FSC Biology Book MCQs Chapter 8 Class 12

Created on March 02, 2022 By Ali Durrani

Biology Ch 8 Class 12 Variation and Genetics

1 / 66

A cross between F1 hybrids and either of parents is called:

NUMS 2014
Biology
Variation and Genetics
Elimination Tool:

A) Back cross

B) Test cross

C) Reverse cross

D) None of the above

2 / 66

An X to A ratio of ⅖ in drosophila produces:

NUMS 2020
Biology
Variation and Genetics
Elimination Tool:

A) Sterile female

B) Fertile female

C) Male

D) Both male and female but sterile

3 / 66

Red green colour blindness is a sex linked recessive trait. If a colour blind man marries a woman with normal vision whose father is colourblind and they have a daughter, what is the probability that she will be colourblind?

NTS 2015 SMBBMC
Biology
Variation and Genetics
Elimination Tool:

A) 0

B) 0.25

C) 0.50

D) 0.75

4 / 66

Haemophilia occurs because of:

PMC 2020
Biology
Variation and Genetics
Elimination Tool:

A) Mutation in an autosome

B) Mutation of Y-chromosome

C) Mutation of X-chromosome

D) Deficiency of iron

5 / 66

Filial is a Latin word. It means

Spring

Issue

Progeny

Descendent

6 / 66

A continuously varying trait is encoded by alleles of two or more different genes located at different loci, all influencing the same trait in an additive way is known as ________.

NUMS 2019 Cancelled
Biology
Variation and Genetics
Elimination Tool:

A) Poly traits

B) Polygenes

C) Multiple alleles

D) Different gene

7 / 66

A pure breeding plant with the dominant phenotype of character P and the recessive phenotype of character Q was crossed with another pure breeding plant with the recessive phenotype of chracter P and the dominant phenotype of Q. The offspring of this cross were crossed with a double homozygous recessive for P and Q and the following results were obtained:

22 were phenotypically dominant for P and recessive for Q
5 were phenotypically dominant for both P and Q
4 were phenotypically recessive for both P and Q
24 were phenotypically recessive for P and dominant for Q
Which one of the following types of Inheritance is illustrated by these results?

Biology
Variation and Genetics
NTS 2013
Elimination Tool:

A) gene linkage of P and Q

B) Independent segregation of P and Q

C) Mendelian dihybrid Inheritance

D) multiple alleles

8 / 66

______ alleles both have an effect on the phenotype of a heterozygous organism:

PMC 2020
Biology
Variation and Genetics
Elimination Tool:

A) Dominant

B) Recessive

C) Co-dominant

D) Multiple

9 / 66

When phenotype of a heterozygote is in between the phenotypes of both the homozygous parents, it is called:

NUMS 2016
Biology
Variation and Genetics
Elimination Tool:

A) Incomplete dominance

B) Epistasis

C) Pleiotropy

D) Codominance

10 / 66

Two animals are mated. One is homozygous dominant for one character and homozygous recessive for another. The other animal Is heterozygous for both characters. How many phenotypes are expected in the offspring of this cross?

NTS 2016 DUHS & JSMU
Biology
Variation and Genetics
Elimination Tool:

A) 1

B) 2

C) 3

D) 4

11 / 66

In birds the male is the homogametic sex. A male bird showing the recessive trait was mated with a female showing the dominant trait of a characteristic governed by a pair of alleles which are sex-linked. What is the probability that the male offspring will show the dominant trait?

NTS 2014
Biology
Variation and Genetics
Elimination Tool:

A) 0

B) 0.25

C) 0.50

E) 1.00

12 / 66

Hybrid black Guinea pigs are crossed with each other. The resulting offsprings will be:

NUMS 2016
Biology
Variation and Genetics
Elimination Tool:

A) All black

B) All white

C) 3 black, 1 white

D) 3 white, 1 black

13 / 66

______ are more common in human males than females.

NUMS 2016
Biology
Variation and Genetics
Elimination Tool:

A) X-linked dominant traits

B) X-linked recessive traits

C) Y-linked dominant traits

D) Autosomal linked recessive traits

14 / 66

Which of the following X – linked recessive trait that is more common in human males than in human females?

