Xque
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Some genetics questions
Studying for my exam I have come upon some problems:
Regarding a population of Hardy-Weinberg equilibrium, how can I tell 2pq or p by q? Let me give an example. I have this exercise that tells me to
calculate the frequency of the gene r, based on the fact that in the given H-W equilibrium population there is a frequence of 16% rr persons.
As I was typing this, a solution came to my mind, is it correct?
rr = q^2 => q = 0.4, p= 1 - q = 0,6
p^2 + 2pq + q^2 = 0,36 + 0,48 + 0,16, thus the frequence of q is 0,16 + 0,24 = 0,40 or 40%
Another exercise asks, could individuals of a homogametic sex be used for making a linkage map (in the previous exercise we did a map of some
x-chromosome genes based on the phenotypes of heterogametic individuals). Depending on the lack of facts in this exercise, my answer is no, at least
you can't do it based on the phenotypes since it is not the direct expression of single genes, unlike the males we mapped earlier. If I am to take
Barr bodies into consideration, my answer might be different.
You talk the talk, but do you walk the walk?
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futurenobellaureate
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Please be more specific in your questions.
Sciguy
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Twilight
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For the first question:
Frequency of dominant gene (presumably R) = p
Frequency of recessive gene r = q
So yes, the frequency of rr = 0.16 = q^2, which directly implies that q = 0.40 = frequency of r. The other steps are not necessary.
For the second question, I am not sure, but it should be possible to do a linkage map on a homogametic sex because the actual genotypes themselves do
not matter as much as the fact that the phenotype of the offspring differs from that of the parents (implying a "recombinant" genotype). Besides, how
would one map an autosomal chromosome if the linkage map analysis may only be performed on genes that have only one copy in the organism (so to
speak)?
Feel free to correct me, as my AP biology knowledge is quite rusty.
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