24) (5 pts) Two mutant genes in Drosophila melanogaster produce the recessive tr

Question

24) (5 pts) Two mutant genes in Drosophila melanogaster produce the recessive traits crossveinless wings (cv) and singed bristles (sn). Given the following genetic map, if the F1 dihybrid flies (produced from a cross of pure-breeding males with singed bristles and pure-breeding females with crossveinless wings) for these two genes were crossed to flies that expressed both recessive phenotypes, what outcome would you predict in terms of the phenotypic classes seen and numbers of progeny in each class (assume a population size of 1000)?

Explanation / Answer

The map distance between the genes is 10 mu, which means the proportion of recombinants is 10%.

Consider the genotype of F1 population as follows:

Pure breeding males with singed bristles: cv+ - sn (wild type for wings)

Pure breeding females with crossveinless wings: cv – sn+

A cross between these two genotypes gives the following offspring: cv+ - sn/ cv – sn+

The offspring are now crossed to flies recessive for both the phenotypes. The phenotype of these recessive flies is: cv – sn/ cv – sn

A cross between these two parents will give the following offspring

Gametes

cv - sn

Offspring phenotype

proportion

cv+ - sn (parental)

cv+ - sn / cv - sn

Normal wings, singed bristles

45% = 450

cv – sn+(parental)

cv – sn+ / cv - sn

Crossveinless wings, normal bristles

45% = 450

cv+ - sn+ (recombinant)

cv+ - sn+ / cv - sn

Wild type

5% = 50

cv – sn (recombinant)

cv – sn / cv - sn

Crossveinless wings, singed bristles

5% = 50

Gametes

cv - sn

Offspring phenotype

proportion

cv+ - sn (parental)

cv+ - sn / cv - sn

Normal wings, singed bristles

45% = 450

cv – sn+(parental)

cv – sn+ / cv - sn

Crossveinless wings, normal bristles

45% = 450

cv+ - sn+ (recombinant)

cv+ - sn+ / cv - sn

Wild type

5% = 50

cv – sn (recombinant)

cv – sn / cv - sn

Crossveinless wings, singed bristles

5% = 50

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