Question

1. spongebob’s aunt is famous around town for her itty, bitty stubby nose! she recently met a cute squarepants fellow who also has a stubby nose, which is a recessive trait. would it be possible for them to have a child with a regular long nose? why or why not? create a punnett square to help you answer this question.

1. if spongebob’s aunt described in #7 wanted children with long noses, what type of fellow would she need to marry in order to give her the best chances? create a punnett square to help you answer this question.

t. trimpe 2003 http://sciencespot.net/

Explanation:

Question 7

Step1: Define Traits and Genotypes

Let the recessive trait (stubby nose) be represented by \( r \), so the genotype for stubby nose is \( rr \) (since it's recessive, both parents must have \( rr \) to express the trait). The dominant trait (long nose) is \( R \), so the genotype for long nose is \( RR \) or \( Rr \).

Step2: Create Punnett Square

Both parents (SpongeBob's aunt and her fellow) have stubby noses, so their genotypes are \( rr \) (mother) and \( rr \) (father). The Punnett square will have all cells as \( rr \) (since \( r \times r = rr \) for both rows and columns).

Step3: Analyze Offspring Genotypes

All offspring will have the genotype \( rr \), which expresses the recessive trait (stubby nose). So, no offspring will have a long nose because both parents only contribute the recessive allele \( r \), so the child can only have \( rr \) genotype.

Step1: Determine Aunt's Genotype

From Question 7, SpongeBob's aunt has a stubby nose (recessive trait), so her genotype is \( rr \).

Step2: Identify Best Mate's Genotype

To maximize the chance of having children with long noses (dominant trait, \( R \)-), the best mate would be someone with the dominant genotype that can pass on \( R \). The genotype \( RR \) (homozygous dominant) is best because when crossed with \( rr \), all offspring will be \( Rr \) (long nose, since \( R \) is dominant). If the mate was \( Rr \), only 50% of offspring would be \( Rr \) (long nose), and 50% \( rr \) (stubby nose).

Step3: Create Punnett Square (for \( rr \times RR \))

	\( R \)	\( R \)
\( r \)	\( Rr \)	\( Rr \)

All offspring are \( Rr \), which expresses the long nose trait (since \( R \) is dominant over \( r \)).

Step4: Compare with Other Genotypes

If the mate was \( Rr \) (heterozygous), the Punnett square \( rr \times Rr \) would be:

	\( R \)	\( r \)
\( r \)	\( Rr \)	\( rr \)

Here, 50% are \( Rr \) (long nose) and 50% are \( rr \) (stubby nose). So, a mate with \( RR \) (homozygous dominant) gives a 100% chance of long - nosed offspring, which is better than a mate with \( Rr \).

Answer:

No, it is not possible for them to have a child with a regular long nose. Because both parents have the recessive genotype \( rr \) (for stubby nose, a recessive trait), all their offspring will inherit \( r \) from both parents, resulting in the \( rr \) genotype (stubby nose). The Punnett square for \( rr \times rr \) shows all offspring as \( rr \).

Question 8