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東京大学 情報理工学系研究科 コンピュータ科学専攻 2016年8月実施 専門科目II 問題6

Author

祭音Myyura (co-authored with GPT 5.6 SOL)

Description

On generation and activation of protein YY we consider the following three models. In what follows we designate the name of protein and its concentration by the same symbol.

In the first model, protein YY is produced at a constant speed β\beta and its degradation rate α\alpha is also constant (here α,β\alpha,\beta are nonnegative real values). That is, we describe the change of the concentration of protein YY by

dYdt=βαY.\frac{dY}{dt}=\beta-\alpha Y.

Answer the following questions.

(1) Find the steady-state concentration of protein YY.

(2) We set the concentration of protein YY at time t=0t=0 to 0. Under this setting, express the concentration of protein YY as a function over time tt (here t0t\ge0).

In the second model we suppose that mm types of transcription factors control the concentration of protein YY. Let X1,,XmX_1,\ldots,X_m be the concentrations of mm transcription factors. We then describe the change of the concentration of protein YY by

dYdt=j=1mXjγjαY,\frac{dY}{dt}=\prod_{j=1}^{m}X_j^{\gamma_j}-\alpha Y,

where α,γ1,,γm\alpha,\gamma_1,\ldots,\gamma_m are unknown parameters.

Answer the following questions.

(3) Consider a steady state. Express the value of logY\log Y in terms of X1,,XmX_1,\ldots,X_m and α,γ1,,γm\alpha,\gamma_1,\ldots,\gamma_m.

(4) Consider the following experiment: we measure the steady-state concentration of YY for fixed values of X1,,XmX_1,\ldots,X_m. Using different values of X1,,XmX_1,\ldots,X_m, we repeat this experiment nn times. Let (Yi,Xi,1,,Xi,m)(Y_i,X_{i,1},\ldots,X_{i,m}) be the ii-th experimental data (where i=1,,ni=1,\ldots,n). Give an experimental condition for this series of experiments to uniquely determine the parameters α,γ1,,γm\alpha,\gamma_1,\ldots,\gamma_m.

As the third model we consider the following. Proteins X1X_1 and X2X_2 are activated by phosphorylation; and activated X1X_1 and X2X_2 phosphorylate protein YY. We describe this model by

dYpdt=3Xp,1Y0+2Xp,2Y0Yp,\frac{dY_p}{dt}=3X_{p,1}Y_0+2X_{p,2}Y_0-Y_p,

where: Xp,1X_{p,1} and Xp,2X_{p,2} are the concentrations of phosphorylated X1X_1 and phosphorylated X2X_2, respectively; and Y0Y_0 and YpY_p are the concentrations of non-phosphorylated YY and phosphorylated YY, respectively. Here we assume that Xp,1X_{p,1} and Xp,2X_{p,2} are constant over time, and that Y0+Yp=CY_0+Y_p=C (CC is a constant).

Answer the following question.

(5) Find a condition on Xp,1X_{p,1} and Xp,2X_{p,2} so that, in a steady state, Yp/CY_p/C is greater than 0.5.

题目描述

考虑蛋白质 YY 的生成与活化模型。以下用同一符号表示蛋白质名称及其浓度。

第一模型为

dYdt=βαY,\frac{dY}{dt}=\beta-\alpha Y,

其中生成速率 β\beta、降解率 α\alpha 为非负常数。

(1)求 YY 的稳态浓度。

(2)设 Y(0)=0Y(0)=0,求 t0t\ge0 时的 Y(t)Y(t)

第二模型中,mm 种转录因子的浓度为 X1,,XmX_1,\ldots,X_m,并有

dYdt=j=1mXjγjαY,\frac{dY}{dt}=\prod_{j=1}^{m}X_j^{\gamma_j}-\alpha Y,

其中 α,γ1,,γm\alpha,\gamma_1,\ldots,\gamma_m 未知。

(3)在稳态下,用 X1,,Xm,α,γ1,,γmX_1,\ldots,X_m,\alpha,\gamma_1,\ldots,\gamma_m 表示 logY\log Y

