Question

In chickens, the allele for feathered legs (F) is dominant over the allele for bare legs (f). In a population of chickens on a farm, 240 chickens have feathered legs (FF or Ff), and 80 chickens have bare legs (ff). How many chickens can be expected to be heterozygous for feathered legs (Ff)? Assume that the conditions of Hardy-Weinberg equilibrium apply to the population.

Answer 1

Answer:

0.5

Explanation:

According to Hardy-Weinberg equilibrium p2+2pq+q2=1 (p+q=1)

p2 is frequency of the dominant homozygous genotype

2pq is the frequency of the heterozygous genotype

q2 is the frequency of the recessive homozygous genotype

In the example above 80 chickens have bare legs (ff)-recessive homozygous which means that the frequency of that genotype is 80/240+80=0.25 (q2), frequency of the recessive allele is$\sqrt{0.25}$=0.5. This means that the frequency of the dominant allele (p) is 1-0.5=0.5

So, the frequency of the heterozygous genotype (2pq) is 2*0.5*0.5=0.5

The frequency of the dominant homozygous genotype is p2=0.25

Answer 2

Hardy-Weinberg equilibrium referred to constant genetic variation in population from one generation to other in absence of disturbing factors. Here, heterozygous for feathered legs will be 160.

What is equation for Hardy-Weinberg equilibrium ?

The basic requirements for Hardy-Weinberg equilibrium are no mutation, no gene flow, no selection etc.

Hardy-Weinberg equilibrium can be calculated as

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

Here,

• p=dominant true breed frequency.
• 2pq=heterozygous frequency.
• q= recessive true breed frequency.

Thus, according to given scenario, the  heterozygous for feathered legs is expected to be 160.

For more information about Hardy-Weinberg equilibrium, visit:

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