Lecture 7 Inbreeding and Crossbreeding - University of...
Transcript of Lecture 7 Inbreeding and Crossbreeding - University of...
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Lecture 7Inbreeding andCrossbreeding
Bruce Walsh lecture notesSynbreed course
version 3 July 2013
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Inbreeding• Inbreeding = mating of related individuals
• Often results in a change in the mean of a trait
• Inbreeding is intentionally practiced to:
– create genetic uniformity of laboratory stocks
– produce stocks for crossing (animal and plantbreeding)
• Inbreeding is unintentionally generated:
– by keeping small populations (such as is found atzoos)
– during selection
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Genotype frequencies under inbreeding
• The inbreeding coefficient, F– We (and the literature) interchangeably use f
as well
• F = Prob(the two alleles within an individualare IBD) -- identical by descent
• Hence, with probability F both alleles in anindividual are identical, and hence ahomozygote
• With probability 1-F, the alleles arecombined at random
4q2 + Fpq(1-F)q2FqA2A2
(1-F)2pq(1-F)2pq0A2A1
p2 + Fpq(1-F)p2FpA1A1
frequencyAlleles not IBDAlleles IBDGenotype
p A1
A2q
F
F
A1A1
A2A2
p
p A1A1
A2A1
q
A2A1q
A2A2
Alleles IBD
1-F
1-F
Random mating
Alleles IBD
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Changes in the mean under inbreeding
µF = µ0 - 2Fpqd
Using the genotypic frequencies under inbreeding, the population mean µF under a level of inbreeding F isrelated to the mean µ0 under random mating by
Genotypes A1A1 A1A2 A2A2
0 a + d 2a
freq(A1) = p, freq(A2) = q
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For k loci, the change in mean is
• There will be a change of mean value dominance is present (d not zero)
• For a single locus, if d > 0, inbreeding will decrease the mean valueof the trait. If d < 0, inbreeding will increase the mean
• For multiple loci, a decrease (inbreeding depression) requires directional dominance --- dominance effects di tending to be positive.
• The magnitude of the change of mean on inbreeding depends on gene frequency, and is greatest when p = q = 0.5
µF = µ0 ! 2Fk!
i=1
pi qi di = µ0 ! B F
Here B is the reduction in mean under complete inbreeding (F=1) , where
B = 2!
pi qi di
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Inbreeding Depression and Fitnesstraits
Inbred Outbred
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Define ID = 1-µF/µ0 = 1-(µ0-B)/µ0 = B/µ0
0.02Thorax length
0.02 (0.03, 0.01)Wing length
-.005 (-0.001, 0)Sternopleural bristles
0.077 (0.06, 0.05, 0)Abdominal bristles
-0.10Female weight
0.085 (0.1, 0.07)Male weight
0.18Male longevity
0.11 (0.22, 0)Male fertility
0.905 (0.97, 0.84)Competitive ability
0.773 (0.92, 0.76, 0.52)Male mating ability
0.603 (0.96, 0.57, 0.56, 0.32)Female reproductive rate
0.417 (0.81, 0.35, 0.18)Female fertility
0.442 (0.66, 0.57, 0.48, 0.44, 0.06)Viability
Lab-measured ID = B/µ0Drosophila Trait
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Why do traits associated with fitnessshow inbreeding depression?
• Two competing hypotheses:– Overdominance: Genetic variance for fitness is caused by loci
at which heterozygotes are more fit than both homozygotes.Inbreeding decreases the frequency of heterozygotes,increases the frequency of homozygotes, so fitness isreduced.
– Dominance:: Genetic variance for fitness is caused by raredeleterious alleles that are recessive or partly recessive; suchalleles persist in populations because of recurrent mutation.Most copies of deleterious alleles in the base population are inheterozygotes. Inbreeding increases the frequency ofdeleterious homozygotes, reducing fitness.
