00:01
Now to answer this question, let's talk about hardingwemeter equilibrium.
00:03
Remember that according to hardingwember, you have that 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 alleles in the population that are recessive alils.
00:15
In this case, we're using the example of p .a .j.
00:18
You also have that p square plus 2pq plus q plus q is equal to 1, where p square is a number of individuals or organisms in the population a that have the genotype homocygous lominant, 2pq is the same but for heterozygose and qsquare is for homozygose.
00:39
Let's answer these questions.
00:40
It says, question one.
00:42
This hard -de -weamer equation and they have p -square plus 2 -p -q plus q -square is equal to 1 means.
00:48
And the answer for this question is option a that says the assume of genotype frequencies equals to 1 because this p -square, 2 -pq and q -square, we present the frequencies for the genotypes in this.
01:01
Population.
01:01
So this is the answer for question a.
01:04
Then it says, question number two says, in the hardy whimary equation, b squared plus pq plus q squared equal to 1, what does that term q represent? it means they're asking us for the frequency for the excessive allele.
01:16
And the answer for this question is going to be option b, that says the frequency of the excessive allele.
01:21
The next question says, assuming the allele b, well in this case they're using b, okay? so you have here that dominant b and the recessive b.
01:32
There are 40 students in a class and 10 of them have the recessive genotype.
01:38
So you have a total of 40 students and 10 of them have a recessive genotype.
01:47
It means 10 of them have the recessive genotype...