NUMS 2019 Cancelled
Biology
Variation and Genetics
Elimination Tool:

A) Hemophilia

B) Sickle cell Anemia

C) Color Blindness

D) Hypophosphatemic rickets

15 / 66

If we cross round (RR) seeded plants with wrinkled (rr) seed plant, and get F1 round seeded. In F2 generation, what will be genotypic ration of plants try Punnett square method?

NUMS 2019 Cancelled
Biology
Variation and Genetics
Elimination Tool:

B) 1:1

C) 2:1

D) 1:1:2

E) 1:2:1

16 / 66

Baldness is most frequent in

Men

Women

Children

Girls

17 / 66

A child with AB blood group who was born to a mother of the genotype IAIA cannot have the father of the genotype:

NUMS 2014
Biology
Variation and Genetics
Elimination Tool:

A) IA IA

B) IB IB

C) IB IO

D) IB IB and IB IO

18 / 66

Which of the following is monohybrid cross

TT x tt

TTYy x Ttyy

Both of these

None of these

19 / 66

Which statement about dominant and recessive alleles is not correct?

NTS 2016 SMBBMC
Biology
Variation and Genetics
Elimination Tool:

A) A dominant characteristic is seen in the phenotype of a heterozygote.

B) A homozygous genotype may be either dominant or recessive.

C) Recessive phenotypes always have two recessive alleles.

D) The phenotype of a homozygote is always dominant.

20 / 66

The type of gene interaction in which the effect caused by a gene at one locus interfere with the effect caused by another gene at another locus is known as:

NUMS 2018
Biology
Variation and Genetics
Elimination Tool:

A) Pleiotropy

B) Epistasis

C) Polygenic inheritance

D) Gene linkage

21 / 66

In cats, the genes controlling coat-color are co-dominant (incompletely dominant) and are carried on the X chromosomes. When a black female was mated with a ginger male the resulting litter consisted of black male and tortoise-shell female kittens. What phenotypic ratio would be expected in the F2 generation?

NTS 2012
Biology
Variation and Genetics
Elimination Tool:

A) 1 black male : 1 ginger male : 2 tortoise – shell females

B) 1 black male : 1 ginger male : 1 tortoise – shell female : 1 black female

C) 2 black males : 1 tortoise – shell female : 1 ginger female

D) 1 black male : 1 tortoise – shell female : 1 ginger female : 1 black female

22 / 66

A pure breeding tall pea plant was crossed to dwarf plant what will be the frequency of dwarf plants in F1

0.25

0.5

0.75

0

23 / 66

How many genes control Rh blood group system?

One

Two

Three

Four

24 / 66

To check the homozygosity or heterozygosity of apparently dominant trait, we do the test cross by crossing it with:

NUMS 2019 Cancelled
Biology
Variation and Genetics
Elimination Tool:

A) Any Dominant trait

B) Homozygous dominant trait

C) Homozygous Recessive trait

D) Heterozygous dominant trait

25 / 66

Erythroblastosis foetalis occurs when:

NTS 2011
Biology
Variation and Genetics
Elimination Tool:

D) Both mother and baby are Rh-positive

A) Mother is Rh-positive and baby is Rh-negative

B) Mother is Rh-negative and the baby is Rh-positive

C) Both mother and baby are Rh-negative

D) Both mother and baby are Rh-positive

26 / 66

A monohybrid cross yielded 3:1 in F2. What could be mode of inheritance?

Segregation

Independent assortment

Both of these

None of these

27 / 66

Which traits are most likely to affect men than women?

