Chapter 3 - Principles of Inheritance and Variation - 1 File ...

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SECTION - A

Objective Type Questions

(Mendel’s Laws of Inheritance, Inheritance of One Gene, Inheritance of Two Genes)

1. When a pink flowered Antirrhinum plant is test crossed, then phenotypic ratio in resulting progenies is

(1) 1 Red : 1 White (2) 3 Red : 1 White (3) 2 Pink : 1 White (4) 1 Pink : 1 White

Sol. Answer (4)

2. Heterozygous tall and violet flowered pea plants were selfed and total 512 seeds are collected. What will be

total number of seeds for both heterozygous traits?

(1) 128 (2) 256 (3) 384 (4) 64

Sol. Answer (1)

Number of individuals 512

4

TtRr16

4512 128

16

3. Mark the odd one (w.r.t. F2 generation of Mendelian dihybrid cross)

(1) Frequency of TtRR genotype = 12.5% (2) Frequency of ttrr genotype = 6.25%

(3) Frequency of TTRR genotype = 6.25% (4) Frequency of ttRr genotype = 25%

Sol. Answer (4)

2 1ttRr 12.5%

16 8

Level - II

Solutions

Chapter 3

Principles of Inheritance

and Variation

24 Principles of Inheritance and Variation Solutions of Assignment (Level-II)


4. Find out the frequency of AabbCcDdee if parents are AabbCCddEe and AabbccDdee

(1) 0.78% (2) 12.5% (3) 25% (4) 50%

Sol. Answer (2)

AabbCCddEe × AabbCcDdee

AabbCcDdee

2 2 2 812.5

4 4 4 64

5. In incomplete dominance

(1) Dominant trait is completely expressed in F1 generation

(2) Phenotypic and genotypic ratio are different

(3) Two dominant alleles are needed to express the complete dominant trait

(4) F1 individuals have the equal traits of both parents

Sol. Answer (3)

6. Progeny with blood group ‘O’ can not be obtained in cross

(1) A×A (2) A×B (3) O×AB (4) B×B

Sol. Answer (3)

O × AB

ii IAIB

IAi, IBi So no O blood group

7. Which of the following parental combination has produced mutant offspring?

(1) Tt × tt = Tt (2) tt × tt = Tt (3) Tt × Tt = tt (4) TT × tt =Tt

Sol. Answer (2)

tt × tt = Tt

8. Which of the following combination seems to have some linkage in character selected by Mendel?

(1) Stem height and pod colour (2) Flower colour and flower position

(3) Seed shape and seed colour (4) Plant height and pod shape

Sol. Answer (4)

Plant Height and Pod shape

9. A diploid organism is heterozygous for five loci and homozygous for 2 loci, how many types of gametes can

be produced?

(1) 128 (2) 32 (3) 4 (4) 14

Sol. Answer (2)

25Solutions of Assignment (Level-II) Principles of Inheritance and Variation


10. Lesch Nyhan disease is an X-linked recessive disorder that causes neurological damage in human beings. A

survey of 500 mates from a caucasion population revealed that 20 were effected with this disorder. What is

the frequency of the normal allele in this population?

(1) 9.6 (2) 0.8 (3) 0.096 (4) 96

Sol. Answer (2)

2 20 1q 0.2

500 25

q = 0.2

p + q = 1

p = 0.8

11. How many types of zygotic combinations are possible between a cross Aa BB Cc Dd × AA bb Cc DD?

(1) 32 (2) 128 (3) 64 (4) 16

Sol. Answer (4)

Aa BB Cc Dd × AA bb Cc DD

Number of gametes

2n 23 × 21

8 × 2 = 16

12. In F2 generation of a Mendelian dihybrid cross (TTRR × ttrr)

(1) Tall plants and violet flowered plants are obtained in 1 : 1 frequency

(2) Ratio of parental and non-parental plants is 1 : 15

(3) Recombinant plants are obtained in 1 : 1 frequency

(4) More than one option is correct

Sol. Answer (4)

(Two Genes Interaction w.r.t. Post-Mendelism, Sex Determination, Mutation)

13. Morgan hybridised yellow-bodied, white-eyed females to brown-bodied, red-eyed males and intercrossed their

F1 progeny. He observed that

(a) F2 ratio was deviated very significantly from the 9 : 3 : 3 : 1 ratio

(b) Both genes did not segregate independently of each other

(c) Recombinant types are not obtained in F2 generation

(d) Both genes segregate independently of each other

Select the correct set of statements :

(1) (a) & (b) only (2) (b) & (c) only (3) (b) & (d) only (4) (c) & (d) only

Sol. Answer (1)

y+w+

yw

Brown body and red eye are linked gene so F2 ratio deviated from 9 : 3 : 3 : 1.



14. ___(A)___ used the frequency of recombination between gene pairs on the ___(B)___ as a measure of the

distance between genes and mapped their position on the chromosome.

(A) (B)

(1) Morgan Same chromosome

(2) Sturtevant Different chromosomes

(3) Morgan Different chromosomes

(4) Sturtevant Same chromosome

Sol. Answer (4)

15. While solving the problem of sex determination in large number of insects, it was observed that

(1) All eggs lack sex chromosome

(2) Some of the sperms bear the X-chromosome

(3) All eggs as well as sperms bear the X-chromosome

(4) Some of the eggs bear the X-chromosome

Sol. Answer (2)

Insects XX

XO

A + X+

A + X A + O

16. Loss or gain of a segment of DNA results in

(1) Frame-shift muation (2) Point mutation

(3) Polyploidy (4) Chromosomal aberration

Sol. Answer (4)

17. Which one of the following is a physical factor that induce mutation?

(1) Acridines (2) HNO2

(3) UV-rays (4) Base analogue

Sol. Answer (3)

UV-rays and Non-ionised radiation.

18. The chromosome maps are not accurate maps because

(1) Crossing over frequency is higher than recombination frequency

(2) One crossing over interferes and increases the frequency of nearby crossing over

(3) Crossing over frequency decreases towards the ends of chromosome

(4) Heterochromation increases crossing over

Sol. Answer (1)

19. If a agouti mice (CcAa) is crossed with albino mice (ccAA), then how many albino mice are produced in

resulting progeny?

(1) 4 (2) 9 (3) 2 (4) 3

Sol. Answer (3)



CcAa × ccAA

Agouti Albino

CcAA

CA

CcAa

Ca

ccAA

cA

ccAa

ca

cA

Agouti Agouti Albino Alibino

20. Epistasis and dominance are respectively

(1) Intragenic, Intergenic (2) Non-allelic, Extra-allelic

(3) Non-allelic, Interallelic (4) Intergenic, Non-allelic

Sol. Answer (3)

21. In a complimentary gene interaction calculate the number of phenotype and genotype produced in a cross AaBb

× aaBB

(1) 1 phenotype, 2genotypes

(2) 2 phenotypes, 4 genotypes



Sol. Answer (2)

AB

AaBB AaBb

Ab

aaBB

aB

aaBb

ab

aB

1 1 1 1

Phenotype Phenotype

Genotype

22. How many types of gametes will be produced by a O Drosophila having following arrangement of

two genes (y+ and w+) on X-chromosome?

X Y

y+

w+

(1) 2 (2) 4 (3) 1 (4) 8

Sol. Answer (1)

y+w+

YX

23. If interference is complete or cent percent then the frequency of observed double crossover will be

(1) Equal to expected frequency

(2) Greater than expected frequency

(3) Lesser than expected frequency

(4) Zero

Sol. Answer (4)



(Genetic Disorders)

24. In the given pedigree, indicate whether the shaded symbols indicate dominant or recessive allele.

(1) Recessive (2) Codominant

(3) Dominant (4) It can be recessive or dominant both

Sol. Answer (4)

If propositus is Aa

Then it will be dominant pedigree

If is aa then it will be recessive pedigree.

25. In which of the following disorder a single protein that is a part of the cascade of proteins involved in blood

clotting is affected?

(1) Thalassemia (2) Sickle-cell anaemia (3) Haemophilia (4) Phenylketonuria

Sol. Answer (3)

Haemophilia A Blood clot factor VIIIth is absent

Haemophilia B Blood clot factor IXth is absent

26. Mark the correct statement (w.r.t. sickle cell-anaemia)

(1) Homozygous individuals for HbS are apparently unaffected

(2) Heterozygous individuals exhibit sickle-cell trait

(3) Heterozygous individuals are affected as well as carrier

(4) Homozygous individuals for HbA show the diseased phenotype

Sol. Answer (2)

HbA HbS Sicklecell trait

27. The defect sickle-cell anaemia is caused by the ______ of glutamic acid by valine at the 6th position of the

______ globin chain of the haemoglobin molecule.

(1) Substitution, (2) Deletion, (3) Duplication, (4) Translocation,

Sol. Answer (1)

Substitution of Pyrimidine ��⇀↽��

Purine

DNA CTC CAG

RNA GAG GUG

AA Glutanic

acid

Valine



28. A Y-linked gene is responsible for hypertrichosis (long hair on ears). When an affected man marries a normal

woman, what percentage of their daughters would be expected to have hairy ears?

