1. Who is known as the “Father of Genetics”?
a) Charles Darwin
b) Gregor Mendel
c) Thomas Hunt Morgan
d) Watson and Crick
Answer: b) Gregor Mendel
Explanation: Mendel, through his experiments on pea plants (Pisum sativum), formulated the laws of inheritance. He is called the Father of Genetics.
2. Mendel conducted experiments on:
a) Maize
b) Pea (Pisum sativum)
c) Wheat
d) Sunflower
Answer: b) Pea
Explanation: Mendel selected pea plants because they had contrasting characters, short generation time, could self-pollinate, and allowed controlled cross-pollination.
3. Mendel studied inheritance of traits using:
a) Hybridization
b) Pure line breeding
c) Controlled pollination
d) All of these
Answer: d) All of these
Explanation: Mendel used true-breeding plants, hybridization experiments, and carefully controlled pollination to study inheritance.
4. The law of segregation is also known as:
a) Law of purity of gametes
b) Law of dominance
c) Law of independent assortment
d) Law of blending inheritance
Answer: a) Law of purity of gametes
Explanation: It states that alleles segregate during gamete formation and each gamete receives only one allele of a pair.
5. Which Mendelian law explains independent inheritance of traits?
a) Law of dominance
b) Law of segregation
c) Law of independent assortment
d) Law of blending
Answer: c) Law of independent assortment
Explanation: Traits assort independently because alleles of different genes are distributed into gametes randomly.
6. The phenotypic ratio in monohybrid cross (F2 generation) is:
a) 3:1
b) 1:2:1
c) 9:3:3:1
d) 2:1
Answer: a) 3:1
Explanation: In a cross like Tt × Tt, 3 plants are tall and 1 dwarf, giving 3:1 ratio.
7. The genotypic ratio in monohybrid cross (F2 generation) is:
a) 3:1
b) 1:2:1
c) 9:3:3:1
d) 1:1
Answer: b) 1:2:1
Explanation: Genotypes are TT : Tt : tt in ratio 1:2:1.
8. The phenotypic ratio in a dihybrid cross (F2 generation) is:
a) 9:3:3:1
b) 1:2:1
c) 3:1
d) 2:1
Answer: a) 9:3:3:1
Explanation: Crossing RrYy × RrYy produces 9 round yellow, 3 round green, 3 wrinkled yellow, 1 wrinkled green.
9. The parental generation (P) in Mendel’s experiments were:
a) Hybrid plants
b) True-breeding plants
c) Mutants
d) Wild varieties
Answer: b) True-breeding plants
Explanation: Mendel used homozygous plants that consistently produced the same traits.
10. Crossing tall (TT) with dwarf (tt) gives all tall progeny (Tt). This shows:
a) Segregation
b) Dominance
c) Recessiveness
d) Independent assortment
Answer: b) Dominance
Explanation: The tall allele (T) masks dwarf allele (t), proving dominance.
11. When both alleles express equally in heterozygote, it is:
a) Dominance
b) Codominance
c) Recessiveness
d) Epistasis
Answer: b) Codominance
Explanation: In codominance, both alleles are expressed (e.g., blood group IAIB shows both A and B antigens).
12. Incomplete dominance was first reported in:
a) Pea
b) Four o’clock plant (Mirabilis jalapa)
c) Maize
d) Wheat
Answer: b) Four o’clock plant
Explanation: Red (RR) × White (rr) give Pink (Rr) in F1, showing blending of traits.
13. Test cross involves:
a) Crossing F1 with dominant parent
b) Crossing F1 with recessive parent
c) Crossing two F1 individuals
d) Crossing two recessive parents
Answer: b) Crossing F1 with recessive parent
Explanation: Test cross is used to determine whether a plant with dominant phenotype is homozygous or heterozygous.
14. The ratio obtained in test cross of dihybrid (AaBb × aabb):
a) 9:3:3:1
b) 1:2:1
c) 1:1:1:1
d) 3:1
Answer: c) 1:1:1:1
Explanation: Each gamete combination results in equal proportion of four phenotypes.
