Haemoglobin carries more than 20 times its volume of oxygen. It combines so firmly with carbon monoxide that it can no longer combine with oxygen; this causes asphyxiation. After a life of perhaps 120 days, red blood cells are destroyed in the spleen or, in the course of circulation, their haemoglobin is broken into its constituents, including iron, which enters new blood cells formed in the bone marrow. When blood vessels rupture, as in an injury, the red cells are released and escape into tissue, where they are broken down. The haemoglobin is converted into bile pigments, the colour of which is responsible for the appearance of bruises.
Alterations in the structure of haemoglobin can lead to life-threatening illnesses. The most important of these conditions is sickle-cell anaemia, which involves a hereditary change in one of the amino acids that make up haemoglobin. The thalassaemias are a group of hereditary diseases of similar origin.
Hypoxia
Hypoxia is the term for a lack of oxygen. This can be caused by many things, including living and training at high altitudes. In this situation, the body tries to maximize the delivery of oxygen to the tissues, by way of aerobic respiration. This increases the heart rate. After a few days at altitude, water is absorbed from the circulation, concentrating to the Red Blood Cells. Finally, after a few more weeks, the kidneys secrete erythoprotein, which increases production of Red Blood Cells from the bone marrow.
Physical Training
Doing physical exercise or training increases the need for haemoglobin. The total blood volume is increased, probably because blood plasma increases. Thanks to this increase in volume, the Haemoglobin concentratiotn is usually unchanged.
Hypothesis
My null hypothesis is that there will be no significant difference between the results of mock blood A and mock blood B, and no significant difference between mock blood A and mock blood C.
My hypothesis is that Sample C will be quicker at passing through the copper sulphate solution the quickest, since the blood adapts after being at high altitudes for a long time so that it can pass through the body quicker.
I also predict that B will be quicker than A but not C at descending through the copper sulphate solution. This is because sample B was taken after exercise, which affects the density, but A was not so the density will be ordinary, and C was taken after exercise at higher altitudes, making it denser than B, meaning it will pass more quickly.
Method
- First, we mixed some copper sulphate solution. We did this by adding 24g of copper sulphate powder to a litre of water, and shaking this until the powder was completely dissolved, creating copper sulphate solution.
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We then filled a measuring cylinder with this solution until the liquid level was well above 100cm3. We then took a syringe full of Mock Blood Sample A and dropped a drop of blood just above the liquid level.
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As the drop of blood descended through the water, we timed how long it took for the drop to descend between 100cm3 and 10cm3 on a stopwatch.
- We then repeated this another 19 times with sample A in order to collect 20 results. We then did the same experiment 20 times for Sample B and 20 times for Sample C.
To make this a fair test, we used the same copper sulphate solution for each test and the different measuring cylinders for each mock blood sample. We also used a different syringe for each mock blood sample. These measures were taken to prevent cross-contamination and ensure the accuracy of our results.
Results
Analysis of Results
The null hypothesis is that there will be no significant difference between the results of A and B, and the results of A and C.
The Mean
Mean= the sum of the results, divided by the number of the results.
The mean of the tests on sample A was 10.18 seconds
10.03 + 7.60 + 6.47 + 10.29 + 8.84 + 8.89 + 14.16 + 11.78 + 10.84 + 9.94 + 10.64 + 10.95 + 10.70 + 11.07 + 9.80 + 10.26 + 10.31 + 11.09 + 6.50 + 12.58 = 203.6
203.6 divided by 20 = 10.18
The mean of the tests on sample B was 9.03 seconds
9.37 + 11.31 + 8.00 + 8.21 + 9.60 + 9.17 + 9.70 + 10.74 + 12.21 + 10.57 + 9.27 + 8.84 + 10.01 + 8.65 + 8.78 + 9.31 + 8.14 + 8.82 + 6.94 + 8.88 = 180.6
180.6 divided by 20 = 9.03
The mean of the tests on sample C was 7.46 seconds
7.11 + 6.07 + 8.40 + 7.89 + 6.80 + 6.80 + 7.40 + 6.49 + 7.95 + 7.24 + 6.14 + 7.11 + 9.74 + 8.46+ 6.75 + 6.05 + 7.41 + 6.78 + 8.63 + 10.02 = 149.2
149.2 divided by 20 = 7.46
The Standard Deviation
The Standard Deviation is the square root of the sum of the results squared minus the sum of the results squared divided by the number of results, divided by the number of results. [That makes no sense.]
