Production of Fruits, Vegetables, Turf, and Ornamental Plants
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Which is the appropriate pruning tool if a limb to be pruned is less than 0.5 inches in diameter?
bypass hand pruner
hand pruning saw
A hand pruning saw is better used for limbs that are greater than 1.25 inches in diameter.
A chain saw is much too large for pruning; it likely would snag, rip, and tatter such a small limb.
Although a lopper could prune a limb that is less than 0.5 inches in diameter, the bulkiness of its long handles and large pruning head makes it more difficult to maneuver into tight spaces to remove small branches. It is a better choice for limbs between 0.5 and 1.25 inches in diameter.
Question 1 Explanation:
A sharp and properly aligned bypass hand pruner will cleanly cut a limb of less than 0.5 inches in diameter.
Which weather conditions might require the use of an at-transplant fungicide?
Low humidity is not conducive to fungal growth, thus fungicide would not be warranted.
high barometric pressure
Weather effects barometric pressure, and high barometric pressure alone would not warrant an at-transplant fungicide.
low barometric pressure
Weather effects barometric pressure, and low barometric pressure alone would not warrant an at-transplant fungicide.
Question 2 Explanation:
High humidity is conducive to fungal growth, thus fungicide would be warranted as a preventive
You have 3 ft3 of media that contains predominantly peat. You add 1 ft3 of coarse perlite to it. Which of the properties below will increase in the new peat – perlite mixture?
The bulk density of perlite is very low, as demonstrated by its tendency to float to the surface of the substrate when irrigated.
Perlite has no carbon.
Perlite has low water retention compared to peat, thus its addition does not increase the water-holding capacity of the substrate.
Question 3 Explanation:
Perlite is expanded aluminosilicate rock with much air space, thus, it increases the aeration or porosity of the substrate.
A landscaper needs a fertilizer high in nitrogen. What product should be selected?
4 – 12 – 4
This fertilizer is higher in percent phosphorus pentoxide (P2O5) by weight.
4 – 4 – 12
This fertilizer is higher in percent potassium oxide (K2O) by weight.
4 – 4 – 4
A balanced fertilizer is not higher in one macronutrient.
12 – 4 – 4
Question 4 Explanation:
This fertilizer is higher in percent nitrogen by weight.
What determines the size of the transplant hole for a balled and burlapped tree?
caliper of the tree trunk
The caliper of the tree trunk will determine how the tree should be prepared for transplanting, e.g. balled and burlapped vs. tree spade, and the size of the root ball.
proximity to a building
Building proximity does not affect the size of the transplant hole, but it should influence one's choice of plant that would be appropriate for planting near a building.
radius of the tree canopy
The radius of the tree canopy is not directly used to determine appropriate root ball size.
size of the root ball
Question 5 Explanation:
The transplant hole should be dug twice the diameter of the soil ball. The depth of the hole should be about the same as that of the soil ball.
After a long period of consistently warm weather, a cold front causes high winds, subfreezing temperatures and low humidity at a plant nursery. The nursery manager turns on the overhead irrigation system and runs it continuously during the period of sub-freezing temperatures, low humidity and high wind. What is likely to happen to the cold-sensitive, tropical plant species grown in the nursery as a result of these action?
The high level of irrigation will promote fungal disease, which will damage or destroy the crop.
Fungi will not be active at sub-freezing temperatures and low humidity.
Evaporative cooling will occur and the low temperatures will damage or destroy the crop.
A coating of ice will form on the plants, protecting them from cold damage.
Because the humidity is low and the wind is high, the water will evaporate rather than form ice.
As ice forms on the plant leaf surfaces, heat energy released by the water molecules will warm the plants.
If the weather had been calm and the humidity not low, the water would have slowly frozen on the plants, releasing its latent heat of fusion as it transitioned from liquid to solid and, thus, potentially protecting the plant tissues from freezing. However, this was not the case.
Question 6 Explanation:
Because the humidity is low and the wind is high, the water will evaporate. Evaporative cooling will further remove heat from the plant tissues, allowing the plant to freeze.
A flowering potted plant crop currently in the greenhouse has a production schedule with a target market date of November 1. If the current production conditions are continued, the crop will not be ready for the retail market until November 7. Assuming light intensity is not limiting, what changes could be implemented to help ensure quality plant are ready for the November 1 market date?
Increase the nitrogen and potassium fertility levels.
Ammonium level is generally reduced at the end of the crop cycle to improve flower development and postharvest life. Plants are able to store excess nitrates but do not store excess ammonium, thus high levels of ammonium can lead to ammonium toxicity. Excess potassium may accentuate N, Ca, or Mg deficiency.
Increase the pH and raise the night temperatures.
Increasing the pH will reduce the availability of some nutrients. Raising the night temperature will increase respiration and may delay flowering in some crops, e.g. poinsettia. Respiratory rate increases about 2-4✕ for each 10°C (18°F) rise.
Increase the humidity and start night interruption lighting.
Increased humidity can increase disease pressure on the crop. Night interruption lighting will inhibit flower initiation in short day plants. Short day plants (SDP), e.g. poinsettia and mum, are induced to flower when day is shorter than the species’ critical day length. Remember, it’s really night length that the plants are sensing.
Increase the daytime temperature and inject carbon dioxide into the growing area.
Question 7 Explanation:
In order to increase the growth rate, one needs to increase the photosynthetic rate of the plants. Photosynthetic rate approximately doubles in many temperate species for each 10°C (18°F) rise in temperature. Photosynthetic rate can be doubled by increasing CO2 to 0.1%. (Ambient CO2=0.03%)
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