(1) 25% (2) 0% (3) 50% (4) 100%

Sol. Answer (2)

Daughter will not receive Y chromosome from father.

29. A normal woman, whose father had colour blindness, married a normal man. What is the chance of occurrence

of colour blindness in the progeny?

(1) 25% (2) 50% (3) 100% (4) 75%

Sol. Answer (1)

XXC × XY

XX

X

XXC

X YC

XY

X

Y

XC

25% of progeny will be colour blind.

30. Mr. Stevan is suffering from haemophilia and cystic fibrosis. His father is hetrozygous for cystic fibrosis. The

probability of Stevan's sperm having recessive X-linked as well as autosomal allele is

(1)1

4(2)

1

16(3)

1

2(4)

1

8

Sol. Answer (3)

AhAC Xh Y

AC Xh 1

2

AC Y 1

2

31. Select the incorrect one (w.r.t. reciprocal cross)

(1) To know whether the alleles are present on sex chromosomes or autosomes

(2) It is made to eliminate the effect of nuclear traits

(3) Two individuals with contrast genotypes are involved

(4) Results are not changed for autosomal traits

Sol. Answer (2)

32. In Lathyrus odoratus, hybrid blue flowered and long pollen plant is test crossed with homozygous recessive

red flowered and round pollen plant then how many parental types are obtained when genes are present in

cis stage in parents?

(1) 50% (2) 43.7% (3) 87.4% (4) 12.6%

Sol. Answer (3)

BbLl × bb ll

BbLl

BL

Bbll

Bl

bbLl

bL

bbll

Bl

bl

7 1 1 7

14

100 87.4%16



33. Match the following - (w.r.t. Pedigree analysis)

Column - I Column - II

a. Solid symbol (i) Carrier of sex linked trait

b. Horizontal line (ii) Offspring

between symbols

c. Horizontal line (iii) Trait to be studied

above the symbols

d. Dot in centre (iv) Parents

(1) a(iv), b(iii), c(ii), d(i) (2) a(ii), b(iii), c(iv), d(i)

(3) a(iii), b(iv), c(ii), d(i) (4) a(i), b(ii), c(iv), d(iii)

Sol. Answer (3)

34. Select incorrect statement

(A) Linked genes cause absolute lethality

(B) Persons affected by PKU do not show mental disorder

(C) F2 ratio in codominance and incomplete dominance are same

(D) Sex of male Drosophila is dependent on Y-chromosome

(1) (A) & (B) (2) (B) & (C) (3) (A), (B) & (D) (4) All of these

Sol. Answer (3)

35. In phenylketonuria

(1) Break down of phenylalanine is rapid

(2) Accumulation of phenylalanine in body

(3) Chromosomal constitution of patient changes

(4) TSD gene situated on chromosome 15 undergoes mutation

Sol. Answer (2)

Phenylalanine hydroxylase enzyme is absent.

36. Select the odd one out w.r.t. non-allelic gene interactions

(1) Epistasis (2) Duplicate genes

(3) Incomplete dominance (4) Complementary genes

Sol. Answer (3)

Incomplete dominance show interallelic gene interaction.

37. Fruit colour in Cucurbita pepo is an example of

(1) Complementary genes (2) Duplicate genes

(3) Dominant epistasis (4) Polymeric genes

Sol. Answer (3)

38. Complementary genes were demonstrated by Bateson and Punnet in

(1) Capsella (2) Lathyrus odoratus

(3) Antirrhinum (4) Mirabilis

Sol. Answer (2)



39. If dominant alleles of two gene loci produce the same phenotype whether inherit separately or together, it will

be

(1) Recessive epistasis (2) Dominant epistasis

(3) Duplicate genes interaction (4) Inhibitory genes interaction

Sol. Answer (3)

40. A gene which hides the action of another gene is termed as

(1) Co-dominant gene (2) Epistatic gene

(3) Hypostatic gene (4) Lethal gene

Sol. Answer (2)

41. In polymeric gene action, the modified dihybrid phenotypic ratio in F2 generation is

(1) 9 : 3 : 3 : 1 (2) 13 : 3

(3) 9 : 6 : 1 (4) 12 : 3 : 4

Sol. Answer (3)

42. Which of the following genotype of sweet pea plant is related with the production of purple coloured flowers?

(1) CcPp (2) CCpp

(3) ccPP (4) Ccpp

Sol. Answer (1)

CcPp = Purple flower.

43. Select the odd one out w.r.t. polygenic inheritance

(1) Bell-shaped curve is obtained (2) Also called quantitative inheritance

(3) Recessive alleles show cumulative effect (4) Intermediate phenotypes are more frequent

Sol. Answer (3)

44. Select the correct match (w.r.t. dihybrid phenotypic ratio in F2 generation)

(1) Recessive epistasis : 12 : 3 : 1

(2) Dominant epistasis : 9 : 3 : 4

(3) Collaborative gene : 9 : 3 : 3 : 1

(4) Duplicate genes : 9 : 7

Sol. Answer (3)

45. Skin colour in man is controlled by

(1) Three pairs of polygenes (2) Duplicate genes

(3) Six pairs of polygenes (4) Supplementary genes

Sol. Answer (1)

46. Holandric genes are present on

(1) X-chromosomes (2) Y-chromosomes

(3) Sex-chromosomes as well as autosomes (4) Autosomes

Sol. Answer (2)



47. Select the odd one out w.r.t. genic balance theory of sex-determination in Drosophila

(1) Y-chromosome plays no role in sex-determination

(2) Given by C.B. Bridges

(3) If X/A ratio is one, superfemales are produced

(4) If X/A ratio is less than 0.5, supermales are produced

Sol. Answer (3)

X/A ratio = 1 femaleness.

48. Environmental mechanism of sex-determination is seen in

(1) Bonnelia (2) Crepidula

(3) Grasshopper (4) More than one option is correct

Sol. Answer (4)

Bonnelia and Crepidula.

49. Select the correct match

(1) Sex-limited trait – Colour blindness

(2) Sex-limited trait – Express in both sexes

(3) Sex-influenced trait – More frequent in one sex than in the other

(4) Sex-influenced trait – Porcupine skin

Sol. Answer (3)

50. All are sex limited traits, except

(1) Beard in man (2) Porcupine skin

(3) Antlers in male deer (4) Brilliant plumage in peacock

Sol. Answer (2)

Porcupine skin is a Y linked trait.

51. Which of the following chemical is a base analogue?

(1) 5-bromouracil (2) Acridines

(3) Nitrous acid (4) Hypoxanthine

Sol. Answer (1)

52. Cytoplasmic male sterility in maize is due to defective

(1) Mitochondria (2) Lysosome

(3) Golgi body (4) Leucoplast

Sol. Answer (1)

53. Find odd one (w.r.t. dominant traits in humans)

(1) Blue eyes

(2) Brown eyes

(3) Free ear lobes

(4) Myotonic dystrophy

Sol. Answer (1)



54. Which of the following abnormalities is due to X-linked recessive mutation?

(1) Cystic fibrosis

(2) Thalassaemia

(3) Klinefelter's syndrome

(4) Lesch-Nyhan syndrome

Sol. Answer (4)

SECTION - B

Previous Years Questions

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

[NEET-2018]

(1) Only daughters

(2) Only sons

(3) Both sons and daughters

(4) Only grandchildren

Sol. Answer (3)

• Woman is a carrier

• Both son & daughter inherit X–chromosome

• Although only son be the diseased

2. Which of the following characteristics represent ‘Inheritance of blood groups’ in humans? [NEET-2018]

a. Dominance b. Co-dominance

c. Multiple allele d. Incomplete dominance

e. Polygenic inheritance

(1) b, c and e (2) a, b and c (3) a, c and e (4) b, d and e

Sol. Answer (2)

IAIO, IBIO - Dominant–recessive relationship

IAIB - Codominance

IA, IB & IO - 3-different allelic forms of a gene (multiple allelism)

3. Which of the following pairs is wrongly matched? [NEET-2018]

(1) Starch synthesis in pea : Multiple alleles

(2) ABO blood grouping : Co-dominance

(3) T.H. Morgan : Linkage

(4) XO type sex determination : Grasshopper

Sol. Answer (1)

Starch synthesis in pea is controlled by pleiotropic gene.

Other options (2, 3 & 4) are correctly matched.



4. Select the correct statement [NEET-2018]

(1) Franklin Stahl coined the term ‘‘linkage’’

(2) Punnett square was developed by a British scientist

(3) Transduction was discovered by S. Altman

(4) Spliceosomes take part in translation

Sol. Answer (2)

Punnett square was developed by a British geneticist, Reginald C. Punnett.

– Franklin Stahl proved semi-conservative mode of replication.

– Transduction was discovered by Zinder and Laderberg.

