The allele for brown eyes (B) is found within a population at a frequency of 0.7; the allele for blue eyes (b) at a frequency of 0.3. If the conditions for Hardy-Weinberg equilibrium are satisfied, what will be the expected frequency of heterozygous (Bb) individuals?

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To tackle this question, we start by identifying the allele frequencies given: p (brown eyes B) = 0.7 and q (blue eyes b) = 0.3. Under Hardy-Weinberg equilibrium, genotype frequencies in the population are described by p^2 for BB, 2pq for Bb, and q^2 for bb.

Option 1: 0.49. This value corresponds to p^2, the frequency of homozygous dominant individuals (BB). Calculating p^2 = (0.7)^2 = 0.49, which would be the proportion of BB, not Bb. Therefore this option describes a different genotype frequency than asked for.

Option 2: 0.21. This figure does not match any standard Hardy-Weinberg calculation with p = 0.7 and q = 0.3. If you try 2pq, you get 2*(0.7)*(0.3) = 0.42, so 0.21 would be incorrect and suggests a miscalculation or misinterpretation of the 2pq term.

Option 3: 0.42. This exactly matches 2pq, the frequency of heterozygotes (Bb) under Hardy-Weinberg: 2*(0.7)*(0.3) = 0.42. This reflects individuals carrying one B and one b allele.

Option 4: 0.09. This value is q^2, the frequency of homozygous recessive individuals (bb), calculated as (0.3)^2 = 0.09. While correct for bb, it does not answer the question about heterozygotes.

In summary, the heterozygote frequency in a population at Hardy-Weinberg equilibrium with p = 0.7 and q = 0.3 is determined by 2pq, which evaluates to 0.42. The other options correspond to BB (0.49) or bb (0.09) frequencies, or an incorrect calculation (0.21).

BIOL 1009 (001-006/050-055) Final Exam

The allele for brown eyes (B) is found within a population at a frequency of 0.7; the allele for blue eyes (b) at a frequency of 0.3. If the conditions for Hardy-Weinberg equilibrium are satisfied, what will be the expected frequency of heterozygous (Bb) individuals?

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