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Askelto2 Askelto2
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12 years ago
What is the probability that the children of a woman heterozygous for colorblindness an a man with normal color vision will be colorblind? explain... I honestly don't understand it. (:
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datoledo88datoledo88
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12 years ago
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12 years ago
Lets say that B is normal vision and b is colorblindness. The mother is therefore XB/Xb (since she is heterozygous and female). The father is XB/Y (since he has normal vision and male). So if they have children then they will have a 1 in 4 chance that their children will be colorblind. If you draw a Punnet Square for the traits you have:
XBY
XBXB/XBXB/Y
XbXB/XbXb/Y

Since color blindness is X-linked that means that males are more likely to be colorblind because they only carry 1 X gene. So the child that is Xb/Y will be colorblind. The other squares will have normal vision, but the one with XB/Xb is a carrier for colorblindness (like her mother).
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