CBSE Class 10 Science Heredity and Evolution MCQs Set H

Question. Which of the following statement is incorrect?
(a) For every hormone there is a gene.
(b) For every protein there is a gene.
(c) For production of every enzyme there is a gene.
(d) For every molecule of fat there is a gene.
Answer: d

Question. A cross between a tall plant (TT) and short pea plant (tt) resulted in progeny that were all tall plants because
(a) Tallness is the dominant trait.
(b) Shortness is the dominant trait.
(c) Tallness is the recessive trait.
(d) Height of pea plant is not governed by gene ‘T’ or ‘t’.
Answer: a

Question. A zygote which has an X-chromosome inherited from the father will develop into a
(a) Boy.
(b) Girl.
(c) X- chromosome does not determine the sex of a child.
(d) Either boy or girl.
Answer: b

Question. If a round, green seeded pea plant (RRyy) is crossed with wrinkled, yellow seeded pea plant, (rrYY) the seeds produced in F1 generation are :
(a) Round and yellow.
(b) Round and green.
(c) Wrinkled and green.
(d) Wrinkled and yellow.
Answer: a

Question. Two pink coloured flowers on crossing resulted in 1 red, 2 pink and 1 white flower progeny. The nature of the cross will be
(a) Double fertilization.
(b) Self-pollination.
(c) Cross fertilization.
(d) No fertilization.
Answer: c

Question. The maleness of a child is determined by
(a) The X chromosome in the zygote.
(b) The Y chromosome in zygote.
(c) The cytoplasm of germ cell which determines the sex.
(d) Sex is determined by chance.
Answer: b

Question. The number of pair (s) of sex chromosomes in the zygote of humans is:
(a) One.
(b) Two.
(c) Three.
(d) Four.
Answer: a

Assertion and Reason Based MCQs :

Directions : In the following questions, A statement of Assertion (a) is followed by a statement of Reason (R). Mark the correct choice as.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is NOT the correct explanation of A.
(c) A is true but R is false.
(d) A is false and R is true.

Question. Assertion (a): In humans, males play an important role in determining the sex of the child.
Reason (R): Males have two X chromosomes.
Answer: c

Question. Assertion (a): Mendel choose a number of varieties of garden pea as plant material for his experiments.
Reason (R): Garden pea has well defined characters and is bisexual.
Answer: a

Question. Assertion (a): Mendel in his experiment selected only two characters of seed.
Reason (R): He studied single character at one time.
Answer: d

Question. Assertion: A zygote with two X chromosomes develops into a boy.
Reason: If the egg cell carrying an X chromosome fuses with the sperm carrying a Y chromosome, the resulting child would be a boy.
Answer: d

Question. Assertion (a): Mendel proposed the law of inheritance of traits from the first generation to the next generation.
Reason (R): Mendel’s Law of segregation is also known as “Law of purity of gametes.
Answer: b

Question. Assertion: A geneticist crossed a pea plant having violet flowers with a pea plant having white flowers, he got all violet flowers in first generation.
Reason: White colour gene is not passed on to next generation.
Answer: c

Case Based Questions:

I. Read the passage and answer the following questions.
In a cross between plants with purple flowers and plants with white flowers, the offspring of F1 generation all had white flowers. When the F1 generation was self-crossed, it was observed in the F2 generation that out of 100, 75 flowers were white. Make a cross and answer the following questions:

Question. Which of these is not the genotype of F2 progeny?
(a) WW
(b) Ww
(c) ww
(d) Wp

Answer: d

Question. The ratio of ‘White: Purple’ flowers in the F2 generation is:
(a) 3 : 1
(b) 1 : 2
(c) 1 : 3
(d) 2 : 1

Answer: a

Question. The above cross is known as:
(a) Monohybrid cross
(b) Dihybrid cross
(c) Test cross
(d) Back cross

Answer: a

Question. In a monohybrid cross between two heterozygous individuals, percentage of pure homozygous individuals obtained in F1 generation is:
(a) 25%
(b) 50%
(c) 75%
(d) 100%

Answer: b

III. Read the given passage and answer the following questions.
Seema crossed pure breed pea plants having round-yellow seeds with wrinkled green seeds and found that only A-B type of seeds were produced in the F1 generation. When F1 generation pea plants having A-B type of seeds were cross-breed by self pollination, then in addition to the original round yellow and wrinkled green seeds, two new varieties A-D and C-B types of seeds were also obtained.

Question. A-D are __________ and C-B are _______ type of seeds.
(a) Round green and wrinkled yellow respectively
(b) Round yellow and wrinkled green respectively
(c) Wrinkled green and round green respectively
(c) Wrinkled green and round yellow respectively

Answer: a

Question. Which one of these will be produced in maximum number in the F2 generation?
(a) A-B
(b) A-D
(c) Both (a) and (b)
(d) None of these

Answer: a

Question. What are A-B type of seeds?
(a) Round -yellow
(b) Round- green
(c) Wrinkled- yellow
(d) Wrinkled- green

Answer: a

Question. A and B are _____________ traits.
(a) dominant
(b) recessive
(c) Both (a) and (b)
(d) None of these

Answer: a

II. Study the given cross showing self pollination in F1 and answer the following questions from 
RRYY × rryy Parents
(Round Yellow) (Wrinkled Green)
RrYy × ________ F1 generation
(Round Yellow)

Question. The ratio of the combination of characters in the F2 progeny is:
(a) 3 : 1
(b) 1 : 2 : 1
(c) 1 : 1 : 1
(d) 9 : 3 : 3 : 1

Answer: d

Question. A Mendelian experiment consists of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short.
This suggests that the genetic make-up of the tall parent cod depicted as
(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw

Answer: c

Question. The missing blank in the above cross is:
(a) RrYy
(b) RRYY
(c) RryY
(d) rryy

Answer: a

Question. The combination of characters in the F2 progeny are:
(a) Round Yellow: Round Green: Wrinkled Yellow: Wrinkled Green
(b) Round Green : Round Yellow : Wrinkled yellow : Wrinkled Green
(c) Round Yellow : Round Green : Wrinkled Green: Wrinkled yellow
(d) Round Green : Round Yellow : Wrinkled yellow : Wrinkled Green

Answer: a