Final answer:
The F2 offspring from a tetrahybrid cross (AAbbCcDd x AAbbCcDd) are expected to have a 1/256 chance of being homozygous dominant for genotype AABBCCDD using the probability method of genetics.
Step-by-step explanation:
The proportion of F2 offspring expected to have the genotypic combination AABBCCDD in a tetrahybrid cross (AAbbCcDd x AAbbCcDd) is 1/256. The probability method of genetics, which relies on understanding independent assortment and dominance in a dihybrid or a tetrahybrid cross. In the given cross, each gene pair sorts independently. Consequently, we calculate the probability of homozygous dominant for each gene separately and then multiply them together for the overall probability. For gene A, the probabilities for homozygous dominant (AA) is 1/4 and for heterozygous (Aa) is 1/2.
The combined probability of having a dominant allele at A (either AA or Aa) is 1/4 + 1/2 = 3/4. Similarly, this calculation step is applicable for genes B, C, and D. Multiplying these probabilities together for all four genes using the product rule (3/4) × (3/4) × (3/4) × (3/4) gives the proportion of offspring with the dominant phenotype, which is 81/256. However, for the exact homozygous dominant genotype AABBCCDD, each gene must be homozygous dominant (AA/BB/CC/DD), each with a probability of 1/4. Therefore, the final calculation step is (1/4) × (1/4) × (1/4) × (1/4) = 1/256, thus providing plagiarism free content.