Q.In the F2 generation of a Mendelian dihybrid cross the number of phenotypes and genotypes are:

A dihybrid cross tracks two traits simultaneously, each controlled by a different gene with two alleles showing complete dominance. When Mendel crossed pea plants differing in two characters—say, seed shape (round/wrinkled) and seed color (yellow/green)—the F1 generation was uniformly heterozygous for both traits (RrYy) and displayed both dominant phenotypes.

Self-pollinating the F1 dihybrids produces an F2 generation following the famous 9:3:3:1 phenotypic ratio. This ratio arises because each trait segregates independently:

• Four distinct phenotypes appear: 9 showing both dominant traits, 3 showing the first dominant and second recessive, 3 showing the first recessive and second dominant, and 1 showing both recessive traits. …

When Mendel extended his experiments from tracking one trait to tracking two simultaneously, he opened the door to understanding how different characteristics are inherited independently. A dihybrid cross examines the inheritance of two contrasting traits at once—say, seed shape (round vs. wrinkled) and seed color (yellow vs. green) in pea plants.

The classic dihybrid cross begins with two pure-breeding parents that differ in both traits. For instance, one parent might be homozygous dominant for both traits (RRYY—round and yellow), while the other is homozygous recessive (rryy—wrinkled and green). The F₁ generation from this cross is uniformly heterozygous (RrYy), displaying both dominant traits: round and yellow seeds.

The real insight emerges in the F₂ generation, produced by crossing two F₁ individuals (RrYy × RrYy). Because each parent can produce four types of gametes (RY, Ry, rY, ry) in equal proportions, a 4×4 Punnett square reveals sixteen possible offspring combinations.

Counting the phenotypes is straightforward. Any offspring with at least one dominant R allele will have round seeds; any with at least one dominant Y allele will have yellow seeds. This gives us:

• 9 round, yellow (R_Y_)
• 3 round, green (R_yy)
• 3 wrinkled, yellow (rrY_)
• 1 wrinkled, green (rryy)

That's 4 distinct phenotypic classes, even though their frequencies differ.

Counting the genotypes requires more care. We must distinguish every unique allele combination:

• RRYY, RRYy, RRyy (3 genotypes)
• RrYY, RrYy, Rryy (3 genotypes)
• rrYY, rrYy, rryy (3 genotypes) …

Use the general formula instead of listing every box: for n independently assorting gene pairs (each with 2 alleles and complete dominance), the F2 phenotype count is 2^n and the genotype count is 3^n. For a dihybrid cross, n=2, givi …