You are interested in a set of genes regulating digestion in Drosophila. You hav

Question

You are interested in a set of genes regulating digestion in Drosophila. You have two loci (E and F) that both encode digestive enzymes; E is dominant to e and F is dominant to f. You cross a homozygous recessive fly to a homozygous dominant and testcross the F1 progeny. You get the following F2 progeny: Ee Ff 585 Ee ff 202 ee Ff 200 ee ff 566 Perform a chi-square test of independence to determine if these alleles assort independently. Give the chi-square value, df and p-value. What conclusion can you make about the independent assortment of loci E and F?

Explanation / Answer

In case of test cross of the heterozygous progeny of the dihybrid cross, we get the ratio 1:1:1:1.

Total number of progenies = 1553

If the genes were independently assorted then the number of progenies of each genotype would have been equal.

So expected number of progenies - 1553/4 = 388.

We need to find out the Chi-square value.

SO the chi- Square value is 361.

Degree of freedom df = number of alleles- 1

4- 1

3.

For three degree of freedom when the value is 0.05 chi-square value is 7.82.

Our observed value of chi-square is 361 which is way more than the value that should be ideal ie.e 7.82.

So our hypothesis that the alleles are independently assorted is wrong.

Observed o Expected e Deviation d (Observed- expected) Deviation square d2 d2/e EeFf 585 388 197 38809 100.0232 Eeff 202 388 -186 34596 89.16495 eeFf 200 388 -188 35344 91.09278 eeff 566 388 178 31684 81.65979 Total 1553 361.9407

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