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LOD score, or logarithm of odds score, is a statistical test used in genetic linkage analysis. The LOD score compares the probability of obtaining the test data if the two loci are linked to the probability of obtaining the test data if the two loci are not linked. [1]

  1. The most often one is when you have a small number of progeny. In that case, use the method below the heading "When progeny size is small". In case your progeny size is small and you feel that the number of recombinants and non-recombinants accurately represents the genetic distance between the two genes (which is rare in human genetics), use the method under the heading "When progeny size is large". [2]
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Method 1
Method 1 of 2:

When progeny size is small

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  1. Assume you take the genetic distance (r) as 0.1. [3]
  2. The probability of non-recombinations will be 1-r = 1-0.1 = 0.9
  3. (1-r)/2=0.45. The probability of recombinants will be r/2 = 0.1/2 = 0.05 [4]
    • The probability of Data if the two genes are linked is ((r/2)^R)*(((1-r)/2)^NR) where R= # of recombinants and NR = # of non-recombinants. So the probability of Data if the two genes are linked is ((0.05)^2)*((.45)^11)
    • The probability of Data if the two genes are non-linked is (0.25)^(R+NR) = (0.25)^13
    • Now log of odds = (probability of Data if the two genes are linked)/(probability of Data if the two genes are non-linked). So, Log of odds = (((.05)^2)*((0.45)^11)) / ((0.25)^13) = 25.7
    • The LOD score will be log10(25.7) = 1.41.
    • The highest value of LOD score you get is your correct answer. The genetic distance you've assumed is probably close to the actual distance between the two genes of interest.
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Method 2
Method 2 of 2:

When Progeny size is large

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  1. The more data you gather, the better. [5]
  2. For example, suppose alleles A1 and B1 one locus 1 came from one parent, and A2 and B2 on locus 2 came from the other. Progeny inheriting A1B1 or A2B2 is non - recombinant, while those inheriting A1B2 or A2B1 are recombinant.
  3. This is given by ((R/(R+NR))^R)*((1-(R/(R+NR)))^NR), where R = number of recombinants, NR = number of nonrecombinants. [6]
    • For example, in the MNS blood group system, M and S are linked, and N and s are linked. Suppose that in 100 random individuals, 25 have MS haplotype, 30 have Ms haplotype, 6 have NS haplotype, and 39 have Ns haplotype. Since M and S are linked, and N and s are linked, MS and Ns are non - recombinants, and Ms and NS are recombinants.
    • Thus, the number of recombinants = R = 30+ 6 = 36, while the number of nonrecombinants = NR = 25 + 39 = 64. The recombinant frequency is R/(R+NR) = 36/(36+64) = 0.36.
    • The probability of obtaining the results assuming the two loci are linked is therefore ((0.36)^36)*((1-(0.36))^64) = 4.19187538*10^-29.
  4. This is given by 0.5^(NR+R). In the example above, this is 0.5^(64+36) = 0.5^100 = 7.88860905×10^-31.
  5. Divide the probability of obtaining the results assuming the two loci are linked (from Step 3 above), by the probability of obtaining the results assuming the two loci are not linked (from Step 4 above). For our example, this equals 4.19187538*10^-29/7.88860905×10^-31 = 53.14.
  6. For the example, this equals log 53.14 = 1.73.
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      • For interpretation of LOD score, a LOD score greater than 3.0 is considered evidence for linkage. A LOD score of +3 indicates 1000 to 1 odds that the linkage being observed did not occur by chance. On the other hand, a LOD score less than -2.0 is considered evidence to exclude linkage. In the example above with LOD score +1.73, linkage is suggested but not confirmed. Greater number of data points will increase the LOD score if linkage exists for analysis.
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