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alexis_jordyn13

  • 2 years ago

Duchene muscular dystrophy, or DMD, is a recessive, x-linked disease. What would you expect to see in the offspring of a woman who has DMD and a man who does not have the disease? A. All of their sons and none of their daughters would have the disease. B. None of their children would have the disease, but they would all be carriers. C. All of their daughters and none of their sons would have the disease. D.Their daughters would have the disease and their sons would be carriers.

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  1. ataly
    • 2 years ago
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    Hi alexis_jordyn13, you first need to deduce the genotype of parents from their phenotype. then the possible genotype for children and finally their phenotype.

  2. DrAmaQueen
    • 2 years ago
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    Duchenne muscular dystrphy(DMD) is an X linked recessive trait Going back in our genetic lessons we should recall that females have two xs i.e XX(one X from the mother and another from the father) while males are XY(an X from the mother and a Y from the father), dystrophin is a protein present on the short arm of the X chromosome, it is necessary for strenghtening the muscle cells, in the case of DMD this gene is mutated hence no dystrophin is produced, causing weakness and subsequently death of the muscle cells. DMD affects mostly males and in extremely rare cases females who are mainly just carriers of the gene but do not exhibit the trait, this is because females are XX even when they have inherited a faulty X from their mother, they can make up for it with the X from their father which would have a normal dystrophin gene, but with males they have only one X, if this X is affected, they can't make up for it because the Y chromosome does not have the dystrophin gene hence they will manifest DMD. Since the gene is recessive and the mother is infected let me give her XnXn(N represents the normal dystrophin gene and n would represent the mutated dystrophin gene, the mother would be nn because she would have to have the mutation on both genes to be infected) and the father would be XNY since he is neither infected nor a carrier, by drawing the table, we would have a result like this..........................XNXn(female carrier), XnY(infected male), XNXn(female carrier), XnY(infected male).... This would mean that all their male offsprings would be infected with the disease while the females would only be carriers but not infected so A would be a more appropriate answer............ hope its explanatory

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