The lab setup shows four test tubes. Tube 1 contains water only. Tube 2 contains a live snail. Tube 3 contains a live green wate
r plant. Tube 4 contains both a live green water plant and a live snail.
In this setup, which tubes contain at least one organism carrying on cellular respiration?
A) Tubes 1 and 2 only.
B) Tubes 2 and 4 only.
C) Tubes 3 and 4 only.
D) Tubes 2, 3, and 4 only.
In this setup, which tubes contain at least one organism carrying on cellular respiration?
A) Tubes 1 and 2 only.
B) Tubes 2 and 4 only.
C) Tubes 3 and 4 only.
D) Tubes 2, 3, and 4 only.
1 answer:
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Answer:
42,5 mL
Explanation:
We need to use the serial dilution formula beacuse we start with a stock concentrate solution and we need to prepare a new less concentrated one.
<u>DF in the dilution factor, Vi is the initial volume and Vf is the final volume.</u>
The first step is to have the same measurment unit so we need to convert 345 µg to mg.
we know that 1 µg equals 0,001 g, hence:
now the final volume is 0,345 mg protein/ mL and the inital volume is 15mg protein/mL, both of them are in the same unit so we can use the formula
Now since the question said that we already have 1.0mL of the amylase stock solution we need to subtract that 1.0mL to the 43,5 mg protein/mL
So, we need 42,5 mL of diluting buffer if we want a final concentration of 345 µg protein/mL (0.345 mg protein/mL)