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by: Mrs. Sierra Bailey
Mrs. Sierra Bailey
GPA 3.85


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Class Notes
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This 87 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 Staff in Fall. Since its upload, it has received 9 views. For similar materials see /class/191825/npb-100-university-of-california-davis in Neurobiology,Physio & Behavior at University of California - Davis.

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Date Created: 09/08/15
NPE mun Muduie 12 Lalevai inhibiiiun Lateral Inhibition Part I Introduction eoais Undevsiand We enhancemem Madge cunivasis dining Visuai pvucessing eye Reviewing neuvuanaiumy mime Limaus huvseshue main iaievai Undevsiand the neuvai civcuiivy ieading in iaievai inhibiiiun m in in me nevvuus sysiemmai iaievai mimicquot has been mus amuusiy studied We Wiii iuuk ai iaievai mimicquot m inevisuai sysiem Way Fim i dun new inhibi iun Secund a weii knuwn eweei eiiaievai inhibmum cunsisiemwiih us mie m ewieiem mm hencememniedgesisaccumpiishedbyincveasing b fan m m y m uwn eyes Each uHhe imee has is a sum cuiuv yeiihe middie bay appeavsiu be a gvadiamw mnmng mm iigm Wham ii buvdevsihe davk baiiu daMW y yuu cuvevihe suvvuunding baiswiih a uupie pieces nipapev ihe iiiusiunwiii gm away These and nine human psychuphysicaiubsewaiiunsinihe19 cemuvy iead Ems1 Mach mi supevsuni Mach numbeviame in pmpusei aime Visuai sysiem accumpiishes cunivas1 in enhancememihmugh inhibnuvy cunneciiuns beiween adiaeem Visuai vecepiuvsw a aviangemem nuw knuwn as iaievai mimicquot Mmemanau yeavsiaiev H K Hanime cuniivmed Wis pvupusaiwii eiecivuphysiuiugicaiwuvk un me iaievai eye uHhe avihmpud Limauspoyphemus huvseshue main Fm hiswuvk Hanime iaiev shavedihe Nubei pvize m Physiuiugy m Medicine in 1957 E memeae 1916 e K Ha iineimitigm Muduieiz Laievaiinmmimn emmweveex mm NPE mun Mudme 12 Laleva mmnmun Despne Ms cummun name Lunaus 5 nm mam m mm m a mumcean m manna nab but my a chehceva 28ubphymm cmnceraza m fgggggyw a uvpmns n h u u enesma sp 215 and s as a numbevuh h 5n Each ummaudmm cumams an eccenmc 22H neuvumtha depmanzes m vespunse hgmandgenev 2 am rma sthat avesenHume m Huwe quotpm a save amazequot ummanma mmbumy cunnecuuns Whmh uccm m We a eva exuss ave m hm uHa eva mmquot m mudu e yuu Wm mam a mumrum a bash neuvuana mm D 2 L mu us eye and mveshgate huwmat mum yespunuswu ham snmuh owmaudmm Eccenmc new M mevemeccenmc M W muammw s m Mudmmz LalevaunmbmurLSemLAnmvedx mm NPE mun Mudme 12 Laleva mmnmun Part II A Basic 1Dimensional Model Goa s Undevs and the assumpncns m the a eya mmbmcn mudg Genevate a Venn cumammg a new sumums Cveate a dwmensmna mceemm neaves newghbuv ccnnecucns Undevs and huw he mudg s a eded bynme Vuu Wm name a 277mmquot eye mms mudu e C eav y we cc nm have me mu s necessamc Because c1 WS WEWM make a numbev uszmpMymg assumpncns 1 WME each cmmammm s a cump ex um uhvumemus ceHs We Wm vepvesem each cmmammm as a smg e neumn 2 Ms unmessvanab e s yea emc the we neeuency uHhe neumn Huwevev we shumd a eya mmbmcn Fm exammg mu ham unnst 2th mm was uhveuvuna acwuy Fm swmphmty we H vepvesem ham mensny as a unmessvanab e much We amwty A Evevy ceu sends mmbumy ccnnecucnsm Ms newghbuvs The amuum cmnmbmcn s pvupumunaHu the pvesynapu neuvun s acwuy ham and a eva mhmmcn i i Exmlsuon Vi Excwauun l Eccentric Cells pvahem um sc yuu can yevencmem wmxe yuu Wuvk an the pvesem mudu e Vuu H waan pay me waunacm sequence Hyuu havem aheady dune su yuu shumd a su cump ete Graphing M Mndule We wme Back Mudmmz Latevaunmmmn emmmveex mm NPB 100d Module 12 Lateral Inhibition For simplicity our quoteyequot will be 1dimensional a single row of neurons looking out into the world The activities ofthese neurons will be contained in a single vector neuron The light intensities shining down on these neurons will be contained in another similarlysized vector light To specify each of these vectors we39ll first need to specify the size length of our 1dimensional eye in terms of the number of neurons it contains It will be convenient to assign this number to a variable eyesize because we39ll use this value a lot In the white space below use the assignment operator to assign eyesize to be 27 Note that implicitly this is a length ie the eye is 27 ommatidia neurons long but that we are not associating units with this variable This is simply out of convenience and allows us to avoid stating the length in pm of a single ommatidium eyeisize 27 Knowing the size of our eye let us move on to describing the light that shines down on it As mentioned above these light values will be contained in a vector called light To build the light vector we first need to create a range variable to specify the indices of light We39ll call this range variable L because we39ll use it to create light Assign L to be a range variable below starting at 0 having a second value of 1 and ending at eyesize1 M 01 quoteyeisize 7 1 Why is the last value in L equal to eyesize1 and not simply eyesize Our light stimulus will be a bright spot in the middle third of the eye It39s fairly easy to represent numbers as pictures in Mathcad if the numbers range between 0 and 255 So let39s use light levels in that range We39ll assume that the quotbackgroundquot lighting is 100 and at the bright spot the lighting is greater by 50 ie 150 total Lateral nhibitionSemiAnsweredx mcd NPE mun Mudme 12 Laleva mmnmun Cveate memHqug smamsm be uwm pupu ate m hgmvemuvwnh 01252 va ues Ms 39 Tu n w linlnv subssnm Whmh can be yped mmm m 2mm and Msmrumbsy The s symbm can be mm m We Eun ean Tun bav and nutethat N s un y used unce nuHWme 2m n smilme maul mm smL 3y 7 3y 251 H m M 15B m n hgmvame u mu vemu Wham 312012 avgeva ues and haw many 312012127 Ale 01212 anyva ues uthenhan mu m 15m 2m H k HMquot mums mhevways sswsu FwsL as a gvaph smm secund as a pmms Mudmmz Latsvaunmmm smmmvsux mm NPB 100q Module 12 Lateral Inhibition Create a new xy plot below Set the xaxis to be L Set the yaxis to be lightL Set the xaxis to range from 0 to eyesize1 lou I I 140 130 11ghtL 120 110 lUU 0 10 20 Create a new picture by clicking below then going up to menu bar clicking on Insert and then clicking on Picture CtrlT will accomplish the same thing In the empty placeholderthat appears type light A small grey vertical bar will appear Resize the whole picture region so it39s a little bigger and then click on the small barto bring up the Picture Toolbar In the Picture Toolbar click on the Zoom to Window button that looks like a box with crossed arrows inside 2 o lightT You39ll notice that the graph above and the picture you just made sort of disagree as to the orientation of your eye that is does it go leftright or updown Personally I prefer to think of the eye as going leftright so let39s change the picture to reflect that Position the blue editing line at the rightmost end ofthe light in your picture placeholder above Then click on the Matrix Transpose button in the Vector and Matrix Toolbar The Matrix Transpose will append a superscripted Tto light which will convert light from a 1column 27row matrix to a 27column 1row matrix This essentially rotates the picture so that darklight dark runs from left to right as it does in the xy plot above Laterall nhibitionSemiAnsweredx mcd NPE mun Mudme 12 Laleva mmnmun Resize yum mauve SD H mm sumethmg m m We neumm acwmes abuve an acmy mu 5 an appvupnate mma cundnmn m awe neuvuns Asmnhe m uHhe Huwmn N m m m be u mm mm yuu Expert 212 Mudmmz LalevaunmbmurLSemLAnmvedx mm NPB IOOq Module 12 Lateral Inhibition Forthis first model of lateral inhibition we39ll assume that neurons only make connections to their nearest neighbors While this is a fairly simple assumption we nonetheless encounter problems at the boundaries of the eye For example the first neuron neurono has only 1 nearest neighbor to connect to neuron1 On the other end neuronze has a similar problem All the other neurons have 2 nearest neighbors one to the right and one to the left There are a number of ways we could handle this One solution that is easy to implement in Mathcad is to assume that these boundary cells don39t actually see any light that their activity is fixed at zero In this way neurono and neuronze will always remain at their initial condition From there we can concern ourselves with only the neurons that have 2 nearest neighbors To do this we39ll need a new range variable to reference this subset of neuron indices That is we need to be able to reference the 25 elements within the neuron vector that aren39t at the boundaries We39ll call this new range variable x to indicate distance along the eye Assign x below to be a range variable beginning at 1 the second number should be 2 and the last should be eyesize2 213 X 12 eyeisize 7 2 Earlierwe assumed that the amount of inhibition a postsynaptic neuron receives from its neighbors is proportional to that neighbor39s presynaptic activity assumption 4 We now need to specify that constant of proportionality a variable we39ll call synwt for synaptic weight Let us choose a value of 30 03 for synwt Our choice of 30 is fairly arbitrary although if you make the synaptic weight too strong the model will produce negative neuronal activities and results that don39t make a lot of sense Assign synwt below to be 03 synjvt 03 Creating the 1dimensional model is now pretty straightforward The activity of a neuron at point x ie neuronx is the sum ofthree parts 1 The light at that same point ie Iightx 2 MINUS synwt TIMES the activity ofthe neuron one index to the left neuronx1 3 MINUS synwt TIMES the activity ofthe neuron one index to the right neuronx1 Laterall nhibitionSemiAnsweredx mcd NPE mun Mudme 12 Laleva mmnmun mm m mudg nuw by makmg memHqug asswgnmem be uw neumnx ghtx 7 syniw mmn 7 syniw neuranx m nmnx hghlx 7 5mm nmn 7 mm mm same mm be x 5mm yam m be neurnn same mm m vange mm n m eye size71 Rem me gvaph m quotmwnni make u a Me My sumethmg m gvaph at m Ham 5 1U 15 2m 25 when yuu ve dune X m