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Practice calculations for Bioc0036

1.

In the experiment that I performed a stock ovalbumin solution had concentration of 95 mgml^-1.  How many ml of that solution would you need to prepare 25ml of 5 mgml^-1?

Remember, C1V1=C2V2.

Answer: 1.316

Here, C1=95 mgml^-1, C2=5 mgml^-1, V2=25 ml and V1 is the unknown. From there V1=C2V2/C1=1.316 ml

2.

Relationship between absorbance and molar concentration is expressed by Lambert–Beer law: A= elC

Here, A= absorbance, l=path length (standard is 1 cm) and epsilon is an extinction coefficient of the absorbing molecule (chromophore).

What is a molar concentration of a solution which exhibits absorption of 0.86 while the extinction coefficient is 1.08 x 10³ M^-1cm^-1? Enter a number corresponding to concentration expressed with units as Micromolar (µM).

Answer:

796.3

C=A/el

C=0.86/(1.08 x 10³ M^-1cm^-1 x 1cm)=0.0007963 M = 796.3 µM

3.

The relationship between moles and molar concentration is simple:

C=n/V, where n=number of moles, V= volume.

However, in biochemistry, protein concentration is frequently expressed as weight per volume and in mgml^-1 units. That slightly changes the expression above and with the knowledge of the protein's molecular weight we could also express protein concentration in mgml^-1 as:

C=n x MW/V, since n=m/MW, where m=mass (g) and MW=molecular weight (in gmol^-1)

Question:

What is a number of nanomoles in 10 ml solution with the concentration of 13mgml^-1, and the MW of the macromolecule is 32500 gmol^-1?

Answer:

4000

n=CV/MW = 10ml x 0.013gml^-1/32500 gmol^-1

n=4x10^-6=4000x10^-9 mol = 4000 nanomoles

4.

Arrhenius equation, v = Ae-Ea/RT, describes the relationship between a reaction rate and the temperature at which the reaction takes place. Here, A is Arrhenius constant, R is a gas constant (R=8.314 JK^-1mol^-1) and T is temperature in Kelvins and Ea is activation energy.

From that equation we can see that

LnV=Ln(Ae^-Ea/RT ) = LnA + Ln(e^-Ea/RT ) = -Ea/RT +Constant =

LnV=-Ea/R * 1/T + Constant

Thus, if LnV is plotted against 1/T then the slope of the curve would be equivalent to -Ea/R and the activation energy could be calculated.

For a reaction with the activation energy of 76.3 kJmol^-1 what would be the slope value (in Kelvins) of the curve obtained from plotting LnV as a function of 1/T (rounded to one decimal point)?

Answer:

-9177.3

-Ea/R=-76300Jmol^-1/8.314 JK^-1mol^-1 = -9177.3 K

Note: There was a question during the Zoom session, as to why the slope is expressed in Kelvins, this should be obvious from the calculations above: units for activation energy are Jmol^-1 and units for gas constant are JK^-1mol^-1 thus Ea/R (slope) will be in Kelvins since

Jmol^-1/ JK^-1mol^-1 = K

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5.

When measuring a time course of a coloured reaction we will register a change of absorbance per unit of time which can be determined from the slope of the linear portion of the reaction. This would be expressed as deltaA/time.

But if you had determined that there is a change of absorbance of 0.1/min then this should be converted to the change in number of moles of substrate or product per unit of time. And ultimately, if we are interested in enzymatic activity this change of substrate/product concentration per unit of time should be also expressed as a function of a volume and/or amount of enzyme.

Question:

If deltaA=0.1min^-1, and extinction coefficient of the product is 4.14 x 10⁴ M^-1 cm^-1 what is the rate of reaction expressed in terms of micro moles product produced per minute/ml enzyme (rounded up to 3 decimal points)? Reaction was initiated by 0.04 ml of enzyme in the final reaction volume of 2ml.

Answer:

0.121

This is how you should have calculated this:

deltaA=0.1min-1 corresponds to deltaC=0.1/4.14 x 10⁴ M^-1min^-1 = 0.02415 x 10^-4 M min^-1

this corresponds to n=CV=0.02415 x 10^-4 molL^-1 * 2 x 10^-3 L= 0.0483 x 10^-7 mol min^-1

or as a function of enzyme volume the rate would be 0.0483 x 10^-7 mol min^-1 / 0.04 x 10^-3 L

= 0.121 micro moles of product per minute per ml of enzyme!

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If, you know that the enzyme is at C=32 µM what would be the rate of reaction expressed in terms of moles of product per mole of enzyme per unit of time (minute)? Note that this value would be equivalent to kcat.

This is how to calculate:

V= 0.121 micro moles of product per minute per ml of enzyme = 0.121 µmoles min^-1 ml^-1

C= 32 µmoles L^-1 = 32 x 10^-3 µmoles mL^-1,

V= 0.121 µmoles min^-1/32 x 10^-3 µmoles = 3.78 min^-1