– Spliceosome formation is part of post-transcriptional change in eukaryotes

5. A disease caused by an autosomal primary non-disjunction is [NEET-2017]

(1) Down's syndrome (2) Klinefelter's syndrome

(3) Turner's syndrome (4) Sickle cell anemia

Sol. Answer (1)

Down’s syndrome is caused by non-disjunction of 21st chromosome.

6. Thalassemia and sickle cell anemia are caused due to a problem in globin molecule synthesis. Select the

correct statement. [NEET-2017]

(1) Both are due to a qualitative defect in globin chain synthesis

(2) Both are due to a quantitative defect in globin chain synthesis

(3) Thalassemia is due to less synthesis of globin molecules

(4) Sickle cell anemia is due to a quantitative problem of globin molecules

Sol. Answer (3)

Thalassemia differs from sickle-cell anaemia in that the former is a quantitative problem of synthesising too

few globin molecules while the latter is a qualitative problem of synthesising an incorrectly functioning globin.

7. Which one from those given below is the period for Mendel's hybridization experiments? [NEET-2017]

(1) 1856 - 1863 (2) 1840 - 1850

(3) 1857 - 1869 (4) 1870 - 1877

Sol. Answer (1)

Mendel conducted hybridization experiments on Pea plant for 7 years between 1856 to 1863 and his data was

published in 1865.

8. The genotypes of a Husband and Wife are IAIB and IAi. Among the blood types of their children, how many

different genotypes and phenotypes are possible? [NEET-2017]

(1) 3 genotypes ; 3 phenotypes (2) 3 genotypes ; 4 phenotypes

(3) 4 genotypes ; 3 phenotypes (4) 4 genotypes ; 4 phenotypes

Sol. Answer (3)

A B AI I I i

Husband Wife

IA

IB

IA

IA

IA

IA

IB

i I iA

I iB

+



Number of genotypes = 4

Number of phenotypes = 3

IAIA and IAi = A

IAIB = AB

IBi = B

9. Among the following characters, which one was not considered by Mendel in his experiments on pea?

[NEET-2017]

(1) Stem – Tall or Dwarf (2) Trichomes – Glandular or non-glandular

(3) Seed – Green or Yellow (4) Pod – Inflated or Constricted

Sol. Answer (2)

During his experiments Mendel studied seven characters.

Nature of trichomes i.e., glandular or non-glandular was not considered by Mendel.

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

[NEET (Phase-2)-2016]

(1) Inversion (2) Duplication (3) Translocation (4) Crossing-over

Sol. Answer (3)

Translocation is a phenomenon of transfer of a gene segment between non-homologus chromosome, i.e.,

different linkage group.

11. A true breeding plant is [NEET (Phase-2)-2016]

(1) One that is able to breed on its own

(2) Produced due to cross-pollination among unrelated plants

(3) Near homozygous and produces offspring of its own kind

(4) Always homozygous recessive in its genetic constitution

Sol. Answer (3)

True breeding line is one that, having undergone continuous self pollination, shows the stable trait inheritance

and expression for several generations. It is both homozygous recessive as well as dominant in genetic

constitution.

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

being colour-blind is [NEET (Phase-2)-2016]

(1) 0 (2) 0.5 (3) 0.75 (4) 1

Sol. Answer (1)

Colourblindness is X-linked recessive disease and shows criss-cross inheritance.

13. In a testcross involving F1 dihybrid flies, more parental-type offspring were produced than the recombinant-type

offspring. This indicates [NEET-2016]

(1) Both of the characters are controlled by more than one gene

(2) The two genes are located on two different chromosomes

(3) Chromosomes failed to separate during meiosis

(4) The two genes are linked and present on the same chromosome

Sol. Answer (4)

When two genes in a dihybrid cross are situated on the same chromosome, the proportion of parental gene

combinations are much higher than the non-parental or recombinant type.



14. Which of the following most appropriately describes haemophilia? [NEET-2016]

(1) Dominant gene disorder (2) Recessive gene disorder

(3) X-linked recessive gene disorder (4) Chromosomal disorder

Sol. Answer (3)

Haemophilia is X-linked recessive gene disorder. It is a blood clotting disorder and shows criss-cross inheritance.

Father Mother

Son Daughter

15. A tall true breeding garden pea plant is crossed with a dwarf true breeding garden pea plant. When the F1

plants were selfed the resulting genotypes were in the ratio of [NEET-2016]

(1) 3 : 1 : : Dwarf : Tall

(2) 1 : 2 : 1 : : Tall homozygous : Tall heterozygous : Dwarf

(3) 1 : 2 : 1 : : Tall heterozygous : Tall homozygous : Dwarf

(4) 3 : 1 : : Tall : Dwarf

Sol. Answer (2)

Parents - TT

(Tall)

× tt(Dwarf)

F generation1

Tt (Heterozygous tall)

On selfing

F generation2

TT(Tall)

Tt(Tall)

tt(dwarf)

Tt(Tall)

T

t

egg

Pollen T t

Phenotypic ratio = 3 : 1 [Tall : Dwarf]

Genotypic ratio

1 : 2 : 1 [Homozygous tall : Heterozygous tall : Dwarf]

16. Match the terms in Column I with their description in Column II and choose the correct option [NEET-2016]

Column I Column II

(a) Dominance (i) Many genes govern a single character

(b) Codominance (ii) In a heterozygous organism only one allele expresses itself

(c) Pleiotropy (iii) In a heterozygous organism both alleles express themselves fully

(d) Polygenic inheritance (iv) A single gene influences many characters

(1) a(iv), b(iii), c(i), d(ii) (2) a(ii), b(i), c(iv), d(iii)

(3) a(ii), b(iii), c(iv), d(i) (4) a(iv), b(i), c(ii), d(iii)

Sol. Answer (3)



Dominance - Expression of only one allele in heterozygous organism.

Codominance - Side by side full expression of both alleles. F1 resembles both parents.

Pleiotropy - Single gene can exhibit multiple phenotypic expression e.g., Phenyl ketonuria.

Polygenic inheritance - Many genes govern a single character e.g., Human skin colour.

17. Pick out the correct statements : [NEET-2016]

(a) Haemophilia is a sex-linked recessive disease.

(b) Down's syndrome is due to aneuploidy.

(c) Phenylketonuria is an autosomal recessive gene disorder.

(d) Sickle cell anaemia is an X-linked recessive gene disorder.

(1) (a), (b) and (c) are correct (2) (a) and (d) are correct

(3) (b) and (d) are correct (4) (a), (c) and (d) are correct

Sol. Answer (1)

Sickle cell anaemia is an autosomal recessive gene disorder.

18. 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? [Re-AIPMT-2015]

(1) 0.25 (2) 0.5

(3) 1 (4) Nil

Sol. Answer (2)

Father (Colourblind)

Daughter (Carrier)

Grandson [50% Probability (0.5)]

19. The term "Linkage" was coined by [Re-AIPMT-2015]

(1) W. Sutton (2) T.H. Morgan (3) T. Boveri (4) G. Mendel

Sol. Answer (2)

The term "linkage" was coined by T.H. Morgan.

20. A pleiotropic gene [Re-AIPMT-2015]

(1) Controls multiple traits in an individual (2) Is expressed only in primitive plants

(3) Is gene evolved during Pliocene (4) Controls a trait only in combination with another gene

Sol. Answer (1)

The gene which controls multiple traits in an individual.

21. In his classic experiments on pea plants, Mendel did not use [Re-AIPMT-2015]

(1) Flower position (2) Seed colour (3) Pod length (4) Seed shape

Sol. Answer (3)

Mendel did not selected Pod length as a character for study



22. A gene showing codominance has [Re-AIPMT-2015]

(1) Both alleles independently expressed in the heterozygote.

(2) One allele dominant on the other

(3) Alleles tightly linked on the same chromosome

(4) Alleles that are recessive to each other

Sol. Answer (1)

Both alleles are independently expressed in heterozygote during codominance.

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

pedigree.

I

II

III

IV

[Re-AIPMT-2015]

(1) X-linked dominant (2) Autosomal dominant (3) X-linked recessive (4) Autosomal recessive

Sol. Answer (4)

The given pedigree represents inheritance of Autosomal recessive trait.

I

II

III

IV

aa

aa

Aa

Aa

AaAa Aa Aa aa Aa Aa

Aa Aa aa Aa

aa aa Aa

24. Alleles are [AIPMT-2015]

(1) Heterozygotes (2) Different phenotype

(3) True breeding homozygotes (4) Different molecular forms of a gene

Sol. Answer (4)

Alleles are slightly different molecular forms of the same gene.

25. The movement of a gene from one linkage group to another is called [AIPMT-2015]

(1) Crossing over (2) Inversion

(3) Duplication (4) Translocation

Sol. Answer (4)

Translocation is illegitimate crossing over between non-homologous chromosome.

26. Multiple alleles are present [AIPMT-2015]

(1) On non-sister chromatids (2) On different chromosomes

(3) At different loci on the same chromosome (4) At the same locus of the chromosome

Sol. Answer (4)

All alleles of a gene are located on the same loci of chromosome in population.



