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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.

Short Answer

Expert verified
The phenotypic ratios of the offspring are 1/2 (Bald, Short Fingered), 3/10 (Bald, Long Fingered), 1/10 (NonBald, Short Fingered), and 1/10 (NonBald, Long Fingered).

Step by step solution

01

Identify the genotypes of the parents

The father is a heterozygous short-fingered man without the balding trait (Sf, bb). The mother is a short-fingered woman heterozygous for the balding trait (Sf, Bb).
02

Set up a Punnett square

Start by creating a 4x4 Punnett square. On the top row, place the father's gametes (Sb, Sf, bb, bf) and on the left column place the mother's gametes (SB, Sb, Bf, Bb). |[ | Sb | Sf | bb | bf]| |---|---|---|---|---| |SB | | | | | |Sb | | | | | |Bf | | | | | |Bb | | | | |
03

Fill in the Punnett square

For each cell in the Punnett square, combine the corresponding gametes from the father and mother. |[ | Sb | Sf | bb | bf]| |---|---|---|---|---| |SB |SSBb|SSFb|SBBb|SBFb| |Sb |SsBb|SsFb|SbBb|ssf| |Bf |BfSb|BfSf|Bfbb|Bfbf| |Bb |BBBb|BBFb|BbBb|BbFb|
04

Determine the phenotype of each offspring genotype

Now, we'll determine the phenotype for each combination of traits in the Punnett square, taking into account that both traits are dominant in men and recessive in women. |[ | Sb | Sf | bb | bf]| |---|---|---|---|---| |SB |Bald,ShortF|Bald,ShortF|Bald,LongF|Bald,LongF| |Sb |Bald,ShortF|Bald,ShortF|Bald,LongF|Bald,LongF| |Bf |NonBald,ShortF|NonBald,ShortF|NonBald,LongF|NonBald,LongF| |Bb |Bald,ShortF|Bald,ShortF|Bald,LongF|Bald,LongF|
05

Count the phenotypes and calculate the phenotypic ratios

Now count the number of each phenotype in the Punnett square: - Bald, Short Fingered: 10 - Bald, Long Fingered: 6 - NonBald, Short Fingered: 2 - NonBald, Long Fingered: 2 To find the phenotypic ratio, divide each count by the total (20) and express it as a fraction. Bald, Short Fingered (10/20): 1/2 Bald, Long Fingered (6/20): 3/10 NonBald, Short Fingered (2/20): 1/10 NonBald, Long Fingered (2/20): 1/10 Therefore, the phenotypic ratios of the offspring are 1/2 (Bald, Short Fingered), 3/10 (Bald, Long Fingered), 1/10 (NonBald, Short Fingered), and 1/10 (NonBald, Long Fingered).

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Key Concepts

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

Phenotypic Ratio
Phenotypic ratio is a way to represent the observable traits (phenotypes) that result from a genetic cross. It summarizes the proportion of different phenotypes in the offspring. Phenotypes are the expressed traits, like being bald or having short fingers.
In genetic problems, after calculating the genotypes using a Punnett square, you determine the phenotypes to find ratios. For example, in our exercise, offspring were characterized based on balding and finger length traits.
To calculate the phenotypic ratio, count each phenotype and divide by the total number of offspring. For hypotheical offspring:
  • Bald, Short Fingered: 10 out of 20 (1/2)
  • Bald, Long Fingered: 6 out of 20 (3/10)
  • NonBald, Short Fingered: 2 out of 20 (1/10)
  • NonBald, Long Fingered: 2 out of 20 (1/10)
This results in a phenotypic ratio that gives a clear picture of how traits are expected to appear over several offspring.
Punnett Square
A Punnett square is a diagram used to predict the genotype and phenotype probabilities of offspring from a genetic cross. It helps visualize how alleles (gene variants) combine when reproductive cells (gametes) from the parents come together.
The Punnett square is set up by placing the father's possible gametes across the top and the mother's along the side. Each box within the grid is filled by combining the corresponding alleles. This visual aid swiftly shows all possible genetic combinations of the offspring.
For example, in our problem, if a father has gametes represented as (Sb, Sf, bb, bf) and the mother has (SB, Sb, Bf, Bb), you create a 4x4 square and fill it by combining these gametes. This results in a comprehensive map of potential offspring, showcasing each possible genetic combination for the studied traits.
Dominant and Recessive Inheritance
Dominant and recessive inheritance patterns affect how traits are passed from parents to offspring. In simple terms, dominant traits require only one dominant allele to be visible, while recessive traits need two recessive alleles to show.
In the exercise, the traits of baldness and short fingers are sex-influenced. This means that they behave differently in men and women, resembling dominance in one sex and recessiveness in the other. For instance, both are dominant in men but recessive in women.
When dealing with these kinds of traits, it's crucial to understand that a dominant allele masks the presence of a recessive allele in mixed-gender conditions. This knowledge will guide the interpretation of genetic results, ensuring the correct prediction of phenotypes based on the offspring's gender.

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Most popular questions from this chapter

In Drosophila melanogaster, the gene for white eyes, \(\mathrm{w}\), is recessive and X-linked. Red eyes result from the wild-type allele at this locus. (a) On chromosomes, symbolize the genotypes of a white-eyed male, a red-eyed female (both possible types), and a white-eyed female. (b) Diagram a cross between a homozygous red-eyed female and white-eyed male. Carry this to the \(\mathrm{F}_{2}\) generation and give phenotype ratios for each sex. (c) Show a cross between an \(\mathrm{F}_{1}\) female and white-eyed male. Give phenotype ratios. (d) Show a cross between an \(\mathrm{F}_{1}\) female and wild-type male. Also give ratios.

How is sex determination in the parasitic wasp Habrobracon junglandis different from the method characteristic of the insect order Hymenoptece?

Vitamin D resistant rickets is produced by an X-linked dominant allele. Two recessive alleles together will lead to normal bone development. What are the expected results from the following crosses? (a) A normal woman and a man with vitamin D resistant rickets. (b) A normal man and a woman with the condition who has a normal father.

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?

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