00:01
In order to answer this question, i've talked about hard -de -beamer equilibrium.
00:04
It says, what is the first step when determining frequency for either alleles or genotype? so the first step is practically to work with that excessive trait, okay, with a decisive frequency, because remember that according to harding weber, p plus q is equal to 1, where p is the frequency of alleles in the population, that are dominant alleles, and q is the frequency of alias in the population that are recessive alleles.
00:26
Also, you have that p -square plus 2 p -k plus q, plus q, square is equal to 1, where p.
00:32
Square is the frequency of individuals in this population that are homozygous dominant, 2pq is the frequency of individuals in the population that are hydrocygos, and qs, which is a frequency of individuals in the population of that homozygot recessive.
00:46
So if you start working with the dominant phenotype that is that in this case is dominant, okay, dopamine a, then you're going to have to work with these two variables that is p.
00:58
P -square and 2 p -q.
01:00
So you have two variables here that is p -square, p and q.
01:04
So this is very hard to start working with.
01:08
But if you work with the recessive phenotype that is a recessive one, this one here, okay? then you're going to work with q -square.
01:17
So you have only one variable, and this is easier to start working with.
01:21
Okay, so you always have to focus and start working with the some sort of recessive frequency.
01:29
Now it is, what would always be the same for all frequency calculations no matter how the actual number, numbers of alleys or denot types differ.
01:37
And the answer is that p plus q is always going to be equal to 1.
01:41
And also that p squared plus 2 plus n plus q is equal to 1.
01:47
The next question says if the frequency for the dominant ali is 0 .6, what is a frequency of the visceral? so as you can see here, p plus q is equal to 1.
01:56
So if you have that the frequency for the dominant ali, it means the value for p is 0 .6, then the value for q is going to be 0 .4 because 0 .6 plus 0 .4 is 1.
02:09
So this is the answer for this question.
02:11
The frequency for the glycifalil is 0 .4.
02:14
Now it says if the frequency of the homoside good recessive genotype is 0 .64, so you have that the frequency for the homositeal recessive is q squared, so you have that qsquare is equal to 0 .64.
02:27
Against square is the frequency of the dominant alley.
02:32
So, if you have that that u .s.
02:34
Score is equal to 0 .64, then you can apply good square at both sides here and also here in order to get the value for q.
02:42
And you're going to get that q is equal to 0 .8...