Class 12 Biology Topic Wise Line by Line Questions Chapter 4 Principles of Inheritance and Variation

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Mendel’s Laws of Inheritance
1. Genetics is the subject that deals with
A) inheritance B) variation of characteristics
C) reproduction D) both (a) and (b)
2. The basis of heredity is
A) variation B) inheritance
C) mutation D) linkage
3. Humans knew from as early as 8000–1000 BC that one of the causes of variation was
hidden in
A) sexual reproduction B) asexual reproduction
C) vegetative propagation D) none of these
4. Choose the incorrect statement from the following.
A) Humans knew from very early that sexual reproduction is one of the causes of variation.
B) They exploited the variation to obtain plants and animals of desirable characters through
selective breeding.
C) Sahiwal cows were obtained through artificial selection and domestication from ancestral
wild cows.
D) Our ancestors were very well aware about the scientific basis of inheritance of characters and
variation.
5. Which one from the following is the period for Mendel’s hybridization experiments?
A) 1840–1850 B) 1857–1869 C) 1870–1877 D) 1856–1863
6. Who proposed the ‘Laws of Inheritance’ in living organisms?
A) Mendel B) Morgan C) de Vries D) Correns
7. Match Column-I with Column-II and choose the correct answer from the codes given below.
Column-I Column-II
(A) Genetics (1) Process of passing characters from parent to offspring
(B) Inheritance (2) Laws of inheritance
(C) Variation (3) A branch of Biology
(D) Mendel (4) Degree of difference of progeny from their parents
Codes-
A B C D
A) 1 4 2 3
B) 4 2 3 1
C) 3 1 4 2
D) 2 3 1 4
8. Mendel investigated characters in the garden pea plant that were manifested as two
A) linked traits B) opposing traits
C) similar traits D) none of these
9. How many pairs of contrasting characters in pea plants were studied by Mendel in his
experiments?
A) Six B) Eight C) Seven D) Four
10. Which contrasting trait was not studied by Mendel during his experiments?
A) Seed colour B) Leaf colour C) Flower colour D) Stem height
11. Among the following, which one is not a dominating trait?
A) Axial position of flower B) Green colour of pod
C) Violet colour of flower D) Green colour of seed
NCERT LINE BY LINE QUESTIONS

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12. A true-breeding line is one that
A) has undergone continuous self pollination
B) shows stable trait inheritance
C) shows expressions of trait for several generations
D) all of these
13. Match Column-I with Column-II and choose the correct option from the codes given below.
Column-I Column-II
(A) Axial flower (1) Undergone continuous self pollination
(B) Terminal flower (2) Father of genetics
(C) Mendel (3) Dominant trait
(D) True-breeding line (4) Recessive trait
Codes-
A B C D
A) 3 4 2 1
B) 4 3 1 2
C) 1 2 4 3
D) 2 1 3 4
14. Refer to the given figures (A–D) showing traits of pea plant studied by Mendel.Among these,
choose the dominant trait.
A) B B) A C) D D) C
15. Which technique was used by Mendel during his experiments on pea plant?
A) Artificial pollination B) Cross pollination
C) Self-pollination D) All of these
16. Choose the correct statement(s) from the following.
(I) During Mendel’s investigation, statistical analysis and mathematical logic were applied to
problems in Biology.
(II) Mendel investigated characters in the garden pea plant that were manifested as two
opposing traits.
(III) Mendel conducted artificial pollination experiments using several true-breeding pea lines.
(IV) Mendel selected eight true-breeding pea plant varieties as pairs.
A) I and II B) III and IV C) I, II and III D) All of these
17. The contrasting trait(s) selected by Mendel was/were
A) smooth or wrinkled seed
B) yellow or green seed
C) smooth or inflated pods
D) all of these
18. Assertion: Mendel conducted hybridization experiments on garden pea plant.
Reason: He proposed laws of inheritance in living organisms.
A) Both assertion and reason are true and reason is the correct explanation of assertion.
B) Both assertion and reason are true but reason is not correct explanation of assertion.
C) Assertion is true, but reason is false.
D) Both assertion and reason are false.
19. Assertion: Mendel used contrasting traits for his studies.

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Reason: He used Ocimum plant for his experiments.
20. Assertion: Mendel used true-breeding pea lines for his experiments.
Reason: A true-breeding line is one that has undergone continuous selfpollination.
Inheritance of one Gene
21. The first hybrid generation of Mendel’s experiment is known as
A) Filial1 progeny
B) F1-generation
C) Father generation
D) Both (A) and (B)
22. When Mendel crossed true-breeding tall and dwarf plants, in F1-generation all tall plants were
obtained. On self-crossing in the F2 generation, he obtained
A) 1/4th dwarf and 3/4th tall plants
B) 3/4th dwarf and 1/4th tall plants
C) 2/4th dwarf and 2/4th tall plants
D) All dwarf plants
23. During the study of inheritance of one character in F2 generation, Mendel obtained phenotype
in
A) 2 : 1 ratio B) 3 : 1 ratio C) 1 : 2 : 1 ratio D) 1 : 1 : 1 : 1 ratio
24. The ‘factors’ of Mendel are today known as
A) genome B) gene C) DNA D) allele
25. The slightly different forms of the same genes are called
A) genome B) DNA C) allele D) cistron
26. Alleles are
A) true-breeding homozygotes
B) different molecular forms of a gene
C) heterozygotes
D) different phenotype
27. What would be the phenotype of a plant that had a genotype ‘Tt’? Here ‘T’ represent tall trait
while ‘t’ represents dwarf trait.
A) Tall B) Intermediate height C) Dwarf D) None of these
28. In homozygous condition, a particular gene has
A) different alleles on homologous chromosomes.
B) no alleles on homologous chromosomes.
C) same alleles on homologous chromosomes.
D) none of these
29. Tall and dwarf are the two alleles of gene of height. The dominant trait is
A) dwarf B) tall
C) both are equally dominant D) both are recessive
Column-I Column-II
(A) Genes (1) Slightly different forms of the same gene
(B) Alleles (2) Genetic composition of an organism