mumhxl 39 F vn h v That s sumu us we mam expedma he cunespundmg neuvuna ammme a su be s auc me any vepvesemsme 51am ham sumums The mmquot s ave We mm yew Duesthe abuve gvaph vepvesem a s eadyr ate uhveuvuna acthy7 0y anemawe y mums mudg take same We m cunvevge m a smawsmev Mudmmz LalevaunmbmurLSemLAnmvedx mm NPB 100d Module 12 Lateral Inhibition Based on the model structure you39ve seen so far what do you thinK Do you think the above graph is a steadystate of activity Why or why not any answer here is fine so long as you explain your reasoning One way to test whether we39re at a steadystate or not is to rerun the model using the current output as the initial condition If we39re at a steadystate then we should get the same answer again If we39re not at a steadysteadystate then the answer will change To rerun the model with the current output as the initial conditions simply copy your above assignment for neuronx from above and paste it here below Below that copy and paste your graph of neuronx from above neuronX hghtX 7 synfwtneuronxil 7 syniwtneuronx 1 u I I 10039 39 neuronX 80 6039 39 I I 4L 0 10 20 X 10 Laterall nhibitionSemiAnsweredx mcd NPB 100d Module 12 Lateral Inhibition What do you see in your new graph compared to the old graph Did the model output change What does that suggest about whether we were at a steadystate before Is the model at a steadystate now Repeat the above procedure to support your conclusion neuronX hghtX 7 syniwtneuronxil 7 syniwtneuronx 1 u I 100 neuronX 80 60 What39s the problem here Why does the output keep changing Why don39t we arrive at steadystate with the first iteration ofthe model Is this a problem with Mathcad39s calculation of our model Or is this a problem with the way we39ve implemented the model Explain your answer 221 Laterall nhibitionSemiAnsweredx mcd NPB IOOq Module 12 Lateral Inhibition Part III A TimeDependent1Dimensional Model Goal Expand the 1dimensional model to include time The exercises in the previous Part should impress upon you that the lateral inhibition model is affected by time That is it takes several iterations ofthe model and some nonzero amount of time for the output to converge to a single steadystate The reason for this is that the model contains circular references The neural circuit in this case is said to be quotrecurrentquot For example neuron139s activity depends on neuronz39s activity but neuronz39s activity in turn depends on neuron139s activity Thus it is impossible to accurately calculate the steadystate activity of both neuron1 and neuron2 in a single timestep To calculate these steadystate activities we will need to expand the model to include time This will lead us to a model output that is a 2dimension matrix instead of a 1dimensional vector In the matrix each column will represent the activity of a single neuron as it changes in time In turn each row will represent all the neuronal activities at a particulartime With each additional row those activities will come closer and closer to steadystate values In practice the model will converge to a steadystate within about 20 timesteps We don39t need to specify the length of these timesteps but they39re probably pretty short on the order of 10ms or less The reason for this quickness is that all the inhibition happens with graded membrane potentials before spikes are encoded so we don39t have to wait for 2 or more spikes to arrive in the presynaptic neuron before we can infer its activity To specify time in the model begin below by assigning the end time end to be 20 end 20 To create the 2D output matrix for this model we will need two range variables to specify the indices one for the rows and one forthe columns We can already specify the columns with x To specify the rows we39ll create the range variable t for time Assign t below to be a range variable beginning at 1 the second number being 2 and ending at end 302 t12end Here we begin tat 1 rather than zero because the first index zero will contain the initial conditions of the neurons which we will specify shortly To begin our new model we need to first clear out the old neuron vector We can accomplish this by assigning it to be NaN which is a special value in Mathcad that stands for quotNot a Numberquot The effect ofthis assignment will be almost like quotunassigningquot neuron which assures us that none of the previous model output will be carried over to our new model 12 Lateral nhibitionSemiAnsweredx mcd NPB 100d Module 12 Lateral Inhibition Assign neuron below to be NaN neuron NaN To reference elements in a 2dimensional matrix you must use two indices together in the subscript separated by a comma The first index is for the row and the second index is for the column As mentioned above rows in our model will correspond to different times while columns will correspond to different neurons 80 the designation neuron would indicate time index 3 and neuron index 6 Note that this is different from the way you represented neuron indexes earlier When the model output was a single vector then different rows corresponded to different neurons Things are different now Now different columns correspond to different neurons This new scheme is perhaps more natural because the neuron activities really are situated from left to right as in our graphs In addition forthe calculation rules that Mathcad 131 follows this organization is absolutely necessary for the model to give the correct output As before we will assume that the initial activity of each of the neurons is zero Create these initial conditions in the first row of your neuron matrix by assigning neurono to be 0 zero 304 neuronZ 3L 0 Ask Mathcad below what neuron is 1161er11 I l0123456789 You should get a single row of 27 zeroes Lateral nhibitionSemiAnsweredx mcd NPE mun Mudme 12 Laleva mmnmun su pvevmus vuw uhveuvuna acmy we 71 smgxe asswgnmem and subhacuun m m subscnms neuran vx 7 1ng syniw mmn VH 7 syniw mmn M 3m mm 7 hng 7 5mm mum M 7 5mm mum M mesmp Huw duesthe uu put hev cumpave m Me 351 gvaph yuu made m Pan H7 3m nanmm 2n m m am m lt1 n mm m 1 ca cu auunmvyuu an ltmHgt mm 7 mm mnm quotquot m max 7nmz Mudmmz LalevaunmbmurLSemLAnmvedx mm NPB 100q Module 12 Lateral Inhibition Would you say that neuronend x is now a steadystate Why Go up to Exercise 301 and change end to some othervalue try 5 and see how that affects the result of your simulation What is the minimum value of end that gives you less than 1 change in mean activity between the last two timesteps 309 Change end back to 20 Go way back into Part II and copy your picture not graph of LightT and paste it here below Below that copy and paste your graph from Exercise 307 above Adjust the positions and widths ofthe two regions so that they match up so the neuron graph is in register with the light picture 311 lightT l U 100 39 nemonendg 8039 39 60 39 15 Laterall nhibitionSemiAnsweredx mcd NPB100q Module 12 Lateral Inhibition Describe what you see above ignore for now the activity of the first and last neurons How does the activity of the neurons change with changes in the light stimulus Is this result consistent with or contrary to the idea that lateral inhibition enhances edge contrast as discussed in Part 312 What do you think is happening at the boundaries of the model eye That is why is there a sharp increase in neuron activity forthe first and last neurons in your graph above 313 neuron 0 neuron 0 end 0 end eyeislzeel Do you thinkthat similar boundary effects at the border of the eye happen in the Limuus eye in vivo Why or why not 314 16 Laterall nhibitionSemiAnsweredx mcd NPB 100q Module 12 Lateral Inhibition Becausethe neuronal activity represents what a Limuus individual might quotseequot it39s probably worthwhile to make a picture out ofthis neuronal activity To this end Mathcad is quite happy to make a picture out of any vector we give it like we did with the light vector earlier Unfortunately the expression neuron L is not a vector as you might otherwise assume end Show this below by asking Mathcad what neuronend L is Note that although this produces the correct list of numbers there are no indexes associated with them Also note we use L here rather than x because we want to see the activity of all the neurons in the eye even the ones we assumed didn39t see any light 315 1161er11 end L To get around this inconvenience we can simply create a new vector activity and assign it to be neuronend L Do this below You39ll have to come up with the syntax on your own Hint use L If you do it correctly activity will be a vector of 27 values activityL neuronend L Laterall nhibitionSemiAnsweredx mcd NPB 100d Module 12 Lateral Inhibition Copy your picture of light from above and paste it here below Make another copy of this picture right below it Change the placeholder on this second picture to be activityT using the Matrix Transpose button or Ctrl1 If you get an error it may be because you didn39t create your activity vector properly above lightT activityT One thing you might notice in comparing the above two pictures is that the activity picture is overall darkerthan the light picture Indeed the mean value ofthe activity vector is only w 59522 of the mean value of the light vector This is a consequence of the mean ight general inhibition in the model To remove this general inhibition and see the output features a bit better we can normalize the activity vector such that its mean value is equal to the mean value of the light vector Do this below by assigning a new variable normactivity to be activityM The mean mean activity function is builtin to Mathcad and will return the mean value of any array vector or matrix 313 m ean light normiactivity activity mean act1v1ty Laterall nhibitionSemiAnsweredx mcd NPB 100q Module 12 Lateral Inhibition Copy your pair of pictures from above and paste them here below Change the placeholder ofthe activity picture to be normactivityT 319 lightT normiactivityT Take a minute to look at your normactivity picture and think about how lateral