Answer to Question #133611 in Cell Biology for Dennise Claire Tambologan

Question

Calculate the size of an organism that occupies 26 um spaces under scanner, LPO and HPO using stage micrometer and ocular micrometer if the calibration factor is 0.65, 0.75 and 0.85

Answer 1

Total magnification$=4\times10=40×$

Occupied space$=26\mu m$

Calibration factor$=0.65$

size of organism$=\frac{26\mu m}{40\times 0.65}=1\mu m$

calibration factor=$0.75$

size of organism$=\frac{26\mu m}{40\times0.75}=0.87\mu m$

Calibration factor$=0.85$

size of organism$=\frac{0.85\mu m}{40\times0.85}=0.765$

For low power(L.P.O)=$Magnification=10\times10=100\times$

calibration factor=$0.65;$ size of organism =$\frac{26\mu m}{100\times0.65}=0.4\mu m$

calibration factor=$0.75$ ;size of organism=$\frac{26\mu m}{100\times0.75}=0.35\mu m$

calibration factor=$0.85$ ; $\frac{26\mu m}{100\times0.85}=0.31\mu m$

For high power(H.P.O)=magnification=$40\times10=400\times$

calibration=$0.65;$ size of organism=$\frac{26\mu m}{400\times0.65}=0.1\mu m$

Calibration=$0.75;$ size of organism=$\frac{26\mu m}{400\times0.75}=0.087\mu m$

calibration=$0.85;$ size of organism=$\frac{26\mu m}{400\times0.85}=0.077\mu m$