15. Alleles are:
a) Different forms of a gene
b) Different forms of a chromosome
c) Homologous chromosomes
d) Non-coding DNA
Answer: a) Different forms of a gene
Explanation: Alleles represent variants of a gene located at the same locus on homologous chromosomes.
16. If heterozygous tall pea plant (Tt) is selfed, probability of dwarf offspring is:
a) 0%
b) 25%
c) 50%
d) 75%
Answer: b) 25%
Explanation: Cross Tt × Tt gives TT, Tt, Tt, tt → 1/4 are dwarf (tt).
17. A homozygous tall (TT) crossed with heterozygous tall (Tt) gives ratio:
a) 1 TT : 1 Tt
b) 3 Tall : 1 Dwarf
c) 1:2:1
d) All tall
Answer: a) 1 TT : 1 Tt
Explanation: Gametes: TT × Tt → TT and Tt, all tall but genotypic ratio 1:1.
18. Dominant traits studied by Mendel included:
a) Tall stem, round seed, yellow seed
b) Dwarf stem, wrinkled seed, green seed
c) Red flower, white seed
d) Green pod, terminal flower
Answer: a) Tall stem, round seed, yellow seed
Explanation: These were among Mendel’s seven dominant traits in pea.
19. Crossing Aa × Aa produces heterozygotes in what proportion?
a) 25%
b) 50%
c) 75%
d) 100%
Answer: b) 50%
Explanation: Cross Aa × Aa → 25% AA, 50% Aa, 25% aa. Thus, heterozygotes = 50%.
20. Independent assortment occurs because of:
a) Crossing over
b) Random orientation of homologous chromosomes
c) Mutation
d) Gene duplication
Answer: b) Random orientation of homologous chromosomes
Explanation: During metaphase I of meiosis, chromosomes align randomly leading to assortment of alleles.
21. The physical location of a gene on a chromosome is:
a) Allele
b) Genome
c) Locus
d) Codon
Answer: c) Locus
Explanation: A locus is the fixed position where a gene resides on a chromosome.
22. Which of the following is a monohybrid cross?
a) Aa × Aa
b) AaBb × AaBb
c) AaBb × aabb
d) TtRr × TtRr
Answer: a) Aa × Aa
Explanation: Monohybrid cross involves a single trait/gene, hence Aa × Aa.
23. Back cross means:
a) Crossing F1 with F1
b) Crossing F1 with any parent
c) Crossing F1 with dominant parent only
d) Crossing F1 with recessive parent only
Answer: b) Crossing F1 with any parent
Explanation: A back cross can be with either dominant or recessive parent to check genotype or improve traits.
24. Example of recessive trait in Mendel’s pea:
a) Tall stem
b) Round seed
c) Green seed
d) Yellow pod
Answer: c) Green seed
Explanation: Green seed color was recessive, while yellow was dominant.
25. If both alleles are identical, the condition is:
a) Heterozygous
b) Homozygous
c) Hybrid
d) Dominant
Answer: b) Homozygous
Explanation: Homozygous individuals have identical alleles (TT or tt).
26. When a single gene controls more than one trait, the phenomenon is called:
a) Epistasis
b) Pleiotropy
c) Codominance
d) Polygenic inheritance
Answer: b) Pleiotropy
Explanation: In pleiotropy, one gene affects multiple traits. Example: sickle-cell anemia gene affects hemoglobin, RBC shape, and resistance to malaria.
27. Skin color in humans is an example of:
a) Monogenic inheritance
b) Pleiotropy
c) Polygenic inheritance
d) Codominance
Answer: c) Polygenic inheritance
Explanation: Skin color is controlled by multiple genes, each adding a small effect (quantitative inheritance).
28. Interaction of two non-allelic genes where one masks the effect of another is called:
a) Codominance
b) Incomplete dominance
c) Epistasis
d) Polygenic inheritance
Answer: c) Epistasis
Explanation: Example: Coat color in mice, where one gene masks expression of another.
