Protocol Online logo
Top : New Forum Archives (2009-): : -Genetics and Epigenetics-

Calcualting genetic distance - (Jun/30/2013 )

So the question is,

A mutant in the Columbia ecotype of Arabidopsis, pass1, was mapped relative to marker A. For homozygous F2 pass1 plants from a Columbia © x Landsberg (L) cross the results for marker A were 60 CC, 10 CL, and 4 LL. Calculate the genetic distance (cM) between marker A and pass1.

I used the equation: Distance = (Number of recombinant chromosomes)/(Recombinant chromosomes+parental)) x 100 and got 13.51%, however, this does not seem to be the correct answer. Can anyone tell me what I have done wrong here?

This is a multiple choice question and so the answer should be either: 5.4, 18.9, 12.2 or 9.5.


I honestly don't get the crossing scheme there, so I can't tell what is the number of recombinantsm even if it seems to be the CL number on a first sight.

From the options, the right answer should be one of these recombinant numbers: 4, 14, 9, 7 but I can't see right know where to get any of these numbers.


Thanks for the reply Trof. I guess I'll keep plugging at it. Could you though explain or show how you came up with your answers (4, 14, 9, 7) because it might help conceptualize the method to me since none of your answers were the same as mine. So clearly, you used a different approach to the problem.


Well, your equation is a right one for calculating the genetic distance from two linked genes/markers, like A/a and B/b, where it's used to calculate from the percentage in the cross between a dihybride Ab/aB and recessive ab/ab parents, creating parental Ab/ab, aB/ab and recombinant ab/ab and AB/ab genotypes in the offspring. But as I said I failed to fill in the given numbers and crossing schemes into this calculation, so first of all I don't know if the equation used is actually the right one for this case. I can't tell what are the genotypes here from the sentence "results for marker A were 60 CC, 10 CL, and 4 LL" , as this is very different from the usuall settings (mentioned above) I'm used to.

But if we assume it is right, from that equation, the results should be 5.4, 18.9, 12.2 or 9.5 % (for easier calculations I took 0.054, 0.189 and so on, not multiplied by 100). And the divisor in this equation should be always the overall number of the whole offspring so 60+10+4 in this case. From knowing the possible answers and the divisor, you can calculate the number of recombinants supposed for each answer.
From your calculation you obviously took 10 as number of recombinants, to get 0.1351. I assumed the number of recombinants unknown, using the possible answers:
distance (i.e. 0.054) = x/74 -> x=~4

So we are still stuck with the same problem, I don't get the task setting and therefore I'm not sure if the equation is right or not in the first place. If you do understand the setting maybe if you can scheme out the crossing it may shed some light for me.


I am not even sure how I can carry out a F2 test cross for marker A, but I will take a stab at it. However, while pondering this further a thought did strike me; the question states that the mutant, pass1, is in the Colombia wild type and that it is mapped relative to marker A. I assume this means that marker A is a marker from the Colombia strain. Then this line:

For homozygous F2 pass1 plants from a Columbia ( C ) x Landsberg (L) cross the results for marker A were 60 CC, 10 CL, and 4 LL

Since these are results for marker A (which I assumed is a marker in the C genome) does that mean that all the L values are recombinant? Thus the total recombinant value is 14 (10 + 4)? If this is the case, then using the gene mapping equation (which is the only equation I know of that I can utilize) would give me a value of 18.9. Which is an answer on the MCQ. I am just not sure if my interpretation was correct. i.e. assuming that marker A is a fixed region on the Colombia ecotype and that the test cross values of L are therefore recombinations.