Darius Danuseviius and Dag Lindgren Optimize your breeding with
Breeding Cycler Efficient long-term cycling strategy 35 min + 10
min Lithuanian Forest Res. Institute (P15) Swedish University of
Agricultural Sciences Slide 2 How carry breeding ? How to select?
phenotype? clones? progeny? 1$1$ Progeny trial No. 3 I breeding
More gain More diversit y Faster But the budget is so low... Slide
3 Answer is Breeding cycler Deterministic optimizer of one (of many
identical) breeding cycles made in Excel Transparent Interactive
Communistic Slide 4 Basic feature Gain per time Cost Diversity
Complete comparison, as it simultaneously considers: Other things,
e.g. to well see the road Slide 5 It assumes specific long- term
strategy Recurrent cycles of mating, testing and balanced selection
Adaptive environment Testing Within family selection Mating We
consider one such breeding population Breeding population Slide 6
Key-problem: How to deal with relatedness and gene diversity
Solution: Group coancestry (equivalent Status number, Dag Lindgren
et al.) The probability for IBD is group coancestry. f Let's put
all homologous genes in a pool Take 2 (at random with replacement).
Slide 7 Components of Long-Term Breeding Plus trees Long-term
breeding Selection Mating Gain Seed orchard Testing Initiation
Slide 8 Long term breeding goes for many repeated cycles Selection
Mating Testing Long-term breeding Breeding cycler studies what
happens in one complete cycle Slide 9 What happens in one complete
cycle? Long-term breeding The breeding value increases The gene
diversity decreases How to assign a single value to the increase in
BV and the decrease in GD? Slide 10 weighted average of Breeding
Value and Gene Diversity Weighting factor = Penalty coefficient;
(coancestry of 1 means drop in BV dawn to 0) Answer is : Group
merit Lindgren and Mullin 1997 if coancestry of 1= 100% drop in BV,
then coancestry of 0.005= 0.5 % drop in BV Slide 11 We expressed
Group Merit per year to consider 3 key factors: Wei and Lindgren
2001 Genetic gain; Gene diversity; Time. Time factor is missing in
many BPs in EU Slide 12 How to consider the cost? Long-term
breeding Cost of a cycle is depending on number of test plants,
mating techniques, testing strategy etc. Selection Mating Testing
Slide 13 Annual Group Merit progress at a given annual cost for all
BP considers four key factors: Danusevicius and Lindgren 2002
Genetic gain; Gene diversity; Time; Cost. Slide 14 Examples of what
Breeding Cycler can do Which is the best testing strategyWhich is
the best testing strategy What is optimum breeding population
size?What is optimum breeding population size? What is the
influence of the parameters?What is the influence of the
parameters? When to select and what numbers to test ?When to select
and what numbers to test ? Where to allocate resources to
strengthen your breeding plan?Where to allocate resources to
strengthen your breeding plan? How to optimize balance among BP
members?How to optimize balance among BP members? Slide 15 How the
Cycler works (in principle) Long-term breeding Selection age ?
Mating Testing size ? 1. Input Genetic parameters Time components
Cost component 2. Find resource allocation that maximises GM/year?
Test method Clone? Progeny? Size of breeding population? Slide 16
Variables - Genetic parameters Additive variance in test Dominance
variance in test Environmental variance in test Breeding population
size Other Slide 17 How the Cycler works (detail) Insert red
values. The worksheet will calculate the blue values with the
consequences of your choices. Find optimum testing size and testing
time to fit into the budget and annual maximize group merit. Or let
SOLVER find the values which maximise progress in group merit Slide
18 Time and cost components Recombination (cost can be either per
BP member or in total) Cost per tested genotype (it costs to do a
clone or a progeny) Test plant can be economical unit Cycle cost
Under budget constraint Recombination Time for e.g. cloning or
creation of progeny Production of test plants Testing time Cycle
time Slide 19 Variables - Others Rotation time (for J*M
considerations) Annual budget Test method (clonal, progeny or
phenotype) J*M development curve Weighting factor for diversity
versus gain Slide 20 How the Cycler works Results You do almost
nothing input the parameters and look for result Inputs Slide 21
J-M correlation is important Choice can be made of J-M function
including custom, Lambeth and Dill 2001 (genetic) is our favourite.
Slide 22 Constraints and limitations of breeding cycler Sh in sh
out= the inputs must be chosen with care The inputs may need
adjustment from the most evident for considering factors not
considered in the math Breeding heads for an area and get
information from a number of sites, this is considered by
modification to the variance components Breeding heads for
improvement in many characters, we set goal as one character value
for forestry. The observation is an index of observations, and J*M
has to be adjusted. Slide 23 Constraints and limitations of
breeding cycler - continued Plant cost is seen as independent of
age of evaluation. This can cause problems for some type of
comparisons. It is rather easy to add many type of considerations
to the EXCEL sheet, the problem is that it makes it too complex for
the user journal papers. Slide 24 Cycling will accumulate gain.