27. An abnormal human baby with 'XXX' sex chromosomes was born due to [AIPMT-2015]

(1) Fusion of two sperm and one ovum (2) Formation of abnormal sperms in the father

(3) Formation of abnormal ova in the mother (4) Fusion of two ova and one sperm

Sol. Answer (3)

Due to non-disjunction of X-chromosomes in mother

A + XX (egg) × A + X (sperm)

28. How many pairs of contrasting characters in pea plants were studied by Mendel in his experiments?

[AIPMT-2015]

(1) Seven (2) Five (3) Six (4) Eight

Sol. Answer (1)

7 pairs of contrasting characters in pea plant were studied by Mendel in his experiment.

29. Fruit colour in squash is an example of [AIPMT-2014]

(1) Recessive epistasis (2) Dominant epistasis

(3) Complementary genes (4) Inhibitory genes

Sol. Answer (2)

Dominant epistasis is the phenomenon of masking or supressing the expression of a gene by a dominant non-

allelic gene.

e.g., fruit colour in Cucurbita pepo (Summer squash)

30. A man whose father was colour blind marries a woman who had a colour blind mother and normal father. What

percentage of male children of this couple will be colour blind ? [AIPMT-2014]

(1) 25% (2) 0% (3) 50% (4) 75%

Sol. Answer (3)

X Y+

X+

X

Y

X+

Xc

X+

Xc

X+

Xc

X+

Xc

X+

Y Xc

Y

Colourblind male = 50%

31. A human female with Turner's syndrome: [AIPMT-2014]

(1) Has 45 chromosomes with XO (2) Has one additional X chromosome

(3) Exhibits male characters (4) Is able to produce children with normal husband

Sol. Answer (1)

Turner's syndrome is caused due to the absence of one of the X chromosomes i.e., 45 (44 + XO).

32. In a population of 1000 individuals 360 belong to genotype AA, 480 to Aa and the remaining 160 to aa. Based

on this data, the frequency of allele A in the population is [AIPMT-2014]

(1) 0.4 (2) 0.5 (3) 0.6 (4) 0.7

Sol. Answer (3)

According to Hardy Weinberg principle, p2 + 2pq + q2 = 1; (p + q)2 = 1

(AA) p2 = 360 out of 1000 individual or p2 = 36 out of 100



q2 = 160 out of 1000 or q2 = 16 out of 100

so q = 0.16 = 0.4. As p + q = 1

so, p is 0.6.

33. If both parents are carriers for thalessemia, which is an autosomal recessive disorder, what are the chances

of pregnancy resulting in an affected child? [NEET-2013]

(1) 50% (2) 25% (3) 100% (4) No chance

Sol. Answer (2)

ATA × ATA

AA : AAT : ATAT

1 : 2 : 1

25%

34. The incorrect statement with regard to Haemophilia is [NEET-2013]

(1) It is a recessive disease

(2) It is a dominant disease

(3) A single protein involved in the clotting of blood is affected

(4) It is a sex-linked disease

Sol. Answer (2)

35. 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

is an example of [NEET-2013]

(1) Incomplete dominance (2) Partial dominance (3) Complete dominance (4) Codominance

Sol. Answer (4)

Both IA and IB give its expression.

36. Which Mendelian idea is depicted by a cross in which the F1 generation resembles both the parents?

[NEET-2013]

(1) Law of dominance

(2) Inheritance of one gene

(3) Co-dominance

(4) Incomplete dominance

Sol. Answer (3)

37. Which of the following statements is not true of two genes that show 50% recombination fequency ?

[NEET-2013]

(1) The genes are tightly linked

(2) The genes show independent assortment

(3) If the genes are present on the same chromosome, they undergo more than one crossovers in every

meiosis

(4) The genes may be on different chromosomes

Sol. Answer (1)



38. F2 generation in a Mendelian cross showed that both genotypic and phenotypic ratios are same as 1 : 2 : 1.

It represents a case of [AIPMT (Prelims)-2012]

(1) Monohybrid cross with complete dominance

(2) Monohybrid cross with incomplete dominance

(3) Co-dominance

(4) Dihybrid cross

Sol. Answer (2)

39. 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?

[AIPMT (Prelims)-2012]

(1) 25% (2) 50% (3) 100% (4) Zero percent

Sol. Answer (4)

X Y X XC

X Y X XC

Gametes :

X

X X C

X YC

X X

X Y

X Y

XC

X

100% daughter = Normal but 50% carrier

50% son – Colourblind

50% son – Normal

40. A test cross is carried out to [AIPMT (Mains)-2012]

(1) Determine the genotype of a plant at F2

(2) Predict whether two traits are linked

(3) Asses the number of alleles of a gene

(4) Determine whether two species or varieties will breed successfully

Sol. Answer (1)

TT × tt

Tt

Cross with recessive parent

41. Represented below is the inheritance pattern of a certain type of traits in humans. Which one of the following

conditions could be an example of this pattern? [AIPMT (Mains)-2012]

Female

Mother Father

Male

Daughter Son

(1) Phenylketonuria (2) Sickle cell anaemia (3) Haemophilia (4) Thalassemia

Sol. Answer (3)

Haemophilia is sex linked character.



42. Which one of the following is a wrong statement regarding mutations? [AIPMT (Mains)-2012]

(1) Deletion and insertion of base pairs cause frame-shift mutations

(2) Cancer cells commonly show chromosomal aberrations

(3) UV and Gamma rays are mutagens

(4) Change in a single base pair of DNA does not cause mutation

Sol. Answer (4)

Point mutation can cause mutation.

43. Which one of the following conditions correctly describes the manner of determining the sex in the given

example? [AIPMT (Prelims)-2011]

(1) Homozygous sex chromosomes (XX) produce male in Drosophila.

(2) Homozygous sex chromosomes (ZZ) determine female sex in Birds.

(3) XO type of sex chromosomes determine male sex in grasshopper

(4) XO condition in humans as found in Turner Syndrome, determine female sex

Sol. Answer (3)

AA – XO – A + X

A + O

44. When two unrelated individuals or lines are crossed, the performance of F1 hybrid is often superior of both its

parents. This phenomenon is called [AIPMT (Prelims)-2011]

(1) Metamorphosis (2) Heterosis (3) Transformation (4) Splicing

Sol. Answer (2)

45. Test cross in plants or in Drosophila involves crossing [AIPMT (Mains)-2011]

(1) The F1 hybrid with a double recessive genotype

(2) Between two genotypes with dominant trait

(3) Between two genotypes with recessive trait

(4) Between two F1 hybrids

Sol. Answer (1)

46. ABO blood groups in humans are controlled by the gene I. It has three alleles – IA, IB and i. Since there are

three different alleles, six different genotypes are possible. How many phenotypes can occur?


(1) Three (2) One (3) Four (4) Two

Sol. Answer (3)

47. Which one of the following cannot be explained on the basis of Mendel’s Law of Dominance?


(1) Factors occur in pairs

(2) The discrete unit controlling a particular character is called a factor

(3) Out of one pair of factors one is dominant, and the other recessive

(4) Alleles do not show any blending and both the characters recover as such in F2 generation

Sol. Answer (4)



48. The genotype of a plant showing the dominant phenotype can be determined by [AIPMT (Prelims)-2010]

(1) Back cross (2) Test cross

(3) Dihybrid cross (4) Pedigree analysis

Sol. Answer (2)

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


(1) Tightly linked genes on the same chromosome show very few recombinations

(2) Tightly linked genes on the same chromosome show higher recombinations

(3) Genes far apart on the same chromosome show very few recombinations

(4) Genes loosely linked on the same chromosome show similar recombinations as the tightly linked ones

Sol. Answer (1)

Tightly linked gene will show less recombination.