inhibition processed your light input 320 Lateral nhibitionSemiAnsweredx mcd NPB 100q Module 12 Lateral Inhibition Part IV SecondNearest Neighbors Goal Expand the model to include secondnearest neighbors So far we have restricted our model to only nearest neighbors In reality connections within the Limulus eye extend some distance beyond nearest neighbors up to onethird of the distance across the eye in some cases Barlow R 1969 Inhibitory Fields in the Limulus Lateral Eye J Gen Physiol 541383396 As you will see in this Part such extended connections appear to quotsmooth outquot the neuronal responses in space which is desirable in processing the visual scene Here you will modify your existing model to incorporate 2 additional neighbors th e quotsecondnearestquot neighbors This will mean that a total of 4 cells inhibits each neuron in this Part39s model two to the left and two to the right Clear out your neuron matrix by assigning it below to be NaN neuron NaN Create the initial conditions for your model below by assigning neurono L to be 0 zero neuronZ L 0 When we used nearest neighbors previously we discussed how problems at the boundary ofthe eye led us to assume that the first and last neuron did not see any light We will continue with that assumption but because we39re now including inhibitory connections from secondnearest neighbors we will need to assume that a total of 4 neurons do not see any light That is easily accomplished by simple redefining our range variable x Assign x below to be a range variable beginning at 2 with a second value of 3 and ending at eyesize3 X 23 eyeisize 7 3 Lateral nhibitionSemiAnsweredx mcd NPE mun Mudme 12 Laleva mmnmun m n nth m ms am curmecuuns s essmn m suengm uhnuve neavby curmecuuns Tu venemm We Wm un y14 Nhe suengm um must neay newghbuv ParHH mm h MW mesame x 1 neumn V 1ng 7 syniw neumn i M neuran i vm 7 2mm neumn i H neuran i 4m 1 WW quotmm H Wm x MUM hng 7 5mm mmr i vxi maer 4 m mm mp nm 455 1ng 12 mn mmnm x 2D 6D 40 IEI 2U Mudmmz Lamaunmmm emmmveux mm NPB 100q Module 12 Lateral Inhibition Checkthat this represents a steadystate by copying the provided math region from Part III and pasting it here below 406 7 T ltendgt T ltem 1gt neuron 7 neuron m 0874 1 ml T ltend71gt J me 116111 01 1 Does the above graph of neuronend x represent a steadystate Why or why not 407 Copy your assignments for activity and normactivity from Part III and paste them here below Below that copy and paste your pair of pictures of IightT and normactivityT 40a activityL neuronend L mean1ight norm actlv1t actlv1ty meanactivity lightT lll normiactivityT 22 Lateral nhibitionSemiAnsweredx mod NPB 100q Module 12 Lateral Inhibition Describe what you see in the output graphs and pictures from this Part39s model How does the output compare to what you saw in Part III What advantage if any does there appear to be in including connections beyond nearest neighbors 409 23 Lateral nhibitionSemiAnsweredx mcd NPB 100q Module 12 Lateral Inhibition Part V Model Dynamics Goal Understand how the steadystate output of the model develops in time From the work you did in Parts amp ll you should have a basic understanding of how and why the lateral inhibition model takes some time to develop to a steadystate output Using your model from Part IV you will now investigate the development ofthat steadystate in more detail Let39s begin by examining some of the rows of the neuron matrix Just to run a check what does a single row from the neuron matrix represent What does a single column represent Copy your light picture and neuronend x graph from Part IV above and paste them here below Set the yaxis on your graph to range from 5 to 155 neuronendy x neuronl 7x neuronz X To the above graph add yaxis traces for neural activity at times t 0 t 1 and t 2 24 Lateral nhibitionSemiAnsweredx mcd NPB 100q Module 12 Lateral Inhibition 503 Describe what you see in the graph above Do the changes in neural activities follow any general pattern in time 504 Why is neurono x all zeros 505 Why does neuron1 x look exactly like the light vector we used for input 506 Why are the activities in neuronzyx so much less than the steadystate neuronend x 507 e12 LateralnhibitionSemiAnswereclx mcd NPB 100d Module 12 Lateral Inhibition At this point you probably have a general idea of how the activity of the model oscillates in time However this may be hard to see when looking atjust a few plots in a graph To get a better idea of what we39re seeing let39s make a movie To set up a movie in Mathcad you must set up a region or set of regions that will be controlled by a special builtin Mathcad variable called FRAME all caps In our movie we39d like to see how a graph of neuront X changes as t changes Before we start to deal with FRAME let39s first set up the graph we want to make a movie of 1 Copy your light picture and neuronendx graph from Part IV not the multiplot graph you just made and paste them here below 2 Set the yaxis of your graph to range from 5 to 155 3 Change the yaxis to be neurontx 4 Adjust your light picture39s position if need be so it still lines up with your neuron graph 02 t FRAME lightT 150 39 39 100 neurontvX 50 0 10 20 X Your graph will look like a mess That39s OK Mathcad is plotting all rows of your neuron matrix one on top of another But that39s good That39s what we want for now Now add an assignment fortto be FRAME in all caps above your light picture 503 Your graph will be reduced to a single plot where implicitly t 0 Again this is OK Every good movie starts at the beginning The instructions below for making your movie may be easier to follow if you print them out first 26 Laterall nhibitionSemiAnsweredx mcd NPE mun Mudme 12 Laleva mmnmun chck un Tnnk Animminn Thwswm hung up a meme buxwheve yuu can de ne FRAME Fm FRAME 1 Set me e be u the dEVauH 2 Sena asZ the same as um and vanab e 3 SetM e be 2 FumesSec m u m1 memee yuuvl asswgnme se me muuse m dvaw a bux avuund hem yum neurnm x gvaph and the ligle mauve m 2 mesh Oneemshee Mame Wm upen a new WmdDW cumammg yum muwe Chck my We may buI un m 015 new 0 ee yu Wmduwmwa ch yum muwe u ve saus ed Wm yum meme ch un SaveAs m We FRAME mama bux Gwe yum w ame m quotd u may Vuu H need m submmhwsme a ung w m yuuvcummemd mudu e m Mhe Mme space be uw drzgrzndrdmp yum muwe aw me Tu u we 1 Usmg hewmduwsexp menupenmem devcumammgme awme yuujus1 made 2 Le hck my me we me n e W5 Mame Wmduw and ve ease n m We vegmn be uw 5m mum x Maeuxe12Lateyaxxnh n z anxDaynam aw Vuu canwew yum muwe Wm Mame nuw by enumeehekmg my me abuve mspxay Muwe Mudmmz Lateyauquotmemememmw eyee x mm NPB 100d Module 12 Lateral Inhibition controls will automatically appear DON39T DELETE your avi file The above appearance of your movie is only a linkto yourfile not a quotrealquot copy ofthe movie You still need your avi file to view the movie and you39ll still need to turn in your avi file along with this module to receive full credit foryour work Describe what you see in your movie How does neuronal activity of the eye change through time 507 Reset t below by assigning it to be a range variable beginning at 1 the second number being 2 and ending at end 508 539 12end These past three Parts have illustrated the following features of our lateral inhibition model 1 In response to static input the model takes some time to converge to a static output 2 Including connections between more distant ommatidia reduce spatial quotnoisequot compared to strictly nearestneighbor models 3 Spatial changes in luminance ie edges are exaggerated Image contrast near edges is enhanced This last result enhancement of edge contrast is clearly something lateral inhibition does However it is not clear that edge detection is really a major goal early in the visual system For example we know precious little about the higher objectextracting brain areas that might use such edge information The last two parts of this module will look at lateral inhibition from a somewhat different perspective efficient coding We will first point out that neurons have limits to the amount of visual information they can encode And then given these limitations you will see how lateral 28 Laterall nhibitionSemiAnsweredx mcd NPB IOOq Module 12 Lateral Inhibition inhibition contributes to the efficient coding of visual information Part VI Limits Of Neuronal Bandwidth Goals Understand why neuronal bandwidth is limited Explore how lateral inhibition makes efficient use of neuronal bandwidth Visual systems must perform over a wide range of light intensities In bright sunlight our visual scene might be composed of 100 million photons falling on the eye each millisecond Yet under darkadapted conditions humans can perceive a dim 1ms flash composed of less than 10 photons Hecht et al 1942 Energy Quanta and Vision J Gen Physiol 251819840 Thus our dynamic range of light inputs is somewhere on the order of 107 The ability of a photosensitive neuron to communicate light levels to higher brain areas is limited For example if a brain neuron quotlistensquot to action potentials from the eye over a 200ms time period 10 incoming spikes could mean anywhere from just over 7 45Hzto just under ms 7 11 200ms 55Hz a uncertainty range of 10Hz Further if we assume that neuronal firing rates max out at 200Hz this suggests that a 200Hz 10Hz photosensitive neuron has only 20 discernable light output quotlevelsquot to work with This is what we mean when we say that neuronal bandwidth is limited Clearly the neuron cannot dedicate a separate output level to singlephoton increments Such exquisite sensitivity would force it to ignore large parts of the visual scene And similarly if each level is dedicated to millionphoton increments fine details such as object texture would have to be ignored Thus the early components of the visual system must deal with the competing goals of sensitivity and wide dynamic range to make efficient use of limited bandwidth There are a number of mechanisms by which eyes make efficient use of limited neuronal bandwidth