5. As the lab instructor for Cell Biology, you need to make a solution of DCMU,

Question

5. As the lab instructor for Cell Biology, you need to make a solution of DCMU, an inhibitor of photosynthetic electron transport. You want a final concentration of 2x10-4M in a final volume of 2mls, and this needs to be achieved using a 10ul shot from the micropipette. How much DCMU should you weigh out to dissolve in 10mls of alcohol? The MW of DCMU is 400g/mole.

4. You are given 1ml of a 30mM stock solution of NADPH. You are told to dilute it so that its absorbance in the Spec20 equals 0.60. The molar extinction coefficient for NADPH is 7105L/mole·cm and the path length is 1.6cm. The volume of the tube is 10mls. How many microliters of stock NADPH do you transfer?

2. Show how you would construct a serial dilution if you were given a 1M solution and you need exactly 10mls of a 20 nanomolar solution.

3. How many microliters of a stock solution of protein that is 21mg/ml would you transfer to get 6mls of a 0.04mg/ml solution?

Explanation / Answer

Ans 5)

Formula: N1 × V1 = N2 × V2                                              N = Normality ; V = Volume

0.0002 (2x10-4M) × 2ml =   N2 × 10ml                                     N2 = Required moles of DCMU

0.0002 × 2 / 10 = N2

N2 = 0.00004 M

So,

1 M = 400 g dissolved in 1000ml

So 0.00004 M is 0.00016gms of DCMU dissolved in 10 mls of alcohol.

Ans 6)

A = c l                              - Beer - Lamberts Law

= molar extinction coefficient

c = concentration

l = path length                                                 

0.6 = 7105 × c × 1.6

c = 0.00005277973 Mole/L

c = 0.05277973 mM OR 0.05 mM

Original stock = 30 mM   Required concentration = 0.05 mM

So by formula = required / Given

= 0.05 / 30

= 0.001 ul of NADPH from the Original stock should be transferred.

Ans 2 )

Assuming given 1M solution is prepared in 1 Litre(1000ml)

N1 × V1 = N2 × V2

1M × V1(ml) = 0.00000002M × 10 ml

V1 = 200 ml

In this case 2 ml can be dissolved in 8 ml to get 20nM concentration and final volume 10ml.

Ans 3)

N1 × V1 = N2 × V2

21 × V1 (ml ) = 0.04 × 6ml

V1 (ml) = 0.0115

So, 11.5 microliters of a stock solution of protein needs to transfer.

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