12 posts
• Page **1** of **1**

Be the first to discuss this question!

- admin
**Posts:**1065**Joined:**Tue Jun 29, 2010 7:47 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!!

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!!

- miri
**Posts:**10**Joined:**Mon Apr 23, 2012 3:26 pm

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!!

- miri
**Posts:**10**Joined:**Mon Apr 23, 2012 3:26 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.

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.

- goldstanda3269
**Posts:**1722**Joined:**Wed Aug 25, 2010 10:59 pm

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!

- goldstanda3269
**Posts:**1722**Joined:**Wed Aug 25, 2010 10:59 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?

- laura.cp897117
**Posts:**27**Joined:**Mon Mar 02, 2015 7:42 pm

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 ).

- goldstanda3269
**Posts:**1722**Joined:**Wed Aug 25, 2010 10:59 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?

the value mentinoned in question is for extra cellular value right?

- sukij
**Posts:**2**Joined:**Sun Sep 20, 2015 9:45 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 ).

- goldstanda3269
**Posts:**1722**Joined:**Wed Aug 25, 2010 10:59 pm

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

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

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

Thanks

- HunterEd
**Posts:**45**Joined:**Fri Jan 17, 2020 6:10 pm

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

- goldstanda3269
**Posts:**1722**Joined:**Wed Aug 25, 2010 10:59 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.

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.

- HunterEd
**Posts:**45**Joined:**Fri Jan 17, 2020 6:10 pm

12 posts
• Page **1** of **1**

Return to Section III: Reasoning in Biological and Physical Sciences

Users browsing this forum: No registered users and 1 guest