A geneticist is studying the inheritance of a trait in pea plants. He crosses a true-breeding purple-flowered plant with a true-breeding white-flowered plant and observes the following offspring: 100 purple-flowered, 100 white-flowered. Which of the following is the most likely ratio of genotypes in the offspring?

Core Concept

PILLAR 1 — MOLECULAR/CONCEPTUAL MECHANISM

Step-by-Step Analysis

The transmission of flower-color alleles from parent to offspring in pea plants (Pisum sativum) begins with meiosis, a reductional division that segregates homologous chromosomes—and their resident alleles—into haploid gametes. During meiosis I, homologous chromosomes pair along the metaphase plate and are pulled to opposite poles by spindle microtubules attached to kinetochore proteins. This physical separation ensures that each gamete receives exactly one allele at each locus. When a parent is homozygous (e.g., PP or pp), every gamete carries the same allele; when heterozygous (e.g., Pp), half the gametes carry P and half carry p, a direct consequence of the random alignment of homologous tetrads at metaphase I (independent assortment). The anthocyanin pigment pathway in pea flowers involves enzymes such as chalcone synthase and dihydroflavonol 4-reductase, whose transcription is governed by alleles at the P locus. The functional P allele produces a protein with correct folding geometry at the active site, enabling catalysis of leucoanthocyanidin to anthocyanidin—the purple pigment precursor. The recessive p allele often contains a nonsense or frameshift mutation that introduces a premature stop codon, truncating the enzyme and eliminating its catalytic function. In a Pp heterozygote, one functional copy suffices for pigment production (dominance), whereas pp homozygotes produce no functional enzyme and thus bear white flowers.

Why Other Options Are Wrong

PILLAR 2 — STEP-BY-STEP LOGIC

A cross between a true-breeding purple-flowered plant (PP) and a true-breeding white-flowered plant (pp) would, under standard Mendelian expectations, produce all Pp heterozygotes—200 purple-flowered offspring, zero white. The observed 1:1 phenotypic ratio of 100 purple to 100 white immediately signals a deviation from the predicted F1 uniformity of a simple PP × pp cross. The 1:1 ratio is the diagnostic signature of a test cross: a heterozygous parent (Pp) crossed with a homozygous recessive (pp). In such a cross, the heterozygous parent produces P-bearing and p-bearing gametes in equal proportion (50% each) because of Mendel's law of segregation. The homozygous recessive parent produces only p-bearing gametes. Fertilization thus yields Pp and pp zygotes in equal frequency—half purple (Pp) and half white (pp). Therefore, the genotype ratio among the 200 offspring is 100 Pp : 100 pp, corresponding to option B. The most plausible explanation is that the purple-flowered parent was not truly homozygous PP but was actually heterozygous Pp, despite the geneticist's assumption that it was true-breeding. Alternatively, the white-flowered parent could have been misidentified, but the simplest reconciliation of the data with Mendelian principles is a Pp × pp cross.

PILLAR 3 — DISTRACTOR ANALYSIS

Option A (100 PP, 100 pp) proposes equal numbers of both homozygous genotypes. This would require both parents to be heterozygous (Pp × Pp), yielding a 1:2:1 genotype ratio from which only the homozygous classes are (selectively) reported. However, PP homozygotes would express purple flowers and pp homozygotes would express white, matching the 1:1 phenotype count—but the problem asks for the most likely genotype ratio given a cross between two plants, one purple and one white. A Pp × Pp cross would yield 50% Pp heterozygotes (purple), which are conspicuously absent from option A. A student choosing A has likely confused the F2 dihybrid or monohybrid F2 ratio with the test-cross scenario presented here, or has selectively deleted the heterozygous class without justification.

Option C (75 PP, 25 Pp, 25 pp) presents a genotype distribution totaling only 125 offspring, not the 200 observed. This option appears to be a distorted fragment of a 3:1 F2 phenotypic ratio (which would be 150 purple : 50 white in a Pp × Pp cross) combined with an incorrect partitioning of the purple class into PP and Pp genotypes. A student selecting C likely conflated the F2 phenotypic ratio with the genotypic ratio and then applied faulty arithmetic. The stimulus data (100 purple, 100 white) do not approximate a 3:1 phenotypic ratio, so this option fails on both numerical and Mendelian grounds.

Option D (50 PP, 50 Pp) lists only purple-flowered genotypes, yielding 100 purple and zero white offspring. This directly contradicts the observed 100 white-flowered plants. A student choosing D may have reasoned backward from only the purple-flowered offspring, attempting to determine their genotypic composition (which would indeed split into PP and Pp in a 1:1 ratio if the purple parent were Pp × PP), while entirely ignoring the white-flowered class. This reflects a common error: answering a question about all offspring using only partial data. The presence of 100 white-flowered plants demands that pp homozygotes appear among the offspring, which option D omits entirely.

In summary, only option B (100 Pp, 100 pp) reconciles the observed 1:1 phenotypic ratio with Mendelian segregation, identifying the cross as most consistent with a heterozygous purple parent (Pp) crossed with a homozygous recessive white parent (pp).