5. Your Favorite Gene (YFG) is cloned into pAMP and E. coli is transformed with
ID: 1089005 • Letter: 5
Question
5. Your Favorite Gene (YFG) is cloned into pAMP and E. coli is transformed with 0.2 ug of intact pAMP/YFG recombinant according to the protocol above. Using the information below, calculate the number of molecules of YFG that have been cloned in the entire culture when the culture enters the stationary phase 200 minutes after inoculation. a. A transformation efficiency equal to 10 colonies per microgram of intact pAMP is achieved. b. pAMP has an average copy number of 100 molecules per transformed cell c. Following heat shock (step 12) the entire 250 1 of cell suspension is used to inoculate 25 ml of fresh LB broth. The culture is incubated, with shaking, at 37°C Transformed cells enter the log phase 60 minutes after inoculation and then begin to replicate on an average of once every 20 minutes. d. 6. The above transformation protocol is used with 10 1 of intact plasmid DNA at nine different concentrations. The following numbers of colonies are obtained when 100 1 of transformed cells are plated on selective medium: 0.00001 /! 0.00005 g/1 0.0001 gjul 0.0005 gjul 0.001 gjul 0.005 g/MI 0.01 g/1 4 colonies 12 colonies 32 colonies 125 colonies 442 colonies 542 colonies 507 colonies 475 colonies 516 colonies 0.1 Hg/ul a. Calculate the transformation efficiency at each concentration. b. Plot a graph of DNA mass versus colonie Plot a graph of DNA mass versus transformation efficiency What is the relationship between the mass of DNA transformed and transformation efficiency c. d. e. At what point does the transformation reaction appear to be saturated f. What is the true transformation efficiency?Explanation / Answer
Solution:
1ug –> 10^6 colonies
0.2ug – >10^6 * 0.2 = 2*10^5 colonies are transformed with the given amount of gene construct.
The copy number of the construct is 100 molecules per every transformed cell
The question is to find out how many molecules of the construct are present 200 minutes after transformation.
If the log phase started 60 minutes after inoculation, the remaining 140 minutes are used by the bacterial cells to multiply.
In 140 minutes, 2*10^5 colonies are present
In 120 minutes, half of it should be present=> 10^5
In 100 minutes, 0.5 *10^5 colonies
In 80 minutes, 0.25*10^5 colonies
In 60 minutes, 0.125*10^5 colonies
In 40 minutes, 0.0625*10^5 colonies
In 20 minutes, 0.03125*10^5 colonies have to be present
Or 3125 colonies are present 20 minutes after attaining log phase
Initially, 1562.5 colonies are present.
It is a calculation done from several experiments that one ml of an aerated overnight culture of E.coli is added to 20ml of fresh broth, it would contain after incubation of 1.75 hours, log phase consisting of 2*10^8cells/ml.
1562.5 colonies of E.coli here are supposed to have 2*10^8 cells
If 1ml of culture is added to 20ml of broth to get 2*10^8cells/ml
250ul is added to 25ml of broth to get (2*10^8 *0.25*25)/20 = 6.25*10^7cells/ml
If 100 molecules are formed from a single transformed cell
6.25*10^7 cells will have 6.25*10^9 molecules of pAMP/YFG.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.