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Chapter 1: Problem 2

Probabilities of sequences. Assume that the four bases A, C, T, and G occur with equal likelihood in a DNA sequence of nine monomers. (a) What is the probability of finding the sequence AAATCGAGT through random chance? (b) What is the probability of finding the sequence AAAAAAAAA through random chance? (c) What is the probability of finding any sequence that has four A's, two T's, two G's, and one C, such as that in (a)?

Short Answer

Expert verified
a) \frac{1}{262144}, b) \frac{1}{262144}, c) \frac{1890}{262144} \approx 0.0072

Step by step solution

01

Title - Probability of a Specific Sequence

For part (a), begin by understanding that each base (A, C, T, G) has an equal likelihood of occurring. Since there are 4 possible bases, the probability of any specific base occurring at a given position in the sequence is \(\frac{1}{4}\). To find the probability of a specific 9-base sequence like AAATCGAGT, multiply the probabilities for each individual base: \[ \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{4^9} \] Calculate this value.
02

Title - Evaluating the Probability

Compute \(\frac{1}{4^9}\) step by step: \[ 4^9 = 262144 \ \frac{1}{4^9} = \frac{1}{262144} \] So, the probability is \(\frac{1}{262144}\).
03

Title - Probability of All Same Bases

For part (b), the process is similar. The probability of finding the sequence AAAAAAAAA is determined the same way because the chances of getting an 'A' at each position is \(\frac{1}{4}\). Thus, the calculation remains \[ \frac{1}{4^9} = \frac{1}{262144} \] as before.
04

Title - Probability of Any Sequence with Specific Base Composition

For part (c), consider the total number of unique sequences that can be formed with 4 A's, 2 T's, 2 G's, and 1 C. This is calculated using combinatorial counting: \[ \frac{9!}{4! 2! 2! 1!} = \frac{362880}{24 \cdot 2 \cdot 2 \cdot 1} = 1890 \] Each of these 1890 sequences has an individual probability of \frac{1}{262144}, so the total probability of any sequence with the specified base composition is \[ 1890 \times \frac{1}{262144} = \frac{1890}{262144} \] Simplify to get the final result: \[ \frac{1890}{262144} \approx 0.0072 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculation in Genetics
In genetics, calculating the probability of specific DNA sequences is fundamental in understanding genetic variation and mutation rates. Let's break down these concepts:

Every DNA sequence consists of four bases: Adenine (A), Cytosine (C), Thymine (T), and Guanine (G). Since each base can occur with equal likelihood, the probability of any base appearing at a given position in a sequence is \(\frac{1}{4}\).

For instance, to find the probability of a 9-base sequence like AAATCGAGT, you multiply the individual probabilities for each base:
\[ \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{4^9} = \frac{1}{262144} \]

By calculating these probabilities, geneticists can predict the likelihood of specific sequences appearing by random chance.
Combinatorial Counting in DNA Sequences
Combinatorial counting helps us determine the number of different sequences possible with a specific base composition. This is crucial when we want to know the likelihood of sequences with repetitive or certain patterns.

For example, if we want to find sequences with four A's, two T's, two G's, and one C in a series of 9 bases, we use the formula involving factorials:
\[ \frac{9!}{4! 2! 2! 1!} = \frac{362880}{24 \times 2 \times 2 \times 1} = 1890 \]

This tells us that there are 1890 unique sequences that can be made with this specific composition of A's, T's, G's, and C's.

Next, each of these sequences occurs with a probability of \(\frac{1}{262144}\). So, to find the total probability of any sequence with the given composition, you multiply this by the number of unique sequences:
\[ 1890 \times \frac{1}{262144} = \frac{1890}{262144} \]

Simplifying this gives us approximately 0.0072, indicating that there's a 0.72% chance of randomly generating a sequence with this base composition.
Base Composition in Sequences
Understanding base composition is vital for analyzing and predicting DNA behavior and functions. Each DNA sequence's base composition can significantly influence its stability, replication, and translation processes.

In genetic studies, when we talk about base composition, we often refer to the proportion of each base (A, C, T, G) within a sequence. For instance, a sequence might have a high proportion of adenine or a specific ratio of all four bases.

