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Table 8.1 shows that Turner syndrome occurs when an individual inherits one X chromosome but lacks a second sex chromosome. Can Turner syndrome be due to nondisjunction during oogenesis, spermatogenesis, or both? If a
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- Human sex chromosomes are XX for females and XY for males. a. With respect to an X-linked gene, how many different types of gametes can a male produce? b. If a female is homozygous for an X-linked allele, how many different types of gametes can she produce with respect to this allele? c. If a female is heterozygous for an X-linked allele, how many different types of gametes can she produce with respect to this allele?arrow_forwardA boy with Klinefelter syndrome (47,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 boy has patches of normal skin and patches of abnormal skin. Which of the following statemnets likely explains these observations? The father contributed the extra X chromosome in the son as a result of a non-disjunction in meiosis I during spermatogenesis. The mother contributed the extra X chromosome in the son as a result of a non-disjunction in meiosis II during oogenesis. The mother contributed the extra X chromosome in the son as a result of a non-disjunction in meiosis I during oogenesis. The father contributed the extra X chromosome in the son as a result of a non-disjunction in meiosis II during spermatogenesis. Either parent could have contributed to the extra X chromosome in the son as a results of disjunction in either meiosis I or meiosis II during…arrow_forwardButterflies have an X-Y sex-determination system that is different from that of flies or humans. Female butterflies may be either XY or X0, while butterflies with two or more X chromosomes are males. This photograph shows a tiger swallowtail gynandromorph, which is half male (left side) and half female (right side). Given that the first division of the zygote divides the embryo into the future right and left halves of the butterfly, propose a hypothesis that explains how nondisjunction during the first mitosis might have produced this unusual-looking butterfly. Question is also in the picture.arrow_forward
- A boy with Klinefelter syndrome (47,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.arrow_forwardA common kind of red-green blindness in humans is caused by the presence of a sex-linked recessive gene c, whose normal allele is c+. Using these genes, what are the possible genotypes and their corresponding phenotypes in males and females? Can two colorblind parents produce a normal son? (b) A normal daughter? (c) Can two normal parents produce a colorblind son? (d) a colorblind daughter?arrow_forwardAssume that the incidence of an X-linked recessive disorder is 1 in 200 male births. What is the frequency of the mutant allele?arrow_forward
- In humans, baldness (Xᵇ) is a recessive sex-linked trait. Two people with normal hair (Xᴮ) have a bald son. What are the genotypes of the parents?arrow_forwardIn C. elegans, lon-2 and unc-2 are recessive mutations that are 8 map units apart on the X chromosome. An hermaphrodite who is Lon and Unc is mated to a wild-type male. An F1 hermaphrodite is mated to a wild-type male. What are the expected percentages of the different phenotypes among the male progeny?arrow_forwardThe 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?arrow_forward
- In humans, Widow’s peak (W) is dominant over a continuous hairline (w), and short fingers (F) are dominant over long fingers (f). Two individuals with widow’s peaks and short fingers have a child with a continuous hairline and long fingers. Determine the genotype of the parents.arrow_forwardRed–green color blindness is a human X-linked recessive disorder. A young man with a 47,XXY karyotype (Klinefelter syndrome) is color blind. His 46,XY brother is also color blind. Both parents have normal color vision. Where did the nondisjunction that gave rise to the young man with Klinefelter syndrome take place? Assume that no crossing over took place in prophase I of meiosis.arrow_forwardVermillion eye color in Drosophila is a sex-linked recessive trait. What phenotype would be found in this progeny of a cross between a vermillion female and a wild type male?arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage LearningHuman Biology (MindTap Course List)BiologyISBN:9781305112100Author:Cecie Starr, Beverly McMillanPublisher:Cengage Learning