In: Biology
What is the probability that the first offspring from this cross: Aa bb Dd Ee x AA bb Dd Ee will be a son with the genotype: Aa bb DD ee Type your answer as a decimal and round to 6 decimal places.
To solve this problem we'll use the concept of product of independent probabilities.
Probability that Aa will be present in the son,P(Aa), will be determined using a Punnet square.
A | a | |
A | AA | Aa |
A | AA | Aa |
Therefore P(Aa) = 2/4 = 1/2
Similarly P(bb) = 1 , as both the parents have bb genotype.
P(DD) =
D | d | |
D | DD | Dd |
d | Dd | dd |
Thus P(DD) = 1/4
similarly for P(ee) = 1/4 (Punnet square will be same as that of Dd x Dd)
Thus the P(Aa bb DD ee) = (1/2).(1).(1/4)2
= 1/32
= 0.031250