PLANT BREEDING
AGR 3204
GENETICS AND VARIABILITY
IN CROP PLANTS
Genetics and variability of traits are
grouped by:

Qualitative traits
Traits that show variability that can be classified
into discrete (clear-cut) classes that are easily
identifiable.
Eg. Flower colour, fruit shape, stem colour etc.

Quantitative traits
Traits that show variability in continuous form, and
could only be identified through measurements.
They do not show any particular discrete form.
Eg. Sugar content, plant height, fruit size, number
of fruits per plant etc.
QUALITATIVE TRAITS
Controlled by few number of genes,
1-3 loci (major gene)
 Mostly expressed in dominant/
recessive forms
 Absence / very minimum influence of
environment on their expression

Example of Qualitative Trait
Red(RR)
Red (Rr)
White (rr)
EFFECT OF SELECTION ON QUALITATIVE TRAITS
A) Recessive traits
Only expressed in the homozygous form
in composition of segregating generation (e.g.
F2)
RR
2Rr
No. of Loci(n)
Rate of Recessive
Homozygous Individuals
(1/4)n
1
2
3
1/4
1/16
1/64
rr
EFFECT OF SELECTION ON
QUALITATIVE TRAITS
A) Recessive Traits(cont.)


Can be recognized and selected in one
generation only – but required an appropriate
minimum population size (big enough) to detect
its presence.
Dominant gene can be eliminated in one
generation of selection.
EFFECT OF SELECTION ON
QUALITATIVE TRAITS
B) Dominant traits

Expressed in the homozygous (RR) and heterozygous
(Rr) forms.
RR
2Rr
rr
EFFECT OF SELECTION ON QUALITATIVE TRAITS
B) Dominant Trait (Cont.)
 More difficult to select for dominant traits
– need more than one generation of selection.
 Example to select Red Petunia with red
flowers
 Colour
of petunia flower: Red (RR, Rr) and white (rr).
 F2 – ¾ red (RR, Rr) and ¼ white.
EFFECT OF SELECTION ON QUALITATIVE TRAITS
B) Dominant Traits (Cont.)
 Methods:1.
Select for plant with red flower- 1/3 RR & 2/3 Rr
2.
Selfed pollinate plants with red flowers and grow seeds
from them.


3.
4.
Selfing of Rr – gives progenies that are ¾ red & ¼ white flowers
Selfing of RR – gives all progenies with red flowers (RR).
Select only the plants that give progenies with all red
flowers. Discard the plants with progenies segregating for
the flower colour.
To select dominant gene or to eliminate the recessive gene
requires two generations.
First generation: selection
Second generation: progeny testing
QUANTITATIVE TRAIT
Most of the economically valuable characters.
 Controlled by many genes - polygenes.
 Each gene has cumulative contribution to the
expression of the character.
 Expression of quantitative genes usually
influenced by environment effects.

PHENOTYPIC VARIATION (VP) OF QUANTITATIVE
TRAITS

1.
2.
3.
Consist of:
Genetic Variance (VG)
Environmental Variance (VE)
Variance Due to Interaction between Genetic
and Environment (VGE)
Therefore: VP = VG + VE + VGE
HERITABILITY


DEFINITION: Contribution of genetic component to a
certain character, compared to that of the
environment
Heritability (%) = VG / VP X 100
VG
VG + VE + VGE

X 100
Heritability calculated based on all genetic factors over
phenotypic variance is called BROAD-SENSE HERITABILITY
MAJOR COMPONENT OF GENETIC EFFECT

1.
2.
3.
Genetic effect are divided to 3 components:
VA – Additive variance: Indicates the number
of favorable alleles needed for a particular
locus
VD – Dominance variance: Interaction
between alleles within the same locus
VI – Epistasis: Interaction among genes of
different loci
Therefore: VG = VA + VD + VI
MAJOR COMPONENTS OF GENETIC EFFECTS


Ratio of additive variance over phenotypic variance is
called NARROW-SENSE HERITABILITY
Narrow-sense Heritability =
VA
VP
x 100
Narrow-sense heritability is more meaningful because:

Additive effect are transmitted to the next generation

Dominance (interaction between alleles within the same locus)
and epistasis (interaction between loci) varied between
generations.

Epistasis effects are usually small and could be neglected.
EXAMPLE
Consider plant height controlled by one locus
A/a
 A=45 cm and a= 15cm


Additive effect: AA = 90cm, Aa = 60 cm aa=30cm
aa=30cm

Aa = 60cm
Dominance effect: AA = Aa = 90cm
aa=30cm
M
AA = 90cm
aa=30cm
AA = Aa = 90cm
EXAMPLE
GENE
EFFECTS
No Dominance
(Completely
additive)
Complete
Dominance
FEMALE
PARENT
MALE
PARENT
AVERAGE
HEIGHT OF
PROGENIES
(F1)
AA (90 cm)
aa (30 cm)
Aa (60 cm)
AA (90 cm)
aa (30 cm)
Aa (90 cm)
METHODS TO DETERMINE GENETIC VARIANCE
COMPONENTS AND HERITABILITY

Crosses between 2 homozygous parents
Parent P1 x Parent P2
(A1A1)
(A2A2)
F1
A1A2
F2
1(A1A1) 2(A1A2) 1(A2A2)
METHODS TO DETERMINE GENETIC VARIANCE
COMPONENTS AND HERITABILITY (Cont.)

