In horses black is dependent upon a dominate allele (B) and chestnut upon its re

Question

In horses black is dependent upon a dominate allele (B) and chestnut upon its recessive (b). The trotting gait is due to allele (T) dominate to the allele (t) which determines the pacing gait. These genes are on different chromosomes

a) If a homozygous black pacer is mated to a chestnut trotter what will the F1 generation be like?

b) What will be the result if two F1 individuals are crossed? Draw a punnett square are derive the genotypic and phenotypic ratio.


So Im not sure how to set up the table exactly but here are my thoughts maybe with the answer ill get it

BB TT B T

bb BBbb bbTT one posibilty ?? or is it more like this?? b Bb bt

tt BBtt ttTT t Bt Tt

or do i need a bigger table to answer this question.?? My problem is is that they are homozgous so does that mean that one horse has both the dominate genes and the other has both recessive??

Like my first guess at what it is supposed to be or my second guess at the gametes was correct? But the outcomes from the offspring dont make sense a one horse will be black with unknown trot and the other horse will have the trotting gate with unknown color and so I am a bit confused on how to properly set up this table and answer the question

Explanation / Answer

Your first idea is correct. The first table would look like this:

Bt

Bt

bT

Bb Tt

Bb Tt

bT

Bb Tt

Bb Tt

Then a cross of the F1 generation would look like this:

BT

Bt

bT

bt

BT

BB TT

Bb Tt

Bb TT

Bb Tt

Bt

BB Tt

BB tt

BbTt

Bbtt

bT

Bb TT

Bb Tt

bbTT

bbTt

bt

BbTt

Bbtt

bbTt

bbtt

The genotypic ratio would then be: 1:2:2:1:4:2:1:2:1

and the phenotypic ratio would be: 9:3:3:1

Does that make sense?

Bt

Bt

bT

Bb Tt

Bb Tt

bT

Bb Tt

Bb Tt

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