UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter
9
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
PRINCIPLES OF INHERITANCE
AND VARIATION
P Generation
Red
RR
Meiosis I - anaphase
Meiosis II - anaphase
Meiosis II - anaphase
Germ cells
Germ cells
White
rr
×
Gametes
Meiosis I - anaphase
r
R
F1 generation
Pink
Rr
Gametes ½ R ½ r
Eggs ½ R
½ r
F2 generation
½ R Sperms
RR
½ r
Rr
Rr
rr
Independent assortment of chromosomes
Introduction....

Gregor Johann Mendel an Austrian monk, discovered the basic principles of heredity

He conducted experiments on the Pea plant

In 1850 Mendel began to research the transmission of Hereditary traits in plant
hybrids.

He published the results of his studies under the title ‘Experiments on plant Hybrids’

In 1900, Hugo de Vries, Karl correns and Tschermak these three workers rediscovered
Mendel’s work independently

After rediscovery of Mendel’s work, he is rightly known as Father of Genetics.
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CHAPTER AT A GLANCE

INTRODUCTION
 PRACTICE TEST - I

9.1 MENDEL’S EXPERIMENTS

GENETIC TERMINOLOGY

MENDEL’S EXPERIMENTAL PLANT
 PRACTICE TEST - II

9.2 INHERITANCE OF ONE GENE (MONOHYBRID CROSS)
 PRACTICE TEST - III

9.3 DEVIATIONS FROM MENDELIAN CONCEPT OF
DOMINANCE
 PRACTICE TEST - IV

9.4 INHERITANCE OF TWO GENES (DIHYBRID CROSS)


PRACTICE TEST - V
9.5 CHROMOSOMAL THEORY OF INHERITANCE
 PRACTICE TEST - VI

9.6 LINKAGE AND RECOMBINATION

9.7 MUTATIONS
 PRACTICE TEST - VII
QUESTION BANK

EXERCISE - I
CHECK YOUR MEMORY

EXERCISE- II
SHARPEN YOUR REFLEXES

EXERCISE-III
THINK TWICE BEFORE YOU CHOOSE

EXERCISE - IV SIMPLE MATCHING

EXERCISE - V

EXERCISE - VI TRUE OR FALSE

EXERCISE - VII PREVIOUS MEDICAL ENTRANCE
MULTIPLE MATCHING
QUESTIONS
100
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Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
SYNOPSIS
PRACTICE TEST-I
INTRODUCTION
 Genetics is the science deals with the inheritance
as well as the variation of characters from parents
to offspring.
 Inheritance is the process by which characters
are passed from parents to offspring.
 The entire body of molecular biology was a
consequent development with major contributions
from Watson, Crick, Nirenberg, Khorana,
Kornbergs (father and son), Benzer, Monod,
Brenner, etc.
 Mendel factors (now a days genes), those controls
inheritance patterns represent the genetic basis of
inheritance, understanding the structure of genetic
material and the structural basis of genotype and
phenotypic conversions.
 Gregor Johann Mendel is an Austrian monk,
discovered the basic principles of heredity through
experiments on the pea plant.
 Around 1850, Mendel began to research the
transmission of hereditary traits in plant hybrids.
He published the results of his studies under the title
“Experiments on Plant Hybrids”. The
observations he made while growing peas in his
monastery’s garden became the foundation of
modern genetics and the study of heredity.
 “Like begets like” and “as you show so you
reap” because an organism gives birth to organisms
similar to it only.
Eg: An elephant always gives birth to baby elephant
and mango seed forms only plant producing
mangoes.
 Children in a family show similarities to their
parents because certain characters are passed
from the parents to the offspring or progeny
during sexual reproduction. This is called heredity.
However siblings are not identical to parents and
they show differences which are called variations.
 Between 8000 - 1000 BC humans recognised
that causes of variations are in sexual
reproduction.
 They used the variations present in wild populations
for improvement of plants and animals by selective
breeding.
Eg: Sahiwal cows in Punjab.
Ongole bulls in Andhra Pradesh.
 Though ancestors know about heredity and
variation yet they had no idea of scientific basis of
heredity.
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1. Sexual reproduction brings about
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
1) Polyploidy
2) Aneuploidy
3) Euploidy
4) Genetic recombination
Genetics deals with
1) Heredity and Variations 2) Mutations
3) Variations
4) Mendel’s laws
Laws of inheritance are the base to the
scientific explanation of
1) Heredity
2) Variations
3) Eugenics
4) 1 and 2
During sexual reproduction, characters are
transmitted through
1) Vegetative cells
2) Gametes
3) Propagules
4) Clones
The unit of heredity is
1) Chromosome 2) DNA and RNA molecules
3) Nucleus
4) Gene
The most popularised person considered in the
field of genetics is
1) Engler
2) Schwann
3) Johanssen
4) Mendel
Example of an animal developed by artificial
selection and domestication in north India
1) Drosophila
2) Mule
3) Sahiwal cows
4) Ongole bulls
The entire body of molecular biology was a
consequent development with major
contributions from
1) Benzer 2) Brenner 3) Nirenberg 4) 1,2 and 3
The process of transmission of parental
characters to the offspring is
1) Variations
2) Inheritance
3) Mendelism
4) Hybridization
“Characteral resemblances among same
progeny and also with their parents is possible”.
The reason is the difference in characters
among same progeny is due to
1) Variations
2) Gene manipulation
3) Cloning of genes
4) Lack of alleles
Who discovered the basic principles of heredity
through experiments on the pea plant.
1) Mendel
2) Morgan
3) Hugo de Vries
4) Correns
Mendel published the results of his studies
under the title
1) Experiments on animals 2) Enquiry into plants
3) Historia plantarum
4) Experiments on plant hybrids.
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Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
13. Between 8000 - 1000 BC humans recognised
that causes of variations are found in.
1) Asexual reproduction 2)Vegetative reproduction
3) Fission
4) Sexual reproduction
14. Mendel conducted experiments on a plant
belongs to family
1) Solanaceae
2) Fabaceae
3) Liliaceae
4) Cucurbitaceae
15. Example of an animal, developed by artificial
selection and domestication in AP
1) Sahiwal cows
2) Mules
3) Ongole bulls
4) Drosophila
Gregor Johann Mendel
 Mendel was born on 22 July 1822 in Austria.
 In 1856–57, he started his historical experiments
of heredity on garden pea plant [Pisum sativum]
and continued upto 1863 for 7 years.
 The results of his experiments were published in
the science Journal in 1866 entitled “Experiments
on plant Hybrids”.
 His experiments were ingenious. He is the first
person to use statistical analysis and
mathematical logic for interpreting the results.
The use of larger sample size brought credibility to
the data collected.
 He prepared the basic frame work of rules
governing inheritance as the characters in pea are
maintained as opposite traits.
By using true breeding line/pure line he conducted
artificial cross pollination experiments.
Genetic Terminology
 Gene: A gene is the unit of DNA responsible for
the appearance and inheritance of a character in
the present day literature and Mendel called it as
element or factor.
 Allele is one of the two members of a gene pair.
Eg: T or t
 Allelomorphs or Alleles are a pair of genes that
control the two alternatives of the same character.
 Gene Locus: The position of a gene on
chromosome is called locus. The alleles of a gene
are present at the same gene locus on homologous
chromosome.
 Heterozygous: The organism in which both the
genes of a character are unidentical is said to be
heterozygous. Such organisms do not breed true
on self fertilization eg. Tt.
 Homozygous: The organism in which both the
genes of a character are identical is said to be
homozygous or genetically pure for that character.
e.g TT or tt.
 Dominant and recessive gene: A heterozygote
possesses two contrasting genes or alleles but
102




only one of the two is able to express itself, while
the other remains hidden.
The gene which is expressed in F1 hybrid is
known as dominant gene.
The gene which is not expressed in F1 hybrid
is known as recessive gene.
Genotype: The genetic constitution of an organism
is called genotype.
OR
The sum total of genes inherited from both the
parents irrespective of whether they are expressed
or not is called genotype (Johanssen).
Phenotype: The physical appearance of an
individual is called phenotype.
purelines: The homozygous organisms which
produce offspring of only one type are called
purelines. (or)
Progeny produced from a single homozygous
parent by repeated self pollination.
Genome: Total set of genes in the haploid set of
chromosomes is called genome.
9.1 Mendel’s Experiments
 Mendel selected common garden pea (Pisum
sativum) for his experiments, because the plant
shows the presence of several contrasting traits,
hence it was useful in his experiments.
 He selected fourteen true breeding pea plant
varieties which were similar except for one
character with contrasting traits.
 Mendel conducted artificial pollination/ cross
pollination experiments using several true breeding
pure lines (A truebreeding line is one that, having
undergone continuous self pollination showing the
stable trait inheritance and expression for several
generations).
 Mendel selected 14 true breeding pea plant varieties.
A true breeding line is one that having undergone
continuous self pollination, shows the stable trait
inheritance and expression for several generations.
Table : Characters of Garden Pea picked up
by Mendel
Characters
Stem Height
Flower Colour
Flower position
Pod shape
Pod colour
Seed shape
Seed colour
Dominant trait
Recessive trait
Tall (T)
Violet (W)
Axial (A)
Dwarf (t)
White (w)
Terminal (a)
Inflated (I)
Green (G)
Round (R)
Constricted (i)
Yellow (g)
Yellow (Y)
Wrinkled (r)
Green (y)
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
16
17
18.
19.
20.
21.
22.
23.
24.
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
3) alternative forms of a gene
Advantages of selecting garden pea plant by Mendel
4) alternative forms of a character
for his hybridization experiments are
25.
Mendel conducted hybridisation experiments on
1.It is an annual plant that has well defined
1) Pigeon Pea
2) Garden Pea
characters.
3) Wild Pea
4) Sweet Pea
2. It can be grown and crossed easily.
3. It has bisexual flowers containing both female and 26. Sexuality of the flowers of pea plant
1) Staminate flower
2) Bisexual flower
male sex organs.
3) Pistillate flower
4) 1 and 3
4. It can be self fertilized conveniently.
5. It has a short life cycle and produces large no of 27. Offsprings produced from a cross made between
two pure lines is called
offsprings.
1) hybrids
2) mutants
3) P generation
4) F2 generation
PRACTICE TEST-II
28. Which one of the following pairs is not of a
contrasting character
Genotype is
1) Tall and dwarf
2) Axial and terminal
1) Genetic composition of species
3)
Green
and
yellow
4) Round and yellow
2) Genetic composition of plastids
29.
Inbreeding
is
also
known
as
3) Genetic composition of germ cells
1) cross fertilization
2) self fertilization
4) Genetic composition of an individual
3)
cross
pollination
4) vegetative fertilization
Alleles of a gene are found on
30.
In
Pisum
sativum,
the
dwarfness of plant is a
1) non homologous chromosomes
....
character.
2) homologous chromosomes
1) dominant
2) recessive
3) same chromosome 4) any chromosomes
3)
co-dominant
4) incomplete dominant
The external appearance of an individual
31.
Mendel
selected
pea
plant
for his experiments
1) genotype
2) phenotype
because
it
has
3) recessive
4) dominant
1) short life cycle
Organism with two copies of the same allele is
2) large number of offsprings
1) homozygous for that trait
3) self fertilized conveniently
4) 1, 2 and 3
2) homologous for the allele
32.
Which
of
the
following
characters
is not
3) heterozygous for that trait
considered in pea plant by Mendel
4) heterologous for the allele
1) Pod shape
2) Leaf shape
The F1 generation is
3) Plant height
4) Pod colour
1) homozygous
2) heterozygous
33. Genes of today’s world were Mendel’s
3) recessive trait 4) dominant for both the alleles
1) genes only
2) factors
On inbreeding, the homozygous parents will
3) chromosomes
4) plasmid
produce
34. Which of the following principle was not
1) all similar offsprings
proposed by Mendel.
2) 25% similar and 75% dissimilar
1) segregation
2) blending inheritance
3) 75% similar and 25% dissimilar
3) dominance
4) independent assortment
4) 50% similar and 50% dissimilar
35. In Pisum sativum, the dominant character of
Organisms phenotypically similar but
seed is
genotypically different are due to
1) violet
2) round
1) heterozygosity
2) monozygosity
3) wrinkled
4) green
3) multizygosity
4) homozygosity
36. Number of vegetative characters considered
Allele is called
by Mendel in pea plant
1) a pair of chromosomes
1) seven
2) three
2) a pair of sex chromosomes
3) fourteen
4) one
3) a pair of gene of contrasting character
9.2 Inheritance of one Gene
4) a pair of vegetative chromosomes
(or) Monohybrid Cross
An allelomorph is
 The cross made between two parents differing in
1) total number of genes for a trait
one contrasting trait is called monohybrid cross.
2) total number of genes on chromosome
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Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION

Mendel crossed pure tall and dwarf pea plants to
F2 Genotypic ratio 1 : 2 : 1.
study the inheritance of one gene or one factor.
F2 parental combinations to recombination ratio 1 : 1.
 Crossing or hybridization involves selection of
parents, emasculation of flowers of female parent,  F2 dwarf plants are homozygous because they form
artificial pollination, fertilization.
100% dwarf in F3 , F4 or onward generation.
 He found only tall plants and no dwarf plants in F1
or first filial generation. Only one parental trait is  Among F2 tall plants out of three one plant breeds
true and the remaining two form both tall and dwarf
expressed. The same occurred in the case of all
in next generation on selfing. Hence among tall 1/3
contrasting characters.
are homozygous 2/3 are heterozygous.
 The character expressed in F1 is called dominant
 The binomial expression of monohybrid genotypic
and unexpressed is called recessive. Selfing of F1
ratioof homozygous tall, heterozygous tall and
1 1 1
2
3
1
homozygous dwarf is : :   ax  by  .
plants produced tall and dwarf in
and
4 2 4
4
4
proportions. No plant was in between tall and dwarf  The expression is expanded as given below
 1/ 4TT  1/ 2Tt  1/ 4tt
and so there was no blending of traits in either F1
Pure homozygous

dwarf pea plant
or F2 generations.
tall pea plant
tt  P1Parental gen.
TT
Mendel proposed that characters are governed by
factors or elements or units now called genes. They
t  Gametes
T
occur in different forms and denoted by alphabetical
symbols. They occur in pairs because one is
 F gen.
contributed by male and another by female during
Tt
(1st filial gen.)
Heterozygous tall or hybrid tall
fertilization.
(Monohybrid)
Selfing
Different forms of the same gene are called alleles.
Presence of identical alleles of same trait is called
homozygous and dissimilar alleles is called
t
T
heterozygous.
TT
Tt
T
Pure or
Hybrid or
All F1 progeny of pure tall and dwarf is
Homozygous tall Heterozygous tall  F gen.
(2nd filial gen.)
Tt
tt
t
heterozygous. The cross between TT and tt is called
Hybrid tall
Dwarf
monohybrid cross.
In heterozygous condition (Tt) the phenotype is that 9.2.1(a) Back cross and Test cross:
 Back cross: It is a cross between F1 individual
of dominant allele.
and with either one of its parents. In a back
Pure tall and dwarf plants genotypes are TT and tt
respectively. As one allele is transmitted to a gamete,
cross with dominant parent, the phenotypic
parent with TT produces all gametes carrying
ratio of tall and dwarf plants is 1:0 and the





Results of selfing can be
represented by Checkar
board or Punnet-square
1
allele only. Parent with tt produces all gametes
carrying
allele only..
 The hybrids are phenotypically tall but have factors
governing contrasting traits. Hence they are Tt or
heterozygous. F1 hybrid produce gametes carrying
and

2
genotypic ratio of tall and dwarf is1:1.
Tall
Tall
TT
T
X
T
T
t
T
T
.
TT
Parent
T t
gametes
t
Tt
The F2 is raised by selfing F1 . In F2 all possible
phenotypes and genotypes probabilities are
T
TT
Tt
calculated by punnettt square developed by
Reginald C Punnett.
 Test cross: The cross made between F1 with
 F1 phenotypic and genotypic ratio of tall to dwarf 1 : 0.
recessive parent is called test cross.
F2 phenotypic ratio 3 : 1.
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Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
 Monohybrid test cross ratio 1 : 1.
 The law of segregation postulates that alleles in
 Test cross is conducted to find out the genotype
heterozygous condition co exist and do not blend
of unknown parent.
or mix and they separate and enter into different
 When tested plant is homozygous the ratio of
gametes during meiosis.
dominant phenotype to recessive phenotype is 1 : 0.

