Question

1. A large protein of 1205 amino acids was produced errors of average 8 mistakes/200 amino acids.

Calculate probability to produce protein with:

1. No mutation
2. Less than 2 mutations
3. Exactly 3 mutations

Give mathematical expression and the number used in these expressions.

Answer 1

We can model the problem as a Poisson experiment as it satisfies the below properties -

• The experiment results in outcomes that can be classified as successes (no mutation) or failures (mutation).
• The average number of mutations (μ) that occurs in a specified region (number of amino acids) is known.
• The probability that a success (mutations) will occur is proportional to the size of the region ((number of amino acids).
• The probability that a success (mutations) will occur in an extremely small region (negligible number of amino acids) is virtually zero.

Let X be the number of mutation in a large protein of 1205 amino acids.

Mutation rate = (8 / 200) * 1205 = 48.2

Then X ~ Poisson(μ = 48.2) and the probability mass function of X is,

P(X = x) = exp(-μ) μ^x /x!

=> P(X = x) = exp(-48.2) 48.2^x /x!

a.

No Mutation (X = 0)

P(X = 0) = exp(-48.2) 48.2⁰ /0! = exp(-48.2) = 1.17 x 10^-21

b.

Less than 2 mutations   (X < 2)

P(X < 2) = P(X = 0) + P(X = 1)  = exp(-48.2) 48.2⁰ /0! + exp(-48.2) 48.2¹ /1! = 49.2 exp(-48.2) = 5.74 x 10^-20

c.

Exactly 3 mutations  (X = 3)

P(X = 3) = exp(-48.2) 48.2³ /3! = exp(-48.2) = 2.18 x 10^-17