00:01
Okay, so this question asks about true statements regarding the hardy -winberg expression p -square plus 2pq plus q squared.
00:09
So i have a picture here that obviously describes the equation that we are talking about.
00:15
So p -squared is the frequency of the homozygous dominant genotype, and p is the frequency of the dominant allele.
00:24
2 -pq demonstrates the frequency of the hemoglobinate.
00:30
Heterocygous genotype, and then q squared here is the frequency of the homozygous recessive genotype.
00:37
And q being here, the frequency of the recessive allele.
00:47
And so as we look down at the options that are stated, we can eliminate and see which ones are true and which ones aren't true regarding this expression.
01:02
So a says, knowing either p squared or q squared, you can calculate all other frequencies.
01:09
So if we know either p or q, then we can calculate the frequency of p squared, 2 pq, and q squared.
01:17
So the values of these three respective values, the values of these three expressions.
01:23
This is true because if we know p squared, which is the frequency of the homozygous dominant genotype, right? so we say, let's say that equals a number.
01:34
If we take the square root of that, then we get the frequency of p, which is the dominant allele.
01:45
And another important fact to note is that p plus q is equal to 1 because the frequency of the dominant and the recessive will have to equal 1.
01:54
So if we find p, then we can find q by subtracting 1, and then we could find the frequency of the homozygous recessive q squared and 2pq.
02:02
And the same goes if we know q squared, we could find p via this expression and then find p squared into bq.
02:11
So a is true...