1. QTL Mapping Quantitative Trait Loci (QTL): A chromosomal segments that contribute to variation in a quantitative phenotype

2. Maize Teosinte tb-1/tb-1 mutant maize

3. Mapping Quantitative Trait Loci (QTL) in the F2 hybrids between maize and teosinte

4. Nature 432, 630 - 635 (02 December 2004) The role of barren stalk1 in the architecture of maize ANDREA GALLAVOTTI(1,2), QIONG ZHAO(3), JUNKO KYOZUKA(4), ROBERT B. MEELEY(5), MATTHEW K. RITTER1,*, JOHN F. DOEBLEY(3), M. ENRICO PÈ(2) & ROBERT J. SCHMIDT(1) 1 Section of Cell and Developmental Biology, University of California, San Diego, La Jolla, California 92093-0116, USA 2 Dipartimento di Scienze Biomolecolari e Biotecnologie, Università degli Studi di Milano, 20133 Milan, Italy 3 Laboratory of Genetics, University of Wisconsin, Madison, Wisconsin 53706, USA 4 Graduate School of Agriculture and Life Science, The University of Tokyo, Tokyo 113-8657, Japan 5 Crop Genetics Research, Pioneer-A DuPont Company, Johnston, Iowa 50131, USA * Present address: Biological Sciences Department, California Polytechnic State University, San Luis Obispo, California 93407, USA

5. Effects of ba1 mutations on maize development Mutant Wild type No tassel Tassel

6. A putative QTL affecting height in BC Sam- HeightQTL ple(cm, y) genotype 1184Qq (1) 2185Qq (1) 3180Qq (1) 4182Qq (1) 5167qq (0) 6169qq (0) 7165qq (0) 8166qq (0) If the QTL genotypes are known for each sample, as indicated at the left, then a simple ANOVA can be used to test statistical significance.

7. Suppose a backcross design ParentQQ (P1) x qq (P2) F1 Qq x qq (P2) BCQq qq Genetic effect a* 0 Genotypic value  +a* 

8. QTL regression model The phenotypic value for individual i affected by a QTL can be expressed as, y i =  + a* x* i + e i where is the overall mean, x* i is the indicator variable for QTL genotypes, defined as x* i =1 for Qq 0 for qq, a* is the “real” effect of the QTL and e i is the residual error, e i ~ N(0,  2 ). x* i is missing

9. Data format for a backcross Sam- Height Marker genotype QTL ple(cm, y) M1M2Aaaa 1184Mm (1)Nn (1)½ ½ 2185Mm (1)Nn (1) ½ ½ 3180Mm (1)Nn (1) ½ ½ 4182Mm (1)nn (0) ½ ½ 5167mm (0)nn (1) ½ ½ 6169mm (0)nn (0) ½ ½ 7165mm (0)nn (0) ½ ½ 8166mm (0)Nn (0) ½ ½ Observed dataMissing dataComplete data = +

10. Two statistical models I - Marker regression model y i =  + ax i + e i where x i is the indicator variable for marker genotypes defined as x i = 1 for Mm 0 for mm, a is the “effect” of the marker (but the marker has no effect. There is the a because of the existence of a putative QTL linked with the marker) e i ~ N(0,  2 )

11. Heights classified by markers (say marker 1) MarkerSampleSampleSample groupsizemeanvariance Mm n 1 = 4m 1 =182.75s 2 1 = mm n 0 = 4m 0 =166.75s 2 0 =

12. The hypothesis for the association between the marker and QTL H 0 : m 1 = m 0 H 1 : m 1  m 0 Calculate the test statistic: t = (m 1 –m 0 )/  [s 2 (1/n 1 +1/n 0 )], where s 2 = [(n 1 -1)s 2 1 +(n 0 -1)s 2 0 ]/(n 1 +n 0 –2) Compare t with the critical value t df=n1+n2-2 (0.05) from the t-table. If t > t df=n1+n2-2 (0.05), we reject H 0 at the significance level 0.05  there is a QTL If t < t df=n1+n2-2 (0.05), we accept H 0 at the significance level 0.05  there is no QTL

