Q.A population of snakes lived in a desert with brown sand. Study the drawings given below showing the change in the population from 'one' to 'two' over time and answer the question that follows. Brown snakes and Grey snakes are represented by alleles A/a (Dominant/recessive). [Figure: Population-one and Population-two (Migration of Birds) — drawings of snakes on desert sand, with birds shown migrating in over Population-two]
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Start your 14-day free trial to unlock the full solution →Using Hardy-Weinberg equilibrium, the recessive allele frequency is 0.3, so homozygous dominant frequency is 49% and heterozygous frequency is 42%. The mechanism is natural selection (specifically, directional selection due to bird predation).
The Concept: Hardy-Weinberg and Selection
This problem connects two big ideas in evolution. First, the Hardy-Weinberg principle gives us a mathematical baseline: in a non-evolving population, allele and genotype frequencies stay constant from generation to generation. The formula is:
p^2 + 2pq + q^2 = 1
where p = frequency of dominant allele (A), q = frequency of recessive allele (a), and p + q = 1.
Second, when a population does change — as shown in the drawings, where brown snakes become more common after birds arrive — some evolutionary force is at work. The birds are predators that can see grey snakes more easily against the brown sand. That's natural selection in action.
The key insight: we're told the recessive trait (grey snakes) has a frequency of 9% in population-one. That's the phenotype frequency, not the allele frequency. Since grey is recessive, every grey snake must be homozygous recessive (aa). So q^2 = 0.09.
Step-by-Step Solution
Part (a): Finding genotype frequencies
1. Find the recessive allele frequency q.
If 9% of snakes are grey (aa), then:
q^2 = 0.09
Taking the square root:
q = sqrt(0.09) = 0.3
A common mistake is to confuse the phenotype frequency (9%) with the allele frequency. Remember: q^2 is the frequency of the genotype aa, not the frequency of allele a. Always take the square root.
2. Find the dominant allele frequency p.
Since p + q = 1:
p = 1 - q = 1 - 0.3 = 0.7
3. Find the frequency of homozygous dominant snakes (AA).
This is p^2:
p^2 = (0.7)^2 = 0.49
So 49% of the snakes are homozygous dominant (brown).
4. Find the frequency of heterozygous snakes (Aa).
This is 2pq:
2pq = 2 x 0.7 x 0.3 = 0.42
So 42% of the snakes are heterozygous (also brown, since A is dominant).
You can check your work: p^2 + 2pq + q^2 = 0.49 + 0.42 + 0.09 = 1.00. Everything adds up perfectly.
Part (b): The mechanism of evolution …
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