Question

in creatures without color have this trait, it must inherit follow the directions below to use punnett square models to predict the probability that offspring will have albinism in each situation described. refer to the passing on genes sidebar on page 17 for additional help. scenario 1 you are a biologist who works with a species of highly endangered lizards at a zoo. you are planning a breeding program to produce baby lizards that will be released into the wild to boost the animal’s population. you have one female lizard and two male lizards at the zoo. the female lizard, named lisa, is an albino. one male, called lenny, isn’t albino, but he is a carrier of albinism with one recessive gene and one dominant gene. 1. complete the punnett square on the right to show the likely pigment traits of offspring produced by breeding lisa with lenny. punnett square diagram with lenny and lisa labels 2. use the equations below to determine the probability that lisa and lenny will have offspring that have albinism (aa) = number of offspring with trait ÷ 4 × 100 = ____% have traditional coloring (aa or aa) = number of offspring with trait ÷ 4 × 100 = ____%

Explanation:

Part 1: Complete the Punnett Square

Step 1: Determine Genotypes

• Lisa is albino, so her genotype is \( aa \) (recessive for albinism).
• Lenny is a carrier (not albino but has one recessive gene), so his genotype is \( Aa \) (one dominant, one recessive).

Step 2: Set Up Punnett Square

• Lisa’s alleles: \( a \) (top row, left) and \( a \) (top row, right).
• Lenny’s alleles: \( A \) (left column, top) and \( a \) (left column, bottom).

	\( a \)	\( a \)
\( a \)	\( aa \)	\( aa \)

Part 2: Calculate Probabilities

For Albinism (\( aa \)):

Step 1: Count Offspring with \( aa \)

From the Punnett square, there are 2 offspring with \( aa \) (albinism) out of 4 total.

Step 2: Apply Probability Formula

Probability (\%) for albinism:
\( \frac{\text{Number of } aa \text{ offspring}}{4} \times 100 = \frac{2}{4} \times 100 = 50\% \)

For Traditional Coloring (\( AA \) or \( Aa \)):

Step 1: Count Offspring with \( Aa \)

From the Punnett square, there are 2 offspring with \( Aa \) (traditional coloring) out of 4 total.

Step 2: Apply Probability Formula

Probability (\%) for traditional coloring:
\( \frac{\text{Number of } Aa \text{ offspring}}{4} \times 100 = \frac{2}{4} \times 100 = 50\% \)

Final Answers

• Probability of offspring with albinism: \( \boldsymbol{50\%} \)
• Probability of offspring with traditional coloring: \( \boldsymbol{50\%} \)

Answer:

Part 1: Complete the Punnett Square

Step 1: Determine Genotypes

• Lisa is albino, so her genotype is \( aa \) (recessive for albinism).
• Lenny is a carrier (not albino but has one recessive gene), so his genotype is \( Aa \) (one dominant, one recessive).

Step 2: Set Up Punnett Square

• Lisa’s alleles: \( a \) (top row, left) and \( a \) (top row, right).
• Lenny’s alleles: \( A \) (left column, top) and \( a \) (left column, bottom).

	\( a \)	\( a \)
\( a \)	\( aa \)	\( aa \)

Part 2: Calculate Probabilities

For Albinism (\( aa \)):

Step 1: Count Offspring with \( aa \)

From the Punnett square, there are 2 offspring with \( aa \) (albinism) out of 4 total.

Step 2: Apply Probability Formula

Probability (\%) for albinism:
\( \frac{\text{Number of } aa \text{ offspring}}{4} \times 100 = \frac{2}{4} \times 100 = 50\% \)

For Traditional Coloring (\( AA \) or \( Aa \)):

Step 1: Count Offspring with \( Aa \)

From the Punnett square, there are 2 offspring with \( Aa \) (traditional coloring) out of 4 total.

Step 2: Apply Probability Formula

Probability (\%) for traditional coloring:
\( \frac{\text{Number of } Aa \text{ offspring}}{4} \times 100 = \frac{2}{4} \times 100 = 50\% \)

Final Answers

• Probability of offspring with albinism: \( \boldsymbol{50\%} \)
• Probability of offspring with traditional coloring: \( \boldsymbol{50\%} \)