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Homework Help: Eadie Scatchard Plot

  1. Nov 24, 2011 #1

    I'm having trouble plotting this data

    Is there a way to attach the hw assignment so that you guys can view it?

    Anyways, I'll type it for now:

    We need to determine the dissociation constant for L (ligand) and the number of L binding sites on P using a scatchard plot.

    here is the problem:

    L is available in a radioactive form with an activity of 10^9 disintegrations per second per micromole of L. You set up a series of equilibrium dialysis experiments and add to each side:

    There's a drawing of a dialysis membrane with the left side where you add one micromolar of protein and to the right you add various quantities of L.

    It then says that after equilibration, you withdraw 10 microliters from the left and right side and determine how many disintegrations per second from each in 6 experiments with different amounts of radioactive L added: these are all in DPS=disintegrations per second)

    25,000 10,000
    44,000 20,000
    60,000 30,000
    100,000 60,000
    135,000 90,000
    200,000 150,000

    Our professor gave us the equation that b/f= -1/ksb+n[E]t/ks

    So the y intercept is Sbound/Sfree while the x intercept is Sbound.

    So I figured that the left side was the Sbound and the Sfree so in order to get the Sbound I had to subtract the right side which represents the Sfree. So the data is

    bound free
    15,000 10,000
    24,000 20,000
    30,000 30,000
    40,000 60,000
    45,000 90,000
    50,000 150,000

    And now he wants it in molar for x......so I was thinking that you take the S bound and change it to molar so you have 15,000* [10^9 disintegrations/(second *μmol)/ 10 microliters].....I did that and got

    Sbound S bound/Sfree
    1.5E+24 0.15
    2.4E+24 0.12
    3E+24 0.1
    4E+24 0.066666667
    4.5E+24 0.05
    5E+24 0.033333333

    and for my graph i got y=.2 ???? So i know there's something wrong. Also i know for the units like the micromole divided by 10 microliters i dont get molar right? I get 10^9 disintegrations/(second*umol*10*microliters)?
  2. jcsd
  3. Nov 24, 2011 #2


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    Not followed everything but there appears to be a tenfold factor error in your column of Sbound/Sfree. Also if E+24 means X1024 I'm pretty doubtful you can ever get such a factor here whatever it represents.

    For figs if you can scan them pics can be put up by loading them into tinypic.com where there is a website or a messageboard size option on the 'resize' button. You just copy what it gives you and paste it here and it comes up - there used to be a delay here for checks but I think there isn't any more.

    But don't forget to label figs with what they are, and the units as well as numbers.
  4. Nov 24, 2011 #3

    I don't have a scanner with me now......actually the assignment is in pdf format. I don't know if that makes a difference? But yea....E24 mean 10^24....

    I'm confused if I'm doing unit conversions wrong?

    But I think our professor wants the x axis in molar and the y axis shouldnt have any units because Sbound/Sfree should cancel.
  5. Nov 25, 2011 #4


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    1μmole/10μliters = 0.1 mol/L

    I have just skimmed both posts so I can be missing something, but it nicely fits the tenfold difference.
  6. Nov 27, 2011 #5


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    Somebody just posted on this thread but seems to have deleted his post.

    I do not see the difficulty.

    The problem seems to have been designed for didactic purposes. The results are quoted in round tens of thousands dpm. The specific activity dpm/μmole is a round number, 109, which you might or might not aim to achieve in a real experiment and then might or might not succeed exactly. The measurements have all come as round numbers of tens of thousands.

    10,000 dpm = 104dpm = 104/109 = 10-5μmoles

    but this is in 10μl so it is 10-6μmoles/μl i.e. 10,000 dpm means 1μM solution.

    So it is particularly easy to plot the results in μM units.

    The slope is -1/ks, or if you rotate the graph, use the axes the other way round the slope is ks.

    This gives a ks in the region of 3μM. Just plot the results Sbound against Sfree directly and you will see whether this looks a reasonable result.

