Question

In a mythical insect called the Sorefly, the dominant allele, L. is responsible for resistance to a common insecticide called Loritol. Another dominant allele, M, is responsible for the ability of Soreflies to hiss like rattlesnakes. A true-breeding mute (cannot hiss), Loritol-resistant Sorefly, was mated to a true-breeding hissing, Loritol-sensitive Sorefly, and the hissing, Loritol-resistant female progeny were testcrossed with true-breeding wild-type (i.e., mute Loritol-sensitive) males. Of the 455 total progeny produced, 139 were mute and Loritol-resistant, 127 could hiss and were Loritol-sensitive; whereas 111 could hiss and were Loritol-resistant, and 107 were mute and Loritol- sensitive. Perform a Chi-square test for independent assortment to determine whether or not these genes may be linked. Show all your work (scratch paper & turn in), but indicate your final calculate Chrsquare value balow • Calculated Chi-square value:

Answer 1

If these two genes were assorting independently, then the cross Mm Ll x mm ll would produce progeny of the phenotype ML, Ml, mL, and ml in a 1:1:1:1 ratio.

To calculate the Chi-square statistic:

Progeny	Observed (O)	Expected (E)	O - E	(O - E)2	(O - E)2/E
Hissing, Loritol Resistant (M L)	111	121	-10	100	0.826
Hissing, Loritol Sensitive (M l)	127	121	6	36	0.298
Mute, Loritol Resistant (m L)	139	121	18	324	2.678
Mute, Loritol Sensitive (m l)	107	121	-14	196	1.62
Total	484				$\chi ^{2}$ = 5.422

At 3 degrees of freedom, the Chi-square value at 95% confidence level is 7.815. Since this value is less than the critical value, the two alleles are assorting independently.