Segregation of genes is a universal phenomenon
Eg: Violet  White flowered plants of Pea = All
in all organisms reproducing by normal sexual
Violet.
method.
 When tested plant is heterozygous the ratio of dominant
phenotype to recessive phenotype is 1 : 1.
Eg : Violet  White = 50% Violet : 50% white
Homozygous
recessive
Homozygous
recessive
ww
w
ww
w
w
?
W
Ww Ww
if
if
WW
Ww
W
Ww Ww
w
W
Ww Ww
Result
w
ww
All flowers are violet
Interpretation Unknown flower is
homozygous
(WW)
ww
Half of the flowers are violet and
half of the flowers are white
Unknown flower is
heterozygous
(Ww)
Diagrammatic representation of a test cross.
 Monohybrid cross and its reciprocal cross gives
same results, based on it Mendel proposed
law of dominance and segregation.
9.2.1(b) Law of Dominance :
(i) Characters are controlled by discrete units
called factors.
(ii) Factors occur in pairs.
(iii) In a dissimilar pair of factors pertaining to a
character one member of the pair dominates
(dominant) the other (recessive).
The law of dominance is used to explain the
expression of only one of the parental characters in
a monohybrid cross in the F1 and the expression of
both in the F2 . It also explains the proportion of 3 :
1 obtained at the F2 .
 When dissimilar alleles of same trait are present
(heterozygous) one allele has phenotypic expression
is called dominant. This is called law of
dominance.
9.2.2 Law of segregation or Law of
purity of gametes
Mendel’s Law of segregation states that “the two
alleles of a gene when present together in a
heterozygous state, do not fuse or blend in any
way, but remain distinct and segregate during meiosis
or in the formation of gametes so that each meiotic
product or gamete will carry only one of them”.
 This law is based on the fact that the alleles do not
show any blending and that both the characters are
recovered as such in the F2 generation though one
of these is not seen at the F1 stage.
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37. A pure tall Pea plant is crossed with pure dwarf
Pea plant. The progeny is self pollinated. The
ratio of true breeding tall Pea plants to true
breeding dwarf Pea plants shall be
1) 2 : 1 2) 1 : 1
3) 3 : 1 4)
1:2
38. In a cross 45 tall and 14 dwarf plants were
obtained, genotype of parents was
1) TT × TT
2) TT × Tt
3) Tt × Tt
4) TT × tt
39 The offspring of a cross between two individuals
differing in at least one set of characters is
called
1) polyploid 2) mutant 3) hybrid 4) variant
40. F1 hybrids contain
1) two different genes of a contrasting pair of
characters
2) single gene of only one character
3) two genes of same character
4) single gene for all characters
41 What will be the genetic constitution of the
offspring of a cross of individuals heterozygous
(Zz) for an allele?
1) All ZZ
2) All zz
3) 1/2 ZZ : 1/2 zz 4) 1/4 ZZ : 1/2 Zz : 1/4 zz
42. An offspring of two homozygous parents
differing from one another by alleles at only
one gene locus is
1) back cross
2) monohybrid
3) dihybrid
4) trihybrid
43. If F1 is selfed for red colour, the genotypic ratio
of the offsprings would be
1) 1 : 2 : 1 2) 3 : 1
3) 1 : 1 4) 9 : 7
44. Law of dominance and recessiveness was the
result of
1) Back cross
2) Incomplete dominance
3) Dihybrid cross 4) Monohybrid cross
45. In reciprocal Mendelian crosses, the ratio of
the offsprings will be
1) similar in phenotypic ratio
2) similar in genotypic ratio
3) both (1) and (2)
4) different in phenotypic ratio
46. Cross between hybrid and its parent is
1) Back cross
2) Reciprocal cross
3) Monohybrid cross 4) Dihybrid cross
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Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
47. Heterozygous tall (Tt) and dwarf (tt) Pea plants
are crossed. The ratio in the offspring shall be
1) All tall
2) All dwarf
3) 3 tall : 1 dwarf
4) 1 tall : 1 dwarf
48. tt crossed with Tt. What will be characteristic
of offspring
1) 75% recessive
2) 50% recessive
3) 25% recessive
4) All dominant
49. Test cross is a cross between
1) Hybrid X Dominant parent
2) Hybrid X Recessive parent
3) Hybrid X Hybrid
4) Recessive & recessive
50. What is the process of crossing a recessive
phenotype with its own hybrid called
1) Back cross
2) Test cross
3) Reciprocal cross
4) Out cross
51. In which position female gametes are written
on Punnett's checker board
1) Outside the board on upper side
2) Outside the board on left side
3) Outside the board on right side
4) Outside the board any where
52. Law of dominance is proved by
1) Back cross
2) Incomplete dominance
3) Monohybrid cross 4) Dihybrid cross
53. The possibilities of genotype combination in F2
generation can be determined by
1) Punnett's checker board method
2) Forked line method
3) Multiplying the types of gametes 4) 1, 2 and 3
54. In which position male gametes are written on
Punnett's checker board
1) Outside the board on upper side
2) Outside the board on right side
3) Outside the board on left side
4) Outside the board any where
55. The characters that appear in the first
generation of pea plant are called
1) recessive characters
2) dominant characters
3) both recessive and blended characters
4) incomplete dominant
56. The appearance of a hidden character in some
offsprings in F2 generation obeys the law of
1) purity of gametes
2) dominance
3) independent assortment 4) co-dominance
57. When one member of a pair of allelic genes
express itself as a whole, it is a case of
1) dominance
2) co-dominance
3) incomplete dominance 4) pleomorphism
58. Mendel’s law of segregation is applicable
during
1) formation of gametes 2) fusion of gametes
3) formation of zygote 4) formation of PEN
9.3-Deviations from
Mendelian Concept of Dominance
106
9.3.1 Incomplete Dominance
 In monohybrid cross when F1 is intermediate
(between the parents), does not resemble either
of the parents is called incomplete dominance.
 In dog flower (Antirrhinum - Snapdragon)
crossing of pure red (RR) and pure white (rr)
results in pink (Rr).
 Similar like Mendelian monohybrid cross, even
during incomplete dominance F2 genotypic ratio
is 1(RR) : 2(Rr) : 1(rr). But phenotypic ratio is
not 3 : 1, it is 1Red : 2Pink : 1White.
 During incomplete dominance F2 progeny both
phenotypic and genotypic ratios are same
i.e.1 : 2 : 1.
P Generation
Red
RR
White
rr
×
Gametes
R
r
F1 generation
Pink
Rr
Gametes ½ R ½ r
Eggs ½ R
½ r
F2 generation
½ R Sperms
RR
½ r
Rr
Rr
rr
Figure: Results of monohybrid cross in the
plant snapdragon where one allele is
incompletely dominant over the other allele.
9.3.2 Co-dominance:
 In monohybrid cross when F1 hybrids resembles
both the parents is called co-dominance.
 Good examples are different types of red blood
cells that determine ABO blood grouping in human
beings and seed coat pattern and size in lentil
plants.
 Lentil [Lens culinaris ssp. culinaris (Medik)
Williams] is a major grain legume (pulse) crop in
North America.
 A cross between pure-breeding spotted (having
few big irregular patches) lentils and pure-breeding
dotted (having several small circular dots) lentils
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UNIT - III :GENETICS
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Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
produce heterozygotes that are both spotted and  It depends as much on the gene product and the
dotted. The F1 show the phenotypic features of
production of a particular phenotype from this
both parents(the alleles are termed co-dominant).
product. It also depends on particular phenotype
 Self-pollination of the spotted/dotted F1 generation
that we choose to examine, in case more than
produces F2 progeny in the ratio of 1 spotted: 2
one phenotype is influenced by the same gene.
spotted & dotted : 1 dotted.
9.3.3 Concept of dominance
 Once again, because the heterozygotes can be  Every gene contains the information to express a
distinguished from both homozygotes, the
particular trait. In a diploid organism, there are
phenotypic and genotypic ratios coincide.
two copies of each gene, i.e., as a pair of alleles.
 Now, these two alleles need not always be
identical, (eg. a heterozygote). One of them may
Parents
CSCS
CDCD
be different due to some changes that it has
undergone which modifies the information that the
particular allele contains.
Gametes
CD
CS
 Dominance is due to the work of a gene or what
a gene does.
 Gene contains the information to express a trait
through an enzyme.
 In heterozygous condition each allele expresses to
produce one specific enzyme and so two enzymes
F1(all identical)
CSCD ×
CSCD
are produced.
 The enzyme produced by the normal allele.
converts substrate into product so the same
CS CD
F2
phenotype would be the result.
CS S S S D
 Theoretically, the modified allele could be
CC CC
responsible for the production of,
D
C S D D D
(i) the normal/less efficient enzyme, or
CC CC
S S
(ii) a non-functional enzyme, or
1 C C (spotted):
S D
(iii) no enzyme at all
2 C C (spotted & dotted):
D D

The phenotype/trait will only be dependent on the
1 C C (dotted)
functioning of the unmodified allele. The
 Pleiotropy: Occasionally, a single gene
unmodified (functioning) allele, which represents
product may produce more than one effect so
the original phenotype is the dominant allele and
that a single gene may be related to more than one
the modified allele is generally the recessive
character. Such phenomenon is called Pleiotropy.
allele.
 Eg: Starch synthesis in pea seeds is controlled by
one gene. It has two alleles B and b.
 Large starch grains (Starch synthesised effectively)
PRACTICE TEST-IV
- BB
 Smaller starch grains(lesser efficiency in starch
59. When F1 in monohybrid cross is intermediate
synthesis) - bb.
between the parents phenotypically and does
 After maturation of the seeds, BB seeds are
not resemble them it is due to
round where bb seeds are wrinkled.
1)
Dominance
2) Heterozygous
 Heterozygotes (Bb) produce round seeds and
3)
Incomplete
dominance
4) Recessive
starch grains with intermediate size.
 So B seems to be the dominant allele. But, the 60. The F1 progeny of a cross between pure red
starch grains produced are of intermediate size in
and pure white in dog flower
seeds of heterozygotes (Bb). So if starch grain
1) Red 2) White 3) Pink 4) Yellow
size is considered as the phenotype, then from 61. In incomplete dominance genotypic ratio
this angle, the alleles show incomplete dominance.
1) 1 : 1 : 1
2) 1 : 2 : 1
 Therefore, dominance is not an autonomous feature
3) 3 : 1 : 1
4) 2 : 1 : 1
of a gene or the product that it has information for.
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UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
seed shape also segregated in 3:1 ratio; just like
62. In incomplete dominance phenotypic ratio
1) 1 : 1 2) 3 : 1 3) 1 : 2 : 1 4) 1 : 0
in a monohybrid cross.
63. Dominance is mainly depends on
 In F2 there were four phenotypes i.e, yellow,,
1) Structure of the gene
round, yellow wrinkled, green round and green
2) The product of the gene
wrinkled. In the proportion of 9 : 3 : 3 : 1.
3) Phenotype produced by the product
 Yellow round and green wrinkled are parental
4) 1 or 2 or 3
combination. Yellow wrinkled and green round
64. The genotype causing wrinkled seeds in pea
are recombinants.
plant
P generation
1) BB 2) bb
3) Bb
4) 1,2,3
Wrinkled green
Round yellow
65. The genotype of seeds in pea phenotypically
rr
yy
RR YY
similar to dominant but with intermediate
×
ry
Gametes
RY
size starch grains in seeds is
1) bb 2) Bb
3) BB
4) yy
66. How many different types of gametes can be
Round yellow
F generation
Rr Yy
formed by F1 progeny, resulting from the
Selfing
following cross: AA BB CC  aa bb cc
1) 3
2) 8
3) 27
4) 64
RY
RY
In
Lentil
the
pure-breeding
spotted seeds
67.
Gametes
Gametes
rY
rY
bear
RRYY
1) Few small irregular patches
Ry
Ry
RrYY
RrYY
2) Several small circular patches
ry
ry
RRYy
rrYY
RRYy
3) Few small circular patches
4) Few big irregular patches
RrYy
RrYy
RrYy
RrYy
F generation
68. The phenomenon of a gene responsible for
rrYy
RRyy
rrYy
more than one function is
Rryy
Rryy
1) Dominance
2) Pleiotropy
3) Incomplete dominance 4) Co-dominance
rryy
69. Seed coat pattern and size in Lentil plants
Phenotypic ratio: round yellow: round green: wrinkled yellow: wrinkled green
are good example for
9
3
3
1
1) complete dominance 2) co-dominance
 In dihybrid cross F2 progeny
3) Incomplete dominance4) dominance
phenotypic ratio : 9 : 3 : 3 : 1
9.4 Inheritance of two Genes (Digenotypic ratio: 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1
hybrid Cross)
 Probability of the F2 Genotypes and
 Cross conducted between parents different in two
Phenotypes of a Dihybrid Cross : The nine
characters is called dihybrid cross.
genotypes of F2 generation for the two genes are
 Crossing of true breeding yellow, round seeded
the products of three genotypes for each gene as
pea plant with green, wrinkled is an example of
shown below :
dihybrid cross or inheritance of two genes.
Gene 1 Gene 2 Probabilities Phenotypes
 F1 plants of dihybrid cross are similar to dominant
of genotypes
phenotype i.e., yellow and round seeds.
1/4 RR × 1/4 YY 1/16 RRYY Round Yellow
 Similar to monohybrid cross, even in dihybrid
2/4 Yy 2/16 RRYy Round Yellow
cross F1 hybrids expresses only dominant
1/4 yy 1/16 RRyy Round Green
phenotype.
 YYRR and yyrr are the genotypes of pure yellow,
2/4 Rr × 1/4 YY 2/16 RrYY Round Yellow
round and green, wrinkled parents respectively.
2/4 Yy 4/16 RrYy Round Yellow
 The gametes of parents, during the formation of
1/4 yy 2/16 Rryy Round Green
F1 are (YR) and (yr). So the genotype of F1 is
1/4 rr × 1/4 YY 1/16 rrYY Wrinkled Yellow
YyRr.
2/4 Yy 2/16 rrYy Wrinkled Yellow
 When Mendel self pollinated the F1 plants, he
1/4 yy 1/16 rryy
Wrinkled Green
found that 3/4th of F2 plants had yellow seeds
and 1/4th had green. The yellow and green
colour segregated in 3:1 ratio. Round and wrinkled
1
2
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UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
25%.

The
combination of gametes (Egg and pollen
The probabilities of the F2 phenotypes of a selfed
grain) results in four types of phenotypes and
F1 dihybrid is a product of the probabilities of the
nine types of genotypes.
separate selfed F1 monohybrid phenotypes i.e; 3/

Among
the four types of plants(phenotypes)
4 : 1/4
produced in dihybrid cross, 9 are double dominant
Gene 1
Gene 2
Probabilities
phenotype, 6 with one dominant and one recessive
of Phenotype
and one is double recessive.
3/4 Round × 3/4 Yellow 9/16 Round Yellow
 Among the six single dominant each dominant trait
1/4 Green 3/16 Round Green
is found in 3. i.e., colour dominant 3, shape
dominant 3.
1/4 Wrinkled × 3/4 Yellow 3/16 Wrinkled Yellow

Mendel’s
independent assortment will not apply
1/4 Green 1/16 Wrinkled Green
to two genes on the same chromosome or when
These 9 : 3 : 3 : 1 ratios for the pairs of traits are
the genes are linked.
independent of one another and that combinations
Relation among pairs of independent alleles,
turn up as expected according to chance.
gametes, F2 genotypes and F2 phenotypes
 F2 phenotypic ratio 9 : 3 : 3 : 1.
when dominance is present.
F2 genotypic ratio 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1.
No. of
No. of No. of
No. of
Number of types of phenotypes = 4
Heterozygous kinds of types of F2 types of F2
Number of types of genotypes = 9
pairs
gametes Genotypes Phenotypes
Ratio of double heterozygous to double
1
2
3
2
homozygous (4 : 4) 1 : 1
Number of offspring phenotypically similar to
9
4
2
4
F1 = 9
Number of offspring genotypically similar to
3
8
27
8
F1 = 4
4
16
81
16
In F2 the ratio between parental types to
10
1024
59059
1024
recombinants (phenotypically) = 10 : 6 (5 : 3)
F2 Phenotypes
 Each parent forms four types of gametes
n
2n
3n
. The punnett square has 16 boxes.
 Based upon dihybrid cross results Mendel
proposed the law of independent assortment.
9 Genotypes
9.4.1-Law of Independent Assortment:
YYRR
1
YYRr, YYRr
2
YyRR, YyRR
3
YyRr, YyRr, YyRr, YyRr
4
YYrr
1
Yyrr, Yyrr
2
yyRR
1
yyRr, yyRr
2
yyrr
1
The law states that when a pair of contrasting
traits are combined in a hybrid (RrYy) the
segregation (or) separation of pair of characters
is independent of other pair of characters.
 Segregation of factors within Rr and Yy and also
between the pairs are independent during meiosis
(Rr and Yy)
 Segregation within Rr and Yy is 50% and results
in 4 types of gametes.
.
1
 The frequency of each type of gamete is
or
4
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2n
Genotypic Phenotypic
ratio
ratio
9 Yellow,
round
3 Yellow,
wrinkled
3 Green,
round
1 Green,
wrinkled
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UNIT - III :GENETICS
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
PRACTICE TEST-V
70. When parents used in a cross are different in a
pair of contrasting characters the cross is
called
1) Monohybrid cross
2) Dihybrid cross
3) Trihybrid cross
4) Test cross
71. In pea plant the progeny of a cross between
yellow round and green wrinkled is called
1) Monohybrid
2) Trihybrid
3) Dihybrid
4) Nullisomic
72. The phenotype of F1 in dihybrid cross is
1) similar to dominant
2) similar to recessive
3) intermediate to parents 4) dissimilar to parents
73. Which of the following type of gametes is not
formed in Mendels dihybrid cross
1)
YR
2)
Yr
3)
yR
4)
rr
74. Phenotypic ratio of F2 in dihybrid cross
1) 9 : 3 : 3 : 1
2) 1 : 1 : 1 : 1
3) 6 : 3
4) 9 : 7
75.Genotypic ratio of dihybrid cross
1) 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1
2) 1 : 3 : 3 : 4 : 1 : 2 : 1 : 2 : 1
3) 1 : 1 : 1 : 1
4) 9 : 3 : 3 : 1
76.The percentage of gametes with (Yr) genotype
formed from F1
1) 50% 2) 25% 3) 75%
4) 100%
77. Percentage of genotypes with YyRr in F2
generation
1) 6.25%
2) 12.5% 3) 18.75% 4) 25%
78. Which among the given genotypes in F2 of
dihybrid cross respresent 6.25%
1) yyRR 2) YyRr 3) yyRr
4) YYRr
79. A dihybrid condition is
1) tt Rr 2) Tt rr 3) tt rr
4) Tt Rr
80. Gametes of AaBb individual can be
1) Aa, Bb
2) AB, ab
3) AB, ab, aB 4) AB, Ab, aB, ab
81. Mendel's law based on results of dihybrid
cross is
1) Segregation
2) Polygenic inheritance
3) Independent assortment
4)Dominance
82. Mendel's law of independent assortment can
be demonstrated by
1) Test cross
2) Back cross
3) Monohybrid cross 4) Dihybrid cross
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SR.BOTANY EAMCET - VOL - I
83. In the F2 generation of dihybrid cross, number of
double dominant homozygous individuals is/are
1) 1
2) 2
3) 3
4) 4
84. In F2 generation of dihybrid cross what is the
possibility for heterozygosity for two
characters
1) 1/16 2) 8/16 3) 4/16
4) 9/16
85. In a cross, offsprings were 25% yellow round,
25% yellow wrinkled, 25% green round, 25%
green wrinkled. Of which cross these results
were obtained
1) YYRR x yyrr
2) YyRr x YyRr
3) YyRR x yyrr
4) YyRr x yyrr
86. How many types and in what ratio gametes will
be formed in a plant which is heterozygous for
two characters ?
1) Four types, 9:3:3:1 2) Two types, 3:1
3) Three types, 1:2:1 4) Four types, 1:1:1:1
87. In a typical dihybrid cross in pea plants with
TTaa and ttAA, 560 pea plants are formed in
F 2 generation. The expected number of
progeny homozygous for only tallness is
1) 280
2) 60
3) 120
4) 140
88. When two hybrids rrTt and Rrtt are crossed,
the phenotypic ratio of offspring shall be
1) 3 : 1
2) 1 : 1 : 1 : 1
3) 1 : 1
4) 9 : 3 : 3 : 1
9.5 Chromosomal Theory
of Inheritance
 Mendel laws was not accepted by contemporary


