NEET Senior Batches

The Leader In Chemistry ..... Biology- Test No. : 21

• Topic : Principles of Inheritance and Variation•
Date Time No. Of MCQs Marks

RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *
26/02/2024 60 mins. 100 360 (+4, –1)

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Section-C: Biology 7. Mutation [Old Page. 88]
1) Results in alteration of DNA sequaence
Section-A
2) Change phenotype but not genotype
1. In a few insects, Henking observed a specific
3) Creates discontinuous variations
nuclear structure only in 50% of the sperms,
whereas the other 50% lack this structure. He 4) More than 1 correct
termed this structure as. [Old Page. 87] Ans.(4)
1) Barr body 2) X-bod 8. Chromosomal aberration is quite common in
3) Y-body 4) Drum - stick [Old Page. 91]
1) Cancer cells 2) Skin cells
Ans.(2)
3) Meristematic cells 4) Muscle cells
2. If we change the position of genes on
Ans.(1)
chromosomes it will be included in
9. Choose the odd option w.r.t sickle cell anaemia.
[Old Page. 91]
[Old Page. 90]
1) Polyploidy
1) It is a result of point mutation.
2) Genomic mutation 2) GUG in the mRNA is replaced by GAG.
3) Chromosomal Mutation 3) HbAHbs individuals apparently unaffected.
4) Gene Mutation 4) Mutant haemoglobin molecule undergoes
Ans.(3) polymerisation under high oxygen tension.
3. The sum total of genes in a population is known Ans.(2)
as 10. Select the incorrect match.
1) Gene bank 2) Genome [Old Page. 91, 92]
3) Linkage group 4) Gene pool 1) Phenylketonuria - Autosomal recessive trait
Ans.(4) 2) Down’s syndrome - Trisomy of allosome
3) Klinefelter’s syndrome - 47, XXY
4. Chemical or physical factors that induce mutation
are referred to as. [Old Page. 88] 4) Turner’s syndrome - 45 with XO
Ans.(2)
1) Mutagens 2) Mutants
11. If a mother is carrier for colourblindness and
3) Carcinogens 4) Statins
father is colourblind, then what percent of the
Ans.(1) children will be boys with normal vision?
5. Sex of progeny is determined by female parent [Old Page. 89]
in [Old Page. 87] 1) 50% 2) 75%
1) Grasshopper 2) Bird 3) 25% 4) 0%
3) Drosophila 4) Human Ans.(3)
Ans.(2) 12. In phenylketonuria, the affected individual
6. Genes for which of the following characters in lacks an enzyme that converts the amino acid A
Drosophila were found to be very tightly linked into B. Identify A and B. [Old Page. 91]
by Morgan? [Old Page. 84] A B
1) Eye colour and wing size 1) Phenylalanine Tyrosine
2) Tyrosine Phenylalanine
2) Body colour and wing size
3) Phenylalanine Valine
3) Wing size and eye colour
4) Glutamic acid Phenylalanine
4) Body and eye colour
Ans.(1)
Ans.(4)
Prof. Motegaonkar S. R. M.Sc. Che. Gold Medalist, SET/NET--JRF, GATE, BARC,TIFR Qualified 1
13. If the haplod number of chromosome in an 19. Observe the given pedigree.
organism is 20, what will be the chromosome
number present in its trisomy?
[Old Page. 92]

RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *
1) 11 2) 21
3) 41 4) 39 Which of the following disorder can have such
Ans.(3) kind of pedigree ? [Old Page. 89]