To illustrate this, let's consider the sequence AAAAAAAAA. The probability of this occurring in a 9-base sequence is the same as any other specific sequence because each position still has a \(\frac{1}{4}\) chance of being an 'A', calculated as:
\[ \frac{1}{4^9} = \frac{1}{262144} \]

Similarly, for a sequence with mixed bases like AAATCGAGT, the computation remains consistent with the same approach. This emphasizes that regardless of the base mix, the fundamental calculation of probability remains hinged on the equal likelihood of each base appearing at any position.

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Most popular questions from this chapter

Combining independent probabilities. You have applied to three schools: University of California at San Francisco (UCSF), Duluth School of Mines (DSM), and Harvard (H). You guess that the probabilities you'll be accepted are \(p(\mathrm{UCSF})=0.10, p(\mathrm{DSM})=0.30\), and \(p(\mathrm{H})=0.50\). Assume that the acceptance events are independent. (a) What is the probability that you get in somewhere (at least one acceptance)? (b) What is the probability that you will be accepted by both Harvard and Duluth?

Predicting the rate of mutation based on the Poisson probability distribution function. The evolutionary process of amino acid substitution in proteins is sometimes described by the Poisson probability distribution function. The probability \(p_{s}(t)\) that exactly \(s\) substitutions at a given amino acid position occur over an evolutionary time \(t\) is $$ p_{s}(t)=\frac{e^{-\lambda t}(\lambda t)^{s}}{s !}, $$ where \(\lambda\) is the rate of amino acid substitution per site per unit time. Fibrinopeptides evolve rapidly: \(\lambda_{F}=9.0\) substitutions per site per \(10^{9}\) years. Lysozyme is intermediate: \(\lambda_{L} \approx 1.0\). Histones evolve slowly: \(\lambda_{H}=0.010\) substitutions per site per \(10^{9}\) years. (a) What is the probability that a fibrinopeptide has no mutations at a given site in \(t=1\) billion years? (b) What is the probability that lysozyme has three mutations per site in 100 million years? (c) We want to determine the expected number of mutations \(\langle s\rangle\) that will occur in time \(t\). We will do this in two steps. First, using the fact that probabilities must sum to one, write \(\alpha=\sum_{s=0}^{\infty}(\lambda t)^{s} / s !\) in a simpler form. (d) Now write an expression for \(\langle s\rangle\). Note that $$ \sum_{s=0}^{\infty} \frac{s(\lambda t)^{s}}{s !}=(\lambda t) \sum_{s=1}^{\infty} \frac{(\lambda t)^{s-1}}{(s-1) !}=\lambda t \alpha $$ (e) Using your answer to part (d), determine the ratio of the expected number of mutations in a fibrinopeptide to the expected number of mutations in histone protein, \(\langle s\rangle_{\mathrm{fib}} /\langle s\rangle_{\mathrm{his}}[6]\).

Evolutionary fitness. Suppose that the probability of having the dominant allele (D) in a gene is \(p\) and the probability of the recessive allele \((\mathbf{R})\) is \(q=1-p\). You have two alleles, one from each parent. (a) Write the probabilities of all the possibilities: DD, DR, and RR. (b) If the fitness of \(\mathrm{DD}\) is \(f_{\mathrm{DD}}\), the fitness of \(\mathrm{DR}\) is \(f_{\mathrm{DR}}\), and the fitness of \(R R\) is \(f_{R R}\), write the average fitness in terms of \(p\).

Predicting compositions of independent events. Suppose you roll a fair six- sided die three times. (a) What is the probability of getting a 5 twice from all three rolls of the dice? (b) What is the probability of getting a total of at least two 5 's from all three rolls of the die?

The Maxwell-Boltzmann probability distribution function. According to the kinetic theory of gases, the energies of molecules moving along the \(x\) direction are given by \(\varepsilon_{x}=(1 / 2) m v_{x}^{2}\), where \(m\) is mass and \(v_{x}\) is the velocity in the \(x\) direction. The distribution of particles over velocities is given by the Boltzmann law, \(p\left(v_{x}\right)=e^{-m v_{x}^{2} / 2 k T}\). This is the Maxwell-Boltzmann distribution (velocities may range from \(-\infty\) to \(+\infty\) ). (a) Write the probability distribution \(p\left(v_{x}\right)\), so that the Maxwell-Boltzmann distribution is correctly normalized. (b) Compute the average energy \(\left\langle\frac{1}{2} m v_{x}^{2}\right\rangle\). (c) What is the average velocity \(\left\langle v_{x}\right\rangle\) ? (d) What is the average momentum \(\left\langle m v_{x}\right\rangle\) ?
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