Download Lecture 5 Artificial Selection

Lecture 8
Short-Term Selection
Response
R=
2
h
S
Applications of Artificial Selection
• Applications in agriculture and forestry
• Creation of model systems of human diseases and
disorders
• Construction of genetically divergent lines for
QTL mapping and gene expression (microarray)
analysis
• Inferences about numbers of loci, effects and
frequencies
• Evolutionary inferences: correlated characters,
effects on fitness, long-term response, effect of
mutations
Response to Selection
• Selection can change the distribution of
phenotypes, and we typically measure this
by changes in mean
– This is a within-generation change
• Selection can also change the distribution
of breeding values
– This is the response to selection, the change in
the trait in the next generation (the betweengeneration change)
The Selection Differential and
the Response to Selection
• The selection differential S measures the
within-generation change in the mean
– S = m* - m
• The response R is the between-generation
change in the mean
– R(t) = m(t+1) - m(t)
(A) Parental Generation
Truncation selection
uppermost fraction
p chosen
S
mp
(B) Offspring Generation
R
m*
Within-generation
change
Between-generation
change
mo
The Breeders’ Equation: Translating S into R
Recall the regression of offspring value on midparent value
µ
yO = š P + h
2
(
P f + Pm
° šP
2
Ž
)
Averaging over the selected midparents,
E[ (Pf + Pm)/2 ] = m*,
Likewise, averaging over the regression gives
E[ yo - m ] = h2 ( m* - m ) = h2 S
Since E[ yo - m ] is the change in the offspring mean, it
represents the response to selection, giving:
R = h2 S
The Breeders’ Equation
• Note that no matter how strong S, if h2 is
small, the response is small
• S is a measure of selection, R the actual
response. One can get lots of selection but
no response
• If offspring are asexual clones of their
parents, the breeders’ equation becomes
– R = H2 S
• If males and females subjected to
differing amounts of selection,
– S = (Sf + Sm)/2
– An Example: Selection on seed number in plants -pollination (males) is random, so that S = Sf/2
Break for Problem 1
Response over multiple generations
• Strictly speaking, the breeders’ equation only holds for
predicting a single generation of response from an
unselected base population
• Practically speaking, the breeders’ equation is usually pretty
good for 5-10 generations
• The validity for an initial h2 predicting response over several
generations depends on:
– The reliability of the initial h2 estimate
– Absence of environmental change between generations
– The absence of genetic change between the generation in
which h2 was estimated and the generation in which
selection is applied
The selection differential is a function of both
the phenotypic variance and the fraction selected
50% selected
Vp = 4, S = 1.6
20% selected
Vp = 4, S = 2.8
20% selected
Vp = 1, S = 1.4
(C)
(A)
(B)
S
S
S
The Selection Intensity, i
As the previous example shows, populations with the
same selection differential (S) may experience very
different amounts of selection
The selection intensity i provided a suitable measure
for comparisons between populations,
S
S
i= p
=
æp
VP
Selection Differential Under
Truncation Selection
Mean of truncated distribution
Z
E [z j z > T ] =
1
•
•
T° š
pT = ¡
æ
( )
z p(z)
æ ¢pT
dz = š +
ºT
ºT
P ( z > T)
Height of standard
the unit normal
density function
deviations
Hence, S = sPhenotypic
pT/pT
at the truncation point T
Likewise, i = S/ s = pT/pT µ
Ž
t
1
i ' 0:8 + 0:41 ln ( ° 1
p
Change in the Variance
"
1+
pT ¢(z ° š )=æ
°
ºT
µ
pT
ºT
Ž2#
æ2
)
Selection Intensity Versions of the
Breeders’ Equation
R =
h2 S
=
h2
S
æp = i h 2 æp
æp
2s =write
2 ) s
We canhalso
this
Since
(s2A/s
P
P as
P = sA(sA/sP) = h sA
R = i h sA
Since h = correlation between phenotypic and breeding
R = i PA A
values, h = rPA
r s
Response = Intensity * Accuracy * spread in Va
When we select an individual solely on their phenotype,
the accuracy (correlation) between BV and phenotype is h
Accuracy of selection
More generally, we can express the breeders
equation as
R = i ruAsA
Where we select individuals based on the index u
(for example, the mean of n of their sibs).
