Learning Module -
Period 7
Estimation of DNA
yield from phage and E. coli XL-1 Blue
Your gel picture
should look something like this, although the lane order may be different:
Do not make any
marks on the picture itself. After affixing the picture in your notebook,
proceed to add labels and a legend.
Each lane should be numbered from left to right (with the sample wells at the top) starting with Lane 1. Each major band should be identified with a letter and a Legend describing each lane must be placed below or next to the picture:
Note: In the above species identification, Form I is CCC supercoiled and Form II, IV are relaxed circular (covalently closed or nicked)
By comparison of
fluorescence intensities, estimate the amount of DNA in each of your
preparations:
For pTrc99A, Lane 1 (2 µL) has about twice the amount of DNA as the
Standard in Lane 8 (200 ng).
Therefore, the concentration in the mini-prep is: 400 ng / 2µL = 200 ng / µL
and the yield is: 200 ng / µL x 30 µL = 6000 ng, or 6 µg
For ZAP/topAcysB, Lane 5 (4 µL) has about the same amount of DNA as
Lane 10 (100 ng): 100 ng / 4 µL = 25 ng / µL
and Lane 4 (2 µL) has about 3/4 the amount of DNA as in Lane 9 (50 ng): (50 ng
x 0.75) / 2 µl = 18.75 ng / µL
Therefore the concentration in the mini-prep is: (25 ng / µL + 18.75 ng / µL) /
2 = 21.9 ng / µL
and the yield is: 21.9 ng / µL x 60 µl = 1313 ng or 1.31 µg
From the concentrations, the volume of each mini-prep needed for use in the subcloning (Experiment 8) can be calculated, and from the yields, a comparison of the actual yield to the expected DNA yields can be made. If one assumes 100% recovery of both DNAs, the % viability of the phage can be calculated and the copy number of pTrc99A can be determined. These topics should be addressed in the Discussion section for Period 7.