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University Degree: Microbiology

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  1. Dermatology and Microbiology - The Growth of Nails.

    The nails will grow when the top layer of cells are transformed into nail cells by keratinisation. The cells within the matrix will be divided up, the upper ones will become thickened and toughened through the keratinisation process. As more cells are produced the old ones are pushed outwards and flattened, they then become transparent and form part of the nail plate. The matrix also determines the shape and thickness of a nail so the longer the matrix is the thicker the nail will be. If however the matrix is damaged it can result in temporary loss of the nail or permanent damage to the nail plate.

    • Word count: 900
  2. Mechanism of Insulin Secretion From Pancreatic Beta-cells

    An increase in cAMP concentration, without glucose, is not sufficient to stimulate insulin secretion. Glucose therefore leads to an increased intracellular concentration of cAMP that is in turn thought to promote insulin secretion by depolarising the cell, that is, by making the resting potential become more positive. IF 15mM of glucose is added to a beta-cell within an isolated Islet of Langerhans, within about a minute its membrane potential is found to change from its resting potential of about -60mV to -30mV. This depolarization is a consequence of the decrease in the membrane's permeability to potassium ions that is observed in the presence of glucose.

    • Word count: 939
  3. Describe the Concept and Discuss the Importance of Homeostasis

    This explains the reason why more lizards are caught by kookaburras on cold cloudy days than on sunny hot days. The regulatory processes that are the basis of homeostasis depend solely on the use of negative feedback systems, which work by analysing sensory information about a particular variable (e.g. pH, salinity, temperature etc), and using this information to respond to the stimulus and bring the variable back to the equilibrium state. Control systems are said to consist of receptors, which measure the factor being controlled, an amplifier, which makes the signal larger, and a negative feedback system which applies the amplified error signal back to reduce the error.

    • Word count: 2390
  4. Qualification and quantification of microorganisms in soil role of microorganisms in the nitrogen cycle.

    There may be hundreds of millions to billions of microbes in a single gram (Table 1). The most numerous microbes in soil are the bacteria followed in decreasing numerical order by the actinomycetes, the fungi, adapted for exploiting the three-dimensional pore network of the soil, soil algae and cyanobacteria (photosynthetic microbes which can add small amounts of carbon to soil) and soil protozoa (unicellular soil organisms that decompose organic materials as well as consume large numbers of bacteria) (Sylvia et al., 1998). Microbial Group No./Gram of soil Bacteria 100,000,000 - 1,000,000,000 Fungi 100,000 - 1,000,000 Algae and Cyanobacteria 1000 - 1,000,000 Protozoa 1000 - 100,000 Table 1: Numbers of Microbes in Soil Source: Sylvia, Fuhrmann, Hartel and Zuberer, 1998.

    • Word count: 5376
  5. "So many peptides, so few grooves" - compare the ways in which specific antigen recognition is accomplished by MHC molecues, by T cell receptors and by antibody molecules.

    One way in which the molecular interaction between peptide and MHC has been studied is by introducing mutations in an immunogenic peptide. With this technique, it is possible to identify certain residues that make contact with the MHC molecule and others that are essential for TCR recognition. Together with the evidence presented above, this has lead to the idea that MHC molecules only recognize certain anchor residues of the bound peptide. Typically, a MHC I molecule has been found to have six pockets (A-F)

    • Word count: 1570
  6. The theory behind this experiment is that of rate laws and rate expressions. Arsenious acid, the reducing agent present in this experiment, reacts with iodine at the rate it is formed.

    however the most frequently type of rate law for this reaction is of the form; d [IO3-] =k([IO3-] m,[I-]n,[H+}p,.) dt (Equation 3) where m, n, p,... are determined by experiment, and each exponent in the equation is the order of the reaction with respect to the corresponding species, and the algebraic sum of the exponents is the overall order of the reaction. The equation for the reaction of arsenious acid and I3- : H3AsO3 + I3- + H2O?H AsO42- + 3I- +4H+ (Equation 4) While the equation for the overall reaction, up to the time of the starch end point, can be written as: IO3- + 3H3AsO3 ?

