An Investigation into the Effects of a Single Factor on the Rate of Transpiration of 2 Different Species of Plant
An Investigation into the Effects of a Single Factor on the Rate of Transpiration of 2 Different Species of Plant. The aim of this investigation is to determine and compare how variation of a named factor affects the transpiration rates of 2 different species of plant. The process of transpiration is the process in which water vapour is lost from within the plant and goes into the atmosphere. It occurs only when stomata are open on the leaves of the plant which allows the water vapour to pass through going to the lower water potential outside the leaf.(1) The actual rate of the transpiration can be effected by many different factors including light intensity, wind speed, temperature and humidity of air surrounding plant. In this investigation I am going to look into how one of these factors affects the rate of transpiration of 2 different species of plant. Factor (variant) = wind speed Plants = holly and privet To get a fair investigation then the other factors must be kept as constants i.e. they must remain the same throughout the experiment whilst taking results as if they are not the results may be inaccurate and inconsistent in respect to what they should be. I think that as wind speed increases the rate of transpiration will increase, this is because the wind intensity will be getting rid of the vapour lost from the plant quicker and therefore the water potential
Reaction of Catalase and Hydrogen Peroxide
PLANNING SECTION Aim: - To find the optimum temperature for the enzyme catalyst. Hypothesis: - I hypothesize that as the temperature increases the rate of enzyme catalysed reaction increases. Enzymes Enzymes are complex protein molecules produced by living organisms. They catalyse a vast range of chemical reactions without themselves being chemically changed at the end of the reaction. These biological catalysts are important because they speed up the rate of reaction they catalyse that would otherwise be too slow to support life. For example Catalase catalysis the decomposition of hydrogen peroxide into water and oxygen. 2H2O2 ? 2H2O + O2 Enzymes bind temporarily to one or more of the reactants of the reaction they catalyse. In doing so, they lower the amount of activation energy needed and thus speed up the reaction. Cofactors are non-proteins components required by enzymes for their efficient functioning. An enzyme-cofactor complex is called a holoenzyme. An enzyme without its cofactor is called apoenzyme. The Chemical Structure Of Enzymes All enzymes are globular proteins. Proteins consist of long chains of amino acids. In a globular protein the amino acid chain is folded and wound into spherical or globular shape. While the enzyme molecule is normally larger than the substrate molecule it acts upon, only a small part of the enzyme molecule actually comes into
How do Cytotoxic T cells recognise and kill virally infected cells?
How do Cytotoxic T cells recognise and kill virally infected cells? * Cytotoxic t- cells are developed from precursors in the thymus. * They develop from naïve CD8 t cells (called thus due to presence of CD8 molecule which is important in antigen binding). * They recognise antigens via t-cell receptors (TCRs) on the cell surface - a single cell may have up to 300000 identical TCR's - these are similar to immunogloulin molecules - each TCR has an alpha and a beta region T cells have different amino acid sequences in the variable region of the TCR receptors, this is what causes specificity. This specificity is crucial as it allows cytotoxic t cells to recognise and bind to the many different foreign antigens which might invade our bodies. Naïve cytotoxic t cells -T cells which have not yet met their specific antigens are called naïve t cells. -naïve CD8 t cells will later develop into cytotoxic t cells if they are activated. Naïve t cells are released from the thymus and circulate through the blood stream and through lymphoid organs through a process of migration. As they migrate they become temporarily bound to antigen presenting cells (APCs), usually dendritic cells. This binding is caused by adhesion molecules. All cells have major histocompatibility (MHC) complexes on their surfaces. Class 1 MCH molecules express a sample portion of the proteins
This experiment is to investigate the properties of some carbohydrates, including monosaccharides and disaccharides , physically, functionally and sensually.
