In each of the scenarios, an individual heterozygous for the indicated gene or genes is crossed with an identical heterozygous individual. Predict the number of different phenotypes that would be observed among the progeny of such a cross. Scenario 1 : R is completely dominant to r.
Scenario 2: R is codominant with r.
Scenario 3: R and S are completely dominant to rand s, respectively; R and S are linked by 15 m. u. and do not interact with each other.
Scenario 4: Genes R and S perform the same function; R and S are completely dominant to r and s, respectively; and R and S are unlinked.
Scenario 5: R and S are completely dominant to r and s, respectively; R and S are linked by 10 m. u. and show dominant epistasis.

Explanation:

Scenario 2: R is codominant with r.

Scenario 3: R and S are completely dominant to rand s, respectively; R and S are linked by 15 m.u. and do not interact with each other.

Scenario 4: Genes R and S perform the same function; R and S are completely dominant to r and s, respectively; and R and S are unlinked.

Scenario 5: R and S are completely dominant to r and s, respectively; R and S are linked by 10 m.u. and show dominant epistasis.

essentially, one bacterial cell splits into two daugurer cells. bacterial offspring don't need or get care from their parents because they are in a sense their own parents. that said, some bacteria can and do depend on other bacteria especially in harsh environments such as bacterial microfilms in hot water tanks.