Question

A boy with Klinefelter syndrome \((47, \mathrm{XXY})\) is born to a mother who is phenotypically normal and a father who has the X-linked skin condition called anhidrotic ectodermal dysplasia. The mother's skin is completely normal with no signs of the skin abnormality. In contrast, her son has patches of normal skin and patches of abnormal skin. (a) Which parent contributed the abnormal gamete? (b) Using the appropriate genetic terminology, describe the meiotic mistake that occurred. Be sure to indicate in which division the mistake occurred. (c) Using the appropriate genetic terminology, explain the son's skin phenotype.

Short answer

Answer: The mother contributed the abnormal gamete due to a meiotic mistake involving nondisjunction. The son's skin phenotype can be explained by X-inactivation and mosaicism.

Step-by-step solution

(Determine the involved chromosomes)

To analyze the parents' gametes and the son's chromosomes, we need to determine the involved chromosomes. In this case, the boy has Klinefelter syndrome, which means he has an extra X chromosome (47, XXY). We are also given that the skin condition, anhidrotic ectodermal dysplasia, is X-linked, which means it is a gene located on the X chromosome.

(Parent contribution)

As the boy has Klinefelter syndrome, we know that one of the parents contributed an extra X chromosome. Since the mother is phenotypically normal and the father has the X-linked skin condition, we can conclude that the mother contributed the extra X chromosome and the father contributed the Y chromosome, together with his X chromosome carrying the skin condition. So, the answer for part (a) is that the mother contributed the abnormal gamete.

(Meiotic mistake description)

Now, we need to find which meiotic division produced the mistake and explain it using genetic terminology. The mistake occurred in the mother's meiosis, leading to a gamete with an extra X chromosome. Nondisjunction is a process in which homologous chromosomes or sister chromatids fail to separate properly during meiosis. As the mother contributed the extra X chromosome, we can infer that there was a nondisjunction event in the mother's meiosis. Considering this, the answer for part (b) should be that the meiotic mistake was nondisjunction, and it occurred in the mother's meiosis. The question does not provide enough specific information to pinpoint whether the mistake occurred in meiosis I or meiosis II.

(Explaining the son's skin phenotype)

The son's skin phenotype can be explained by the process of X-inactivation and the concept of mosaicism. In mammalian cells, one of the X chromosomes in every female cell is inactivated, leading to a "mosaic" pattern of gene expression from the X chromosomes. This process ensures that females and males have similar levels of X-linked gene expression. In the son's case, he has two X chromosomes – one from the father carrying the gene for anhidrotic ectodermal dysplasia and one normal X chromosome from the mother. During early embryonic development, X-inactivation occurs, and in some cells, the father's X chromosome is inactivated, leading to patches of normal skin. In other cells, the mother's normal X chromosome is inactivated, resulting in patches of abnormal skin. Therefore, the answer for part (c) is that the son's skin phenotype can be explained by X-inactivation and mosaicism.