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Testing Natural Selection vs. Genetic Drift in Phenotypic Evolution Using Quantitative Trait Locus Data
H. Allen Orraa Department of Biology, University of Rochester, Rochester, New York 14627
Corresponding author: H. Allen Orr, Department of Biology, University of Rochester, Rochester, New York 14627, haorr{at}urhep.pas.rochester.edu (E-mail).
Communicating editor: P. D. KEIGHTLEY
| ABSTRACT |
|---|
Evolutionary biologists have long sought a way to determine whether a phenotypic difference between two taxa was caused by natural selection or random genetic drift. Here I argue that data from quantitative trait locus (QTL) analyses can be used to test the null hypothesis of neutral phenotypic evolution. I propose a sign test that compares the observed number of plus and minus alleles in the "high line" with that expected under neutrality, conditioning on the known phenotypic difference between the taxa. Rejection of the null hypothesis implies a role for directional natural selection. This test is applicable to any character in any organism in which QTL analysis can be performed.
TO determine if a genetic difference is adaptive, we typically consider its phenotypic consequences. We might ask, for instance, if different alleles at PGI in Colias affect flying time or if AdhF survives better than AdhS among Drosophila in wine cellars. Here I consider the reverse possibility: Can we show that a phenotypic difference is adaptive by looking at the genes involved?
The idea is simple. Imagine that two plant varieties differ in height by 50 mm and that (an unusually powerful) quantitative trait locus analysis reveals that this difference is due to 50 quantitative trait loci (QTL). QTL analysis also reveals the direction and magnitude of each factor's phenotypic effect. Although many QTL in the tall line are likely to be plus ("tall") factors, some may be minus ("short"). Indeed, QTL studies have shown that mixtures of plus and minus factors in the high line are common (![]()
![]()
![]()
![]()
The problem, however, is more subtle than it first appears. Because one variety is taller than the other, it is obvious that the tall line must contain some plus factors. But because the mere existence of the phenotypic difference did not convince us of a role for natural selection, the necessary existence of some plus alleles in the high line cannot do so either. Instead, we must ask if the ratio of plus-to-minus alleles in the high line is more extreme than expected under neutrality given the known phenotypic difference.
The problem can be restated as follows. Under neutral evolution, we predict some distribution of phenotypic differences between taxa after a period of divergence (![]()
![]()
![]()
I sketch such a test here. As we will see, given the phenotypic difference separating two taxa, the number of QTL, and the approximate distribution of QTL effects, we can find the probability that the observed number of plus factors would show up in the high line under the null hypothesis of neutrality.
| THE NULL MODEL |
|---|
Our null hypothesis is that the observed phenotypic difference between two taxa is neutral. In particular, we assume that all phenotypes in the neighborhood of the observed two populations have equal fitness (e.g., ![]()
The phenotypic difference between our two taxa is due to n loci. At any locus, genotypic values, G, are assigned as follows:
The quantities a and d can take different values at different loci. When d = 0, there is no dominance, and the heterozygote is intermediate between the homozygotes. The test described below makes no assumption about dominance.
Populations P1 (the high line) and P2 (the low line) are assumed to be homozygous at all n loci. The ith locus in P1 might carry a plus allele (of effect G1i = a) or a minus allele (G1i = -a). With no epistasis, the mean phenotypes are
1 =
ni=1Gli and
2 =
ni=1G2i and the phenotypic difference between populations is
1 -
2 = 2
nG1i .
Because QTL analysis is imperfect, detecting only a subset of the genes causing a phenotypic difference, in practice n will refer to the number of factors detected in an actual QTL analysis of two particular lines. Similarly, we will usually replace
1 -
2 with R, the phenotypic difference between two lines that are homozygous for the appropriate alleles at the n loci actually found in QTL analysis: R = 2
nG1i .
Assume that the absolute value of QTL effects, |a|, for a trait are drawn from some probability density f(|a|). The choice of distribution does not matter for our purposes, as long as it can be written down. In practice, the distribution used will be decided by best fit to the data.