PMC 2020
Biology
Variation and Genetics
Elimination Tool:

A) X linked recessive

B) X linked dominant

C) Autosomal dominant

D) Autosomal recessive

28 / 66

The genes which do not follow law of independent assortment

Crossed genes

Linked genes

Recessive genes

Dominant genes

29 / 66

The gene which controls ABO group has how many alleles in an individual

One

Two

Three

Four

30 / 66

Read the passage below to answer the question:

Haemophilia is a disorder in which blood fails to clot. Saad, a male haemophiliac, marries Sara, a normal woman and together they have four children, two boys (Ahmed and Ali) and two girls (Alia and Ayesha). None of the children display the symptoms of Haemophilia.

Ahmed, Ali, Ayesha, and Ali all marry normal individuals and have children. None of Ahmed’s or Ali’s children, male or female, display symptoms of haemophilia, but the sons of Alia and Ayesha display symptoms of haemophilia while the daughters of Alia and Ayesha do not.

Which of the following best explains the reason that Ahmed, Ali, Ayesha and Alia do not display symptoms of haemophilia, even though their father, Saad, is a haemophiliac.

NTS 2010
Biology
Variation and Genetics
Elimination Tool:

A) Haemophilia is an X-linked disorder, and Saad can only pass on his Y chromosome

B) Haemophilia is an X-linked disorder, and even though Alia and Ayesha receive a hemophiliac X-chromosome from Saad, Sara gave them a normal X-chromosome.

C) Haemophilia is a Y -linked disorder, and therefore cannot be displayed in females

D) Haemophilia is a Y-linked disorder ,and Ahmed and Ali must have received an X Chromosome from Saad

31 / 66

The total number of genes present in a particular population, at a particular time is called:

NUMS 2019 Cancelled
Biology
Variation and Genetics
Elimination Tool:

A) Gene

B) Genome

C) Genotype

D) Gene Pool

32 / 66

A pea plant with yellow seed was crossed to a plant having green seeds. What will happen in F1; if plants are true breeding

All seeds will be yellow

Half of seeds will be yellow

All the seeds will be green

Both will be present in ratio of 1:2:1

33 / 66

The disease, which is caused by defect in a single gene or pair of genes is referred to as:

NTS 2016 SMBBMC
Biology
Variation and Genetics
Elimination Tool:

A) Unifactorial

B) Multifactorial

C) Down syndrome

D) Turner syndrome

34 / 66

The law of dominance is illustrated in the garden pea by:

NTS 2010
Biology
Variation and Genetics
Elimination Tool:

B) Heterozygous tall x heterozygous tall

C) Homozygous tall x homozygous tall

D) Pure short x pure short

E) Homozygous tall x pure short

35 / 66

In the pea plants, the factors round and yellow seeds (RRYY) is dominant over wrinkled and green seeds (rryy). Keeping this in mind, identify the seed that has the genotype round and green.

NTS 2011
Biology
Variation and Genetics
Elimination Tool:

A) Rryy

B) rrYy

C) RrYy

D) RRYy

36 / 66

In question number 2 what will be frequency of dwarf plant in F2

0.25

0.5

0.75

0

37 / 66

Mendel concluded that each organism has two hereditary factors for each trait, now called:

NTS 2015 SMBBMC
Biology
Variation and Genetics
Elimination Tool:

A) Chromatids

B) Alleles

C) Chromosomes

D) None of the above

38 / 66

How many pairs of homologous chromosomes are present in Pisum sativum

Five pairs

Six pairs

Seven pairs

Eight pairs

39 / 66

Read the passage below to answer the question:

Hemophilia is a disorder in which blood fails to clot. Saad, a male hemophiliac, marries Sara, a normal woman and together they have four children, two boys (Ahmed and Ali) and two girls (Alia and Ayesha). None of the children display the symptoms of Hemophilia. Ahmed, Ali, Ayesha, and Ali all marry normal individuals and have children. None of Ahmeds or Ali’s children, male or female, display symptoms of hemophilia, but the sons of Alia and Ayesha display symptoms of hemophilia while the daughters of Alia and Ayesha do not.

