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ild ild
wrote...
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12 years ago
Jamal is colour blind. Anna is not colour blind. Anne is pregnant with their first child. Will their child be colour blind? What will be the child?s genotype and phenotype? Make a Punnett Square using B for not colour blind and b for colour blind.
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wrote...
12 years ago
Jamal's genotype is bY, but you can't make a definite Punnett square, because Anna's genotype is not known.  Is she BB or Bb?
wrote...
12 years ago
first of all you have to consider that color blindness can be an x-linked mutation, thus if the kid might not be color blind because the mother is not. The other things we don't know if the genes are homozygous or heterozygous. Since we don't know I will assume that colour blindness is heterozygous Bb and not color blind is homozygous BB.
you will "multiply" BB*Bb=BB, BB, Bb, Bb.
There fore I would say that the kid can have a 50% chance of caring (genotype) the trait but it will probably never express the trait (phenotype) because the mother is not a carrier, and the dominant trait is the not color blind.
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