biologists of those days, because
1.Mendel proposed that heredity is under the
control of factors (now called genes). The factors
controlling a trait occur in pairs. They do not blend
or mix. They segregate or separate and enter into
different gametes (It is not accepted by
contemporaries due to appearance of continuous
variations in nature).
2.Lack of wide publicity due to shortage of easy
communication.
3.Rejection of mathematical approach to explain
biological phenomena.
4.Failure of Mendel to show physical existence of
genes or their composition are the reasons for not
accepting Mendel’s theory of heredity.
Hugo de Vries, Correns and Van Tschermak
rediscovered independently Mendel’s laws.
Microscopy helped to observe chromosome
(coloured bodies) doubling and division just before
each cell division. Chromosome movement during
cell division had been worked out (both in mitosis
and meiosis).
NARAYANAGROUP
UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
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Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
Walter Sutton and Theodore Boveri proposed  Anaphase-I of Meiosis acts as cytological
evidence to Mendel’s theory of segregation.
chromosomal theory of inheritance (i.e,
Meiosis I - anaphase
Meiosis I - anaphase
chromosomes are physical basis of heredity) after
observing close similarity, behaviour of
chromosomes was parallel to the behaviour of
genes predicted by Mendel.
They explained Mendel laws by using the
movement of chromosomes during cell division.
Genes are located on chromosomes and alleles of
Meiosis II - anaphase
Meiosis II - anaphase
a gene are at homologous site (locus) on
homologous chromosomes.
Sutton and Boveri explained pairing (synopsis)
and separation of homologous chromosomes
(disjunction) would lead to separation of factors
they carried during meiosis.
Germ cells
Germ cells
During Metaphase of meiosis-I, the two
chromosome pairs can align at the metaphase
plate independently of each other.
Hence they united the knowledge of chromosomal
segregation and Mendelian principles. This is
called chromosomal theory of inheritance.
G1
G2
Meiosis I
anaphase
Meiosis II
anaphase
Germ cells
Bivalent
meiosis and germ cell formation in a cell with four chromosomes.
Can you see how chromosomes segregate when germ cells are formed?
Chromosomes
Genes
i) Occur in pairs
i) Occur in pairs
ii) Separate during ii) Separate at
gametogenesis
gamete formation
such that only one
and only one of
of each pair is
each pair is
transferred to a
transmitted to a
gamete.
gamete (as one
iii) Independent pairs
on a
segregate
chromosome).
independently of iii) One pair
each other
segregates
independently of
iv) They retain their
another pair
number, structure
iv)
They
retain their
and individuality
number, structure
throughout their
and individuality
life and from one
throughout their
generation to the
life and from one
next.
NARAYANAGROUP
Independent assortment of chromosomes
 TH Morgan through his experiments on
Drosophila verified chromosomal theory of
inheritance and demonstrated that cause of
variation lies in sexual reproduction and proposed
linear arrangement of genes on chromosomes.
 Short life span, production of many offspring by
single mating, clear and easy recognition of male
and female insects and easy observation of
variations, helped Morgan’s work on Drosophila.
 Any structural change in chromosome (deletion ,
addition) brings structural or functional changes in
organisms. This is an additional evidence that
genes are on chromosomes.
 Morgan proved that certain characters are sex
linked and associates with 'X’ or 'Y' chromosomes.
PRACTICE TEST-VI
89. Mendel’s work remained unrecognised for
many years because
1) Shortage of easy publicity in Mendel’s days
2) Biologists of the day did not accept Mendel’s
explanation as the cause of continuous variations.
3) Mendel could not provide for the physical
existence of his factors controlling traits.
4) 1,2,3
90. Mendel’s laws were rediscovered by
1) Sutton and Boveri 2) TH Morgan and Bridges
3) Waldeyer and Hofmeister
4) de Vries, Correns and Tschermak.
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UNIT - III :GENETICS
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
91. Behaviour of chromosomes is parallel to
behaviour of
1) nucleus 2) genes 3) centriole 4)spindle fibres
92. Parallelism between chromosomes and genes
was explained by
1) Mendel
2) Johanssen
3) Sutton & Boveri
4) Correns
93. Genes are located on
1) Cells
2) Tissues
3) Chromosomes
4) Nucleolus
94. Pairing of homologous chromosomes occurs
during
1) Pachytene
2) Zygotene
3) Leptotene
4) Diakinesis
95. Chromosomal theory of inheritance is proposed
by
1) Watson and Crick 2) Sutton and Boveri
3) Morgan and Mendel 4) Hershey and Chase
96. Which of the following occur in pairs
1) Chromosomes
2) Genes
3) Both 1 and 2
4) RNA
97. Morgan worked on Drosophila because
1) Short life span
2) Production of many offspring by single mating
3) Clear and easy recognition of male and female
insects
4) 1,2,3
98. Cytological evidence to Mendel’s theory of
segregation is
1) Anaphase-I
2) Anaphase-II
3) Metaphase-II
4) Metaphase-I
99. Sutton and Boveri explained Mendel’s laws
by using
1) Movement of chromosomes during cell
division.
2) Movement of nuclei during cell division
3) Movement of cell organelles during cell division
4) Movement of proteins during cell division
100. Who united the knowledge of chromosomal
segregation with Mendelian principles
1) Boveri only
2) Sutton & Boveri
3) Morgan
4) Sutton only
9.6 Linkage and Recombination


An organism has many traits each controlled by a
factor or gene but has limited chromosomes.
Hence a chromosome has many genes.
Morgan and his co workers through their
experiment on Drosophila hypothesized that certain
hereditary characters are sex linked. This is
revealed by monohybrid cross, where the F1
results of normal cross and reciprocal cross are
different.
112
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SR.BOTANY EAMCET - VOL - I
Like Mendel, T.H. Morgan carried out dihybrid
crosses in Drosophila (fruit fly) to study the
heredity of sex linked characters.
Grey body (brown), Red eye and long winged
insects are normal and called wild.
Yellow body, white eye and miniature winged are
mutants.
Wild are dominant and shown '+' as superscript.
All these characters are sex linked. Hence normal
cross and reciprocal cross results will be different.
In his first dihybrid cross T.H. Morgan crossed
brown bodied, red eyed males W Y   with
yellow bodied, white eyed female mutant (wy)
and inter crossed their F1 progeny..
In F1 all males are white eyed and yellow bodied
(wy) and all females are wild type heterozygous
brown bodied, red eyed colour (W+wY+y).
This indicated that Y W  genes of parent male
are not on "Y" chromosome but only on 'X"
chromosome.
In F2 Morgan found 4 types of insects. 2 male
types and 2 female types. Among them parental
combination 98.7% and recombinant types 1.3%.
This is a deviation from 9 : 3 : 3 : 1 ratio of G.J.
Mendel’s dihybrid ratio.
Morgan coined the name linkage and gave
linkage as the reason for the appearance of more
parental combinations. When two genes W Y 
or yw are on the same chromosomes, the
proportion of parental combinations are much
higher than nonparental combination or
recombination.
The physical association of genes on a
chromosome is called linkage. On the same
chromosome some genes are very tightly linked
some are loosely linked.
The strength of linkage depends on the distance
between genes.
The farther are the genes, the weaker is the
linkage, the nearer are the genes the stronger is
the linkage (inversely related).
T.H. Morgan conducted second dihybrid cross
between white miniature winged female  wild
male (grey bodied, large winged). In F1 all
females were wild and all males are white eyed,
miniature wings. This indicated that genes are
only on X chromosome and not on Ychromosome.
NARAYANAGROUP
UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
Cross A
Cross B
w
+
y w+
yw
Parental
yw
Yellow,white
+
w
m
w m
white,
miniature
Wild type

m
+

Wild type
Chromosome Mapping:

yw
yw
w
F1 generation
+
+
+
w
yw
Wild type
Yellow,white
m
+
m
Wild type
w
m
+
+
yw
Wild type
yw
Yellow,white
Recombinant Parental
Recombinant
types(1.3%) type(62.8%) types(37.2%)
+
yw
white
+
+
+
w m
Wild type
yw
w m
Yellow
White,
miniature

White
miniature
Gametes
Parental
type(98.7%)

w+ m
Miniature
w m+
white

+
+
yw
yw
Wild type


white
+
w m+

w m
Wild type
+
w m
w m
miniature
yw
w m
w m
yw
yw
w m
w m
White,
miniature
White
+
Yellow
Figure shows linkage: 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 alleles are represented with (+) sign in
superscript.
Note: The strength of linkage between y and w is
higher than w and m.
Wild type: has dominant phenotypes for both
characters (brown body and red eye)
White: had dominant phenotype(wild colour) for
body colour but recessive phenotype for eye
colour.
Yellow: had recessive phenotype for body colour
(yellow) but dominant phenotype ( wild character)
for eye colour(red).
He intercrossed the F1 to form the F2 with four
types of insects (2 female type + 2 male types).
He recorded 62.8% parental combination and
37.2% recombination.
This proved that genes for wing size and body
colour are loosely linked.
Morgan gave recombination as the reason for the
higher percentage of new combinations.
NARAYANAGROUP

+
yw
Yellow,white

+
yw
Alfred Sturtevant student of Morgan mapped
the positions of genes on a chromosome.
He used the frequency of recombination in crosses
(percentage) between the gene pairs as a measure
of the distance between genes and mapped and
determined their position on a chromosome.
The genetic maps are starting points in the
sequencing of whole genome as was done in the
case of the Human Genome Sequencing Project
(HGSP)
The generation of non-parental gene
combinations is recombination.
9.7 Mutations
F2 generation
yw
Crossing over in pachytene stage of Meiosis and
disjunction in Anaphase-I gives cytological
evidence for recombination.
In Drosophila for those given characters crossing
over occurs in females only.
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Mutation is a phenomenon which results in
alteration of genes (DNA sequences) and
consequently results in changes in the genotype
and the phenotype of an organism.
In addition to recombination, mutation is another
phenomenon that leads to variation in DNA.
Mutation were first noticed by Hugo de Vires in
the plant Oenothera lamarckiana (Lamarck’s
evening primrose).
One DNA helix runs continuously from one end
to the other in each chromatid in a highly
supercoiled form. Therefore loss (deletions) or
gain (insertion/duplication) of a segment of DNA
results in alteration in chromosomes.
Since genes are known to be located on
chromosomes, alteration in chromosomes results
in abnormalities or aberrations. Chromosomal
aberrations are commonly observed in cancer
cells.
In addition to the above, mutations also arise due
to change in a single base pair of DNA. This is
known as point mutation. A classical example of
such a mutation is sickle cell anemia.
Deletions and insertions of base pairs of DNA
cause frame shift mutations.
There are many chemical and physical factors
that induce mutations. These are referred to as
mutagens. UV radiations can cause mutations
in organisms it is a mutagen.
Significance of Mutations

Mutations generate a large amount of variability
in a population from which a breeder can select
113
UNIT - III :GENETICS
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
the desirable types. Hence improved varieties of
crop plants with several desirable characters can
be obtained after careful selection and hybridization.
H ots
Red iscove ry of M en del’s W ork (the birth of
Gen etics)
It w as in 1900 that three scie ntists
indepe ndently rediscovered the principles of
heredity already w orked out by M e ndel. T hey
were :–
i)Karl Correns (G ermany – W orked on maiz e
Zea m ays)
ii)H ugo de Vries(H olland – Worked on
evening primrose Oenother a).
iii)E rich von T scherm ak (A ustria – Wo rked
on flow ering plants).

