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<p>The share of sequences that shows at least one mutation in \(L_{Sequence}\) bp is \(p_{M}=\) <span id="amount_mutated"></span> % of sequences</p> | <p>The share of sequences that shows at least one mutation in \(L_{Sequence}\) bp is \(p_{M}=\) <span id="amount_mutated"></span> % of sequences</p> | ||
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Revision as of 09:55, 19 September 2017
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From 2014.igem.org
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Number of mutations and mutated sequences
Expected number of mutations in a single sequence: $$p_{m} = \frac{N_{mutations}}{L_{Sequence}} = N_{generations} * r_{mutation} = t_{total} * \Phi * r_{mutation}$$
The expected share of sequences that shows at least one mutation in \(L_{Sequence}\) bp is the probability that \(L_{sequence}\) basepairs stay unchanged when \(\frac{N_{mutations}}{L_{Sequence}}\) mutations are expected: $$p_{M} = \frac{N_{mutated}}{N_{Sequences}} = 1 - p(N_{mutations}=0) = 1 - (1-p_{m})^{L_{Sequence}} $$
With this equation we can also calculate the number of sequences \(N_{Sequences}\) that have to be sequenced in order to find a mutated one with a probability of \(p(N_{mutated} > 0)\). $$ N_{Sequences} = \frac{p(N_{mutated} > 0)}{p_{M}} $$
The probability to find at least one mutated sequence under the given conditions is $$p(N_{mutated}>0) = 1 - (1-p_{M})^{N_{sequences}}$$ which gives $$N_{Sequences} = \frac{ln(1-p(N_{mutated}>0))}{ln(1-p_{M})}$$
Set \(\Phi\) to zero to use the number of generations for the calculation. If \(\Phi\) and the number of generations are given, \(\Phi\) is used.
Consider \(L_{Sequence}\) as the number of basepairs that is expected to be mutated. If half of the sequence you are interested in, is highly conserved choose a lower \(L_{Sequence}\).
\(p_{m} =\) %(bp/bp).
\(N_{mutations} =\) bp per sequence.
The share of sequences that shows at least one mutation in \(L_{Sequence}\) bp is \(p_{M}=\) % of sequences