Materi Kuliah Metode Pemuliaan Tanaman Fakultas Pertanian UGM

1. HERITABILITY

2. Remember this?
VP = VG + VE + VGxE

3. BB
Bb
bb
p2
2pq
q2 0 g22 g12 g11
-a
d
a

4. Populasi cross pollinated
Frequensi Assign value
p2 a
2pq d
q2 -a
Rerata = p2
a+2pqd+q2
(-a) = (p-q)a+2pqd
Varian nilai genotipe = p2
a+2pqd+q2
(-a)–{(p-q)a+2pqd}2
= 2pq{a+(q-p)d}2
+4p2
q2
d2

5. Varian aditif dan varian dominan
• Var aditif = σ2
A=pqα2
= pq{a+(q-p)d}2
• Var dominan = σ2
D
=
4p2
q2
d2

6. This is due to genes whose effects are
additive.
Such genes are the chief genetic cause
for the resemblance among relatives.

7. The genetic variance that results from
the actions of such genes is called the
additive genetic variance (VA).

8. VA measures the amount of variance
produced by each allele taken
individually, then summing up all
these individual effects.

9. VA is the part of genetic variation that
plant and animal breeders can utilize
in artificial selection.
They often refer to it as the breeding
value of an individual.

10. Think of the ears of corn.
Breeders could use artificial selection
to produce corn varieties with longer
or shorter ears.

11. The additive genetic variance is also
the component of genetic variation
that natural selection acts on to
change phenotypes over time.

12. When we start putting genes together
into pairs to make genotypes, we have
to allow for possible non-additive
interactions between the alleles of a
gene.

13. Such interactions also may contribute
to the overall genetic variance and are
also a component of it.

14. Such intra-genic interactions give rise
to the dominance variance (VD).

15. As an extreme case, consider a
population made up of dominant
homozygotes and heterozygotes.

16. As an extreme case, consider a
population made up of dominant
homozygotes and heterozygotes.
Clearly, in this case there is genetic
variation – all individuals are not
genetically identical.

17. As an extreme case, consider a
population made up of dominant
homozygotes and heterozygotes.
Yet, there is no phenotypic
variance!

18. Lastly, we also know that genes can
interact with other, nonallelic, genes.

19. The component of variance resulting
from inter-genic interactions is called
the interaction variance (VI).

20. Putting these sources of genetic
variance together, we have the
following equation for genetic
variance:

21. VG = VA + VD + VI
If we now include this into our
equation for phenotypic variance,
we have:

22. VP = VA + VD + VI + VGxE
So, here’s our completed
breakdown of phenotypic variance
into its components.

23. VP = VA + VD + VI + VGxE
Partitioning the phenotypic
variance in this way is useful in
helping us think about the sources
of variation in our population.

24. This is the concept that we refer to as
“heritability”.

25. H = VG / VP
This is the so-called “broad-sense
heritability”.
Your textbook uses a different
symbol for it.

26. H = VG / VP
It can range from 0 to 1.
0 means that none of the variation
seen is due to genetic differences.

27. H = VG / VP
A value of 1 means that all of the
observed variation is due to genetic
differences.

28. Broad sense heritability estimate
• Two pure lines are crossed to produce F1
and F2 population offspring.
• Variations from all generations should be
estimated in the same growing environment
• H = (VF2- VE)/VF2
where
VE = (VF1 + VP1 + VP2)/3

29. H = VG / VP
Although this measure is probably
what we intuitively think of as
“heritability”, it’s not very useful.

30. What would be more useful is a
measure of heritability based on just
the additive genetic variance.
Because this would be useful to
plant and animal breeders wanting
to use artificial selection.

31. What would be more useful is a
measure of heritability based on just
the additive genetic variance.
And for evolutionary biologists
wanting to model how natural
selection affects quantitative traits.

32. What would be more useful is a
measure of heritability based on just
the additive genetic variance.
And because VA allows us to
predict the phenotype of offspring
from the phenotype of the parent.

33. h2
= VA / VP
This type of heritability is called
“narrow-sense” heritability.

