recombination frequency problem

Question

Three loci C, D and E are located on the same chromosome in this order. We found that the frequency of recombinants between C and D is 10% and that between D and E it is 20%. Assuming that crossing over occurs randomly on the chromosome, what is the expected frequency of recombinants between C and E?

I know that it should be less than 30% (10%+20%) but the answer given is 26%. I don't understand how they figured that out.

Edit :

The frequency of recombination is <30% because of the chance that a crossover happens between C and D & also between D and E. The probability of that to happen is 2% (10/100 X 20/ 100) So, why is it 26% instead of 28% ?

Another posssibility : Am I totally off the track ?

Answer 1

I got $26$% as the answer.

To get a recombination between C and E, there are two possible mechanisms:-

1. C and D produce a recombinant, but D and E remain linked, therefore the final genotype will be a recombinant considering C and E(Chiasmata between C and D). Here $P_1=P_{CD}\times P'_{DE}$ where $P$ is the probability of recombination and $P'=1-P$ is the probability of linkage.Hence $P_1=0.08$
2. C and D remain linked but D and E undergo recombination (Chiasmata between D and E). Here, $P_2=P'_{CD}\times P_{DE}=0.18$

If there is a chiasmata between both C and D, and D and E, the resultant will not have a recombination in C and E locus due to double crossing over. If there is no crossing over, there will be no recombination at either loci.

The net probability of recombination $P_{CE}$ will be the sum of these two, as any of these possibilities can occur to give a recombination between C and E. $P_{CE}=P_1+P_2=0.26$

And Biogirllajja, your calculation (the Edit) was almost correct, but you have to subtract $2$% from both $10$% and $20$%, because, both of them include the probability of crossover at one loci, including the case where the other loci has also crossed over. Hence, you need to subtract it from both, and then add to get $26$%. Tell me if you want me to expand on my explanation or provide some clarifications.

If we simply add the map distances, we get the actual map distance which disregards the possibility of double recombinations. All map units greater than 10cM have significant chances of double crossovers between them, and hence, the observed recombination frequency would be less than the expected 0.1. Hence, for distances greater than 10cM, the observed frequency of recombination will not be the same as (map distance/100).