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
The answer is asexual reproduction. This type of reproduction does not involve fusion of gametes. The offspring arise from a single organism and inherit the genes of that parent only. Consequently, there is little variation from parent compared to sexual reproduction.
B. can start in many different places on a sequence at the same<span> time.</span>
Describing how is the cells group is the process of multiplication of cell or the so-called cell division, mitosis. This mechanism of each cell produces another cell that binds together and produces again, repeatedly, to be a larger system called now the tissue as explained earlier. How? Through cell division called, mitosis. Excerpt meiosis –cell division specifically in the sex cells.
