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1. SCALING TEST
GPB 621 – PRINCIPLES OF QUANTITATIVE GENETICS
Class - 6
Dr. K. SARAVANAN
Professor
Department of Genetics and Plant Breeding
Faculty of Agriculture
Annamalai University

2. Dr. K. Saravanan, GPB, AU
SCALING TEST
• A test for additivity.
• Suggested by Mather.
• It is made by using mean measurements of the traits of various
generations.
.

3. Dr. K. Saravanan, GPB, AU
• If the gene effects involved in a hybrid is additive only, then F1 value
should not differ significantly from the mid-parental mean.
i.e.,
= P1 + P2 = 2F1
= 2F1 = P1 + P2 (or) 2F1 – (P1 + P2) = 0
The value obtained 2F1 – (P1 + P2) should not differ significantly from zero
.
1
2
1
2





F
P
P

4. Dr. K. Saravanan, GPB, AU
• The same principle is applicable to the segregating generations like F2
or back cross generation either P1 nor P2.
• This we refer to “Adequacy of the scale”
• Hence,
.
4
2 2
1
1
2



 


P
F
P
F
4
2 2
2
1
3



 


P
F
P
F

1
P
to
BC
2
1
1
1


 

P
F
B

2
P
to
BC
2
2
1
2


 

P
F
B
0
)
2
(
4 2
1
1
2 







P
F
P
F
0
)
2
(
4 2
2
1
3 







P
F
P
F
0
)
(
2 1
1
1 





P
F
B
0
)
(
2 2
1
2 





P
F
B

5. Dr. K. Saravanan, GPB, AU
• Mather worked out this scaling for other segregating generation as
below
• A = (For Back cross to P1)
• B = (For Back cross to P2)
• C = (For F2 generation)
• D = (For F3 generation)
The above formulae are valid on the assumption that the epistatis is
absent
.
0
)
(
2 1
1
1 





P
F
B
0
)
(
2 2
1
2 





P
F
B
0
)
2
(
4 2
1
1
2 







P
F
P
F
0
)
2
(
4 2
2
1
3 







P
F
P
F

6. Dr. K. Saravanan, GPB, AU
• Conversely if A. B. C. or D. significantly deviated from zero for a trait
then it demonstrates the “inadequacy” of the scale, that is some non
additive gene effects like epistasis are present.
.

7. Dr. K. Saravanan, GPB, AU
How to test whether these values are significant or not?
• By using the variances of the various generations for the respective means,
tests of significance are made.
• The variances of the various generations is arrived by squaring the standard
errors of the means.
.
Variance of F2 = VF2
Variance of B1 = VB1
Variance of B2 = VB2
Variance of P1 = VP1
Variance of P2 = VP2
Then the variance of the quantities A.B.C and D determined as
Variance of A = VA = 4VB1+VP1+VF1
Variance of B = VB = 4VB2+VP2+VF1
Variance of C = VC = 16VF2+4VF1+VP1+VP2
Variance of D = VD = 16VF3+4VF2+VP1+VP2

8. Dr. K. Saravanan, GPB, AU
• The standard errors are found as the square roots of the corresponding
variances
i.e., SE of A = √𝑉𝐴
SE of B = √𝑉𝐵
SE of C = √𝑉𝐶
SE of D = √𝑉𝐷
.
Then significant can be tested as
usual
• i.e., if values of A, B, C or D do
not differ by twice the SE then
the values can be considered
that they are not significant from
zero

9. Dr. K. Saravanan, GPB, AU
• The adequacy or otherwise of this model has generally no
relationship with the range of variation of the crosses.
• The model which is adequate for one cross may not be adequate for
another which covers the same range or variation and still in-
adequate for other crosses covering wider ranges of variation.
• If inadequacy of the scale is noted in a cross, it may be then reworked
using
1. Logarithmic transformations (or)
2. Joint scaling test.
The analysis may also be extended directly to estimate parameters of epistatic
components.
.

