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Lecture 7:
Inbreeding and
Crossbreeding
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
A1 A1
p
F
p
q
A1
q
A1 A2
1-F
Random
Alleles
IBDmating
1-F
A2
A1 A1
p
A2 A1
q
A2 A2
F
A2 A2
Genotype
Alleles IBD
Alleles not IBD
frequency
A1A1
Fp
(1-F)p2
p2 + Fpq
A2A1
0
(1-F)2pq
(1-F)2pq
A 2A 2
Fq
(1-F)q2
q2 + Fpq
Changes in the mean under inbreeding
Genotypes
A1A1
0
A1A2
a+d
A2A2
2a
freq(A1) = p, freq(A2) = q
Using the genotypic frequencies under inbreeding, the
population mean mF under a level of inbreeding F is
related to the mean m0 under random mating by
mF = m0 - 2Fpqd
For k loci, the change in mean is
š F = š 0 ° 2F
Xk
pi qi di = š 0 ° B F
i= 1
X
Here B is the reduction in mean under
B= 2
complete inbreeding (F=1) , where
pi qi di
• 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
Break for Problem 1
Inbreeding Depression and Fitness
traits
Inbred
Outbred
Define ID = 1-mF/m0 = 1-(m0-B)/m0 = B/m0
Drosophila Trait
Lab-measured ID = B/m0
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 B
In 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 mF on F,
mF = m0 - BF
m0
mF
m0 - B
0
F
1
If epistasis is present, this regression is non-linear,
with CkFk for k-th order epistasis
Minimizing the Rate of Inbreeding
• Avoid mating of relatives
• Maximum effective population size Ne
• Ne maximized with equal representation
– 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
reduces variation
withinbetween
each population
Inbreeding increases
the variation
populations
(i.e., variation in the means of the populations)
1/4
0
F = 3/4
1
Variance Changes Under Inbreeding
General
F=1
F=0
Between lines
2FVA
2VA
0
Within Lines
(1-F) VA
0
VA
Total
(1+F) VA
2VA
VA
Line Crosses: Heterosis
When inbred lines are crossed, the progeny show an increase in mean
for characters that previously suffered a reduction from inbreeding.
P1over
P2 average value of the
x the
This increase in the mean
parents is called hybrid vigor or heterosis
F1
H F1
š P1 + š P 2
= š F 1 F°2
2
A cross is said to show heterosis if H > 0, so that the
F1 mean is average than the average of both parents.
Expected levels of heterosis
If pi denotes the frequency of Qi in line 1, let pi + di denote
the frequency of Qi in line 2.
The expected amount of heterosis becomes
Xn
H F1 =
(±pi ) 2 d i
i= 1
• 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 | di| = 1). H = 0 if d = 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.
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,
HF 2 = š F 2
š P 1 + š P2
(±p) 2 d
HF 1
°
=
=
2
2
2
Since random mating occurs in the F2 and subsequent
generations, the level of heterosis stays at the F2 level.
Agricultural importance of heterosis
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.
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
Break for Problem 2
Crossbreeding in Animals
Individual heterosis:
Enhanced performance in a hybrid individual
Maternal heterosis:
Enhanced maternal performance, e.g.,
increased litter size and higher survival
rates of offspring
Maternal heterosis is often comparable,
and can be greater than, individual heterosis
hI
n
hM
n
Birth weight
3.2%
42
5.1%
12
Weaning weight
5.0%
56
6.3%
27
Preweaning growth rate
5.3%
19
Postweaning growth rate
6.6%
10
Yearling weight
5.2%
18
-2.0%
4
Ovulation rate
Fertility
2.6%
20
8.7%
30
Prolificacy
2.8%
20
3.2%
31
Birth-weaning survival
9.8%
29
2.7%
25
Lambs per ewe
5.3%
20
11.5%
25
Lambs reared per ewe
15.2%
20
14.7%
25
Total weight lambs/ewe
17.8%
24
18.0%
25
Carcass traits
0%
7
25
Maternal and individual heterosis effects can be combined by
using crossbred (hybrid) dams.
For example, for total weight of lambs rear per
mated ewe has an 18% individual heterotic advantage
in a crossbred offspring and an addition 18% advantage
(from maternal heterosis) when crossbred
ewes are used in place of purebred ewes.
This combining of maternal and individual heterotic effects
is one reason why three-way crosses are common in animal
breeding, generally by crossing a male from line A with
a hybrid female (from a B x C cross).
