NEET AIPMT Biology Chapter Wise Solutions – Principles of Inheritance and Variation
1. A gene showing codominance has
(a) alleles that are recessive to each other
(b) both alleles independently expressed in the heterozygote
(c) one allele dominant on the other
(d) alleles tightly linked on the same chromosome. (AIPMT 2015)
2. In his classic experiments on pea plants, Mendel did not use
(a) seed shape
(b) flower position
(c) seed colour
(d) pod length. (AIPMT 2015)
3. 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 ?
(d) 1 (AIPMT 2015)
4. A pleiotropic gene
(a) controls a trait only in combination with another gene
(b) controls multiple traits in an individual
(c) is expressed only in primitive plants
(d) is a gene evolved during Pliocene. (AIPMT 2015)
5. In the following human pedigree, the filled symbols represent the affected individuals. Identify the type of given pedigree.
(a) Autosomal recessive
(b) X-linked dominant
(c) Autosomal dominant
(d) X-linked recessive (AIPMT 2015)
6. The term “linkage” was coined by
(a) G Mendel
(b) W. Sutton
(c) T.H. Morgan
(d) T. Boveri. (AIPMT 2015)
7. Alleles are
(a) different molecular forms of a gene
(c) different phenotype
(d) true breeding homozygotes. (AIPMT 2015, Cancelled)
8. Multiple alleles are present
(a) at the same locus of the chromosome
(b) on non-sister chromatids
(c) on different chromosomes
(d) at different loci on the same chromosome. (AIPMT 2015, Cancelled)
9. A man with blood group ‘A’ marries a woman with blood group ‘B’. What are all the possible blood groups of their offsprings?
(a) A, B, AB and O
(b) O only
(c) A and B only
(d) A, B and AB only (AIPMT 2015, Cancelled)
10. How many pairs of contrasting characters in pea plants were studied by Mendel in his experiments?
(d) Six (AIPMT 2015, Cancelled)
11. An abnormal human baby’with ‘XXX’ sex chromosomes was bom due to
(a) fusion of two ova and one sperm
(b) fusion of two sperms and one ovum
(c) formation of abnormal sperms in the father
(d) formation of abnormal ova in the mother. (AIPMT 2015, Cancelled)
12. Fmit colour in squash is an example of
(a) recessive epistasis
(b) dominant epistasis
(c) complementary genes
(d) inhibitory genes. (AIPMT 2014)
13. 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?
(d) 75% (AIPMT 2014)
14. A human female with Turner’s syndrome
(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. . (AIPMT 2014)
15. Select the incorrect statement with regard to haemophilia.
(a) It is a dominant disease.
(b) A single protein involved in the clotting of blood is affected.
(c) It is a sex-linked disease.
(d) It is a recessive disease. (NEET 2013)
16. Which of the following cannot be detected in a developing foetus by amniocentesis?
(a) Down’s syndrome
(c) Klinefelter’s syndrome
(d) Sex of the foetus (NEET 2013)
17. If both parents are carriers for thalassaemia, which is an autosomal recessive disorder, what are the chances of pregnancy resulting in an affected child?
(c) No chance
(d) 50% (NEET 2013)
18. If two persons with ‘AB’ blood group marry and have sufficiently large number of children, these children could be classified as ‘A’ blood group: ‘AB’ blood group : ‘B’ blood group in 1 : 2 : 1 ratio. Modern technique of protein electrophoresis reveals presence of both ‘A’ and ‘B’ type proteins in ‘AB’ blood group individuals. This in an example of
(a) partial dominance
(b) complete dominance
(d) incomplete dominance. (NEET 2013)
19. Which idea is depicted by a cross in which the F1 generation resembles both the parents?
(a) Inheritance of one gene
(c) Incomplete dominance
(d) Complete dominance (NEET 2013)
20. Which one is the incorrect statement with regard to the importance of pedigree analysis?
(a) It confirms that DNA is the carrier of genetic information.
(b) It helps to understand whether the trait in question is dominant or recessive.
(c) It confirms that the trait is linked to one of the autosome.
(d) It helps to trace the inheritance of a specific trait. (Karnataka NEET 2013)
21. Down’s syndrome in humans is due to
(a) three ‘X’ chromosomes
(b) three copies of chromosome 21
(d) two ‘Y’ chromosomes. (Karnataka NEET 2013)
22. A normal-visioned man whose father was colour¬blind, marries a woman whose father was also colour-blind. They have their first child as a daughter. What are the chances that this child would be colour-blind?
(b) zero percent
(d) 50%. (Prelims 2012)
23. F2 generation in a Mendelian cross showed that both genotypic and phenotypic ratios are same as 1 : 2 : 1. It represents a case of
(b) dihybrid cross
(c) monohybrid cross with complete dominance
(d) monohybrid cross with incomplete dominance. (Prelims 2012)
24. Represented below is the inheritance pattern of a certain type of trait in humans. Which one of the following conditions could be an example of this pattern?
(b) Sickle cell anaemia
(d) Thalassemia. (Main 2012)
25. Read the following four statements (A – D).
(A) In transcription, adenosine pairs with uracil.
(B) Regulation of lac operon by repressor is referred to as positive regulation.
(C) The human genome has approximately 50,000 genes.
(D) Haemophilia is a sex-linked recessive disease.
How many of the above statements are right?
(d) One. (Main 2012)
26. A test cross is carried out to
(a) determine the genotype of a plant at F2
(b) predict whether two traits are linked
(c) assess the number of alleles of a gene
(d) determine whether two species or varieties will breed successfully. (Main 2012)
27. When two unrelated individuals or lines are crossed, the performance of F1 hybrid is often superior to both its parents. This phenomenon is called
(d) metamorphosis (Prelims 2011)
28. Which one of the following conditions correctly descibes the manner of determining the sex?
(a) homozygous sex chromosomes (ZZ) determine female sex in birds
(b) XO type of sex chromosomes determine male sex in grasshopper
(c) XO condition in humans as found in Turner’s syndrome, determines female sex
(d) homozygous sex chromosomes (XX) produce male in Drosophila. (Prelims 2011)
29. Which one of the following conditions of the zygotic cell would lead to the birth of a normal human female child?
(a) two X chromosomes
(b) only one Y chromosome
(c) only one X chromosome
(d) one X and one Y chromosome (Main 2011)
30. Test cross in plants or in Drosophila involves crossing
(a) between two genotypes with recessive trait
(b) between two F1 hybrids
(c) the F1 hybrid with a double recessive genotype
(d) between two genotypes with dominant trait. (Main 2011)
31. Which one of the following symbols and its representation, used in human pedigree analysis is correct?
32. ABO blood groups in humans are controlled by the gene I. It has three alleles – and IA ,IB i. Since there are three different alleles, six different genotypes are possible. How many phenotypes can occur?
(d) two.(Prelims 2010)
33. Select the correct statement from the ones given below with respect to dihybrid cross.
(a) tightly linked genes on the same
chromosomes show higher recombinations
(b) genes far apart on the same chromosome show very few recombinations
(c) genes loosely linked on the same
chromosome show similar recombinations
(d) tightly linked genes on the same
chromosome show very few recombinations. (Prelims 2010)
34. The genotype of a plant showing the dominant phenotype can be determined by
(a) test cross
(b) dihybrid cross
(c) pedigree analysis
(d) back cross. (Prelims 2010)
35. Which one of the following cannot be explained on the basis of Mendel’s Law of Dominance?
(a) the discrete unit controlling a particular character is called a factor
(b) out of one pair of factors one is dominant, and the other recessive
(c) alleles do not show any blending and both the characters recover as such in F2 generation.
(d) factors occur in pairs. (Prelims 2010)
36. ABO blood grouping is controlled by gene I which has three alleles and show co-dominance. There are six genotypes. How many phenotypes in all are possible?
(d) five (Main 2010)
37. Study the pedigree chart of a certain family given below and select the correct conclusion which can be drawn for the character
(a) the female parent is heterozygous
(b) the parents could not have had a normal daughter for this character
(c) the trait under study could not be colour blindness
(d) the male parent is homozygous dominant (Main 2010)
38. A cross in which an organism showing a dominant phenotype is crossed with the recessive parent in order to know its genotype is called
(a) monohybrid cross
(b) back cross
(c) test cross
(d) dihybrid cross (Main 2010)
39. In Antirrhinum two plants with pink flowers were hybridized. The F2 plants produced red, pink and white flowers in the proportion of 1 red, 2 pink and 1 white. What could be the genotype of the two plants used for hybridization? Red flower colour is determined by RR, and white by rr genes.
(d) rr (Main 2010)
40. Select the incorrect statement from the following.
(a) galactosemia is an inborn error of metabolism
(b) small population size results in random genetic drift in a population
(c) baldness is a sex-limited trait
(d) linkage is an exception to the principle of independent assortment in heredity (Prelims 2009)
41. Sickle-cell anaemia is
(a) caused by substitution of valine by glutamic acid in the beta globin chain of haemoglobin
(b) caused by a change in a single base pair of DNA
(c) characterized by elongated sickle like RBCs with a nucleus
(d) an autosomal linked dominant trajt (Prelims 2009)
42. The genetic defect-adenosine deaminase (ADA) deficiency may be cured permanently by
(a) administering adenosine deaminase activators
(b) introducing bone marrow cells producing ADA into cells at early embryonic stages
(c) enzyme replacement therapy
(d) periodi infusion of genetically engineered lymphocytes having functional ADA cDNA (Prelims 2009)
43. The most popularly known blood grouping is the ABO grouping. It is named ABO and not ABC, because “O” in it refers to having
(a) overdominance of this type on the genes for A and B types
(b) one antibody only – either anti – A or anti – B on the RBCs
(c) no antigens A and B on RBCs
(d) other antigens besides A and B on RBCs (Prelims 2009)
44. Study the pedigree chart given below. What does it show?
(a) inheritance of a condition like phenylketonuria as an autosomal recessive trait
(b) the pedigree chart is wrong as this is not possible
(c) inheritance of a recessive sex-linked disease like haemophilia
(d) inheritance of a sex-linked inborn error of metabolism like phenylketonuria (Prelims 2009)
45. Which one of the following conditions in humans is correctly matched with its chromosomal abnormality/linkage?
(a) erythroblastosis foetalis – X-linked
(b) Down’s syndrome – 44 autosomes + XO
(c) Klinefelter’s syndrome – 44 autosomes + XXY
(d) colour blindness – Y-linked. (Prelims 2008)
46. A human male produces sperms with the genotypes AB, Ab, aB, and ab pertaining to two diallelic characters in equal proportions. What is the corresponding genotype of this person?
(d) AaBb. (2007)
47. 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?
