Question

This question is about the Hamming distance (HD).

Given a reference DNA sequence and a set of candidate sequences, you are going to find out which candidate(s) has/have the Hamming distance smaller or equal to a threshold k.

e.g.

Reference DNA: AATGCGC

Candidate 1: AATCCCC

Candidate 2: TTTGCTC

Candidate 3: AATAAAA

If the Hamming distance threshold k = 3:

HD1 = 2

HD2 = 3

HD3 = 4

Your result should return a list of all the qualified candidate numbers and their Hamming distances in format:

[[candidate number, Hamming distance]].

For the result of the example above:

[[1, 2], [2, 3]]

Here begins the question.

Reference sequence: AGAAACTCTCTGGCCTAAAG

Candidate 1: ACTTAGGTCTCTAAGCCCTC
Candidate 2: AGAAACGTTATGTGGACGTT
Candidate 3: AGTCTGACTCTGATCCAAAG
Candidate 4: AGTCTGCCTTGGCCATTAGC

Candidate 5: GTAAGCTAACCCCGCCAGCA

Candidate 6: AGAAACTCTCTTGTCTAAAG

Candidate 7: AGAAATGATGTCGCCTAAAG

Candidate 8: GCGAATGCGACTGGCGAGGT
Threshold = 11.

Submit the codes you have written to solve the previous question (the one of Hamming distance).

If you are using another language other than Python3, please note which language it is.

Answer 1

• Below is the detailed explanation of the problem mentioned above in python3 with code and it's output.
• For better understanding please read the comments mentioned in the code.
• CODE:

#function which returns list of id's with hamming distance of candidates whose hamming distance is less than equal to the threshold k
def find(reference, candidates,threshold):
#answer list
ans=[]
#initialize id to 1
id=1
#iterate for all candidates
for s in candidates:
#to scalculate hamming distance of this candidate
hamming=0
#calculate hamming distance
for i in range(len(s)):
if s[i]!=reference[i]:
hamming+=1
#if hamming distance is less than threshold
if hamming<=threshold:
ans.append([id,hamming])
#move to next candidate
id+=1
#return answer
return ans

• INPUT/OUTPUT:

1. #test case 1
   reference="AATGCGC"
   candidates=["AATCCCC", "TTTGCTC", "AATAAAA"]
   threshold=3
   print(find(reference,candidates,threshold))

   [[1, 2], [2, 3]]

2. #test case 2
   reference="AGAAACTCTCTGGCCTAAAG"
   candidates=["ACTTAGGTCTCTAAGCCCTC", "AGAAACGTTATGTGGACGTT", "AGTCTGACTCTGATCCAAAG", "AGTCTGCCTTGGCCATTAGC", "GTAAGCTAACCCCGCCAGCA", "AGAAACTCTCTTGTCTAAAG", "AGAAATGATGTCGCCTAAAG", "GCGAATGCGACTGGCGAGGT"]
   threshold=11
   print(find(reference,candidates,threshold))

   [[2, 11], [3, 8], [6, 2], [7, 5]]

• For better understanding and clarity below are the screenshot attached of the code and it's input/output.

CODE and INPUT/OUTPUT

So if you have any doubt regarding this solution please feel free to ask it in the comment section below and if it is helpful then please upvote this solution, THANK YOU.