The mutant FMR-1 allele that causes fragile X syndrome is considered to be X-linked dominant withincomplete penetrance and variable expressivity.Why do most females heterozygous for one mutantand one normal allele have at least some symptomsof the disease?
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The mutant FMR-1 allele that causes fragile X syndrome is considered to be X-linked dominant with
incomplete penetrance and variable expressivity.
Why do most females heterozygous for one mutant
and one normal allele have at least some symptoms
of the disease?
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- In 1952, an article in the British Medical Journalreported interesting differences in the behavior ofblood plasma obtained from several people who suffered from X-linked recessive hemophilia. Whenmixed together, the cell-free blood plasma from certain combinations of individuals could form clots inthe test tube. For example, the following table showswhether clots could form (+) or not (−) in variouscombinations of plasma from four people withhemophilia:1 and 1 − 2 and 3 +1 and 2 − 2 and 4 +1 and 3 + 3 and 3 −1 and 4 + 3 and 4 −2 and 2 − 4 and 4 −What do these data tell you about the inheritance ofhemophilia in these individuals? Do these data allowyou to exclude any models for the biochemical pathway governing blood clotting?The Xg cell-surface antigen is coded for by a gene located on the X chromosome. No equivalent gene exists on the Y chromosome. Two codominant alleles of this gene have been identified: Xg1 and Xg2. A woman of genotype Xg2/Xg2 bears children with a man of genotype Xg1/Y, and they produce a son with Klinefelter syndrome of genotype Xg1/Xg2Y. Using proper genetic terminology, briefly explain how this individual was generated. In which parent and in which meiotic division did the mistake occur?Duchenne muscular dystrophy is a recessive disorder caused by a rare,loss-of-function allele that is located on the X chromosome in humans. Anunaffected woman (i.e., without disease symptoms) who is heterozygousfor the X-linked allele causing Duchenne muscular dystrophy has childrenwith a man with a functional (non-disease-causing) allele. What is theprobability that this couple will have an unaffected son?
- 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.Analysis of X-Linked Dominant and Recessive Traits In the eighteenth century, a young boy with a skin condition known as ichthyosis hystrix gravior was identified. The phenotype of this disorder includes thickening of skin and the formation of loose spines that are sloughed off periodically. This man married and had six sons, all of whom had the same condition. He also had several daughters, all of whom were unaffected. In all succeeding generations, the condition was passed on from father to son. What can you theorize about the location of the gene that causes ichthyosis hystrix gravior?In 1995, doctors reported a Chinese family in whichretinitis pigmentosa (progressive degeneration of theretina leading to blindness) affected only males. Allsix sons of affected males were affected, but all of thefive daughters of affected males (and all of thechildren of these daughters) were unaffected.a. What is the likelihood that this form of retinitispigmentosa is due to an autosomal mutationshowing complete dominance?b. What other possibilities could explain the inheritance of retinitis pigmentosa in this family? Whichof these possibilities do you think is most likely?
- Figure 1-15 shows the family tree, or pedigree, for LouiseBenge (Individual VI-1) who suffers from the diseaseACDC because she has two mutant copies of the CD73gene. She has four siblings (VI-2, VI-3, VI-4, and VI-5)who have this disease for the same reason. Do all of the10 children of Louise and her siblings have the samenumber of mutant copies of the CD73 gene, or mightthis number be different for some of the 10 children?A boy with Klinefelter syndrome (47,XXY) is born to a motherwho is phenotypically normal and a father who has the X-linkedskin condition called anhidrotic ectodermal dysplasia. The mother’sskin is completely normal with no signs of the skin abnormality.In contrast, her son has patches of normal skin and patchesof abnormal skin. Question: Using the appropriate genetic terminology, describe themeiotic mistake that occurred. Be sure to indicate in whichdivision the mistake occurred.The maternal-effect mutation bicoid (bcd) is recessive. Inthe absence of the bicoid protein product, embryogenesis isnot completed. Consider a cross between a female heterozygousfor the bicoid mutation (bcd+/ bcd-) and a homozygousmale(bcd-/ bcd-). How is it possible for a male homozygous for the mutationto exist?
- A boy with Klinefelter syndrome (47,XXY) is born to a motherwho is phenotypically normal and a father who has the X-linkedskin condition called anhidrotic ectodermal dysplasia. The mother’sskin is completely normal with no signs of the skin abnormality.In contrast, her son has patches of normal skin and patchesof abnormal skin. Question: Which parent contributed the abnormal gamete?Female fruit flies homozygous for the X-linked white-eye alleleare crossed to males with red eyes. On very rare occasions, an offspringof such a cross is a male with red eyes. Assuming these rareoffspring are not due to a new mutation in one of the mother’s Xchromosomes that converted the white-eye allele into a red-eyeallele, explain how a red-eyed male arises.Two male mice, which we will call male A and male B, are bothphenotypically normal. Male A was from a litter that containedhalf phenotypically normal mice and half dwarf mice. The motherof male A was known to be homozygous for the normal Igf2 allele.Male B was from a litter of eight mice that were all phenotypicallynormal. The parents of male B were a phenotypically normal maleand a dwarf female. Male A and male B were put into a cage withtwo female mice that we will call female A and female B. FemaleA is dwarf, and female B is phenotypically normal. The parents ofthese two females were unknown, although it was known that theywere from the same litter. The mice were allowed to mate witheach other, and the following data were obtained:Female A gave birth to three dwarf babies and four normal babies.Female B gave birth to four normal babies and two dwarf babies.Which male(s) mated with female A and female B? Explain.