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===Simple example=== The quasispecies concept can be illustrated by a simple system consisting of 4 sequences. Sequences [0,0], [0,1], [1,0], and [1,1] are numbered 1, 2, 3, and 4, respectively. Let's say the [0,0] sequence never mutates and always produces a single offspring. Let's say the other 3 sequences all produce, on average, <math>1-k</math> replicas of themselves, and <math>k</math> of each of the other two types, where <math>0\le k\le 1</math>. The '''W''' matrix is then: :<math>\mathbf{W}= \begin{bmatrix} 1&0&0&0\\ 0&1-k&k&k\\ 0&k&1-k&k\\ 0&k&k&1-k \end{bmatrix} </math>. The diagonalized matrix is: :<math>\mathbf{W'}= \begin{bmatrix} 1-2k&0&0&0\\ 0&1-2k&0&0\\ 0&0&1&0\\ 0&0&0&1+k \end{bmatrix} </math>. And the eigenvectors corresponding to these eigenvalues are: :{| class="wikitable" |- ![[Eigenvalue]] !! [[Eigenvector]] |- |1-2k || [0,-1,0,1] |- | 1-2k || [0,-1,1,0] |- | 1 || [1,0,0,0] |- | 1+k || [0,1,1,1] |} Only the eigenvalue <math>1+k</math> is more than unity. For the n-th generation, the corresponding eigenvalue will be <math>(1+k)^n</math> and so will increase without bound as time goes by. This eigenvalue corresponds to the eigenvector [0,1,1,1], which represents the quasispecies consisting of sequences 2, 3, and 4, which will be present in equal numbers after a very long time. Since all population numbers must be positive, the first two quasispecies are not legitimate. The third quasispecies consists of only the non-mutating sequence 1. It's seen that even though sequence 1 is the most fit in the sense that it reproduces more of itself than any other sequence, the quasispecies consisting of the other three sequences will eventually dominate (assuming that the initial population was not homogeneous of the sequence 1 type).{{citation needed|date=August 2013}}
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