HBC1011 Biochemistry I Trimester 1, 2018/2019

1. FACULTY OF INFORMATION SCIENCE & TECHNOLOGY
(FIST)
HBC1011 - BIOCHEMISTRY I
Trimester 1, 2018/19
PRACTICAL 1: GENERAL LABORATORY PROCEDURES
NAME : MOHAMED REDA
ID : 1151303266

2. Exercise I:
Complete the following conversions:
a. 1 ml = 1000 µl
b. 0.006 µl = 6 x 10-6 ml
c. 0.0089 μg of DNA = 8900 pg of DNA
d. 8.95 x 10-12 g of DNA = 8.95 x 10-6 μg of DNA
e. The mass of green fluorescent protein (GFP) is 28000 Da which is equivalent to 28
kDa.
f. The mass of myoglobin is 0.00017 kDa which is equivalent to 0.17 Da.
Exercise II:
1. How many grams of alanine (MW=89.1) are present in 100 ml of the 1mM alanine
solution? How many milligrams of alanine are present in 100 ml?
1 mM of alanine = 1 x 10-3 M
Mass of alanine = 1 x 10-3 M x 89.1
= 0.0891 g/L
Mass of alanine in 100 ml = 0.0891 g/L (100/1000)
= 0.00891 g
Mass of alanine in milligrams= 0.00891g x 103
= 8.91 mg
2. A solution of alanine, concentration w/v= 5 g/L. How many grams of alanine would be
present in 2 liters of this solution? And in 10 ml?
Mass of alanine in 2 liters = ( 5 ) 2
= 10 g
Mass of alanine in 10 ml = ( 5 ) 0.01
= 0.05 g
3. Convert the concentration units of 0.1 M glucose to the units of mg of glucose in 1 ml
(mg/ml).
MW of glucose, C6H12O6 = 6(12) + 12(1) + 6(16)
= 180 g
Mass of glucose = 0.1M x 180 g x 0.001L
= 0.018 g
Concentration of glucose in mg/ml = (0.018 g x 103 )/ ml
= 18 mg/ml

3. 4. How would you prepare 150 ml of 60mM calcium chloride (CaCl2)? Given that the
molecular weight for CaCl2 is 110.98.
Mass of CaCl2 = 0.06 M x 110.98 g/mol
= 66.588 g/L
Mass of CaCl2needed = 66.588 g/L x 0.025 L
= 1.66 g
Excercise III:
1. Prepare a linear dilution table as below. Assume that the stock solution is of 2.0 M and
you require 20 ml diluted samples of final concentrations 1.66 M, 1.33 M, 1.0 M, 0.66
M, 0.33 M, and 0.00 M.
Stock
Concentration
(M1)
Final
Concentration
(M2)
Volume of the
Stock (ml)
Volume of Diluent
(ml)
Final Volume
(ml)
2.0 1.66 16.6 3.4 20
2.0 1.33 13.3 6.7 20
2.0 1.00 10.0 10.0 20
2.0 0.66 6.6 13.4 20
2.0 0.33 3.3 16.7 20
2.0 0.00 0 20.0 20
2. The following serial dilution was carried out:
1.0 ml of Solution A is diluted to a final volume of 10 ml Solution B
1.0 ml of Solution B is diluted to a final volume of 10 ml Solution C
1.0 ml of Solution C is diluted to a final volume of 10 ml Solution D
Assume that Solution A has a concentration of 2.0M. Calculate the concentrations of
Solution B, C and D
A : B = 1 : 10 A : C = 1 : 100 A : D = 1 : 1000

4. Concentration of solution B = 2.0 M x (1/10)
= 0.2 M
Concentration of solution C = 2.0 M x (1/100)
= 0.02 M
Concentration of solution D = 2.0 M x (1/1000)
= 0.002 M
3. How would you prepare a 600ml of 18mM of EDTA from a 50M EDTA stock solution?
Please show your workings and finally state how much of distilled water needed to
dilute the 50M EDTA.
M1V1 = M2V2
V2 = (0.6 L x 0.018 M) / 50
= 2.16 x 10-4 L
distilled water needed= 0.6 L - 2.16 x 10-4 L
= 0.599 L
= 599 ml
Excercise IV:
For each of the following questions, state how many unit volumes of material to be diluted and
diluent needed to prepare the following solutions?
a. 1:25 of a 10M sodium chloride stock solution to make a total volume of 500ml. What
is the final concentration of sodium chloride after dilution?
Final concentration of NaCl= 1/25 x 10
= 0.4 M
M1V1 = M2V2
V1 = (0.4M x 0.5L)/ 10M
= 0.02L
0.5L - 0.02L = 0.48L
Hence, to prepare 500 ml of 1X NaCl from 25X NaCl, we need 0.02L of 25X NaCl
diluted with 0.48 L of distilled water.
b. a solution of 150mM is known to have been diluted 50X. What is the actual
concentration of this solution?
The ratio is 1 : 50
actual concentration of this solution= 0.15M / 50
= 0.003 M
c. How would you prepare a 2X Buffer A from 40X Buffer A stock solution.

5. V1C1 = V2C2
V1 = (2X x 1L)/40X
= 0.05L
1L - 0.05L = 0.95L
Hence, to prepare a litre of 2X Buffer A from 40X Buffer A, we need 0.05L of 40X
Buffer A diluted with 0.95L of distilled water.
Excercise V:
1. What are the differences between positive displacement and air displacement pipettes?
List them in a table.
Positive displacement Air displacement
volume range 1 μL to 10 mL volume range 0.2 μL to 10 mL
piston is direct contact with the aspirated
liquid
piston is not contact with the aspirated
liquid
no need to discard tips after each pipetting
operation
tips are discard after each pipetting
operation
use for solution that oily, viscous and
volatile liquid
use for aqueous solution
2. List ways to increase pipetting accuracy.
- Use consistent force when installing the pipette tip
- Pre-wet the pipette tip to flush any residual substances from the pipette tip and
prevent cross-contamination
- Aspirate at 90 degrees, dispense at 45 degrees
- When aspirating, place the tip just below the surface to ensures no air is being
aspirated
- Release plunger slowly to avoid air bubbles
- Dispense against the side of the receptacle or into the liquid already in the
receptacle
- Touch off at the very end of your receptacle to pull the residual amount out of
the tip