What is recombination frequency?
Enter the recombinant offspring and the total offspring of a cross. The calculator gives the recombination frequency, the map distance in centimorgans, and whether the genes look linked, or the recombinant offspring to expect for genes a given distance apart.
- Answer
- RF = 5% (5 cM)
RF = 5% (5 cM) from 1,000 offspring.
- Recombination frequency (%)
- 5
- Recombination frequency (fraction)
- 0.05
- Map distance (cM)
- 5
- Recombinant offspring
- 50
- Parental offspring
- 950
- Each recombinant class
- 25
- Linkage
- linked: below 50% recombination
Answer: RF = 5% (5 cM). RF = 5% (5 cM) from 1,000 offspring.
How to calculate
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.
Example with the default inputs (Find Recombination frequency, Recombinant offspring 50, Total offspring 1,000): RF = 5% (5 cM) from 1,000 offspring.
Method: RF = recombinant offspring ÷ total offspring; map distance (cM) = RF × 100; expected recombinants = total offspring × map distance ÷ 100.
- 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
Each example is checked against the calculator on every build.
- Find Recombination frequency, Recombinant offspring 50, Total offspring 1,000 gives Recombination frequency (%) 5, Map distance (cM) 5, Parental offspring 950, Answer 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
- Find Recombination frequency, Recombinant offspring 391, Total offspring 2,300 gives Recombination frequency (%) 17, Recombination frequency (fraction) 0.17, Answer 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
- Find Expected recombinants, Map distance (cM) 17, Total offspring 2,300 gives Recombinant offspring 391, Parental offspring 1,909, Each recombinant class 195.5, Answer 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
- Find Recombination frequency, Recombinant offspring 500, Total offspring 1,000 gives Recombination frequency (%) 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
How it works
Recombination frequency (OpenStax Biology 2e §13.1).
RF = recombinant offspring ÷ total offspring
The page shows RF as a fraction and as a percent (RF × 100). The map distance is the percent in map units: 1 cM for each 1% recombination.
- Recombination frequency: type the recombinant offspring (both recombinant classes added together) and the total offspring. Recombinants are whole numbers from 0 to 1,000,000,000; the total is a whole number from 1 to 1,000,000,000.
- Expected recombinants: type the map distance in cM (any number from 0 to 50) and the total offspring. Expected recombinants = total × distance ÷ 100. Each of the two recombinant classes is expected to hold half of them.
The page also gives the parental offspring (total − recombinants) and half the recombinants (one recombinant class).
Linkage, from the recombination frequency:
| RF | The page says |
|---|---|
| below 50% | linked: below 50% recombination |
| exactly 50% | unlinked: 50% recombination, as for independent assortment |
| above 50% | above 50%: the larger classes are usually the parental ones |
Rules the page checks. The recombinant offspring cannot be more than the total offspring; if they are, the page gives no answer. A result that is not 0 but smaller than 2⁻¹⁰²² (about 2.2 × 10⁻³⁰⁸, from a map distance typed that small) has no answer, because a number cannot keep its digits.
How results are written. Every result is worked exactly and rounded once. In the sentence, numbers have 6 significant figures with trailing zeros dropped and thousands commas (RF = 33.3333%); a number below 0.000001 is written m × 10ⁿ, in the sentence and the rows (1 recombinant of 1,000,000,000 offspring: RF = 1 × 10⁻⁷% (1 × 10⁻⁷ cM)). The percent row is in percent (%). The sentence reads “RF = 5% (5 cM)” or “391 recombinant offspring expected”, then “from 1,000 offspring.” with the total typed.
Assumptions. The counts come from a cross that shows recombination directly, such as a test cross. One map unit is 1% recombination, which holds for genes close together.
Worked examples by hand
Test cross (OpenStax). 50 recombinant offspring of 1,000: RF = 50 ÷ 1,000 = 0.05 = 5%, so the genes are 5 cM apart; 950 offspring are parental.
Morgan’s fruit flies. Recombinants 206 + 185 = 391; total 965 + 944 + 206 + 185 = 2,300. RF = 391 ÷ 2,300 = 0.17 = 17% (17 cM).
Expected recombinants for 17 cM. 2,300 × 17 ÷ 100 = 391 recombinant offspring (195.5 in each recombinant class) and 2,300 − 391 = 1,909 parental offspring.
Unlinked genes. 500 recombinants of 1,000: RF = 50%, the value for independent assortment.
Other questions people ask
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.