29. The ABO blood group in humans shows:
a) Dominance
b) Codominance and multiple alleles
c) Incomplete dominance
d) Epistasis
Answer: b) Codominance and multiple alleles
Explanation: IA and IB are codominant, i is recessive; thus, 3 alleles control ABO system.
30. Incomplete dominance results in:
a) Blending of traits in heterozygote
b) Expression of both traits fully
c) Complete dominance
d) Suppression of traits
Answer: a) Blending of traits in heterozygote
Explanation: Red × White flower in Mirabilis gives pink.
31. Example of codominance in humans is:
a) Sickle-cell anemia
b) ABO blood group
c) Albinism
d) Cystic fibrosis
Answer: b) ABO blood group
Explanation: Both IA and IB express equally in blood group AB.
32. Lethal alleles were first studied in:
a) Mice
b) Pea
c) Maize
d) Drosophila
Answer: d) Drosophila
Explanation: Inheritance of yellow body color in Drosophila led to discovery of lethal alleles.
33. Which cross gives 9:7 phenotypic ratio?
a) Duplicate genes
b) Complementary genes
c) Codominance
d) Lethal genes
Answer: b) Complementary genes
Explanation: Both genes are required for expression; absence of either masks the trait.
34. Which cross gives 15:1 phenotypic ratio?
a) Duplicate genes
b) Complementary genes
c) Epistasis
d) Codominance
Answer: a) Duplicate genes
Explanation: Either of two dominant genes can produce the trait.
35. Which cross gives 9:3:4 phenotypic ratio?
a) Recessive epistasis
b) Dominant epistasis
c) Codominance
d) Incomplete dominance
Answer: a) Recessive epistasis
Explanation: Example: Coat color in mice.
36. Which cross gives 12:3:1 ratio?
a) Duplicate genes
b) Recessive epistasis
c) Dominant epistasis
d) Complementary genes
Answer: c) Dominant epistasis
Explanation: Presence of one dominant allele masks expression of another gene.
37. Which cross gives 13:3 ratio?
a) Suppressor gene action
b) Recessive epistasis
c) Duplicate gene
d) Dominant epistasis
Answer: a) Suppressor gene action
Explanation: One gene suppresses the effect of another.
38. Who discovered linkage?
a) Mendel
b) Morgan
c) Bateson and Punnett
d) Watson and Crick
Answer: c) Bateson and Punnett
Explanation: They observed deviation from 9:3:3:1 ratio in sweet pea (Lathyrus).
39. Who explained linkage using Drosophila?
a) Mendel
b) Thomas Hunt Morgan
c) Bateson
d) Darwin
Answer: b) Thomas Hunt Morgan
Explanation: Morgan studied X-linked genes in Drosophila and explained linkage.
40. Linkage is:
a) Independent assortment of genes
b) Tendency of genes to be inherited together
c) Random mutation
d) Codominance
Answer: b) Tendency of genes to be inherited together
Explanation: Genes close on the same chromosome are inherited together.
41. The strength of linkage between two genes depends on:
a) Distance between genes on chromosome
b) Type of gametes
c) Type of dominance
d) DNA replication
Answer: a) Distance between genes on chromosome
Explanation: Closer genes show stronger linkage.
42. Crossing over occurs in:
a) Mitosis
b) Prophase I of meiosis
c) Metaphase II
d) Anaphase I
Answer: b) Prophase I of meiosis
Explanation: Homologous chromosomes exchange segments during pachytene.
43. The exchange of genetic material between non-sister chromatids is called:
a) Mutation
b) Crossing over
c) Independent assortment
d) Recombination
Answer: b) Crossing over
44. The unit of recombination frequency is:
a) Centromere
b) Map unit (centimorgan)
c) Base pair
d) Chromatid
Answer: b) Map unit (centimorgan)
Explanation: 1% recombination = 1 centimorgan (cM).
45. Maximum recombination frequency possible between two genes is:
a) 25%
b) 50%
c) 75%
d) 100%
Answer: b) 50%
Explanation: When genes are very far apart, they assort independently (like unlinked).
