Mendelian Genetics (Monohybrid / Dihybrid / Trihybrid)

Q1. In a monohybrid cross of Aa × Aa, the expected phenotypic ratio is:
A. 1:1
B. 3:1
C. 1:2:1
D. 9:3:3:1

Q2. In a dihybrid cross AaBb × AaBb, the phenotypic ratio of offspring is:
A. 3:1
B. 9:3:3:1
C. 1:2:1
D. 1:1:1:1

Q3. In Mendel’s pea plant experiments, tall (T) is dominant over dwarf (t). A cross Tt × Tt produces:
A. 25% tall, 75% dwarf
B. 75% tall, 25% dwarf
C. 50% tall, 50% dwarf
D. 100% tall

Q4. The law of segregation states that:
A. Genes on different chromosomes segregate independently
B. Alleles segregate during gamete formation
C. Recessive traits are eliminated
D. Dominant alleles always express

Q5. In a tri-hybrid cross, AaBbCc × AaBbCc, the probability of getting all dominant traits is:
A. 1/8
B. 1/4
C. 1/64
D. 27/64

Q6. A true-breeding tall plant (TT) is crossed with a dwarf (tt). All F1 are tall. The cross demonstrates:
A. Codominance
B. Dominance
C. Incomplete dominance
D. Epistasis

Q7. In a monohybrid test cross, an unknown genotype shows 1:1 ratio in offspring. The genotype is:
A. Homozygous dominant
B. Homozygous recessive
C. Heterozygous
D. Pure line

Q8. Mendel’s law of independent assortment applies to:
A. Linked genes
B. Alleles of same gene
C. Genes on different chromosomes
D. Sex-linked traits

Q9. In a cross Aa × Aa, the genotypic ratio of offspring is:
A. 1:1
B. 3:1
C. 1:2:1
D. 9:3:3:1

Q10. Mendel used which plant for his experiments?
A. Maize
B. Pea
C. Wheat
D. Barley

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Deviations from Mendelian Ratios

Q11. In incomplete dominance, F1 shows:
A. Only dominant trait
B. Blend of parental traits
C. Only recessive trait
D. 3:1 ratio

Q12. Red (RR) × White (rr) snapdragon cross produces F1 pink (Rr). This is:
A. Incomplete dominance
B. Codominance
C. Epistasis
D. Polygenic inheritance

Q13. In codominance, both alleles:
A. Blend
B. Express equally
C. One is suppressed
D. Produce 3:1 ratio

Q14. ABO blood group shows:
A. Incomplete dominance
B. Codominance
C. Epistasis
D. Polygenic inheritance

Q15. Bombay phenotype occurs due to:
A. O allele masking AB
B. Mutation in H gene
C. Dominance of A allele
D. Recessive epistasis

Q16. In co-dominance, heterozygote phenotype shows:
A. Only dominant allele
B. Both alleles expressed equally
C. Blend of alleles
D. Only recessive allele

Q17. In snapdragon, red × white → pink, F2 ratio is:
A. 1:2:1
B. 3:1
C. 1:1
D. 9:3:3:1

Q18. Incomplete dominance differs from codominance because:
A. One allele dominates
B. Heterozygote shows blending
C. Alleles expressed equally
D. No variation occurs

Q19. Multiple alleles mean:
A. Only 2 alleles exist
B. More than 2 alleles for a gene
C. Polygenic inheritance
D. Mendelian dominance

Q20. ABO blood group is controlled by:
A. Two alleles
B. Three alleles
C. Four alleles
D. Single dominant gene

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Pleiotropy & Gene Interaction

Q21. Pleiotropy is when:
A. One gene influences multiple traits
B. Multiple genes influence one trait
C. Alleles blend
D. Genes segregate independently

Q22. Epistasis occurs when:
A. One gene masks effect of another gene
B. All genes express equally
C. Traits blend
D. Alleles segregate independently

Q23. Bombay phenotype is an example of:
A. Pleiotropy
B. Epistasis
C. Codominance
D. Incomplete dominance

Q24. Polygenic inheritance results in:
A. Continuous variation
B. Discrete variation
C. 3:1 ratio
D. 9:3:3:1 ratio

Q25. Skin color in humans is an example of:
A. Codominance
B. Polygenic inheritance
C. Incomplete dominance
D. Multiple alleles

Q26. Eye color in humans is controlled by:
A. Single gene
B. Multiple genes
C. Codominant alleles only
D. Sex-linked gene

Q27. Coat color in mice showing epistasis is due to:
A. One gene masking another
B. Polygenic inheritance
C. Multiple alleles
D. Pleiotropy

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Sex-linked & Chromosomal Disorders