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(C) Genotype (3) Physical appearance of an organism
(D) Phenotype (4) Unit of inheritance
Codes–
A B C D
A) 4 1 2 3
B) 1 4 3 2
C) 3 2 4 1
D) 2 3 1 4
31. A cross that is performed for the study of a single character is
A) dihybrid cross B) test cross
C) monohybrid cross D) back cross
32. The given figure is the diagrammatic representation of a monohybrid cross. In the figure, some
plants are mentioned as A and B. What will be the genotype of these plants?
A) A – tt, B – Tt B) A – Tt, B – tt C) A – TT, B – TT D) A – Tt, B – Tt
33. Choose the incorrect statement about Mendel’s monohybrid cross.
A) The recessive parental trait is expressed without any blending in F2 generation.
B) The alleles of parental pair segregate from each other and both alleles are transmitted to a
gamete.
C) The segregation of alleles is a random process.
D) There is a 50% chance of a gamete containing either allele.
34. The production of gametes by the parents the formation of zygotes, the F1 and F2 plants, can be
understood by using
A) Wenn diagram B) Pie diagram C) A pyramid diagram D) Punnett square
35. Select the correct statement.
A) Franklin Stahl coined the term ‘linkage’.
B) Punnett square was developed by a British scientist.
C) Spliceosomes take part in translation.
D) Transduction was discovered by SAltman.
36. In the test cross, organism whose genotype is to be determined, is crossed with the
A) recessive parent B) dominant parent
C) both parents one by one D) none of these
37. On crossing two tall plants, in F1-generation few dwarf offspring were obtained. What would
be the genotype of the both the parent?
A) TT and Tt B) Tt and Tt C) TT and TT D) TT and tt

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38. Based on his observations of monohybrid cross, Mendel proposed which law of inheritance?
A) Law of dominance B) Law of segregation
C) Law of independent assortment D) Both (A) and (B)
39. According to Mendel, characters are controlled by discrete units called
A) genes B) factors C) alleles D) allelomorph
40. Choose the incorrect statement about law of dominance.
A) It is used to explain the expression of only one of the parental characters in a monohybrid
cross in F1-generation.
B) It does not explain the expression of both parental characters in F2-generation.
C) It also explains the proportion of 3: 1 obtained in F2-generation.
D) It states that characters are controlled by discrete units called factors.
Column-I Column-II
(A) First law of inheritance (1) Law of segregation
(B) Second law of inheritance (2) 3: 1
(C) Monohybrid cross (3) Law of dominance
(D) Test cross (4) 1: 1
Codes-
A B C D
A) 3 1 2 4
B) 1 3 4 2
C) 2 3 1 4
D) 4 2 3 1
42. The second law of inheritance, i.e., law of segregation is based on the fact that
A) alleles do not show any blending.
B) both characters are recovered as such in F2 generation.
C) one allele dominates the other allele.
D) Both (A) and (B)
43. The factor controlling any character is discrete and independent. It was concluded on the basis
of
A) results of F3-generation of a cross.
B) observations of a cross made between the plants having two contrasting traits where
offspring shows only one trait without any blending.
C) self-pollination of F1-offspring.
D) cross pollination of parental generations.
44. In Antirrhinum (Snapdragon), a red flower was crossed with a white flower and in F1
generation, pink flowers were obtained. When pink flowers were selfed, the F2 generation
showed white, red and pink flowers. Choose the incorrect statement from the following
A) The experiment does not follow the principle of dominance.
B) Pink colour in F1 is due to incomplete dominance.
C) Ratio of F2 is ¼ (Red): 2/4 (Pink): ¼ (white).
D) Law of segregation does not apply in this experiment.
45. It was being observed that sometimes, the F1 shows a phenotype that does not resemble either
of the two parents and remains in between the two. It can be explained by
C) Law of incomplete dominance D) None of these
46. The genotypic ratio obtained in incomplete dominance is
A) 3 : 1 B) 1 : 1 : 2 C) 2 : 1 : 1 D) 1 : 2 : 1
47. In case of co-dominance, the F1 progeny

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A) resembles either of the two parents B) is in between of parents
C) resembles both the parents D) none of these
48. A person of AB blood group has IA and IB genes. It is an example of
A) pleiotropy B) segregation
C) co-dominance D) None of these
49. In a marriage between male with blood group A and female with blood group B, the progeny
had either blood group AB or B. What could be the possible genotype of parents?
A) IAi (Male); IBi (Female)
B) IAi (Male); IBIB (Female)
C) IAIA (Male); IBIB (Female)
D) IAIA (Male); IBi (Female)
50. A person has ‘O’ blood group. His mother has ‘A’ while father has ‘B’ blood group. What
would be the genotype of mother and father?
A) Mother is homozygous for ‘A’ blood group and father is heterozygous for ‘B’ blood group.
B) Mother is heterozygous for ‘A’ blood group and father is homozygous for ‘B’ blood group.
C) Both mother and father are homozygous for ‘A’ and ‘B’ blood groups respectively.
D) Both mother and father are heterozygous for ‘A’ and ‘B’ blood groups respectively.
51. Which of the following characteristics represent ‘inheritance of blood groups’ in humans?
(I) Dominance
(II) Co-dominance
(III) Multiple dominance
(IV) Incomplete dominance
(V) Polygenic inheritance
A) II, III and V B) I, II and III C) II, IV and V D) I, III and V
52. A man with blood group ‘A’ marries a woman with blood ‘B’. What are all possible blood
groups of their offsprings?
A) A, B and AB only B) A, B, AB and O
C) O only D) A and B only
53. The genotypes of a husband and wife are IAIB and IAi. Among the blood types of their
children, how many different genotypes and phenotypes are possible?
A) 3 genotypes: 4 phenotypes
B) 4 genotypes: 3 phenotypes
C) 4 genotypes: 4 phenotypes
D) 3 genotypes: 3 phenotypes
54. Multi alleles are present
A) at different loci on the same chromosome
B) at the same locus of the chromosome
C) on non-sister chromatids
D) on different chromosome
Column-I Column-II
(A) Dominance (1) ABO blood group
(B) Codominance (2) Appearance of pink flowers in snapdragon in F1 generation
(C) Incomplete dominance (3) Starch synthesis in pea seeds
(D) Pleiotropy (4) Appearance of violet flowers in F1 generation in garden pea
Codes-
A B C D
A) 4 1 2 3
B) 1 4 3 2
C) 3 2 4 1

