Chapter Five Exercises
1. Re-compute the pattern of linkage disequilibrium at the human LDL locus.
Download the file consisting of the complete genotypes for 71 individuals at 88 SNPs in the LPL locus, derived from Figure 5.5. The genotypes are coded 0 for homozygotes of the common allele, 1 for heterozygotes, 2 for homozygotes of the more rare allele. Missing data is indicated as a point. Use appropriate software provided by your instructor (for example, GDA) to verify that the pattern of linkage disequilibrium is similar to that shown in Figure 5.4.
2. Carry out a hypothetical case-control association study from the LDL locus.
Download the Hypothetical Case-Control data file, consisting of the genotypes of 500 "affected" and "unaffected" individuals at five of the SNPs in the human LDL locus (based on the documented levels of linkage disequilibrium). The five biallelic SNPs are numbers 7, 18, 42, 51, and 70. The alleles are coded with the letters A, B, C, D, and E respectively.
Compute the case-control trend test and genotype test statistics for each of the five SNPs, and draw conclusions about the likely effect of LPL haplotype variation on the hypothetical disease status. Before doing this exercise, PLEASE NOTE THE FOLLOWING ERROR IN BOX 5.2: The squared term in the denominator of the trend test chi-square formula is misplaced outside the square bracket. It should be immediately inside the square bracket so that it is only the last (n1 + 2n2) term that is squared. The correct formula should be:
c2 = N[N(r1 + 2r2) - R(n1 + 2n2)2] / RS[N(n1 + 2n2) - (n1 + 2n2)2]
Also, the next sentence describing this formula should read:
"For the example given in the text, if none of the 302 alleles in the affected individuals were in homozygotes for the A allele (r0 = 127, r1 = 24, r2 = 0), while 9 of the 382 alleles in the unaffecteds were in homozygotes (s0 = 114, s1 = 68, r2 = 9), then c2 = 32.8."
This minor correction of the r0 and s0 values results in a c2 value of 26.4 instead of 27.8 in the next paragraph.
3. Carry out a hypothetical transmission-disequilibrium association study.
Now suppose that a follow-up replication study was performed in 500 families with one affected and one unaffected child, focusing solely on SNP 51. Perform TDT and S-TDT tests on the dataset in the Hypothetical Family file, and compare the power of the two tests. PLEASE NOTE THE FOLLOWING ERROR IN BOX 5.3: The formula for the Family-based TDT also has a mis-placed bracket, such that the numerator should be
b - [(b+c)/2]
rather than
[b-(b+c)]/2 .
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