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A dominant allele \(H\) reduces the number of body bristles that Drosophila flies have, giving rise to a "hairless" phenotype. In the homozygous condition, \(H\) is lethal. An independently assorting dominant allele \(S\) has no effect on bristle number except in the presence of \(H\) in which case a single dose of \(S\) suppresses the hairless phenotype, thus restoring the hairy phenotype. However, \(S\) also is lethal in the homozygous (S/S) condition. a. What ratio of hairy to hairless flies would you find in the live progeny of a cross between two hairy flies both carrying \(H\) in the suppressed condition? b. When the hairless progeny are backcrossed with a parental hairy fly, what phenotypic ratio would you expect to find among their live progeny?

Short Answer

Expert verified
a. The ratio is 2 hairy: 1 hairless among live progeny. b. All progeny are hairy (1:0).

Step by step solution

01

Determine the Genotypes of Parents

Each hairy fly has a genotype of Hs/hs. The 'H' is the dominant allele for the hairless trait, suppressed by 'S,' and each fly is heterozygous for both traits due to the lethal condition of homozygous dominant alleles.
02

Set up a Punnett Square for Hs/hs x Hs/hs

Create a Punnett square considering each parent can contribute either Hs, hs, or Hs alleles. This results in a 2x2 Punnett square filled with potential combinations.
03

Calculate Phenotypic Outcomes from the Punnett Square

Analyze the resulting genotypes: (1) Hs/hs - hairy fly; (2) hs/hs - hairy fly without H expression; (3) HS/HS, HS/hs, and hs/hs - lethal or non-viable. Count the number of viable combinations.
04

Determine Ratio of Hairy to Hairless Progeny

From viable genotypes (only Hs/hs and hs/hs give rise to a phenotype), determine the ratios: 2 hairy (Hs/hs) : 1 hairless (hs/hs) as the hh combination is non-viable.
05

Consider Hairless Progeny Backcross

Cross hairless progeny (hs/hs) with heterozygous hairy (Hs/hs) yielding offspring of genotypes hs/Hs and hs/hs.
06

Phenotypic Ratio for Backcross Progeny

Both genotypes (hs/Hs and hs/hs) produce a hairy phenotype due to suppression by 'S,' so with equal probability, the phenotypic ratio is 1 hairy : 0 hairless given 'S' presence.

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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 visual tool used in genetics to predict the possible genotypes of offspring from two parent organisms. It's like a genetic calculator that shows how alleles from each parent combine. In our exercise with Drosophila flies, the Punnett Square helps us understand crosses involving the "hairless" phenotype controlled by the allele pair \( H \) and \( S \). Each parent in our example carries the genotypes \( Hs/hs \). The Punnett Square reveals the possible genetic outcomes for their progeny.
When you set up a Punnett Square for this cross, consider each parent can pass an allele combination of \( Hs \) or \( hs \). This forms a 2x2 matrix when combined:
  • Hs from Parent 1 can pair with Hs from Parent 2
  • Hs from Parent 1 with hs from Parent 2
  • hs from Parent 1 with hs from Parent 2
  • And the reverse combinations
By filling in the Punnett Square, we predict the genotypes and, subsequently, the phenotypes of the offspring.
Lethal Alleles
Lethal alleles in genetics are mutations that cause death when present in a certain combination. In our Drosophila case, both the alleles \( H \) and \( S \) demonstrate lethality in their homozygous form. The allele \( H \) is lethal when present as \( HH \), meaning two dominant alleles lead to non-viability—you don't see these flies. Similarly, the \( S \) allele is lethal when you have \( S/S \) genetic makeup.
This means crosses must avoid homozygous combinations of these alleles. Lethal alleles drastically affect phenotypic ratios because they reduce the number of viable offspring. In our exercise, since \( HS/HS \) also leads to lethal results, we only consider viable offspring genotypes like \( Hs/hs \) and \( hs/hs \) in subsequent calculations.
Drosophila Genetics
Drosophila, or fruit flies, are a classic model organism in genetics due to their simple genetic makeup and fast life cycle. In this exercise, Drosophila are used to demonstrate genetic concepts like inheritance, alleles, and lethal traits. The flies express a hairless phenotype, with a significant twist when conditions of the environment interact with their genome, impacting traits like number of bristles.
The interaction between alleles \( H \) and \( S \) exemplifies how multiple genes can affect a phenotype, often observed in real-world scenarios. In Drosophila genetics, these interactions often complicate the straightforward Mendelian ratios, providing students with a nuanced look at heredity. By studying such genetic crosses in Drosophila, students gain a richer understanding of inheritance and phenotypic expression.
Phenotypic Ratios
In genetics, phenotypic ratios tell us the observable traits of the offspring in relation to one another. They are predicted by looking at the genotypic combinations from a Punnett Square. In Drosophila crosses, these ratios can illustrate complex genetic interactions, particularly when lethal alleles are involved.In our exercise, crossing hairy flies (\( Hs/hs \)) results in a 2:1 ratio of hairy to hairless viable offspring. The component of lethality (\( HH \) and \( S/S \)) eliminates some genotype combinations, affecting expected phenotypic ratios observed in upcoming generations. Furthermore, upon backcrossing the hairless progeny with a parental hairy fly, all observed phenotypes were hairy due to the suppressing effect of \( S \), resulting in a 1:0 phenotypic ratio. These ratios guide predictions and illustrate inheritance patterns in the flies, emphasizing the uniqueness of complex genetic influences at work.

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Most popular questions from this chapter

A plant believed to be heterozygous for a pair of alleles \(B / b\) (where \(B\) encodes yellow and \(b\) encodes bronze) was selfed, and, in the progeny, there were 280 yellow and 120 bronze plants. Do these results support the hypothesis that the plant is \(B / b ?\)

Most flour beetles are black, but several color variants are known. Crosses of pure-breeding parents produced the following results (see table) in the \(\mathrm{F}_{1}\) generation, and intercrossing the \(\mathrm{F}_{1}\) from each cross gave the ratios shown for the \(\mathrm{F}_{2}\) generation. The phenotypes are abbreviated Bl, black; Br, brown; Y, yellow; and \(\mathrm{W}\), white. $$\begin{array}{clll} \text { Cross } & \text { Parents } & \mathrm{F}_{1} & \mathrm{F}_{2} \\ \hline 1 & \mathrm{Br} \times \mathrm{Y} & \mathrm{Br} & 3 \mathrm{Br}: 1 \mathrm{Y} \\ 2 & \mathrm{Bl} \times \mathrm{Br} & \mathrm{Bl} & 3 \mathrm{Bl}: 1 \mathrm{Br} \\ 3 & \mathrm{Bl} \times \mathrm{Y} & \mathrm{Bl} & 3 \mathrm{Bl}: 1 \mathrm{Y} \\\ 4 & \mathrm{W} \times \mathrm{Y} & \mathrm{Bl} & 9 \mathrm{Bl}: 3 \mathrm{Y}: 4 \mathrm{W} \\ 5 & \mathrm{W} \times \mathrm{Br} & \mathrm{Bl} & 9 \mathrm{Bl}: 3 \mathrm{Br}: 4 \mathrm{W} \\ 6 & \mathrm{Bl} \times \mathrm{W} & \mathrm{Bl} & 9 \mathrm{Bl}: 3 \mathrm{Y}: 4 \mathrm{W} \\ \hline \end{array}$$ a. From these results, deduce and explain the inheritance of these colors. b. Write the genotypes of each of the parents, the \(\mathrm{F}_{1}\) and the \(\mathrm{F}_{2}\) in all crosses.

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