Eugenics and Euphenics are branches of human
genetics. The former deals with application of
genetics knowledge for improvement of human
rays and the later deals with phenotypic
improvement of humans after birth.
101.Sex linked inheritance in Drosophila studied
by
1) T.H. Morgan
2) G.J. Mendel
3) de Vries
4) Hofmeister
102.Which of the following is representing wild
female pure parent for red eyed
W+
2)
1)
3)
+
W
W
X
Y
X
W
X
W
+
+
X
4)
Y
SR.BOTANY EAMCET - VOL - I
105. When a dihybrid cross is conducted between
brown bodied red eyed males (wild) with yellow
bodies white eyed female (m) in F1 all males
are white eyed because Y W  genes of
parents are on
1) X chromosome only 2) Y chromosome only
3) both X and Y chromosomes
4) Autosomes
106. Parental types are 98.7 and recombinant types
are 1.3% in a dihybrid cross F2 . This indicates
1) Independent assortment
2) Higher percentage of crossing over
3) 100% linkage
4) Linkage in most of the genes
107. Term linkage coined by
1) G.J. Mendel
2) Bateson
3) Johanssen
4) T.H. Morgan
108. Physical association of a genes on a
chromosome is called
1) Dominance
2)
Independent
assortment
3) Linkage
4) Crossing over
109. Strength of linkage is inversely proportional
to
1) distance between locii
2) distance between telomeres
3) distance between chromatids
4) distance between pairs of chromosomes
110. Starting point in gene sequencing is
1) Chromosome mapping 2) Crossing over
3) Linkage
4)Chromosome tracing
EXERCISE - I
W
(Check Your Memory)
111. Mendel was a/an
103. When white eyed female is crossed with wild
1) Austrian 2) Australian 3)American 4)German
(red eye) in the F1
112. A gene is a
1) All are wild
2) All are white eyed
1) Complete DNA molecule 2) Specific part of a
DNA molecule 3) Heterochromatic part of
3) All males are red eyed
DNA 4) Set of ribonucleotides
4) All females are red eyed
104. Which of the following is not a mutant character 113. Genome is a
1) diploid set of chromosomes
2) haploid set
in Drosophila
of chromosomes
1) White eye
2) Miniature wing
3) set of genes lying closely on a chromosome
3) Yellow body
4) Brown body
4) set of genes present in homo logo us
chromosomes
114
NARAYANAGROUP
UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
114. The unit of heredity is
1) Chromosome
2) Nucleus
3) DNA and RNA molecules 4) Gene
115. The segregation of the paired genes occur
during
1) interphase between two meiotic divisions
2) metaphase of first meiotic division
3) anaphase of first meiotic division
4) interkinesis
116. Offsprings are not exactly similar to parents
because of
1) variations produced by crossing over at the time
of gamete formation
2) variations produced by chance distribution of
chromosome
3) both 1 and 2
4) neither 1 nor 2
117. Which one of the following pairs is not of a
contrasting character
1) Tall and dwarf
2) Axial and terminal
3) Green and yellow 4) Round and oval
118. F1 hybrids contain
1) two different genes of a contrasting pair of
characters
2) single gene of only one character
3) two genes of same character
4) single gene for all characters.
119. The process of removing stamens from flowers
during hybridization is
1) crossing 2) selfing 3) emasculation 4) bagging
120. Which is the source of variations
1) Mutation
2) Recombination
3) Deletion
4) 1, 2 and 3
121. A Phenotypically round, yellow pea plant is test
crosssed with rryy, in the obtained progeny the
ratio of round yellow & round green is 1:1 then
the genotype of the plant tested is
1) RRyy 2) RrYY 3) RRYy 4) RrYY
122. When true breeding yellow, wrinkled seeded
pea plant is crossed to true breeding green,
round seeded pea plant. The progeny will be
1) All yellow, round
2) All yellow, wrinkled
3) All green, round4) All green, wrinkled
123. The entire body of molecular biology was a
consequent development with major
contribution from
1) Watson, Crick, F.W.Went, Skoog,
Kornberg, Benzer, Monod, Brenner.
2) Watson, Crick, F.W.Went, Skoog,
Kornberg, Calvin, Monod, Brenner.
3) Watson, Crick, F.W.Went, Skoog,
Kornberg, Calvin, Monod, Mitchell.
4) Watson, Crick, Nirenberg, Khorana,
Kornberg, Benzer, Monod, Brenner.
NARAYANAGROUP
124. It is exception to Mendel’s laws
1) Purity of gametes 2) Dominance
3) Linkage and crossing over
4) Independent assortment
125. Mendel’s work on inheritance of two genes
states that
1) In fertilization the combining of sperm and eggs
is random
2) In meiosis, crossing over creates genetically
diverse gametes
3) In any dihybrid cross, it is possible to get any
combination of phenotypes
4) During gametes formation, gene pairs are
transmitted independently of each other
126. One of Mendel’s pure strains of pea plants had
green peas. How many different kinds of eggs
could such a plant produce with regard to pea
colour
1) 1
2) 2 3) 4
4) 8
127. In Pisum sativum, ___ form of seed is dominant
character.
1) wrinkled 2) round 3) yellow 4) green
128. The characters that appear in the first
generation are called
1) recessive characters 2) dominant characters
3) both recessive and blend characters
4) 2 and 3
129. Mendelian laws of heredity include
1) gene linkage, segregation and independent
assortment
2) gene linkage, dominance and independent
assortment
3) segregation, dominance and independent
assortment
4) segregation, independent assortment and
recombination
130.An offspring of two homozygous parents
differing from one another by alleles at only
one gene locus is
1) back cross
2) monohybrid
3) dihybrid
4) trihybrid
131.A pure tall pea plant can be differentiated from
hybrid tall by
1) Treating with GBA
2) Measuring and
comparing height
3) Selfing and noticing that all progeny is tall
4) Selfing and noting that all progeny is short
132.A testcross is done to find out
1) the genotype of an individual by examining the
phenotypes of its offsprings from a particular mating
2) the genotype of an individual for testing for its
DNA content
3) whether a mating is fertile
4) whether two
species can interbreed
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UNIT - III :GENETICS
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
133.The phenomenon of incomplete dominance was
observed in
1) Pea 2) Snapdragon 3) Lentil 4) Maize
134.In co-dominance, genotypes of F1 hybrids
1) C S C D
2) C S C S
3) C DC D
4) Both C S C S and C DC D
135.Pink coloured flowers are obtained from a cross
between a red flower pea plant and white flower
pea plant. The appearance of this pink colour
character is known as
1) co-dominance
2) complete dominance
3) incomplete dominance 4) segregation
136.Emasculation is
1) done for obtaining mixed breeds
2) done for obtaining hybrids
3) done for cloning
4) done for autogamy
137.The phenomenon of segregation and
independent assortment occurs in
1) meiosis I during gamete formation
2) zygote division
3) segregation during meiosis I and independent
assortment of characters during fusion
4) fusion of male and female gametes
138.A monohybrid cross between two plants, one
having 24 cm long internodes and the other
having 12 cm long internodes, produced F1
hybrids all having 18 cm long internodes. This
is a case of
1) co-dominance 2) incomplete dominance
3) incomplete dominance 4) multiple allelism
139.When a red (RR) flowered plant is crossed with
white (WW) flowered plant. all the off spring
of F1 generation are pink. It indicates that ‘R’
gene is
1) Hybrid
2) Incompletely dominant
3) Complete dominance 4) Recessive
140.If a dwarf variety of pea plant was treated with
GA, it grew as tall as the pure tall pea plant.
On selfing the phenotypic ratio is likely to be
1) all dwarf
2) all tall
3) 50% tall
4) 75% tall; 25% dwarf
141.Rarely observed phenotype in population is
called
1) wild type
2) mutant type
3) variant type
4) recombinant type
142.Heterozygosity can be ascertained by
1) back cross
2) reciprocal cross
3) blending inheritance 4) multiple crossing
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SR.BOTANY EAMCET - VOL - I
143 Allele is called
1) a pair of contrasting characters
2) pair of homologous chromosomes
3) one of the alternative forms of a gene
4) physical appearance of an individual
144.In Antirrhinum, when a cross is made between
red and white flowers all resultant flowers are
1) red 2) pink 3) white
4) both 1 and 2
145.A red and tall dominant character hybrid plant
(Rr Tt ) is crossed with recessive white dwarf
plant (rr tt). What will be the ratio of respective
four combinations : red tall, red dwarf, white
tall and white dwarf plants in the next
generation ?
1) 9 : 3 : 3 :1
2) 15 :1: 0 : 0
3) 9 : 3 : 4 : 0
4) 1 : 1 : 1 : 1
146.‘Gametes are never hybrid’. This is a
statement of law of
1) dominance
2) independent assortment
3) segregation
4) random fertilization
147.In a dihybrid test cross of Pisum 300 yellow
round seeds, 302 yellow wrinkled seeds, 302
green round seeds and 299 green wrinkled
seeds are formed, the percentage of
recombinants is
1) 25 %
2) 75 % 3) 50 % 4) 12.50 %
148.If heterozygous tall plant is crossed with a
homozygous dwarf plant, then the percentage
of dwarf plant is
1) 25%
2) 50% 3) 75% 4) 100%
149.Mendel started his experiments on pea plants
at
1) Brno 2) Olmutz 3) Australia 4) San Francisco
150.In dihybrid crosses the F1 heterozygous plants
are self fertilized to produce F2 generation and
if offsprings are computed in punnett square
the phenotypic F2 ratio as per Mendel’s
independent assortment will yield.
1) 9 : 7 2) 9 : 3 : 3 :1
3) 9 : 6 :1 4) 12 : 3 :1
151.Number of linkage groups in Pisum sativum
1) 2
2) 4
3) 7
4) 8
152.The commonly used animal in genetics is
1) Escherichia 2) Drosophila
3) Neurospora 4) Chlorella
153.Which one of the following pairs consists of
both recessive traits in Pisum sativum as
studied by Mendel
1) Dwarf plant and axial flower position
2) Constricted pod shape and round seed shape
3) Green seed colour and terminal flower position
4) White flower and yellow seed colour
NARAYANAGROUP
UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
154.Ratio between number of chromosomes and
linkage groups in the meiocyte of Pisum
1) 2 : 1
2) 1 : 1 3) 1 : 2
4) 7 : 1
155.The percentage of yr gametes in YyRr is
1) 25%
2) 50% 3) 75% 4) 100%
156.Phenotype of an organism is the result of
1) genotype and environment interactions
2) mutations and linkages
3) cytoplasmic effects and nutrition
4) environmental changes and sexual dimorphism
157.Test cross involves
1) crossing between two genotypes with dominant trait
2) crossing between two genotypes with recessive
trait
3) crossing between two F1 hybrids
4) crossing the F1 hybrid with recessive genotype
158.Mendel crossed tall, round seeded plant with dwarf
, wrinkled seeded plant. In F2 generation he saw
1/16 were dwarf and wrinkled seeds. What did he
conclude from this
1) Short and wrinkled show linkage
2) Tall and wrinkled never combine
3) They show independent assortment
4) Tall and dwarf are blended
159.A true breeding plant producing red flowers is
crossed with a pure plant producing white
flowers. Allele for red colour of flower is
dominant. After selfing the plants of first filial
generation, the proportion of plants producing
white flowers in the progeny would be
1) 3/4
2) 1/4 3) 1/3
4) 1/2
160.Gene mutations is caused by
1) linkage
2)reproduction
3) changes in sequence of chromosome
4) changes in sequence of nitrogenous bases
161.Chromosomal aberrations are commonly
observed in
1) Meristems
2) Cancer cells
3) Parenchyma cells
4) all dead cells
162.Mutations result in
1) change in phenotypes
2) change in genetic constitution
3) better varieties
4) 1, 2 and 3
163.Mutations bring about …. Change
1) small
2) large 3)negligible 4)1,2 and 3
164.Point mutations affect DNA’s
1) phosphate group
2) nitrogen base molecule
3) sugar molecule
4) 1, 2 and 3
165.A normal gene sequence in a chromosomal
sequence is ABCDEFGH, if is changed to
ACGH. This is a case of
NARAYANAGROUP
1) frame-shift mutations 2) deletions
3) inversions
4) duplication
166.The ratio of yellow and green seeded progeny
in F2 generation of a dihybrid cross according
to Mendel is
1) 1 : 3 2) 3 : 1
3) 1 : 1 4) 9 : 7
167.The ratio between number of chromosomes in
a diploid compliment of Pisum to the number
of truebreeding lines selected by Mendel for
his experiments
1) 1 : 7 2) 2 : 1
3) 1 : 14 4) 1 : 1
168 Morgan was the first man to
1) rediscover Mendel’s work
2) give association of a particular gene with a
particular chromosome.
3) carryout controlled breeding experiments
4) observed recombination of chromosomes
169.F1 hybrid of a dihybrid cross is crossed with its
dominant parental genotype and also with its
recessive parental genotype. How many types
of total genotypes and common types of
genotypes are found in the resultant progeny
1) 7 & 1
2) 6 & 2
3) 6 & 1 4) 4 & 2
170.From a cross between RRYy X rrYy plants,
the progeny plants with genotypes RrYY:
RrYy:Rryy : rryy will be obtained in the
following ratio
1) 1 : 1 : 1 : 1
2) 1 : 2 : 1 : 0
3) 1 : 3 : 1 : 0
4) 1 : 1 : 1 : 0
171.Mendel formulated the law of purity of gametes
on the basis of
1) Monohybrid cross 2) Dihybrid cross
3) Test cross
4) Back cross
172.Behaviour of chromosomes during meiosis was
linked to the Mendelian principles by
1) Devries & Correns
2) Bateson
3) Sutton & Boveri
4) Morgan
173.Genotypic ratio of round and yellow seeded pea
plants in F2 generation of Mendelian dihybrid
cross is
1) 9 : 3 : 3 :1
2) 1: 2 : 2 : 4
3) 1: 2 : 2 :1
4) 1: 2
174.Genotypic ratio of round and green seeded
plants in F2 generation of Mendelian dihybrid
cross is
1) 9 : 3 : 3 :1
2) 1: 2 : 2 : 4
3) 1: 2 : 2 :1
4) 1: 2
175.Among the progeny round and yellow are 3,
round and green are 1, wrinkled and yellow are
3 and wrinkled and green are 1, find out the
genotypes of the parents that produce the
above combinations
1) Rr Yy x rr Yy
2) Rr Yy x rr yy
3) Rr Yy x Rr Yy
4) Rr yy x rr Yy
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UNIT - III :GENETICS
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
176.What is the ratio of F2 progeny of Mendel’s
dihybrid cross which are double heterozygous
to double homozygous to single homozygous
1) 4 : 6 : 6
2) 4 : 8 : 4
3) 4 : 4 : 8
4) 4 :1: 4
177.The set of genotypes of offspring which
together constitute 8/16th of the total
offspring of the cross Rr Yy  Rr Yy
1) RR YY, rr yy, RR Yy, rr Yy
2) RR YY, Rr Yy, RR yy, rr yy
3) RR yy, Rr yy, rr Yy, rr yy
4) RR Yy, Rr YY, rr Yy, Rr yy
178.In a F2 generation of a dihybrid cross among
the recombinants the ratio between
heterozygous recombinants to homozygous
recombinants is
1) 1:1
2) 1: 2 3) 4 : 2 4) 3 :1
179.All the genes located on a chromosome are
called.
1) Multiple genes
2) Multiple alleles
3) Linked genes
4) Pleiotropic genes
180.If pink flowered snapdragon plant is crossed
with a white flowered snapdragon plant the
ratio of the pink flowered and white flowered
off spring is expected to be
1) 3 :1
2) 2 :1
3) 1:1 4) 1: 2
181.Condition in which two alleles in a heterozygote
show the individual expression of each allele
in the phenotype is
1) Complete dominance 2) Partial dominance
3) Co-dominance
4) No dominance
182.Tallness (T) is completely dominant over
dwarfness(t). Red colour (R ) of flower is
incompletely dominant over white(r ), the
heterozygote being pink. Plant having
genotype of TtRr is self – pollinated. What
would be the percent of plants with dwarf and
pink character in its progeny?
1) 6.25% 2) 50% 3) 25% 4) 12.5%
183.Which pair of characters among the given are
pleiotropic in Pisum
1) Seed shape
2) Pod colour
3) Flower colour
4) Size of starch grain
184.Incorrect statement among the following is
1) Mendel was first to consider results of
hybridization in terms of single traits
2) Mendel compared his experimental proportions
with mathematical models
3) Mendel’s experiments laid foundation for modern
genetics
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SR.BOTANY EAMCET - VOL - I
4) Mendel was the first person to perform
hybridization experiments
185.F2 progeny can be obtained from F1 hybrid by
this method
1) by crossing F1 hybrid and P1 recessive parent
2) by crossing F1 hybrid and P1 dominant parent
3)by crossing F1 hybrid of any two P1 generations
4) by selfing F1 hybrids
186.Checker board is also called
1) Mendel’s square
2) Punnett’s square
3) Bateson’s square
4) Morgan’s square
187.In a monohybrid cross, the F2 progeny show 3/
4 proportion of tall plants and 1/4 proportion
of dwarf plants. From the group of tall plants
1) 1/3 Proportion are heterozygous
2) 1/3 Proportion are homozygous
3) All are heterozygous
4) 2/3 Proportion are homozygous
188.Scientist who did not discover the Mendel’s
paper presented on his experiments
1) Carl Correns
2) Morgan
3) Tschermak
4) Hugo de Vries
189.The correct genotypes of the parents used in
P1 generation of a monohybrid cross based
on the length of the stem are
1) TT  Tt
2) Tt  Tt
3) Tt  tt
4) TT  tt
190.The famous Indian breed Sahiwal cows belong to
1) Ongole
2) Karnataka
3) Punjab
4) Kerala
191.Number of true breeding pea plant varieties
selected by Mendel
1) 70
2) 30
3) 14
4) 7
192.Find the mismatch from the following
Character - Dominant trait
1. Pod colour Yellow
2. Flower position Axial
3. Pod shape Inflated
4. Seed colour
Yellow
193.Selecting garden pea plant by Mendel is an
advantage because1
1) It is a perennial plant with well defined
characteristics
2) It is monoecious with both male & female flowers
3) It has papillionaceous corolla
