In: Biology
Assume you are studying some species of flower that comes in three different colors: red, pink, and white. You know that color is determined by a single locus with two alleles with incomplete dominance. You count 124 red flowers, 62 pink flowers, and 30 white flowers. What are the allele frequencies and is the population in Hardy-Weinberg equilibrium? Calculate the chi squared test statistic and p-value.
Estimation of allele frequencies :
Phenotype |
Genotype |
Freequency |
Allele R |
Allele r |
Total |
Red |
RR |
124 |
248 |
0 |
248 |
Pink |
Rr |
62 |
62 |
62 |
124 |
White |
rr |
30 |
0 |
60 |
60 |
Total |
Total |
216 |
310 |
122 |
432 |
Allele R |
= 310 / 432 |
= 0.72 |
Allele r |
= 122 / 432 |
= 0.28 |
Expected frequencies =
Red |
RR |
= 0.72 *0.72 = 0.515 |
*216 = 111 |
Pink |
Rr |
= 2 * 0.72 * 0.28 = 0.405 |
*216 = 88 |
White |
rr |
= 0.28 *0.28 = 0.080 |
*216 = 17 |
Null hypothesis: The observed values are not deviating from the expected values.
Category |
RR |
Rr |
rr |
|
Observed values |
124 |
62 |
30 |
|
Exprected Values |
111 |
88 |
17 |
|
Deviation |
13 |
-26 |
13 |
|
D^2 |
163.1533 |
652.6133 |
163.1533 |
|
D^2/E |
1.47 |
7.45 |
9.47 |
18.39 |
X^2 |
18.39 |
|||
Degrees of freedom |
1 |
Inference: The calculated chisquare value i.e. 18.39 is greater than the table value i.e. 3.84 at 1 DF and 0.05 probability, hence the null hypothesis is rejected which means the popuation is not in HW equilibrium.