Question
Mendel crossed pure-breeding round yellow pea plants (RRYY) with pure-breeding wrinkled green pea plants (rryy). What is the phenotypic ratio in the F₂ generation, and why does each class appear in that proportion?
This is a direct NEET question type — expect it worded as “ratio of round yellow to round green to wrinkled yellow to wrinkled green in F₂.”
Solution — Step by Step
Parent 1 (round yellow): RRYY → produces only one gamete type: RY
Parent 2 (wrinkled green): rryy → produces only one gamete type: ry
Both parents are homozygous, so each produces exactly one gamete type. F₁ is always RrYy.
F₁ plant is RrYy. During meiosis, R/r and Y/y segregate independently (Law of Independent Assortment — the two genes are on different chromosomes).
The four possible gametes from RrYy are: RY, Ry, rY, ry — each produced with equal frequency (25% each).
Cross F₁ × F₁: RrYy × RrYy
| RY | Ry | rY | ry | |
|---|---|---|---|---|
| RY | RRYY | RRYy | RrYY | RrYy |
| Ry | RRYy | RRyy | RrYy | Rryy |
| rY | RrYY | RrYy | rrYY | rrYy |
| ry | RrYy | Rryy | rrYy | rryy |
Total 16 combinations.
Remember: capital letter = dominant allele present → dominant phenotype expressed.
- Round Yellow (R_Y_): 9 boxes — RRYY, RRYy, RrYY, RrYy (×4 positions) = 9
- Round Green (R_yy): RRyy, Rryy = 3
- Wrinkled Yellow (rrY_): rrYY, rrYy = 3
- Wrinkled Green (rryy): rryy = 1
Phenotypic ratio = 9 : 3 : 3 : 1
Mendel’s Law of Independent Assortment: genes controlling different characters assort independently into gametes. This works here because the R/r and Y/y loci are on non-homologous chromosomes.
Why This Works
The 9:3:3:1 ratio is actually just two 3:1 ratios multiplied together. Round:wrinkled alone gives 3:1; yellow:green alone gives 3:1. When the two traits are independent, we multiply the probabilities: (3/4 × 3/4) : (3/4 × 1/4) : (1/4 × 3/4) : (1/4 × 1/4) = 9:3:3:1.
This multiplicative relationship is the entire logic of independent assortment. If the genes were linked (on the same chromosome), we’d get a skewed ratio — that’s how Morgan discovered linkage groups later.
For NEET, also remember: out of 16 offspring, 1 is RRYY (pure round yellow, same as the original parent) and 1 is rryy (pure wrinkled green). These are the parental combinations recovered in F₂.
Alternative Method — Forked Line / Branch Diagram
Instead of a 4×4 Punnett square (tedious in an exam), use the forked-line method:
Round (3/4) ——— Yellow (3/4) → Round Yellow = 3/4 × 3/4 = 9/16
└── Green (1/4) → Round Green = 3/4 × 1/4 = 3/16
Wrinkled (1/4) — Yellow (3/4) → Wrinkled Yellow = 1/4 × 3/4 = 3/16
└── Green (1/4) → Wrinkled Green = 1/4 × 1/4 = 1/16Multiply fractions across each branch, then read the ratio. This saves 2–3 minutes in NEET — absolutely worth practicing.
In any dihybrid cross where both parents are RrYy, you can skip the Punnett square entirely. Just split each trait independently (3:1 each), then multiply. Forked-line method works for trihybrid too — just add a third branch.
Common Mistake
Students often list only 4 gamete types from RrYy but then assign them unequal frequencies — writing RY as “more common” because R and Y are both dominant. Wrong. Each gamete (RY, Ry, rY, ry) is produced at exactly 25% frequency. Dominance affects phenotype, not gamete proportion. This error leads to an incorrect ratio like 6:2:3:1 instead of 9:3:3:1.
A second trap: confusing genotypic ratio with phenotypic ratio. The genotypic ratio in F₂ is 1:2:1:2:4:2:1:2:1 (nine genotypic classes). NEET almost always asks for phenotypic ratio — 9:3:3:1 — so read the question carefully before writing your answer.