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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Solution Summary
Disease probability is examined. The chances the child will have the condition are determined.
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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 :)
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