Question

In Drosophila, an \(\mathrm{X}\) -linked recessive mutation, scalloped (sd), causes irregular wing margins. Diagram the \(F_{1}\) and \(F_{2}\) results if (a) a scalloped female is crossed with a normal male; (b) a scalloped male is crossed with a normal female. Compare these results to those that would be obtained if the scalloped gene were autosomal.

Short answer

In the X-linked inheritance problem, the F2 generation exhibits a 1:1:1:1 ratio for each phenotype (normal females, scalloped females, normal males, and scalloped males). If the scalloped gene were autosomal, the F2 generation would show a 3:1 ratio of normal to scalloped phenotypes.

Step-by-step solution

Cross (a): Scalloped Female x Normal Male

Firstly, let us consider the cross between a scalloped female and a normal male. Scalloped females have the genotype: \(X_{sd} X_{sd}\) Normal males have the genotype: \(X^{+} Y\) Create a Punnett square to predict the offspring genotypes and phenotypes: |\(X_{sd}\)|\(X_{sd}\)| |-|-| |\(X^{+}\)|\(X^{+} X_{sd}\)|\(X^{+} X_{sd}\)| |\(Y\)|\(X_{sd}Y\)|\(X_{sd}Y\)| The F1 generation will have the following genotypes and phenotypes: - \(X^{+} X_{sd}\) (normal females): 2 - \(X_{sd} Y\) (scalloped males): 2

F1 Cross: Normal Female x Scalloped Male

Now, let us cross the F1 offspring, which have the following genotypes: Normal females: \(X^{+} X_{sd}\) Scalloped males: \(X_{sd} Y\) Create a Punnett square to predict the offspring genotypes and phenotypes: |\(X^{+}\)|\(X_{sd}\)| |-|-| |\(X_{sd}\)|\(X^{+} X_{sd}\)|\(X_{sd} X_{sd}\)| |\(Y\)|\(X^{+} Y\)|\(X_{sd} Y\)| The F2 generation will have the following genotypes and phenotypes: - \(X^{+} X_{sd}\) (normal females): 1 - \(X_{sd} X_{sd}\) (scalloped females): 1 - \(X^{+} Y\) (normal males): 1 - \(X_{sd} Y\) (scalloped males): 1

Cross (b): Scalloped Male x Normal Female

Now, let us consider the cross between a scalloped male and a normal female. Scalloped males have the genotype: \(X_{sd} Y\) Normal females have the genotype: \(X^{+} X^{+}\) Create a Punnett square to predict the offspring genotypes and phenotypes: |\(X^{+}\)|\(X^{+}\)| |-|-| |\(X_{sd}\)|\(X^{+} X_{sd}\)|\(X^{+} X_{sd}\)| |\(Y\)|\(X^{+} Y\)|\(X^{+} Y\)| The F1 generation will have the following genotypes and phenotypes: - \(X^{+} X_{sd}\) (normal females): 2 - \(X^{+} Y\) (normal males): 2

See if autosomal: Scalloped Female x Normal Male

- If the scalloped gene were autosomal, the following results would be obtained for F1 and F2 when a scalloped female is crossed with a normal male: Scalloped female: \(A_{sd} A_{sd}\) Normal male: \(A^{+} A^{+}\) Create a Punnett square to predict the offspring genotypes and phenotypes: |\(A_{sd}\)|\(A_{sd}\)| |-|-| |\(A^{+}\)|\(A^{+} A_{sd}\)|\(A^{+} A_{sd}\)| |\(A^{+}\)|\(A^{+} A_{sd}\)|\(A^{+} A_{sd}\)| F1 generation: - \(A^{+} A_{sd}\) (heterozygous, normal offspring): 4 For the F2 generation with an autosomal gene, the results would show a 3:1 ratio of normal to scalloped phenotypes, regardless of the parental genotypes. In conclusion, in the X-linked inheritance problem, the F2 generation exhibits a 1:1:1:1 ratio for each phenotype, while in the autosomal case, the F2 generation shows a 3:1 ratio of normal to scalloped phenotypes.