QuestionMay 31, 2025

Problem 3 DNA Methylation DNA methylation is a process by which methyl chemical groups are attached to the DNA. Thus each gene in the genome can be methylated or unmethylated.Suppose that 80% of methylated genes remain methylated and 50% of unmethylated genes remain unmethylated each generation. (a) Express methylation model with the recursion x_(t+1)=Ax_(t) (b) Calculate the eigenvalues of A. (c) What is the long-term behavior of x_(t) (d) Using the initial condition x_(0)=[} a 1-a ] , express the solution to the recursion in terms of the eigenvectors and eigenvalues of A.

Problem 3 DNA Methylation DNA methylation is a process by which methyl chemical groups are attached to the DNA. Thus each gene in the genome can be methylated or unmethylated.Suppose that 80% of methylated genes remain methylated and 50% of unmethylated genes remain unmethylated each generation. (a) Express methylation model with the recursion x_(t+1)=Ax_(t) (b) Calculate the eigenvalues of A. (c) What is the long-term behavior of x_(t) (d) Using the initial condition x_(0)=[} a 1-a ] , express the solution to the recursion in terms of the eigenvectors and eigenvalues of A.
Problem 3 DNA Methylation DNA methylation is a process by which methyl chemical groups are attached to the DNA. Thus each gene in the genome can be methylated or unmethylated.Suppose that 80% of methylated genes remain methylated and 50% of unmethylated genes remain unmethylated each generation. (a) Express methylation model with the recursion x_(t+1)=Ax_(t) (b) Calculate the eigenvalues of A. (c) What is the long-term behavior of x_(t) (d) Using the initial condition x_(0)=[} a 1-a ] , express the solution to the recursion in terms of the eigenvectors and eigenvalues of A.

Solution
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Answer

(a) x_{t+1} = \begin{bmatrix} 0.8 & 0.5 \\ 0.2 & 0.5 \end{bmatrix} x_t ### (b) Eigenvalues: \lambda_1 = 1, \lambda_2 = 0.3 ### (c) Long-term behavior: Stabilizes at eigenvector associated with \lambda_1 = 1 ### (d) Solution: x_t = c_1 v_1 + c_2 (0.3)^t v_2, where c_1, c_2 from initial condition x_0 Explanation 1. Define the matrix A The transition matrix A is defined based on the given probabilities. Methylated genes remain methylated with probability 0.8, and unmethylated genes become methylated with probability 0.5. Thus, A = \begin{bmatrix} 0.8 & 0.5 \\ 0.2 & 0.5 \end{bmatrix}. 2. Calculate eigenvalues of A Find eigenvalues by solving \det(A - \lambda I) = 0. For A = \begin{bmatrix} 0.8 & 0.5 \\ 0.2 & 0.5 \end{bmatrix}, solve (0.8-\lambda)(0.5-\lambda) - 0.1 = 0. This simplifies to \lambda^2 - 1.3\lambda + 0.3 = 0. Solve using quadratic formula: \lambda = \frac{1.3 \pm \sqrt{1.69 - 1.2}}{2} = \frac{1.3 \pm 0.7}{2}. Eigenvalues are \lambda_1 = 1, \lambda_2 = 0.3. 3. Determine long-term behavior Long-term behavior depends on dominant eigenvalue. Since \lambda_1 = 1 is the largest, system stabilizes at eigenvector associated with \lambda_1. 4. Express solution in terms of eigenvectors and eigenvalues Using initial condition x_0 = \begin{bmatrix} a \\ 1-a \end{bmatrix}, express x_t as linear combination of eigenvectors. Let v_1 and v_2 be eigenvectors for \lambda_1 and \lambda_2. Solution: x_t = c_1 \lambda_1^t v_1 + c_2 \lambda_2^t v_2. Coefficients c_1 and c_2 determined by initial condition.

Explanation

1. Define the matrix A<br /> The transition matrix $A$ is defined based on the given probabilities. Methylated genes remain methylated with probability 0.8, and unmethylated genes become methylated with probability 0.5. Thus, $A = \begin{bmatrix} 0.8 & 0.5 \\ 0.2 & 0.5 \end{bmatrix}$.<br /><br />2. Calculate eigenvalues of A<br /> Find eigenvalues by solving $\det(A - \lambda I) = 0$. For $A = \begin{bmatrix} 0.8 & 0.5 \\ 0.2 & 0.5 \end{bmatrix}$, solve $(0.8-\lambda)(0.5-\lambda) - 0.1 = 0$. This simplifies to $\lambda^2 - 1.3\lambda + 0.3 = 0$. Solve using quadratic formula: $\lambda = \frac{1.3 \pm \sqrt{1.69 - 1.2}}{2} = \frac{1.3 \pm 0.7}{2}$. Eigenvalues are $\lambda_1 = 1$, $\lambda_2 = 0.3$.<br /><br />3. Determine long-term behavior<br /> Long-term behavior depends on dominant eigenvalue. Since $\lambda_1 = 1$ is the largest, system stabilizes at eigenvector associated with $\lambda_1$.<br /><br />4. Express solution in terms of eigenvectors and eigenvalues<br /> Using initial condition $x_0 = \begin{bmatrix} a \\ 1-a \end{bmatrix}$, express $x_t$ as linear combination of eigenvectors. Let $v_1$ and $v_2$ be eigenvectors for $\lambda_1$ and $\lambda_2$. Solution: $x_t = c_1 \lambda_1^t v_1 + c_2 \lambda_2^t v_2$. Coefficients $c_1$ and $c_2$ determined by initial condition.
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