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Efficient long-term cycling strategy

Efficient long-term cycling strategy. DaDa work 2001-2003. Contents of 1 h. Introduction and our studies (5 min.) Main finding (2 min) Testing strategy: optimization and timing (50 min): Single-stage strategies compared, Two-stage strategies compared,

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Efficient long-term cycling strategy

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  1. Efficient long-term cycling strategy DaDa work 2001-2003

  2. Contents of 1 h • Introduction and our studies (5 min.) • Main finding (2 min) • Testing strategy: optimization and timing (50 min): • Single-stage strategies compared, • Two-stage strategies compared, • Amplified case: Progeny testing versus Pheno/Progeny. • Main finding separately for pine and spruce (5 min.)

  3. 4: BP size optimised Seminar 2004.03.02 3: Ph/Prog amplified (pine), effect of J-M. 1-2: Best testing strategy The Road to this semianr Hungry shark Breeding cycler

  4. Main findings: cloning is the best strategy

  5. Main findings • Clonal test is superior (use for spruce) • Progeny testing not efficient • For Pine, use 2 stage Pheno/Progeny • Pine flowers not needed before age ~ 10-15

  6. General M&M

  7. Basic advantage of our approach Cost Gain per time Diversity Other things, e.g. to well see the road Is a complete comparison as it simultaneously considers:

  8. The long-term program Adaptive environment Mating Breeding population Within family selection Testing We consider one such breeding population Recurrent cycles of mating, testing and balanced selection

  9. Benefit = Group Merit/Year Diversity loss was set to be as important as gain Gain Diversity Time

  10. Low Main High lower reasonable bound typical for Pine or spruce higher reasonable bound Main inputs and scenarios Genetic parameters Time components Cost components While testing an alternative parameter value, the other parameters were at main scenario values

  11. The time and cost explained Production of sibs (4 years) Cutting of ramets Rooting of ramets (1 year) Nursery Establishment, maintenance and assessments Field trial Transportation Crossing Established in 5 years after seed harvest Recombination cost=20, Time=4 Genotype depend. cost=2 (per ortet) Plant dependent cost=1 (per ramet) Time before Testing time Lag Mating time • Cost per test plant = 1 ’cost unit’, all the other costs expressed as ratio of this 1. • Such expression also helped to set the budget constraint corresponding to the present-day budget

  12. All these costs should fit to a present-day budget • Budget estimate is taken from pine and spruce breeding plan ~ test size expressed per year and BP member. • ~ 10 ’cost units’ for pine, 20- for spruce. Budget constraint

  13. Why budget constraint per BP member and year? • Because costs expressed per BP member = easier to handle • Gain efficiency should be assessed per unit of time • Optimization= optimum combination of testing time and testing size to obtain max GM/Year and to satisfy the budget constraint (use Solver)

  14. The Relativity theory holds for the Cycler as well… It optimizes “your case” • What if budget is such • What if costs are such • What if we reduce them • What if heritably is such • What if J-M correlation is • So, interpretation should consider that everything is relative to each other

  15. Single-stage testing strategies

  16. Objective: compare strategies based on phenotype, clone or progeny testing Phenotype testing Clone or progeny testing N=50 N=50 (…n) (…n) (…n) (…n) OBS: Further result on numbers and costs- for one of these families (…m) (…m) (…m) (…m) (…m) (…m) (…n), (…m) and selection age were optimized

  17. Parameters- for reference

  18. CVa at mature age • CVa=14 % is based on pine tests in south Sweden Jansson et al (1998), • 1/2 of additive var in pop is within full sib families, • Our program is balanced= gain only from within full-sib selection, • Thus, CVa within fam= CVa in pop divided by the square root of 2, thus a CV = 10%, which we use here (even if not quite correct). CVa within =sqrt(2/2)=sqrt(2)/sqrt(2)= 2/sqrt(2)

  19. 0.6 Annual Group Merit, % 0.5 0.4 Clone 0.3 Phenotype 0.2 Progeny 0.1 0.0 0 0.1 0.2 0.3 0.4 0.5 0.6 Narrow-sense heritability Results-clonal best, progeny worst Test 26 clones with 21 ramet (18/15  budget), select at age 20 Test 182 phenotypes; select at age 15, ( budget: 86, for 17 years) (second best) Test 11 female parents with 47 progeny each; select at age 34 ( budget: 8/34, 40 years) At all the scenarios, Clonal was superior, except high h2.

