Plz Help with mendilian genetics 1. aristaless-3 [al3] autosomal mutant Drosophi

Question

Plz Help with mendilian genetics

1. aristaless-3 [al3] autosomal mutant Drosophila melanogaster flies have no Aristae (antennae) Female flies that are homozygous for this mutation are sterile. Another autosomal mutation, aristapedia [ap] produces flies that have abnormal development of antennae, where they develop into deformed legs. Aristapedia is homozygous lethal. If you were to design a vial such that it contained 1 homozygous purebred wild type [wt] female, 1 heterozygous aristaless-3 and aristapedia mutant male, 1 homozygous aristaless-3 (homozygous wildtype at the other locus) female, and 1 homozygous aristaless-3 male (homozygous wildtype at the other locus), (a) what is the phenotype ratio expected in this vial in the next generation? (b) If you had a vial with a heterozygous aristaless-3 and aristapedia mutant male, and one heterozygous aristaless-3 and aristapedia mutant female, what is the expected phenotype ratio in the vial in the surviving offspring in the next generation?

Explanation / Answer

Solution :

aristaless - al3 = AL = A

aristapedia - ap3 = AP = P

a]The vial has the follwoing :1 homozygous purebreed wild type [wt]female(AApp)

                                            1 hetreozygous aristaless-3 and aristapedia mutant male(AaPp)

                                            1 homozygous aristaless-3 female(AAPp)

                                            1 homozygous aristaless-3 male(AAPp)

AApp x AaPp

Phenotypic ratio obtained :1:1:1:1

b]The vial has the following:heterozygous aristaless-3 and aristapedia mutant male(AaPp)

                                            heterozygous aristaless-3 and aristapedia mutant female(AaPp)

AaPp x AaPp

Phenotypic ratio obtained : 9:3:3:1

AP Ap aP ap Ap AAPp AApp AaPp Aapp Ap AAPp AApp AaPp Aapp Ap AAPp AApp AaPp Aapp Ap AAPp AApp AaPp Aapp

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