00:01
Given is a single nucleotide polymorphism that is associated with the disease, and the mutant allele is present at the frequency of 3 % in the population.
00:13
And according to hardy -wainburg equation, if p is the frequency of normal allele and q is the frequency of mutant allule, then p plus q is equals to 1.
00:29
Squaring both sides so we have p plus q squared so we will have p squared plus q squared plus two p q is equals to one in this p squared is the frequency of homozygous dominant individual q squared is the frequency of homozygous recessive individuals and two p q is the frequency of heterozygotes or the carrier and according to the question, your q is equals to 3 % or 0 .03.
01:06
Therefore, your p is equals to 1 minus q.
01:10
So p is equals to 1 minus 0 .03, so our p is 0 .97.
01:17
According to hardy -wainburg equation, the frequency of homozygose recessive individual would be q squared.
01:25
And the frequency of effective individual is 0 .03 squared, or this would be equivalent to 0 .0009 or 0 .09 % the frequency of the carrier that is the 2pq, so therefore 2 times 0 .03 times 0 .097, we have 0 .0582 or 5 .82 or 5.
01:57
The ratio to affected individuals can be calculated.
02:04
So the percentage of carriers, that is 5 .82%.
02:12
The percentage of affected individuals is 0 .09%.
02:21
So the ratio for this one, you will just divide 5 .82 by 0 .09.
02:28
We have 1 is to 64 .67.
02:35
If the snp was present on x chromosome, then there will be a slight difference in calculation of frequency of affected and carrier individuals...