Yourwork in this Part should impress upon you that lateral inhibition is one ofthese mechanisms It is difficult to explain what lateral inhibition quotdoesquot for the visual system In this Part we will strive for an empirical explanation of lateral inhibition39s effects feeding various light inputs to your current model and examining the results You will use these observations to infer the effects of lateral inhibition and why lateral inhibition it is superiorto alternative encoding schemes In this Part we are concerned with quantifying the effects of lateral inhibition To that end it will be useful to talk about light and neuronal activity in concrete terms Let us begin to use units in our assignments The units for light incident on a surface ie illuminance are qu Ix And while we recognize that lateral inhibition occurs before proximal to action potential generation let us now for convenience associate the units of Hertz Hz with neuronal activities We will begin simply by generating three light inputs that are of uniform intensity across the eye Eventually we39ll graph these vectors all together so we39ll need to give them distinct names Without much thought let39s call them IightA lightB and lightC Laterall nhibitionSemiAnsweredx mcd NPE mun Mudme LalevaHnmmtmn Use yum vange mama L m make We quot9th vectuv cumammg va ues u 12m Du me same m am hghtAL 12m nng 751x ngmcL 251x m metevs y ZuuxamheHae mum R22 Cum Mamasnea annulans Mar Sm BE 2537254 Vuu Wuum nd swm av ham wavexswuhm a tymnHy m huuse at mam m 7 mm PanVabuve and paste n have be uw hghmx 1th i mm 7 Change We y awsm be 1kth Add addnmna View quotaces m X nng ugch ch 5n 7 and Ix Ix 7 7 7 7 7 7 7 7 7 7 Vuuv maph mum uuk n7 v sumemmq hke m when vuu ve 5 m 15 2m 25 dune X am Mudmmz LalevaunmbmurLSemLAnmvedx mm NPB IOOq Module 12 Lateral Inhibition Now we39ll reinitialize the neuron matrix and re run the model Clear out your current neuron matrix by assigning it to be NaN neuron NaN Initialize your neuron matrix by assigning neurono L to be 0H Don39t leave out the units m L OHZ Copy your neurontx assignment from way back in Part IV Exercise 404 and paste it here below In the assignment change light to be lightAX This will do two things X 1 Set the input light vectorto be IightA instead of light 2 Implicitly convert light input to firing frequency such that in the absence of lateral inhibition 1x of input elicits 1H2 of neuronal activity as output This assumption that neuronal output varies linearly with illuminance is a gross simplification In truth neuronal output typically varies linearly with the logarithm of illuminance but we will gloss over that detail 605 neuront X 7 1g X 1X 7 syniwtneurontilaxi1 neurontilax 7 4syniwtneurontil3X72 neurr Eventually we39ll want to graph all together the outputs generated from IightA IightB and IightC To that end you39ll be creating neuronA neuronB and neuronC to store the respective outputs One way to accomplish this for neuronA would be to edit the above assignment changing neuron to neuronA in the five locations it appears However that39s a little tedious especially since we39ll have to do it all over again for neuronB and neuronC Instead let39s run the model with the neuron matrix and then post hoc assign neuronA to be the last row in the neuron matrix the activities at time t end This will make neuronA a vector containing the steadystate activities Conveniently this will allow us to graph the steadystate output by simply specifying neuronAx much like IightAx instead of neuronAendYX Assign neuronAL below to be neuronend L neuronAL neuron end L Laterall nhibitionSemiAnsweredx mcd NPE mun Mudme 12 Laleva mmnmun mm PanVabuve and pad n have be uw Chem 012 vrawsm be nzumnA Hz Vuuv maph 5mm uuk sume hmg hke m X am 15n mmmxmn H 5n n 1m 2m Wm a cuup e sentences abumwhat yuu 522 m yum gvaph u neurnnA m Much mum umpm mama m neumn A s unwmm acvuss x m ummanma Tm seems vanauun s aHhe buundanes We mscussed m buundavy 21121 b21uvew nhasm u th mm y y an nm mum ummauma mum buundanes vespemwe hgmvecmvs Ahhuugh yuuwun meedm yemmamme neurnn mamagan Cupy yuu asswgnmem m neu n m Exev E as have and paste n have be uw VWhm 012 r quotw m Use a asswgnmem change hgmAMu be ligth Evevylhmg 2 52 shm d be me am Mudmmz LalevaunmbmurLSemLAnmvedx mm NPB 100d Module 12 Lateral Inhibition Hz 1 neuront X hghLBXEj 7 syniwtneurontilaxi1 neurontilax 7 Zsyniwtneurontil3X72 neurr As you did with neuronA above assign neuronBL below to be neuronend L 610 neuronB neuron L end L Copy your graph of neuronA from above and paste it here below Change the yaxis to display neuroan instead Hz 150 39 39 neuronBx 100 Use similar techniques in the space below to 1 Run the model with lightc as an input 2 Assign the output ofthe model at t end to neuronC 3 Make a graph of neuronC Hz 1 neuron hghtC 7 syn wt neuron neuron syn wt neuron neurr tX X 1X t71x71 t71x17 4 t71x72 neuronC neuron L end L e12 LaterallnhibitionSemiAnsweredx mcd NPE mun Mudme 12 Laleva mmnmun 15n mumncxmn H 5n n 1m 2m awn hghtAX nzumnAX neuvunvemuvs Fuvexamp e change mm Mudmmz LalevaunmbmurLSemLAnmvedx mm NPB 100q Module 12 Lateral Inhibition 150 39 39 39 neuronAX Hz 100 39 39 neuroan I Hz nu neuroan 50 I Hz n I l 0 10 20 x Considering your graphs above write a few sentences about how neuronal activity relates to illuminance when passed through your lateral inhibition model You can ignore what39s happening at the boundaries of the eye 614 You may notice that the neuronal activities are suppressed compared to the light input That is in the end 1lx does not yield 1H2 Forthese uniform light stimuli what does 1lx generate in terms of Hz You can use the builtin mean function to answer this for neuronA and lightA by asking Mathcad what m 39 mean1ightA Change the output units to Do the same for neuronB and lightB and neuronC and lightC Don39t forget to answer the question in words meanneuronA 7 Hz 5 mean1ightA 1x Lateral nhibitionSemiAnsweredx mcd NPB 100q Module 12 Lateral Inhibition meanneuronB 7 0 5 E meanligth 7 39 1x meanneuronC 7 E meanlightC 39 1x Notice two things 1 Lateral inhibition has scaled the neural output frequency such that each 05Hz of activity corresponds to 1lx of illuminance whereas in Exercise 605 we assumed a 1Hz1lx relationship If we assume that neural firing frequency is limited to say 200Hz then lateral inhibition has increased the dynamic range of the neurons39 response to light input Now rather than reporting illuminance in the range 0 200lx the neurons can report illuminance in the range 0 400lx All else being equal this is a better arrangement There are other ways to scale the output frequency that do not require lateral inhibition For example we could just assume at the outset that 1lx of illuminance generates 05Hz of neural activity and eliminate the connections between ommatidia N Given this last point what is the benefit of using lateral inhibition over a more simple scaling technique To be sure there are benefits in boosting edge contrast as we saw in Parts I amp IV But in the context of efficient coding are there any obvious advantages of using lateral inhibition We will answerthat question presently Part VII Fine Stimulus Details Texture Goals Generate a light stimulus with fine details Observe how stimulus details are lost when the input image is simply scaled Observe how stimulus details are preserved when lateral inhibition is used One ofthe effects of lateral inhibition is to increase the dynamic range of lightresponsive neurons This is good because we would like our neurons to report with their limited firing frequency range as large a range of visual stimuli as possible However assuming we are free to set a scaling factor between light input Ix and neuron output Hz it39s not clear why we need lateral inhibition We could set 1lx to correspond to 1H2 or 05Hz or even 01Hz In each case we would be expanding the dynamic range ofthe neuron allowing it to use its limited range of firing frequencies to respond to greater and greater ranges of input values However higherlevel neurons are limited in their ability to distinguish firing frequencies coming from lightsensitive neurons see Part VI To account for this let us assume that firing frequencies 5Hz are interpreted as having been generated from the same light intensity This is of minor consequence in detecting uniform light levels such as those in Part VI However if we introduce some fine details to the stimulus the benefits of lateral inhibition soon become clear You will create a new light vector IightD This vectorwill be based on IightA but with a small cosine function added to it giving it a quotbumpyquot appearance These slight variations in input signal 36 Lateral nhibitionSemiAnsweredx mcd NPB 100d Module 12 Lateral Inhibition might correspond to the texture of an object being viewed Assign IightDL below to be lightAL 4cosL7r1x That39s Ix lux the units for illuminance at the end not onex You can get the 1 symbol from the Greek Symbol Toolbar or by typing pctrlg 701 ligh DL lightAL 4 cosL7r1x To get an idea of what this looks like copy your picture of lightT from Exercise 502 and paste it here below Change the lightT designation to be IightDT instead 702 ligh DT In addition make a graph of lightD below Start by copying your graph of neuronC from Exercise llgh D 612 above and pasting it here below Then change the yaxis to be IX X instead 150 39 39 lighth 100 1X 50 39 Now that we have an input signal let39s forget about lateral inhibition for minute Instead let39s assume that light Ix is encoded in the firing rate of neurons through some simple scaling function Forthe sake of example we ll assume a relationship of 05Hz for every 1lx This will give us the same dynamic range as what we saw with lateral inhibition using uniform inputs Part VI 37 Lateral nhibitionSemiAnsweredx mcd NPB 100d Module 12 Lateral Inhibition We39ll first look at what this simple scaling process produces from lightD a result we39ll call scalingD Then later we39ll run the same lightD