00:01
Here we have given an example of a population which is in hardy -winberg equilibrium.
00:06
And the hardy -winberg equation is just like this, that p -square plus 2 p -q plus q -plus q -square is equal to 1.
00:17
And in that particular population, the data given saying that the 38 percentage of the individuals are homozygous recessive for a specific trait or character and the total population is of 14 ,500 individuals.
00:47
So now we have to calculate the percentage of homozygous dominant individuals and the heterozygous individuals.
00:53
So here we'll always start with the homozygous recessive percentage if it is given here and it is given here that equals to the q square.
01:06
We are assuming that homozygous recessive alleles as the q square, the homozygous dominant alleles as the p square and the heterozygous alils as the 2pq.
01:17
So if we have given the homozygous recessive percentage of individuals then it will be is equal to the q square.
01:26
So the decimal of 38 % will be 0 .5%.
01:31
And according to that the q is equal to 0 .38 under roots and ultimately the value of q will be 0 .616.
01:48
If we focus to find the p values so it is now straightforward that point p we can use this.
01:55
P plus q is equal to one formula and thus we already have the q values.
02:01
So we'll put here as 0 .616 is equal to 1 and the p is equal to 0 .38.
02:12
So this is how we got the p and q values.
02:15
Now the homozygous dominant individuals are represented by p square...