Question

DNA synthesis. Suppose that upon synthesizing a molecule of DNA, you introduce a wrong base pair, on average, every 1000 base pairs. Suppose you synthesize a DNA molecule that is 1000 bases long. (a) Calculate and draw a bar graph indicating the yield (probability) of each product DNA, containing 0,1 , 2 , and 3 mutations (wrong base pairs). (b) Calculate how many combinations of DNA sequences of 1000 base pairs contain exactly 2 mutant base pairs. (c) What is the probability of having specifically the 500 th base pair and the 888 th base pair mutated in the pool of DNA that has only two mutations? (d) What is the probability of having two mutations side-by-side in the pool of DNA that has only two mutations?

Short answer

0 mutations: 0.3679, 1 mutation: 0.3679, 2 mutations: 0.184, 3 mutations: 0.0612.\Binomial coefficient for 2 mutations: 499500.\ Probability for 500th and 888th: 10^{-6}. Probability of adjacent: 9.99 \times 10^{-4}.

Step-by-step solution

Understand the Mutation Probability

The probability of introducing a wrong base pair at any position is \(\frac{1}{1000}\). The probability of having the correct base pair is thus \(1 - \frac{1}{1000} = \frac{999}{1000}\).

Use Binomial Distribution (Part a)

The number of mutations (wrong base pairs) follows a binomial distribution \(B(n, p)\) where \(n = 1000\) and \(p = \frac{1}{1000}\). The probability of having \(k\) mutations is given by \(\binom{n}{k} p^k (1 - p)^{n - k}\).

Calculate Probabilities for k = 0, 1, 2, 3 (Part a)

Calculate the probabilities for \(k = 0, 1, 2, 3\) mutations:

Draw Bar Graph (Part a)

Create a bar graph with the x-axis representing the number of mutations (0, 1, 2, 3) and the y-axis representing the corresponding probabilities (0.3679, 0.3679, 0.184, 0.0612).

Calculate the Combinations (Part b)

The number of ways to choose 2 positions out of 1000 for mutations is given by the binomial coefficient \(\binom{n}{k}\). Here \(n = 1000\) and \(k = 2\), so \(\binom{1000}{2} = \frac{1000!}{2!(1000-2)!} = \frac{1000 \times 999}{2 \times 1} = 499500\).

Specific Mutation Combination Probability (Part c)

The probability of having the 500th and 888th base pairs mutated is \(\left(\frac{1}{1000}\right)^2 = 10^{-6}\).

Adjacent Mutations Probability (Part d)

The number of ways to place 2 adjacent mutations in 1000 bases is \(999\). The probability of having two specified bases mutated is \(\left(\frac{1}{1000}\right)^2 = 10^{-6}\). Thus, the total probability is \(999 \times 10^{-6} = 9.99 \times 10^{-4}\).

Key concepts

Binomial Distribution

A binomial distribution is useful for modeling the number of successes in a fixed number of independent trials. Each trial has only two outcomes: success or failure. In our DNA synthesis problem, a success is defined as introducing a wrong base pair.
The probability of introducing a wrong base pair at any position in the DNA is given by \(p = \frac{1}{1000}\).
Conversely, the probability of inserting a correct base pair is \(1 - p = \frac{999}{1000}\).
For a DNA strand of length 1000 base pairs (trials), we model the number of wrong base pairs (mutations) using the binomial distribution \(B(n, p)\), where \(n = 1000\) and \(p = \frac{1}{1000}\). The formula for the probability of exactly \(k\) mutations is:
\[P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \]
This equation helps calculate the likelihood of encountering exactly \(k\) wrong base pairs out of 1000.

Mutation Frequency

Mutation frequency is the rate at which mutations occur within a given DNA sequence. In the exercise, the mutation frequency is 1 mutation per 1000 base pairs.
To calculate this rate, we consider how often a wrong base pair is introduced compared to the length of the DNA strand. Given that each base pair has a \(\frac{1}{1000}\) chance of being incorrect, longer sequences are likely to have more mutations.
For example, if we synthetize 1000 base pairs, we expect on average 1 mutation. The actual occurrence varies, so it's essential to consider the distribution, which isn't equal for all base pairs but follows the binomial distribution.
Specific mutation frequencies in different segments allow us to analyze region-specific mutation patterns and offer insight into DNA stability and mutagenesis mechanisms.

Combinatorial Calculations

Combinatorial calculations are crucial for determining the number of ways to achieve a specific number of mutations. For instance, choosing 2 mutations out of 1000 base pairs involves selecting positions where these mutations occur. This is calculated using the binomial coefficient, often represented as \(\binom{n}{k}\).
In our problem, \(n = 1000\) and \(k = 2\), so:
\[\binom{1000}{2} = \frac{1000!}{2!(1000-2)!} = \frac{1000 \times 999}{2 \times 1} = 499500 \]
This number tells us there are 499500 unique ways to place exactly 2 mutations within a DNA strand of 1000 bases.

Adjacent Mutations

Adjacent mutations refer to mutations that occur next to each other in a sequence. In the context of this problem, we want to know the probability of having two side-by-side mutations within 1000 bases.
First, we determine the possible placements for such pairs. If two mutations are placed adjacent, there are \(1000 - 1 = 999\) possible pairs.
The probability of having two specific adjacent mutations can be found using:
\[\left(\frac{1}{1000}\right)^2 = 10^{-6} \]
Thus, the total probability for any two adjacent positions is:
\[999 \times 10^{-6} = 9.99 \times 10^{-4} \]
This suggests that although the event of adjacent mutations is rare, it is not impossible.