Question

1.) What is the concentration of a solution of PCR primers if the absorption of 260 nm (A260) is 0.478 the path length is 1.0 cm and the extinction coefficient (absorptivity) is 185 mM^-1 cm ?

a. 0.39 mN

b. 3.9 x 10^-3 M

c. 2.6 x 10^-3 mmol/L

d. 2.6 mmol/L

2.) What is the concentration of a stock solution of PCR primers if the absorption of a diluted solution at 260 nm (A260) is 0.528, the path length is 1.0 cm, the extinction coefficient (absorptivity) is 180 mM^-1 cm^-1, and the diluted solution read in the spectrometer was made by adding 10 uL of the stock solution to 100 uL of TE buffer.

a. 3.2 x 10^-2 mmol/L

b. 2.9 x 10^-5 mmol/L

c. 2.9 x 10^-1 mmol/L

d. 2.6 x 10^-4 mmol/L

3.) An alternate set of units to express a concentration of 3.9 x 10^-3 mM is

a. 3.9 x 10^-3 mmol/L

b. 3.9 x 10^-6 umol/uL

c. 3.9 uM

d. All of the above

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Answer 1

1)

Given

1) Molar extinction coefficient of a substance =185 mM ^-1 cm ^-1

2) concentration of a substance = ?

3) unit cell length =1.0 cm

4) Absorbance = 0.478

From Beer's we can write A = b c

where A= absorbance, b = unit cell length,c= concentration.

Putting given values in above formula, we get

0.478 = 185 mM ^-1 cm ^-1 1 cm x c

c = 0.478 / 185 mM ^-1 cm ^-1 1 cm = 2.6 10 ^-03 mM ^-1 = 2.6 10 ^-03 mmol / L

ANSWER : Concentration of Solution of PCR primers = 2.6 10 ^-03 mmol / L

2)

Given

1) Molar extinction coefficient of a substance =180 mM ^-1 cm ^-1

2) concentration of a substance = ?

3) unit cell length =1.0 cm

4) Absorbance = 0.528

We have, A = b c

where A= absorbance, b = unit cell length,c= concentration.

Putting given values in above formula, we get

0.528 = 180 mM ^-1 cm ^-1 1 cm x c

c = 0.528 / 180 mM ^-1 cm ^-1 1 cm = 2.9 10 ^-03 mM ^-1 = 2.9 10 ^-03 mmol / L

Here volume of diluted solution = 10 uL + 100 uL = 110 uL

Concentration of diluted solution = 2.9 10 ^-03 mmol / L

Volume of stock solution = 10 uL

Now, we can use dilution formula to calculate concentration of stock solution.

C stock x V stock = C dilute x V dilute

  C stock = C dilute x V dilute / V stock =  2.9 10 ^-03 mmol / L 110 uL / 10 uL = 0.0319 mmol / L

  C stock = 3.2 10 ^-02 mmol / L

ANSWER : Concentration of stock Solution of PCR primers =3.2 10 ^-02 mmol / L

3)

We have 3.9 10 ^-03 mM = 3.9 10 ^-03 mmol / L

We have relation , 1 uM = 1000 mM

Therefore, 3.9 10 ^-03 mM = 3.9 10 ^-03 mM   ( 1 uM / 1000 mM ) = 3.9 uM

We have , 1 L = 10,00,000 uL

Therefore, 3.9 uM = 3.9 micromol / L x ( 1 L / 10 ⁶ uL) = 3.9 x 10 ^-06 uM / uL

ANSWER : All of the above.