(a) 9 : 1
(b) 1 : 3
(d) 50:50. (2007)
48. Inheritance of skin colour in humans is an example of
(a) point mutation
(b) polygenic inheritance
(d) chromosomal aberration. (2007)
49. In the hexaploid wheat, the haploid (n) and basic (x) numbers of chromosomes are
(a) n = 21andx = 21
(b) n = 21andx=14
(c) n= 21andx=7
(d) n = 7andx=21. (2007)
50. A common test to find the genotype of a hybrid is by
(a) crossing of one F2 progeny with female parent
(b) studying the sexual behaviour of F1 progenies
(c) crossing of oneF1 progeny with male parent
(d) crossing of one F2 progeny with male parent. (2007)
51. Test cross involves
(a) crossing between two genotypes with dominant trait
(b) crossing between two genotypes with recessive trait
(c) crossing between two F1 hybrids
(d) crossing the F1 hybrid with a double recessive genotype (2006)
52. Both sickle cell anaemia and Huntington’s chorea are
(a) virus-related diseases
(b) bacteria-related diseases
(c) congenital disorders
(d) pollutant-induced disorders (2006)
53. If a colourblind woman marries a normal visioned man, their sons will be
(a) all colourblind
(b) all normal visioned
(c) one-half colourblind and one-half normal
(d) three-fourths colourblind and one-fourth normal (2006)
54. Cri-du-chat syndrome in humans is caused by the
(a) trisomy of 21st chromosome
(b) fertilization of an XX egg by a normal Y-bearing sperm
(c) loss of half of the short arm of chromosome 5
(d) loss of half of the long arm of chromosome 5 (2006)
55. Sickle cell anaemia has not been eliminated from the African population because
(a) it is controlled by dominant genes
(b) it is controlled by recessive genes
(c) it is not a fatal disease
(d) it provides immunity against malaria. (2006)
56. In Mendel’s experiments with garden pea, round seed shape (RR) was dominant over wrinkled seeds (rr), yellow cotyledon (YY) was dominant over green cotyledon (yy). What are the expected phenotypes in the F2 generation of the cross RRYY x rryy?
(a) round seeds with yellow cotyledons, and wrinkled seeds with yellow cotyledons
(b) only round seeds with green cotyledons
(c) only wrinkled seeds with yellow cotyledons
(d) only wrinkled seeds with green cotyledons (2006)
57. How many different kinds of gametes will be produced by a plant having the genotype AABbCC?
(d) nine (2006)
58. Phenotype of an organism is the result of
(a) genotype and environment interactions
(b) mutations and linkages
(c) cytoplasmic effects and nutrition
(d) environmental changes and sexual dimorphism (2006)
59. Which one of the following is an example of polygenic inheritance?
(a) skin colour in humans
(b) flower colour in Mirabilis jalapa
(c) production of male honey bee
(d) pod shape in garden pea (2006)
60. In order to find out the different types of gametes produced by a pea plant having the genotype AaBb it should be crossed to a plant with the genotype
(d) aaBB. (2005)
61. G-6-P dehydrogenase deficiency is associated with haemolysis of
(d) RBCs. (2005)
62. A man and a woman, who do not show any apparent signs of a certain inherited disease, have seven children (2 daughters and 5 sons). Three of the sons suffer from the given disease but none of the daughters affected. Which of the following mode of inheritance do you suggest for this disease?
(a) sex-linked dominant
(b) sex-linked recessive
(c) sex-limited recessive
(d) autosomal dominant. (2005)
63. A woman with 47 chromosomes due to three copies of chromosome 21 is characterized by
(c) Turner’s syndrome
(d) Down’s syndrome. (2005)
64. Haemophilia is more commonly seen in human males than in human females because
(a) a greater proportion of girls die in infancy
(b) this disease is due to a Y-linked recessive mutation
(c) this disease is due to an X-linked recessive mutation
(d) this disease is due to an X-linked dominant mutation. (2005)
65. Which of the following is not a hereditary disease?
(a) cystic fibrosis
(d) cretinism. (2005)
66. A woman with normal vision, but whose father was colour blind, marries a colour blind man. Suppose that the fourth child of this couple was a boy. This boy
(a) may be colourblind or may be of normal vision
(b) must be colour blind
(c) must have normal colour vision
(d) will be partially colour blind since he is heterozygous for the colourblind mutant allele. (2005)
67. In a plant, red fruit (R) is dominant over yellow fruit (r) and tallness (T) is dominant over shortness (t). If a plant with RRTt genotype is crossed with a plant that is rrtt,
(a) 25% will be tall with red fruit
(b) 50% will be tall with red fruit
(c) 75% will be tall with red fruit
(d) all the offspring will be tall with red fruit. (2004)
68. A nutritionally wild type organism, which does not required any additional growth supplement is known as
(d) prototroph. (2004)
69. A male human is heterozygous for autosomal genes A and B and is also hemizygous for hemophilic gene h. What proportion of his sperms will be abh?
(d) 1/4. (2004)
70. Lack of independent assortment of two genes A and B in fruit fly Drosophila is due to
(d) crossing over. (2004)
71. One of the parents of a cross has a mutation in its mitochondria. In that cross, that parent is taken as a male. During segregation of F2 progenies that mutation is found in
(a) one-third of the progenies
(b) none of the progenies
(c) all the progenies
(d) fifty percent of the progenies. (2004)
72. A normal woman, whose father was colour-blind is married to a normal man. The sons would be
(a) 75% colour-blind
(b) 50% colour-blind
(c) all normal
(d) all colour-blind. (2004)
73. The recessive genes located on X-chromosome humans are always
(c) expressed in males
(d) expressed in females. (2004)
74. Which one of the following traits of garden pea studied by Mendel was a recessive feature ?
(a) axial flower position
(b) green seed colour
(c) green pod colour
(d) round seed shape (2003)
75. The genes controlling the seven pea characters studied by Mendel are now known to be located on how many different chromosomes ?
(d) four (2003)
76. Two crosses between the same pair of genotypes or phenotypes in which the sources of the gametes are reversed in one cross, is known as
(a) test cross
(b) reciprocal cross
(c) dihybrid cross
(d) reverse cross (2003)
77. Pattern baldness, moustaches and beard in human males are examples of
(a) sex-linked traits
(b) sex limited traits
(c) sex influenced traits
(d) sex determining traits (2003)
78. Which one of the following conditions though harmful in itself, is also potential saviour from a mosquito borne infectious disease ?
(b) sickle cell anaemia
(c) pernicious anaemia
(d) leukaemia (2003)
79. Down’s syndrome is caused by an extra copy of chromosome number 21. What percentage of offspring produced by an affected mother and a normal father would be affected by this disorder ?
(d) 25% (2003)
80. In Drosophila, the sex is determined by
(a) the ratio of number of X-chromosome to the sets of autosomes
(b) X and Y chromosomes
(c) the ratio of pairs of X-chromosomes to the pairs of autosomes
(d) whether the egg is fertilized or develops parthenogenetically (2003)
81. Which of the following is an example of pleiotropy ?
(c) sickle cell anaemia
(d) colour blindness. (2002)
82. There are three genes a, b, c. Percentage of crossing over between a and b is 20%, b and c is 28% and a and c is 8%. What is the sequence of genes on chromosome
(a) b, a, c
(b) a, b, c
(c) a, c, b
(d) none of these. (2002)
83. On selling a plant of F1-generation with genotype “AABbCC”, the genotypic ratio in F2-generation will be
(c) 9 : 3 : 3 : 1
(d) 27:9:9:9:3:3:3:1. (2002)
84. A gene is said to be dominant if
(a) it expresses it’s effect only in homozygous state
(b) it expresses it’s effect only in heterozygous condition
(c) it expresses it’s effect both in homozygous and heterozygous condition.
(d) it never expresses it’s effect in any condition. (2002)
85. A diseased man marries a normal woman. They get three daughters and five sons. All the daughters were diseased and sons were normal. The gene of this disease is
(a) sex linked dominant
(b) sex linked recessive
(c) sex limited character
(d) autosomal dominant. (2002)
86. Which of the following is a correct match?
(a) Down’s syndrome – 21st chromosome
(b) Sickle cell anaemia – X-chromosome
(c) Haemophilia – Y-chromosome
(d) Parkinson’s disease – X and Y chromosome. (2002)
87. Two nonallelic genes produces the new phenotype when present together but fail to do so independently then it is called
(c) non complementary gene
(d) complementary gene. (2001)
88. A and B genes are linked. What shall be genotype of progeny in a cross between AB/ab and ab/ab
(a) AAbb and aabb
(b) AaBb and aabb
(c) AABB and aabb
(d) none of the above. (2001)
89. When dominant and recessive alleles express itself together it is called
(d) pseudo dominance. (2001)
90. Ratio of complementary genes is
(a) 9 : 3 : 4
(b) 12 : 3 : 1
(c) 9 : 3 : 3 : 4
(d) 9 : 7. (2001)
91. Independent assortment of genes does not takes place when
(a) genes are located on homologous chromosomes
(b) genes are linked and located on same chromosome
(c) genes are located on non-homogenous chromosome
(d) all the above. (2001)
92. Sickle cell anaemia induce to
(a) change of’ amino acid in α-chain of haemoglobin
(b) change of amino acid in β-chain of haemoglobin
(c) change of amino acid in both α and β chains of haemoglobin
(d) change of amino acid either α or βchains of haemoglobin. (2001)
93. Number of Barr bodies in XXXX female is
(d) 4. (2001)
94. Male XX and female XY sometime occur due to
(b) transfer of segments in X and Y chromosome
(d) hormonal imbalance. (2001)
95. Probability of four sons to a couple is
(d) 1/32. (2001)
96. Due to the cross between TTRr x ttrr the resultant progenies show what percent of tall, red flowered plants
(d) 100%. (2000)
97. According to Mendelism, which character shows dominance?
(a) terminal position of flower
(b) green colour in seed coat
(c) wrinkled seeds
(d) green pod colour. (2000)
98. Mongolian Idiocy due to trisomy in 21st chromsome is called
(a) Down’s syndrome
(b) Turner’s syndrome
(c) Klinefelter’s syndrome
(d) Triple X syndrome. (2000)
99. Erythroblastosis foetalis is caused when fertilization takes place between gametes of
(a) Rh– female and Rh+ male
(b) Rh+ female and Rh– male
(c) Rh+ female and Rh+ male
(d) Rh– female and Rh– male. (2000)
100. In Drosophila the XXY condition leads to femaleness whereas in human beings the same condition leads to Klienfelter’s syndrome in male. It proves
(a) in human beings Y chromosme is active in sex determination
(b) Y chromosome is active in sex determination in both human beings and Drosophila .
(c) in Drosophila Y chromosome decides femaleness
(d) Y chromosome of man have genes for syndrome. (2000)
101. A gene pair hides the effect of another gene. The phenomenon is called
(d) mutation. (1999)
102. In hybridization, Tt × tt gives rise to the progeny of ratio
(a) 2 : 1
(d) 1:2. (1999)
103. A marriage between normal visioned man and colour blind woman will produce offspring
(a) colour blind sons and 50% carrier daughter
(b) 50% colourblind sons and 50% carrier daughter
(c) normal males and carrier daughters
(d) colour blind sons and carrier daughters. (1999)
104. The problem, due to Rh– factor arises when the blood of two (Rh+ and Rh–) mix up
(a) during pregnancy
(b) in a test tube
(c) through transfusion
(d) both (a) and (c). (1999)
105. Haemophilic man marries a normal woman. Their offsprings will be
(a) all haemophilic
(b) all boys haemophilic
(c) all girls haemophilic
(d) all normal. (1999)
106. In human beings, multiple genes are involved in the inheritance of
(a) sickle-cell anaemia
(b) skin colour
(d) phenylketonuria. (1999)
107. When a single gene influences more than one trait it is called
(d) none of these. (1998)
108. If Mendel had studied the seven traits using a plant with 12 chromosomes instead of 14, in what way would his interpretation have been different?
