where:
Given that the DNA is 5250 base pairs long and there are 10.5 base pairs per turn of the DNA helix:
\[ Tw = \frac{5250}{10.5} = 500 \] Step 2: Considering Supercoils (\( Wr \)).The plasmid has 10 negative supercoils:
\[ Wr = -10 \] Step 3: Calculating Linking Number (\( Lk \)). \[ Lk = Tw + Wr = 500 - 10 = 490 \] Conclusion:Explanation: The linking number for this plasmid is **490**, accounting for the number of helical turns and the negative supercoiling present. This value represents the total number of times the strands are intertwined, including both twists and supercoils.