mixing

**labrat** · June 21st 2007, 06:17 AM

1.If 0.5ml of serum is mixed with 1.0ml of water and a glucose test is performed resulting in a reading of 200mg/dl from the instrument, what is the final reported result?

2.If a 1:5 dilution is made with 0.5 ml of serum, how much water is used?

3.You are given a series of ten tubes, each of which contain 5ml of diluents. To the first tube is added 1ml of serum, and a serial dilution using 1ml is carried out in the remaining tubes. what is the serum concentration in tubes 4 and 8?

dratiffarid · June 21st 2007, 08:44 AM

Patient has diabetes mellitus.
Anything more than 140 mg/dL is considered as diabetes mellitus.

**labrat** · June 21st 2007, 02:52 PM

yes that is true but i need to try to work this problem. Thank you very, very much for your reply. It is informativeand is relative, but not the answer needed in this particular problem.

**topsquark** · June 22nd 2007, 04:42 AM

Originally Posted by labrat

1.If 0.5ml of serum is mixed with 1.0ml of water and a glucose test is performed resulting in a reading of 200mg/dl from the instrument, what is the final reported result?

Ummm... 200 mg/dL? Reported result of what, specifically?

-Dan

**topsquark** · June 22nd 2007, 04:44 AM

Originally Posted by labrat

2.If a 1:5 dilution is made with 0.5 ml of serum, how much water is used?

$\frac{1}{5} = \frac{0.5~mL}{x}$

So
$x = 5 \cdot (0.5~mL) = 2.5~mL$

-Dan

**topsquark** · June 22nd 2007, 04:54 AM

Originally Posted by labrat

3.You are given a series of ten tubes, each of which contain 5ml of diluents. To the first tube is added 1ml of serum, and a serial dilution using 1ml is carried out in the remaining tubes. what is the serum concentration in tubes 4 and 8?

The first tube contains 5 mL of dilutent (solvent?) and 1 mL serum. We take 1 mL of solution from this so we have
$\frac{1}{5} = \frac{x}{1~mL}$

$x = 0.2~mL$
we are taking 0.2 mL of serum from tube 1.

We mix this into 5 mL of dilutent in tube 2. Then take 1 mL of solution from this and put it into tube 3, etc.

Each time we do this we are reducing the amount of serum by a factor of 5. So the amount of serum s, in each tube t is going to be of the form:
$s_t = (1~\text{mL serum}) \left ( \frac{1}{5} \right ) ^t$

So in tube 4 we have
$s_4 = (1~\text{mL serum}) \left ( \frac{1}{5} \right ) ^4 = 0.0016~mL$

So the concentration in tube 4 is $\frac{0.0016~\text{mL serum}}{5~\text{mL dilutent}}$

and in tube 8 we have
$s_8 = (1~\text{mL serum}) \left ( \frac{1}{5} \right ) ^8 = 2.56 x 10^{-6}~mL$

So the concentration in tube 8 is $\frac{2.56 x 10^{-6}~\text{mL serum}}{5~\text{mL dilutent}}$

-Dan

**labrat** · June 24th 2007, 11:51 AM

Thank you so very much. You have been a big help. Im in my first semester of college and struggling.I really appriciate your help so much.

**topsquark** · June 24th 2007, 05:12 PM

Originally Posted by labrat

Thank you so very much. You have been a big help. Im in my first semester of college and struggling.I really appriciate your help so much.

No problem! Glad I was able to help.

-Dan

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