50. A cross in which an organism showing a dominant phenotype is crossed with the recessive parent in order

to know its genotype is called [AIPMT (Mains)-2010]

(1) Monohybrid cross (2) Back cross (3) Test cross (4) Dihybrid ross

Sol. Answer (3)

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

for the character [AIPMT (Mains)-2010]

(1) The female parent is heterozygous

(2) The parents could not have had a normal daughter for this character

(3) The trait under study could not be colourblindness

(4) The male parent is homozygous dominant

Sol. Answer (1)

52. 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? [AIPMT (Mains)-2010]

(1) Six (2) Three (3) Four (4) Five

Sol. Answer (3)



53. The fruit fly Drosophila melanogaster was found to be very suitable for experimental verification of chromosomal

theory of inheritance by Morgan and his colleagues because: [AIPMT (Mains)-2010]

(1) It reproduces parthenogenetically

(2) A single mating produces two young flies

(3) Smaller female is easily recognisable from larger male

(4) It completes life cycle in about two weeks

Sol. Answer (4)

54. In Antirrhinum two plants with pink flowers were hybridized. The F1 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. [AIPMT (Mains)-2010]

(1) rrrr (2) RR (3) Rr (4) rr

Sol. Answer (3)

Antirrhinum majus

RR × rr

F1

Rr

RR : Rr : rr

1 : 2 : 1F2

55. Which one of the following symbols and its representation, used in human pedigree analysis is correct?


(1) = male affected (2) = mating between relatives

(3) = unaffected male (4) = unaffected female

Sol. Answer (2)

56. Point mutation involves [AIPMT (Prelims)-2009]

(1) Deletion (2) Insertion

(3) Change in single base pair (4) Duplication

Sol. Answer (3)

Sickele cell anaemeia

57. Study the pedigree chart given below:

What does it show? [AIPMT (Prelims)-2009]

(1) Inheritance of a condition like phenylketonuria as an autosomal recessive trait

(2) The pedigree chart is wrong as this is not possible

(3) Inheritance of a recessive sex-linked disease like haemophilia

(4) Inheritance of a sex-linked inborn error of metabolism like phenylketonuria



Sol. Answer (1)

Aa Aa

aa Aa Aa

A–

aa

A–

aa A–

58. Select the incorrect statement from the following: [AIPMT (Prelims)-2009]

(1) Galactosemia is an inborn error of metabolism

(2) Small population size results in random genetic drift in a population

(3) Baldness is a sex-limited trait

(4) Linkage is an exception to the principle of independent assortment in heredity

Sol. Answer (3)

Baldness is sex influenced character.

59. Haploids are more suitable for mutation studies than the diploids. This is because [AIPMT (Prelims)-2008]

(1) All mutations, whether dominant or recessive are expressed in haploids

(2) Haploids are reproductively more stable than diploids

(3) Mutagens penetrate in haploids more effectively than diploids

(4) Haploids are more abundant in nature than diploids

Sol. Answer (1)

60. Which one of the following conditions in humans is correctly matched with its chromosomal abnormality/

linkage? [AIPMT (Prelims)-2008]

(1) Down syndrome - 44 autosomes + XO

(2) Klinefelter syndrome - 44 autosomes + XXY

(3) Colour blindness - Y-linked

(4) Erythroblastosis foetalis - X-linked

Sol. Answer (2)

61. In the hexaploid wheat, the haploid (n) and basic (x) numbers of chromosomes are [AIPMT (Prelims)-2007]

(1) n = 21 and x = 7 (2) n = 7 and x = 21

(3) n = 21 and x = 21 (4) n = 21 and x = 14

Sol. Answer (1)

n = 7

6n = 42

n = 21

x 7 (Monoploid)



62. Inheritance of skin colour in humans is an example of [AIPMT (Prelims)-2007]

(1) Codominance (2) Chromosomal aberration

(3) Point mutation (4) Polygenic inheritance

Sol. Answer (4)

63. A common test to find the genotype of a hybrid is by [AIPMT (Prelims)-2007]

(1) Crossing of one F1progeny with male parent

(2) Crossing of one F2 progeny with male parent

(3) Crossing of one F2 progeny with female parent

(4) Studying the sexual behaviour of F1 progenies

Sol. Answer (1)

Tt × tt or +

64. Two genes R and Y are located very close on the chromosomal linkage map of maize plant. When RRYY and

rryy genotypes are hybridized, the F2 segregation will show [AIPMT (Prelims)-2007]

(1) Higher number of the parental types (2) Higher number of the recombinant types

(3) Segregation in the expected 9: 3: 3: 1 ratio (4) Segregation in 3:1 ratio

Sol. Answer (1)

65. In pea plants, yellow seeds are dominant to green. If a heterozygous yellow seeded plant is crossed with a

green seeded plant, what ratio of yellow and green seeded plants would you expect in F1 generation?


(1) 3 : 1 (2) 50 : 50 (3) 9 : 1 (4) 1 : 3

Sol. Answer (2)

Yy × yy

Yy : yy

50 50

66. A human male produces sperms with genotypes AB, Ab, aB and ab pertaining to two diallelic characters in

equal proportions. What is the corresponding genotype of this person? [AIPMT (Prelims)-2007]

(1) AABB (2) AaBb (3) AaBB (4) AABb

Sol. Answer (2)

AaBb

AB, Ab, aB, ab

67. Which one of the following is the most suitable, medium for culture of Drosophila melanogaster ?


(1) Moist bread (2) Agar agar (3) Ripe banana (4) Cow dung

Sol. Answer (3)

68. Phenotype of an organism is the result of [AIPMT (Prelims)-2006]

(1) Mutations and linkages

(2) Cytoplasmic effects and nutrition

(3) Environmental changes and sexual dimorphism

(4) Genotype and environment interactions

Sol. Answer (4)



69. In which mode of inheritance do you expect more maternal influence among the offspring ?


(1) Autosomal (2) Cytoplasmic (3) Y-linked (4) X-linked

Sol. Answer (2)

70. How many different kinds of gametes will be produced by a plant having the genotype AABbCC?


(1) Three (2) Four (3) Nine (4) Two

Sol. Answer (4)

AABbCC

2n 21 271. Which one of the following is an example of polygenic inheritance? [AIPMT (Prelims)-2006]

(1) Flower colour in Mirabilis jalapa

(2) Production of male honey bee

(3) Pod shape in garden pea

(4) Skin colour in humans

Sol. Answer (4)

72. 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? [AIPMT (Prelims)-2006]

(1) Only round seeds with green cotyledons

(2) Only wrinkled seeds with yellow cotyledons

(3) Only wrinkled seeds with green cotyledons

(4) Round seeds with yellow cotyledons and wrinkled seeds with yellow cotyledons

Sol. Answer (4)

RRYY rryy×

RY ry

RrYy

RY Round yellow

y Round green

rY

Y Wrinkled yellow

73. Test cross involves [AIPMT (Prelims)-2006]

(1) Crossing between two genotypes with recessive trait

(2) Crossing between two F1 hybrids

(3) Crossing the F1 hybrid with a double recessive genotype

(4) Crossing between genotypes with dominant traits

Sol. Answer (3)

74. If a colourblind woman marries a normal visioned man, their sons will be [AIPMT (Prelims)-2006]

(1) All normal visioned (2) One-half colourblind and one-half normal

(3) Three-fourths colourblind and one-fourth normal (4) All colourblind

Sol. Answer (4)



75. Cri-du-chat syndrome in humans is caused by the [AIPMT (Prelims)-2006]

(1) Fertilization of an XX egg by a normal Y-bearing sperm

(2) Loss of half of the short arm of chromosome 5

(3) Loss of half of the long arm of chromosome 5

(4) Trisomy of 21st chromosome

Sol. Answer (2)

76. 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 are

affected. Which of the following mode of inheritance do you suggest for this disease?


(1) Autosomal dominant (2) Sex-linked dominant (3) Sex-limited recessive (4) Sex-linked recessive

Sol. Answer (4)

XXC × XY

XCY / XX / XCX / XY

Sex linked recessive

77. At a particular locus, frequency of ‘A’ allele is 0.6 and that of ‘a’ is 0.4. What would be the frequency of

heterozygotes in a random mating population at equilibrium? [AIPMT (Prelims)-2005]

(1) 0.16 (2) 0.48 (3) 0.36 (4) 0.24

Sol. Answer (2)

78. A woman with normal vision, but whose father was colour blind, marries a colourblind man. Suppose that the

fourth child of this couple was a boy. This boy: [AIPMT (Prelims)-2005]

(1) Must have normal colour vision

(2) Will be partially colourblind since he is heterozygous for the colourblind mutant allele

(3) Must be colourblind

(4) May be colourblind or may be of normal vision

Sol. Answer (4)

XXC X XCY

XCY or XY

79. Haemophilia is more commonly seen in human males than in human females because


(1) This disease is due to an X-linked dominant mutation

(2) A greater proportion of girls die in infancy

(3) This disease is due to an X-linked recessive mutation

(4) This disease is due to a Y-linked recessive mutation

Sol. Answer (3)

80. A woman with 47 chromosomes due to three copies of chromosome 21 is characterized by


(1) Down syndrome (2) Triploidy (3) Turner syndrome (4) Super femaleness

Sol. Answer (1)

2n + 1 21 chromosome



81. 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 [AIPMT (Prelims)-2005]

(1) aaBB (2) AaBb (3) AABB (4) aabb

Sol. Answer (4)

AaBb × aabb

Dihybrid test cross

82. Which of the following is not a hereditary disease? [AIPMT (Prelims)-2005]

(1) Cretinism (2) Cystic fibrosis (3) Thalassaemia (4) Haemophilia

Sol. Answer (1)

83. The salivary gland chromosomes in the dipteran larvae, are useful in gene mapping because


(1) These are much longer in size (2) These are easy to stain

(3) These are fused (4) They have endoreduplicated chromosomes

Sol. Answer (4)

84. Genetic variation in a population arises due to

(1) Mutations only (2) Recombination only

(3) Mutations as well as recombination (4) Reproductive isolation and selection

Sol. Answer (3)

85. Which one is the incorrect statement with regards to the importance of pedigree analysis?

(1) It helps to trace the inheritance of a specific trait

(2) It confirms that DNA is the carrier of genetic information

(3) It helps to understand whether the trait in question is dominant or recessive

(4) It confirms that the trait is linked to one of the autosome

Sol. Answer (2)

It is used to study inheritance of character.

