A Drosophila female homozygous mutant for two recessive genes ( a and b ) is cro

ID: 187236 • Letter: A

Question

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 2 analysis to determine if genes a
and b assort independently. Give the value for 2, 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.

2 = 5.48, DF = 3, prob < 0.05, reject, linked

2 = 8.45, DF = 3, 0.1 < prob < 0.9. accept, not linked

2 = 4.58, DF = 4, 0.1< prob< 0.5, accept, not linked

2 = 5.48, DF = 3, 0.1 < prob < 0.5, accept, not linked

a b

497

+ +

460

a +

465

+ b

426

Total

1848

Explanation / Answer

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 = (O - E)2 / E

Where O = Observed frequency

E = Expected frequency

Phenotype

(O-E)

(O-E)^2

(O-E)^2/E

A_B_

497

462

1225

2.651515

A_bb

460

462

-2

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.