Question

In a population of 50000 individuals, if the frequency of recessive allele for a character is $40\%$, then find the number of organisms having dominant character?

A. 30000

B. 4200

C. 42000

D. 36000

Answer 1

In Hardy-Weinberg equilibrium for a gene in a population, it is not evolving; the allele frequency stays the same across the generations. Hardy-Weinberg has five assumptions: random mating, no mutation, infinite population size, no gene flow, and no selection.

Complete answer:

According to Hardy-Weinberg principle Here frequency recessive allele given is $40\%$ hence $q=0.4, p=0.6$

frequency of recessive genotype will be (aa) $q^2= 0.16$

Hence the total number of individuals having recessive trait $= 50000 \times 0.16= 8000$

The number of individuals with dominant trait is $50000 - 8000= 42000$

or frequency of individuals having dominant trait will be $p^2 + 2pq = 0.36+ 2(0.6 \times 0.4)= 0.84$

number of individuals having dominant traits will be $= 50000 \times 0.84 = 42000$

Hence the correct answer is option C, $42000.$

Additional information:

The Hardy-Weinberg principle is a theory which states that in the absence of disturbing factors, genetic variation in a population will remain constant from one generation to the next. In a large population with no destructive conditions, when mating is random, the law predicts that both genotype and allele frequencies will stay constant since they are in balance.

A variety of forces can disrupt the Hardy-Weinberg equilibrium, including mutations, natural selection, nonrandom mating, genetic drift, and gene flow. For example, by the introduction of new alleles into a population, mutations disturb the equilibrium of allele frequencies. Similarly, the Hardy-Weinberg equilibrium is broken by natural selection and nonrandom mating because they result in shifts in gene frequencies.

Note: Remember the basic formulas:

${p^2} + 2pq + {q^2} = 1$ with $p + q = 1$

Throughout the community, p = frequency of the dominant allele.

$q =$ The recessive allele rate in the population.

${p^2}$ = Percentage of dominant homozygous individuals.

${q^2}$ = percentage of recessive homozygous individuals.

$2pq$ = percentage of individuals heterozygous.