Answer: Father's phenotype = non-roller
Father's genotype = rr (homozygous for non-rolling)
Mother's phenotype: tongue-roller
Mother's genotype: Rr ( heterozygous for tongue-rolling)
In a cross between a man who is a non-roller (rr), and a woman who is heterozygous for tongue-roller (Rr), the possible genotypes of the children are 2 Rr and 2 rr. Therefore, the probability of having a child who is a non-roller is 50/100 = 1/2
The probability of having a child who is a tongue-roller is 50/100 = 1/2
Explanation:
The mother is heterozygous for tongue rolling, having a genotype of Rr will manifest outwardly as a tongue-roller because the gene for being a tongue-roller (R) is dominant over the gene for being a non-roller (r). In a cross between the man and the woman, 50% of the children will be heterozygous for tongue-rolling (Rr) while 50% will be homozygous for non-tongue rolling (rr).
See the attached punnet square for more information