(a) he would not have discovered the law of independent assortment
(b) he would have discovered sex linkage
(c) he could have mapped the chromosome
(d) he would have discovered blending or incomplete dominance. (1998)
109. Crossing over in diploid organism is responsible for
(a) segregation of alleles
(b) recombination of linked alleles
(c) dominance of genes
(d) linkage between genes. (1998)
110. How many different types of genetically different gametes will be produced by a heterozygous plant having the genotype AABbCc?
(d) four. (1998)
111. A woman with two genes for haemophilia and one gene for colour blindness on one of the ‘X’ chromosomes marries a normal man. How will the progeny be?
(a) 50% haemophilic colour-blind sons and 50% normal sons
(b) 50% haemophilic daughters (carrier) and 50% colour blind daughters (carrier)
(c) all sons and daughters haemophilic and colour-blind
(d) haemophilic and colour-blind daughters. (1998)
112. Mental retardation in man, associated with sex chromosomal abnormality is usually due to
(a) moderate increase in Y complement
(b) large increase in Y complement
(c) reduction in X complement
(d) increase in X complement. (1998)
113. Albinism is known to be due to an autosomal recessive mutation. The first child of a couple with normal skin pigmentation was an albino. What is the probability that their second child will also be an albino?
(d) 25%. (1998)
114. A fruit fly is heterozygous for sex-linked genes, when mated with normal female fruit fly, the males specific chromosome will enter egg cell in the proportion
(d) 2:1. (1997)
115. A person with the sex chromosomes XXY suffers from
(b) Klinefelter’s syndrome
(c) Down’s syndrome
(d) Turner’s syndrome. (1997)
116. Genetic identity of a human male is determined by
(b) cell organelles
(d) nucleolus. (1997)
117. The polygenic genes show
(a) different karyotypes
(b) different genotypes
(c) different phenotypes
(d) none of these. (1996)
118. When two dominant independently assorting genes react with each other, they are called
(a) collaborative genes
(b) complementary genes
(c) duplicate genes
(d) supplementary genes. (1996)
119. In which of the following diseases, the man has an extra X-chromosome?
(a) Turner’s syndrome
(b) Klinefelter’s syndrome
(c) Down’s syndrome
(d) haemophilia. (1996)
120. A person whose father is colour blind marries a lady whose mother is daughter of a colour blind man. Their children will be
(a) all sons colour blind
(b) some sons normal and some colour blind
(c) all colour blind
(d) all daughters normal. (1996)
121. A genetically diseased father (male) marries with a normal female and gives birth to 3 carrier girls and 5 normal sons. It may be which type of genetic disease?
(a) sex-influenced disease
(b) blood group inheritance disease
(c) sex-linked disease
(d) sex-recessive disease. (1996)
122. When two genetic loci produce identical phenotypes in cis and trans position, they are considered to be
(a) multiple alleles
(b) the parts of same gene
(d) different genes. (1995)
123. The phenomenon, in which an allele of one gene suppresses the activity of an allele of another gene, is known as
(d) inactivation. (1995)
124. The most striking example of point mutation is found in a disease called
(a) Down’s syndrome
(b) sickle cell anaemia
(d) night blindness. (1995)
125. An abnormal human male phenotype involving an extra X-chromosome (XXY) is a case of
(a) Edward’s syndrome
(b) Klinefelter’s syndrome
(d) Down’s syndrome. (1995)
126. The genes, which remain confined to differential
region of Y-chromosome, are
(a) autosomal genes
(b) holandric genes
(c) completely sex-linked genes
(d) mutant genes. (1994)
127. A child’s blood group is ‘O’. The parent’s blood groups cannot be
(a) A and B
(b) A and A
(c) AB and O
(d) B and O. (1994)
128. Albinism is a congenital disorder resulting from the lack of which enzyme?
(b) xanthine oxidase
(d) ffuctokinase. (1994)
129. The colour blindness is more likely to occur in males than in females because
(a) the Y-chromosome of males have the genes for distinguishing colours
(b) genes for characters are located on the sex- chromosomes
(c) the trait is dominant in males and recessive in females
(d) none of the above. (1994)
130. Of both normal parents, the chance of a male child becoming colour blind are
(b) possible only when all the four grand parents had normal vision
(c) possible only when father’s mother was colour blind
(d) possible only when mother’s father was colour blind. (1993)
131. Mr. Kapoor has Bb autosomal gene pair and d allele sex-linked. What shall be proportion of Bd in sperms
(d) 1/8. (1993)
132. Which of the following is suitable for experiment on linkage
(a) aaBB x aaBB
(b) AABB x aabb
(c) AaBb x AaBb
(d) AAbb x AaBB. (1993)
133. Of a normal couple, half the sons are haemophiliac while half the daughters are carriers. The gene is located on
(a) X-chromosome of father
(b) Y-chromosome of father
(c) one X-chromosome of mother
(d) both the X-chromosomes of mother. (1993)
134. Two dominant nonallelic genes are 50 map units apart. The linkage is
(a) cis type
(b) trans type
(d) absent/incomplete. (1993)
135. A polygenic inheritance in human beings is
(a) skin colour
(c) colour blindness
(d) sickle cell anaemia. (1993)
136. Mendel studied inheritance of seven pairs of traits in Pea which can have 21 possible combinations. If you are told that in one of these combinations, independent assortment is not observed in later studies, your reaction will be
(a) independent assortment principle may be wrong
(b) Mendel might not have studied all the combinations
(c) it is impossible
(d) later studies may be wrong. (1993)
137. Sex is determined in human beings
(a) by ovum
(b) at time of fertilization
(c) 40 days after fertilization
(d) seventh to eight week when genitals differentiate in foetus. (1993)
138. A child of O-group has B-group father. The genotype of father will be
(d) IBI°. (1992)
139. An allele is dominant if it is expressed in
(a) both homozygous and heterozygous states
(b) second generation
(c) heterozygous combination
(d) homozygous combination. (1992)
140. In a cross between AABB x aabb, the ratio of F, genotypes between AABB, AaBB, Aabb and aabb would be
(a) 9 : 3 : 3 : 1
(b) 2 : 1 : 1 : 2
(c) 1 : 2:2: 1
(d) 7 : 5 : 3 : 1. (1992)
141. Segregation of Mendelian factors (no linkage, no crossing over) occurs during
(a) anaphase I
(b) anaphase II
(d) metaphase I. (1992)
142. An organism with two identical alleles is
(d) homozygous. (1992)
143. A colour blind mother and normal father would have
(a) colour blind sons and normal/carrier daughters
(b) colour blind sons and daughters
(c) all colour blind
(d) all normal. (1992)
144. Down’s syndrome is due to
(a) crossing over
(c) sex-linked inheritance
(d) nondisjunction of chromosomes. (1992)
145. In human beings 45 chromosomes/ single X/XO abnormality causes
(a) Down’s syndrome
(b) Kinefelter’s syndrome
(c) Turner’s syndrome
(d) Edward’s syndrome. (1992)
146. A man of A-blood group marries a women of AB blood group. Which type of progeny would indicate that man is heterozygous A?
(d) B. (1991)
147. Multiple alleles control inheritance of
(b) colour blindness
(c) sickle cell anaemia
(d) blood groups. (1991)
148. The contrasting pairs of factors in Mendelian crosses are called
(a) multiple alleles
(d) paramorphs. (1991)
149. First geneticist/father of genetics was
(a) De Vries
(d) Morgan. (1991)
150. Mendel’s last law is
(c) independent assortment
(d) polygenic inheritance. (1991)
151. A dihybrid condition is
(c) independent assortment
(d) polygenic inheritance. (1991)
152. Blue eye colour is recessive to brown eye colour. A brown eyed man whose mother was blue eyed marries a blue-eyed women. The children will be
(a) both blue eyed and brown eyed 1 : 1
(b) all brown eyed
(c) all blue eyed
(d) blue eyed and brown eyed 3:1. (1991)
153. The allele which is unable to express its effect in the presence of another is called
(d) recessive. (1991)
154. RR (Red) Antirrhinum is crossed with white (WW) one. Offspring RW are pink. This is an example of
(b) incomplete dominance
(d) supplementary genes. (1991)
155. A colour blind girl is rare because she will be bom only when
(a) her mother and maternal grand father were colour blind
(b) her father and maternal grand father were colour blind
(c) her mother is colour blind and father has normal vision
(d) parents have normal vision but grand parents were colour blind. (1991)
156. Cross between AaBB and aaBB will form
(a) 1 AaBB :1 aaBB
(b) All AaBB
(c) 3AaBB :1 aaBB
(d) lAaBB : 3aaBB. (1990)
157. In a genetic cross having recessive epistasis, F2 phenotypic ratio would be
(a) 9 : 6 : 1
(b) 15 : 1
(c) 9 : 3: 4
(d) 12 : 3 : 1. (1990)
158. ABO blood group system is due to
(a) multifactor inheritance
(b) incomplete dominance
(c) multiple allelism
(d) epistasis. (1990)
159. tt mates with Tt. What will be characteristic of offspring?
(a) 75% recessive
(b) 50% recessive
(c) 25% recessive
(d) All dominant. (1990)
160. In Down’s syndrome of a male child, the sex complement is
(d) XXY (1990)
161. Haemophilia is more common in males because it is a
(a) recessive character carried by Y-chromo- some
(b) dominant character carried by Y-chromo- some
(c) dominant trait carried by X-chromosome
(d) recessive trait carried by X-chromosome. (1990)
162. Which one is a hereditary disease?
(d) phenylketonuria. (1990)
163. Both husband and wife have normal vision though their fathers were colour blind. The probability of their daughter becoming colour blind is
(d) 75%. (1990)
164. Bateson used the terms coupling and repulsion for linkage and crossing over. Name the correct parental of coupling type alongwith its cross over or repulsion
(a) coupling AABB, aabb; Repulsion AABB, a abb
(b) coupling AAbb, aaBB; Repulsion AaBb, aabb
(c) coupling aaBB, aabb; Repulsion AABB, aabb
(d) coupling AABB, aabb : Repulsion AAbb, aaBB. (1990)
165. A normal green male Maize is crossed with albino female. The progeny is albino because
(a) trait for a albinism is dominant
(b) the albinos have biochemical to destroy plastids derived from green male
(c) plastids are inherited from female parent
(d) green plastids of male must have mutated. (1989)
166. Two linked genes a and b show 20% recombination, the individuals of a dihybrid cross between ++/++ × ab/ab shall show gametes
(a) ++ 80 : ab : 20
(b) ++ 50 : ab : 50
(c) ++ 40 : ab 40 : + a 10 : + b : 10
(d) ++ 30 : ab 30 : + a 20 : + b : 20 (1989)
167. Which contribute to the success of Mendel?
(a) qualitative analysis of data
(b) observation of distinct inherited traits
(c) his knowledge of biology
(d) consideration of one character at one time. (1988)
168. A family of five daughter only is expecting sixth issue. The chance of its beings a son is
(d) 100%. (1988)
169. Haploids are able to express both recessive and dominant alleles/mutations because there are
(a) many alleles for each gene
(b) two alleles for each gene
(c) only one allele for each gene in the individual
(d) only one allele in a gene. (1988)
1. (b) :
The phenomenon of expression of both the alleles in a heterozygote is called codominance. The alleles which do not show dominance-recessive relationship and are able to express themselves independently when present together are called codominant alleles. As a result the heterozygous condition has a phenotype different from either of homozygous genotypes, e.g., alleles for blood group A ( IA) and for blood group B ( IB) are codominant so that when they come together in an individual, they produce blood group AB.
