a)A couple plan on having six children. What is the probability that they will h

Question

a)A couple plan on having six children. What is the probability that they will have 3 boys and 3 girls?
(Answer in reduced fraction format)

b)Consider the cross AabbCcDD x AaBbccDd. What proportion of the progeny will have the same phenotype as the first parent (ie A_bbC_D_)?

c)Black skin colour is dominant to pink skin colour in pigs. Two heterozygous black pigs are crossed. What is the probability that their first and second offspring will have black skin?

d)A male and female are each heterozygous for both cystic fibrosis (CF) and phenylketonuria (PKU). Both conditions are autosomal recessive and they assort independently of one another. What proportion of the children of this couple will have neither condition?

Explanation / Answer

G- girl, B- boy P(G,B)= probability of boy and girl

Each possible birth order has P=2/16. That is, P(G,G,B)=P(G,B,G)=P(B,G,G)=2/16.
So, P(2G,1B)= 6/16 and P(1G,2B)= 6/16.

2/16 + 6/16 + 6/16 + 2/16 = 1 (100%)

b)Consider the cross AabbCcDD x AaBbccDd. What proportion of the progeny will have the same phenotype as the first parent (ie A_bbC_D_)?

1/64

c)Black skin colour is dominant to pink skin colour in pigs. Two heterozygous black pigs are crossed. What is the probability that their first and second offspring will have black skin?

offspring with a 3:1ratio of black to pink pigs

First child = P(B:p) = P(3:1) therefore probaility for first child black -1/3

Second child = P(B:p) = therefore probaility for second child black -2/3

Cross between heterozygous for cystic fibrosis and phenylketonuria

           CcPp XCcPp

1. Offspring will not have cystic fibrosis condition = 12/16

2. Offspring will not have phenylketonuria condition = 12/16

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