For the purpose of this problem assume that in humans eye co

For the purpose of this problem assume that in humans eye color is controlled by a single gene and that the allele for brown eyes is dominant to the blue-eye allele. Two brown-eyed parents planning a pregnancy already have a blue-eyed daughter. PLEASE NOTE: gender (being male or female) is a SECOND trait (like a second Punnett square), but no need for a Punnett square because the chance that an offspring will be male or female is always 1/2. Assign letters to each allele: BROWN allele = & BLUE allele = The parent\'s genotypes are: x Eye-color Punnett square for these two parents (show fractions): What is the probability that these parents will have a brown-eyed boy? (the child has brown eyes AND is male) For the same parents as in (a), what is the probability that they their next children might be (in this order): two blue-eyed girls and the third is a brown-eyed boy? (please show all math/work) For the same parents as in (a), what is the probability that their next children might be: two brown-eyed daughters or three blue-eyed sons?

Solution

Let us assign letters to the brown eye allele as \'B\' and the blue eyes allele as \'b\'

Given that brown is dominant over blue, and the both parents have brown eyes.

So the possibility that the genotype of the parents could be one of either \'BB\' or \'Bb\'.

Note that, it could be \'Bb\', because they have a daughter with blue eyes. So that means, both the parents have one allele of each of the recessive Blue allele.

Hence, the parents genotype is \'Bb\' and \'Bb\'. The cross becomes Bb * Bb which yields, BB, Bb, Bb, bb.

BB=Bb=Bb=Brown eyed bb= blue eyed

(a) Probability of these parents having a BROWN EYED BOY --

Probabality of having a brown eyes child = 3/4 ( as only 1 chance in 4 chances gives you a blue eyes child)

Probablity of having a boy= 1/2

Combining both these probabilities, the fina probability is 3/8.

(b) Probability of getting Two blue eyed girls and the third brown eyed boy in that order --

The simple trick for this is -

P= (p(blue-eye)*p(girl))*( p(blue-eye)*p(girl))*(p(brown-eye)*p(boy))

They want this particular combination and not any of these individual ones, hence we multiply the proabability.

we know that p(blue eye ) is 1/4 and p(brown eye) is 3/4 [from the prunnete square of Bb and Bb]

and p(girl)=p(boy)=1/2

substituting these values, P = 3/512

(c) Probability of getting two brown eyed daughters or three blue eyes sons--

Note that here the parents do not mind either the brown eyed daughters or the blue eyed sons, so we add up the probabilites.

P= (p(brown)p(girl))^2 + (p(blue)p(boy))^3

P= (3/4*1/2)^2 + (1/4*1/2)3

P= 9/64 + 1/512 = 73/512.