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arcanos616p53hw1
06.05.2020 •
Biology
Both Martha and her mother are normal (not color-blind or anemic) and Martha's mother is not a carrier for either disorder. Martha's father is color-blind and anemic. George and Martha marry and have a daughter with normal phenotype named Kate. Question A. If George and Martha have another child, what is the probability the child will be a son with normal phenotype?
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Ответ:
1/4
Explanation:
The probability of having a son with normal phenotype would be 1/4.
Both color blindness and anemia are X-linked and are linked together. The genes for X-linked disorders are found on the sex chromosomes with the male being XY and female being XX. The male child gets his X chromosome from the mother and Y from the father while the female child gets one of her X chromosome form the father and the other from the mother.
For recessive disorders, two recessive alleles are needed by the female on the two X chromosome to become affected while only one allele is needed on one X chromosome of the male to become affected.
Assuming the allele for color blindness is a and that of anemia is b.
George is color blind but not anemic, the genotype would be![X^{aB}Y](/tpl/images/0647/7853/caeec.png)
Martha is normal but her dad is color blind and anemic, this means that Martha is a carrier for both disorders with genotype![X^{AB}X^{ab}](/tpl/images/0647/7853/12644.png)
Crossing the two genotypes
offspring:![X^{AB}X^{aB}, X^{aB}X^{ab}, X^{AB}Y, X^{ab}Y](/tpl/images/0647/7853/97fad.png)
Probability of having a son = 1/2
Probability of producing a normal phenotype child = 1/2
Hence,
Probability of a son with normal phenotype = 1/2 x 1/2 = 1/4.
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