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Principles Of Inheritance And Variation

Total questions: 102

Worksheet time: 51mins

Name
Class
Date
1.

Genetics is the branch of biology which deals with

a)

variation

b)

inheritance

c)

Both (a) and (b)

d)

study of characters

2.

The inheritance of characters from parents to offspring is

a)

variation

b)

heredity

c)

inheritance

d)

resemblance

3.

The tendency of offspring to differ from their parents is called

a)

variation

b)

heredity

c)

inheritance

d)

resemblance

4.

Mendel’s hybridisation experimental material was

a)

Pisum sativum

b)

Lathyrus odoratus

c)

Oryza sativa

d)

Mirabilis jalapa

5.

Which one from those given below is the period of Mendel’s hybridisation experiments?

a)

1856-1863

b)

1840-1850

c)

1857-1869

d)

1870-1877

6.

Mendel investigated characters in garden pea plant manifested in two traits which were

a)

similar

b)

non-zygote

c)

identical

d)

opposite

7.

A true breeding line is characterised by the presence of

a)

stable trait inheritance due to the continuous selfpollination

b)

varying traits in different generations due to the cross pollination

c)

single trait in all generations due to allogamy

d)

varying trait inheritance in a single generation due to geitonogamy

8.

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

a)

12

b)

13

c)

14

d)

15

9.

Out of 7 contrasting trait pairs selected by Mendel, how many traits were dominant and recessive?

a)

7 and 7

b)

8 and 6

c)

6 and 8

d)

5 and 9

10.

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

a)

Stem – Tall or Dwarf

b)

Trichomes – Glandular or Non-glandular

c)

Seed – Green or Yellow

d)

Pod – Inflated or Constricted

11.

Which is correct about traits choosen by Mendel for his experiment on pea plant?

a)

Terminal pod was dominant

b)

Constricted pod was dominant

c)

Green coloured pod was dominant

d)

Tall plants were recessive

12.

What contributed to Mendel’s success?

I. Selection of pureline pea varieties.

II. Knowledge of history.

III. Selecting one character at a time.

IV. Statistical analysis and mathematicallogic.

Choose the correct option

a)

I, II, III and IV

b)

II and III

c)

I, III and IV

d)

II, III and IV

13.

The first hybrid progeny obtained by Mendel were called

a)

F1 -progeny

b)

F0 -progeny

c)

F2 -progeny

d)

F3 -progeny

14.

F1 -progeny of a cross between pure tall and dwarf plant is always

a)

tall

b)

short

c)

intermediate

d)

None of these

15.

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

a)

F4

b)

F2

c)

F1

d)

F0

16.

The Mendel crossed true breeding tall and dwarf plant varieties in his experiment. Tallness was the dominant character and dwarfness was recessive. The recessive character appeared in

a)

F1

b)

F2

c)

F3

d)

F2 and F3

17.

How did Mendel obtained recessive (dwarf) character in F2 -generation?

a)

By self-pollinating F1

b)

By self-pollinating F2

c)

By cross-pollinating F1

d)

By cross-pollinating F2

18.

The proportion of plants that were dwarf and tall, respectively in F2 -generation of Mendel’s experiment was

a)

14\frac{1}{4} th and 34\frac{3}{4} th

b)

34\frac{3}{4} th and 14\frac{1}{4} th

c)

23\frac{2}{3} rd and 13\frac{1}{3} rd

d)

13\frac{1}{3} rd and 43\frac{4}{3} rd

19.

Mendel crossed tall and dwarf plants. In F2 -generation both the tall and dwarf plants were produced. This shows

a)

blending of characters

b)

atavism

c)

non-blending of characters

d)

intermediate characters

20.

During his experiments, Mendel used the term factor for

a)

genes

b)

traits

c)

characters

d)

qualities

21.

Genes which codes for a pair of contrasting characters are

a)

recessive character

b)

dominant character

c)

alleles

d)

alternative gene

22.

