Chapter 10 – Factors Affecting the Rate of
Genetic Change
I. Elements of the Key Equation for Genetic Change
II. The Key Equation in More Precise Terms
III. Trade-Offs Among Elements of the Key Equation
IV. Comparing Selection Strategies Using the Key Equation
V. The Key Equation in Perspective
VI. Male Versus Female Selection
Chapter 10 – Factors Affecting the Rate
of Genetic Change

Learning Objective: To understand the
factors (heritability, intensity of selection,
phenotypic variation, and generation
interval) that affect the predicted amount of
genetic progress, and how certain factors
can be manipulated by the breeder.
I. Elements of the Key Equation for Genetic
Change
The Key Equation: Four Factors
Accuracy of Selection – A measure of the
strength of the relationship between true
breeding values and their predictions for a trait
under selection (largely a function of
heritability).
Selection Intensity – A measure of how
“choosy” breeders are in deciding which
individuals are selected.
Genetic Variation – Variability of breeding
values within a population for a trait under
selection. This is the key to genetic progress.
Generation Interval – The average amount of
time required to replace one generation with
the next.
II. The Key Equation in More Precise Terms
^ i σ
rBV,BV
BV
∆BV/t = ________
L
where ∆BV/t = predicted rate of genetic change per unit
of time
^
rBV,BV
= accuracy of selection
i
= selection intensity
σBV
= genetic variation
L
= generation interval
∆BV/t =
^ iσ
rBV,BV
BV
________
L
where ∆BV/t = The predicted change in mean breeding
value of a population due to selection. Only BV is
passed on from selected parents to offspring.
^
rBV,BV
= The strength of the relationship between
true breeding values and their predictions for a trait
under selection. Accuracy ranges from 0 (no
information) to 1 (lots of information).
i
= selection intensity (phenotypic
difference between the mean selection criterion of those
individuals selected to be parents and the population
mean = the “Standardized Selection Differential”).
σBV
= genetic variation (the standard deviation
of breeding values).
L
= generation interval (the average age of
parents when their selected offspring are born).
L
= generation interval (the average age of
parents when their selected offspring are born).
Generally, the larger the size of the species, the longer
the generation interval.
TABLE 10.2 Common generation intervals (modified)
Species
Horses
Dairy cattle
Beef cattle
Sheep/Goats
Swine
Chickens/Rabbits
Generation Interval
(years)
8 to 12
4 to 6
4 to 6
3 to 5
1.5 to 2
1 to 1.5
Source: http://www.lfba.lv/en/industry/history
III. Trade-Offs Among Elements of the Key
Equation
∆BV/t =
^ iσ
rBV,BV
BV
________
L
Accuracy vs. Generation Interval – To wait for more progeny
records to enhance accuracy would increase the generation interval.
A strategy is to choose the better young sires with the highest
accuracies and use for only one year.
Accuracy vs. Intensity – Likewise, to wait for more progeny records
to enhance accuracy for only a few sires would decrease the intensity
of selection (i.e., the percentage selected). One solution is to test
more sires that have fewer progeny records.
Intensity vs. Generation Interval – Being too choosy usually results
in a longer generation interval. However, the real trade-off is between
selecting females vs. males (e.g., a lower replacement rate dictates
that animals remain longer in the population).
Intensity vs. Risk – A problem may exist if very few sires are used
that have low accuracies, whereby their breeding values turn out to
be poorer than what was initially reported.
Truncation selection – Only need to know the proportion (p) of offspring saved as replacements.
Truncation selection – Only need to know the proportion (p) of offspring saved as
replacement.
___
__ __
_
SCS - SC
i = _______
σSC
Mathematical Proof of the Key Equation:
Given the following parameters for a sheep population being selected for yearling weight
(YW, lbs): μ = 130 Mean of selected males = 190 and females = 158 lbs
h2 = 0.36
σP = 30 lbs
L = 3 years
1) R = h2 x SD
= 0.36 x (174 – 130)
R = 15.84 lbs
_
2) R = h2 i σP
= 0.36 x 1.465 x 30
{SD (Selection Differential) = 44 lbs
_
{i = “Standardized Selection Differential”
_
= i = SD/σP ; 44/30 = 1.465
R = 15.84 lbs
3)
ˆ
_
rBV, BV i σBV
∆BV/t = __________
L
{ = h2 x σ2P = σ2BV = 0.36 x 302 = 324 (√ 324 = 18)
_
i σBV
= √.36 x 1.465 x 18
______________
______________________
= √ h2
L
3
= 15.84
______
3
= 5.28 lbs/yr
IV. Comparing Selection Strategies Using the Key Equation
Mathematical proof of i : = “Standardized Selection Differential”
Determine the intensity of selection values (i) for each sex.
Top 1% (p) of males; i = 2.67
Top 10% (p) of females; i = 1.76
Mean of i = (2.67+1.76)/2 = 2.215
Mean of selected males: μ + iσ
5.0 + 2.67(√4.0) = 10.34 kg
Mean of selected females: μ + iσ
5.0 + 1.76(√4.0) = 8.52 kg
__
Mean of selected offspring : SCS
__
SCS = 10.34 + 8.52 = 9.43 kg
2
___
__ __
_
SCS - SC
i = _______ = 9.43 – 5.0 = 2.215
σSC
√4
i = “Standardized Selection Differential”
IV. Comparing Selection Strategies Using the Key Equation
{∆BV = R, genetic “response” to selection}
Mean of i = (2.67+1.76)/2 = 2.215
∆BV
=
=
New mean =
=
√.35 x 2.215 x √ 1.4
1.55 kg
5.0 + 1.55
6.55 kg
R/t = 1.55/1.5 = 1.034 kg/year
{σ2BV = h2 x σ2p}
= .35 x 4
= 1.4
σBV = √ 1.4
= 1.1832
IV. Comparing Selection Strategies Using the Key Equation

