1. 1
Name: Abah Simon Peter
Class: PhD Student
Course Code: WACI702 (Quantitative Genetics in Plant Breeding)
Assignment: #2
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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