Probability of two recessive traits
Two AaBb parents have a 1 in 16 chance, 0.0625 or 6.25%, of an offspring that shows both recessive traits, because 1/4 at each locus multiplies to 1/16. No need to redraw the sixteen-cell grid. Free in your browser.
- Free
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- Your data stays on your device
Run — free
Worked example
- Parent1
- AaBb
- Parent2
- AaBb
- Cross
- AaBb × AaBb
- Trait count
- 2
- Loci
- A, B
- Per trait recessive probability
- 0.25
When two parents are each heterozygous at the same two unlinked loci, classic Mendelian genetics asks how often an offspring will express both recessive phenotypes at once. Students meet this as the double-recessive corner of a sixteen-cell Punnett square; tutors and exam writers need the same fraction without redrawing the grid every time. This calculator accepts two dihybrid heterozygous genotypes such as AaBb and AaBb, checks that each locus is truly heterozygous and that both parents share the same two letters, then applies the product rule under independent assortment: one-fourth chance of the recessive phenotype at each locus multiplies to one-sixteenth for both together.
Why two independent recessive traits multiply to 1/16
A monohybrid heterozygous cross Aa times Aa produces the familiar genotypic ratio one AA to two Aa to one aa, so the recessive phenotype aa appears in exactly one quarter of offspring when dominance is complete. A dihybrid parent that is heterozygous at two unlinked loci, written AaBb, forms four gamete types AB, Ab, aB, and ab in equal frequency under independent assortment. Crossing two such parents fills a four-by-four Punnett square with sixteen equally likely zygote combinations.
Only one of those sixteen cells is aabb, the genotype that expresses both recessive traits when each recessive allele is fully masked in heterozygotes. The same count falls out of the product rule without drawing the grid: the probability of aa is one fourth, the probability of bb is one fourth, and independence multiplies those events to one sixteenth. This capability encodes that textbook identity for any pair of locus letters you supply, as long as both parents are dihybrid heterozygotes at the same two loci.
It does not model linkage, incomplete dominance, epistasis, or lethal genotypes; those extensions change the joint phenotype frequencies and belong to specialized tools. For the standard independent recessive case that drives most secondary and introductory college exercises, the answer is always the fixed dyadic rational one sixteenth, returned here as probability 0.0625, fraction 1/16, and percent 6.25 so clients can display whichever form their lesson expects.
Which genotypes you can enter and what you get
Each value must be a four-letter dihybrid genotype such as AaBb or RrYy. Letters may be spaced for readability but are stripped before parsing. Each consecutive pair of characters is one locus: the two alleles must use the same base letter in opposite case so the locus is heterozygous (Aa is valid, AA and aa are rejected).
The two loci on one parent must use different letters. Both parents must share the same two locus letters, though the order of loci may differ and is normalized to parent1 order in the output. Example: parent1 AaBb and parent2 AaBb yields probability one sixteenth for the aabb phenotype under independent assortment.
What you can do with it
Dihybrid Punnett homework checks
Students submit two dihybrid heterozygous genotypes and verify that the double-recessive phenotype probability is exactly 1/16 under independent assortment.
LMS and exam auto-grading
Mark items that ask for the joint recessive probability of an AaBb × AaBb cross with a deterministic oracle that rejects non-heterozygous parents.
FAQ
What is the probability of two recessive traits from AaBb × AaBb?
Under independent assortment and complete dominance, P(aa and bb) = (1/4)×(1/4) = 1/16 (0.0625, or 6.25%).
Which genotypes does the calculator accept?
Both parent1 and parent2 must be four-letter dihybrid heterozygotes such as AaBb or RrYy: each locus pair opposite case of the same letter, two distinct locus letters, same letters on both parents.
Why are AABb or Aabb rejected?
This capability models two dihybrid heterozygous parents only.
Does the tool model genetic linkage?
No. It assumes independent assortment of two unlinked loci. Linked genes, epistasis, and incomplete dominance are out of scope and would change the joint phenotype frequencies.
Can I use letters other than A and B for the genes?
Yes. Any pair of locus letters works, such as RrYy, as long as both parents are heterozygous at the same two loci. The order of the loci may differ between parents; it is normalized before the cross.
Related tools
For developers — API access
Everything on this page is available programmatically. This section is for teams who want to wire it into their own systems; everyone else can just use the tool above.
API endpoint
Prefer to automate it? One authenticated POST creates the task; the result comes back by webhook or a signed link. The same capability also runs here on the web, by email and from Telegram — and soon from our app too.
