Question

1. make a restriction map of the isolated plasmid by determining the relative positions of the restriction sites. the best approach is to realize that the double digest is really an extension of the single digests. for example, the double digest is analogous to taking the bands you see in the a restriction digest and digesting these bands with the b enzyme. the smaller bands you see in the double digest should add up to the bands seen in the a digest alone.

a. determine which bands of the double digest come together to form the small a restriction fragment.

b. determine which bands of the double digest make up the large a fragment.

c. draw the two a restriction fragments in the space provided, and indicate where the b restriction sites fall on the fragments. indicate the size of the fragments that are produced in the double digest.

Explanation:

Step1: Analyze A digest bands

A digest has bands at 3 Kb and 1 Kb.

Step2: Analyze double digest bands

Double digest (A and B) has bands: 1.75, 1.25, 0.75, 0.25 (for small A? Wait, small A is 1 Kb. Wait, no: small A fragment is 1 Kb. Let's check which double digest bands sum to 1 Kb. Wait, 0.75 + 0.25 = 1 Kb? Wait no, 0.75 + 0.25 = 1? 0.75 + 0.25 = 1? Yes. Wait, but also 1.25? No. Wait, the small A restriction fragment is 1 Kb. So we need to find double digest bands that add up to 1 Kb. Let's list double digest bands: 1.75, 1.25, 0.75, 0.25? Wait no, the double digest bands are 1.75, 1.25, 0.75, 0.25? Wait the image shows: A and B digest has 1.75, 1.25, 0.75, 0.25? Wait no, looking at the gel: A and B digest bands (from top to bottom? Wait origin is at the top. So bands in A and B digest: 1.75, 1.25, 0.75, 0.25? Wait no, the gel: Markers are 4,3,2,1,0.5 Kb. A digest: 3,1. B digest: 2.5,1.5. A and B digest: 1.75,1.25,0.75,0.25? Wait no, maybe I misread. Wait A digest: two bands, 3 Kb and 1 Kb (so total plasmid size is 3 + 1 = 4 Kb? Because it's a plasmid, circular, so single digest with A gives two bands, so total size 3 + 1 = 4 Kb. B digest: two bands, 2.5 and 1.5 Kb (2.5 + 1.5 = 4 Kb, matches). Double digest: let's sum the bands. 1.75 + 1.25 + 0.75 + 0.25 = 4 Kb (1.75+1.25=3, 0.75+0.25=1; 3+1=4). Now, small A fragment is 1 Kb (the 1 Kb band in A digest). So which double digest bands sum to 1 Kb? 0.75 + 0.25 = 1 Kb. Wait, but also, the large A fragment is 3 Kb. Which double digest bands sum to 3 Kb? 1.75 + 1.25 = 3 Kb (1.75 + 1.25 = 3).

Part a: Small A (1 Kb)

The small A restriction fragment is 1 Kb. The double digest bands that add up to 1 Kb are 0.75 Kb and 0.25 Kb (since 0.75 + 0.25 = 1).

Part b: Large A (3 Kb)

The large A restriction fragment is 3 Kb. The double digest bands that add up to 3 Kb are 1.75 Kb and 1.25 Kb (since 1.75 + 1.25 = 3).

Part c: Drawing the map

• Plasmid is circular, 4 Kb.
• A restriction sites: cut the plasmid into 3 Kb and 1 Kb fragments.
• B restriction sites: cut the plasmid into 2.5 Kb and 1.5 Kb fragments.
• In the 3 Kb A fragment (large A), B cuts it into 1.75 Kb and 1.25 Kb (since 1.75 + 1.25 = 3).
• In the 1 Kb A fragment (small A), B cuts it into 0.75 Kb and 0.25 Kb (since 0.75 + 0.25 = 1).

So the restriction map: Plasmid (4 Kb circular). A sites: divide into 3 and 1. B sites: in 3 Kb A fragment, B cuts at 1.75 from one end (so 3 Kb becomes 1.75 and 1.25); in 1 Kb A fragment, B cuts at 0.75 from one end (so 1 Kb becomes 0.75 and 0.25). So the order (circular) could be: A site, then B site (1.75 from A), then A site, then B site (0.75 from A), and back to start.

Answer:

Part a:

The bands in the double digest (A and B) that form the small A restriction fragment (1 Kb) are 0.75 Kb and 0.25 Kb (since \( 0.75 + 0.25 = 1 \) Kb).

Part b:

The bands in the double digest (A and B) that form the large A restriction fragment (3 Kb) are 1.75 Kb and 1.25 Kb (since \( 1.75 + 1.25 = 3 \) Kb).

Part c:

• Large A fragment (3 Kb): Contains B restriction sites that produce fragments of 1.75 Kb and 1.25 Kb (from the double digest).
• Small A fragment (1 Kb): Contains B restriction sites that produce fragments of 0.75 Kb and 0.25 Kb (from the double digest).
• The plasmid is circular (total size \( 3 + 1 = 4 \) Kb, matching \( 2.5 + 1.5 = 4 \) Kb from B digest). The restriction map shows:
• A sites divide the plasmid into 3 Kb and 1 Kb segments.
• B sites within the 3 Kb segment split it into 1.75 Kb and 1.25 Kb; within the 1 Kb segment, split it into 0.75 Kb and 0.25 Kb.

(For drawing, represent the circular plasmid with A sites (separating 3 Kb and 1 Kb) and B sites at \( 1.75 \) Kb from one A site (in the 3 Kb segment) and \( 0.75 \) Kb from the other A site (in the 1 Kb segment), labeling the fragment sizes from double digest: 1.75, 1.25, 0.75, 0.25.)