Discovering the Lowest Concentration Of Lead Ions Needed to Cause the Loss of Partial Permeability of Red Onion Cell Membranes

Authors Avatar

Geoff Cunningham

Student Number 0919

Centre Number 22133

Discovering the Lowest Concentration Of Lead Ions Needed to Cause the Loss of Partial Permeability of Red Onion Cell Membranes

Aim:

                

To discover the lowest concentration of lead ions needed to cause the loss of partial permeability of red onion cell membranes.

Prediction:

        I predict the lowest concentration of lead ions to cause the loss of partial permeability in red onion cell membranes to be 0.4M

Background Knowledge

Lead ions reduce the partial permeability of cell membranes. The principle mechanism by which lead does this is by reduction in pore size and by reducing the solubility of the phospholipid bilayer, the principle material from which cell membranes made. Pores are proteins which allow uninhibited access to water molecules to and from the cell. These proteins contain thiol groups (-SH), to which lead ions have a particular affinity. A reduction in solubility reduces the permeability of the cell membrane to water; a less polar phosphate group on the head of the phospholipid repels the polar water molecules. This reduces the rate at which water can travel by osmosis through the cell membrane, and hence the cell is less likely to become plasmolysed through water loss when it has suffered poisoning by lead ions, and in addition the chance of deplasmolysation in distilled water is also reduced. (Knowledge derived from References 1 and 4)

Join now!

Method

List of Equipment

        7 clean evaporating dishes

        A supply of 1M lead nitrate

        A supply of 1M glucose solution

        A 10 ml syringe

        A stopclock

        Microscope

        Red onion

        Scalpel

        Ceramic Tile

        A beaker

        Slides and Coverslips

        Forceps

1. Preparing Lead Nitrate Concentrations

        Six concentrations of lead nitrate should be prepared, 0.2M, 0.3M 0.4M 0.5M, 0.6M and 0.7M. To do this, a supply of 1M lead nitrate is needed. Then, using a syringe to measure amounts, the concentrations above should be prepared according to the table below.

        20ml of each concentration of lead nitrate ...

This is a preview of the whole essay