Red = RR, Rr
Purple = rr
Genotype of F1 parents = Rr and Rr
Null hypothesis : Both the parents were heterozgyotes.
Rr × Rr
Expected ratio = 3 (red) : 1 (purple)
Expected numbers = 696.75 (red), 232.25 (purple)
Phenotype | #observed | #expected | (O-E)^2/E |
Red | 705 | 696.75 | 0.0977 |
Purple | 224 | 232.25 | 0.293 |
Calcuated chi sqaure = 0.0977 + 0.293 = 0.3907
Degrees of freedom = 2 - 1 = 1
p value = 0.1 to 0.9
p value is more than 0.05. So, we will accept null hypothesis.
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