    • Word count: 994
  7. An investigation to show the effects of temperature on the rate of diffusion

    If the concentration is greater between the two regions, then the rate of diffusion is also increased. 2} The distance over which diffusion takes place, the shorter the distance between the two regions of different concentration the greater the rate of diffusion. The rate can then be said directly proportional to the diameter of the organism. 3} the area over which diffusion takes place. The larger the surface area then the more molecules or ions can travel at any one time, therefore the faster the rate of diffusion. 4} Variations in the cells membrane may affect diffusion as well.

    • Word count: 1460
  8. Searle's solution to the mind-body problem.

    Let us keep these two notes in mind (one, rejection of Cartesian vocabulary, and two, scientific world-view) as we consider the two theses put forth by BN. According to Searle, the two theses of BN work within a micro-macro schema that is compatible with the current scientific world-view. Using the micro-macro model, we can formulate the theses (combined) in another way by saying that minds and mental states are the higher-level phenomenon of the lower-level phenomenon that is physical brain processes.

    • Word count: 974
  9. Measure cellular respiration in living and dormant pea seeds at two different temperatures.

    Each step in cellular respiration insures that the energy in the glucose bonds is conserved. The amount of cellular respiration reflects the amount of activity in the cell. In order to produce energy and continue living, cells perform cellular respiration. Thus, cells that perform cellular respiration more are also more active. There are three ways that the amount of cellular respiration can be measured: the consumption of oxygen, the production of carbon dioxide, and the release of energy. Hypothesis: If a respirometer only contains germinating peas and is placed in a temperature of twenty five degrees Celsius, then this respirometer will show the highest amount of oxygen consumption.

    • Word count: 2043
  10. Bacterial Resistance to Antibiotics.

    aureus strains are resistant to penicillin, ampicillin, and the antipseudomonas penicillins (Neu 1065). Staphylococcus aureus can cause deadly infections of the skin, heart valves, blood, and bones, and in 1997, three Americans were reported to be infected with strains of it that were resistant to our last line of defense- the antibiotic vancomycin. Methicillin, a semisynthetic penicillin, was created to respond to this challenge. However, in the 1980s, methicillin-resitant S. aureus (MRSA) became a problem as well. Now, MRSA is resistant to all penicillins, cephalosporins, carbapenems, and penems due to a gene that produces a protein capable of bonding to and deactivating the penicillin (Neu 1065). Antibiotics are mostly derived from natural products.

    • Word count: 1891
  11. Investigation into the effect of Temperature on the action of the Enzyme Lipase.

    This method shows that enzymes with their precise structure fit perfectly into their substrate to break it down. If their structure changes just slightly then they will be unable to fit precisely into their substrate's shape and they will be useless. Background Information and secondary sources used: Biological Studies by Henry Wilkinson, published by Nelson, page number 46. Encarta '97 Prediction In the experiment I shall be testing six different temperatures: 20?, 30?C, 40?C, 50?C, 60?C and 70?C. This leaves a range of 50?C and I feel that this is a wide enough variety to obtain a valid conclusion.

    • Word count: 3024
  12. An antigen is anything which having invaded a host, causes the host to generate an immune response against itself. In this case the virus causing smallpox in humans and cowpox in cows.

    It is these biochemical clusters which trigger the immune response. Immunological memory is the term used to describe how the adaptive immune system appears to remember all pathogens it has encountered previously, and changes behaviour as a consequence of earlier experience. It relates only to epitopes that have been encountered previously. Small lymphocytes are T and B cells found circulating in the blood and lymphatic system and stationary in the lymph nodes, spleen and other lymphoid organs. Each has receptor molecules on its surface with a unique structure and specificity that binds to just one epitope shape.

    • Word count: 921
  13. The aim is to compare brewing techniques in school and industry.