The standard deviation for the Sample A test was 44.37359125
(1.) Sum of results squared = 203.6 x 203.6 = 41452.96
(2.) (1.) divided by number of results= 41452.96 divided by 20 = 2072.648
(3.) (1.) minus (2.) = 41452.96 – 2072.648 = 39380.312
(4.) (3.) divided by number of results = 39380.312 divided by 20 = 1969.0156
(5.) The square root of (4.) = 44.37359125
The standard deviation for the Sample B test was
(1.) Sum of results squared = 180.6 x 180.6 = 32616.6
(2.) (1.) divided by number of results= 32616.6 divided by 20 = 1630.818
(3.) (1.) minus (2.) = 32616.6 – 1630.818 = 30985.782
(4.) (3.) divided by number of results = 30985.782 divided by 20 = 1549.2891
(5.) The square root of (4.) = 39.3610099
The standard deviation for the Sample C test was
(1.) Sum of results squared = 149.2 x 149.2 = 22260.64
(2.) (1.) divided by number of results= 22260.64 divided by 20 = 1113.032
(3.) (1.) minus (2.) = 22260.64 – 1113.032 = 21147.608
(4.) (3.) divided by number of results = 21147.608 divided by 20 = 1057.3804
(5.) The square root of (4.) = 32.51738612
The t-Test
To compare sample A (x1) with sample B (x2)
To calculate the value of t
(1.) Mean 1 minus Mean 2 = 10.18 – 7.463 = 2.717
(2). S1 squared divided by number of results = 2.3732 divided by 20 = 0.11866
(3). S2 squared divided by number of results = 49.43911 divided by 20 = 2.4719555
(4.) (2) plus (3) = 0.11866 + 2.4719555 = 2.59060155
(5.) (1) divided by the square root of (4) = 1.688065632
T = 1.688065632 degrees of freedom = number of results in test A + number of results in test B – 2 = 20 + 20 – 2 = 38
Critical value= 2.024
2.024 > 1.688065632
The critical value is greater than the t value so there is a significant difference between the two sets of results.
To compare sample A (x1) with sample C (x2)
To calculate the value of t
(1.) Mean 1 minus Mean 2 = 10.18 – 9.179 = 1.001
(2). S1 squared divided by number of results = 2.3732 divided by 20 = 0.11866
(3). S2 squared divided by number of results = 2.177179 divided by 20 = 0.10885895
(4.) (2) plus (3) = 0.11866 + 0.1088595 = 0.2275195
(5.) (1) divided by the square root of (4) = 2.098576306
T = 2.098576306 degrees of freedom = number of results in test A + number of results in test C– 2 = 20 + 20 – 2 = 38
Critical value= 2.024
2.024 < 2.098576306
The critical value is lower than the t value so there is not a significant difference between the two sets of results.
Conclusion
The t-test clearly shows that there is a significant difference between the results of mock blood A and mock blood C. This shows that mock blood C is much denser than mock blood A, making it better for carrying oxygen.
At higher altitudes, there is less oxygen. The body, therefore, creates more and more red blood cells, and these carry more haemoglobin, allowing for more oxygen to be carried around the body. The presence of more red blood cells makes the blood denser, or heavier, making it travel through the body faster, allowing for faster and, therefore, more efficient delivery of the blood around the body. Mock blood C is denser since it represents blood taken from a normal, healthy adult after he has spent three months undergoing aerobic training at high altitude. In these conditions, the number of red blood cells would increase dramatically, thus making the blood denser.
The t-test also shows that there isn’t a significant difference between mock blood A and mock blood B, although B is slightly denser than A, since it needs to carry more oxygen.
Exercise creates a larger need for oxygen, since oxygen is being used up much quicker by aerobic respiration. Like at higher altitudes, more red blood cells are created, but not on a scale even close to that of red blood cell production at high altitudes. This will make the blood slightly denser, in order to provide a faster and more efficient delivery of oxygen to the areas that need it, such as the limbs. This is shown by mock blood B, which, though it didn’t become as dense as mock blood C, it did become denser than A.
Evaluation
There were a few faults, mostly unavoidable, with the implementation of the experiment. Firstly, the drops of mock blood dispersed into random shapes as they entered the copper sulphate solution. The different shapes descended through the solution at different speeds. There is no way of controlling this, so the results can never be entirely accurate. Also, there was no way of measuring the distance between the needle and the liquid level before we dropped the blood drop into the solution so there is no way of accurately making sure that the experiment was carried out exactly the same way each time. Finally, though I used the same solution and measuring cylinder for each blood sample, the later results in each experiment could have been affected by the piles of previous blood drops lying at the bottom of the measuring cylinder.
These three faults provided the largest margin of error, but were unavoidable. Aside from these, the implementation was as accurate and fair as I could make it. If I were to do the experiment again, I would, time permitting, find ways of minimalizing these margins of error, e.g. finding some way to measure the distance between the syringe needle and the liquid level.