Q28. Hemophilia is:
A. Autosomal recessive
B. X-linked recessive
C. Y-linked
D. Codominant

Q29. Color blindness in humans is:
A. Autosomal dominant
B. X-linked recessive
C. Y-linked
D. Polygenic

Q30. Turner syndrome (XO) affects:
A. Males
B. Females
C. Both
D. None

Q31. Klinefelter syndrome (XXY) is:
A. Female disorder
B. Male disorder
C. Both
D. None

Q32. Down syndrome occurs due to:
A. Monosomy
B. Trisomy 21
C. Turner syndrome
D. Polyploidy

Q33. Y-linked traits are passed:
A. Father → daughter
B. Father → son
C. Mother → son
D. Mother → daughter

Q34. Barr body is:
A. Inactive X chromosome in females
B. Active X chromosome
C. Y chromosome
D. Autosomal gene

Q35. Duchenne muscular dystrophy is:
A. Autosomal recessive
B. X-linked recessive
C. Y-linked
D. Autosomal dominant

Q36. SRY gene determines:
A. Female sex
B. Male sex
C. Both
D. None

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Mutation

Q37. Point mutation affects:
A. One nucleotide
B. Whole gene
C. Chromosome
D. Genome

Q38. Sickle cell anemia is caused by:
A. Chromosomal mutation
B. Point mutation in β-globin gene
C. Deletion of gene
D. Polygenic trait

Q39. Frameshift mutation occurs due to:
A. Substitution
B. Insertion/deletion
C. Translocation
D. Duplication

Q40. Silent mutation:
A. Changes amino acid
B. Does not change amino acid
C. Causes disease
D. Causes deletion

Q41. Mutagens include:
A. Chemicals
B. Radiation
C. Viruses
D. All of the above

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Pedigree Analysis

Q42. In pedigree, a shaded square represents:
A. Unaffected male
B. Affected male
C. Affected female
D. Carrier female

Q43. A circle represents:
A. Male
B. Female
C. Carrier male
D. Dominant trait

Q44. Autosomal dominant trait appears in:
A. Every generation
B. Skips generations
C. Only males
D. Only females

Q45. Autosomal recessive trait appears:
A. Every generation
B. Skips generations
C. Only males
D. Only females

Q46. X-linked recessive trait more common in:
A. Females
B. Males
C. Both equally
D. None

Q47. Carrier female symbol:
A. Shaded circle
B. Half-shaded circle
C. Shaded square
D. Half-shaded square

Q48. Pedigree helps to:
A. Trace inheritance patterns
B. Identify carriers
C. Predict offspring phenotype
D. All of the above

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Gene Interaction / Polygenic Traits

Q49. Epistasis can alter:
A. Mendelian ratios
B. Gene expression
C. Both A & B
D. None

Q50. Coat color in Labrador dogs is an example of:
A. Epistasis
B. Polygenic trait
C. Codominance
D. Incomplete dominance

Q51. Polygenic inheritance produces:
A. Discontinuous variation
B. Continuous variation
C. 3:1 ratio
D. 1:2:1 ratio

Q52. Quantitative traits include:
A. Height
B. Skin color
C. Weight
D. All of the above

Q53. Interaction of multiple genes is called:
A. Pleiotropy
B. Epistasis
C. Polygenic inheritance
D. Codominance

Q54. Epistasis can cause ratios like:
A. 9:3:3:1
B. 9:3:4
C. 3:1
D. 1:2:1

Q55. Multiple alleles differ from polygenic traits because:
A. Polygenic controlled by many genes
B. Multiple alleles by single gene
C. Both A & B
D. None

Q56. ABO blood group shows:
A. Multiple alleles + codominance
B. Polygenic inheritance
C. Epistasis
D. Pleiotropy

Q57. Bombay phenotype is due to:
A. H gene mutation masking ABO
B. Polygenic inheritance
C. Codominance
D. Epistasis

Q58. Traits influenced by environment + genes:
A. Polygenic
B. Quantitative
C. Multifactorial
D. All of the above

Q59. Epistatic gene suppresses:
A. Allele of same gene
B. Allele of different gene
C. Both
D. None

Q60. Trihybrid cross with epistasis may deviate from:
A. 9:3:3:1 ratio
B. 3:1 ratio
C. 1:2:1 ratio
D. None

A. Assertion–Reason MCQs (15)

Instructions:
A. Both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.
B. Both Assertion and Reason are correct, but Reason is NOT the correct explanation of Assertion.
C. Assertion is correct, but Reason is incorrect.
D. Assertion is incorrect, but Reason is correct.

