B) The figure below shows a pedigree for a rare trait in humans. Affected indivi
ID: 193938 • Letter: B
Question
B) The figure below shows a pedigree for a rare trait in humans. Affected individuals are indicated by the filled-in symbols. The woman lt2 is pregnant, with lI-3 being the father. Her unborn child is represented by the "?". Q1.ls the trait inherited as a dominant trait or as a recessive trait? What is the best evidence for your answer? (6) Q2. What is the probability that both ll.2 and I1.3 are heterozygous for the trait? (6) Q3. What is the overall probability that the child will be unaffected by the trait and be a girl? (6) 1 4 4 1Explanation / Answer
1) Recessive trait - because, I-1&2 do not express the phenotype and II-1 expresses the phenotype. Hence, I-1&2 should be a carrier for II-1 to inherit the disorder. Also II-3&4 do not express the phenotype when I-3 expresses the phenotype.
2) For II-2 there is a 50% chance the person is heterozygous for the trait, Because, when both parents are carriers, 50% of the offsprings will be carrier, 25% will be normal and 25% will be affected. But for II-3, since one of the parent is recessive for both the alleles, 100% of the offsprings will be carriers.
3) Since the probability of II-2 being a carrier is 50%, then it is expressed as 2/4 - 1/2 and II-3 is 100% - 1. When II-2 and II-3 cross, the probability ratio for an unaffected offspring will be 1/2*1/4 = 1/8. And the probability for whether the offspring is male or female is 1/2. So, 1/8*1/2 = 1/16. So there is an 1/16 or 0.0625 chance that the offspring is female and unaffected.
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