In drosphila the mutant black (b) has a black body and the wild type has a grey

Question

In drosphila the mutant black (b) has a black body and the wild type has a grey body ; the mutant vestigial (vg) has wings that are much shorterand crumpled compared with the long wings of the wild type . In the following cross the true breeding parents are given together with the counts of off spring of p females * black and vestigial males p black and. Normal * grey and vestigial.....f1 females * black vestigial males progeny: ...grey normal 283 , grey vestigial 1294, black normal1418...black vestigial241. ?from this data , calculate the map distance between the black and vestigial genes.

Explanation / Answer

Gene Mapping Example continued? 1st step: identify the parental and recombinantsP: black, normal x grey, vestigialF1: females x black, vestigialProgeny: 283 grey, normal 1,294 grey, vestigial 1,418 black, normal 241 black, vestigial Gene Mapping Example continuedP: black, normal x grey, vestigialF1: females x black, vestigialProgeny: 283 grey, normal 1,294 grey, vestigial 1,418 black, normal 241 black, vestigialThe parentals are the same phenotype as parents ? Grey, normal ? Black, vestigial 10. Gene Mapping Example continued2nd step: Identify recombinants and figureout percentage of progenyP: black, normal x grey, vestigialF1: females x black, vestigialProgeny: 283 grey, normal 1,294 grey, vestigial 1,418 black, normal 241 black, vestigial . Gene Mapping Example continuedRecombinants : grey, vestigial black, normalP: black, normal x grey, vestigialF1: females x black, vestigialProgeny: 283 grey, normal 1,294 grey, vestigial 1,418 black, normal 241 black, vestigial Gene Mapping Example continuedTo find percentage of recombination take each recombinantand divide by the total offspring and multiply by 100.P: black, normal x grey, vestigialF1: females x black, vestigialProgeny: 283 grey, normal 1,294 grey, vestigial 1,418 black, normal 241 black, vestigial . Gene Mapping Example continuedProgeny: 283 grey, normal 1,294 grey, vestigial 1,418 black, normal 241 black, vestigial Total 3236Recombinants : grey, vestigial (1,294/3236) x 100 = 40% black, normal (1,418/3235) x 100 = 43.8 % . Gene Mapping Example continuedTotal recombination: 40% + 43.8% = 83.8%Progeny: 283 grey, normal 1,294 grey, vestigial 1,418 black, normal 241 black, vestigial Total 3236Recombinants : grey, vestigial (1,294/3236) x 100 = 40% black, normal (1,418/3235) x 100 = 43.8 % Gene Mapping Example continued Change the percentage to map units 83.8% = 83.8 map units The black and vestigial genes are located 83.8 map units apart.

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