Which of the following individuals are heterozygous for hemophilia:

NTS 2010
Biology
Variation and Genetics
Elimination Tool:

A) Saad, Ahmed, and Ali

B) Ahmed, Ali, Alia, and Ayesha

C) Saad and Sara

D) Alia and Ayesha

40 / 66

Read the passage below to answer the question:

Hemophilia is a disorder in which blood fails to clot. Saad, a male hemophiliac, marries Sara, a normal woman and together they have four children, two boys (Ahmed and Ali) and two girls (Alia and Ayesha). None of the children display the symptoms of Hemophilia. Ahmed, Ali, Ayesha, and Ali all marry normal individuals and have children. None of Ahmeds or Ali’s children, male or female, display symptoms of hemophilia, but the sons of Alia and Ayesha display symptoms of hemophilia while the daughters of Alia and Ayesha do not.

If one of Ali’s daughters marries a normal man,what is the probability that one of their children will display symptoms of hemophilia:

NTS 2010
Biology
Variation and Genetics
Elimination Tool:

A) 0%

B) 25%

C) 50%

D) 75%

41 / 66

A trait whose alleles are present in both males and females but prominent more in one sex than other is a:

NUMS 2016
Biology
Variation and Genetics
Elimination Tool:

A) Sex-linked trait

B) Sex limited trait

C) Sex influenced trait

D) X-linked trait

42 / 66

The position of a gene on chromosome is called

Habitat

Position

Locus

Location

43 / 66

An Rh negative woman is married to an Rh positive man whose father was also Rh negative. What are the chances that their child will be affected?

NTS 2016 SMBBMC
Biology
Variation and Genetics
Elimination Tool:

A) 25%

B) 50%

C) 75%

D) 100%

44 / 66

If a heterozygous individual shows the complete effect of both alleles, the type of inheritance would be

Complete dominance

Non dominance

Incomplete dominance

Co-Dominance

45 / 66

If the seed is heterozygous yellow (Yy). In a test cross, what will be the probability of appearance of seed colors?

NUMS 2019 Reconduct
Biology
Variation and Genetics
Elimination Tool:

A) 0 % yellow, 100 % green

B) 25 % yellow, 75 % green

C) 50 % yellow, 50 % green

D) 75 % yellow, 25 % green

46 / 66

If we cross round yellow seeded plant with green wrinkled seeded plants, what will be phenotypic ratio at F2 generation?

NUMS 2019 Cancelled
Biology
Variation and Genetics
Elimination Tool:

A) 9:3:3:1

B) 9:4:3:2

C) 3:9:3:1

D) 8:3:3:2

47 / 66

Which of the following does not belong to the same linkage group?

NUMS 2015
Biology
Variation and Genetics
Elimination Tool:

A) Sickle cell anemia

B) Albinism

C) Leukemia

D) Gout

48 / 66

A man with blood group “A” marries a woman of blood group “B”. Both are heterozygous. What is the off-springs having phenotype “O”

Zero

25%

50%

75%

49 / 66

Phenotype for both TT and Tt is tall, so the allele T is known as:

NUMS 2019 Cancelled
Biology
Variation and Genetics
Elimination Tool:

A) Dominant

B) Recessive

C) Homozygous

D) Heterozygous

50 / 66

The trait “Kernel colour in corn” in controlled by how many pairs of genes

One pair

Two pairs

Three pairs

Many pairs

51 / 66

In which of the following cases, genotypic and phenotypic ratio will remain same in F2 generation

Law of independent assortment

Law of Segregation

Test cross

Incomplete dominance.

52 / 66

Genotype of blood group ‘AB’ in human is ______.

