### Assignment on Quantitative Genetics

• 1. 1 Name: Abah Simon Peter Class: PhD Student Course Code: WACI702 (Quantitative Genetics in Plant Breeding) Assignment: #2 --------------------------------------------------------------------------------------------------------------- Question 1: The following are the days to flowering of individual plants from two inbred lines of sunflower (P1 and P2) and their F1 generation. For each generation, calculate: a. Mean b. Variance c. Variance of mean d. Standard error of mean Inbred P1: 98, 93, 100, 96, 86, 102, 94, 101, 90, 99, 87, 100, 98, 96, 86, 99, 91, 93, 92, 91. Inbred P2: 46, 38, 32, 43, 37, 32, 44, 32, 36, 40, 42, 35, 40, 38, 48, 36, 39, 41, 40, 36 F1 (P1 x P2): 83, 90, 85, 80, 81, 90, 81, 90, 85, 89, 86, 78, 90, 83, 83, 86, 90, 88, 82, 90, 90, 88, 82, 85, 90, 87, 75, 76, 83, 79, 82, 93, 95, 83, 88, 90, 87, 94, 85, 102 Solution: 1a. Calculationfor Inbred P1 Generation: (a) Mean (𝑥̅) = ∑ 𝑥 𝑛 Where 𝑥̅ = Sample mean of P1 x = summation of individual value n = sample size Therefore, 𝑥̅ = 98+93+100+⋯+91 20 = 94.6 (b) Variance (𝑆2 ) = ∑ 𝑥2 − (∑ 𝑥)2 𝑁 𝑁−1 Where S2 = Sample Variance of P1 x2 = sum of square of P1 individual (∑x)2 N = Mean square of P1 N = sample size Therefore, 𝑆2 = 179468−178983.2 19 = 484.8 19 = 25.52
• 2. 2 (c) Variance of mean ( 𝑆 𝑥̅ 2) = 𝑆2 𝑛 Where 𝑆 𝑥̅ 2 = mean variance of P1 S2 = sample variance n = sample size Therefore, 𝑆 𝑥̅ 2 = 25.52 20 = 1.28 (d) Standard error of mean ( 𝑆 𝑥̅ ) = √ 𝑆2 𝑛 Where 𝑆 𝑥̅ = mean standard error of P1 S2 = sample variance n = sample size Therefore, 𝑆 𝑥̅ = √ 𝑆 𝑥̅ 2 = √1.28 = 1.13 1b. Calculationfor Inbred P2 Generation: (a) Mean (𝑥̅) = ∑ 𝑥 𝑛 Where 𝑥̅ = Sample mean of P2 x = summation of individual value n = sample size Therefore, 𝑥̅ = 46+38+32+⋯+36 20 = 38.8 (b) Variance (𝑆2 ) = ∑ 𝑥2 − (∑ 𝑥)2 𝑁 𝑁−1 Where S2 = Sample Variance of P2 x2 = sum of square of P1 individual (∑x)2 N = Mean square of P2 N = sample size Therefore, 𝑆2 = 30413−30031.25 19 = 381.75 19 = 20.09 (c) Variance of mean ( 𝑆 𝑥̅ 2) = 𝑆2 𝑛 Where 𝑆 𝑥̅ 2 = mean variance of P2 S2 = sample variance
• 3. 3 n = sample size Therefore, 𝑆 𝑥̅ 2 = 20.09 20 = 1.01 (d) Standard error of mean ( 𝑆 𝑥̅ ) = √ 𝑆2 𝑛 Where 𝑆 𝑥̅ = mean standard error of P2 S2 = sample variance n = sample size Therefore, 𝑆 𝑥̅ = √ 𝑆 𝑥̅ 2 = √1.01 = 1.00 1c. Calculationfor F1 (P1xP2)Generation: (a) Mean (𝑥̅) = ∑ 𝑥 𝑛 Where 𝑥̅ = Sample mean of F1 x = summation of individual value n = sample size Therefore, 𝑥̅ = 83+90+85+⋯+102 40 = 86.1 (b) Variance (𝑆2 ) = ∑ 𝑥2 − (∑ 𝑥)2 𝑁 𝑁−1 Where S2 = Sample Variance of F1 x2 = sum of square of P1 individual (∑x)2 N = Mean square of P1 N = sample size Therefore, 𝑆2 = 297662−296528.4 39 = 1133.6 39 = 29.07 (c) Variance of mean ( 𝑆 𝑥̅ 2) = 𝑆2 𝑛 Where 𝑆 𝑥̅ 2 = mean variance of F1 S2 = sample variance n = sample size Therefore, 𝑆 𝑥̅ 2 = 29.07 40 = 0.73