Call it from your stack
curl -X POST https://api.kit.forhosting.com/bio/punnett-probability-two-recessive \
-H "Authorization: Bearer $KIT_KEY" \
-H "Content-Type: application/json" \
-d '{"parent1":"AaBb","parent2":"AaBb"}'const res = await fetch("https://api.kit.forhosting.com/bio/punnett-probability-two-recessive", {
method: "POST",
headers: {
"Authorization": `Bearer ${process.env.KIT_KEY}`,
"Content-Type": "application/json"
},
body: JSON.stringify({
"parent1": "AaBb",
"parent2": "AaBb"
})
});
const { task_id } = await res.json();import os, requests
res = requests.post(
"https://api.kit.forhosting.com/bio/punnett-probability-two-recessive",
headers={"Authorization": f"Bearer {os.environ['KIT_KEY']}"},
json={
"parent1": "AaBb",
"parent2": "AaBb"
},
)
task_id = res.json()["task_id"]<?php
$res = file_get_contents("https://api.kit.forhosting.com/bio/punnett-probability-two-recessive", false, stream_context_create([
"http" => [
"method" => "POST",
"header" => "Authorization: Bearer " . getenv("KIT_KEY") . "\r\nContent-Type: application/json",
"content" => '{"parent1":"AaBb","parent2":"AaBb"}',
],
]));
$task = json_decode($res, true);body := bytes.NewBufferString(`{"parent1":"AaBb","parent2":"AaBb"}`)
req, _ := http.NewRequest("POST", "https://api.kit.forhosting.com/bio/punnett-probability-two-recessive", body)
req.Header.Set("Authorization", "Bearer "+os.Getenv("KIT_KEY"))
req.Header.Set("Content-Type", "application/json")
res, _ := http.DefaultClient.Do(req)Example request
{
"parent1": "AaBb",
"parent2": "AaBb"
}Example response
{
"task_id": "tsk_a1b2c3d4e5f6a1b2c3d4e5f6",
"type": "bio.punnett_probability_two_recessive",
"status": "queued",
"_links": {
"result": "/tasks/tsk_…/result"
}
}The API is asynchronous: the call returns a task_id immediately and the result arrives by webhook. Polling is capped at 1 req/s per task.
Pricing
Published price — no tokens, no invented credits. A failed task is never charged.
Errors
| HTTP | Code | Meaning |
|---|---|---|
401 | unauthorized | Missing or invalid API key. |
402 | insufficient_balance | Your balance doesn't cover the task price. |
404 | unknown_type | That task type doesn't exist. |
429 | rate_limited | Too many requests. Use the webhook instead of polling. |
Technical notes
The response returns the exact probability, the fraction 1/16, a percent form, the normalized cross string, and a short formula you can show beside homework or LMS feedback.
Input contract, genotype rules, and response fields
POST a JSON body with two required string fields: parent1 and parent2. Aliases p1, mother, genotype1 and p2, father, genotype2 are accepted. On success the body echoes normalized parent1 and parent2 strings, a cross field like AaBb times AaBb, trait_count equal to two, the loci letter list, per_trait_recessive_probability 0.25 with fraction 1/4, the joint probability 0.0625 with fraction 1/16 and percent 6.25, a phenotype_target naming the double-recessive genotype, and a formula string restating the product rule. Missing fields, non-string values, wrong length, homozygous loci, mixed letters inside a locus, or mismatched locus sets fail with invalid_input and are not billed.
Classroom, exam, and content uses for a double-recessive API
Biology teachers and tutoring platforms use a two-recessive-traits calculator so learners can confirm the double-recessive corner of a dihybrid Punnett square without arithmetic slips when they only need the joint recessive probability rather than all sixteen genotypes. Keep requests small, pin the exact 0.0625 and 1/16 fields in tests, and you get reproducible Mendelian answers for homework checks, content embeds, and edge workers without standing up a full genetics simulation stack. Exam-prep generators and LMS auto-graders need a deterministic oracle that accepts standard genotype strings, rejects homozygous or monohybrid inputs with a clear invalid_input signal, and never invents a different fraction when the cross is the classic heterozygous dihybrid case. Research notebooks and teaching demos can call the endpoint when sketching expected counts for a planned litter size by multiplying the returned probability by n outside the API.
Integration use cases
SEO lessons and tutoring widgets
Embed a free browser calculator for two recessive traits that matches the paid API byte-for-byte on the same pure Mendelian product-rule module.
Technical questions
What does the two recessive traits calculator cost?
The browser tool is free. API requests use prepaid KIT balance at $0.002 per successful call; invalid input is not billed.