Dihybrid Cross Punnett Square Calculator
Generate a 4x4 dihybrid cross Punnett square for two genes. Calculate genotypic and phenotypic ratios with detailed offspring analysis.
Formula
Phenotypic Ratio (AaBb x AaBb) = 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb
In a dihybrid cross between two heterozygotes, each parent produces 4 gamete types. The 4x4 Punnett square yields 16 offspring combinations. With independent assortment and complete dominance, the phenotypic ratio is 9:3:3:1. Different parental genotypes produce different ratios.
Worked Examples
Example 1: Classic Mendel Dihybrid Cross (AaBb x AaBb)
Problem:Cross two organisms heterozygous for both traits. What are the expected phenotypic and genotypic ratios?
Solution:Parent gametes: AB, Ab, aB, ab (each parent) 16 offspring combinations in 4x4 Punnett square Phenotypic ratio: 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb Genotypic ratio: 1 AABB : 2 AABb : 2 AaBB : 4 AaBb : 1 AAbb : 2 Aabb : 1 aaBB : 2 aaBb : 1 aabb
Result:Phenotypic ratio 9:3:3:1 with 4 phenotypes and 9 genotypes
Example 2: Test Cross (AaBb x aabb)
Problem:Cross a dihybrid individual with a homozygous recessive individual to determine the dihybrid's gamete production.
Solution:AaBb gametes: AB, Ab, aB, ab aabb gametes: ab only Offspring: AaBb, Aabb, aaBb, aabb Each genotype appears in equal proportions (1:1:1:1) All 4 phenotypes equally likely at 25% each
Result:Phenotypic ratio 1:1:1:1 confirming independent assortment
Frequently Asked Questions
What is a dihybrid cross?
A dihybrid cross is a genetic cross between two organisms that are heterozygous for two different genes. This means each parent carries two different alleles for each of the two traits being studied. The classic example is Mendel's cross of pea plants that were heterozygous for both seed shape (Rr) and seed color (Yy). A dihybrid cross produces a 4x4 Punnett square with 16 possible offspring combinations, which typically yields the famous 9:3:3:1 phenotypic ratio when both genes assort independently.
How do you set up a dihybrid Punnett square?
First, determine the genotypes of both parents for two genes. Next, figure out the possible gametes each parent can produce by combining one allele from each gene. For a dihybrid cross, each parent produces 4 types of gametes. Place one parent's gametes along the top of a 4x4 grid and the other parent's gametes along the side. Fill in each cell by combining the gamete from the column with the gamete from the row. This produces 16 offspring genotype combinations that you can then analyze for genotypic and phenotypic ratios.
What is the expected ratio from a dihybrid cross of two heterozygotes?
When both parents are heterozygous for both genes (AaBb x AaBb), the expected phenotypic ratio is 9:3:3:1 under Mendelian inheritance with independent assortment. This means 9 out of 16 offspring show both dominant traits, 3 show the first dominant and second recessive, 3 show the first recessive and second dominant, and 1 shows both recessive traits. The genotypic ratio is more complex with 9 distinct genotypes. This 9:3:3:1 ratio was one of Mendel's key discoveries and supports the law of independent assortment.
What is independent assortment?
Independent assortment is Mendel's second law, which states that alleles of different genes sort independently of one another during gamete formation. This occurs because different genes are located on different chromosomes, and chromosomes segregate randomly during meiosis I. However, genes located close together on the same chromosome tend to be inherited together (genetic linkage) and do not assort independently. Independent assortment is what produces the 9:3:3:1 ratio in dihybrid crosses and is a fundamental principle of genetics.
When does a dihybrid cross NOT produce a 9:3:3:1 ratio?
Several factors can alter the expected 9:3:3:1 ratio. Genetic linkage occurs when two genes are on the same chromosome and close together, causing them to be inherited together more often than expected. Epistasis is when one gene affects the expression of another, producing modified ratios like 9:7, 12:3:1, or 9:3:4. Incomplete dominance or codominance at either locus changes the number of distinguishable phenotypes. Lethal alleles can eliminate certain genotypes from the offspring pool, also modifying the expected ratios.
Is there a way to get dihybrid probabilities without drawing all 16 cells?
Yes, the forked-line or product method. Because unlinked loci assort independently, treat each gene as its own monohybrid cross and multiply the results. Aa x Aa gives 3/4 dominant and 1/4 recessive, and Bb x Bb does the same, so A_B_ is 3/4 x 3/4 = 9/16, A_bb and aaB_ are each 3/4 x 1/4 = 3/16, and aabb is 1/4 x 1/4 = 1/16. That reproduces 9:3:3:1 exactly, and it scales to any number of genes.
What is a test cross and what does it reveal?
A test cross mates the individual under investigation with a fully recessive partner, aabb in this two-gene case. That parent contributes only ab gametes, so each offspring's phenotype is determined entirely by the allele set it received from the other parent, and the offspring classes read out that parent's gamete proportions directly. A genuine dihybrid produces four phenotypes in 1:1:1:1 proportions. A skew away from 1:1:1:1, with parental combinations over-represented, is the classic signature of linkage.
How do I test whether real offspring counts fit 9:3:3:1?
Apply a chi-square goodness-of-fit test: chi-square = sum of (observed - expected)^2 / expected, with 3 degrees of freedom for four phenotype classes and a critical value of 7.815 at the 0.05 level. Mendel's 556 F2 seeds gave 315, 108, 101 and 32 against expectations of 312.75, 104.25, 104.25 and 34.75, which comes to roughly 0.47. That sits far below 7.815, so the deviation is comfortably explained by sampling chance.
How does the square grow for three or more genes?
Each heterozygous locus doubles the gamete types, so n loci give 2^n gametes per parent and a grid of 4^n cells. Two genes mean 4 gametes and 16 cells, three genes mean 8 gametes and 64 cells, and four genes mean 16 gametes and 256 cells. The phenotypic ratio is the 3:1 expansion raised to the same power, so a trihybrid cross of two triple heterozygotes yields 27:9:9:9:3:3:3:1 across 64 offspring. Past three genes, the product method beats drawing a grid.
Why does the calculator list nine genotypes but only four phenotypes?
Genotype counts every allele combination, while phenotype counts only what is visible under complete dominance. From AaBb x AaBb the nine genotypes fall out as 1 AABB : 2 AABb : 1 AAbb : 2 AaBB : 4 AaBb : 2 Aabb : 1 aaBB : 2 aaBb : 1 aabb, which sums to 16. AA and Aa are indistinguishable in appearance, so they collapse into the A_ shorthand used in the phenotype column, leaving four classes in 9:3:3:1.
References
Background & Theory
History
Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer ยท Editorial policy
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