4) It can be self fertilized conveniently
194.A cross was made between tall and dwarf
plants. If F1 plants were selfed, the tall and
dwarf plants appeared in 3:1 ratio in F2
generation. This phenomenon is known as
1) Linkage
2) Segregation
3) Hybridization
4) Crossing over
NARAYANAGROUP
UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
195.Which of the following characters of pea was
not studied by Mendel
1) Length of stem
2) Colour of plant
3) Shape of pod
4) Colour of pod
196.Which term represents a pair of contrasting
characters
1) Homozygous
2) Heterozygous
3) Complementary genes 4) Allelomorphs
197.The trait that is expressed in homozygous and
heterozygous conditions
1) Co-dominant trait 2) Dominant trait
3) Recessive trait 4) Incomplete dominant trait
198.Monohybrid test cross ratio is
1) 1:3 2) 3:1
3) 1:1 4) 2:1
199.Mendel’s principle of segregation was based
on the separation of alleles in the garden pea
during
1) Embryonic development 2) Seed formation
3) Gamete formation
4) Pollination
200.Progeny are phenotypically and genotypically
uniform in
1) F2 generation
2) F3 generation
3) F4 generation
4) F1 generation
201.A cross between a plant with any of the parents
is called
1) Test cross
2) Back cross
3) Hybrid cross 4) Inbreed cross
202.Total types of gametes produced in monohybrid
back cross of Tt and TT plants
1) One
2) Two 3) Three 4) None
203.The genotypic ratio and phenotypic ratio of back
cross and test cross of a Mendelian
monohybrid respectively
1) 3:1 and 1:1
2) 1:1 and 1:1
3) 1:3 and 3:1
4) 1:2 and 1:1
204.Mendel’s law of segregation is based upon the
ratio of
1) 1:2 2) 3:1 3) 9:3:3:1
4) 1:1
205.When a tall pea plant is crossed to dwarf pea
plant in the progeny all tall plants are produced,
the tall plant is
1) Only heterozygous 2)Recessive alleles absent
3) Only homozygous
4) May be homozygous or heterozygous
206.In a typical monohybrid cross % of F2 progeny
resembling the parents genotypically
1) 100 2) 75
3) 50
4) 25
207.In a typical monohybrid cross% of F2 progeny
resembling the dominant parent phenotypically
1) 100 2) 75
3) 50
4) 25
NARAYANAGROUP
208.The ratio of homozygous and heterozygous
organisms in the F2 progeny of monohybrid
cross
1) 1:1 2) 3:1
3) 1:3
4) 1:2
209.The type of cross conducted to know the
genotype of tall progeny
1) Back cross
2) Test cross
3) Monohybrid cross 4) Selfing of
210.Estimate the number of tall individuals in the
progeny of 2000 individuals obtained by selfing
the heterozygous plant in monohybrid cross
1) 1000 2) 1500
3) 500
4) 2000
211.Number of homozygous tall plants among 1000
individuals obtained by monohybrid test cross
1) 500 2) 1000
3) 750
4) Zero
212.Morgan hybridized yellow bodied, white eyed
female Drosophila to brown bodied, red eyed
male Drosophila and intercrossed their
F1 progeny. He observed that
1. F2 ratio was deviated very significantly from the
9:3:3:1 ratio
2. Both the genes segregate independently of each
other
3.Recombinant types are not obtained F2 generation
4. Parental types are not obtained in F2 generation
213.Which of the following phenomenon leads to
variation in DNA
1. Linkage, Mutation 2.Recombination, Linkage
3. Mutation, Recombination
4. Aneuploidy, Linkage
214.Loss or gain of a segment of DNA results in
1. Frame shift mutation
2. Point mutation
3. Polyploidy
4. Linkage
215.Sickle cell anemia disorder arises due to
1. Duplication of segment of DNA
2. Substitution in a single base of DNA
3. Deletion of a segment of DNA
4. Duplication of in a base pair of RNA
216.Which one of the following is a physical
mutagen
1. Colchicine
2.UV-rays
3. Formaldehyde
4. DES
217.Mutations first noticed in Oenothera
lamarckiana by
1. Alfred Sturtevant
2. Morgan
3 Hugo de Vries
4. Muller
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UNIT - III :GENETICS
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
218.The law of independent assortment states that
:
1) In fertilization , the combining of sperm and eggs
is random.
2) In meiosis, crossing over creates genetically
diverse gametes
3) In any dihybrid cross, it is possible to get any
combination of phenotypes.
4) During gametes formation, gene pairs are
transmitted independently of each other
219.In which type of mutation the length of the DNA
increases
1. Deletion
2. Substitution of nitrogen base in DNA
3. 1 and 2 4. Insertion
220.Absence of independent segregation or
deviation from 9:3:3:1 ratio in F2 indicates
1. Linkage 2. Dominance
3. Crossing over 4. Absence of sexual reproduction
221.The number of linkage groups in a cell having
40 pairs of chromosomes is
1. 20
2. 40 3. 30 4. 80
222.Number of linkage groups in a species is equal
to the number of chromosomes in a
1. Genotype 2. Genome
3. Zygote
4. Spore mother cell
223.In a dihybrid cross between yellow round
seeded plants and green wrinkled seeded
plants, the number of F2 plants with ‘YyRr’
genotype
1. 1 out of 16
2. 2 out of 16
3. 4 out of 16
4. 3 out of 16
224.The organisms which has undergone a mutation
is called
1. Mutagen 2. Mutant 3. Pure line4. Inbred line
225.In F2 progeny of dihybrid cross the expected
genotypic proportions of individuals that are
homozygous for both dominant characters is
1. 3/16
2. 9/16 3. 12/16
4. 1/16
226.If linkage was known at the time of Mendel
which of the following law of Mendel was
unable to explain
1. Law of dominance
2. Law of independent assortment
3. Law of segregation
4. Law of purity of gametes
227.From a cross YYRr x yyRr, the genotypes
YyRR, YyRr, Yyrr, yyrr will be obtained in the
following ratio
1. 1:1:1:1 2. 1:2:1:0 3. 1:3:1:0 4. 1:1:1:0
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SR.BOTANY EAMCET - VOL - I
228.A and B genes are completely linked. What
should be the genotype of progeny in a cross
between AB/ab and ab/ab
1. AAbb and aabb
2. AaBb and aabb
3. ABB and aabb
4. 1 and 3
229.A plant of unknown genotype with yellow seeds
and purple flowers is crossed to a plant with
green seeds and white flowers. The offspring
all have yellow seeds, but some have purple
flowers and some have white flowers what is
the genotype of yellow seeded, purple flowered
plant
1) YyPp
2) Yypp 3) YYpp 4) YYPp
230.Study the following with regard to pleiotropy.
The starch synthesis in pea plants is controlled
by a single gene. It has two alleles B and b.
BB homozygotes produced large starch grains
as compared to that produced by bb
homozygotes. After maturation it was observed
that BB seeds were round and bb were
wrinkled. When they were crossed, the
resultant heterozygotes produced round seeds,
but starch grains produced are of intermediate
size in Bb seeds.
According to the above information, correct
statements of the following
1) The dominance is an autonomous feature of a
gene or for the information that has.
2) If starch grain size is considered as the phenotype,
the alleles show co-dominance
3) Dominance is not an autonomous feature of a
gene or the product that has the information for ; it
depends as much on the gene product and the
production of a particular phenotype from this
product
4) F2 progeny shows round and wrinkled seeds in
1 : 1 ratio
231.Total population of 800 individuals which
formed F2 population (9:3:3:1) of a cross
between yellow round and green wrinkled. Find
the number of plants with yellow and wrinkled
seeds
1) 150 2) 400 3) 800 4) 300
232.The recombinant phenotypic ratio in F2
obtained from parental cross having genotypes
TTRR x ttrr will be
1) 9:3:3:1
2) 3:1 3) 1:2:1 4) 3:3
233.In a cross between AABB x aabb, the ratio of
genotypes in F2 generation between AABB,
AaBB, Aabb, aabb would be
1) 2 : 1 : 1 : 2
2) 1 : 2 : 1 : 2
3) 1 : 2 : 2 : 1
4) 1 : 2 : 2 : 4
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UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
234.Dominant genes for tallness is T and for yellow
colour is Y. A plant with heterozygous for both
traits is selfed. Then the ratio of pure
homozygous dwarf and green offspring would
be
1) 1/4
2) 2/3
3) 3/16 4) 1 /16
235.Experimental verification of the chromosome
theory of inheritance was given by
1) Sutton & Boveri
2) Alfred Sturtevant
3) Thomas Hunt Morgan 4) Hugo de Vries
236.In Mendel’s dihybrid cross, percentage of
genotypes obtained for the character yellow in
heterozygous condition is
1) 25
2) 50
3) 75 4) 6.25
237.There are three genes a, b, c. Percentage of
crossing over between a & b is 20, b & c is 28
and a & c is 8. Correct sequence of genes on
chromosome is
1) a, b, c
2) a, c, b 3) b, a, c
4) b, c, a
238.The exception to Mendel’s law of independent
assortment is
1) Linkage
2) Recombination
3) Co-dominance 4) Incomplete dominance
239.In a cross between tall plant with round seeds
[ TTRR ] with a dwarf plant with wrinkled seeds
[ ttrr ]. The F1 generation consists of tall plants
with round seeds. What would be the proportion
of dwarf plants with wrinkled seeds in F2
generation
1) Zero 2) 1 / 2
3) 1 / 4
4) 1 / 16
240.In the law of Mendelian genetics, independent
assortment of alleles is due to
1)Location of different alleles on same chromosome
2) Location of allelic pairs on a particular part of
homologous chromosome
3) Location of allelic pairs on different homologous
pairs of chromosomes
4) Both 1 and 2
241.Genetic map is one that
1) Shows stages of cell division
2) Establishes various stages seen in gene evolution
3) Establishes sites of genes on a chromosome
4) Shows distribution of various chromosomes
in a cell
242.Two genes situated very close on the
chromosome show
1) High crossing over
2) Hardly no crossing over
3) No crossing over
4) Many times crossing over
243.F1 plant is crossed with one of the parents. In
the resultant progeny, phenotype is uniform &
genotypic ratio is 1 : 1. It is a type of
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1) Reciprocal cross
2) Back cross
3) Test cross
4) Pureline
244.Read the following and identify the wrong
statement
1) During Mendel’s investigation, statistical q q
analysis & mathematical logic were applied to
problems of biology for the 1st time.
2) Mendel selected 14 true breeding pea varieties
3) Mendel was the 1st to conduct hybridization
experiments
4) Mendel’s experiments had large sampling size
which gave greater credibility to the data collected
245.When a round seeded pea plant is crossed with
a wrinkled seeded pea plant, progeny produced
were with round seed & wrinkled seeds. Round
seeded parent pea plant in this cross is
1) Homozygous
2) Homozygous or Heterozygous
3) Heterozygous 4) Hemizygous
246.A dihybrid plant on self pollination produced
600 plants with 4 types of genotypes. How
many seeds will have genotype Yy Rr
1) 50 2) 100
3) 150
4) 200
247.Which one of the following is incorrectly
matched in relation to characters studied by
Mendel in Pisum
1) Seed characters-2 2) Flower characters-2
3) Vegetative character-1 4) Fruit characters-3
248. An allele is dominant if it is expressed in
1) Both homozygous and heterozygous states
2) Second generation
3) Heterozygous combination only
4) Homozygous combination only
249 Alleles are
1) Genes found on allosomes
2) Pair of genes governing a specific character like
height
3) Genes present on same chromosome
4) Genes present on Y chromosome
250.Two individuals with similar external
appearance but different genetic make up have
the similar
1) Genotype
2) Phenotype
3) Heterozygote
4) Homozygote
251.Genetic recombination is caused by
1)Fertilisation and mitosis 2)Fertilisation and meiosis
3) Mitosis and meiosis
4) Somatic division
252.Pure line is connected with development of
1) Recombination
2) Heterozygosity
3) Homozygosity and self assortment
4) Heterozygosity and linkage
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UNIT - III :GENETICS
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
253.If a homozygous dominant red flowered plant
is crossed with a homozygous recessive white
flowered plant, the offspring would be
(according to law of dominance)
1) half red flowered
2) half white flowered
3) all red flowered
4) half pink flowered
254.When two hybrids undergo crossing then
percentage of recessive is
1) 25
2) 100 3) 50
4) 75
255. When a heterozygous tall plant is crossed with
a homozygous dwarf , then the percentage of
recessive is
1) 25
2) 50
3) 100
4) 75
256. If a pure tall pea plant is raised in nutrient
deficient media such that it grows to the size of
a dwarf plant and is then selfed, the progeny in
normal media will be
1) dwarf
2) all tall
3) 50% tall 50% dwarf 4)75% tall and 25% short
257.Which of the following, follows Mendel’s law
of segregation
1) Each gamete receives one gene for an allele
2) Genetic material is equally distributed during
mitosis
3) Genetic material enters into one gamete only
4) 1, 2and3
258.When parents ‘A’ and ‘B’ were crossed, F2
progeny was produced with three fourth similar
features in phenotype of ‘B’ and one fourth
possessed contrasting traits. If the traits being
considered here are for height with ‘T’ for tall
and ‘t’ for short. What will be the possible
genotype of ‘A’ and one fourth of F2
1) tt and Tt
2) Tt and tt
3) Tt and Tt
4) tt and tt
259.Monohybrid cross F1 progeny Tt is inbred.
What will be the types of F2 genotypes
1) 3
2) 2
3) 4
4) 1
260.A true breeding plant producing red flowers is
crossed with a pure plant producing white
flowers. Allele for red colour of flower is
dominant. After selfing the plants of first filial
generation, the proportion of plants producing
white flowers in the progeny would be
1) 3/4
2) 1/4
3) 1/3
4) 1/2
261. A farmer planted 200 seeds collected from a
single F1 pea plant, which produced 140 tall
and 40 short plants in F2 generation. The
genotypes of these offsprings are most likely
1) TT and Tt only
2) TT, Tt, tt
3) Tt and tt only
4) TT and tt only
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SR.BOTANY EAMCET - VOL - I
262.If a cross is made between AaTT and aatt,
where ‘A’ stands for red dominant and ‘T’ for
tall dominant. What will be the percentage of
red tall plants
1) 25%
2) 50% 3) 75% 4) 100%
263.Heterozygous tall and red flowered pea plants
were selfed and total 2000 seeds are collected.
What is the total number of seeds for both
heterozygous traits
1) 250
2) 500 3) 1250 4) 750
264.Total 512 seeds are collected from the cross
WwYy  WwYy. Find the number of plants
produced with first dominant and second
recessive trait
1) 288
2) 96
3) 32
4) 320
265.The term genome is used for
1) Diploid set of chromosomes
2) Polyploid set of chromosomes
3) Triploid set of chromosomes
4) Haploid set of chromosomes
266.The principle of independent assortment of
characters is proved by
1) The appearance of tall and dwarf plants in F2
population
2) The appearance of yellow and green in the ratio
3) 1 and also the appearance of round and
wrinkled seeded plants in the ratio of 3 : 1
3) The appearance of round and wrinkled seeds
plants in the F2 population
4) The observation that F1 progeny is tall
267.Which types of gametes does RrYy produce
1) RY, Ry, rY, ry
2) RY, RY, Ry, RY
3) RY, rY, rY, ry
4) RY, Ry, rY, rY
268.In a dihybrid cross, pure homozygous offspring
in the F2 progeny will be
1) 1/2 2) 1/4
3) 1/8
4) 1/16
269.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) Data insufficient
270.Genes ‘A’ and ‘B’ are present in different
chromosomes. A female AaBb was crossed with
a male AAbb. The genotypes of gametes will be
1) Female AB, Ab, aB, ab; male Ab
2) Female AA, bb, AB, ab; male Ab
3) Female Aa, Bb; male AA, bb
4) Female AB, ab; male AA, bb
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UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
271. In pea plants round seed is dominant over
wrinkled seed. A plant heterozygous for round
seeds was crossed with a plant with wrinkled
seeds. Which one of the following progenies
agrees with expected result
1) 99 round and 300 wrinkled
2) 301 round and 100 wrinkled
3) 305 round and 301 wrinkled
4) 200 round and 100 wrinkled
272.What proportion of the offspring of a cross
AABBcc  AaBbCc will be completely
heterozygous if all genes segregate individually
1) 1/4
2) 1/16 3) 1/2
4) 1/8
273.If a dwarf pea plant was treated with Gibberellic
acid, it grew as tall as the pure tall pea plant.
If this treated plant is crossed with a pure tall
plant then the phenotypic ratio is likely to be
1) all dwarf
2) 50% dwarf 50% tall
3) 75% tall 25% dwarf 4) all tall
274. RR(red) is crossed with rr(white). All the Rr
offspring are pink. This is an indication that R
gene is
1) Hybrid
2) Recessive
3) Incompletely dominant 4) Mutant
275. In the case of incomplete dominance F2
generation has
1) Genotype ratio 3 : 1 2) Phenotypic ratio 3 : 1
3) Genotypic ratio equal to phenotypic ratio
4) Genotypic ratio is not equal to phenotypic ratio
276. Co - dominance is observed in
1) Pisum sativum
2) Lens culinaris
3) Lathyrus odoratus 4)Oenothera lamarckiana
277. In a cross between a pure tall pea plant with
green pod and a pure short plant with yellow
pod, how many short plants out of 16 you would
expect in F2 generation
1) 3
2) 9
3) 4
4) 1
278.In Mendel’s experiment how many different
kinds of seeds are produced from a short plant
with wrinkled seeds (ttrr)
1) 9
2) 4
3) 2
4) 1
279.In garden pea, yellow colour of cotyledons is
dominant over green and round shape of seed
is dominant over wrinkled. When a plant with
yellow and round seeds is crossed with a plant
having yellow and wrinkled seeds, the progeny
