In a certain population, the genotype of an individual can be represented by the following Punnett square: | Rr | Rr || rR | rR |. What is the most likely genotype of the offspring from a cross between two heterozygous individuals?

Core Concept

PILLAR 1 — MOLECULAR/CONCEPTUAL MECHANISM

Step-by-Step Analysis

The transmission of alleles from parent to offspring during sexual reproduction depends on the chromosomal mechanics of meiosis, a reductional division that halves the diploid (2n) chromosome complement to produce haploid (1n) gametes. During Meiosis I, homologous chromosomes—each consisting of two sister chromatids joined at the centromere by cohesin protein complexes—pair up along the metaphase plate. The R allele and the r allele occupy corresponding loci on maternally and paternally derived homologues. Spindle microtubules, composed of α- and β-tubulin heterodimers, attach to kinetochore protein complexes assembled at the centromeric regions, exerting pulling forces that segregate the homologous chromosomes to opposite poles. This physical separation ensures that each gamete receives exactly one allele at the R/r locus, producing a 50% probability of carrying R and a 50% probability of carrying r in each haploid cell. The independent assortment of chromosomes on the metaphase plate further randomizes which allele any single gamete inherits. When two heterozygous (Rr) individuals undergo gametogenesis, each parent contributes an R-bearing sperm or egg with probability 0.5, and an r-bearing sperm or egg with probability 0.5. Fertilization restores the diploid state by combining two haploid nuclei, and the zygote's genotype at the R/r locus depends entirely on which two alleles—drawn independently from each parent—unite.

Why Other Options Are Wrong

PILLAR 2 — STEP-BY-STEP LOGIC

Constructing a standard Punnett square for an Rr × Rr cross places the maternal gametes (R and r) along one axis and the paternal gametes (R and r) along the other. The four resulting cells yield the genotypic ratios: one RR, two Rr, and one rr. Expressed as probabilities, RR = 0.25, Rr = 0.50, and rr = 0.25. Notice that the homozygous outcomes RR and rr each occur with identical frequency—one-quarter of all offspring—because each results from the fusion of two identical gametes (R + R or r + r), and each gamete type has an independent probability of 0.5. The heterozygous Rr appears twice because it arises from two distinct fertilization events: maternal R combining with paternal r, and maternal r combining with paternal R. Thus, while Rr is the single most frequent genotype, the question specifically asks about the offspring genotype pattern, and the critical insight is that the two homozygous classes, RR (Option A) and rr (Option C), are produced at precisely equal frequencies. Option D correctly identifies this equivalence: both homozygous conditions arise with equal probability, a direct consequence of the symmetric segregation of alleles during meiosis I anaphase.

PILLAR 3 — DISTRACTOR ANALYSIS

Option A (RR) attracts students who focus exclusively on the dominant phenotype or who mentally collapse the Punnett square to 'half dominant, half recessive' without distinguishing homozygous from heterozygous classes. The specific flaw is conflating phenotypic frequency (75% show the dominant trait) with genotypic frequency (only 25% are RR). A student who selects A has not recognized that rr offspring are equally probable.

Option B (Rr) is perhaps the most seductive distractor because it is numerically the single most common genotype at 50%. Students who choose B misread the question as asking for the single most likely individual genotype rather than recognizing the structural equivalence between the two homozygous outcomes. The underlying conceptual error is treating 'most likely genotype' as a request for the modal class rather than evaluating whether two classes share identical probabilities.

Option C (rr) functions as a mirror image of Option A. Students who select rr may be reasoning from a recessive-focused perspective—perhaps because many AP Biology exam scenarios highlight recessive disorders—and similarly fail to recognize that both homozygous classes arise with the same 25% probability, making neither uniquely correct on its own. Selecting C in isolation commits the same categorical oversight as selecting A.