Required practicals

• Describe a method a student could use to prepare colour standards and use them to give data for the total anthocyanin extracted

  -use a known concentration of pigment solution

  -prepare a dilution series

  -compare results with dilutions to give concentration
• REQUIRED PRACTICAL 1

  
  

  Investigation into the effect of a named variable on the rate of an enzyme-controlled reaction.

  
  

  METHOD

  (1) Make two control samples:

  Take two flat bottomed tubes.

  ● Add 5cm3 of milk suspension to each tube.

  ● Add 5cm3 of distilled water to one tube- this control will indicate the absence of enzyme activity.

  ● Add 5cm3 of hydrochloric acid to the other- this control indicates the colour of a completely hydrolysed sample.

  (2) Take three test tubes and measure 5cm3 milk into each. Place in water bath at 10°C for 5 minutes to equilibrate.

  (3) Add 5cm3 trypsin to each test tube simultaneously and start the timer immediately.

  (4) Record how long it takes for the milk samples to completely hydrolyse and become colourless.

  (5) Repeat steps 2-3 at temperatures of 20°C, 30°C, 40°C and 50°C.

  (6) Find the mean time for the milk to be hydrolysed at each temperature and use this to work out the rate of reaction.
• REQUIRED PRACTICAL 1

  
  

  Investigation into the effect of a named variable on the rate of an enzyme-controlled reaction.

  
  

  CONCLUSION

  Milk contains a protein called casein which, when broken down, causes the milk to turn colourless. Trypsin is a protease enzyme which hydrolyses the casein protein.

  
  

  As the temperature increases from 10°C, kinetic energy increases so more enzyme-substrate complexes form. This means that the rate of reaction increases up to the optimum temperature.

  
  

  At temperatures beyond the optimum, bonds in the enzyme tertiary structure break, which changes the shape of the active site. This means that the substrate and enzyme are no longer complementary.
• REQUIRED PRACTICAL 2

  
  

  Preparation of stained squashes of cells from plant root tips; set-up and use of an optical microscope to identify the stages of mitosis in these stained squashes and calculation of a mitotic index

  
  

  METHOD

  (1) Heat 1 mol dm-3 HCl at 60°C in a water bath.

  (2) Cut a small sample of the root tip using a scalpel.

  (3) Transfer root tip to HCl and incubate for 5 minutes.

  (4) Remove from HCl and wash sample in cold distilled water and remove the very tip using a scalpel.

  (5) Place tip on a microscope slide and add a few drops of stain (e.g. toluidine blue O). This makes the chromosomes visible and will therefore show which cells are undergoing mitosis.

  (6) Lower the cover slip down carefully onto the slide. Make sure there are no air bubbles in the slide which may distort the image, and that the coverslip doesn't slide sideways which could damage the chromosomes. (7) Place under a microscope and set the objective lens on the lowest magnification.

  (8) Use the coarse adjustment knob to move the lens down to just above the slide.

  (9) Use the fine adjustment knob to carefully re-adjust the focus until the image is clear (you can use a higher magnification if needed).

  (10) To calculate mitotic index, cells undergoing mitosis must be counted (cells with chromosomes visible), as well as the total number of cells.
• REQUIRED PRACTICAL 2

  
  

  Preparation of stained squashes of cells from plant root tips; set-up and use of an optical microscope to identify the stages of mitosis in these stained squashes and calculation of a mitotic index

  
  

  MITOTIC INDEX


  Plant cells undergo mitosis at shoot and root tips in areas called meristems. Cells in the meristems are totipotent and retain the ability to differentiate. The mitotic index of a sample is the ratio of cells undergoing mitosis to the total number of cells in a sample. To find the mitotic index, cells from the meristem must be viewed under an optical microscope.
• REQUIRED PRACTICAL 3

  
  

  Production of a dilution series of a solute to produce a calibration curve with which to identify the water potential of plant tissue

  
  

  METHOD

  (1) Make a series of dilutions of 1M sucrose solution. These should be at 0.0, 0.2, 0.4, 0.6, 0.8 and 1.0M sucrose. Dilute using distilled water.

