Answer:
Option C.
Step-by-step explanation:
Total population = 1000 snails
AA = 160 snails
Aa = 480 snails
aa = 360 snails
Frequency of each type.
[tex]AA=\dfrac{160}{1000}=0.16[/tex]
[tex]Aa=\dfrac{480}{1000}=0.48[/tex]
[tex]aa=\dfrac{360}{1000}=0.36[/tex]
Now, we get
[tex]A^2=0.16\Rightarrow A=\sqrt{0.16}=0.4[/tex]
he frequency of the A allele in this population is 0.4.
Therefore, the correct option is C.