input through the lateral inhibition model a result we39ll call neuronD Comparing the two outputs will give us some insight into the advantages of using lateral inhibition over simple scaling Implement the scaling process below by assigning scalingD to be ligh D O39ISHZ X Note that we don39t have to use a range variable here although we could if we really wanted 704 05H scalingD ligh DTZ Make a graph of scalingD below Start by copying your graph of lightD from Exercise 703 above scalingDX and pasting it below Then change the yaxis to be 705 15039 39 39 39 scalinng 100 HZ WW 5039 I l I 0 10 20 Assuming everything went as planned scalingD should have a lower mean value than lightD but still look bumpy As mentioned at the beginning of this Part we will now assume that higher level neurons have difficulty distinguishing between firing rates of similar frequency i5HZ To implement this assumption we will simply round the values within scalingD to the nearest 10Hz 38 Laterall nhibitionSemiAnsweredx mcd NPB 100d Module 12 Lateral Inhibition Mathcad has a builtin function called Round that will do this Round accepts two inputs The First input is the value or set of values that you wish to round the scalingD vector in our case The second input is the precision you wish to round to 10Hz in our case Copy your graph of scalingD from above and paste it here below To simulate how higher level Roundsca1ingDX IOHZ neurons quotseequot the output of this scaling process change the yaxis to be Hz 706 150 39 39 Roundsca1inng10Hz 100 Hz 50 I I I 0 10 20 X What do you see in your quotRoundedquot graph above What is this supposed to represent What does this imply about using simple scaling to increase the dynamic range of lightsensitive neurons 707 Having examined the result of simple scaling let us now return to your lateral inhibition model The IightD input vector is still available to you you don39t have to create it again And you39ve already run the lateral inhibition model several times eg in Part VI 80 in this section you39ll be largely left on you own In the space below use your lateral inhibition model from Part VI to process the IightD vector The Laterall nhibitionSemiAnsweredx mcd NPB 100q Module 12 Lateral Inhibition model should be as you used in Part VI with a relationship not a 0sz relationship The X output should be called neuronD Make a graph ofthis neuronD vector that you create Hz 1 neuront X hgh DXEj 7 syniwtneurontil xil neurontilax 7 syn7wtneuron neun 4 t71X72 neuronD neuron L end L neuroanlOO Hz Now use the Round function to make a graph below showing neuronD rounded to the nearest 10Hz much like you did in Exercise 706 with the scalingD vector 15039 39 39 ROMHWDX IOHZ 100 Hz 50 What do you see in the Rounded graph of neuronD How does this compare to the Rounded graph of scalingD that you made earlier e12 LateralnhibitionSemiAnsweredx mcd NPB 100d Module 12 Lateral Inhibition We have discussed how increasing the dynamic range of lightsensitive neurons is desirable Would you say that lateral inhibition is a good mechanism for achieving this Why Module 12 Lateral Inhibition Last Modified December 7 2007 400pm Help Improve This Module Completing this section is not required but will be helpful for future development ofthis module Your answers here will not affect your grade About how much time did you spend completing this module Were there any particular parts of this module that you found too difficult too confusing or too tedious How might these sections be improved Lateral nhibitionSemiAnsweredx mcd NPB 100q Module 12 Lateral Inhibition Were there any particular parts that you found especially useful or helpful in understanding the material from NPB 100 Other suggestions kudos criticism Laterall nhibitionSemiAnsweredx mcd NPB 100q Module 12 Lateral Inhibition e12 LateranhibitionSemiAnsweredx mod NPB 100q Module 12 Lateral Inhibition mm x2 e12 LateranhibitionSemiAnsweredx mod NPB 100q Module 12 Lateral Inhibition mm x2 mm x2 e12 LateranhibitionSemiAnsweredx mod NPB 100q Module 12 Lateral Inhibition mm x2 e12 LateranhibitionSemiAnsweredx mod NPE mun Mudme m Hamemev HalfCenter Model Part I Introduction R ewew EEnUa pa em genevamvs she MEN m e m mum acuvny Rewemhehsheemeymeee Neuva EDMYD Mmusde muvemem can he bmken eeWh chm e a basal ganglia mmuum dESgnEd u genevate s1eyemypee umpms 77 52mm panequot genevatuvs CPGS memqem spmmm Ewdence m CPGs m the spma emu e1 mamma swas ms1 puuenh by Gvaham EmWh m 1911 based Em abs 3 cats EVEHWNM cu vd ehmmshhg shy a ehmhg eehheh he ubsevved that We y shh eehusehehem dhmb News was pvesewed when the ca was p aced huh suppun eh a muvmgheadmm Based Em mesa ubsevvatmns EmWh pvupused the EX S EHEE e quotand eheuuwuhm We spma emu that 5 summen uv hasm she appmpnate mum cunth Muxlt s The ame be uw shews EMG musde vecuvdmgs m a spma ca The Mame excess uVextensuv eehhseheh 5 typmaHuHhe wshmg gen enhe shhhsx 00121 gens sueh ashumng m gaHupmg shew mme eeusx dmy cyc es U e equd eehhseheh hmes meme my a quotmm J W e new He 5 sew Muememjah CemeLSem Anmvedxmm NPE mun Mudme m Hamemev Tu assuunuunms cemva pansm gsnsyamn Gyaham pvupused a hammer mudeLwhmh 5 Shaw be uw W hypmheuca vuhage vecuvdmgshum A uHhe ceHs m We mvcun mtemeumn 1 and 2 vespedwe y m m dwagvamywhmh make smamyy synapses am my mm m M hm n Hence We name hawrcemev H I mun mum s 7 if nan mnmrm F Wswm n ms mudu e wewwmucus an m Mu excnatuvy memeumns 1 and 2 m We magvam genevauun HuwevELwe Wm need m mm mm mudg a makemese ceHs pmduce buys1s av asuun putarma s We Wm mssuss mummy m Pan m Fm swmphcny we Wm assume tha whenevenhe smamyy memeuvuns ave vwnthew vespecwe A su m swmphmty We Wm assume thanhe mmbumy memeuvuns Hue essenuaHyUackme m n rm mm m m w n mmnsumnsw and 2 as dwec y mmbmng each mm and ehmmates any need m exphm y mudg ms mmbumy memeuvuns These assumpnuns aHuW usm veduce m hawssmsymuasu as dammed abuvsJu a mum swmp e mudg mvu vmg un ylwu cestnh yempvuca mmbumy synapns sunnssuuns Muammnjam 2 CemeLSemwAnmvedxmm NPB 100d Module 10 HalfCenter When we are finished each ofthese 2 cells will have the following currents 1 Voltagegated Na part of the Hodgkin amp Huxley model and contributes to the regenerative upstroke of the action potential and its termination 2 Voltagegated K part ofthe Hodgkin amp Huxley model and contributes to both the resting membrane potential and termination ofthe action potential 3 Leak a predominantly Kseective resting conductance Contributes to the resting membrane potential 4 Injected input depolarizing presumably from commandcontrol neurons that drive the CPG 5 Ca2activated K a slowlyactivating hyperpolarizing current that will terminate bursts of action potentials 6 GABA CIbased synaptic current that will be the mechanism by which the two cells inhibit each other The voltagegated Na and K conductances in the Hodgkin amp Huxley model rely on the opening and closing of quotgatesquot The activation gates forthe Na an K conductances are represented by m and n respectively In addition the Na conductance has an inactivation gate represented by h The rates at which m n and h change are each determined by a pair of voltagedependent rate functions on and 3 V thout getting into what that means exactly those six functions are simply provided for your use here 105E7m 4x1072 104E7m 55x1072 0cmE7m 111E7m 7 67 102E7m4x107 2 1 7 e7102E7m55x1072 7 Eim 7 Eim 2x107 2 1 amplt 10 2 0LhE7m 27e BmE7m 108e 39 7 Eim 7 2 BnE7m 555eamplt10 3 10 BhE7m 1 67 102E7m35x107 2 This module will make use of Mathcad39s ability to solve differential equations If you have not yet completed the minimodule on differential equations you should do so now 101 Welcome back Module10Half 3 CenterSemiAnsweredxmcd NPB IOOq Module 10 HalfCenter Part II Single Neuron No Bursting Goals Create a single nonbursting neuron using the Hodgkin amp Huxley model Graph Em in time showing repetitive action potential firing This section assumes that you have already completed the Action Potential module previously or are otherwise familiarwith the Hodgkin amp Huxley HH model of the action potential Throughout this module you will use Odesolve solve blocks to solve ordinary differential equations Odesolve does not allow units so we will have to be careful to use a consistent set of implicit units The safest way to do this is to consistently use base units Volts Farads Seimens Molar etc You can always scale your results solutions to mV and ms in your graphs when neccessary We will begin by defining the necessary constants for the HH model The first ofthese is membrane capacitance c m In past modules we39ve dealt with c m in one of two ways In the Synapses module we explicitly determined the c m for a 20me diameter spherical cell SOpF which led to calculating membrane current with implicit units of Amps In contrast the Action Potential model arbitrarily assumed a axonal section with a surface area of 1cm2 to conveniently um assign cm to be the specific capacitance of biological membranes However this led to cm2 membrane currents being represented as a current densities with implicit units of Amis cm At this point we are simply trying to recreate the HH model so we can use it in the context of a halfcenter model From the previous Action Potential module we already have parametervalues for Na W and leak maximal conductances GNa GK and Gleak respectively that are known to quotworkquot That is if we39re willing to work with current densities and scale everything to an abitrary surface area of 1cm2 per cell then we can use the parameter values from the Action Potential module straight away 6 7 F Forthis reason alone let us assume a cm value of 1 X 10 2 cm Use the assignment operator below to assign cm to be 10396 without units Next we39ll need to know the reversal potentials of the Na W and leak conductances In the past we39ve assumed certain ion concentrations on the inside and outside of the cell and then used the Nernst or GHK equation to calculate the reversal potential Here we won39t bother We39ll skip