Question

Q1 i got 3.7x10^5 is that right?

i dont know how to calculate Q2 or 3

The diagram below represents a haemocytometer grid with shaded cells indicating those stained with trypan blue and unshaded cells which are unstained by trypan blue. To prepare this slide you mixed 20 wl of the cell suspension with 20 l of trypan blue. O 0o . ooo Q ° 0 O 008 il 0.0 0.0 Oo 110 0 1 0 ooo 0.0 % VOLO DO 0 OOO Koo ooºoo 0 20 o 14 Millal oo. Use the diagram above to answer the following questions. 5. Showing your calculations, determine the concentration of viable cells per ml in your sample. 6. Showing all calculations, what is the percentage of viable cells in your sample? 7. Showing all calculations, what volume of this sample would you require in order to remove 5x10° viable cells?

Answer 1

Haemocytometer o Include cells touching middleline on top and left are counted Tolol & Enclude cells touching middle line on bottom Estimating cell density activity and right Trypan Blue is a stain that selectively stains dead cells Clustained cells are not stained viable. Chamber #Live Cells | #Deadeels #Totalceus Quadrant 1 19 = 20 Suadrant 2 23 i 2 Suadrant 3 22 3 Suadrant 4 ] 18 2 23 4 25 07w + 0 A Total amount of sample used = 204L 420u1. Tryphan Average #cells from the given quadrants blue. Average #Average # Average cells # Live cells Dead cells 20 +23+25+23 19+19+22 1+ 4 +3+5 4. +18 4 = 3:25 = 91 2 22.75 = 19.5 Total Dination factor = volume = 40 ul = 2 Volume of sample zou 4

NO #concentration of viable cells per ml.: - 19.5 - Average & dilution x 2x104. I factor & 104 live x dilution factor to cells the pet ml : 19.5 = 3.96 105 viable cells/ml. 140 x 10-3 # Percentage of viable cells in the viabletteelts 100 de = 19.5 +100 – 85.71% Total cell 22.75 # Volume of this sample required to remove 5x105 viable. cells. -x 100 =

Final volume = Meas weed cell Initial volume density cell Recommended density = 3.9x105 .42yb 200LC volume of samuple taken = 15.6ML