14. Which of the following is true for 1) Sickle cell anaemia 2) Haemophilia
phenylketonuria and down’s syndrome? 3) Colourblindness 4) Myotonic dystrophy
[Old Page. 91, 92] Ans.(1)
1) Both of them are X - linked recessive traits 20. Assertion : A true breeding line shows the stable
2) Both are caused due to gene mutation trait inheritance and expression for several
generations. [Old Page. 70]
3) Mental retardation occurs in both of them
Reason : A true breeding line is one that have
4) Reduction in hair and skin pigmentation in both undergone continuous cross-pollination.
the cases
1) If both assertion and reason are true and the
Ans.(3) reason in the correct explanation of the assertion
15. How many single heterozygous will be prduced 2) If both assertion and reason are true, but reason
in F2 generation of a dihybrid cross of Mendel? is not the correct explanation of the assertion.
[Old Page. 79] 3) If assertion is true, but reason is false.
1) 8 / 16 2) 2 / 16 4) If both assertion and reason are false
3) 1 / 16 4) 4 / 16 Ans.(3)
Ans.(1) 21. Which of the following Mendelian disorder is a
16. How many true breeding pea plant varieties did quantitative problem of synthesizing few  -
Mendel select as pairs, which were similar globin molecules? [Old Page. 91]
except in one character with contrasting traits? a. Sickle cell anaemia b. Haemophilia
[Old Page. 70] c. Thalassemia
1) 4 2) 2 1) a & c 2) Only c
3) 14 4) 8 3) Only a 4) a, b and c
Ans.(3) Ans.(2)
17. A woman has an X-linked condition on one of 22. Which of the following cross will give recessive
her X chromosomes. This chromosome can be progeny in F1 generation? [Old Page. 73]
inherited by [Old Page. 89]
1) TT  tt 2) Tt  TT
1) Only daughters
3) TT  TT 4) tt  tt
2) Only sons
Ans.(4)
3) Only grandchildren
23. The ability of a gene to have multiple phenotypic
4) Both sons and daughters effect is known as. [Old Page. 85]
Ans.(4) 1) Epistasis 2) Pleiotropy
18. A genetic disorder is transferred from a 3) Linkage 4) Polygenic inheritance
phenotypically normal but carrier female to only
some of the male progeny. The disease could be Ans.(2)
[Old Page. 90] 24. A gamete will contain [Old Page. 75]
1) Phenylketonuria 1) Both the alleles of a gene
2) Cystic fibrosis 2) Multiple alleles of a gene
3) Haemophilia 3) No allele
4) Sickle cell anaemia 4) Only one alleles of a gene
Ans.(3) Ans.(4)

Prof. Motegaonkar S. R. M.Sc. Che. Gold Medalist, SET/NET--JRF, GATE, BARC,TIFR Qualified 2
25. Statement I : In snapdragon, F1 plants do not have 30. Select the correct match.[Old Page. 90]
red or whte flowers.[Old Page. 76] 1) Haemophilia – Y linked
Statement II : It is intermediate inheritance with 2) Phenylketonuria – Autosomal dominant trait
neither of the two alleles of a gene being dominant
3) Sickle cell anaemia – Autosomal recessive trait,
over each other.

RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *
chromosome - 11
1) Both statement I and statement II are true. 4) Thalassemia – X linked
2) Both statement I and statement II are false Ans.(3)
3) Statement I is true but statement II is false 31. Statement - I : The XX-XO type of chromosomes
4) Statement I is false but statement II is true constitution for sex determination is seen in few
Ans.(1) insects. [Old Page. 85]
Statement - II : Henking observed a specific
26. If a pink flowered snapdragon plant is crossed
with homozygous white flowered plant, the nuclear structure X-body through spermatogenesis
in birds.
offspring would be [Old Page. 76]
1) Both statement I and statement II are true.
1) All red flowered
2) Both statement I and statement II are false
2) All pink flowered
3) Statement I is true but statement II is false
3) 50 % white flowered
4) Statement I is false but statement II is true
4) 50 % red flowered
Ans.(2)
Ans.(3)
32. Broad palm with single palm crease is visible in
27. Match the columns and select the correct option. person suffering from [Old Page. 92]
[Old Page. 90, 92] 1) Turne’s syndrome

