× Didn't find what you were looking for? Ask a question
Top Posters
Since Sunday
w
5
a
3
j
2
a
2
t
2
u
2
r
2
j
2
j
2
l
2
d
2
y
2
New Topic  
smag smag
wrote...
Posts: 39
Rep: 1 0
12 years ago
A recessive trait appears in 81% of the individuals in a population that is in Hardy-Weinberg equilibrium.  What percent of the population in the next generation is expected to be homozygous dominant?
Read 461 times
1 Reply

Related Topics

Replies
wrote...
12 years ago
In Hardy-Weinberg equilibrium, (p + q)^2 = p^2 + 2pq + q^2 = 1; and p + q = 1
The frequency of recessive trait, q^2(aa) = 0.81
The frequency of recessive allele, q(a) = SQRT 0.81 = 0.9
The frequency of dominant allele, p(A) = 1 - 0.9 = 0.1

Percentage of homozygous dominant in the population in the next generation would be
p^2 (AA) = 0.1^2 = 0.01
New Topic      
Explore
Post your homework questions and get free online help from our incredible volunteers
  1397 People Browsing
Related Images
  
 766
  
 68
  
 197
Your Opinion