Dogs have a variety of genes that govern coat colour. There are at least eleven identified gene series $(A,\, B,\, C, \,D, \,E, \,F, \,G, \,M, \,P, \,S, \,T) $ that influence coat colour in dog.

$A$ dog inherits one gene from each of its parents. The dominant gene gets expressed in the phenotype. For example, in the $B$ series, a dog can be genetically black or brown.

Let us assume that one parent is homozygous black $(BB)$, while the other parent is homozygous brown $(bb)$

$bb$	$BB$
		$B$	$B$
	$b$	$Bb$	$Bb$
	$b$	$Bb$	$Bb$

In this case, all the offsprings will be heterozygous $(Bb)$.

Since black $(B)$ is dominant, all the offsprings will be black. However, they will have both $B$ and $b$ alleles.

If such heterozygous pups are crossed, they will produce $25\%$ homozygous black $(BB)$, $50\%$ heterozygous black $(Bb)$, and $25\%$ homozygous brown $(bb)$ offsprings.

$\begin{array}{ccc} & B & b \\ B & B B & B b \\ b & B b & B b\end{array}$