Thread: Chlorophyll-binding protein

Hi all,

I'm a third year undergraduate doing his final-year project. I'm really confused about something and could really do with some quick help as deadline is soon. I have worked out the concentration of chlorophyll within a solution (3 mg/ml). Within this solution the chlorophyll is bound to protein (LHC-II). Each monomeric LHC-II protein binds 14 chlorophyll molecules. Using this information how do i calculate the total LHC-II (protein) concentration of the solution.

Any help will be much appreciated

Originally Posted by whusean

Hi all,

I'm a third year undergraduate doing his final-year project. I'm really confused about something and could really do with some quick help as deadline is soon. I have worked out the concentration of chlorophyll within a solution (3 mg/ml). Within this solution the chlorophyll is bound to protein (LHC-II). Each monomeric LHC-II protein binds 14 chlorophyll molecules. Using this information how do i calculate the total LHC-II (protein) concentration of the solution.

Any help will be much appreciated

Hi Whusean,

So the ratio is

Chlorophyll : LHC-II
14 : 1
3mg/ml : x

Remember that cconcentration = moles/ volume i.e. c = n/V

And moles = mass/molar mass i.e. n = m/M

The molecular weight of chlorophyll-a is 893.49 and the molecular weight of chlorophyll-b is 907.47. The molecular weight of LHC-II is ~29.6kD (type 1) or 27.5kD (type 2) or 26.0kD (type 3).

Does your solution contain chlorophyll-a or -b, or both? Which type of LHC-II does your solution contain? What volume of solution do you have?

Best wishes,

Tri~

Thanks for the reply.

the molecular weight of the LHC-II monomers i used was 29.6kD.

In order to find out the concentration I put a pigment mixture (chlorophyll a and b, and carotenoids) into a UV/Vis spectrophotometer. I think the volume of pigment is used was about 800 microlitres. I then used the spectra to calculate the total chlorophyll conc.

Sorry, could you possibly explain how to calculate the conc in a slightly different way if that is possible.

I appreciate the reply, thanks

Hiya,

Since the total chlorophyll concentration of your solution is known and you know the volume of the solution, it should be possible to calculate the number of moles of chlorophyll in the solution (n = c x V). The overall mass of chlorophyll can be calculated by multiplying the number of moles of chlorophyll in the solution by the molar mass of chlorophyll. It should be possible to work out the overall copy number of chlorophyll molecules in the solution using Avogadro's constant. The total number of LHC-II molecules in the solution will be 14 times as great. So, if the molecular copy number and Mw of LHC-II are known, and the volume of solution is known, I think that it should be possible to extrapolate the overall LHC-II concentration.

I think.

Mathematics definitely isn't my strong point, as you will have noticed. Other members: can you help?

Thanks,

Tri~

But because i have chlorophyll a and chlorophyll b in my solution....what molecular mass shall i use (a or b).

So far ive calculated: moles = conc x vol = 3 mg/ml x 0.8 ml (800 uL) = 2.4 mol

mass = moles x Mw = 2.4 x molar mass of chlorophyll ??? which chlorophyll

Once i have worked that out, you talk about working out a copy number. I have not come across this term before could you explain.

Im pretty rubbish with the maths side of biology as well, i find it very petty

Hi Whusean,

So, here's what I get:

Chlorophyll: LHC-II
14 : 1
3mg/ml : x mg/ml
Volume = 0.8 ml

Mw chlorophyll-a = 893.49 g mol^-1
Mw chlorophyll-b = 907.47 g mol^-1
Mean Mw chlorophylls a+b = 900.48 g mol^-1
Mw LHC-II = 29.6kDa

Remember that
n = m/M
n = c x V
1 Da = 1.660 538 921(73)x10^-27kg
NA (Avogadro's constant) = N (copy number)/n (moles)

mass (chlorophyll) = (3mg/ml)*(0.8ml) = 2.4mg total
n = m/M
n = 0.0024g/900.48 g mol^-1
n = 0.0000026652452 moles chlorophyll

N (chlorophyll) = (6.02214129(27)x10^23 mol^-1)*(0.0000026652452 moles) = 1 605 048 317 409 047 004 molecules chlorophyll total
(1 605 048 317 409 047 004)/14 = 114 646 308 386 360 500 molecules LHC-II

1Da = 1.660 538 921(73) x10^-27 kg
29.6kDa = (29.6x10^3)*(1.66053892173x10^-27) kg = 0.000 000 000 000 000 000 000 049 151 952 08 kg

(0.000 000 000 000 000 000 000 049 151 952 08 kg)*N(LHC-II)
(0.000 000 000 000 000 000 000 049 151 952 08 kg)*(114 646 308 386 360 500) = 0.00000563508985595529342160484 kg LHC-II = 5.64 mg LHC-II in 800ul

5.64 x 1.25 = 7.05 mg ml^-1


It really wouldn't surprise me if I have got this wrong, though. Can anyone else verify or nullify please?


Thanks,

Tridimity