Suppose you perform a serial dilution of Amylase from a stock solution with a concentration of 500ug/mL. In five tests tubes you add 4mL of water. Then you aliquot 1mL of the stock solution into test tube 1 and vortex. From there you aliquot 1mL from test tube 1 into test tube 2. You continue to mix this way into each consecutive test tube. What is the final concentration in test-tube 3?

The final concentration in test-tube 3 is = 4ug/mL

Explanation:

Dilution factor = V initial / V final

Concentration of Stock solution = 500 ug/mL

5 test tubes are prepared with 4 mL of water

1 ml of stock solution is added to the 1st test tube.

therefore, the total volume of 1st test tube is = 4 ml + 1 ml = 5ml

∴ the concentration of 1st Test tube →

 C = m/V   [m= mass (solute) , v= volume]

     =500 ug / 5 ml  = 100 ug/ml

From 1st test tube, 1 ml is taken out and that is added to the 2nd  test tube.

∴ the concentration of 2nd Test tube = 100 ug / 5ml

                                                                = 20 ug/ml

From  2nd test tube, 1 ml is taken out and that is added to the 3rdd  test tube.

∴ the concentration of 3rd Test tube = 20 ug / 5ml

                                                              = 4 ug/ml

The final concentration in test-tube 3 is = 4ug/mL

c.more precipitation,but only when emission levels.

stimulus is a change in the environment that causes an organism to react. the response is the reaction to the stimulus. some examples of stimuli for zebras are drought, to many predators, and bad feeding grounds such as dead grass or plants. zebras respond to stimuli just like all other living organisms