Question

scenario 2
the second male at the zoo, called leo, is not albino or a carrier for albinism. he has two dominant traditional coloring...

1. complete the punnett square on the right to show the likely offspring traits of offspring produced by breeding lisa with leo.

punnett square diagram with leo (top) and lisa (side) labels, empty cells

1. use the equations below to determine the probability that lisa and leo 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:

To solve this problem, we first need to determine Leo's and Lisa's genotypes. Since Leo is not an albino and not a carrier, his genotype is \( AA \) (homozygous dominant). We assume Lisa's genotype (usually, if we're dealing with albinism which is a recessive trait, and if we're calculating probabilities, Lisa's genotype would likely be \( Aa \) or \( AA \); but let's assume the standard case where Lisa is a carrier or has a certain genotype. Wait, the problem says "Leo is not albino or a carrier, he has two dominant traditional coloring genes" – so Leo is \( AA \). Let's assume Lisa's genotype (maybe the problem implies Lisa's genotype, but since it's a Punnett square, let's complete it.

Step 1: Determine Genotypes

Leo: \( AA \) (so his gametes are \( A \) and \( A \))
Assume Lisa's genotype (let's say, for example, if we're looking at albinism, and if Lisa is a carrier, her genotype would be \( Aa \), but the problem might have Lisa's genotype as \( Aa \) or \( AA \). Wait, the Punnett square has Leo's gametes as two \( A \)s (since he's \( AA \)). Let's assume Lisa's gametes (maybe the problem has Lisa's genotype, but let's proceed.

Step 2: Complete the Punnett Square

Leo's gametes: \( A \) (top left), \( A \) (top right)
Lisa's gametes: Let's assume Lisa's genotype (maybe the problem has Lisa's genotype, but since it's a Punnett square, let's say Lisa's gametes are, for example, \( A \) and \( a \) (if she's a carrier \( Aa \)). But the problem says "Leo is not albino or a carrier, he has two dominant traditional coloring genes" – so Leo is \( AA \). Let's assume Lisa's genotype (maybe the problem is about albinism, where albinism is \( aa \), and traditional coloring is \( A \) dominant.

If Leo is \( AA \) (gametes \( A, A \)) and Lisa is, say, \( Aa \) (gametes \( A, a \)):

	\( A \) (Leo)	\( A \) (Leo)
\( a \) (Lisa)	\( Aa \)	\( Aa \)

Now, for the probability calculations:

For Albinism (\( aa \)):

Number of offspring with \( aa \): 0 (since all are \( AA \) or \( Aa \))
Probability: \( \frac{0}{4} \times 100 = 0\% \)

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

Number of offspring with \( AA \) or \( Aa \): 4 (all)
Probability: \( \frac{4}{4} \times 100 = 100\% \)

Final Answers

• Probability of offspring with albinism (\( aa \)): \( 0\% \)
• Probability of offspring with traditional coloring (\( AA \) or \( Aa \)): \( 100\% \)

Answer:

To solve this problem, we first need to determine Leo's and Lisa's genotypes. Since Leo is not an albino and not a carrier, his genotype is \( AA \) (homozygous dominant). We assume Lisa's genotype (usually, if we're dealing with albinism which is a recessive trait, and if we're calculating probabilities, Lisa's genotype would likely be \( Aa \) or \( AA \); but let's assume the standard case where Lisa is a carrier or has a certain genotype. Wait, the problem says "Leo is not albino or a carrier, he has two dominant traditional coloring genes" – so Leo is \( AA \). Let's assume Lisa's genotype (maybe the problem implies Lisa's genotype, but since it's a Punnett square, let's complete it.

Step 1: Determine Genotypes

Leo: \( AA \) (so his gametes are \( A \) and \( A \))
Assume Lisa's genotype (let's say, for example, if we're looking at albinism, and if Lisa is a carrier, her genotype would be \( Aa \), but the problem might have Lisa's genotype as \( Aa \) or \( AA \). Wait, the Punnett square has Leo's gametes as two \( A \)s (since he's \( AA \)). Let's assume Lisa's gametes (maybe the problem has Lisa's genotype, but let's proceed.

Step 2: Complete the Punnett Square

Leo's gametes: \( A \) (top left), \( A \) (top right)
Lisa's gametes: Let's assume Lisa's genotype (maybe the problem has Lisa's genotype, but since it's a Punnett square, let's say Lisa's gametes are, for example, \( A \) and \( a \) (if she's a carrier \( Aa \)). But the problem says "Leo is not albino or a carrier, he has two dominant traditional coloring genes" – so Leo is \( AA \). Let's assume Lisa's genotype (maybe the problem is about albinism, where albinism is \( aa \), and traditional coloring is \( A \) dominant.

If Leo is \( AA \) (gametes \( A, A \)) and Lisa is, say, \( Aa \) (gametes \( A, a \)):

	\( A \) (Leo)	\( A \) (Leo)
\( a \) (Lisa)	\( Aa \)	\( Aa \)

Now, for the probability calculations:

For Albinism (\( aa \)):

Number of offspring with \( aa \): 0 (since all are \( AA \) or \( Aa \))
Probability: \( \frac{0}{4} \times 100 = 0\% \)

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

Number of offspring with \( AA \) or \( Aa \): 4 (all)
Probability: \( \frac{4}{4} \times 100 = 100\% \)

Final Answers

• Probability of offspring with albinism (\( aa \)): \( 0\% \)
• Probability of offspring with traditional coloring (\( AA \) or \( Aa \)): \( 100\% \)