Choose the incorrect match.

a)

Phenotype – Physical appearance of an organism

b)

Genotype – Expressed genes

c)

Homozygous – Identical alleles of a gene present at the same locus

d)

Heterozygous – Genes of an allelic pair are not same

23.

Number of gametes produced by a homozygous and a heterozygous individuals of genotype AA and Aa, respectively are

a)

1 and 2

b)

2 and 3

c)

3 and 5

d)

many

24.

A cross in which parents differ in a single pair of contrasting character is called

a)

monohybrid cross

b)

dihybrid cross

c)

trihybrid cross

d)

tetrahybrid cross

25.

The phenotypic ratio of a monohybrid cross in F2 -generation is

a)

3 : 1

b)

1 : 2 : 1

c)

2 : 1 : 1

d)

9 : 3 : 3 : 1

26.

The genotypic ratio of a monohybrid cross in F2 -generation is

a)

3 : 1

b)

1 : 2 : 1

c)

2 : 1 : 1

d)

9 : 3 : 3 : 1

27.

F2 -generation in a Mendelian cross showed that both genotypic and phenotypic ratios are same as 1 : 2 : 1. It represents a case of

a)

codominance

b)

dihybrid cross

c)

monohybrid cross with complete dominance

d)

monohybrid cross with incomplete dominance

28.

If the male plant has the genotype TT and the female plant has the genotype tt then they contribute pollen and egg, respectively with

a)

T and T gametes

b)

tt and TT gametes

c)

TT and tt gametes

d)

T and t gametes

29.

Graphical representation to calculate the probability of all possible genotype of an offspring in genetic cross is called

a)

Bunett square

b)

Morgan square

c)

Punnett square

d)

Mendel square

30.

Test cross involves a cross between

a)

recessive F1 -plant and dominant F2 -plant

b)

recessive F2 -plant and dominant F3 -plant

c)

dominant F2 -plant and recessive parent plants

d)

dominant F2 -plant and heterozygous parent plants

31.

Mendel performed test cross to know the

a)

genotype of F1

b)

genotype of F2

c)

genotype of F3

d)

genotype of F4

32.

When alleles of two contrasting characters are present together and one of the character expresses itself during the cross while the other remains hidden gives the

a)

aw of purity of gametes

b)

law of segregation

c)

law of dominance

d)

law of independent assortment

33.

The allele which expresses itself in both homozygous and heterozygous condition is called

a)

dominant allele

b)

recessive allele

c)

incomplete dominant allele

d)

split allele

34.

3:1 ratio in F2 -generation is explained by

a)

law of partial dominance

b)

law of dominance

c)

law of incomplete dominance

d)

law of purity of gametes

35.

The law of dominance is applicable in inheritance of

a)

seed colour in pea

b)

flower colour in Mirabilis jalapa

c)

starch grain size in pea

d)

roan coat colour in cattles

36.

Mendel’s principle of segregation means that the germ cells always receive

a)

one pair to alleles

b)

one quarter of the genes

c)

either one allele of father or one allele of mother

d)

any pair of alleles

37.

The law based on fact that the characters do not show any blending and both the characters are recovered as such in F2 -generation although one character was absent in F1 -progeny, is

a)

law of purity of gametes

b)

law of independent assortment

c)

law of incomplete dominance

d)

law of dominance

38.

The types of gametes formed by the genotype RrYy are

a)

RY, Ry, rY, ry

b)

RY, Ry, ry, ry

c)

Ry, Ry, Yy, ry

d)

Rr, RR, Yy, YY

39.

In law of independent assortment how many factors are involved (for a dihybrid cross)

a)

1

b)

2

c)

3

d)

4

40.

In Mendel’s experiments with garden pea, round seed shape (RR) was dominant over wrinkled seeds (rr) and yellow colour (YY) was dominant over green colour (yy). What are the expected phenotypes in the F1 -generation of the cross RRYY × rryy?

a)

Only round seeds with yellow cotyledons

b)

Only wrinkled seeds with yellow cotyledons

c)

Only wrinkled seeds with green cotyledons

d)

Round seeds with yellow cotyledons and wrinkled seeds with yellow cotyledons

41.

