Answer:
Four
Explanation:
Let the allele for black fur be B (the alternate will be b), the allele for active behavior be A (the alternate will be a), and the allele for light eyes be L (the alternate will be l).
<em>Heterozygous dominant for black fur = Bb</em>
<em>Heterozygous dominant for active behavior = Aa</em>
<em>Homozygous recessive for light eyes = ll</em>
Hence, the genotype for a male that is heterozygous for the dominant traits of black fur and active behavior but is homozygous recessive for light eyes will be BbAall.
Bb Aa ll
Possible Gametes
<em>BAl, Bal, bAl, and bal</em>
Therefore, the number of different types of gametes produced by the male mouse is four.