Question

1. the presence of freckles (f) is dominant to no freckles (f). jake and his mom have freckles, but his sister and dad do not have freckles. complete the punnett square that shows jakes family. dad phenotype probability (%) freckles: no freckles:

Explanation:

Step1: Determine parents' genotypes

Since Dad has no freckles, his genotype is \( ff \). Mom has freckles and has a child (sister) with no freckles (\( ff \)), so Mom must be \( Ff \).

Step2: Complete Punnett square

The Punnett square for \( Ff \) (Mom) \( \times ff \) (Dad):

	\( f \)	\( f \)
\( f \)	\( ff \)	\( ff \)

Step3: Calculate probabilities

For freckles (\( Ff \)): There are 2 out of 4 offspring with \( Ff \). Probability \(=\frac{2}{4}\times100\% = 50\%\)
For no freckles (\( ff \)): There are 2 out of 4 offspring with \( ff \). Probability \(=\frac{2}{4}\times100\% = 50\%\)

Answer:

Freckles: \( 50\%\)
No Freckles: \( 50\%\)