00:01
Hello students, in this video we will discuss about population.
00:06
In a population that is in hardy -weinberg equilibrium, the frequency of the homozygous recessive genotype q square is given as 0 .81.
00:37
According to the hardy -weinberg equilibrium equation, p square plus 2pq plus q square equals to 1, where p square represents the frequency of the homozygous dominant genotype, 2pq represents the frequency of the heterozygous genotype, and q square represents the frequency of the homozygous recessive genotype.
01:31
Given that q square is equals to 0 .81, we can calculate q, the frequency of the recessive allele, by taking the square root of q square, that is q equals to 0 .81, q equals to 0 .9.
02:05
Since p plus q equals to 1, we can determine the frequency of the dominant allele p.
02:13
So p equals to 1 minus q, p equals to 1 minus 0 .9...