Part 5: Hardy–Weinberg Principle (25 MCQs)
Part 5: Hardy–Weinberg Principle (25 MCQs)
Q101. The Hardy–Weinberg principle deals with:
a) Genetic drift
b) Genetic equilibrium in a population
c) Speciation
d) Adaptive radiation
Answer: b
- b) Correct: H–W principle → allele frequencies remain constant in a population if evolutionary forces are absent.
Q102. Hardy–Weinberg equilibrium assumes:
a) Random mating
b) No mutation
c) No migration, selection, or drift
d) All of the above
Answer: d
- d) Correct: Conditions = no selection, no mutation, no migration, large population, random mating.
Q103. Hardy–Weinberg equation is expressed as:
a) p + q = 1
b) p² + 2pq + q² = 1
c) p² + q² = 1
d) 2pq = 1
Answer: b
- b) Correct: Genotype frequencies = p² + 2pq + q² = 1.
Q104. In Hardy–Weinberg principle, p stands for:
a) Frequency of dominant allele
b) Frequency of recessive allele
c) Frequency of heterozygotes
d) Gene pool size
Answer: a
- a) Correct: p = dominant allele frequency, q = recessive allele frequency.
Q105. In Hardy–Weinberg equation, 2pq represents:
a) Homozygous dominant
b) Homozygous recessive
c) Heterozygous frequency
d) Mutant frequency
Answer: c
- c) Correct: 2pq = heterozygote genotype frequency.
Q106. If p = 0.6, then q = ?
a) 0.2
b) 0.4
c) 0.6
d) 0.8
Answer: b
- b) Correct: p + q = 1 → q = 1 – 0.6 = 0.4.
Q107. If q = 0.3, what is the frequency of homozygous recessive (q²)?
a) 0.3
b) 0.09
c) 0.7
d) 0.49
Answer: b
- b) Correct: q² = (0.3)² = 0.09 (9%).
Q108. If p = 0.7, frequency of homozygous dominant (p²) is:
a) 0.49
b) 0.21
c) 0.91
d) 0.30
Answer: a
- a) Correct: p² = (0.7)² = 0.49.
Q109. If heterozygotes are 48%, then 2pq = 0.48. What is pq?
a) 0.24
b) 0.48
c) 0.12
d) 0.96
Answer: a
- a) Correct: 2pq = 0.48 → pq = 0.24.
Q110. Which of the following disturbs Hardy–Weinberg equilibrium?
a) Large population size
b) Random mating
c) Natural selection
d) No migration
Answer: c
- c) Correct: Selection alters allele frequencies.
Q111. If a population is not in Hardy–Weinberg equilibrium, it means:
a) No evolution is occurring
b) Evolution is occurring
c) Random mating is happening
d) Population is infinite
Answer: b
- b) Correct: Deviation from H–W → evolutionary forces are acting.
Q112. A population has p = 0.8, q = 0.2. Frequency of heterozygotes?
a) 0.16
b) 0.32
c) 0.64
d) 0.04
Answer: b
- b) Correct: 2pq = 2 × 0.8 × 0.2 = 0.32.
Q113. If 9% of a population is homozygous recessive, what is q?
a) 0.09
b) 0.3
c) 0.7
d) 0.81
Answer: b
- b) Correct: q² = 0.09 → q = √0.09 = 0.3.
Q114. With q = 0.3, p = ?
a) 0.3
b) 0.5
c) 0.7
d) 0.9
Answer: c
- c) Correct: p = 1 – q = 0.7.
Q115. If q² = 0.16, frequency of heterozygotes (2pq)?
a) 0.48
b) 0.32
c) 0.16
d) 0.36
Answer: a
- a) Correct: q = 0.4, p = 0.6 → 2pq = 0.48.
Q116. Which is NOT a factor in Hardy–Weinberg equilibrium?
a) Large population
b) Migration
c) Random mating
d) No mutation
Answer: b
- b) Correct: Migration changes allele frequencies → equilibrium is disturbed.
Q117. The term “gene pool” refers to:
a) All alleles in a population
b) All mutations
c) Only dominant alleles
d) Only recessive alleles
Answer: a
- a) Correct: Gene pool = total alleles of all genes in a population.
Q118. If allele frequencies remain constant, it indicates:
a) Evolution is occurring
b) No evolution is occurring
c) Genetic drift is active
d) Mutation is frequent
Answer: b
- b) Correct: Constant allele frequencies = genetic equilibrium (no evolution).
Q119. Small populations deviate from Hardy–Weinberg due to:
a) Genetic drift
b) Random mating
c) Natural selection
d) Mutation
Answer: a
- a) Correct: Drift affects small populations strongly, disturbing equilibrium.
Q120. If 49% are homozygous recessive, then q = ?
a) 0.49
b) 0.7
c) 0.3
d) 0.09
Answer: b
- b) Correct: q² = 0.49 → q = √0.49 = 0.7.
Q121. If q = 0.7, what is heterozygote frequency?
a) 0.21
b) 0.42
c) 0.49
d) 0.91
Answer: b
- b) Correct: p = 0.3, so 2pq = 2 × 0.3 × 0.7 = 0.42.
Q122. Which of the following is a real-life application of Hardy–Weinberg principle?
a) Calculating frequency of genetic disorders
b) Explaining fossils
c) Determining mutations in viruses
d) Studying embryology
Answer: a
- a) Correct: Used to estimate genetic disorder frequencies in populations.
Q123. If in a population, p = 0.5, q = 0.5, heterozygotes = ?
a) 0.25
b) 0.5
c) 0.75
d) 0.1
Answer: b
- b) Correct: 2pq = 2 × 0.5 × 0.5 = 0.5.
Q124. Deviation from Hardy–Weinberg principle indicates:
a) Population is evolving
b) Population is stable
c) No mutations
d) No selection
Answer: a
- a) Correct: Deviation = evolution is happening.
Q125. Hardy–Weinberg equilibrium proves that:
a) Evolution is not always happening
b) Evolution is a random process
c) Acquired traits are inherited
d) All mutations are harmful
Answer: a
- a) Correct: H–W law shows allele frequencies can remain constant → no evolution under ideal conditions.
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