The options are missing iin this question. The complete question is as follows.
Suppose 2 parents, a father with the genotype AaBbCcDcee and a mother with the genotype aabbCcDDEe, wanted to thave children. Assume each locus follows Mendelian inheritance patterns for dominance. What proportion of the offspring will ahve each of the following characteristics? Round your answers to two decimal places. (A) Same genotype as the father. (B) Phenotypically resemble the father. (C) same genotype as the mother. (D) Phenotypically resemble the mother. (E) Phenotypically resemble neither parent.
Answer: (A) P = 0.03125; (B) P=0.09375; (C) P=0.03125; (D) P=0.09375; (E) P=0.8125
Explanation: This is a polyhibrid crossing. As the punnet square, which shows each possibility for the offspring, is too big, we can separate each genotype and cross it:
For <u>allele A:</u> Aa x aa
a a
A Aa Aa
a aa aa
The probabilities are P (Aa) = 0.5 and P(aa) = 0.5
For <u>allele B:</u> Bb x bb
b b
B Bb Bb
b bb bb
The probabilities are P(Bb) = 0.5 and P(bb) = 0.5
For <u>allele C:</u> Cc x Cc
C c
C CC Cc
c Cc cc
Probabilities: P(CC) = 0.25; P(Cc) = 0.5; P(cc) = 0.25
For <u>allele D:</u> Dd x DD
D D
D DD DD
d Dd Dd
Probabilities: P(DD) = 0.5; P(Dd) = 0.5
For <u>allele E:</u> ee x Ee
E e
e Ee ee
e Ee ee
Probabilities: P(Ee) = 0.5 P(ee) = 0.5
(A) For the child to have the same genotype as the father, the child has to be AaBbCcDdee, so the probability is:
P1 = P(Aa).P(Bb).P(Cc).P(Dd).P(ee) = 0.5*0.5*0.5*0.5*0.5 = 0.03125
(B) For the child to phenotypically resemble the father, it has to be dominant for the allele A, because A is dominant and a is recessive; dominant for B; dominant for C; dominant for D and recessive for 3. This is a 'or' case, because the offspring can be AA and Aa at the same time. So, the probability is:
P = [P(Aa)+P(Aa)]*[P(Bb)+P(Bb)]*[P(CC)+P(Cc)]*[P(DD)+P(Dd)]*[P(ee)+P(ee)]
P = (0.25+0.25)*(0.25+0.25)*(0.25+0.25+0.25)+(0.25+0.25+0.25+0.25)*(0.25+0.25)
P2 = 0.09375
(C) To has the same genotype of the mother, it has to be: aabbCcDDEe.
So, the probability is:
P3 = P(aa).P(bb).P(Cc).P(DD).P(Ee) = 0.5*0.5*0.5*0.5*0.5 = 0.03125
(D) To phenotypically resemble the mother, it has to be recessive for A and B; dominant C, D and E:
P = [P(aa)+P(aa)]*[P(bb)+P(bb)]*[P(CC)+P(Cc)]*[P(DD)+P(Dd)]*[P(Ee)+P(Ee)]
P = (0.25+0.25)*(0.25+0.25)*(0.25+0.25+0.25)+(0.25+0.25+0.25+0.25)*(0.25+0.25)
P4 = 0.09375
(E) For the offspring to phenotypically resemble neither parent, it has to be what the parents aren't, so:
P = 1 - (P2+P4)
P = 1 - (2*0.9375)
P5 = 0.8125
