Genetic control method for tea geometrids
A genetic control and tea geometrid technology, which can be applied to botany equipment and methods, horticulture, plant protection, etc., can solve the problem of unsatisfactory control effect
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Embodiment 1
[0015] Example 1 Prevention and control of tea geometrid (population A) in Hangzhou area (release males of population B to population A)
[0016] (1) Prediction of tea geometrid control plots
[0017] 1. Prediction of occurrence: In the plots that need to be controlled, at the 2nd instar larvae blooming stage (middle stage), use the 5-point sampling method, take a 1m-long tea row at each point, check and record the number of larvae on the surface of the tea bushes, according to The area of 1m long tea row is 1.5m 2 , 1 mu covers an area of 667m 2 , to calculate the population density. Such as 5 points (7.5m 2 ) the total number of larvae is 70, then the number of 2nd instar larvae per mu is 6225 (70 / 7.5×667). In general, the survival rate (field) of 2nd instar larvae to pupa is 50%, and the survival rate to adult is 40%. 6225 heads × 50% = 3113 heads; 6225 heads × 40% = 2490 heads, that is, in this tea garden, the number of pupae occurrences is 3113 heads, and the num...
Embodiment 2
[0031] Example 2 Control of tea geometrid (population B) in Quzhou area (release the males of population A to population B)
[0032] (1) Prediction of tea geometrid control plots
[0033] 1. Prediction of occurrence: In the plots that need to be controlled, at the 2nd instar larvae blooming stage (middle stage), use the 5-point sampling method, take a 1m-long tea row at each point, check and record the number of larvae on the surface of the tea bushes, according to The area of 1m long tea row is 1.5m 2 , 1 mu covers an area of 667m 2 , to calculate the population density. Such as 5 points (7.5m 2 ) the total number of larvae is 70, then the number of 2nd instar larvae per mu is 6225 (70 / 7.5×667). In general, the survival rate (field) of 2nd instar larvae to pupa is 50%, and the survival rate to adult is 40%. 6225 heads × 50% = 3113 heads; 6225 heads × 40% = 2490 heads, that is, in this tea garden, the number of pupae occurrences is 3113 heads, and the number of adults...
Embodiment 3
[0047] Example 3 Tea geometrid control in a large area (20 mu) of tea gardens (such as releasing the male insects of population B to control population A)
[0048] (1) Prediction of tea geometrid A population in the control area
[0049] 1. Occurrence prediction: Visually check whether the occurrence of tea geometrid larvae varies greatly, and investigate in 3 to 4 plots respectively. Refer to Example 1 for the investigation and calculation method of each plot.
[0050] 2. Prediction of occurrence period: In spring and autumn, there are 4 instars of tea geometrid larvae. The duration of the 2nd, 3rd, 4th instar and pupae are 4, 3, 5 days and 8.5 days respectively. ) to pupal stage is 10 days (2+3+5); to adult stage is 18.5 days (2+3+5+8.5 days). For example, the field survey (the middle stage of the second instar) is June 15th, then the pupal stage is June 25th, and the adult stage is July 3rd to 4th (also the egg-laying stage).
[0051] (2) Collection and rearing of tea geo...
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