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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Genetics
Genetics is the study of heredity and how traits are passed down from one generation to the next. This field helps us to understand the genetic code that determines everything from eye color to susceptibility to certain diseases. Genes, the basic units of heredity, are sections of DNA that carry the instructions for making proteins – the building blocks of life. In genetic problems, such as those illustrated with Punnett squares, traits are often expressed in terms of alleles. Alleles are different forms of a gene, where dominant alleles mask the expression of recessive ones in heterozygous pairings.

This is crucial when predicting the phenotypes of offspring from a genetic cross. For example, in the exercise given, the scalloped-winged trait is denoted by the recessive allele 's', while normal wings are denoted by the dominant 'S'. Genetics help us use tools like Punnett squares to visually represent and predict how these genetic traits are likely to be inherited by the next generation. When creating small models of inheritance as seen in the exercise, the segregation of these alleles during gamete formation plays a key role in the resulting genetic combinations.
Drosophila
Drosophila melanogaster, commonly known as the fruit fly, is a favored organism for genetics studies. With over 100 years of research backing it, Drosophila offers several advantages for genetic experiments. It has a simple genetic structure, a short life cycle, and produces a large number of offspring. This makes it ideal for observing various hereditary patterns and genetic mutations over a short period. Researchers can easily manipulate its genome and observe the resulting phenotypes.

In the exercise example, one mutation of interest is the ebony mutation, which affects the color of the body. Being a recessive trait, it's denoted by the allele 'e'. When investigating this type of mutation, geneticists often look into how different allele combinations result in different observable characteristics, or phenotypes. Thus, Drosophila serves as a powerful model for exploring complex genetic phenomena, including autosomal inheritance and the impact of genetic mutations.
Autosomal Inheritance
Autosomal inheritance refers to the transmission of genes located on the autosomes, which are the non-sex chromosomes. In humans, these chromosomes are numbered from 1 to 22. Unlike sex-linked traits, autosomal traits are not affected by the organism's sex, meaning they can appear in any offspring, regardless if they are male or female.

In the context of the exercise involving Drosophila, autosomal inheritance is illustrated through the transmission of the ebony mutation located on chromosome 3. Individuals must carry two copies of the recessive allele, 'ee', for the ebony phenotype to manifest. The study of such inheritance patterns can help predict how traits controlled by autosomal genes will be passed from parent to offspring.

The exercise solution applies the Punnett square method and reveals the phenotypic ratios of offspring based on these autosomal genes. By understanding basic principles of autosomal inheritance, such as dominance and recessivity, we can predict how combinations of autosomal alleles can influence the phenotypes of subsequent generations.

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Most popular questions from this chapter

In Drosophila, an X-linked recessive mutation, scalloped (sd), causes irregular wing margins. Diagram the \(F_{1}\) and \(\mathrm{F}_{2}\) 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 with those that would be obtained if the scalloped gene were autosomal.

When summer squash plants (Cucurbita pepo) with discshaped fruits are crossed to ones with long fruits, the \(\mathrm{F}_{1}\) generation all have disc-shaped fruits. When the \(F_{1}\) plants are crossed to each other, the \(\mathrm{F}_{2}\) produce spherical fruits as well as exhibit the two parental strains. The phenotypic ratio is 9: 6: 1 (disc-shaped:spherical:long). (a) Which type of gene interaction is this an example of? (b) Explain the phenotypes observed in terms of the number of gene pairs involved and by designating genotypes for all the fruit shapes in the cross. (Use dashes where required.)

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?

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.

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

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