By y.c. pong Introduction: Carbohydrates are important in human diet. This experiment is to investigate the properties of some carbohydrates, including monosaccharides and disaccharides , physically, functionally and sensually. Expt 1 relative solubility of carbohydrates, glucose, galactose and lactose Aim: To determine the relatives solubility of glucose, galactose and lactose by measure their saturation point at room temperature and pressure. Principle and procedure: By adding carbohydrates ( glucose , galactose and lactose)into a known volume of water, until no more sugar can be dissolve by water. The solution is considered to be saturated. Solubility means the amount of substance that can be dissolve in a given amount of solvent.so the saturation point can represent the solubility, expressed in mg/ml . By do the test of glucose, galactose and lactose, their relative solubility can be compared. Data: st Mass of beaker (g) Mass of beaker + water(g) Final mass (g) Saturation point (mg/ml) Glucose 36.84 45.35 50.70 628.7 Galactose 36.60 46.93 48.70 71.4 lactose 38.99 48.34 48.78 47.1 2nd Mass of beaker (g) Mass of beaker + water(g) Final mass (g) Saturation point (mg/ml) Glucose 36.84 45.17 50.58 649.5 Galactose 36.60 46.05 48.01 207.4 lactose 39.06 50.62 51.01 33.7 Table of experimental average saturation point st
Enzymes uses in industry and medicine bio cw
The Uses of enzymes in industry and medicine Enzymes are Biological catalysts which allow the chemical reactions of metabolism to take place so therefore controlling the speed of the reaction. They are found in all living cells and are divided into two main groups, intracellular and extra cellular. Intracellular are found and work inside the cells, therefore are secreted inside the cell membrane, from where they control metabolism. Enzymes are complex globular proteins. Their long peptide chains of amino acids linked by peptide bonds are wound, folded and bonded into a precise 3D structure, owing their activity to this particular shape. They are compounds of high molecular weight. Enzymes are truly amazing molecules as they are used within industry and medicinal purposes as they are thermostable as they can work at high and low temperatures in contrast to catalysts which can only work at high temperatures as this is very useful for the industrial processes to produce products with less heat so less energy and to make it cost efficient in industrial terms. (1) Enzymes can also work at many different ranges of pH as well as temperature changes and this is due to when they are immobilized in the industrial and medicine process. (3) This essay will now look at particular enzymes in industry and medicine within their main uses. An enzyme that is very important in the
Statistical Technique.
Statistical Technique The Mann-Whitney U Test will be used as differences between two treatments (two carbohydrates, sucrose and glucose) are being investigated and measurements have been taken at least six times with each treatment. Random sampling will therefore be useful and the data summarised using the median. Results Table: This table shows the volume of gas collected in five minutes in cm from each carbohydrate . Starch Sucrose Lactose Glucose 0 4 0 4 0 3 0 6 0 3 0 6 0 3 0 4 0 2 0 5 0 3 0 4 As neither starch nor lactose produced any result they shall not be investigated in the statistical test. Null Hypothesis: there is no difference between the volume of gas produced by sucrose or by glucose. Data set A: Rankings 7.5 3.5 3.5 3.5 3.5 Data set A: Observations 4 3 3 3 2 3 Data set B: Rankings 7.5 1.5 1.5 7.5 0 7.5 Data set B: Observations 4 6 6 4 5 4 Ra = 22.5 Na = 6 Nb = 6 Rb = 55.5 Ra + Rb = [(Na + Nb)(Na + Nb + 1)] = Ra + Rb = 78 2 Ua = NaNb + Nb(Nb + 1) - Rb = Ua = 36 + 21 - 55.5 = 1.5 2 Ub = NbNa + Na(Na + 1) - Ra = Ub = 36 + 21 - 22.5 = 34.5 2 Na = 6 Nb = 6 CV = 5 U CV 1.5 < 5 As the smallest U value is less than the critical value the null hypothesis can be rejected. Therefore it has been statistically
Sam and Sarah planned and carried out an experiment to see the effect of pH on the activity of the enzyme amylase which breaks down starch into sugar.
