Answer:
A) Brown-Brown ,Brown-Blue, Blue-Brown, Blue-Blue B) 1/4 =0,25 C)3/4=0,75
Step-by-step explanation:
Lets mother's "BROWN" is "BROWN-M",
mother's "BLUE" is " BLUE-M"
Lets father's "BROWN" is "BROWN-F" and
father's "BLUE " is "BLUE-F"
The kid can have the genotype as follows (list of possible outcomes) :
1. BROWN-M>BROWN-F ( received BROWN as from mother as from father)
2. BROWN-M>BLUE-F ( Received BROWN from mother and BLUE from father)
3. BLUE-M>BROWN-F ( Received BLUE from mother and Brown from father)
4. BLUE-M>BLUE-F ( Received BLUE as from mother as from father)
b) As we can see in a) only 1 outcome from 4 is BLUE-BLUE. So the probability of BLUE-BLUE genotype is
P(BLUE>BLUE)=1/4=0.25
c) As we know that if the child has at least one brown allele, that color will dominate and the eyes will be brown.
It means that outcomes BROWN-BROWN, BROWN-BLUE and BLUE-BROWN determine brown color of eye. So the number of these outcomes is 3. Total amount of outcomes is 4.
So probability that eyes are brown is P(Brown eyes)=3/4 =0.75