34. h2
= VA / VP
It is the proportion of the total
phenotypic variance which is due
to the additive genetic variance.

35. h2
= VA / VP
This is the type of heritability
almost always measured by
quantitative geneticists in model
organisms.

36. h2
= VA / VP
We can use the fact that relatives
resemble each other for additive
traits, to estimate h2
.

37. Parent-Offspring Regression
The mean of the parental values (“mid-
parental value”) is plotted (regressed)
against the mean of offspring values from
each family.
Or the value of one parent is regressed
against the value of an offspring.

38. Parent-Offspring Regression
The slope of the regression line is the
narrow-sense heritability.
y = a + bx
h2
= b

39. Halfsib mating design
Sumber
ragam
db KT E(MS)
Jantan m-1 M3 σ2
e+ rσ2
fm+ rfσ2
m
Betina/jantan (f-1)m M2 σ2
e+ rσ2
fm
Sesatan Sisa M1 σ2
e
Total
σ2
m = (M3-M2)/rf σ2
A = 4σ2
M

40. Varian halfsib
• Var HS = ¼ σ2
A
• Var Fullsib = ½ σ2
A+¼ σ2
D

41. heritability: no selection
heritability:
between generations:
slope regression line
Parents-Offspring
covP(P,O)/VP(P)
=
(1/2 VG(P))/(1/2VP(P))
(parental mean
on x-axis) TRAIT VALUE PARENTS
TRAIT VALUE OFFSPRING
mean
mean

42. Regressions of mean parental wing length on
mean offspring wing length in Drosophila.

43. Heritability in the broad sense (HB)
0 1
No genetic
influence
Complete
genetic
influence
Most
traits

44. Beak size in song
sparrows
HN = 0.98
Parent-offspring regression

46. Flower size in
alpine skypilots
HN = 0.50

47. TYPICAL HERITABILITY
ESTIMATES - LOW
Calving interval .05
Female fertility .04
Cutting ability .12
Weaning weight .10
Hatchability .10
Number born .15

48. TYPICAL HERITABILITY
ESTIMATES - MEDIUM
Weaning weight .30
Milk yield .35
Racing time .35
Feed conver. .35
500 d. egg production .25
Birth weight .30

49. TYPICAL HERITABILITY
ESTIMATES - HIGH
Mature weight .50
Butterfat % .55
Canon bone
circumference
.45
Loin eye area .50
Egg size .45
Fleece weight .40

50. Phenotypic selection in parents …
selected
parents
have a
different
mean trait
value than
the
population
TRAIT VALUE PARENTS
TRAIT VALUE OFFSPRING
mean
before selection
mean
without
selection
mean selected group

51. leads to a genetic selection response.
selected
parents have
a different
mean trait
value
and a different
mean value
in their
offspring
TRAIT VALUE PARENTS
TRAIT VALUE OFFSPRING
mean
before selection
mean
without
selection
mean selected group
mean of offspring
of selected group

52. a simple prediction of the selection
response …
response R
TRAIT VALUE PARENTS
TRAIT VALUE OFFSPRING
differential S
slope h2

53. Heritability estimation
Selection response = R
Selection difference = S
• R = h2 × S h2 = R/S
• Heritability is the part of the parents’
phenotypic deviation, which can be
transferred to their offspring

54. follows from the definition of ‘slope’.
response R
TRAIT VALUE PARENTS
TRAIT VALUE OFFSPRING
differential S
slope h2
slope = R/S
h2
= R/S
R = h2
S

55. PREDICTION
OF
GENETIC SELECTION RESPONSE
FROM
PHENOTYPIC
SELECTION DIFFERENTIAL
R = h2
S
NO GENETICS NECESSARY

56. Response to Selection
Remember that it is the additive
genetic variation which can
respond to selection, either natural
or artificial.