2. (d) :
Characters of garden pea plant selected by Mendel were plant height (tall/dwarf), flower position (axial/ terminal), pod colour (green/ yellow), pod shape (inflated/constricted), flower colour (violet/white), seed shape (smooth/wrinkled) and seed colour (yellow/ green).
3. (b) :
When a colour blind man (XCY) marries a normal woman (XX), all of their daughters are carriers and all of their sons are normal, as shown in following figure:
when the Carrier daughter Normal son When the carrier daughter (XXC) is married to a normal man, the probability of their son being colour blind is 0.25, as shown in following figure:
daughter son daughter blind son From above crosses, it is clear that the probability of occurrence of colour blindness in the grandson of a colour blind man and a normal woman is 0.25.
4. (b) :
The ability of a gene to have multiple phenotypic effects because it influences a number of characters simultaneously is known as pleiotropy. The gene having a multiple phenotypic effect because of its ability to control expression of two or more characters is called pleiotropic gene. In human beings pleiotropy is exhibited by syndromes called sickle cell anaemia and phenylketonuria.
Autosomal recessive traits are the traits which are caused by recessive autosomal genes when present in homozygous condition. The given pedigree can be explained as:
As the trait appears only in homozygous recessive individuals (aa), therefore it is an autosomal recessive trait.
Linkage is the phenomenon of certain genes staying together during inheritance through generations without any change or separation due to their being present on the same chromosome. Linkage was first suggested by Sutton and Boveri (1902-1903) when they propounded the famous “chromosomal theory of inheritance.” Bateson and Punnett (1906) while working on sweet pea found that the factors for certain characters do not show independent assortment. However, it was Morgan (1910) who clearly proved and defined linkage on the basis of his breeding experiments in fruit fly (Drosophila melanogaster).
Genes are the units of inheritance and contain the information that is required to express a particular trait in an organism. Alternating forms of a single gene which code for a pair of contrasting traits are known as alleles. For example, two alleles determine the height of pea plant (tall and dwarf).
The man has blood group A, thus its genotype can either be IAIA or IAI°. Similarly, woman can either have IBIB , or IBI° genotype. Thus, their offspring can have any of the blood groups A ( IAIA or IAI°). B ( IBIB or IBI°), AB (IAIB) or O (I°I°).
10. (b) :
Characters studied by Mendel are given as:
The abnormal baby has chromosome, thus it must have been an extra X produced by fusion of abnormal XX ovum with a normal X sperm. Abnormal XX sperm is not possible because, males have XY genotype, and if produce abnormal sperms, then XY sperms and O sperms will be produced. If fusion of multiple gametes have occurred (either two ova with one sperm or two sperms with one ovum), then the human baby will have triploid genotype not the trisomy of sex chromosomes.
12. ( b) :
A dominant epistatic allele suppresses the expression of a non-allelic gene whether the latter is dominant or recessive. The gene which suppresses the expression of a nonallelic gene is known as epistatic gene. The gene or locus which is suppressed by the presence of nonallelic gene is termed as hypostatic gene. In summer squash or Cucurbita pepo, there are three types of fruit colour— yellow, green and white. White colour is dominant over other colours while yellow is dominant over green. Yellow colour is formed only when the dominant epistatic gene is represented by its recessive allele (w). When the hypostatic gene is also recessive (y), the colour of the fruit is green, i.e.,
W – Yy, W – yy —> White
wwyy —> Green
13. (c) :
It is given that the man had colour blind father, i.e., man’s genotype would be XY.
Now, the woman had a colourblind mother and normal father, thus her genotype would be XCX. A cross between them can be represented as below.
Therefore, 50% of male children of this couple will be colourblind.
A human female with Turner’s syndrome has single sex chromosome i.e., 44 + X0 (45). Such females are called sterile females with rudimentary ovaries. Other associated phenotypes of this condition are short stature, webbed-neck, broad chest, lack of secondary sexual characteristics and sterility. Thus, any imbalance in the copies of the sex chromosomes may disrupt the genetic information necessary for normal sexual development.
Haemophilia is sex-linked disease which is also known as bleeder’s disease as the patient will continue to bleed even from a minor cut since he or she does not possess the natural phenomenon of blood clotting due to absence of antihaemophiliac globulin or factor VIII (haemophilia – A) and plasma thromboplastin factor IX (haemophilia-B, Christmas disease) essential for it. As a result of continuous bleeding the patient may die of blood loss. It is genetically due to the presence of a recessive sex linked gene h, carried by X-chromosome. A female becomes haemophiliac only when both of her X-chromosomes carry the gene (XhXh). However, such females generally die before birth because the combination of these two recessive alleles is lethal. A female having only one allele for haemophilia (XXh) appears normal because the allele for normal blood clotting present on the other X-chromosome is dominant. Such females are known as carriers. In case of males, a single gene for the defect is able to express itself as the Y-chromosome is devoid of any corresponding allele (XhY).
16. (b) :
Amniocentesis is a foetal sex determination test in which amniotic fluid containing foetal cells which surrounds the developing embryo is extracted and cells are tested for chromosomal pattern to identify genetic disorders, if any. Jaundice is not a chromosomal disorder thus cannot be tested by amniocentesis.
17. (a) :
Thalassaemia is an autosomal recessive blood disorder. In the given case, both the partners are unaffected carriers for the gene i.e., have heterozygous genotype Tt. Persons homozygous for the autosomal recessive gene of (3-thalassaemia suffer from severe haemolytic anaemia. Heterozygous persons are also not normal, but show the defect (thalassaemia minor).
The phenomenon of expression of both the alleles in a heterozygote is called codominance. Alleles for blood group A ( IA) and B(IB) are codominant so when they come together in an individual, they produce blood group AB characterized by presence of both antigens A and B over the surface of erythrocytes.
In codominance, both the alleles are able to express themselves independently when present together resulting in a phenotype that is intermediate between both the parental homozygous phenotypes, thereby resembling both of them. E.g., roan coat colour in cattle is a result of co-dominance of alleles for white and red coat colour.
Pedigree analysis is a system of analysis by following the movement and distribution of certain genetic traits in many generations of a family. Pedigree analysis cannot confirm that DNA is the carrier of genetic information because it is an analysis system. For DNA based experiments, molecular biology techniques are used.
21. (b) :
Down’s syndrome is the trisomy of 21st chromosome in man. Down’s syndrome is characterized by short stature, warty skin, protmding tongue, slanting eyes, with folded eyelids. The affected person’s face presents a typical mongoloid look. Hence it is also called as mongoloid idiocy. It occurs due to the phenomenon of non-disjunction. Non-disjunction occurs when a pair of homologous chromosomes do not separate in meiosis but migrate to the same pole of the cell resulting in an uneven number of chromosomes in the daughter cells (45 in one and 47 in other) This numerical abnormality results in trisomy (2n + 1) and monosomy (2n – 1). Non-disjunction is more common in sex chromosomes.
22. (b) :
In the given condition the chances of child to be colour-blind is zero percent. It can be understood by the given cross :
23. (d) :
The inheritance of flower colour in the dog flower (snapdragon or Antirrhinum sp.) is a good example which shows incomplete dominance. In a cross between true-breeding red-flowered (RR) and true- breeding white-flowered plants (rr), the F1 (Rr) was pink. When the F2 was self-pollinated the F1 resulted in the following ratio, 1 (RR) Red : 2 (Rr) Pink : 1 (rr) White. Here the genotype ratios were 1 : 2 : 1 as in any Mendelian monohybrid cross, but the phenotype ratios had changed from the 3 : 1 dominant: recessive ratio to 1:2:1.
25. (a) :
Transcription is the process of transferring of information stored in DNA to MRNA through the synthesis of RNA over the template DNA. As RNA is a polymer of ribonucleotide and is made up of pentose ribose sugar, phosphoric acid and nitrogenous bases (A, U, G and C), so during base pairing ribonucleoside triphosphate come to lie opposite to the nitrogen bases of DNA template and form complementary pair; uracil opposite adenine, adenine opposite thymine, cytosine opposite guanine and guanine opposite cytosine. Haemophilia is a sex-linked recessive disease, which shows its transmission from unaffected carrier female to some of the male progeny, hence shows criss-cross
inheritance pattern. It is also known as bleeder’s disease as the patient continues to bleed from a minor cut as he does not possess the natural phenomenon of blood clotting due to absence of antihaemophiliac globulin or factor VIII and plasma thromboplastin factor IX. It is genetically, due to recessive sex-linked gene h carried by X chromosome. A female becomes haemophiliac only when both its X chromosomes carry the gene (XhXh) whereas in case of males, a single gene for the defect is able to express itself (XhY) due to presence of a recessive sex-linked gene h carried by X chromosome. The repressor of the operon is synthesized (all-the- time – constitutively) from the i gene. The repressor protein binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon. In the presence of an inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase to access the promoter and transcription proceeds. Hence, regulation of lac operon by repressor is referred to as negative regulation. Human genome is said to have approximately 3 x 109 bp and approximately 20,000 – 25,000 genes.
26. (a) :
Test cross is performed to determine the genotype of F2 plant. In a typical test cross an organism showing dominant phenotype and whose genotype is to be determined is crossed with one that is homozygous recessive for the allele being investigated, instead of self-crossing. The progenies of such a cross can easily be analyzed to predict the genotype of the test organism. Given ahead is an illustration of test cross.
The increased vigour displayed by the offspring from a cross between genetically different parents is called heterosis. Hybrids from crosses between different crop varieties (F1 hybrids) are often stronger and produce better yields than the original varieties.
28. (b) :
XO type of sex chromosomes determine male sex in grasshoppers. This type of sex-determination comes under XX-XO type. Its common examples are cockroaches, grasshoppers and bugs. The female has two homomorphic sex chromosomes XX and is homogametic. It produces similar eggs, each with X-chromosome. The male has one chromosome only and is heterogametic. It produces 2 types of sperms : gynosperms with X and androsperms without X. Fertilization of an egg by X-bearing sperm yields female offspring, and by no X sperm yields male offspring.
In humans, the female has a pair of X chromosome (homogametic composition) and the male has XY chromosomes (heterogametic composition). Therefore, two normal X chromosomes in zygotic cell lead to the birth of a normal human female child.