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D) 2 3 1 4
56. ABO blood grouping is a good example of
A) incomplete dominance
B) mutation
C) multiple alleles
D) pleiotropy
57. Sometimes a single gene product may produce more than one effect. This phenomenon is
known as
A) mosaicism B) pleiotropy C) multiple allelism D) polygeny
58. Starch synthesis in pea seeds is an example of
A) multiple allelism B) incomplete dominance
C) co-dominance D) pleiotropy
59. Pea seeds having Bb genotype produce starch grains of
A) large size B) small size
C) intermediate size D) they do not produce starch.
60. Choose the incorrect statement from the following about pleiotropy.
A) In pleiotropy, a single gene produces more than one effect.
B) Starch synthesis in pea seeds is controlled by one gene.
C) Pea seeds having BB genotypes, produce small starch grains.
D) bb homozygotes of pea produce wrinkled seeds.
61. Assertion: The law of dominance is used to explain the expression of only one of the parental
characters in a monohybrid cross.
Reason: It also explains the proportion of 3: 1 obtained at F2 generation.
62. Assertion: The pink flower of dog plant show incomplete dominance.
Reason: In pink flowers, both alleles are expressed equally.
63. Assertion: A person having IAIB genotype has AB blood group.
Reason: IA and IB alleles are co-dominant
Inheritance of two Genes
64. Crosses that are performed to study two contrasting characters at a time are called
A) monohybrid cross
B) dihybrid cross
C) test cross
D) back cross
65. The phenotypic ratio obtained by Mendel in his dihybrid cross was
A) 1 : 2 : 1 : 2 B) 3 : 2 : 2 : 1 C) 9 : 3 : 3 : 1 D) 2 : 3 : 1 : 2
66. The third law of inheritance proposed by Mendel is
A) Law of dominance
B) Law of independent assortment

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C) Law of incomplete dominance
D) Law of segregation
67. The ratio 9: 3: 3: 1 of a dihybrid cross denotes that
A) it is a multigenic inheritance.
B) the alleles of two genes are interacting with each other.
C) it is a case of multiple allelism.
D) the alleles of two genes are segregating independently.
68. The numbers of phenotypes and genotypes in F2 generation of a Mendelian dihybrid
cross are
A) phenotypes 4: genotypes 16
B) phenotypes 4: genotypes 8
C) phenotypes 9: genotypes 4
D) phenotypes 4: genotypes 9
69. Mendel’s law of independent assortment is true for the genes situated on the
A) same chromosome
B) non-homologous chromosomes
C) homologous chromosomes
D) extra nuclear genetic element
70. Genes A and B are linked. The F1 heterozygote of a dihybrid cross involving these genes is
crossed with homozygous recessive parental type (aabb). What would be the ratio of offspring
in the next generation?
A) 1: 1 B) 1: 1: 1: 1 C) 9: 3: 3: 1 D) 3: 1
71. Mendel’s work remained unrecognized for many years. Find out the true reason for the same.
(I) Mendel’s concept of genes was not accepted by his contemporaries as an explanation for the
continuous variation seen in nature.
(II) The approach of using mathematics was new and unacceptable by other biologists.
(III) He could not provide any physical proof for the existence of factors.
(IV) Communication was not easy in those days and his work could not be widely published.
A) I and II B) II and III C) III and IV D) All of these
72. Mendel’s results on the inheritance of characters were rediscovered by:
A) de Vries B) Correns C) von Tschermak D) all of these
73. Among the following, who noted that the behaviour of chromosomes was parallel to the
behavior of genes?
A) Walter Sutton B) Theodore Boveri C) Von Tschermak D) Both (A) and (B)
74. Refer to the given figure showing meiosis and germ cell formation in a cell with four
chromosomes. Which law of Mendel can be effectively explained by this figure?
C) Law of independent assortment D) All of these
75. The chromosomal theory of inheritance was proposed by
A) Sutton B) Boveri C) Morgan D) Both (A) and (B)

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Column-I Column-II
(A) Mendel (1) Rediscovery of Mendel’s law
(B) Correns, Tschermak and Vries (2) Worked on Drosophila melanogaster
(C) Sutton and Boveri (3) Law of independent assortment
(D) T. H. Morgan (4) Chromosomal theory of inheritance
Codes-
A B C D
A) 3 1 4 2
B) 1 4 3 2
C) 2 3 1 4
D) 4 2 3 1
77. Morgan performed his experiments on
A) Garden pea B) Drosophila C) Snapdragon D) None of these
78. When two genes are located on the same chromosome, the proportion of parental gene
combination is
A) higher than non-parental
B) lower than non-parental
C) equal to non-parental
D) None of these
79. Genes which are present on the same chromosome
A) do not form any linkage group.
B) affect the phenotype by forming interactive groups.
C) form a linkage group.
D) form different groups depending upon their relative distance.
80. The term used to describe the generation of nonparental gene combination is
A) linkage B) recombination C) mutation D) none of these
81. Which type of relationship is found between the distance of genes and percentage of
recombination?
A) Inverse B) Parallel C) Direct D) None of these
82. Among the following which will not cause variations among siblings?
A) Linkage B) Independent assortment of genes
C) Crossing over D) Mutation
Column-I Column-II
(A) Linkage (1) Non-parallel gene combination
(B) Recombination (2) Genetic map
(C) Sturtevant (3) Unit of distance between genes
(D) Centimorgan (4) Physical association of genes
Codes-
A B C D
A) 3 2 4 1
B) 2 3 1 4
C) 4 1 2 3
D) 1 4 3 2
84. What map unit (centimorgan) is adopted in the construction of genetic maps?
A) A unit distance between two expressed genes, representing 10% cross over.
B) A unit distance between two expressed genes, representing 100% cross over.
C) A unit distance between genes on chromosomes, representing 1% cross over.
D) A unit distance between genes on chromosomes, representing 50% cross over.
85. The concept of genetic map was given by