rua = the accuracy of using the measure u to
predict an individual's breeding value =
correlation between u and an individual's BV
Example: Estimating an individual’s BV with
progeny testing. For n progeny,
ruA2 = n/(n+a), where a = (4-h2)/h2
Mathematica program
Overlapping Generations
Lx = Generation interval for sex x
= Average age of parents when progeny are born
The yearly rate of response is
Ry =
im + if
Lm + Lf
h2sp
Trade-offs: Generation interval vs. selection intensity:
If younger animals are used (decreasing L), i is also lower,
as more of the newborn animals are needed as replacements
Generalized Breeder’s Equation
Ry =
im + if
Lm + Lf
ruAsA
Tradeoff between generation length L and
accuracy r
The longer we wait to replace an individual, the more
accurate the selection (i.e., we have time for progeny
testing and using the values of its relatives)
Finite sample correction for i
When a finite number of individuals form the next
generation, previous expressions overestimate i
Suppose M adults measured, of which N are selected
for p = N/M
E[i] depends on the expected value of the order statistics
Rank the M phenotypes as z1,M > z2,M . .. > zM,M
N
N
zk,M = kth order
1 1 Xstatistic in a sample
1 X of M
¦
E(i ) =
æ
!
(
N
E (zk; M ) ° š
k= 1
=
N
E (zk0 ; M )
k= 1
Standardized order statistics zk0 ; M = (zk ; M ° š )=æ
Expected Selection intensity
M = infinity
M = 100
M = 50
M = 20
M = 10
.01
.1
1
p = N/M, the proportion selected
Finite population size results in infinite M expression overestimating
the actual selection intensity, although the difference is small unless
N is very small.
Burrows' approximation (normally-distributed
trait)
•
E ( i( M ; N ) ) ' i °
1° p
2p(M + 1)
•
1
i
Bulmer’s approximation:
uses infinite population result, with p replaced by
N + 1=2
pe =
M + N =(2M )
Selection on Threshold
Traits
Assume some underlying continuous value z, the
liability, maps to a discrete trait.
z<T
character state zero (i.e. no disease)
z>T
character state one (i.e. disease)
Alternative (but essentially equivalent model) is a
probit (or logistic) model, when p(z) = Prob(state one | z)
Frequency of trait
Liability scale
Mean liability before selection
Selection differential
on liability scale
Mean liability after selection q * - q is the
t
t
(but before reproduction)
selection differential
on the phenotypic scale
Frequency of character state on
Mean liability in
in next
next generation
generation
Steps in Predicting Response to Threshold Selection
i) Compute initial mean m0
Hence,
-choose
m>00)
is =a aP(z
unit
variable
We can
scale
the >liability
P(trait)
= zP(z
- normal
m >where
-m) random
= P(U
-m)
z has variance is
ofaone
and
a threshold T = 0
unit
normal
Define z[q] = P(UU< z
)
=
q.