    The wort is sent from the whirlpool via a wort cooler or paraflow to a fermentation vessel. As the wort leaves the paraflow, yeast is added, as well as oxygen to allow the yeast to multiply before fermentation begins. Fermentation. In brewing, fermentation is the conversion of sugar into carbon dioxide gas (CO2) and ethyl alcohol. Enzymes within a yeast cell carry out this process. It is in fact a complex series of conversions that bring about the conversion of sugar to CO2 and alcohol.

    • Word count: 1294
  14. Experiment (polygenic traits), fingerprint patterns will be examined

    Looking at the general population, the fingerprint ridge patterns tend to be: arch, 5.0%; radial loop 5.4%; ulnar loop, 63.5%; and whorl, 26.1%. (Also see, (Bio molecules, cells and genetics, unit handbook: 61BL0020), Polygenic traits: fingerprint ridge count Page: 20-22). Materials and Procedure (See, (Bio molecules, cells and genetics, unit handbook: 61BL0020), Polygenic traits: fingerprint ridge count Page: 22-23). Results Table of class results No. Of students Loops Whorls Arches Total Ridge count Gender Radial Ulnar (TRC) (M/F) 1 1 6 6 0 185 M 2 3 3 3 0 138 F 3 4 0 0 6 18 M 4

    • Word count: 691
  15. AS Chemistry - CourseworkOpen Book Paper 1. There are two types of rubbers, natural and synthetic rubber. Natural rubber is predominantly

    Below is the reaction of a synthetic rubber being polymerised. H H H H H2C CH2 CH2 H2C Butadiene Poly(cis-1,3-butadiene) There are many similarities between natural and synthetic rubber. Firstly it is evident that both contain cis molecules 2 .Discuss how the structures of natural and vulcanised rubber determine their properties and describe how vulcanising rubber leads to an improvement in its properties for use in car tires 3.Natural rubber has a thermoplastic nature and consequently it is not suitable for being used independently to construct a tyre. Therefore it is necessary that a tyre must be have specific chemicals and synthetic fibres added in order to improve the suitability of rubber being used as car tyres.

    • Word count: 849
  16. Should we think of Fleming as a hero? IntroductionPenicillin has been such an important discovery because it has saved so many lives

    dish flew in through the open window, when Fleming found the dish with the penicillin mould in it he was showing someone the work he had been doing recently and it was in the sink ready for him to wash it up. When Fleming found the dish he noticed that there was no germs growing around the structure of the mould, Fleming's individual skill was very important in the observation of the mould because he managed to notice that there were no germs when he was just skimming through his work showing it to one of his friends, when he managed to notice that there were no germs near the mould.

    • Word count: 601
  17. The Building Blocks of Biology

    Then you should describe the experiment you would do to test your hypothesis. Be sure to include controls, and describe the variables that would remain constant and the one that you want to test. (Remember, you can only test one variable at a time.) Include the type of data you would collect. The next section would be your expected results. Then you need to state what your conclusion would be depending on those results. Finally, do you accept or reject your hypothesis? Introduction: During summer last year, I had a rash, more like pimples only on my right cheek.

    • Word count: 623
  18. With reference to membrane proteins, discuss the important role of water in defining the shape and function of biomolecules.

    Due to the polar nature and the ability to form hydrogen bond make water a very versatile solvent. This is very important role of waters in biomolecules since most of the chemical reactions that occur in organism require material dissolved in water. The high H-bonding capacity allow it to act a non-electrolytes and dissolve polar compound such as glucose in the body. Water can also act as electrolytes due to the high capacity of hydration shells and high dielectric constant. This allows inorganic ionic material such as NaCl to become hydrated and dissolve in water. It can also act as polyelectrolytes where particles are in a colloidal suspension by hydrated surface ionic charge and an example of this is globular proteins.

