We observed some California Lupine sampling 1000 individuals

We observed some California Lupine, sampling 1000 individuals. You found the following genotype frequencies: AA = 0.080, Aa = 0.280; aa = 0.640. From these data, calculate the allele frequencies of the A and a alleles in this population. Can you reject the null hypothesis that the population is in Hardy-Weinberg Equilibrium? (e.g., is the chi-square value you have calculated greater than 3.84?) A botanist is investigating a population of plants whose petal color is controlled by a single gene whose two are incompletely dominant. She finds 170 plants that are homozygous brown, 21 plants petals are purple-brown, 340 plants that are homozygous purple. Is this population in HWE? How do you know?

Solution

7: Allele frequency
Frequency of A (p)= 0.080 + 1/2 (0.280)= 0.22
Frequency of a (q)= 0.640 + 1/2 (0.280)=0.78
Expected genotype frequency
AA= (0.22)²= 0.0484
Aa= 2 x 0.22 x 0.78= 0.3432
aa= (0.78)²= 0.6084
Expected numbers
AA = 0.0484 x 1000= 48.4
Aa =0.3432 x 1000= 343.2
aa= 0.6084 x 1000= 608.4
chi square= (80 - 48.4)²/ 48.4 + (280- 343.2)²/ 343.2 + (640-608.4)²/ 608.4= 20.63 + 11.63 + 1.64 = 33.9
Since calculated chi square value > 3.84= Reject null hypothesis; not in HWE