Heredity and Evolution Class 10 MCQ Chapter 9 Science

Heredity and Evolution Class 10 MCQ with Solution

1.The ultimate source of variations is :

(A) Natural selection

(B) Sexual reproduction

(D) Hormonal actions

Ans. (C)

Sol. Inheritable variations formed mostly in germinal cells which are either already present in the ancestors or develop anew due to mutations.

2. An allele is said to be dominant if it is :

(A) Expressed only in heterozygous condition

(B) Expressed only in homozygous condition

(D) Expressed in F₂ generation

Ans. (C)

Sol. Dominant Factor or Allele is one of a pair of alleles which can express itself whether present in homozygous or heterozygous state.

3. The branch dealing with inheritance of traits and variation is called :

(A) Genetics

(B) Mutations

(D) Evolution

Ans. (A)

Sol. The branch dealing with inheritance of traits and variation is called Genetics.

4. An individual with constrasting alleles is called :

(A) Homozygous (pure)

(B) Monoecious

(D) Dioecious

Ans. (C)

Sol. Heterozygous (hybrid) is an individual which contains the two contrasting factors of a character or two different alleles of a gene on its homologous chromosomes. It is not pure and is called hybrid for that character.

5. An individual with two identical members of a pair of factors is called :

(A) Heterozygote

(B) Hybrid

(D) Pureline

Ans. (C)

Sol. Homozygote is an individual which contains the identical allele of a gene or factors of a character on its homologous chromosomes. It is pure for that character.

6. A pair of genes representing alternate expressions of the same character is called :

(A) Alleles or allelomorphic pair

(B) Nonallelic alleles

(D) Heterozygous genes

Ans. (A)

Sol. The two mendelian factors which occur on the same locus in the two homologous chromosomes of an individual and control the expression of a chracter are called alleles or allelomorphs.

7. If 75% of the offspring show dominant character, the parents are :

(A) Both hybrids

(B) One dominant and one recessive

(D) Both dominant

Ans. (A)

Sol. If hybrids (Tt × Tt) are crossed then phenotypic ratio of next generation is 3 : 1. In this 75% of the offsprings show dominant character.

8. Continuity of life is made possible by :

(A) Sexual reproduction

(B) Mutations

(D) Meiosis

Ans. (A)

Sol. Sexual reproduction is the process of development of next organisms through the formation and fusion of gametes.

9. Which one of the following best describes a gene?

(A) A triplet of nucleotide bases

(B) A specific length of a single staranded DNA

(D) The nucleotide sequence of a chromatid

Ans. (C)

Sol. A gene is defined as a unit of inheritance composed of a segment of DNA or chromosome situated at a specific locus which carries coded information associated with a specific function and DNA is double stranded.

10. Genes are not found in pairs in :

(A) Sperm cell

(B) Fertilized egg

(D) Epithellal cell

Ans. (A)

Sol. Sperm cells are male gametes and these have half number of genes as compared to diploid cell.

11. The pairs of characters used by Mendel during his experiment were :

(A) Ten

(B) Six

(D) Two

Ans. (C)

Sol. Mendel restricted his experiments to one or few pairs of contrasting traits in each experiment and in all 7.

12. The monohybrid genotypic ratio 1 : 2 : 1 in F₂ generation indicates :

(A) Segregation

(B) Independent assortment

(D) Incomplete dominance

Ans. (A)

Sol. Hereditary characters are determined by particular factors. Those units occur in pair in an individual. On formation of germ cell. They become segregated so that only one of pair transmitted through any one gamete. This law of purity of gametes or segregation.

13. Genotype is :

(A) Genetic composition of species

(B) Genetic composition of plastids

(D) Genetic composition of an individual.

Ans. (D)

Sol. The genetic complement of an organism is genotype.

14. Sexually reproducing organisms contribute in their offspring :

(A) All of the genes

(B) One half of their genes

(D) Double the number of genes

Ans. (B)

Sol. During sexual reproduction meiosis takes place and number of chromosomes are halved.

15. Male sex is determined in human zygote is by :

(A) Nutrition of mother

(B) Strength of father

(D) Chromosome composition of sperm

Ans. (D)

Sol. The sperm may have either X or Y chromosomes determining the sex. A sperm has 22 autosomes & either an X or Y with them. So, if y fuses with egg containing x, it will result in male child.

