Question

In: Biology

Given: 11 ml of splenocytes with concentration of 9.5 x 10^6 cells/ml diluting buffer= RPMI/2% FCS...

Given:

  • 11 ml of splenocytes with concentration of 9.5 x 10^6 cells/ml
  • diluting buffer= RPMI/2% FCS

What needs to be made?

  • 4 plates with 96 wells each

What are the requirements?

  • Each well will contain - 4 x 10^6 cells with the volume of 105 microlitres

How do you create a cell sample given the information above?

Solutions

Expert Solution

Given the concentration of splenocytes = 9.5 *106 cells/mL

We are asked to prepare a cell concentration of 4 * 106 cells in 105 uL volume.

The per mL concentration required would then be ; 1000 / 105 * 4 * 106 = 9.5 * 4 * 106 = 38  * 106 = 3.8 * 107 cells per mL

So, we need a concentration of 3.8 * 107 cells per mL so that when we aliquot 105 uL we get a concentration of 4 * 106 cells.

Now we need to prepare this into 4 plates with 96 wells, so 4 * 96 = 384 wells * 105 uL = 40320 uL or 40.32 mL.

We need a total of 40.32 mL, let us make it rounded to about 45 mL with a cell concentration of 3.8 * 107 cells per mL so that we can aliquot 105 uL in 4 plates.

So, for a total of 42 mL, the cell concentration required would be: 42 * 3.8 * 107 = 159.6 *107 = 1.6 *109 cells per 42 mL.

We know the original cell concentration to be: 9.5 *106 cells/mL

The total volume is given as 11 mL, so the total number of cells in the original culture = 9.5 * 106 * 11 = 105.6 * 106 = 1.1 * 108 cells / 11 mL .

This appears to be less than that calculated for the amount of cells to be transferred into 4 plates.

Let us find the amount required per plate:

96 * 105 = 10080 uL = 10.1 mL, let us round it off to 15 mL of the desired concentration.

To get 15 mL of the desired diluted culture, the number of cells needed would be : 15 * 3.8 * 107 = 57 *107 = 5.7 * 108 cells.

We can see that even this is less than the number of cells available in the original culture!


Related Solutions

You are given a vial of cells at a density of 6.8 x 10^6 cells/ml. You...
You are given a vial of cells at a density of 6.8 x 10^6 cells/ml. You are asked to seed a 12 well plate, which requires 2 x 10^5 cells per well. Because every well requires a final media volume of 1mL, your professor asks you to dilute your cells with media so that you can directly add 1ml of diluted cells per well in the plate. How you would dilute your cells so that each well receives 2 x...
An organic acid concentration in a hard sample was determined by diluting a 15.00 mL sample...
An organic acid concentration in a hard sample was determined by diluting a 15.00 mL sample of 200.00 mL solution. A 20.00 mL portion of the diluted sample reacted with 40.0 mL of 0.05417 M NaOH completed. The excess OH- was back titrated with 5.21 mL of 0.02154 M HCL. The concentration of the organic acid in the orginal sample is: Answer is 1.370 M
Calculate the concentration of buffer components present in 217.00 mL of a buffer solution that is...
Calculate the concentration of buffer components present in 217.00 mL of a buffer solution that is 0.270 M NH4Cl and 0.270 M NH3 after the addition of 2.10 mL of 6.0 M HNO3. [NH3] = ___ M [NH4+] = ____ M
Given the following data: x 2 8 5 12 9 y 6 11 7 14 10...
Given the following data: x 2 8 5 12 9 y 6 11 7 14 10 a) draw/graph a scatter plot. b) by hand, find the correlation coefficient r. c) by hand, find b0 and b1. d) write the regression equation.
Solve for x -2 < (1/6)(10-x) < 2
Solve for x -2 < (1/6)(10-x) < 2
11 mL of 0.0100 M HCl are added to 25.0 mL of a buffer solution that...
11 mL of 0.0100 M HCl are added to 25.0 mL of a buffer solution that is 0.010 M HC2H3O2 and 0.08 M C2H3O2-.  What is the pH of the resulting solution?
A beaker with 2.00×10^2 mL of an acetic acid buffer with a pH of 5.000 is...
A beaker with 2.00×10^2 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.60 mL of a 0.400 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740
A beaker with 1.10×10^2 mL of an acetic acid buffer with a pH of 5.000 is...
A beaker with 1.10×10^2 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 mol L−1. A student adds 7.00 mL of a 0.490 mol L−1 HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760. Please show where all numbers comes from.
The final volume of buffer solution must be 100.00 mL and the final concentration of the...
The final volume of buffer solution must be 100.00 mL and the final concentration of the weak acid must be 0.100 M. Based on this information, what mass of solid conjugate base should the student weigh out to make the buffer solution with a pH=5.5?
The final volume of buffer solution must be 100.00 mL and the final concentration of the...
The final volume of buffer solution must be 100.00 mL and the final concentration of the weak acid must be 0.100 M. Based on this information, what mass of solid conjugate base should the student weigh out to make the buffer solution with a pH=7.0 (in grams)? Based on this information, what volume of acid should the student measure to make the 0.100 M buffer solution (in mL)? My weak acid is Sodium dihydrogen phosphate monohydrate, Ka= 6.23X10^-8, 2.00M and...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT