To determine which individuals carry the gene for disease A, we need to identify those who possess at least one dominant allele 'A'. The gene for having the disease is represented by a capital 'A', while the gene for not having the disease is represented by a lowercase 'a'.
Here is the Punnett square for heritability, showing the genotypes for both parents and their children:
[tex]\[
\begin{tabular}{|l|c|c|}
\hline
& \begin{tabular}{c}
MOM \\
Aa
\end{tabular}
& \begin{tabular}{c}
DAD \\
Aa
\end{tabular} \\
\hline
CHILD 1
& a
& a
\\
\hline
CHILD 2
& A
& a
\\
\hline
CHILD 3
& A
& A
\\
\hline
CHILD 4
& a
& a
\\
\hline
\end{tabular}
\][/tex]
Now let's analyze each individual:
Parents:
- Mom: Has the genotype 'Aa' – Contains the dominant allele 'A', so Mom is a carrier of the gene for disease A.
- Dad: Has the genotype 'Aa' – Also contains the dominant allele 'A', so Dad is a carrier of the gene for disease A.
Children:
- Child 1: Has the genotype 'aa' – Does not have the dominant allele 'A', so Child 1 does not carry the gene for disease A.
- Child 2: Has the genotype 'Aa' – Contains the dominant allele 'A', so Child 2 is a carrier of the gene for disease A.
- Child 3: Has the genotype 'AA' – Contains two dominant alleles 'A', so Child 3 carries the gene for disease A.
- Child 4: Has the genotype 'aa' – Does not have the dominant allele 'A', so Child 4 does not carry the gene for disease A.
Based on this analysis, the individuals who have the gene that carries the disease are:
- Both Mom and Dad
- Child 2
- Child 3
Thus, the individuals who are carriers of the gene for disease A are:
- Mom
- Dad
- Child 2
- Child 3