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A) de Vries B) Morgan C) Sturtevant D) Mendel
86. Assertion: Mendel proposed the law of independent assortment on the basis of results of
dihybrid cross.
Reason: When two pairs of traits are combined in a hybrid, segregation of one pair of characters
is independent of the other pair of characters.
A) Both assertion and reason are true and the reason is the correct explanation of assertion.
B) Both assertion and reason are true but the reason is not the correct explanation of assertion.
C) Assertion is true but reason is false.
87. Assertion: The chromosomal theory of inheritance was proposed by T. H. Morgan.
Reason: Morgan worked on garden pea plants to give this theory.
88. Assertion: Morgan coined the term linkage to describe the physical association of genes on a
chromosome.
Reason: Linkage shows more non-parental type combination of genes.
Sex Determination
89. X-body was discovered by
A) Mendel B) Morgan C) Henking D) de Vries
90. In XO type of sex determination, who does possess the X chromosome?
A) Female B) Male
C) Sometimes female and sometimes male D) None of these
91. X-chromosome is designated as
A) autosome B) sex chromosome
C) somatic chromosome D) none of these
92. Which type of sex determination is found in grasshoppers?
A) XX – XY type B) XX – XO type
C) ZZ – ZW type D) None of these
93. In a specific taxon of insects, some possess 17 chromosomes while others have 18 chromosomes.
These 17 and 18 chromosomes bearing organisms are [NCERT Exemplar]
A) All males B) All females
C) Females and males, respectively D) Males and females, respectively
94. In Drosophila, males possess
A) XO chromosomes B) XX chromosomes
C) XY chromosomes D) YY chromosomes
Column-I Column-II
(A) X-body (1) Autosomes
(B) X and Y chromosome (2) Henking
(C) Somatic chromosome (3) Grasshopper
(D) XO-types of sex determination (4) Allosomes
Codes-
A B C D
(a) 2 4 1 3

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(b) 4 2 1 3
(c) 3 1 4 2
(d) 1 3 2 4
96. XY type of sex determination is found in
A) Drosophila B) humans C) grasshopper D) both (A) and (B)
97. Choose the incorrect statement about XY type of sex determination.
A) Both males and females have same number of chromosomes.
B) The counter part of X chromosome is distinctly smaller and called Y chromosome.
C) Males and females possess different number of autosomes.
D) This type of sex determination is found in Drosophila.
98. Male heterogamety is found in
A) grasshopper B) Drosophila C) humans D) all of these
99. In female heterogamety, females
A) one type of gametes B) two types of gametes
C) three types of gametes D) none of these
100. ZZ/ZW type of sex determination is the characteristics feature of [NCERT Exemplar]
A) platypus B) snails C) peacock D) cockroach
101. Among the following, which has a different mechanism of sex determination?
A) Birds B) Humans C) Drosophila D) None of these
102. Refer to the given figure which is followed by few statements. Choose the incorrect statement
about it.
A) It shows male heterogamety.
B) Both possess same types of autosomes.
C) The sex of progeny is determined by females.
D) This type of sex determination is different from humans.
103. In humans, sex is determined by
A) females B) males C) environmental factors D) none of these
Column-I Column-II
(A) XO-type (1) Drosophila
(B) XY-type (2) Grasshopper
(C) ZZ-ZW type (3) Birds
(4) Humans progeny from their parents
Codes-
A B C
(a) 1,4 2 3
(b) 2 1,4 3
(c) 3,2 1 4
(d) 4 3 2,1
105. Match the items of Column I with Column II.
Column-I Column-II
(A) XX-XO method of sex determination (1) Turner’s syndrome
(B) XX-XY method of sex determination (2) Female heterogametic
(C) Karyotype-45 (3) Grasshopper

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(D) ZW-ZZ method of sex determination (4) Female homogametic
Codes-
A B C D
A) 4 2 1 3
B) 2 4 1 3
C) 1 4 2 3
D) 3 4 1 2
106. Select the incorrect statement.
A) Male fruit fly is heterogametic.
B) In male grasshoppers, 50% of sperms have no sex chromosome.
C) In domesticated fowls, sex of progeny depends on the type of sperm rather than egg.
D) Human males have one of their sex chromosome much shorter than the other.
107. Assertion: Grasshoppers show male heterogamety.
Reason: Male grasshoppers produce two types of gametes.
108. Assertion: In fruitfly, sex of progeny is decided by females.
Reason: Females produce two types of gametes.
109. Assertion: Birds show female heterogamety.
Reason: In birds, the sex of progeny is determined by males.
Mutation and Genetic Disorders
110. The phenomenon which results in alteration of DNA sequences is
A) mutation B) transpiration C) transcription D) translation
111. Chromosomal aberrations are commonly observed in
A) cardiac cells B) cancer cells C) skeletal cells D) none of these
112. A classical example of point mutation is
A) gout B) night blindness C) sickle cell anaemia D) Turner’s syndrome
113. The factors that cause mutations are called
A) mutagens B) teratogens C) allergens D) none of these
114. An analysis of traits in several of generations of a family is called
A) mutation B) pedigree analysisC) genetic map formation D) none of these
115. In a pedigree analysis, the given symbol represents
A) affected individuals B) mating
C) consanguineous mating D) unspecified sex

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116. Pedigree analysis is used to study the inheritance pattern of a gene over generations. The
character that is studied in the pedigree analysis is equivalent to [NCERT Exemplar]
A) Mendelian trait B) Maternal trait
C) Polygamic trait D) Quantitative trait
117. Mendelian disorders are mainly determined by alteration or mutation in the
A) chromosomes B) single gene
C) array of genes D) none of these
118. Among the following which one is a Mendelian disorder?
A) Haemophilia B) Sickle cell anaemia
C) Cystic fibrosis D) All of these
119. Choose the incorrect statement about Mendelian disorders.
A) These are usually caused by mutation in a single gene.
B) These disorders are transmitted to the offspring according to the laws of inheritance.
C) Mendelian disorders are always sex linked.
D) The trait in question can be dominant or recessive.
120. A genetic disease transmitted from a carrier female that is phenotypically normal to only some
male progeny is [NCERT Exemplar]
A) sex-linked dominant B) sex-linked recessive
C) autosomal dominant D) autosomal recessive
121. Refer to the given pedigree analysis. It is related to the analysis of
A) autosomal dominant trait B) autosomal recessive trait
C) sex-linked dominant trait D) sex-linked recessive trait
122. Haemophilia is a/an
A) sex-linked recessive disease B) sex-linked dominant disease
C) autosomal recessive disease D) autosomal dominant disease
123. The possibility of a female becoming a haemophilic is
A) extremely high B) extremely rare
C) equal to a male D) none of these
124. Haemophilia A and B are due to deficiencies of respectively clotting factor
A) VIII and IX B) IX and VIII C) VII and IX D) X and VII
125. Sickle cell anaemia is a/an
A) sex-linked recessive disease B) sex-linked dominant disease
C) autosomal recessive disease D) autosomal dominant disease
126. In sickle cell anaemia, valine replaces glutamic acid. This valine is coded by the triplet [NCERT
Exemplar]
A) AAG B) GGG C) GUG D) GAA
127. Sickle Cell Anaemia (SCA) is transferred from parents to offspring when
A) father is affected and mother is normal.

