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Chapter 4: Problem 17

In rats, the following genotypes of two independently assorting autosomal genes determine coat color: A third gene pair on a separate autosome determines whether or not 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 \(c c\) genotype results in albino rats regardless of the \(A\) and \(B\) alleles present. Determine the \(F_{1}\) phenotypic ratio of the following crosses: (a) \(A A b b C C \quad \times \quad\) aaBBcc (b) \(A a B B C C \quad \times \quad A A B b c c\) (c) \(A a B b C c \quad \times \quad\) AaBbcc (d) \(A a B B C c \quad \times \quad A a B B C c\) (e) \(A A B b C c \quad \times \quad A A B b c c\)

Short Answer

Expert verified
Answer: The phenotypic ratio of this cross is 1:0, with all progenies having the same colored phenotype.

Step by step solution

01

(a) Cross: \(A A b b C C \times aaBBcc\)

First, let's perform a Punnett square with the given genotypes of the two rats. Since both of the rats' genotypes for Gene A are homozygous (AA and aa), the progeny will maintain their respective genotype. For Gene B, both rats are homozygous but different (bb x BB), and Gene C, one rat is homozygous dominant and the other one is homozygous recessive (CC x cc). The resulting offspring genotype will be \(AaBbCc\). Since C is present, color will be expressed. The \(F_{1}\) phenotypic ratio is 1:0, since all progenies will share the same phenotype.
02

(b) Cross: \(A a B B C C \times A A B b c c\)

Perform a Punnett square for each gene pair. For Gene A: All offspring will be Aa. For Gene B: Half the offspring will be BB, and the other half will be Bb. For Gene C: All offspring will be Cc. As C is present, color is expressed. The \(F_{1}\) phenotypic ratio will be 1 (A_) : 1 (a_) for Gene A and 2 (B_) : 0 (bb) for Gene B, resulting in a 2:0 overall ratio.
03

(c) Cross: \(A a B b C c \times A a B b c c\)

Perform a Punnett square for each gene pair. For Gene A: The ratio is 1 (AA) : 2 (Aa) : 1 (aa). For Gene B: The ratio is 1 (BB) : 2 (Bb) : 1 (bb). For Gene C: One-fourth of the offspring will be cc (albino), and the remaining three-fourth will be C_. This results in a 9:3:4 phenotypic ratio when considering the three genes together: 9 colored with dominant A and B, 3 colored with dominant A and recessive b, and 4 albino.
04

(d) Cross: \(A a B B C c \times A a B B C c\)

Perform a Punnett square for each gene pair. For Gene A: The ratio is 1 (AA) : 2 (Aa) : 1 (aa). For Gene B: All offspring will be BB. For Gene C: One-fourth of the offspring will be cc (albino), and the remaining three-fourth will be C_. This results in a 3:1 phenotypic ratio: 3 colored with dominant A and B and 1 albino.
05

(e) Cross: \(A A B b C c \times A A B b c c\)

Perform a Punnett square for each gene pair. For Gene A: All offspring will be AA. For Gene B: The ratio is 1 (BB) : 2 (Bb) : 1 (bb). For Gene C: One-fourth of the offspring will be cc (albino), and the remaining three-fourth will be C_. This results in a 6:2 phenotypic ratio: 6 colored with dominant A and various combinations of B and 2 albino.

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Most popular questions from this chapter

In a cross in Drosophila involving the X-linked recessive eye mutation white and the autosomally linked recessive eye mutation sepia (resulting in a dark eye), predict the \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) results of crossing true-breeding parents of the following phenotypes: (a) white females \(\times\) sepia males (b) sepia females \(\times\) white males Note that white is epistatic to the expression of sepia.

Discuss how temperature influences phenotypic expression.

While vermilion is X-linked in Drosophila and causes the eye color to be bright red, brown is an autosomal recessive mutation that causes the eye to be brown. Flies carrying both mutations lose all pigmentation and are white-eyed. Predict the \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) results of the following crosses: (a) vermilion females \(\times\) brown males (b) brown females \(\times\) vermilion males (c) white females \(\times\) wild-type males

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 thatKerrys were true breeding. Dexters, on the other hand, were not true breeding and never produced as many offspring as Kerrys. Provide a genetic explanation for these observations.

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