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Q1.
Assertion: Incomplete dominance produces F1 with intermediate phenotype.
Reason: Heterozygote expresses blend of both alleles.

Q2.
Assertion: ABO blood group shows codominance.
Reason: IA and IB alleles express equally in AB phenotype.

Q3.
Assertion: Hemophilia is more common in males.
Reason: It is an X-linked recessive trait.

Q4.
Assertion: Polygenic traits show continuous variation.
Reason: Multiple genes contribute cumulatively to phenotype.

Q5.
Assertion: Bombay phenotype lacks H antigen.
Reason: Mutation in H gene prevents expression of ABO antigens.

Q6.
Assertion: Sickle cell anemia is a result of mutation.
Reason: It involves a point mutation in β-globin gene.

Q7.
Assertion: Epistasis can alter Mendelian ratios.
Reason: One gene can mask or modify effect of another gene.

Q8.
Assertion: Pleiotropic genes influence multiple traits.
Reason: Single gene produces multiple phenotypic effects.

Q9.
Assertion: Trihybrid cross produces 27:9:9:9:3:3:3:1 phenotypic ratio.
Reason: Independent assortment of three genes leads to 8 phenotypic classes.

Q10.
Assertion: X-linked recessive traits skip generations.
Reason: Carrier females transmit allele to sons without being affected.

Q11.
Assertion: Mendel’s law of segregation is universal.
Reason: Each gamete receives only one allele of a gene.

Q12.
Assertion: Linked genes do not assort independently.
Reason: Genes on same chromosome are inherited together unless crossing over occurs.

Q13.
Assertion: Point mutation may not affect phenotype.
Reason: Silent mutation does not change amino acid sequence.

Q14.
Assertion: Polygenic inheritance traits show multifactorial influence.
Reason: Environmental factors also influence expression of these traits.

Q15.
Assertion: Pedigree analysis helps trace inheritance.
Reason: It shows patterns of trait transmission across generations.

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B. Difficult / Calculation / Diagram-Based MCQs (30)

Q1. A monohybrid cross of Aa × Aa produces offspring. Probability of getting homozygous dominant genotype:
A. 25%
B. 50%
C. 75%
D. 100%

Q2. Dihybrid cross AaBb × AaBb, probability of obtaining offspring with genotype AABB:
A. 1/16
B. 1/4
C. 1/8
D. 1/2

Q3. In a tri-hybrid cross AaBbCc × AaBbCc, probability of abc genotype:
A. 1/64
B. 1/8
C. 1/16
D. 27/64

Q4. In a pedigree, a trait is present in every generation. This suggests:
A. Autosomal dominant
B. Autosomal recessive
C. X-linked recessive
D. X-linked dominant

Q5. Hemophilia appears in 50% of sons from carrier mother (XHXh) and normal father. Probability:
A. 25%
B. 50%
C. 75%
D. 100%

Q6. If red flower (RR) × white flower (rr) gives pink (Rr), F2 phenotypic ratio:
A. 1:2:1
B. 3:1
C. 1:1:1
D. 9:3:3:1

Q7. Two heterozygotes for AB blood group (IAi × IBi) produce offspring. Probability of O blood group:
A. 25%
B. 50%
C. 0%
D. 100%

Q8. Probability of obtaining male child in one pregnancy:
A. 25%
B. 50%
C. 75%
D. 100%

Q9. Probability of getting a homozygous recessive (aa) in monohybrid Aa × Aa cross:
A. 25%
B. 50%
C. 75%
D. 100%

Q10. In Drosophila, cross AaBb × AaBb for linked genes with 20% recombination frequency. Probability of parental phenotype:
A. 40%
B. 60%
C. 80%
D. 20%

Q11. In a cross AaBbCc × AaBbCc, probability of all dominant traits:
A. 27/64
B. 1/64
C. 1/8
D. 1/16

Q12. Color blindness is X-linked recessive. Carrier female (XBXb) × normal male (XBY). Probability of color-blind sons:
A. 0%
B. 25%
C. 50%
D. 100%

Q13. Down syndrome due to trisomy 21. Probability in child if one parent carries Robertsonian translocation:
A. 1/4
B. 1/2
C. 1/16
D. 1/3

Q14. Cross AaBbCc × aabbcc. Probability of offspring AaBbCc:
A. 1/8
B. 1/4
C. 1/16
D. 1/64

Q15. Pedigree shows X-linked dominant trait. Affected father × normal mother. Probability daughter affected:
A. 0%
B. 25%
C. 50%
D. 100%

Q16. Probability of getting 2 sons and 1 daughter in 3 children:
A. 1/8
B. 3/8
C. 1/2
D. 1/4

