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GS-1 Section III Question 83

GS-1 Section III Question 83

Postby admin » Tue Aug 10, 2010 3:43 am

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GS-1 Section III Question 83 (BOOK)

Postby miri » Sat Aug 18, 2012 11:48 pm

The book equivalent of this is using the same table is:

If the concentration of potassium outside a mammalian motor neuron were changed to 0.55mol/L, what would be the predicted change in the equilibrium potential?
A 12mv
B 120 mV
C 60 mV
D 600mV

My reasoning (below does not get me the correct answer). Can anyone point out where I am going wrong?

1) we are given this equation:
E(k) = 60.log ([k+] out / [K+] in)

2) according to the values in the table, [k+] would decrease by a factor of 10 (from 5.5 to 0.55)

3) assuming that [k]o remains the same, then Ek = 60.log(-)10 = -60mV

4) If we compare the original E(k) and new E(k) then the difference is 30.

I don't see how this gets a predicted change in the EQ potential of 120!!
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Re: GS-1 Section III Question 83 (BOOK)

Postby miri » Sun Aug 19, 2012 12:05 am

miri wrote:The book equivalent of this is using the same table is:

If the concentration of potassium outside a mammalian motor neuron were changed to 0.55mol/L, what would be the predicted change in the equilibrium potential?
A 12mv
B 120 mV
C 60 mV
D 600mV

My reasoning (below does not get me the correct answer). Can anyone point out where I am going wrong?

1) we are given this equation:
E(k) = 60.log ([k+] out / [K+] in)

2) according to the values in the table, [k+] would decrease by a factor of 10 (from 5.5 to 0.55)

3) assuming that [k]o remains the same, then Ek = 60.log(-)10 = -60mV

4) If we compare the original E(k) and new E(k) then the difference is 30.

I don't see how this gets a predicted change in the EQ potential of 120!!


Further to this - I have just realised that the E(k) equation is given in mmol whereas the question says the change is to 0.55 mol/L.

So does this mean that:
(a) [k] is increased by a factor 10^3
(b) E(k) = 60.log ([k+] out / [K+] in) = 60*log10^3 = 60*3 = 180...
(c) HELP please!!
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Re: GS-1 Section III Question 83

Postby goldstanda3269 » Wed Nov 21, 2012 5:23 pm

You were getting closer with your second post.

If [K+]o = 0.55 mol/L = 550 mmol/L then this is 100 times the value of [K+]o in Table 1 which is only 5.5 mmol/L. Thus 60log(100) = 60log 10^2 = 120 mV.
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Re: GS-1 Section III Question 83

Postby goldstanda3269 » Thu Nov 22, 2012 4:31 am

Now let’s pretend that you did not notice how the problem could be solved easily and you had 2 other things that you won’t have for the GAMSAT: lots of time and a scientific calculator. If you did, then you would do the long calculation using the approximation of the Nernst equation given in the passage, using the data in Table 1 where [K+]o = 0.55 mol/L and [K+]i = 150 mol/L and your scientific calculator would estimate Ek to be -86 (not far from the real value of -90 in Table 1). If you then used [K+]o = 0.55 mol/L = 550 mmol/L in the same equation, your calculator would suggest about +34. And so, the difference would be 34 – (-86) = 120 mV, success! But, there would not be enough time to do the other questions!
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Re: GS-1 Section III Question 83

Postby laura.cp897117 » Fri Feb 26, 2016 10:06 pm

Can you please explain the logic of using Ek = 60log(100) as the equation? I understand that the concentration has increased x100. But I don't quite understand how we can sub 100 for [K+]o / [K+]i?
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Re: GS-1 Section III Question 83

Postby goldstanda3269 » Mon Feb 29, 2016 3:35 am

laura.cp897117 wrote:Can you please explain the logic of using Ek = 60log(100) as the equation? I understand that the concentration has increased x100. But I don't quite understand how we can sub 100 for [K+]o / [K+]i?


