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For the following crosses, determine as accurately as possible the genotypes of each parent, the parent in whom nondisjunction occurs, and whether nondisjunction takes place in the first or second meiotic division. Both color blindness and hemophilia, a
A man and a woman who each have
A man who is color blind and a woman who is wild type have a son with Jacob syndrome (XYY) who has hemophilia.
A
A man who is color blind and has hemophilia and a woman who is wild type have a daughter with triple X syndrome (XXX) who has hemophilia and normal color vision.
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- 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 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_forwardA color-blind man marries a woman with normal vision whose father was color-blind. Remember that color-blindness is an X-linked recessive trait. Hint: see figure 12.7 in book. A) What is the probability that their first child will be a color-blind daughter? B) What is the probability that their first son will be color-blind?arrow_forwardAssuming no other abnormalities, classify each of the following human sex chromosome complements as to whether or not individuals with that complement will be morphologically male or female. Three human sex chromosome complements will be classified as male and three will be classified as female. XXY XYY XXX XO XY XXarrow_forward
- Men have XY (or YX) chromosomes and women have XX chromosomes. X-linked recessive genetic diseases (such as juvenile retinoschisis) occur when there is a defective X chromosome that occurs without a paired X chromosome that is not defective. Represent a defective X chromosome with lowercase x, so a child with the xY or Yx pair of chromosomes will have the disease and a child with XX or XY or YX or xX or Xx will not have the disease. Each parent contributes one of the chromosomes to the child. Complete parts a through d below. a. If a father has the defective x chromosome and the mother has good XX chromosomes, what is the probability that a son will inherit the disease? nothing (Type an integer or a decimal. Do not round.) b. If a father has the defective x chromosome and the mother has good XX chromosomes, what is the probability that a daughter will inherit the disease? nothing (Type an integer or a decimal. Do not round.) c. If a mother has one defective x…arrow_forwardIn the following cross, imagine that you have a female fly that has two Xs and one Y due to a nondisjunction event in her mother's germ cells. Draw out what the possible gametes are for both the female and the male and also a Punnett square showing the genotypes, phenotypes, and sex of the possible flies as a result of this cross. You do not need to provide the probabilities of each of these. Red-eyed wi C Ở Red-eyed wt XX Y X Y Meiosisarrow_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 their first child will be a brachydactylous girl? ¼ 1/2 1/8 3/4 2/3arrow_forward
- Draw a Punnett square for the dihybrid cross. There are two known alleles of gene occupying a specific locus in the X chromosome. The gene in question codes for a transcription factor involved in digit development. The mutant allele is dominant and gives rise to an additional but non-functioning little finger (polydactyly) on both hands. A couple have had their DNA sequenced at the region of interest, the male exhibits polydactyly because of the mutation, the female is homozygous wild type at the same locus and therefore has the wild type phenotype. Both have green eyes. In this story; eye colour shows a monogenic autosomal inheritance pattern and the allele for brown eyes shows incomplete dominance with that for blue eyes, the heterozygote phenotype is green eyes. The genes for eye colour and polydactyly show no linkage.arrow_forwardHemophilia is caused by an X-linked recessive mutation in humans. If a man whose paternal uncle (father's brother) was a hemophiliac marries a woman whose brother is also a hemophiliac, what is the probability that their first child will have hemophilia? (Assume that no other cases of hemophilia exist in the pedigree.) 1/3 0 1/8 0 1/4 1/2arrow_forwardConsider a couple: a woman who is homozygous for a recessive mutation that causes X-linked colorblindness, and a man with full color vision (he does not carry a copy of the mutation). a) What is the probability that a son of this couple will be colorblind? b) What is the probability that a daughter of the couple will be colorblind?arrow_forward
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