How to count point mutations

Drashti Shah

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Bioinformatics with Rust

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2 min read

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Dec 7, 2023

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A Rust function that returns count of point mutations (hamming distance)

Arguments

• seq1: a string slice that holds a DNA sequence
• seq2: another string slide that holds a DNA sequence

Example

let seq1 = "ACGT";
let seq2 = "ACCT";
let count = count_point_mutations(seq1, seq2).unwrap();

// count holds 1

Code

fn main() {
    let seq1: &str = "GAGCCTACTAACGGGAT";
    let seq2: &str = "CATCGTAATGACGGCCT";
    let count = count_point_mutations(seq1, seq2).unwrap();
    println!("{}", count)
}

fn count_point_mutations(seq1: &str, seq2: &str) -> Option<i32> {
    if seq1.len() == seq2.len() {
        let mut distance = 0;
        for (x, y) in seq1.chars().zip(seq2.chars()) {
            if x != y {
                distance += 1;
            }
        }
        Some(distance)
    } else {
        None
    }
}

Some notes on the code

• I need to use .unwrap() because count_point_mutations returns an Option
• The zip function takes 2 iterators and returns a new iterator that yields pairs of elements from both input iterators
• If the inputs sequences are not of the same length, the code will panic. You can use if let instead.
• distance stands for hamming distance. Naming this variable count is probably a wiser choice.

fn main() {
    let seq1: &str = "GAGCCTACTAACGGGAT";
    let seq2: &str = "CATCGTAATGACGGCCT";
    if let Some(count) = count_point_mutations(seq1, seq2) {
        println!{"{}", count};
    } else {
        println!("The input sequence lengths are not same.");
    }
}

Next Steps

• Play with this code in the Rust Playground
• Solve a Rosalind challenge