In cross between pure breeding pea plants having yellow round (YYRR) and green wrinkled (yyrr) seeds, find out the total seeds (plants) having yellow colour in F2 -generation.

a)

12

b)

10

c)

14

d)

11

42.

In a cross between plants having yellow round (YYRR) and green wrinkled (yyrr) seeds, what will be the ratio between seeds having yellow and green seed colour?

a)

3 : 2

b)

3 : 1

c)

9 : 7

d)

7 : 9

43.

Total number of round seed in the F2 -generation of a cross between plants having pure yellow round and pure green wrinkled seeds is

a)

9

b)

12

c)

11

d)

1

44.

Ratio observed in dihybrid cross (phenotypically)

a)

3 : 1

b)

1 : 2 : 1

c)

9 : 7

d)

9 : 3 : 3 : 1

45.

The number of different genotypes observed in the F2 -generation of a dihybrid cross are

a)

9

b)

12

c)

4

d)

6

46.

Mendel’s result on inheritance of characters were rediscovered by

a)

de Vries

b)

Correns

c)

von Tschermak

d)

All of these

47.

The literal meaning of chromosome is

a)

painted body

b)

coloured body

c)

doubling body

d)

thread-like body

48.

The concept of chromosome movement during meiosis to explain Mendel’s laws was used by

a)

Sutton and Boveri

b)

Malthus

c)

Correns

d)

Morgan

49.

The chromosomes as well as genes occur in pair and the two alleles of a gene pair are located on

a)

homologous chromosomes

b)

non-homologous chromosomes

c)

single chromosome

d)

All of the above

50.

Who proposed the chromosomal theory of inheritance?

a)

Sutton and Mendel

b)

Boveri and Morgan

c)

Morgan and Mendel

d)

Sutton and Boveri

51.

Experimental evidences of chromosomal theory of inheritance was given by

a)

S Boveri

b)

TH Morgan

c)

de Vries

d)

W Sutton

52.

Morgan’s experimental organism was

a)

Drosophila melanogaster

b)

Mangifera indica

c)

Mirabilis jalapa

d)

Drosophila indica

53.

Both chromosome and gene (Mendelian factors) whether dominant or recessive are transmitted from generation to generation in

a)

changed form

b)

unaltered form

c)

altered form

d)

disintegrated form

54.

Choose the incorrect pairing among the following.

a)

Sutton and Boveri – Chromosome theory

b)

Walter and Boveri – Behaviour of chromosome during cell divisions

c)

TH Morgan – Mutation

d)

Henking – Barr bodies

55.

Linked genes that were observed by Morgan were present on

a)

X-chromosome

b)

different chromosome

c)

heterologous chromosome

d)

paired chromosome

56.

Strength of the linkage between the two genes is

a)

proportionate to the distance between them

b)

inversely proportionate to the distance between them

c)

depend on the chromosomes

d)

depend upon the size of chromosomes

57.

In Morgan’s experiment, white and yellow genes were linked tightly, while white and miniature wing were loosely linked. The per cent recombination shown by these genes were

a)

50% each

b)

72% and 8.3%, respectively

c)

0.3% and 53%, respectively

d)

1.3% and 37.2%, respectively

58.

In a test cross involving F1 dihybrid flies, more parental-type offspring were produced than the recombinant type offspring. This indicates

a)

chromosomes failed to separate during meiosis

b)

the two genes are linked and present on the same chromosome

c)

Both of the characters are controlled by more than one gene

d)

the genes are located on two different chromosomes

59.

The frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes was explained by

a)

Gregor J Mendel

b)

Alfred Sturtevant

c)

Sutton-Boveri

d)

TH Morgan

60.

What map unit (centi Morgan) is adopted in the construction of genetic maps?

a)

A unit of distance between two expressed genes representing 100% cross over

b)

A unit of distance between genes on chromosomes, representing 1% cross over

c)

A unit of distance between genes on chromosomes, representing 50% cross over

d)

A unit of distance between two expressed genes representing 10% cross over

61.

