Please Help! Question Below!
The wing color of a certain species of moth is controlled by alleles at a single locus. Gray (G) is dominant to white (g). A scientist studied a large population of these moths, tracking the frequency of the G allele over time, as shown in the figure below.
Assuming that the population was in Hardy-Weinberg equilibrium for this gene, what percentage of the moth population was homozygous recessive (gg) in 1975?
A) 25% B) 36% C) 64% D) 75%
@amistre64 @Nnesha @Compassionate @mathmate
define: Hardy-Weinberg equilibrium
p + q = 1?
I think the answer is C by the way.
the wiki defines it as: allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences.
Okay
that simply means that the ratio remains consistent .... how to apply that i dont know yet, need to scour the google some more
Okay please do. I really need this :D
ok, i see the p+q=1 in that f(AA) = p^2 f(aa) = q^2 f(Aa) = 2pq which are the terms of the expansion of: (p+q)^2 = 1
Okay I understand so far.
Gg is a geno type? G and g are alleles? or how do we define those words?
Yes that is how we define it in my class as well.
im trying to determine how we go from a frequency an allele of G, to determining a genotype of (gg)
a pundit square is like G g G GG Gg g Gg gg g g G Gg Gg g gg gg g g G Gg Gg G Gg Gg
Yes
So would my answer be correct?
i cannot verify your answer, i simply do not have enough experience with this stuff to confirm or deny it. it 'seems' like a good answer; but can you back up its reasoning? even then, i cant verify it.
Okay thank you so much! @amistre64
good luck with it
i wonder \[(G+g)^2=1\] \[GG+2Gg+\underbrace{gg}_{\text{solution ?}}=1\]
if G is between .4 and .5 g is between .5 and .6 gg, or g^2, is between .25 and .36 but of course both of those extremes are solution options, and i dont see an option between them so this idea might not hold any value
http://www.k-state.edu/parasitology/biology198/answers1.html ok, the idea is good ... at least according to this
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