16. Base pairs found in one turn of DNA spiral are:

(A) 12

(B) 11

(D) 8

Ans. (C)

Sol. A turn of DNA has 10 base pairs, adjacent base pairs seperated by 3.4 Å.

17. Who ascertained the number of human chromomes as 46.

(A) T.H Morgan

(B) Bareson and purmnet

(D) Turner

Ans. (C)

Sol. Tijo & Levan gave the number of chromosomes in human as 46.

18. Haemophilia is a genetic disorder in which :

(A) Blood fails to coagulate after an injury

(B) There is delayed coagulation of blood

(D) Blood cell count falls

Ans. (B)

Sol. Haemophilia also called bleeders disease, blood clotting is deficient because of lack of necessary substrate, thromboplastin.

19. Auto somes are :

(A) Sex chromosomes

(B) Chromosomes except sex chromosome

(D) Unit membrane bound cell organella

Ans. (B)

Sol. Chromosomes other than sex chromosomes are called autosomes.

20. The person with Turner’s syndrome has :

(A) 45 autosomes and x sex chromosome

(B) 44 autosomes and xxy sex chromosome

(D) 44 autosomes and x sex chromosomes

Ans. (D)

Sol. The chromosome number 2A = 45, chromosomal formula is 44+XO. Turner syndrome is a sterile female.

21. Who proved that DNA is the genetic material :

(A) Griffith

(B) Watson and Crick

(D) Nirenberg

Ans. (C)

Sol. Hershey & Chase performed the experiment to give DNA as genetic material.

22. Which of the following statements is false :

(A) Recessive allele is unable to express it’s effect in heterozygous state

(B) Dominant allele express itself in heterozgous state

(D) Heterozygous individual show hybrid vigour

Ans. (C)

Sol. Homozygous individuals produce only a single type of gamete.

23. The haploid set of chromosomes or genes of an organism is called :

(A) Genome

(B) Karyotype

(D) None of the above

Ans. (A)

Sol. The haploid set of genes of an organism is genome.

24. What do you mean by ‘phenotype’ :

(A) Obervable morphological or physiological characters of an individual

(B) Genetic constitution of an individual

(D) Both morphological as well as genetic characters of an individual

Ans. (A)

Sol. Phenotype is external appearance (morphological expression) which is product of genotype and environment.

25. Organisms phenotypically similar but genetically different are :

(A) Heterozygous

(B) Homozygous

(D) Multizygous

Ans. (A)

Sol. Individuals having different alleles on homologous chromosomes are heterozygous.

26. Heterozygous organism for two genes shall be :

(A) RRYY

(B) RrYY

(D) RRYy

Ans. (C)

Sol. Individuals having different alleles on homologous chromosomes are heterozygous.

27. Sex chromosomes of man are best depicted as :

(A) Homologous

(B) Heterologous

(D) None of the above

Ans. (B)

Sol. Males have a X & Y chromosome, both are different. So, condition is heterologous.

28. The frequency of mutant gene in a population is expected to increase if that gene is :

(A) Dominant

(B) Recessive

(D) Favourable selected

Ans. (B)

Sol. Sudden writable changes occurring in an organism are mutations neutral or even harmful mutatious are not lost from gene because they are recessive and survive in heterozygous individuals.

29. Parents are heterozygous albino. What will be the first three children?

(A) Some normal, heterozygous and albino

(B) All normal

(D) No norma

Ans. (A)

Sol. The children produced are some heterozygous, while other are normal but albino.

30. Mutation is :

(A) A change that is inherited

(B) A change, which affects the parents only but never inherited

(D) A factor responsible for plant growth

Ans. (A)

Sol. Sudden writable changes occurring in an organism are mutations neutral or even harmful mutatious are not lost from gene because they are recessive and survine in heterozygous individuals.

31. The following ratio is generally constant for a given species :

(A) A + C / T + G

(B) A + G / C + T

(D) G + C / A + T

Ans. (D)

Sol. A+T/G+C is constant for a given species.

