Hope that helps. -UF aka Nadia
Two brown-eyed parents have produced three blue-eyed sons . This means that the parents have heterogyzous genotype with one blue allele. Here, brown color is dominant over the blue color of the eye.
Let, the brown color eye allele be represented by "B" and the blue color allele be represented by allele "b"
So the genotype of parents would be "Bb"
If a cross is carried out between two parents with genotype "Bb" the following offspring will be produced -
Bb x Bb
BB, Bb, Bb, bb
So out of the four offspring three children have brown color eye.
Thus the probability of a child having brown color eye is
Option D,they both carry a recessive allele for blue eyes.
A recessive allele is expressed only when the recessive allele are in pairs.
In this case, if "B" represents brown color of eye and "b" represents blue color of eye, then the genotype of blue eyed sons would be "bb"
Thus, both the parents must be a carrier of recessive allele so that each of them can contribute one blue color allele to their son.
Hence option D is correct.
They are both heterozygous.
They both carry a recessive allele for blue eyes.