  20. GM/Y digits after comma are important • If for Clone GM/Y=0.25%; cycle= 30 years then • Cycle GM=8 % (gain 8.5 - 0.5 div loss) • Thus GM/Y reduction by 0.03 (10%) = Cycle gain reduction by 1% • Loss of Cycle gain by 1% = important loss

  21. Optimum 18(15) How flat are the optima (clone)? h2=0.1, lower budget, at optimum testing time 0.30 • Clone number (ramet per clone) = 12(22)-24 (14) • Less ramets at optimum clone number is sensitive: no > than 5, (not shown) • If problems with cloning, better-> clones with < ramets • If h2 is higher , see next 0.25 0.20 Annual Group Merit , % GM/Y by Pheno 0.15 0.10 4(59) 10(25) 15(18) 20(14) 30(10) 40(8) Clone no (ramets per clone) 17 18 20 22 23 25 Test time

  22. If not enough cuttings, better more clones with less ramets, rather than to reduce ramet number at optimum clone number Budget=20, h2=0.1, Cycling cost=20, time 4, Tbefore=5, Cg=2, J-M corr by L(2001), c=100 0.450 0.436 0.440 0.430 0.420 0.410 This line marks loss of GM/Y > 0,03 0.400 0.390 0.380 0.370 0.360 0.350 Variation in these outlined numbers will 0.340 not cause marked loss of benefit 0.330 57(6) 70(5) 39(9) 50(7) 45(8) 22(16) 24(15) 26(14) 28(13) 30(12) 32(11) 36(10) 19(19) 20(18) 21(18) Optimum for clone number (ramet no per clone) GM/Y by Phenotype=0,275 testing time 12 12 12 12 12 13 13 13 14 14 15 15 15 15 17

  23. Higher h2 = more clones and less ramets Clone no/ramet no 0.50 46/5 Optimum then is between 18/15 and 30/10 0.40 28/9 18/15 0.30 GM/Y, % 0.20 13/23 0.10 Spruce plan 40/15 Ola’s thesis, paper I, Fig. 9= 40 cl with 7 ram at test size 280 0.00 0 0.1 0.2 0.3 0.4 0.5 Narrow-sense heritability Budget= 10

  24. The optimal testing time 18-20 • No effect to test longer than 18-20 years • These 18-20 years with conservative J-M function (Lambeth 1980) • With Lambeth 2001, about 15-17 years 0.30 Clone strategy 0.25 0.20 Annual Group Merit, % 0.15 0.10 0.05 0.00 15 16 17 18 19 20 21 22 23 24 25 Testing time, years Figure with optimum at main scenario parameters (budget=10) clones/ramets 18/15

  25. How realistic are the optima? • Optima depends on budget, h2, J-M correlation- how realistic are they? • Budget is the present-day allocation. Increase will result in more gain. But we test how to optimise the resources we have. • h2 =0,1 seems to be reasonable • J-M functions taken from southerly pines, it affects the timing with stand. error of 2 years (7-10-12).

  26. Why Phenotype ≥Progeny ? • Drawbacks of Progeny: long time and high cost (important to consider for improvement) • Phenotype generates less gain but this is compensated by cheaper and faster cycles.

  27. Dominance seems to matter little 0.6 0.5 Dominance would not markedly affect superior performance of clonal testing Annual Group Merit, % 0.4 Clone 0.3 0.2 Phenotype Progeny 0.1 0.0 0 25 50 75 100 125 Dominance variance (% of additive)

  28. 0.30 0.25 0.20 0.15 0.10 0.05 0 3 6 9 12 15 18 Delay before establishment of selection test (years) On Genotype cost Tbefore 0.30 Clone 0.25 Clone 0.20 Progeny 0.15 Phenotype Progeny 0.10 0.05 0 1 2 3 4 5 6 Cost per genotype Expensive genotypes are of interest only if it would markedly shorten T before for Progeny or improve cloning

  29. Recombinatin cost and total budget 0.3 0.30 Clone Clone 0.25 Phenotype 0.2 0.20 Phenotype Annual Group Merit , % 0.15 0.1 Progeny Progeny 0.10 0.0 0.05 0 5 10 15 20 25 10 20 30 40 50 60 Budget per year and parent Recombination cost Important factors; what happens if they fluctuate? Phenotype get more attractive at low budget, strategy choice not depending on recombination cost

  30. Conclusions • Clonal testingis the best breeding strategy • Phenotype2nd best, except very low h2 or high budget • Superiority of the Phenotypeover Progenyis minor = additional considerations may be important (idea of a two-stage strategy).