Q17. Cross Aa × Aa. Probability of getting at least one heterozygous offspring in two children:
A. 3/4
B. 7/16
C. 1/2
D. 9/16

Q18. In snapdragon, red × white → pink. Probability of red in F2:
A. 25%
B. 50%
C. 75%
D. 100%

Q19. ABO blood group cross: IAi × IBi. Probability of AB:
A. 25%
B. 50%
C. 75%
D. 0%

Q20. A cross AaBbCc × AaBbCc. Probability of recessive for all three traits:
A. 1/8
B. 1/64
C. 1/16
D. 1/4

Q21. Epistasis 9:3:4 ratio indicates:
A. Dominant epistasis
B. Recessive epistasis
C. Duplicate gene interaction
D. Polygenic inheritance

Q22. Polygenic trait shows 64% tall, 36% short. Environmental influence shifts distribution. This is:
A. Multifactorial inheritance
B. Mendelian inheritance
C. X-linked inheritance
D. Codominance

Q23. Probability of 2 heterozygotes in F2 from Aa × Aa cross:
A. 1/16
B. 1/4
C. 1/2
D. 3/4

Q24. In tri-hybrid cross, probability of exactly 2 dominant traits:
A. 27/64
B. 9/16
C. 3/8
D. 1/4

Q25. Linked genes, recombination frequency 10%. Probability of recombinant gametes:
A. 10%
B. 20%
C. 90%
D. 50%

Q26. Pedigree: male shows trait, father unaffected. Suggests:
A. Autosomal dominant
B. X-linked recessive
C. Autosomal recessive
D. Y-linked

Q27. Sickle cell trait F1 heterozygote × heterozygote. Probability of affected offspring:
A. 1/4
B. 1/2
C. 3/4
D. 100%

Q28. Probability of 3 children all female:
A. 1/8
B. 1/4
C. 1/2
D. 1/16

Q29. Cross of blood group IAIB × IAi. Probability of AB:
A. 25%
B. 50%
C. 75%
D. 0%

Q30. In a population, mutant allele frequency = 0.1. Probability of homozygous recessive (q²):
A. 0.01
B. 0.1
C. 0.2
D. 0.9

A. Standard / Conceptual MCQs – 70 Answers

Q.No	Answer	Explanation
1	B	Monohybrid Aa × Aa gives 3:1 phenotypic ratio (dominant:recessive).
2	B	Dihybrid AaBb × AaBb gives 9:3:3:1 phenotypic ratio.
3	B	Tt × Tt produces 3 tall :1 dwarf.
4	B	Alleles segregate during gamete formation (Law of Segregation).
5	D	Trihybrid all dominant traits probability = 27/64.
6	B	True-breeding tall × dwarf shows dominance.
7	C	Test cross with 1:1 ratio indicates heterozygous genotype.
8	C	Independent assortment applies to genes on different chromosomes.
9	C	Aa × Aa genotypic ratio is 1:2:1.
10	B	Mendel’s experiments used pea plants.
11	B	Incomplete dominance gives F1 with blended phenotype.
12	A	Pink F1 results from blending → incomplete dominance.
13	B	Codominance: both alleles expressed equally.
14	A	ABO shows multiple alleles with codominance.
15	B	Bombay phenotype due to mutation in H gene, masking ABO antigens.
16	B	Heterozygote expresses both alleles equally in codominance.
17	A	F2 ratio in incomplete dominance = 1:2:1 (red:pink:white).
18	B	Incomplete dominance → heterozygote shows intermediate trait.
19	B	Multiple alleles: more than 2 alleles exist for a gene.
20	B	ABO controlled by IA, IB, i (three alleles).
21	A	Pleiotropy: one gene affects multiple traits.
22	A	Epistasis: one gene masks effect of another gene.
23	B	Bombay phenotype shows epistatic interaction of H gene.
24	A	Polygenic inheritance → continuous variation.
25	B	Skin color influenced by multiple genes → polygenic.
26	B	Eye color controlled by multiple genes.
27	A	Coat color in mice shows epistasis (one gene masking another).
28	B	Hemophilia is X-linked recessive, affects mostly males.
29	B	Color blindness is X-linked recessive.
30	B	Turner syndrome (XO) affects females.
31	B	Klinefelter syndrome (XXY) affects males.
32	B	Down syndrome due to trisomy 21.
33	B	Y-linked traits pass father → son.
34	A	Barr body = inactive X chromosome in females.
35	B	Duchenne muscular dystrophy is X-linked recessive.
36	B	SRY gene on Y chromosome determines maleness.
37	A	Point mutation affects single nucleotide.
38	B	Sickle cell caused by point mutation in β-globin gene.
39	B	Frameshift mutation caused by insertion/deletion.
40	B	Silent mutation does not change amino acid.
41	D	Mutagens include chemicals, radiation, viruses.
42	B	Shaded square = affected male.
43	B	Circle represents female.
44	A	Autosomal dominant traits appear in every generation.
45	B	Autosomal recessive traits often skip generations.
46	B	X-linked recessive traits more common in males.
47	B	Half-shaded circle = carrier female.
48	D	Pedigree analysis traces inheritance, predicts traits, identifies carriers.
49	C	Epistasis can alter Mendelian ratios and gene expression.
50	A	Labrador coat color is example of epistasis.
51	B	Polygenic inheritance → continuous variation.
52	D	Quantitative traits include height, skin color, weight.
53	C	Interaction of multiple genes = polygenic inheritance.
54	B	Epistasis can produce 9:3:4 ratio.
55	B	Multiple alleles involve single gene; polygenic involves many genes.
56	A	ABO blood group = multiple alleles + codominance.
57	D	Bombay phenotype shows epistasis of H gene.
58	C	Traits influenced by genes + environment = multifactorial/polygenic.
59	B	Epistatic gene suppresses effect of another gene.
60	A	Trihybrid cross with epistasis deviates from 9:3:3:1 ratio.
61	–	Placeholder for numbering continuation.

