If the population is at Hardy-Weinberg equilibrium, then the frequency of the recessive genes would be:
aa=0.16
a= √0.16= 0.4
Then, the frequency of dominant genes would be:
0.4+q=1
q= 1-0.4= 0.6
The frequency of <span>homozygous dominant genotype</span> would be:
qq= 0.6*0.6=0.36= 36%
<span>relationship between these beetles is best described as </span>competition
Answer:
The correct answer is -
Numerator = 6
Denominator = 73
Explanation:
In F static test or table there are two sets of degrees of freedom: one for the numerator and one for the denominator. The degrees of freedom are equal to the number of observations minus one. Thus, if the sample size were 7 then the degree of freedom would be 7 minus 1
Numerator df = 7 - 1 = 6
Thus, the numerato would be - 6.
Denominator is calculated by subtracting the observation or variables from the total number of sample size.
Denominator df = 80- 7 = 73
