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Chapter 4: Problem 27

Two mothers give birth to sons at the same time at a busy urban hospital. The son of mother 1 has hemophilia, a disease caused by an X-linked recessive allele. Neither parent has the disease. Mother 2 has a son without hemophilia, despite the fact that the father has hemophilia. Several years later, couple 1 sues the hospital, claiming that these two newborns were swapped in the nursery following their birth. As a genetic counselor, you are called to testify. What information can you provide the jury concerning the allegation?

Short Answer

Expert verified
In conclusion, the inheritance patterns of hemophilia in both the couple's situations do not support the allegation that the newborns were swapped in the hospital. Couple 1's son has a 25% chance of having hemophilia, while Couple 2's son has a 100% chance of not having hemophilia. These patterns indicate that the given scenarios are possible without involving a newborn swap.

Step by step solution

01

Understanding Hemophilia Inheritance

Hemophilia is a sex-linked recessive disease, which means that it is carried on the X chromosome. Males have one X and one Y chromosome (XY), and females have two X chromosomes (XX). A male offspring receives the X chromosome from his mother and the Y chromosome from his father, while a female offspring receives an X chromosome from both parents. If a male has the hemophilia allele on his X chromosome, he will have the disease since there is no corresponding allele on the Y chromosome to mask its effect. In contrast, females can be carriers of the disease without manifesting any symptoms if they have only one X chromosome with the disease allele.
02

Analyzing the First Couple's Situation

Couple 1 has a son with hemophilia, and neither parent has the disease. For the son to have hemophilia, he must have received the X chromosome with the hemophilia allele from his mother, who must be a carrier (XHXh). The father has a normal X chromosome (XHY). The inheritance pattern can be illustrated as follows: Mother (XHXh) x Father (XHY) - XHXH (daughter, non-carrier) - XHXh (daughter, carrier) - XHY (son, non-hemophiliac) - XhY (son, hemophiliac) In this case, there is a 25% chance of having a son with hemophilia, making their situation possible without any evidence of a newborn swap.
03

Analyzing the Second Couple's Situation

Couple 2 has a son without hemophilia, even though the father has the disease. The father has an X chromosome with the hemophilia allele (XhY) and the mother has two normal X chromosomes (XHXH). The inheritance pattern can be illustrated as follows: Mother (XHXH) x Father (XhY) - XHXh (daughter, carrier) - XHY (son, non-hemophiliac) In this case, there is a 100% chance of having a son without hemophilia and a 100% chance of having a daughter who is a carrier, which also makes this situation possible without evidence of a newborn swap.
04

Conclusion

Based on the information provided and the inheritance patterns of hemophilia, both the first and second couple's situations are possible without any indication of a newborn swap. It is essential to inform the jury about these inheritance patterns in this case, which do not support the allegation that the newborns were switched in the hospital.

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Most popular questions from this chapter

The specification of the anterior-posterior axis in Drosophila embryos is initially controlled by various gene products that are synthesized and stored in the mature egg following oogenesis. Mutations in these genes result in abnormalitics of the axis during embryogenesis, illustrating maternal effect. How do such mutations vary from those involved in organelle heredity that illustrate extranuclear inheritance? Devise a set of parallel crosses and expected outcomes involving mutant genes that contrast maternal effect and organelle heredity.

As in the plants of Problem \(6,\) color may be red, white, or pink and flower shape may be personate or peloric. Determine the \(\mathrm{P}_{1}\) and \(\mathrm{F}_{1}\) genotypes for the following crosses: (a) red, peloric \(\times\) white, personate \(\mathrm{F}_{1}:\) all pink, personate (b) red, personate \(\times\) white, peloric \(\mathrm{F}_{1}:\) all pink, personate (c) pink, personate \(\times\) red, peloric $F_{1:}\left\\{\begin{array}{l}1 / 4 \text { red, personate } \\ 1 / 4 \text { red, peloric } \\ 1 / 4 \text { pink, personate } \\ 1 / 4 \text { pink, peloric }\end{array}\right.$ (d) pink, personate \(x\) white, peloric $\mathbf{F}_{1:}\left\\{\begin{array}{l}1 / 4 \text { white, personate } \\ 1 / 4 \text { white, peloric } \\ 1 / 4 \text { pink, personate } \\ 1 / 4 \text { pink, peloric }\end{array}\right.$ (e) What phenotypic ratios woud result from crossing the \(\mathrm{F}_{1}\) of (a) to the \(F_{1}\) of \((b) ?\)

Three gene pairs located on separate autosomes determine flower color and shape as well as plant height. The first pair exhibits incomplete dominance, where color can be red, pink (the heterozygote), or white. The second pair leads to the dominant personate or recessive peloric flower shape, while the third gene pair produces either the dominant tall trait or the recessive dwarf trait. Homozygous plants that are red, personate, and tall are crossed with those that are white, peloric, and dwarf. Determine the \(F_{1}\) genotype(s) and phenotype(s). If the \(F_{1}\) plants are inter. bred, what proportion of the offspring will exhibit the same phenotype as the \(\mathrm{P}_{1}\) plants?

The following genotypes of two independently assorting autosomal genes determine coat color in rats: \(A-B-(\text { gray }) ; A-b b\) (yellow) \(; a a B-\) (black); \(a a b b\) (cream) A third gene pair on a separate autosome determines whether any color will be produced. The \(C C\) and \(C c\) genotypes allow color according to the expression of the \(A\) and \(B\) alleles. However, the ce genotype results in allbino rats regardless of the \(A\) and \(B\) alleles present. Determine the \(F_{1}\) phenotypic ratio of the following crosses: (a)AAbbCC \(\times\) aaBBcc; (b) \(A a B B c c \times A A B b c c\) (c) \(A a B b C c \times A a B b c c\)

Horses can be cremello (a light cream color), chestnut (a reddish brown color), or palomino (a golden color with white in the horse's tail and mane). Of these phenotypes, only palominos never breed true. The following results have been observed: cremello \(\times\) palomino \(\longrightarrow 1 / 2\) cremello \(1 / 2\) palomino chestnut \(\times\) palomino \(\longrightarrow 1 / 2\) chestnut \(1 / 2\) palomino palomino \(\times\) palomino \(\longrightarrow 1 / 4\) chestnut \(1 / 2\) palomino \(1 / 4\) cremello (a) From these results, determine the mode of inheritance by assigning sene symbols and indicating which genotypes yield which phenotypes. (b) Predict the \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) results of many initial matings between cremello and chestnut horses.
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