86. In our society women are blamed for producing female children. Choose the correct answer for the sex-

determination in humans

(1) Due to some defect in the women

(2) Due to some defect like aspermia in man

(3) Due to the genetic make up of the particular sperm which fertilizes the egg

(4) Due to the genetic make up of the egg

Sol. Answer (3)

Next generation sex is determined by male.

87. Down’s syndrome in humans is due to

(1) Two ‘Y’ chromosomes (2) Three ‘X’ chromosomes

(3) Three copies of chromosome 21 (4) Monosomy

Sol. Answer (3)

2n + 1 21 Chromosome



88. The variation/difference in the offsprings of a species from their parents constitutes an important component

of

(1) Genetics (2) Speciation (3) Species fixation (4) Heredity

Sol. Answer (1)

89. If two pea plants having red (dominant) coloured flowers with unknown genotypes are crossed, 75% of the

flowers are red and 25% are white. The genotypic constitution of the parents having red coloured flowers will

be

(1) Both homozygous (2) One homozygous and other heterozygous

(3) Both heterozygous (4) Both hemizygous

Sol. Answer (3)

90. Walter Sutton is famous for his contribution to

(1) Genetic engineering (2) Totipotency

(3) Quantitative genetics (4) Chromosomal theory of inheritance

Sol. Answer (4)

91. A polygenic trait is controlled by 3 genes A, B and C. In a cross AaBbCc × AaBbCc, the phenotypic ratio of

the offsprings was observed as

1 : 6 : x : 20 : x : 6 : 1

What is the possible value of x?

(1) 3 (2) 9 (3) 15 (4) 25

Sol. Answer (3)

Polygenic inheritance – 3 genes

1 : 6 : 15 : 20 : 15 : 6 : 1

92. The chromosome constitution 2n – 2 of an organism represents

(1) Monosomic (2) Nullisomic (3) Haploid (4) Trisomic

Sol. Answer (2)

Nullisomic 2n – 2

93. Mendel's principle of segregation means that the germ cells always receive

(1) One pair of alleles (2) One quarter of the genes

(3) One of the paired alleles (4) Any pair of alleles

Sol. Answer (3)

94. Absence of one sex chromosome causes

(1) Turner's syndrome (2) Klinefelter's syndrome

(3) Down's syndrome (4) Tay-Sach's syndrome

Sol. Answer (1)

AA – XO

Absence of X chromosome.



95. Chimera is produced due to

(1) Somatic mutations (2) Reverse Mutations

(3) Lethal mutations (4) Pleiotropic mutations

Sol. Answer (1)

Mutation in different tissue.

96. Haploids are more suitable for mutation studies than the diploids. This is because

(1) All mutations, whether dominant or recessive are expressed in haploids

(2) Haploids are reproductively more stable than diploids

(3) Mutagens penetrate in haploids more effectively than diploids

(4) Haploids are more abundant in nature than diploids

Sol. Answer (1)

97. Which one of the following conditions in humans is correctly matched with its chromosomal abnormality/

linkage?

(1) Down’s syndrome – 44 autosomes + XO (2) Klinefelter’s syndrome – 44 autosomes + XXY

(3) Colour blindness – Y-linked (4) Erythroblastosis foetalis – X-linked

Sol. Answer (2)

Klinefelter's syndrome – Trisomy of sex chromosome.

98. The genes, which remain confined to differential region of Y-chromosome, are

(1) Autosomal genes (2) Holandric genes

(3) Completely sex-linked genes (4) Mutant genes

Sol. Answer (2)

99. The colour blindness is more likely to occur in males than in females because

(1) The Y-chromosome of males have the genes for distinguishing colours

(2) Genes for characters are located on the X-chromosomes

(3) The trait is dominant in males and recessive in females

(4) None of these

Sol. Answer (2)

Since carry one X chromosome so X-Linked character will give its expression.

100. Albinism is a congenital disorder resulting from the lack of the enzyme

(1) Tyrosinase (2) Xanthine oxidase (3) Catalase (4) Fructokinase

Sol. Answer (1)

Loss of pigment caused by Tyrosinase.

101. An abnormal human male phenotype involving an extra Y-chromosome (XYY) is a case of

(1) Edward’s syndrome (2) Jacob syndrome (3) Intersex (4) Down’s syndrome

Sol. Answer (2)

Trisomy of Y-Chromosome



102. The phenomenon, in which an allele of one gene suppresses the activity of an allele of another gene, is known

as

(1) Epistasis (2) Dominance (3) Suppression (4) Inactivation

Sol. Answer (1)

103. Barr body in mammals represents

(1) All the heterochromatin in male and female cells

(2) The Y-chromosome in somatic cells of male

(3) All the heterochromatin in female cells

(4) One of the two X-chromosomes in somatic cells of females

Sol. Answer (4)

One of X-chromosome change into heterochromatin in +

.

104. When two dominant independently assorting genes react with each other producing effect jointly they are called

(1) Collaborative genes (2) Complementary genes

(3) Duplicate genes (4) Supplementary genes

Sol. Answer (2)

Both gene complement each other to give new expression.

105. 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 genetical disease?

(1) Sex-influenced disease (2) Blood group inheritance disease

(3) Sex-linked disease (4) Sex-limited disease

Sol. Answer (3)

XDY × XY

5 Normal son (XY), 3(XXD) carrier daughter

106. A person whose father is colour blind marries a lady whose mother is daughter of a colour blind man. Their

children will be

(1) All sons colour blind

(2) Some sons normal and some daughters colour blind

(3) All sons and daughters colour blind

(4) All daughter normal

Sol. Answer (4)

XY × XX / X XC

All normal daughter

107. In which of the following disease, the individual has one less X-chromosome?

(1) Turner’s syndrome (2) Klinefelter’s syndrome

(3) Bleeder’s disease (4) Down’s syndrome

Sol. Answer (1)

AA – XO (2n – 1) Monosomic



108. H.J. Muller had received Nobel Prize for

(1) His studies on Drosophila for genetic study

(2) Proving that the DNA is a genetic material

(3) Discovering the linkage of genes

(4) Discovering the induced mutations by X-rays

Sol. Answer (4)

109. The polygenic genes show

(1) Different karyotypes (2) Different genotypes (3) Different phenotypes (4) None of these

Sol. Answer (3)

Skin shade of human being.

110. Foetal sex can be determined by examining cells from the amniotic fluid by looking for

(1) Chiasmata (2) Kinetochore

(3) Barr bodies (4) Autosomes

Sol. Answer (3)

+ contain 1 barr body

contain 0 barr body

111. A fruit fly is hemizygous for sex-linked genes, mated with normal female fruit fly, the males specific chromosome

will enter egg cell in the proportion

(1) 3 : 1 (2) 7 : 1

(3) 1 : 1 (4) 2 : 1

Sol. Answer (3)

XAY × XX

XAX × XY

1 : 1

112. Genetic identity of a human male is determined by

(1) Sex-chromosome (2) Cell organelles (3) Autosome (4) Nucleolous

Sol. Answer (1)

Y-Chromosome

113. Different forms of a gene located at the same locus of chromosomes are called

(1) Multiple alleles (2) Polygenes (3) Oncogenes (4) None of these

Sol. Answer (1)

They are mutated genes occupy same gene locus on homologous chromosome.

114. After crossing two plants, the progenies are found to be male sterile. This phenomenon is found to be maternally

inherited and is due to some genes which reside in

(1) Mitochondria (2) Cytoplasm (3) Nucleus (4) Chloroplast

Sol. Answer (1)



115. 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?

(1) 50% (2) 75% (3) 100% (4) 25%

Sol. Answer (4)

AA : AA : A Aa a a

1 2 1

25%

AAa

× AAa

116. How many different types of genetically different gametes will be produced by a heterozygous plant having the

genotype AABbCc?

(1) Six (2) Nine (3) Two (4) Four

Sol. Answer (4)

AA Bb Cc

2n (2)2 4

117. When a single gene influences more than one traits it is called

(1) Pseudodominance (2) Pleiotropic (3) Epistasis (4) None of these

Sol. Answer (2)

Sickle cell anemia

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

(1) Moderate increase in Y complement (2) Large increase in Y complement

(3) Reduction in X complement (4) Increase in X complement

Sol. Answer (4)

Increase in X complement

119. Loss of a X-chromosome in a particular cell, during its development, results into

(1) Gynandromorphs (2) Meta female (3) Triploid individual (4) Myotonic dystrophy

Sol. Answer (1)

Which contain both and +

character.

120. 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 not have discovered the law of independent assortment

(2) He would have discovered sex linkage

(3) He could have mapped the chromosome

(4) He would have discovered blending or incomplete dominance

Sol. Answer (1)

121. 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?

