Top 100+ Principles of Inheritance and Variation - Biology Questions and Answers For NEET 2024 Exam Preparation

Q1. Ratio of complementary genes is

1. 9 : 3 : 3 : 4
2. 12 : 3 : 1
3. 9 : 3 : 4
4. 9 : 7

Answer: (d)

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Q2. α-thalassemia in humans is controlled by

1. HBA1 gene on chromosome 12
2. HBA1 and HBA2 genes on chromosome 16
3. HBA2 gene on chromosome 11
4. HBA1 and HBA2 genes on chromosome 9

Answer: (b)

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Q3. Independent assortment of genes does not take place when

1. genes are linked and located on same chromosomes
2. genes are located on homologous chromosomes
3. genes are located on non homologous chromosomes
4. All the above

Answer: (a)

Independent assortment of genes takes place only when they are located on separate non-homologous chromosomes. Where two or more than two genes are located on same chromosome, independent assortment will not be possible.

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Q4. If a child has O type of blood group and the father has B type, then the genotype of the father will be

1. $I^AI^B$
2. $I^OI^O$
3. $I^OI^B$
4. $I^BI^B$

Answer: (c)

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Q5. Two linked genes a and b show 20% recombination. the individuals of a dihybrid cross between ++/ ++ × ab/ab shall show gametes

1. ++ : 50 : : ab : 50
2. ++ : 80 : : ab : 20
3. ++ : 40 : : ab : 40 : : + a : 10 : : + b : 10
4. ++ : 30 : : ab : 30 : : + a : 20 : : + b : 20.

Answer: (c)

Two linked genes a and b show 20% recombination. The individuals of a dihybrid cross between + +/+ + × ab/ab shall show gametes + + : 40 : : ab : 40 : : + a : 10 : : + b : 10.

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Q6. Select the correct statement from the ones given below with respect to dihybrid cross.

1. Genes far apart on the same chromosome show very few recombinations.
2. Tightly linked genes on the same chromosome show higher recombinations.
3. Genes loosely linked on the same chromosome show similar recombinations as the tightly linked ones.
4. Tightly linked genes on the same chromosome show very few recombinations.

Answer: (d)

When two allelic pair are used for crossing, it is called dihybrid cross. Linkage is the inheritance of genes of same chromosome together and capacity of these genes to retain their parental combination in subsequent generation.

The strength of linkage between two genes is inversely proportional to the distance between the two. This means, two linked genes show higher frequency of recombination if the distance between them is higher and lower frequency if the distance is smaller.

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Q7. Which of the given pedigree shows inheritance of autosomal recessive gene. What is the genotype of given parents?

1. Aa, AA
2. AA, aa
3. aa, aa
4. Aa, Aa

Answer: (b)

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Q8. Study the pedigree chart of a certain family given below and select the correct conclusion which can be drawn for the character.

1. The trait under study could not be colour blindness.
2. The parents could not have had a normal daughter for this character.
3. The female parent is heterozygous.
4. The male parent is homozygous dominant

Answer: (c)

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Q9. The percentage of crossing over will be more if

1. Linked genes are located close to each other
2. Linked genes are located far apart from each other
3. Genes are not linked
4. Genes are located in a different cell

Answer: (b)

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Q10. If a colour-blind man marries a woman who is homozygous for normal colour vision, the probability of their son being colour-blind is

1. 0.5
2. 0
3. 0.75
4. 1.

Answer: (b)

Genotype of colour blind man – X$^c$Y

Genotype of women homozygous – XX

for normal woman

Hence, there is zero (0) probability of their son to be colour-blind.

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Q11. Select the correct statement from the ones given below with respect to dihybrid cross.

1. Genes loosely linked on the same chromosome show similar recombinations.
2. Genes far apart on the same chromosome show very few recombinations.
3. Tightly linked genes on the same chromosomes show higher recombinations.
4. Tightly linked genes on the same chromosome show very few recombinations

Answer: (d)

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.

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Q12. RR (red) Antirrhinum is crossed with WW (white) one. Offspring RW are pink. This is an example of

1. hybrid
2. incomplete dominance
3. dominant-recessive
4. supplementary genes

Answer: (b)

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Q13. Which one of the following is an example of polygenic inheritance?

1. Pod shape in garden pea
2. Production of male honey bee
3. Skin colour in humans
4. Flower colour in Mirabilis jalapa

Answer: (c)

Polygenic inheritance is the inheritance of traits which are dependent on the no. of genes such as the skin colour of human beings, eg. AABB is black AaBB in neither dark nor black. AaBb is again wheat is Aabb is light and aabb is white colour.

