Consider a true-breeding plant, \(A A B B C C\), crossed with another true- breeding plant, aabbcc, whose resulting offspring are AaBbCc. If you cross the \(\mathrm{F}_{1}\) generation, and independent assortment is operational, the expected fraction of offspring in each phenotypic class is given by the expression \(N ! / M !(N-M) !\) where \(N\) is the total number of alleles (six in this example) and \(M\) is the number of uppercase alleles. In a cross of \(A a B b C c \times A a B b C c,\) what proportion of the offspring would be expected to contain two uppercase alleles?

First, we need to understand that we have two true-breeding plant genotypes: AABBCC and aabbcc. When crossed, we obtain an F1 generation plant with genotype AaBbCc. Our goal is to find the proportion of the offspring with two uppercase alleles when we cross two F1 generation plants.

The rule of Independent Assortment states that genes for different traits assort independently during the formation of gametes. In this case, each allele pair segregates independently from other allele pairs during gamete formation. Therefore, we can compute the probability of each pair of alleles separately and multiply them afterward.

For each pair of alleles, there are three possibilities: uppercase uppercase (e.g. AA), uppercase lowercase (Aa or aA), and lowercase lowercase (aa). The probability of each possibility occurring in the F1 generation is: - \(P(AA) = \frac{1}{4}\) - \(P(Aa) = P(aA) = \frac{1}{2}\) - \(P(aa) = \frac{1}{4}\) The same probabilities apply to the other allele pairs (BB, Bb, bb, CC, Cc, cc) since their inheritance is identical due to the rule of independent assortment.

To find the probability of the offspring having exactly two uppercase alleles, we need to consider all possible combinations that lead to this outcome: AABbcc, aaBBcc, and aaBbCC. Calculating the probabilities based on previous step results: - \(P(AABbcc) = P(AA) \times P(bb) \times P(cc) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}\) - \(P(aaBBcc) = P(aa) \times P(BB) \times P(cc) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}\) - \(P(aaBbCC) = P(aa) \times P(Bb) \times P(CC) = \frac{1}{4} \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{32}\) To get the total probability of offspring having exactly two uppercase alleles, we add the probabilities of all three cases: \(P(2\,uppercase\,alleles) = P(AABbcc) + P(aaBBcc) + P(aaBbCC) = \frac{1}{64} + \frac{1}{64} + \frac{1}{32} = \frac{4}{64} = \frac{1}{16}\) The proportion of offspring with two uppercase alleles is \(\frac{1}{16}\).