Consider a species in which dark brown fur is dominant relative to light brown fur.

Explanation:

To interpret this pedigree, let’s start with information that we already know:

Brown is recessive, which means brown individuals must have the phenotype BB. In this pedigree, brown individuals are filled in.

Black is dominant, which means black individuals must have at least one B allele. Their phenotype could be either BB or BB. In this pedigree, black individuals are not filled in.

Figure 5 shows the same pedigree, but with information about the individual’s phenotype filled in.

The shaded individual, who is a brown female puppy, must have the phenotype BB. If she had any B alleles, she would be black because the black allele is dominant over the brown allele.

In order for the brown puppy to have the phenotype BB, she must have gotten two “b” alleles: one from each of her parents. We know that her parents are both black (because they are unshaped), which means they must have a least one “B” allele. This means that both parents must be heterogeneous: BB.

The three black puppies must have at least one “B” allele in order for them to be black in color. However, we can’t tell whether they are homologous dominant (BB) or heterogeneous (BB) since both of those phenotype would result in black color. One way to represent this on a pedigree is B-, meaning that the second allele could be either B or b.

i had this on my unit activity, i can you if you are a plato student. are you a plato student?

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%