Imagine that you are a genetic counselor, and a couple planning to start a family comes to you for information. Charles was married once before, and he and his first wife had a child with cystic fibrosis. The brother of his current wife, Elaine, died of cystic fibrosis. What is the probability that Charles and Elaine will have a baby with cystic fibrosis? (Neither Charles, Elaine, nor their parents have cystic fibrosis.)

Short Answer

Expert verified

The probability that Charles and Elaine's child will have cystic fibrosis is 1/6.

Step by step solution

01

Cystic Fibrosis

Cystic Fibrosis (CF) is a genetic disorder that is caused due to a defect in the CFTR gene. It affects the lungs, and individuals have breathing difficulties due to the build-up of mucus. It is an autosomal recessive disease.

An individual must inherit two copies of the defective gene to be diseased with CF. However, individuals with one copy are the carriers of the disease.

02

The genotypes of Charles (father) and Elaine (mother)

Let C be the normal allele and c be the allele for cystic fibrosis. The genotype of Charles would be Cc as he had a diseased child with his first wife. The genotype of Elaine can be either CC or Ccbecause her parents did not have the disease, but her brother had CF.

As a result, the woman has 1/3 chance of being CC as her parents were not diseased. However, she has a 2/3 chance of being a carrier (Cc) for the trait as her brother died of CF.

03

The probability of having a diseased child

The father is a carrier (Cc) for CF. If the mother is also a carrier (Cc) for CF, then the chances that Charles and Elaine would have a diseased child is ¼ (1/2 × ½). However, if the mother is normal (CC), there are zero chances that the child would have CF.

This means the probability of the child being diseased depends on the genotype of the mother. Therefore, the probability that the child has CF can be calculated as(23×14)+(13×0)=16

Thus, the probability that the child will have CF is 1/6.

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Most popular questions from this chapter

Karen and Steve each have a sibling with sickle cell disease. Neither Karen nor Steve nor any of their parents have the disease, and none of them have been tested to see if they carry the sickle-cell allele. Based on this incomplete information, calculate the probability that if this couple has a child, the child will have sickle-cell disease.

Joan was born with six toes on each foot, a dominant trait called polydactyly. Two of her five siblings and her mother, but not her fathers, also have extra digits. What is Joan’s genotype for the number of digits character? Explain your answer. Use D and d symbolize the alleles for this character.

The pedigree below traces the inheritance of alkaptonuria, a biochemical disorder. Affected individuals, indicated hereby the colored circles and squares, are unable to metabolize a substance called alkapton, which colors the urine and stains body tissues. Does alkaptonuria appear to be caused by a dominant allele or a recessive allele? Fill in the genotypes of the individuals whose genotypes can be deduced. What genotypes are possible for each of the other individuals?

Hemochromatosis is an inherited disease caused by a recessive allele. If a woman and her husband, who are both carriers, have three children, what is the probability of each of the following?

(a) All three children are of normal phenotype.

(b) One or more of the three children have the disease.

(c) All three children have the disease.

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(Note: It will help to remember that the probabilities of all possible outcomes always add up to 1.)

In tigers, a recessive allele of a particular gene causes both an absence of fur pigmentation (a white tiger) and a cross-eyed condition. If two phenotypically normal tigers that are heterozygous at this locus are mated, what percentage of their offspring will be cross-eyed? What percentage of cross-eyed tigers will be white?

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