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In turkeys, the gene for short wattle \((\ell)\) is \(\mathrm{X}\) -linked recessive. Its wild-type allele (L) is responsible for producing a long wattle. In I turkeys, like in all birds, the female is the heterogametic sex, possessing an \(\mathrm{X}\) and a \(\mathrm{Y}\) chromosome. The male has two \(\mathrm{X}\) chromosomes. The sex of the female can be reversed to male if the one functional ovary is destroyed or removed. Assuming that such a reversal can yield a fertile male, what will be the phenotypic ratio of a cross between a short-wattled reversed male and a long-wattled female?

Short Answer

Expert verified
The phenotypic ratio of a cross between a short-wattled reversed male and a long-wattled female will depend on the genotype of the long-wattled female. If she is homozygous dominant (LL), all offspring will have a long wattle (100% long-wattled). If she is heterozygous (Ll), the phenotypic ratio will be 1:1, meaning 50% long-wattled and 50% short-wattled offspring.

Step by step solution

01

We are given a short-wattled reversed male turkey and a long-wattled female turkey. The short wattle gene (l) is X-linked recessive, so the male turkey will have the ll genotype. The female will be a long-wattled turkey, so she will have the L phenotype. Since females have two X chromosomes, she can either be homozygous dominant (LL) or heterozygous (Ll). We are not told the female's exact genotype, so we will consider both possibilities. #Step 2: Set up Punnett squares for both female genotypes#

We will create a Punnett square for each possible genotype of the female turkey. For the homozygous dominant female, we'll use (LL), and for the heterozygous female, we'll use (Ll). Homozygous dominant female (LL) cross: ``` l ------------- L | Ll | ------------- L | Ll | ``` Heterozygous female (Ll) cross: ``` l ------------- L | Ll | ------------- l | ll | ``` #Step 3: Determine the phenotypic ratios for each cross#
02

Now, we will determine the phenotypic ratios based on the Punnett squares. Homozygous dominant female (LL) cross: All offspring will have the Ll genotype, so there will be a 100% long-wattled phenotype. Heterozygous female (Ll) cross: - 50% offspring will have the Ll genotype (long-wattled phenotype) - 50% offspring will have the ll genotype (short-wattled phenotype) #Step 4: Write the final answer#

The phenotypic ratio will depend on the genotype of the long-wattled female. If she is homozygous dominant (LL), all offspring will have a long wattle (100% long-wattled). If she is heterozygous (Ll), the phenotypic ratio will be 1:1, meaning 50% long-wattled and 50% short-wattled offspring.

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Key Concepts

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

Punnett square
The Punnett square is a fundamental tool in genetics used to predict the genotypes and phenotypes of offspring from a genetic cross. It helps visualize how alleles from each parent can combine during reproduction. In this exercise, we used the Punnett square to determine possible outcomes for the offspring of a reversed hermaphroditic male turkey and a female turkey, both carrying genes for wattle length. The male turkey, having two X chromosomes, possesses the recessive allele \( ll \). For the female turkey, we considered two scenarios: homozygous dominant \( LL \) and heterozygous \( Ll \). By slotting the alleles into the Punnett square, we can clearly see potential genetic combinations for the offspring. This graphical representation ensures we systematically cover all possibilities for the traits being studied, such as the length of the wattle in turkeys.
genetic cross
A genetic cross involves breeding two individuals to study the inheritance pattern of specific traits. In the given exercise, the cross is between a short-wattled reversed male turkey and a long-wattled female turkey. The gene of interest is X-linked with two alleles: recessive \( l \) and dominant \( L \). The reversed male turkey possesses the genotype \( ll \), meaning he carries two recessive alleles that result in a short wattle. Meanwhile, the female turkey can either be homozygous dominant \( LL \) or heterozygous \( Ll \). Each mating scenario examines how their genetic information gets passed down to their offspring. The process of genetic crossing, depicted via Punnett squares, helps deduce potential genotypes and the associated phenotypes among their progeny. The insights gained through such a cross not only predict trait inheritance but also highlight the role of different alleles in the expression of traits.
phenotypic ratio
Phenotypic ratio is a vital concept in genetics that signifies the ratio of different observable traits among the offspring of a genetic cross. In our turkeys' exercise, the phenotypic ratio is calculated after analyzing the resulting genotypes from the Punnett squares. For a homozygous dominant female \( LL \) crossed with a short-wattled male \( ll \), all offspring inherit at least one dominant allele, leading to 100% long-wattled phenotypes. Conversely, when the female is heterozygous \( Ll \), the offspring distribution changes: 50% have the long-wattled \( Ll \) genotype, and 50% possess the short-wattled \( ll \) genotype. Thus, it creates a phenotypic ratio of 1:1 for long and short wattles. Understanding phenotypic ratios is crucial as it connects the dots between genetic compositions and their physical manifestations, aiding predictions about the prevalence of traits in future generations.

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Most popular questions from this chapter

Explain the mechanism of the genetic determination of sex in humans.

Pattern baldness and short index fingers are both sex-influenced. Both are dominant in men and recessive in women. Determine the phenotypic ratios of the \(\mathrm{F}_{1}\) generation resulting from a cross between a heterozygous short-fingered man without the balding trait and a short-fingered woman heterozygous for the balding trait.

In man, the gene (i) for ichthyosis is sex-linked and recessive to the gene (I) for normal skin. Ichthyosis is a disease which produces scaley skin. Diagram on chromosomes the genotypes of the following crosses and summarize the expected phenotypic ratios: (a) a normal man, and a woman with ichthyosis; (b) a man with ichthyosis and a normal heterozygous woman; (c) a man with ichthyosis and a normal homozygous woman.

Consider the gene for vermilion eye color (v) in the fruit fly, Drosophila melanogaster, to be X-linked and recessive to its wild-type allele ( \(\mathrm{v}^{+}\) ), which produces dull red eye color. The heterogametic sex (XY) in the fruit fly is the male. A female with wild-type eyes is crossed with an unknown male and the following progeny are observed: $$ \begin{array}{|l|l|} \hline \text { Females } & \text { Males } \\ \hline 64 \text { wild-type } & 42 \text { wild-type } \\ \hline 57 \text { vermilion } & 59 \text { vermilion } \\ \hline \end{array} $$ (a) What is the genotype of the female parent? (b) What is the genotype of the unknown male parent?

Chromosomal sex in mammals has been shown to be strongly \(\mathrm{Y}\) chromosome determined. How was this shown to be true?

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