Question

1. the axial flower position is dominant over the terminal flower position in pea plants. fill in the punnett square and determine the expected genotypic and phenotypic ratios from crossing heterozygous and homozygous dominant parents.

genotypes: ________________________ genotypic ratio: ______
phenotypes: _______________________ phenotypic ratio: ______

Explanation:

Step 1: Determine Parent Genotypes and Gametes

• Heterozygous parent: \( Aa \) (produces gametes \( A \) and \( a \)).
• Homozygous dominant parent: \( AA \) (produces gametes \( A \) only).

Step 2: Fill Punnett Square

• Combine gametes from each parent into the Punnett square cells as shown above.

Step 3: Determine Genotypes of Offspring

• From the Punnett square, the offspring genotypes are \( AA \) (two cells) and \( Aa \) (two cells).

Step 4: Calculate Genotypic Ratio

• Count the number of each genotype: \( AA = 2 \), \( Aa = 2 \). Ratio \( AA:Aa = 2:2 = 1:1 \).

Step 5: Determine Phenotypes of Offspring

• Since \( A \) is dominant, both \( AA \) and \( Aa \) give axial flower position. So all 4 offspring have axial flowers.

Step 6: Calculate Phenotypic Ratio

• Number of axial flowers: 4, terminal flowers: 0. Ratio axial:terminal \( = 4:0 = 1:0 \).

Answer:

Punnett Square Filling:

Let the dominant allele for axial flower position be \( A \) and the recessive allele for terminal flower position be \( a \). The heterozygous parent has genotype \( Aa \) and the homozygous dominant parent has genotype \( AA \).

The gametes produced by the \( AA \) parent are all \( A \), and the gametes produced by the \( Aa \) parent are \( A \) and \( a \).

Filling the Punnett square:

	\( A \) (from \( AA \))	\( A \) (from \( AA \))
\( a \) (from \( Aa \))	\( Aa \)	\( Aa \)

Genotypes:

The genotypes of the offspring are \( AA \) and \( Aa \).

Genotypic Ratio:

The number of \( AA \) offspring is 2 and the number of \( Aa \) offspring is 2. So the genotypic ratio is \( AA:Aa = 2:2 = 1:1 \) (simplified).

Phenotypes:

Since \( A \) is dominant (axial flower position), both \( AA \) and \( Aa \) genotypes result in axial flower position. So all offspring have axial flower position.

Phenotypic Ratio:

All offspring show the axial flower phenotype. So the phenotypic ratio is axial:terminal \( = 4:0 = 1:0 \) (or just all axial, so we can say axial flower position : terminal flower position \( = 4:0 \) which simplifies to \( 1:0 \) or we can express it as all axial, so the ratio of axial to terminal is \( 4:0 \) or \( 1:0 \)).