Talk:BE.109:DNA engineering/Examine candidate clones
- In the future, I think it would be better to use 1 ng DNA instead of 5. The colonies were far too crowded to count on the 5 ng plate, which makes the homework assignment for tomorrow impossible.
- Definitely better to use less DNA.
- Could also plate less of transformation mix...
- To complete the calculation, you can roughly estimate the crowded plate as 5000 colonies, though this could easily be 2X off in either direction. (Thanks to Yoon Sung for catching the error in the homework related to the plasmid's name...it should be pCX-EGFP) And if you'd like to see the class data again, here it is -- Natalie
- For future reference, another option is to count the colonies in a sector of the plate (i.e. 1/8th or 1/16th ... draw lines on the plate like pie slices to divide into sectors) and then multiply by the number of sectors. This should lead to a reasonable approximation to the number of colonies. Of course, this only works if your colonies are actually distinguishable from one another. If you have a continuous lawn of cells, this won't work. Regardless, go with Natalie's suggestion for your homework assignment. -- Reshma (your Mod 3 TA chiming in).
|no ligase control||1||7||5||0||0||0|
|bkb + ligase control||1||19||8||0||1||114|
|complete ligation mixture||21, 22||800, 850||75, 80||7, 13||65||89, 105|
|no ligase control||400||0||1?||0||0||3|
|bkb + ligase control||0||8||0||1||0||3|
|complete ligation mixture||220, 250||22, 28||0, 0||0, 0||6, 5||44, 44|