Backcross 1 to parent P1 (BC1P1)
Parent P1 x Parent P2
(A1A1)
(A2A2)
F1
(A1A2)
BC1P1
1(A1A1) 1(A1A2)
METHODS TO DETERMINE GENETIC VARIANCE
COMPONENTS AND HERITABILITY (Cont.)

Backcross 1 to parent P2 (BC1P2)
Parent P1 x Parent P2
(A1A1)
(A2A2)
F1
A1A2
BC1P2
1(A2A2) 1(A1A2)
METHODS TO DETERMINE GENETIC VARIANCE
COMPONENTS AND HERITABILITY (Cont.)

All populations are planted at the same time in
the same environment
Population
Variance
VP1
VP2
VF1
VF2
VBC1P1
VBC1P2
Expected
Variance
Components
VE
VE
VE
VA+VD+VE
½VA+VD+VE
½VA+VD+VE
Genotype
A 1A 1
A 1A 1
A1A1
A 1A 2
2A1A2
A1A2
A 1A 2
A 2A 2
A 2A 2
A 2A 2
METHODS TO DETERMINE GENETIC VARIANCE
COMPONENTS AND HERITABILITY (Cont.)
1. Environmental Variance (VE)
VE = (VP1 + VP2 + VF1)/3
2. Phenotypic Variance (VP)
VP = VG + VE = VA + VD + VE = VF2
3. Genetic Variance (VG)
VG = VP - VE
= VF2 –[(VP1 + VP2 + VF1)/3]
METHODS TO DETERMINE GENETIC VARIANCE
COMPONENTS AND HERITABILITY (Cont.)
4. Additive Variance (VA)
2VF2 = 2VA + 2VD + 2VE
VBC1P1 + VBC1P2 = VA + 2VD + 2VE
VA = 2VF2 - (VBC1P1 + VBC1P2 )
5. Dominance Variance (VD)
VD = VG - VA
= {VF2 –[(VP1 + VP2 + VF1)/3]} {2VF2 - (VBC1P1 + VBC1P2 )}
METHODS TO CALCULATE HERITABILITY
1. Based on P1, P2, F1, dan F2 Population Variation
Broad-sense Heritability (HB) = VG/VP
= VF2 –[(VP1 + VP2 + VF1)/3]
VF2
2. Based on F2 , BCP1 & BCP2 Population Variation
Narrow-sense Heritability (HN) = VA/VP
= 2VF2 - (VBC1P1 + VBC1P2 )
VF2
METHODS TO CALCULATE HERITABILITY (Cont.)
3. Parent (X) to Offsprings (Y) Regression Method
Y
Y= a + bX
X
High Heritability value = character from the
parent is highly inherited by the offsprings
METHODS TO CALCULATE HERITABILITY (Cont.)
3. Parent (X) to Offspring (Y) Regression Method

Arrangement of parent and offspring data
Female
Parent
Male
Parent
Parent
Average
Offspring
X1
X1
X1
Y1
X2
X2
X2
Y2
X3
X3
X3
Y3
.
.
.
.
.
.
.
.
Xn
Xn
Xn
Y4
SX
SX
SX
SY
Parent- Offspring Regression (bxy)
=
Sxy - {(SxSy)/n}
2
2
Sx - {(Sx) /n}
where: y = offspring value
x = parent value


If X is the value of one of the parent (male or female):
Narrow-sense Heritability (HN) = 2b
If X is the average value of the parents:
Narrow-sense Heritability(HN) = b
METHODS TO CALCULATE HERITABILITY (Cont.)
4. Components in Analysis of Variance (ANOVA)
Method
Source of
variation
d.f.
Replication
Genotype
Error
r-1
g-1
(r-1)(g-1)
Mean
squares
M1
M2
Expected Mean
Squares
se2 + rsg2
se2
METHODS TO CALCULATE HERITABILITY (Cont.)
Computation of Variance Components:
VG = sg2 = (M1 – M2)/r
= (se2 + rsg2 - se2)/r
= rsg2/r
= sg2
VE = se2 = M2
Broad-sense Heritability (HB) = VG /(VG + VE )
Genetic Advance From Selection

From heritability value, genetic advance from
selection can be estimated:
XS=14 t/ha
Original
population
XO =10 t/ha
Selected
parent
Progenies
(offspring) of
selected parents
XE = ?
Genetic Advance From Selection
(Cont.)
Original population
Selected population
Progenies of Selected population
Genetic Advance From Selection
(Cont.)

Computation of Genetic Advance (GA) and
population mean of progenies of selected
population (XE):Consider the Heritability (H) = 60%
GA
= (XS- XO)H
= (14 – 10)0.6
=2.4 t/ha
XE = XO + (XS - XO)H
= 10 + 2.4 t/ha
= 12.4 t/ha