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NCERT Exemplar · Q28

Q.In a certain population, the frequency of three genotypes is as follows:
Genotypes: BB Bb bb
frequency: 22% 62% 16%
What is the likely frequency of B and b alleles?

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The frequency of an allele is calculated by summing the frequency of homozygous individuals for that allele and half the frequency of heterozygous individuals.

In population genetics, we often distinguish between genotype frequencies and allele frequencies. Genotype frequency refers to the proportion of individuals in a population that possess a specific genotype (e.g., BB, Bb, or bb). Allele frequency, on the other hand, refers to the proportion of a specific allele (e.g., B or b) within the gene pool of a population. Understanding how to derive allele frequencies from observed genotype frequencies is fundamental.

Consider a gene with two alleles, B and b. Each individual in the population carries two alleles for this gene.

  • An individual with genotype BB carries two B alleles.
  • An individual with genotype Bb carries one B allele and one b allele.
  • An individual with genotype bb carries two b alleles.

To find the frequency of an allele, say B, we need to count all the B alleles present in the population and divide by the total number of alleles for that gene in the population. Since each individual has two alleles, the total number of alleles is twice the number of individuals.

Let's denote the frequency of allele B as $p$ and the frequency of allele b as $q$.

Given the genotype frequencies:

  • Frequency of BB ($f_{BB}$) = $22% = 0.22$
  • Frequency of Bb ($f_{Bb}$) = $62% = 0.62$
  • Frequency of bb ($f_{bb}$) = $16% = 0.16$

To calculate the frequency of allele B ($p$):

Every individual with the BB genotype contributes two B alleles to the gene pool.

Every individual with the Bb genotype contributes one B allele to the gene pool.

Therefore, the total proportion of B alleles in the population can be found by taking the full frequency of BB individuals and adding half the frequency of Bb individuals (since only one of the two alleles in Bb is B).

$p = f_{BB} + \frac{1}{2} f_{Bb}$

Substituting the given values:

$p = 0.22 + \frac{1}{2} (0.62)$

$p = 0.22 + 0.31$

$p = 0.53$ …

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