the calculation and assign these parameters to be a physiological voltage Module10Half 4 CenterSemiAnsweredxmcd NPE mun Mudme m Hawcamev Use We asswgnmem upmmy wqu asswgn E Numb nnsz Ejezkm be 111155 The mphcn unns have aveVuHs but am mdude my 2m E7K G m each mum cunductances Pecamhat c y N and s a cunvememway m thmk u ave m knuwn Usemassmnmem upenuw bE uwmassmn 67mm be n1zn 67K m be nms Gilezktu be 3 x m A The mphcn unns have ave but dun t mama my 2m GiNa n 12D 67K n n25 aim 3 x mquot Let us nuw cvea e me mums thatwm cumpute me vanuus Na K and eak cunems WE H u Hm m y Ms cundudance duesn change Tm mphestha Ms amua cundudance gjeak s equaHu Ms mawma cundudance Gilezk mm a mummy m UeakEim be uw and asswgn m be aim 57m 7 571m 2m gamut Gileak 57m 7 E 1er Nancemame Llezkmnmmn amms Eim as an mpm Tm s necessavymv cumpmmgthe Nwmn y and n Wh ch a ec s aim Muumemjam CEMELSemwAnwEvedxmm NPB 100d Module 10 HalfCenter Create a function for KEmn below and assign it to be GiKn4E7m 7 EiK IfKE7mn GiKn4E7m 7 EiK The final current function for this Part is Na which will need three input parameters Em forthe driving force and both m and h which affect gNa Create a function for NaEmmh below and assign it to be GiNam3hEim 7 EiNa IiNaE7mmh GiNam3hE7m 7 EiNa Let us now define time We39ll want to run a simulation long enough to show repetitive action potential firing so something on the order of 200ms should be fine We39ll also need to tell Mathcad how detailed of a solution in time to our ODEs we want We39ll do this by specifying a quotstepsizequot that we will use in two ways 1 in our ODEsolve statement at the end ofthe Solve Block and 2 in our assignment for t time for use in making graphs In the white space below assign stepsize to be 2 X 10 4 This will give our solutions a resolution of about 5 datapoints perms 207 stepsize 2 X 107 4 Make two assignments below 1 For end to be 02 2 Fortto be a range variable beginning at 0 having a second value of stepsize and ending with end 208 end 02 t 05tepsize end We don39t really have to define t right now The subsequent ODESolve Solve Block will work fine without it the end assignment however is necessary However we WILL need twhen we start making graphs Module10Half 6 CenterSemiAnsweredxmcd NPB 100q Module 10 HalfCenter In this Part we are only concerned with modeling a single cell Arbitrarily we will model cell quot1quot from the diagram presented in Part Later when we model both cells we will need to keep track ofthe Em n m and h for each cells separately 80 let us prepare forthat now with this single cell by labeling these quantities Em1 n1 m1 and hi respectively Below you will find most ofthe solve blockthat you need to solve this single cell model Note that the initial conditions for Em1 n1 m1 and hi have been provided for you These are quasisteadystate values obtained by running the model for a long time like 1 second using the parameters you39re about to use ie the ones you set above The first ODE is for calculating Em1 As in other modules the first derivative of voltage is proportional to the injected current minus the three membrane currents K Na and leak In the past linj represented current experimentally injected into the neuron via an electrode We can assume that this is still the case or alternatively that this represents excitatory input from commandcontrol neurons that drive the CPG It really makes no difference we39re not going to complicatethe model by trying to explicitly represent this input The last three ODEs compute n m and h as dictated by their respective 0c and 5 rate constants Again we39re not going to talk about those much but they39re a necessary part ofthe HH model Given E7m10 70063 1110 0353 m10 007 1110 0513 cimiEimHt 17mg 7 17KE7m1tn1t 7 IfNaE7m1t m1t h1t 7 IileakE7m1t 1110 unltEm1lttgtgtlt1 7 mm A 6nltEm1lttgtgtn1lttgt imut amltEm1lttgtgtlt1 7 mm A BmltEm1lttgtgtm1lttgt h1t ozhE7m1t1 7 1110 7 BhE7m1th1t t To finish the solve block you39ll need to make the Odesolve statement In this module we39ll want to quotparameterizequot the solve block so we can easily make multiple graphs from a single solve block This is a fancy way of saying that you39re going to use the Odeolve statement as an assignment to a function Throughout this module you39ll want to use linj as the input parameter This will allow you to change how hard you quotdrivequot the cell or circuit and observe how the model responds Forthis first model of a single nonbursting neuron we39ll assign the Odesolve output to the function HHinj for quotHodgkin amp Huxleyquot becuase the model is essentially theirs Module10Half 7 CenterSemiAnsweredxmcd NPB 100d Module 10 HalfCenter Eiml 111 end Create the function HHIInj below and assign it to be Odesolve tend You m1 step51ze L hi can create the vector using the Vector and Matrix Toolbar T Eiml end n1 HHIiinj Odesolve tend m1 step51ze hl Within the Odesolve statement we39ve specified a number of instructions to Mathcad The first is a vector of variables that we want solutions to This includes Eiml but also the gating variables for l K and l Na Second we istt time as our independent variable Third we specify the duration ofthe simulation which is equal to end 02 which you specified earlier Based on the the initial conditions listed earlier in the solve block like quotE m10 063 Mathcad infers that the simulation is supposed to start at t 0 Last we specify the number of datapoints our solutions should contain This indicates the resolution ofthe solution with more datapoints showing us more details but also taking longer to compute and graph In this module we will vary the duration of the simulation somewhat end but we39d like to keep the resolution constant Specifying for the number of datapoints step51ze does this nicely Mathcad will calculate on the flyjust how many datapoints to use for our solutions The default number of datapoints for an Odesolve solution is 1000 Odesolve will do this if you leave out the last parameter In the preent model 1 X 103 so we39re step51ze pretty much sticking to the default value for now One last thing you39ll need to change is the method by which Odesolve obtains its solution By default Odesolve is set to the quotFixedquot algorithm We need to change that to quotAdaptivequot You don39t really need to know what the difference is between these algorithms lt39s sufficient to understand that the default algorithm doesn39t always work In your Odesolve statement above RIGHTclick on the word Odesolve A small popup menu will appear with a check mark next to the word quotFixedquot LEFTclick on the word quotAdaptivequot instead 210 OK now let39s take a look at this HHIinj function you made Module10Half 8 CenterSemiAnsweredxmcd Mudme m Hawcamev NPE mun AskMamcadwha HHMw emw ThwsaskMathcadmvuntheswmu auunwnhUmseHuD 22m Vuu 5mm get an anwevthat uuks m m KUm ess gt Um ess muss PIX10 Kquot KUm es KUm essji 2M Um ess aUm ess muess aUm ess Hm NJ gt Um ess aUm ess Um ess aUm ess accepts same unmess mum and gamma same unmess uutput Thuse A unmmns 312012 sumnunsvm E7m1 quot1 m1 and h1Jespedwe y The sumuuns ave m m case D Em vuhage as an umpm mum unns uvvms gvaph HH dwec y Vuu musmm asswgn une u hvesuwng uncuunsm anumwaname Tm s eas y dune by usmgme subscnm u mm 52012 asswgnmem upmmy wqu 52 E7m1 m be HHEID Thwswm mam a Eimm mmmn mm vuu can nuW maph Nu wcetha even thuuuh m vemun memes a E mm11uncuunnm svmax have mama m mm 212 57m HHEIU Mudmem H am CEMELSemwAnwEvedxmm NPE mun Mudme m Hawcamev Cveate a newgvaph be uw Senhexrawsmbe L Senhe yamm be E 0 Makeme maph aswme asme name 5 Vuu can see Whafs numu un aHhe eaywumes E71111 763 7 7 mV 7 rm 7 SD 15 2 n m 39 5n mm 15 n I 39njn y m m m n vuhage 5 amy ame amund reamvo 2 Am ves mg membvane pu erma Theve s same smaH mm Hm neumn pmenuaxs Let s nuw 522 mm happenswhen We mquot up heme L39 j Cnpy We asswgnmen uv EJM mm abuve and paste n have be uw Change We va ue DVLW msm We pavemhesesm be 2 x m 6 ms12ad mm The mphed um have ave Am ZPS m n Muumemjam CEMELSemwAnwEvedxmm NPB 100q Module 10 HalfCenter Copy your graph of Em1 from above and paste it here below 215 U I I I E7m1t mV 7 50 I I I lUU 0 50 100 150 200 t ms Describe what you see in your graph Does this amount of injected current produce repetitive action potentials 216 Copy your assignment for Em1 andthe graph of Em1 and paste them both here below Change the value of inj in your Em1 assignment to be 3 X 10 6 217 7 6 W HH3 gtlt 10 0 u I I I E7m1t mV 7 50 39 I I I lUU 0 50 100 150 200 t ms Module lO Half 11 CenterSemiAnsweredxmcd NPB 100q Module 10 HalfCenter Now what do you see in your graph Copy your assignment and graph for Em1 them both here below Change the value of inj in your Em1 assignment to be 13 X 107 6 219 7 6 W HH13gtlt 10 0 E7m1t mV t ms How did this change your graph Explain what you think is going on here 220 Module10Half CenterSemiAnsweredxmcd NPB 100q Module 10 HalfCenter Copy your assignment and graph for Em1 them both here below Change the value of inj in your Em1 assignment to be 150 X 107 6 221 7 6 W HH150 X 10 0 U I I I E7m1t mV 7 50 I I I lUU 0 50 100 150 200 t ms How did this change your graph Explain what you think is going on here 222 Module10Half 13 CenterSemiAnsweredxmcd NPE mun Mudme m Hamemev Part III Bursting Goa s R ewew vu e uhmvaceHmav Caz m yegu a mg exee neaw aewny Extend HVH mudeHu mcmde mueceHu avC 239 Extend HVH mudg e m a Cahamwa ee w cunductance m m nne shuwmg bummg acwn n ms am yuu Wm an n n 7H ude e mcmde bummg behavmv Tha s when appmpnate y yu v n2uvun wwne a sequence Mammy pmennaxe and than qme even than n hem a ev a penud qumesenceJhe neuvun MM 3 me a se uenee e1 mudg Huwevemu assuve armphase benawenwe Wm need u add vempvuca mmbmun whmh We Wm u a ev m Pan w Caz Vuu may knqumm synapucUansmwssmmmat depmanzatmn uhveuvuns upens 393 Wm M v mneceHu av ca mum than causesvesm e msmn and ve ease uMeumUansmMev Heweven mmeases m mneceHu av Caz 09 can have mnen ewems asweH ena He s m We p asma membvane n b WS 5 an 012 pnesynapne eeu e We ene Wnn mgh 032 