  31. Let’s do it in 2 stages?

  32. Phenotype/Progeny strategy Mating Mating Stage1: Phenotype test and pre-selection Reselection Reselection based on based on performance of performance of the progeny the progeny Stage 2: Stage 2: .Sexual .Sexual propagation of propagation of pre pre - - selected selected individuals individuals Testing of Testing of the progeny the progeny

  33. Alternative Parameters Main scen ario scenarios 2 ( s 1 - Additive variance ) A Dominance variance, % of the additive variance in 25 0; 100 2 ( s BP ) D 2 s 88 0; 38; 94 Environmental variance, % of total variance ( ) E s 10 5; 20 Additive standard deviation at mature age ( ), % Am Diversity loss per cycle , % 0.5 0.25;1; 5 Rot ation age, years 60 10; 20; 120 1 ( phenotype ) 3; 5 ( phenotype ) 5 ( clone; 3; 7 ( clone; Time before establishment of the selection test phenotype/clone ) phenotype/clone ) ( T ), years BEFORE 17 ( progeny; 5; 7 ( progeny; phenotype/progeny ) phenotype/progen y ) Recombination cost ( C ), $ 30 - RECOMB 0.1 ( clone ), 1; 5 ( clone ), Cost per genotype ( Cg ), $ 1 ( progeny ) 0.1; 5 ( progeny ) Cost per plant ( Cp ), $ 1 0.5; 3 Budget per year and parent (the constraint) 10 5; 20; 50 Group Merit Gain per year ( GMG/Y ) T o be maximized Values- study 2

  34. Clone= Phenotype/Clone= no need for 2 stages. • Phenotype/Progeny is 2nd best = best for Pine • If Progeny initiated early, may~ Phenotype/Progeny= need for a amplification • Phenotype/Progeny is shown with a restriction for Phenotype selection age > 15 Results: two-stage 2nd best 0.30 0.25 Clone Pheno/Progeny 0.20 0.15 Progeny Phenotype 0.10 1 3 5 7 9 11 13 15 17 Delay before establishment of selectiontest (years) arrows show main scenario

  35. Budget cuts = switching to Phenotype tests in Pine 0.3 If budget is cut by half = simple Phenotype test Clone Pheno/Progeny Annual Group Merit, % 0.2 Phenotype Progeny 0.1 0 5 10 15 20 Budget per year and parent (%)

  36. Budget cuts for Pheno/Progeny 5 Budget = resources reallocated on cheaper Phenotype test 32 Stage 2 Progeny 4 Genetic gain, % 17 3 Stage 1 Phenotype 5(44) 5(72) 2 Budget=10 Budget=5 Testing time 10 (stage 1) and 14 (stage 2) little affected by the budget

  37. Why Pheno/Progenywas so good? • It generated extra gain by taking advantage of the time before the candidates reach their sexual maturity • This was more beneficial than single-stage Progeny test at a very early age • Question for the next study: is there any feasible case where Progeny can be better?

  38. Progeny test with and without phenotypic pre-selection • Is there any realistic situation where Progeny testing is superior over Pheno/Progeny(reasonable interactions and scenarios) • What and how flat is the optimum age of pre-selection for Pheno/Progeny? (when do we will need flowers?) Progeny test Phenotypetest Pre-selection age?

  39. Simply- where best to invest? • Phenotype-based pre-selection Early flowering induction

  40. Time and cost components CPer CYCLE = Crecomb + n (CG + m CP), Tcycle = Trecomb + TMATING + TLAG + Tprogtest TMATINGage of sufficient flowering capacity to initiate progeny test (for 2-stage strategy it corresponds to the age of phenotypic pre-selection TLAG is crossing lag for progeny test (polycross, seed maturation, seedling production)

  41. Main scenario Alternative Interactive Parameters values scenario values* scenario values 2 ( s 1 - - Additive variance ) A Dominance variance, % of the additive variance 25 0; 100 - 2 ( s ) D 2 Narrow - sense heritability ( h ) (obtained by 0.1 0.01; 0.5 0.01 2 s changing ) E Additive standard deviation at mature age % 10 - - Diversity loss per cycle , % 0.5 - - J - M gen etic correlation function L (2001) L(1980); G(2000) L(1980) Age of mating for progeny test (age of sufficient 3 to 25 by 1 - - flowering capacity for progeny testing), years Crossing lag for progeny test (crossing; seed 3 5; 8 - maturation, seedling production), years Rotation age ( RA ), years 50 20; 30; 80 80 Recombination cost ( C ), $ 30 0; 100 - RECOMB Cost per genotype ( Cg ), $ 1 0.1; 10 - Cost per plant ( Cp ), $ 1 0.1; 2 - Budget per year and parent, $ (the constraint) 10 5; 20 - Annual progre ss in Group Merit ( GM/Y ) To be maximized 1 Parameters study 3

