Question

In: Biology

A Drosophila female homozygous mutant for two recessive genes (a and b) is crossed to a...

A Drosophila female homozygous mutant for two recessive genes (a
and b) is crossed to a wild-type male. The F1 are all wild-type. An F1 female is crossed to a double-mutant male and the following phenotypes were seen in the F2:

a b	497
+ +	460
a +	465
+ b	426
Total	1848

Perform a “best fit” or apriori χ² analysis to determine if genes a
and b assort independently. Give the value for χ², DF and Probability; as well as your decision. Are these genes linked? Use the method you used in Statistics lab and round to two decimal places throughout.

		χ² = 5.48, DF = 3, prob < 0.05, reject, linked
		χ² = 8.45, DF = 3, 0.1 < prob < 0.9. accept, not linked
		χ² = 4.58, DF = 4, 0.1< prob< 0.5, accept, not linked
		χ² = 5.48, DF = 3, 0.1 < prob < 0.5, accept, not linked

Expert Solution

Given that the female is homozygous mutant for the two recessive genes a and b. Means, its genotype is, aa bb. The genotype of a pure breed Wild-type male is, AA BB. Cross between these two will have the progeny with the following genotypes.

aa bb* AA BB -----> Aa Bb (All wild-type) -------> F1

Cross between F1 female and double-mutant male (aa bb) will have the progeny with the following genotypes.

Aa Bb* aa bb ----> Aa Bb (++) (1/4, all wild-type), Aa bb (+b) (1/4, mutant for b), aa Bb (a+) (1/4, mutant for a), aa bb (ab) (1/4, double mutant)

The given phenotypes are in approximately equal ratios similar to that of the expected phenotypes.

CHI - SQUARE (X²):

X² = Σ(O - E)² / E

Where O = Observed frequency

E = Expected frequency

Phenotype	O	E	(O-E)	(O-E)^2	(O-E)^2/E
A_B_	497	462	35	1225	2.651515
A_bb	460	462	-2	4	0.008658
aa A_	465	462	1.5	2.25	0.155172
aa bb	426	462	-36	1296	2.805195
					2.815345

The degrees of freedom (DF) = n-1 = 4-1 = 3

The calculated P value is p < 0.421856, which is not significant at p< 0.05. So, we accept the null hypothesis.

The difference between the expected an observed phenotypic ratios is not significant and the genes are assorting independently.

gladiator answered 1 year ago

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