Choose the incorrect pair with respect to sex determination in different organisms.

a)

Grasshopper = XO type

b)

Birds = ZZ-ZW type

c)

Drosophila = XX-XO type

d)

Human = XX-XY type

62.

In XX and XY type of sex-determination

a)

males are heterogametic

b)

females are isogametic

c)

Both (a) and (b)

d)

None of the option is correct

63.

Male heterogamety is seen in

a)

Humans

b)

Grasshopper

c)

Drosophila

d)

All of these

64.

Choose the incorrect pair amongst the following

a)

Male bird – Homogametic

b)

Female bird – Heterogametic

c)

Male Drosophila – Heterogametic

d)

None of the above

65.

The chromosomal denotation for heterogametic female and homogametic males are

a)

ZW and ZZ

b)

ZO–ZZ

c)

XX–XO

d)

Both (a) and (b)

66.

The number of chromosomes in females and males honeybees are

a)

32

b)

16

c)

32 and 16, respectively

d)

16 and 32, respectively

67.

The unfertilised eggs in honeybees develop into

a)

males

b)

queen

c)

worker

d)

Both (a) and (c)

68.

In honeybees, male and female gametes are produced through

a)

mitosis

b)

mitosis and meiosis, respectively

c)

meiosis

d)

meiosis and mitosis, respectively

69.

Mutation is a phenomena which results in alteration in sequences of

a)

DNA

b)

RNA

c)

proteins

d)

Both (a) and (b)

70.

Mutation may result in

a)

change in genotype

b)

change in phenotype

c)

change in metabolism

d)

All of these

71.

Chromosomal abberation is commonly found in the

a)

cancer cells

b)

normal cells

c)

healthy cells

d)

autosomal cells

72.

Point mutation arises due to the change in

a)

single base DNA

b)

single base pair of DNA

c)

segment of DNA

d)

double base pair of DNA

73.

If there are four different types of nitrogenous bases (A, T, G and C) then how many different types of transitions and transversion are possible?

a)

Transition = 8, Transversion = 4

b)

Transition = 4, Transversion = 4

c)

Transition = 8, Transversion = 4

d)

Transition = 4, Transversion = 8

74.

Sickle-cell anaemia is a classical example of

a)

frame-shift mutation

b)

point mutation

c)

Both (a) and (b)

75.

Frame-shift mutation arises due to

a)

deletion of base pair of DNA

b)

insertion of base pair of DNA

c)

Both (a) and (b)

d)

change in single base pair of DNA

76.

Mutagens are

a)

chemical agents which cause change in DNA

b)

physical agents which cause mutation

c)

Both (a) and (b)

d)

None of the above

77.

Analysis of traits of several generation of a family in the form of diagram is called

a)

gene analysis

b)

chromosome analysis

c)

allele analysis

d)

pedigree analysis

78.

Pedigree analysis is very important in human beings because

a)

it helps genetic counselors to avoid disorders

b)

it shows origin of traits

c)

it shows the flow of traits in family

d)

All of the above

79.

Colour blindness in humans

a)

results in defect in either red or green cone of eyes

b)

is caused due to the mutation in gene found on X-chromosome

c)

affects males more frequently than females

d)

All of the above

80.

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

a)

Only grand children

b)

Only sons

c)

Only daughters

d)

Both (b) and (c)

81.

A normal-visioned man whose father was colourblind, marries a woman whose father was also colourblind. They have their first child as a daughter. What are the chances that this child would be colourblind?

a)

100%

b)

0%

c)

25%

d)

50%

82.

A man whose father was colourblind marries a woman, who had a colourblind mother and normal father. What percentage of male children of this couple will be colourblind ?

a)

25%

b)

0%

c)

50%

d)

75%

83.

A normal woman whose father was colourblind, marries a normal man. What kinds of children can be expected and in what proportion ?

a)

All daughters normal, 50% of sons colourblind

b)

All daughters normal, all sons colourblind

c)

50% daughters colourblind, all sons normal

d)

All daughters colourblind, all sons norma

84.

