Answer:
b).
The expected ratio of the above progeny would be 1:1:1:1.
Phenotype |
Observed(O) |
Expected (E) |
O-E |
(O-E)2 |
(O-E)2/E |
Gray body / long wings |
965 |
575.00 |
390.00 |
152100.00 |
264.52 |
black body / vestigial wings |
944 |
575.00 |
369.00 |
136161.00 |
236.80 |
gray body/vestigial wings |
206 |
575.00 |
-369.00 |
136161.00 |
236.80 |
black body / long wings |
185 |
575.00 |
-390.00 |
152100.00 |
264.52 |
Total |
2300 |
2300.00 |
1002.65 |
X^2 value = 1002. 65
Degrees of freedom = Number of phenotypes – 1
Df = 4-1=3
Critical value = 7.815
The chi-square value of 1002.65 is greater than the critical value of 7.815. We can reject null hypothesis and the two genes are linked.
c). Based on the above results, it is clear that if the two genes are present on the same chromosome, thery are said to be linked. Testcross ration of linked genes and genes assorted independently assorting genes are different.
(b) Morgan actually obtained the following results: 965 offspring were gray body/long wings; 944 were black...
o a phenotypically wildtype fruit fly of unknown origin to a fly homozygou #1) and vg (vestigial wings mutations the following offspring were obtained: pts) In a cross of a phenotypicall ck body, fr . wildtype - 12 black body, normal wings -34 normal body, vestigial wings 38 black body, vestigial wings - 16 a. Are genes b and vg linked or not? what would you have seen if the opposite were true? b. If they are not linked, prove...