Expert: Dana Krempels, Ph.D. - 7/11/2007

Question
Dana I would to take the time to say my Thank you for the quick respond on my previous question. THANK YOU VERY MUCH
My second question is and I hope you can help me on this..

Righthandedness is dominant to left-handedness. Brown eyes is dominant to blue eyes. A homozygous right-handed, heterozygous brown-eyed man marries a blue-eyed, left-handed woman. They have 4 children.
A) What is the probability that the 2nd child was a girl    who was right-handed and had blue eyes?
B) What is the probability that the last child was a boy with brown eyes and left handed?

well I drew a Punnett square and bloged in llbbRRBb (16boxes). A)The probability that the 2nd child was a girl who was right-handed and had blue eyes is 12.5%
B) The probability that the last child was a boy with brown eyes and left handed is 0%.

I am not confident with this answer but if my concept is correct but I am wording differently may I please ask you to correct me the answer and how I should worded if my answer is correct? Thank you in advance


Answer
Dear Farduus,

If we designate the handedness gene as "R" and the eye color gene as "B", then the genotype of the man is RRBb, and the woman is rrbb.  The Punnett Square won't have 16 squares, but rather only two:  The man can make the following gametes:

RB and Rb

The woman can make only rb

The resulting offspring can ONLY be RrBb (right handed, brown eyes) or Rrbb (right handed, blue eyes).  There's a 50% chance of either (combined) phenotype.

There's also a 50% chance that any birth will be a boy or a girl.  

Using the Product Rule to determine the combined probability of the Rrbb genotype occurring *with* the female sex, you get 0.5 x 0.5, or 25%.  

Your answer to B is correct.

I hope this helps!

Dana