Dihybrid cross probabilities (two genes)
A dihybrid cross follows inheritance of two independent genes at the same time.
Here the genes are A/a (gene 1) and B/b (gene 2), with complete dominance at each locus
(A dominant over a, and B dominant over b).
Core assumptions (Mendelian)
• Each gene segregates into gametes according to the parent’s genotype.
• Independent assortment means allele choice at A/a is independent of allele choice at B/b (default in this calculator).
• No linkage modeling here (linkage requires recombination frequency).
Genotypes and phenotype classes
For each locus there are three possible genotypes:
\(AA, Aa, aa\) at the A locus and \(BB, Bb, bb\) at the B locus.
A combined dihybrid genotype is a concatenation such as \(AaBb\) or \(AAbb\).
Phenotype classes under complete dominance:
- A_B_: at least one A and at least one B (dominant at both loci)
- A_bb: at least one A, but bb (dominant at A, recessive at B)
- aaB_: aa, but at least one B (recessive at A, dominant at B)
- aabb: aa and bb (recessive at both loci)
Gamete probabilities for each parent
First determine allele probabilities at each locus from the parent genotype:
\[
\begin{aligned}
AA &\Rightarrow P(A)=1,\;P(a)=0 \\
Aa &\Rightarrow P(A)=\tfrac{1}{2},\;P(a)=\tfrac{1}{2} \\
aa &\Rightarrow P(A)=0,\;P(a)=1
\end{aligned}
\]
\[
\begin{aligned}
BB &\Rightarrow P(B)=1,\;P(b)=0 \\
Bb &\Rightarrow P(B)=\tfrac{1}{2},\;P(b)=\tfrac{1}{2} \\
bb &\Rightarrow P(B)=0,\;P(b)=1
\end{aligned}
\]
Under independent assortment, each parent produces four gamete types:
\(AB, Ab, aB, ab\). Their probabilities are products:
\[
\begin{aligned}
P(AB) &= P(A)\cdot P(B) \\
P(Ab) &= P(A)\cdot P(b) \\
P(aB) &= P(a)\cdot P(B) \\
P(ab) &= P(a)\cdot P(b)
\end{aligned}
\]
Punnett approach (4×4)
A 4×4 Punnett square lists Parent 1 gametes on one side and Parent 2 gametes on the other.
Each cell is an offspring genotype from combining the two gametes, and its probability is
the product of the corresponding gamete probabilities:
\[
P(\text{cell}) = P_1(\text{gamete})\cdot P_2(\text{gamete})
\]
The calculator can show this 4×4 square (optional). It is useful for learning, but the
probability method below is usually faster.
Probability approach (clean method)
A key shortcut is that the phenotype can be found by combining results from each locus.
For the A locus:
\[
\begin{aligned}
P(aa) &= P_1(a)\cdot P_2(a) \\
P(A\_) &= 1 - P(aa)
\end{aligned}
\]
For the B locus:
\[
\begin{aligned}
P(bb) &= P_1(b)\cdot P_2(b) \\
P(B\_) &= 1 - P(bb)
\end{aligned}
\]
Then multiply A-part by B-part to get each phenotype class (labels shown as plain text):
\[
\begin{aligned}
P(A\_)\cdot P(B\_) &\Rightarrow \text{A\_B\_} \\
P(A\_)\cdot P(bb) &\Rightarrow \text{A\_bb} \\
P(aa)\cdot P(B\_) &\Rightarrow \text{aaB\_} \\
P(aa)\cdot P(bb) &\Rightarrow \text{aabb}
\end{aligned}
\]
This is exactly what the calculator visualizes as a probability tree:
first split by A_ vs aa, then split each branch by B_ vs bb.
Phenotype ratio
Ratios summarize expected outcomes. The classic dihybrid cross
\(AaBb \times AaBb\) produces:
Ratio: 9 : 3 : 3 : 1 (for A_B_ : A_bb : aaB_ : aabb)
The calculator reports an exact ratio when the probabilities match multiples of \(1/16\).
Otherwise it reports an approximate simplified ratio.
Worked example: AaBb × AaBb
Show solution
For each locus in \(Aa \times Aa\), the recessive genotype probability is
\(P(aa)=\tfrac{1}{2}\cdot\tfrac{1}{2}=\tfrac{1}{4}\), so \(P(A\_)=\tfrac{3}{4}\).
Similarly, \(P(bb)=\tfrac{1}{4}\) and \(P(B\_)=\tfrac{3}{4}\).
\[
\begin{aligned}
\tfrac{3}{4}\cdot\tfrac{3}{4} &= \tfrac{9}{16} \\
\tfrac{3}{4}\cdot\tfrac{1}{4} &= \tfrac{3}{16} \\
\tfrac{1}{4}\cdot\tfrac{3}{4} &= \tfrac{3}{16} \\
\tfrac{1}{4}\cdot\tfrac{1}{4} &= \tfrac{1}{16}
\end{aligned}
\]
Therefore:
A_B_ = 9/16, A_bb = 3/16, aaB_ = 3/16, aabb = 1/16
→ ratio 9:3:3:1.
Batch CSV mode
Batch mode lets you evaluate many crosses quickly. Each row supplies two parent genotypes
(for example \(AaBb\), \(aabb\)). The output is a CSV table of phenotype probabilities and ratios for every row.
CSV suggestion
Header: p1,p2
Rows: AaBb,AaBb • AaBb,aabb • AABb,AaBb
Common pitfalls
- Gene order doesn’t matter: \(AaBb\) and \(BbAa\) represent the same genotype.
- Dominant phenotype hides genotype: A_ includes both \(AA\) and \(Aa\); B_ includes both \(BB\) and \(Bb\).
- Linkage: if genes are linked, independent assortment fails and you need recombination frequency.
- Sampling variation: observed offspring counts can deviate from expected probabilities, especially with small sample sizes.
