BCH2011 The Pre-Practical exercise

Students must come to the practical class with written answers to the following QUESTIONS. If NOT completed and submitted before class commences 50% of the total possible marks for the weeks exercise will be deducted. Demonstrators will check these during class and return to students. The Pre-Practical exercise must be submitted with the report’. Both the exercise below and the Assay of Glucose require the preparation of a standard curve. A standard curve, which is determined using a standard solution, relates data for two variables, e.g. absorbance and concentration. These data are connected with a curve of best fit. Consider the data below for a 2.0 mM solution of glucose, which has been diluted to cover a range of concentrations appropriate for a planned experiment.

Tube Number 1 2 3 4 5 6 Volume of 2.0 mM glucose (ml) - 0.2 0.4 0.6 0.8 1.0 Volume of water (ml) 1.0 0.8 0.6 0.4 0.2 - Glucose (µmoles/tube) - 0.4 0.8 1.2 1.6 2.0 Absorbance at 500 nm 0.05 0.30 0.76 1.01 1.23 1.35

In this graph, the linear region of the graph where the Beer-Lambert Law is obeyed lies between 0 and 1.2 µmol of glucose per tube. Continued next page Tube Number 1 2 3 4 5 6 Volume of 2.0 mM glucose (ml) - 0.2 0.4 0.6 0.8 1.0 Volume of water (ml) 1.0 0.8 0.6 0.4 0.2 - Glucose (µmoles/tube) - 0.4 0.8 1.2 1.6 2.0 Absorbance at 500 nm 0.05 0.30 0.76 1.01 1.23 1.35 Linear region Pre-Practical Exercise continued The axes of a standard curve need to be labelled, with the units specified, where appropriate.

Thus, in the current example, the x-axis is entitled Glucose and the units are µmoles/tube and the y-axis is entitled Absorbance at 500 nm; no units are specified for the latter as absorbance has no units. Each axis also needs to be marked into divisions (the scale) to allow the data to be plotted and read accurately. These divisions usually involve multiples 1, 2, 5 or 10, or 0.1, 0.2, 0.5 and 1.0, etc. and not 3, 7, etc. which do not allow data to be plotted accurately using the grid of graph paper or read accurately from the graph. Furthermore, the scale should be chosen so that the resulting curve has a gradient of 45° with respect to the x-axis and y-axis; this minimises errors due to scaling.

Exercise The haemoglobin (Hb) content of blood is normally within the following ranges: males 130-180 g per litre females 110-160 g per litre. It can be ascertained by treating a blood sample with a reagent containing an excess of ferricyanide and cyanide, which converts both Hb and oxygenated Hb into cyanomet-Hb. The latter can be determined conveniently from its absorbance at 540 nm by comparison with a standard curve. In order to construct a standard curve Tubes 1-5 (see below) were prepared using a standard solution containing 0.5g Hb per litre. At the same time a 0.01 ml sample of blood was diluted with water and treated with ferricyanide/cyanide reagent (Tube 6).

All tubes had a total volume of 5ml. The absorbance at 540 nm of Tubes 1-6 was measured and is recorded in the protocol below. Tube Number 1 2 3 4 5 6 (sample) Volume of standard Hb solution (ml) - 1.0 2.0 3.0 4.0 - Volume of cyanide reagent (ml) 1.0 1.0 1.0 1.0 1.0 1.0 Volume of water (ml) 4.0 3.0 2.0 1.0 3.99 Volume of blood (ml) - - - - 0.01 Absorbance at 540 nm 0.025 0.090 0.160 0.230 0.290 0.210 Amount of Hb (mg/tube) Determine the amount of Hb in Tubes 1-5 and enter these data in the protocol. Using the data for Tubes 1-5, construct a standard curve by plotting the absorbance at 540 nm versus the amount of Hb (mg/tube). Answer the following questions about the blood sample: The amount of Hb in Tube 6 = . . . . . . . . . mg The amount of Hb in the 0.01 ml blood sample = . . . . . . . . . mg The concentration of Hb in the blood sample = . . . . . . . . . g per litre Is the Hb content normal? . . . . . . . . .

