Answer:
2/3
Explanation:
This question involves a single gene coding for the possession or not of Cystic Fibrosis (CF). Since, the disorder is inherited as a recessive pattern, it means the C allele (no CF) is dominant over c allele (CF). This means that only an individual with (cc) genotype can be affected.
However, if a cross between two unaffected parents produced a child with CF, it means both parents are heterozygous or carriers of the trait i.e. Cc genotype. Thus, using a punnet square, a cross between them will give rise to four possible offsprings with CC, Cc, Cc and cc genotypes.
Two children were given birth to, with one having the disorder (cc) and one not having. The CC, Cc, and Cc genotypes will not be affected but Cc will be a carrier. Therefore, the probability of having the unaffected children be a carrier is 2 out of the normal 3 children i.e. 2/3.
The answer is c.
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Answer:
An electron microscope
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