Where is the limit? Slide 25 Example of what Breeding cycler can do
studies by Dag and Darius Slide 26 Optimizing the balance 3:
Ph/Prog amplified (pine), effect of J-M. Seminar 2009 Best testing
strategy Contents Breeding cycler shark Slide 27 Main findings
Clonal test is superior (use for spruce)Clonal test is superior
(use for spruce) Progeny testing not efficientProgeny testing not
efficient For Pine, use 2 stage Pheno/ProgenyFor Pine, use 2 stage
Pheno/Progeny If 2stage is used, pine flowers not needed before age
~ 10-15If 2stage is used, pine flowers not needed before age ~
10-15 Optimizing the balance may give 50% more gainOptimizing the
balance may give 50% more gain Slide 28 Main inputs and scenarios
While testing an alternative parameter value, the other parameters
were at main scenario values Low lower reasonable bound Genetic
parameters Time components Cost components Main typical for Pine or
spruce High higher reasonable bound Slide 29 Time compnents Slide
30 Cost per test plant = 1 cost unit, all the other costs expressed
as ratio of this 1. 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 Such expression also helped to set the budget
constraint corresponding to the present-day budget The time and
cost explained Established in 5 years after seed harvest Field
trial Establishment, maintenance and assessments Cutting of ramets
Rooting of ramets (1 year) Transportation Crossing Recombination
cost=20, Time=4 Plant dependent cost=1 (per ramet) Genotype depend.
cost=2 (per ortet) Nursery Production of sibs (4 years) Mating time
Time before Testing time Lag Slide 31 All these costs should fit to
an annual budget Budget estimate is taken from pine and spruce
breeding plan ~ test size expressed per year and BP member (can be
all BP). ~ 10 cost units for pine, 20- for spruce. Budget
constraint When it is annual, you can optimize time Slide 32
Single-stage testing strategies Slide 33 Objective: compare
strategies based on phenotype, clone or progeny testing (n)
Phenotype testing N=50 (n), (m) and selection age were optimized
Clone or progeny testing N=50 (n) (m) OBS: Further result on
numbers and costs- for one of these families Slide 34
Results-clonal best, progeny worst At all the scenarios, Clonal was
superior, except high h 2. 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) Annual
Group Merit, % 0.0 0.1 0.2 0.3 0.4 0.5 0.6 00.10.20.30.40.50.6
Narrow-sense heritability Phenotype Clone Progeny Slide 35 0.10
0.15 0.20 0.25 0.30 4(59)10(25)15(18)20(14)30(10)40(8) Clone no
(ramets per clone) Annual Group Merit, % How flat are the optima
(clone)? No marked effect: 12 clones with 22 ramets or 30 clones
with 10 ramets. This means: If problems with cloning, better->
clones with clones with < ramets GM/Y by Pheno h 2 =0.1, lower
budget, at optimum testing time Optimum 18(15) 17 18 20 22 23 25
Test time Slide 36 Higher h 2 = more clones and less ramets Spruce
plan 40/15 Olas thesis, paper I, Fig. 9= 40 cl with 7 ram at test
size 280 0.00 0.10 0.20 0.30 0.40 0.50 00.10.20.30.40.5
Narrow-sense heritability GM/Y, % 13/23 18/15 28/9 46/5 Clone
no/ramet no Optimum then is between 18/15 and 30/10 Budget= 10
Slide 37 Clone strategy 0.00 0.05 0.10 0.15 0.20 0.25 0.30
1516171819202122232425 Testing time, years Annual Group Merit, %
The optimal testing time 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 Figure with optimum at main
scenario parameters (budget=10) clones/ramets 18/15 Slide 38 Why
Phenotype Progeny ? Drawbacks of Progeny: long time and high cost
(important to consider for improvement) 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. Phenotype generates less gain but this is
compensated by cheaper and faster cycles. Slide 39 On Genotype cost
Tbefore 0.05 0.10 0.15 0.20 0.25 0.30 0123456 Cost per genotype
0.05 0.10 0.15 0.20 0.25 0.30 0369121518 Delay before establishment
of selection test (years) Expensive genotypes are of interest only
if it would markedly shorten T before for Progeny or improve
cloning So it pays off to make expensive cloning Clone Progeny
Phenotype Clone Progeny Slide 40 Conclusions Clonal testing is the
best breeding strategy Phenotype 2nd best, except very low h 2
Superiority of the Phenotype over Progeny is minor = additional
considerations may be important (idea of a two-stage strategy).
