Question

Given below is the restriction site of a restriction endonuclease Pst I and the cleavage sites on a DNA molecule.

Choose the option that gives the correct resultant fragments.

• A. 5' C – T – G C – A – G 3' and 3' G – A – C – G – T C 5'
• B. 5' C – T G – C – A – G 3' and 3' G – A – C – G T – C 5'
• C. 5' C – T – G – C A – G 3' and 3' G – A – C – G T – C 5'
• D. 5' C – T – G – C – A G 3' and 3' G A – C – G – T – C 5'

Hint:

Always follow the arrows literally! Draw a vertical line where the arrow tells you to cut on each strand, then connect them horizontally across the middle to see your physical sticky fragments clearly.

Answer 1

The Correct Option is D

Concept: Restriction endonucleases are molecular scissors that recognize specific palindromic nucleotide sequences in double-stranded DNA. They cleave the phosphodiester backbone at precise locations.
• If the enzyme cuts right in the middle of the recognition sequence, it creates blunt ends.
• If it cuts staggered away from the center but between the same two bases on both strands, it creates single-stranded overhangs called sticky ends.

Step 1: Identify the exact position of the cleavage arrow on the top strand ($5' \rightarrow 3'$).
The endonuclease $Pst\,\text{I}$ targets the sequence $5'\text{-CTGCAG-}3'$. Looking at the positional arrows provided in the problem statement: The downward arrow ($\downarrow$) is positioned right before the final Guanine base (G) at the $3'$ end of the top sequence: \[ 5'\text{—C—T—G—C—A} \downarrow \text{G—}3' \] This means the top strand is severed between Adenine (A) and Guanine (G). The resulting left piece retains $5'\text{-CTGCA}$ while the isolated right piece is $\text{G-}3'$.

Step 2: Identify the position of the cleavage arrow on the bottom strand ($3' \rightarrow 5'$).
The upward arrow ($\uparrow$) is located right after the initial Guanine base (G) at the $3'$ start of the lower sequence, which corresponds to cutting between Guanine (G) and Adenine (A): \[ 3'\text{—G} \uparrow \text{A—C—G—T—C—}5' \] This cleaves the strand, leaving an isolated $3'\text{-G}$ fragment on the left, and an $\text{ACGTC-}5'$ fragment on the right side.

Step 3: Assemble the physical fragments generated.
When the physical split occurs, the DNA molecule separates cleanly into two staggered sections:
• Fragment 1: \[ \begin{aligned} 5' &\text{ C—T—G—C—A } 3' 3' &\text{ G } 5' \end{aligned} \]
• Fragment 2: \[ \begin{aligned} 5' &\text{ G } 3' 3' &\text{ A—C—G—T—C } 5' \end{aligned} \] Combining the structural look shows that the cut separates the sequence into a large block containing $5'\text{-CTGCA-}3'$ paired with $3'\text{-G-}5'$, and a second block of $5'\text{-G-}3'$ over $3'\text{-ACGTC-}5'$. This matches the staggered split presented in Option (D).