Under our null hypothesis of neutrality, note that:
- At each locus, each line has a 1/2 chance of fixing the plus allele by genetic drift.
- The distribution of fixed allelic effects reflects the distribution of mutations available. If alleles of small effect are more common than those of large effect, that is because small mutations arise more often than large ones.
These two facts let us model neutral phenotypic evolution as an unbiased random walk in which some step sizes are more common than others.
| QTL SIGN TEST |
|---|
General model:
Our question is simple: Given all the ways of neutrally evolving a phenotypic difference of R or more when drawing QTL from our distribution of effects, how often does one see the observed number of plus factors or more in the high line? (We consider differences of R or more because, presumably, we would have performed QTL analysis in either case.) If cases of neutral evolution yielding differences of R or more usually involve the observed number of plus factors, we have no reason to reject the null hypothesis. But if differences of R or more rarely involve such an extreme number of plus factors, we reject the null hypothesis.
To build our test, we must find the probability that n+ = 1, 2, ..., n plus factors will show up in the high line under neutrality conditioned on a line difference of R. Our critical probability, P, is just the total probability of observing at least n+obs plus factors in the high line by chance given R. In symbols,
![]() |
(1) |
If P < 0.05, we reject the null hypothesis.
By Bayes' theorem, the probability of finding n+ = i plus alleles in the high line is
![]() |
(2) |
The probability P(2
G1
R | n+ = j) is slightly more difficult to find and is derived in the Appendix 1 It is
![]() |
(3) |
+ and
- give the mean effects of the j plus factors and n - j minus factors, respectively, residing in the high line. f(j)+ (
+) and f(n-j)- (
-) are the sampling distributions of these means, and F(j)+(
+) is the cumulative distribution function of f(j)+(
+) .
Substituting into Equation 1, the critical probability P is
![]() |
(4) |
The sampling distributions in Equation 3 and Equation 4 depend on the distribution of QTL effects. If, for instance, QTL effects are gamma distributedas often assumed (![]()
2. In particular, with gamma distributed effects, that is, f(|a|) =
e-
|a|
, the density of sample means is
![]() |
(5) |
- to refer to the absolute value of the mean of the minus factors.) With gamma distributed QTL effects, one can therefore calculate the critical probability P exactly by numerical integration. The biologically important point is simple. Knowing only the phenotypic difference separating two taxa, the number of QTL, and the approximate distribution of QTL effects, we can find the probability that some number of plus factors would show up in the high line by chance. The ratio of plus-to-minus factors residing in the high line, in other words, can serve as a footprint for natural selection.
Although we assumed for illustration that QTL effects are gamma distributed, the QTL sign test can be performed for any distribution of QTL effects. Exact calculation of P is possible whenever the sampling distribution of means is known (![]()
A C program that calculates P for gamma distributed QTL is available from the author (see Appendix 1). This program also remedies a potential problem with the above simple approach: it allows one to set a threshold for QTL detection. Thus, although QTL are drawn from a distribution having estimated scale (
) and shape (ß) parameters, factors having an effect smaller than T are assumed to be undetectable and so are ignored. Use of a truncated distribution surely better captures the realities of QTL mapping. Although I ignore all parameter estimation problems in this note, this program can also obviously be used to assess the effect on P of variation in
and ß about their estimated values.
Equal effects:
It is worth considering a variation on the above model that might be appropriate in certain cases. Imagine that evolution is constrained to build phenotypes from factors of equal effect (cf. ![]()
- =
+ . For any combination of j plus factors and n - j minus factors, we can obtain the observed R only if
j =
. But R > 0 only when j > n/2. In other words, one cannot obtain the observed phenotypic difference of R unless most plus factors reside in the high line.
Thus, the probability of seeing n+obs or more plus factors in the high line conditioned on R > 0 equals the probability of seeing n+obs or more plus factors conditioned on the majority of plus factors residing in the high line, and
![]() |
(6) |
In the fortunate case where n is large, we can use a normal approximation:
![]() |
(7) |
If P < 0.05, we reject the neutral null hypothesis. A minimum of n = 6 factors must be detected by QTL analysis to reject the null hypothesis.