46. If two genes show 20% recombination, distance between them is:
a) 2 cM
b) 20 cM
c) 50 cM
d) 200 cM
Answer: b) 20 cM
47. The first genetic map was prepared by:
a) Mendel
b) Morgan
c) Sturtevant
d) Watson
Answer: c) Sturtevant
Explanation: A student of Morgan, he mapped genes in Drosophila using recombination frequencies.
48. Coupling and repulsion hypothesis was given by:
a) Bateson and Punnett
b) Morgan
c) Mendel
d) Sturtevant
Answer: a) Bateson and Punnett
Explanation: They observed linked genes inherited together (coupling) or separately (repulsion).
49. A test cross with linked genes shows:
a) 9:3:3:1
b) 1:2:1
c) Deviation from Mendelian ratio
d) 3:1
Answer: c) Deviation from Mendelian ratio
Explanation: Linked genes do not assort independently.
50. Linkage reduces:
a) Mutation
b) Independent assortment
c) Crossing over
d) Gene expression
Answer: b) Independent assortment
Explanation: Genes located close together on the same chromosome tend to be inherited together.
51. The physical basis of heredity is:
a) Genes
b) Chromosomes
c) DNA
d) Protein
Answer: b) Chromosomes
Explanation: Chromosomes carry genes, the units of heredity. Sutton and Boveri (1902) proposed the chromosomal theory of inheritance.
52. Genes are located on:
a) Centromere
b) Chromosomes
c) Ribosomes
d) Mitochondria only
Answer: b) Chromosomes
Explanation: Genes are specific DNA sequences present at loci on chromosomes.
53. Who proposed the chromosomal theory of inheritance?
a) Mendel
b) Sutton and Boveri
c) Morgan
d) Watson and Crick
Answer: b) Sutton and Boveri
Explanation: They correlated Mendel’s factors with chromosomes.
54. In humans, the number of chromosomes is:
a) 23
b) 46
c) 44
d) 48
Answer: b) 46
Explanation: Humans have 23 pairs = 46 chromosomes (22 autosome pairs + 1 sex chromosome pair).
55. The term “linkage group” refers to:
a) Chromosome
b) Gene cluster
c) Alleles
d) Genome
Answer: a) Chromosome
Explanation: Each chromosome carries one linkage group (all genes inherited together).
56. Mutation is:
a) Change in phenotype only
b) Sudden heritable change in genetic material
c) Gene recombination
d) Environmental adaptation
Answer: b) Sudden heritable change in genetic material
57. Mutations that occur in gametes are called:
a) Somatic mutations
b) Germinal mutations
c) Chromosomal aberrations
d) Induced mutations
Answer: b) Germinal mutations
Explanation: They are heritable and passed to offspring.
58. Mutations occurring in body cells are:
a) Germinal mutations
b) Somatic mutations
c) Induced mutations
d) Chromosomal mutations
Answer: b) Somatic mutations
Explanation: Somatic mutations are not inherited by offspring.
59. Mutation in a single nucleotide is:
a) Point mutation
b) Frameshift mutation
c) Chromosomal mutation
d) Polyploidy
Answer: a) Point mutation
Explanation: Example: Sickle cell anemia caused by substitution of A → T in HBB gene.
60. Sickle cell anemia is caused by substitution of:
a) Glutamic acid → Valine
b) Valine → Glutamic acid
c) Glycine → Serine
d) Alanine → Glycine
Answer: a) Glutamic acid → Valine
Explanation: At 6th position of β-globin chain, glutamic acid is replaced by valine.
61. Frameshift mutation is caused by:
a) Substitution of base
b) Addition or deletion of base
c) Inversion of segment
d) Duplication of gene
Answer: b) Addition or deletion of base
Explanation: Alters reading frame, changing all downstream codons.
62. A mutation that changes one codon to a stop codon is:
a) Missense mutation
b) Nonsense mutation
c) Silent mutation
d) Neutral mutation
Answer: b) Nonsense mutation
Explanation: Leads to premature termination of protein synthesis.