(1) 50% haemophilic colour-blind sons and 50% haemophilic sons

(2) 50% haemophilic daughters and 50% colour blind daughters

(3) All sons and daughters haemophilic and colour-blind

(4) Haemophilic and colour-blind daughters



Sol. Answer (1)

XhXhc Woman

122. In human beings, multiple genes are involved in the inheritance of

(1) Sickle cell anaemia (2) Skin colour (3) Colour blindness (4) Phenylketonuria

Sol. Answer (2)

Polygenic inheritance

123. Haemophilic man marries a normal woman. Their offsprings will be

(1) All haemophilic (2) All boys haemophilic (3) All girls haemophilic (4) All normal

Sol. Answer (4)

XhY × XX

All normal

124. A marriage between normal visioned man and colour blind woman will produce offspring

(1) Colour blind sons and 50% carrier daughters

(2) 50% colourblind sons and 50% carrier daughters

(3) Normal males and carrier daughters

(4) Colour blind sons and carrier daughters

Sol. Answer (4)

XY × XCXC

X YC

X XC

125. In hybridization, Tt × tt gives rise to the progeny in the ratio

(1) 2 : 1 (2) 1 : 2 : 1 (3) 1 : 1 (4) 1 : 2

Sol. Answer (3)

Tt × tt

Tt : tt

1 : 1

126. According to Mendelism, which character shows dominance?

(1) Terminal position of flower (2) Green colour in seed coat

(3) Wrinkled seeds (4) Green pod colour

Sol. Answer (4)

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

(1) 50% (2) 75%

(3) 25% (4) 100%

Sol. Answer (1)

TTRr × ttrr

TtRr

TR

Ttrr

Tr

tr

1 1



128. In Drosophila, the XXY condition leads to femaleness whereas in human beings the same condition leads to

Klienfelter’s syndrome in male. It proves

(1) In human beings, Y chromosome is active in sex determination

(2) Y chromosome is active in sex determination in both human beings and Drosophila

(3) In Drosophila, Y chromosome decides femaleness

(4) Y chromosome of man has genes for syndrome

Sol. Answer (1)

129. Independent assortment of genes does not take place when

(1) Genes are located on homologous chromosomes

(2) Genes are linked and located on same chromosome

(3) Genes are located on non-homogenous chromosomes

(4) All of these

Sol. Answer (2)

130. Mendel obtained wrinkled seeds in pea due to deposition of sugars instead of starch. It was due to which

enzyme?

(1) Amylase (2) Invertase

(3) Diastase (4) Absence of starch branching enzyme

Sol. Answer (4)

131. Ratio of complementary genes is

(1) 9 : 3 : 4 (2) 12 : 3 : 1 (3) 9 : 3 : 3 : 4 (4) 9 : 7

Sol. Answer (4)

9 : 7

132. When both parental alleles are expressed together, it is called

(1) Co-dominance (2) Dominance (3) Incomplete dominance (4) Pseudodominance

Sol. Answer (1)

133. A and B genes are linked. What shall be genotype of progeny in a cross between AB/ab and ab/ab?

(1) AAbb and aabb (2) AaBb and aabb (3) AABB and aabb (4) aaBB × Aabb

Sol. Answer (2)

A

B

a

b ×

a

b

a

b

A

B

a

b

1

a

b

a

bA

B

a

b

a

b

: 1

134. Probability of four sons to a couple is

(1) 1/4 (2) 1/8 (3) 1/16 (4) 1/32

Sol. Answer (3)

1 1 1 1 1

2 2 2 2 16



135. If recombination frequency between AB genes is 20% and BC gene is 40% and interference is 30% in the

case of double cross over then what will be coincidance under this condition?

(1) 2.4 (2) 8 (3) 5.6 (4) 0.7

Sol. Answer (4)

C + I = 1

0.3 + 0.7 = 1

136. Male XX and female XY sometime occur due to

(1) Deletion (2) Transfer of segments in X and Y chromosome

(3) Aneuploidy (4) Hormonal imbalance

Sol. Answer (2)

137. Number of Barr body in XXXX female is

(1) 1 (2) 2 (3) 3 (4) 4

Sol. Answer (3)

Barr Body = Number of X-chromosome – 1

4 – 1 =3

138. Extranuclear inheritance occurs in

(1) Killer Paramoecium (2) Killer Amoeba (3) Euglena (4) Hydra

Sol. Answer (1)

139. Which of the following is correct match?

(1) Down’s syndrome - 21st chromosome (2) Sickle cell anaemia – X-chromosome

(3) Haemophilia – Y-chromosome (4) Parkinson’s disease – X & Y chromosome

Sol. Answer (1)

140. How many genome types are present in a typical green plants cell?

(1) More than five (2) More than ten (3) Two (4) Three

Sol. Answer (3)

Cell 2n

141. Which of the following is an example of sex linked disease?

(1) AIDS (2) Colour blindness (3) Syphilis (4) Gonorrhoea

Sol. Answer (2)

AIDS – Viral

Syphilis, Gonorrhoea Spirochetes

142. Which of the following is an example of pleiotropy?

(1) Haemophilia (2) Thalassemia

(3) Sickle cell anaemia (4) Colour blindness

Sol. Answer (3)

One gene control multiple feature.



143. A gene is said to be dominant if

(1) It expresses its effect only in homozygous state

(2) It expresses its effect only in heterozygous condition

(3) It expresses its effect both in homozygous and heterozygous condition

(4) It never expresses it’s effect in any condition

Sol. Answer (3)

Tt Tall

144. On selfing a plant of F1-generation with genotype ‘’AABbCC’’, the genotypic ratio in F

2-generation will be

(1) 1 : 2 : 1 (2) 1 : 1

(3) 9 : 3 : 3 : 1 (4) 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1

Sol. Answer (1)

ABC

AbC

AABbCC

2 type gamete

145. A diseased mans 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

(1) Sex linked dominant (2) Sex linked recessive

(3) Sex limited character (4) Autosomal dominant

Sol. Answer (1)

XDY × XX

(3. ) X X : XY (5 Sons)D

+

+

146. Down’s syndrome is caused by an extra copy of chromosome number 21. What percentage of offspring is

produced by an affected mother and a normal father?

(1) 100% (2) 75% (3) 50% (4) 25%

Sol. Answer (3)

147. Which one of the following discoveries resulted in a Nobel Prize?

(1) X-rays induce sex-linked recessive lethal mutations

(2) Cytoplasmic inheritance

(3) Recombination of linked genes

(4) Genetic engineering

Sol. Answer (1)



148. 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

(1) Test cross (2) Reciprocal cross

(3) Dihybrid cross (4) Reverse cross

Sol. Answer (2)

+

149. The genes controlling the seven pea characters studied by Mendel are now known to be located on how many

different chromosomes?

(1) Seven (2) Six (3) Five (4) Four

Sol. Answer (4)

Chromosome Number 1, 4, 5, 7

150. Which one of the following traits of garden pea studied by Mendel was a recessive feature ?

(1) Axial flower position (2) Green seed colour

(3) Green pod colour (4) Round seed shape

Sol. Answer (2)

151. Moustaches, beard and horseness in voice in human males are examples of

(1) Sex-linked traits (2) Sex limited traits

(3) Sex differentiating traits (4) Sex determining traits

Sol. Answer (2)

152. In Drosophila, the sex is determined by

(1) The ratio of number of X chromosomes to the sets of autosomes

(2) X and Y chromosomes

(3) The ratio of X-chromosomes to the pairs of autosomes

(4) Whether the egg is fertilized or develops parthenogenetically

Sol. Answer (1)

Gene balance theory X

= ratioA

153. 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

(1) One-third of the progenies

(2) None of the progenies

(3) All the progenies

(4) Fifty percent of the progenies

Sol. Answer (2)

Mitochondrial inheritance is cytoplasmic inheritance



154. Lack of independent assortment of two genes A and B in fruit fly Drosophila is due to

(1) Repulsion (2) Recombination

(3) Linkage (4) Crossing over

Sol. Answer (3)

Two genes are present on same chromosome.

155. What kind of evidence suggested that man is more closely related with chimpanzee than with other hominoid

apes?

(1) Evidence from DNA from sex chromosomes only

(2) Comparison of chromosomes morphology only

(3) Evidence from fossil remains, and the fossil mitochondrial DNA alone

(4) Evidence from DNA extracted from sex chromosomes, autosomes

Sol. Answer (4)

156. The recessive genes located on X-chromosome of humans are always

(1) Lethal (2) Sub-lethal

(3) Expressed in males (4) Expressed in females

Sol. Answer (3)

Hemizygous condition in XCY

157. A male human is heterozygous for autosomal genes A and B and is also hemizygous for haemophilic gene

h. What proportion of his sperms will be abh?