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Estimating BIn many cases, lines cannot be completely inbred due to either time constraints and/or because in many species lines near complete inbreeding are nonviable
In such cases, estimate B from the regression of µF on F,
µF = µ0 - BF
0
µ0
1
µ0 - B
If epistasis is present, this regression is non-linear, with CkFk for k-th order epistasis
µF
F
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Minimizing the Rate of Inbreeding
• Avoid mating of relatives
• Maximize effective population size Ne
• Ne maximized with equal representation– Ne decreases as the variance of contributed offspring
increases
– Contribution (number of sibs) from each parent as equal aspossible
– Sex ratio as close to 1:1 as possible
– When sex ratio skewed (r dams/sires), every male shouldcontribute (exactly) one son and r daughters, while everyfemale should leave one daughter and also with probability 1/rcontribute a son
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Variance Changes Under Inbreeding
Inbreeding reduces variation within each population
Inbreeding increases the variation between populations(i.e., variation in the means of the populations)
F = 0
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F = 1/4
F = 3/4
F = 1
Between-group variance increases with F
Within-group variance decreases with F
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Variance Changes Under Inbreeding
VA2VA(1+F) VATotal
VA0(1-F) VAWithin Lines
02VA2FVABetween lines
F = 0F = 1General
The above results assume ONLY additive variancei.e., no dominance/epistasis. When nonadditivevariance present, results very complex (see WL Chpt 10).
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Mutation and Inbreeding• As lines lose genetic variation from drift, mutationintroduces new variation
• Eventually these two forces balance, leading to an equilibrium levelof genetic variance reflecting the balance between loss from drift,gain from mutation
Symmetrical distribution of mutational effects
Assuming:
Strictly neutral mutations
Strictly additive mutations
VM = new mutation variation each generation, typically VM = 10-3 VE
VA = VG = 2NeVM
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Between-line Divergence
The between-line variance in the mean (VB) in generationt is
For large t, the asymptotic rate is 2VMt
VB = 2VM [ t! 2Ne(1 ! e!t/2Ne)]
Implications: Two identical lines will have theirdifference in means eventually (approximately)following a normal distribution with mean 0 andvariance 2VMt, e.g., µ(1) - µ(2) ~ N(0, 2VMt)
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Line Crosses: Heterosis
When inbred lines are crossed, the progeny show an increase in meanfor characters that previously suffered a reduction from inbreeding.
This increase in the mean over the average value of theparents is called hybrid vigor or heterosis
A cross is said to show heterosis if H > 0, so that the F1 mean is larger than the average of both parents.
! µP1 + µP2HF1 = µF1 2
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Expected levels of heterosis
If pi denotes the frequency of Qi in line 1, let pi + !pi denotethe frequency of Qi in line 2.
• Heterosis depends on dominance: d = 0 = no inbreeding depression and no
Heterosis. As with inbreeding depression, directional dominance is required for heterosis.
• H is proportional to the square of the difference in gene frequency between populations. H is greatest when alleles are fixed in one population and
lost in the other (so that |!pi| = 1). H = 0 if !p = 0.
• H is specific to each particular cross. H must be determined empirically,
since we do not know the relevant loci nor their gene frequencies.
The expected amount of heterosis becomes
HF1 =n!
i=1
(!pi)2 di
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Heterosis declines in the F2
In the F1, all offspring are heterozygotes. In the F2, random mating has occurred, reducing the frequency of heterozygotes.
As a result, there is a reduction of the amount of heterosis in the F2 relative to the F1,
Since random mating occurs in the F2 and subsequentgenerations, the level of heterosis stays at the F2 level.
HF2 = µF2 !µP1 + µP2
2=
(!p)2 d2
=HF1
2
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Agricultural importance of heterosis
6 x 106 ha15 x 10643012Rice
6 x 106 ha7 x 106305060Sunflower
9 x 106 ha13 x 106194048Sorghum
13 x 106 ha55 x 106101565Maize
Annual landsavings
Annualadded yield:
tons
Annualadded yield:
%
% yieldadvantage
% plantedas hybrids
Crop
Crosses often show high-parent heterosis, wherein the F1 not only beats the average of the two parents (mid-parent heterosis), it exceeds the best parent.
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Hybrid Corn in the US
Shull (1908) suggested objective of corn breeders should be to find and maintain the best parentallines for crosses
Initial problem: early inbred lines had low seed set
Solution (Jones 1918): use a hybrid line as the seed parent, as it should show heterosis for seed set
1930’s - 1960’s: most corn produced by double crosses
Since 1970’s most from single crosses
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A Cautionary Tale1970-1971 the great Southern Corn Leaf Blight almostdestroyed the whole US corn crop
Much larger (in terms of food energy) than the great Irishpotato blight of the 1840’s
Cause: Corn can self-fertilize, so to make hybrids either haveto manually detassle the pollen structures or use genetic tricksthat cause male sterility.