Answer:

Solution

Tube number	1	2	3	4	5	6 sample
Volume of standard Hb solution (ml)	-	1.0	2.0	3.0	4.0	-
Volume of cyanide reagent (ml)	1.0	1.0	1.0	1.0	1.0	1.0
Volume of water (ml)	4.0	3.0	2.0	1.0		3.99
Volume of blood (ml)	-	-	-	-		0.01
Absorbance at 540 nm	0.025	0.090	0.160	0.230	0.290	0.210
Amount of Hb (mg/tube)	0.000 g	0.0005 g	0.001 g	0.003 g	0.012 g

Determine the amount of Hb in Tubes 1-5 and enter these data in the protocol.

Volume of standard Hb taken = 1.0 ml

Amount of Hb taken = volume of standard Hb taken * standard solution containing Hb

= 1.0 ml * (g/ml) = 0.0005g

Therefore, total mass of Hb in tube 2 = 0.0005g

Tube3

= 2.0 ml * (g/ml) = 0.001g

Therefore, total mass of Hb in tube 3 = 0.001g

Tube 4

= 3.0 ml * (g/ml) = 0.003g

Therefore, total mass of Hb in tube 4 = 0.003g

Tube 5

= 4.0 ml * (g/ml) = 0.012g

Therefore, total mass of Hb in tube 5 = 0.012g

Note: the addition of any amount of water does not affect the total amount of Hb in the tubes because it remains constant, however, it reduces Hb in final solution.

Also the volume of distilled water added to tube 2 = 3 ml

Total volume of tube 2 = 1.0 ml (Volume of standard Hb solution) + 1.0 ml [Volume of cyanide reagent (ml)] + 3.0 ml (Volume of water (ml)] = 5.00 ml

Total volume of tube 3 = 5.0 ml

Total volume of tube 4 = 5.0 ml

Total volume of tube 5 = 5.0 ml

Now, using C₁V₁ = C₂V₂ ………..(i)

C₁ = concentration

V₁ = volume of initial solution 1, standard Hb

C₂ = concentration

V₂ = volume of final solution 2, final solution

Putting the values in equation 1

Tube 2

1ml * (g/ml) = 5.0 ml * C₂

0.0005g = 5.0 * C₂

C₂ = 0.0001 g/ml

Tube 3

2ml * (g/ml) = 5.0 ml * C₂

0.001g = 5.0 * C₂

C₂ = 0.0002 g/ml

Tube 4

3ml * (g/ml) = 5.0 ml * C₂

0.003g = 5.0 * C₂

C₂ = 0.0006 g/ml

Tube 5

4ml * (g/ml) = 5.0 ml * C₂

0.012g = 5.0 * C₂

C₂ = 0.0024 g/ml

Amount of Hb

= C₂V₂

Tube 2

0.0001* 5= 0.0005 g

Tube 3

0.0002 * 5 = 0.001 g

Tube 4

0.0006 * 5 = 0.003 g

Tube 5

0.0024 * 5 = 0.012 g

From the graph

The amount of Hb in tube 6 = 1.6 mg/tube
The amount of Hb in the 0.01 ml blood sample

Note that 1.6 mg = 5.0 ml of solution

? = 0.01 ml of the blood in the solution

= 0.0032 mg

The concentration of Hb in the blood sample
=

= 0.32 g/l

Is the Hb content normal

No, it is not within the ranges given.

Buy BCH2011 The Pre-Practical exercise Answers Online

Talk to our expert to get the help with BCH2011 The Pre-Practical exercise Answers to complete your assessment on time and boost your grades now

The main aim/motive of the management assignment help services is to get connect with a greater number of students, and effectively help, and support them in getting completing their assignments the students also get find this a wonderful opportunity where they could effectively learn more about their topics, as the experts also have the best team members with them in which all the members effectively support each other to get complete their diploma assignments. They complete the assessments of the students in an appropriate manner and deliver them back to the students before the due date of the assignment so that the students could timely submit this, and can score higher marks.Â The experts of the assignment help services at urgenthomework.com are so much skilled, capable, talented, and experienced in their field of programming homework help writing assignments, so, for this, they can effectively write the best economics assignment help services.

Get Online Support for BCH2011 The Pre-Practical exercise Assignment Help Online

Not the Exact Question you were looking for ? Post your question for assignment help and get instant help on your homework and assignment questions from our experts