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# Neurobiology NPB 100

UCD

GPA 3.85

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This 4 page Class Notes was uploaded by Mrs. Sierra Bailey on Tuesday September 8, 2015. The Class Notes belongs to NPB 100 at University of California - Davis taught by Mitchell Sutter in Fall. Since its upload, it has received 79 views. For similar materials see /class/191815/npb-100-university-of-california-davis in Neurobiology,Physio & Behavior at University of California - Davis.

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Date Created: 09/08/15

Homework 1 Key 1 Problem 1 Given The beaker below is divided by a semipermeable membrane that is only permeable to K and has 100 times the amount of potassium acetate on the left side than on the right side A Under these conditions what will the voltage of the right side of the beaker be relative to the left side To solve this problem we will calculate the Nemst potential for potassium We K K use the Nemst equation rather than the GHK equation because otassium is the z 39 K K 0711 permeable lon ER 2 E lnl 57510g10 C ZF K K We used In K1KR because we always put the concentration 39of he side for which we want to know the voltage in the denominator At 17 C for positive monovalent ions like potassium 23RTZF575 mV and the equation reduces to the following Ek5logff gt Ek575log100 EREk575mV2 115 mV You now perform an experiment where you want to set clamp the voltage of the right side at a speci c value In this particular experiment we are going to clamp the right side of the beaker at 17 mV relative to the left side of the beaker You do this by putting two electrodes in the right side of the beaker see gure to left One electrode measures the voltage This value is fed back to electronic equipment which then calculates how much current must be injected through the second clamping electrode to hold the right side voltage at 17 mV relative to the left side Before beginning the experiment the right side is sitting at the voltage you calculated in part A You then turn the voltage clamp on and within a millisecond the clamp is generating current which clamps the right side s voltage at 17 mv B Shortly after clamping at 17 mV before a large change in the concentration gradient can occur what is the magnitude and direction of the electrical force acting on potassium ions The electrical eld force drives positive potassium ions to the left side of the beaker because the right side ofthe beaker is clamped at positive 17 mV relative to the left Under these conditions positive potassium ions will be repelled by the positive voltage and driven away from the right towards the left side of the beaker by electrical forces The size ofthe electrical force can be calculated with Felemca 12W So Felemical 11171 l7 arbitrary force units afu C Shortly after clamping at 17 mV before a large change in the concentration gradient can occur what is nght slde of beaker CLAMFED the magnitude and direction of the chemical force acting on KT The chemical force will tend to drive K ions from the high concentration of K on the left side to the lower concentration of K on the right side of the beaker s e an r NOTCLAMPED The size of the chemical force can be calculated with Fchemicall 575 llogIomIml 17 C or for the beaker Fchemicall 575 llogIRigmILe l 575 log 1001 115 afus Note in the chemical force equation it does not matter if you use Ingridhm or LegLight because these will give the same number with a different sign because we are getting an min a absolute value the sign of the number does not matter af rm 115m 115339quot D Shortly after clamping at 17 mV before alarge change in Elecmcal force on K 1151 quotan the concentration gradient can occur what is the net force driving Kl Net almamo ve on K 0 sin 98 Bin Homework 1 Key 2 Adding the electrical and chemical forces yields the net force on Kl If we add 115 afu toward the right and 17 afu toward the left Fuel 98 afu toward the right side for K see Fig to right This question shows how a voltage clamp experiment works Important principles used in this problem are also described in your book pp 119121 For part E the experiment is a little different We are now using a membrane that is permeable to both K and acetate E multiple choice Ifthe investigator lets this beaker stay clamped at 17 mV for a long time 1 there will be no signi cant changes in the ionic concentrations on either side of the beaker 2 the concentration gradient will break down until there is no potasium on the left side 3 the concentration gradient will break down until the there is twice as much postassium on the left side as the right 4 the concentration gradient will break down until the potasium is equal on both sides 5 the concentration gradient will break down until the there is 15 times as much potassium on the right side as the left Explain your answer The answer is 3 There are two different methods to solve this one Method 1 Trial and Error If we allow the concentration gradient to change since there are no pumps a new equilibrium will be established by the concentration gradient slowly changing We have voltage clamped at 17 mv and as calculated before this will create an electrical force of 17 afus to the left therefore for equilibtium to be established the chemical force must be equal and opposite or 17 mv to the right We must use the chemical force equation chhemicall 575 llOgQKRigmlKLe Dl 1700 For each choice one should plug in for KRKL For choice 1 KRKL 1100 9 115 afus choice For choice 2 KRKL 00 which creates an in nite chemical force For choice 3 KRKL 05 9 17 afus which is correct For choice 4 KRKL 1 9 0 afus For choice 5 KRKL 15 9 101 afus but in the wrong direction since the chemcial force would go R 9 L and it should go L 9 R to oppose the electrial force Method 2 solve the chemical force equationlfwe allow the concentration gradient to change a new equilibrium will be established We have voltage clamped at 17 mv and as calculated before this will create an electrical