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Using the streak plate method, compare the effect of two different brands of toothpaste on the growth of five different bacteria.

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Introduction

Biology experiment and investigation AIM: Using the streak plate method, compare the effect of two different brands of toothpaste on the growth of five different bacteria. HYPOTHESIS: The toothpaste containing higher levels of alcohol ingredients will have a greater effect of inhibition on the growth of bacteria than the toothpaste containing low levels of alcohol ingredients. This means that Colgate toothpaste will work better than the Sainsbury's own brand. METHOD: Wipe all work surface areas with 'Milton' using cotton wool, and wash hands with soap. Heat the inoculating loop in the Bunsen burner, on the blue flame, until the loop is red hot. Hold the loop upright to prevent any bacteria falling onto the hand. Leave the loop to cool for about ten seconds. Next, collect an agar plate with one bacteria already growing on the agar jelly. Remove the lid slightly from the plate containing the bacteria. Once the loop has cooled, insert it into the plate avoiding contact with the stem of the loop and the top of the plate. Wipe the loop over the slime until there is some in the loop. Using a prepared set agar plate, open the lid and swipe the loop over the set jelly without breaking it. First swipe the loop down the middle only a third of the way. Then swipe it across so the jelly is covered. Move the petri dish around whilst swiping the loop in a zigzag motion. Then repeat for last third in a wider zigzag motion until the loop ends up where the streaking started. Flame the loop again. Replace the lid and add cellotape at three different points around the petri dish. Label the Petri dish with its contents and put in the incubator at 30�C. Once the 5 types of bacteria incubated for 24 hours and allowed to grow, each one is needed to be transferred into sterile distilled water. ...read more.

Middle

The Mann Whitney U Test is used to test the difference between two sets of data. If the U value is lower than or equal to the critical value then it is positive to reject the null hypothesis and accept that there is a significant difference between the two sets of data. INTRODUCTION: I will test two different brands of toothpaste on five different gram-positive bacteria and analyse the results of growth. I have chosen to use gram-positive bacteria rather than gram-negative because of the effect that the toothpaste will have on the gram-positive. The experiment will work much better, this is because of the structure of a gram-positive bacteria. From a previous gram staining experiment I know that the two bacterial cells look very different following staining with the Gram stain. Gram-positive cells are purple and gram-negative are red. The basis for this differential reaction relates to the cell wall. The knowledge gained from gram staining helps me with the scientific background to apply to my experiment. Looking at the two different electron micrograph pictures below I can see that the gram-negative cell has an additional layer and from the external view of the cell the outside is convoluted. (Not really obvious in the below photo). The gram-positive wall is much thicker then the gram-negative and from the external view has a smoother appearance. The gram staining experiment helps us realise that there is a difference in structure of the two bacteria which can then lead to certain effects. In both gram-positive and negative cells they both have peptidoglycan in common. Most cell walls contain peptidoglycan; it is a thick rigid layer. It is composed of an overlapping lattice of two sugars that are cross linked by amino acid bridges. The two sugars are N-acetyl glucosamine (NAG) and N-acetyl muramic acid (NAM). A distinguishing factor among gram-positive bacteria is that roughly 90% of their cell wall is comprised of peptidoglycan and a gram-positive bacteria can have more than twenty layers of peptidoglycan stacked together to form the cell wall which causes it to be very thick. ...read more.

Conclusion

58 1 Sainsbury's 2. 253.5 2 Sainsbury's 3. 341 3 Sainsbury's 4. 370.5 4 Sainsbury's 5. 433 5 Sainsbury's 6. 439.5 6 Colgate 7. 569.5 7 Colgate 8. 624 8 Colgate 9. 673 9 Colgate 10. 699.5 10 Colgate Rank 1 1 2 3 4 5 Sainsbury's 58 253.5 341 370.5 433 Rank 2 6 7 8 9 10 Colgate 439.5 569.5 624 673 699.5 Sum of rank 1= 15 Sum of rank 2= 40 Formulae to calculate U1 and U2 R1 (Rank 1) R2 (Rank 2) n1 = number of average Sainsbury's values n2 = number of average Colgate values Sainsbury's toothpaste: U1 = n1 x n2 + n2 (n2+1) - R1 = 5 x 5 + 5 (5+1) - 15 = 25 + 5 (6) - 15 2 = 25 +15 -15 U1 = 25 Colgate toothpaste: U1 = n1 x n2 + n2 (n2+1) - R2 = 5 x 5 + 5 (5+1) - 40 = 25 + 5 (6) - 40 2 = 25 +15 - 40 U1 = 0 The smallest U value is taken U = 0 When the value of n1 is 5 and the value of n2 is 5, the critical value is 2. The smallest U value, 0 is less than the critical value of 2. The null hypothesis is rejected. "There is no significant difference between the effects on the toothpaste with a high alcohol content on bacterial growth compared to the toothpaste with a low alcohol content." In order to eliminate terms like certainly or probably, and state whether or not there is a significant difference, the critical value is used as a cut-off point along the scale of overlap. As the smallest U value is under the critical value it means there is no overlap and it is possible to be certain there is a significant difference. The null hypothesis is rejected at the p=0.005 or 5% level. P< 0.005 This means there is a 5% confidence level that there is only a 5% chance that the null hypothesis is incorrect. ...read more.

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