Final answer:
The number of possible genotypes in the offspring from the cross AaBbCCDd x aaBBccDd is 8 different genotypes, by calculating the independent assortments of each allele and considering homozygosity where applicable.
Step-by-step explanation:
Number of Genotypes in Offspring
To determine how many different genotypes are possible in the offspring of the cross AaBbCCDd x aaBBccDd, where genes are independently assorting, we must consider each gene pair separately:
- For the A allele, there are 2 possibilities (A from the first parent, a from the second).
- For the B allele, there is 1 possibility (B is homozygous in both parents).
- For the C allele, there is 1 possibility (C is homozygous in the first parent and absent in the second).
- For the D allele, there are 4 possibilities (D or d from the first parent combined with D or d from the second).
Therefore, the total number of genotypes is the product of the possibilities for each gene: 2 (A) x 1 (B) x 1 (C) x 4 (D) = 8 different genotypes.
The possible genotypes from this cross, taking each gene into account, are:
- AaBbCcDd
- AaBbCcdd
- aaBbCcDd
- aaBbCcdd
However, since the original question specifies independent assortment and fixed homozygous states for the B and C genes, the actual variety comes only from the alleles A and D, hence simplifying the genotypic diversity to variants of A and D alleles only.