Question

Using the same example in the book, work the Hardy-Weinberg Equation for the following The sample size is 3200 plants. Keep in mind, flowering plants are diploid. (6 Pts.)

There are:                                                                                                                                                                                         

1980 Blue flowers                 1000 Teal flowers                220 Green flowers

What does this all mean? What are the numbers doing for you?

Answer 1

Blue = TT = 1980

Teal = Tt = 1000

Green = tt = 220

Total T alleles = 2*1980+1000 = 4960

Total t alleles = 2*220+1000 = 1440

Total alleles = 70+90 = 56400

Frequency of dominant allele = p(T) = 4960/6400 = 0.775

Frequency of recessive allele = p(t) = 1440/6400 =0.225

p^2 = 0.775*0.775 = 0.601

2pq = 2* 0.775*0.225 =0.349

q^2 = 0.225*0.225 = 0.050

Expected values:

TT = 0.601*3200 = 1923

Tt = 0.349*3200 = 1116.8

tt = 0.050*3200 = 160

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
TT	1980	1923.2	56.8	3226.2400	1.6775
Tt	1000	1116.8	-116.8	13642.2400	12.2155
tt	220	160	60	3600.0000	22.5000
Total	3200	3200			36.3930

Chi-square value = 36.39

Degrees of freedom = number of genotype – 1

Df = 3-1=2

Ther critical X^2 value is 5.99

The X^2 value of 36.39 is greater than the critical value of 5.99. So we can reject the null hypothesis. The observed values are not consistent with the expected values and the population is not in Hardy-weinberg equilibrium