(N/A) To determine the phenotypic distribution,we consider a cross between a heterozygous male (genotype $Bb$) and a homozygous recessive female (genotype $bb$).
$1$. The male produces two types of gametes: $B$ and $b$.
$2$. The female produces only one type of gamete: $b$.
$3$. Using a Punnett square:
| Gametes | $b$ |
| :--- | :--- |
| $B$ | $Bb$ |
| $b$ | $bb$ |
Resulting genotypes in the $F_{1}$ generation are $Bb$ (heterozygous) and $bb$ (homozygous recessive) in a $1:1$ ratio.
Phenotypically,this results in $50\%$ individuals showing the dominant trait and $50\%$ individuals showing the recessive trait,giving a phenotypic ratio of $1:1$.