When there is found independent assortment, then the each progeny are found in 1/4 proportions of total. 1/4×(650)=162.5
Chi square table:
Phenotype | observed flies (O) | Expected flies (E) | (O-E)^2/E |
vg+ qu+ | 230 | 162.5 | 28.03846 |
vg qu | 224 | 162.5 | 23.2753846 |
vg qu+ | 97 | 162.5 | 26.4015385 |
vg+ qu | 99 | 162.5 | 24.8138462 |
Total | 650 | 650 | 102.5292 |
The answer upto one decimal place is 102.5
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The chi square vakue from chi square table at degree of freedom n-1=4-1=3 and at p =0.05 is 7.815
As our chi square experimental value is more tham chi square value from table.
That's why answer is:
No, the X^2 = value indicates that the observed progeny are significantly different from what would be expected with independent assortment of genes.
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