Although this test ignores all information from QTL analysis about the actual sizes of the factors involved, generously allowing factors to assume whatever
is required to explain R (for a given j), it has one obvious merit: it requires only that we know n and n+obs. Although this test is not as biologically realistic as the one described above, it is preferable toand more conservative thana simple sign test, which fails to condition on the necessary existence of plus factors in the high line.
| EXAMPLES |
|---|
I now consider three QTL data sets to demonstrate the use of the QTL sign test.
Tomato fruit mass:
In one of the best-known QTL analyses, ![]()
2a = 2.27 [units are log10 (grams)]. [I have averaged across the two California environments considered by ![]()
Looking across the several characters studied, ![]()
0.10.
When evolution is neutral and involves factors of mean effect of 1/
0.10, there is a small probability of finding all 11 plus alleles in the high line by chance alone: P = 0.02. We thus reject the neutral null hypothesis. This probability remains essentially unchanged for similar values of mean QTL effect and for any realistic QTL effect threshold (0 < T < 0.025, where 0.025 is close to the heterozygous effect of the smallest QTL actually seen).
This result is, of course, hardly surprising. Tomato fruit mass has obviously been subjected to strong directional artificial selection.
Maize grain weight:
Factors affecting grain weight between two inbred lines of maize were studied by ![]()
![]()
![]()
![]()
![]()
The magnitude of QTL effects is approximately exponentially distributed for most traits studied (see Figure 3 in ![]()
3.33. (This value is computed using only those QTL of significant effect; if all QTL are used, a slightly different value is obtained.)
Conditioning on the phenotypic difference explained by the QTL, n+ = 11 or more factors would result by chance alone P = 0.19 of the time. This result does not qualitatively depend on the precise value of the mean of QTL effects nor on any realistic threshold value (0
T
1.5). We cannot, therefore, reject the neutral null hypothesis.
Posterior lobe area in Drosophila:
![]()
![]()
86% of the species difference in the posterior lobe area. (I have assumed no dominance and have re-expressed ![]()
All eight QTL have roughly similar effects, ranging between |a| = 2.5% and 8% of the species difference (see also ![]()
![]()
![]()
| CONCLUSIONS |
|---|
Evolutionary biologists have long desired some method for determining if a phenotypic difference between taxa reflects adaptation or neutrality. Ironically, quantitative trait locus analysis was never intended as such a method. It is clear, however, that QTL data do provide information on the roles of natural selection vs. genetic drift in phenotypic evolution. Here I have suggested a test to extract this information. Although the power of the test is obviously constrained by the present power of QTL analyses, there is every reason to believe that future analyses will uncover more factors as well as provide less biased estimates of their effects. In principle, then, QTL analyses may routinely provide the information required to test for the action of directional natural selection.
It is important to understand what the QTL sign test does not test. In particular, we do not ask if a phenotypic difference can or cannot be explained by natural selection. As ![]()
To see why failure to reject the null hypothesis does not imply no role for selection, imagine that natural selection "built" the high line by fixing a major plus factor (which overshoots the optimum somewhat), followed by several smaller compensatory minus factors. It is very unlikely that we can reject the neutral null hypothesis in this case. Such a pattern is simply too common under neutral evolution. Regardless of any intuitions we may have about how selection acts, it will always be easier to reject the null hypothesis when more, rather than fewer, plus factors reside in the high line. Intuitively, it might also seem that selection would sometimes fix minus factors in the high line: linked minus factors could hitchhike to fixation during strong selection for a major plus factor. Although this could well occur, it should not complicate the present test: if natural selection could not separate the undesirable minus factor from the linked desired plus factor, it seems very unlikely that our F2 QTL analysis could do so.