Almost 85% of US corn in 1970 had Texas cytoplasm Tcms, amtDNA encoded male sterilty gene
Tcms turned out to be hyper-sensitive to the fungusHelminthosporium maydis. Result over a billion dollarsof crop loss
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Crossing Schemes to Reduce theLoss of Heterosis: Synthetics
Take n lines and construct an F1 population bymaking all pairwise crosses
Allow random mating from the F2 on to produce asynthetic population
F2 = F1 !F1 ! P
n H/n
HF2 = HF1
"1! 1
n
#Only 1/n of heterosislost vs. 1/2
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Schemes to Reduce the Loss of Heterosis:Rotational Crossbreeding
Originally suggested for pig populations
Suppose we have three “pure” lines, A, B, C
A B
AxB C
AxBxC AEach generation, crossa crossbred dam with a sirefrom the next line in the sequence
Dam
Sire
Use of a crossbred dam exploitsmaternal heterosis (better moms)
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The expected mean value under a two-way rotation:
R2 = zAB !zAB ! P 2
3, where P2 =
zA + zB
2
Key: Heterosis advantage divided by 3, not by 2 as in F2
The expected mean value under a three-way rotation:
$R3 = SC3 !zAB ! P 3
7, where SC3 =
zAB + zAC + zBC
3
1/7th of heterosis is lost
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Under a 4-way rotation, the order matters:
$R(A,B,C,D)4 = SC4 !
SCna
1!5
P4 , where SCna =zAC + zBD
2
Mean of all six pair-wise crosses
1/15 ofheterosis is
lost
Mean of crosses ofnonadjacent lines
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61.764.664.468.964.412-18 m weight gain
295.3276.6296.6315.7274.918-month weight
233.6212.3232.2246.8210.512-month weight
181.4170.1178.3180.5154.2Weaning weight
BCSRF1PTrait
Note that F1 > R > S > P
$
For the 2-breed synthetic,$S2 = 180.5! 8 . ! 11 0 5 54.2
2= 167.4
For weaning weight
$R2 = 180.5! 180.5! 154.23
= 171.7
3R2 = F1
F1 ! P 2
For a 2-way rotation:
-
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Individual vs. Maternal Heterosis
• Individual heterosis– enhanced performance in a hybrid individual
• Maternal heterosis– enhanced maternal performance (such as
increased litter size and higher survival ratesof offspring)
– Use of crossbred dams
– Maternal heterosis is often comparable, and canbe greater than, individual heterosis
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Individual vs. Maternal Heterosis in Sheep traits
5.7%3.2%2.5%Prolificacy
35.8%18.0%17.8%Total weightlambs/ewe
29.9%14.7%15.2%Lambs rearedper ewe
12.5%2.7%9.8%Birth-weaningsurvival
11.3%6.3%5.0%Weaning weight
8.3%5.1%3.2% Birth weight
totalMaternal HIndividual HTrait
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Estimating the Amount ofHeterosis in Maternal Effects
µA = µ + gIA + gM
A + gM !
A
Contributions to mean value of line A
Individualgeneticeffect(BV)
Maternalgenetic
effect (BV)
Grandmaternalgenetic effect
(BV)
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µAB = µ +gI
A + gIB
2+ gM
B + gM!
B + hIAB
Consider the offspring of an A sire and a B dam
Individual geneticvalue is the
average of bothparental lines
Maternal andgrandmaternal
effectsfrom the B mothers
Contributionfrom (individual)
heterosis
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µBA = µ +gIA + gI
B
2+ gM
A + gM!
A + hIAB
Now consider the offspring of an B sire and a A dam
Maternal and grandmaternalgenetic effects for B line
µAB = µ +gI
A + gIB
2+ gM
B + gM!
B + hIAB
Difference between the two line means estimatesdifference in maternal + grandmaternal effectsin A vs. B
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µAB + µBA
2µAA + µBB
2= hI
AB
Hence, an estimate of individual heteroic effects is
µBA ! µAB
% &!
%gMB + gM!
B
&= gM
A + gM!
A
Likewise, an estimate of maternal/grandmaternaleffects is given by
How about estimation of maternal heteroic effects?
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µC·AB =2gI
C + gIA + gI
B
4+
hICA + hI
CB
2+
gMA + gM
B
2+ hM
AB + gM!
B +rIab
2
The mean of offspring from a sire in line C crossed toa dam from a A X B cross (B = granddam, AB = dam)
Average individual genetic value(average of the line BV’s)
New individualheterosis of C x AB
cross
Genetic maternal effect(average of maternal BV for both
lines) Grandmaternalgenetic effect
Maternal geneticheteroic effect
“Recombinational loss” --- decay of the F1
heterosis in the F2
µC·AB !µCA + µCB
2= hM
AB +rIab
2.
One estimate (confounded) of maternal heterosis