Dihybrid Cross Practice Problems With Answers

Mastering dihybrid crosses is essential for understanding Mendelian genetics and predicting the inheritance of two traits simultaneously. This guide walks you through the logic behind dihybrid crosses, provides a step‑by‑step method, offers several practice questions, and supplies detailed solutions with explanations. Here's the thing — whether you are preparing for a biology exam, teaching a genetics unit, or simply curious about how traits combine, working through dihybrid cross practice problems with answers builds confidence and sharpens problem‑solving skills. By the end, you’ll be able to set up Punnett squares, interpret genotypic and phenotypic ratios, and apply the principles of independent assortment to real‑world scenarios.

How to Solve a Dihybrid Cross Problem

A dihybrid cross examines the inheritance of two different genes, each with two alleles. The classic example uses pea plant traits: seed shape (round R vs. wrinkled r) and seed color (yellow Y vs. green y). Because the genes assort independently, the gamete combinations follow the FOIL method (First, Outer, Inner, Last) or a simple 2 × 2 grid for each parent Worth keeping that in mind. Which is the point..

Short version: it depends. Long version — keep reading.

Step‑by‑Step Procedure

1. Identify the parental genotypes – Write each parent’s genotype for both traits (e.g., RrYy).
2. Determine possible gametes – List all allele combinations that can be passed on. For a heterozygote (RrYy), the gametes are RY, Ry, rY, ry.
3. Set up a 4 × 4 Punnett square – Place one parent’s gametes across the top and the other’s down the side.
4. Fill in the squares – Combine the alleles from the row and column to obtain each offspring’s genotype.
5. Phenotype classification – Convert each genotype to its observable trait using dominance rules (e.g., R_ = round, yy = green).
6. Count phenotypes – Tally the number of each phenotype to derive the phenotypic ratio (typically 9:3:3:1 for two heterozygotes).
7. Interpret the ratio – Relate the observed ratio to Mendel’s law of independent assortment.

Tip: If one parent is homozygous for a trait, reduce the square size accordingly (e.g., a 4 × 2 grid).

Practice Problems

Below are four dihybrid cross scenarios. But attempt each before checking the answers. Write down the parental genotypes, list gametes, draw the Punnett square, and note the phenotypic ratio.

Problem 1

In pea plants, tall stems (T) are dominant to short stems (t), and purple flowers (P) are dominant to white flowers (p). Cross a plant that is heterozygous for both traits (TtPp) with a plant that is homozygous recessive for both traits (ttpp). What are the expected phenotypic ratios of the offspring?

Problem 2

A fruit fly has red eyes (R) dominant to white eyes (r) and normal wings (N) dominant to vestigial wings (n). Cross a male RrNn with a female rrnn. Determine the phenotypic ratio of the progeny.

Problem 3

In a certain plant, seed texture is either smooth (S) or wrinkled (s), with smooth dominant. Seed color is either yellow (Y) or green (y), with yellow dominant. If two plants both heterozygous for both traits (SsYy) are crossed, what fraction of the offspring will be smooth‑seeded and yellow‑seeded?

Problem 4

A cattle breed shows two independently assorting traits: coat color (black B dominant to brown b) and presence of horns (horned H dominant to polled h). A bull with genotype BbHh is mated with a cow of genotype bbhh. List all possible genotypes of the calves and indicate which phenotypes correspond to each genotype.

Answers and Explanations

Answer to Problem 1

Parental genotypes: TtPp × ttpp

Gametes:

• TtPp → TP, Tp, tP, tp
• ttpp → tp only

Punnett square (4 × 1):

Phenotypes:

• TtPp → tall, purple
• Ttpp → tall, white
• ttPp → short, purple
• ttpp → short, white

Each appears once, so the phenotypic ratio is 1 : 1 : 1 : 1 (tall‑purple : tall‑white : short‑purple : short‑white).

Answer to Problem 2

Parental genotypes: RrNn × rrnn

Gametes:

• RrNn → RN, Rn, rN, rn
• rrnn → rn only

Punnett square (4 × 1):

Phenotypes:

• RrNn → red eyes, normal wings
• Rnnn → red eyes, vestigial wings
• rrNn → white eyes, normal wings
• rrnn → white eyes, vestigial wings

Again, each phenotype occurs once → 1 : 1 : 1 : 1 ratio.

Answer to Problem 3

Parental genotypes: SsYy × SsYy

Gametes for each parent: SY, Sy, sY, sy (four each).

**Punnett square