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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 Shas 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. 2:1 hairy to hairless. b. 1:1 hairy to hairless.

Step by step solution

01

Determine the Genotypes of the Parental Flies

The hairy phenotype in the suppressed condition suggests that the flies have one dose of both the dominant alleles, i.e., H (dominant for hairless) and S (suppressor for hairless) while being heterozygous for both alleles to be viable. Therefore, the genotypes of the hairy parental flies are HhSs.
02

Set up the Punnett Square for the Cross

We perform a dihybrid cross of HhSs x HhSs. The possible gametes of each parent are HS, Hs, hS, and hs.
03

Use the Punnett Square to Determine Offspring Genotypes

From the Punnett square, the offspring genotypes are: 1 HHSS, 2 HHSs, 1 HHss, 2 HhSS, 4 HhSs, 2 Hhss, 1 hhSS, 2 hhSs, 1 hhss.
04

Identify Lethal Genotypes

HH and SS in homozygous forms are lethal. Therefore, genotypes HHSS, HHSs, HHss, HhSS, hhSS are non-viable.
05

Calculate Viable Offspring for Phenotypic Ratio

From viable offspring (HhSs, Hhss, hhSs, hhss), identify: HhSs = Hairy, Hhss = Hairless, hhSs = Hairy, hhss = Hairless. Thus, we get 4 hairy and 2 hairless.
06

Determine Phenotypic Ratio of Hairy to Hairless

The phenotypic ratio of viable offspring is 4 hairy: 2 hairless, simplifying to 2:1.
07

Cross Hairless With Parental Hairy

Hairless (Hhss) with parental hairy (HhSs) results in possible genotypes: HhSs, Hhss, hhSs, hhss for viable combinations.
08

Calculate Phenotypic Ratio for Backcross

For viable offspring, categorize HhSs and hhSs as hairy, Hhss and hhss as hairless, yielding 2 hairy: 2 hairless, simplifying to a ratio of 1:1.

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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 useful diagram that helps visualize the possible genotypes of offspring from a genetic cross. Imagine it as a simple grid. Each parent's potential gametes are placed along the top and side of the grid. With Drosophila crosses such as HhSs x HhSs, each parent can produce four types of gametes: HS, Hs, hS, and hs. You place these gametes across the grid and systematically fill in the squares to see all possible offspring combinations. This tool makes it clear which genotypes are viable and which phenotypes these will manifest as based on dominance and recessiveness of alleles.
Drosophila Genetics
Drosophila melanogaster, commonly known as the fruit fly, is a vital model organism in genetics. Researchers often choose Drosophila due to its short life cycle and simple genetics. In our exercise, we consider dominant alleles, such as H for 'hairless' and S for 'suppressor'. This cross elucidates the importance of allele interaction within the species. For example, the presence of one S allele can suppress the hairless phenotype caused by H, showcasing how genetic interactions go beyond simple dominance-recessiveness and can lead to interesting phenotypic outcomes.
Allele Interaction
Alleles don't work in isolation. They interact in both expected and unexpected ways. Here, we have two dominant alleles: H, which induces a hairless phenotype, and S, which can suppress this phenotype, restoring hairiness. Such interactions suggest epistatic relationships, where the expression of one gene masks or modifies the expression of a second gene. Additionally, both H and S in a homozygous form (HH or SS) are lethal, illustrating a critical allele relationship where viability is concerned. This gives students insightful examples of how allele interaction complicates genetic predictions compared to simple Mendelian inheritance.
Phenotypic Ratio
Phenotypic ratios reveal the observable characteristics produced by certain genetic crosses. For instance, from a cross between two Drosophila flies with genotypes HhSs, some offspring are non-viable due to lethal allele combinations, impacting the expected ratios. In our example, removing the lethal and non-viable combinations, the phenotypic ratio of hairy to hairless flies is calculated as 2:1. In contrast, backcrossing hairless offspring with a hairy parent changes this ratio. After recalculating, we get an expected phenotypic ratio of 1:1. Understanding these ratios help students predict the distribution of traits within a population based on parental genotypes.

