Correction for my last post:

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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.

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