You are interested in a set of genes regulating digestion in Drosophila. You hav
ID: 193290 • Letter: Y
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
The paretnal crossis, EE FF* ee ff à Ee Ff à F1 generation
The testcross of F1 progeny is, Ee Ff* ee ff à Ee Ff (1/4), Ee ff (1/4), ee Ff (1/4), ee ff (1/4)
So, the expected genotypic ratio is, 1:1:1:1
The total number of progeny is = 585+202+200+566 = 1553.
Thus, the expected number of each genotype is, 1553/4 = 388.25
CHI - SQUARE (X2):
X2 = (O - E)2 / E
Where O = Observed frequency
E = Expected frequency
P
O
E
(O-E)
(O-E)^2
(O-E)^2/E
Ee Ff
585
388.25
196.75
38710.56
99.70525
Ee ff
202
388.25
-186.25
34689.06
89.34723
ee Ff
200
388.25
-188.25
35438.06
91.2764
ee ff
566
388.25
177.75
31595.06
81.37814
1553
1553
361.707
Thus, the chi-square value is, 361.707
The degrees of freedom is = n – 1 = 4-1 = 3
The P-Value is < 0.00001. The result is significant at p < 0.05.
Conclusion: Thus, the difference between expected and observed values is significant, so the genes are not following the independent assortment.
P
O
E
(O-E)
(O-E)^2
(O-E)^2/E
Ee Ff
585
388.25
196.75
38710.56
99.70525
Ee ff
202
388.25
-186.25
34689.06
89.34723
ee Ff
200
388.25
-188.25
35438.06
91.2764
ee ff
566
388.25
177.75
31595.06
81.37814
1553
1553
361.707
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