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.
Answer:
what exactly do you need help with? like whats the question?
Yes I would, the map show that the continents were originally in one placement whereas now they are way different placement then before.
The abdominal wall muscle that is most superficial of the oblique muscles is the rectus abdominis (or A)
Pond over growing is the answer because that takes much more time