(4)每次实验固定 X1,,XmX_1,\ldots,X_m,测量 YY 的稳态浓度;采用不同的 X1,,XmX_1,\ldots,X_m 值重复 nn 次实验,第 ii 次观测为 (Yi,Xi,1,,Xi,m)(Y_i,X_{i,1},\ldots,X_{i,m}),其中 i=1,,ni=1,\ldots,n。给出能唯一确定所有参数的实验条件。

第三模型中,X1,X2X_1,X_2 经磷酸化而活化,活化后的 X1,X2X_1,X_2 使蛋白质 YY 磷酸化:

dYpdt=3Xp,1Y0+2Xp,2Y0Yp,\frac{dY_p}{dt}=3X_{p,1}Y_0+2X_{p,2}Y_0-Y_p,

其中 Xp,1,Xp,2X_{p,1},X_{p,2} 分别为磷酸化的 X1,X2X_1,X_2 的浓度,Y0,YpY_0,Y_p 分别为未磷酸化和已磷酸化的 YY 的浓度。Xp,1,Xp,2X_{p,1},X_{p,2} 不随时间变化,且 Y0+Yp=CY_0+Y_p=C,其中 CC 为常量。

(5)求稳态下 Yp/C>0.5Y_p/C>0.5Xp,1,Xp,2X_{p,1},X_{p,2} 应满足的条件。

Kai

(1)

令导数为 00。当 α>0\alpha>0 时,

Yss=βα.\boxed{Y_{\rm ss}=\frac\beta\alpha}.

α=0,β>0\alpha=0,\beta>0,则无有限稳态;若 α=β=0\alpha=\beta=0,则任意非负常数浓度均为稳态。

(2)

解一阶线性方程并代入 Y(0)=0Y(0)=0,得

Y(t)=βα(1eαt)(α>0).\boxed{Y(t)=\frac\beta\alpha\left(1-e^{-\alpha t}\right)}\qquad(\alpha>0).

α=0\alpha=0,则 Y(t)=βtY(t)=\beta t

(3)

稳态满足

Y=1αj=1mXjγj.Y=\frac1\alpha\prod_{j=1}^{m}X_j^{\gamma_j}.

在各 Xj>0X_j>0α>0\alpha>0 时取对数:

logY=logα+j=1mγjlogXj.\boxed{\log Y=-\log\alpha+\sum_{j=1}^{m}\gamma_j\log X_j}.

(4)

θ=(logα,γ1,,γm)T,\boldsymbol\theta=(-\log\alpha,\gamma_1,\ldots,\gamma_m)^{\mathsf T},

则实验给出线性方程

logYi=(1,logXi,1,,logXi,m)θ.\log Y_i=(1,\log X_{i,1},\ldots,\log X_{i,m})\boldsymbol\theta.

因此须有 nm+1n\ge m+1,所有浓度均为正,并且设计矩阵

M=(1logX1,1logX1,m1logXn,1logXn,m)M=\begin{pmatrix} 1&\log X_{1,1}&\cdots&\log X_{1,m}\\ \vdots&\vdots&&\vdots\\ 1&\log X_{n,1}&\cdots&\log X_{n,m} \end{pmatrix}

满足 rankM=m+1\operatorname{rank}M=m+1。这正是唯一确定 θ\boldsymbol\theta,进而唯一确定 α,γ1,,γm\alpha,\gamma_1,\ldots,\gamma_m 的充要条件。

(5)

设总浓度 C>0C>0,使比例 Yp/CY_p/C 有定义。令 k=3Xp,1+2Xp,2k=3X_{p,1}+2X_{p,2}。稳态时

0=k(CYp)Yp,0=k(C-Y_p)-Y_p,

所以

YpC=kk+1.\frac{Y_p}{C}=\frac{k}{k+1}.

于是

YpC>12    3Xp,1+2Xp,2>1.\boxed{\frac{Y_p}{C}>\frac12\iff 3X_{p,1}+2X_{p,2}>1}.