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(B) $1$. The pedigree shows an autosomal recessive trait because an affected female $(F_0)$ produces carrier daughters,and an affected male $(F_2)$ is

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$1/2$

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(B) $1$. The pedigree shows an autosomal recessive trait because an affected female $(F_0)$ produces carrier daughters,and an affected male $(F_2)$ is produced from the mating of a carrier female and an unaffected male (implying the father was also a carrier or the trait is recessive).$2$. Let the normal allele be $A$ and the disease allele be $a$. The affected individuals are $aa$.$3$. In the $F_2$ generation,the parents are a carrier female $(Aa)$ and an affected male $(aa)$.$4$. The cross is $Aa 	imes aa$. The possible genotypes of the offspring are $Aa$ (carrier) and $aa$ (affected) in a $1:1$ ratio.$5$. The probability of a child being a carrier $(Aa)$ is $1/2$.$6$. The probability of a child being affected $(aa)$ is $1/2$.$7$. The question asks for the probability of a child having no disease and being a carrier. Since a carrier $(Aa)$ is phenotypically normal (has no disease),the probability is $1/2$.

With the help of the given pedigree,find out the probability for the birth of a child having no disease and being a carrier (has the disease mutation in one allele of the gene) in the $F_3$ generation.

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