) C
4) D
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108. Assertion : The presence of introns is reminiscent of antiquity & the process of splicing represents
the dominance of RNA world.
Reason : The split gene arrangement represents an advanced feature of the genome.
1) A
2) B
3) C
4) D
109. Assertion : Genetic codes are unambiguous & specific.
Reason : Some amino acids are coded by more than one codon.
1) A
2) B
3) C
4) D
110. Assertion : t RNAs are specific for each amino acid but there are no t-RNAs for stop codons.
Reason : The secondary structure of t-RNA has been depicted that looks like a clover-leaf.
1) A
2) B
3) C
4) D
111. Assertion : An mRNA also have some additional sequences that are not translated & are referred as
UTR.
Reason : The UTRs are present at both 5' end & at 3' end and they have no specific function.
1) A
2) B
3) C
4) D
112. Assertion : Inhibition of transcription is the only method of regulation of gene expression in all organisms.
Reason : Transcription is the first step of gene expression.
1) A
2) B
3) C
4) D
113. Assertion : It is the metabolic, physiological or environmental conditions that regulate the expression
of genes.
Reason : The genes in a cell are expressed to perform a particular function or a set of functions.
1) A
2) B
3) C
4) D
114. Assertion : The development & differentiation of embryo into adult organisms are the result of coordinated regulation of expression of several sets of genes.
Reason : The expression of all genes are not required simultaneously during development of organism.
1) A
2) B
3) C
4) D
115. Assertion : In prokaryotes, control of the rate of transcriptional initiation is the predominant site for
control of gene expression.
Reason : The regulatory protein can act both positively (activators) & negatively (repressors)
1) A
2) B
3) C
4) D
116. Assertion : Lac operator is present only in lac operon & it interact specifically with lac repressor only.
Reason : Each operon has its specific operator & specific repressor.
1) A
2) B
3) C
4) D
117. Assertion : .-The activity of RNA polymerase at a promotor is regulated by interaction with accessory
protein, which affect its ability to recognise start sites.
Reason : The accessibility of promotor region of prokaryotic DNA is in many cases is regulated by
interaction of proteins with operators.
1) A
2) B
3) C
4) D
118. Assertion : Gene regulation in prokaryotes is comparatively simple than eukaryotes.
Reason : In most of prokaryotic operons the genes present in the operon are needed together to function in the same or related metabolic pathway.
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1) A
2) B
3) C
4) D
119. Assertion : In lac operon, a polycistronic structural gene is regulated by a common promoter & regulatory genes.
Reason : Such arrangement is very common in bacteria & is referred as operon.
1) A
2) B
3) C
4) D
120. Assertion : The gene i codes for the repressor of the lac operon.
Reason : The y-gene codes for permease, which increases permeability of the cell to  -galactosidase.
1) A
2) B
3) C
4) D
121. Assertion : Lactose is the substrate for the enzyme p-galactosidase & it regulates switching on & off
of the operon.
Reason : A very low level of expression of lac operon is always present in cell all the time.
1) A
2) B
3) C
4) D
122. Assertion : Regulation of lac operon by repressor is referred to as positive regulation.
Reason : Regulation of lac operon can also be visualised as regulation of enzyme synthesis by its
product.
1) A
2) B
3) C
4) D
123. Assertion : HGP was closely associated with the rapid development of a new area in biology called as
Bioinformatics.
Reason : The enormons amount of data generated in HGP necessitated the use of high speed computational devices for data storage & analysis.
1) A
2) B
3) C
4) D
124. Assertion : About 3300 books, each of 1000 pages are required to store the sequences of human
genome in typed form.
Reason : Human genome is said to have approximately 9 billion base pairs.
1) A
2) B
3) C
4) D
125. Assertion : Addressing the ELSI, in one of the goal of HGP.
Reason : The HGP was a 13 year long project & was completed in 2003.
1) A
2) B
3) C
4) D
126. Assertion : Many non-human model organisms have also been sequenced in HGP.
Reason : Their sequences can be applied towards solving challenges in health care, agriculature,
energy production etc.
1) A
2) B
3) C
4) D
127. Assertion : ESTs are the sequences in DNA that expressed as RNA.
Reason : Sequence Annotation is the blind approach of sequencing entire codeing sequence of genome.
1) A
2) B
3) C
4) D
128. Assertion : For sequencing, the total DNA from a cell is isolated and converted into random fragments.
Reason : DNA is a long polymer & there are technical limitations in sequencing very long pieces of
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DNA.
1) A
2) B
3) C
4) D
129. Assertion : BAC & YAC are the common vectors used in HGP.
Reason : In HGP, sequencing was done by automated DNA sequencers that worked on methods of F. Sanger.
1) A
2) B
3) C
4) D
130. Assertion : The sequencing of chromosome-1 was completed at last in May-2006.
Reason : Chromosome-1 in the longest chromosome with maximum number of genes.
