Monohybrid cross (Punnett probabilities)
A monohybrid cross tracks inheritance of a single gene with two alleles:
A (dominant) and a (recessive). Under complete dominance, any genotype
containing at least one A shows the dominant phenotype.
Assumptions (Mendelian, complete dominance)
• Each parent produces gametes with allele proportions determined by their genotype.
• Fertilization is random: offspring allele combinations follow product probabilities.
• No linkage, no mutation, no selection, no Hardy–Weinberg step (not needed here).
Genotypes and phenotypes
Possible genotypes for one gene with alleles A/a:
- AA (homozygous dominant) → dominant phenotype
- Aa (heterozygous) → dominant phenotype
- aa (homozygous recessive) → recessive phenotype
Gametes produced by each genotype
A Punnett square is built from the gamete allele probabilities of each parent.
For a single parent genotype:
\[
\begin{aligned}
AA &\Rightarrow P(A)=1,\; P(a)=0 \\
Aa &\Rightarrow P(A)=\frac{1}{2},\; P(a)=\frac{1}{2} \\
aa &\Rightarrow P(A)=0,\; P(a)=1
\end{aligned}
\]
These gamete proportions are exactly what the calculator displays as gamete chips
(for example: A 50%, a 50% for a heterozygous parent).
How the Punnett square gives probabilities
Let Parent 1 have gamete probabilities \(P_1(A)\), \(P_1(a)\),
and Parent 2 have \(P_2(A)\), \(P_2(a)\).
Each cell in the 2×2 Punnett square is a product:
\[
\begin{aligned}
P(AA) &= P_1(A)\cdot P_2(A) \\
P(Aa) &= P_1(A)\cdot P_2(a) + P_1(a)\cdot P_2(A) \\
P(aa) &= P_1(a)\cdot P_2(a)
\end{aligned}
\]
The calculator computes these values and shows them in:
(1) the Punnett square cells, and (2) the genotype probability bar chart.
Phenotype probabilities (complete dominance)
Under complete dominance, AA and Aa are dominant phenotype, and aa is recessive.
\[
\begin{aligned}
P(\text{dominant phenotype}) &= P(AA)+P(Aa) \\
P(\text{recessive phenotype}) &= P(aa)
\end{aligned}
\]
The calculator shows this as a separate phenotype bar chart.
Expected ratios
Ratios are a compact way to express expected outcomes.
When probabilities match classic Mendelian quarters, the common results are:
- Aa × Aa → genotype ratio 1 : 2 : 1 (AA : Aa : aa) and phenotype ratio 3 : 1 (dominant : recessive)
- Aa × aa → genotype ratio 0 : 1 : 1 and phenotype ratio 1 : 1
- AA × aa → genotype ratio 0 : 1 : 0 and phenotype ratio 1 : 0
Note: If a parent is entered as “unknown” or “dominant phenotype, unknown carrier,” the probabilities may not be neat quarters.
The calculator still reports a simplified (or approximate) integer ratio.
When a parent is entered as phenotype (unknown carrier)
If a parent shows the dominant phenotype, their genotype is either AA or Aa.
The calculator lets you choose a prior carrier probability:
\(p = P(Aa \mid \text{dominant phenotype})\).
\[
\begin{aligned}
P(AA \mid \text{dominant}) &= 1 - p \\
P(Aa \mid \text{dominant}) &= p \\
P(aa \mid \text{dominant}) &= 0
\end{aligned}
\]
From that genotype mixture, it computes gamete probabilities and then offspring probabilities exactly as above.
Worked examples
Example 1: Aa × Aa (classic 3:1 phenotype)
Gametes: each parent produces A and a with 50% each.
\[
\begin{aligned}
P(AA)&=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4} \\
P(Aa)&=\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{2} \\
P(aa)&=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}
\end{aligned}
\]
\[
\begin{aligned}
P(\text{dominant})&=\frac{1}{4}+\frac{1}{2}=\frac{3}{4} \\
P(\text{recessive})&=\frac{1}{4}
\end{aligned}
\]
Genotype ratio: 1 : 2 : 1 | Phenotype ratio: 3 : 1
Example 2: Aa × aa (classic 1:1 phenotype)
Parent 1 gametes: A 50%, a 50%. Parent 2 gametes: a 100%.
\[
\begin{aligned}
P(AA)&=\frac{1}{2}\cdot 0 = 0 \\
P(Aa)&=\frac{1}{2}\cdot 1 + \frac{1}{2}\cdot 0 = \frac{1}{2} \\
P(aa)&=\frac{1}{2}\cdot 1 = \frac{1}{2}
\end{aligned}
\]
\[
\begin{aligned}
P(\text{dominant})&=\frac{1}{2} \\
P(\text{recessive})&=\frac{1}{2}
\end{aligned}
\]
Genotype ratio: 0 : 1 : 1 | Phenotype ratio: 1 : 1
Batch CSV mode
Batch mode is for quickly evaluating many crosses. Each CSV row contains two parental genotypes:
AA, Aa, or aa. The output table reports the probabilities for each row.
CSV suggestion
Header: p1,p2
Rows: Aa,Aa • Aa,aa • AA,Aa
Common pitfalls
- Genotype vs phenotype: dominant phenotype does not distinguish AA from Aa.
- Order doesn’t matter: Aa × aa gives the same probabilities as aa × Aa.
- Small sample sizes: real offspring counts can deviate from expected probabilities by chance.
FAQ: Why is the Punnett square still 2×2 if a genotype is “unknown”?
The square is based on gamete allele probabilities (A vs a), which always have two possibilities.
When a parent is “unknown,” the calculator first forms a mixture over AA/Aa/aa using your priors,
then converts that mixture into \(P(A)\) and \(P(a)\). The Punnett square remains 2×2 because it models
allele combinations, not the parent’s internal uncertainty directly.
