Classically, the linkage between two loci can be measured in centimorgans (cM), which represents the percent chance that these two loci will recombine an odd number of times (generating a recombinant genotype).

Due to independent assortment, markers on different chromosomes are expected to recombine 50% of the time. As I understand, if one conducts a classic experiment and counts various progeny, then deduces the apparent linkage, values of 50 cM or more are interpreted as "impossible to determine whether on separate chromosomes or just recombines very often". See (source), with grammar corrected:

The final point that we need to make regards the maximum distance that we can measure. Because of the way in which the calculations are performed, we can never have more (than) 50% recombinant gametes. Therefore the (maximum) distance that two genes can be apart and still measure that distance is just less (than) 50 cM. If two genes are greater than 50 cM apart, then we can not determine if they reside on the same chromosome or are on different chromosomes.

Wikipedia gives an analytical solution and also remarks that (d is physical distance):

The probability of recombination is approximately d/100 for small values of d and approaches 50% as d goes to infinity.

However, what happens if the linkage is greater than 50 cM? It seems like classical experiments aside, such a situation is possible in reality.

For instance, yeast chromosome IV is 1530 kb long, and averages 0.31 cM/kbp. Let's take two genes on Chr.IV:

  • DNF2 is at about +631 kb.
  • TOM1 is at about +1370 kb.

The physical distance between these is about 734 kb. According to the cM/kbp value, the centimorgan distance is $734\text{ kb} \cdot 0.31\text{ cM}/\text{kb} = 228 \text{ cM}$ . How to interpret this 228?

If I mate a haploid yeast which is DNF2 TOM1 with another which is dnf2 tom1 (lowercase indicating minor allele rather than deletion), then sporulate them, what is the chance of getting spores with DNF2 tom1 or dnf2 TOM1 genotypes?

I realize that the cM/kbp value is only a simplification and that in practice linkage is a more complex phenomenon. Nevertheless, it seems plausible that several crossover events can happen on Chr.IV since it is so big. This includes the possibility of 1, 3, 5 and more crossovers which would create a hybrid (assuming they all happen between these two loci), as well as the possibility of 2, 4, etc crossovers which would create a non-hybrid spore (at least as far as our chosen markers are concerned).


2 Answers 2


Understanding the statistics we use when talking about recombination rate is an important question that is unfortunately too often dismissed in an intro course to evolutionary biology or population genetics and that is misunderstood by many. I have a partial understanding of it but hopefully, I'll manage to write a decent answer.

#Short answer

A recombination rate and a genetic distance (in centiMorgan) are two different things. While the recombination rate is bounded between 0 and 0.5, the genetic distance is bounded between 0 and infinity. There is a one-to-one function from the genetic distance to the recombination rate. For small recombination rate, the recombination rate and the genetic distance take very similar values but for greater values, the recombination rate is much lower than the genetic distance.

#Long answer

There is a tiny bit of math below. These equations are mainly for curiosity as one can understand the answer without understanding the maths behind it.

Definitions of $r$ and $M$

You are getting confused between two different statistics

  • The rate of recombination $r$ between two loci
  • $r$ is the probability for two loci to remain in the same gamete after recombination occurred. This probability cannot be greater than 0.5 ($0 \le r \le 0.5$).
  • The distance in Morgans $M$ (or more commonly in centiMorgans) between two loci. $M$ is the expected number of cross-over that occurs between the two loci.

Morgans and centiMorgans

You will note that I talk in Morgans rather than centiMorgans, which is not typical in the literature, but it helps at conveying the intuition of what it means. If $150$ centiMorgans $=$ $1.5$ Morgans. For two loci distant by 1.5 Morgans, the expected number of cross-over between the two loci is 1.5 (by definition). Below are some more explanations on these two definitions with some drawing :)

Case study with loci A and B

While $M$ and $r$ are closely related, they are not exactly the same thing. Consider the following sequence with the loci A and B


Let's assumed the two loci are very far apart and $M=2$. The probability of having exactly $k$ crossovers is therefore given by a Poisson distribution with rate $M=2$

$$P(k) = \frac{e^{-M}M^k}{k!}$$

Let's say for a given case that $k=1$ (a single cross-over occurred). This cross over is represented by a "/" below


Here clearly the two sequences at loci A and B will be separated. Let's say now that $k=2$ (two cross-over occurred).


Here, even if crossovers have occurred, the two sequences at loci A and B will remain together. Only the sequence in between the two cross-overs will come from the homologous chromosome.

Relationship between $r$ and $M$ - in words

You might see it coming from the previous section. The probability $r$ of these two loci A and B to be separated via recombination is the probability of an odd number of recombination events to occur between them (knowing that $M$ is the expected number of cross-overs).