NUMS 2015
Biology
Variation and Genetics
Elimination Tool:

A) IA IB

B) IB Io

C) IA IA

D) IoIo

53 / 66

When both the alleles of a gene pair are same, the organism is said to be:

PMC 2020
Biology
Variation and Genetics
Elimination Tool:

A) Heterozygous

B) Genotype

C) Homozygous

D) Phenotype

54 / 66

Which of the following condition is hybrid A. TT B. Tt C. tt D. All of these

TT

Tt

tt

All of these

55 / 66

In ‘Complete Dominance’:

PMC 2020
Biology
Variation and Genetics
Elimination Tool:

A) Different alleles of a gene are both expressed in heterozygous condition

B) One allele (R) is completely dominant over the other (r) and the presence of the recessive allele is functionally hidden, so the heterozygote (Rr) has the same round phenotype as (RR) homozygote.

C) The phenotype of the heterozygote is intermediate between phenotypes of the two homozygotes

D) Gene mutations may produce many different alleles of a gene

56 / 66

Continuous variations in a population were first observed by:

NUMS 2015
Biology
Variation and Genetics
Elimination Tool:

A) Mendel

B) Correns

C) Nilsson

D) Darwin

57 / 66

What is the risk of color blind baby boy in a family when mother is color blind but father is normal?

NUMS 2019 Cancelled
Biology
Variation and Genetics
Elimination Tool:

A) 25%

B) 50%

C) 75%

D) 100%

58 / 66

A pea plant has purple colored flowers (dominant trait). To check the genotype, which of the following cross is performed

NUMS 2018
Biology
Variation and Genetics
Elimination Tool:

A) Test cross with homozygous white flowered plant

B) Test cross with homozygous purple flowered plant

C) Test cross with heterozygous purple flowered plant

D) Hybird cross

59 / 66

Linked genes with closer loci ______ Mendel’s law of independent assortment as they _____ assort independently.

NUMS 2019 Reconduct
Biology
Variation and Genetics
Elimination Tool:

A) Don’t obey: cannot

B) Don’t obey: can

C) obey : cannot

D) obey : can

60 / 66

Pick the correct option about Drosophila.

NUMS 2015
Biology
Variation and Genetics
Elimination Tool:

A) The Male is larger with a pointed abdomen

B) The female has sex combs on its front legs

C) It has a generation time of just eight weeks

D) Salivary gland cells have giant chromosomes in their nuclei

61 / 66

A pure breeding tall plant was crossed to dwarf plant. What would be probability of “Tt” genotype in F2

0.25

0.5

0.75

0

62 / 66

Ahmed’s blood CANNOT clot properly; he was diagnosed with hemophilia A. He passed this trait to his grandson through his:

NUMS 2019 Reconduct
Biology
Variation and Genetics
Elimination Tool:

A) Wife

B) Son

C) Daughter

D) Brother

63 / 66

What CANNOT be true for a man having B+ve blood group?

NUMS 2019 Reconduct
Biology
Variation and Genetics
Elimination Tool:

A) His mother’s blood group is BB

B) His mother’s blood group is BO

C) His father’s blood group is AO

D) His father’s blood group is AA

64 / 66

In nature, Garden pea is

Self-fertilized

Cross fertilized

Cross pollinated

None of these

65 / 66

J. Seiler in 1914 discovered which type of sex determination in moths?

NUMS 2015
Biology
Variation and Genetics
Elimination Tool:

A) XO – XX

B) XY – XX

C) ZZ – ZW

D) None of the given options

66 / 66

Which of the following characters of pea plant is dominant?

Axial flowers

Yellow pods

White flowers

Wrinkled seeds

Your score is

The average score is 59%

0%

PakLearningSpot LMS

FSC Biology Book MCQs Chapter 8 Class 12

Learn Anytime Anywhere

PakLearningSpot

2 responses to “FSC Biology Book MCQs Chapter 8 Class 12”

FSC Biology Book MCQs Chapter 8 Class 12

Learn Anytime Anywhere

PakLearningSpot

Share this:

2 responses to “FSC Biology Book MCQs Chapter 8 Class 12”