Theoretical Considerations
Power Analysis of QTL Mapping with CSS vs. Crosses
We calculate the number of progeny required to detect a QTL using a CSS panel versus using traditional crosses, under straightforward and commonly used assumptions about the QTLs.
Suppose that host Strain A and donor Strain B differ with respect to a quantitative trait, owing to allelic differences at a modest number of loci, L1, L2, .…, Lk. We will assume that the loci are unlinked, that alleles contribute additively without dominance or epistasis, and that environmental noise is additive and normal. The phenotype of a mouse or inbred strain derived from Strains A and B is therefore given by:
Y = a1d1 + a2d2 + …. akdk + e,
where ai is the number of copies (0, 1, or 2) and di is the effect of each copy of the B allele at locus Li , and e is a normal variable with variance Ve.
The sample needed to detect the effect of an explanatory variable X on a quantity Y is given by a basic statistical formula:
N = (Ca)2[Vres / Vexp], 1
where Vexp is the variance explained by X, Vres is the residual variance unexplained by X, a is the required significance level, and Ca is the threshold such that Prob(|Z|>Ca) = a for a standard normal variable Z.
A CSS panel typically has greater power than crosses for two reasons that are directly related to Eq. (1). First, a given QTL will usually explain a greater proportion of the total variance because a CSS eliminates the variance resulting from segregation of QTLs on other chromosomes. Second, a lower significance level a suffices because the problem of multiple testing is less severe in comparing a limited number of CSSs than in a survey of many markers across the entire genome.
Progeny required with F2 intercrosses. Specifically, suppose that one wishes to detect a locus L1 in an F2 intercross. The relevant quantities are
Vexp = 1/2 (d12) = V1
Vres = 1/2 (d22 + d32. … + dk2) + VE = VG* + VE,
where VG* denotes the genetic variance due to the segregation of the other k-1 QTLs. A stringent significance level a is needed to correct for searching the entire genome to achieve a genome-wide false positive rate of 5%. According to ref (23), the appropriate level a is:
a = 5.2 x 10-5
(Ca)2 = 65.5
From (1), the number (NF2) of intercross progeny required to detect a QTL is therefore:
NF2= 65.5 ([VG* + VE] / V1) 2
Progeny required with a panel of CSSs. With a CSS p1nel, we will compare a sample of n mice from strain A to a sample of n mice from each of 20 CSS (ignoring chromosome Y and mitochondria). We thus use a total of 21n mice and perform 20 comparisons. The nominal significance level a needed to ensure a overall false positive rate of 5% after correcting for multiple hypothesis testing is therefore a = 0.0025, which corresponds to (Ca)2 = 9.1.
With locus L1 as the only major QTL on chromosome j, the comparison between CSSj and Strain A is
Vexp = d12= 2V1
Vres = VE.
(N.B. The genetic variance is twice as large as for the F2 intercoss, because the CSS involves homozygous mice only whereas the intercross involves both homozygotes and intermediate heterozygotes.)
By (1), the number of mice required for each pairwise comparison is then
2n = 9.1 (VE / 2V1)
and the total number NCSS of mice to be phenotyped is
NCSS = 21n = 47.8 (VE / V1) 3
Comparison between CSS and F2 intercross. Comparison of equations (2) and (3) shows that NF2 is always larger than NCSS, with the ratio being:
NF2/NCSS = 1.37 [1 + (VG*/VE)]
The F2 intercross thus requires a relative increase of at least 37% in the number of intercross mice to identify the presence of a QTL on a chromosome, compared to using CSS panels. This 37% increase occurs when there are no other QTLs of substantial effect (that is, VG*/VE ~ 0). The increase is much larger when there are additional segregating QTLs contributing to the genetic variance. CSS panels thus become increasingly advantageous as the genetic complexity of the trait increases.
In short, a CSS panel can detect a QTL with a given effect with fewer mice than a cross, or a QTL with weaker effects with the same number of mice as in a cross. A recent theoretical study (28) compared CSSs and segregating populations, with similar results as above. This study showed that the heritability of a given QTL is usually several times greater, thresholds for declaring significant and suggestive linkage were 20-30 times greater, and the sample size needed to detect linkage is smaller.
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