Carly Mckenzie 17/10/01 Investigation on Enzymes Introduction Sam and Sarah planned and carried out an experiment to see the effect of pH on the activity of the enzyme amylase which breaks down starch into sugar. They decided to investigate 5 different pH's 2,4,6,8 and 10. . The used a measuring cylinder to measure out 5cm3 of starch and placed it in a test tube. They then added acid or alkali until they got to pH2. They checked this using indicator paper 2. They placed the test tube in a rack on the bench and added 5cm3 of amylase which they had also measured using a measuring cylinder. 3. Immediately, they took out a drop of the mixture and placed it on a spotting tile which had 2 drops of Iodine in it. They repeated this every 30 seconds until they knew all of the starch had been broken down, so they stopped. 4. They then repeated the experiment at the different pHs'. 5. They did the experiment 3 times for each pH Aim Sam and Sarah planned to discover the effects of pH on the activity of the enzyme amylase which breaks down starch into sugar. Prediction I predict that the pH that the enzyme will work best at is 8 because amylase is mostly found in the mouth and salivary glands. This area is normally slightly alkaline, so the pH that this enzyme would most likely
Technical documentation - Design specification.
Technical Documentation Design Specification The system is required to calculate the weekly income including bonuses and tips for each driver, and also the weekly Income for the Boss. The system should be able to tell automatically which drivers are to be taxed and which ones shouldn't. It should be able to compare which drivers earned well and which drivers didn't earn so well. Printouts of work and income details of each driver should be available from the simple click of a button. Navigating around the necessary spreadsheets in the workbook should be made simple with the use of fully working buttons that run macros that replace tab clicking. Hardware Required Minimum Specification for PC: Pentium 200 Mhz 64 MB Memory SVGA Monitor 5 MB of free HD space Software Required Windows 95 Operating System (or above) Microsoft Excel Opening and configuring spreadsheet To open the spreadsheet, open the file harveys.xls To configure the lay out of the spreadsheet, click on the ON or OFF buttons on the Calculations worksheet. OFF Button runs GridOff macro The OFF button switches off the Standard and Formatting Toolbars, sheet tabs, column and row headings. The Off button also changes the font colour of the hidden cells (i.e. cell link values) white, on the Calculations worksheet ON Button runs GridOn macro The ON button switches on the Standard and Formatting
Lipids - what are they?
Lipids Lipids are fats, oils and waxes, organic compounds containing carbon, hydrogen and oxygen. The same three elements are involved in the structure of carbohydrates, but the amount of oxygen in the molecule present is much less than in carbohydrates. Lipids are insoluble in water but soluble in organic solvents such as acetone and ether. They are relatively small molecules compared to the polysaccharides, but because they are insoluble they tend to join together to form globules. Lipids can be split into the following They are glycerol, fatty acids, oils, fats, waxes, phospholipids and triglycerides (ester). The properties of the fats are, * Fats are insoluble in water but soluble in organic solvents. * Behave as water hating molecules ( hydrophobic molecules) * They are relatively small in size. * Fats are solids at room temperature Oils * Insoluble in water but soluble in organic solvents * They are relatively small in size * Behave as water hating molecules ( hydrophobic molecules) * Oils are liquids at room temperature Wax * Insoluble in water but soluble in organic solvents * Behave as water hating molecules ( hydrophobic molecules) * They are relatively large molecules * Wax is a solid at room temperature. Naturally occurring fats, oils and wax are esters formed by condensation reaction between glycerol (an alcohol) and organic acids known as
permeability of beetroot membranes
An investigation into how the concentration of a solution affects the rate of osmosis in a potato Definitions of important terms: Osmosis: Osmosis is the movement of water through a partially permeable membrane from a low concentration into a high concentration. This continues until both solutions are of equal concentration. Osmosis is a natural occurrence that can be simulated with an artificial membrane such as a visking tube. The rate of osmosis can be determined by the difference in water potential between the two substances on either side of the partially permeable membrane. Water potential (symbol??): Water molecules possess kinetic energy, which means that they are continually moving around when in a gas or liquid state. The higher the concentration of water molecules in a system, the greater the total kinetic energy of water molecules in that system and also the higher its water potential. Water potential is a measure of the energy available in an aqueous solution to cause the migration of water molecules across a semi-permeable cell membrane during osmosis. Water moves from areas of high (less negative) to areas of low (more negative) potential. Pure water is given the value zero. Aim of My Experiment: I am going to study whether or not the amount of water that may be taken into a potato cell by osmosis is affected by the concentration of a solution on one
