# Question #f71f9

Oct 17, 2017

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 $\frac{X}{X f}$ . 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 = $\frac{236}{653}$= 0.367140....
fpq = $\frac{313}{653}$= 0.47932....
fqq= $\frac{104}{653}$=0.15926.....