Problem 10.1: Calculate the rate of genetic change in feed conversion in a
swine population given the following:








Heritability of feed conversion (h2) = .35
Phenotypic standard deviation (σp) = .2 lb/lb
Accuracy of male selection (rBVm,BVm) = .8
Accuracy of female selection (rBVf,BVf) = .5
Intensity of male selection (im) = -2.4
Intensity of female selection (if) = -1.5
Generation interval for males (lm) = 1.8 years
Generation interval for females (lf) = 1.8 years
IV. Comparing Selection Strategies Using the Key
Equation
(Problem 10.1):
∆BV/t =
^ iσ
rBV,BV
BV
_________
L
^ i +r
^ i)σ
(rBVm,BV
m
BV
,BV
BV
m
f
f f
= _______________________
Lm + Lf
= (.8(-2.4) + .5(-1.5))(.118) = -.088 lb/lb/year
1.8 + 1.8
IV. Comparing Selection Strategies Using the Key Equation
Problem 10.6: A rancher runs a closed herd of breeding cattle. He normally
keeps and breeds the top 3% of his bull calves based on individual performance for
yearling weight (YW). His sires average three years of age when their offspring are
born. He is studying two female replacement strategies:


a.
Saving the top 20% of his heifers based on YW (Lf = 6.2 years)
Saving the top 60% based on YW (Lf = 3.2 years; cull more older cows)
If h2YW = .5, and σPYW = 60 lb, calculate the expected annual rate of
genetic change in YW for each strategy. Strategy 1:
^ i +r
^
(rBVm,BV
BVf,BVf if) σBV
m m
∆BV/t = _______________________ = (.7071(2.27) + .7071(1.4))(42.4264) = 11.97 lb/yr
Lm + Lf
3.0 + 6.2
b.
What elements of the key equation is the rancher experimenting with?
c.
What element appears to be more important?
IV. Comparing Selection Strategies Using the Key Equation
A rancher runs a closed herd of breeding cattle. He normally keeps and breeds the
top 3% of his bull calves based on individual performance for yearling weight (YW).
His sires average three years of age when their offspring are born. He is studying
two female replacement strategies:


a.
Saving the top 20% of his heifers based on YW (Lf = 6.2 years)
Saving the top 60% based on YW (Lf = 3.2 years ; cull more older cows)
If h2YW = .5, and σPYW = 60 lb, calculate the expected rate of genetic change in
YW for each strategy. Strategy 2:
^ i +r
^
(rBVm,BV
BVf,BVf if) σBV
m m
∆BV/t = _______________________ = (.7071(2.27) + .7071(.64))(42.4264) = 14.08 lb/yr
Lm + Lf
3.0 + 3.2
b.
What elements of the key equation is the rancher experimenting with?
if and Lf
c.
What element appears to be more important?
The i affects L (i.e., by culling more cows that are aged or less productive, if
increases, which potentially can translate into more genetic progress).