A given genotype is either dominant with genotype AA, hybrid with genotype Aa or recessive with genotype aa. Offspring inherits one gene randomly from each parent. One parent has dominant genes, with genotype AA. The possibilities are:
Parent 1 Type AA\Parent 2 Type AA | A | A |
A | AA | AA |
A | AA | AA |
Parent 1 Type AA\Parent 2 Type Aa | A | a |
A | AA | Aa |
A | AA | Aa |
Parent 1 Type AA\Parent 2 Type aA | a | A |
A | aA | AA |
A | aA | AA |
Parent 1 Type AA\Parent 2 Type aa | A | A |
A | Aa | Aa |
A | Aa | Aa |
If the unknown parent has genotype Aa, the probability of the offspring having genotype Aa is 1/2.
If the unknown parent has genotype aa, the probability of the offspring having genotype AA is 0.
Carrying on in this way we can construct the table
Parent 2 Type\Offspring Type | AA | Aa | aa |
AA | 1/2 | 1/2 | 0 |
Aa | 1/4 | 1/2 | 1/4 |
aa | 0 | 1/2 | 1/2 |
\[G= \left( \begin{array}{ccc} 1/2 & 1/2 & 0 \\ 1/4 & 1/2 & 1/4 \\ 0 & 1/2 & 1/2 \end{array} \right)\]
.Solving the equation
\[ \left( \begin{array}{ccc} 1/2 & 1/2 & 0 \\ 1/4 & 1/2 & 1/4 \\ 0 & 1/2 & 1/2 \end{array} \right) \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}x\\y\\z\end{pmatrix}\]
Gives the solution
\[x=1/4, \; y=1/2, \; z=1/4\]
so a quarter of the offspring will have genotype AA, half the offspring will have genotype Aa and a quarter will have genotype aa.