In HORSES, BLACK IS DEPENDANT UPON A DOMINANT GENE ,B, and chestnut upon its recessive allele, b,.The trotting gait is due to a dominant gene, T, 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.
Please solve this genetics question.

Dear Student,
 
In this question we are given certain information

Black is dominant and defined by B
Chestnut is recessive and defined by b

Trotting is define by dominant T
Pacing is defined by recessive t


Now look at the condition given

We have

Homozygous black pacer which can be defined by BBtt
and
Homozygous  chestnut trotter which can be defined by bbTT

If we cross these two organisms by Punnet square method we get.



So the F1 generation will be 
heterozygous Black Trotter

 

  • 2
Both the offsprings will be blacks trotters (BbTt).
  • 0
What are you looking for?