Answer:
C. 1/4
Explanation:
From the question, we can deduce that
R = abililty to roll tongue
r = inability to roll tongue
F = having Freckles
f = no Freckles
So, if a woman heterozygous for both traits , (i.e RrFf ) married a man with no freckles who couldn't roll tongue (i.e rrff )
The question ask us what is the probability that they would have a freckled, and tongue-rolling child. (i.e RrFf )
The probability is 1/4
The self cross of each parents goes thus,
RrFf = RF, Rf, rF, rf
rrff = rf, rf, rf, rf
Lets look at the dihybrid cross between both parents in order to determine the probability of the child that will have a freckled, and tongue-rolling traits.
RF, Rf, rF, rf × rf, rf, rf, rf
= RrFf, Rrff, rrFf, rrff
= RrFf, Rrff, rrFf, rrff
= RrFf, Rrff, rrFf, rrff
= RrFf, Rrff, rrff, rrff
RrFf (i.e freckled, tongue-rolling child) = 4/16
=1/4
i also attahed a file, that clearly depicts the cross using a punnet square.
