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Punnett square for taster vs. nontaster heritability:

\begin{tabular}{|l|c|c|}
\hline
& \begin{tabular}{c}
MOM \\
[tex]$T n$[/tex]
\end{tabular}
& \begin{tabular}{c}
DAD \\
[tex]$n n$[/tex]
\end{tabular} \\
\hline
CHILD 1 & [tex]$T$[/tex] & [tex]$n$[/tex] \\
\hline
CHILD 2 & [tex]$n$[/tex] & [tex]$n$[/tex] \\
\hline
CHILD 3 & [tex]$T$[/tex] & [tex]$n$[/tex] \\
\hline
CHILD 4 & [tex]$n$[/tex] & [tex]$n$[/tex] \\
\hline
\end{tabular}

Key:
[tex]$T = \text{"taster"}$[/tex] \\
[tex]$n = \text{"nontaster"}$[/tex] \\

Recall that the trait of being a taster is a dominant trait. Which of the children in the family are tasters? Check all that apply.

A. Child 1 \\
B. Child 2 \\
C. Child 3 \\
D. Child 4



Answer :

GinnyAnswer
To determine which of the children are tasters, we need to recall that the trait for being a taster is dominant. In other words, if a child has at least one "T" allele, they will exhibit the taster phenotype.

Let's analyze the given data for each child:

- Child 1: Genotype "Tn"
- Since this child has one "T" allele, they are a taster.

- Child 2: Genotype "nn"
- This child has two "n" alleles, meaning they do not have the dominant "T" allele, so they are a nontaster.

- Child 3: Genotype "Tn"
- Similar to Child 1, this child also has one "T" allele, meaning they are a taster.

- Child 4: Genotype "nn"
- Identical to Child 2, this child has two "n" alleles and thus is a nontaster.

Given this analysis:
- Child 1 is a taster.
- Child 2 is a nontaster.
- Child 3 is a taster.
- Child 4 is a nontaster.

Therefore, the children who are tasters are:
- Child 1
- Child 3

The correct response is:
- Child 1
- Child 3

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