How do I work this out??

Question

can someone plz explain this to me? my answer was 0.3 but i got it wrong..thank u (:

Researchers examining a particular gene in a fruit fly population discovered that the gene can have either of two slightly different sequences, designated A1 and A2. Further tests showed that 70% of the gametes produced in the population contained the A1 sequence. If the population is at Hardy-Weinberg equilibrium, what proportion of the flies carries both A1 and A2?

Answer 1

You need these formulas:
p(squared) + 2pq + q(suared) = 1
p + q = 1

p = A1 = 70% = 0.7

using the 2nd formula:
0.7 - q = 1
q = 0.3

Your question asks what percentage of the population are heterozygotes.

Now we use the first formula and square both p and q

p = 0.7
p2 = 0.49

q = 0.3
q2 = 0.09

p2 + 2pq + q2 = 1

0.49 + 2pq + 0.09 = 1
2pq = 0.42

Your answer is 42% contain both alleles (are heterozygotes).

We also could have:

2pq = A1A2 (heterozygotes)

2 x 0.7 x 0.3 = 0.42

Answer 2

OK, let's go from the start:
In the entire population, 70% of the gametes are A1, so 30% are A2 (that was probably where you got your 0.3).
Now let's set up a combination box:

gametes........A1(0.7) .. A2(0.3)

A1(0.7).......A1A1 (0.49)...A1A2 (0.21)
A2(0.3).......A2A1 (0.21)...A2A2 (0.09)

So, homozygous A1A1 individuals constitute about 49% of the population (70% X 70%)
homozygous A2A2 individuals constitute about 9% of the population (30% X 30%)
and heterozygous (A1A2 and A2A1) individuals make up about 42% of the population ((70% X 30%) + (30% X 70%)).
Does it add up? 49% + 9% + 42% = 100%, so it looks OK!
Now this only applies at population equilibrium, where the gene percentages are stable.
Hope that's a little clearer now!