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Chapter 3: Problem 13

Albinism, lack of pigmentation in humans, results from an autosomal recessive gene (a). Two parents with normal pigmentation have an albino child. (a) What is the probability that their next child will be albino? (b) What is the probability that their next child will be an albino girl? (c) What is the probability that their next three children will be albino?

Short Answer

Expert verified
Answer: The probability of their next child being an albino girl is 12.5%.

Step by step solution

01

(Identify genotypes)

Since both parents have normal pigmentation but have produced an albino child, we can determine that both of them are heterozygous carriers of the albinism gene. This means their genotypes are Aa x Aa, where "A" represents the dominant allele for normal pigmentation and "a" is the recessive allele for albinism.
02

(Create a Punnett square)

To determine the probability of different genotypes for the next child, we can use a Punnett square to list all possible combinations of alleles from both parents. The Punnett square is as follows: | A | a ----------- A | AA | Aa ----------- a | Aa | aa
03

(a) Calculate the probability of an albino child)

The probability of an albino child is determined by the probability of inheriting the recessive "aa" genotype. According to the Punnett square, there is one "aa" genotype out of four possible genotypes, so the probability of their next child being albino is 1/4 or 25%.
04

(b) Calculate the probability of an albino girl)

To determine the probability of the next child being an albino girl, we first need to consider the gender probability. The probability of having a girl is 1/2 or 50%. Therefore, the probability of an albino girl is the product of the probabilities of being a girl and being albino: (1/2) * (1/4) = 1/8 or 12.5%
05

(c) Calculate the probability of three consecutive albino children)

To determine the probability of their next three children all being albino, we need to calculate the probability of three consecutive albino children. This probability can be calculated by taking the probability of one albino child and raising it to the power of 3 (because there are three independent events): (1/4) * (1/4) * (1/4) = (1/4)^3 = 1/64 or 1.56%

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Most popular questions from this chapter

The basis for rejecting any null hypothesis is arbitrary. The researcher can set more or less stringent standards by deciding to raise or lower the \(p\) value used to reject or not reject the hypothesis. In the case of the chi- square analysis of genetic crosses, would the use of a standard of \(p=0.10\) be more or less stringent about not rejecting the null hypothesis? Explain.

Two true-breeding pea plants were crossed. One parent is round, terminal, violet, constricted, while the other expresses the respective contrasting phenotypes of wrinkled, axial, white, full. The four pairs of contrasting traits are controlled by four genes, each located on a separate chromosome. In the \(\mathrm{F}_{1}\) only round, axial, violet, and full were expressed. In the \(\mathrm{F}_{2},\) all possible combinations of these traits were expressed in ratios consistent with Mendelian inheritance. (a) What conclusion about the inheritance of the traits can be drawn based on the \(\mathrm{F}_{1}\) results? (b) In the \(\mathrm{F}_{2}\) results, which phenotype appeared most frequently? Write a mathematical expression that predicts the probability of occurrence of this phenotype. (c) Which \(\mathrm{F}_{2}\) phenotype is expected to occur least frequently? Write a mathematical expression that predicts this probability. (d) In the \(F_{2}\) generation, how often is either of the \(P_{1}\) phenotypes likely to occur? (e) If the \(F_{1}\) plants were testcrossed, how many different phenotypes would be produced? How does this number compare with the number of different phenotypes in the \(\mathrm{F}_{2}\) generation just discussed?

Mendel crossed peas having round green seeds with peas having wrinkled yellow seeds. All \(\mathrm{F}_{1}\) plants had seeds that were round and yellow. Predict the results of testcrossing these \(\mathrm{F}_{1}\) plants.

Tay-Sachs disease (TSD) is an inborn error of metabolism that results in death, often by the age of \(2 .\) You are a genetic counselor interviewing a phenotypically normal couple who tell you the male had a female first cousin (on his father's side) who died from TSD and the female had a maternal uncle with TSD. There are no other known cases in either of the families, and none of the matings have been between related individuals. Assume that this trait is very rare. (a) Draw a pedigree of the families of this couple, showing the relevant individuals. (b) Calculate the probability that both the male and female are carriers for TSD. (c) What is the probability that neither of them is a carrier? (d) What is the probability that one of them is a carrier and the other is not? [Hint: The \(p\) values in (b), (c), and (d) should equal \(1 .]\)

The autosomal (not X-linked) gene for brachydactyly, short fingers, is dominant to normal finger length. Assume that a female with brachydactyly in the heterozygous condition is married to a man with normal fingers. What is the probability that (a) their first child will have brachydactyly? (b) their first two children will have brachydactyly? (c) their first child will be a brachydactylous girl?
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