showed segregation for all the four characters.
The probability of obtaining green round seeds
in the progeny of the cross is
1) 1/8 2) 1/4 3) 1/16
4) 3/16
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280.A true breeding plant producing red flower is
crossed with a pure plant producing white
flower. Allele for red colour of flower is
dominant. After selfing the plants of first filial
generation, the proportion of plants producing
white flower in the progeny would be
1) 3/4
2) 1/4 3) 1/3
4) 1/2
281.How many types of gametes may be produced
by genotype DdEeFf
1) 27
2) 8
3) 3
4) 6
282.How many pairs of contrasting characters in
pea seed were chosen by Mendel
1) 2
2) 3
3) 4
4) 7
283.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 F1 generation
1) 50 : 50
2) 9 : 1 3) 1 : 3 4) 3 : 1
284.A pea plant parent having violet coloured
flowers with unknown genotype was crossed
with a plant having white coloured flowers, in
the progeny 50% of the flowers were violet and
50% were white.The genotypic constitution in
the parent having violet coloured flowers was
1) Homozygous
2) Merozygous
3) Heterozygous
4) Hemizygous
285.A white flowered (rr) Snapdragon was crossed
with red coloured (RR). If 120 plants are
produced in F2 generation, the result would be
1) 30 red, 60 pink and 30 white
2) 60 red, 30 pink and 30 white
3) 30 red, 30 pink and 60 white
4) 90 red, 0 pink and 30 white
286.Cytological evidence for chromosomal theory
of inheritance is found in
1) Prophase
2) Metaphase-I
3) Anaphase-I
4) Telophase-I
287.Heritable variations occur due to
1) asexual reproduction 2) vegetative propagation
3) sexual reproduction 4) 1 and 2
288.Linear arrangement of genes on chromosomes
proposed by
1) T.H. Morgan
2) Sutton and Boveri
3) Sturtevant
4) G.J. Mendel
289.Sutton and Boveri proposed parallelism
between chromosomes and genes because
1) both are in the nucleus
2) both segregate at the time of gametogenesis
3) both show independent segregation always
4) 1 and 3
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UNIT - III :GENETICS
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
290.Separation of genes due to segregation of
chromosomes is called
1) cell theory
2) cell lineage
3) chromosomal theory of heredity
4) mutations theory
291.In a grey bodied Drosophila character 'a' has
68.2% parental combination character 'b' has
98.7% parental combination. This indicates
1) 'a' and 'b' genes are at the same place as that of
grey body gene
2) 'a' is nearer grey body gene and 'b' is far away
3) 'a' is far away and 'b' is near grey body gene
4) a ,b are equidistantly placed with respect to grey
body gene
292.Crossing over occurs during
1) Anaphase-I
2) Diplotene
3) Pachytene
4) Diakinesis
293.When genes are located on 'X' chromosome
crossing over does not occur in
1) Male
2) Female
3) 1 and 2
4) 1 or 2
294.Gene position on a chromosome were mapped
by
1) G.J. Mendel
2) Bateson
3) Sturtevant
4) G.H. Shall
295.Number of Mendelian recombinants in F2
dihybrid cross is
1) 10
2) 6
3) 9
4) 16
296.Mendel to pea :: T.H. Morgan to
1) Zea
2) Pisum
3) Lathyrus
4) Drosophila
297.Absence of independent segregation or
deviation from 9 : 3 : 3 : 1 ratio in F2 indicates
1) Linkage
2) Crossing over
3) Dominance
4) Absence of sexual reproduction
298.When linkage is present which of the following
cannot be observed in a dihybrid cross
1) Dominant trait in F1
2) Recessive in F2 characters
3) 1 : 1 : 1 : 1 ratio in test cross
4) Dominant trait in F2
299.Of white (eye colour gene) and yellow (body
colour gene) are tightly linked but white and
miniature wing genes are loosely held because
1) recombination percentages is lesser in the former
but more in the later
2) recombination percentage equal in both of them
3) parental combinations are lesser in the former
and more in the later
4) crossing over absent in both of them.
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SR.BOTANY EAMCET - VOL - I
300.Which of the following may not apply when the
gene controlling two different traits are located
on the same chromosome
1) Law of segregation 2) Incomplete dominance
3) Law of dominance
4) Law of independent assortment
301.How many linkage group are there in nucleoid
of bacteria
1) One 2) Two 3) Four
4) None
302.If distance between gene on chromosome is
more, then gene shows
1) Weak linkage
2) Strong linkage
3) Less crossing
4) Both1 and 3
303.The number of linkage groups in a cell having
10 pairs of chromosomes are
1) 5
2) 10
3) 15
4) 20
304.Complete linkage is found in
1) Birds
2) Snakes
3) Female Drosophila 4) Male Drosophila
305.A phenomenon which works opposite to the
linkage is
1) Independent assortment 2) Crossing over
3) Segregation
4) Mutation
306.Which of the following will not result in
variations among siblings
1) Independent assortment 2) Crossing over
3) Linkage
4) Mutation
307.Number of linkage groups in a species equals
to number of chromosomes in a
1) Spore mother cell
2) gamete
3) zygote
4) sporophyte cell
308.The position of a gene on a given chromosome is
1) Allele
2) Gene locus
3) Linkage
4) Chaismata
309.How many types of gametes will be produced
from a microspore mother cell with Bb and Ll
genotype when there is 100% linkage between
the characters denoted by 'B' and 'L' genes
1) 1
2) 2
3) 4
4) 6
310.Two genes R and Y are located very close on
chromosome linkage map of a maize plant.
When RRyy and rryy genotypes are hybridized.
Then the F2 segregation will show
1) higher number of recombination types
2) segregation in the expected 9 : 3 : 3 : 1 ratio
3) segregation in 3 : 1 ratio
4) higher number of parental types
311.Find odd one out with reference to 100%
linkage
1) 100% parental combination in F2
2) F2 phenotypic ratio 3 : 1
3) Dihybrid test cross ratio 1 : 1
4) Linked genes tend to separate frequently
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UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
312.In what ratio the parental phenotypes will
appear in F2 generation of dihybrid cross
1) 3/16 : 3/16
2) 4/16 : 4/16
3) 9/16 : 1/16
4) 12/16 : 4/16
313.On which characters the Mendel's dihybrid
experiment was based
1) Green seed coat, round seed x yellow seed coat,
wrinkled seed
2) Tall plant, red flower x dwarf plant, white flower
3) Yellow cotyledons, round seed x green
cotyledons, wrinkled seed
4) Tall plant, yellow flower x dwarf plant, pink flower
314.Independent assortment means
1) separation of characters of one parent
2) non-separation of characters of one parent
3) combination of parental characters
4) separation of parental characters
315.Mendel’s law of independent assortment
1) will not apply to two genes on the same
choromosome
2) will apply only if two genes are on the same
choromosome
3) will apply to all genes
4) has now been completely disproved
316.If it is imagined in pea plants that genes for
controlling seed coat colour and shape are
present on the same chromosome very closely;
performing dihybrid experiments with these
characters Mendel couldn’t have been able to
arrive at the idea of
1) independent assortment 2) dominance
3) segregation
4) incomplete dominance
317.From a cross PPCC  ppCC following
genotypic ratio will be obtained in F1
generation
1) 1 PpCC: 3ppCC
2) 3 PpCC:1 ppCC
3) all PpCC: No ppCC 4) 1 PpCC:1 ppCC
318.Which of the following explains, how progeny
can possess the combinations of traits that none
of the parent possessed
1) Chromosome theory 2) Polygenic inheritance
3) Law of segregation
4) Law of independent assortment
319.Who observed that the behaviour of
chromosomes at meiosis can serve as the
cellular basis of both segregation and
independent assortment
1) Sutton and Boveri 2) Banden and Boveri
3) W. Flemming
4) Boveri and Braue
320.A red and tall dominant character hybrid plant
is crossed with recessive white dwarf plant
(RrTt  rrtt). What will be the ratio of
respective four combinations : red tall, red
dwarf, white tall and white dwarf plants in the
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next generation
1) 9 : 3 : 3 : 1
3) 9 : 3 : 4 : 0
1) 15 : 1 : 0 : 0
4) 4 : 4 : 4 : 4
EXERCISE-II
(Sharpen Your Reflexes)
321.Test plant used by Mendel is (1) a member of
(2) family in it yellow seed coat is dominant
because crossing of pure yellow and pure green
gives 100% (3). 1,2,3 respectively are
1) Pea, Fabaceae, Yellow
2) Lathyrus, Fabaceae, Long pollen
3) Drosophila, Diptera - Vestigial wing
4) Antirrhinum, Verbinaceae, Yellow
322.Taking the symbols on certain chromosomes
present in lentils such as for spotted - CSCS
, dotted CDCDand spotted & dotted CSCD,
find out the ratios obtained in F2 generation
and identify the dominance respectively
I. Spotted, dotted & spotted, dotted are in ratio of
1 : 1 : 1 respectively and is incomplete dominance.
II. Spotted, Spotted & dotted, Dotted are in ratio
of 1 : 2 : 1 respectively and is co-dominance.
III. Dotted, spotted, spotted & dotted are in ratio
of 1 : 1 : 2 respectively and is incomplete dominance.
IV. Spotted, dotted, spotted & dotted are in ratio
of 1 : 1 : 2 respectively and is incomplete dominance.
1) 1 & II are correct 2) Except II all are incorrect
3) I, II, III are correct 4) IV alone correct.
323.Study the following statements
I) A homozygous tall plant is crossed with
homozygous dwarf (TT  tt), produces 100% tall
progeny
II) A heterozygous tall selfed, produces tall and dwarf
progeny successively 75%, 25%
III) A heterozygous tall crossed with homozygous
dwarf, tall and dwarf are produced in 1 : 1.
IV) A heterozygous tall crossed with homozygous
tall, produces tall and dwarf in 100 : 0 ratio.
1) I and II are correct 2) II and III are correct
3) I, II, III, IV are correct 4) I and IV are correct
324.When YYRR is crossed with yyrr the genotype
of F1 (1). Number of types of gametes in F1
(2). Number of phenotypes in F2 (3).
1) yyRR - 4 - 9
2) yyRR - 4 - 9
3) YyRr - 4 - 4
4) YyRr - 4- 10
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UNIT - III :GENETICS
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
325.Find the correct ideas of homozygous plants
A) They are formed by repeated self fertilization
B) They have two copies of same gene.
C) They form two types of gametes each containing
different copy of the same gene.
D) They are formed by cross fertilization.
1) A, B, D are correct 2) A and D only correct
3) A and B are only correct 4) A, B, C are correct
326.The examples for co-dominance are
A) ABO blood grouping in human beings
B) Seed coat pattern in lentil plants
C) Inheritance of flower colour in Antirrhinum
1) B only 2) A only 3) A & B only 4) A,B&C
327.The progeny obtained by crossing pure/true
breeding plants is
A) a hybrid
B) inbred
C) Heterozygous
D) Homozygous
The correct pair is
1)A and B 2)C and D 3)A and C 4)A and D
328.Find the combination of dominant characters
studied by Mendel
A) Round shaped yellow seeds
B) Wrinkled seeds axial flowers
C) Round shaped green seeds
D) Axial flowers on long stems
1)A and D 2)A and B 3)C and D 4)A and C
329.Mendel in his experiments conducted
A) Natural pollination B) Artificial pollination
C) Vegetative propagation D) Apomixes
1)A and C 2)A and D 3)A and B 4)B and D
330.When same path of characters are passed to
several generations the possible combination
of reasons
I) Homozygosity II) Vegetative propagation
III) Linkage of genes (100%)
IV) Crossing over
1) I and II only
2) I, II, III only
3) II and III only
4) IV only
331.Linkage decreases the frequency of
I) Hybrid
II) Recessive allele
III) Dominant allele
IV) Recombinants
1) I and II
2) I and IV
3) I and III
4) II and IV
332.Both genotypic phenotypic ratio of progeny is
1:2:1 in case of
I. Incomplete dominance
II. Co-dominance
III. Monohybrid back cross
IV. Monohybrid test cross
1) I & III
2) All 3) I, III & III 4) I & II
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SR.BOTANY EAMCET - VOL - I
EXERCISE - III
Think Twice Before you choose
(‘Assertion’ and ‘Reason’ type Questions)
1) A is correct and R is explanation to A.
2) A is correct and R is not an explanation to A
3) A is true and R is wrong.
4) A is wrong and R is true.
333.(A) : Test cross is conducted to know the genotype
of F1 individual.
(R) : In a test cross dominant phenotypic organism
is crossed with heterozygous one.
334.(A) :Mendel selected garden pea plant for his
hybridization experiments
(R) :Morphology of useful parts in garden pea are
seeds
335.(A) : A cross between pure-breeding spotted Lentil
and pure-breeding dotted Lentils produce
heterozygotes that are both spotted and dotted.
(R) : The (R) for this heterozygosity is that the spotted
allele and dotted allele are co-dominant.
336.(A): If a single gene is related to more than one
character, that single gene product may produce
more than one effect.
(R): Production of more than one character by one
single gene is called polymorphism.
337.(A): Chromosomal theory of inheritance was
proposed by Sutton and Boveri.
(R): The knowledge of chromosomal segregation
when united with Mendelian principles it is called
chromosomal theory of inheritance.
338.(A): The physical association of genes on a
chromosome is called Linkage.
(R): The generation of non-parental gene
combinations is described as recombination
339.(A): The frequency of recombination between gene
pairs on the same chromosome was used as a
measure of the distance between genes on the
chromosome.
(R): Several dihybrid crosses in Drosophila were
carried out by Morgan.
340.(A): The alteration in chromosomes by deletion or
duplication of a segment of DNA results in
chromosomal aberrations.
(R): Chromosomal aberrations are commonly
observed in cancer cells.
341.(A): Sickle cell anemia is a classical example for
point mutation.
(R): The change in a single base pair of DNA is
called point mutation.
NARAYANAGROUP
UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
342.(A): Ultra violet radiation is a mutagen.
(R):He was lucky in selecting seven characters in
(R) : U.V radiation can cause mutations in
wild pea.
organisms.
355.(A) : Mutations generate a large amount of variations
343.(A): In dihybrid cross conducted by Mendel, the
in a population.
pairs of factors are located in two pairs of
(R) : A breeder can select all the variations caused
homologous chromosomes.
by mutations.
(R) :Each homologous pair of chromosomes
356.(A) : Linkage leads to more parental types.
bears two pairs of factors for same gene.
(R) : Linkage reduces the number of types of
344.(A): Hybrids produced by hybridization process
variable gametes.
always carry desired genes only.
357.(A) : Mendel did not recognise the linkage
(R): Hybridization process involves crossing two
phenomenon in his experiments.
plants with desirable characters.
(R) : The factors he considered were located on
345.(A): The linked genes tend to get inherited together.
different chromosomes.
(R): Crossing over never occurs between two
homologous chromosomes of a bivalent at the sites
EXERCISE - IV
in between two weak linked genes.
346.(A) : Pea plant shows stable trait inheritance and
(Simple matching Questions)
expression for several generations
358.Correct
match the following
(R) : Pea plant is a self pollinated member of
List
–
I
List–II
Solanaceae.
In
Monohybrid
cross
347.(A) : In Mendel’s monohybrid cross F1 plant
A) Phenotypic ratio of F2 progeny I) TT
produce 2 types of female gametes and 2 types of
B) Genotypic ratio of F2 progeny II) TT, tt
male gametes
C) Genotypes of Parents
III) 3 : 1
(R) : F1 Plants of normal monohybrid cross are in
D) Genotype of F1 hybrid
IV) Tt
heterozygous (Tt) condition.
V) 1 : 2 : 1
348.(A) : Linkage is inversely proportional to distance
A
B
C
D
A
B
C D
between genes on a chromosome.
1)
II
V
III
IV
2)
III
IV
II V
(R) : Closed genes linked strongly where as farther
3)
III
V
II
IV
4)
IV
II
V III
genes linked weakly.
349.(A) : In Mendel’s monohybrid cross, the probability 359. Match the following
List – I
List – II
of obtaining dwarf plants in F1 is zero.
A) Genetic make up of
I) Contrasting traits
(R) : F1 individuals of monohybrid cross show
an individual
alleles for tallness only.
B) The alternative forms II) Allelomorphs
350.(A) : In co-dominance, the phenotypic and genotypic
of a Character
ratios are same.
C)
The two alternative
III) Genotype
(R) : Heterozygotes can be distinguished from
forms of a gene
homozygotes.
D) Pairs of contrasting
IV) Alleles
351.(A) : Loss or gain of a segment of DNA results in
characters
alternation in chromosomes.
(R) : One DNA helix runs continuously from one The correct match is
1) A-II, B-I, C-IV, D-III
end to the other in each chromatid in a highly super
2) A-II, B-IV, C-I, D-III
coiled form .
3) A-III, B-I, C-IV, D-II
352.(A): Mendel observed recombinants along with
4) A-III, B-IV, C-I, D-II
parental combinations during dihybrid experiment.
(R): Characters studied by Mendel for his each 360.Study the following columns and identify the
correct match
dihybrid cross experiment were on a separate
Column – I
Column – II
chromosomes but not on same chromosome.
A
Monohybrid
cross
p. T and t
353.(A) : All test cross are back crosses.
B
Test
cross
q. TT
(R) :A back cross in which recessive parent is
C
Alleles
r. TT X tt
involved is test cross.
D
Homozygous
tall
s. tt
354.(A):Mendel gave postulates like “principles of
t. Tt X tt
segregation and principles of independent
A
B
C
D
A B C D
assortment” after studying seven pairs of contrasting
1) r p t
c
2) r
t
p q
traits in garden pea.
3) r t