13. Why can the t-test probe a QTL? Assume a backcross with two genes, one marker (alleles M and m) and one QTL (allele Q and q). These two genes are linked with the recombination fraction of r. MmQqMmqqmmQqmmqq Frequency(1-r)/2r/2r/2(1-r)/2 Mean effectm+amm+am Mean of marker genotype Mm: m 1 = (1-r)/2 (m+a) + r/2 m = m + (1-r)a Mean of marker genotype mm: m 0 = r/2 (m+a) + (1-r)/2 m = m + ra The difference m 1 – m 0 = m + (1-r)a – m – ra = (1-2r)a

14. The difference of marker genotypes can reflect the size of the QTL, This reflection is confounded by the recombination fraction Based on the t-test, we cannot distinguish between the two cases, - Large QTL genetic effect but loose linkage with the marker - Small QTL effect but tight linkage with the marker

15. Example: marker analysis for body weight in a backcross of mice _____________________________________________________________________ Marker class 1Marker class 0 ___________________________________________ Markern1m1 s 2 1 n1m1 s 2 1 tP value _____________________________________________________________________________ 1 Hmg1-rs134154.20111.816247.3263.673.754 <0.01 2 DXMit574255.21104.126146.5156.124.99 <0.01 3 Rps17-rs114355.30101.986046.3054.385.231 <0.000001 _____________________________________________________________________

16. Marker analysis for the F2 In the F2 there are three marker genotypes, MM, Mm and mm, which allow for the test of additive and dominant genetic effects. GenotypeMeanVariance MM:m 2 s 2 2 Mm:m 1 s 2 1 mm:m 0 s 2 0

17. Testing for the additive effect H0: m 2 = m 0 H1: m 2  m 0 t 1 = (m 2 –m 0 )/  [s 2 (1/n 2 +1/n 0 )], where s 2 = [(n 2 -1)s 2 2 +(n 0 -1)s 2 0 ]/(n 1 +n 0 –2) Compare it with t df=n2+n0-2 (0.05)

18. Testing for the dominant effect H0: m 1 = (m 2 + m 0 )/2 H1: m 1  (m 2 + m 0 )/2 t 2 = [m 1 –(m 2 + m 0 )/2]/  {[s 2 [1/n 1 +1/(4n 2 )+1/(4n 0 )]], where s 2 = [(n 2 -1)s 2 2 +(n 1 -1)s 2 1 +(n 0 -1)s 2 0 ]/(n 2 +n 1 +n 0 –3) Compare it with t df=n2+n1+n0-3 (0.05)

19. Example: Marker analysis in an F2 of maize ______________________________________________________________________________________________ Marker class 2Marker class 1Marker class 0AdditiveDominant ________________________________________ Mn 2 m 2 s 2 2 n 1 m 1 s 2 1 n 0 m 0 s 2 0 t 1 P t 2 P _______________________________________________________________________________________________ 143 5.24 2.4486 4.27 2.9342 3.11 2.766.10 <0.0010.38 0.70 2 48 4.82 3.1589 4.17 3.2634 3.54 2.843.28 0.001 -0.05 0.96 3 42 5.01 3.2392 4.14 3.1837 3.57 2.683.71 0.0002 -0.57 0.57 _______________________________________________________________________________________________

20. II – QTL regression model based on markers (interval mapping) Suppose gene order Marker 1 – QTL – Marker 2 y i =  + a*z i + e i where a* is the “real” effect of a QTL, z i is an indicator variable describing the probability of individual i to carry the QTL genotype, Qq or qq, given a possible marker genotype, e i ~ N(0,  2 )

21. Indicators for a backcross Sam- Height MarkersThree-locusQTLMarkerQTL|marker ple (cm, y i ) 12genotypex* i x i z i 11841111111 1 P(1|11)  1 101 21851 111111 1P(1|11)  1 101 31801111111 1P(1|11)  1 101 41821011011 0P(1|10)  1-  100 5167010110 1P(1|01)  0010 6169000100 0P(1|00)  0 0000 7165000100 0P(1|00)  0 0000 8166000100 0P(1|00)  0 0000

