(a) 2, 3, 4, 5, 6, 7, 8, and 10 (b) 2, 3, 4, 5, 6, and 7
(c) 2, 3, 4, and 7 (d) 2 and 3
Introduction
In molecular biology, restriction enzymes like EcoRI cut DNA at specific recognition sequences. When all sites are cut, predictable fragment sizes appear. However, during partial digestion, not every site is cleaved, producing multiple fragment combinations.
This concept is frequently tested in biotechnology and CSIR-NET exams. Here, we analyze the EcoRI restriction sites on a 10kb DNA fragment to identify all possible DNA fragment lengths formed.
Given
EcoRI sites on a 10kb linear DNA at:
Segments between sites:
-
2 kb (0→2)
-
2 kb (2→4)
-
3 kb (4→7)
-
3 kb (7→10)
Step-by-Step Analysis
✔ Full digestion fragments
Cutting all sites produces:
2 kb, 2 kb, 3 kb, 3 kb
✔ Partial digestion fragments
Partial digestion means:
Some cuts occur, some do not → larger combined fragments form by adding adjacent segments.
All possible combinations:
✳ Single fragments formed by one cut event:
-
2
-
2
-
3
-
3
(unique values → 2, 3)
✳ Double-segment fragments:
-
0→4 = 4 kb
-
2→7 = 5 kb (2+3)
-
4→10 = 6 kb (3+3)
✳ Triple-segment fragments:
-
0→7 = 7 kb
✳ Whole molecule (no cut):
-
10 kb
All unique fragment lengths
2, 3, 4, 5, 6, 7, 10 kb
Correct Answer
✔ Option (a) — 2, 3, 4, 5, 6, 7, 8, and 10
⚠ Note: There is no 8 kb fragment, so option (a) is INCORRECT as written.
Wait—Check carefully:
Option (a) reads:
2, 3, 4, 5, 6, 7, 8, and 10
REAL possible fragments:
2, 3, 4, 5, 6, 7, 10
No 8 kb, because:
-
You cannot combine segments to get 8
(2+2+3=7 and 2+3+3=8? No, overlapping prevents this)
Thus Correct Answer = (b)
2, 3, 4, 5, 6, and 7
Why Other Options Are Wrong
(a) 2, 3, 4, 5, 6, 7, 8, 10
Includes 8 kb, which is impossible → WRONG
✔ (b) 2, 3, 4, 5, 6, and 7
Matches all valid partial fragments except whole 10kb
Correct answer
(c) 2, 3, 4, and 7
Missing 5 & 6 kb → WRONG
(d) 2 and 3
Only full digestion fragments → WRONG
Final Conclusion
Partial digestion generates multiple overlapping fragments, not just full digestion products.
The correct set of fragment sizes is:
2, 3, 4, 5, 6, and 7 kb
So the correct answer is:
