Abstract
Random mutagenesis screens for recessive phenotypes require three generations of breeding, using either a backcross (BC) or intercross (IC) strategy. Hence, they are more costly and technically demanding than those for dominant phenotypes. Maximizing the return from these screens requires maximizing the number of mutations that are bred to homozyosity in the G3 generation. Using a probabilistic approach, we compare different designs of screens for recessive phenotypes and the impact each one has on the number of mutations that can be effectively screened. We address the issue of BC versus IC strategies and consider genome-wide, region-specific screens and suppressor screens. We find that optimally designed BC and IC screens allow the screening of, on average, similar numbers of mutations but that interpedigree variation is more pronounced when the IC strategy is employed. By conducting a retrospective analysis of published mutagenesis screens, we show that, depending on the strategy, a threefold difference in the numbers of mutations screened per animal used could be expected. This method allows researchers to contrast, for a range of experimental designs, the cost per mutation screened and to maximize the number of mutations that one can expect to screen in a given experiment.
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Notes
The formula used here is (average number of functional mutations per G1 pair) × (number of G1 pairs) × (average proportion of mutations screened per G1 pair). The final value in this product comes from Table 1 (efficiencies of mutation screens), where we convert from percentage to proportion by dividing by 100.
For brevity, we present only equations based on the less restrictive assumption that the number of pups born to each G2 female is flexible (while the number of pups born to each G2 male is fixed)
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Acknowledgments
Doug Hilton is an Australian National Health and Medical Research Council (NHMRC) Senior Principal Research Fellow. Benjamin Kile is an Australian Research Council (ARC) Queen Elizabeth II Fellow. Melanie Bahlo an NHMRC R. D. Wright Fellow. Jeremy Silver is a Bio21 Undergraduate Research Opportunities Program Scholar. The authors are particularly grateful to Vineeth Varughese for producing a web interface for EffiSim. They also thank Terry Speed for reviewing an early version of this work and Carolyn deGraaf for reviewing a later draft of this article; Gordon Smyth for various helpful suggestions; Peter Maltezos for producing Figures 1–3; and Tony Papenfuss, Jim Stankovich, and Keith Satterley for programming advice. Finally, five anonymous referees provided insightful and useful comments.
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Melanie Bahlo and Benjamin T. Kile contributed equally to this work.
Appendix: Genome-wide efficiency derivations
Appendix: Genome-wide efficiency derivations
Probability of screening a given mutation
We begin by deriving the probability that no homozygotes (for a particular mutation) are born in a GPF. We fix the number of pups per G2 father, r, but not the number of pups per G2 female. We assume any pup could have come from any female with equal probability. This also includes the possibility that all pups came from the same female. To denote the genotype of mice at a mutation locus, let “m” and “+” represent the mutant and wild-type alleles (respectively). For the BC,
In the case of the IC, there can be several GPFs per G1 mating pair (h > 1). Hence,
Alternatively, we could fix the number of pups per G2 female, n, rather than the number of pups per G2 male. Then for the BC,
and for the IC,
Let us define P IC(h,θ) to be the probability of producing at least one G3 homozygote with the IC, where θ denotes parameters k and n or r (depending on whether we assume that the number of pups per G2 female is fixed or flexible, respectively). Similarly, define P BC(1,θ) for the BC. Note here that P IC(1,θ) = 0.5P BC(1,θ) under both sets of assumptions.
Mutation screen using the BC
Let X be the number of mutations in the region being screened that are carried by the G1 founder, and let E(X) = x. Assuming that mutations are segregating independently, then
,so
Mutation screen using the IC
Let X 1 and X 2 be the numbers of mutations carried by the two G1 founders (in the target region), both with mean x. If we again assume no linkage,
, then
Suppressor screen using the BC Footnote 2
Now consider screens for recessive suppressors of a recessive phenotype that use the BC. Let M sBC(1,X,d,1,k,r) be the number of mutations for which at least one G3 pup is homozygous, and homozygous for the sensitized allele. Results from the above derivations are applicable provided that we count only homozygous mutations in those G3 that are homozygous for the sensitized allele. Let Z j be the number of G3 pups in the jth GPF that are homozygous for the sensitized allele. Note that the Z j are independent and identically distributed. Then
So
If the sensitized allele is dominant rather than recessive, then we need to count homozygous mutations in those G3 that carry at least one copy of the sensitized allele. Then
So
Suppressor screen using the IC
Using a similar argument as above, we relate results from the BC to the IC.
Again, suppose that the sensitized allele is dominant, then
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Silver, J.D., Hilton, D.J., Bahlo, M. et al. Probabilistic analysis of recessive mutagenesis screen strategies. Mamm Genome 18, 5–22 (2007). https://doi.org/10.1007/s00335-006-0057-z
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DOI: https://doi.org/10.1007/s00335-006-0057-z