In horses black is dependent upon a dominant gene, B, and chestnut upon its recessive allele, b.?

The trotting gait is due to a dominant allele T, and the pacing gait to its recessive allele, t. If a homozygous black pacer is mated to a homozygous chestnut trotter, what will be the appearance of the F1 generation?

In horses black is dependent upon a dominant gene, B, and chestnut upon its recessive allele, b.?
BBtt x bbTT = BbTT





The result is purely heterozygous. The F1 offspring will all be heterozygous black trotters.
Reply:He's write but we wrote the equation wrong.


If you draw a Punnet Square it is easy to see that:


BBtt x bbTT = BbTt


The F1 Generation is purely heterozygous Black Trotter.

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