Question

1. in humans, being a tongue roller (r) is dominant over non - roller (r). a man who is a non - roller marries a woman who is heterozygous for tongue rolling.

father’s phenotype ____ mother’s phenotype ____
father’s genotype ____ mother’s genotype ____
what is the probability of this couple having a child who is a tongue roller? ______

1. brown eyes in humans are dominant to blue eyes. a brown - eyed man, whose mother was blue - eyed, marries a brown - eyed woman whose father had blue eyes.

what is the probability that this couple will have a blue - eyed child? ______
answer the following questions by completing the punnett square.

Explanation:

Step1: Determine genotypes and phenotypes for question 17

• Since non - roller (r) is recessive, father's phenotype is non - roller and genotype is \( rr \).
• Mother is heterozygous for tongue rolling, so her genotype is \( Rr \) and phenotype is tongue roller.
• Using Punnett square:

	\( R \)	\( r \)
\( r \)	\( Rr \)	\( rr \)

• Probability of tongue roller (\( Rr \)) is \( \frac{1}{2} \)

Step2: Determine genotypes for question 18

• Let \( B \) (brown) be dominant and \( b \) (blue) be recessive.
• Man's mother was blue - eyed (\( bb \)), so man (brown - eyed) has genotype \( Bb \).
• Woman's father was blue - eyed (\( bb \)), so woman (brown - eyed) has genotype \( Bb \).
• Using Punnett square:

	\( B \)	\( b \)
\( b \)	\( Bb \)	\( bb \)

• Probability of blue - eyed (\( bb \)) is \( \frac{1}{4} \)

Answer:

17.
Father’s phenotype: non - roller
Mother’s phenotype: tongue roller
Father’s genotype: \( rr \)
Mother’s genotype: \( Rr \)
Probability of tongue - roller child: \( \frac{1}{2} \)

1. Probability of blue - eyed child: \( \frac{1}{4} \)