Question

Four bases \((\mathrm{A}, \mathrm{C}, \mathrm{T}, \text { and } \mathrm{G})\) appear in DNA. Assume that the appearance of each base in a DNA sequence is random. a. What is the probability of observing the sequence AAGACATGCA? b. What is the probability of finding the sequence GGGGGAAAAA? c. How do your answers to parts (a) and (b) change if the probability of observing A is twice that of the probabilities used in parts (a) and (b) of this question when the preceding base is G?

Short answer

The probabilities of observing the sequences AAGACATGCA and GGGGGAAAAA under equal probability of each base are approximately \(9.5367 \times 10^{-7}\). When the probability of observing A is twice as likely when the preceding base is G, the probability for AAGACATGCA increases to approximately \(7.6294 \times 10^{-6}\) and for GGGGGAAAAA decreases to approximately \(2.3842 \times 10^{-6}\).

Step-by-step solution

Determine the probability of each base appearing in the DNA sequence

Since each base's appearance is random, they have an equal probability of occurring in the sequence. We have four bases, A, C, T, and G. Because there are four options and each has an equal chance of being chosen, the probability of each base appearing is 1/4.

Calculate the probability of observing the sequence AAGACATGCA

To find the probability of observing this specific sequence, we multiply the probability of each base appearing in the positions it appears in the sequence. The probability of AAGACATGCA is, therefore, the product of the probabilities of each base appearing in the sequence in order: \(P(\text{AAGACATGCA}) = P(A) \times P(A) \times P(G) \times P(A) \times P(C) \times P(A) \times P(T) \times P(G) \times P(C) \times P(A)\) Since the probability of each base appearing is 1/4, our equation becomes: \(P(\text{AAGACATGCA}) = (1/4)^{10}\) \(P(\text{AAGACATGCA}) \approx 9.5367 \times 10^{-7}\)

Calculate the probability of observing the sequence GGGGGAAAAA

Similarly, to find the probability of the second sequence, we multiply the probability of each base appearing in the positions in the sequence. The probability of GGGGGAAAAA is: \(P(\text{GGGGGAAAAA)} = P(G) \times P(G) \times P(G) \times P(G) \times P(G) \times P(A) \times P(A) \times P(A) \times P(A) \times P(A)\) Since the probability of each base appearing is 1/4, our equation becomes: \(P(\text{GGGGGAAAAA}) = (1/4)^{10}\) \(P(\text{GGGGGAAAAA}) \approx 9.5367 \times 10^{-7}\)

Calculate the new probabilities if the probability of observing A is twice that when the preceding base is G

If the probability of observing A is twice that when the preceding base is G, we need to adjust our initial probabilities. The new probabilities are: \(P(A|G) = 1/2\) \(P(C|G) = 1/4\) \(P(T|G) = 1/4\) \(P(G|G) = 1/4\) We will now recalculate the probabilities for sequences AAGACATGCA and GGGGGAAAAA using these new probabilities for the cases where the preceding base is G. For AAGACATGCA: \(P(\text{AAGACATGCA}) = P(A) \times P(A) \times P(G|A) \times P(A|G) \times P(C|A) \times P(A|C) \times P(T|A) \times P(G|T) \times P(C|G) \times P(A|C)\) The probability of AAGACATGCA is now: \(P(\text{AAGACATGCA}) = (1/4)^2 \times (1/4) \times (1/2) \times (1/4) \times (1/4) \times (1/4) \times (1/4) \times (1/4) \times (1/4)\) \(P(\text{AAGACATGCA}) \approx 7.6294 \times 10^{-6}\) For GGGGGAAAAA: \(P(\text{GGGGGAAAAA)} = P(G) \times P(G) \times P(G|G) \times P(G|G) \times P(G|G) \times P(A|G) \times P(A) \times P(A) \times P(A) \times P(A)\) The probability of GGGGGAAAAA is now: \(P(\text{GGGGGAAAAA}) = (1/4)^5 \times (1/2) \times (1/4)^4\) \(P(\text{GGGGGAAAAA}) \approx 2.3842 \times 10^{-6}\) The probabilities of observing sequences AAGACATGCA and GGGGGAAAAA are now different when the probability of observing A is twice that when the preceding base is G. The probability of AAGACATGCA increased and the probability of GGGGGAAAAA decreased.

Key concepts

Base Appearance Probability

In DNA sequences, we have four nucleotides or bases: Adenine (A), Cytosine (C), Thymine (T), and Guanine (G). A crucial aspect of analyzing DNA is determining the probability of each base's appearance. If we assume that each base appears randomly and with equal likelihood, then each base has a probability of 1/4, because there are four bases in total.
This assumes a uniform distribution, where no base is more or less likely to appear. In practice, real DNA may not have a perfectly uniform distribution. However, understanding this basic assumption is important for foundational studies in probabilistic DNA sequence analysis.
If conditions change, such as one base being more likely after another, probabilities need reassessment to capture these dependencies effectively. This forms the basis for deeper probabilistic methods in sequence analysis.

Conditional Probability

Conditional probability allows us to determine the likelihood of an event happening based on the occurrence of a previous event. In the context of DNA sequence analysis, this concept helps address situations where the probability of a base appearing might depend on the base before it. For example, if the probability of observing 'A' is higher when the preceding base is 'G', this dependence needs expression using conditional probability.
The notation \( P(A|G) \) means the probability of 'A' given G has occurred. Incorporating conditional probabilities changes the computation of sequence likelihoods substantially.
Applying this to sequences, if 'A' is twice as likely after 'G', then \( P(A|G) = 1/2 \), while the probability of other bases like 'C', 'T', and 'G' remain at 1/4. This modifies sequence probability outcomes, enhancing our ability to analyze and predict genetic sequences within more realistic settings.

DNA Sequence Analysis

DNA sequence analysis involves calculating the probabilities of specific sequences appearing, which is fundamental in genetics, bioinformatics, and evolutionary studies. In the exercises, we found probabilities of sequences like AAGACATGCA or GGGGGAAAAA by multiplying the probabilities of individual bases appearing in exact positions.
Standard calculations assume a uniform probability across all bases, but in more complex models, one might operate under non-uniform or conditional base probabilities. This complexity allows for better modeling of natural DNA, reflecting patterns or preferences in a sequence.

• Simple cases treat each base as independent.
• Complex cases incorporate dependencies, like conditional probabilities.

Sequence analysis impacts understanding mutations, hereditary diseases, and evolutionary biology, providing insights into how genetic material evolves and functions.

Random Sequence Generation

When generating random DNA sequences, each base is typically chosen based on its probability of occurrence. If the base probability is uniform, sequences can be generated by selecting each base randomly with a 1/4 probability of being either A, C, T, or G.
In practice, random sequence generation mimics natural variability, assisting simulations, and modeling in genetic research. When probabilities shift based on conditions, such as after a preceding base, this generates more biologically realistic sequences.

• Uniform probability leads to equal chances for each base.
• Conditional models reflect biases found in actual DNA.

By considering base dependencies in generation processes, researchers can create sequences that better mimic real-world genetic structures, thus aiding studies related to genetics and evolutionary biology.