Answer:
In Labrador dogs, black coat is dominant to chocolate, normal vision is dominant to progressive retinal atrophy (PRA), and normal hip joint is dominant to hip dysplasia. All these genes assort independently. Two dogs that are heterozygous for alleles of all three genes are crossed. Using rules of probability (not a Punnett square), what is the chance that the first pup born to these dogs will be chocolate, have normal vision, and have normal hip joints?
BbVvHh x BbVvHh= BBVVHH, BbVvHh, BbVvHh, bbvvhh
Bb= black coat dominant
Vv= Normal vision dominant
Hh= Normal hip join dominant
probability of having a first born of these dogs will be chocolate, have normal vision and have normal hip joint is 0
Explanation:
As the punette square gives 3:1 phenotype having three black coat, normal vision and normal hip joint and one chocolate, progressive retina altropy and hip dysplasia
In prophase 1:
Chromosomes become visible, crossing-over occurs, the nucleolus disappears, the meiotic spindle forms, and the nuclear envelope disappears.
In metaphase 1:
The pairs of chromosomes (bivalents) become arranged on the metaphase plate and are attached to the now fully formed meiotic spindle. The centrioles are at opposite poles of the cell.
Stimulation of the sympathetic nervous system is required
