An experiment to determine the amount of urea in a specimen of urine. Introduction. Metabolism produces a number of toxic by-products, particularly the nitrogenous wastes that result from the breakdown of proteins and nucleic acids. Amino (NH 2) groups are the result of such metabolic reactions and can be toxic if ammonia (NH 3) is formed from them. Ammonia tends to raise the pH of bodily fluids and interfere with membrane transport functions. To avoid this the amino groups are converted into urea, which is less toxic and can be transported and stored to be released by the excretory system.

Urea is the result of two amino groups being joined to a carbonyl (C = O) to form CO (NH 2) 2, the process of which is called the ornithine cycle and takes place in the liver. The ornithine cycle was developed by Hans Krebs in 1932 and is similar to the Krebs cycle through the use of. One of the steps in the cycle the breakdown of arginine into ornithine and urea, a reaction catalyzed by the enzyme. (See below) (Fig 1.

0) Arginine Orthinine Urea Urease is the enzyme which catalyses the hydrolysis of urea according to the following equation: (NH 2) 2 CO (aq) + 3 H 2 O (l)  CO 2 (g) + 2 NH 3 (g) The acidic ammonium carbonate is formed because the carbon dioxide dissolves in water to produce carbonic acid (H 2 CO 3), which immediately reacts with ammonia to form the ammonium carbonate. This is shown by the following equation: 2 NH 3 (g) + H 2 CO 3 (aq)  (NH 4+) 2 CO 3 (aq) The resulting solution can then be titrated against hydrochloric acid with methyl orange as the indicator in order to determine how much urea was present initially. The point of neutralization using a methyl orange indicator is determined using the following colour changes.  Acid  Red.  Neutral  Yellow.

 Alkali  Orange. Enzymes are nearly all made up of globular proteins. The structure of enzymes can be divided into three categories: 1. The primary structure, which is the sequence of amino acids. 2. The secondary structure, which is the coiling of the protein into an alpha helix 3.

The tertiary structure, which is the 3 D shape into which the protein is folded. This shape gives the enzyme its properties and specificity. The shape is held together by ionic bonds, disulphide bridges and the weaker hydrogen bonds. Method. Six urea solutions were prepared an placed in conical flasks one of which was of unknown concentration. The flasks were sealed to prevent CO 2 and NH 3 gases from escaping and then placed in a water bath at 35 oC for 1 hour.

The temperature was kept at 35 oC as each enzyme has an optimum temperature at which it works best, so it was important that the temperature remained constant for the duration of the reaction. After 1 hour all the flasks were removed. A burette was washed first with distilled water to remove any impurities. Then with HCl to prevent the acid from being neutralised by the remaining water, as this would increase the pH of the acid and give a less accurate title.

It will also remove any impurities not dealt with by the water. The burette was then carefully filled to the top with HCl. A 10 cm 3 puppet was used to place portions of the urea solution into a beaker, into which a few drops of methyl orange were placed to act as an indicator. The beaker was arranged on top of a white tile so that the end-point of the titration could be determined more accurately. At the start of the titration the solution was yellow and at the end-point it turned red. This process was repeated for each solution, and the volume needed to completely neutralise 10 cm 3.

Each time the procedure was repeated 3 times and the average title would be calculated. Results. (Fig 2. 0) Concentration of urea (g/100 cm 3) Volumes of 0. 1 M HCl Required (cm 3) Mean volume of 0.

1 M HCl required (cm 3) (1 dp) 0. 32 9. 2, 9. 3, 9. 3 9. 30.

48 10. 9, 11. 0, 10. 9 10.

91. 0 13. 1, 13. 5, 13. 6 13. 42.

0 15. 3, 15. 2, 15. 7 15. 44. 0 18.

8, 18. 4, 18. 5 18. 6 Unknown 11.

4, 11. 0, 11. 1 11. 2 Discussion. My average title for the unknown sample was 11. 2 cm 3.

When I applied this to the graph I found the concentration of urea to be 0. 58 g/100 cm 3. Figure 2. 2 clearly shows that as the concentration of urea increases, the volume of HCl required for neutralization also increases. This is to be expected as there are more moles of urea being hydrolyzed, which would mean more HCl would be required.

However the curve definitely indicates that the rate of increase decreases as the amount of urea present initially rises. This means that the concentration of urea is not directly proportional to the amount of ammonium carbonate produced, as the equations shown earlier suggested. This leads me to believe that it is the activity of the enzyme, which has caused this result. It may be due to availability of the active site on the enzyme, i.

e. because the Urease causes the release of CO 2, which dissolves in the water forming carbonic acid. This in turn reacts with ammonia forming ammonium carbonate, which is then titrated with HCl. So the availability of the active site would dictate how much ammonium carbonate would be produced, and hence the volume of HCl required to neutralise it. In which case I would expect to see the graph level off if solutions of greater concentration were used.

Temperature could not have been a factor as this was kept constant through the use of a water bath at 35 oC. The pH is a likely factor as throughout the reaction there are acids being produced. The precise 3-dimensional shape of the enzyme is partly the result of hydrogen bonding. However these bonds may be broken by the concentration of hydrogen ions (H+) which are present when acids are in solution.

The CO 2 being produced formed carbonic acid, which then went on to form ammonium carbonate. Both of which are relatively weak acids, however it need not be strong to affect the enzyme as pH is a logarithmic scale and a change of one pH point represents a tenfold change in the H+ concentration. By breaking the hydrogen bonds, which give shape to the enzyme, any pH change can effectively denature it. Every enzyme has an optimum pH at which it functions at its best and in this case the enzyme is likely to have been affected by the change in pH that occurred. The enzyme Urease has fallen victim to what could either be toxic accumulation, or a kind of feedback inhibition, which resulted from its own waste products. Limitations.

As described the fact that the pH changed during the course as a result of the products formed, could have affected the activity of the enzyme. If I were to repeat the experiment I would try using a buffer solution to overcome this problem. The time could have been a factor as the enzyme may not have had sufficient time to catalyst the reactions. Again if I were to repeat the experiment I would leave the reaction for 2 or more hours. Judging the end-point with the naked eye is not sufficiently accurate and human error is more than likely, especially when the colour change is so gradual. The best way to overcome this in future investigations would be to use a colorimeter.

This would give more accurate titration readings. Biological significance. When patients undergo pregnancy and other urine tests they are asked for an early morning sample. This is because no water has been drunk for a very long time and a great deal of reabsorption has occurred in the kidneys. This is due to an increase in ADH (anti diuretic hormone) secretion from the pituitary gland. Such reabsorption has caused the urine to be more concentrated with urea than normal.

During the day water will taken in so the reabsorption of water in the kidneys will be unnecessary. This will result in a decrease in the concentration of urea during the day. Urease can also be used by plants, which take in urea from the soil and obtain the nitrogen from its decomposition to CO 2 and NH 3. The plant then uses this to make nucleic acids or amino acids.