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Annex 1. Cases and levels of comparison of regeneration capacity of different explant types - hypothetical examples

Objective: To study the shoot regeneration capacity of a source organ using two types of explants:

1) conventional explant (conv); 2) tTCL explant (tTCL)

From the source organ, two conventional explants or 20 tTCL explants can be prepared, that is, 10 tTCL explants can be prepared theoretically from one source organ.

Two parameters are observed at the end of the experiments for both explant types:

(1) average number of shoots per regenerating explant (SN);

(2) percentage of explants which regenerated compared to explants prepared (R%)

Example 1 / Example 2 / Example 3 / Example 4 / Example 5
SN / R% / SN / R% / SN / R% / SN / R% / SN / R%
conventional explant / 12 / 100 / 12 / 100 / 12 / 100 / 12 / 70 / 6 / 80
tTCL explant / 6 / 100 / 6 / 70 / 6 / 30 / 6 / 100 / 8 / 100

According to the two parameters observed at the end of the experiments, there are two cases or levels of comparison:

Case I – per regenerating explant base: SN on tTCL type of explant is compared to the SN of the conventional type of explant; that is, comparison of the shoot yield on one regenerating explant.

Case II – per organ base (GCF); SN that regenerates from the source organ using tTCL and conventional explant types; that is, comparison of the shoot yield from the source organ using different types of explants.

In our first four hypothetical examples (Examples 1-4), making a Case I comparison, we can conclude that by using a tTCL explant the number of shoots we could obtain is half that that can be obtained by using a conventional explant, because:

SNtTCL / SNconv. = 6/12 = 0.5 (see the table below), and in Example 5 we can conclude that by using tTCL explants 1.3-fold more shoots can be obtained compared to the use of a conventional explant.

Example 1 / Example 2 / Example 3 / Example 4 / Example 5
Level of comparison / Case I / Case II / Case I / Case II / Case I / Case II / Case I / Case II / Case I / Case II
comparison of RC of tTCL and conventional explants / 0.5 / 5 / 0.5 / 3.5 / 0.5 / 1.5 / 0.5 / 40 / 1.3 / 10.7

However, when making a comparison on a source organ basis, that is, Case II comparison, we have to also take the number of explants that can be prepared from the source organ and the percentage of explants which regenerated from the explants prepared into consideration for both explant types.

In this case, as can be seen in the table above in our first four hypothetical examples (Examples 1-4), we can conclude that the number of shoots can be obtained using a tTCL explant is higher than that that can be obtained by using a conventional explant according to the following calculation:

, where n = the number of tTCL explants that can be theoretically prepared from a source organ.

Example 1: ; Example 2: ; Example 3: , Example 4:

In Example 1 the shoot number that can be obtained is 5-fold, in Example 2, 3.5-fold, in Example 3, 1.5-fold and in Example 4, 40-fold higher by using a tTCL explant than the shoot number obtained using a conventional explant, depending on R%s.

In our Example 5, the essence of the conclusion remains the same, i.e., that more shoots can be produced by using a tTCL explant than a conventional explant. However, according to the Case II comparison:

Example 5 , the conclusion should be modified because the efficacy of tTCLs is 10.7-fold higher instead of 1.3-fold as calculated considering the Case I comparison.

This calculation is the GCF itself, that is,(Eq. 1).


Annex 2 (modified (that is, wording and mathematical notation) from http://answers.yahoo.com/question/index?qid=20090611090428AAgXEwo, and notes therein)

Surface area = A = 2 R π h

where R = radius of a sphere; h = height of a dome.

R is determined from Pythagoras's rule: R2 = r2 + (R - h)2

where r = radius of a circular foundation.

R2 = r2 + R2 - 2Rh + h2

R = (r2 + h2) / (2 h)

r is found from circumference L:

r = L/(2 π) = 19.18/(2 π) = 3.05

R = (3.052 + 11.12) / (2 × 11.1) = 5.97

Surface area:

A = 2 × 5.97 × π × 11.1 = 416.37


Annex 3 (modified (that is, wording and mathematical notation) from http://mathcentral.uregina.ca/qq/database/QQ.09.99/wilkie1.html, and notes therein)

A B C

When a sphere of radius R is placed inside a cylinder, with the cylinder just touching the equator, and cut off at the height of the top and bottom of the sphere (see cutaway view in A), then the area of the curved part of the cylinder = 2πR×2R=4πR2, found by slicing the cylinder surface and rolling it out as a rectangle. The cylinder surface is exactly the area of the sphere. When a small horizontal slice is made through the diagram (B), this cuts a rectangle out of the rolled out cylinder and a slightly distorted rectangle out of the sphere. In B (cross-sectional view), hc = height of the slice on the cylinder; hs = length of the arc on the sphere cut out by the slice; r = radius of the distorted rectangle on the sphere; R = radius of the sphere.

