Restriction Mapping Analysis of a Plasmid

The question asks us to determine the correct statement based on the restriction digestion profiles of a 4.8 kb plasmid using four different enzymes: E1, E2, E3, and E4.

First, let's analyze the given digestion data:

• Single Digests:
  • E1: 4.8 kb
  • E2: 4.8 kb
  • E3: 4.8 kb
  • E4: 4.8 kb

  When a circular plasmid is digested with an enzyme that cuts once, it produces a linear fragment equal to the size of the plasmid. The fact that each enzyme individually produces a 4.8 kb fragment indicates that each enzyme (E1, E2, E3, and E4) has exactly one recognition site on the 4.8 kb plasmid.

• Double Digests:
  • E1 + E2: 500 bp + 4.3 kb. Since $500\ \text{bp} = 0.5\ \text{kb}$, the fragments are $0.5\ \text{kb}$ and $4.3\ \text{kb}$. $0.5 + 4.3 = 4.8\ \text{kb}$. This means the distance between the E1 site and the E2 site is either $0.5\ \text{kb}$ or $4.3\ \text{kb}$ around the circle.
  • E2 + E3: 400 bp + 4.4 kb. Since $400\ \text{bp} = 0.4\ \text{kb}$, the fragments are $0.4\ \text{kb}$ and $4.4\ \text{kb}$. $0.4 + 4.4 = 4.8\ \text{kb}$. The distance between E2 and E3 sites is either $0.4\ \text{kb}$ or $4.4\ \text{kb}$.
  • E3 + E4: 2 kb + 2.8 kb. $2 + 2.8 = 4.8\ \text{kb}$. The distance between E3 and E4 sites is either $2\ \text{kb}$ or $2.8\ \text{kb}$.
  • E1 + E4: 1.9 kb + 2.9 kb. $1.9 + 2.9 = 4.8\ \text{kb}$. The distance between E1 and E4 sites is either $1.9\ \text{kb}$ or $2.9\ \text{kb}$.

  Double digests produce two fragments because each enzyme has one site. The two sites divide the circular plasmid into two segments, whose lengths are given by the fragment sizes.

Constructing the Restriction Map

We can use the double digest data to construct a map showing the relative positions of the restriction sites on the circular plasmid. Let's place the E1 site at an arbitrary position, say 0 kb, on the circle.

From the E1 + E2 digest, E2 is either $0.5\ \text{kb}$ or $4.3\ \text{kb}$ away from E1. Let's tentatively place E2 at $0.5\ \text{kb}$.

From the E2 + E3 digest, E3 is either $0.4\ \text{kb}$ or $4.4\ \text{kb}$ away from E2 (which is at $0.5\ \text{kb}$). If E3 is $0.4\ \text{kb}$ away from E2, its position could be $0.5 + 0.4 = 0.9\ \text{kb}$ or $0.5 - 0.4 = 0.1\ \text{kb}$ (or wrapping around, $0.5 + 4.4 = 4.9\ \text{kb} = 0.1\ \text{kb}$ on a $4.8\ \text{kb}$ plasmid). So, E3 is potentially at $0.1\ \text{kb}$ or $0.9\ \text{kb}$ from the E1 site (at 0).

From the E1 + E4 digest, E4 is either $1.9\ \text{kb}$ or $2.9\ \text{kb}$ away from E1 (at 0). So, E4 is at $1.9\ \text{kb}$ or $2.9\ \text{kb}$.

From the E3 + E4 digest, the distance between E3 and E4 is $2\ \text{kb}$ or $2.8\ \text{kb}$.

Let's test the possibility that the sites are relatively close in the order E1 - E2 - E3, with distances derived from the smaller fragments in the double digests:

• E1-E2 distance: $0.5\ \text{kb}$
• E2-E3 distance: $0.4\ \text{kb}$

If the order is E1 - E2 - E3, then the distance E1 to E3 along this path is $0.5 + 0.4 = 0.9\ \text{kb}$. The sites would be at E1(0), E2(0.5), E3(0.9).

