Predict the probability (using a Punnett square) of producing an offspring affected by a single gene X-linked mutation given the genotype of the parents. Compare these crosses in two cases: where the mutant allele is “A” vs where the mutant allele is “a” (XAXA x XaY) (XAXa x XAY) (XAXa x XaY) (XaXa x XAY) (XaXa x XaY). Include gender in the phenotype.
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Predict the probability (using a Punnett square) of producing an offspring affected by a single gene X-linked mutation given the genotype of the parents. Compare these crosses in two cases: where the mutant allele is “A” vs where the mutant allele is “a” (XAXA x XaY) (XAXa x XAY) (XAXa x XaY) (XaXa x XAY) (XaXa x XaY). Include gender in the
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- Normal vision (XA) in humans is dominant to color blindness (Xa) and is X-linked. A man with normal vision, whose father was colorblind, marries a colorblind woman. What are the chances that a son will be colorblind? What are the chances that a daughter will be colorblind? The determiner for brown eyes (B) is dominant to blue eyes (b) and is not X-linked. A colorblind man with brown eyes, whose mother was blue-eyed, marries a blue-eyed woman having normal vision, whose father was colorblind. Show the expected phenotypes ratio of their children involving eye color, color blindness, and sex.On average, what proportion of X-linked genes in the first individual are the same (inherited from a common ancestor) as those in the second individual? a. A male and his mother b. A female and her mother c. A male and his father d. A female and her father e. A male and his brother f. A female and her sister g. A male and his sister h. A female and her brotherSuppose gene B is X-linked and is embryonically lethal when homozygous or hemizygous recessive. A man marries a woman who is heterozygous for this gene. They want to have three kids – one girl and two boys. Using a Punnett square, answer the following: What is the probability that they will have a son that dies before birth? _______________________ What is the probability that they will have a daughter who has the same genotype as her mother? _________________ One of their daughters eventually has a child with a man. One of their sons dies before birth. What was the genotype of this daughter? _______________________________ please show me how to get the answer and explain how you got and use a punnet square
- Duchenne muscular dystrophy is an X-linked, recessive disorder in which muscles waste away early in life, resulting in death in the teens or twenties. A man and woman in their late thirties have five children—three boys (ages 1, 3, and 10 years) and two girls (ages 5 and 7 years). The oldest, boy shows symptoms of the disease. What are the probabilities that their other children will develop the disease? Give only typing answer with explanation and conclusionWhich of the following statements is false? Dosage compensation is accomplished in humans by stimulation of gene expression from the single X chromosome. Dosage compensation is accomplished in humans by inactivating a female X chomosome. An individual with Turner Syndrome has no Barr bodies. An individual with Klinefelter syndrome generally has one Barr body. A typical XX human female has one Barr body.Calico is a coat color found in cats, which is caused by a X-linked, co-dominant alleleFemale cats: XBXB= black; XOXO = orange; and XBXO = calico Male cats: XBY = black; orange XOY = orange What is the outcome of a black male crossed with an orange female? A. All offspring are black B. All female cats are calico, male cats are orange C. All female cats are calico, male cats are orange D. 1/4 female calico, 1/4 female black, 1/4 male black, 1/4 male orange
- Choose correct option and do explain. Considering an X-linked dominant trait, if an affected woman and an unaffected man decide to have children, which of the answer choices is possible for their children? a. All of their sons are expected to show the dominant trait. b. Their daughters are expected be heterozygous for the gene. c. Their daughters are not expected to show the dominant trait. d. Their sons are expected to be heterozygous for the gene. e. All their children, whether male or female, are expected to show the dominant trait.The following X-linked recessive traits are found in fruit flies:vermilion eyes are recessive to red eyes, miniature wings are recessiveto long wings, and sable body is recessive to gray body. A cross wasmade between wild-type males with red eyes, long wings, and graybodies and females with vermilion eyes, miniature wings, and sablebodies. The heterozygous female offspring from this cross, whichhad red eyes, long wings, and gray bodies, were then crossed tomales with vermilion eyes, miniature wings, and sable bodies. Thefollowing data were obtained for the F2 generation (including bothmales and females):1320 vermilion eyes, miniature wings, sable body1346 red eyes, long wings, gray body102 vermilion eyes, miniature wings, gray body90 red eyes, long wings, sable body42 vermilion eyes, long wings, gray body48 red eyes, miniature wings, sable body2 vermilion eyes, long wings, sable body1 red eyes, miniature wings, gray bodyA. Calculate the map distances separating the three genes.B. Is…The following X-linked recessive traits are found in fruit flies:vermilion eyes are recessive to red eyes, miniature wings are recessiveto long wings, and sable body is recessive to gray body. A cross wasmade between wild-type males with red eyes, long wings, and graybodies and females with vermilion eyes, miniature wings, and sablebodies. The heterozygous female offspring from this cross, whichhad red eyes, long wings, and gray bodies, were then crossed tomales with vermilion eyes, miniature wings, and sable bodies. Thefollowing data were obtained for the F2 generation (including bothmales and females):1320 vermilion eyes, miniature wings, sable body1346 red eyes, long wings, gray body102 vermilion eyes, miniature wings, gray body90 red eyes, long wings, sable body42 vermilion eyes, long wings, gray body48 red eyes, miniature wings, sable body2 vermilion eyes, long wings, sable body1 red eyes, miniature wings, gray bodyWhat information do you know based on the question and your…
- The following X-linked recessive traits are found in fruit flies:vermilion eyes are recessive to red eyes, miniature wings are recessiveto long wings, and sable body is recessive to gray body. A cross wasmade between wild-type males with red eyes, long wings, and graybodies and females with vermilion eyes, miniature wings, and sablebodies. The heterozygous female offspring from this cross, whichhad red eyes, long wings, and gray bodies, were then crossed tomales with vermilion eyes, miniature wings, and sable bodies. Thefollowing data were obtained for the F2 generation (including bothmales and females):1320 vermilion eyes, miniature wings, sable body1346 red eyes, long wings, gray body102 vermilion eyes, miniature wings, gray body90 red eyes, long wings, sable body42 vermilion eyes, long wings, gray body48 red eyes, miniature wings, sable body2 vermilion eyes, long wings, sable body1 red eyes, miniature wings, gray bodyWhat topic in genetics does this question address?The following X-linked recessive traits are found in fruit flies:vermilion eyes are recessive to red eyes, miniature wings are recessiveto long wings, and sable body is recessive to gray body. A cross wasmade between wild-type males with red eyes, long wings, and graybodies and females with vermilion eyes, miniature wings, and sablebodies. The heterozygous female offspring from this cross, whichhad red eyes, long wings, and gray bodies, were then crossed tomales with vermilion eyes, miniature wings, and sable bodies. Thefollowing data were obtained for the F2 generation (including bothmales and females):1320 vermilion eyes, miniature wings, sable body1346 red eyes, long wings, gray body102 vermilion eyes, miniature wings, gray body90 red eyes, long wings, sable body42 vermilion eyes, long wings, gray body48 red eyes, miniature wings, sable body2 vermilion eyes, long wings, sable body1 red eyes, miniature wings, gray bodyAnalyze data. Make a drawing. Make a calculation.Jane is heterozygous for both X-linked traits like her mother. Her father is normal for both X-linked traits. James has a mother who suffers from G6PDD but not from fragile X-syndrome. His father does not exhibit any X-linked disorder but has amelogenesis imperfecta. What are the genotypes of the following: Jane: ______________________ James: ______________________ Jane’s mother: _________________ James’ mother: ________________ Jane’s father: __________________ James’ father: _________________