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Chapter 4: Problem 14

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

Expert verified
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

01

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
02

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
03

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
04

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.

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Most popular questions from this chapter

Students taking a genetics exam were expected to answer the following question by converting data to a "meaningful ratio" and then solving the problem. The instructor assumed that the final ratio would reflect two gene pairs, and most correct answers did. Here is the exam question: "Flowers may be white, orange, or brown. When plants with white flowers are crossed with plants with brown flowers, all the \(\mathrm{F}_{1}\) flow ers are white. For \(\mathrm{F}_{2}\) flowers, the following data were obtained: 48 white 12 orange 4 brown Convert the \(\mathrm{F}_{2}\) data to a meaningful ratio that allows you to explain the inheritance of color. Determine the number of genes involved and the genotypes that yield each phenotype." (a) Solve the problem for two gene pairs. What is the final \(\mathrm{F}_{2}\) ratio? (b) A number of students failed to reduce the ratio for two gene pairs as described above and solved the problem using three gene pairs. When examined carefully, their solution was deemed a valid response by the instructor, Solve the problem using three gene pairs.

Labrador retrievers may be black, brown, or golden in color (see the chapter opening photograph on \(\mathrm{p} .53\) ). Although each color may breed true, many different outcomes occur if numerous litters are examined from a variety of matings, where the parents are not necessarily true-breeding. The following results show some of the possibilities. Propose a mode of inheritance that is consistent with these data, and indicate the corresponding genotypes of the parents in each mating. Indicate as well the genotypes of dogs that breed true for each color. (a) black \(\times\) brown \(\longrightarrow\) all black (b) black \(\times\) brown \(\longrightarrow \quad 1 / 2\) black \(1 / 2\) brown (c) black \(\times\) brown \(\longrightarrow \quad 3 / 4\) black \(1 / 4\) golden (d) black \(\quad \times\) golden \(\longrightarrow \quad\) all black (e) black \(\times\) golden \(\longrightarrow \quad 4 / 8\) golden 318 black \(1 / 8\) brown (f) black \(\times\) golden \(\longrightarrow \quad 2 / 4\) golden \(1 / 4\) black \(1 / 4\) brown (8) brown \(\times\) brown \(\longrightarrow \quad 3 / 4\) brown \(1 / 4\) golden (h) black \(\times\) black \(\longrightarrow 9 / 16\) black \(4 / 16\) golden \(3 / 16\) brown

In Dexter and Kerry cattle, animals may be polled (hornless) or horned. The Dexter animals have short legs, whereas the Kerry animals have long legs. When many offspring were obtained from matings between polled Kerrys and horned Dexters, half were found to be polled Dexters and half polled Kerrys. When these two types of \(\mathrm{F}_{1}\) cattle were mated to one another, the following \(\mathrm{F}_{2}\) data were obtained: \(3 / 8\) polled Dexters 3/8 polled Kerrys \(1 / 8\) horned Dexters \(1 / 8\) horned Kerrys A geneticist was puzzled by these data and interviewed farmers who had bred these cattle for decades. She learned that Kerrys were true-breeding. Dexters, on the other hand, were not truebreeding and never produced as many offspring as Kerrys. Provide a genetic explanation for these observations.

The following genotypes of two independently assorting autosomal genes determine coat color in rats: \(A-B-(\text { gray }) ; A-b b\) (yellow) \(; a a B-\) (black); \(a a b b\) (cream) A third gene pair on a separate autosome determines whether any color will be produced. The \(C C\) and \(C c\) genotypes allow color according to the expression of the \(A\) and \(B\) alleles. However, the ce genotype results in allbino rats regardless of the \(A\) and \(B\) alleles present. Determine the \(F_{1}\) phenotypic ratio of the following crosses: (a)AAbbCC \(\times\) aaBBcc; (b) \(A a B B c c \times A A B b c c\) (c) \(A a B b C c \times A a B b c c\)

As in the plants of Problem \(6,\) color may be red, white, or pink and flower shape may be personate or peloric. Determine the \(\mathrm{P}_{1}\) and \(\mathrm{F}_{1}\) genotypes for the following crosses: (a) red, peloric \(\times\) white, personate \(\mathrm{F}_{1}:\) all pink, personate (b) red, personate \(\times\) white, peloric \(\mathrm{F}_{1}:\) all pink, personate (c) pink, personate \(\times\) red, peloric $F_{1:}\left\\{\begin{array}{l}1 / 4 \text { red, personate } \\ 1 / 4 \text { red, peloric } \\ 1 / 4 \text { pink, personate } \\ 1 / 4 \text { pink, peloric }\end{array}\right.$ (d) pink, personate \(x\) white, peloric $\mathbf{F}_{1:}\left\\{\begin{array}{l}1 / 4 \text { white, personate } \\ 1 / 4 \text { white, peloric } \\ 1 / 4 \text { pink, personate } \\ 1 / 4 \text { pink, peloric }\end{array}\right.$ (e) What phenotypic ratios woud result from crossing the \(\mathrm{F}_{1}\) of (a) to the \(F_{1}\) of \((b) ?\)
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