翻译Rab GTPases serve as master regulators of membrane trafficking. They can be activated by guanine nucleotide exchange factors (GEF) and be inactivated by GTPase-activating proteins (GAPs). The roles of some GAPs have been explored in Saccharomyces cerevisiae, but are largely unknown in filamentous fungi. Here, we investigated the role of GAP Gyp3 gene, an ortholog of S. cerevisiae Gyp3, in an entomopathogenic fungus, Metarhizium acridum.

结合素GTP酶是膜货运的主要调节因子，它们可以通过环腺苷酸交换因子（GEF）激活，并被GTP酶激活蛋白（GAP）抑制。在酵母菌中，一些GAP的作用已经被探索，但在丝状真菌中却大多未知。在这里，我们研究了昆虫病原真菌Metarhizium acridum中GAP Gyp3基因的作用，它是酵母菌Gyp3的同源基因。

相关问题

用c++解决Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points specializing in the same pair of currencies. Each point has its own exchange rates, exchange rate of A to B is the quantity of B you get for 1A. Also each exchange point has some commission, the sum you have to pay for your exchange operation. Commission is always collected in source currency. For example, if you want to exchange 100 US Dollars into Russian Rubles at the exchange point, where the exchange rate is 29.75, and the commission is 0.39 you will get (100 - 0.39) * 29.75 = 2963.3975RUR. You surely know that there are N different currencies you can deal with in our city. Let us assign unique integer number from 1 to N to each currency. Then each exchange point can be described with 6 numbers: integer A and B - numbers of currencies it exchanges, and real RAB, CAB, RBA and CBA - exchange rates and commissions when exchanging A to B and B to A respectively. Nick has some money in currency S and wonders if he can somehow, after some exchange operations, increase his capital. Of course, he wants to have his money in currency S in the end. Help him to answer this difficult question. Nick must always have non-negative sum of money while making his operations. Input The first line contains four numbers: N - the number of currencies, M - the number of exchange points, S - the number of currency Nick has and V - the quantity of currency units he has. The following M lines contain 6 numbers each - the description of the corresponding exchange point - in specified above order. Numbers are separated by one or more spaces. 1 ≤ S ≤ N ≤ 100, 1 ≤ M ≤ 100, V is real number, 0 ≤ V ≤ 103. For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10-2 ≤ rate ≤ 102, 0 ≤ commission ≤ 102. Let us call some sequence of the exchange operations simple if no exchange point is used more than once in this sequence. You may assume that ratio of the numeric values of the sums at the end and at the beginning of any simple sequence of the exchange operations will be less than 104. Output If Nick can increase his wealth, output YES, in other case output NO.

这道题目可以用贝尔曼-福德算法（Bellman-Ford Algorithm）来解决。算法的主要思想是通过动态规划的方式来计算从起点到各个顶点的最短路径。在这个问题中，我们可以将起点设置为 Nick 持有的货币，然后通过不断地进行货币兑换来尝试增加他的财富。

具体来说，我们可以定义一个一维数组 d，其中 d[i] 表示从起点到货币 i 的最大收益。初始时，我们将 d[S] 设置为 V，表示 Nick 初始持有的货币数量。然后，我们可以通过遍历所有的兑换点来更新数组 d。对于每个兑换点，我们可以根据其提供的兑换率和佣金计算出从 A 货币到 B 货币的汇率和佣金，然后使用这些值来更新数组 d。具体来说，我们可以使用以下公式来更新 d：

d[B] = max(d[B], (d[A] - C) * R)

其中 C 是佣金，R 是汇率。这个公式的意思是，我们将从 A 货币到 B 货币的汇率应用到当前持有的 A 货币数量上，减去佣金后得到实际能够兑换成的 B 货币数量，然后将这个数量与原本持有的 B 货币数量进行比较，取较大值作为更新后的 d[B] 值。

我们可以重复执行上述更新操作，直到没有任何一次更新能够使 d 数组发生变化为止。如果此时 d[S] 大于 V，那么说明 Nick 可以增加他的财富，输出 YES，否则输出 NO。

下面是完整的 C++ 代码实现：

fit.score Call: coxph(formula = Surv(OS.time, OS) ~ ., data = my.cli.2) coef exp(coef) se(coef) z p NRP1 0.5326 1.7033 0.3492 1.525 0.127 RAB31 -0.5530 0.5752 0.3731 -1.482 0.138 LOX 0.3780 1.4593 0.2487 1.520 0.129 ###哪列是变量的系数

在fit.score的结果中，第3列（coef）是变量的系数，表示每个预测变量对生存时间的影响大小。具体来说，coef的正负号表示变量对生存时间的影响方向（正表示增加该变量会增加生存时间，负表示增加该变量会减少生存时间），coef的绝对值表示变量对生存时间的影响大小（绝对值越大，影响越明显）。例如，在这个结果中，NRP1的系数为0.5326，表示每单位NRP1的增加，生存时间增加1.7033倍；RAB31的系数为-0.5530，表示每单位RAB31的增加，生存时间减少0.5752倍；LOX的系数为0.3780，表示每单位LOX的增加，生存时间增加1.4593倍。