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This question already has an answer here:

Centimorgan's aren't physical distances. The analogy that I could come up with is if you had an array with 10 items,

A = [a, b, c, d, e, f ,g ,h ,i ,j, k]

The distance of a in the array is closer to b than k, ie A[0] is closer to A[1] than A[10] so that the centimorgan of A[0] and A[1] is less than A[0] and A[10]?

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marked as duplicate by David, terdon, Bryan Krause♦, Charles, user24284 Jan 7 '18 at 10:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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I personally like to talk in Morgan (M), instead of centiMorgan (cM), because the Morgan has a more intuitive meaning than one-hundredth of its value.

Morgan

If two loci are at distance 1M, it means the expected number of cross-over happening between these two loci is 1. If they are at 0.1M, then the expected number of crossover is 0.1. If they are at 1.4M, then the expected number of crossover is 1.4.

centiMorgan

As the typical distance of interest is much lower than the Morgan, we typically talk in centiMorgan.

Rate of recombination

If in a given meiosis, there is 0 crossover between two loci, then the two alleles from the same haplotype will end up in the same gamete. If there is 1 crossover, then the two alleles from the same haplotype end up in different gametes. If there are 2 crossovers between two loci, then the two alleles from the same haplotype will end up in the same gamete.

In short, if the number of crossovers between two loci (which expected value is the number of Morgans) is even, then the two alleles from the same haplotype will end up in the same gamete.

At very low Morgan (say 0.000001M), the probability of having an odd number of crossovers is low and the recombination rate is therefore low. As the number of Morgans increases, this probability of having an even number of crossover approaches 0.5. The recombination rate can therefore never be greater than 0.5.

Physical distance

Note that as the recombination rate per nucleotide varies throughout the genome (and among genders and among lineages), the distance in Morgan between two loci is not a simple function of their physical distance in number of nucleotides. The two are obviously correlated though.

Human genome

In the human genome, the average per nucleotide recombination rate is of the order $5\cdot 10^{-8}$ (Wang et al. 2012). This means that 1 cM is about 1.5 millions of nucleotides.

Highly Related posts

This question has actually been answered several times already. You'll find more information, some drawings and some math at the posts

For a more introductory post, have a look at

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1 centimorgan is equivalent to the percent recombination frequency for 2 genes during the crossing over step of meiosis. You can calculate it with

(number of recombinant offspring)/(total number of offspring) * 100%

and the result will be in cM (centimorgans).

If you have many offspring that are recombinant for the 2 genes, you can see that you'll get a high percent recombination and therefore linkage (not physical) distance. Since there are many recombinant offspring, it can be inferred that the 2 genes have a lower likelihood of translocating together, since they are likely to be physically distant.

In the case where you have very few offspring that are recombinant, you'll get a low percent recombination and linkage distance because 2 genes that are close together have a greater likelihood of being translocated together during crossing over.

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  • $\begingroup$ This is actually wrong. You are confusing a distance in cM with a probability of recombining. They are not quite the same (also they are good approximation for closely linked loci). Please have a look at my answer or to the two posts I linked at the end of my answer. $\endgroup$ – Remi.b Nov 29 '17 at 19:04
  • $\begingroup$ Oops, my mistake. Isn't looking at recombinant offspring a "good" metric for inferring probability of recombining, though? $\endgroup$ – jidiculous Nov 29 '17 at 19:07
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    $\begingroup$ Yes it is but it is not the issue here. The distance in cM is not equal to the probability (in percent) of recombination. It is approximatively equal for low distance but start to get widely different for higher distances. $\endgroup$ – Remi.b Dec 12 '17 at 1:40
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The Morgan is a unit of how likely two traits will be conveyed together in offspring - also known as linkage.

Centimorgan's are not physical distances because the chromosome has specific variations in how well it crosses over when it recombines.

For instance in men and women there are more crossovers and this makes the genes appear to be further apart as they don't translate together in Meiosis...

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The recombination frequency of two genes on the same chromosome is expressed in centimorgans (cM).

1% recombination frequency is a distance of 1 centimorgan (cM) on the gene map.

1 centimorgan roughly corresponds to one million base pairs of DNA. The actual distance between two genes on a chromosome may vary, but this serves as a useful approximation to estimate the distance between two genes on a chromosome, in terms of the number of base pairs.

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  • $\begingroup$ 1% recombination frequency is a distance of 1 centimorgan (cM) on the gene map is wrong. It is a good approximation but is not correct. Please have a look at my answer for more info. $\endgroup$ – Remi.b Dec 12 '17 at 1:38

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