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B. Assertion–Reason MCQs – Answers + Explanation

Q.No	Answer	Explanation
1	A	Both correct; F1 shows intermediate trait due to blending of alleles.
2	A	Both correct; IA and IB co-expressed in AB blood group.
3	A	Both correct; X-linked recessive trait manifests more in males.
4	A	Both correct; multiple genes cumulatively influence phenotype.
5	A	Both correct; mutation in H gene masks ABO expression.
6	A	Both correct; point mutation in β-globin causes sickle cell.
7	A	Both correct; one gene can mask expression of another.
8	A	Both correct; pleiotropy affects multiple traits from one gene.
9	A	Both correct; 8 phenotypic classes arise from independent assortment.
10	A	Both correct; X-linked recessive traits skip generations via carrier females.
11	A	Both correct; gametes receive one allele from each parent.
12	A	Both correct; linked genes inherited together unless crossing over occurs.
13	A	Both correct; silent mutation does not change amino acid sequence.
14	A	Both correct; environmental factors affect polygenic traits.
15	A	Both correct; pedigree traces inheritance patterns.

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C. Difficult / Calculation / Diagram-Based MCQs – Answers + Explanation

Q.No	Answer	Explanation
1	A	Homozygous dominant (AA) probability in Aa × Aa = 25%.
2	A	Dihybrid AaBb × AaBb → AABB = 1/16.
3	A	Trihybrid AaBbCc × AaBbCc → abc (all recessive) = 1/64.
4	A	Trait in every generation → autosomal dominant.
5	C	Carrier XHXh × XHY → 50% sons affected.
6	A	F2 phenotypic ratio in incomplete dominance = 1:2:1 (red:pink:white).
7	A	IAi × IBi → O blood group = 25%.
8	B	Probability male child = 50%.
9	A	Aa × Aa → aa probability = 25%.
10	B	Parental phenotypes = 1 − recombination frequency = 60%.
11	A	Probability all dominant = 27/64.
12	C	XBXb × XBY → 50% sons color-blind.
13	D	Probability varies; depends on translocation; typically 1/3.
14	A	AaBbCc × aabbcc → AaBbCc = 1/8.
15	D	Affected daughter from affected father × normal mother = 100%.
16	B	Two sons + one daughter = 3/8 probability.
17	B	At least one heterozygote = 1 − probability both homozygous = 7/16.
18	A	F2 red = 25% in incomplete dominance cross.
19	A	IAIB × IAi → AB = 25%.
20	B	Recessive for all three traits = 1/64.
21	B	9:3:4 ratio indicates recessive epistasis.
22	A	Polygenic traits influenced by genes + environment → multifactorial inheritance.
23	B	Two heterozygotes in two offspring = 1/4.
24	C	Exactly 2 dominant traits = 3/8 probability.
25	A	Recombinant gametes = recombination frequency = 10%.
26	C	Trait appears in male; father unaffected → X-linked recessive.
27	A	Heterozygote × heterozygote → affected offspring = 1/4.
28	A	Three female children probability = 1/8.
29	A	IAIB × IAi → AB probability = 25%.
30	A	Mutant allele frequency q = 0.1 → q² = 0.01.

Disclaimer:

All MCQs and answers are for educational purposes only. They are based on NCERT and NEET syllabus and do not represent official NEET questions.