force of 17 afus to the left therefore for equilibtile to be established the chemical force must be equal and opposite or 17 afus to the right We must solve for the concentration ratio of KRightKLe inthe chmical force equation chhemicall 575 llOgQKRigmlKLe Dl 17 C Fea anging 9 chhemical 5751 110gKRigmK1enll raising both sides to the power of 10 9 10lFChemlcal57395l the ratio We put the ratio around the right side of the equation because 10chhemical57395l will give us the ratio t cannot distinguish if it s left to right or right to left This is because by using the absolute value of lFChemical57395l in 10lFChemical57395l we can now get eltherKRigmKLe or KLe Kigm Therefore we must rst solve for the ratio and then determine which side has the higher concentration First we solve for the ratio We know for equilibrium chhemicall should be 17 afus so 101757 5 the ratio 2 We can reason that there will be more potassium on the left side because this will make Fchemical go to the right and counteract the electrical force which drives K to the left Therefore KLe lKigm 2 and there will be twice as much K on the left When performing voltage clamps on real cells one must always carefully monitor the health of the cell and make sure concentrations do not change Homework 1 Key 3 Problem 2 The point of this problem is to understand how two opposing forces create a membrane potential and where all the terms came from in the Nemst equation Given the chemical work required to prevent ionsfrom moving down their concentration gradient is represented by the equation below hint another way ofthinking about this work is that it is the energy associated with a chemical concentration gradient AriWWW AnRTln I 11 1r where R gas constant T absolute temperature I1 the concentration of I on the left side of a semipermeable membrane in a beaker IL the concentration of I on the right side of a semipermeable membrane in a beaker An the number of moles of the molecule Given the work required to prevent ionsfrom moving toward a voltage E is represented by the equation below hint another way ofthinking about this work is that is the energy required to move charges AVVelectricul ZFE where z charge of the ion F Faraday s constant E is the potential difference voltage Given the above information derive the equation for the equilibrium potential Hint this should look like the Nemst equation Make sure to show your work and eXplainyour answer When the work required to prevent ions from moving down their concentration gradient is equal amp opposite to the work required to prevent ions from moving towards a voltage the system is said to be in electrochemical equilibrium 9 Therefore the rst step of this question requires that you make the chemical work equal amp opposite to the electrical work This is actually the most difficult part of the question From this point forward we will simply rearrange variables to solve for the voltage that must exist across the membrane when the system is in electrochemical equilibrium AnRT1nI1 AnZFE lsieftrrangfingEterms amp 39AV39RT 1n Ill E Elm Il ll 0 or 39 AnzF Ir 2F Hr Chemical Work Electrical Work Now do you see why the chemical force only depends on concentrations of ions and the electrcial force only depends on the voltage Homework 1 Key 4 Problem 3 A special single cell organism that lives in a natural mineral water spring has the following concentrations and permebilities associated with it Permeability outside in Cl39 042 21 3 K 024 10 760 Na 006 07 365 Br 39 l 04 08 other non permeable ions for osmotic balance and charge neutrality A What is the concentration of nonpermable anions inside the cell if the concentration of nonpermeable cations is 30 mM and all anions and cations inside the cell are monovalent The total ofpositive charge inside the cell is 760 mM K 365 mM Na 30 mM of nonpermeable cations l 155 mM of positive charge The total of negative charge is 3mM C139 8 mM Br39 x mM of nonpermeable anions We know the net positive charge will almost exactly balance the net negative charge except for about a 00000001 mM difference causing about 100 mV voltage across the cell membrane Therefore 3 mM C139 08mM Br39 x mM 1155 mM solving for x we get x 11512 mM ofnonpermeable anions B To maintain the chloride gradient this cell has a special chloride pump Which way does the pump have to move chloride to work Explain the logic of your answer and show all math to maximize partial credit note if this were a test question it would be worth about 20 points out of 100 You would only get 1 point for guessing 19 points would be based on your explanation and math This is the kind of question where we DO NOT lead you to the answer but to get credit you have to go through the logic of the whole problem Solution You might want to work backwards l The pump will have to counteract the net force acting on chloride so we need to calculate the net force 2 To calculate the net force we need to calculate the chemical and electrical force Now to solve this we work back 3 to calculate the Fchemical on C139 we need to know the Cl39concentrations and the temperature 4 to calculate the Felemical we need to know the voltage across the cell membrane and the ion s charge 5 we were not given the voltage but we can calculate the resting membrane voltage using the GHK Now lets work back from step 5 to step 1 5 calculate resting membrane potential RMP using GHK We use the GHK equation to nd the RMP if multiple ions are permeable and therefore contribute to the RMP Also remember that positive ions have outside concentrations in the numerator and negative ion have inside concentrations in the numerator p51 Clix pk K10 pm N010 pBriBrii F 75116710 Pk K11 PM N011 PBJBVL 170 P51 C1 PkK0 PM N010 PBJBVL ER 575 log a C110 pk K11 MalVal 103 Brig 042302410006 071 08 E 57510g 575 10 0021 9637mV R 04221024760006365104 g 4 Calculate electrical force lzVl on C139 964 arbitrary force units to the outside negative ion repelled out by negative intemal voltage ER

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