Question

steps of setting up a punnett square

• establish the genotype of each parent.
• place one parent genotype across the top
• place the other on the side.
• fill in the punnett square.

1. show the cross between a round (rr)

plant and a wrinkled (rr) plant.
rr × rr

1. identify the phenotypes and

their percentages: remember,
each box is worth 25%.
__ % are round
__ % are wrinkled.

Explanation:

Step1: Fill the Punnett Square

The parent with genotype \( RR \) will produce gametes with allele \( R \) (both gametes, since it's homozygous dominant). The parent with genotype \( rr \) will produce gametes with allele \( r \) (both gametes, since it's homozygous recessive). When we fill the Punnett Square:

• The top row has \( r \) and \( r \), and the left column has \( R \) and \( R \).
• The first cell (top - left) is \( R \times r = Rr \).
• The second cell (top - right) is \( R \times r = Rr \).
• The third cell (bottom - left) is \( R \times r = Rr \).
• The fourth cell (bottom - right) is \( R \times r = Rr \).

So all four cells in the Punnett Square have the genotype \( Rr \).

Step2: Determine Phenotypes

The allele \( R \) (round) is dominant over \( r \) (wrinkled). So any genotype with at least one \( R \) allele will show the round phenotype. Since all four genotypes (\( Rr \)) have the dominant \( R \) allele, all offspring will have the round phenotype.

To find the percentage:

• Total number of cells in the Punnett Square is 4.
• Number of cells with round phenotype (all 4, since \( Rr \) is round) is 4.
• Percentage of round plants \(=\frac{4}{4}\times100 = 100\% \).
• Number of cells with wrinkled phenotype: Since the wrinkled phenotype is only shown when the genotype is \( rr \) (homozygous recessive), and there are 0 cells with \( rr \) genotype.
• Percentage of wrinkled plants \(=\frac{0}{4}\times100 = 0\% \).

Answer:

100% are round, 0% are wrinkled.