In test cross, F1individual showing a dominant phenotype is crossed with homozygous recessive parent. The progenies of such a cross can easily be analysed to predict the genotype of the test organism.
A record of inheritance of certain genetic traits for two or more generations presented in the form of a diagram or family tree is called pedigree. In a pedigree a square represents the male, a circle the female, solid (blackened) symbol shows the trait under study or affected individual; unaffected or normal individual by an open or clear symbol and a cross or shade (of any type) in the symbol signifies the carrier of a recessive allele. Words can also be used in place of symbols. Parents are shown by horizontal line while their offsprings are connected to it by a vertical line. The offsprings are then shown in the form of a horizontal line below the parents and numbered with arabic numerals.
The three alleles IA Is and i of gene IB in ABO blood group system can produce six different genotypes and four different phenotypes as shown below :
Linkage is the phenomenon of certain genes staying together during inheritance through generations without any change or separation due to their being present on the same chromosome. Linked genes occur in the same chromosome. Strength of the linkage between two genes is inversely proportional to the distance between the two i.e., two linked genes show higher frequency of crossing over (recombination) if the distance between them is higher and lower frequency if the distance is small.
Refer answers 30.
35. (c) :
According to Mendel’s law of dominance, in heterozygous individuals a character is represented by two contrasting factors called alleles or allelomorphs which occur in pairs. Out of the two contrasting alleles, only one is able to express its effect in the individual. It is called dominant factor or dominant allele. The other allele which does not show its effect in the heterozygous individual is called recessive factor or recessive allele. The option (c) in the given question cannot be explained on the basis of law of dominance. It can only be explained on the basis of Mendel’s law of independent assortment, according to which in a dihybrid cross, the two alleles of each character assort independently (do not show any blending) of the alleles of other character and separate at the time of gamete formation. Both the characters are recovered as such in F2 generation producing both parental and new combinations of traits.
36. (c) :
Refer answer 32.
38. (c) :
Refer answer 30.
The given situation is an example ofincomplete dominance where phenotype found in F1 generation do not resemble either of the two parents. The genotype of the two plants used for cross will be
The incomplete dominance of dominant allele (here ‘R’) over recessive allele (here ‘r’) could be due to mutations (insertion, deletion, substitution or inversion of nucleotides). The mutant allele generally produces a faulty or no product. This modification in the product may lead to incomplete dominance of the (unmodified) wild type dominant allele.
40. (c) :
Baldness is a sex influenced trait. The dominance of alleles may differ in heterozygotes of the two sexes. This phenomenon is called “sex influenced dominance”. Gene products of heterozygotes in the two sexes may be influenced differentially by sex hormones.
41. (b) :
Sickle-cell anaemia is an autosomal hereditary disorder in which erythrocytes become sickle shaped. It is caused by the formation of abnormal haemoglobin called haemoglobin-S. Hamoglobin-S is formed when 6th amino acid of β-chain, i.e., glutamic acid is replaced by valine due to substitution. It occurs due to a single – nucleotide change (A —> T) in the P-globin gene of coding strand. In the normal β-globin gene the DNA sequence is CCTGAGGAG, while in sickle-cell anaemia, the sequence is CCTGTGGAG
ADA deficiency can be permanently cured if the isolated gene from bone marrow cells producing ADA is introduced into cells at early embryonic stages.
In ABO blood group O refers to O blood group. It has no antigen (A and B) on RBCs.
44. (a) :
This chart shows inheritance of an autosomal recessive trait like phenylketonuria. An autosomal recessive trait may skip a generation. It appears in case of marriage between two heterozygous individuals (Aa x Aa = 3 Aa + 1 aa), a recessive individual with hybrid (Aa x aa = 2 Aa + 2 aa) and two recessive (aa x aa = all aa). Phenylketonuria is an inborn, autosomal, recessive metabolic disorder in which homozygous recessive individual lacks the enzyme phenylalanine hydroxylase. The heterozygous individuals are normal but carriers.
45. (c) :
Klinefelter’s syndrome is a genetic disorder in which there are three sex chromosomes, XXY, rather than the normal XX or XY. The number of autosomes are normal i.e., 44. Affected individuals are apparently male but are tall and thin, with small testes, failure of normal sperm production (azoospermia), enlargement of the breasts (gynaecomastia), and absence of facial and body hair.
46. (d) :
As sperms produced are with genotypes AB, Ab, aB, ab (two diallelic character) the person must be heterozygous for both genes. So his genotype will be AaBb.
47. (d) :
Yellow (Y) seeds are dominant to green (y). So a heterozygous yellow seeded plant will have the genotype of (Yy) and a green seeded plant will have genotype of (yy). When these two plants are crossed, the F1 generation will have the ratio of yellow : green as 50 : 50. It is shown as
48. (b) :
Polygenic (or Quantitative) inheritance is that type of inheritance in which the complete expression of a trait is controlled by two or more genes in which a dominant allele of each gene contributes only a unit fraction of the trait and total phenotypic expression is the sum total of a additive or cumulative effect of all the dominant alleles of genes/polygenes. Human skin colour is an example of such polygenic inheritance which is controlled by three pairs of polygenes A, B and C. Negro/black colour is due to presence of all the six dominant contributing alleles AABBCC. Very light colour or white colour is due to presence of all six recessive noncontributing alleles aabbcc.
49. (c) :
Hexaploid wheat is a result of allopolyploidy induced by doubling the chromosome number of the hybrid produced by crossing two different plants. In hexaploid wheat triticale 2n = 6x = 42. So x stands for basic chromosome number and n for haploid chromosome number. So, n = 21 and x = 7 for hexaploid wheat.
50. (c) :
A common test to find the genotypes of a hybrid is by crossing of one F1 progeny with male parent.
51. (d) :
Refer answers 30.
52. (c) :
A congenital disorder is a medical condition that is present at birth. Congenital disorders can be a result of genetic abnormalities, the intrauterine environment, or unknown factors. Sickle cell disease [a group of genetic disorders caused by sickle haemoglobin (Hbs). Hbs molecules tend to clump together, making red blood cells sticky, stiff, and more fragile, and causing them to form into a curved, sickle shape.] and Huntington’s chorea (an inherited disorder characterized by degenerative changes in the basal ganglia structures, which ultimately result in a severely shrunken brain and enlarged ventricles, abnormal body movements called chorea, and loss of memory) are congenital disorders.
Colourblindness is a recessive sex-linked trait.
All sons will be colourblind and all daughters will be carriers.
54. (c) :
Cri du chat syndrome, also called deletion 5p syndrome, (or 5p minus), is a rare genetic disorder. Cri du chat syndrome is due to a partial deletion of the short arm of chromosome number 5. The name of this syndrome is French for “cry of the cat,” referring to the distinctive cry of children with this disorder. The cry is caused by abnormal larynx development, which becomes normal within a few weeks of birth. Infants with cri du chat have low birth weight and may have respiratory problems. Some people with this disorder have a shortened lifespan, but most have a normal life expectancy.
55. (d) :
Sickle cell anaemia is an autosomal hereditary disorder in which the erythrocytes become sickle shaped. The disorder or disease is caused by the formation of an abnormal haemoglobin called haemoglobin-S. As found out by Ingram (1958), haemoglobin-S differs from normal haemoglobin-A in only one amino acid – 6th amino acid of (β-chain, glutamic acid, is replaced by valine. This is the major effect of the allele. During conditions of oxygen deficiency 6-valine forms hydrophobic bonds with complementary sites of other globin molecules. It distorts their configuration. As a result, erythrocytes having haemoglobin-S become sickle-shaped. Carriers of the sickle cell anaemia gene are protected against malaria because of their particular haemoglobin mutation; this explains why sickle cell anaemia is particularly common among people of African origin. The malarial parasite has a complex life cycle and spends part of it in red blood cells and feeds on haemoglobin. Both sickle-cell anaemia and thalassemia are more common in malaria areas, because these mutations convey some protection against the parasite. In a carrier, the presence of the malaria parasite causes the red blood cell to rupture, making the Plasmodium unable to reproduce. Further, the polymerization of Hb affects the ability of the parasite to digest Hb in the first place. Therefore, in areas where malaria is a problem, people’s chances of survival actually increase if they carry sickle cell anaemia. Thus, sickle-cell anaemia is a potential saviour from malaria.
56. (a) :
Since round seed shape is dominant over wrinkled seed shape and yellow cotyledon is dominant over green cotyledon so RRYY individuals is round
57. (a) :
The plant having genotype AABbCC is heterozygous for only one character B. Number of gametes = 2X, where n is the heterozygosity.
Since n = 1 so 2 gametes will be formed. Those are ABC and AbC.
So the two types of gametes will be ABC and AbC.
58. (a) :
The external manifestation, morphological or physiological expression of an individual with regard to one or more characters is called phenotype. For recessive genes, phenotype and genotype are similar. For dominant genes, the phenotype is same for both homozygous states. Phenotype is influenced by environment as well as age. A child definitely differs from adolescent, the latter from adult and an adult from aged one. Many phenotypes are determined by multiple genes. Thus, the identity of phenotype is determined by genotype and environment.
59. (a) :
Refer answer 48.
60. (c) :
A test cross involving the crossing of F1 individual with the homozygous recessive parent. It is done to find out homozygous and heterozygous individuals. So AaBb, should be crossed with aabb.
61. (d) :
Glucose-6-phosphate dehydrogenase (G-6- PD) deficiency is a group of hereditary abnormalities (X linked disorder) in which the activity of the erythrocyte enzyme G-6-PD is markedly diminished leading to haemolysis.
62. (b) :
Traits governed by sex-linked recessive genes are : (a) produce disorders in males more often than in females, (b) express themselves in males even when represented by a single allele because Y-chromosome does not carry any corresponding alleles, (c) seldom appear in both father and son. (d) fail to appear in females unless their father also possesses the same and the mother is a carrier, (e) female heterozygous for the trait function as carrier, (f) female homozygous for the recessive trait transfer the trait to all the sons.
Take the example of colour blindness which is a recessive sex-linked trait. In the question, as man and woman do not show any signs of disease, so man must be normal and woman must be carrier.
63. (d) :
Down’s syndrome is caused by the presence of an extra chromosome number 21. Both the chromosomes of the pair 21 pass into a single egg. Thus, the egg possesses 24 chromosomes instead of 23 and offspring has 47 chromosomes (45 + XY in males, 45 + XX in males) insted of 46. Turner’s syndrome is formed by the union of an abnormal O egg and a normal X sperm or a normal egg and an abnormal O sperm. The individual has 45 chromosomes (44 + X) instead of 46. Female with more than two X chromosomes is called superfemale. Triploidy is a condition in which an organism has three times (3n) the haploid number (n) of chromosomes.