Now let's place E4 relative to E1 and E3. E1-E4 distance is $1.9\ \text{kb}$ or $2.9\ \text{kb}$. E3-E4 distance is $2\ \text{kb}$ or $2.8\ \text{kb}$.

If we place E4 at $2.9\ \text{kb}$ from E1 (at 0), let's check the distance to E3 (at 0.9). Distance E3 to E4 = $|2.9 - 0.9| = 2.0\ \text{kb}$. This matches one of the fragment sizes from the E3 + E4 digest ($2\ \text{kb}$ and $2.8\ \text{kb}$).

So, a possible map configuration with E1 at 0 is:

• E1 at $0\ \text{kb}$
• E2 at $0.5\ \text{kb}$ (distance E1-E2 is $0.5\ \text{kb}$)
• E3 at $0.9\ \text{kb}$ (distance E2-E3 is $0.9 - 0.5 = 0.4\ \text{kb}$)
• E4 at $2.9\ \text{kb}$ (distance E3-E4 is $2.9 - 0.9 = 2.0\ \text{kb}$)

Let's verify the remaining distances and the total size:

• Distance E1-E2: $0.5 - 0 = 0.5\ \text{kb}$. Matches E1+E2 fragment ($0.5\ \text{kb}$, $4.3\ \text{kb}$).
• Distance E2-E3: $0.9 - 0.5 = 0.4\ \text{kb}$. Matches E2+E3 fragment ($0.4\ \text{kb}$, $4.4\ \text{kb}$).
• Distance E3-E4: $2.9 - 0.9 = 2.0\ \text{kb}$. Matches E3+E4 fragment ($2.0\ \text{kb}$, $2.8\ \text{kb}$).
• Distance E4-E1 (wrapping around the circle): $4.8\ \text{kb} - (2.9 - 0) = 4.8 - 2.9 = 1.9\ \text{kb}$. Matches E1+E4 fragment ($1.9\ \text{kb}$, $2.9\ \text{kb}$).

The order of recognition sites around the circle appears to be E1 - E2 - E3 - E4 - E1, with approximate distances:

The sum of these adjacent distances is $0.5 + 0.4 + 2.0 + 1.9 = 4.8\ \text{kb}$, which is the size of the plasmid.

The order of sites around the circular plasmid is E1, E2, E3, E4 in this arrangement. This means E2 is located between E1 and E3 (along the shorter path).

Evaluating the Statements

Let's check each given statement based on our constructed map (E1 - E2 - E3 - E4):

1. E1 and E4 have two recognition sites each in the plasmid.

   This is incorrect. As determined from the single digests, each enzyme cuts the plasmid only once, meaning they each have only one recognition site.

2. E2 recognition sequence is located between E1 and E3.

   According to our map (E1 - E2 - E3 - E4), E2 is indeed located between E1 and E3 along one segment of the plasmid circle (the segment with distances 0.5 kb and 0.4 kb). This statement is consistent with the map.

3. The sequence of recognition sites after digestion of the plasmid with E3 is: E4 - E2 - E1.

   Digesting with E3 linearizes the plasmid. The linear sequence starting from E3 would be E3 - E4 - E1 - E2 or E3 - E2 - E1 - E4 (depending on which side you traverse). The sequence E4 - E2 - E1 is not a contiguous segment in our determined circular order (E1 - E2 - E3 - E4). This statement is incorrect.

4. The sequence of recognition sites after digestion of the plasmid with E2 is: E1 - E3 - E4.

   Digesting with E2 linearizes the plasmid. The linear sequence starting from E2 would be E2 - E3 - E4 - E1 or E2 - E1 - E4 - E3. The sequence E1 - E3 - E4 is not a contiguous segment in our determined circular order (E1 - E2 - E3 - E4). This statement is incorrect.

Based on the restriction mapping derived from the digestion data, the statement that "E2 recognition sequence is located between E1 and E3" is correct.