  (2) Measure 5cm3 of each dilution into separate test tubes.

  (3) Use a cork borer to cut out six potato chips and cut down the sections into identically sized chips. Dry each chip using a paper towel to remove excess water but do not squeeze.

  (4) Weigh each before the start of the experiment.

  (5) Place a potato chip in each test tube (one per sucrose concentration) and leave for 20 minutes.

  (6) Remove each potato chip, dry gently using paper towel, and weigh them in turn.

  (7) Calculate the percentage change in mass for each sucrose solution.
• REQUIRED PRACTICAL 3

  
  

  Production of a dilution series of a solute to produce a calibration curve with which to identify the water potential of plant tissue

  
  

  CALLIBRATION CURVE and WATER POTENTIAL


  Calibration curves are graphs used to determine an unknown concentration of a sample by comparing the unknown to a set of standard samples with known concentrations - they are also known as standard curves.

  
  

  A dilution series can be used to create a set of samples with known concentrations. A calibration curve can be used to determine an unknown water potential in a potato sample.

  
  

  Water potential is the tendency of water to diffuse from one area to another. Water molecules move from areas of high water potential to areas of low water potential by osmosis. The water potential is determined by the concentration of solutes. The movement of water in and out of cells is related to the relative concentration of solutes either side of the cell membrane.
• REQUIRED PRACTICAL 3

  
  

  Production of a dilution series of a solute to produce a calibration curve with which to identify the water potential of plant tissue

  
  

  CONCLUSION

  Potato chips in lower concentrations of glucose solution will increase in mass, whilst those in the higher concentrations of glucose solution will decrease in mass.

  
  

  In the dilute glucose solutions, the solution has a higher water potential than the potato, so water passively moves via osmosis to the area of lower water potential (the potato). This causes the potato to increase in mass.

  
  

  In concentrated glucose solutions, water will move out of the potato, thus the potato will decrease in mass
• REQUIRED PRACTICAL 4

  
  

  Investigation into the effect of a named variable on the permeability of cell-surface membranes.

  
  

  METHOD

  (1) Cut beetroot into 6-10 identical cubes using a scalpel.

  (2) Wipe/rinse to clean off any pigment released as a result.

  (3) If investigating temperature: place each of the cubes of beetroot in an equal volume of distilled water (5-15ml).

  (4) Place each test tube in a water bath at a range of temperatures (30-80°C).

  (5) If investigating concentration of solvents: create a dilution series of ethanol using distilled water. Ethanol concentrations should range from 0-100% ethanol.

  (6) Leave the samples for 20 minutes - the pigment will leak out of the beetroot.

  (7) Set the colorimeter to a blue filter and zero using a cuvette with distilled water.

  (8) Filter each sample into a cuvette using filter paper.

  (9) Measure the absorbance for each solution. A higher absorbance indicates higher pigment concentration, and hence a more permeable membrane.
• REQUIRED PRACTICAL 4

  
  

  Investigation into the effect of a named variable on the permeability of cell-surface membranes.

  
  

  CONCLUSION

  As the temperature increases, the permeability of the cell-surface membrane also increases. This is because the proteins in the membrane denature as the heat damages the bonds in their tertiary structure. This creates gaps in the membrane, so it is easier for molecules to pass through it.

  
  

  At low temperatures, phospholipids have little energy and are packed closely together to make the membrane rigid. This causes a decrease in permeability and restricts molecules from crossing the membrane

  
  

  NB: At very low temperatures, ice crystals can form which pierce the cell membrane and increase the permeability.

  
  

  Ethanol causes the cell-surface membrane to rupture, releasing the betalain pigment from the cell. Higher concentrations of ethanol will cause more disruption to the membrane and more gaps will form. Thus, as concentration of ethanol increases, the permeability of the cell-surface membrane also increases.