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Chapter 4: Problem 26

Another recessive mutation in Drosophila, ebony \((e),\) is on an autosome (chromosome 3) and causes darkening of the body compared with wild-type flies. What phenotypic \(F_{1}\) and \(F_{2}\) male and female ratios will result if a scalloped-winged female with normal body color is crossed with a normal-winged cbony male? Work out this problem by both the Punnett square method and the forked-line method.

Short Answer

Expert verified
The phenotypic ratios for the F1 generation are 100% normal-winged and normal body color. For the F2 generation, the ratios are 50% normal-winged and normal body color, and 50% normal-winged and ebony body color.

Step by step solution

01

Determine the genotypes of the parents

Let's denote the scalloped-winged trait as s and the normal wing trait as S (this is dominant). The female has scalloped wings (ss) and normal body color (EE), so her genotype is ssEE. The male has normal wings (SS) and the ebony mutation (ee), so his genotype is SSee.
02

Determine the possible gametes of each parent

The female can produce only one type of gamete: sE. The male can produce only one type of gamete as well: Se.
03

Use the Punnett square method to cross the gametes and find \(F_{1}\) generation phenotypes

Since both parents can produce only one type of gamete, so the Punnett square will only contain one cell: |:-: | :-: | |sE |Se | The \(F_{1}\) generation will have the genotype sESe which represents normal-winged (S) and normal body color (E) offspring.
04

Use the forked-line method to cross the gametes and confirm the \(F_{1}\) generation phenotypes

With the forked-line method, we represent the possible gametes from each parent as branches off a starting point, and then cross the resulting branches (in this case, there's only one combination): Female (sE) x Male (Se) -> sESe As we can see, this confirms the same \(F_{1}\) generation phenotype as the Punnett square method.
05

Determine the possible gametes for the \(F_{1}\) generation to find the \(F_{2}\) generation phenotypes

The \(F_{1}\) generation has the genotype sESe. Since they are heterozygous for both traits, they can produce four different gametes: sE, sE, sE, and sE.
06

Use the Punnett square method to cross the \(F_{1}\) gametes and find the \(F_{2}\) generation phenotypes

Constructing a Punnett square using the gametes from the \(F_{1}\) generation: | | sE | se | | --- | --- | --- | | Se | sESe| seSe| | Se | sESe| seSe| From the Punnett square, we can find the \(F_{2}\) phenotypes: - 2 normal-winged (S) and normal body color (E) offspring (sESe). - 2 normal-winged (S) and ebony body color (e) offspring (seSe).
07

Use the forked-line method to cross the \(F_{1}\) gametes and confirm the \(F_{2}\) generation phenotypes

Using forked-line method for the \(F_{2}\) generation: sESe x seSe Possible branches from this cross: 1. sESe (normal-winged, normal body color) 2. seSe (normal-winged, ebony body color) This confirms the same \(F_{2}\) generation phenotypes as the Punnett square method. In conclusion, the phenotypic ratios for the \(F_{1}\) generation are 100% normal-winged and normal body color. For the \(F_{2}\) generation, the ratios are 50% normal-winged and normal body color, and 50% normal-winged and ebony body color.

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Most popular questions from this chapter

The \(A\) and \(B\) antigens in humans may be found in water-soluble form in secretions, including saliva, of some individuals \((\text {Se/Se and Se/se})\) but not in others (se/se). The population thus contains "secretors" and "nonsecretors." (a) Determine the proportion of various phenotypes (blood type and ability to secrete) in matings between individuals that are blood type \(A B\) and type \(O,\) both of whom are \(S e / s e\) (b) How will the results of such matings change if both parents are heterozygous for the gene controlling the synthesis of the H substance \((H h) ?\)

A husband and wife have normal vision, although both of their fathers are red- green color-blind, an inherited X-linked recessive condition. What is the probability that their first child will be (a) a normal son? (b) a normal daughter? (c) a color-blind son? (d) a color- blind daughter?

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 a unique species of plants, flowers may be yellow, blue, red, or mauve. All colors may be true breeding, If plants with blue flowers are crossed to red- flowered plants, all \(\mathrm{F}_{1}\) plants have yellow flowers. When these produced an \(\mathrm{F}_{2}\) generation, the following ratio was observed: \(9 / 16\) yellow: \(3 / 16\) blue: \(3 / 16\) red: \(1 / 16\) mauve In still another cross using true-breeding parents, yellow-flowered plants are crossed with mauve-flowered plants. Again, all \(\mathrm{F}_{1}\) plants had yellow flowers and the \(\mathrm{F}_{2}\) showed a 9: 3: 3: 1 ratio, as just shown. (a) Describe the inheritance of flower color by defining gene symbols and designating which genotypes give rise to cach of the four phenotypes. (b) Determine the \(F_{1}\) and \(F_{2}\) results of a cross between truebreeding red and true-breeding mauve-flowered plants.

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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