A student sets up five plates. The cells with no DNA give 0 colonies…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A student sets up five plates. The cells with no DNA give colonies. The uncut vector gives colonies. The cut vector without ligase gives colonies. The cut vector with ligase and no insert gives colonies. The cut vector with ligase and the insert gives colonies.
- Which plate shows that the cells are competent?
- What share of the uncut vector's colonies does the cut vector without ligase give? What does it show?
- What share of the colonies on the last plate could be an empty vector?
- About how many colonies on the last plate probably carry the insert?
Need a reminder? The key ideaShow
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Given. Counts: , , , , .
Plan. Compare each plate with the plate it controls.
Step 1. (a) The uncut vector, with colonies. The cells took the DNA up well.
Step 2. (b) per cent. It shows that the digest was nearly complete. A little uncut vector is left.
Step 3. (c) The background of the vector that closes on itself is colonies. per cent of the last plate could be an empty vector.
Step 4. (d) colonies probably carry the insert.
Answer: (a) the uncut vector plate. (b) per cent, so the digest was nearly complete. (c) per cent. (d) about .
Check. The plate with no DNA shows no colonies, so the antibiotic worked.
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