Question

Please help !!

Answer 1

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

Answer 2

Answer:

what is the probability of this couple having a child who is a non - roller ?

bb X Cc= bC, bc, bC and bc

The probability  of the couple having a child who is a non-roller is 1/2

Explanation:

homozygous bb= non-roller father

heterozygous Cc= roller

From the crossing above we have equal halves of both roller and non-roller

1/2 roller

1/2 non-roller while the becomes heterozygous and roller is also heterozygous