b******v
发帖数: 1493 | 1 第13题我的想法是这样的，不知道对不对
P(exactly one accept) = N*p*(1-p)^{N-1}
logP = logN + log(p/(1-p)) + Nlog(1-p)
d(logP)/dN = 0 ==> N = 1/log(1/(1-p)) |
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b******v
发帖数: 1493 | 2 第13题我的想法是这样的，不知道对不对
P(exactly one accept) = N*p*(1-p)^{N-1}
logP = logN + log(p/(1-p)) + Nlog(1-p)
d(logP)/dN = 0 ==> N = 1/log(1/(1-p)) |
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j*****h
发帖数: 252 | 4 门外汉请教：知道化合物的水溶性怎么评估？LogP参数能评估它的水溶性？LogP的大小
和水溶性大小相关么？
感谢！ |
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m*********1
发帖数: 42 | 5 整天比来比去,换个口味做个题:设psi(x)为Chebyshev psi函数， 即psi(x)=sigma_{p=<
x}logp [logx/logp], 其中p泛指素数， [y]指小等于y之最大整数。证明或证伪：对任
意正数epsilon, psi(x)-x=o(x^{1/2+epsilon})。 |
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b**u
发帖数: 2761 | 6 Yup, agree. If the compound is too lipophilic, it will somehow remain in the
membrane. I think I've seen
a graph of Permeability vs LogP. There is a range of optimum LogP.
therapeutic |
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U******m
发帖数: 423 | 7 当然是有一个最优的logp了，经常是一个二次曲线的关系。
Raj经常喜欢搞极端，我说的是相对于先导结构，增加logP,他就非要理解成增加到很大
，搞得got stuck in membrane.
哼哼... |
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E*******F
发帖数: 2165 | 8 这个很随意的
比如你可以定义logP(red) = log(0.5) + c*log(red_count/total_count + 0.5)
c是一个系数
还要保证概率小于一，最后得有个配分 |
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R*******N
发帖数: 7494 | 9 概而言之,对后验概率P(Q,V|O),
(Q是参数，V是变量，O是观察结果)
E-step计算P(V｜O,Q),
M-step计算argmax logP(O,Q)=log{SIGMA(P(O,Q,V)},
如此迭代即得到大后验概率的Q,
Q是自由变量(free variable),
之所以引入log概率，
我认为可能是Bayes公式的缘故,
可参考:
Dempster, A., Laird, N., and Rubin, D. (1977). Maximum likelihood from incom-
plete data via the EM algorithm. Journal of the Royal Statistical Society,
Series B, 39(1):1–38. |
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w***y
发帖数: 2537 | 10 "has not passed windows Logp testing to verify its compatibility with
windows XP"
how to solve it? the webcamera doesn't work w/o the driver setup properly
Thanks |
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w***y
发帖数: 2537 | 11 "has not passed windows Logp testing to verify its compatibility with
windows XP"
how to solve it? the webcamera doesn't work w/o the driver setup properly
Thanks |
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j*****h
发帖数: 252 | 12 谢谢nanomotor1.
我查了下，这个octanol-water partition coefficient是化合物分别在octanol和水相
的溶解度的比值，所以想要知道一个小分子的溶解度，通过这个值能不能知道，还是要
通过实验方法指导。
用pH值打个比方，pH=3,知道溶液的[H]是10^(-3)mol....如果一个化合物的XLogP3是3
，能知道它在水里面的溶解度值么？不知道LogP有没有什么公式求到分子的溶解度？
感谢！！ |
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d*****h
发帖数: 61 | 13 1. 可以用logP 和 melting point 估计，很粗略。
2. 根据结构用ACDlabs算
3. 实验测（即可用hplc测也可目测估计） |
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x******g
发帖数: 318 | 14 [p^a/q^b]能取遍所有的自然数
p,q为固定的正整数,且logp(q)不是有理数;a,b可以取任意的自然数. |
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e*******y
发帖数: 73 | 15 Is there always a prime in the intervals
(n^2,(n+1)^2),
(n,n+n^{1/2}),
(N^2+N+2)/2<=M<=(N^2+3*N+2)/2(in conjecture A),
OR
(N^2+N)/2<=M<=(N^2+3*N)/2(in Conjecture B)? {{{note that
we only consider it, for sufficiently large n}}}
It is just the same thing(Opperman conjecture)!
It also equlvalent to
p'-p << p^{1/2}logp,
where p',p are consecutive primes. And we can easily obtain
under Riemann hypothesis. |
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b*********f
发帖数: 2125 | 17 1, 你的给药剂量是多少？iv 1mpk ？ po 10mpk？给个Cmax没有意义
2，貌似溶解度问题，HPbCD 最多用10%， 好好做一下formulation，10%Cremophor EL
不work？
3，本身分子permeability不好， F《30% ，引入亲水基团，降logP |
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s******r
发帖数: 30 | 18 Thank you very much for your infor. I also read of this unbound ratio
parameter. Seems so much to learn in CNS drug discovery, but very
interesting.
btw, NP-12 has been discontinued due to lack of efficacy in Phase II. But I
really wonder its in vitro PK properties, considering its high lipophilicity
, high logP, low ligand efficiecy....
Thanks. |
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