3 years ago
What is the probability of producing a child that will phenotypically resemble either one of the two parents in the following four crosses? How many phenotypically different kinds of progeny could potentially result from each of the four crosses?
Aa Bb Cc Dd * aa bb cc dd
1 Answers
Best Answer
caleighgrace13 Staff answered 3 years ago
caleighgrace13 Staff answered 3 years ago
One parent has the genotype Aa Bb Cc Dd while hte other has aa bb cc dd. If an ofFSPring looks like the heterozygous parent, it must inherit dominant alleles (A,B,C, and D) from the first parent, while the recessive alleles will be inherited from the second parent. The probability for this to occur is (1)/(2)* (1)/(2)*(1)/(2)*(1)/(2)=(1)/(16).
If an ofFSPring looks like the homozygous recessive parent, it must inherit recessive alleles (a,b,c, and d) from the first parent. The probability is (1)/(2) * (1)/(2) * (1)/(2) * (1)/(2)=(1)/(16).
Thus, the possibility for a child to look like either of his parents is (1)/(16)+(1)/(16).
The number of possible phenotypes is 2 for each of the 4 trails (for example Aa and aa for the first trait), which makes it 16 (2 2 2*2).