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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?

Short Answer

Expert verified
It is genetically impossible for two albino dogs (aa) to produce black puppies.

Step by step solution

01

Understand Albino Phenotype

The albino phenotype is caused by an autosomal recessive allele. For a dog to express an albino phenotype, it must have two recessive alleles (aa). This means both copies of the gene must carry the recessive allele responsible for albinism.
02

Identify Possible Parental Genotypes

Since both the albino poodle and the supposed albino stud are from purebred lines, both must have the genotype aa (homozygous recessive) to express the albino trait.
03

Determine Genotypic Possibilities for Offspring

Each parent can only pass on a recessive allele (a) to their offspring. When two albino parents (aa) mate, each offspring receives one allele from each parent, resulting in an offspring genotype of aa.
04

Analyze Phenotypic Outcome for Offspring

The genotype aa results in an albino phenotype. Therefore, it is not genetically possible for two albino parents, each providing only recessive alleles (a), to produce offspring with a black coat.
05

Conclusion on Possible Genotypes

Given the genetic model, the breeding of two homozygous recessive (aa) dogs can only produce offspring that are also homozygous recessive (aa), and therefore, albino.
06

Formulate Testimony as an Expert Witness

Based on genetic principles, it is impossible for two purebred albino (aa) dogs to produce black puppies. Thus, my testimony would support the claim that a mix-up occurred with the stud male.

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Key Concepts

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

Autosomal Recessive Traits
In genetics, traits that are inherited in an autosomal recessive manner require the presence of two recessive alleles for their expression. Autosomal refers to any chromosome that is not a sex chromosome, meaning these traits are not linked to the individual's sex. Recessive, on the other hand, implies that the trait is only visible when both inherited alleles are recessive. Let's look at the albino trait in dogs, specifically in poodles as in our scenario. An albino poodle would possess an "aa" genotype, where "aa" represents two recessive alleles responsible for albinism. Each allele comes from one parent, and only dogs with "aa" genotype will exhibit the albino characteristic, which affects pigmentation in their skin and fur. When we observe a purebred albino poodle, we can confidently conclude that it must carry two recessive alleles. If its partner is also a purebred albino, it too must carry the same recessive genotype. In the given exercise, both the female and male dogs have the "aa" genotype, ensuring that their phenotypic expression is albino. They can only pass the recessive allele "a" to their offspring.
Phenotypic Expression
Phenotypic expression refers to the visible traits manifested in an organism. It's the result of the interaction between an organism's genotype and its environment. In simpler terms, it's how the genetic instructions within an organism are translated into observable characteristics. For the recessive albino trait in our dog scenario, the phenotypic expression would be a lack of pigmentation, resulting in a white coat. This trait is only expressed when the dog receives a recessive allele from each parent. Remember, both parents in this case are "aa", indicating all their offspring should also appear albino. However, the unexpected appearance of black puppies is a discrepancy. If both parents genuinely are "aa" albinos, black coats cannot appear as phenotypic expressions, as black coat color suggests the presence of at least one dominant allele. This discrepancy implies that one of the parents must have a different genotype than expected, or a different dog was used in the mating process.
Genotype Analysis
Genotype analysis involves examining an organism's genetic makeup to understand the specific alleles it carries for certain traits. It allows us to predict the potential inheritance patterns and the probable phenotypic outcomes in offspring. In the context of our exercise, we explored the dogs' genotypes to predict the coat color of their offspring. The analysis starts by identifying the parental genotypes. Both dogs were believed to be homozygous recessive, expressed as "aa", according to the original claim made by the woman and breeder. When mating occurs between two "aa" dogs, each offspring's genotype is expected to be "aa", leading to an albino phenotype. This is because each parent can only contribute a recessive allele. Hence, producing black puppies—requiring a dominant allele—is an anomaly. The genotype analysis becomes critical here to assess whether there could've been a mistake with the breeding pair or if one of the dogs was mischaracterized, suggesting a different genetic makeup than initially supposed.

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Most popular questions from this chapter

If a man of blood-group AB marries a woman of bloodgroup A whose father was of blood-group \(\mathrm{O}\), to what different blood groups can this man and woman expect their children to belong?

Mice of the genotypes \(A / A ; B / B ; C / C ; D / D ; S / S\) and \(a / a ; b / b ; c / c ; d / d ; s / s\) are crossed. The progeny are intercrossed. What phenotypes will be produced in the \(\mathrm{F}_{2}\) and in what proportions? (The allele symbols stand for the following: \(A=\) agouti, \(a=\) solid (nonagouti) \(; B\) \(=\) black pigment, \(b=\) brown; \(C=\) pigmented, \(c=\) albino; \(D=\) nondilution, \(d=\) dilution (milky color); \(S=\) unspotted, \(s=\) pigmented spots on white background.)

A plant thought to be heterozygous for two independently assorting genes \((P / p ; Q / q)\) was selfed, and the progeny were \\[ 88 \quad P /-; Q /- \\] \(25 p / p ; Q /-\) \\[ 32 P /-; q / q \\] \(14 \mathrm{p} / \mathrm{p} ; \mathrm{q} / \mathrm{q}\) Do these results support the hypothesis that the original plant was \(P / p ; Q / q ?\)

For a period of 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 inconvenience the rabbits. a. When rabbits showing the typical 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 appears to be 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 showing 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 do you suppose there were only 39 of them? c. What additional experimental evidence might you collect to support or disprove your answers to 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. Can you suggest why, if you are permitted an analogy with the condition in rabbits? e. Again by analogy with rabbits, what phenotypes and genotypes might be expected among the children of a man and woman who both show the Pelger anomaly? (Problem 26 is from A. M. Srb, R. D. Owen, and R. S. Edgar, General Genetics, 2nd ed. W. H. Freeman and Company, \(1965 .\))

The production of eye-color pigment in Drosophila requires the dominant allele \(A\). The dominant allele \(P\) of a second independent gene turns the pigment to purple, but its recessive allele leaves it red. A fly producing no pigment has white eyes. Two pure lines were crossed with the following results: \(\mathrm{P} \quad\) red-eyed female \(\times\) white-eyed male \\[ 1 \\] \(\mathrm{F}_{1}\) purple-eyed females red-eyed males \\[ \mathrm{F}_{1} \times \mathrm{F}_{1} \\] \(\mathrm{F}_{2} \quad\) both males and females: \(\frac{3}{8}\) purple eyed \(\frac{3}{8}\) red eyed \(\frac{2}{8}\) white eyed Explain this mode of inheritance and show the genotypes of the parents, the \(\mathrm{F}_{1},\) and the \(\mathrm{F}_{2}\).

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