Yes, there is a step in between which, after some times of seeing these types of problems, you will begin to skip, but you are right to ask for it!

First a reminder, since the following rule is essential to understanding what happens next (GAMSAT Math section 3.7):

log(MN) = log M + log N

Now the equation we are given:

Ek = 60 log ( [K+]o / [K+]i )

If [K+]o = 0.55 mol/L = 550 mmol/L then this is 100 times the value of [K+]o in Table 1 which is only 5.5 mmol/L, and so we have:

Ek = 60 log ( 100 [K+]o / [K+]i ) = 60 log ( 100 ) + 60 log ( [K+]o / [K+]i )

So now it is evident that the original Ek which is 60 log ( [K+]o / [K+]i ) has been changed by (increased by) the amount 60 log ( 100 ).
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Re: GS-1 Section III Question 83

Postby sukij » Thu Mar 17, 2016 10:37 pm

why are we NOT taking the value of 150mmol/L from the table as it is the inside cell value ..which is what {K+}i is supposed to be?
the value mentinoned in question is for extra cellular value right?
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Re: GS-1 Section III Question 83

Postby goldstanda3269 » Thu Mar 17, 2016 11:48 pm

sukij wrote:why are we NOT taking the value of 150mmol/L from the table as it is the inside cell value ..which is what {K+}i is supposed to be?
the value mentinoned in question is for extra cellular value right?


We are doing a mathematical manipulation of the equation in a way to avoid using the original values (i.e. original inside and outside concentrations). If you use those original values, you will triple your work and get the same answer (eventually!). So to simplify the calculation, we focus on the change. This is commonly done in GAMSAT math. This is the calculation that we put above:

Now the equation we are given:

Original: Ek = 60 log ( [K+]o / [K+]i )

If [K+]o = 0.55 mol/L = 550 mmol/L then this is 100 times the value of [K+]o in Table 1 which is only 5.5 mmol/L, and so we have:

Ek = 60 log ( 100 [K+]o / [K+]i ) = 60 log ( 100 ) + 60 log ( [K+]o / [K+]i )

So now it is evident that the original Ek which is 60 log ( [K+]o / [K+]i ) has been changed by (increased by) the amount 60 log ( 100 ).
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Re: GS-1 Section III Question 83

Postby HunterEd » Sat Mar 06, 2021 1:17 am

If the equation is Ek = 60 log ( [K+]o / [K+]i )

why are we using log(MN) = log M + log N

Should it not be log(M/N) = log M - log N ?

Thanks
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Re: GS-1 Section III Question 83

Postby goldstanda3269 » Sat Mar 06, 2021 3:51 am

HunterEd wrote:If the equation is Ek = 60 log ( [K+]o / [K+]i )

why are we using log(MN) = log M + log N

Should it not be log(M/N) = log M - log N ?

Thanks


I can't help but quote the overused: "Just because you can do something, does not mean that you should do it!"

If you have the equation for velocity: v = d/t

If distance d is constant, and the event takes 4 times longer, how does it affect velocity v?

v = d/(4t) = d/t (1/4)

d/t is the original v, now we now that v is a 1/4 of its original value.

But, we did we not do this:

v = d/(4t) = d x 1/4 x 1/t

we would be applying the rule: d/t = d x 1/t and then randomly place 1/4 anywhere, which is also permitted for multiplication. The equation is true but unhelpful. It is better to keep the original equation together so we can see how it was changed in the process.

Ek = 60 log ( 100 [K+]o / [K+]i ) = 60 log ( 100 ) + 60 log ( [K+]o / [K+]i )

Ek = 120 + 60 log ( [K+]o / [K+]i ) = 120 + original Ek
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Re: GS-1 Section III Question 83

Postby HunterEd » Sat Mar 06, 2021 11:47 pm

Oh, wow. Silly error on my part. I was taking Ko to be 100 as opposed to realising its 100 (times) its original value.

so I was imagining this EK= 60 log 100/Ki as opposed to Ek= 60 log 100(Ko/Ki)

Thank you again for your explanation.
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