# What are Centimorgans?

If you had your DNA completed by Ancestry.com, they now show how many centimorgans you share with the other people in your DNA profile. This is most interesting. My daughter who I have always though to be MOST like my side of the family has 3454 centimorgans in common with me. That is equal to 72% of her total centimorgans. What happened to the l/2 from father and l/2 from Mother. Sounds to me like it is possible to inherit more from one parent than the other. One thing that has been stated is that the number 3454 centimorgans is the amount of DNA that we share in common that was acquired from a recent ancestor. Since men only have 2809 centimorgans then her remaining dna = 47% of his total centimorgans. There is confusing information on the web with regard to centimorgans and how to calculate this. If anyone knows for sure, I would appreciate hearing from you.

From Wikipedia:

In genetics, a **centimorgan** (abbreviated **cM**) or **map unit** (**m.u.**) is a unit for measuring genetic linkage. It is defined as the distance between chromosome positions (also termed loci or markers) for which the expected average number of intervening chromosomal crossovers in a single generation is 0.01. It is often used to infer distance along a chromosome. However, it is not a true physical distance.

The number of base-pairs to which it corresponds varies widely across the genome (different regions of a chromosome have different propensities towards crossover) and it also depends on if the meiosis where the crossing-over takes place is a part of oogenesis (formation of female gametes) or spermatogenesis (formation of male gametes).

One centimorgan corresponds to about 1 million base pairs in humans on average. The relationship is only rough as the physical chromosomal distance corresponding to one centimorgan varies from place to place in the genome, and also varies between men and women since recombination during gamete formation in females is significantly more frequent than in males. Morton et al. calculated that the female genome is 4782 centimorgans long, while the male genome is only 2809 centimorgans long.^{[3]} *Plasmodium falciparum* has an average recombination distance of ~15 kb per centimorgan: markers separated by 15 kb of DNA (15,000 nucleotides) have an expected rate of chromosomal crossovers of 0.01 per generation. Note that non-syntenic genes (genes residing on different chromosomes) are inherently unlinked, and cM distances have no meaning between them.

Because genetic recombination between two markers is detected only if there are an odd number of chromosomal crossovers between the two markers, the distance in centimorgans does not correspond exactly to the probability of genetic recombination. Assuming Haldane’s map function, in which the number of chromosomal crossovers is distributed according to a Poisson distribution,^{[4]} a genetic distance of *d* centimorgans will lead to an odd number of chromosomal crossovers, and hence a detectable genetic recombination, with probability

- {\displaystyle \Pr[{\text{recombination}}|{\text{linkage of }}d{\text{ cM}}]=\sum _{k=0}^{\infty }\Pr[2k+1{\text{ crossovers}}|{\text{linkage of }}d{\text{ cM}}]}
- {\displaystyle {}=\sum _{k=0}^{\infty }e^{-d/100}{\frac {(d/100)^{2\,k+1}}{(2\,k+1)!}}=e^{-d/100}\sinh(d/100)={\frac {1-e^{-2d/100}}{2}}\,,}

where sinh is the hyperbolic sine function. The probability of recombination is approximately *d*/100 for small values of *d* and approaches 50% as *d* goes to infinity.

The formula can be inverted, giving the distance in centimorgans as a function of the recombination probability:

- {\displaystyle d=50\ln \left({\frac {1}{1-2\Pr[{\text{recombination}}]}}\right)\,.}

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