Codon count on a given mrna sequence in python

Question

This is the code that I have been trying to use but it does not work properly:

mrna = input("Please enter mRNA sequence: ")
start = mrna.find('AUG')
if start != -1:
    while start + 2 < len(mrna):
        codon = mrna[start:start + 3]
        if codon == "UAA": 
            break
        print(codon)
        start += 3

Expected start codon: AUG

Expected stop codon: UAA or UAG or UGA If for example:

input = "AUGUGA"
output = 1

input = "GAGAUGUUGGUUUAA"
output = 3

I don't really know what's wrong.

Answer 1

You aren't displaying numbers anywhere in your code. The number of codons is the difference between the end and the start index, floor divided by 3.

You can use a generator to help you check the codons.

mrna = input("Please enter mRNA sequence: ")
start = mrna.find('AUG')
if start != -1:
    end, last = next((x, mrna[x:x + 3] for x in range(start + 3, len(mrna) - 3, 3) if mrna[x:x + 3] in ('UAA', 'UAG', 'UGA')), (len(mrna), 'end'))
    print(f'{(end - start) // 3} codons between AUG and {last}')
else:
    print('AUG not found')

The generator (x, mrna[x:x + 3] for x in range(start + 3, len(mrna) - 3, 3) if mrna[x:x + 3] in ('UAA', 'UAG', 'UGA')) steps over all codons from index start + 3 to the end and yields codons that match anything in ('UAA', 'UAG', 'UGA') , along with the index. next normally returns the next (first) element of an iterator. With the second argument, it returns the extra argument as a sentinel instead of raising a StopIteration when the iterator runs out. // is the truncation division operator, so it will work correctly even if len(mrna) is not a multiple of 3 away from start .