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--- chapter_22 [2024/11/25 18:23] – [LOD score formula for known phase] mike
+++ chapter_22 [2024/11/25 18:34] (current) – [LOD score formula for unknown phase] mike
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 In mouse experiments, the phase of all offspring are usually known. Now, let's return to a human example where it is more common that the phase is unknown:
-<figure Fig3>
+<figure Fig4>
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 $$ \begin{aligned} \text{LOD} &=\log{\frac{\frac{\theta^{R1} \cdot (1-\theta)^{NR1}+\theta^{R2} \cdot (1-\theta)^{NR2}}{2}}{(\frac{1}{2})^{R+NR}}} \\
 &=(T-1)\log{2}+\log{\theta^{R1} \cdot (1-\theta)^{NR1}+\theta^{R2} \cdot (1-\theta)^{NR2}}\end{aligned}$$
+In this example, there are a total of 5 informative chromosomes (R1=1, NR1=4; R2=4, NR2=1). Using the formula, we can calculate that at θ=0.25 we get a max LOD score of 0.25 - this does not cross the threshold of LOD=3.3 and therefore is not evidence of linkage. When the phase is unknown, LOD scores will be substantially lower than if the phase were known. If we knew the phase in our example to be phase 1, we could calculate at θ=0.25, LOD=0.403. That's still not significant but it's higher.
+One last important point on LOD scores is that they are additive. This is because probabilities (odds) are multiplicative, but since we're using the logarithm of odds, we can use addition. In our example above, we were able to calculate the LOD score for linkage between $D$ and $m$ for this family as LOD=0.25 at θ=0.25. If we obtained data from another family that resulted in LOD=0.24 at θ=0.25, we could then combine the data for the two families as LOD=0.25+0.29=0.54. This makes analyzing the data much easier and intuitive; the more data you have, the stronger case you have for linkage.