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6. everyone in squidwards family has light blue skin, which is the domi…

Question

  1. everyone in squidwards family has light blue skin, which is the dominant trait for body color in his hometown of squid valley. his family brags that they are a \purebred\ line. he recently married a nice girl who has light green skin, which is a recessive trait. create a punnett square to show the possibilities that would result if squidward and his new bride had children. use b to represent the dominant gene and b to represent the recessive gene. purebred = homozygous

a. list the possible genotypes and phenotypes for their children.
b. what are the chances of a child with light blue skin? ____%
c. what are the chances of a child with light green skin? ____%
d. would squidwards children still be considered purebreds? explain!

Explanation:

Step1: Determine the genotypes of Squidward and his bride

Since Squidward's family is pure - bred (homozygous) for the dominant light - blue skin trait, his genotype is \(BB\). His bride has light - green skin (recessive trait), so her genotype is \(bb\).

Step2: Create the Punnett square

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

Step3: Analyze genotypes and phenotypes (for part A)

All of their children will have the genotype \(Bb\). Since \(B\) (light - blue) is dominant over \(b\) (light - green), all children will have the phenotype of light - blue skin.

Step4: Calculate probabilities (for parts B and C)

Since all 4 boxes in the Punnett square are \(Bb\) (light - blue), the probability of a child with light - blue skin (\(P(Bb)\)) is \(\frac{4}{4}=1 = 100\%\). The probability of a child with light - green skin (\(P(bb)\)) is \(\frac{0}{4}=0 = 0\%\)

Step5: Determine pure - bred status (for part D)

Pure - bred means homozygous (\(BB\) or \(bb\)). Since all children have the genotype \(Bb\) (heterozygous), they are not pure - bred.

Answer:

A. Genotype: \(Bb\) for all children; Phenotype: All have light - blue skin.
B. \(100\%\)
C. \(0\%\)
D. No. Because their genotype is \(Bb\) (heterozygous), not homozygous (\(BB\) or \(bb\)).