What is the probability that a family of five children will
consistof a blue, a brown, a blue, a blue and a brown -eyed child
in thisorder. assume the parents are those in problem 1.
revised below:
1) two brown eyed parents are each heterozygous for a recessiveblue
eyed gene.

The problem tells you that both parents are brown eyes with heterozygous genes.

That tells you both have the recessive allele but are phenotypically brown.

If brown is E, and blue is e then both parents are Ee for eye color.

If you draw a Punnett table with the cross you'll find:

Ee x Ee = EE Ee eE ee

You have 25% chances of EE, phenotypically brown, 50% Ee heterozygous dominant for brown, and 25% recessive for blue.

phenotypically talking, you have 75% chances of having a brown eye child and 25% chances of having a blue eye child.

when it is at an intermediate population density

the correct answer to this question is #5