Haldane's Mapping Function

Citations:
   Crow, J.F. and W.F. Dove. 1990. Anecdotal, historical and critical commentaries on genetics. Genetics 125:669-671.
   Liu, B.H. 1997. Statistica Genomics: Linkage, Mapping, and QTL Analysis. CRC Press. Pgs. 18-19. Pgs. 328-329.
   Haldane, J.B.S. 1919. J. Genet. 8:299-309.

Synopsis
1) Haldane's mapping function adjusts the observed proportion of recombinant gametes for unobserved double crossovers so that map distances are additive.
2) When observed recombination is less than r = 10% (r<10%), we do not need to use Haldane's mapping function. This is because when loci are located close together the amount of double crossovers within such a small interval is negligible.
3) When r = 50% two loci are inherited independently and the distance between them in cM is infinite. This means that when two loci are inherited independently, we can not determine how many cM there are between the two loci.
4) When two loci are separated by 50cM, the two loci are not inherited independently. A 50 cM map distance between two loci is the equivalent of 32% recombination or r = 0.32.

Derivation of Haldane's Mapping Function:

Let r = observed recombination in crossover units (c.u); d = actual recombination in cM l = number of crossover events = 2d

Distribution of the number of events     Probability of that number of events
0                                            e^-l = e^-2d
1                                            le^-l
2                                            (l²/2)e^-l
3                                            (l³/3!)e^-l

If e^-l is the probability of zero crossover events, then 1 - e^-l is the probability of at least one crossover event. The probability of at least one recombinant gamete is r = (½)(1 - e^-l), because for each crossover event only one-half of the gametes will be recombinant types. This is because only two of four strands are involved in a crossover. Therefore, there is a coefficient of one-half in front of the (1 - e^-l) term. The exponent of l is explained by the fact that the probability of a crossover is 2d.

Determining the map distance using Haldane's mapping function is based on the observed proportion of recombinant gametes adjusted for the number of unobservable double crossovers. Solving for d:
2r = 1 - e^-l = 1 - Prob.( zero crossover events) where the probability of zero crossover events equals e^-l. We can put the equation 2r = 1 - e^-l in words. The probability of at least one recombinant event equals one minus the probability of no crossover events. We can solve for l.

       2r = 1 - e^-l
       e^-l = 1 - 2r
       ln( e^-l) = ln(1 - 2r)
       -l = ln(1 - 2r)

Now 2d = l because d is a measure of the proportion of actual recombinant gametes and l is a measure of the proportion of actual crossover events. d = l/2 means that each crossover event results in only one-half recombinant gametes because only 2 of the 4 strands participate in a crossover event.

       -l = ln(1 - 2r)
       -2d = ln(1 - 2r)
       d = -(1/2)ln(1 - 2r)

The above equation provides a solution for d in M for an observed recombination of r crossover units. For short chromosome segments the map distance (d) equals the recombination fraction and d = r. For example, 4 c.u. = 4 cM = 8% crossover events. When r < 10 c.u. then r = d. However, consider three loci, each separated by 20 c.u.

Let r_AC = observed recombination between the A and C loci;
     r_AB = observed recombination between the A and B loci;
     r_BC = observed recombination between the B and C loci.

1 - r_AC = (1 - r_AB)(1 - r_BC)
r_AC = r_AB + r_BC - 2r_AB r_BC
r_AC = 0.2 + 0.2 - 2(0.2)(0.2) = 0.32

The reason is as follows:
1 - 2 r_AC = the probability of no crossover event between the A and C loci;
1 - 2 r_AB = the probability of no crossover event between the A and B loci;
1 - 2 r_BC = the probability of no crossover event between the B and C loci.

The probability of no crossover event between the A and C loci is the probability of no crossover event between the A and B loci multiplied by the probability of no crossover event between the B and C loci.

