Section 7: Example Growth Regulator Calculations
- Determine the number of ml of Bonzi (paclobutrazol) required per liter for a 10 ppm solution.
Bonzi contains 4000 mg a.i./l or 4 mg a.i./ml
10 ppm = 10 mg/l
Therefore, we need to add enough of the Bonzi to have 10 mg a.i. per liter.
1 ml contains 4 mg a.i.
Therefore, 2.5 ml contains 10 mg a.i.
If we add 2.5 ml of Bonzi to 997.5 ml water (final volume of 1 liter) we will have a 10 ppm solution.
- We need enough 10 ppm Sumagic (uniconazole) solution to cover 1000 ft2 of bench space using the 2 quarts per 100 ft2 rate. Determine the ml of Sumagic required per liter and per gallon for a 10 ppm solution. Then determine the number of quarts required, the total volume of solution required and the total amount of Sumagic required for that volume.
We need 10 mg/L and Sumagic contains 500 mg a.i./L or 0.5 mg a.i./ml
Therefore, 20 ml contains 10 mg a.i. and 20 ml Sumagic added to 980 ml of water (total volume of 1 liter) will provide yield a 10 ppm solution.
There are 3.78 liters/gallon. Therefore, if we needed 10 mg a.i./L, we need 37.8 mg a.i./gallon.
37.8 mg a.i./0.5 mg a.i. per ml = 75.6.
Therefore, we need 75.6 ml/gallon.
Or, if we needed 20 ml for a liter, we need 20 ml x 3.78 = 75.6 ml/gallon
There are 4 quarts per gallon. Therefore, we need 75.6 ml & divide; 4 = 18.9 ml/quart.
At 2 quarts per 100 ft2, we would need 20 quarts to cover 1000 ft2. At 18.9 ml/quart. We would need to add 378 ml/20 quarts to cover the required area with a 10 ppm Sumagic solution.
- You want to apply 0.25 mg a.i. A-Rest (ancymidol) per container and you have 2000 containers. Determine the total volume of solution required and the total amount of A-Rest required.
Some publications recommend applying 3.4 fl. oz. per container while others recommend 8 fl. oz. per container. The goal is to nearly saturate the substrate without having runoff. Remember that you can multiply fl. oz. by 0.03 to convert to liters or 30 to convert to ml.
If we want to apply 100 ml per container, we need a total volume of 100 ml x 2000 = 200,000 ml or 200 liters.
2000 containers x 0.25 mg a.i./container = 500 mg a.i. required
A-Rest = 264 mg a.i./L
Therefore, 500 mg a.i. / 264 mg a.i/L = 1.9 liters
Therefore, add 1.9 liters A-Rest to 198.1 liters of water (total volume of 200 liters) and apply 100 ml per container.
You are now at the end of Unit 10: Plant Growth Regulators. Click here to take the self-exam.
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