Column - I Column - II
2) Klinefelter’s syndrome
3) Thalassemia
a. Down's syndrom i. Sex influenced trait
4) Down’s syndrome
b. Haemophilia ii. Holandric trait
Ans.(4)
c. Pattern Baldness iii. Trisomy
33. Who used the frequency of recombination
d. Hypertrichosis iv. Sex-linked trait
between gene pairs on the same chromosome as
1) a-iii, b-i, c-ii, d-iv 2) a-iii, b-iv, c-i, d-ii a measure of the distance between genes ?
3) a-iii, b-ii, c-i, d-iv 4) a-iii, b-ii, c-iv, d-i [Old Page. 83]
Ans.(2) 1) Sutton and Boveri 2) T.H Morgan
28. Law of independent assortment 3) Sturtevant 4) Mendel
[Old Page. 80] Ans.(3)
1) Is a universal phenomenon 34. An individual suffereing from a particular
genetic disorder has
2) Results in new linkage group
a. Gynaecomastia
3) Can be observed for genes present on same
b. Feminine pitched voice
chromosome
What will be the chromosome complement of
4) Occurs only when genes are present on non-
such individual? [Old Page. 92]
homologous chromosomes
1) 45 + XY 2) 44 + XXY
Ans.(4)
3) 44 + XO 4) 45 + XY
29. If the size of starch grain in pea is considered as
phenotype. The Bb allele will show Ans.(2)
35. A child of blood group ‘O’ cannot have parents
[Old Page. 78]
of blood group [Old Page. 77]
1) Complete dominance
1) AB and O
2) Incomplete dominance
2) A and B
3) Co-dominance 3) B and B
4) Pleiotropism 4) O and O
Ans.(2) Ans.(1)

Prof. Motegaonkar S. R. M.Sc. Che. Gold Medalist, SET/NET--JRF, GATE, BARC,TIFR Qualified 3
 Section-B 41. Statement I : Mendel studied seven pairs of
36. Select the option in which the combination will contrasting traits in pea plants and proposed the
result in Turner ’s syndrome (A represent Laws of Inheritance. [Old Page. 70]
autosomes i.e.,22) [Old Page. 92] Statement II : Seven characters examined by
Mendel in his experiment on pea plants were seed

RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *
1) Egg (A + 0) × sperm (A + 0)
shape and colour, flower colour, pod shape and
2) Egg (A + 0) × sperm (A + X)
colour, flower position, and stem height.
3) Egg (A + X) × sperm (A + XY)
In the light of the above statements, choose the
4) Egg (A + XY) × sperm (A + X) correct answer from the options given below.
Ans.(2) 1) Statement I is correct, but statement II is incorrect
37. Match column I with column II. 2) Statement I is incorret, but statement II is correct
[Old Page. 76, 77, 78] 3) Both statement I and statement II are correct
Column - I Column - II 4) Both statement I and statement II are incorrect
a Incomplete i ABO blood grouping Ans.(3)
dominance 42. The two genes of a chromosome lie at adjacent
b Co-dominance ii Snapdragon gene locus; they show / are [Old Page. 84]
c Multiple allele iii Garden peas 1) Maximum recombination
d Dominance iv Fruit-flies 2) More crossing over
3) No linkage
1) a-ii, b-iv, c-iii, d-i 2) a-i, b-ii, c-iv, d-iii
3) a-iii, b-iv, c-i, d-ii 4) a-ii, b-i, c-i, d-iii 4) Tightly linked
Ans.(4) Ans.(4)
38. When Morgan crossed yellow bodied and white 43. Which of the following characteristics represent
eyed female to wild type male, in F1 progeny ‘Inheritance of blood groups’ in humans?
[Old Page. 83] [Old Page. 77, 75, 78, 85]
1) All the males were of wild type a. dominance b. Co-dominance
2) All females were identical females of parental c. Multiple allele d. Incomplete dominance
generation e. Polygenic inheritance
3) Females were heterozygous for both the 1) b, c, and e 2) a, b and c
characters 3) b, d and e 4) a, c and e
4) Males were yellow bodied and red eyed Ans.(2)
Ans.(3) 44. A test cross is done to find out
39. Mark the pedigree which shows the inheritance of [Old Page. 74]
disorder like haemophilia. [Old Page. 89] 1) The genotype of an individual by examining
the phenotypes of its offspring from a particular
1) 2) mating
2) The genotype of an individual by testing its
DNA content
3) 4) 3) Whether a mating is fertile
4) Whether two species can interbed
Ans.(4)
Ans.(1)
40. Identify the chromosomal disorder on the basis
of given features. [Old Page. 92] 45. Statement I : Dominance is not an autonomous
feature of a gene or the product that it has
a. Congenital heart disease
information for. [Old Page. 76]
b. Broad plam and characteristic palm crease
Statement II : Dominance depends much on the
c. Physical, psychomotor and mental
gene product and the production of a particular
development is retarded. phenotype from this product.
1) Turner’s syndrome
1) Both statement I and statement II are true.
2) Edward syndrome
2) Both statement I and statement II are false
3) Klinefelter’s syndrome
3) Statement I is true but statement II is false
4) Down’s syndrome
4) Statement I is false but statement II is true
Ans.(4)
Ans.(1)