P(U
> z[1-q] ) = q
[q]
For
example,
suppose
of the pop shows the
General
result:
m = - z5%
[1-q]
trait. P(U > 1.645) = 0.05, hence m = -1.645
ii) The frequency qt+1 of the trait in the next
generation is just
qt+1 = P(U > - mt+1 ) = P(U > - [h2S + mt ] )
= P(U > - h2S - z[1-q] )
iii) Hence, we need to compute S, the selection
differential on liability
Let pt = fraction of individuals chosen in
generation t that display the trait
> 0; š t )
š t = (1 °- pt ) E (z jz < 0; š t ) + pt E (zjz •
When
z isThis
normally
distributed,
this
reduces
to z > 0
This
fraction
does
not
display
trait,
hence
fraction
displaythe
the
trait,
hence
z<0
' (š t ) pt °- qt
*
St = š ° š t =
qt
1 °- qt
Height of the unit normal density function
At the point mt
Hence, we start at some initial value given h2 and
m0, and iterative to obtain selection response
2.25
100
2.00
90
S
Selection differential S
1.75
q
80
70
1.50
60
1.25
50
1.00
40
0.75
30
0.50
20
0.25
10
0.00
0
0
5
10
15
Generation
20
25
q, Frequency of character
Initial frequency of q = 0.05. Selection only on adults
showing the trait
15 Minute Break
Permanent Versus Transient
Response
Considering epistasis and shared environmental values,
the single-generation response follows from the
midparent-offspring regression
R = h2 S +
S
æz2
µ
(
æA2 A
2
+
æA2 A A
4
Ž
)
+ ¢¢¢+ æ(E s i r e ; E o ) + æ(E da m ; E o )
Breeder’s
Equation of response
Permanent
component
Response from shared
Responsecomponent
from epistasis
Transient
of response --- contributes
environmental
to short-term response.
Decays awayeffects
to zero
over the long-term
Response with Epistasis
The response after one generation of selection from
an unselected base population with A x A epistasis is
µ
R= S
(
h2 +
æ2A A
2 æz2
Ž
)
The contribution to response from this single generation
after t generations of no selection is
µ
R (1 + ø) = S
(
h2 + (1 °
2
ø æA A
c)
2æz2
Ž
)
2 S) is due to changes in
Response
from
additive
effects
(h
c is the average (pairwise) recombination between loci
allele frequencies and hence is permanent. Contribution
involved in A x A
from A x A due to linkage disequilibrium
Contribution to response from epistasis decays to zero as
linkage disequilibrium decays to zero
Why unselected base population? If history of previous
selection, linkage disequilibrium may be present and
the mean can change as the disequilibrium decays
More generally, for t generation of selection followed by
t generations of no selection (but recombination)
R(t + ø) = t h 2 S + (1 ° c) ø R A A (t )
Maternal Effects:
Falconer’s dilution model
z = G + m zdam + e
Direct effect
geneticpassed
effectfrom
on character
Maternal
dam to offspring is just
G presence
= A + Dm+of
I. the
E[A]
= (Aphenotypic
Adam)/2value
sire +effects
The
of
the
maternal
means that response
A fraction
dam’s
is not necessarily linear and time lags can occur in response
m can be negative --- results in the potential for
a reversed response
Parent-offspring regression under the dilution model
In terms of parental breeding values,
Ada m
As i r e
E (zo j Ada m ; As i r e ; zda m ) =
+
+ m zda m
2
2
Regression of BV on phenotype
A = š A + bA z ( z ° š z ) + e
2 2/(2-m)
With
aeffects,
covariance
BV
The maternal
resulting
slope becomes
bAz b=azhbetween
With no effects,
maternal
= h2
and maternal effect arises, with æA ;M = m æA2 =( 2 ° m )
The response thus becomes
µ
¢ š z = S da m
(
h2
2° m
Ž
)+ S
+m
s ir e
h2
2° m
Response to a single generation of selection
Recovery of genetic response after
initial maternal correlation decays
0.10
0.05
(in terms of S)
Cumulative Response to Selection
