In question 5 I didn\'t understand how we got 0.16 tried to use the equation but

Question

In question 5 I didn't understand how we got 0.16 tried to use the equation but didn't get the same answer. 5) Cystic fibrosis occurs 2% of the time in a large population. Two parents who are from this population and are heterozygous carriers of the recessive disease cystic fibrosis produce a phenotypically normal child. This child breeds with a phenotypically normal individual from the same population. What is the probability that their first child will be affected by the disease? A. 0.50 B. 0.42 C. 0.04 D, 0.16 0.01 6) If 600 offspring resulted from a cross of two A/a, B/b individuals, how many of the offspring would have the parental phenotype? C oepr A 338 121 C. 600 D. 452 E. 267

Explanation / Answer

Answer 5: A. 338 (As explained below)

The parents have dominant A and B alleles and so all children with dominant A and B alleles will show the same phenotype as their parents. There are 9 such offsprings with dominant A and B alleles out 16 i.e. 56.25% have the phenotype of parents.

So, of the 600 offspring, there will 56.25% of 600 i.e. 338 with parental phenotype.

AB Ab aB ab AB AABB AABb AaBB AaBb Ab AABb AAbb AaBb Aabb aB AaBB AaBb aaBB aaBb ab AaBb Aabb aaBb aabb

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