64. (c) :
Flaemophilia is a sex linked disease in which the patient continues to bleed even from a minor cut since he or she does not possess the natural phenomenon of blood clotting. Haemophilia (= hemophilia) is genetically due to the presence of a recessive gene h, carried by Y-chromosome. A female becomes haemophiliac only when both its X- chromosomes carry the gene (XhXh). However, such females generally die before birth because the combination of these two recessive alleles is lethal. A female having only one allele for haemophilia (XXh) appears normal because the allele for normal blood clotting present on the other X-chromosome is dominant. Such females are known as carriers. In case of males, a single gene for the defect is able to express itself as theY-chromosome is devoid of any corresponding allele (XhY) in which an organism has three times (3n) the haploid number (n) of chromosomes.
65. (d) :
Cretinism occurs due to hyposecretion of thyroid hormones. Haemophilia is a sex linked recessive trait. Cystic fibrosis is also a recessive autosomal disorder resulting in mucus clogging in lungs. Thalassemia involves a gene mutation in the polypeptide chains of haemoglobin.
Colourblindness is a recessive sex-linked trait. Since the woman’s father was colour blind. She should be carrier of the colour blind gene (XCX). When she marries to colour blind man their progeny could be
67. (b) :
Since red fruit colour is dominant over yellow fruit colour and tallness is dominant over shortness
These are produced in 1 : 1 ratio
Prototroph is the nutritionally wild strain which is unable to grow in the minimal medium unless additional nutrients were added to the medium. Phenotype is the kind of organism produced by the reaction of a given genotype with the environment. Holotype is one of the original type used to describe a new species.
The male human is heterozygous for autosomal gene A and B and also hemizygous for hemophilic gene h then his genotype will be AaBbXhY because hemophilia is a sex linked trait that is present on X- chromosoine. So the total number of gametes will be abXh, abY, ABXh, ABY, AbXh,AbY, aBXh, aBY. So the proportion of abXh sperm will be 1/8.
70. (c) :
Mendel’s law of independent assortment states that when the parent differs from each other in two or more pairs of contrasting characters, the inheritance of one pair of factor is independent of the other. For the character to assort independently they should be located on separate non homologous chromosomes. Genes present on the same chromosome show linkage. It means that these characters remain together and thus low numbers of combinations are formed. This phenomenon is called linkage such genes are called linked genes. So A and B are linked genes.
Mutation is a sudden alteration of the chemical structure of a gene or the alteration of its position on the chromosome by breaking and rejoining of the chromosome. It has occured in male parent. But organelles like mitochondria, chloroplast etc. are a part of cytoplasmic inheritance.
Cytoplasmic inheritance is the passage of traits from parents to offspring through structures present inside the cytoplasm of contributing gametes. Plasma genes occur in plastids, mitochondria, plasmids and some special particles like kappa particles, sigma particles, etc. In higher organisms cytoplasmic inheritance is called maternal inheritance because the zygote receives most of its cytoplasm from the ovum. Therefore, cytoplasmic inheritance is usually unparental.So none of the progeny will show mutation.
72. (b) :
In question, where the genotype of the other parent is not mentioned then that should be considered normal. Colourblindness is a recessive sex- linked trait (i) To find out the genotype of a woman.
Her father is colour-blind \ his genotype is XCY and her mother is normal so her genotype is XX.
So, woman is carrier
(ii) When this woman marries normal man 50% of the sons would be colourblind.
Normal girl Normal boy Carrier Colourblind girl boy
73. (c) :
The recessive genes located on X-chromosome of humans are always expressed in males e.g. colour blindness is a recessive sex-linked trait in which the eye fails to distinguish red and green colours. The gene for the normal vision is dominant. The normal gene and its recessive allele are carried by X-chromosomes. In females colour blindness appears only when both the sex chromosomes carry the recessive gene (XCXC). The females have normal vision but function as carrier if a single recessive gene for colour blindness is present (XXC). However, in human males the defect appears in the presence of a single recessive gene (XCY) because Y- chromosome of male does not carry any gene for colour vision.
74. (b) :
Mendel worked on garden pea and choose
seven characters for this. Green seed colour is the recessive character in his experiment and the dominant character for seed colour is yellow. Axial flower position, green pod colour and round seed shape are all dominant characters. .
Mendel worked on garden pea. He chose only those characters which showed consistent results. He worked on seven characters. These characters showed complete independent assortment despite the seven characters choosen by him were present on four chromosomes -1,4, 5 and 7.
76. (b) :
A reciprocal cross means that the same two parent are used in two experiments in such a way that if
in one experiment A is used as the female parent and B is used as the male parent then in the other experiment A will be used as the male parent and B as the female parent. Thus the sources of gametes are reversed. When the F1 individuals obtained in a cross is crossed with the recessive parent is called a test cross. When inheritance of two pairs of contrasting character is studied simultaeneously it is called dihybrid cross.
77. (c) :
Sex influenced traits are autosomal traits that are influenced by sex. If a male has one recessive allele, he will show that trait, but it will take two recessive alleles for the female to show that same trait e.g. pattern baldness, moustaches and beard in males. Sex linked traits are those traits the determining genes of which are found on the sex chromosomes. Sex limited traits are the traits which are expressed in a particular sex though their genes also occur in the other sex e.g., milk secretion in mammalian females.
78. (b) :
In sickle cell anaemia, the shape of the RBC change (sickle shaped) in comparison to normal one. The sickle cells (RBCs) cannot pass through capillaries and thus clog them.
On the other hand, the mosquito borne disease is malaria. Here, the main phase of sexual cycle, in affected mosquito occurs in normal RBC. But as in sicke cell anaemia, RBCs become deformed, the mosquito cycle cannot be continued here. Thus sickle cell anaemia is a potential saviour from malaria.
79. (c) :
Down’s syndrome is the example of autosomal aneuploidy. Here, an extra copy of chromosome 21 occurs. As it is an autosomal disease, the offsprings produced from affected mother and normal father should be 50%.
80. (c) :
According to genic balance theory of sex determination the ratio between the number of X-chromosomes and number of complete sets of autosomes will determine the sex. The X-chromosome is believed to carry female tendency genes, while autosomes carry male tendency genes. Both these sets of genes start functioning and there has to be a balance between them for an individual to become male or female. If the ratio between X and A is 1.0 it will be a female individual and when it is 0.5, it would be male.
81. (c) :
Pleiotropic gene is such a gene which has a wider effect on phenotype i.e., it controls several phenotypic traits. Sickle cell anaemia is considered to be caused by one such pleiotropic gene. Sickle cell anaemia is caused due to a mutation in β globin gene of haemoglobin. Single point mutation leads to one change of amino acid.
But this small point mutation & change in the resulting β globin loads to sickling of RBCs, haemolytic anaemia, pain in joints, splenomegaly etc.
Moreover carrier of sickle cell anaemia gene remains protected from malaria. Thus the gene is naturally selected in tropical and subtropical Asia and Africa.
Crossing over and recombination frequencies : One crossing over produces 50% recombinant types. Therefore, frequency of crossing over would be double the frequency of recombinants. 1% recombinants (1% cross-over) means crossing over in 2% meiocytes. Linkage/ Cross over/ Chromosome maps is a graphic representation of relative positions/ order and relative distances of genes in a chromosome in the form of line like a linear road map depicting different places and their relative distances without giving exact mileage. It is based on Morgan’s hypothesis (1911) that frequency of crossing over/recombination between two linked genes is directly proportional to the physical distance between the two. 1 map unit or centrimorgan is equivalent 1% recombination between two genes. Percentage of crossing over between a and b is 20% so they are 20 map distance apart and b and c are 28 map distance apart. So that correct sequence of genes on chromosomes will be as
Selfing is the process of fertilization with polar or male gametes of the same individual. AABbCC will produce two type of gametes ABC and AbC. Thus in F2 generation three genotypes will be obtained. These are AABBCC, AABbCC and AAbbCC in the ratio of 1 : 2 : 1. Phenotypically AABBCC and AABbCC are same. So the phenotypic ratio in F2 generation will be 3 : 1.
84. (c) :
Dominant factor is an allele or Mendelian factor which expressess itself in the hybrid (heterozygous) as well as in homozygous state. It is denoted by capital letter (T or D for tallness in pea, R or W for round seeds in pea).
A dominant gene in genotype will be expressed even in the presence of a corresponding recessive gene.
85. (a) :
In the inheritance pattern of sex chromosomes X-chromosome of father always passes to daughter and X-chromosome of mother passes to son. As the father is diseased and all the girls inherit it, it is obvious the disease is sex-linked. The mother is not a carrier (as evident from the fact that no son is diseased). Thus the gene is dominant and expresses even in heterozygous condition.
86. (a) :
Down’s syndrome (Mongolian Idiocy, Mongolism) is caused by the presence of an extra chromosome number 21. The infant has a characteristic face with a flat nasal bridge, brushfield spots, protuding tongue, small ears and cardiac lesions. All affected children have mental retardation (moderate), short stature and autoimmune abnormalities. All males are infertile and females have reduced fertility.
Sickle cell anaemia is not a sex linked (i.e., X linked) disease but an autosomally inherited recessive trait. Hemophilia is X-linked but not holandric/Y-linked. Parkinson’s disease is a degenerative disease due to absence of neurotransmitter substance “dopamine” from brain. It is not at all hereditary.
87. (a) :
Epistasis is the phenomenon of suppression of phenotypic expression of gene by a nonallelic gene which shows its own effect. The gene which masks the effect of another is called epistatic gene while the one which is suppressed is termed hypostatic gene. Epistasis is of three types – dominant, recessive and dominant-recessive. Two or more different pairs of alleles, with a presumed cummulative effect governing quantitive traits like size, pigmentation etc., are called polygenes.
88. (b) :
The tendency of potential combinations to remain together, which is expressed in terms of low frequency of recombinations (new combinations) is called linkage. Genes present on same chromosomes show linkage. These genes are called linked genes. Since A and B genes are linked they will be passed on together in the progeny.
89. (a) :
Dominance is the phenomenon in which one member of a pair of allelic genes expresses itself as a whole (complete dominance) or in part (incomplete dominance). Codominance is when the two genes neither show dominant-recessive relationship nor show intermediate condition, but both of them express themselves simultaneously. This has been reported in roan character of cattle (i.e., patches of 2 different colours on the skin). Pseudodominance is when a recessive allele of normal homologous chromosome will behave like a dominant allele and will be phenotypically expressed.
90. (d) :
If two genes present on different loci produce the same effect when present alone but interact to form a new trait when present together, they are called complementary genes. The F2 ratio is modified to 9 : 7 instead of 9 : 3 : 3 : 1.
91. (b) :
According to law of independent assortment, the two factors of each trait assort at random and independent of the factors of other traits at the time of meiosis and get randomly as well as independently rearranged in the offspring. Principle of law of independent assortment is applicable to only those factors or genes which are present on different chromosomes.
92. (b) :
Refer answer 41.
93. (c) :
Barr body is a mass of condensed sex chromatin in the nuclei of normal female somatic cells due to inactive X chromosome. Thus Barr body represents the inactivated X-chromosome. Whenever the number of X-chromosomes is two or more than two, the number of barr bodies is one less than the number of X-chromosomes. Therefore, the number of barr bodies in XXXX female is three.
