Incomplete Genetics multiple choice questions, solution provided.

Parent 1: Homozygous black ($BB$)
Parent 2: Homozygous white ($bb$)

The gametes from $BB$ will all be $B$.
The gametes from $bb$ will all be $b$.

Crossing $BB \times bb$:

All offspring ($100\%$) will have the genotype $Bb$.
Since black color ($B$) is dominant over white color ($b$), all offspring will have a black phenotype.

The correct option is C.
C) 100 percent

5. Assuming that both parent plants in the diagram below are homozygous. Why would all of the f1 generation have yellow phenotypes? It is because:

One parent is yellow pea seed, and the other is green pea seed. Both are stated to be homozygous.
Since all F1 generation offspring are yellow, this indicates that yellow is the dominant trait.

Let $Y$ represent the allele for yellow seeds and $y$ represent the allele for green seeds.
Since the parents are homozygous and all F1 are yellow, the yellow parent must be $YY$ and the green parent must be $yy$.

$YY \times yy \rightarrow$ all offspring are $Yy$.
All $Yy$ individuals will express the dominant yellow phenotype.

The F1 generation ($Yy$) is self-crossed to produce the F2 generation.
$Yy \times Yy \rightarrow 1 YY : 2 Yy : 1 yy$
This results in a 3:1 phenotypic ratio of yellow to green, which matches the diagram.

The reason all F1 generation have yellow phenotypes is because yellow is dominant over green.

The correct option is B.
B) Yellow is dominant over green