Meaning of overhang / watson_overhang in single stranded sequences

[manulera]

This is an edge case, but came up in the tests. We should think about what overhang should represent in single stranded sequences. The below code shows the problem

from pydna.dseq import Dseq

seqs = [
    Dseq.from_full_sequence_and_overhangs("AAAA", 0, 3),
    Dseq.from_full_sequence_and_overhangs("AAAA", -3, 0),
    Dseq.from_full_sequence_and_overhangs("AAAA", 0, 4),
]

for seq in seqs:
    print(repr(seq))
    print("ovhg:", seq.ovhg)
    print("watson_ovhg:", seq.watson_ovhg())
    print()

prints:

Dseq(-4)
AAAA
T
ovhg: 0
watson_ovhg: 3

Dseq(-4)
AAAA
   T
ovhg: -3
watson_ovhg: 0

Dseq(-4)
AAAA

ovhg: 0
watson_ovhg: -4

What the last case should return is not clear to me. If we think that the "previous step" is coming from ovhg = 0 (first sequence), then it makes sense to return 0, but if we think the previous step is ovhg = -3, then it feels like ovhg should be -4.

[manulera]

Maybe it should return None? Not sure if that would break things

[manulera]

This also applies to what should three_prime_end and five_primer_end return

[manulera]

I propose the following, which makes all tests pass, but seems to break .melt (there are no unit tests for this function, but it's used in USER)

from pydna.dseq import Dseq

seq = Dseq("EEEE")
print(repr(seq))
print(seq.ovhg)
print(seq.five_prime_end())
print(seq.three_prime_end())

print("--------------------------------")
seq = Dseq("FFFF")
print(repr(seq))
print(seq.ovhg)
print(seq.five_prime_end())
print(seq.three_prime_end())

returns

Dseq(-4)
AAAA

None
('single', 'aaaa')
('single', 'aaaa')
--------------------------------
Dseq(-4)

TTTT
None
('single', 'tttt')
('single', 'tttt')

[manulera]

To reproduce the error of .melt

from pydna.dseq import Dseq

seq = Dseq("GFGGATTaaaCGGCGEg")

print(repr(seq))

stuffer, insert, stuffer = seq.melt(1)

print(repr(insert))

[BjornFJohansson]

I agree that ovhg should return None for a sequence that has only one strand.

[BjornFJohansson]

from pydna.dseq import Dseq

seq1 = Dseq("EEEE")
print(repr(seq1))
# Dseq(-4)
# AAAA
#
print(seq1.ovhg)
print(seq1.five_prime_end())
print(seq1.three_prime_end())

print("--------------------------------")

seq2 = Dseq("FFFF")
print(repr(seq2))
# Dseq(-4)

# TTTT
print(seq2.ovhg)
print(seq2.five_prime_end())
print(seq2.three_prime_end())

print("--------------------------------")

seq3 = Dseq("XXXX")
print(repr(seq3))
# Dseq(-4)
# TTTT
#
print(seq3.ovhg)
print(seq3.five_prime_end())
print(seq3.three_prime_end())


"""

Seq2 and 3 return the same thing in the example above.

I suggest 

seq2.five_prime_end()
('lower', 'tttt') for seq2

seq3.five_prime_end()
('upper', 'tttt') for seq3

"""