Question

the ability to roll ones tongue is a dominant trait (shown below). if a dad who is a homozygous dominant tongue-roller (tt) has a child with a mom who is a homozygous recessive non-tonue-roller (tt), what is the phenotypic probability that they could have a child that cannot roll its tongue? work the punnett square and then choose the correct answer based on your results.

Explanation:

Step1: Set up Punnett Square

Dad's genotype: \( TT \), so he can only pass on \( T \) alleles. Mom's genotype: \( tt \), so she can only pass on \( t \) alleles. The Punnett Square will have \( T \) (from dad) in both top cells and \( t \) (from mom) in both side cells.

	\( T \)	\( T \)
\( t \)	\( Tt \)	\( Tt \)

Step2: Analyze Genotypes and Phenotypes

All offspring have the genotype \( Tt \). Since \( T \) (tongue - rolling) is dominant, any offspring with at least one \( T \) allele will be a tongue - roller. A non - tongue - roller would need the genotype \( tt \), but none of the offspring have this genotype.

Step3: Calculate Probability

The number of offspring with the non - tongue - rolling phenotype (\( tt \)) is 0 out of 4 possible offspring. So the probability is \( \frac{0}{4}=0 \).

Answer:

The phenotypic probability that they could have a child that cannot roll its tongue is \( 0 \) (or \( 0\% \)).