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Using Subcellular Fractionation to Separate Organelles and Study Their Properties

Info: 5728 words (23 pages) Essay
Published: 8th Feb 2020 in Biology

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Abstract:

Differential centrifugation is one of the most common initial steps taken in protein purification from cells or tissues according to (Lodish, H., et al., 2016.); to study the function and structure of a protein. The aim of this study is to use differential centrifugation to produce a fraction enriched in nuclei and mitochondria from a rat liver homogenate, obtained using a motor driven pestle. To assess if a good separation has occurred, the recovery of the mitochondria in the fractions will be measured using the activity of succinate dehydrogenase (SDH) as a marker. SDH is an enzyme found only in mitochondria. Thus, if any activity is detected in a cell fraction, this would mean some mitochondria are present in that fraction. Indicating that the centrifugation has not occurred well and the fractions are impure. Overall, the specific SDH activity of the Mitochondria Fraction was high as expected and the Homogenate Fraction Total SDH Activity. However the % recovery of SDH was the highest in the SF this was unexpected and could have been due to inaccuracies in carrying out the protocol. Overall majority of the results conclude what was intended, slight changes to laboratory equipment and a change in some of the methods used would mean less contamination and purer fraction could be obtained avoiding results such as finding SDH activity in the supernatant.

Results:

The results in table 1 were obtained by using differential centrifugation. The speed of the centrifuge is an important factor in determining if separation will occur. A low g force is used to separate molecules with a high density in this case nuclei, whereas a high g force is used to separate soluble organelles/proteins such as the mitochondria.

The samples were stored at -20oC to prevent SDH losing its catalytic activity, and the different fractions were suspended into an isotonic buffer to prevent damage to organelles due to osmotic imbalance.

To obtain a series of pure fractions, centrifugation of the homogenate was done twice at  1500 x g for 10 min to sediment the nuclei from the other organelles such as whole cells, endomembrane organelles, and the mitochondria producing supernatant I and II. The same principle was used to obtain the mitochondrial fraction by combining Supernatant I and II, but at 20,000 x g force producing supernatant III / IV, which were combined together to produce supernatant fraction (SF). To obtain the final total volume both NF (10mL) and MF (8mL) were suspended in isotonic buffer.

 

 

TABLE 1: Total volume for each cellular fraction

Fraction  

Total Volume (mL) 

Homogenate (H)

Exact volume of the homogenate remaining

AFTER taking 3 X 1 mL H aliquots.

11.5ml

Nuclei fraction (NF) before taking the aliquots

9.45ml

Mitochondrial Fraction (MF) before taking the aliquots

9ml

Supernatant fraction (SF)

FROM Combined supernatants lll and IV

26ml

 

 

 

 

 

 

 

 

 

 

Table 1. Total volume for each of the fraction. Homogenate volume is lower as 3 x 1mL of aliquots were taken to be later used to measure the protein content of each subcellular fraction and to measure the succinate dehydrogenase activity (SDH) by using red formazan essay. Conversely the SF fraction has a high volume as Supernatant III and IV were combined.

 

Measuring protein content in each subcellular fraction using a biuret assay

To measure the protein content in each of the subcellular concentrations, a Biuret assay was used. In this the proteins form a blue complex with alkaline Cu2+, this colour has an absorbance peak at 550nm. Although the reaction is specific for peptide bonds as stated by… the reaction is not sensitive and thus can only be used with crude extracts, which contain large amounts of protein. The biuret reaction also requires alkaline condition to achieve this 0.1M of NaOH was used. Known amounts of bovine serum albumin (BSA), were reacted with biuret reagent, the absorbance values were measured to create a standard calibration graph to calculate the protein concentration in each of the subcellular concentrations as can be seen below in table 3.

Table 2. Averaged absorbance values for each BSA protein amount

Values for BSA standard curve

Protein amount (mg)

12 

16 

Absorbance at 550nm 

0.089-0.089= 

0 (Blank) 

0.0925-0.089= 

0.0035 

0.104-0.089= 

0.015 

0.147-0.089= 

0.058 

0.138-0.089= 

0.049 

0.168-0.089= 

0.079 

0.185-0.089= 

0.096 

 

 

 

TABLE 3: Protein amount in homogenate and subcellular fractions 

Homogenate 

Nuclei 

Mitochondria 

Supernatant 

A 

Average absorbance (550nm) 

0.206-0.089=

 0.117 

0.1017-0.089 = 0.0127 

 

0.173- 0.089= 0.084 

0.188- 0.089=

0.099 

B 

Protein amount in sample’s aliquot (mg

18.406 mg 

2.188 mg 

13.328 mg 

15.672 mg 

C 

Protein concentration in fraction (mg/mL

18.406/0.1=184.062 

2.188/0.4= 5.468 

13.328/0.4= 33.320 

15.672/0.4= 39.180 

D 

Protein amount in fraction’s Total Volume (mg

184.062* 11.5ml=2116.719  

5.468*9.45ml= 51.673 

33.320*9ml= 299.880 

39.180*26ml=1018.680 

H 

Percentage recovery 

100% 

2.441 

14.167 

48.125 

OVERALL PERCANTAGE RECOVERY 

64.734% 

Calculations used to obtain the figures in table 3:

To find out the protein amount in sample’s aliquot (mg) the equation y = mx + c was used from the graph created from table 2 as can be seen above.