10. Dr. K. Saravanan, GPB, AU
Problem 1
• Work out scaling test for given population. Data per plant yield in
gram of the hybrid Reba B.50 x Laxmi.
.
Generation Mean ± SE
P1 7.20 ± 2.47
B1 20.42 ± 3.50
F1 24.07 ± 2.64
F2 21.66 ± 4.60
B2 19.25 ± 3.71
P2 9.20 ± 2.70

11. Dr. K. Saravanan, GPB, AU
• A = 2B1 – (F1 + P1)
= (2 X 20.42) – (24.07 +7.20)
= 9.57
• B = 2B2 – (F1 + P2)
= (2 X 19.25) – (24.07 + 9.20)
= 5.23
• C = 4F2 – (2F1 +P1 +P2)
= (4 X 21.66) – ((2 X 24.07) + 7.20 + 9.20))
= 22.10
.
Generation Mean ± SE
P1 7.20 ± 2.47
B1 20.42 ± 3.50
F1 24.07 ± 2.64
F2 21.66 ± 4.60
B2 19.25 ± 3.71
P2 9.20 ± 2.70

12. Dr. K. Saravanan, GPB, AU
Working of SE
• VA = 4VB1+VP1+VF1
= 4(3.50)2 + (2.47)2+ (2.64)2
= 62.0705
SE =√𝑽𝑨
= 62.0705 = 7.88
Test of Significant
t =
𝑨
𝑺𝑬𝑨
=
𝟗.𝟓𝟕
𝟕.𝟖𝟖
= 1.2144
.
Generation Mean ± SE
P1 7.20 ± 2.47
B1 20.42 ± 3.50
F1 24.07 ± 2.64
F2 21.66 ± 4.60
B2 19.25 ± 3.71
P2 9.20 ± 2.70

13. Dr. K. Saravanan, GPB, AU
.
Generation Mean ± SE
P1 7.20 ± 2.47
B1 20.42 ± 3.50
F1 24.07 ± 2.64
F2 21.66 ± 4.60
B2 19.25 ± 3.71
P2 9.20 ± 2.70
• VB = 4VB2+VP2+VF1
= 4(3.71)2 + (2.70)2+ (2.64)2
= 69.3160
SE =√𝑽𝑩
= 69.3160 = 8.33
Test of Significant
t =
𝑩
𝑺𝑬𝑩
=
𝟓.𝟐𝟑
𝟖.𝟑𝟑
= 0.6278

14. Dr. K. Saravanan, GPB, AU
.
Generation Mean ± SE
P1 7.20 ± 2.47
B1 20.42 ± 3.50
F1 24.07 ± 2.64
F2 21.66 ± 4.60
B2 19.25 ± 3.71
P2 9.20 ± 2.70
• VC = 16VF2+4VF1+VP1 +VP2
= 16(4.60)2 + 4(2.64)2 +
(2.47)2+ (2.70)2
= 379.8293
SE =√𝑽𝑪
= 379.8293 = 19.48
Test of Significant
t =
𝑪
𝑺𝑬𝒄
=
𝟐𝟐.𝟏𝟎
𝟏𝟗.𝟒𝟖
= 1.1345

15. Dr. K. Saravanan, GPB, AU
Result :
• A = 9.57 ± 7.88
• B = 5.23 ± 8.33
• C = 22.10 ± 19.48
.
• if values of A, B, C or D do not
differ by twice the SE then the
values can be considered that
they are not significant from zero
• t – Table value = 1.96
Scale SCALE VARIANCE SE CALCULATED t VALUE
A 9.57 62.0705 7.88 1.2144 Non Significant
B 5.23 69.3160 8.33 0.6278 Non Significant
C 22.10 379.8293 19.48 1.1345 Non Significant

16. Dr. K. Saravanan, GPB, AU
• Problem 2
• Work out scaling test for given sesame seed yield data.
.
Generation MEAN VARIANCE
P1 18.09 0.23
B1 28.72 0.25
F1 30.05 0.15
F2 42.63 0.32
B2 36.71 0.29
P2 26.05 0.21

17. Thank q
Dr. K. Saravanan, GPB, AU