94. (b) :
Male XX and female XY sometimes occur due to transfer of segments in X and Y chromosomes. Deletion is the loss of an intercalary segment of a chromosome which is produced by a double break in the chromosomes followed by the union of remaining parts. Aneuploidy is a condition of having fewer or extra chromosomes than the normal genome number of the species.
96. (a) :
The cross can be represented as :
Tall Red TtRr – 50%
Tall White Ttrr – 50%
Mendel worked on garden pea. He chose only those characters which showed consistent results. He worked on seven characters which are given as:
From the above table it is confirmed that green pod colour is dominant character.
98. (a) :
Refer answer 86.
If fertilization takes place between gametes of Rh female and Rh– male then the resulting foetus’ blood is RhThis is a serious problem. If the Rlr child does not suffer (although the Rh’ blood of the foetus stimulates the formation of anti Rh factors are not produced in the mother’s blood to harm the foetus). But in second pregnancy (with Rh+ foetus), the anti Rh factors of the mother’s blood destroy the foetal red blood corpuscles. This is called erythroblastosis foetalis. New bom may survive but it is often anaemic.
100. (a) :
in Drosophila. XXY condition leads to femaleness. Cytological examination have suggested that Y-chromosome does not play any role in determination of sex in Drosophila. In human being, XXY is phenotypically male with underdeveloped testes gynecomastia and often mental retardation. It is caused by the union of a non-disjunct XX egg and sperm and a normal X egg and abnormal XY sperm. This indicates that in human being Y chromosome is active in sex determination.
101. (c) :
Refer answer 87.
102. (c) :
Crossing of individuals having dominant phenotype with its homozygous recessive is a test cross, which can be represented as:
Tt (Tall) tt (Dwarf) Test cross progeny
Thus ratio of progeny is = I : 1
103. (d) :
Colourblindness is the inability of certain human beings to distinguish red from green colour. It is produced by a recessive gene which lies on X chromosome. A marriage between normal visioned man (XY) and colourblind women (XCXC), results in colourblind sons (XCXC) and carrier daughters (XXC).
104. (d) :
A protein named as rhesus antigen, is present on the surface of red blood corpuscles in many persons. Depending on the race, 85 to 99 percent of the white population have this rhesus antigen (also called Rh factor) and are called Rh positive (Rh+). Others who do not have this factor are known as Rh negative (Rh ). Both Rh+ and Rh– individuals are phenotypically normal. The problem arises during blood transfusion and pregnancy.
The first blood transfusion of Rh+ blood to the person with Rh+ blood causes no harm because the Rh– person develops anti Rh factors or antibodies in his/her blood. In second blood transfusion of Rh blood to the Rh–person, the latter’s anti Rh factors attack and destroy the red blood corpuscles of the donor. If father’s blood is Rh– and mother’s blood is Rh+, the foetus’ blood is Rh–. This is a serious problem. If the Rh’ blood of mother has not earlier come in contact with Rh+ blood through transfusion, her first child does not suffer (although the Rh+ blood of the foetus stimulates the formation of anti Rh factors or antibodies in the mother’s blood yet enough anti Rh factors are not produced in the mother’s blood to harm the foetus). But in second pregnancy (with Rh+foetus), the anti Rh factors of the
mother’s blood destroy the foetal red blood corpuscles. This is called erythroblastosis foetalis. New bom may survive but it is often anaemic.
105. (d) :
Haemophilia is a defect of blood which prevents its clotting. It is caused by a recessive gene located in the X-chromosome. When a haemophilic man (XhY) marries a normal woman (XX), produces carrier girls (XXh) and normal boys (XY), i.e. all their offsprings will be normal.
In human beings, inheritance of skin colour is due to multiple genes. This type of inheritance is called quantitative inheritance. Quantitative inheritance is a type of inheritance controlled by one or more genes in which the dominant alleles have cumulative effect with each codominant allele expressing a part or unit of the trait, the full trait being shown only when all the dominant alleles are present. A few instances of such inheritance are skin colour in human beings, human intelligence, height in human beings etc. Sickle-cell anaemia and phenylketonuria are due to recessive genes (autosomal) in homologous condition. Colourblindness is produced by a recessive gene lying on X-chromosome.
107. (b) :
The ability of a gene to have multiple phenotypic effect because it influences a number of characters simultaneously is known as pleiotropy. The phenomenon by which a gene suppresses the phenotypic expression of a nonallelic gene is called epistasis.
108. (a) :
According to principle of independent assortment, the two factors of each trait assort at random and independent of the factors of other traits at the time of meiosis and get randomly as well as independently rearranged in the offspring. Priniciple of independent assortment is applicable to only those factors or genes which are present on different chromosomes. Chromosome have hundreds of genes which show linked inheritance or linkage. Linkage is the phenomenon of certain genes (present on the same chromosome) to remain together and get inherited through generations. The seven characters that Mendel choose were present on 14 chromosomes and so they did not show linkage but if present on 12 chromosomes they would have shown linkage and the principle of independent assortment would not have been discovered.
109. (b) :
Crossing over is the reciprocal exchange of segments between nonsister chromatids of a pair of homologous chromosomes. It results in recombination of genes. The nonsister chromatids in which exchange of segments takes place are known as cross-overs or recombinants while other chromatids not involved in exchange of segments are called noncross-over or parental types. Single crossing-over at one point between two nonsister chromatids resulting in two cross-over and two parental types. Double crossing over at two points in a tetrad of chromatids.
110. (d) :
A heterozygous individual is the one having two contrasting mendelian factors or alleles for a character.
Number of gametes = 2n = 22 = 4.
where n is the number of gene in heterozygous form. The four gametes formed will be —>ABc, AbC, Abe and ABC,
111. (b) :
Haemophilia is a defect of blood w’hich prevents its clotting. Colour blindness is the inability of certain human beings to distinguish red from green colour. Both these diseases are produced by a recessive gene which lies on the X-chromosomes. A woman having both gene for haemophilia on one X-chromosome and gene for colourblindness on another X-chromosome will have genotype XhXC.
XhXC x XY
Thus, progeny includes 50% haemophilic daughters (carrier) and 50% colour blind daughters (carrier).
112. (d) :
In humans, sex chromosomal abnormality is due to gene carried on X-chromosome. Increase in X- complement leads to, Klinefelter’s syndrome. Klinefelter’s syndrome, 47, XXY or XXY syndrome is a condition caused by a chromosome aneuploidy. Affected males have an extra X sex chromosome. It is formed by the union of an XX egg and normal Y sperm or normal X egg and abnormal XY sperm. Affected males are almost always effectively sterile, although advanced reproductive assistance is sometimes possible and some degree of language learning impairment and mental retardation may be present. In adults, possible characteristics vary widely and include little to no signs of affectedness, a lanky, youthful build and facial appearance, or a rounded body type with some degree of gynecomastia (increased breast tissue). It is the second most common extra chromosome condition, and is named after Dr. Harry Klinefelter, an endocrinologist at Massachusetts General Hospital, Boston, Massachusetts, who first described it in 1942. The condition exists in roughly 1 out of every 500 to 1,000 males.
113. (d) :
Albinism is caused by the absence of the enzyme tyrosinase which is essential for the synthesis of the pigment from dihydroxy-phenyalanine. The gene for albinism (a) does not produce the enzyme tyrosinase but its normal allele (A) does. Thus, only homozygous individual (aa) is affected by this disease. Albinos (individuals with albinism) lack dark pigment melanin in the skin, hair and iris. Although albinos have poor vision yet they lead normal life. On the basis of principles of simple recessive inheritance, the probability of albinic child from a normally pigmented parents, will be 1/4 or 25 %.
114. (c) :
The female Drosophila possesses two homomorphic sex chromosomes, (XX) and the male Drosophila contains two heteromorphic sex chromosomes (XY). The differential or non-homologous region of Y-chromosome is mostly heterochromatic. The female parent produces only one type of eggs (22 + X). The male parent produces two types of gametes (22 + Y) and (22 + X). They are produced in equal proportions. As the two types of sperms are produced in equal proportions, there are equal chances of getting a male or female fly in a particular mating.
115. (b) :
Klinefelter’s syndrome is an aneuploid condition with 3 sex chromosomes (trisomy). Klinefelter’s syndrome is caused by XXY genotype. This genotype results from the union of a nondisjunct XX egg and a normal Y sperm or normal X egg and abnormal XY sperm. The individual has 47 chromosomes (2n +1).
The person having Klinefelter’s syndrome is a sterile male with small testes, unusually long legs, obesity, and sparse body hair, and with many female characteristics such as breasts. The victim usually has normal intelligence. The more the X chromosomes, the greater is the mental defect. Klinefelter’s syndrome occurs about once in 2000 live births.
Turner’s syndrome is formed by the union of an abnormal O egg and a normal X sperm or a normal egg and an abnormal O sperm. The individual has 45 chromosomes (44 + X). Down’s syndrome is caused by the presence of an extra chromosome number 21. Gynandromorphism is a phenomomenon by which one part of the body of an animal is male and the other part is female.
116. (a) :
Sex chromosomes are those chromosomes whose presence, absence or particular form determines the sex of the individual in unisexual or dioecious organisms, e.g., XX – XY. XY method (XX – XY). It occurs in mammals and many insects with females having homomorphic XX sex chromosomes and males possessing heteromorphic XY – chromosomes. In human being the Y-chromosome is straight and acrocentric (centromere subterminal near one end). It is about 2 mm in length and is thus the shortest of all chromosomes. X-chromosome is 5.0 – 5.5 mm in length. It is metacentric (centromere in the middle).
Despite differences in morphology, XY chromosomes synapse during zygotene. They have two parts, homologous and differential. Homologous regions of the two take part in synapsis.
Refer answer 48.
118. (b) :
Complementary genes are those nonallelic genes which independently show a similar effect but produce a new trait when present together in the dominant form. Supplementary genes are a pair of nonallelic genes, one of which produces its effect independently in the dominant state while the dominant allele of the second gene is without any independent effect but is able to modify the effect of the former to produce a new trait. Duplicate genes are independent genes producing the same or similar effect.
119. (b) :
Refer answer 115.
120. (d) :
In question where the genotype of the other parent is not mentioned then that should be considered as normal.
About cases show that if mother (lady) is carrier then options (a) and (c) are not true. Option (b) is true and option (d) all daughters normal (though phenotypically) is also true. If mother is normal then options (a), (b) and (c) are not true and option (d) is true so from the cases, it is concluded that option (d) is true.
121. (c) :
It is a sex-linked disease. It is a disease which occurs due to gene responsible for determining the character concerned and is carried on a sex chromosome. In humans, this is always X-chromosome e.g. colourblindness. Here a genetically diseased father (male) marries with a normal female (homozygous). Colour blindness is recessive to normal vision so that the sons produced would be normal, but daughters will be heterozygous (normal phenotype), which means that these daughters will be carriers of this trait.
