itte formProbe bal Cin eability to 1. You have prepared a cell suspension of mou
ID: 146721 • Letter: I
Question
itte formProbe bal Cin eability to 1. You have prepared a cell suspension of mouse lymph node cells (10o ml total in your tube) and novw perform a 1/10 dilution, wish to determine the concentration of cells using a hemacytometer. You and put a sample in the hemacytometer. You count two of the large (1mm x 1mm) squares and find a total of 147 live cells and 15 dead cells. (3 pts) r a. What is your concentration of live lymph node cells per milliter 2 b. How many total live lymph node cells are present in your tebe? c. What is the percent viability of your ceils? 142- 0.967 xoo- 0.7%) 2. You have counted a 1/20 dilution of a cell preparation. After counting 4 of the tiniest squares (each with an area of 1/400 mm2) on the hemacytometer, you have a total of 284 live cells and 2 dead cells a. What is the percent viability of your preparation? 284 :(-19.3% x100 b What is the sonsentration (number of cells/ml) of live cells in your preparation? 20 c. If your suspension is in a volume of 0.5 mL, how many total cells do you have? Viabl Cells 0Explanation / Answer
As you counted In two squares so we take the average of the cells counted which is 147/2 and multiply it by 10000 and then again multiply by 10 to account for dilution . The answer therefore is 73.5 * 105 cells/ml.
As we have 10 ml cell suspension so the total number of cells in tube is 73.5 * 106
73.5*105 + 7.5 * 105 = 81*105 . This is the total number of cells calculated by live cells plus dead cells . Now we divide the live cells by the total and the answer is 0.907 which is 90.7%.
In the seconds question we use the same formula as above .
The number of live cells is 284/4 * 10000*20= 14200000 and dead cells is 2/4 * 10000* 20=100000. Total number of cells is 14300000. Therefore percentage viability is 14200000/1430000= 0.993 *100=99.3%.
In 0.5 ml we have 14200000*0.5 = 7100000 cells .
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