Final answer:
The question seems to have an inconsistency in the lettering of the dominant and recessive alleles, but based on standard convention, a Punnett square for two homozygous parents (one with AA and one with aa) would result in 100% Aa heterozygous offspring, all displaying the dominant phenotype.
Step-by-step explanation:
No option is given that shows two homozygous parents producing heterozygous offspring. However, to set up a correct Punnett square for two homozygous parents, with dominant amethyst (A) and recessive white (a), you will place AA on one side for the homozygous dominant parent and aa on the other side for the homozygous recessive parent. The resulting offspring would all be heterozygous (Aa).
In classical Mendelian genetics, when two homozygous organisms with different alleles are crossed (AA x aa), the result is 100% heterozygous (Aa) offspring. This occurs because each parent can only contribute one type of allele to their gametes—dominant (A) from the homozygous dominant parent and recessive (a) from the homozygous recessive parent. When these gametes combine during fertilization, the resulting offspring are all Aa, displaying the dominant phenotype. Do note, since the question mentions the symbols 'a' and 'a' for dominant and recessive traits, respectively, ensure consistent symbol usage when solving. It is crucial to maintain the convention of using uppercase for dominant alleles and lowercase for recessive alleles, which implies that 'A' should represent the dominant amethyst and 'a' the recessive white trait in a conventional Punnett square.