A man who cannot roll his tongue marries a woman who can. The woman's mother cannot roll her tongue. What are the genotypes and phenotypes of their children?

Here's the genetic cross for the F1 generation: Step 1: Define the alleles. Let R represent the dominant allele for tongue rolling. Let r represent the recessive allele for non-rolling. Step 2: Determine the genotypes of the parents. The man cannot roll his tongue, so his genotype is rr*. The woman can roll her tongue, but her mother cannot. Since her mother's genotype must be rr, the woman inherited an r allele from her mother. As the woman can roll her tongue, she must have an R allele from her father. Therefore, the woman's genotype is Rr*. Step 3: Set up the Punnett square for the cross between the man (rr) and the woman (Rr). |c|c|c| Gametes & R & r \\ r & Rr & rr \\ r & Rr & rr \\ Step 4: Determine the F1 generation genotypes and phenotypes. Genotypes: Rr: 2 out of 4, or 50% rr: 2 out of 4, or 50% Phenotypes: Rollers (Rr): 2 out of 4, or 50% Non-rollers (rr): 2 out of 4, or 50% The F1 generation will have a 1:1 ratio of heterozygous rollers (Rr) to homozygous non-rollers (rr). Phenotypically, there will be a 1:1 ratio of rollers to non-rollers.