    The horizontal (Sbound) intercept gives you the total molarity of binding sites, the amount bound at saturation. The concentration of protein was not mentioned, which you would need to see the number of binding sites per protein molecule. (But maybe this was a round number too and it is a hexamer? :wink:)
    Last edited: Nov 27, 2011
  7. Nov 27, 2011 #6
    Thank you for clearing up the units for me. Now I got a linear regression line that actually makes sense. Our professor told us in picture that he put one micromolar of protein so I'm guessing that's the E total I need to solve for the number of binding sites?...anyways, when i solved that i got .66 moles? am I supposed to come out with a whole number
  8. Nov 28, 2011 #7
    I'm probably in your class UCD MCB123

    We can help each other out if you like, but I sort of don't do my homework until the last minute, as you can see.

    Unit Conversion
    You want Mol/L for concentration of S,
    but you must start with DPS.
    So DPS/(Activity of DPS), where Activity = 10^9 DPS/uMol gives you 10^-9 uMol.
    Now take uMol and divide it by 10uL; 10uL because that's how much you extracted in the experiment.
    Now you have uMol/10uL. u cancels out, gives you Mol/10L
    And the final conversion is (1/10^9)(1/10) Mol/L = 10^-10 Mol/L

    For your confirmation:
    Ks = 3 Mol^-1
    n = 6

    Edit: I also noticed this is my first post.... Yay for me :D
  9. Nov 28, 2011 #8


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    Slight reserve on that last phrase. Putting yourself in the shoes of the experimenter he hopes it will come out approximately a whole number which it should if he has not made any errors, if what is in the dialysis cell is what he thinks ìt is, if the protein is pure, if its concentration is what he thinks it is, if it is 100% active in binding and not a bit denatured or maybe got some other stuff tightly bound to it. Then if he gets approximately a whole number he can allow himself to believe that that whole number is what it is.

    You must have made some other mistake, put something upside down or whatever. But rather than an 'external' "am I supposed to come out with" anything you will get less stress and not less errors but less uncorrected errors and be more on top if you do your own checks for reasonable answers. I mean 0.66 is not totally unreasonable in itself - it might be 0.5 with a fairly big error or 1 with a bigger error. But look at the last result in your second table. That is telling you in that experiment 5μmoles of ligand are bound to what you now essentially tell us is 1μmole protein. So you know your number should be at least 5. If you'd done this
    you'd have known it too.

    When you have suspicions it is probably easier to sniff out your error or it may be easier to do it all again from scratch.
    Last edited: Nov 28, 2011
  10. Nov 28, 2011 #9
    at ieatpasta: Lol, yup, mcb 123 with Wilson? I don't know how much of help I would be to you because honestly, I am like sooooooo lost in this class....

    On a more serious note: I checked my math and I got 6 binding sites too. But i think you have to look at your units again. I think it's 3 micromolar for the Ks and its not inverse.
  11. Nov 28, 2011 #10
    at epenguin, yea it was my math that was wrong. It turns out that the answer was 6. :)
  12. Nov 28, 2011 #11
    You're right, but it's not micro. Sb/Sf = no units.
    1/Ks * Mol/L + (1+Mol/L)/Ks = 1
    1/Ks * (Mol/L + Mol/L) = 1
    Ks = Mol/L = M
  13. Nov 29, 2011 #12
    I divided everything by 10E-10 to get it all in M. Then I made the Bound/Free vs Bound plot...but got a slope of -0.3. Shouldn't it be -3 so that Ks is 3? Anyone know what I did wrong? Any help would be greatly appreciated!
  14. Nov 29, 2011 #13
    It might be a typo you put, but you divide by 10^10, and not 10^-10. Your final units of Sbound should be 10^-6

    Sbound/Sfree should be from 0 to 1.5 and Sbound should be in units of 10^-6 Mol/L If you forget to put all Sbound in 10^-6, you're slope will come out as -0.333333 instead of 333333

    It is 10^-6 because:
    Sbound+Sfree - Sfree = Sbound
    25000 - 10000 = 15000

    15000/10^10 = 1.5 x 10^-6 is your first value of Sbound.
  15. Nov 29, 2011 #14
    ok yea that makes sense. Like I have 3 micromolar instead of 3 X10^-6. Like I just thought it would be easier to plot the values without all the extra zeros
  16. Nov 29, 2011 #15
    thanks! so i had a slope of 333333 the first time i did it, but how does this give us a Kd of 3? would only make sense if it was in micromolar, but didnt the micro cancel out?
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