# Investigation in to the effect of bile salts on the digestion of fat.

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Introduction

By Julie-Anne Miskelly Variables Controlled variables are the variables, which should be kept constant to ensure a fair test. Controlled variables are: - The same temperature (37oC) The same amount of time taken (10mins) The same volume of lipase The same volume of milk The same volume of sodium carbonate The same variety of milk The same equipment used From my preliminary experiment I found that these controlled variables worked well and feel that I should use the same controlled variables in this experiment. Independent variable is the variable you change in the experiment. Independent variable is: - The concentration of bile salts. The concentrations of bile salts I will use will be 0%, 1%, 2%, 3%, 4% and 5%. I decided to pick these as I wanted a wide range of concentrations of bile salts to see what effect the concentrations will have on the rate of reaction. Dependant variable is the variable you measure in the experiment. Dependant variable is: - The pH of the mixture Prediction I predict that as the concentration of bile salts increases, also the rate of reaction will increase the activity of lipase in the solution. This will in turn make the amount of fats emulsified into smaller droplets increase. Therefore a larger surface area for the enzyme lipase to break down the triglycerides to fatty acids and glycerol. This is because the more surface area the more active sites for the substrate concentration to get to for the enzyme-substrate complex to take place. Therefore the reaction will take place quicker as there is more enzyme -substrate complex taking place. Preliminary Experiment I carried out a preliminary experiment, firstly to see what variables I will use and change and also to get to grips with the equipment that I will use in the actual experiment. In the preliminary experiment the method was:- * Collect apparatus and set up apparatus. ...read more.

Middle

Any difference has occurred only by chance. difference in mean xx - xy Now t = standard error of mean Sx2 + Sy2 nx ny I can then determine exact level of significance for nx + ny - 2 degrees of freedom. To see if the means of X and Y are significantly different or not. I will now carry out the statistical tests on 0% and 5% bile at 1 minute. 0% Bile at 1 min 5% Bile at 1 min X X2 9.07 82.26 8.98 80.64 9.02 81.36 9.08 82.45 9.40 88.36 9.27 85.93 8.94 79.92 9.22 85.01 9.39 88.17 9.52 90.63 ?x= 91.89 ?x2= 844.73 Mean of x = 91.89 ? 10 Mean of y = 87.13 ? 10 = 9.189 = 8.713 ?x2 - (?x)2 ?y2 - (?y)2 n n Sx = -------------- Sy = -------------- n-1 n-1 844.73 - (91.89)2 760.77 - (87.13)2 10 10 Sx = ------------------- Sy = -------------------- 10 -1 10 - 1 844.73 - 844.38 760.77 - 759.16 Sx = -------------------- Sy = --------------------- 9 9 Sx = 0.20 Sy = 0.42 difference in mean xx - xy Now t = standard error of mean Sx2 + Sy2 nx ny t = 9.189 - 8.713 (0.20)2 + (0.42)2 10 10 t = 0.476 0.147 t = 3.24 Is this significant? Degrees of freedom = nx + ny -2 = 10 +10 - 2 = 18 The value 3.24 falls between p= 0.01 and p = 0.02 (see t table) The level of rejection in biological investigations is normally taken at p =0.05. Therefore there is a significant difference between the two sets of results so the null hypothesis is rejected because p = 0.01. difference in mean xx - xy Now t = standard error of mean Sx2 + Sy2 nx ny t = 8.602 - 7.361 (0.44)2 + (0.15)2 10 10 t = 1.241 0.147 t = 8.44 Is this significant? ...read more.

Conclusion

= 569.36 - 568.82 + 509.56 -508.94 18 s = 0.54 + 0.62 18 s = 0.06 mean of X - mean of Y mean of X - mean of Y Now t = standard error of mean 1 1 nx ny (?x)2 (?y)2 ?x2 - -------- + ?y2 - ------- s = nx ny nx + ny - 2 (72.70)2 (70.34)2 s = 528.70 - --------- + 494.89 - -------- 10 10 10 + 10 - 2 s = 528.70 - 528.53 + 494.77 -494.77 18 s = 0.17 + 0.12 18 s = 0.16 mean of X - mean of Y mean of X - mean of Y Now t = standard error of mean 1 1 nx ny (?x)2 (?y)2 ?x2 - -------- + ?y2 - ------- s = nx ny nx + ny - 2 (71.31)2 (69.64)2 s = 508.63 - --------- + 485.08 - -------- 10 10 10 + 10 - 2 s = 508.63 - 508.51 + 485.08 -484.97 18 s = 0.12 + 0.11 18 s = 0.013 mean of X - mean of Y mean of X - mean of Y Now t = standard error of mean 1 1 nx ny (?x)2 (?y)2 ?x2 - -------- + ?y2 - ------- s = nx ny nx + ny - 2 (69.06)2 (68.57)2 s = 477.07 - --------- + 470.27 - -------- 10 10 10 + 10 - 2 s = 477.07 - 476.93 + 470.27 - 476.93 18 s = 0.14 + 6.66 18 s = 0.38 mean of X - mean of Y mean of X - mean of Y Now t = standard error of mean 1 1 nx ny Student's t values exceeded with probability p d.f. P = 0.1 0.05 0.02 0.01 0.02 0.001 1 2 3 4 6.314 2.920 2.353 2.132 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 40 60 120 ? ...read more.

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