Compare the sizes of b. cereus m. luteus and r. rubrum

Scenario 5 1) take 10 red and 10 black beans and place them, mixed, on the table. record the starting phenotype # and frequencies (% of your total population) of your starting population in the table provided (generation 0). 2) act as a predator. “capture” as many organisms as you can until you have reduced the population to three organisms. put them aside. at this point, the predators die. 3) the remaining organisms each produce 2 clonal offspring. multiply your organisms accordingly and allow them to mix on the table. calculate and record the resultant phenotype # and frequencies (% of your total population) of your population in the table provided (generation 1). 4) repeat the reproduction event, allowing each of your organisms to produce 2 clonal offspring. calculate and record the resultant phenotype # and frequencies (% of your total population) of your population in the table provided (generation 2). 5) repeat the reproduction event, allowing each of your organisms to produce 2 clonal offspring. calculate and record the resultant phenotype # and frequencies (% of your total population) of your population in the table provided (generation 3).

What best explains whether bromine (br) or neon (ne) is more likely to form a covalent bond? bromine forms covalent bonds because it has seven valence electrons, but neon has eight valence electrons and already fulfills the octet rule. bromine forms covalent bonds because it has many electron shells, but neon has only two electron shells and is tightly bound to its electrons. neon forms covalent bonds because it can share its valence electrons, but bromine has seven valence electrons and can gain only one more electron. neon forms covalent bonds because it has only two electron shells, but bromine has many electron shells and will lose electrons in order to fulfill the octet rule.

Physical characteristics of down syndrome