If the dominant allele makes up $73 \%$ of the gene pool, then the recessive allele must make up the remaining $27 \%$ of the gene pool.
Let $p$ represent the frequency of the dominant allele and $q$ represent the frequency of the recessive allele. According to the Hardy-Weinberg principle, $p+q=1$.
We know that $p=0.73$, so we can solve for $q$:
\begin{align*}
p+q &=1 \\
q &=1-p \\
q &=1-0.73 \\
q &=0.27
\end{align*}
Now we can use the Hardy-Weinberg equation to determine the frequency of homozygous dominant individuals in the population:
\begin{align*}
p^2 + 2pq + q^2 &= 1 \\
p^2 &= (0.73)^2 \\
p^2 &= 0.5329 \\
\end{align*}
Therefore, the frequency of homozygous dominant individuals in the population is $p^2 = 0.5329$, or approximately $53 \%$.
So, about $53 \%$ of the population is homozygous dominant.