Genetics problem, please help?

In horses, the trotting gait, T, is dominant to the pacing gait, t, and black, B, is dominant to chestnut, b. What are the genotypes of teh corss that would produce 3/8 black trotting, 3/8 black pacing, 1/8 chestnut trotting, and 1/8 chestnut pacers?

Can someone explain how to solve it?

Genetics problem, please help?
A homozygously dominant horse would have the genotype TTBB--it would be a black trotter. A homozygously recessive horse would have the genotype ttbb--it would be a chestnut pacer, and we would be able to recognize it at a glance. This is not exactly the same as the 9:3:3:1 ratio you would expect from crossing a pair of heterozygous horses (TtBb). One-quarter of these horses are chestnut (bb) and three-quarters are black (BB or Bb); half of them are trotters and the other half are pacers. This means that the parents of the foals are both heterozygous for color (Bb) while one of them is heterozygous for gait (Tt) and the other is homozygous recessive (tt). I hope this helps.
Reply:OK lets start with the trotting and pacing :



so



horse A mates with Horse B



we know that T is dominant and t is recessive, therefore

for a horse to get this gene they must inherit a copy from both parents, there is only a 1 in 4 chance of this occuring.



here are the possibilities



Tt Tt



gives you



Tt tT tt TT



only one of these can result in a horse that paces..

this is tt





The same also applies with the brown and the chestnut colouring, there is only a 1 in 4 chance of having a horse with chestnut colouring :



Bb Bb



Bb BB BB bb



only bb will result in a chestnut horse...



so :



the genotype of each parent would be :

Bb Bb

Tt Tt







hope this helps
Reply:wouldnt this go in chemistry? not sure.





help me:



http://answers.yahoo.com/question/index;...
Reply:A cross between BbTt and Bb tt would produce the final ratios that you are interested in.



You can break it down into the individual groups by noticing that 3/4 of the final cross are Black, and 1/4 is chestnut. This cross is easier to figure out. Similarly, you've got 1/2 and 1/2 trotting versus pacing. Also, you know that each parent much carry the recessive gene in at least one copy, because otherwise, the recessive phenotype will never show up. You also know that at least one parent (but not necessarily both) needs to carry the dominant trait.



Once you have the individual traits, you can pair them up and draw a Punnett Square to check yourself. Here's some Punnett Square examples: http://www.athro.com/evo/gen/punexam.htm...