1

Lecture 7Inbreeding andCrossbreeding

Bruce Walsh lecture notesSynbreed course

version 3 July 2013

2

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

3

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

5

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

6

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

7

Inbreeding Depression and Fitnesstraits

Inbred Outbred

8

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

9

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.

10

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

11

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

12

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

13

F = 1/4

F = 3/4

F = 1

Between-group variance increases with F

Within-group variance decreases with F

14

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).

15

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

16

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)

17

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

18

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

19

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

20

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.

21

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

22

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

23

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

24

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)

25

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

26

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

27

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:

-

28

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

29

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

30

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)

31

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

32

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

33

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?

34

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