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B) father is normal and mother is carrier.
C) father is normal and mother is affected.
D) both mother and father are carrier.
E)
Column-I Column-II
(A) Myotonic dystrophy (1) Autosomal recessive
(B) Sickle cell anaemia (2) Sex-linked recessive
(C) Haemophilia (3) Sex-linked dominant
(D) Rett syndrome (4) Autosomal dominant
Codes-
A B C D
(a) 4 1 2 3
(b) 4 2 3 1
(c) 3 4 1 2
(d) 2 3 4 1
129. Thalassemia and sickle cell anaemia are caused due to a problem in globin molecule synthesis.
Select the correct statement.
A) Both are due to a quantitative defect in globin chain synthesis.
B) Thalassemia is due to less synthesis of globin molecules.
C) Sickle cell anaemia is due to quantitative problem of globin molecules.
D) Both are due to qualitative defect in globin chain synthesis.
130. The person suffering from phenylketonuria disease lacks enzyme
A) phenylalanine hydroxylase B) phosphates
C) enolase D) none of these
131. Phenylketonuria is an inborn error in which affected individual lacks an enzyme that converts
A) phenylalanine into tyrosine B) tyrosine into phenylalanine
C) glutamic acid into valine D) valine into glutamic acid
132. Phenylketonuria is a/an
A) autosomal dominant trait B) autosomal recessive trait
C) sex-linked dominant trait D) sex-linked recessive trait
133. If a colourblind man marries a women who is homozygous for normal colour vision, the
probability of their son being colour blind is
A) 0.75 B) 1 C) 0 D) 0.5 25.
134. The chromosomal disorders are
A) absence of one or more chromosomes
B) excess of one or more chromosomes
C) abnormal arrangement of chromosomes
D) all of these
135. Condition of having 2n ± 1 or 2n ± 2 chromosomes is called [NCERT Exemplar]
A) polyploidy B) aneuploidy C) allopolyploidy D) monosomy
136. An increase in a whole set of chromosomes in an organism is called
A) aneuploidy B) linkage C) polyploidy D) none of these
137. Condition (2n + 1) of chromosomes is known as
A) trisomy B) monosomy C) polyploidy D) haploidy
Column-I Column-II
(A) Deletion (1) Loss of a gene or a segment of chromosome
(B) Duplication (2) A segment of chromosome is turned around 180° within a
chromosome

16
(C) Inversion (3) Presence of a gene or segment of chromosome more than once
(D) Translocation (4) Exchange of segments between two homologous chromosomes
Codes-
A B C D
A) 1 3 2 4
B) 4 2 3 1
C) 3 1 4 2
D) 2 4 1 3
Column-I Column-II
(A) Aneuploidy (1) An increase in whole set of chromosomes
(B) Polyploidy (2) 2n + 1
(C) Trisomy (3) Gain or loss of a chromosome
(D) Monosomy (4) 2n − 1
Codes-
A B C D
A) 1 3 4 2
B) 3 1 2 4
C) 4 2 3 1
D) 2 4 1 3
140. A disease caused by an autosomal primary nondisjunction is
A) Klinefelter’s syndrome B) Turner’s syndrome
C) Sickle cell anaemia D) Down’s syndrome
141. Refer to the given figure. It is showing the characteristic features of
A) Down’s syndrome B) Turner’s syndrome
C) Klinefelter’s syndrome D) None of these
142. The disease caused by the trisomy of chromosome number 21 is
A) Turner’s syndrome
B) Haemophilia
C) Klinefelter’s syndrome
D) Down’s syndrome
143. An abnormal human baby with ‘XXX’ sex chromosomes was born due to
A) formation of abnormal ova in the mother.
B) fusion of two ova and one sperm.
C) fusion of two sperms and one ovum.
D) formation of abnormal sperms in the father.
144. What is the genetic disorder in which an individual has an overall masculine development,
gynaecomastia and is sterile?
A) Turner’s syndrome
B) Klinefelter’s syndrome

17
C) Edward’s syndrome
D) Down’s syndrome
145. In which genetic condition, each cell in the affected person, has three sex chromosomes XXY?
A) Turner’s syndrome B) Thalassemia
C) Kleinfelter’s syndrome D) Phenylketonuria
146. A disorder caused due to the absence of one of the X chromosomes is
A) Turner’s syndrome B) Down’s syndrome
C) Klinefelter’s syndrome D) Edward’s syndrome
147. Assertion: The possibility of a female becoming a haemophilic is extremely rare.
Reason: For being haemophilic, the mother of such a female has to be at least carrier and the
father should be haemophilic.
148. Assertion: Aneuploidy is the gain or loss of chromosomes.
Reason: It is caused due to the failure of cytokinesis after telophase stage of cell division.
149. Assertion: Klinefelter’s syndrome is caused due to the presence of an additional copy
of X-chromosome. Reason: Such individuals are sterile.
1. Select the correct statement : [2018]
(a) Franklin Stahl coined the term “linkage”.
(b) Punnett square was developed by a British scientist.
(c) Transduction was discovered by S. Altman.
(d) Spliceosomes take part in translation.
2. Which of the following pairs is wrongly matched? [2018]
(a) Starch synthesis in pea : Multiple alleles
(b) ABO blood grouping : Co-dominance
(c) T.H. Morgan : Linkage
(d) XO type sex determination : Grasshopper
3. Select the correct match. [2018]
(a) Ribozyme - Nucleic acid (b) F2 × Recessive parent - Dihybrid cross
(c) G. Mendel – Transformation (d) T.H. Morgan - Transduction
4. Which of the following characteristics represent ‘Inheritance of blood groups’ in humans? [2018]
A. Dominance B. Co-dominance
C. Multiple allele D. Incomplete dominance
E. Polygenic inheritance
(a) B, C and E (b) A, B and C (c) A, C and E (d) B, D and E
5. A woman has an X-linked condition on one of her X chromosomes. This chromosome can be inherited
by : [2018]
(a) Only daughters (b) Only sons (c) Both sons and daughters (d) Only grandchildren
NEET PREVIOUS YEARS QUESTIONS