       (1 - 2 r_AC) = (1 - 2r_AB)(1 - 2r_BC)
       1 - 2 r_AC = 1 - 2r_BC - 2r_AB + 4r_ABr_BC
       2 r_AC = 2 r_BC + 2 r_AB - 4r_AB r_BC
       r_AC = r_BC + r_AB - 2r_ABr_BC

In our example, r_BC = 0.2 = r_AB then,
r_AC = r_BC + r_AB - 2r_ABr_BC
      = 0.2 + 0.2 - 2(0.2)(0.2)
      = 0.32 c.u. = 32%

The point here is that the observed recombination between the A and C loci does not add up to the sum of the observed recombination between the A and B loci plus the observed recombination between the B and C loci.This is why we need to use Haldane's mapping function, to make recombination additive between loci.

Let us convert r_AC, which is the observed proportion of recombination gametes into the actual proportion of recombinant gametes.

d = -(½) ln(1 - 2r)
   = -(½)ln(1 - 2r_AC)
   = -(½)ln(1 - 0.64)
   = -(½)ln(0.36)
   = -(½)ln(-1.022)
   = 0.51 M

When loci are closely linked, it is not necessary to convert recombination units to cM. Let r = 0.1. We can convert this to d = 11 cM, which is approximately equal to 10 c.u. This means that for closely linked loci the distances are additive.

We will now convert our observed recombination values to actual recombination values.
Previously in our example, r_AB = 0.2 = r_BC; r_AC = 0.32
We have shown that:
        r_AC = r_BC + r_AB - 2r_ABr_BC and d = -(½)ln(1 - 2r)

d_AB = -(½)ln(1 - 2r_AB) = -(½)ln(1 - 0.4) = 0.255 M = 25.5 cM
d_BC = -(½)ln(1 - 2r_BC) = -(½)ln(1 - 0.4) = 0.255 M = 25.5 cM
d_AC = -(½)ln(1 - 2r_AC) = -(½)ln(1 - 0.64) = 0.511 M =51.1 cM

Haldane's mapping function converts recombination fractions to map distances, which are now additive.
              r_AB + r_BC = r_AC, but d_AB + d_BC = d_AC

Why was r_AB = 0.2 converted to d_AB = 25.5 cM? Because there are double crossovers between the A and B loci which we cannot detect. There is no locus between the A anc B loci which would permit us to identify double crossovers between A and B. Double crossovers between the A and B loci do not result in observable recombination between the A and B loci.

We have shown that the average recombination for 2, 3 and 4-strand double crossovers is 50% recombination. Double crossovers should be counted as two crossover events, but the recombination fraction appears the same as for a single crossover event. Thus, the relative frequency of crossover events is underestimated by the observable recombination fraction. We can use the function d = -(½)ln(1 - 2r) to convert observed recombination (r) to map distance in centiMorgans (d).

Independent Loci Example:

Let r = 0.5, which means that the two loci are inherited independently.
    d = -(½)ln(1 - 2r)
       = -(½)ln(1 - 1)
       = -(½)ln(0)
       = infinity

Let r = 0.49
    d = -(½)ln(1 - 2r)
       = -(½)ln(1 - 0.98)
       = 1.956 M = 195.6 cM

50cM Example:

Now we will use the formula to convert map distance to observed recombination. We have to convert d from cM to Morgans to use the formula.
Let d = 50 cM = 0.5 M
    d = -(½)ln(1 - 2r)
    0.5 M = -(½) ln (1 - 2r)
    -1 = ln( 1 - 2r)
    e^-1 = 1 - 2r
    1 - e^-1 = 2r
    (½) (1 - e ^-1) = r
    (½) (1 - 1/e) = r
    ½ (1 - 0.368) = r = 0.316 = 31.6 c.u.

We can see that 50cM = 31.6 c.u. A 50 cM map distance does not equal independence between loci. 50 c.u. = 50% recombination and 50 cM = 32 c.u.

Example of Loci Closely Linked:

When there is less than 10% recombination between loci, it is not necessary to use Haldane's mapping function to convert to cM. There are so few double crossovers between closely linked loci that the observed and actual recombination are approximately equal.
Let r_DE = 6 c.u. and r_EF = 8 c.u.,
then r_DF = r_EF + r_DE = 6 + 8 = 14 c.u. = 14 cM