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Q14. tt mates with Tt. What will be characteristic of offspring?

1. 25% recessive
2. 50% recessive
3. 75% recessive
4. All dominant

Answer: (b)

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Q15. In a plant, red fruit (R) is dominant over yellow fruit (r) and tallness (T) is dominant over shortness (t). If a plant with RRTt genotype is crossed with a plant that is rrtt,

1. 75% will be tall with red fruit
2. 50% will be tall with red fruit
3. 25% will be tall with red fruit
4. all of the offspring will tall with red fruit

Answer: (b)

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Q16. Punnett square is used to know the

1. probable result of a cross
2. outcome of a cross
3. types of gametes
4. result of meiosis

Answer: (a)

Punnett square is a checker-board used to show the result of a cross between two organisms. It was devised by geneticist, R.C. Punnett (1927). It depicts both genotype and phenotype of the progeny.

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Q17. Taylor conducted the experiments to prove semi conservative mode of chromosome replication on

1. Drosophila melanogaster
2. Vicia faba
3. E. coli
4. Vinca rosea

Answer: (b)

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Q18. An abnormal human baby with ‘XXX’ sex chromosomes was born due to :

1. fusion of two ova and one sperm
2. formation of abnormal ova in the mother
3. fusion of two sperms and one ovum
4. formation of abnormal sperms in the father

Answer: (b)

A human baby having abnormality with ‘XXX’ sex chromosomes is born due to evolution of abnormal ova inmother’s ovary. This is caused due to non-disjunction of X chromosome in the mother.

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Q19. Experimental evidences of chromosomal theory of inheritance was given by

1. TH Morgan
2. S Boveri
3. de Vries
4. W Sutton

Answer: (a)

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Q20. The ratio of phenotypes in $F_{2}$ of a monohybrid cross is

1. 1 : 2 : 1
2. 3 : 1
3. 9 : 3 : 3 : 1
4. 2 : 1

Answer: (b)

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Q21. Sex determination in grasshoppers, humans, and Drosophila is similar because

1. males have one X-chromosome and females have two X-chromosomes.
2. females are hemizygous.
3. all males always have one Y-chromosome in all three species.
4. the ratio of autosomes to sex chromosomes is the same in all three organisms.

Answer: (a)

Sex determination in grasshoppers, humans, and Drosophila is similar because males have one X-chromosome & females have two X-chromosomes. In these three species, females have two X-chromosomes and males have one X-chromosome.

The ratio of X-chromosomes to autosomes is important (and different in each organism) in Drosophila and grasshoppers, but not in humans. In all three species, males have one Y-chromosome, but the Y-chromosome is required for male fertility, not for Drosophila to be male (in Drosophila, male flies can be XO).

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Q22. Due to the cross between TTRr × ttrr the resultant progenies show what percent of tall, red flowered plants

1. 75%
2. 50%
3. 25%
4. 100%.

Answer: (b)

The cross can be represented as :

Tall Red TtRr - 50%

Tall White Ttrr - 50%

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Q23. Which one of the following conditions correctly describes the manner of determining the sex in the given example?

1. XO condition in human as found in Turner syndrome, determines female sex.
2. XO type of sex chromosomes determines male sex in grasshopper.
3. Homozygous sex chromosomes (ZZ) determines female sex in birds.
4. Homozygous sex chromosomes (XX) produces male in Drosophila.

Answer: (b)

In grasshopper the males lack a Y-sex chromosome and have only an X-chromosome. They produce sperm cells that contain either an X chromosome or no sex chromosome, which is designated as O.

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Q24. Mendel did not recognize the linkage phenomenon in his experiments because

1. Characters he studied were located on different chromosomes.
2. There were many chromosomes to handle.
3. He did not have powerful microscope.
4. He studied only pure plants.

Answer: (a)

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Q25. 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?

1. Blood group inheritance disease
2. Sex-influenced disease
3. Sex-linked disease
4. Sex-recessive disease

Answer: (c)

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Q26. Which cross shows very tight linkage?

1. Cross B
2. Cross A
3. Both (a) and (b)
4. None of these

Answer: (b)

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Q27. A common test to find the genotype of a hybrid is by

1. studying the sexual behaviour of $F_1$ progenies
2. crossing of one $F_2$ progeny with female parent
3. crossing of one $F_1$ progeny with male parent
4. crossing of one $F_2$ progeny with male parent.

