Within my last post, I started the graph at F1 and it was reflecting something other than the original F1 from the self-pollination image.
This is more accurate:
Being that our example begins with YY and yy
one is yellow, the other green making it 50/50 percentage wise.
The recessive nature is shown in F2 (second generation) where 25% are green which the new graph now reflects.
If the F1 example had a heterozygote Yellow (Yy), we would start with
Yy and yy
which would mean 50% yellow as well.
In that case, F2 would be the same with F3 being 5/8 and F4 being 11/16 green.
In such case, a graph would show green percentages approaching 75% which is the same as the presence of the genes regardless of phenotype in the beginning F1, the same as the graph I am using shows green approaching 50% as is the frequency of the genes in the YY and yy example using homozygotes.