Hi! :)
Make a = q, aa (homozygous recessive) = q^2,
A = p, AA (homozygous dominant) = p^2, and
2pq = heterozygous
This was derived from p + q = 1
Therefore all a in pop (q) = 20% = .20
And all A in pop (p) = 80% = .80
Since the disease is homozygous recessive (affected), then aa = qq or q×q = q^2
q^2 = (.20)^2 = .040 = 4%
Answer: 4%
C the ovary is being pointed at
I think the answer is coilibscillosis
<h3><em>False! I believe.</em></h3><h3><em>This is because, in a low preasure aea the air becomes more lighter. Like High, high up in the earths atmophere is has little air, so it would behard to breath.</em></h3><h3><em>Hope this helps, sorry if not tho </em></h3>
It’s really all of the above but i think the best answer is C :)) hope this helped !!