18
6. A disease caused by an autosomal primary non-disjunction is [2017]
(a) Klinefelter's syndrome (b) Turner's syndrome
(c) Sickle cell Anaemia (d) Down's syndrome
7. Thalassemia and sickle cell anemia are caused due to a problem in globin molecule synthesis. Select the
correct statement. [2017]
(a) Both are due to a quantitative defect in globin chain synthesis.
(b) Thalassemia is due to less synthesis of globin molecules.
(c) Sickle cell anemia is due to a quantitative problem of globin molecules.
(d) Both are due to a qualitative defect in globin chain synthesis.
8. Which one from those given below is the period for Mendel's hybridisation experiments? [2017]
(a) 1840 - 1850 (b) 1857 - 1869 (c) 1870 - 1877 (d) 1856 - 1863
9. Among the following characters, which one was not considered by Mendel in his experiments on pea?
[2017]
(a) Trichomes – Glandular or non-glandular (b) Seed – Green or Yellow
(c) Pod – Inflated or Constricted (d) Stem – Tall or Dwarf
10. The genotypes of a husband and wife are IAIB and IAi. Among the blood types of their children, how
many different genotypes and phenotypes are possible? [2017]
(a) 3 genotypes ; 4 phenotypes (b) 4 genotypes ; 3 phenotypes
(c) 4 genotypes ; 4 phenotypes (d) 3 genotypes ; 3 phenotypes
11. Pick out the correct statements. [2016]
1. Haemophilia is a sex-linked recessive disease.
2. Down’s syndrome is due to aneuploidy.
3. Phenylketonuria is an autosomal recessive gene disorder.
4. Sickle cell anaemia is a X-linked recessive gene disorder.
(a) 1 and 4 are correct (b) 3 and 4 are correct (c) 1, 3 and 4 are correct (d) 1, 2 and 3 are correct
12. Which of the following most appropriately describes haemophilia? [2016]
(a) Recessive gene disorder (b) X - linked recessive gene disorder
(c) Chromosomal disorder (d) Dominant gene disorder
13. A cell at telophase stage is observed by a student in a plant brought from the field. He tells his teacher
that this cell is not like other cells at telophase stage. There is no formation of cell plate and thus, the
cell is containing more number of chromosomes as compared to other dividing cells. This would result
in:
[2016]
(a) Aneuploidy (b) Polyploidy (c) Somaclonal variation (d) Polyteny
14. In a testcross involving F1 dihybrid flies, more parentaltype offspring were produced than the
recombinant-type offspring. This indicates: [2016]
(a) the two genes are located on two different chromosomes.
(b) chromosomes failed to separate during meiosis.
(c) the two genes are linked and present on the same chromosome.
(d) both of the characters are controlled by more than one gene.
15. Match the terms in Column-I with their description in Column-II and choose the correct option. [2016]

19
(a) A – II, B – I, C – IV, D – III (b) A – II, B – III, C – IV, D – I
(c) A – IV, B – I, C – II, D – III (d) A – IV, B – III, C – I, D – II
16. A tall true breeding garden pea plant is crossed with a dwarf true breeding garden pea plant. When the
F1 plants were selfed, the resulting genotypes were in the ratio of [2016]
(a) 1 : 2 : 1 :: Tall homozygous : Tall heterozygous : Dwarf
(b) 1 : 2 : 1 :: Tall heterozygous : Tall homozygous : Dwarf
(c) 3 : 1 : : Tall : Dwarf (d) 3 : 1 : : Dwarf : Tall
17. A pleiotropic gene: [2015]
(a) is a gene evolved during Pliocene.
(b) controls a trait only in combination with another gene.
(c) controls multiple traits in an individual.
(d) is expressed only in primitive plants.
18. An abnormal human baby with ‘XXX’ sex chromosomes was born due to [2015]
(a) formation of abnormal ova in the mother. (b) fusion of two ova and one sperm.
(c) fusion of two sperms and one ovum. (d) formation of abnormal sperms in the father.
19. A colour blind man marries a woman with normal sight who has no history of colour blindness in her
family. What is the probability of their grandson being colour blind? [2015]
(a) 1 (b) 0 (c) 0.25 (d) 0.5
20. In the following human pedigree, the filled symbols represent the affected individuals. Identify the type
of given pedigree. [2015]
(a) X- linked recessive (b) Autosomal recessive(c) X-linked dominant (d) Autosomal dominant
21. The term ‘linkage’ was coined by : [2015]
(a) T. Boveri (b) G. Mendel (c) W. Sutton (d) T.H. Morgan
22. A gene showing co-dominance has [2015]
(a) alleles tightly linked on the same chromosome.
(b) alleles that are recessive to each other.
(c) both alleles independently expressed in the heterozygote.
(d) one allele dominant on the other.
23. Multiple alleles are present [2015]
(a) at different loci on the same chromosome.(b) at the same locus of the chromosome.
(c) on non-sister chromatids. (d) on different chromosomes.
24. Alleles are [2015]
(a) true breeding homozygotes. (b) different molecular forms of a gene.
(c) heterozygotes. (d) different phenotype.
25. Which is the most common mechanism of genetic variation in the population of sexually reproducing
organism? [2015]
(a) Chromosomal aberrations (b) Genetic drift (c) Recombination (d) Transduction
26. How many pairs of contrasting characters in pea plants were studied by Mendel in his experiments?
[2015]
(a) Six (b) Eight (c) Seven (d) Five
27. In his classic experiments on pea plants, Mendel did not use [2015]
(a) Pod length (b) Seed shape (c) Flower position (d) Seed colour
28. Person with blood group AB is considered as universal recipient because he has [2014]
(a) both A and B antigens on RBC but no antibodies in the plasma.
(b) both A and B antibodies in the plasma.
(c) no antigen on RBC and no antibody in the plasma.
(d) both A and B antigens in the plasma but no antibodies.
29. A human female with Turner’s syndrome [2014]
(a) has 45 chromosomes with XO (b) has one additional X chromosome
(c) exhibits male characters (d) is able to produce children with normal husband
30. A man whose father was colour blind marries a woman who had a colour blind mother and normal
father. What percentage of male children of this couple will be colour blind? [2014]
(a) 25% (b) 0% (c) 50% (d) 75%
31. Fruit colour in squash in an example of [2014]

