Anhidrotic ectodermal dysplasia is an X-linked recessive disorder in humans characterized by small teeth, no sweat glands, and sparse body hair. This trait is usually seen in men, but women who are heterozygous carriers of the trait often have irregular patches of skin with few or no sweat glands (see the illustration below). Q. Explain why women who are heterozygous carriers of a recessive gene for anhidrotic ectodermal dysplasia have irregular patches of skin lacking sweat glands.
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Anhidrotic ectodermal dysplasia is an X-linked recessive disorder in humans characterized by small teeth, no sweat glands, and sparse body hair. This trait is usually seen in men, but women who are heterozygous carriers of the trait often have irregular patches of skin with few or no sweat glands (see the illustration below).
Q. Explain why women who are heterozygous carriers of a recessive gene for anhidrotic ectodermal dysplasia have irregular patches of skin lacking sweat glands.
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- Anhidrotic ectodermal dysplasia is an X-linked recessive disorder in humans characterized by small teeth, no sweat glands, and sparse body hair. This trait is usually seen in men, but women who are heterozygous carriers of the trait often have irregular patches of skin with few or no sweat glands (see the illustration below). Q. Why does the distribution of the patches of skin lacking sweat glands differ among the females depicted in the illustration, even between the identical twins?Anhidrotic ectodermal dysplasia is an X-linked recessive disorder in humans characterized by small teeth, no sweat glands, and sparse body hair. This trait is usually seen in men, but women who are heterozygous carriers of the trait often have irregular patches of skin with few or no sweat glands (see the illustration below). a. Explain why women who are heterozygous carriers of a recessive gene for anhidrotic ectodermal dysplasia have irregular patches of skin lacking sweat glands. b. Why does the distribution of the patches of skin lacking sweat glands differ among the females depicted in the illustration, even between the identical twins?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.
- 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 conclusionRed-green color blindness is inherited as an X-linked recessive (Xc). If a color-blind man marries a woman who is heterozygous for normal vision, what would be the expected phenotypes of their children with reference to this character? In your answer, specify in your phenotype descriptions the gender of the children. (For example, don’t just say 75% of the children would be colorblind – you would instead say 100 % of the daughters would be colorblind and 50% of the sons would be colorblind. Note that this is not a correct answer; it is just to give you an idea of how to explain the correct phenotypes of the cross.)___Suppose 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
- Let’s suppose that two different X-linked genes exist in mice,designated with the letters N and L. Gene N exists in a dominant,normal allele and in a recessive allele, n, that is lethal. Similarly,gene L exists in a dominant, normal allele and in a recessive allele,l, that is lethal. Heterozygous females are normal, but males thatcarry either recessive allele are born dead. Explain whether or notit would be possible to map the distance between these two genesby making crosses and analyzing the number of living and deadoffspring. You may assume that you have strains of mice in whichfemales are heterozygous for one or both genes.Consider 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?Color blindness in humans is controlled by an X-linked completely recessive allele (Xc), while breast cancer is controlled by an autosomal completely dominant allele, B. A color blind male, who is a heterozygote carrier for breast cancer has three children/n with a normal eyed female (whose mother was color blind), who is homozygote recessive for the breast cancer allele. What is the probability that out of three children, 2 will be color blind males, and not show breast cancer, and one will be a color blind female, who shows breast cancer?