32. Barbara McClintock’s work showed that in maize :

(A) RNA transcribed from genes form mature mRNA

(B) Genes are located at specific sites of chromosomes

(D) Genes are exchanged between chromosomes during chiasmata formation

Ans. (C)

Sol. Barbara McClintock’s gave the concept of transposons i.e. genes movement within or one to another chromosome.

33. Mendel formulated the law of purity of gametes on the basis of :

(A) Test cross

(B) Back cross

(D) Dihybrid cross

Ans. (C)

Sol. Herediteray characters are determined by particular factors. Those units occur in pair in an individual. On formation of germ cell. They become segregated so that only one of pair transmitted through any one gamete. This law of purity of gametes or segregation.

34. Mutations are commonly :

(A) Dominant

(B) Codominant

(D) Incomplete

Ans. (C)

Sol. Sudden heritable changes occurring in an organism are mutations neutral or even harmful mutations are met lost from gene because they are recessive and survive in heterozygous individuals.

35. Which of the following constitute the seven contrasting pair of characters noticed by G.J Mendel in Pea :

(A) Height of the plant

(B) Colour of seed coat

(D) All of the above

Ans. (D)

Sol. The traits of pea studied by mendel are plant height flower colour, flower position, pod colour, pod shape, seed colour, seed shape.

36. Medel’s law were rediscovered by :

(A) Correns

(B) Hugodevaries

(D) All of the above

Ans. (A)

Sol. Mendels laws were rediscovered by three scientist Hugo de vries, Tschermak & Correns simultaneously.

37. Law of mendel which is not completely applicable :

(A) Law of Dominance

(B) Law of segregation

(D) Co-dominance

Ans. (C)

Sol. Due to the phenomenon of linkage law of independent is not applicable.

38. Which of the following characters of pea were not studied by Mendel :

(A) Length of plant

(B) Colour of plant

(D) Colour of pod

Ans. (B)

Sol. The traits of pea studied by Mendel are plant height, flower colour, flower position, pod colour, pod shape, seed colour, seed shape.

39. 1 : 1 : 1 : 1 ratio is obtained when :

(A) Yy Rr is crossed with yyrr

(B) YyRr is crossed with YYrr

(D) None of these

Ans. (A)

Sol. Back cross to recessive parent is test cross. A cross of F₁, with recessive parent is test cross. During mohybrid cross, ratio is 1:1 & dihybrid 1:1:1:1.

40. Mendel’s laws of independent assortment is :

(A) Applicable to all the genes on all the chromosomes

(B) Applicable to all the dominant genes

(D) Is not applicable when genes are on the same chromosomes

Ans. (D)

Sol. Due to the phenomenon of linkage, (the closure the distance between two genes, they tend to remain linked) law of independent assortment is not applicable.

41. In mendelian crosses the genes responsible for different characters separate through :

(A) Fusion

(B) Aggregation

(D) Distribution

Ans. (C)

42. A pure red flowered Pea plant is crossed with pure white flowered Pea plant. The offspring will be after F₁ generation :

(A) All red flowered

(B) Half red flowered

(D) Half pink flowered

Ans. (A)

Sol. During monohybrid cross the F₁, always has dominant characters and all are phenotypically similar to dominant but are heterozygous.

43. Mendal’s laws of heredity can be explained with the help of :

(A) Mitosis

(B) Meiosis

(D) Both A and B

Ans. (B)

Sol. Meiosis helps in making number of choromosomes half, thus aid in explaining mendel’s laws.

44. A recessive trait in Garden Pea is :

(A) Wrinkled seeds

(B) Tall stem

(D) Coloured seed coat

Ans. (A)

Sol. The recessive traits in pea plant are dwarf (Height), white (flower colour), Terminal (Flower Position), Yellow (Pod Colour), Constricted (Pod Shape), Green (Seed Colour), wrinkled (seed shape).

45.

(A) a – Tall Red, b – Tall Red, c – Dwarf Red, d – Dwarf White

(B) a – Tall Red, b – Tall White, c – Tall Red, d – Dwarf Red

(D) None of these

Ans. (B)

Sol. Tall & Red are Dominant traits, this is cross between Tall Red and dwarf white. (Dihybrid cross)

46. Which one of the following produced only one type of gamete :

(A) AA, aa, YYrr

(B) aaBB, AABB, aabb

(D) All correct

Ans. (D)

Sol. All are in homozygous conditions. So, all will produce similar gametes.