s
q
4) t
r
q s
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UNIT - III :GENETICS
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
SR.BOTANY EAMCET - VOL - I
361.Study the following and identify the correct
dihybrid cross
C) 5 : 3
match.
III) Dihybrid phenotypic ratio
D) 1 : 3 : 1
Character of pea Recessive trait
IV) Dihybrid test cross ratio
E) 1 : 2 : 1
A. Pod shape
I. Wrinkled
The correct combination is
B. Pod colour
II. Yellow
1) I-E, II-C, III-D, IV-A
C. Seed shape
III. Constricted
2) I-B, II-C, III-A, IV-D
D. Seed colour IV. Green
3) I-E, II-C, III-B, IV-A
V. Grey
4) I-A, II-B, III-C, IV-D
1) A - IV, B - V, C - II, D - I
365. Column - I
Column - II
2) A - III, B - II, C - I, D - IV
Cross
Phenotypic ratio
3) A - II, B - III, C - I, D - IV
I) Rr  Rr
A) ! : 1 : 1 : 1
4) A - I, B - II, C - IV, D - III
II) Rr  rr
B) 9 : 3 : 3 : 1
362.Match the following and find the correct
III) RrYy  RrYy
C) 1 : 1
combination.
IV) RrYy  rryy
D) 3 : 1
I) Homozygous A) Forms more than one type The correct combination is
of gametes
1) I-D, II-C, III-B, IV-A
II) Phenotype
B) Appears in F1 of a cross
2) I-A, II-B, III-C, IV-D
3) I-A, II-D, III-B, IV-C
between pure parents.
4) I-B, II-C, III-D, IV-A
III) Dominant
C) Physical appearance of a
366.
List - I
List - II
character
trait
A) T.H. Morgan
I)Chromosomal maps
IV) Heterozygous D) Arises by self fertilization
1) I-D, II-C, III-B, IV-A
B) G.J. Mendel
II) Linkage
C) Sturtevant
III) Checker board
2) I-C, II-D, III-A, IV-B
D) Reginald C Punnett
IV) Laws of heredity
3) I-A, II-B, III-C, IV-D
The
correct
combination
is
4) I-B, II-C, III-A, IV-D
1) A-II, B-IV, C-I, D-III
363.Match the following and find the correct
combination
2) A-III, B-II, C-IV, D-I
Character
Phenotype
3) A-II, B-III, C-I, D-IV
I) Flower position
A) Axil
4) A-II, B-I, C-III, D-IV
II) Colour of flower
B) Violet
367.List - I
List - II
III) Seed shape
C) Round
I) G.J. Mendel
A) Chromosomal theory
of inheritance
IV) Pod colour
D) Green
II) Hugo de Vries
B) Linkage in Drosophila
1) I-A, II-C, III-D, IV-B
III) T.H. Morgan
C) Mutations
2) I-C, II-A, III-D, IV-B
3) I-A, II-B, III-C, IV-D
IV) Sutton and Boveri D) Law of dominance
The correct combination is
4) 1-D, II-B, III-C, IV-A
1) I-D, II-C, III-B, IV-A
364. List - I
List - II
2) I-A, II-B, III-C, IV-D
I) Incomplete dominance
A) 1 : 1 : 1 : 1
3) I-B, II-C, III-D, IV-A
genotypic ratio in F2 of
4) I-D, II-A, III-B, IV-C
dog flowers
II) Ratio of parental types
B) 9 : 3 : 3 : 1
to recombinant types in
the phenotypes of F2
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NARAYANAGROUP
UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
EXERCISE - V
368. Character
Dominant Trait
Recessive Trait
I. Shape of the seed
Round
Wrinkled
II. Colour of the pod
Green
Yellow
III.Position of the flower Terminal
Axial
IV Colour of the seed
Yellow
Green
1) I, II and IV only
2) I and II only
3) I and IV only
4) I and III only
369. Which of the following combinations has all the dominant characters with reference to contrasting
traits of Garden pea?
Colour of
Shape of Colour of
Position of
Colour of
unripe pod Seed
Cotyledon
flowers
Flower
1) Yellow
Round
Green
Terminal
White
2) White
Wrinkled Yellow
Axial
Green
3) Green
Round
Yellow
Axial
Violet
4) Green
wrinkled
Yellow
Terminal
White
370.Study the following
Phenotypes
No.in F2 progeny
Number of Genotypes for different traits
I) Yellow, round
9
homozygous for yellow = 4
II) Yellow, wrinkled
3
homozygous for wrinkled = 0
III) Green, round
3
homozygous for round = 1
IV) Green, wrinkled
1
homozygous for green = 3
The correct combination is
1) I, II, IV
2) II, IV 3) I, III
4) III, IV
371.Study the following
I) Hybridisation
Recombination
and gametes
II) Mutation
Variations
Genetic integrity
III) Linkage
Stability of genes
New varieties
IV) Crossing over
Evolution
New species
The correct combination is
1) I, II
2) I, IV
3) II, III
4) III, IV
372.Study the following
Phenotype
Genotype
Nature of the trait
I) Homozygous tall
TT
Dominant
II) Heterozygous tall
Tt
Recessive
III) Homozygous dwarf tt
Recessive
IV) F1 hybrid
TT
Dominant
The correct combination is
1) II, III
2) I, IV
3) II, IV 4) I, III
373.Study the following
I) Garden pea
Law of dominance
Mendel
II) Lentil
Co-dominance
Flower colour
III) Snapdragon
Incomplete dominance
Seed coat patterns
IV) Fruit fly
Linkage
Morgan
The wrong combination is
1) II, IV
2) II, III
3) I, IV 4) III, IV
NARAYANAGROUP
129
UNIT - III :GENETICS
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
374.
375.
376.
377.
378.
SR.BOTANY EAMCET - VOL - I
Study the following
Genotype
Phenotype
Probability
I) RRyy
Round, green
1/16
II) RrYy
Round, yellow
4/16
III) RRYy
Wrinkled, yellow
2/16
IV) rryy
Wrinkled, yellow
1/16
The correct combination is
1) I, II, IV
2) II, III, IV
3) I, II, III
Study the following
Type of cross
Genotype
Genotypic ratio
I) Monohybrid cross
TT x tt
1:2:1
II) Back cross
Tt x TT
3:1
III) Dihybrid cross
RRYY x rryy
1:1:1:1
IV) Test cross
Tt x tt
1:1
The correct combination is
1) I, II, III
2) I, IV
3) II, III, IV
Study the following
I) Chromosome
Cell division
Two genomes
II) DNA
High molecular weight
Stable
III) Gene
Segment of DNA
Contrasting traits
IV) Allele
Alternate forms of same gene Unit of inheritance
The wrong combination is
1) I, II
2) II, III
3) I, III
Study the following combinations
Genotypes
Numbers
Probability
I) YYRR
1
1/4 x 1/4
II) YyRR
2
3/4 x 1/4
III) Yyrr
2
1/4 x 3/4
IV) YyRr
4
1/2 x 1/2
The correct combination is
1) I, IV
2) II, III
3) III, IV
Study the following
I) Dihybrid test cross
1:1:1:1
Genotypic ratio
II) Co-dominance
1:2:1
Phenotypic ratio
III) Incomplete dominance
1:1
F1 progeny
IV) Monohybrid cross
3:1
F2 genotypic ratio
The correct combination is
1) I, II
2) III, IV
3) II, IV
4) I, III
4) I, III
4) III, IV
4) I, III
4) II, III
3) Pasteur - Inheritance of acquired characters
4) de Vries - Natural selection
EXERCISE - VI
380.In Pisum sativum there are 14 chromosomes.
How many pairs with different chromosomal
(QUESTION FROM PREVIOUS
composition can be prepared? [BHU 2005]
MEDICAL EXAMINATIONS )
1) 214
2)27
3) 7
4)14
379.Which one of the following scientist’s name is
381.Source
of
Mendelian
recombination
is
correctly matched with the theory put forth by
[D.P.M.T. 2001]
him [CBSE-2008]
1)
Linkage
2)
Independent
assortment
1) Mendel - Theory of pangenesis
3)
Mutations
4)
Dominant
traits
2) Weiseman - Theory of continuity of germplasm
130
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UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
382.A recessive trait in Garden Pea is
[Karnataka 1999]
1) Wrinkled seeds
2) Tall stem
3) Round seeds
4) Coloured seed coat
383.Which is correct about traits chosen by Mendel
[C.B.S.E. 2000]
1) Terminal pod is dominant
2) Constricted pod is dominant
3) Green coloured pod is dominant
4) Tall plants are recessive
384.Which one is not a dominant trait amongst
seven Pea traits chosen by Mendel
[ManipaI 2000]
1) Flower Colour–Purple
2) Pod Colour–Yellow
3) Shape of Seed–Round 4) Flower–Axial
385.Main reason for success of Mendel was
[B.V. 2000]
1) He employed statistical analysis
2) He performed statistical experiments
3) His choice of Pisum sativum
4) He was the first to work on a plant
386.Correct reason for Mendel’s success was
[B.H.U. 2001]
1) He repeated each experiment several times
2) Traits chosen by him had genes far apart so that
linkage was absent
3) He kept record of all experiments
4) He used statistical techniques
387.In Pea, wrinkling of seeds is due to starch
because of the absence of enzyme
[C.B.S.E. 2001]
1) Amylase
2) Invertase
3) Branching enzyme 4) Diastase
388.Who amongst the following scientists
rediscovered Mendel's work ?
[M.P.P.M.T. 2002]
1) T.H. Morgan
2) W. Bateson
3) E. Strasburger
4) Von Tschermak
389.Emasculation is related to [Manipal 2002]
1) Pure line
2) Homozygocity
3) Clonal selection
4) Hybridisation
390.Which one of the following traits of Garden Pea
studied by Mendel was a recessive feature
[C.B.S.E. 2002]
1) Axial flower position 2) Green seed colour
3) Green pod colour 4) Round seed shape
391.Which trait was not incorporated by Mendel
for his experiments?
[RPMT 2005]
1) Colour of Pea seed 2) Colour of Pea flower
3) Colour of Pea Plant 4) Colour of Pea pod
392.The phenotypic ratio of a monohybrid cross is
[MHCET 2003]
NARAYANAGROUP
1) 9 : 3 : 3 : 1
2) 1 : 2 : 1
3) 3 : 1
4) 1 : 2 : 3
393. In a cross 50% individuals were tall and 50%
were dwarf. So the cross was between
[MHCET 2005]
1) Tt  tt
2) Tt  TT
3) TT  tt
4) Tt  Tt
394.Crosses between F1 offsprings and either of
their parents are known as[AFMC 2006]
1) reciprocal cross 2) back cross
3) dihybrid cross 4) polyhybrid cross
395.Hetrozygous purple flower is crossed with
recessive white flower. The progeny has the
ratio
[DPMT 2003]
1) all purple 2)all white
3) 50% purple and 50% white
4) 75% purple and 25% white
396.The crossing of an organism with a double
(homozygous) recessive in order to determine
whether it is homozygous or heteorzygous for
a character under consideration is known as
[AMU 1998, KEREALA PMT 2001,2002,
GGSIPU 2003]
1) back cross 2) test cross
3) reciprocal cross 4) dihybrid cross
397. A cross between hybrid and a recessive parent
(Tt  tt) gives a ratio of [Kerala PMT 2004]
1) 1 : 1 2) 2 : 1
3) 3 : 1 4) 4 : 1
398. Test cross involves [AIPMT 2006]
1) crossing between two genotypes with dominant
trait
2) crossing between two genotypes with recessive
trait
3) crossing between two F1 hybrids
4) crossing the F1 hybrid with a double recessive
genotype
399 .A cross between the heterozygous F1 hybrid
and the double recessive homozygous parent
is known as [KCET 2007]
1) heterosis
2) test cross
3) reciprocal cross
4) inbreeding
400.Identify the wrong statement (EAMCET 2012)
1) The probability of homozygous dwarf progeny
formed in a cross TT  TT is zero.
2) The heterozygous tall progeny formed in a cross
TT  Tt is 0.5.
3) The homozygous dwarf formed in TT  Tt is
0.75.
4) The homozygous tall progeny formed in a cross
involving TT  TT is 1.0
131
UNIT - III :GENETICS
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
401.In dihybrid cross the F1 heterozygous plants
are self fertilized to produce an F2 generation
and if offsprings are computed in Punnett
square the phenotypic F2 ratio as per Mendel’ss
independent assortment will yield
[AFMC 2000, Orissa JEE 2003]
1) 9 : 7
2) 9 : 3 : 3 : 1
3) 9 : 6 : 1
4) 12 : 3 : 1
402.In a cross between individuals with genotypes
TtRr, if the resulting number of offsprings is
16, identify the number of genotypes with TtRr
and TtRR among them [EAMCET 2003]
1) 1 and 2 2) 2 and 3 3) 3 and 1 4) 4 and 2
403.The product of cross between pure tall, green
seeds and pure dwarf, yellow seeds is crossed
with pure dwarf and green seeds then number
of phenotype produced are
[BVP Pune 2003]
1) 2
2) 6
3) 4
4) 1
404.The ratio of 1 : 1 : 1 : 1 is obtained from a
cross between the parents [Kerala PMT 2003]
1) RRYY  rryy 2) RRYy  rrYy
3) RrYY  Rryy 4) RrYy  rryy
405.Mendelian recombination is due to
[DYPatil Pune2004]
1) Independent assortment 2) segregation
3) Mutation
4) Linkage
406.What will be ratio of combination of parental
to non-parental in F2 generation where F1 is
produced by pure breeding yellow round seeds
crossed with green wrinkled seed?
[DY Patil Pune 2004]
1) 12 : 4
2) 9 : 1 3) 5 : 3 4) 16 : 1
407.Number of male gametes having both dominant
genes for YYRr per 1000 is
[DY Patil Pune 2004]
1) 500 2) 250 3) 750
4) 1000
408.The ratio of homozygous dominant and
homozygous recessive in dihybrid cross is
[BVP Pune 2004]
1) 1 : 16 2) 1 : 8 3) 16 : 1
4) 1 : 1
409.A dihybrid for qualitative trait is crossed with
homozygous recessive individual of its type, the
phenotypic ratio is [Orissa JEE 2005]
1) 1 : 2 : 1 2) 3 : 1
3) 1 : 1 : 1 : 1 4) 9 : 3 : 3 : 1
410.The percentage of yr gametes in YyRr is
[Maharashtra CET 2005]
1) 25% 2) 50% 3) 75%
4) 100%
411. In order to find out the different types of
gametes produced by a pea plant having the
132
SR.BOTANY EAMCET - VOL - I
genotype AaBb, it should be crossed to a plant
with the genotype
[AIPMT 2005]
1) aaBB 2) AaBb 3) AABB 4) aabb
412.Which genotype represents a true hybrid
condition?
[CPMT 2005]
1) TTrr 2) TtRr 3) TTRr
4) ttrr
413.In progeny of dihybrid cross in F2 generation,
how many plants are homozygous recessive?
[Harayana PMT 2005]
1) One
2) two 3) Three
4) Four.
414.A dihybrid plant on self pollination, produced
400 plants with 4 types of genotype. How many
seeds will have genotype TtRr?
[CPMT 2006]
1) 200
2) 100 3) 50
4) 150
415.Cross between plants having YYRR and yyrr
will yield
[Harayana PMT 2006]
1) yellow and round seeds
2) yellow and wrinkled seeds
3) green’s round seeds
4) green and wrinkled seeds
416.The ratio of F2 phenotype generation in
Mendel’s dihybrid cross between yellow and
green seeds was [DY Patil 2006]
1) 3 : 1
2) 2 : 1 3) 1 : 1
4) 12 : 4
417.A double homozygous tall pea plant with green
cotyledons is crossed with pure dwarf plant with
yellow cotyledons. The offspring are crossed
with pure dwarf plant with green cotyledons.
How many different phenotypes will be
obtained in this progeny? [DY Patil Pune 2007]
1) 16
2) 2
3) 4 4) 1
418.Dihybrid genotypic testcross ratio is
[CMC Vellore 2007]
1) 1 : 12 2) 1 : 1 : 1 : 1 3) 3 : 1 4) 1 : 1
419.If dominant alleles or recessive alleles tend to
remain in different plants, then this could be
[CMC Vellore 2007]
1) Independent assortment 2) Segregation
3) Dominance4) by chance
420.When a dihybrid cross is fitted into a Punnett
square with 16 boxes, the maximum number of
different phenotypes available are
[Kerala PMT 2008]
1) 8 2) 4
3) 2 4) 16
421.Mendel crossed tall and rounded seeds with
short and wrinkled. In F2 generation he saw 1/
16 were short and wrinkled seeds. What did
he conclude from this? [HP PMT 2008]
1) Short and wrinkled show linkages
2) Tall and wrinkled never combine
3) They show independent assortment
4) None of the above
NARAYANAGROUP
UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
422.The number F2 plants found similar to F1
phenotype and genotypes respectively in a total
of 128 plants in F2 progeny of a Mendelian
dihybrid cross [EAMCET 2011]
1) 70, 30 2) 72, 32 3) 24, 12 4) 40,70
423.In garden pea yellow is dominant over green,
round is dominant over wrinkled yellow round
is crosses with yellow wrinkled. Progeny shows
segregation of all the four types. The
probability of green round in this crosses
[EAMCET 2009]
430.Any change in the structure of the
chromosome will change the structure of gene.
Hence it is true statement.
424.How many different types of gametes can be
formed by F1 progeny, resulting from the
following cross: AA BB CC  aa bb cc
[AIIMS 2004]
1) 3 2) 8
3) 27 4) 64
425.A self-fertilizing trihybrid plant forms
[APMT 2004]
1) 8 different gametes and 64 different zygotes
2) 4 different gametes and 16 different zygotes
3) 8 different gametes and 16 different zygotes
4) 8 different gametes and 32 different zygotes
426.How many different kinds of gametes will be
produced by a plant having the genotype
AABbCC?
[APMT 2006]
1) Two 2) Three 3) Four 4) Nine
427.A dwarf pea plant treated with gibberellic acid,
becomes tall, then crossed with pure
homozygous tall pea plants, then what will be
the phenotypic ratio of tall pea plants in F1
progeny? [BVP Pune 2007]
1) 100% dwarf plants 2) 50% tall plants
3) 100% tall plants4) all tall pea plants.
428.The number of genotypes produced when
individual of genotype YyRrTt are cross with
each other?
[MGIMS Wardha2006; DYPatil Pune 2004]
1) 4 2) 64
3) 28 4) 27
429.A homozygous sweet pea plant with blue flowers
(RR) and long pollen (  R0 R0  is crossed with a
homozygous plant having red flowers (rr) and
round pollen  r0 r0  . The resultant hybrid is
test crossed. Which of the following genotype
does not appear in its progeny
[EAMCET 2009]
1) Rrrr0 2) RrR0 r0 3) Rrr0 r0
4) rrR0 r0
NARAYANAGROUP
430.All genes located on the same chromosome
[PMT 2007]
1) Form different groups depending upon their
relative distance.
2) Form one linkage groups,
3) will not form any linkage group.
4) form interactive group.
431.Distance between the genes and percentage
of recombination shows
[PMT 2008]
1) direct relationship
2) an inverse relationship
3) a parallel relationship 4) no relationship
432.The first attempt to show linkage in plants was
done on
[PMT 2002]
1) Pisum sativum
2)Lathyrus odoratus
3) Oenothera lamarkiana 4) Zea mays
433.When two genes are situated very close to each
other in a chromosome then
[CPMT 2007]
1) the percentage of crossing over between them is
very high
2) hardly any crossing over detected
3) no crossing over takes place between them.
4) no double cross over can take place between
them.
434.Bacteria and blue green algae contain
[AIIMS 2003]
1) one linkage group
2) two linkage groups
3) many linkage groups 4) no linkage group.
435.A cross between two tall garden pea plants
produced all tall plants. The possible genotypes
of the parents are
[EAMCET -2014]
I) TT,TT II) TT,Tt
III) Tt, tt
IV) Tt, Tt
The correct answer is
1) III, IV
2) I, IV
3) I, II 4) II, III
436.Tall (T) is completely dominant over dwarf (t).
Red flower colour (R) is incompletely dominant
over white (r), the heterozygote being pink. Plant
having genotype of Tt Rr is self pollinated. What
would be the proportion of plants with dwarf and
pink characters in its progeny ?
[EAMCET-2014]
1)
2
16
2)