Slide 41 Lets do it in 2 stages? Slide 42 Phenotype/Progeny
strategy Stage 1 Phenotype select at age 10 (15 only 3% GM lost)
Stage 2 Progeny test select at ca 10 (70) (30) Slide 43 arrows show
main scenario 0.10 0.15 0.20 0.25 0.30 1357911131517 Delay before
establishment of selection test (years) Phenotype/ProgenyIf Progeny
initiated early, may~ Phenotype/Progeny = need for a amplification
Phenotype/Progeny is shown with a restriction for Phenotype
selection age > 15 Clone = Phenotype/Clone = no need for 2
stages.Clone = Phenotype/Clone = no need for 2 stages.
Phenotype/Progeny is 2nd best = best for PinePhenotype/Progeny is
2nd best = best for Pine Clone Progeny Phenotype Pheno/Progeny
Results: two-stage 2nd best Slide 44 Budget cuts = switching to
Phenotype tests in Pine If budget is cut by half = simple Phenotype
test 0.1 0.2 0.3 05101520 Budget per year and parent (%) Annual
Group Merit, % Clone Progeny Phenotype Pheno/Progeny Slide 45 Why
Pheno/Progeny was so good? It generated extra gain by taking
advantage of the time before the candidates reach their sexual
maturity 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 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? Question for the next study: is there any
feasible case where Progeny can be better? Slide 46 Results: 2
stage is better 2 stage strategy was better under most reasonable
values 2 stage strategy was better under most reasonable values
Main scenario 0.0 0.3 0.6 0510152025 Annual Group Merit (%) Age of
mating for progeny test (years) No marked loss would occur if
mating is postponed to age 15No marked loss would occur if mating
is postponed to age 15 Pheno/Progeny Progeny Slide 47 When the loss
from optimum is important? Rotation age = 20 0.0 0.3 0.6 0510152025
When early testing is advantageous h 2 is high but then Phenotype
alone is better 0.0 0.3 0.6 0510152025 Plant cost= 0.1 0.0 0.3 0.6
0510152025 Rotation is short Plants are cheap h 2 = 0.5 0.0 0.2 0.4
0.6 0.8 0510152025 Pheno/Progeny Progeny Annual Group Merit (%) Age
of mating for progeny test (years) Slide 48 Main findings - spruce
(18) (15)(15)(15) (15)(15) (15)(15)(15)(15) (15)(15) Clonal test by
far the best Select at age 15 (20) depending on J-M correlation If
higher h 2 more clones less ramets Present plans: size 40/15,
selection age: 10 years With L(2001), Cycle time~ 21 Gain=8.2 %
GM/Y= 0,34% Slide 49 Main findings- P ine Use 2 stage Pheno/Progeny
strategy Stage 1 Phenotype select at age 10 (15 only 3% GM lost)
Stage 2 Progeny test select at ca 10 (70) (30) With L(2001), Cycle
time~ 27 Gain=8 % GM/Y= 0,27% Slide 50 Conclusions Under all
realistic values, Pheno/Progeny better than Progeny Sufficient
flowering of pine at age 10 is desirable, but the disadvantage to
wait until the age of 15 years was minor, If rotation short, h 2
high, testing cheap, delays from optimum age could be important
Slide 51 Research needs- Faster cloning Slide 52 Optimizing the
Balance: restrict grandparental but relax parental contributions
Slide 53 () Cycle 1 SPM parental balance (almost current Swedish
pine program) Founder selection Mating of founders Select and mate
2 best sibs to create 2 families () Cycle 2 Select and mate 2 best
sibs to create 2 families () Cycle 3 Select and mate 2 best sibs to
create 2 families Slide 54 () Cross e.g. 4 best sibs in the 2 best
families 2 nd rank family () 1 st rank family () 3 rd rank family
() n th rank family () Multiple SPMs 2 nd rank family () 1 st rank
family () 3 rd rank family () n th rank family () Founders Green
trees show pedigree Slide 55 Multiple SPMs Pedigrees BP third
generation 2 nd rank family () 1 st rank family Trees selected for
crossing in the 3 rd generation () Pedigree to later breeding
population Founders Slide 56 Note that retrospectively SPM and
multiple SPM give identical pedigrees, thus identical increase of
coancestry.. Slide 57 10 5 14 0.0 0.1 0.2 0.3 0.4 0.5 0.6
0510152025 Low budget High budget 2 Medium budget Families &
parents cost nothing Number of parents per selected family Annual
progress (%) 2=phenotypic Slide 58 Slide 59 Conclusions Multiple
SPM strategy is VERY promising and can beat conventional single SPM
strategy with 20-70 % gain. Variants of strategy 5 which are still
more efficient can be constructed. Multiple SPM strategy is VERY
promising and can beat conventional single SPM strategy with 20-70
% gain. Variants of strategy 5 which are still more efficient can
be constructed. Slide 60 The end
