You can get BB, Bb, bB, and bb
Pr=1/4
a) Suppose both parents of an individual are Bb. What is the probability that a randomly selected child of this couple will have blue eyes?
I know how to calculate this using a punnett square, but this is for stats class...so i'm not sure how i'm supposed to work this out. do i use a venn diagram? my professor wasn't very clear explaining this. thanks!
Pr=1/4
According to the Law of Segregation, when two parents with the gene pair Bb reproduce, each parent randomly passes on one of their two genes to their child. In this case, since both parents are Bb, each parent has one dominant B gene and one recessive b gene.
To determine the probability of a child having blue eyes, you need to calculate the chances of the child inheriting two recessive b genes (bb). The probability of receiving a b gene from the first parent is 1 out of 2, and the same applies to the second parent. To find the overall probability, you multiply the individual probabilities together.
So, using the multiplication rule, the probability of having a child with blue eyes is:
1/2 (probability of receiving b gene from the first parent) * 1/2 (probability of receiving b gene from the second parent) = 1/4 or 25%.
Therefore, there is a 25% chance that a randomly selected child of this couple will have blue eyes. You don't need to use a Venn diagram for this specific question; rather, understanding the laws of genetics and using probability calculations will yield the answer.