Onee anwa ethese charme s aHuW w e em hypevpmanzmg We neumn Bunny 3 buvs1 Mammy pmennawe CaZ L mcveaseswnh eaen depmanzauun unm Cahamwa ee w cunductances Caz s pumped um mm M Asa nM Ca Meememjan CemeLsemwAnmvedxmcd NPB 100q Module 10 HalfCenter The figure above is an experimental recording of a bursting neuron from Apysia a sea hare mollusc Voltage is shown in the lower trace while intracellular calcium is simultaneously recorded in the upper trace with a calcium sensitive dye Note that Ca2i increases during each burst of action potentials and then decays back to baseline between bursts We will use a Ca2activated K conductance to induce bursting in our quotneuronquot This will require us to create an additional differential equation that tracks Ca2i To be as simplistic as possible we won39t explicitly model the activation of voltagegated Ca2 channels and the subsequent Ca2 current Ca To relate Ca to Ca2i we would need to suddenly deal with such things as the geometry and volume of our cell which we have so far managed to avoid Let us continue to avoid these details and create a more empirical and simple representation of how Em affects Ca2i You probably already have an intuitive if only general feel for how Em and Ca2i are related When Em is at rest around 63mV Ca2i is low about 100nM When Em is depolarized Ca2i increases More depolarization means Ca2i increases faster and to higher steadystates perhaps up to tens or even hundreds of HM if we sustain the depolarization long enough Let us begin first by ignoring time That is we39ll simply ask if Em is a certain value at t 00 what is Ca2i We39ll call this Ca2i value Cainf because it is Ca2i at infinite time V thout too much rigamarole the following function for Cainf as a function of E m meets our intuition fairly well E7m005 CainfE7m 107 7e 0015 To get an idea of what Cainf looks like make a graph of CainfEm against Bi below You39ll In need to define Em as a range variable first ranging from 007 to 005 is probably fine 301 Eim 7007 70069 005 1x10 4 8x10 I I 1x107 6x10 5 CainfE7m 1x107 5 CainfE7m 4X10 5 1x10 2x10 5 1x107 8 39 39 u 39 7 50 0 50 7 50 0 50 Eim Eim mV mV Module10Half 15 CenterSemiAnsweredxmcd NPE mun Mudme m Hamemev Nme tha Cam Eim has mphcn mu av um s My hm because M 57 a buHHn Ma hcad umLWe hem buthev pu mg n m We denummamvwhen We make gvaphs Duesthe Czin Eimj when match um mmmun as hsted abuve abum hew steadyr ate CaZ L s yea eme E m7 Why mwhy new Tu get ah ea e What gumg Em nEaVrE mV yuu can change M the y aws u he Em a wee seaxe Nme tha m 7M1EIEIn mz Cain Wen deawmh tha atevwhen we make the coach 032 1 eehemahee Shea Use the ass gnmem upevatuv be uwm 52 GJC n he 9 x mquot The mphcn uhus have ave s GiKCa 9 x m WE H avbmavhy assume tha gjlt0a vanes hheahywuh 0321 heyeaneuus1 Cz Wuh g3lt0a GJC when Ca ma 1 am As yuu H see m yum atev swmu auuns Ca Wun t eve exceed m 6 suthwsassumptmn spmbab hhe Thws suggesmhe uhewmg mdmn m LKCa Ca LKCsE7mCa 7K0 57m 7 Egg m39 6 CVEa E W5 ass gnmem be uw Ca LKC E7mC GiKCa 57m 7 Fin m Meememjah CemeLsemwAnmvedxmm NPE mun Mudme m Hamemev sun Wen keep me same mmpuva vesuhmun 5 we Wun t change slaysize 15 Tn n n w m y H 77 Stan a secund va ue s slepsize aswa ue s end am we 14 El5tepaze and m nm Wm Mm n u u n onesewe atemenLWE H cveate a new ene nenn swatch m a pn ma Gwen 157mm ems nxnnzsz ml r 7 h1nn513 7m 7 IKE7mltn1t e 7N5E7mltmlth1t e Ueaka m e LKCs a m gem in e magma n e um e New mo in e magma n e m e enema mm gm 7 enema 1 7 mo 7 magma mo 1 Lch w eenne mm p ay mne n51 ODE abuven We ene ma ca cu ates E7m1 Append ma 7 sznmmn H v Jn and Ca useme su uuunsthat onesewe wm pmduce w E7m1 m and Cz1myespeewew When yuu ve dune the new ODE enema uuk hke W5 7m 7 Uqum nm e LNsUEimlO mm mo 7 I lea EimlOD 7 17mm a cmrEmlt at 0 Meememjam CemeLsemwAnwEvedxmcd NPE mun Mudme m Hamemev rquot y m Secund 3211 Use We Eun ean equaws swgnl 9 be uwm senhe mma Ca me szme be Sn x m because Cmnf zj 42m 1 9 cum Sn x m 9 cun5nxm 9 FuHhe ODE quaH We Hassumethat C2 meveases E0510 5 pusmve M32 5 essman Cain t C2 decveases E0510 5 negawe M32 5 wave man Cain t C2 s1ays the same ECMO s zem WCz Cain t a a Cain we when ICmnf 7 Cal s ang vwmm eenmemu ancv the uHuwme oDEwm accumphsh mesa assummmns 30511 Cmn EimlOD e cuojjm t Use the Eun ean equaxs swgnl m cveate W5 ODE be uw E0510 c mnfE7mlt e cu 1n t 39 The m m e ec detevmmmg the qmcknesswnh Whmh C2 appmaches Cam m essence ms s equauun saus esmem As yuu mu m Pan us yuu Wm nuw msh W5 su ve b uckwnh an Odesu ve smemem asswgned m 3 nj as an mpm u u hwsmndmn musesquot j Muememjam CEMELSemwAnwEvedxmm NPB 100d Module 10 HalfCenter Eiml Cal Create the function Burstslinj below and assign it to be Odesolve n1 tendiq step51ze m1 hl can create the vector using the Vector and Matrix Toolbar You39ll notice that the appearace of terms in the vector doesn39t have to correspond to the order in which the ODEs appear in the solve block Because ofthis we39ll group the variables we39re actually interested in Em1 and Ca1 at the top for convenience 310 Eiml Cal d Bursts iinj Odesolv n1 tend en step51ze m1 hl As mentioned in Part II we can39t directly graph the Burstslinj function However we can assign it to another variable and then graph that variable Let39s do that first with Em1 and then with Cal Use the assignment operator below to set Em1 to be Bursts6 X 10 60 using the subscript operator D This value of inj should give us some very brief bursts Depending on the speed of your computer it may take Mathcad some time to compute the solutions Just be patient 311 E m1 Bursts6 X 107 60 m U I 0 E7m1t mV 7 50 W W I I lUU 0 500 1x103 15x103 t ms Module lO Half 19 CenterSemiAnsweredxmcd NPB 100q Module 10 HalfCenter Copy one of your Em1t vs tgraphs from Part II and paste it above 312 Describe what you see in the graph Explain what you think is happening with Ca1 and gKCa and how this is affecting Em1 To confirm our intuition we39d now like to graph Cal on the same graph that we39ve plotted Em1 Use the assignment operator below to set Ca1 to be Bursts6 gtlt 10 61 using the subscript operator Again it may take Mathcad a while to compute this solution Cal Bursts6 X 107 61 Copy your graph of E m1t from above and paste it here below Enable the secondary yaxis and Ca1t 9 on the secondary yaxis The 10 implies that the units here are nM although M 9 place 10 isn39t a builtin mathcad unit so we39re just sort of pretending Make the graph more narrow so that it fits on the page Set the secondary yaxis to range from 0 to 1000 315 0 I I 1 103 39800 0 Em1t 39600 Ca1t mV 9 750 400 10m x W LK 200 39 1 r l lUU U 0 500 1x103 15x103 t ms Module lO Half 20 CenterSemiAnsweredxmcd NPE mun Mudme m Hamemev cm ury 75m n E71111 mV 7 v u x v I ma 40D 5m 2m mun nun 1 n Aswe mu m Pan w nuw M s see Wm happens e W5 eeu when We mquot up meme I nj 5 me eehmeax de a We shumd eeamm muuuh m makmu We WEI sEpava e assmnmems E m1 and cm abuvewe ve essermaHv askmu Mame e mme swmu atmn Mme me my each meme Because We swmu atmn aheadvtakes mm a m emme m mm 5 sun ewe paw Them m Eus39s6 x M M s say run And Wen asswgn E7m1 and Cz1meyms emquot se m We space be uw cveate sumethmg hkemws mesmeskxm J 5 ml quotu m run 316 mu e Emb x m 2 Nude matwe usedvecmvsm asswgn hem E7m1and Cz1m a smg e s1ep WME 2 23371LW5 s m necessavy We eeum have de ned E7m1 and Cz1meyms enun Wm sepavate vegmns Mudmem H am 21 CEMELSemwAnwEvedxmm NPB 100d Module 10 HalfCenter Copy your graph of Em1t and Ca1t from above and paste it here below You should get the same graph as above because inj is set to the same value 317 EM Cal 10gt mV 10 9 t ms Now we can turn up thejuice Change the run assignment above such that inj is 9 X 107 6 318 What do you see in the graph now How is this different from when inj was 6 X 10 6 319 Make a new simulation below by copying and pasting the two assignments involving run from above Change inj forthis new simulation to be 13 X 10 6 320 Iun Bursts13gtlt 107 6 MAM Module10Half 22 CenterSemiAnsweredxmcd Module 10 HalfCenter NPB 100q Copy your graph of Em1t from above and paste it here below 321 N I 1 103 800 E7m1t 0 600 Ca1t mv 39 400 10 9 MW ii Iiiiilliii i I 200 I l lUU U 1x103 15x103 How did increasing inj to 13 X 10 6 affect what you see in the graph What do you think is going on 321 Module10Haf CenterSemiAnsweredxmcd NPB 100d Module 10 HalfCenter Part IV HalfCenter Goals Use Bursting HH model to simulate two cells simultaneously Extend the Bursting model to include reciprical inhibition Graph Em in time for both cells showing antiphase bursting behavior Having created a Bursting model we will now extend the model to include two cells with inhibitory synaptic connections between them Arbitrarily we39ll assume that GABA is neurotransmitter at these inhibitory synapses GABA binds to GABAA receptors which are ligandgated Cl39selective channels For simplicitly we will drop the A subscript and refer to these as simply GABA receptors Once completed these two cells will show antiphase bursting behavior That is one cell will fire bursts while the other quiet and vice versa As you39ll eventually see this antiphase behaviorwill involve longer bursts that you saw in Part III Consequently we39ll need to extend the duration of the simulations Assign end below to be 2 Then below that reassign tto be a range variable as before in Parts ll amp start at 0 second value is stepsize last value is end 401 m 2 t 05tepsize end Before addressing the GABA connections between the two cells let us simply expand our Bursting model from Part III to include two cells To do this copy you solve block from Part III and paste it here You will need the 9 regions around and including the given statement You will also need to separately copy the two regions controlling Ca its initial condition and the ODE When you39re done you39ll have a total of 11 regions below Do not copy your Odesolve statement Given E7m10 70063 n10 0353 m10 007 h10 0513 cimE7m1t 17in 7 IiKE7m1tn1t 7 IfNaE7m1tm1th1t 7 IileakE7m1t 7 IfKCa t nl t 0mE7m1t1 7 n1t 7 BnE7m1tn1t t