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Ch. 29 Development and Inheritance
Martini - Fundamentals of Anatomy & Physiology 12th Edition
Martini, Nath, Bartholomew12th EditionFundamentals of Anatomy & PhysiologyISBN: 9780137854011Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoMartini, Nath, Bartholomew 12th EditionCh. 29 Development and InheritanceProblema 25
Capitolo 28, Problema 25

Hemophilia A, a condition in which blood does not clot properly, is a recessive trait located on the X chromosome (Xʰ). Suppose that a woman who is heterozygous for this trait (XXʰ) has children with a normal male (XY). What is the probability that the couple will have daughters with hemophilia? What is the probability that the couple will have sons with hemophilia?

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Identify the genotypes of the parents: The woman is heterozygous for hemophilia A, so her genotype is X X\^h (one normal X chromosome and one X chromosome with the hemophilia allele). The man is normal, so his genotype is X Y.
Determine the possible gametes each parent can produce: The woman can produce eggs carrying either X or X\^h. The man can produce sperm carrying either X or Y.
Set up a Punnett square to find all possible combinations of the offspring's sex chromosomes: Combine the woman's gametes (X and X\^h) with the man's gametes (X and Y) to get the genotypes of daughters and sons.
Analyze the daughters' genotypes: Daughters receive one X chromosome from each parent. The possible daughter genotypes are X X (normal) and X\^h X (carrier). Since hemophilia is recessive and X-linked, daughters must have two X\^h alleles (X\^h X\^h) to express the disease. Determine the probability of this genotype from the Punnett square.
Analyze the sons' genotypes: Sons receive the X chromosome from the mother and the Y chromosome from the father. The possible son genotypes are X Y (normal) and X\^h Y (hemophiliac). Since males have only one X chromosome, the presence of X\^h means the son will have hemophilia. Determine the probability of this genotype from the Punnett square.

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X-linked Recessive Inheritance

X-linked recessive traits are caused by genes located on the X chromosome. Males (XY) are more likely to express the trait because they have only one X chromosome, while females (XX) must inherit two copies of the recessive allele to express the trait. Females with one affected X are carriers but usually do not show symptoms.
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X-Linked Inheritance

Genotype and Phenotype of Parents

Understanding the parents' genotypes is essential: the mother is heterozygous (XXʰ), meaning she carries one normal and one affected X chromosome, while the father is normal (XY). This determines the possible combinations of sex chromosomes and alleles their children can inherit, influencing the probability of affected offspring.
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Genotype & Phenotype

Probability of Inheritance in Offspring

Calculating the probability involves combining the inheritance patterns of sex chromosomes and the recessive allele. Daughters inherit one X from each parent, while sons inherit the X from the mother and Y from the father. This affects the likelihood of daughters or sons having hemophilia based on the mother's carrier status.
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Percorso guidato
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X-Linked Inheritance
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Indicate the type of inheritance involved in each of the following situations. 

(a) Children who exhibit the trait have at least one parent who also exhibits it. 

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Distinguish between the following paired terms: 

(b) heterozygous and homozygous, 

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