This strategy exploits maternal heterosis in the female,
with the sire line often chosen for its contribution to
some production trait.
Synthetics and Rotational
Crossbreeding
To maximally exploit heterosis, we ideally would use only
F1 individuals, as the heterotic advantage decreases
in the F2
Problem: In (most) large farm animals, the number of
offspring is on order of the number of dams. Thus, to
produce n triple cross hybrid offspring requires on the
order of 2n dams (the granddam to produce the
B x C dam, and the dam herself in the A X (B X C) cross).
One partial solution around this problem:
Synthetics: n parental lines are chosen and a
random-mating population formed by first making all
n(n-1)/2 pairwise intercrosses between the lines
F1 ° P
F2 = F 1 °
n Average of the founding lines
Mean of the F1 from all pairwise crosses
This is equivalent to
HF 2
•
•
1
= H 1°
n
(
)
The larger n (number of founding lines), smaller decrease
in H
Second Solution: Rotational Crossbreeding
AxB
cross dam back to A sire
dam
(A
x B) x A
cross dam back to B sire
dam
((A
x B) x A) x B
cross dam back to A sire
dam
And so on ….
The expected mean value under a two-way rotation:
R 2 = zA B
zA B ° P 2
°
;
3
where
zA + z B
P2 =
2
Key:mean
Heterosis
advantage
dividedrotation:
by 3, not by
The expected
value under
a three-way
z 2 as
° in
P F2
z
+ z
+ z
b3 = SC 3 °
R
AB
3
7
;
where
S C3 =
AB
AC
BC
3
Under
a 4-way
rotation,isthe
1/7th
of heterosis
lostorder matters:
SCna ° P4
zA C + zB D
( A ;B ; C ; D )
b
R4
= S C4 °
; where S C n a =
15
2
Mean of
all of
sixheterosis
pair-wise
1/15
Meaniscrosses
oflost
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
64.4
68.9
64.4
64.6
61.7
12-18 m weight gain
Note that F1 > R > S > P
For a 2-way rotation:
For weaning weight
F1 ° P 2
b
R 2 = F1 °
3
For the 2-breed synthetic,
180:5 ° 154:2
Rb 2 = 180:5 °
= 171:7
3
180:5 ° 154:2
b
S 2 = 180:5 °
= 167:4
2
Break for Problem 3
Estimating the Amount of
Heterosis in Maternal Effects
Contributions to mean value of line A
šA = š +
I
gA
+
M
gA
+
M
gA
0
Grandmaternal
genetic
IndividualMaternal
genetic
effect
genetic(BV)
effect
(BV) effect (BV)
Consider the offspring of an A sire and a B dam
š AB = š +
I
gA
+
2
I
gB
+
M
gB
+
M0
gB
+
I
hAB
Maternal and grandmaternal effects
Contribution
from
(individual)
Individual
genetic
value
is
the
average
of
both
parental
from
the
B
mothers
Now consider the offspring ofheterosis
an B sire and a A dam lines
šBA
gAI + g IB
M
M0
= š +
+ gA + gA + h IA B
2
Individual genetic and heterotic effects as in A x B cross
Maternal and grandmaternal genetic effects for B line
Hence, an estimate of individual heteroic effects is
š AB + š B A
2
š AA + š B B
I
= hAB
2
Likewise, an estimate of maternal/grandmaternal
effects is given by
š B A ° š AB
•
¥ •
¥
0
0
M
= gAM + gAM ° gM
+
g
B
B
(
) (
)
How about estimation of maternal heteroic effects?
The mean of offspring from a sire in line C crossed to
a dam from a A X B cross (B = granddam, AB = dam)
š C ¢A B
2gIC + gAI + gBI
h IC A + h IC B
gAM + gBM
r aI b
M
M0
=
+
+
+ hA B + gB +
4
2
2
2
Average individual genetic value
Maternal
genetic
heteroic
effect
New
individual
Genetic
heterosis
maternal
Grandmaternal
of
C
x
effect
AB
cross
genetic
effect
“Recombinational
loss”
--decay
of
the
F
heterosis
in
(average of the line BV’s)
1
(average of maternal BV for both lines)
The F2
One estimate (confounded) of maternal heterosis
š C .¢A B
I
š CA + š CB
r
ab
°
= hM
+
AB
2
2