1. GENETICS AND VARIABILITY IN CROP PLANTS

2. 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.

3. 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

4. Example of Qualitative Trait Red(RR)Red (Rr)White (rr)

5. EFFECT OF SELECTION ON QUALITATIVE TRAITS A) Recessive traits Only expressed in the homozygous form in composition of segregating generation (e.g. F 2 ) RR2Rrrr No. of Loci(n)Rate of Recessive Homozygous Individuals (1/4) n 123123 1/4 1/16 1/64

6. 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

7. B) Dominant traits  Expressed in the homozygous (RR) and heterozygous (Rr) forms. RR2Rrrr EFFECT OF SELECTION ON QUALITATIVE TRAITS

8. 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).  F 2 – ¾ red (RR, Rr) and ¼ white.

9. 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.  Selfing of Rr – gives progenies that are ¾ red & ¼ white flowers  Selfing of RR – gives all progenies with red flowers (RR). 3. Select only the plants that give progenies with all red flowers. Discard the plants with progenies segregating for the flower colour. 4. To select dominant gene or to eliminate the recessive gene requires two generations. First generation: selection Second generation: progeny testing EFFECT OF SELECTION ON QUALITATIVE TRAITS

10. 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.

11. PHENOTYPIC VARIATION (V P ) OF QUANTITATIVE TRAITS  Consist of: 1. Genetic Variance (V G ) 2. Environmental Variance (V E ) 3. Variance Due to Interaction between Genetic and Environment (V GE ) Therefore: V P = V G + V E + V GE

12. HERITABILITY  DEFINITION: Contribution of genetic component to a certain character, compared to that of the environment  Heritability (%) = V G / V P X 100 V G  Heritability calculated based on all genetic factors over phenotypic variance is called BROAD-SENSE HERITABILITY V G + V E + V GE X 100

13. MAJOR COMPONENT OF GENETIC EFFECT  Genetic effect are divided to 3 components: 1. V A – Additive variance: Indicates the number of favorable alleles needed for a particular locus 2. V D – Dominance variance: Interaction between alleles within the same locus 3. V I – Epistasis: Interaction among genes of different loci Therefore: V G = V A + V D + V I

14.  Ratio of additive variance over phenotypic variance is called NARROW-SENSE HERITABILITY  Narrow-sense Heritability = V A x 100 V P 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. MAJOR COMPONENTS OF GENETIC EFFECTS

15. 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  Dominance effect: AA = Aa = 90cm aa=30cm aa=30cmAa = 60cmAA = 90cmaa=30cmMAA = Aa = 90cm

16. EXAMPLE GENE EFFECTS FEMALE PARENT MALE PARENT AVERAGE HEIGHT OF PROGENIES (F 1 ) No Dominance (Completely additive) Complete Dominance AA (90 cm) aa (30 cm) Aa (60 cm) Aa (90 cm)

17. METHODS TO DETERMINE GENETIC VARIANCE COMPONENTS AND HERITABILITY  Crosses between 2 homozygous parents Parent P 1 x Parent P 2 F1A1A2F1A1A2 F 2 1(A 1 A 1 ) 2(A 1 A 2 ) 1(A 2 A 2 ) (A1A1)(A1A1)(A2A2)(A2A2)

18. METHODS TO DETERMINE GENETIC VARIANCE COMPONENTS AND HERITABILITY (Cont.)  Backcross 1 to parent P 1 (BC 1 P 1 ) Parent P 1 x Parent P 2 F 1 (A 1 A 2 ) BC 1 P 1 1(A 1 A 1 ) 1(A 1 A 2 ) (A1A1)(A1A1) (A2A2)(A2A2)

19. METHODS TO DETERMINE GENETIC VARIANCE COMPONENTS AND HERITABILITY (Cont.)  Backcross 1 to parent P 2 (BC 1 P 2 ) Parent P 1 x Parent P 2 F1A1A2F1A1A2 BC 1 P 2 1(A 2 A 2 ) 1(A 1 A 2 ) (A1A1)(A1A1)(A2A2)(A2A2)

