Eye Color Each of two parents has the genotype brown>blue, which consists of the pair of alleles that determine eye color, and each parent contributes one of those alleles to a child. Assume that if the child has at least one brown allele, that color will dominate and the eyes will be brown. (The actual determination of eye color is more complicated than that.) a. List the different possible outcomes. Assume that these outcomes are equally likely. b. What is the probability that a child of these parents will have the blue>blue genotype? c. What is the probability that the child will have brown eyes?

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

hello from mrbilldoesmath!

answer:

the domain is the set { -2, 0, 2, 3}

discussion:

the domain is the set of x values in the ordered pairs (x,y). taking the x, or first component, of each ordered pair gives   -2, 0, 2, 3.

you,

mrb