Y = average absorbance of each of the fractions taken at 550nm

Example: Homogenate

Y = (0.117 + 0.0013) / 0.0064

X = 18.48mg

To work out the protein concentration in fraction, divide the protein amount in sample aliquot by the volume taken to produce the reaction tubes to be used for the Biuret assay.

Example:

For the homogenate 0.1mL was used

Protein amount in homogenate aliquot 18.48mg divided by 0.1mL volume taken

= 184.062 mg/mL

To calculate protein amount in fraction’s total volume, multiply it by total volume of fraction found in table 1

= 184.062 * 11.5ml=2116.719  mg

To calculate the percentage of protein recovery in comparison to the initial homogenate (HF)

  1. HF = 100%
  2. % of MF = (299.880 mg / 2116.719 mg) x 100% = 14.1672%
  3. Rounded to 3dp = 14.167%

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

To obtain reasonable values of SDH activity, the fractions were diluted with phosphate extraction buffer to the approximate protein concentration given in the first column of Table 4. To get the suggested concentration given in table 4 for each of the fractions in a volume of 2000 μL, the fold dilution was calculated using the equation C1 *V1 = C2*V2 as shown below.

TABLE 4. Dilution of Fractions in PBS 

Homogenate

Or

Fraction

Volume of homogenate or fraction required (μL)

Volume of diluent required (μL) to make up to 2000 µL total

Actual concentration used (mg/mL)*

Homogenate:

to approx. 5 mg/mL

184.062/5=36.812 fold

2000/36.812=54.329

Rounded up: 54 μL

2000-54.329= 1945.67

Rounded up:

1946 μL

V1 = 54 microliter

C1=184.062mg/ml

V2=2ml(2000 μL)

C1*V1=C2*V2

4.969mg/mL

Nuclear fraction:

to approx. 2 mg/mL

5.468/2 = 2.734

fold

2000/2.734 = 731.528

Rounded up: 732 μL

2000 – 731.528 = 1268.472

Rounded up:

1268 μL

V1=732microliter

C1=5.468mg/ml

V2= 2ml(2000 μL)

C1*V1=C2*V2

2.001 mg/mL

Mitochondrial fraction:

to approx. 1 mg/mL

33.320/1 = 33.32

Fold

2000/33.32 = 60.024

Rounded up: 60 μL

2000 – 60.024 = 1939.976

Rounded up to:

1940 μL

V1=60

C1=33.320

V2=2ml

C1*V1=C2*V2

Supernatant fraction:

use undiluted

N/A

N/A

Value from Table 3, row C

39.180 mg/ml

Fold dilution was calculated using the equation C1 *V1 = C2*V2 as shown below.

Example: Nuclear fraction (NF)

  1. The protein concentration in fraction (mg/mL) from table 3 column C is divided by the approximate protein concentration given in the first column of Table 4 mg/mL. In this case
  2. 5.468 (mg/mL) / 2 (mg/mL) = 2.734 concentration fold
  3. To work out the suggested concentration in a volume of 2000 μL:
  4. 2.734 / 2000 (μL) = 731.528 μL
  5. Rounded up to 732 μL

To work out the volume of diluent required (μL) to make up to 2000 µL total

  1. 2000 (μL) – 731.528 (μL) = 1268.472
  2. Rounded up to 1268

Actual concentration used (mg/mL)*

V1=732 μL

C1=5.468mg/ml

V2= 2ml (2000μL)

C1*V1=C2*V2  

C2 = (C1*V1) / V2

= (5.458 (mg/mL) * 732 (μL) / 2000 (μL)

= 2.001 mg/mL

 

For TABLE 5: Determining the amount of Formazan content at an absorbance of 490nm  

In this experiment Succinate dehydrogenase is being used as a marker enzyme, it is found in the citric acid cycle in the inner membrane of mitochondria as stated by … it catalyses the conversion of succinate oxidation into fumarate. This reaction happens simultaneously however as both reagents are colourless, we cannot see the reaction taking place that’s why a coupled reaction is used as shown below.

