Respuesta :
We solve the problem using the Hardy-Wineberg equation.
Let us start with the homozygous recessive percentage first,
q² = 35/100 = 0.35
We have the frequency for short stem (s) as q = √0.35 = 0.59
Solving for the frequency of long stem (S) by using the fact that,
p + q = 1
∴ p = 1 - q = 1 - 0.59 = 0.41
So the long stem (S) has a frequency of 0.41.
Now we can calculate 2pq which is the heterozygous alleles (Ss) frequency.
∴ 2pq = 2 x 0.59 x 0.41 = 0.48
Hence 48% of the population is heterozygous for long stem.