Prof. Motegaonkar S. R. M.Sc. Che. Gold Medalist, SET/NET--JRF, GATE, BARC,TIFR Qualified 4
46. Consider the following characteristics. 50. Match column I with column II.
a. Monosomy of X chromosome [Old Page. 73, 70, 81, 83, 85]
b. Karyotype is 45; XO
c. Presence of barr bodies Column - I Column - II

RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *
d. Sterile females with rudimentary ovaries. a. Reginald C.Punnett i. Sex determination
e. Lack of secondary sexual characters. b. Mendel ii. Genotype
Which of the given statements are related to c. Sutton and Boveri iii. probability
Linkage and
Turner’s syndrome? [Old Page. 92] recombination
1) a, b, c and d 2) Only a, d and e d. T.H. Morgan iv. Laws of inheritance
3) All except c 4) a, b, c, d and e e. Henking v. Parallel behaviour of
Ans.(3) genes and
47. A dihybrid cross was conducted by Morgan chromosomes
between the following parents of fruit fly.
1) a-ii, b-iv, c-iii, d-i, e-v 2) a-iv, b-v, c-i, d-ii, e-iii
3) a-ii, b-iv, c-v, d-iii, e-i 4) a-v, b-iii, c-ii, d-i, e-iv

[Old Page. 84]

Ans.(3)
Section-D: Biology
The progeny possible in F1 generation is. Section-A
51. Chromosomal theory of inheritance was
1) 2) proposed by [Old Page. 81]
1) Watson and Crick
2) Sutton and Boveri
3) 4) 3) Bateson and Punnet
4) T.H. Morgan
Ans.(2)
48. Statement I : Mendel gave postulates such as Ans.(2)
‘principles of segregation and principles of 52. The number of contrasting characters studied by
independent assortment’ after studying seven Mendel for his experiments were
pairs of contrasting traits in garden pea. [Old Page. 70]
Statement II : He was lucky in selecting seven 1) 7 2) 14
characters in pea that were located on seven
3) 4 4) 2
different chromosomes. [Old Page. 75]
1) Both statement I and statement II are true. Ans.(1)
2) Both statement I and statement II are false 53. The production of gametes by the parents,
3) Statement I is true but statement II is false formaton of zygotes, the F1 and F2 plants, can be
understood from a diagram called.
4) Statement I is false but statement II is true
[Old Page. 73]
Ans.(1)
49. Assertion : In human, the gamete contributed by 1) Bullet square
the male determines whether the child produced 2) Punch square
will be male or female. 3) Punnett square
Reason : Sex in human in a polygenic trait 4) Net square
depending upon a cumulative effect of some genes
on X-chromosome and some on Y-chromosome. Ans.(3)
[Old Page. 86] 54. Distance between the genes and percentage of
1) If both assertion and reason are true and the recombination shows. [Old Page. 83]
reason in the correct explanation of the assertion 1) A direct relationship
2) If both assertion and reason are true, but reason 2) An inverse relationship
is not the correct explanation of the assertion.
3) A parallel relationship
3) If assertion is true, but reason is false.
4) No relationship
4) If both assertion and reason are false
Ans.(2)
Ans.(3)