57. Response to Selection
The amount of phenotypic change
per generation can be used to
estimate h2
.

58. G = YO-Y = 0 G G
D = YP-Y DD
D
G
hHN == 2Y = parental mean
(dotted lines)

60. Oil content
in corn
Are there limits to
the response to
natural selection?
Oil content
in corn

61. Causes of Limits to Response to Selection
(1) Exhaustion of genetic variation (VG = 0). There is
no longer any genetic variation remaining to improve
the phenotype under selection.
(2) Counteracting natural selection. Most genes
affect more than one trait – pleiotropy. Therefore,
selection on one trait will usually affect many other
traits. There will almost always be a trade-off. For
example, larger elephant seals may be more
dominant, but they will be less able to move on land.

63. So, with suitable organisms we can
obtain good estimates of narrow-sense
heritability.

65. Genetic advance in one
generation selection
Gs = (i)√Vp(h2
)
(i) is average of selected
individuals in standard
deviation units
√Vp is phenotypic
variation in parental
population
h2 is narrow sense
heritability in fraction
Pollen control factor
If only the female is known
the advance will be only half

66. Tandem
Independent Culling Levels
Selection Index
most efficient method
should cause maximum
improvement in
overall
objective
Multiple trait selection

67. G1
G2
G3
P1
P2
P3
rG rP
I
h
h
h
H
H = aggregate genotype
I = selection index

68. Index equations
a = relative economic weights
change in profit per 1 unit change in
a trait
h2
= heritability
rG = genetic correlations
rP = phenotypic correlations
σP = standard deviation
b = index weights

69. where:
I = selection index
bn = index weights for each
trait in the index
Pn = phenotypic
measurement for each trait in
the index.
nn PbPbPbI ⋅⋅⋅++= 2211

70. PnPnGnnPnPPnPnP rhhaharbb σσσσσσ 111
2
1
2
1111
2
11 +⋅⋅⋅+=+⋅⋅⋅+
22
1111
2
111 PnnnPnPnGnPnnnPPnP harhhabrb σσσσσσ +⋅⋅⋅+=+⋅⋅⋅+
. . . .
. . . .
. . . .
Index equations

71. σ2
P = Var P
h2
σ2
P = σ2
G = Var G
rP σ1 σ2 = σP12
= Cov P12
h1h2rG σ1 σ2 = σG12
= Cov G12

72. b1Var P1 + b2Cov P12 = a1 Var G1 + a2 Cov G12
b2Cov P12 + b2Var P2 = a1 Cov G12 + a2 Var G2
To solve for b: Pb =Ga
[P][b] = [G][a]
and b = P-1
Ga,
where
P = matrix of phenotypic variances and covariances
b = vector of index weights
G = matrix of genetic variances covariances
a = vector of economic values.

73. Var P1 Cov P1P2 Cov P1P3 b1 Var G1 Cov G1G2 Cov G1G3 a1
CovP2P1 Var P2 Cov P2P3 b2 = Cov G2G1 VarG2 Cov G2G3 a2
Cov P3P1 Cov P3P2 Var P3 b3 Cov G3G1 Cov G3G2 Var G3 a3
Obtain the index weights by solving for the b vector
(with 3 traits in objective and 3 traits in index)
may be more traits in objective than in index

74. Parameter estimates (NSIF)
Trait Heritability st dev EconValue
Number born alive(L) 0.10 2.50 13.50
21-day litter weight(W) 0.15 16.0 0.50
Days to 250 pounds(D) 0.30 13.0 -0.12
Backfat probe (B) 0.40 .10 -15.00
Feed efficiency (F) 0.30 0.25 -13.00

75. Phenotypic and genetic correlations
Genetic Correlations
L W D B F
L - 0.12 -0.2 0.0 -0.2
W 0.0 - 0.0 0.0 -0.3
Phenotypic D 0.1 0.0 - -0.2 0.65
Correlations
B 0.0 0.0 -0.18 - 0.33
F - - 0.6 0.25 -

76. SPI = 100 +6.5(L) + W
MI = 100 + 6(L) + .4(W) - 1.6(D) - 81(B)
TI = 100 -1.7(D) - 168(B)
SPI = Sow Productivity Index
MI = Maternal Index
TI = Terminal Index
constructed so that mean = 100 and
st dev = 25