q^2 = 1/900 --> q = 1/30

Carrier frequency per Hardy-Weinberg formula is 2pq.

Now, for rare autosomal recessive diseases the carrier frequency of 2pq ≈ 2q.

2q = 2(1/30) = 2/30 = 1/15

Answer = 1/15

Just as a note for anybody else who was WTF at how 2(29/30)(30/30) = 1/15...a lot of question banks round 29/30 (or any similarly large fraction) out to 1

p^2 + 2pq + q^2 = 1 p+q= 1

q^2= 1/900 q= 1/30 p= 29/30 which rounds to 1

They are asking for carrier frequency which is 2pq.

If u are not goot in math like me xD. I did it this way: They gave me p^2=1/900 so p=1/30.

Now I need to find P, so use p+q=1. p= 1-(1/30)= 1-0.03= 0.97

So, 2pq= 2(0.97)(0.03)= 0.064 = (1/15) :D I got it right, this is my way to do it cause im not good at fractions.

Here's how I did it

X is the frequency a parent will be a carrier of the disease. 1/4 is the probability the parents will have a child with the disease. Therefore

X * X * 1/4 = 1/900

X = 1/15