1
``
16
3)
9
16
4)
3
16
437.Assume that blue flower of a plant is dominant
character over the white. When a blue flowered
plant is crossed with white flowered plant, the
progeny showed 50% of plants with blue flowers
and 50% of plants with white flowers. The
genotypes of blue and white parents
respectively are
[EM -2013]
1) BB,bb
2) Bb,bb
3) bb,bb
4) BB,Bb
133
UNIT - III :GENETICS
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
EXERCISE - VII
TRUE/ FALSE STATEMENTS
438.Mendel’s genes are now a days factor’s
439.The entire body of molecular biology was a
consequent development with major contribution of
Krebs, Calvin, Crick and Mendel
440.One of the causes of variation was hidden in a sexual
reproduction
441.Natural selection and breeding are the key process
in the development of Sahiwal Cows in Punjab and
Ongole bulls in A.P.
442.Mendelian general rules of inheritance were not
substantiated ideas
443.Pea plant self fertilized inconveniently
444.According to Mendelian rules, F1 is always similar
to one of the parent
445.Emasculation is essential for obtaining F 2generation
in Mendelian monohybrid cross
446.Mendel was an Australian monk
447.The phenotype height is represented with the
genotypes TT or Tt or tt
448.The genotype of F2 dwarf plants was homozygous
459.Mendel found that dwarf plants of F2 continued to
generate dwarf plants in F3 and F4
450.Checker Board developed by British chemist
Punnett
451.Recessive trait may not be in heterozygous
452.During formation of gametes, the paired unit factors
separate randomly with equal probability due to
meiosis
453.Generally gametes are pure for their factor they
posses
454.F1 hybrid is crossed with parental type having
dominant trait in homozygous condition, the no. of
recombinant individuals is 50%
455.Pleiotropy means a single character influenced by
many genes
456.Mutation useful in crop improvement
457.Alternative form a character is called allele
458.In co-dominance the phenotypic ratio of F2 progeny
is 3 : 1.
459.Drosophila is morphologically easily distinguishable
460.According to concept of linkage, two alleles of
a gene located on same chromosome.
461.Mendel proposed his law of independent assortment
based on results on dihybrid cross.
462.Alleles of same gene present on similar locations of
homologous chromosomes.
463.The probability of double homozygous individuals
in F2 generation Mendel dihybrid cross is 1/4.
134
SR.BOTANY EAMCET - VOL - I
464.Externally it is possible to distinguish between the
plants with the genotypes TT and Tt.
465.In a test cross, dominant phenotype is crossed with
recessive parent instead of self fertilization.
466.Genes can never be mutated
467.Dominance is not an autonomous feature of a gene.
468.Linkage is inversely proportional to distance
between the genes of the same chromosome.
469.Mutations first discovered by Hugo de Vries
470.The term linkage coined by T.H. Morgan.
471.RBC that determining ABO blood group and seed
coat pattern example for incomplete dominance
472.In dihybrid cross of Morgan, the proportion of
parental gene combination was much higher than
the non parental type
473.Chromosomal theory of inheritance proposed by
Watson and Crick.
474.Mendel dihybrid test cross ratio is 1 : 1 : 1 : 1
475.Co - dominance is seen in Snapdragon.
476.Generally gametes are pure for the factors they
possess
477.Number of monohybrid traits in typical dihybrid is
8
478.Several dihybrid crosses in Drosophila were carried
out by Sutton and Boveri
479.Number of phenotype in monohybrid test cross is
2 and genotype also 2.
480.Number of double dominant individuals in Mendel
dihybrid F2 progeny is 1.
481.Chromosomal aberrations are commonly observed
in cancer cells.
482.Factors are discrete which controls characters.
483.Unit factors occurs in paired condition.
484.Sickle cell anemia is a classical example for point
mutation
485.The change in a single base pair of DNA is called
point mutation
486.Deletions and insertions of base pairs of DNA cause
frame shift mutations.
487.UV radiations is an example of chemical mutagen
488.Desirable mutations helps in crop improvement
489.Any change in the structure of the chromosome will
change the structure of gene.
490.Number types of gametes produced from Tt
genotype is two
491.Sutton and Boveri noted parallelism between
behaviour of chromosomes and genes.
492.Number of Homozygous tall individuals in F 2
generation of monohybrid cross is 1.
NARAYANAGROUP
UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
KEY
PRACTICE TEST-I
1) 4
7) 3
13) 4
2)1
8)4
14) 2
3)4
4) 2 5) 4
9) 2 10) 1 11)1
15) 3
6) 4
12) 4
PRACTICE TEST-II
16) 4
22) 1
28)4
34) 2
17)2
23) 3
29) 2
35) 2
18) 2 19) 1 20) 2 21) 1
24) 3 25) 2 26) 2 27)1
30) 2 31) 4 32) 2 33) 2
36) 4
PRACTICE TEST-III
37)2
43) 1
49)2
55) 2
38)3
44) 4
50)2
56) 1
39) 3
45) 3
51) 2
57) 1
40) 1 41) 4 42) 2
46)1 47)4 48)2
52) 3 53) 4 54) 1
58)1
PRACTICE TEST-IV
59) 3
65)2
60)3
66)2
61)2
67)4
62)3
68)2
63)3 64)2
69) 2
PRACTICE TEST-V
70) 2
76) 2
82) 4
88) 2
71) 3
77) 4
83) 1
72) 1 73) 4 74) 1 75) 1
78) 1 79) 4 80) 4 81) 3
84) 3 85) 4 86) 4 87)4
PRACTICE TEST-VI
89) 4
95) 2
90) 4
96) 3
91) 2 92) 3 93) 3 94) 2
97) 4 98) 1 99) 1 100)4
PRACTICE TEST-VII
101) 1 102) 3
107)4 108) 3
103) 4 104) 4 105) 1 106)4
109) 1 110) 1
QUESTION BANK (EI - EVII)
CHECK YOUR MEMORY (E–I)
111)1
117)4
123)4
129)3
135)3
141)2
147)3
153)3
159)2
165)2
171)1
177)4
183)4
189)4
195)2
201)2
207)2
213)3
219)4
225)4
112)2
118)1
124)3
130)2
136)2
142)1
148)2
154)1
160)4
166)2
172)3
178)3
184)4
190)3
196)4
202)3
208)1
214)1
220)1
226)2
NARAYANAGROUP
113)2
119)3
125)4
131)3
137)1
143)3
149)1
155)1
161)2
167)4
173)2
179)3
185)4
191)3
197)2
203)2
209)2
215)2
221)2
227)2
114)4
120)4
126)1
132)1
138)2
144)2
150)2
156)1
162)4
168)2
174)4
180)3
186)2
192)1
198)3
204)2
210)2
216)2
222)2
228)2
115)3
121)3
127)2
133)2
139)2
145)4
151)3
157)4
163)4
169)1
175)1
181)3
187)2
193)4
199)3
205)1
211)4
217)3
223)3
229)4
116)1
122)1
128)2
134)1
140)1
146)3
152)2
158)3
164)2
170)2
176)3
182)4
188)2
194)2
200)4
206)3
212)1
218)4
224)2
230)3
231)1
237)3
243)2
249)2
255)2
261)2
267)1
273)4
279)1
285)1
291)3
297)1
303)2
309)2
315)1
232)4
238)1
244)3
250)2
256)2
262)2
268)2
274)3
280)2
286)3
292)3
298)3
304)3
310)4
316)1
233)3
239)4
245)3
251)2
257)1
263)2
269)1
275)3
281)2
287)3
293)1
299)1
305)2
311)4
317)3
234)4
240)3
246)3
252)3
258)4
264)2
270)1
276)2
282)1
288)1
294)3
300)4
306)3
312)3
318)4
235)3
241)3
247)4
253)3
259)1
265)4
271)3
277)3
283)1
289)2
295)2
301)1
307)2
313)3
319)1
236)2
242)2
248)1
254)1
260)2
266)2
272)4
278)4
284)3
290)3
296)4
302)1
308)2
314)3
320)4
SHARPEN YOUR REFLEXES(E-II)
321)1 322)2 323)3 324)3 325)3 326)3
327)3 328)1 329)3 330)2 331)2 332)4
THINK TWICE BEFORE YOU
CHOOSE (E-III)
333)3
339)2
345)3
351)1
357)1
334)2
340)2
346)3
352)1
335)1
341)1
347)1
353)1
336)3
342)1
348)1
354)3
337)2
343)3
349)3
355)3
338)2
344)4
350)1
356)1
SIMPLE MATCHING QUESTIONS
(E– IV)
358)3
364)1
359)3
365)1
360)2 361)2 362)1 363)3
366)1 367)1
MULTIPLE MATCHING QS (E– V)
368)1
373)2
378)1
369)3
374)1
370)3 371)2 372)3
375)2 376)3 377)1
QUESTIONS FROM PREVIOUS
MEDICAL EXAMINATIONS (E-VI)
379)2
385)1
391)3
397)1
403)4
409)3
415)1
421)3
427)4
433)3
380)2
386)2
392)3
398)4
404)4
410)1
416)1
422)2
428)4
434)1
381)2
387)3
393)1
399)2
405)1
411)4
417)3
423)2
429)1
435) 3
382)1
388)4
394)2
400)3
406)3
412)2
418)2
424)2
430)2
436)1
383)3
389)4
395)3
401)2
407)1
413)1
419)2
425)1
431)1
437)2
384)2
390)2
396)2
402)4
408)4
414)2
420)2
426)1
432)2
135
UNIT - III :GENETICS
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
TRUE OR FALSE STATEMENT (E-VII)
438) F
443) F
448) T
453) T
458) F
463) T
468) T
473) F
478) F
483) T
488) T
439) F
444) T
449) T
454) T
459) T
464) F
469) T
474) T
479) T
484) T
489) T
440) T
445) F
450) F
455) F
460) F
465) T
470) T
475) F
480) T
485) T
490) T
441) F
446) F
451) T
456) T
461) T
466) F
471) F
476) T
481) T
486) T
491) T
442) F
447) T
452) T
457) F
462) T
467) T
472) T
477) T
482) T
487) F
492) T
HINTS FOR EXCERSISE - III
333. In a test cross F1 organism is crossed with
recessive parent hence R is false.
334.both are true but not explains.
335.In Lentil plant the heterozygous F1 individual show
the phenotypic features of both the parents which
means that neither “spotted” nor “dotted” allele is
dominant or recessive to other and the alleles are
co-dominant, hence R explains A.
336.Production of more than one character by one single
gene is called pleiotrophism.
337.Both are true but are not related.
338.Both linkage and recombination are individual
definitions but are not related.
339.Both are true but are not related
340.Assertion explains what are chromosomal aberrations
and the reason explains where they commonly
occurred so they are not related.
341.Sickle cell anemia is because of change occurs in a
single base pair of DNA hence it is point mutation.
342.U.V radiation can cause mutations in organisms
hence U.V radiation is a mutagen.
343.In a dihybrid cross conducted by Mendel, fortunately
the pairs of factors are located in two pairs of
homologous chromosomes and generally each
homologous pair of chromosome bears only one pair
of factors or alleles for one gene hence R is false.
344.Hybrids produced by hybridization process may
have undesirable genes along with the desired genes,
hence A is false.
345.Crossing can occur in between two weakly linked
genes so R is false.
346.Pea plant is self pollinated member of Fabaceae
but not Solanaceae hence R is false.
347.In Mendel monohybrid cross F1 individual are in
heterozygous condition (Tt). They produce gametes
like female T, female t & male T, male t. Hence R
explains A.
348.If distance between genes on a chromosome is less
then linkage is more, if distance is more then linkage
136
SR.BOTANY EAMCET - VOL - I
is weak so Linkage is inversely proportional to
distance between genes.
349.F1 individuals of monohybrid cross show alleles for
tallness (T) and for dwarfness (t), but because
complete dominance only tall phenotype is
expressed. So the probability of obtaining dwarf
plants in F1 is zero. Hence R is false.
350.As the heterozygotes show the phenotype that is
inbetween the two parents , so the phenotypic and
genotypic ratios are same. Hence R explains A.
351.Usually chromosomes are made up of DNA and
histone proteins. One chromatid
(anaphase chromosome) is made upof only one
DNA. if loss or gain in that segment causes
alternation of chromosomes, so R explains A.
352.Fortunately characters selected by Mendel for his
hybridization experiments in pea were on different
chromosomes, but not two characters selected by
Mendel for a dihybrid cross on a same
chromosome. So the characters are
independently assorted and results in the formation
of recombinants, So R explains A.
353.Back cross involves the cross between F1 individual
with either dominant or recessive parent, where as
test cross is a type of back cross in which only
recessive parent is involved.hence test cross is also
a type of back cross.
354.Mendel conducted hybridization experiments in
garden plant ( Pisum sativum) but not in Wild pea
( Lathyrus sativus), hence R is false.
355.A breeder can select only desirable mutations, but
not undesirable mutations.
356.Usually The linked genes tend to get inherited together
so they reduces the possibility of gametes production,
so they produce more number of parental combinations.
357.As the factors selected by Mendel present on
different chromosomes so they are independently
assorted, so Mendel didn’t find any linkage in pea
plant. Hence R explains A.
HINTS FOR TRUE OR FALSE
STATEMENTS (E-VII)
379.Mendel’s factors were today’s genes hence the
statement is false.
380.Krebs and Calvin are related to physiology but not
molecular biology hence it is false statement
381.One of the causes of variation was hidden in sexual
reproduction hence it is true statement.
382.Artificial selection (but not natural selection) and
breeding are the key process in the development of
Sahiwal Cows in Punjab and Ongole bulls in A.P ,
hence it is a false statement
383.Mendelian general rules of inheritance were
substantiated ideas, hence it is a false statement
384.Pea plants are self fertilized conveniently hence
it is false statement.
NARAYANAGROUP
UNIT - III :GENETICS
SR.BOTANY EAMCET - VOL - I
Chapter-9 : PRINCIPLES OF INHERITANCE AND VARIATION
385.According to Mendelian rules, F1 is always similar
to one of the parents, hence it is true statement
386.Emasculation is not required for obtaining F2
generation in Mendelian monohybrid cross because
they are produced by self pollination
387.Mendel was an Austrian but not Australian monk
hence it is a false statement.
388.The phenotype height is represented with the
genotypes TT or Tt or tt hence it is true statement.
389.In order to express recessive character, both alleles
must be recessive. Hence the statement is true.
390.If dwarf plants with ‘tt’ genotype are self crossed
they are going to produce only dwarf plants hence
the statement is true.
391.Checker Board developed by British geneticist but
not chemist, hence it is false
statement.
392.Recessive trait may not be in heterozygous because
it requires two recessive alleles, hence
it is a
true statement.
393.During formation of gametes, the paired unit factors
separate randomly with equal probability due to
meiosis hence it is true statement.
394.Generally gametes are pure for their factor they
posses hence it is true statement.
395.F1 hybrid is crossed with parental type having
dominant trait in homozygous condition, the number
of recombinant individuals is 0% hence it is true
statement.
396.Pleiotropy means a single gene is responsible for
many characters,hence it is false statement.
397.Mutations can generate large number of variability
with in the population. Hence a breeder can use
desirable mutations, hence it is true statement.
398.Alternative form a character is called contrasting
traits but not alleles, hence it is a false statement.
399.In co-dominance the phenotypic ratio of F2 progeny
is 1 : 2 : 1 but not 3 : 1, hence it is a false statement.
400.Drosophila is morphologically easily distinguishable
based on sex, because females are larger than males.
Hence it is a true statement.
401.According to concept of linkage, two or more
genes of a chromosome are linked but not two
alleles of same gene. Hence it is a false statement.
402.Mendel proposed his law of independent assortment
based on results on dihybrid cross, hence it is a true
statement.
403.Alleles of same gene are present on similar locations
of homologous chromosomes. Hence it is true
statement.
404.The probability of double homozygous individuals
in F2 generation Mendel dihybrid cross is 1/4. Hence
it is a true statement.
405.Externally it is not possible to distinguish between
the plants with the genotypes TT and Tt because
the genotype, an organism can not be predicted by
physical appearance. Hence it is false statement .
406.In a test cross, dominant phenotype is crossed
NARAYANAGROUP
with recessive parent instead of self fertilization
hence it is true statement.
407.Genes can be mutated hence it is false statement.
408.Dominance is not an autonomous feature of a gene,
because it depends upon the phenotype produced
by the product. Hence it is a false statement.
409.Linkage is inversely proportional to distance
between the genes of the same chromosome. Hence
it is true statement.
410.Mutations first discovered by Hugo de Vries. Hence
it is a true statement.
411.The term linkage coined by T.H. Morgan. hence it
is true statement.
412.RBC that determining ABO blood group and seed
coat pattern example for co - dominance. Hence it
is a false statement.
413.In dihybrid cross of Morgan, the proportion of
parental gene combination was much higher than
the non parental type because he found linkage in
Drosophila, hence it is true statement.
414.Chromosomal theory of inheritance proposed by
Sutton and Boveri but not Watson and Crick. Hence
it is a false statement.
415.Mendel dihybrid test cross ratio is 1 : 1 : 1 : 1.
Hence it is true statement.
416.Co - dominance is not seen in Snapdragon. Hence
it is a false statement.
417.Generally gametes are pure for the factors they
possess. Hence it is true statement.
418.Number of monohybrids in typical dihybrid is 8.
Hence it is a true statement.
419.Several dihybrid crosses in Drosophila were carried
out by T.H.Morgan. Hence it is a false statement.
420.Number of phenotypes in monohybrid test cross is
2 and genotypes also is 2. hence it is true a statement.
421.Number of double dominant individuals(RRYY) in
Mendel dihybrid F2 progeny is 1. Hence it is true
statement.
422.Chromosomal aberrations are commonly observed
in cancer cells. Hence it is true statement.
423.Factors are discrete units which controls characters.
Hence it is true statement.
424.Unit factors occurs in paired condition. Hence it is
true statement.
425.Sickle cell anemia is a classical example for point
mutation. Hence it is true statement.
426.The change in a single base pair of DNA is called
point mutation. Hence it is a true statement.
427.Deletions and insertions of base pairs of DNA cause
frame shift mutations. Hence it is true statement.
428.UV radiations is an example of physical mutagen
but not chemical mutagen, Hence it is a false
statement.
429.Desirable mutations helps in crop improvement.
Hence it is true statement.
137