1. Genotypic value is not transferred from parent to
offspring; genes are.
Need a value that reflects the genes that an
individual carries and passes on to it’s offspring
Breeding Value
Empirically: An individual’s value based on the mean deviation of its progeny
from the population mean.
Theoretically: An individual’s value based on the sum of the average effects
of the alleles/genes it carries.

2. Average Effect of an Allele
Type of Values and Freq Mean value Population Average
gamete of gametes of genotypes mean effect of
gene
A1A A1A A2A2
a d a
A p q pa + qd a(p-q) + 2dpq q[a+d(q-p)]
A p q qa + pd a(p-q) + 2dpq -p[a+d(q-p)]
average effect of An:
an = mean deviation from the population mean of individuals that received An from one parent, if the other parent’s allele chosen randomly
a1 = pa + qd - [ a (p – q) + 2dpq ]
population mean .
f (A1)
f (A2)
a1 = q [ a + d (q – p)]
a2 = –p [ a + d (q – p)]

3. When there are only two alleles at a locus
Average effect of a gene substitution
A1A1
A1A2
A2A2 +a d -a
(a - d)
(d + a)
p(a - d) q(d + a)
a = a + d(q - p)
a
pa

4. Average Effects Frequency q (A2 orTHa2) 0.0 0.3 0.5 0.7 1.0
: THa
: THa
 
d = -15; a = 25

5. Theoretically: An individual’s value based on the sum of the
average effects of the alleles/genes it carries.
Genotype
Breeding Value
A1A1
A1A2
A2A2
2a1 = 2qa
a1 + a2 = (q - p)a
2a2 = -2pa

6. Breeding Values - THa example
A2 or Tha Pop Mean A1A1 A1A A2A2
q =
q =
q =
q =
q =

7. = = + (A) G = A + D Sum of average effects across loci +
Breeding Value
(A)
A1A A1A A2A2
2a1 a1 + a a2
B1B B1B B2B2
2a1 a1 + a a2 +
(breeding values)
(breeding values)
G = A + D = +
Genotypic
Value
Additive effects
of genes
Dominance
deviation

8. Partitioning the phenotypic value
genotypic value
breeding
value
deviations from population mean
phenotypic value
of individual
genotypic value
G = A + D
breeding value
dominance
deviation
P = G
Pop
Mean
G = G1 + G2 + I12
interaction
two-locus:
P = A1 + D1 + A2 + D2 + I12
d=3/4a, q = 1/4

9. Environmental effects on phenotypes
P = A + D
One locus, two alleles
One locus, two alleles + environmental variation
environmental
deviation
P = A + D + E

10. Amount of genetic variation in a population depends on
# of genotypes, genotypic value, and gene frequencies.
More variation
Less variation
0.75
0.75
p = 0.5
p = 0.9
0.50
0.50
0.25
0.25 9 10 11 9 10 11
A1A1 A1A2 A2A2
A1A1 A1A2 A2A2
Mean
Mean

11. Components of phenotypic variation
f (A1)
a = d = 0.07
P = A + D + I + E
Variance partitioning:
VP = VG VE
VP = VA + VD + VI + VE .
total
phenotypic
variance
additive
genetic
dominance
interaction
(epistatic)
environmental
• Phenotypic variation can be decomposed into additive genetic and other variation
• Relative contributions of different sources depend on allele frequencies

12. TH alpha Example VA = 2pq[a + d (q - p)]2 VD = (2pqd)2 q = 0.3 443.31
403.62
39.69
q = 0.7
191.24
151.62
39.69
Genotypic, VG
Additive, VA
Dominance, VD
d = -15; a = 25
(178.7)
(158.7)

13. Environment and genotype can interact
Different environmental effects across genotypes result in
G x E interaction variance
VP = VA + VD + VI + VE + VGxE
genetic x
environmental
interaction
variance
VP = VA + VD + VI + VE

14. Heritability
h2 = VA/Vp
Proportion of total phenotypic variance attributed to
variation in breeding values.
Expresses the extent to which genes are transmitted
from parents to offspring

15. Detecting the genetic component of phenotypic variation
Resemblance between parents and offspring indicates degree to which a trait is inherited
Drosophila wing length
offspring (O)
bOP
parental average (P)
heritability (narrow sense):
h2 = VA VP
h2 =
bOP
Fraction of variation in parents that
is explained by variation in their genes (VA).
• heritability can also be estimated with offspring–midparent regression

16. Bill depth
in the
Ground Finch
1976
h2 = 0.9
1978

17. Heritability is often
measured in the lab;
estimates are higher
than would be
expected in nature.
Almost all characteristics
in almost all species are
genetically variable to
some extent.