1. Consider this cross:

                    Male = AaBbCcDdEe   x  Female = AaBBccDdEe

If we assume complete dominance at each locus, and  assume segregation and independent assortment then we can answer the following questions:

        1) What fraction of sperm will contain the 'a' allele?
                (just consider this locus for the male.  He is Aa.)

                so, if segregation works properly, then he should produce two types of gametes with respect to this locus; those carrying 'A' and those carrying 'a', and they should be produced in equal frequency.  So, the fraction of gametes carrying 'a' should be  1/2.

        2) What fraction of the female gametes should contain the 'aBc' genes?

                Well, use the same logic as above, but for each locus SEPARATELY:
                A locus - 1/2 the gametes will contain 'a' (fraction = 1/2)
                B locus - All of the gametes MUST contain a 'B' (fraction = 1)
                C locus - All of the gametes must contain a 'c' (fraction = 1)

                Now, multiply these separate fractions together = 1/2 x 1 x 1 = 1/2 for 'aBc'

        3) How many different genetic combinations can the female make in her gametes, with respect to these loci?

                Answer with respect to each locus, then multiply:

                A locus - 2 types, 'A' or 'a'
                B locus - 1 type, all have 'B'
                C locus - 1 type, all have 'c'
                D locus - 2 types, 'D' or 'd'
                E locus - 2 types, 'E' or 'e'

                Multiply answers together = 8 different gene combinations in gametes.

        4) OK, now how about the offspring?  How many different genotypes and phenotypes are possible in the offspring?

                A locus:  Aa x Aa = 3 possible genotypes, 2 phenotypes
                B locus:  Bb x BB = 2 possible genotypes, 1 phenotype
                C locus:  Cc x cc = 2 possible genotypes, 2 phenotypes
                D locus:  Dd x Dd = 3 possible genotypes, 2 phenotypes
                E locus:  Ee x Ee = 3 possible genotypes, 2 phenotypes

                So, the number of differnt genotypes possible = 3 x 2 x 2 x 3 x 3 = 81
                And, the number of possible phenotypes = 2 x 1 x 2 x 2 x 2 = 16

        5) What is the probability of producing an offspring with the Aa genotype?

                Well, just answer the question for this locus.

                A locus:  Aa x Aa  =  1/2 the offspring will be Aa

        6) What is the probability of producing an offspring that has the AB phenotype?

                A locus:  Aa x Aa =  3/4 will have the A phenotype
                B locus:  Bb x BB =  all will have the B phenotype
                3/4 x 1 = 3/4 will have the AB phenotype.

        7)  What fraction will have the aBCdE phenotype?

                A locus:  Aa x Aa = 1/4 will be a
                B locus:  Bb x BB = all will be B
                C locus:  Cc x cc = 1/2 will be c
                D locus:  Dd x Dd = 1/4 will be d
                E locus:  Ee x Ee = 3/4 will be E
                ABcdE = 1/4 x 1 x 1/2 x 1/4 x 3/4 = 3/128.

        8) Consider this cross:  AaBbCc   X  AaBBCC

                Assume the following:
                    A is incompletely dominant to a
                    B and b are codominant
                    C and c exhibit overdominance.
                    These genes assort independently

                    - How many gametes can the male make? (8)   Female? (2)

                    - How many phenotypes are possible in the offspring?
                        3 X 2 X 2 = 12

                    - What fraction will be ABC? 1/4 (only AA) X 1/2 (only BB) x 1/2 (only CC) = 1/16