Question

in clowns, a red nose is dominant to a blue nose. bozo has a red nose and so do both of his parents. his brother twisty has a blue nose. what are the genotypes of the parents?
○ aa x aa
○ aa x aa
○ aa x aa
○ aa x aa
○ aa x aa
question 17
josh has a big second toe, which is a dominant trait. he is homozygous dominant for the trait. his wife, irma, has a short second toe. is it possible for them to have a child with a short second toe like irma?
○ yes
○ no
○ no answer text provided.
○ no answer text provided.
question 18
pku (phenylketonuria) is a dominant disease, where just having 1 copy of the dominant pku allele will result in the individual having the disease. krystal does not have pku, but both her parents do. what are all their genotypes?
○ krystal: pp | parents: pp x pp
○ krystal: pp | parents: pp x pp
○ krystal: pp | parents: pp x pp
○ krystal: pp | parents: pp x pp

Explanation:

Question about Clown Nose Genetics (First Question on the Right)

1. Understand Dominance: Red nose (A) is dominant over blue nose (a).
2. Analyze Phenotypes: Bozo and both parents have red noses (dominant), brother has blue nose (recessive, genotype \( aa \)).
3. Determine Parent Genotypes: For the brother to have \( aa \), both parents must contribute an \( a \) allele. Since parents have red noses (dominant), they can't be \( aa \) (would be blue). So parents must be heterozygous (\( Aa \)) to pass \( a \) and show red (dominant). Wait, no—wait, Bozo has red, parents have red, brother has blue (\( aa \)). So parents must each have at least one \( a \) (to give to brother) and at least one \( A \) (to be red). So parents are \( Aa \times Aa \)? Wait, no, the options: Let's re - check. Wait, the options are AA x Aa, AA x aa, Aa x Aa, Aa x aa, aa x aa. Wait, brother has \( aa \), so each parent must give an \( a \). So parents can't be AA (can't give \( a \)). So possible parent genotypes: both must have \( a \) allele. So if parents are \( Aa \times Aa \), they can have a child with \( aa \) (brother). But wait, Bozo has red nose—could be \( AA \) or \( Aa \). But since brother is \( aa \), parents must be \( Aa \) each? Wait, no, the options: Wait the first option is AA x Aa, second AA x aa, third Aa x Aa, fourth Aa x aa, fifth aa x aa. Wait, aa x aa would have all blue noses, so eliminate. AA x aa: children would be \( Aa \) (all red), no blue - nosed children, but brother has blue nose, so eliminate. Aa x aa: children would be \( Aa \) (red) and \( aa \) (blue) in 1:1. But parents: if one is \( Aa \) (red) and one \( aa \) (blue), but the problem says both parents have red noses, so \( aa \) parent would be blue—so eliminate. AA x Aa: children would be \( AA \) or \( Aa \) (all red), no blue - nosed children, eliminate. Wait, that can't be. Wait, maybe I made a mistake. Wait, the problem says "Bozo has a red nose and so do both of his parents. His brother Twisty has a blue nose." So brother is \( aa \), so parents must each have an \( a \) allele (to pass to brother) and be red (so at least one \( A \)). So parents must be \( Aa \times Aa \) (because \( Aa \times Aa \) can produce \( AA \), \( Aa \), \( aa \) children—so red - nosed parents (\( Aa \)) can have red - nosed (\( AA \), \( Aa \)) and blue - nosed (\( aa \)) children). But the option is "Aa x Aa"? Wait, the third option is "Aa x Aa". Wait, but let's check the options again. Wait the options are:

• AA x Aa

• AA x aa

• Aa x Aa

• Aa x aa

• aa x aa

Wait, if parents are \( Aa \times Aa \), then:

• Genotypes of children: \( AA \) (25%), \( Aa \) (50%), \( aa \) (25%). So brother (aa) is possible, parents are red (Aa), Bozo (red) could be AA or Aa. So the correct option should be Aa x Aa? But wait, the problem says "Bozo has a red nose and so do both of his parents"—so parents are red, which \( Aa \) is red. So the answer is Aa x Aa (third option). Wait, but earlier I thought maybe I messed up. Let's re - do:

1. Brother has blue nose (\( aa \)) → both parents must have at least one \( a \) allele (to pass to him).

1. Parents have red noses → they must have at least one \( A \) allele (since red is dominant).

1. So possible parent genotypes: \( Aa \) (has \( A \) and \( a \)) or \( aa \) (but \( aa \) is blue, so parents can't be \( aa \)). So parents must be \( Aa \) (since they have red noses and can pass \( a \)).

1. So parents are \( Aa \times Aa \), which is option C (Aa x Aa).

1. Genotype of Josh: Big second toe is dominant, Josh is homozygous dominant (\( TT \) - let's use \( T \) for dominant, \( t \) for recessive). So Josh's genotype: \( TT \).

1. Genotype of Irma: Short second toe is recessive, so her genotype: \( tt \).

1. Possible Offspring Genotypes: When Josh (\( TT \)) and Irma (\( tt \)) have a child, the child will get one \( T \) from Josh and one \( t \) from Irma. So all children will have genotype \( Tt \), which is the dominant (big toe) phenotype.

1. Conclusion: Since all children will have \( Tt \) (big toe), they cannot have a child with short second toe (recessive, \( tt \)). So the answer is no.

1. Understand PKU Inheritance: PKU is dominant, and having 1 dominant allele (\( P \)) causes the disease. Krystal does not have PKU, so her genotype must be \( pp \) (homozygous recessive, no dominant allele).

1. Determine Parent Genotypes: Both parents have PKU (so they have at least one \( P \) allele) but Krystal (\( pp \)) got one \( p \) from each parent. So each parent must have a \( p \) allele (to pass to Krystal) and at least one \( P \) allele (to have PKU). So parents' genotypes: \( Pp \) (heterozygous dominant—has \( P \) (causes PKU) and \( p \) (passed to Krystal)).

1. Check the Options: The options are Krystal: pp | Parents: Pp x Pp (first option), Krystal: pp | Parents: Pp x pp (second), Krystal: pp | Parents: PP x pp (third), Krystal: pp | Parents: pp x pp (fourth). Let's analyze:

• Option 1: Parents \( Pp \times Pp \). Possible children: \( PP \) (PKU), \( Pp \) (PKU), \( pp \) (no PKU). Krystal is \( pp \) (no PKU), parents are \( Pp \) (have PKU) – this fits.

• Option 2: Parents \( Pp \times pp \). Parent \( pp \) does not have PKU (since PKU is dominant, \( pp \) has no dominant allele), but the problem says both parents have PKU—so eliminate.

• Option 3: Parents \( PP \times pp \). Parent \( PP \) has PKU, parent \( pp \) does not (no dominant allele) – eliminate (parents must both have PKU).

• Option 4: Parents \( pp \times pp \). Neither parent has PKU (no dominant allele) – eliminate.

Answer:

Aa x Aa

Question 17 (Josh and Irma's Toe)