20
(a) Recessive epistasis (b) Dominant epistasis
(c) Complementary genes (d) Inhibitory genes
32. What map unit (Centimorgan) is adopted in the construction of genetic maps ? [NEET-2019]
(1) A unit of distance between two expressed genes, representing 10% cross over
(2) A unit of distance between two expressed genes, representing 100% cross over
(3) A unit of distance between genes on chromosomes, representing 1% cross over
(4) A unit of distance between genes on chromosomes, representing 50% cross over
33. The frequency of recombination between gene pairs on the same chromosome as a measure of the
distance between genes was explained by : [NEET-2019]
(1) T.H. Morgan (2) Gregor J. Mendel (3) Alfred Sturtevant (4) Sutton Boveri
34. In Antirrhinum (Snapdragon), a red flower was crossed with a white flower and in F1 generation,
pink flowers were obtained. When pink flowers were selfed, the F2 generation showed white, red
and pink flowers. Choose the incorrect statement from the following : [NEET-2019]
(1) This experiment does not follow the Principle of Dominance
(2) Pink colour in F1 is due to incomplete dominance.
(3) Ratio of F2 is 1/4 (Red) : 2/4 (Pink) :1/4 (White)
(4) Law of Segregation does not apply in this experiment.
35. What is the genetic disorder in which an individual has an overall masculine development,
gynaecomastia, and is sterile ? [NEET-2019]
(1) Turner's syndrome (2) Klinefelter's syndrome
(3) Edward syndrome (4) Down's syndrome
36. Select the incorrect statement. [NEET-2019]
(1) Male fruit fly is heterogametic.
(2) In male grasshoppers, 50% of sperms have no sex-chromosome.
(3) In domesticated fowls sex of progeny depends on the type of sperm rather than egg.
(4) Human males have one of their sex-chromosome much shorter than the other.
37. The production of gametes by the parents, the formation of zygotes, the F1 and F2 plants, can be
understood using :- [NEET-2019 ODISSA]
(1) Pie diagram (2) A pyramid diagram (3) Punnet square (4) Wenn diagram
38. Match the items of column I with column II [NEET-2019 ODISSA]
Column I Column II
(a) XX-XO method of sex determination (i) Turner's syndrome
(b) XX-XY method of sex determination (ii) Female heterogametic
(c) Karyotype-45 (iii) Grasshopper
(d) ZW-ZZ method of sex determination (iv) Female homogametic
Select the correct option from the following :
(1) a-ii, b-iv, c-i, d-iii (2) a-i, b-iv, c-ii, d-iii (3) a-iii, b-iv, c-i, d-ii (4) a-iv, b-ii, c-i, d-iii
39. In which genetic condition, each cell in the affected person has three sex chromosomes XXY ?
[NEET-2019 ODISSA]
(1) Thalassemia (2) Kleinfelter's Syndrome
(3) Phenylketonuria (4) Turner's Syndrome
40. How many true breeding pea plant variation did Mendel select as pairs, which were similar except in
one character with contrasting traits? [NEET-2020]
1) 8 2) 4 3) 2 4) 14
41. Experimental verification of the chromosomal theory of inheritance was done by [NEET-2020]
1) Morgan 2) Mendel 3) Sutton 4) Boveri
42. Select the correct match [NEET-2020]
(1) Thalassemia - X linked
(2) Haemophilia - Y linked
(3) Phenylketonuria - Autosomal dominant triat
(4) Sickle cell anaemia - Autosomal recessive triat, chromosome-11
43. The best example for pleiotropy is :- [NEET-2020 COVID]
(1) Skin colour (2) Phenylketoneuria (3) Colour Blindness (4) ABO Blood group

21
44. Chromosomal theory of inheritance was proposed by : [NEET-2020 COVID]
(1) Sutton and Boveri (2) Bateson and Punnet (3) T. H. Morgan (4) Watson and Crick
45. The number of contrasting characters studied by Mendel for his experiments was:
[NEET-2020 COVID]
(1) 14 (2) 4 (3) 2 (4) 7
46. Mutations in plant cells can be induced by: [NEET-2021]
(1) Infrared rays (2) Gamma rays (3) Zeatin (4) Kinetin
47. The production of gametes by the parents formation of zygotes, the 1
F and 2
F plants, can be understood
form a diagram called [NEET-2021]
1)Punch square 2) Punnett square 3) Net square 4) Bullet square
48. In a cross between a male and female, both heterozygous for sickle cell anaemia gene, what percentage
of the progeny will be diseased? [NEET-2021]
1) 75% 2) 25% 3) 100% 4) 50%
49. The recombination frequency between the genes a & c is 5%, b & c is 15%, b & d is 9%, a & b
is 20%, c & d is 24% and a & d is 29%. What will be the sequence of these genes on a linear
chromosome?
1) a, d, b, c 2) d, b, a, c 3) a, b, c, d 4) a, c, b, d
50. If a colour blind female marries a man whose mother was also a colour blind, what are the
chances of her progeny having colour blindness?
1) 25% 2) 50% 3) 75% 4) 100%
NCERT LINE BY LINE QUESTIONS – ANSWERS
1) D 2) B 3) A 4) D 5) D 6) A 7) C 8) B 9) C 10) B
11) D 12) D 13) A 14) A 15) D 16) C 17) D 18) B 19) C 20) B
21) D 22) A 23) B 24) B 25) C 26) B 27) A 28) C 29) B 30) A
31) C 32) A 33) B 34) D 35) B 36) A 37) B 38) D 39) B 40) B
41) A 42) D 43) B 44) D 45) C 46) D 47) C 48) C 49) B 50) D
51) B 52) B 53) B 54) B 55) A 56) C 57) B 58) D 59) C 60) C
61) B 62) C 63) A 64) B 65) C 66) B 67) D 68) D 69) C 70) B
71) D 72) D 73) D 74) D 75) D 76) A 77) B 78) A 79) C 80) B
81) C 82) A 83) C 84) C 85) C 86) B 87) D 88) C 89) C 90) A
91) B 92) B 93) D 94) C 95) A 96) D 97) C 98) D 99) B 100) C
101) A 102) A 103) B 104) B 105) D 106) C 107) A 108) D 109) C 110) A
111) B 112) C 113) A 114) B 115) C 116) A 117) B 118) D 119) C 120) D
121) A 122) A 123) B 124) A 125) C 126) C 127) D 128) A 129) B 130) A
131) A 132) B 133) C 134) D 135) B 136) C 137) A 138) A 139) B 140) D
141) A 142) D 143) A 144) B 145) C 146) A 147) A 148) C 149) B
NEET PREVIOUS YEARS QUESTIONS-ANSWERS
1 (b) 2 (a) 3 (a) 4 (b) 5 (c) 6 (d) 7 (b) 8 (d) 9 (a) 10 (b)
11 (d) 12 (b) 13 (b) 14 (c) 15 (b) 16 (a) 17 (c) 18 (a) 19 (c) 20 (b)
21 (d) 22 (c) 23 (b) 24 (b) 25 (c) 26 (c) 27 (a) 28 (a) 29 (a) 30 (c)
31 (b) 32 (3) 33 (3) 34 (4) 35 (2) 36 (3) 37 (3) 38 (3) 39 (2) 40 (4)
41 (1) 42 (4) 43 (2) 44 (1) 45 (1) 46 (2) 47 (2) 48 (2)

