BioTech LabReport Essay

Submitted By tevanmore
Words: 1198
Pages: 5




Calculating Transformation Efficiencies:

1. Consider a transformation experiment with the following characteristics
a. 15 ng of DNA was used in the transformation.
b. The final volume at recovery was 110µl.
c. The plating volume was 100µl.
d. The experiment produced 200 transformants from the 100µl plating volume.

CALCULATE, the Transformation Efficiency for this experiment.
The unit for Transformation Efficiency is “Number of Transformants per µg of DNA”.

Show your work for INCLUDING ALL UNITS. MAKE SURE YOUR FINAL ANSWER HAS THE CORRECT UNIT DESIGNATION. All entries must be hand-written, use pencil or ink. Express you answer in scientific notation, round the base number to the nearest 1/10th.

Transformation Efficiency = _________________________

Show Work:

2. The photo below shows a pre & post ligation gel pattern. Answer the questions below based on these data in these gels

Which Lane shows the expected pre-ligation gel pattern, which Lane shows the expected post-ligation gel pattern

2a. Pre-Ligation Gel Pattern: Lane #_____1_______

2b. Post-Ligation Gel Pattern: Lane #_____2_______

Explain what the 2 bands in Lane 1 and the single band in Lane 2 represent. Include the name of the molecule and the estimated size. Answer must be contained in a complete sentence (50% penalty if not in sentence form)

Lane 1 – First (top) band:

The top band represents the 3000 base pair plasmid vector. It contains the EcoR1 and PvuII restriction sites and has termini compatible with the kanr DNA fragment.

Lane 1 – Second band:

The second band is the kanr fragment that is about 1300 base pairs in length (1301). This is ligated with T4 DNA ligase which catalyzes the bonding.

Lane 2 band:

This band is the construct or recombinant plasmid that is 4300 base pairs in length. The construct contains the gene of interest that is to be inserted in the bacteria strain.

GEL #1

Pvu2/Cla 1

GEL #2

Pvu2/Cla 1

3. Answer the following questions based on the 2 gel patterns on the previous page.

3a. Which Gel (#1 or #2) represents a 3’ to 5’ insert? ____#2______

3b. EXPLAIN your answer in 3a.
Explain Expected Data and then compare to the Observed Data
REFERENCE ONLY THE LANES THAT ARE RELEVANT TO THE EXPLANATION. Deductions for referencing non-relevant lanes.
Suggestion: Draw restriction maps for Expected data and reference in comparison to Observed data – DO NOT USE/INSERT ANY FIGURES CREATED BY THE INSTRUCTOR INCLUDING THOSE ON HOMEWORK OR THE CLASS WEBSITE

The expected data is that the insert would be 5’ to 3’ as shown in Gel #1. When observing the Pvu2/Cla1 lane in Gel #1 the top band is the vector while the lower band is the fragment between the two restriction sites. When the fragment is inserted 3’ to 5’ we have the observed data which is different in the Pvu2/Cla1 lane. When the fragment is 3’ to 5’ the Pvu2 and Cla1 restriction sites are much closer than they are in the 5’ to 3’ orientation. This results in a smaller fragment and a larger plasmid vector. This explains why the top band in the Gel #2 is slightly higher than the top band in Gel #1. In this way whether the insert is 5’ to 3’ or 3’ to 5’ will alter the lengths of the bands and therefore result in the different gel electrophoreses. The two ways the kanr insert is situated can therefore be determined by the size of the fragments.
The data below is a DNA Fingerprint for a paternity case.

Lane 1: Standard DNA fragments
Lane 2: Mother’s DNA
Lane 3: Child’s DNA cut
Lane 4: Potential Father 1’s DNA
Lane 5: Potential Father 2’s DNA

Based on the data, which potential father is NOT EXCLUDED from further paternity investigations in this case? (Write you answer on the line below)…