22. Conditional probabilities (  1|i or  0|i ) of the QTL genotypes (missing) based on marker genotypes (observed) Marker QTL genotype GenotypeFreq.Qq(1)qq(0) 11½(1-r)(1-r 1 )(1-r 2 )/r 1 r 2 / (1-r)  1(1-r)  0 10½r(1-r 1 )r 2 /rr 1 (1-r 2 )/r  1-  = 1-r 1 /r   = r 1 /r 01½r r 1 (1-r 2 )/r (1-r 1 )r 2 /r   1 -  00½(1-r)r 1 r 2 / (1-r 1 )(1-r 2 )/ (1-r)  0 (1-r)  1 r is the recombination fraction between two markers r 1 is the recombination fraction between marker 1 and QTL r 2 is the recombination fraction between QTL and marker 2 Order Marker 1–QTL–Marker 2

23. Interval mapping with regression approach Consider a marker interval M 1 -M 2. We assume that a QTL is located at a particular position between the two markers (r 1 and  are fixed) With response variable, y i, and dependent variable, z i, a regression model is constructed as y i =  + a*z i + e i Statistical software, like SAS, can be used to estimate the parameters ( , a*,  2 ) for a particular QTL position contained in the regression model Move the QTL position every 2cM from M 1 to M 2 and draw the profile of the F value. The peak of the profile corresponds to the best estimate of the QTL position. F-value M 1 M 2 M 3 M 4 M 5 Testing position

24. Interval mapping with maximum likelihood Linear regression model for specifying the effect of a putative QTL on a quantitative trait Mixture model-based likelihood Conditional probabilities of the QTL genotypes (missing) based on marker genotypes (observed) Normal distributions of phenotypic values for each QTL genotype group Log-likelihood equations (via differentiation) EM algorithm Log-likelihood ratios The profile of log-likelihood ratios across a linkage group The determination of thresholds Result interpretations

25. Linear regression model for specifying the effect of a QTL on a quantitative trait y i =  + a*z i + e i, i = 1, …, n (latent model) a* is the (additive) effect of the putative QTL on the trait, z i is the indicator variable and defined as 1 when QTL genotype is Qq and 0 when QTL genotype is qq, e i  N(0,  2 ) Observed data:y i and marker genotypes M Missing data: QTL genotypes Parameters:  = ( , a*,  2,  =r 1 /r) Observed marker genotypes and missing QTL genotypes are connected in terms of the conditional probability (  1|i or  0|i ) of QTL genotypes (Qq or qq), conditional upon marker genotypes (11, 10, 01 or 00).

26. Mixture model-based likelihood without marker information L(y|  ) =  i=1 n [½f 1 (y i ) + ½f 0 (y i )] Sam- Height ple(cm, y) QTL genotype 1184Qq (1) 2185Qq (1) 3180Qq (1) 4182Qq (1) 5167qq (0) 6169qq (0) 7165qq (0) 8166qq (0)

27. Mixture model-based likelihood with marker information L(y,M|  ) =  i=1 n [  1|i f 1 (y i ) +  0|i f 0 (y i )] Sam- Height Marker genotype QTL ple(cm, y) M1M2Aaaa 1184Mm (1)Nn (1)½ ½ 2185Mm (1)Nn (1) ½ ½ 3180Mm (1)Nn (1) ½ ½ 4182Mm (1)nn (0)½ ½ 5167mm (0)nn (1)½ ½ 6169mm (0)nn (0)½ ½ 7165mm (0)nn (0)½ ½ 8166mm (0)Nn (0) ½ ½

28. Conditional probabilities of the QTL genotypes (missing) based on marker genotypes (observed) L(y,M|  ) =  i=1 n [  1|i f 1 (y i ) +  0|i f 0 (y i )] =  i=1 n1 [1 f 1 (y i ) + 0 f 0 (y i )]Conditional on 11   i=1 n2 [(1-  ) f 1 (y i ) +  f 0 (y i )]Conditional on 10   i=1 n3 [  f 1 (y i ) + (1-  ) f 0 (y i )]Conditional on 01   i=1 n4 [0 f 1 (y i ) + 1 f 0 (y i )]Conditional on 00

