Question #f71f9

1 Answer
Oct 17, 2017

Answer:

p= 0.6001
q= 0.399
#p^2#= pp = 235.864 --> 236
pq= 313.177 --> 313
#q^2#=qq=103.958 --> 104

Explanation:

We know that our Population X is 653 individuals. Of this population 104 have floppy ears (Xf). We know that p (short) is dominant over q (floppy).

Basic equations for Hardy Weinberg:
#p^2# + 2pq + #q^2# = 1
p + q = 1

#q^2# is the same as the # X / (Xf)# . solving this gives 0.15926
Taking the square root of this number gives us the frequency
q= 0.399

Now we can enter this into p+q=1
This means that p= 1 - q which solves to 0.6001

Now that we now both p and q we can start calculating the frequencies of the genotypes
#p^2# for type pp gives 0.6001 * 0.6001 = 235.864 Round this number up to 236

2pq gives 2 * 0.6001 * 0.399 = 313.177 which gets rounded up to 313

#q^2# for type qq gives 0.399 * 0.399= 103.958 which gets rounded up tp 104

To find the frequency of each genotype we divide the total population by each genotypes' population.
fpp = #236/653#= 0.367140....
fpq = #313/653#= 0.47932....
fqq= #104/653#=0.15926.....