Lab 4: Transformant analysis

Question 1

What were the 10 plates used?

Answer 1

• Selective Ligation (500 ul)
• Selective Ligation (100 ul)
• Non-selective ligation (10^-5)
• Non selective ligation (10^-6)
• Selective Positive control
• Non-selective positive control (10^-5)
• Non selective positive control (10^-6)
• Selective negative control
• Non-selective negative control (10^-5)
• Non selective negative control (10^-6)

Question 2

In Cv = LB/(D x V), what is used in the calculation?

Answer 2

Cv: concentration of viable cells (CFU/mL)
LB: number of colonies on one LB plate
D: the dilution used on that LB plate
V: the volume plated on that LB plate

Question 3

In Te = Nt/Ns x 100, what is used in the calculation?

Answer 3

Te: transformation efficiency
Nt: number of transformed cells that survive on selective positive control
Ns: number of cells plated on selective media

Question 4

In Ns = Cv x V, what is used in the calculation?

Answer 4

Ns: number of cells plated on selective media
Cv: concentration of viable cells
V: volume plated

Question 5

Should the transformation efficiency be very high or very low? Why?

Answer 5

Very low. We are trying to force DNA through the cell membrane and cells go out of their way to avoid that

Question 6

What are the phenotypes we expected to see on the selective ligation plates?

Answer 6

Blue, white and non-fluorescent, and white and fluorescent

Question 7

What are the phenotypes we expected to see on the non-selective ligation plates?

Answer 7

These were the viability controls, no X-gal was present so we should only have white non-fluorescent

Question 8

What are the phenotypes we expected to see on the selective positive control plate?

Answer 8

Blue only

Question 9

What are the phenotypes we expected to see on the non-selective positive control plates?

Answer 9

Viability controls, so only white non-fluorescent

Question 10

What are the phenotypes we expected to see on the selective negative control plate?

Answer 10

Nothing, there should be no growth

Question 11

What are the phenotypes we expected to see on the non-selective negative control plates?

Answer 11

Viability controls, so white non-fluorescent

Question 12

What are two ways to verify the presence of a recombinant plasmid when there isn’t a very obvious phenotype like fluorescence?

Answer 12

PCR and a diagnostic digest

Question 13

What are the 4 reagents required for PCR?

Answer 13

Template DNA, primers, dNTPs, Taq polymerase

Question 14

Why do we use Taq polymerase for PCR instead of E. coli polymerase?

Answer 14

Taq polymerase is from a thermophilic bacteria and doesn’t denature at high temperatures, so it doesn’t require pipetting in new polymerase after every cycle

Question 15

What are the two controls required for PCR?

Answer 15

Positive and negative controls

Question 16

What is the template DNA used for the positive control in PCR? What does this control tell us? What should we see?

Answer 16

Something we know contains the sequence we want to amplify, in this case the pGFPuv plasmid. It tells us the procedure is working correctly and we didn’t miss a reagent. We should see a band.

Question 17

What is the template DNA used for the negative control in PCR? What does this control tell us? What should we see?

Answer 17

No DNA, just water. It tells us there was no contamination. We should see no product

Question 18

What is a master mix? Why do we use it?

Answer 18

All components of the PCR reaction except the template DNA are mixed together in one batch, then that is added to the correct template for that reaction. It accounts for pipetting error and makes sure each reaction has the same amount of reagent