Lecture 13:Inbreeding and Heterosis

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 plant

breeding)• Inbreeding is unintentionally generated:

– by keeping small populations (such as is found at zoos)

– during selection

Genotype frequencies under inbreeding

• The inbreeding coefficient, F• F = Prob(the two alleles within an

individual are IBD) -- identical by descent

• Hence, with probability F both alleles in an individual are identical, and hence a homozygote

• With probability 1-F, the alleles are combined at random

Genotype Alleles IBD Alleles not IBD frequency

A1A1 Fp (1-F)p2 p2 + Fpq

A2A1 0 (1-F)2pq (1-F)2pq

A2A2 Fq (1-F)q2 q2 + Fpq

p A1

A2q

F

F

A1A1

A2A2

1-F

1-Fp

p A1A1

A2 A1

q

A2A1q

A2A2

Alleles IBDRandom mating

Changes in the mean under inbreeding

F = 0 - 2Fpqd

Using the genotypic frequencies under inbreeding, the population mean F under a level of inbreeding F isrelated to the mean 0 under random mating by

Genotypes A1A1 A1A2 A2A2

0 a+d 2a

freq(A1) = p, freq(A2) = q

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 value of 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

πF=π0°2FkXi=1piqidi=π0°BFHere B is the reduction in mean under complete inbreeding (F=1) , whereB=2Xpiqidi

Inbreeding Depression and Fitness traits

Inbred Outbred

Define ID = 1-F/0 = 1-(0-B)/0 = B/0

Drosophila Trait Lab-measured ID = B/0

Viability 0.442 (0.66, 0.57, 0.48, 0.44, 0.06)

Female fertility 0.417 (0.81, 0.35, 0.18)

Female reproductive rate 0.603 (0.96, 0.57, 0.56, 0.32)

Male mating ability 0.773 (0.92, 0.76, 0.52)

Competitive ability 0.905 (0.97, 0.84)

Male fertility 0.11 (0.22, 0)

Male longevity 0.18

Male weight 0.085 (0.1, 0.07)

Female weight -0.10

Abdominal bristles 0.077 (0.06, 0.05, 0)

Sternopleural bristles -.005 (-0.001, 0)

Wing length 0.02 (0.03, 0.01)

Thorax length 0.02

Why do traits associated with fitness show inbreeding depression?

• Two competing hypotheses:– Overdominance Hypothesis: 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 is reduced.

– Dominance Hypothesis: Genetic variance for fitness is caused by rare deleterious alleles that are recessive or partly recessive; such alleles persist in populations because of recurrent mutation. Most copies of deleterious alleles in the base population are in heterozygotes. Inbreeding increases the frequency of homozygotes for deleterious alleles, so fitness is reduced.

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 F on F,

F = 0 - BF

0

0

1

0 - B

If epistasis is present, this regression is non-linear, with CkFk for k-th order epistasis

F

F

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

as possible – Sex ratio as close to 1:1 as possible– When sex ratio skewed (r dams/sires ), every male should

contribute (exactly) one son and r daughters, while every female should leave one daughter and also with probability 1/r contribute a son

Variance Changes Under Inbreeding

F = 0F = 1/4F = 3/4F = 1

Inbreeding reduces variation within each populationInbreeding increases the variation between populations(i.e., variation in the means of the populations)

Variance Changes Under Inbreeding

General F = 1 F = 0Between lines

2FVA 2VA 0

Within Lines (1-F) VA 0 VA

Total (1+F) VA

2VA VA

Line Crosses: Heterosis

P1 P2

F1

x

F2

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 average than the average of both parents.

HF1=πF1°πP1+πP22

Expected levels of heterosis

If pi denotes the frequency of Qi in line 1, let pi + i 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 andlost in the other (so that | i| = 1). H = 0 if = 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 becomesHF1=nXi=1(±pi)2di

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=πF2°πP1+πP22=(±p)2d2=HF12

Agricultural importance of heterosis

Crop % planted as hybrids

% yield advantage

Annual added

yield: %

Annual added

yield: tons

Annual land

savings

Maize 65 15 10 55 x 106 13 x 106

ha

Sorghum 48 40 19 13 x 106 9 x 106 ha

Sunflower 60 50 30 7 x 106 6 x 106 ha

Rice 12 30 4 15 x 106 6 x 106 ha

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.

