[2] For the hemocytometer problem below: a) What is the concentration of the cel

Question

[2] For the hemocytometer problem below: a) What is the concentration of the cells that are in the tube that has been added to the hemocytometer? (be SURE to include all calculations b) If the ube from part 2a was the 4 tube in a series of S-fold serial dilutions, what would the concentration of cells be in the undiluted tube? e) Your lab partner took the undiluted sample (determined in 2b) and set up a series of 6,4-fold serial dilutions. What would you predict the concentration of cells to be in their 4 tube.

Explanation / Answer

2a. No of cells in the tube= number of cells in all 4 corners /4 x 10000 × dilution factor

Number of cells in the 1st block=~ 80

In second block=~80

In 3rd block=~118

In 4th block =~88

Average number of cells = 80+80+118+88 / 4 = 366/ 4= 91.5 cells

Cell count= 91.5 × 10000 × 1 =

If dilution factor is 1 then number of cells will be = 91.5 × 10^4=

9.15 × 10^5 cell are present in the tube.

Q2 b when 4th tube of dilution is taken, the dilution factor will be 5×5×5 =125

then number of cells will be = 91.5 × 10^4 × 125 =

11437.5 x 10^4 = 1.14 × 10^8 cells in undiluted tube.

25. If diluted in 6 tube, the number of cells in the 4th tube will be = 91.5 × 10^4 × 64 = 5856 × 10^4 = 5.8 × 10^7 cells in undiluted tube

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