Answer: (c)

A common test to find the genotypes of a hybrid is by crossing of one $F_1$ progeny with male parent.

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Q28. Linkage group is

1. non-linearly arranged group of linked gene
2. linearly arranged group of linked gene
3. non-linearly arranged group of unlinked gene
4. non-linearly arranged group of single gene

Answer: (b)

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Q29. Failure of segregation of chromatid during cell division cycle results in the gain or loss of chromosome which as called

1. hypopolyploidy
2. aneuploidy
3. hyperpolyploidy
4. polyploidy

Answer: (b)

Aneuploidy occurs when the chromatids fail to segregate during cell division, resulting in gain or loss of a chromosome.

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Q30. In the F2 -generation of a Mendelian dihybrid cross the number of phenotypes and genotypes are

1. phenotypes-9, genotypes-4
2. phenotypes-4, genotypes-16
3. phenotypes-4, genotypes-8
4. phenotypes-4, genotypes-9

Answer: (d)

Mendel's dihybrid cross is depicted below

Thus, from the cross, it can be determined that the number of phenotypes and genotypes in the F2 -generation of a Mendelian dihybrid cross is 4 and 9, respectively.

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Q31. Distance between the genes and percentage of recombination shows

1. an inverse relationship
2. a direct relationship
3. a parallel relationship
4. no relationship

Answer: (b)

The distance between the genes and percentage of recombination shows a direct relationship. This can be explained as when genes are close together, or are linked exhibit low recombination frequencies. And when they are far apart the recombination is high.

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Q32. In a plant, red fruit (R) is dominant over yellow fruit (r) and tallness (T) is dominant over shortness (t). If a plant with RRTt genotype is crossed with a plant that is rrtt,

1. 75% will be tall with red fruit
2. 50% will be tall with red fruit
3. 25% will be tall with red fruit
4. all the offspring will be tall with red fruit

Answer: (b)

Since red fruit colour is dominant over yellow fruit colour and tallness is dominant over shortness.

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Q33. A colour blind mother and normal father would have

1. all colour blind
2. colour blind sons and daughters
3. colour blind sons and normal/carrier daughters
4. all normal

Answer: (c)

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Q34. Which of the following is a correct match?

1. Sickle cell anaemia - X-chromosome
2. Down's syndrome - 21 st chromosome
3. Haemophilia - Y-chromosome
4. Parkinson's disease - X and Y chromosome.

Answer: (b)

Down's syndrome (Mongolian Idiocy, Mongolism) is caused by the presence of an extra chromosome number 21. Sickle cell anaemia is not a sex linked (i.e., X linked) disease but an autosomally inherited recessive trait.

Haemophilia is X-linked but not holandric/Y-linked. Parkinson's disease is a degenerative disease. It is not at all hereditary.

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Q35. Which one of the following conditions correctly describes the manner of determining the sex?

1. XO type of sex chromosomes determine male sex in grasshopper.
2. Homozygous sex chromosomes (ZZ) determine female sex in birds.
3. XO condition in humans as found in Turner's syndrome, determines female sex.
4. Homozygous sex chromosomes (XX) produce male in Drosophila.

Answer: (a)

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 : gymnosperms with X and angiosperms without X. Fertilisation of an egg by X-bearing sperm yields female offspring and by no X sperm yields male offspring.

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Q36. Linkage in Drosophila was first discovered by

1. Bateson and Punnett
2. Morgan
3. Sturtevant
4. Bridges

Answer: (b)

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Q37. Exchange of genetic material between chromatids of homologous chromosomes during meiosis is called

1. Chiasmata
2. Synapsis
3. Transformation
4. Crossing over

Answer: (d)

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Q38. Which of the following is not a hereditary disease?

1. Thalassemia
2. Cystic fibrosis
3. Haemophilia
4. Cretinism

Answer: (d)

Cystic fibrosis : It is a common disorder of the Caucasian race in which thick and more salty mucus blocks the respiratory tract. The homozygous recessive condition produces the defective protein which regulates chloride transport channel.

Cretinism : In this disorder the physical growth, mental growth and sexual growth in children is retarded. Such a dwarf and sterile child is called a cretin. It is due to hyposecretion of thyroid hormones.

Thalassemia : Due to defective production of a or b chains of haemoglobin, autosomal recessive.

Haemophilia : Sex linked disorder due to defective recessive gene.

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Q39. Who coined the term linkage?

1. Tschermak
2. Mendel
3. Sturtevant
4. T. H. Morgan

Answer: (d)

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Q40. A disease caused by an autosomal primary non-disjunction is

1. Turner’s Syndrome
2. Klinefelter’s Syndrome
3. Sickel Cell Anemia
4. Down’s Syndrome

Answer: (d)

Down ’s syndrome is caused by non-disjunction of $21^{st}$ chromosome i.e. Trisomy.

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Q41. Two nonallelic genes produces the new phenotype when present together but fail to do so independently then it is called

1. polygene
2. epistasis
3. non complementary gene
4. complementary gene.

Answer: (b)

Epistasis is the phenomenon of suppression of phenotypic expression of gene by a non-allelic 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.

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Q42. 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?

1. He would have discovered blending or incomplete dominance.
2. He could have mapped the chromosome.
3. He would not have discovered the law of independent assortment.
4. He would have discovered sex linkage.

Answer: (c)

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Q43. Which of the following is incorrect regarding ZW-ZZ type of sex determination?

1. 1 : 1 sex ratio is produced in the offsprings
2. Females are homogametic and males are heterogametic
3. It occurs in birds and some reptiles
4. All of these

Answer: (b)

In ZW-ZZ type of sex determination, the male has two homomorphic sex chromosomes (ZZ) and is homogametic, and the female has two heteromorphic sex chromosomes (ZW) and is heterogametic. There are, thus, two types of eggs: Z and W, and only one type of sperms, i.e., each with Z.

Fertilization of an egg with Z chromosome by a sperm with Z chromosome gives a zygote with ZZ chromosomes (male). Fertilization of an egg with W chromosome by a sperm with Z chromosome yields a zygote with ZW chromosomes (female). This mechanism operates in some vertebrates (fishes, reptiles and birds).

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Q44. A true breeding plant is

1. Near homozygous and produces offspring of its own kind
2. Produced due to cross-pollination among unrelated plants
3. Always homozygous recessive in this genetic constitution
4. One that is able to breed on its own

Answer: (a)

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Q45. If a colour-blind man marries a woman who is homozygous for normal colour vision, the probability of their son being colour-blind is

1. 0.75
2. 0.5
3. 0
4. 1

Answer: (c)

Genotype of colour blind man – $X^c$ Y

Genotype of woman homozygous – XX for normal woman

Hence, there is zero (0) probability of their son to be colour-blind.

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Q46. How many true breeding pea plant varieties did Mendel select as pairs, which were similar except in one character with contrasting traits?

1. 14
2. 2
3. 4
4. 8

Answer: (a)

Mendel conducted artificial pollination or cross pollination experiments using several true-breeding pea lines. He selected 14 true-breeding pea plant varieties, as pairs which were similar except for one character with contrasting traits. Some of the contrasting traits selected were smooth or wrinkled seeds, yellow or green seeds, inflated (full) or constricted green or yellow pods and tall or dwarf plants.

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Q47. In a family, father had a trait but mother did not. All their sons and daughter had this trait. The same trait was found in some grand daughters, though daughter were married to the normal persons. Choose the correct pedigree chart for this condition.

1. (a)
2. (b)
3. (c)
4. (d)

Answer: (a)

Figure is as the case is that of criss-cross inheritance. It is a type of sex-linked inheritance, where a parent passes the traits to the grand child of the same sex through offspring of the opposite sex. The father passes the traits to grandson through his daughter (diagynic), while the mother transfers traits to her grand daughter through her son (dia-andric)

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Q48. A certain road accident patient with unknown blood group needs immediate blood transfusion his doctor friend at ones offers his blood what was the blood group of the donor?

1. Blood group B
2. Blood group A
3. Blood group O
4. Blood group AB

Answer: (c)

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Q49. Phenotype of an organism is the result of

1. mutations and linkages
2. genotype and environment interactions
3. cytoplasmic effects and nutrition
4. environmental changes and sexual dimorphism.

Answer: (b)

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.

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Q50. Identify the wrong statement with reference to the gene 'I' that controls ABO blood groups.

1. When $I^{A}$ and $I^{B}$ are present together, they express same type of sugar.
2. A person will have only two of the three alleles.
3. The gene (I) has three alleles.
4. Allele i does not produce any sugar

Answer: (a)

ABO blood groups are controlled by the gene I. The gene (I) has three alleles $I^A, I^B$ and i. The alleles $I^A$ and $I^B$ produce a slightly different form of the sugar while allele i does not produce any sugar. Because humans are diploid organisms, each person possesses any two of the three I gene alleles. When $I^A$ and $I^B$ are present together they both express their own types of sugars,because of co-dominance.

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Q51. Match the following columns.

Codes A B C D

1. 4 3 2 1
2. 1 2 3 4
3. 1 3 2 4
4. 2 1 4 3

Answer: (c)

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Q52. Which one of the following conditions though harmful in itself, is also potential saviour from a mosquito borne infectious disease ?

1. Sickle cell anaemia
2. Thalassaemia
3. Pernicious anaemia
4. Leukaemia

Answer: (a)

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Q53. Sickle cell anaemia is caused by the substitution of Glutamic acid (Glue) by ________ at the sixth position of the beta globin chain of the haemoglobin molecule.

1. Gly (Glycine)
2. Asn (Aspargine)
3. Arg (Argenine)
4. Val (Valine)

Answer: (d)

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Q54. The "Cri-du-Chat" syndrome is caused by change in chromosome structure involving

1. inversion
2. duplication
3. deletion
4. translocation

Answer: (c)

Cri-du-chat/cat cry syndrome is due to the deletion of a large part of the small or one of the 5th chromosome.

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Q55. Which contribute to the success of Mendel?

1. Observation of distinct inherited traits
2. Qualitative analysis of data
3. His knowledge of biology
4. Consideration of one character at one time

Answer: (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.

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Q56. Which is the most common mechanism of genetic variation in the population of sexually reproducing organism?

1. Genetic drift
2. Chromosomal aberrations
3. Recombination
4. Transduction

Answer: (c)

The most common cause of variations is recombination in the organism which are reproduced by sexual way.

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Q57. A woman has an X-linked condition on one of her X-chromosomes. This chromosome can be inherited by

1. Only sons
2. Only grand children
3. Only daughters
4. Both (a) and (c)

Answer: (d)

In the given problem, the woman has an X-linked condition and she can transmit the carrier allele to both her son and daughter. The resulting son will affected because X-linked disorder always affect males as males contain a single X-chromosome. The daughter offspring will be a carrier, but not diseased because females are affected by X-linked disorder in homozygous recessive condition only, i.e. two recessive alleles are required.

Hence, out of the 4 offspring possible 25% of sons are diseased and 25% are normal. Similarly, 50% daughters normal out of which half are carriers.

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Q58. Mental retardation in man associated with sex chromosomal abnormality is usually due to

1. increase in size of Y-chromosome.
2. increase in size of X-chromosome.
3. increase in number of Y-chromosome.
4. increase in number of X-chromosome.

Answer: (d)

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Q59. The mechanism that causes a gene to move from one linkage group to another is called

1. translocation
2. duplication
3. inversion
4. crossing-over

Answer: (a)

Translocation is a chromosomal abnormality caused by rearrangement of parts between nonhomologous chromosomes. It may cause a gene to move from one linkage group to another.

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Q60. Study the pedigree chart given below:

What does it show?

1. Inheritance of a condition like phenylketonuria as an autosomal recessive trait
2. Inheritance of a sex-linked inborn error of metabolism like phenylketonuria
3. Inheritance of a recessive sex-linked disease like haemophilia
4. The pedigree chart is wrong as this is not possible

Answer: (a)

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Q61. Which one of the following conditions correctly describes the manner of determining the sex in the given example?

1. XO type of sex chromosomes determine male sex in grasshopper
2. Homozygous sex chromosomes (ZZ) determine female sex in birds.
3. XO condition in human as found in Turner syndrome, determines female sex.
4. Homozygous sex chromosomes (XX) produce male in Drosophila.

Answer: (a)

In grasshopper the males lack a Y-sex chromosome and have only an X-chromosome. They produce sperm cells that contain either an X chromosome or no sex chromosome, which is designated as O.

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Q62. 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?

1. Sickle cell anaemia
2. Phenylketonuria
3. Haemophilia
4. Thalassemia

Answer: (c)

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Q63. Which of the following will not result in variations among siblings?

1. Crossing over
2. Independent assortment of genes
3. Linkage
4. Mutation

Answer: (c)

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Q64. 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?

1. AABb
2. AaBB
3. AABB
4. AaBb.

Answer: (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.

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Q65. A gene showing co-dominance has

1. one allele dominant on the other
2. both alleles independently expressed in the heterozygote
3. alleles that are recessive to each other
4. alleles tightly linked on the same chromosome

Answer: (b)

The phenomenon of expression of both the alleles in a heterozygote is called co-dominance. The alleles which do not show dominance-recessive relationship and are able to express themselves independently when present together are called co-dominant alleles.

As a result the heterozygous condition has a phenotype different from either of homozygous genotypes, e.g., alleles for blood group A ($I^A$) and for blood group B ($I^B$) are codominant so that when they come together in an individual, they produce blood group AB.

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Q66. ZZ/ZW type of sex determination is seen in

1. Snails
2. Platypus
3. Cockroach
4. Peacock

Answer: (a)

In ZZ/ZW case, the female has heteromorphic (ZW) sex chromosomes. Thus peacock shows ZZ/ZW sex determination type.

In Platypus the sex determination is of XX-XY type. In snails the sex determination is environmentally induced, while in cockroaches it is of XX-XO types.

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Q67. A gene is said to be dominant if

1. it expresses its effect only in heterozygous condition
2. it expresses its effect only in homozygous state
3. it expresses its effect both in homozygous and heterozygous condition
4. it never expresses its effect in any conditions

Answer: (c)

A dominant gene would lead to the expression of its phenotype irrespective of the fact whether its allelic gene is dominant or recessive.

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Q68. A pleiotropic gene

1. controls multiple traits in an individual
2. controls a trait only in combination with another gene
3. is expressed only in primitive plants
4. is a gene evolved during Pliocene.

Answer: (a)

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.

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Q69. A tall true breeding garden pea plant is crossed with a dwarf true breeding garden pea plant. When the $F_1$ plants were selfed the resulting genotypes were in the ratio of

1. 3 : 1 : : Dwarf : Tall
2. 3 : 1 : : Tall : Dwarf
3. 1 : 2 : 1 : : Tall homozygous : Tall heterozygous : Dwarf
4. 1 : 2 : 1 : : Tall heterozygous : Tall homozygous : Dwarf.

Answer: (c)

When a tall true breeding garden pea plant is crossed with a dwarf true breeding garden pea plant and the $F_1$ plants were selfed the resulting genotypes were in the ratio of 1 : 2 : 1

i.e., Tall homozygous : Tall heterozygous : Dwarf

It can be illustrated as given below:

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Q70. Which of the following is genetically dominant in man?

1. Rh positive
2. Colour blindness
3. Haemophilia
4. Albinism

Answer: (a)

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Q71. Red green colourblindness is a sex linked trait. Which of the given statements is not correct regarding colourblindness?

1. Males can be carriers of the trait
2. Homozygous recessive condition is required for the expression of colourblindness in females
3. It is more common in males than in females
4. Colourblind women always have colourblind father and always produce colourblind son.

Answer: (a)

Since colourblindness is a sex-linked recessive trait and males just have one X chromosome, they can never be the carriers. Males will always express the disease/phenotype.

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Q72. Person having genotype $I^A I^B$ would show the blood group as AB. This is because of

1. codominance
2. pleiotropy
3. segregation
4. incomplete dominance

Answer: (a)

ABO blood grouping in humans is an example of codominance.

When $I^A$ and $I^B$ are present together, both express equally and produce the surface antigens A and B. Thus, a person having genotype $I^A I^B$ would show the blood group as AB.

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Q73. The movement of a gene from one linkage group to another is called

1. crossing over
2. translocation
3. inversion
4. duplication.

Answer: (b)

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Q74. Which one from those given below is the period for Mendel's hybridisation experiments?

1. 1870-1877
2. 1857-1869
3. 1840-1850
4. 1856-1863

Answer: (d)

Mendel carried out hybridisation experiments on garden pea for 7 years from 1856-1863.

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Q75. Match the following columns.

Codes A B C

1. 2 3 1
2. 2 1 3
3. 3 2 1
4. 1 3 2

Answer: (a)

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Q76. How many pairs of true breeding varieties were selected by Mendel for his experiment on pea plant?

1. 13
2. 12
3. 14
4. 15

Answer: (c)

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Q77. A fruit fly heterozygous for sex-linked genes, is mated with normal female fruit fly. Male specific chromosome will enter egg cell in the proportion

1. 3 : 1
2. 2 : 1
3. 1 : 1
4. 7 : 1

Answer: (c)

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Q78. Down's syndrome is due to

1. sex-linked inheritance
2. linkage
3. crossing over
4. non-disjunction of chromosomes

Answer: (d)

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Q79. Select the correct statement from the ones given below with respect to dihybrid cross.

1. Genes far apart on the same chromosome show very few recombinations
2. Tightly linked genes on the same chromosome show higher recombinations
3. Genes loosely linked on the same chromosome show similar recombinations as the tightly linked ones
4. Tightly linked genes on the same chromosome show very few recombinations

Answer: (d)

Linkage is the inheritance of genes of same chromosome together and capacity of these genes to retain their parental combination in subsequent generation. The strength of linkage between two genes is inversely proportional to the distance between the two.

This means, two linked genesis how higher frequency of recombination if the distance between them is higher and lower frequency if the distance is smaller.

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Q80. Which one from those given below is the period of Mendel's hybridisation experiments?

1. 1840-1850
2. 1856-1863
3. 1857-1869
4. 1870-1877

Answer: (b)

Mendel performed his hybridisation experiment on Pisum sativum (garden pea) for 7 years between 1856-1863.

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Q81. It is said that Mendel proposed that the factor controlling any character is discrete and independent. This proposition was based on the

1. observations that the offspring of a cross made between the plants having two contrasting characters shows only one character without any blending
2. results of $F_3$ -generation of a cross
3. self-pollination of $F_1$ offsprings
4. cross-pollination of $F_1$ -generation with recessive parent

Answer: (a)

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Q82. Which Mendelian idea is depicted by a cross in which the $F_{1}$ generation resembles both the parents?

1. Inheritance of one gene
2. Law of dominance
3. Co-dominance
4. Incomplete dominance

Answer: (c)

In co-dominance, $F_{1}$ generation resemble both the parents. e.g., Blood group inheritance.

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Q83. Multiple alleles control inheritance of

1. sickle cell anaemia
2. colour blindness
3. phenylketonuria
4. blood groups

Answer: (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 - $I^A I^B$ and $I^O$. A person is having only two of these three alleles and blood type can be determined.

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Q84. 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 chance that this child would be colour blind?

1. zero percent
2. 100%
3. 25%
4. 50%

Answer: (a)

If a normal visioned man marries a woman whose father was also colour blind. Then his wife would be carried of this disease if her mother was normal. This trait passed into children but daughters produce by this couple are carrier not the colour blind. 50% of sons would be colour blind.

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Q85. Which contribute to the success of Mendel?

1. His knowledge of biology
2. Observation of distinct inherited traits
3. Qualitative analysis of data
4. Consideration of one character at one time

Answer: (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.

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Q86. Extranuclear inheritance occurs in

1. mitochondria and ribosome
2. chloroplast and mitochondria
3. peroxisome and ribosome
4. chloroplast and lysosome

Answer: (b)

In eukaryotic cells, two cytoplasmic organelles, mitochondria and chloroplast of green plants, contain their own genetic materials.

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Q87. If a colour blind woman marries a normal visioned man, their sons will be

1. all normal visioned
2. all colour blind
3. one-half colour blind and one-half normal
4. three-fourths colour blind and one-fourth normal.

Answer: (b)

Colour blindness is a recessive sex-linked trait.

All sons will be colour blind and all daughters will be carriers.

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Q88. Mental retardation in man, associated with sex chromosomal abnormality is usually due to

1. reduction in X complement
2. large increase in Y complement
3. moderate increase in Y complement
4. increase in X complement

Answer: (d)

In humans, sex chromosomal abnormality is due to gene carried on X-chromosome. Increase in X-complement leads to Klinefelter's syndrome. The individiuals of Klinefelter's syndrome has 47 chromosomes (44 + XXY), this 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).

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Q89. According to Mendel's observation, which generation of progeny always represents the phenotype of the dominant parent?

1. $F_2$
2. $F_4$
3. $F_1$
4. $F_0$

Answer: (c)

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Q90. The amino acid Tryptophan is the precursor for the synthesis of:

1. Thyroxine and Triiodothyronine
2. Melatonin and Serotonin
3. Estrogen and Progesterone
4. Cortisol and Cortisone

Answer: (b)

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Q91. Mendel selected pea as material for his experiments because

1. the flowers are self-pollinated.
2. it is an annual plant with comparatively short life cycle.
3. the number of seeds produced is quite large.
4. all of the above.

Answer: (d)

Mendel selected Garden pea as material for his hybridization experiments because of the following reasons:

1. Hybridizationor crossing in pea is easy.
2. It has bisexual flowers.
3. It has a number of well defined contrasting characters.
4. It shows predominantly self - fertilization.
5. It has a short life span.

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Q92. In the following human pedigree, the filled symbols represent the affected individuals. Identify the type of given pedigree.