63. Down’s syndrome is caused by:
a) Trisomy of chromosome 21
b) Monosomy of X
c) Trisomy of 18
d) Deletion of chromosome 5
Answer: a) Trisomy of chromosome 21
64. Turner’s syndrome is:
a) 47, XXY
b) 45, XO
c) 47, XYY
d) 46, XY
Answer: b) 45, XO
Explanation: Females with single X chromosome, sterile.
65. Klinefelter’s syndrome is:
a) 47, XXY
b) 45, XO
c) 47, XYY
d) 46, XY
Answer: a) 47, XXY
Explanation: Males with extra X chromosome, show sterility and feminine characters.
66. Cri-du-chat syndrome results from:
a) Trisomy of chromosome 18
b) Deletion of short arm of chromosome 5
c) Trisomy of chromosome 13
d) Translocation of chromosome 21
Answer: b) Deletion of short arm of chromosome 5
Explanation: Causes cat-like cry in infants.
67. Who discovered transformation in bacteria?
a) Griffith
b) Avery, MacLeod, McCarty
c) Hershey and Chase
d) Watson and Crick
Answer: a) Griffith
Explanation: In 1928, Griffith showed transfer of genetic material in Streptococcus pneumoniae.
68. Avery, MacLeod and McCarty proved that the transforming principle was:
a) Protein
b) RNA
c) DNA
d) Lipid
Answer: c) DNA
69. Hershey and Chase experiment with bacteriophage showed:
a) Protein is genetic material
b) DNA is genetic material
c) RNA is genetic material in all organisms
d) Chromosomes carry only proteins
Answer: b) DNA is genetic material
Explanation: DNA (labelled with 32P) entered bacteria, proteins (labelled with 35S) did not.
70. In some viruses, genetic material is:
a) DNA only
b) RNA only
c) Either DNA or RNA
d) Both DNA and RNA
Answer: c) Either DNA or RNA
Explanation: Viruses may have DNA (bacteriophages) or RNA (HIV, influenza).
71. The double helix model of DNA was proposed by:
a) Hershey and Chase
b) Avery and MacLeod
c) Watson and Crick
d) Franklin and Wilkins
Answer: c) Watson and Crick (1953)
72. The base pairing rule of DNA was given by:
a) Watson
b) Chargaff
c) Crick
d) Franklin
Answer: b) Chargaff
Explanation: A = T, G = C, purines equal pyrimidines.
73. In DNA, adenine pairs with thymine by:
a) One H-bond
b) Two H-bonds
c) Three H-bonds
d) Covalent bond
Answer: b) Two H-bonds
74. In DNA, guanine pairs with cytosine by:
a) One H-bond
b) Two H-bonds
c) Three H-bonds
d) Covalent bond
Answer: c) Three H-bonds
75. The backbone of DNA is made up of:
a) Nucleotide bases only
b) Sugar and phosphate
c) Ribose and uracil
d) Hydrogen bonds
Answer: b) Sugar and phosphate
Explanation: Deoxyribose sugars linked by phosphate groups form the backbone, while bases project inward.
76. The three stop codons are:
a) AUG, UAA, UAG
b) UAA, UAG, UGA
c) AUG, UGA, UGG
d) UAA, UGA, GUG
Answer: b) UAA, UAG, UGA
Explanation: These codons do not code for amino acids and terminate protein synthesis.
77. The start codon for protein synthesis is:
a) UAA
b) AUG
c) UGA
d) UAG
Answer: b) AUG
Explanation: AUG codes for methionine and serves as initiation codon.
78. The genetic code is:
a) Ambiguous
b) Non-overlapping, nearly universal
c) Random
d) Different for each organism
Answer: b) Non-overlapping, nearly universal
Explanation: The same codons specify the same amino acids in almost all organisms.
79. Who proved that one gene codes for one enzyme?
a) Griffith
b) Beadle and Tatum
c) Watson and Crick
d) Mendel
Answer: b) Beadle and Tatum
Explanation: In Neurospora crassa, they showed “one gene–one enzyme” hypothesis.