(1) 1/8 (2) 1/32 (3) 1/16 (4) 1/4

Sol. Answer (1)

Aa Bb XhY

1 1 1 1

2 2 2 8

158. A self-fertilizing trihybrid plant forms

(1) 8 different gametes and 64 different zygotes (2) 4 different gametes and 16 different zygotes

(3) 8 different gametes and 16 different zygotes (4) 8 different gametes and 32 different zygotes

Sol. Answer (1)

159. 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?

(1) b, a, c (2) a, b, c (3) a, c, b (4) None of these

Sol. Answer (1)

b20 a 8 c

28%

Gene sequence b – a – c



160. The linkage map of X-chromosome of fruit fly has 66 units, with yellow body gene (y) at one end and bobbed

hair (b) gene at the other end. The recombination frequency between these two genes (y and b) should be

(1) 66% (2) > 50% (3) 50% (4) 100%

Sol. Answer (3)

y b

Recombination frequency will be less than 50%.

161. 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) 25% will be tall with red fruit



(4) All the offsprings will be tall with red fruit

Sol. Answer (2)

RRTt × rrtt

RT Rt rt

RT

RrTt

Rt

Rrttrt

+

50 50

162. A normal woman, whose father was colour blind is married to a normal man. The sons would be

(1) 75% colour blind (2) 50% colour blind

(3) All normal (4) All colour blind

Sol. Answer (2)

XXC × XY

X Y, XYC

Colour blind

163. De Vries gave his mutation theory on organic evolution while working on

(1) Pisum sativum

(2) Drosophila melanogaster

(3) Oenothera lamarckiana

(4) Althea rosea

Sol. Answer (3)

164. Triticale, the first man-made cereal crop, has been obtained by crossing wheat with

(1) Barley (2) Rye (3) Pearl millet (4) Sugarcane

Sol. Answer (2)

Rye Secale cereal



165. Normally DNA molecule has A-T, G-C pairing. However, these bases can exist in alternative valency status owing

to rearrangements called

(1) Frame-shift mutation (2) Tautomerisational mutation

(3) Analog substitution (4) Point mutation

Sol. Answer (2)

A T Tautomerisational mutation

G C Tautomerisational mutation

A T Amino Imino group

166. The most striking example of point mutation is found in a disease called

(1) Down’s syndrome (2) Sickle cell anaemia

(3) Edward syndrome (4) Night blindness

Sol. Answer (2)

167. Identify the one, which causes gene mutation

(1) Granoson (2) Colchicine (3) Crossing over (4) X-rays

Sol. Answer (4)

Ionised radiation

168. The mutations are mainly responsible for

(1) Increasing the population rate (2) Maintaining genetic continuity

(3) Constancy in organisms (4) Variation in organisms

Sol. Answer (4)

169. Which of the following is the main category of mutation?

(1) Somatic mutation (2) Genetic mutation

(3) Heterosis (4) None of these

Sol. Answer (2)

170. Change in sequence of nucleotide in DNA is called

(1) Mutagen (2) Mutation

(3) Recombination (4) Translation

Sol. Answer (2)

171. When a cluster of genes show linkage behaviour they

(1) Do not show a chromosome map (2) Show recombination during meiosis

(3) Do not show independent assortment (4) Induce cell division

Sol. Answer (3)

Linked genes will not show independent assortment.



172. Genetic map is one that

(1) Establishes sites of the genes on a chromosome

(2) Establishes the various stages in gene evolution

(3) Shows the stages during the cell division

(4) Shows the distribution of various species in a region

Sol. Answer (1)

Linear arrangement of gene on chromosome.

173. In mutational event, when adenine is replaced by guanine, it is a case of

(1) Frame shift mutation (2) Transcription

(3) Transition (4) Transversion

Sol. Answer (3)

Replaced byA G Transition

Purine Purine��⇀↽��

174. The most likely reason for the development of resistance against pesticides in insects damaging a crop is

(1) Random mutations (2) Genetic recombination

(3) Directed mutations (4) Acquired heritable changes

Sol. Answer (1)

175. When two genetic loci produce identical phenotypes in cis and trans position, they are considered to be

(1) Multiple alleles (2) The parts of same gene

(3) Pseudoalleles (4) Different genes

Sol. Answer (3)

When two adjacent genes occupy different gene locus but performer regulate same function.

176. What base is responsible for hot spots for spontaneous point mutations?

(1) 5-bromouracil (2) 5-methylcytosine

(3) Guanine (4) Adenine

Sol. Answer (2)

177. Nucleus of a donor embryonal cell/somatic cell is transferred to an enucleated egg cell. Then after the formation

of organism, what shall be true?

(1) Organism will have extranuclear genes of the donor cell

(2) Organism will have extranuclear genes of recipient cell

(3) Organism will have extranuclear genes of both donor and recipient cell

(4) Organism will have nuclear genes of recipient cell

Sol. Answer (2)

Organism contain Extra chromosome DNA of recepient cell



178. Genes for cytoplasmic male sterility in plants are generally located in

(1) Chloroplast genome (2) Mitochondrial genome

(3) Nuclear genome (4) Cytosol

Sol. Answer (2)

179. Extranuclear inheritance is the consequence of presence of genes in

(1) Mitochondria and chloroplasts

(2) Endoplasmic reticulum and mitochondria

(3) Ribosomes and chloroplast

(4) Lysosomes and ribosomes

Sol. Answer (1)

SECTION - C

Assertion-Reason Type Questions

1. A : Turner's syndrome generally does not occur in males.

R : Foetus with 44 + YO complement generally dies.

Sol. Answer (1)

Because absence of X chromosome

2. A : Sickel cell anaemia occurs due to the point mutation.

R : mRNA produced from Hb(s) gene has GAG instead of GUG.

Sol. Answer (3)

GUG Valine

GAG Glutamine

3. A : Holandric traits are passed from one generation to the next generation.

R : These traits appear more frequently in one sex than in other.

Sol. Answer (3)

Hoalandric genes are Y linked.

4. A : Dominance is not an autonomous feature of a gene.

R : It depends as much on the gene product and the production of a particular phenotype from this product.

Sol. Answer (1)

5. A : The posssibility of a female becoming a haemophilic is extremely rare.

R : Mother of such a female has to be carrier and father should be haemophilic.

Sol. Answer (1)

It is X linked recessive

6. A : Polyploids with odd number of chromosomes are propagated vegetatively.

R : Seed formation is absent due to meiotic abnormality.

Sol. Answer (1)

7. A : Pseudoalleles are actually closely linked genes.

R : These can be identified easily as both affect different characters, providing separate phenotypes.

Sol. Answer (3)

These genes occupy different location on a chromosome but they regulate same feature.



8. A : The heterozygotic female for haemophilia may transmit the disease to sons.

R : Such traits show criss-cross inheritance.

Sol. Answer (2)

9. A : Non-allosomic genic determination of sex is found in bacteria.

R : Sex is dependent on some environmental factors in prokaryotes.

Sol. Answer (3)

Sex is regulated by F-plasmid or fertility genes.

10. A : Crossing over is exchange of genetic material between non-homologous chromosomes.

R : It produces new linkages.

Sol. Answer (4)

Exchange of genetic material between homologous chromosome.

11. A : Mendel gave postulates like "principles of segregation and principles of independent assortment" after

studying seven pairs of contrasting traits in garden pea.

R : He was lucky in selecting seven characters in pea that were located on seven different chromosomes.

Sol. Answer (3)

12. A : Test cross is the tool for knowing linkage between genes.

R : Monohybrid test cross gives two phenotypes and two genotypes.

Sol. Answer (2)

13. A : Myotonic dystrophy is caused by recessive mutant pleiotropic gene.

R : Gene mutation leads to more synthesis of fibrillin proteins.

Sol. Answer (4)

Caused by Dominant genes.

14. A : In snapdragon, F1 plants do not have red or white flowers.

R : It is intermediate inheritance with neither of the two alleles of a gene being dominant over each other.

Sol. Answer (1)

Because of incomplete dominance they allele cannot give its expression completely in figuration.

15. A : en block inheritance of all genes located on the same chromosome may occur in some organisms.

R : Dihybrid test cross will have only two phenotypes.

Sol. Answer (1)

16. A : Morgan's cross III was conducted in Drosophila to locate genes on chromosome for white eye colour.

R : The cross was done between red eyed hybrid female and white eyed male.

Sol. Answer (3)

It was a reciprocal cross so +

white x red eye.

17. A : Antlers in male deer are sex influenced traits.

R : These are controlled by autosomal genes which are influenced by the sex of bearer.

Sol. Answer (4)

Antler of deer is sex limited feature.



18. A : One drum stick per nucleus is present in the neutrophil of normal female.

R : It is absent in the neutrophil of male.

Sol. Answer (2)

19. A : Blood group phenotype is controlled by presence or absence of antigens present on surface coating of

RBC.

R : These antigens are of three types and found in the oligosaccharides rich head regions of a glycophorin.

Sol. Answer (2)

20. A : XO type sex determination is found in large number of insects.

R : Some of the sperms bear the X-chromosome whereas some do not.

Sol. Answer (2)

XX – XO – Type is found in some insects.

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