1. Autosomal recessive
2. X- linked recessive
3. X-linked dominant
4. Autosomal dominant

Answer: (a)

Autosomal recessive is a type of disorder in which two copies of an abnormal gene must be found for the disease in the affected person.

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Q93. In honeybees, male and female gametes are produced through

1. mitosis and meiosis, respectively
2. mitosis
3. meiosis
4. meiosis and mitosis, respectively

Answer: (a)

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Q94. Thalassemia in humans

1. can transmit from parents to offspring when both parents are unaffected carriers (heterozygous)
2. is an autosome linked recessive blood disorder
3. caused due to the mutation or deletion of one of the α or β-globin chain
4. All of the above

Answer: (d)

Thalassemia is also an autosome linked recessive blood disease transmitted from parents to the offspring when both the partners are unaffected carrier for the gene (or heterozygous).

The defect could be due to either mutation or deletion which ultimately results in reduced rate of synthesis of one of the globin chains (α and β-chains) that make up haemoglobin. This causes the formation of abnormal haemoglobin molecules resulting into anaemia which is characteristic of the disease.

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Q95. Which one of the following conditions correctly describes the manner of determining the sex?

1. XO condition in humans as found in Turner's syndrome, determines female sex.
2. XO type of sex chromosomes determine male sex in grasshopper.
3. Homozygous sex chromosomes (ZZ) determine female sex in birds.
4. Homozygous sex chromosomes (XX) produce male in Drosophila

Answer: (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. Fertilisation of an egg by X-bearing sperm yields female offspring and by no X bearing sperm yields male offspring.

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Q96. If a genetic disease is transferred from a phenotypically normal but carrier female to only same of the male progeny, the disease is

1. autosomal recessive
2. Autosomal dominant
3. sex-linked dominant
4. sex-linked recessive

Answer: (d)

Most sex-linked (X-linked) conditions are recessive. This means that person having two X-chromosomes (females), both copies of a gene (i,e., one on each X-chromosome), must have a change or mutation whereas in a person with one X- chromosomes (males),only one copy or a gene must have a mutation. A female with a mutation in one copy of a gene on the X chromosome is said to be a ‘carrier’ for an X- linked condition.

For X-linked recessive disorders, and unaffected carrier mother who has a mutation in a gene on the X-chromosome can transfer either the X- chromosome with this mutation or a normal X-chromosome to her children. The pattern of inheritance of a condition directly or indirectly due to a dominant faulty gene located on autosome is known as autosomal dominant inheritance.The condition caused directly or indirectly due to a recessive faulty gene copy on autosome is known as autosomal recessive inheritance.

Rare trait that is caused by single abnormal gene on the X-chromosome is called sex-linked dominant.

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Q97. The contrasting pairs of factors in Mendelian crosses are called

1. alloloci
2. allelomorphs
3. multiple alleles
4. paramorphs

Answer: (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.

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Q98. At a particular locus, frequency of allele A is 0.6 and that of allele a is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium?

1. 0.24
2. 0.16
3. 0.36
4. 0.48

Answer: (d)

In a stable population, for a gene with two alleles, 'A' (dominant) and 'a' (recessive), if the frequency of 'A' is p and the frequency of 'a' is q, then the frequencies of the three possible genotypes (AA, Aa and aa) can be expressed by the Hardy-Weinberg equation:

$p^2 + 2pq + q^2$ = 1

where $p^2$ = Frequency of AA (homozygous dominant) individuals $q^2$ = Frequency of aa (homozygous recessive) individuals

2pq = Frequency of Aa (heterozygous) individuals

so, p = 0.6 and q = 0.4 (given)

∴ 2pq (frequency of heterozygote) = 2 × 0.6 × 0.4 = 0.48.

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Q99. Human skin colour is controlled by several gene pairs. Let us assume here that there are just three gene pairs on different chromosomes and that for each pair there are two alleles-an incompletely dominant one that codes for melanin deposition and an incompletely recessive one that codes for no melanin deposition. If a very dark-skinned person marries a very light-skinned woman, what will be the chance that their offspring will have very dark skin?

1. 1/4
2. 0
3. 5/8
4. 9/64

Answer: (b)

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Q100. A character which is expressed in a hybrid is called

1. recessive
2. dominant
3. co-dominant
4. epistatic

Answer: (b)

A character which is expressed in a hybrid is called dominant character. It is an inherited character expressed by a dominant gene in the $F_{1}$ generation.

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