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METHOD FOR ESTIMATING COMMERCIAL YIELDS OF WATER MELON BREEDING STOCK

机译:西瓜育种商品商业产量的估算方法

摘要

the invention u043eu0442u043du043eu0441u0438u0442u0441u00a0 to plant breeding, in particular to a selection u0430u0444u0431u0443u0437u0430. the purpose of u0438u0437u043eu0431u0440u0435u0442u0435u043du0438u00a0 - accelerating the evaluation of productivity u0441u0435u043bu0435u043au0446u0438u043eu043du043du043eu0433u043e material. the u043fu0440u043eu0432u043eu0434u00a0u0442 by counting number of standard size and u0432u0437u0432u0435u0448u0438u0432u0430u043du0438u00a0 one the largest fruit on u0434u0435u043bu00a0u043du043au0435.u0438u0441u043fu043eu043bu044cu0437u0443u00a0 installed high u043au043eu0440u0440u0435u043bu00a0u0446u0438u043eu043du043du0443u044e u0441u0432u00a0u0437u044c between product productivity and mass of single fruit with the largest u0434u0435u043bu00a0u043du043au0438 (u0433u0443u0445 1 + 0.63 - 0.69) and quantities om of standard size (u0433u0443u0445u0408 + 0.87, 0.91, rely on crop productivity formula 0.35 x.+ 2,60 (xg / n) - 1.16, where - u0442u043eu0432u0430u0440u043du0430u00a0 productivity one u0440u0430u0441u0442u0435u043du0438u00a0, kg; x is one of the largest fruit on u0434u0435u043bu00a0u043du043au0435, kg;, x is the number of times of the standard measure; n is the number of these plants. 1, table 2). and ss (l
机译:本发明可以用于植物育种,尤其是选择到品种0430 u0444 u0431 u0443 u0437 u0430。 u0438 u0437 u043e u0431 u0440 u0435 u0442 u0435 u043d u0438 u00a0的目的-加速对生产率的评估 u0441 u0435 u043b u0435 u0435 u043a u0446 u0438 u043e u043d u043d u043e u0433 u043e材料。 u043f u0440 u043e u0432 u043e u0434 u00a0 u0442通过计算标准尺寸和 u0432 u0437 u0432 u0435 u0448 u0438 u0432 u0430 u043d u0438 u00a0一个最大的水果在 u0434 u0435 u043b u00a0 u043d u043a u0435上。 u0438 u0441 u043f u043e u043b u044c u0437 u0443 u00a0安装了高 u043a u043e u0440 u0440 u0435 u043b u00 u0446 u0438 u043e u043d u043d u0443 u044e u0441 u0432 u00a0 u0437 u044c在产品生产率和具有最大 u0434 u0435 u043b u00a0 u043d u043a u0438( u0433 u0443 u0445 1 + 0.63-0.69)和标准尺寸的数量om( u0433 u0443 u0445 u0408 + 0.87,0.91,取决于作物生产力公式0.35 x。+ 2,60(xg / n) -1.16,其中- u0442 u043e u0432 u0430 u0440 u043d u0430 u00a0生产力一个 u0440 u0430 u0441 u0442 u0435 u043d u0438 u00a0,公斤; x是最大的果实之一 u0434 u0435 u043b u00a0 u043d u043a u0435,kg; x是标准量的次数; n是这些植物的数量1 ,表2)。和ss(l

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