Answer:
a. 
= 
b. i. 0.75
ii. 0.000061
iii. 0.012
iv. 0.17
c. 0.67
Explanation:
a. The expansion of the binomial (p + q)7 would be such that:
=
b. Both couples are heterozygous:
Aa x Aa
AA Aa Aa aa
Since A is dominant over a,
probability of having mole (aa) = 1/4
probability of not having moles = 3/4
<em>Therefore, the probability of the first child not having moles </em>= 3/4 or 0.75
ii. Let the probability of not having mole = p and the probability of having mole = q. From the binomial expansion:
=
<em>Probability that all of the children will have moles </em>= 
since p = 3/4 and q = 1/4
= 
= 0.000061
iii. <em>Probability that the first two children will have no moles and the last five will have moles</em> = 
= 
= 0.012
iv. <em>Probability that 4 will have no moles and 3 will have moles out of the 7 children</em> = 
= 
= 0.17
c. <em>Probability that the child born without moles is a carrier of the a-allele = probability of heterozygou</em>s.
From the cross in (b), the genotypes of those born without moles are AA and 2Aa. Therefore, the probability of not having moles and be Aa is:
= 2/3 or 0.67