Relationship between $r$ and $M$ - in equation

Let's calculate first the probability $p_{even}$ that an even number of cross-overs occur. This probability is just

$$p_{even} = \sum_{k = 0}^{\infty}{e^{-M}M^{2k} \over (2k)!}$$

, where I just added the constant $2$ before $k$ at both the numerator and denominator. With some algebra and trig, one can show that

$$p_{even} = \sum_{k = 0}^{\infty}{e^{-M}M^{2k} \over (2k)!} = e^{-M}\sum_{k = 0}^{\infty}{M^{2k} \over (2k)!} = $$

$$e^{-M} \: cosh(M) = e^{-M}\left ( \frac{e^{M} + e^{-M}}{2}\right ) = {1 + e^{-2M}\over 2} $$

As, $r = p_{odd} = 1 - p_{even}$,

$$r = 1 - {1 + e^{-2M}\over 2} = {1 - e^{-2M}\over 2}$$

Here we go! We have our relationship between $r$ and $M$! Let's graph it

#Relationship between $r$ and $M$ - on a graph

I just graphed the above equation in Mathematica (Plot[y = (1 - Exp[-2 M ])/2, {M, 0, 5}])

enter image description here

Here is the same graph but zoomed on lower values of $r$ and $M$ (Plot[y = (1 - Exp[-2 M ])/2, {M, 0, 0.1}])

enter image description here

We clearly see from the graph that for low values of $M$, as $M$ increases $r$ increase quasi linearly ($M≈r$). For greater values of $M$, $r$ still increases but slower and slower until reaching an asymptote / plateau at $\frac{1}{2}$. $r$ is indeed bounded between 0 (when $M=0$) and $\frac{1}{2}$ (when $M=\infty$).

Note, the fact that the sum of probabilities of every even $k$ in a Poisson distribution is always lower or equal to 0.5 is an interesting math fact in itself!

  • $\begingroup$ From a deleted comment by @emre : You should start with $p_{even}=\sum_{k=0}^{\infty}\frac{2}{2}\frac{d^{k}e^{-d}}{(k)!}$. Because, simply, $\sum_{k=0}^{\infty}\frac{d^{2k}e^{-d}}{(2k)!}\not=\sum_{k=0}^{\infty}\frac{d^{2k}e^{-d}}{2(k)!}$ $\endgroup$
    – AliceD
    Commented Sep 6, 2017 at 18:25
  • $\begingroup$ @Remi.b As AliceD pointed out, some of your math was wrong. I modified it to be valid, and to still demonstrate the equality in question by using common trig identities of $cosh(x)$. $\endgroup$
    – user22020
    Commented Sep 20, 2017 at 3:12

Just to be clear, markers on different linkage groups (chromosomes) do not recombine 50% of the time. It is just that unlinked mutations co-segregate 50% of the time.

This is a direct quote from Strickberger's textbook "Genetics" 3rd ed. 1985 p 397:

> After many tests involving numerous sex-linked genes, the entire X chromosome of D. melanogaster was mapped and found to have a length of 68 map units. This, however, does not mean that there is a recombination frequency of 68 percent in a linkage experiment between yellow and bobbed which are at the extreme ends of the X chromosome. As explained previously, no more than 50 percent recombination is expected between any two loci, since only two of four chromatids in a meiotic tetrad are involved in any particular crossover point. In fact, the actual recombination value observed between yellow and bobbed may be even less than 50 percent in a linkage experiment for the simple reason that not every X chromosome bivalent may have a crossover in that particular interval. <<

1 percent recombination = 1 map unit

  • 1
    $\begingroup$ Now that we know what this 68 does NOT mean. Can you explain what it means and how to interpret it? $\endgroup$
    – Remi.b
    Commented Feb 12, 2016 at 1:10
  • $\begingroup$ If you are trying to use meiotic recombination between visible markers to construct a genetic map of a chromosome then you will need multiple linked genetic markers. Once you order the markers with respect to each other then you do additional mapping crosses to get accurate recombination rates (cM) between the pairs of adjacent linked markers. Then you sum those distances up to get the total length of the linkage group in map units. $\endgroup$
    – mdperry
    Commented Feb 12, 2016 at 2:27
  • $\begingroup$ So if you have know that two genes are tightly linked to the two telomeres: TEL-A----Z-TEL, and you have a couple more well-placed genes between them: TEL-A-----D----------H---------Z-TEL. If A is 18 cM from D, and D is 20 cM from H, and H is 30 cM from Z then the total distance from A to Z is 18 + 20 + 30 = 68 map units. If however you mapped the distance between A and Z using just those two markers you would calculate that they co-segregate 50 percent of the time, and may be unlinked. So if you know they are linked but you get 50 cM then you know you need more markers for your map. $\endgroup$
    – mdperry
    Commented Feb 12, 2016 at 2:35
  • $\begingroup$ Yes, that's exactly what I would be suggesting to include in the answer :) $\endgroup$
    – Remi.b
    Commented Feb 12, 2016 at 5:25
  • $\begingroup$ You're killing me! (by which I mean LOL). Sorry, I wasn't trying to be deliberately dense or obtuse. I was confused because I know you have a rep of like 10X mine, and that you are an expert on evolution (among many other things), but for some reason my frame of mind when I wrote those comments was "he genuinely doesn't know, so I will try to explain it explicitly" and I didn't clue in at all to the StackExchange mantra subtext of improving my existing answer. I am almost totally off of Bio, I just found it too frustrating. So once my initial enthusiasm wore off, I drifted away. $\endgroup$
    – mdperry
    Commented Feb 12, 2016 at 12:24

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