In: Biology
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 |
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.