h2 = 0.11, m = -0.13 (litter size in mice)
0.00
Reversed response in 1st
generation largely due to
negative maternal correlation
masking genetic gain
-0.05
-0.10
-0.15
0
1
2
3
4
5
6
Generation
7
8
9
10
Cumulative Response (in units of S)
Selection occurs for 10 generations and then stops
1.5
m = -0.25
1.0
m = -0.5
0.5
m = -0.75
0.0
h2 = 0.35
-0.5
-1.0
0
5
10
Generation
15
20
Gene frequency changes under selection
Genotype
A1A1
A1A2
A2A2
Fitnesses
1
1+s
1+2s
Additive fitnesses
Let q = freq(A2). The change in q from one
generation of selection is:
sq(1 ° q)
¢D q =
' sq(1 ° q)
1 + 2sq
when
j 2sq j < < 1
In finite population, genetic drift can overpower
selection. In particular, when 4N j s j < < 1
e
drift overpowers the effects of selection
Strength of selection on a QTL
Have to translate from the effects on a trait under
selection to fitnesses on an underlying locus (or QTL)
Suppose the contributions to the trait are additive:
Genotype
Contribution to
Character
A1A1
0
A1A2
a
A2A2
2a
For a trait under selection (with intensity i) and
phenotypic variance sP2, the induced fitnesses
are additive with s = i (a /sP )
Thus, drift overpowers
selection on the QTL when
4N e j a i j
4N e j s j =
<< 1
æP
More generally
Genotype
A1A1
A1A2
A2A2
Contribution to trait
0
a(1+k)
2a
Fitness
1
1+s(1+h)
1+2s
Change in allele frequency:
¢ q ' 2q(1 ° q)[1 + h(1 ° 2q)]
D
Selection coefficients for a QTL
s = i (a /sP )
h=k
15 Minute Break
Changes in the Variance under Selection
The infinitesimal model --- each locus has a very small
effect on the trait.
Under the infinitesimal, require many generations
for significant change in allele frequencies
However, can have significant change in genetic
variances due to selection creating linkage disequilibrium
Underpositive
linkage linkage
equilibrium,
negative
linkagedisequilibrium,
disequilibrium,f(AB)
f(AB)><f(A)f(B),
f(A)f(B),
With
freq(AB
gamete)
= freq(A)freq(B)
so that
AB gametes
are
less frequent
more
frequent
Changes in VA with disequilibrium
Starting
Under
the
from
infinitesimal
an unselected
model,
base
disequilibrium
population, only
a single
disequilibrium
Va + d a disequilibrium
A (1) =
changes theofV
generation
additive
selection
variance.
generates
contribution d to the additive variance
variance
VP (1) =genetic
VA (1)Additive
+ VD +genetic
VE = V
P + d in the
Additive
variance
after
one generation ofunselected
selection base population. Often
the
additive
genic
variance
Changes in VA changes
variance
VAcalled
(1) the
Vaphenotypic
+
d
h 2 (1) =
VP (1)
=
VP + d
The
amount
generated by a single generation
Changes
in Vdisequilibrium
A and VP change the heritability
of selection is
Within-generation change
4
h
the variance
d = æ2O ° æ2P =
±æz2 An
Aindecrease
increase
in
in the
the variance
variance
2
*
generates d >< 0 and hence
positive
negativedisequilibrium
disequilibrium
A “Breeders’ Equation” for Changes in Variance
d(0) = 0 (starting with an unselected base population)
Within-generation
d(t)
h4 (t)
change in the variance
d(t + 1) =
+
±æz ( t )
-- akin to S
2
2
New disequilibrium
generated
by
Many
forms
of
selection
(e.g.,
truncation)
satisfy
) disequilibrium
Va +thed(tresponse
) from recombination
A (tmeasures
d(t+1)
-V
d(t)
in
selection
Decay
in previous
2
selection
that
is
passed
onto
the
hon(tthe
) = variance =
measuring the mean)
2
VP (tnext
) 2(akin
VPto+R d(t)
generation
æz (t ) = (1 ° k)æz (t)
k <> 0. Within-generation
d(t)
h4 (t) reduction
invariance.
variance.
2increase in
d(t + 1) =
° k
ænegative
(t)
zpositive
disequilibrium, d <> 0
2
2
Break for Problem 2