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# Excluded as a possibility, therefore the final probability of the woman having either aa or Aa genotypes is 1/3 and 2/3, respectively

 tarix 27.06.2016 ölçüsü 20 Kb.
 Albinism is a somatic recessive condition resulting from the inability to produce the dark pigment melanin in skin and hair. A man and woman with normal skin pigmentation have two children. The man has one albino parent; the woman has parents with normal pigmentation, but an albino brother. What is the probability that at least one of the children is albino? What is the probability of both children being albino? Using alleles A and a, since one of the man’s parents is albino, he must be a carrier (Aa) of the recessive allele (probability = 1). The woman has an albino brother, which means both her parents must be carriers (Aa). However, the woman (who is not albino) could have either an AA or Aa genotype. In the woman’s case the aa (albino) genotype must be excluded as a possibility, therefore the final probability of the woman having either AA or Aa genotypes is 1/3 and 2/3, respectively. Case 1: If the woman is Aa, the probability of at least one of the two children being albino from the cross (Aa × Aa) equals the probability that one child is albino (1/4) × the probability that the other child is normal (3/4) × 2 (must multiply by 2 because either birth order is possible for this outcome), plus the probability that both children are albino, (1/4) × (1/4) = 1/16. Therefore, the final probability that at least one child is albino equals 2(3/4)( 1/4) + (1/16) = 6/16 + 1/16 = 7/16 (43.75%). Case 2: If the woman is AA, the probability of at least one of the two children being albino from the cross (Aa × AA) is zero. As noted above, if Case 1 is true, then the probability of both children being albino (aa) equals (1/4) × (1/4) = 1/16 (6.25%).

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