Question

try this one:
a homozygous yellow seeded plant is crossed
with a green seeded plant.

1. establish the genotypes

of the parents:

1. set up the square
2. determine the phenotypic percentages:

____ % yellow
____ % green

Explanation:

Question 14

In genetics, for seed color, yellow is often dominant (let's assume allele \( Y \) for yellow, \( y \) for green). A homozygous yellow - seeded plant has two dominant alleles, so its genotype is \( YY \). A green - seeded plant, being recessive, must be homozygous recessive with genotype \( yy \) (since if it had a dominant allele, it would show the dominant phenotype).

To set up the Punnett square, we take the alleles from each parent. The homozygous yellow - seeded parent (\( YY \)) can only produce gametes with the \( Y \) allele. The green - seeded parent (\( yy \)) can only produce gametes with the \( y \) allele. So we place the alleles of one parent along the top and the alleles of the other parent along the side.

	\( Y \)	\( Y \)
\( y \)	\( Yy \)	\( Yy \)

Step 1: Analyze the genotypes from the Punnett square

From the Punnett square, all the offspring have the genotype \( Yy \). Since \( Y \) (yellow) is dominant over \( y \) (green), all the offspring will show the yellow phenotype.

Step 2: Calculate the percentages

The total number of offspring genotypes in the Punnett square is 4 (from the 2x2 square). The number of yellow - phenotypic offspring is 4, and the number of green - phenotypic offspring is 0. To find the percentage, we use the formula \( \text{Percentage}=\frac{\text{Number of a particular phenotype}}{\text{Total number of offspring}}\times100 \). For yellow: \( \frac{4}{4}\times 100 = 100\% \), for green: \( \frac{0}{4}\times100=0\% \).

Answer:

\( YY \) \( \times \) \( yy \)

Question 15