Crossing Schemes to Reduce the Loss 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 populationF2=F1°F1°PnH/nHF2=HF1µ1°1n∂( ) Only 1/n of heterosis

lost vs. 1/2

Schemes to Reduce the Loss of Heterosis: Rotational

CrossbreedingOriginally suggested for pig populationsSuppose 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

The expected mean value under a two-way rotation:

R2 = zAB °zAB ° P 2

3; where P2 =

zA +zB

2Key: Heterosis advantage divided by 3, not by 2 as in F2

The expected mean value under a three-way rotation:

bR3 = SC3 °zAB ° P 3

7; where SC3 =

zAB +zAC +zB C

3

1/7th of heterosis is lostUnder a 4-way rotation, the order matters:

bR (A;B ;C;D)4 = SC4 °

SCna ° P4

15; where SCna =

zAC + zB D

2Mean of all six pair-wise crosses1/15 of heterosis is lostMean of crosses of nonadjacent lines

Trait P F1 R S BC

Weaning weight 154.2 180.5 178.3 170.1 181.4

12-month weight 210.5 246.8 232.2 212.3 233.6

18-month weight 274.9 315.7 296.6 276.6 295.3

12-18 m weight gain 64.4 68.9 64.4 64.6 61.7

Note that F1 > R > S > P

3bR2 = F1 °

F1 ° P 2

For a 2-way rotation:

For the 2-breed synthetic,

bS2 = 180:5 °180:5 ° 154:2

2= 167:4

b

For weaning weight

R2 = 180:5 °180:5 ° 154:2

3= 171:7

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 rates of offspring)

– Use of crossbred dams– Maternal heterosis is often comparable, and

can be greater than, individual heterosis

Individual vs. Maternal Heterosis in Sheep traits

Trait Individual H

Maternal H total

Birth weight 3.2% 5.1% 8.3%

Weaning weight

5.0% 6.3% 11.3%

Birth-weaning survival

9.8% 2.7% 12.5%

Lambs reared per ewe

15.2% 14.7% 29.9%

Total weight lambs/ewe

17.8% 18.0% 35.8%

Prolificacy 2.5% 3.2% 5.7%

Estimating the Amount of Heterosis in Maternal Effects

πA = π +gIA +gM

A +gM 0

A

Contributions to mean value of line A

Individual genetic effect (BV)Maternal genetic effect (BV)Grandmaternal genetic effect (BV)

πA B = π +gI

A +gIB

2+gM

B +gM0

B +hIA B

πB A = π +gI

A +gIB

2+gM

A +gM0

A +h IAB

Consider the offspring of an A sire and a B dam

Individual genetic value is the average of both parental linesMaternal and grandmaternal effectsfrom the B mothersContribution from (individual)

heterosisNow consider the offspring of an B sire and a A dam

Individual genetic and heterotic effects as in A x B crossMaternal and grandmaternal genetic effects for B line

πAB + πB A

2πAA + πB B

2= h I

AB

Hence, an estimate of individual heteroic effects is

πB A ° πAB =≥gM

A +gM0

A

¥°

≥gM

B +gM0

B

¥( ( ))

Likewise, an estimate of maternal/grandmaternaleffects is given by

How about estimation of maternal heteroic effects?

πC¢AB =2gI

C +gIA +gI

B

4+

h ICA +h I

CB

2+

gMA +gM

B

2+hM

AB +gM 0

B +r Iab

2

πC¢AB °πC A + πC B

2= hM

A B +r I

ab

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 individual heterosis of C x AB crossGenetic maternal effect

(average of maternal BV for both lines)Grandmaternal genetic effectMaternal genetic heteroic effect“Recombinational loss” --- decay of the F1 heterosis in

The F2

One estimate (confounded) of maternal heterosis