Let us assign the
recessive allele for freckle to be f and
the dominant F. According to the statement, we deduce that ff = 0.04.
The frequency of the recessive f allele is, therefore, is
0.2 (square root of 0.04 ).
Hardy-Weinburg p + q = 1
P + 0.2 = 1; p = 1- 0.2 = 0.8
Therefore using the Hardy-Weinburg equation of a population
in equilibrium
P2+2pq+q2=1
Heterozygous individuals are;
2pq = 2*0.8*0.2 = 0.32
This is 32% of the population
<span>humus; Increase..................................................</span>
The protein and non-protein
Answer:
Explanation:
Nonspecific defenses include physical and chemical barriers, the inflammatory response, and interferons. Physical barriers include the intact skin and mucous membranes. ... An example of such a substance is lysozyme, an enzyme present in tears that destroys the cell membranes of certain bacteria.
