Efficient long-term cycling strategy

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
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 MM
7
Basic advantage of our approach
Is a complete comparison as it simultaneously
considers
8
The long-term program
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
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

• 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
(n), (m) and selection age were optimized
17
Parameters- for reference
Parameters Main scenario Alternative scenarios
Additive variance (sA2 ) 1
Dominance variance, of the additive variance in BP (sD2) 25 0 100
Narrow-sense heritability (h2) (obtained by changing sE2) 0.1 0.05 0.5
Additive standard deviation at mature age (sAm), 10 5 20
Diversity loss per cycle, 0.5 0.251
Rotation age, years 60 10 120
Time before establishment of the selection test (TBEFORE), years 1 (phenotype) 3 5 (phenotype)
Time before establishment of the selection test (TBEFORE), years 5 (clone) 3 7 (clone)
Time before establishment of the selection test (TBEFORE), years 17 (progeny) 5 7 (progeny)
Recombination cost (CRECOMB), 30 15 50
Cost per genotype (Cg), 0.1 (clone), 1 5 (clone),
Cost per genotype (Cg), 1 (progeny) 0.1 5 (progeny)
Cost per plant (Cp), 1 0.5 3
Cost per year and parent (constraint) 10 5 20
Group Merit Gain per year (GMG/Y) To be maximized To be maximized
18
CVa at mature age

• CVa14 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
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/Y0.25 cycle 30 years then
• Cycle GM8 (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
How flat are the optima (clone)?
h20.1, 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 gt than 5, (not shown)
• If problems with cloning, better-gt clones with lt
  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
GM/Y by Phenotype0,275
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 Olas 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
(budget10) 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
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 testing is the best breeding strategy
• Phenotype 2nd best, except very low h2 or high
  budget
• Superiority of the Phenotype over Progeny is
  minor additional considerations may be
  important (idea of a two-stage strategy).

31
Lets 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
.Sexual
Stage 2
.Sexual
propagation of
propagation of
pre
-
selected
pre
-
selected
individuals
individuals
Testing of
Testing of
the progeny
the progeny
33
Values- study 2
34
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
selection test (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
Budget10
Budget5
Testing time 10 (stage 1) and 14 (stage 2) little
affected by the budget
37
Why Pheno/Progeny was 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
Phenotype test
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
TMATING age 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
Parameters study 3
42
J-M correlation functions

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
Main scenario
0.6

• 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

• 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
Rotation age 20
0.6
0.6
Plant cost 0.1
0.6
2
0.8
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
5
10
15
20
25
0
5
10
15
20
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
Crossing lag 5
0.5
Crossing lag 8
0.5
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)

• 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.

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)

Interactive scenario
0.06
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
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 gt 25
years and For the two-stage, pre-selection not at
its optimum age (10 years)
Main (h0.1, budget10), 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?
Cycle, years GM/year, GM/cycle 2 cycle s of Pheno
Phenotype 20 0,152 3,04 6,08
Pheno/Prog 40 0,181 7,26

Answer is No 7,26 is gt 6,08
51
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, h2 high, testing cheap, delays
  from optimum age could be important

52
Our main findings
53
Main findings- spruce
Clonal test by far the best
If higher h2 more clones less ramets Present
plans size 40/15, selection age 10 years
Select at age 15 (20) depending on J-M correlation
With L(2001), Cycle time 21 Gain8.2 GM/Y
0,34
54
Main findings- Pine
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)
(70)
With L(2001), Cycle time 27 Gain8 GM/Y 0,27
55
Research needs- Faster cloning
56
Research needs (a PhD thesis)

• Faster, better cloning embryogenesis, rooting,
  C-effects (especially for pine)
• Sufficient flowering at age 10 (15) for pine
• Documentation of flowering in breeding stock
• How sexual maturation, flowering abundance are
  related to breeding value?

57
In breeding, thanks to Dag there may be less risk
to enter a wrong way ...
58
The end
The end