• 4. 4 (d) Standard error of mean ( 𝑆 𝑥̅ ) = √ 𝑆2 𝑛 Where 𝑆 𝑥̅ = mean standard error of F1 S2 = sample variance n = sample size Therefore, 𝑆 𝑥̅ = √ 𝑆 𝑥̅ 2 = √0.73 = 0.85 Summary of the calculation Generation Mean Variance Variance of mean Standard error of mean Inbred P1 94.60 25.52 1.28 1.13 Inbred P2 38.80 20.09 1.01 1.00 F1 (P1 x P2) 86.10 29.07 0.73 0.85
• 5. 5 Question 2: The mean and their standard errors of the six basic generations from a cross between two inbreds are as follows: Family Mean Standard error Sample size P1 96.00 1.43 20 P2 40.50 1.44 20 F1 88.20 1.03 40 F2 79.40 1.20 80 BC1.1 93.00 0.74 80 BC1.2 65.20 1.05 80 (a) Determine if there was significant variation in F1, F2 and backcross generations. (b) Estimate VD, VA, VAD and VE and explain what they mean. (c) Estimate the dominance ratio. Solution: We need to derive the variance value for the families. Since 𝑆 = √ 𝜎2 𝑛 Therefore, 𝜎2 = ( 𝑆)2 × 𝑛 ----------------------- (1) Where S = standard error 2 = population variance n = sample size By substituting the value of the given standard error and sample size into equation 1 above. Family Variance Sample size P1 40.90 20 P2 41.47 20 F1 42.44 40 F2 115.20 80 BC1.1 43.81 80 BC1.2 88.20 80 2a) To calculate the significant variation in F1, F2, and the backcross generations We must first calculate the environmental variance (VE) which is the average of the non-segregating variance (VP1, VP2, and VF1). So, VE = 𝑉 𝑃1+𝑉 𝑃2 +𝑉𝐹1 3 = 40.90+41.47+42.44 3 = 124.81 3 = 41.60
• 6. 6 Note that this is applicable if there is no significant difference between the non- segregating families. So, we need to test for significant between the non-segregating families. F-test = 𝐻𝑖𝑔ℎ𝑒𝑠𝑡 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝐿𝑜𝑤𝑒𝑠𝑡 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 , 𝑑𝑓𝐹1, 𝑑𝑓𝑃1 F-test(F1 & P1) = 𝑉𝐹1 𝑉 𝑃1 , 𝑑𝑓(𝐹1), 𝑑𝑓(𝑃1) Where, VF1 = F1 variance, VP1 = P1 variance, df(F1) = degree of freedom for F1, df(P1) = degree of freedom for P1. Therefore, F-test(F1 & P1) = 42.44 40.99 , 39, 19 F-test(F1 & P1) = 1.04, 39, 19 By using excel =FDIST(1.04,39,19) = 0.479 P-value = 0.479, which is greater than 0.05 significant level This showed that there is no significant difference between the non-segregating family, therefore, we can use their mean for VE. (2ai) To calculate the significant variation in F1 generation. F-test(F1) = 𝑉 𝐹1 𝑉 𝐸 , 𝑑𝑓(𝐹1),𝑑𝑓(𝑉 𝐸 ) Where, VF1 = F1 variance, VE = environmental variance, df(F1) = degree of freedom for F1, df(VE) = degree of freedom for VE. Where VF1 = 42.44, VE = 41.60, 𝑑𝑓(𝐹1) = 40 − 1 = 39, 𝑑𝑓( 𝑉 𝐸) = 𝑑𝑓( 𝐹1) + 𝑑𝑓( 𝑃1) + 𝑑𝑓( 𝑃2) = 19 + 19 + 39 = 77 Therefore, F-test(F1) = 42.44 41.60 ,39, 77 F-test(F1) = 1.02, 39, 77 By using excel P-value = FDIST(1.02, 39, 77) = 0.46 P-value = 0.46, which is greater than the significant value of 0.05.
