D)When parents are a carrier and an unaffected O', then on average, 1/2 of the daughters will be carriers and 1/2 of the sons will be affected If the trait is rare then the vast majority of affected individuals will be male which is the hallmark of x-linked traits If)Affected sons inherit the allele from mother Maternal uncles often affected Since inherited only from mother, inbreeding doesn't increase the probability of an affected o Conditional probabilities Consider the following pedigree of a recessive trait female male ? plaffected child)= p(mother carrier and father carrier and affected child) =2/3x2/3×14=1/ However if they have a child that is affected we must reassess the probability that their next child will be affected p(both parents carriers)=1. So, p(next child affected)=1/4 This example shows how probability calculations are based on information. The probability changes not because the parents have changed but because our information about them hasi) When parents are a carrier O and an unaffected O , then on average, 1/2 of the daughters will be carriers and 1/2 of the sons will be affected. If the trait is rare then the vast majority of affected individuals will be male which is the hallmark of X-linked traits. ii) Affected sons inherit the allele from mother • Maternal uncles often affected • Since inherited only from mother, inbreeding doesn’t increase the probability of an affected O . Conditional probabilities Consider the following pedigree of a recessive trait. = female = male ? p(affected child) = p(mother carrier and father carrier and affected child) = 2/3 x 2/3 x 1/4 = 1/9 However, if they have a child that is affected we must reassess the probability that their next child will be affected. p(both parents carriers) = 1. So, p(next child affected) = 1/4 This example shows how probability calculations are based on information. The probability changes not because the parents have changed but because our information about them has