Which of the following most appropriately describes haemophilia?

a)

X-linked recessive gene disorder

b)

Chromosomal disorder

c)

Dominant gene disorder

d)

Recessive gene disorder

85.

In haemophilia, the affected protein is a part of a cascade of protein which is involved in the

a)

formation of RBCs

b)

formation of WBCs and platelets

c)

coagulation of blood

d)

anticoagulation

86.

In sickle-cell anaemia,

a)

Both parents are heterozygous carriers, but are unaffected

b)

Single pair of allele controls the disease

c)

Only Hb Hb s s show diseased phenotype

d)

All of the above

87.

In individual suffering from phenylketonuria,

a)

enzyme phenylalanine hydroxylase is absent

b)

phenylalanine do not convert to tyrosine

c)

phenylpyruvic acid is formed

d)

All of the above

88.

Thalassemia in humans

a)

is an autosome linked recessive blood disorder

b)

can transmit from parents to offspring when both parents are unaffected carriers (heterozygous)

c)

caused due to the mutation or deletion of one of the α or β-globin chain

d)

All of the above

89.

α-thalassemia in humans is controlled by

a)

HBA1 and HBA2 genes on chromosome 16

b)

HBA1 gene on chromosome 12

c)

HBA2 gene on chromosome 11

d)

HBA1 and HBA2 genes on chromosome 9

90.

β-thalassemia in humans is controlled by

a)

HBA2 gene on chromosome 16

b)

HBB gene on chromosome 11

c)

HBA1 gene on chromosome 15

d)

HBA1 and HBA2 gene on chromosome 8

91.

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

a)

aneuploidy

b)

hypopolyploidy

c)

hyperpolyploidy

d)

polyploidy

92.

A cell telophase stage is observed by a student in a plant brought from the field. He tells his teacher that this cell is not like other cells at telophase stage. There is no formation of cell plate and thus the cell is containing more number of chromosomes as compared to other dividing cells. This would result in

a)

polyploidy

b)

somaclonal variation

c)

polyteny

d)

aneuploidy

93.

Non-disjunction in meiosis results in

a)

trisomy

b)

normal diploid

c)

gene mutation

d)

None of these

94.

A disease caused by an autosomal primary non-disjunction is

a)

Down’s syndrome

b)

Klinefelter’s syndrome

c)

Turner’s syndrome

d)

Sickle-cell anaemia

95.

Karyotype of Down’s syndrome has how many chromosomes?

a)

43

b)

46

c)

47

d)

45

96.

I. Short statured body with small round head. II. Furrowed tongue and partially opened mouth. III. Palm is broad with characteristic palm crease. IV. Slow physical, psycomotor and mental development. These are the characters of

a)

Down’s syndrome

b)

Turner’s syndrome

c)

Klinefelter’s syndrome

d)

Edward syndrome

97.

Choose the correct pair.

a)

Gynacoemastia – Development of breasts

b)

Turner’s syndrome – Loss of an X-chromosome in females

c)

Polyploidy – Seen in plants

d)

All of the above

98.

What is the genetic disorder in which an individual has an overall masculine development gynaecomastia and is sterile?

a)

Klinefelter’s syndrome

b)

Edward syndrome

c)

Down’s syndrome

d)

Turner’s syndrome

99.

Klinefelter’s syndrome results from

a)

XX egg and Y from sperm

b)

XX egg and XY sperm

c)

X egg and XY sperm

d)

Both (a) and (c)

100.

In which genetic condition, each cell in the affected person, has three sex chromosomes XXY?

a)

Thalassemia

b)

Klinefelter’s syndrome

c)

Phenylketonuria

d)

Turner’s syndrome

101.

Female suffering from Turner’s syndrome possess

a)

45 + XO

b)

rudimentary ovaries

c)

lack of secondary sexual characters

d)

All of the above

102.

Which of the following are chromosomal disorders.

I. Colour blindness

II. Down’s syndrome

III. Phenylketonuria

IV. Turner’s syndrome

V. Thalassaemia

a)

I, II and III

b)

II, IV and V

c)

III, IV and V

d)

II and IV