This represents a second family. Assume that the phase is known and identical to the first family.

The genotype frequencies are the same, so we can move directly to the lod score:

Z(c)=log[64(1-c)6]

    c

    Z
0 infinite
0.1 1.53
0.2 1.22
0.3 0.88
0.4 0.48
0.5 0

Now, as expected when all progeny exhibit the non-recombinant genotypes, the strongest support is for linkage! If we combine these two families, we simply add the individual lod scores:

    c

    Z
0 infinite
0.1 0.6
0.2 0.8
0.3 0.7
0.4 0.4
0.5 0

Now, the strongest support is for linkage at about the 0.2 frequency of recombination; 20 centiMorgans.

This example is not different from simply having more progeny from the first set of parents. The BIG difference gets to be when one is comparing pedigrees from other situations, in which the phase of the heterozygotes is known, but not the same.

Many students ask me how this differs from the chi-square tests frequently applied to 2 point linkage data. Lod score analysis in this setting really is not very different from comparing the chi-square values for a cross at 20% linkage with these 10 progeny from the same cross assuming no linkage. This is really because chi-square tests are likelihood methods applied to populations derived from defined lines.

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