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Most popular questions from this chapter

You have been given a virgin Drosophila female. You notice that the bristles on her thorax are much shorter than normal. You mate her with a normal male (with long bristles and obtain the following \(\mathrm{F}_{1}\) progeny: \(\frac{1}{3}\) short-bristled females, \(\frac{1}{3}\) long-bristled females, and \(\frac{1}{3}\) long-bristled males. A cross of the \(\mathrm{F}_{1}\) long- bristled females with their brothers gives only long-bristled \(\mathrm{F}_{2}\) A cross of short-bristled females with their brothers gives \(\frac{1}{3}\) short-bristled females, \(\frac{1}{3}\) long-bristled females, and \(\frac{1}{3}\) long-bristled males. Provide a genetic hypothesis to account for all these results, showing genotypes in every cross.

A woman who owned a purebred albino poodle (an autosomal recessive phenotype) wanted white puppies; so she took the dog to a breeder, who said he would mate the female with an albino stud male, also from a pure stock. When six puppies were born, all of them were black; so the woman sued the breeder, claiming that he replaced the stud male with a black dog, giving her six unwanted puppies. You are called in as an expert witness, and the defense asks you if it is possible to produce black offspring from two pure-breeding recessive albino parents. What testimony do you give?

The production of pigment in the outer layer of seeds of corn requires each of the three independently assorting genes \(A, C,\) and \(R\) to be represented by at least one dominant allele, as specified in Problem \(64 .\) The dominant allele \(P r\) of a fourth independently assorting gene is required to convert the biochemical precursor into a purple pigment, and its recessive allele \(p r\) makes the pigment red. Plants that do not produce pigment have yellow seeds. Consider a cross of a strain of genotype \(A / A ; C / C ; R / R ; p r / p r\) with a strain of genotype \(a / a ; c / c ; r / r ; \operatorname{Pr} / \operatorname{Pr}\) a. What are the phenotypes of the parents? b. What will be the phenotype of the \(\mathrm{F}_{1}\) ? c. What phenotypes, and in what proportions, will appear in the progeny of a selfed \(\mathrm{F}_{1}\) ? d. What progeny proportions do you predict from the testcross of an \(\mathrm{F}_{1}\) ?

For several years, Hans Nachtsheim investigated an inherited anomaly of the white blood cells of rabbits. This anomaly, termed the Pelger anomaly, is the arrest of the segmentation of the nuclei of certain white cells. This anomaly does not appear to seriously burden the rabbits. a. When rabbits showing the Pelger anomaly were mated with rabbits from a true-breeding normal stock, Nachtsheim counted 217 offspring showing the Pelger anomaly and 237 normal progeny. What is the genetic basis of the Pelger anomaly? b. When rabbits with the Pelger anomaly were mated with each other, Nachtsheim found 223 normal progeny, 439 with the Pelger anomaly, and 39 extremely abnormal progeny. These very abnormal progeny not only had defective white blood cells, but also showed severe deformities of the skeletal system; almost all of them died soon after birth. In genetic terms, what do you suppose these extremely defective rabbits represented? Why were there only 39 of them? c. What additional experimental evidence might you collect to test your hypothesis in part \(b\) ? d. In Berlin, about 1 human in 1000 shows a Pelger anomaly of white blood cells very similar to that described for rabbits. The anomaly is inherited as a simple dominant, but the homozygous type has not been observed in humans. Based on the condition in rabbits, why do you suppose the human homozygous has not been observed? e. Again by analogy with rabbits, what phenotypes and genotypes would you expect among the children of a man and woman who both show the Pelger anomaly? (Data from A. M. Srb, R. D. Owen, and R. S. Edgar, General Genetics, 2 nd ed. W. H. Freeman and Company, \(1965 .)\)

Wild-type strains of the haploid fungus Neurospora can make their own tryptophan. An abnormal allele \(t d\) renders the fungus incapable of making its own tryptophan. An individual of genotype \(t d\) grows only when its medium supplies tryptophan. The allele \(s u\) assorts independently of \(t d ;\) its only known effect is to suppress the \(t d\) phenotype. Therefore, strains carrying both \(t d\) and \(s u\) do not require tryptophan for growth. a. If a \(t d ; s u\) strain is crossed with a genotypically wildtype strain, what genotypes are expected in the progeny and in what proportions? b. What will be the ratio of tryptophan-dependent to tryptophan-independent progeny in the cross of part \(a ?\)

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