Final answer:
In a cross between two heterozygous individuals for a recessive lethal allele, like sickle cell disease, the phenotypic ratio would be 1:2:0, as the homozygous recessive genotype leads to non-viability in utero.
Step-by-step explanation:
The question concerns the ratio of phenotypes in offspring from a cross between two individuals heterozygous for sickle cell disease, which is an autosomal recessive disorder. Unlike a typical Mendelian 3:1 ratio seen in monohybrid crosses, when dealing with a recessive lethal allele that is expressed in utero, we must adjust our expectations because the homozygous recessive genotype results in non-viability.
In such a cross, the expected genotypic ratio would be 1:2:0, corresponding to homozygous dominant:heterozygous:homozygous recessive. This is because the homozygous recessive genotype, which would be the sickle cell disease in this case, is lethal when expressed in utero and thus, those individuals do not survive to birth. Therefore, they are not present in the genotypic ratio of the surviving offspring. Within this context, the actual phenotype ratio observed would be similar, with the majority of surviving offspring displaying the dominant phenotype (not having the disease) and the rest being carriers (heterozygous).
This scenario is an example of how real-world genetics can differ from textbook examples due to the influence of lethal genes, and also demonstrates the significance of Mendel's large sample size which minimized the effects of random chance on phenotypic ratios.