1) A
2) B
3) C
4) D
131. Assertion : Repeated sequence make up very large portion of the human genome.
Reason : They have no direct coding functions but useful in study of chromosome structure, dynamics &
evolution.
1) A
2) B
3) C
4) D
132. Assertion : DNA fingerprinting is a very quick way to compare the DNA sequences of any two individuals.
Reason : The 0.1% difference in base pairs of all humens make tham unique in their phenotypic appearance.
1) A
2) B
3) C
4) D
133. Assertion : The repetitive DNA are separated from bulk genomic DNA as different peaks during density
gradient centrifugation.
Reason : The bulk DNA forms a minor peak & other major peaks are referred to as satellite DNA.
1) A
2) B
3) C
4) D
134. Assertion : DNA fingerprinting involves identifying differences in some specific regions in DNA called as
repetitive DNA sequences.
Reason : These sequences show high degree of polymorphism & form the basis of DNA fingerprinting.
1) A
2) B
3) C
4) D
135. Assertion : DNA polymorphism arises due to mutations.
Reason : An inheritable mutation which is observed in a population at high frequency, is referred to as DNA
polymorphism.
1) A
2) B
3) C
4) D
136. Assertion : The probability of DNA polymorphism would be higher in non-codeing DNA sequences.
Reason : Mutations in these sequences may not have any immediate effect in an individual's reproductive
ability.
1) A
2) B
3) C
4) D
137. Assertion : The VNTR belongs to a class of satellite DNA referred to as mini-satellite.
Reason : The mini-satellite numbers remains same from chromosome to chromosome in an individual.
1) A
2) B
3) C
4) D
138. Assertion : In DNA fingerprinting, after hybridization with VNTR probe, the autoradiogram gives many
bands of different sizes.
Reason : It differes from individual to individual in a population except fraternal twins.
1) A
2) B
3) C
4) D
139. Assertion : The sensitivity of fingerprinting technique has been increased by the use of PCR.
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Reason : DNA from a single cell is not enough to perform DNA fingerprinting ahalysis.
1) A
2) B
3) C
4) D
140. Assertion : Operon concept is applicable only in prokaryotes.
Reason : Gene expression in prokaryotes is influenced by environmental conditions.
1) A
2) B
3) C
4) D
141. Assertion : Removal of introns is must for transcription of hnRNA to mRNA.
Reason : In bacteria introns are not present.
1) A
2) B
3) C
4) D
142. Assertion : Autopolyploids have more than two copies of a single genome.
Reason : Allopolyploids contain more than one type of genome each with atleast two copies.
1) A
2) B
3) C
4) D
143. Assertion : DNA finger printing is a unique process through which recombinants are identified similar to mother parent.
Reason : Micropropagation is a technique to produce genetically uniform plants.
1) A
2) B
3) C
4) D
144. Assertion : Thymine is one of the DNA bases.
Reason : Complementary base of guanosine is thymine.
1) A
2) B
3) C
4) D
145. Assertion : In a mosaic female cat orange and black patches of fur are present.
Reason : At some place orange gene is suppressed and in some skin cells black gene is suppressed.
1) A
2) B
3) C
4) D
1) 1
11) 2
21) 4
31) 3
41) 2
51) 1
61) 1
71) 4
81) 2
91) 1
101) 1
111) 3
121) 2
131) 2
141) 4
2) 3
12) 3
22) 2
32) 4
42) 4
52) 3
62) 2
72) 1
82) 4
92) 1
102) 1
112) 4
122) 4
132) 2
142) 2
3) 2
13) 2
23) 3
33) 3
43) 3
53) 1
63) 3
73) 3
83) 1
93) 3
103) 1
113) 2
123) 1
133) 3
143) 4
4) 3
14) 3
24) 1
34) 1
44) 3
54) 1
64) 1
74) 1
84) 1
94) 1
104) 2
114) 1
124) 3
134) 1
144) 3
5) 2
15) 2
25) 1
35) 1
45) 1
55) 3
65) 3
75) 1
85) 1
95) 1
105) 1
115) 2
125) 2
135) 2
145) 1
6) 4
16) 4
26) 4
36) 1
46) 4
56) 1
66) 1
76) 3
86) 4
96) 4
106) 2
116) 1
126) 1
136) 1
146
7) 1
17) 1
27) 2
37) 2
47) 2
57) 4
67) 1
77) 2
87) 1
97) 4
107) 4
117) 2
127) 3
137) 3
8) 3
18) 1
28) 3
38) 1
48) 3
58) 1
68) 4
78) 1
88) 1
98) 1
108) 3
118) 1
128) 1
138) 3
9) 1
19) 1
29) 3
39) 2
49) 3
59) 1
69) 4
79) 1
89) 1
99) 3
109) 2
119) 2
129) 2
139) 3
10) 1
20) 2
30) 3
40) 3
50) 1
60) 3
70) 3
80) 1
90) 1
100) 1
110) 2
120) 3
130) 1
140) 2
GENETICS