20. METHODS TO DETERMINE GENETIC VARIANCE COMPONENTS AND HERITABILITY (Cont.)  All populations are planted at the same time in the same environment Population Variance Expected Variance Components Genotype VP 1 VP 2 VF 1 VF 2 VBC 1 P 1 VBC 1 P 2 V E V A +V D +V E ½V A +V D +V E A1A1A1A1A1A1A1A1A1A1A1A1 A 1 A 2 2A 1 A 2 A 1 A 2 A2A2A2A2A2A2A2A2A2A2A2A2

21. METHODS TO DETERMINE GENETIC VARIANCE COMPONENTS AND HERITABILITY (Cont.) 1. Environmental Variance (V E ) V E = (VP 1 + VP 2 + VF 1 )/3 2. Phenotypic Variance (V P ) V P = V G + V E = V A + V D + V E = VF 2 3. Genetic Variance (V G ) V G = V P - V E = VF 2 –[(VP 1 + VP 2 + VF 1 )/3]

22. METHODS TO DETERMINE GENETIC VARIANCE COMPONENTS AND HERITABILITY (Cont.) 4. Additive Variance (V A ) 2VF 2 = 2V A + 2V D + 2V E VBC 1 P 1 + VBC 1 P 2 = V A + 2V D + 2V E V A = 2VF 2 - ( VBC 1 P 1 + VBC 1 P 2 ) 5. Dominance Variance (V D ) V D = V G - V A = {VF 2 –[(VP 1 + VP 2 + VF 1 )/3]} - {2VF 2 - (VBC 1 P 1 + VBC 1 P 2 )}

23. METHODS TO CALCULATE HERITABILITY 1. Based on P 1, P 2, F 1, dan F 2 Population Variation Broad-sense Heritability (H B ) = V G /V P = VF 2 –[(VP 1 + VP 2 + VF 1 )/3] VF 2 2. Based on F 2, BC P1 & BC P2 Population Variation Narrow-sense Heritability (H N ) = V A /V P = 2VF 2 - ( VBC 1 P 1 + VBC 1 P 2 ) VF 2

24. METHODS TO CALCULATE HERITABILITY (Cont.) 3. Parent (X) to Offsprings (Y) Regression Method Y= a + bX High Heritability value = character from the parent is highly inherited by the offsprings Y X

25. 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 Y1 X2 Y2 X3 Y3........ Xn Y4 XX XX XX YY

26. Parent- Offspring Regression (b xy ) =  xy - {(  x  y)/n}  x 2 - {(  x) 2 /n} where: y = offspring value x = parent value  If X is the value of one of the parent (male or female): Narrow-sense Heritability (H N ) = 2b  If X is the average value of the parents: Narrow-sense Heritability(H N ) = b

27. METHODS TO CALCULATE HERITABILITY (Cont.) 4. Components in Analysis of Variance (ANOVA) Method Source of variation d.f.Mean squares Expected Mean Squares Replication Genotype Error r-1 g-1 (r-1)(g-1) M1 M2  e 2 + r  g 2  e 2

28. METHODS TO CALCULATE HERITABILITY (Cont.) Computation of Variance Components: V G =  g 2 = (M1 – M2)/r = (  e 2 + r  g 2 -  e 2 )/r = r  g 2 /r =  g 2 V E =  e 2 = M2 Broad-sense Heritability (H B ) = V G /(V G + V E )

29. Genetic Advance From Selection  From heritability value, genetic advance from selection can be estimated: Original population Selected parent Progenies (offspring) of selected parents X O =10 t/ha X S =14 t/ha X E = ?

30. Original population Selected population Progenies of Selected population Genetic Advance From Selection (Cont.)

31.  Computation of Genetic Advance (GA) and population mean of progenies of selected population (X E ):- Consider the Heritability (H) = 60% GA= (X S - X O )H = (14 – 10)0.6 =2.4 t/ha X E = X O + (X S - X O )H = 10 + 2.4 t/ha = 12.4 t/ha