A line of Drosophila melanogaster with ebony body and sepia eyes (line A) is cro

Question

A line of Drosophila melanogaster with ebony body and sepia eyes (line A) is crossed to a line with tan body and red eyes (line B). The F1 are all tan-bodied, red-eyed. Imagine that An F1 female is crossed to a male from line A. Now imagine that the following data were actually collected from the cross : Body Eyes Tan normal 241 Tan sepia 259 Ebony normal 240 Ebony sepia 260. Diagram this cross.

Explanation / Answer

e: ebony; E: tan body s: sepia: S: red eyes P: eess x EESS F1: EeSs (tan-red) What proportion of gametes produced by line A have the dominant allele for both loci? => 0 (line A has genotype eess) What proportion of the gametes produced by the F1 have the dominant allele for both loci? => 1/2 (for linked) => 1/4 (for unlinked) An F1 female is crossed to a male from line A. If the two genes are unlinked: 250 tan-red 250 tan-sepia 250 ebony-red 250 ebony-sepia Now imagine that the following data were actually collected from the cross above: Body Eyes Tan normal 411 Tan sepia 89 Ebony normal 87 Ebony sepia 413 Are the loci linked? => Yes For the linked genes, the proportion of the gametes would be m : n : n : m ; where 'm' is parental types and 'n' is recombinant types, and m > n. If so, how many map units are they apart ? (89+87)/(411+89+87+413) x 100 = 17.6 mu ~~> 18 mu Do a x^2 test and show your work below: obs: observed exp: expected d = obs - exp --Phenotype--| obs |. exp | d | d^2 | d^2/e | Tan normal -- | 411 | 410 | 1 |..1....| 0.002 Tan sepia ---- |. 89 |... 90 | -1 |.1....| 0.011 Ebony normal |. 87 |... 90 | -3 |.9...| 0.100 Ebony sepia -| 413 | 410 | 3 |..9...| 0.022 The value of Chi-square = 0.002 + 0.011 + 0.100 + 0.022 = 0.135. So, with the degree of freedom 3 (from four classes), the probability, p > 0.95 meaning that the observed data were not signifantly different from the expected data (410 : 90 : 90 : 410).

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