22
NEET PREVIOUS YEARS QUESTIONS-EXPLANATIONS
1. (b) Punnett (British scientists) devised the "Punnett Square" to depict the number and variety of genetic
combinations, and had a role in shaping the Hardy- Weinberg law. Franklin Stahl proved semi-
conservative mode of replication. Transduction was discovered by Zinder and Lederberg. Spliceosome
formation is part of post-transcriptional change in eukaryotes.
2. (a) Starch synthesis in pea is controlled by pleiotropic gene. Pleiotropy occurs when one gene influences
two or more seemingly unrelated phenotypic traits.
3. (a) Ribozyme is a catalytic RNA, which is nucleic acid.
4. (b) IA
IO
, IB
IO
- Dominant-recessive relationship
IAIB - Codominance
IA
, IB
& IO
- Three different allelic forms of a gene (multiple allelism)
5. (c) Woman acts as a carrier. Both son & daughter inherit Xchromosome.
Although only son would be the diseased one.
6. (d) Down’s syndrome is caused by non-disjunction of 21st
chromosome i.e. trisomy.
7. (b) Thalassemia is a quantitative problem of synthesising very few globin molecules while sickle cell
anaemia is a qualitative problem of synthesising an incorrectly functioning globin.
8. (d) Mendel conducted hybridization experiments for 7 years on pea plant between 1856 to 1863 and his
data was published in 1865.
9. (a) During his experiments, Mendel had taken seven characters in a pea plant. In which, nature of
trichomes i.e., glandular or non-glandular was not considered by Mendel.
10. (b) Husband × Wife
Number of genotypes = 4
Number of phenotypes = 3
IA
IA
and IA
i = A
IA
IB
= AB
IB
i = B
11. (d) Sickle cell disease is inherited in an autosomal recessive pattern.
12. (b)
13. (b) This phenomenon is known as polyploidy, wherein the cells contain more than two paired
(homologous) sets of chromosomes. Polyploidy is often seen in the case of plants. The major cause of
polyploidy is the non - disjunction of sister chromatids during meiotic recombination. This condition is
actually useful in development of new crop varieties.
14. (c) When two genes in a dihybrid cross are situated on the same chromosome, the proportion of parental
gene combinations are much higher than the non-parental or recombinant type. This is also called as
incomplete linkage.
15. (b)

23
16. (a)
Phenotypic ratio - 3: 1 (Tall :Dwarf)
Genotypic ratio - 1:2:1 (Homozygous Tall : Heterozygous Tall : Dwarf).
17. (c)
18. (a) A human baby having abnormality with ‘XXX’ sex chromosomes is born due to evolution of
abnormal ova in mother’s ovary. This is caused due to non-disjunction of X chromosome in the mother.
19. (c) XX × XcY
Normal women Colourblind man
The daughters of this couple will have normal eye sight and are carrier, if one of the carrier daughter
marries with normal eyed man.
Only 25% grandson will show colour-blindness.
20. (b) Autosomal recessive is a type of disorder in which two copies of an abnormal gene must be found
for the disease in the affected person.
21. (d) Thomas Hunt Morgan won the Nobel Prize (1933) in physiology or medicine for the function of
chromosomes in heredity.
22. (c) In co-dominance, both alleles are independently expressed in the heterozygote.
23. (b) All alleles of a gene are situated on the same loci of chromosome in organisms.
24. (b) Alleles are defined as alternative form of a same gene.
25. (c) The most common cause of variations is recombination in organisms which reproduce sexually.
26. (c) Seven pairs of contrasting characters were selected in pea plant and studied by Mendel in his
experiment.
27. (a) Mendel did not use pod length for his experiment.

24
28. (a)
29. (a) Turner's syndrome is a chromosomal condition that affects development in females. A human female
with Turner's syndrome has 45 chromosomes with XO. The most common feature of Turner's syndrome
is short stature, which becomes evident by about age 5.
30. (c) Colour blindness is a X-chromosome linked character.
Colour blind male = 50%.
31. (b) Epistasis is the phenomenon of suppression of phenotypic expression of gene by a non-allelic gene
which shows its own effect. A dominant epistatic allele suppresses the expression of a non-allelic gene
whether the latter is dominant or recessive. For example, fruit colour of Summer Squash (Cucurbita
pepo) is governed by a gene which produces yellow colour in dominant state (YY) and green colour in
recessive state (yy).
40. Mendel selected 14 true breeding pea plants
41. Morgan done experimental verification of the chromosomal theory of inheritance
42. Phenylketonuria – Autosomal recessive disorder
Thalassemia – Autosomal recessive disorder
Haemophilia – X linked recessive disorder
Sickle cell anaemia – Autosomal recessive trait associated with chromosome number 11
44. • Several kinds of radiation like gamma rays, Xrays,UV-rays cause mutation.
• These are physical mutagens.
• Such induced mutation in plants is done to develop improved varieties. The first natural cytokinin was
isolated from unripe maize grain known as zeatin. The cytokinin that was obtained from degraded
product of autoclaved herring sperm DNA was kinetin (N6-furfuryl aminopurine). Infrared rays cause
heating effect.
47. Punnett square
48. According to given question;
Total number of affected progenies = 1
* Percentage of diseased/affected progenies=1/4 *100 = 25%
49

25
Recombination frequency is equal to the distance between the genes.
Distance between a & c is 5%, b & c is 15%, b & d is 9%, a &b is 20%, c & d is 24% and a & d is 29%.
Then the sequence of these genes on a linear chromosome will be a, c, b, d
50 Affected parents in x inked recessive inheritance give all progeny affected

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