Question

spooner and wendy both have normal eyes and heterozygous genotypes (ee). spooner e e wendy e ee ee e ee ee phenotype probability (%) normal eyes: heart eyes:

Explanation:

Step1: Count total genotypes

In the Punnett square, there are 4 possible genotypes: \( EE \), \( Ee \), \( Ee \), \( ee \). So total cases \( n = 4 \).

Step2: Count normal eyes genotypes

Normal eyes are associated with genotypes having at least one dominant allele \( E \) (i.e., \( EE \), \( Ee \), \( Ee \)). So number of normal eyes cases \( m_{normal}= 3 \).

Step3: Calculate normal eyes probability

Probability of normal eyes \( P_{normal}=\frac{m_{normal}}{n}\times100\%=\frac{3}{4}\times100\% = 75\% \).

Step4: Count heart eyes genotypes

Heart eyes are associated with genotype \( ee \). Number of heart eyes cases \( m_{heart}= 1 \).

Step5: Calculate heart eyes probability

Probability of heart eyes \( P_{heart}=\frac{m_{heart}}{n}\times100\%=\frac{1}{4}\times100\% = 25\% \).

Answer:

Normal eyes: \( 75\% \)
Heart eyes: \( 25\% \)