Prof. Motegaonkar S. R. M.Sc. Che. Gold Medalist, SET/NET--JRF, GATE, BARC,TIFR Qualified 5
55. Which one of the following symbols represent 61. Morgan utilised Drosophila as an experimental
mating between relatives in human pedigree material for his investigation as.
analysis? [Old Page. 88] A. Female flies are distinguishable from male
flies due to their smally size

RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *
B. It can complete its life cycle in 2 weeks.
1) 2)
C. Single mating produce larger number of
progenies.
3) 4) Choose correct option. [Old Page. 83]
1) Only (A)
Ans.(1)
2) Only (A) and (B)
56. Which of the following is an exception to law of
segregation? [Old Page. 75]
3) Only (B) and (C)
1) Disjunction of chromosomes 4) all (A), (B) and (C)
2) Non-disjunction of non-homologous chromosomes Ans.(3)
3) Incomplete dominance 62. If a colour-blind female marries a man whose
mother was also colour blind, what are the
4) Non-disjunction of homologous chromosomes chances of her progeny having colour blindness?
Ans.(4)
[Old Page. 89]
57. Statement A : The possibility of a female becoming 1) 75 % 2) 100 %
a haemophilic is extremely rare.
3) 25 % 4) 50 %
Statement B : Mother of such female has to be at
least carried and father should be haemophilic Ans.(2)
which is unviable in the later stage of life. 63. Law of segregation and law of independent
[Old Page. 90]
assortment can be explained by _____ movement
of chromosome. [Old Page. 75, 80]
1) Only A is correct
1) Metaphase
2) Only B statement is correct
2) Anaphase I
3) Both A and B statement are correct
3) Anaphase II
4) Both A and B statement are incorrect
4) Crossing over
Ans.(3)
Ans.(2)
58. The family pedigree of queen Victoria shows a
number of haemophilic descendants as she was. 64. Match column I with column II and choose the
correct combination from the options given.
[Old Page. 89]
[Old Page. 77, 75, 88, 83]
1) Haemophilic
2) Carrier of disease haemophilia Column - I Column - II
3) Married to the king affected with haemophilia a ABO blood group i Exception of law of
4) Married to king who was carrier of haemophilia independent
Ans.(2) assortment
59. A man whose mother was colourbind marries a b Law of segregation ii Base pair substitution
normal woman who has colourblind father, then c Linkage iii Multiple allelism
the percentage of their daughter that would be
colourblind is [Old Page. 89] d Mutation iv Monohybrid cross
1) 25 % 2) 75 % 1) a-ii, b-iv, c-iii, d-i 2) a-iii, b-iv, c-i, d-ii
3) 50 % 4) Zero 3) a-iii, b-ii, c-iv, d-i 4) a-iv, b-iii, c-ii, d-i
Ans.(3) Ans.(2)
60. In a cross betwee a male and female, both 65. A husband and wife have normal vision but
heterozygous for sickle cell anaemia gene, what fathers of both of them were colourblind. The
percentage of the progeny will be diseased? probability of their first son to be colourblind
[Old Page. 90] is. [Old Page. 89]
1) 50 % 2) 75 % 1) 25 % 2) 50 %
3) 25 % 4) 100 % 3) 75 % 4) 0 %
Ans.(3) Ans.(1)

Prof. Motegaonkar S. R. M.Sc. Che. Gold Medalist, SET/NET--JRF, GATE, BARC,TIFR Qualified 6
66. From the given pedigree for inheritance of 70. Choose the odd statement w.r.t the given figure.
colourblindness, predict the genotype of [Old Page. 92]
individuals ; I – 2, II – 1 and III – 2.
[Old Page. 89]

RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *
1) X°X, X°X, XX
2) X°X, X°X, X°X
3) XX, XX, XX
4) XX, X°X, XX
Ans.(2)

