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ex8 - k Glycerol not drawn towards either wire 4 Current 34...

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Exercise 8 Analysis 1) The glycerol ensures that the solutions are dense enough so that they can be  inserted into their wells without dispersing into the rest of the buffer. 2) The oxygen (which has a slight negative charge) faces downward toward the anode  and the hydrogen atoms (which have a slight positive charge) face upward towards  the cathode.  3) a) Hydroxyl ion: OH: drawn towards anode b) Hydronium ion: H 3 O +  : drawn towards the cathode c) Tris +  (the buffer): drawn towards the cathode d) Water: not drawn towards either wire e) Chloride ion: Cl - : drawn towards the anode f) NH 3 + -CH 2 -COO : not drawn towards either wire g) NH 2 -CH 2 -COO - : drawn towards the anode h) Bromophenol blue (the tracking dye): drawn towards the anode i) Sodium dodecyl sulfate -  : drawn towards the anode j) Proteins bound to many SDS molecules: drawn towards the anode
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Unformatted text preview: k) Glycerol: not drawn towards either wire 4) Current: 34 mA = .034 amps Average voltage: 80 volts V = IR R = V/I R = 80 volts/ .033 amps R = 2352.94 ohms ( ) Ω Average resistance during electrophoresis: 2.353K Ω 5) V initial / R initial = V final / R final V final = (V initial / R initial )R final V final = (60 volts/ R initial ) (2R initial ) V final = 120 volts 6) a) V = 60 volts I = 34 mA P = V x I P = 60 volts x 34mA P =2040 mW b) P=VI and V=IR 2V=I(2R) 2P= (2V)I So if resistance doubles, then voltage doubles; and when voltage doubles, power also doubles. Heat production increases by a factor of 2 during the electrophoresis run if the resistance doubles....
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