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    Disease probability

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    A woman heterozygous for Huntington diseae reproduces with a man without the condition. Huntington disease is dominant; what are the chances the child will have the condition? The chances the child will not have the condition?

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    Hi there,

    50% chance for the child having the disease
    50% chance for the child not having the disease

    Because a child receives one gene from each parent. The heterozygous affected parent has one normal gene and one affected gene. This parent must give one of the two to his/her child, therefore, the probability is 50%.

    I hope this helps you :)


    The wording in your question leads to this answer...

    I'd be interested in your books ...

    Solution Summary

    Disease probability is examined. The chances the child will have the condition are determined.