1. A particular mutant allele exists in a hypothetical...

1. A particular mutant allele exists in a hypothetical population of 1000 individuals on an island in a frequency of 0.20. Let us refer to this allele as "a". It is recessive to the dominant allele, "A." An analysis of the populations shows the following genotypes: AA -- 0.67 Aa -- 0.30 aa -- 0.03 Is this population in approximate Hardy-Weinberg equilibrium? Show on your answer sheet all calculations that led up to your answer and give the reason or reasons for your answer. Assume that at this particular locus there are only two possible allelic forms. 2. In a natural population we find a genetic disease resulting from a single mutant allele "a," found in the homozygous state. This allele is recessive to a dominant allele, "A." Genetic analysis shows that the disease is found in a frequency of 0.0009 in the population. Assuming that the population is in approximate Hardy-Weinberg equilibrium, what is the estimated frequency of heterozygotes in the population and the estimated frequency of the homozygotes? Assume that at the particular locus we are discussing there are only two possible allelic forms. Show all calculations.

Transcript

00:01 In order to answer this question, let's talk about hardly -weimer equilibrium.

00:04 Question number one says a particular mutant allele exists in a hypothetical population of 1 ,000 individuals on an ice -land in a frequency of 0 .20.

00:14 Like it says, let us refer to this allele as recessive a.

00:18 It is excessive to the dominant allele, dominant a.

00:21 An analysis of the populations show the following genotypes, okay? and, well, they are giving you the frequencies for the genotypes.

00:28 And the question is, is this population? in approximate harding women equilibrium.

00:33 So remember that according to harding to harding women equilibrium, p plus q is equal to 1, and also p -square plus 2pq plus q is equal to 1.

00:41 In this case, the frequency for homozygous dominoleine is 0 .67.

00:48 The frequency for homo -sacogicaseaseif a is 0 .03 and the frequency for the hittrosygos is 0 .30.

00:58 Now, in order to see if this population is in hardly memory equilibrium, we have to find the frequencies for these genotypes but in the next generation because when a population is in hardly one where equilibrium, the frequencies for them is not going to, or are not going to change.

01:15 So in order to find the frequencies for these genotypes in the next generation, you have to group this 0 .67 plus half of these heterozycos in the same here.

01:27 For example, in this case, you have to group like this.

01:30 0 .67 and instead of 0 .30 you're going to get 0 .15 and also here 0 .15 because if you add 0 .15 plus 0 .15 you're going to get 0 .30 that is the frequency for heterocycles okay and you're going to group like this like this so now that you have done this you have to add this value with this one here and the same here and in this case you're going to get 0 .80 2 yep and here you're going to get 0 .48 right in this value here is going to represent the value for p and this value is going to be the value for q now you have to see or you want to see if this population is a highly memory equilibrium so we now have the value for p and for q so if we square this 0 .48 it means if we get the value for q squared we should get 0 .03 okay so let's do that you're going to get 0 .48 and you're going to square this value and if you do that you're going to get 0 .2304 then let's do the same for the value for p p square plus 0 .82 and you're going to square it and here you're going to get 0 .67 to 4 and you have to do the same for 2pq okay so you're gonna you're gonna have you're going to have here 0 .78 70 okay so these are the frequencies for q square p square and 2 pq and as you can see here this frequency for q square for p square and for 2 pq are different for like from these frequencies here that are 0.

03:35 0 .67, 0 .30, and 0 .03.

03:39 So, particularly, this population is not in hard -de -weimber equilibrium.

03:45 And this is the answer for question number one.

03:47 Question number two says, in a natural population, we find a genetic disease resulting from a single mutant allele, recessive a found in the homosygoous state.

03:56 Okay? then it says this allele is recessive to a dominant allele -a.

04:00 Genetic analysis shows that the disease is found in a frequency of 0 .000alalase.

04:05 So again, p plus q is equal to 1, where p is the frequency of alils in the population that are dominant alils, n q is the frequency of alils in the population that are siphales...

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