In peas, round seeds (R) are dominant to wrinkled seeds (r), and yellow seeds (Y) are dominant to green seeds (y). Two plants with the genotype RrYy are crossed. What is the probability that a randomly selected offspring will have wrinkled, green seeds?

Core Concept

PILLAR 1 — MOLECULAR/CONCEPTUAL MECHANISM

Step-by-Step Analysis

The phenotypic ratios observed in a dihybrid cross between two RrYy pea plants emerge from the molecular architecture of meiosis, specifically the independent assortment of homologous chromosome pairs at metaphase I. The R gene encodes starch branching enzyme I (SBEI), a protein that catalyzes the formation of α-1,6-glycosidic bonds in amylopectin within developing seed amyloplasts. When SBEI functions properly (R allele present), branched starch accumulates, the seed retains water via osmotic balance across its tonoplast membrane, and the mature seed appears smooth and round. The r allele contains a transposon insertion in exon 6 of the SBEI gene, producing a truncated, nonfunctional enzyme. Homozygous rr seeds accumulate linear amylose chains with fewer branch points, creating a higher ratio of free sucrose and glucose. This elevated solute concentration generates an osmotic gradient that drives excessive water uptake during early development, followed by disproportionate desiccation during seed maturation—yielding the wrinkled phenotype. Similarly, the Y locus governs a carotenoid biosynthetic pathway enzyme (phytoene synthase) in the seed chloroplast envelope; homozygous yy plants cannot convert geranylgeranyl diphosphate (GGPP) into colored carotenoid pigments, leaving the seed chloroplasts' residual chlorophyll visible through the translucent testa as a green appearance.

Why Other Options Are Wrong

During meiosis I in RrYy heterozygotes, the chromosome carrying R and its homolog carrying r segregate to opposite poles independently of the chromosome pair carrying Y versus y, provided these loci reside on separate chromosomes (or are sufficiently distant for complete recombination between them). At metaphase I, the random orientation of each bivalent at the equatorial plate means that the R-bearing chromosome can migrate toward the same pole as either the Y-bearing or y-bearing chromosome with equal likelihood. This mechanistic independence at the chromosomal level generates four equiprobable gamete genotypes—RY, Ry, rY, and ry—each produced at a frequency of one-quarter.

PILLAR 2 — STEP-BY-STEP LOGIC

To determine the probability of obtaining wrinkled, green (rryy) offspring from an RrYy × RrYy cross, we decompose the dihybrid problem into two monohybrid probabilities and apply the multiplication rule for independent events. For the seed shape locus alone, the cross Rr × Rr produces offspring genotypes in the ratio 1 RR : 2 Rr : 1 rr. Only the rr homozygous recessive class exhibits the wrinkled phenotype, representing one-quarter of all zygotes. For the seed color locus alone, the cross Yy × Yy likewise yields one-quarter yy (green) progeny. Because the R/r and Y/y gene pairs assort independently during meiosis—each bivalent aligning without influence on the other—the probability of any offspring receiving both the rr genotype and the yy genotype equals the product of their individual probabilities: 1/4 × 1/4 = 1/16. Alternatively, one can construct a 4 × 4 Punnett square from the four gamete types (RY, Ry, rY, ry) contributed by each parent. Of the sixteen resulting zygotic cells, only the single rryy cell in the bottom-right corner represents wrinkled, green seeds, confirming the probability of 1/16. This 9:3:3:1 phenotypic ratio (9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green) is the hallmark signature of independent assortment operating on two heterozygous, unlinked loci.

PILLAR 3 — DISTRACTOR ANALYSIS

Option B likely presents 1/4, a value that traps students who recognize the correct phenotype for one locus but fail to multiply probabilities across both loci. These students identify that rr occurs in one-quarter of offspring or that yy occurs in one-quarter, but then stop, neglecting that the joint probability requires multiplication—reflecting a fundamental misunderstanding of independent event probability in genetics.

Option C likely offers 3/16, the individual probability for each of the mixed-phenotype classes (round green or wrinkled yellow). This distractor exploits confusion about which phenotypic class in the 9:3:3:1 ratio corresponds to the double-recessive genotype. Students selecting this answer may be recalling the ratio's numbers without connecting each term to its specific genotype, thereby matching the question's target to the wrong phenotypic category.

Option D likely presents 1/2, ensnaring students who add the single-locus probabilities (1/4 + 1/4) rather than multiplying them. This error reflects a deeper conceptual failure: conflating the word "and" (which demands multiplication of independent probabilities) with "or" (which, for mutually exclusive events, demands addition). These students may reason that since each trait independently appears in one-quarter of progeny, the combined result must be larger than either alone, instinctively adding rather than recognizing that the joint occurrence of two independent rare events is rarer than either event alone.

Option A correctly states 1/16, derived from the product of the two monohybrid recessive probabilities under Mendel's law of independent assortment.