    • Word count: 903
  19. Bounded Buffer Module

    : X size < maxsize buffer' = buffer ? <x?> The new value is appended to the end of sequence buffer. Extracting an item is possible only if the buffer is not empty; the value obtained is the one at the head of buffer: ===BufferOut0 [X]============= UpdateBuffer[X] x! : X buffer = <> buffer' = tail buffer x! = head buffer These schemas represent partial operations: they describe the effect of a succesful insertion and a succesful extraction, respectively. When the operations are not possible, some messages must be produced.

    • Word count: 1507
  20. Cellular Respiration - Fermentation of Corn and Malt Extracts

    This glucose is transformed into sucrose and then taken to other parts of the plant cell. The energy that glucose stores cannot be accessed readily by the cell. The process of cellular respiration transfers the energy stored in glucose bonds to bonds in ATP (adenosine triphosphate) so that it can be used more easily by the cell. As many as 38 ATP molecules can be generated by glucose through cellular respiration (Kenyon 141). Our goal was to consider one method by which cells make ATP using energy that comes from breaking down the chemical bonds in glucose.

    • Word count: 1217
  21. Determine the amount of gas evolved by the fermentation of different sugars by yeast.

    and ethanol. The rates of the chemical reactions of fermentation depend on the action of enzymes. The enzymes present in yeast cells are specific and will only catalyse the fermentation of certain sugars and not others. Yeast fermentation gives the following during reaction: C6H12O6-------> 2CH3CH2OH + 2CO2 + energy Enzymes Glucose ---------------> carbon dioxide + ethanol + energy (ATP) (a gas) Material: 1. 5 - fermentation tubes, labeled 1 through five. 2. 5 large test-tubs 3. constant-temperature water bath (37oC)

    • Word count: 1182
  22. Find out the effect of temperature on enzyme activity.

    1 cm of amylase solution was placed in test-tubes 4, 5, and 6 5. 3 beakers were labeled A, B and C. 6. Beaker A's temperature was about 10oC using ice to water. Beaker B's temperature was 20oC using cold water and beaker C's temperature was about 35oC by heating. 7. Test tubes 1 and 4 were placed in beaker A, test tubes 2 and 5 were placed in beaker B and test tubes 3 and 6 were placed in beaker C.

    • Word count: 1416
  23. Bacterial leaching - The process involved and discusses the advantages and disadvantages of this method of extracting metals from low grade ore.

    Solvent extraction techniques are used to recover the copper ions from the solution by a process of ligand exchange solvent extraction. (Article 2) The ligand, dissolved in an organic solvent such as kerosene is immiscible in water. Cu2+(aq) + 2LH (organic) = CuL2(organic) +2H+(aq) (L represents the ligand) The copper is removed from a low concentration in water to a high concentration in the solution. Mixing the organic solution with a small volume of concentrated acid reverses the process and pushes the Cu2+ ions back into the aqueous solution increasing their concentration.

    • Word count: 1067
  24. Buffer solution making and investigation.

    x x = 0.398/1 100 x 0.398 = 28.46 cm3 salt 1.398 Therefore: 100 - 28.46 = 71.54 acid For the making of the buffer pH 8.8: Equations for the dissociation of ammonia and ammonium chloride respectively: NH3 + H2O - NH2- + H3+O NH2Cl � NH2- + Cl- Using the equation previously stated: 8.8 = 9.3 - log [HA] [A-] 5.2 - 4.8 = - log [HA] [A-] 0.5 = - log [HA] [A-] 10-0.5 = [HA] [A-] 0.316 = [HA] : [A-] CH3COOH + H2O - CH3COO- + H3+O (100 - x)

    • Word count: 1786
  25. The Effect of pH on the Rate of Catalase

    This can lead an enzyme becoming inactive. This is known as denaturing. When an enzyme is denatured the tertiary structure is altered, this affects the structure of the active site. If the active site's structure is changed then, following the 'lock & key' theory, it no longer compliments the structure of the substrate and it can no longer bind. No ES complexes are formed and the reaction is not catalysed. At each pH the amount of oxygen evolved increased at a fairly constant rate, producing a reasonably steady gradient, this suggests the rate of catalysation altered very little during the five minutes.

    • Word count: 1427

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