  42. J-M correlation functions Lambeth (1980) Lambeth & Dill (2001) Gwaze et al. (2000) 1.0 • Lambeth (2001) Main = genetic corrs in 4 series (15 trials) P taeda (296 fams) 0.9 0.8 0.7 • Gwaze et al. (2000)= genetic correlations from 19 trials with 190 fams of P taeda western USA. 0.6 J-M genetic correlation coefficient 0.5 0.4 0.3 0.2 • Lambeth (1980)= phenotypic fam mean corrs from many trials of 3 temperate conifers 0.1 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Ratio selection/rotation age (Q)

  43. Results: 2 stage is better 0.6 Main scenario • 2 stage strategy was better under most reasonable values Annual Group Merit (%) Pheno/Progeny 0.3 • No marked loss would occur if mating is postponed to age 15 Progeny 0.0 0 5 10 15 20 25 Age of mating for progeny test (years)

  44. J-M correlation affects pre-selection age 10 Gain would increase faster if switching to progeny test 12 Gain increases fast by time Lambeth (1980) Lambeth & Dill (2001) Gwaze et al. (2000) 7 • Optimum selection age depends on efficiency of Phenotype to generate enough gain to motivate prolongation of testing for an unit of time. 1.0 0.9 0.8 0.7 0.6 J-M genetic correlation coefficient 0.5 0.4 0.3 • The gain generating efficiency mainly depends on slope of J-M correlation function. 0.2 0.1 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Ratio selection/rotation age (Q) Do we have J-M estimates for spruce and pine?

  45. When the loss from optimum is important? When early testing is advantageous 0.6 Rotation age = 20 Plant cost= 0.1 0.6 0.6 0.8 2 h = 0.5 Pheno/Progeny 0.6 0.3 0.3 0.3 0.4 Annual Group Merit (%) Progeny 0.2 0.0 0.0 0.0 0.0 0 0 5 5 10 10 15 15 20 20 25 25 0 5 10 15 20 25 0 5 10 15 20 25 Age of mating for progeny test (years) h2 is high but then Phenotype alone is better Plants are cheap Rotation is short

  46. Better crossings are motivated 0.5 Crossing lag= 5 0.5 Crossing lag= 8 • Crossing lag and genotype costs had no marked effect = the crosses can be made over a longer time to simultaneously test all pre-selected individuals and their flowering may be induced at a higher cost. 0.4 0.4 Pheno/Progeny 10; 0.26 0.3 0.3 10; 0.25 0.23 0.22 Annual Group Merit (%) 0.2 0.2 Progeny 0.1 0.1 0.0 0.0 0 5 10 15 20 25 0 5 10 15 20 25 Age of mating for progeny test (years)

  47. Progeny is motivated when conditions disfavour Phenotype • These are as for our interactive scenario: • low heritability (0,01), • long rotation (80 y)= less J-M at pre-selection, • weak J-M correlation (L1980) 0.06 Interactive scenario Pheno/Progeny Progeny 0.03 Annual Group Merit (%) 0.00 0 5 10 15 20 25 Age of mating for progeny test (years) But the optima flat and scenario unrealistic

  48. Cycle time~ 27 Gain=8 % GM/Y= 0,27% 2-4 years, at a high cost if feasible Mating Optimum test time and size for pine (for one of the 50 full sib fams) Stage 1: Test 70 full-sibs Select back the best of 5 when progeny- test age is 10 Long-term breeding Stage 2. Progeny-test each of those 5 with 30 offspring Select 5 at age 10 Lag- 3-4 years

  49. What if no pine flowers until age 25? This means, singe stage Phenotype cycle time > 25 years and For the two-stage, pre-selection not at its optimum age (10 years) Main (h=0.1, budget=10), Flowers at age 25 • Pheno/Progeny is still leading • Phenotype with selection age of 25 is better • Progeny is the last • Budget cuts, high h2 will favour Phenotype 0.20 0.179 0.140 0.15 0.135 Annual Group Merit, % 0.10 0.05 0.00 Pheno/Progeny Progeny Phenotype

  50. May be 2 cycles of Phenotype instead of Pheno/Progeny? Answer is No: 7,26 is > 6,08

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