Question

punnett squares practice worksheet

1. in guinea pigs, long hair is dominant to short hair. a purebred short - haired guinea pig is crossed with a hybrid long - haired guinea pig. what are their genotypes? (use l)

pure - bred (homozygous)
hybrid (heterozygous)
set up and complete the punnett square.
what is the probability of the offspring having ll as their genotype?
what is the probability of the offspring being long - haired?
what is the probability of the offspring being short - haired?

1. in rabbits, black fur is dominant to white fur. if you cross a bb male with a bb female, what are the possible genotypes and phenotypes of the offspring? what is the percent chance for each type? what are the genotypic and phenotypic ratios? answer these questions in the table.

punnett square:

genotypes	possibilities	probabilities	ratios
phenotypes

1. in cabbage butterflies, white wings are dominant to yellow wings. if a ww butterfly is crossed with a ww butterfly, what are the possible genotypes and phenotypes of the offspring and the percent chance for each? what are the genotypic and phenotypic ratios? answer these questions in the table.

punnett square:

genotypes	possibilities	probabilities	ratios
phenotypes

Explanation:

Problem 3 Solution:

Step 1: Determine Parental Genotypes

The dominant allele for white wings is \( W \), and the recessive allele for yellow wings is \( w \). The WW butterfly is homozygous dominant, and the ww butterfly is homozygous recessive. So the parents' genotypes are \( WW \) (parent 1) and \( ww \) (parent 2).

Step 2: Set Up Punnett Square

The WW parent can only produce \( W \) gametes, and the ww parent can only produce \( w \) gametes. The Punnett square will have one row (for \( W \) gametes) and one column (for \( w \) gametes), but since both parents are homozygous, the cross is \( W \times w \) for all offspring. The Punnett square cells will all be \( Ww \).

Step 3: Determine Genotypes of Offspring

All offspring will have the genotype \( Ww \) because each gets a \( W \) from the WW parent and a \( w \) from the ww parent.

Step 4: Determine Phenotypes of Offspring

Since white wings (\( W \)) are dominant over yellow wings (\( w \)), the genotype \( Ww \) will express the dominant phenotype, so all offspring will have white wings.

Step 5: Calculate Probabilities and Ratios

• Genotypes: Only \( Ww \) is possible. So the probability for \( Ww \) is \( \frac{4}{4} = 100\% \), and the genotypic ratio is \( 1:0 \) (only \( Ww \), no other genotypes).
• Phenotypes: Only white wings are possible. The probability for white wings is \( 100\% \), and the phenotypic ratio is \( 1:0 \) (white wings : yellow wings).

Filling the Table:

Genotypes	Possibilities	Probabilities	Ratios
Phenotypes
White wings	All offspring	\( 100\% \)	\( 1:0 \)

Answer:

(for Problem 3):

• Genotypes of Offspring: All \( Ww \) (100% chance).
• Phenotypes of Offspring: All white - winged (100% chance).
• Genotypic Ratio: \( 1:0 \) (only \( Ww \)).
• Phenotypic Ratio: \( 1:0 \) (only white wings).

(For the table in the worksheet, fill as above.)