Area of orange rectangle on the cylinder = 2πRhc.

Area of the distorted orange rectangle on the sphere is approx. 2πrhs.

These two areas are approximately equal (proof outlined below*).

Similarly, the volume of a hemi-sphere can be compared with the volume of a cylinder with an inverted cone removed. Consider the hemisphere with the equator at the bottom and the cone with the 'base' at the top. For each horizontal slice, the area of the slice of the sphere (that is, a circle) is exactly the same area of the cylinder outside the cone. Because all slices have the same area, the volumes are equal.

Volume of hemisphere = π R2 ×R - (1/3) π R2 ×R = (2/3) π R3.

*Proof that the areas of the two "rectangular" strips are approximately equal

In C, K is the center of the sphere. A tangent line PA is constructed tangent to the sphere at P and a perpendicular is dropped from A to meet QP at B. Because PA is tangent to the sphere, angle KPA is a right angle.

Because triangle KPQ is a right angle triangle, angleQKP+angleKPQ=90o. However, angleKPQ+angleBPA=90o, and thus angleQKP=angleBPA. Hence the right triangles QKP and BPA are similar. Thus, PQ:KP=AB:PA. But |PQ| = r, |KP| = R, |AB| = hc and |PA| is approximately hs and thus Rhc is approximately rhs. Thus the area of the orange rectangle on the cylinder, 2πRhc and the area of the distorted orange rectangle on the sphere 2πrhs are approximately equal.


Annex 4 Substitutions and simplifications used while calculating equations used in the text.

Eq. 4: Comparison of tTCL and conventional explant in Cymbidium

Substitution into Eq. 3, , using the surface and volume equations from Table 3, such as the epidermal surface and volume of a dome for conventional explant: Aepid = 2 πr2; V = 2/3 πr3, and epidermal surface and explant of a cylinder for tTCL explants: Aepid = 2 πrh; V = 2/3 πr3:

Simplifications in the equation:

Eq. 5: Comparison of lTCL and conventional explant in Cymbidium

Substitution into Eq. 3, , using the surface and volume equations from Table 3, such as the epidermal surface and volume of a dome for conventional explant: Aepid = 2 πr2; V = 2/3 πr3, and epidermal surface and explant of a rectangle-based prism for lTCL explants: Aepid = lw; V = lwh:

Simplifications in the equation:

Eq. 7: Comparison of tTCL and conventional explant in chrysanthemum

Substitution into Eq. 6, , using the surface and volume equations from Table 3, such as the epidermal surface of a half-cylinder for conventional explant: Aepid = 2 πrh, and epidermal surface and explant of a cylinder for tTCL explants: Aepid = 2 πrh:

Simplifications in the equation:

Eq. 9: Comparison of lTCL and conventional explant in chrysanthemum

Substitution into Eq. 8, , using the surface and volume equations from Table 3, such as the epidermal surface and volume of a half-cylinder for conventional explant: Aepid = πrh; V = (r2πh)/2, and epidermal surface and explant of a rectangle-based prism for lTCL explants: Aepid = lw; V = lwh:

Simplifications in the equation:

Eq. 10: Comparison of tTCL and conventional explant in apple

Substitution into, using the surface and volume equations from Table 3, such that the surface and volume of a trapezium-based prism for a conventional explant:

A = h(a+b+c+d)+w(a+c); V = h(w(a+c)/2)), and

the surface and volume of a rectangle-based prism for tTCL explants: A = 2 h(l+w)+ 2 lw; V = lwh:

Annex 5 Changes of k factor of apple cultivar ‘Royal Gala’ after 9-week-long regeneration period under the same (same position of the source leaf and same TDZ content of the medium) and under different experimental protocols (different source leaf position and different TDZ content of the medium, both respectively and in combination)

TDZ content of the medium / Explant type / Position of the source leaf / SN
(shoot number per regenerating explant) / GF
(Geometric Factor) / k
(correction factor) / SR%
(percentage of explants forming shoots) / GCF
(Growth Correction Factor)
0.5 mg/l
(2.27 µM) / conventional / 1st / 10.2 / 0.4712 / 100
2nd / 12.1 / 0.5192 / 100
tTCL / 1st / 5.5 / 1.1 / 22 / 2.9
2nd / 4.1 / 0.7 / 29 / 2.6
5 mg/l
(22.7 µM) / 1st / 5.1 / 1.1 / 70 / 8.8
2nd / 6.5 / 1.0 / 88 / 11.8
* SN and SR% data were published in recent publications (Dobránszki and Teixeira da Silva 2013; Teixeira da Silva and Dobránszki 2013b) in other contexts.

Examples originate from our recent experiments with apple*