Final answer:
In a cross between a true-breeding pea plant with smooth seeds (S) and a true-breeding pea plant with wrinkled seeds (s), the F1 progeny have a 100% probability of having smooth seeds, as smooth seeds are dominant.
Step-by-step explanation:
The question pertains to Mendelian genetics, specifically the inheritance of seed texture in pea plants and the associated probabilities of different genotypes and phenotypes in the progeny. In the classic experiments by Gregor Mendel, smooth seeds (S) were demonstrated to be dominant over wrinkled seeds (s). When considering a cross between two true-breeding pea plants, one with smooth seeds and one with wrinkled seeds, the F1 progeny will all possess the genotype Ss, having received the S allele from the smooth-seeded parent and s allele from the wrinkled-seed parent. Since the S allele is dominant over the s allele, all F1 progeny will display the smooth seed phenotype.
Upon self-crossing the F1 heterozygous plants (Ss), the expected ratio of F2 progeny is typically 3:1 for the dominant to recessive phenotypes. This is because the Punnett square for a cross between two Ss organisms will result in SS, Ss, Ss, and ss genotypes, translating to a 3:1 phenotypic ratio of smooth to wrinkled seeds, or 75% smooth and 25% wrinkled. Thus, if the question asked for the probability that any given F2 offspring will possess smooth seeds, the answer would be 75%.
However, going back to the original question, since the cross is between two true-breeding individuals (SS and ss), all offspring in the F1 generation will be Ss and display smooth seeds, resulting in a 100% probability of F1 progeny with smooth seeds.