80. The updated concept of Beadle and Tatum is:
a) One gene–one protein
b) One gene–one polypeptide
c) One gene–one trait
d) One gene–many enzymes
Answer: b) One gene–one polypeptide
Explanation: Some proteins consist of multiple polypeptide chains encoded by different genes.
81. The central dogma of molecular biology was given by:
a) Watson
b) Crick
c) Mendel
d) Morgan
Answer: b) Crick
Explanation: Information flows DNA → RNA → Protein.
82. Transcription is the process of:
a) RNA → DNA
b) DNA → RNA
c) RNA → Protein
d) DNA replication
Answer: b) DNA → RNA
83. Translation is the process of:
a) RNA → DNA
b) DNA → RNA
c) RNA → Protein
d) DNA replication
Answer: c) RNA → Protein
84. Enzyme involved in transcription:
a) DNA polymerase
b) RNA polymerase
c) Ligase
d) Helicase
Answer: b) RNA polymerase
85. Enzyme that removes supercoiling during DNA replication is:
a) Helicase
b) Topoisomerase (gyrase)
c) Primase
d) Ligase
Answer: b) Topoisomerase
86. Okazaki fragments are formed during:
a) Transcription
b) Leading strand synthesis
c) Lagging strand synthesis
d) RNA splicing
Answer: c) Lagging strand synthesis
Explanation: DNA polymerase works in 5′ → 3′ direction, producing fragments on lagging strand.
87. The functional unit of DNA carrying information for protein synthesis is:
a) Chromosome
b) Gene
c) Codon
d) Nucleotide
Answer: b) Gene
88. Splicing in eukaryotes removes:
a) Exons
b) Introns
c) Codons
d) mRNA cap
Answer: b) Introns
Explanation: Non-coding introns are removed, exons are joined.
89. In prokaryotes, transcription and translation are:
a) Coupled (occur simultaneously)
b) Separate processes
c) Only transcription occurs
d) Only translation occurs
Answer: a) Coupled (occur simultaneously)
90. In eukaryotes, mRNA is first produced as:
a) Mature mRNA
b) Pre-mRNA (hnRNA)
c) tRNA
d) rRNA
Answer: b) Pre-mRNA (hnRNA)
91. The anticodon is found on:
a) DNA
b) mRNA
c) tRNA
d) rRNA
Answer: c) tRNA
Explanation: Anticodon base-pairs with mRNA codon during translation.
92. Ribosomes are sites of:
a) DNA replication
b) RNA transcription
c) Protein synthesis
d) Mutation repair
Answer: c) Protein synthesis
93. The first genetic material was likely:
a) DNA
b) RNA
c) Protein
d) Lipid
Answer: b) RNA
Explanation: RNA world hypothesis suggests RNA acted as both genetic material and catalyst.
94. Who discovered transposons (jumping genes)?
a) Mendel
b) Barbara McClintock
c) Morgan
d) Griffith
Answer: b) Barbara McClintock
95. VNTRs (Variable Number Tandem Repeats) are used in:
a) Protein synthesis
b) DNA fingerprinting
c) Mutation repair
d) RNA editing
Answer: b) DNA fingerprinting
96. Human genome has approximately:
a) 1,000 genes
b) 20,000–25,000 genes
c) 1 million genes
d) 46 genes
Answer: b) 20,000–25,000 genes
97. Human Genome Project was completed in:
a) 1995
b) 2003
c) 2010
d) 1985
Answer: b) 2003
98. A child with blood group O cannot have parents with:
a) A and B
b) A and O
c) B and O
d) AB and O
Answer: d) AB and O
Explanation: O group (ii) requires both parents to contribute i allele; AB parent cannot provide i.
99. Haemophilia is:
a) Autosomal dominant
b) Autosomal recessive
c) X-linked recessive
d) Y-linked
Answer: c) X-linked recessive
Explanation: Males suffer more, females are usually carriers.
100. Color blindness is:
a) Autosomal dominant
b) Autosomal recessive
c) X-linked recessive
d) Y-linked
Answer: c) X-linked recessive
Explanation: The gene is on the X chromosome; more common in males.