47. Mendel, while studying the garden pea (Pisum sativum) observed its seven traits. Which one doesn’t match with its dominant character ?

(A) Flower colour-purple

(B) Flower position-axial

(D) Seed shape-round

Ans. (C)

Sol. The dominant characters are tall (height), purple (flower colour), axial (flower position), green (pod colour), inflated (pod shape), yellow (seed colour), round (seed shape).

48. A red and tall dominant character hybrid plant is crossed with recessive white dwarf plant (RrTt × rr tt). What will be the ratio of respective four combinations : red tall, red dwarf, white tall and white dwarf plants in the next generation :

(A) 9 : 3 : 3 : 1

(B) 15 : 1 : 0 : 0

(D) 4 : 4 : 4 : 4

Ans. (D)

Sol. Back cross to recessive parent is test cross. A cross of F, with recessive parent is test cross. During mohybrid ratio is 1:1 & dihybrid 1:1:1:1.

49. The ratio of phenotypes if F₂ is a monohybrid cross is :

(A) 3 : 1

(B) 1 : 2 : 1

(D) 2 : 1

Ans. (A)

Sol. The phenotypic ratio in monohybrid cross for F₂ is 3:1

50. Meiosis in AaBb will produce gametes :

(A) Aa, Bb

(B) Aa, bb

(D) AB, aB, Ab, ab

Ans. (D)

Sol. Gametes produced are AB, aB, Ab, ab.

51. Male sex is determined in human zygote is by :

(A) Nutrition of mother

(B) Strength of father

(D) Chromosome composition of sperm

Ans. (D)

Sol. The sperm may have either X or Y chromosomes determining the sex. A sperm has 22 autosomes & either an X or Y with them.

52. Out of two alleles of the same gene, one finds morphological expression. The phenomenon is :

(A) Epistasis

(B) Codominance

(D) Incomplete dominance

Ans. (C)

Sol. When an allele fails to express itself in presence of other, the former is recessive while later is dominant.

53. There are 64 codons in genetic code dictionary because :

(A) There are 64 types of RNA’s found in the cell

(B) There are 44 meaning less and 20 codons for amino acids

(D) There are 64 amino acids to be coded

Ans. (C)

Sol. Genetic code appear in triplets, which is made up of different combinations of four nucleotides.

(4³= 4 × 4 × 4 = 64)

54. Eukaryotic chromosome is made of :

(A) DNA

(B) DNA + proteins

(D) RNA

Ans. (B)

Sol. The chromosome in eukaryotes is made up of DNA & proteins called as histones.

55. DNA polymerase enzyme is required for the synthesis of :

(A) RNA from DNA

(B) DNA from DNA

(D) DNA from RNA

Ans. (B)

Sol. DNA polymerase attaches nucleotides to the growing DNA.

56. The two strands of DNA are :

(A) Similar and parallel

(B) Similar but anti-parallel.

(D) Complementary and parallel

Ans. (C)

Sol. The two strands of DNA are antiparallel it means one in 5’-3’ direction while other in 3’-5’ and also complementary. Which means if A pairs with T then T will pair with A. (A = Adenine, T = Thymine).

57. Genes are formed of :

(A) RNA

(B) DNA

(D) DNA & proteins

Ans. (B)

Sol. DNA of some length determining a specific character is Gene.

58. Nucleotide base present in RNA & not in DNA is :

(A) Cytosine

(B) Uracil

(D) Guanine

Ans. (B)

Sol. The DNA is made up of nucleotide purine (Adenine Guanine), pryrimidine (cytosine, thymine) while RNA has uracil instead of thymine.

59. Segment of DNA which can shift positions are :

(A) Transposons

(B) Jumping genes

(D) Both (A) & (B)

Ans. (D)

Sol. The transposons or jumping genes can change there position across a chromosome length.

60. Auto somes are :

(A) Sex chromosomes

(B) Chromosomes except sex chromosome

(D) Unit membrane bound cell organella

Ans. (B)

Sol. Chromosomes other than sex chromosomes are called autosomes.

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