• 7. 7 This implies that there is no significant different among the F1 generation. (2aii) To calculate the significant variationin F2 generation F-test(F2) = 𝑉 𝐹2 𝑉 𝐸 , 𝑑𝑓(𝐹2),𝑑𝑓(𝑉 𝐸 ) Where, VF2 = F2 variance, VE = environmental variance, df(F2) = degree of freedom for F2, df(VE) = degree of freedom for VE. Where VF2 = 115.2, VE = 41.60, 𝑑𝑓(𝐹2) = 80 − 1 = 79, 𝑑𝑓( 𝑉 𝐸) = 𝑑𝑓( 𝐹1) + 𝑑𝑓( 𝑃1) + 𝑑𝑓( 𝑃2) = 19 + 19 + 39 = 77 Therefore, F-test(F2) = 115.2 41.60 ,79, 77 F-test(F2) = 2.77, 79, 77 By using excel P-value = FDIST(2.77, 79, 77) = 0.0000058 P-value < 0.001, which is less than the significant value of 0.001. This implies that there is high significant difference among the F2 generation. (2aiii)To calculate the significant variationin BC1.1 generation F-test(BC1.1) = 𝑉 𝐵𝐶1.1 𝑉 𝐸 , 𝑑𝑓(𝐵𝐶1.1), 𝑑𝑓(𝑉 𝐸) Where, VBC1.1 = BC1.1 variance, VE = environmental variance, df(BC1.1) = degree of freedom for BC1.1, df(VE) = degree of freedom for VE. Where VBC1.1 = 43.81 VE = 41.60, 𝑑𝑓(𝐵𝐶1.1) = 80 − 1 = 79, 𝑑𝑓( 𝑉 𝐸) = 𝑑𝑓( 𝐹1) + 𝑑𝑓( 𝑃1) + 𝑑𝑓( 𝑃2) = 19 + 19 + 39 = 77 Therefore, F-test(BC1.1) = 43.81 41.60 ,79,77 F-test(BC1.1) = 1.05, 79, 77
• 8. 8 By using excel P-value = FDIST(1.05, 79, 77) = 0.42 P-value = 0.42, which is greater than the significant level of 0.05. This implies that there is no significant difference among the BC1.1 generation. (2aiv) To calculate the significant variationin BC1.2 generation F-test(BC1.2) = 𝑉 𝐵𝐶1.2 𝑉𝐸 , 𝑑𝑓(𝐵𝐶1.2), 𝑑𝑓(𝑉 𝐸 ) Where, VBC1.1 = BC1.2 variance, VE = environmental variance, df(BC1.2) = degree of freedom for BC1.2, df(VE) = degree of freedom for VE. Where VBC1.2 = 88.2 VE = 41.60, 𝑑𝑓(𝐵𝐶1.2) = 80 − 1 = 79, 𝑑𝑓( 𝑉 𝐸) = 𝑑𝑓( 𝐹1) + 𝑑𝑓( 𝑃1) + 𝑑𝑓( 𝑃2) = 19 + 19 + 39 = 77 Therefore, F-test(BC1.2) = 88.2 41.60 ,79,77 F-test(BC1.2) = 2.12, 79, 77 By using excel P-value = FDIST(2.12, 79, 77) = 0.00055 P-value = 0.00055, which is less than the significant level of 0.001. This implies that there is high significant difference among the BC1.2 generation. Summary of the Significant variation of F1, F2, and backcross generations Family Mean Variance P-value Remark F1 88.20 42.44 0.46 Not significance F2 79.40 115.20 5.8 x 10-6 Highly significance BC1.1 93.00 43.81 0.42 Not significance BC1.2 65.20 88.20 5.5 x 10-4 Highly significance
