Question

true - breeding pea plants have either spherical seeds or dented seeds with spherical being the dominant phenotype. a genetic cross between two f1 - hybrid pea plants for spherical seeds will yield what percent of plants with a heterozygous genotype in the f2 generation?
○ 0%
○ 75%
○ 25%
○ 50%

Explanation:

Step1: Define Genotypes

Let the dominant allele (spherical seeds) be \( S \) and recessive (dented seeds) be \( s \). F1 - hybrid pea plants for spherical seeds have genotype \( Ss \) (heterozygous).

Step2: Set Up Punnett Square

When crossing two \( Ss \) plants, the Punnett square is:

	\( S \)	\( s \)
\( s \)	\( Ss \)	\( ss \)

Step3: Calculate Heterozygous Proportion

From the Punnett square, the genotypes in F2 are \( SS \) (1), \( Ss \) (2), \( ss \) (1). Total genotypes: \( 1 + 2+ 1=4 \). Heterozygous (\( Ss \)) count is 2. The proportion is \( \frac{2}{4} = 0.5 \) or 50%.

Answer:

50% (corresponding to the option with 50%)