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The question is:

The frequency of a recessive allele a is 0.6. If a population is in Hardy-Weinberg equilibrium, what is the frequency of the dominant phenotype?

This question is a bit confusing to me. I'm not sure how to use my knowledge about Hardy-Weinberg in this problem. I know that when q = 0.6, p = 1 - 0.6 = 0.4. But when they are asking for the dominant phenotype, should it be p^2 or the squareroot of p? I have watched several YT-videos about this, but I still can't get the hang of it. There's not much help in my teacher or books. I would be really grateful if someone could clear up my confusions!

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    $\begingroup$ I was about to recommend the post Solving Hardy Weinberg problems but then realized that you are the OP and that you have accepted the question suggesting that the answer did not help much. The difficulty of this problem is to remember what genotype have the dominant phenotype and then sum their frequencies, just like Alan Boyd did (+1). $\endgroup$ – Remi.b May 14 '17 at 18:49
  • $\begingroup$ Hi! If you are referring to your answer in my previous post - it did help a lot, but these problems still get me confused. You suggested opening a new post with my question, so I did after a while. If you are referring to the answer I got here - also helped, but I'm afraid that when I'm coming across another problem phrased slightly different than this I won't know how to break it down. Anyways I'll check back later $\endgroup$ – Thomas May 14 '17 at 18:58
  • $\begingroup$ Yep. I did not ask for justification. It is all good :) Glad the other post helped. Good luck @Thomas $\endgroup$ – Remi.b May 14 '17 at 19:11
  • $\begingroup$ Ok! Aha - I read the last bit of your comment just now, thanks for the tip. Are you some kind of moderator? Can you think of any good sources to clear up on my confusions about this? My examdate is getting closer and closer $\endgroup$ – Thomas May 15 '17 at 21:56
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p = p(a) = 0.6

q = p(A) = 0.4

Frequency of homozygous recessive, aa = p2 = 0.36

All other genotypes have dominant phenotype therefore the frequency of the dominant phenotype is 1-0.36 = 0.64

(Frequency of homozygous AA = q2 = 0.16

Frequency of heterozygous Aa = 2pq = 0.48)

And if you want a deeper explanation you really should study this answer by Remi.b

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Since the frequency of the dominant allele is 0.4, or 40%, the chance of each allele being the dominant allele is 0.4. Since you are asking what is the probability of BOTH alleles being dominant we use the multiplication rule of probability, 0.4*0.4 = .16.

This is similar to the probability of flipping two coins and having them both be heads. Each coin has a 50% chance, therefore the probability of both being heads is 0.5*0.5 = 0.25.

That is the logic behind the Hardy Weinberg formula, p^2 + 2pq + q^2 = 1.

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    $\begingroup$ “Since you are asking what is the probability of BOTH alleles being dominant” — No. The question is asking after the frequency of the dominant phenotype. That is, any one dominant allele is sufficient. See the other reply for the correct answer. $\endgroup$ – Konrad Rudolph May 14 '17 at 20:43

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