• 9. 9 (2b) To estimate for VD, VA, VAD and VE By using Mather ABC scaling ratio (2bi) Estimationfor value of VD VD = VBC1.1 + VBC1.2 – VF2 – VE Where VD = Dominance variance VBC1.1 = variance due to backcross one = 43.81 VBC1.2 = variance due to backcross two = 88.20 VF2 = variance due to F2 = 115.20 VE = variance due to environment = 41.6 Therefore, VD = 43.81 + 88.20 – 115.20 – 41.6 VD = -24.79 (2bii) Estimationfor value of VA VA = 2VF2 – (VBC1.1 + VBC1.2) Where VA = Additive variance VF2 = variance due to F2 = 115.20 VBC1.1 = variance due to backcross one = 43.81 VBC1.2 = variance due to backcross two = 88.20 Therefore, VA = 2(115.20) – (43.81 + 88.20) VA = 98.39 (2bii) Estimationfor value of VI or VAD VAD = ½ (VBC1.2 - VBC1.1) Where VBC1.1 = variance due to backcross one = 43.81 VBC1.2 = variance due to backcross two = 88.20 Therefore,
• 10. 10 VAD = ½(88.20 – 43.81) VAD = 22.20. (2biii)Estimationfor value of VE VE = 1 3 ( 𝑉𝑃1 + 𝑉𝑃2 + 𝑉𝐹1 ) Where VE = Environmental variance VP1 = variance due to parent one = 40.90 VP2 = variance due to parent two = 41.47 VF1 = variance due to F1 = 42.44 Therefore, 𝑉𝐸 = 40.90 + 41.47 + 42.44 3 𝑉𝐸 = 124.81 3 = 41.60 Summary of the variance component Symbol Variance Variance Value VD Dominance genetic variance -24.79 VA Additive genetic variance 98.39 VAD Interaction genetic variance 22.20 VE Environmental component of variance 41.60 The result from the variance components showed high additive genetic variance (VA) of 98.39%. This implies that large portion of the variance is genetic, then gains can be made from selecting individuals with the metric value we wish to obtain. This analysis also showed a negative dominance genetic variance (VD) of -24.79%. This implies that there is incomplete or partial dominance which favoured the recessive effect of the allele at the locus. The interaction genetic variance (VAD) showed that 22.2% of genetic variance were contributed by non-allelic genes. There was about 41.60% of error variance that was due to environmental component and this account for the environmental variance (VE).
• 11. 11 (2c) Estimationof the dominance ratio Using the Smith (1953) method for Dominance estimation or Potence ratio 𝐷𝑜𝑚𝑖𝑛𝑎𝑛𝑐𝑒 𝑟𝑎𝑡𝑖𝑜 = ( 𝐹1 −𝑀𝑃) 0.5( 𝑃2−𝑃1) , Which ranges from -1 to +1 Where F1 = mean of F1 = 88.20 MP = mid-parent of P1 and P2 = 𝑃1+𝑃2 2 = 96.0+40 .5 2 = 136.5 2 = 68.25 P1 = mean of parent one = 96.00 P2 = mean of parent two = 40.50 Therefore, Dominance ratio = (88.20−68.25) 0.5(40.50−96.00) = 19.95 0.5(−55.5) = 19 .95 −27.75 = −0.72 Dominance ratio = -0.72 This showed that as the dominance ratio moved from 0 to -1, there is increase effect of inbreeding which support pureline or inbred line selectionwhile dominance ratio 0 to +1 showed increase in heterosis which support hybrid selection. The result of the calculation showed a negative effect of dominance ratio of 0.72. which means that there is high inbreeding effect which thence to achieve homozygosity of the allele. -1 +10-0.72 Inbreeding Heterosis