Answer:
0.5
Explanation:
According to Hardy-Weinberg equilibrium p2+2pq+q2=1 (p+q=1)
p2 is frequency of the dominant homozygous genotype
2pq is the frequency of the heterozygous genotype
q2 is the frequency of the recessive homozygous genotype
In the example above 80 chickens have bare legs (ff)-recessive homozygous which means that the frequency of that genotype is 80/240+80=0.25 (q2), frequency of the recessive allele is
=0.5. This means that the frequency of the dominant allele (p) is 1-0.5=0.5
So, the frequency of the heterozygous genotype (2pq) is 2*0.5*0.5=0.5
The frequency of the dominant homozygous genotype is p2=0.25
I believe it is the first one but I am not 100% sure
The answer is phagocytosis which is also known as cell eating. It is a process by which the living cells which are called phagocytes would ingest other cells. It might be an amoeba or a white blood cell. Hope this would help.