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IB DP Biology- D4.1 Natural selection- IB Style Questions For HL Paper 1A -New Syllabus

IBDP Biology HL Paper 1 -All Chapters

Question

What is an outcome of selection?

(A) Directional selection increases the size of the gene pool.
(B) Stabilizing selection increases allele frequencies of phenotypic extremes.
(C) Disruptive selection increases genetic variance in a population.
(D) Sexual selection leads to allopatric speciation.
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathbf{C}} \)

Disruptive selection favours individuals with extreme phenotypes at both ends of the range, increasing genetic variation within the population. This can promote divergence and may contribute to speciation over time.

Question

In a series of studies, Mark investigated sexual and natural selection in Trinidadian guppies (Poecilia reticulata) by varying environmental conditions and predation pressure.

What conclusion was drawn from Mark’s experiments?

A. Guppies have no predators in their native streams.
B. Male guppies show preference for dull-coloured females.
C. Predation pressure determines the colour patterns of guppies in different habitats.
D. Guppies remain unaffected by the presence of other fish species.

▶️ Answer/Explanation
Mark’s research showed that predation plays a key role in guppy colour variation. In areas with high predation, males evolved duller colours for better camouflage, while in low-predation environments, males displayed brighter colours to attract mates. This demonstrates the interaction between natural and sexual selection in evolution.
✅ Answer: (C) Predation influences the colour of guppies in different environments.

Question

The Hardy–Weinberg equation, used to calculate expected genotype frequencies in a population in genetic equilibrium, is:

\(p^{2}+2pq+q^{2}=1\).

The length of the mouthparts in an insect population is controlled by a gene with \(2\) alleles: long mouthparts (dominant) and short mouthparts (recessive). The frequency of the allele for long mouthparts is \(0.72\). What is the expected genotype frequency of individuals with short mouthparts?

(A) \(0.52\)
(B) \(0.40\)
(C) \(0.28\)
(D) \(0.08\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathbf{D}} \)

The frequency of the dominant allele is \(p=0.72\).

Therefore, \(q=1-p=1-0.72=0.28\).

Individuals with short mouthparts have the recessive genotype \(qq\), so their expected genotype frequency is:

\(q^{2}=(0.28)^{2}=0.0784\approx0.08\).

Therefore, the expected genotype frequency is \(0.08\).

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