67. Consider the following statements and select the

correct option. [Old Page. 90, 91] 1) It has overall masculine development.
A. Heterozygous female for haemophilia may 2) Presence of two Barr bodies.
transmit the disease to sons. 3) Such individuls are sexually sterile.
B. A recessive X-linked trait influences malse 4) It is caused due to trisomy of autosomes.
more than females. Ans.(4)
C. Thalassemia is an autosomal recessive blood 71. The probability of the individuals being
disease. homozygous for first character and heterozygous
1) Only B is correct for second character in a dihybrid cross of
2) Only (A) and (C) are correct Mendel is [Old Page. 79]
3) Only (B) and (C) are correct 1) 1 / 8 2) 1 / 2
4) All (A), (B) and (C) are correct 3) 1 / 4 4) 3 / 16
Ans.(4) Ans.(3)
72. Select the correct statement.
68. The modified allele is equivalent to unmodified [Old Page. 83, 73]
allele when it. [Old Page. 77] 1) Franklin Stahl coined the term ‘linkage’.
1) Produce normal functional enzyme 2) Punnett square was developed by a British
2) Produce non-functional enzyme scientist
3) Produce no enzyme at all 3) Spliceosomes take part in translation
4) All except (1) 4) Transduction was discovered by S. Altman
Ans.(1) Ans.(2)
73. Which of the following pairs is wrongly
69. What is the genetic disorder in which an matched? [Old Page. 78, 83, 85]
individual has an overall masculine 1) Starch synthesis in pea : Multiple alleles
development, gynaecomastia, and is sterile? 2) ABO blood grouping : Co-dominance
[Old Page. 92] 3) XO type sex determination : Grasshopper
1) Klinefelter’s syndrome 4) T.H. Morgan : Linkage
2) Edward syndrome Ans.(1)
3) Down’s syndrome 74. Which of the following occurs due to the
4) Turner’s syndrome presence of autosomal-linked dominant trait?
Ans.(1) [Old Page. 89]
1) Haemophilia 2) Thalassemia
3) Sickle cell anaemia 4) Myotonic dystrophy
Ans.(4)

Prof. Motegaonkar S. R. M.Sc. Che. Gold Medalist, SET/NET--JRF, GATE, BARC,TIFR Qualified 7
75. Match column I with column II. 80. XO type of sex determination can be found in
[Old Page. 73, 79] [Old Page. 86]
Column - I Column - II (Ratio) 1) Grasshoppers
(Experiment) 2) Monkeys

RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *
3) Drosophila
a. Typical monohybrid i. 9 : 3 : 3 : 1
4) Birds
phenotypic ratio
Ans.(1)
b. Typical dihybrid ii. 3 : 1
81. Statement I : Colour blindness occurs in about 8
phenotypic ratio
per cent of male and only about 0.4 percent of
c. Typical monohybrid iii. 1 : 2 : 1
females. [Old Page. 89]
genotypic ratio
Statement II : The genes that lead to red-green
d. Typical dihybrid iv. 1 : 1 : 1 : 1 colour blindness are on the X-chromosome.
test cross ratio 1) Both statement I and statement II are true.
1) a-iii, b-i, c-ii, d-iv 2) a-iii, b-iv, c-i, d-ii 2) Both statement I and statement II are false
3) a-ii, b-i, c-iii, d-iv 4) a-ii, b-iv, c-iii, d-i 3) Statement I is true but statement II is false
Ans.(3) 4) Statement I is false but statement II is true
76. Frequency of recombination between gene pairs
Ans.(1)
on same chromosome as a measure of the
distance between genes to map their positions 82. Assertion : In a snapdragon, F1 plants do not have
on chromosome was used for the first time by. red or white flowers.
1) Sutton and Boveri Reason : It is intermediate inheritance with neither
2) Alfred Sturtevent of the two alleles of a gene being dominant over
3) Henking each other. [Old Page. 76]
4) Thomas Hunt Morgan 1) If both assertion and reason are true and the
Ans.(2) reason in the correct explanation of the assertion
77. In drosophila, Morgan found that the genes for 2) If both assertion and reason are true, but reason
eye color and wing ‘s shaped are linked and is not the correct explanation of the assertion.
show recombinationn frequency equal to. 3) If assertion is true, but reason is false.
[Old Page. 84] 4) If both assertion and reason are false
1) 1. 3% 2) 37.2% Ans.(3)
3) 98.7% 4) 62.8% 83. Down’s syndrome results in the gain of extra
Ans.(2) copy of [Old Page. 92]
78. Mendel’s law of segregation states that 1) Chromosome 11 2) Chromosome 12
[Old Page. 75] 3) Chromosome 21 4) Chromosome 6
1) The two factors for the same trait separate in
Ans.(3)
the production of gametes
2) The two different traits will be inherited 84. Gynaecomastia is observed in which genetic
independently of each other disorder. [Old Page. 92]
3) The gametes are produced by meiosis 1) Turner’s syndrome
4) All of these 2) Klinefelter’s syndrome
Ans.(1) 3) Down’s syndrome
79. Assertion : Crossing over between the genes is 4) Jacob’s syndrome
directly proportional to distance. Ans.(2)
Reason : Greater distance between the genes
85. Statement I : The physical association of genes on
reduces the possibility of linkage.
a chromosome is termed as linkage.
[Old Page. 83]
Statement II : T.H. Morgan studied the genes that
1) If both assertion and reason are true and the
ware sex linked. [Old Page. 83]
reason in the correct explanation of the assertion
1) Both statement I and statement II are true.
2) If both assertion and reason are true, but reason
is not the correct explanation of the assertion. 2) Both statement I and statement II are false
3) If assertion is true, but reason is false. 3) Statement I is true but statement II is false
4) If both assertion and reason are false 4) Statement I is false but statement II is true
Ans.(1)
Ans.(1)
Prof. Motegaonkar S. R. M.Sc. Che. Gold Medalist, SET/NET--JRF, GATE, BARC,TIFR Qualified 8
 Section-B 91. Statement I : In birds, the chromosome
composition of the egg determines the sex.
86. If two dihybrid plants (AaBb) are crossed, then
find out the correct combinations of probability Statement II : Female birds are heterogametic.
of following offsprings. [Old Page. 79] [Old Page. 86]

RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *
1) Both statement I and statement II are true.
AaBb aaBb AaBB aabb
1) 8/16 2/16 3/16 1/16
2) Both statement I and statement II are false
2) 4/16 2/16 2/16 1/16 3) Statement I is true but statement II is false
3) 3/16 4/16 1/16 2/16 4) Statement I is false but statement II is true
4) 4/16 1/16 2/16 2/16 Ans.(1)
Ans.(2) 92. State true (T) or false (F) for the following
statements and choose the correct option.
87. Find out the genotype of father and mother in
the given pedigree chart for autosomal dominant A. The characters never blend in heterozygous
disorder. [Old Page. 88] condition.
B. Change in a single base pair of DNA does not
cause mutation
C. Cancer cells commonly shows chromosomal
aberrations.
Father Mother D. Cause of variations are hidden in sexual
1) AA Aa reproduction only. [Old Page. 75]
2) Aa aa 1) A-T, B-F, C-T, D-F 2) A-F, B-T, C-T, D-F
3) aa Aa 3) A-T, B-F, C-F, D-T 4) A-T, B-F, C-F, D-F
4) Aa Aa Ans.(1)
Ans.(2) 93. A gene responsible for disorder
88. Assertion : In a monohybrid cross, the F 2 phenylketonuria is [Old Page. 91]
generation indicates dominant characters. a. A pleiotropic gene
Reason : Dominance occurs only in the b. An autosomal dominant trait
heterozygous state. [Old Page. 71]
c. A chromosome related disorder
1) If both assertion and reason are true and the
d. Lack enzyme which converts tyrosine into
reason in the correct explanation of the assertion
phenylalanine
2) If both assertion and reason are true, but reason
1) a and b 2) a and c
is not the correct explanation of the assertion.
3) b and d 4) c and d
3) If assertion is true, but reason is false.
Ans.(2)
4) If both assertion and reason are false
94. Match column I with column II.
Ans.(3)
[Old Page. 89, 91, 86, 87]
89. The recombination frequency between the genes
a and c is 5% b and c is 15%, b and d is 9%, a and Column - I Column - II
b is 20%, c and d is 24% and a and d is 29%.
a. Mendelian disorder i. Drosophila
What will be the sequence of these genes on a
linear chromosome? [Old Page. 84] b. Chromosomal ii. Grasshopper
1) a, b, c, d 2) a, c, b, d c. disorder
XO type sex iii. Birds
3) a, d, b, c 4) d, b, a, c determination
Ans.(2) d. XY type sex iv. Klinefelter's
90. Which of the following chromosomal determination syndrome
constitution refer to Turner’s syndrome in e. ZW type sex v. Cystic fibrosis
human? [Old Page. 92] determination
1) 44 + XO 1) a-iv, b-v, c-ii, d-i, e-iii
2) 44 + XXY 2) a-iv, b-v, c-iii, d-i, e-ii
3) 44 + XYY
3) a-v, b-iv, c-ii, d-i, e-iii
4) 45 + XYY
4) a-v, b-iv, c-i, d-ii, e-iii
Ans.(1) Ans.(3)

