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At an autosomal gene locus in humans, the allele for brown eyes is dominant over the allele for blue eyes. At another gene locus, located on the X chromosome, a recessive allele produces colorblindness while the dominant allele produces normal color vision. A heterozygous brown-eyed woman who is a carrier of colorblindness has children with a blue-eyed man who is not colorblind. What is the probability that their first child will be a brown-eyed, colorblind male?

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If we name the gene for the eye color with A, then the genotypes for the brown color would be AA, Aa and the genotype for the blue eyes aa.

If we name the gene for the colorblindness with B, then the genotype XbXb is for colorblind woman while XBXb and XBXB are genotypes for the woman with normal vision.

A heterozygous brown-eyed woman who is a carrier of colorblindness has genotype: Aa XBXb and a blue-eyed man who is not colorblind:  aa XBY.

P:  Aa  XBXb  x  aaXBY

F1: (Aa Aa aa aa) ½ or 50% is the probability for the brown eye, ¼ or 25% is the probability that the child would be color blind male (XBXB XBY XbXB XbY)

This means:probability that their first child will be a brown-eyed, colorblind male is ½*1/4=  1/8= 12.5%

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