29. Normal distributions of phenotypic values for each QTL genotype group f 1 (y i ) = 1/(2  2 ) 1/2 exp[-(y i -  1 ) 2 /(2  2 )],  1 =  + a* f 0 (y i ) = 1/(2  2 ) 1/2 exp[-(y i -  0 ) 2 /(2  2 )],  0 = 

30. Differentiating L with respect to each unknown parameter, setting derivatives equal zero and solving the log-likelihood equations L(y,M|  ) =  i=1 n [  1|i f 1 (y i ) +  0|i f 0 (y i )] log L(y,M|  ) =  i=1 n log[  1|i f 1 (y i ) +  0|i f 0 (y i )] Define  1|i =  1|i f 1 (y i )/[  1|i f 1 (y i ) +  0|i f 0 (y i )](1)  0|i =  0|i f 1 (y i )/[  1|i f 1 (y i ) +  0|i f 0 (y i )](2)  1 =(3)  0 =(4)  2 =(5)  =(6)

31. EM algorithm (1) Give initiate values  (0) = (  1,  0,  2,  ) (0), (2) Calculate  1|i (1) and  0|i (1) using Eqs. 1 and 2, (3) Calculate  (1) using  1|i (1) and  0|i (1), (4) Repeat (2) and (3) until convergence.

32. Two approaches for estimating the QTL position (  ) View  as a variable being estimated (derive the log-likelihood equation for the MLE of  ), View  as a fixed parameter by assuming that the QTL is located at a particular position.

33. Log-likelihood ratio (LR) test statistics H0: There is no QTL (  1 =  0 or a* = 0) – reduced model H1: There is a QTL (  1   0 or a*  0) – full model Under H0: L0 = L(y,M|, a*=0, ) Under H1: L1 = L(y,M|^  1, ^  0, ^  2, ) LR = -2(log L0 – log L1)

34. The profile of log-likelihood ratios across a linkage group LR Testing position

35. The determination of thresholds Permutation SampleOriginal12 …1000M1M2QTL 1184165x…xMm (1)Nn (1)? 2185182x…xMm (1)Nn (1)? 3180169x…xMm (1)Nn (1) ? 4182167x…xMm (1)nn (0)? 5167185x…xmm (0)nn (1)? 6169180x…xmm (0)nn (0)? 7165166x…xmm (0)nn (0)? 8166184x…xmm (0)Nn (0)? LRLR1LR2…LR1000 The critical value is the 95th or 99th percentiles of the 1000 LRs

36. Result interpretations A poplar genome project Objectives: Identify QTL affecting stemwood growth and production using molecular markers; Develop fast-growing cultivars using marker-assisted selection

37. Materials and Methods Poplar hybrids F1 hybrids from eastern cottonwood (D)  euramerican poplar (E) (a hybrid between eastern cottonwood  black poplar) Four hundred fifty (450) F1 hybrids were planted in a field trial DNA extraction and marker arrays A total of 560 markers were detected from a subset of F1 hybrids (90) Genetic linkage map construction

38. Profile of the log-likelihood ratios across the length of a linkage group Critical value determined from permutation tests

39. Advantages and disadvantages Compared with single marker analysis, interval mapping has several advantages: The position of the QTL can be inferred by a support interval; The estimated position and effects of the QTL tend to be asymptotically unbiased if there is only one segregating QTL on a chromosome; The method requires fewer individuals than single marker analysis for the detection of QTL Disadvantages: The test is not an interval test (a test that can distinguish whether or not there is a QTL within a defined interval and should be independent of the effects of QTL that are outside a defined region). Even when there is no QTL within an interval, the likelihood profile on the interval can still exceed the threshold (ghost QTL) if there is QTL at some nearby region on the chromosome. If there is more than one QTL on a chromosome, the test statistic at the position being tested will be affected by all QTL and the estimated positions and effects of “QTL” identified by this method are likely to be biased. It is not efficient to use only two markers at a time for testing, since the information from other markers is not utilized.