Question

1. assume that one of squidwards sons, who is heterozygous for the light blue body color, married a girl that was also heterozygous. create a punnett square to show the possibilities that would result if they had children.

a. list the genotypes with the phenotypes of the possible offspring.
b. what are the chances of a child with light blue skin?
c. what are the chances of a child with light green skin?

Explanation:

Part a: Punnett Square and Genotype-Phenotype List

Step 1: Define Alleles

Let's assume the allele for light blue body color is \( B \) (dominant) and for light green is \( b \) (recessive). Both parents are heterozygous, so their genotype is \( Bb \).

Step 2: Set Up Punnett Square

	\( B \)	\( b \)
\( b \)	\( Bb \)	\( bb \)

Step 3: Determine Genotypes and Phenotypes

• \( BB \): Homozygous dominant, phenotype = light blue (since \( B \) is dominant for blue)
• \( Bb \): Heterozygous, phenotype = light blue (dominant allele \( B \) expressed)
• \( bb \): Homozygous recessive, phenotype = light green (recessive allele \( b \) expressed)

Step 4: List Genotypes and Phenotypes

• \( BB \): Light blue
• \( Bb \): Light blue
• \( bb \): Light green

Part b: Chances of Light Blue Skin

Step 1: Count Light Blue Genotypes

From the Punnett square, \( BB \) and \( Bb \) are light blue. There are 3 such genotypes out of 4 total.

Step 2: Calculate Probability

Probability \( = \frac{\text{Number of light blue genotypes}}{\text{Total genotypes}} = \frac{3}{4} = 75\% \)

Part c: Chances of Light Green Skin

Step 1: Count Light Green Genotypes

The \( bb \) genotype is light green. There is 1 such genotype out of 4 total.

Step 2: Calculate Probability

Probability \( = \frac{\text{Number of light green genotypes}}{\text{Total genotypes}} = \frac{1}{4} = 25\% \)

Final Answers

a. Genotypes and Phenotypes:
\( BB \) - Light blue; \( Bb \) - Light blue; \( bb \) - Light green

b. Chances of light blue skin: \( \boldsymbol{\frac{3}{4}} \) (or 75%)

c. Chances of light green skin: \( \boldsymbol{\frac{1}{4}} \) (or 25%)

Answer:

Part a: Punnett Square and Genotype-Phenotype List

Step 1: Define Alleles

Let's assume the allele for light blue body color is \( B \) (dominant) and for light green is \( b \) (recessive). Both parents are heterozygous, so their genotype is \( Bb \).

Step 2: Set Up Punnett Square

	\( B \)	\( b \)
\( b \)	\( Bb \)	\( bb \)

Step 3: Determine Genotypes and Phenotypes

• \( BB \): Homozygous dominant, phenotype = light blue (since \( B \) is dominant for blue)
• \( Bb \): Heterozygous, phenotype = light blue (dominant allele \( B \) expressed)
• \( bb \): Homozygous recessive, phenotype = light green (recessive allele \( b \) expressed)

Step 4: List Genotypes and Phenotypes

• \( BB \): Light blue
• \( Bb \): Light blue
• \( bb \): Light green

Part b: Chances of Light Blue Skin

Step 1: Count Light Blue Genotypes

From the Punnett square, \( BB \) and \( Bb \) are light blue. There are 3 such genotypes out of 4 total.

Step 2: Calculate Probability

Probability \( = \frac{\text{Number of light blue genotypes}}{\text{Total genotypes}} = \frac{3}{4} = 75\% \)

Part c: Chances of Light Green Skin

Step 1: Count Light Green Genotypes

The \( bb \) genotype is light green. There is 1 such genotype out of 4 total.

Step 2: Calculate Probability

Probability \( = \frac{\text{Number of light green genotypes}}{\text{Total genotypes}} = \frac{1}{4} = 25\% \)

Final Answers

a. Genotypes and Phenotypes:
\( BB \) - Light blue; \( Bb \) - Light blue; \( bb \) - Light green

b. Chances of light blue skin: \( \boldsymbol{\frac{3}{4}} \) (or 75%)

c. Chances of light green skin: \( \boldsymbol{\frac{1}{4}} \) (or 25%)