It is also worth noting that rejection of the null hypothesis does not, strictly speaking, allow us to conclude that the analyzed character was the direct target of selection. One can never completely exclude the possibility that the measured character changed as a correlated response to selection (although this seems less plausible for the larger, and sometimes dramatic, character differences often considered in QTL analysis). In any case, rejection of the null hypothesis does demonstrate that the character's evolution was not neutral.
Several previous tests of neutral phenotypic evolution have been proposed (![]()
![]()
![]()
![]()
![]()
The present QTL test, however, does not require information on any quantitative genetic parameters or on time since separation. It requires only that information provided in standard QTL studies: the size of the phenotypic difference distinguishing two taxa, the number of QTL detected, and the direction and magnitude of QTL effects.
Second, rate tests are a priori in the sense that they predict a distribution of phenotypes expected under neutrality. The present test, however, is a posteriori: we begin with an observed phenotypic difference but know nothing of the distribution of differences expected under neutrality. The distinction matters because, in practice, we perform QTL analysis only on characters that are fairly different between taxa. Thus, although a Lande-Lynch-Hill analysis might predict mostly slight differences between taxa separated for a given length of time, these will not be the differences analyzed in QTL studies, as explained earlier.
Last, it is worth noting that the QTL sign test does not depend on any assumptions about dominance or about the shape of the distribution of QTL effects. As long as the parental lines are homozygous for QTL (which will typically be true for the inbred lines used in such analyses), the QTL sign test makes no assumption about heterozygous effect per se. Similarly, athough one must specify some distribution giving reasonably good fit to observed QTL effects, the QTL sign test can be performed no matter what distribution is deemed appropriate.
The most important assumption underlying the QTL sign test involves epistasis. Because the crux of the test involves asking if a given combination of plus and minus factors "adds up" to explaining the observed phenotypic difference, we obviously assume no epistasis. It is difficult to see how this assumption could be relaxed. To the extent, then, that epistasis is common and strong, the QTL sign test is limited. Fortunately, QTL analyses often reveal little epistasis (![]()
Of course, it may prove possible to build related tests of natural selection that are less sensitive to epistasis. And it will surely prove possible to refine the simple test sketched here, increasing its biological realism and addressing the complications of parameter estimation. But the underlying notion characterizing the test seems sound. QTL data must contain information on the role of natural selection in phenotypic evolution. Our task is to devise biologically realistic tests to extract this information.
| ACKNOWLEDGMENTS |
|---|
I thank J. A. COYNE, P. D. KEIGHTLEY, D. PRESGRAVES, and two anonymous reviewers for very helpful comments. I especially thank M. TURELLI for his careful reading of the manuscript (as well as for catching an error). This work was supported by National Institutes of Health grant GM-51932 and by the David and Lucile Packard Foundation.
Manuscript received January 20, 1998; Accepted for publication April 27, 1998.
| APPENDIX 1 |
|---|
Derivation of P(2
G1
R | n+ = j):
We find this probability as follows: f(j)+(
+) is the sampling distribution of means when drawing j plus factors from our distribution of QTL effects. Similarly, f(n-j)-(
-) is the sampling distribution of means when drawing n - j minus factors. Assume that, in some particular case, the n - j minus factors in P1 have a mean effect of
- . Then there is some probability that the mean of the j plus factors will be large enough to give a line difference of 2
G1
R . This probability is
![]() |
(A1) |
+] is the cumulative distribution function of f(j)+(
+). To find the total probability that j plus factors and n - j minus factors will yield a phenotypic difference of at least R, we weight the probability in Equation A1 by the frequency with which
- takes different values. Thus, we get ![]() |
(A2) |
Computer programs:
A computer program that performs the QTL sign test is available from the author. This program assumes that QTL effects are drawn from a gamma distribution. Source code (in C) and a stand-alone program for the Power Macintosh are available and can be downloaded at http://www.rochester.edu/College/bio/orrlab/orrhome.html. The program is a Monte Carlo, that is, it does not calculate probabilities by numerical integration. Although it is easy to calculate the critical P numerically, for example, via Mathematica, the Monte Carlo approach has the advantage of allowing one to easily test the effects of different QTL detection thresholds.
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