Module10Half 24 CenterSemiAnsweredxmcd NPB 100q Module 10 HalfCenter iml OlmE7m1t1 7 m1t 7 BmE7m1tm1t t 110 ozhE7m1t1 7 1110 7 BhE7m1th1t t Ca10 50x107 9 rm t CainfE7m1t 7 Ca1t10 t Just to be nice here39s the rest ofthe solve block forthe running the second cell cell2 Notice that the equations are identical to the first cell except fortheir labeling eg Em2 Caz etc The initial conditions are identical as well except for the E m2 Ratherthan set cell2 to start at E m 0063 as is the case with cell1 cell2 is set to start slightly more depolarized at E m E leak Eileak 70054 Once we have the whole halfcenter model assembled this difference in starting conditions will be necessary to start the antiphase behavior If instead both cells are set to precisely the same initial conditions they will remain on a quotknifeedge of stabilityquot an unstable steadystate where neither cell can get ahead and take control of the first burst The small difference in initial conditions is the little push necessary for one ofthe cells to take charge E7m20 Eileak 1120 0353 m20 007 1120 0513 cimiEimZO Iiinj 7 IiKE7m2tn2t 7 17NaltE7m2lttgt m2t h2t 7 IileakE7m2t 7 17KCalt gum unltEm2lttgtgtlt1 7 ram 7 BnltEm2lttgtgtn2lttgt imam amltEm2lttgtgtlt1 7 m2lttgtgt 7 BmltEm2lttgtgtm2lttgt h2t ozhE7m2t1 7 h2t 7 BhE7m2th2t t Ca20 50 X 10 9 Ca2t CainfE7m2t 7 Ca2t10 t Module10Half 25 CenterSemiAnsweredxmcd NPB 100q Module 10 HalfCenter Finish the solve block by creating the function HalfCenterlinj below and assigning it to be Eiml Odesolv end tend Be sure to get the order of the 139s and 239s correct in the vector m1 stepsize because we39ll rely on that when we make our run statements You39ll notice that for convenience all the Em39s and and Ca39s are grouped together at the top Also notice that this HaIfCenterlinj function like past Odesolve functions accepts l inj as an input llE il39l l 403 E7m2 Ca2 end HalfCenter imJ Odesolve tend step51ze hl m2 llhzl i As you did in Part III let39s use the dummy variable run to avoid having Mathcad do more workthan necessary From there we39ll make assignments for Em1 Ca1 Em2 and Caz so we can graph them Use the assignment operator below to assign run to be HalfCenter9 X 10 6 Note that we used this same value of Linj in Part III to make 3action potential bursts so should see the same results here 404 W HalfCenter9 gtlt 107 6 Module10Half 26 CenterSemiAnsweredxmcd NPE mun Mudme m Hawcamev Makethe uHuwmg asswgnmemm awed him run We sumuuns We Wanna gvaph Em mu m m 57m 39 run on ml mu m u quotH 57m 39 run cu m1 n 3 mm mm3 Muumemjam CEMELSemwAnwEvedxmm NPB 100q Module 10 HalfCenter In the lower copy of your Em1 graph change Em1t to Em2t Also change in the same graph Cal t to Ca2t Change the trace colors on this second graph to remind you that you39re looking at a different cell 407 Describe what you see in the two graphs You should be able to see that Em2 becomes slightly delayed towards the end of the simulation Why don39t the two cells show the exact same timing in their late bursts 408 Now that we have two cells let39s connect them together As mentioned previously we will assume that the cells are connected by GABAergic synapses GABA receptors pass Cl ions and the Nernst potential for CI39 is typically quite negative for neurons To reflect this we will assume that the reversal potential for GABA receptors EGABA is also quite negative Use the assignment operator below to assign EGABA to be 0090 90mV without units EiGABA 70090 Assign the maximal GABA conductance GGABA below to be 3 X 10 4 This value is somewhat arbitrary but should work for our purposes The implicit units are cm GiGABA 3 X 10 4 You may recall from previous modules that we have often used a quotalpha functionquot in time to represent the changes in Po for synaptic receptors due to release binding and removal of neurotransmitter in the synaptic cleft We could use that technique here but it would require us to know the exact time when action potentials are generated in the presynaptic neuron and we don39t have that knowledge a priori To avoid that complication let us simply say that the GABA current IGABA in the postsynaptic cell is proportional to the value of Ca in the presynaptic cell This assumption is actually fairly reasonable Neurontansmitter release really does depend on Ca in the presynaptic neuron And further the quick rise in Ca during a presynaptic action potential followed by a slow Ca decay fairly mimmicks the quick release and slow clearance of neurotransmitter from the synaptic cleft Like gKCa we will arbitrarily assume that gGABA GGABA when presynaptic Ca is 10396 1pM Module10Half 28 CenterSemiAnsweredxmcd NPE mun Mudme m Hawcamev FuHuwmg We Exammg mm LKCa LGABA s vepvesemed svm av y LGAEAE7mpastCapxe 7 GiGAEA C P Eimpastr EiGAEA m 6 m m y w m Eim and Ca va ues quotme hatthese paveme evs ave deswgnamd as Eimpnsl and Carre LGAEAE7mpa5LCapre 7 GiGAEA M Eimwm EiGAEA m 6 E Eim 1 y mquot VFJVZ Caz etc mm abuve and paste n have be uw Vuu H have a mm mm vegmns mdudmg m Gwen s atemem m Gwm 57m 1m 7mm nxnnzsz ml7 h1nn513 aim 5310 7m 7 IKE7mltn1t 7 7N5E7mllmlth1 7 Uuka m 7 Lch in W105an 1 10 5151an n10 Elm MGJHWD 1 HOD Mum mm ah 1051an 1 h10 WEJMOD hm 9 cun7 EIXIEI E0510 7 c 153an 7 Caltl t Muumemjam CEMELSemwAnwEvedxmm NPE mun Mudme m Hawcamev Appendthe abuve ODEVm EJM m subvert aw LGABA ns12ad u usmg me genen mpms Eimn q n r m E Be caveuabuutth21 s and 2mm When yuu ve aunewe new ODE 5mm uuk m m aim SEimIO Ln 7 LIQEimIO mm 7 LNsULmIO mm hlt 7 Lleal iimlo 7 1qu t m y mquot yam F m r cm etc mm abuve and paste n have be uw Vuu H have a ma DMD vegmns m 57mm 7 um ram 7 u 353 mm 7 n In mm 7 u 513 aim 5320 7 17m 7 LIQEimZO mm 7 LNsUEianO mm 4120 7 Jankaimzojj 7 Lch gum mum 1 r n20 r mum n20 gram 7 mum 1 7 mm 7 tnEm1t ma Elma mum 1 7 mo 7 mum hm cum 75m 1 9 ECHO 7 c mama 7 cmmm t Appendthe abuve ODEVm EimZ m subvert aw LGABA ns12ad u usmg me genen mpms Eimn q n r m E Be caveuabuutth21 s and 2mm When yuu ve aunewe new ODE 5mm uuk m m aim SEianO Ln 7 Lm o nzo 7 ma a mZO mo 7 Lhasa ojj 7 1qu t m Muumemjam 3 02mm emmnmvmmm NPB 100q Module 10 HalfCenter Copy your Odesolve statement from above the HalfCenterlinj one and paste it here below Copy the two assignments involving run and paste them below as well 416 llE illl l E7m2 Ca2 end HalfCenter Iimj Odesolve tend step51ze h1 m2 will W w CenterSemiAnsweredxmcd NPB 100d Module 10 HalfCenter Copy your pair of graphs for Em1t and Em2t from above and paste them here below 417 u I I I 1 103 39800 E7m1t 39600 Ca1t mV 750 400 10 9 x 39 T W in W x W x 200 I N N X N l I quot I quotI I nquot lUU U 0 500 1x103 15x103 2x103 t ms U E7m2t Ca2t mV 9 750 quot10 7 lUU What do you see on your graphs Do you see bursting like you did before we added IGABA Are the bursting durations the same as before What about their spacing in time When do the bursts occur in each ofthe ces Do the bursts occur at about the same time in each cell Or are the bursts in antiphase Or do the bursts follow some other pattern Module10Haf 32 CenterSemiAnsweredxmcd NPB 100q Module 10 HalfCenter Let39s see what happens when we turn up the juice inj Copy the two statements involving run from above and paste them here below Change the inj parameterfor HalfCenterinj from 9 X 107 6to 13 X 107 6 419 m HalfCenter13 X 107 6 mow m m 11 lmnsl Copy your pair of graphs for Em1t and Em2t from above and paste them here below 420 U I I I 1 10 800 E7m1t 600 Ca1t V V 7 9 esouill l x J Hill illlllll i400 39 r 200 T I Nquot I N lUU U 0 500 1x103 15x103 2x103 t ms U I I I 1 103 800 E7m2t 600 Ca2t V 79 m 50 MIN NH ill 400 quot10 l 391 I v 200 I hquot I I lUU U 0 500 1x103 15x103 2x103 t ms Module lO Half 33 CenterSemiAnsweredxmcd NPB 100d Module 10 HalfCenter Describe what you see in these graphs How did increasing Iinj change the pattern of bursting in the two cells 421 You used this same value of Iinj 13 X 10 6 back in Part III before we added the second cell How did adding the second cell and its inhibitory connection change the activity ofthe single cell at this level of stimulation ie Iinj 13 X 107 6 422 Suppose this halfcenter model were driving muscle contractions in a cat while walking such that a full period of bursts represented one step How might you change this halfcenter model to make the cat walkfaster Test this by changing your model What seems to work Are there any limitations to this strategy Use the next page to make any graphs you need to support your conclusions Module10Half 34 CenterSemiAnsweredxmcd Module 10 HalfCenter NPB100q 7 6 m HalfCenter14 X 10 m0 W Iun 8 1 533 run3 U I I I 1 103 800 E7m1t I 600 Ca1t mV 400 10 9 vy IIHHX ml ylH m I ll Nquot 200 lUU I I I U 0 500 1x103 15x103 2x103 t ms U I 1 103 800 E7m2t 600 Ca2t mv x 400 10 9 750 l rll I J llrl x KN V200 I I I lUU U 0 500 1x103 15x103 2x103 t ms Module 10 HalfCenter Last Modified November 26 2007 745pm Module IO Half 35 CenterSemiAnsweredxmcd NPB 100d Module 10 HalfCenter Help Improve This Module Completing this section is not required but will be helpful for future development ofthis module Your answers here will not affect your grade About how much time did you spend completing this module Were there any particular parts of this module that you found too difficult too confusing or too tedious How might these sections be improved Were there any particular parts that you found especially useful or helpful in understanding the material from NPB 100 Other suggestions kudos criticism Module10Half 36 CenterSemiAnsweredxmcd NPE mun Mudme m Hawcamev 57m 10 c u 1 mo cuojj Muumemjam CEMELSemwAnwEvedxmm NPB 100q Module 10 HalfCenter E7m1t Cal t Module10Haf 38 CenterSemiAnsweredxmcd NPB 100q Module 10 HalfCenter E7m2t Ca2t Module10Haf 39 CenterSemiAnsweredxmcd NPB 100q Module 10 HalfCenter Eiml t Cal 0 7 IiGABAE7m1 t Ca2t Module10Haf 40 CenterSemiAnsweredxmcd NPE mun Mudme m Hawcamev n10 cuojj 7 LGAEAE7mlt cmg EimZO pm 7 LGAEMme cuo n20 pm 7 LGAEMEJMO cuojj Muumemjam CEMELSemwAnwEvedxmm


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