Final answer:
In the cross aaBbCcDd x AaBbccDD, no offspring will express the dominant phenotype for gene A; approximately 62.5 offspring will express the dominant phenotype for genes B and C each; and all 125 offspring will express the dominant phenotype for gene D.
Step-by-step explanation:
When assessing how many offspring are expected to express the dominant phenotype for each gene in the cross aaBbCcDd x AaBbccDD, we must consider each gene pair separately, assuming independent assortment. For gene A, the resulting dominant phenotype would come from genotypes AA or Aa. Since one parent is homozygous recessive (aa), all offspring from this gene pair will display the recessive phenotype. Moving on to gene B, with one parent heterozygous (Bb) and the other homozygous recessive (bb), a 1:1 ratio of dominant to recessive phenotypes is expected, giving a 50% chance of expressing the dominant phenotype.
For gene C, since one parent is heterozygous (Cc) and the other homozygous recessive (cc), the offspring also have a 50% chance of expressing the dominant phenotype. Lastly, gene D is represented in all offspring as either homozygous dominant (DD) or heterozygous (Dd), both of which express the dominant phenotype, resulting in 100% expression of the dominant allele. Therefore, the expected number of offspring expressing the dominant phenotype for each gene, out of 125 progeny, are: 0 for A, 62.5 for B, 62.5 for C, and 125 for D.