122. (c) :
E.B. Lewis in 1951 reported from a cross of apricot eyed and white eyed flies in Drosophila, he obtained F1 having intermediate eye colour. In F2, he had expected segregation only for apricot and white, but he recovered very low frequency of wild type. Since those alleles behaved as non-alleles, Lewis preferred to call them pseudoalleles and the phenomenon as pseudoallelism. In pseudoallelism in cis position both the mutant alleles are on one chromosome. So the other chromosome will be normal and will be able to produce the end result. But in trans position the sequence of steps involved in synthesis will be interrupted due to mutations on either of the two homologous chromosomes thus leading to a mutant phenotype.
123. (a) :
Refer answer 87.
124. (b) :
Point mutation involves only the replacement of one nucleotide with another. One type of point mutation is missense mutation. These are base changes that alter the codon for an amino acid resulting in its substitution with a different amino acid. For example, mutation of the codon CTT to ATT would result in the replacement of the hydrophobic amino acid leucine with isoleucine, another hydrophobic amino acid. Many other missense mutations have been described which do affect the encoded protein and result in genetic diseases. These include an A to T mutation in the gene for b-globin, one of the polypeptides of’hemoglobin. This mutation changes codon six of the gene from GAG which encodes glutamic acid to GTG which encodes valine. The mutation results in a condition called sickle cell anaemia in which the red blood cells adopt an abnormal sickle shape due to aggregation of the hemoglobin molecules. The abnormal cells are short¬lived, which causes anaemia and become lodged in capillaries, which reduces the blood supply to organs.
125. (b) :
Refer answer 115.
126. (b) :
Despite differences in morphology, the XY chromosomes are homologous and synapse during zygotene. It is because they have two parts, homologous and differential. Homologous regions of the two help in pairing. They carry same genes which may have different alleles. The differential region of Y- chromosome carries only Y-linked or holandric genes, e.g, testis determining factor (TDF). It is perhaps the smallest gene occupying only 14 base pairs. Other holandric genes are hypertrichosis (excessive hairiness) on pinna, porcupine skin, keratoderma dissipatum (thickened skin of hands and feet) and webbed toes. Holandric genes are directly inherited by a son from his father. Chromosomes which control most of the morpho-physiological characters other than sex, are called autosomes. Sex linked genes are those which are found on the sex chromosomes. Mutant genes are formed by a change in the nucleotide type and sequence of a DNA segment representing a gene or a cistron.
127. (c) :
O blood group of a child cannot be obtained from the parents having blood group O x AB. The parents blood groups may be A x O, A x B, B x O, B x A, O x A and O x B.
128. (a) :
Refer answer 113.
130. (d) :
Of both normal parents, the chance of a male child becoming colour blind are possible only when mother’s father was colour blind. It is an example of criss cross inheritance. If a cross is made between two sexes differing in certain characters, in such a way that character of one sex remains hidden in the opposite sex of F1 generation, but it is passed on to the same sex in the F2 generation, it is said to exhibit criss cross inheritance.
131. (c) :
Genotype of Mr. Kapoor will be Bbd hence one fourth of the sperms will have Bd.
132. (b) :
AABB x aabb is suitable for experiment on linkage. Linkage is the tendency for certain genes tend to be inherited together, because they are on the same chromosome. Thus, parental combinations of characters are found more frequently in offspring than non- parental.
133. (c) :
Of a normal couple, half the sons are haemophilic while half the daughters are carries. The gene is located on one X-chromosomes of mother. Cross between a haemophilic carrier female XhX and normal male would yield 50% of the sons being haemophilic and 50% of the daughter are carriers.
134. (d) :
Two dominant nonallelic genes are 50 map units apart. The linkage is absent/incomplete. Chromosome mapping is based on the fact that genes are linearly arranged in the chromosome and frequency of crossing over is directly proportional to the distance between two genes. Dominant genes show cis arrangement. At 50 map units cis is changed to trans and vice-versa hence no fixed linkage is present.
135. (a) :
Refer answer 48.
136. (b) :
Law of independent assortment states that when two individuals differ from each other in two or more pairs of factors, the inheritance of one pair is quite independent of the inheritance of other. Law of independent assortment is applicable to only those factors or genes which are located on different chromosomes.
137. (b) :
Sex is determined in human beings at the time of fertilization. Sex of human body is determined by the karyotype of the zygote or fertilized egg. Sex of the baby depends upon the sperm which fertilizes the ovum.
The child of O-group has B-group father. The genotype of father will be IBI°. The genotype of the child would be I°I° (recessive). Hence, the genotype of the father can only be IBI°.
139. (a) :
An allele is dominant if it is expressed in both homozygous states. Dominant alleles expresses itself in the homozygous as well as heterozygous condition. It is denoted by capital letter.
140. (c) :
In a cross between AABB x aabb, the ratio of F2 genotypes between AABB, AaBB, Aabb and aabb would be 1 : 2 : 2 : 1. Genotype is the genetic make up of an individual. This may refer to just one trait or it may refer to the combination of many or all of the traits of the individuals.
Segregation of Mendelian factors (no linkage, no crossing over) occurs during anaphase I. At anaphase I, actual segregation occurs, but two similar alleles occurs in the dyad chromosome which separate at anaphase II.
142. (d) :
An organism with two identical alleles is homozygous. Homozygous have identical genes at the same locus on each member of a pair of homologous/ chromosomes.
143. (a) :
Refer answer S3.
Refer answer 63.
145. (c) :
In human beings, 45 chromosomes/single XI XO abnormality causes Turner’s syndrome. Individuals having a single X chromosome 2A + XO (45) have female sexual differentiation but ovaries are rudimentary. Other associated phenotypes of this condition are short stature, webbed neck, broad chest, lack of secondary sexual characteristics and sterility. Thus, any unbalance in the copies of the sex chromosomes may disrupt the genetic information necessary for normal sexual development.
146. (d) :
IAI° x IAIB gives us the following genotypes IAIA, I°P, IAIB. Hence, when a man of blood group A marries a women of AB blood group, B progeny would indicate that man is heterozygous A.
147. (d) :
ABO blood group system is due to multiple allelism. A gene can have more than two alleles or allelomorphs, which can be expressed by mutation in wild form in more than one ways. These alleles or allelomorphs make a series of multiple alleles. The mode of inheritance in case of multiple alleles is called multiple allelism. A well known and simplest example of multiple allelism is the inheritance of ABO blood groups in human beings. In human population, 3 different alleles for this character are found – IA IB and I°. A person is having only two of these three alleles and blood type can be determined.
148. (b) :
The contrasting pairs of factors in Mendelian crosses are called allelomorphs. Alleles or allelomorphs are the different forms of a gene, having the same locus on homologous chromosomes and are subject to mendelian (alternative) inheritance.
149. (b) :
An Austrian Monk, Gregor Mendel, developed his theory of in heritance. He formulated the law of Heredity. Therefore, he is called the ‘father of genetics’.
150. (c) :
Mendel’s last law is independent assortment. The principle of independent assortment states that when two individuals differ from each other in two or more pairs of factors, the inheritance of one pair is quite independent of the inheritance of ethers.
151. (d) :
A dihybrid condition is TtRr. Dihybrid conditions involves simultaneous inheritance of two pairs of Mendelian factors or genes.
152. (a) :
Blue eye colour is recessive to brown eye colour. A brown eyed man whose mother was blue eyed marries a blue eyed women. The children shall be both blue eyed and brown eyed 1:1. The brown eyed man will have the genotype Bb and his wife bb. Hence Bb * bb = Bb : bb. Hence, the ratio is 1 : 1.
The allele which is unable to express its effect in the presence of another is called recessive. A member of a pair of alleles that does not show its effect in the phenotype in the presence of any other allele. It is denoted by small letter.
154. (b) :
Refer answer 23.
A colour blind girl is rare because she will be bom only when her father and maternal grand father were colour blind. The genotype of the mother was to be either XCXC or XCX and that of father XCY so that the daughter becomes colour blind.
156. (a) :
Cross between AaBB and aaBB will form 1 AaBB : 1 aaBB. On crossing, AaBB x aaBB gives 50% individuals having genotype AaBB and 50% individuals having genotype aaBB.
157. (c) :
In a genetic cross having recessive epistasis, F2 phenotypic ratio would be 9 : 3 : 4. The recessive epistasis is illustrated by coat colour in mouse, the coat colour is determined by A/a pair, recessive allele b is epistatic over A/a. Thus, in the presence of bb, both A and aa give the same phenotype (albino). The F2 ratio is generally 9:3 : 4.
158. (c) :
Refer answer 147.
159. (b) :
Refer answer 102.
160. (b) :
In Down’s syndrome of a male child, the sex complement is XY. Down syndrome is a relatively common birth defect caused by the presence of an extra chromosome number 21 (three instead of two number 21 chromosomes, or trisomy 21). This chromosome abnormality adversely affects both the physical and irftellectual development of the individual. It is also the first syndrome to be described in humans.
Haemophilia is more common in males because it is a recessive trait carried by X-ehromosome. Haemophilia A is the most common X-linked genetic disease that prevents normal blood clotting when blood vessels are ruptured. It appears in about in 7000 males. In most of the severe cases, an affected person can bleed to death from a bruise or cut. Haemophilia A is caused by the absence of a protein called antihaemophilic factor of factor VIII that is essential for blood clotting.
162. (d) :
Phenylketonuria is a hereditary disease. Phenylketonuria is an inherited error of metabolism caused by a deficiency in the enzyme phenylalanine hydroxylase. It results in mental retardation and is inherited as an autosomal recessive trait. It is a hereditary human condition resulting from the inability to convert phenylaline into tyrosine. This change can be traced to a tiny mutation in a single gene on chromosome 12.
163. (a) :
Both husband and wife have normal vision through their fathers were colour blind, the probability of their daughter becoming colour blind is 0%. The chances of daughter becoming colour blind arises only when the father is also colour blind.
164. (d) :
Bateson and Punnet explained that when two dominants enter from the same parent-they try to remain together, called coupling. When two dominants enter from different parents they try to remain seperate called repulsion. Bateson and Punnett (1906) used the term coupling and repulsion in sweet pea (Lathyrus odoratus) for linkage and crossing over. The correct parental of coupling type along with its cross over or repulsion is coupling AABB, aabb : Repulsion AAbb; aaBB. .
165. (c) :
A normal green male maize is crossed with albino female. The progeny is albino because, plastids are inherited from female parents.
166. (c) :
Two linked genes a and b show 20% recombination. The individuals of a dihybrid cross between + +/+ + x ab/ab shall show gametes + + 40 : ab 40: + a 10: + b: 10.
167. (d) :
Consideration of one character at one time contribute to the success of Mendel. Mendel’s contribution was unique because of his methodological approach to a definite problem, use of clear cut variables and application of mathematics (statistics) to the problem. Using pea plants and statistical methods, Mendel was able to demonstrate that traits were passed from each parent inheritance of genes.
168. (c) :
A family of five daughter only is expecting sixth issue. The chance of its beings a son is 50%. Human have 22 pairs chromosomes which are XX in females and XY in males. So if we cross the parents there is 1 : 1 chance for boy and girl.
169. (c) :
Haploids are able to express both recessive and dominant alleles/ mutations because there are only one allele for each gene in the individual. Hybrid is a organism containing two different alleles or individual containing both dominant and recessive genes of an allelic pairs.