Prof. Motegaonkar S. R. M.Sc. Che. Gold Medalist, SET/NET--JRF, GATE, BARC,TIFR Qualified 9
95. Assertion : Morgan’s cross was conducted in 99. Assertion (A) : Mendel’s law of independent
Drosophila to locate genes on chromosome for assortment does not hold good for the genes that
white eye colour. [Old Page. 83] are located closely on the same chromosome.
Reasion : The cross was done between red-eyed Reason (R) : Closely located genes assort
hybrid female and white-eyed male. independently. In the light of the above statements,

RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *RCC * *
1) If both assertion and reason are true and the choose the correct answer from the options given
reason in the correct explanation of the below. [Old Page. 81]
assertion 1) (A) is correct, but (R) is not correct
2) If both assertion and reason are true, but reason 2) (A) is not correct, but (R) is correct
is not the correct explanation of the assertion. 3) Both (A) and (R) are correct, and (R) is the
3) If assertion is true, but reason is false. correct explanation of (A)
4) If both assertion and reason are false 4) Both (A) and (R) are correct, but (R) is not the
Ans.(3) correct explanation of (A)
96. Statement I : To determine the genotype of a tall Ans.(1)
plant at F1 generation, Mendel crossed that plant 100. Match column I with column II and choose the
with a homozygous tall plant. correct combination from the options given.
Statement II : In test cross, an organism showing [Old Page. 90 to 92]
a dominant phenotype is crossed with the
Column - I Column - II
recessive parents. [Old Page. 74]
a Down syndrome i X-linked
1) Both statement I and statement II are true.
recessive
2) Both statement I and statement II are false
b Phenylketonuria ii Trisomy of
3) Statement I is true but statement II is false allosome
4) Statement I is false but statement II is true c Haemophilia iii Autosomal
Ans.(4) recessive
97. Choose the correct combination of true (T) and d Klinefelter syndrome iv disorder
Additional 21
st

false (F) for the following statements chromosome

[Old Page. 83]
1) a-iv, b-ii, c-iii, d-i
A. The strength of linkage is determined by the
2) a-iii, b-iv, c-i, d-ii
distance between the two linked genes.
3) a-ii, b-i, c-iv, d-iii
B. Strength of linkage in inversely proportional
4) a-iv, b-iii, c-i, d-ii
to the distance between two genes
C. The two genes are said to be linked where Ans.(4)
they fail to show independent assortment.
1) A - T, B - T, C - F
2) A - T, B - T, C - T
3) A - F, B - T, C - T
4) A - T, B - F, C - T
Ans.(2)
98. What map unit (Centimorgan) is adopted in the
construction of genetic maps?
[Old Page. 83]
1) A unit of distance between two expressed
genes, representing 10% cross over
2) A unit of distance between genes on
chromosomes representing 1% cross over
3) A unit of distance between genes on
chromosomes, representing 50% cross over
4) A unit of distance between two expressed
genes, representing 100% cross over
Ans.(2)

Prof. Motegaonkar S. R. M.Sc. Che. Gold Medalist, SET/NET--JRF, GATE, BARC,TIFR Qualified 10