# What is recombination frequency?

Finds the recombination frequency between two linked genes from the recombinant and total offspring counts of a cross, with the map distance in centimorgans, or the recombinant offspring to expect from a map distance.

- Page: https://www.acalculator.org/biology/recombination-frequency-calculator
- JSON spec: https://www.acalculator.org/biology/recombination-frequency-calculator.json
- Version: 921bfbf40d32

## Default answer

Example with the default inputs (Find Recombination frequency, Recombinant offspring 50, Total offspring 1,000): RF = 5% (5 cM) from 1,000 offspring.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| find | Find | The recombination frequency from offspring counts, or the recombinants to expect from a map distance. |
| r | Recombinant offspring | Offspring with a new combination of the two traits, summed over both recombinant classes. |
| cm | Map distance (cM) | The distance between the two genes in map units (centimorgans), from 0 to 50. |
| n | Total offspring | All offspring counted: parental and recombinant classes together. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| result | Answer | The recombination frequency and map distance, or the recombinant offspring to expect. |
| percent | Recombination frequency (%) | Recombinant offspring as a percent of all offspring. |
| fraction | Recombination frequency (fraction) | Recombinant offspring ÷ total offspring, from 0 to 1. |
| cm | Map distance (cM) | The distance in map units: 1 cM for each 1% recombination. |
| recombinants | Recombinant offspring | Offspring with a new combination of the traits. |
| parentals | Parental offspring | Offspring with a parent’s combination: total minus recombinants. |
| perClass | Each recombinant class | Half the recombinants: the two reciprocal recombinant classes are expected in equal numbers. |
| linkage | Linkage | Below 50%: linked genes. 50%: independent assortment. Above 50%: check which classes are parental. |

## Method

RF = recombinant offspring ÷ total offspring; map distance (cM) = RF × 100; expected recombinants = total offspring × map distance ÷ 100.

## Assumptions

- The counts come from a cross that shows recombination directly, such as a test cross of a double heterozygote with a double recessive.
- One map unit (1 cM) is 1% recombination. This holds for genes close together; frequencies do not go above 50%, so map distances add up only over short spans.

## Worked examples

1. find = rf, r = 50, n = 1,000 gives percent = 5, cm = 5, parentals = 950, result = RF = 5% (5 cM), linkage = linked: below 50% recombination. Source: OpenStax, Biology 2e, §13.1 Chromosomal Theory and Genetic Linkage (recombination frequency = recombinant offspring ÷ total; 0.01 = 1 cM; 0 to 0.5), https://openstax.org/books/biology-2e/pages/13-1-chromosomal-theory-and-genetic-linkage: 50/1000 = 0.05, or 5 percent.
2. find = rf, r = 391, n = 2,300 gives percent = 17, fraction = 0.17, result = RF = 17% (17 cM). Source: OpenStax, Biology 2e, §13.1 Chromosomal Theory and Genetic Linkage (recombination frequency = recombinant offspring ÷ total; 0.01 = 1 cM; 0 to 0.5), https://openstax.org/books/biology-2e/pages/13-1-chromosomal-theory-and-genetic-linkage, Morgan’s cross: (206 + 185) ÷ 2,300 = 17 percent.
3. find = exp, cm = 17, n = 2,300 gives recombinants = 391, parentals = 1,909, perClass = 195.5, result = 391 recombinant offspring expected. Source: OpenStax, Biology 2e, §13.1 Chromosomal Theory and Genetic Linkage (recombination frequency = recombinant offspring ÷ total; 0.01 = 1 cM; 0 to 0.5), https://openstax.org/books/biology-2e/pages/13-1-chromosomal-theory-and-genetic-linkage, Figure 13.4: 65.5 − 48.5 = 17 cM.
4. find = rf, r = 500, n = 1,000 gives percent = 50, linkage = unlinked: 50% recombination, as for independent assortment. Source: OpenStax, Biology 2e, §13.1 Chromosomal Theory and Genetic Linkage (recombination frequency = recombinant offspring ÷ total; 0.01 = 1 cM; 0 to 0.5), https://openstax.org/books/biology-2e/pages/13-1-chromosomal-theory-and-genetic-linkage: a frequency of 0.5 means the genes behave as unlinked.

## FAQ

### How do I calculate recombination frequency?

Count the offspring with a new combination of the two traits (both recombinant classes), divide by all the offspring, and multiply by 100. In Morgan’s fruit fly cross, 206 + 185 = 391 of 2,300 offspring were recombinant: 391 ÷ 2,300 = 0.17, or 17%.

### How do I turn recombination frequency into map units?

One map unit, or centimorgan (cM), is 1% recombination (a frequency of 0.01). A recombination frequency of 17% puts the two genes about 17 cM apart.

### Which offspring are the recombinants?

Those that combine the traits differently from either parent. In a test cross of a double heterozygote with a double recessive, the two parental classes are the most common and the two recombinant classes the least common, because crossing over between linked genes is the rarer event.

### Why is recombination frequency never above 50%?

Genes on different chromosomes, or far apart on one chromosome, sort independently, so half the offspring are recombinant: a frequency of 0.5. Linked genes give less than that. A count above 50% usually means the parental and recombinant classes were swapped.

### How many recombinant offspring should I expect?

Multiply the offspring by the map distance and divide by 100. Body color and wing length in fruit flies are 65.5 − 48.5 = 17 cM apart, so about 17 of every 100 offspring, or 391 of 2,300, should be recombinant, about half in each recombinant class.

### Do map distances add up?

For genes close together, yes: that is how Sturtevant built the first chromosome map in 1913. Over long distances the recombination frequency stops growing at 50%, so a frequency near 0.5 cannot tell genes far apart on one chromosome from genes on different chromosomes.

## Sources

- OpenStax, Biology 2e, §13.1 Chromosomal Theory and Genetic Linkage: recombination frequency as the share of recombinant offspring (50/1000 = 0.05), Morgan’s cross (965, 944, 206 and 185 offspring; 17 percent), map units (0.01 = 1 cM), the range from 0 to 0.5, Sturtevant’s map and Figure 13.4 (body color 48.5 cM, wing length 65.5 cM). https://openstax.org/books/biology-2e/pages/13-1-chromosomal-theory-and-genetic-linkage (retrieved 2026-10-19)