Reaction 1:

Succinate + FAD  fumarate + FADH2

The overall reaction is known as an oxidation reaction. This is because when SDH breaks down the succinate into fumarate, 2 electrons are donated/lost to the enzyme bound cofactor flavin adenine dinucleotide (FAD + 2 electrons  FADH2) to produce FADH2.

 

Reaction 2:

FADH2 + INT  FAD + formazan

Since we cannot measure the SDH directly, the formation of formazan is measured instead which forms a deep red compound. This happens when FADH2 reduces tetrazolium salt (INT). This way the Succinate acts as a donor of electrons and INT the final electron acceptor.

 

 

 

 

TABLE 5: Determining the amount of Formazan content at an absorbance of 490nm  

Fraction 

Control 

Test 1 

Test 2 

Average test absorbance values (obtained over a 10 min incubation) 

H 

0.004 

Corrected value 0.000

0.227 -0.004 = 0.223

0.255 – 0.004  = 0.251

0.237

NF 

-0.003 

Corrected value 0.000

0.076 

0.057 

0.0665 

MF 

-0.008 

Corrected value 0.000

0.084 

0.119 

0.1015 

SF 

0.070 

Corrected value 0.000

1.832 

1.7 

1.766 

The control values were subtracted from the test results, and test ½ were added and average was taken.

Example HF = 0.004 subtracted from test 1 (0.227 -0.004 = 0.223) and test 2 (0.255 – 0.004 = 0.251)

(0.223 +0.251) / 2 = 0.237 average absorbance value

As little formation of formazan in the control tubes can be seen even though no sodium succinate was added, this is because INT (tetrazolium salt) is a non-specific electron acceptor, this means it can be reduced by different redox enzymes transferring electrons from many compounds in the cell such as ascorbic acid, hydroquinones, and menadione. However, since SDH is highly specific for succinate the control sample (containing no succinate) will be used to subtract any INT reduction which may produce red formazan due to other enzymes than SDH.

TABLE 6. Calculating Total and Specific activity and % recovery of activity in the fractions in relation to the homogenate

 

Obtained over 10 min incubation period

Homogenate fraction 

Nuclei fraction 

 

Mitochondria fraction   

 

Supernatant fraction  

 

Total Succinate Dehydrogenase (SDH) Activity  (nmoles/min)

9983

168

2998.8

4550

Amount of substrate formed per minute 

(mmoles/per minute)

4.7164

1.3

2

35

% Recovery of Activity (rel.Homogenate)

100%

1.6829

30.0390

     45.5775

Specific SDH Activity

(µmoles/min/mg protein)

(rounded to 3dp)

4.716 X 10-3

3.2512 X 10-3

      0.01

4.477 X 10-3

Calculations:

Example of how to calculate the total SDH activity of HF

  1. Average Absorbance value of homogenate: 0.117     (Obtained over 10min incubation)
  2. Apply Beer lambert law Aλ = ελ L [c]
  3. 0.237= 20100 M-1 cm-1 x 1 cm x [c ] 
  4. 0.237 / 20100 = [c]        
  5. [c]  =1.1791 X 10-5M =  (moles/L)
  6. Convert to µmoles/L by multiplying by 106

(1.1791 X 10-5 moles/L) X (106 µmoles/L) =  11.7910 µmoles/L of formazan in 10min

However we want to find out the total activity of SDH in the whole fraction to calculate that:

Firstly need to calculate per minute:

Note: This would be the concentration of Formazan (11.7910 mM) or the amount of formazan (11.7910 mM) produced and dissolved in 1 litre of ethyl acetate. However, we only dissolved in 4mL

  1. 11.7910µmoles/L of formazan in 10min
  2. (11.7910 μmoles/1000mL) X 4mL = 0.04716µmoles in 10 min (in reaction)

0.04716 (μmoles) x 1000 nmoles/ μmole  = 47.16 nmoles in 10min (in reaction)

  1. 47.16 nmoles in 10min (in reaction)
  2. 47.16 nmoles / 10min = 4.7164nmoles per min (in reaction)

Formazan product was derived from the SDH activity of 0.2mL diluted fraction

  1. 4.7164nmoles per min in 0.2ml         

To find out in 1mL multiply by 5 (mg/mL)  

  1. 4.7164nmoles x 5 (mg/mL) = 23.58nmoles per min in 1mL of diluted homogenate

From table 3, we can find out the protein concentration in fraction mg/mL in this case it is 184.062mg/mL

  1. 184.062 (mg/mL) / 5 (mg/mL) = 36.812 (3.dp) this is the dilution factor used to dilute the sample with extraction buffer to obtain the approximate protein concentration given in first column of table 4.

5mg/mL (Homogenate protein concentration/volume)

  1. 23.58 (nmoles/permin/1mL) X 36.812 = 868.10nmoles/per min in 1mL (undiluted homogenate)
  2. 868.10 (nmoles/permin/mL) X 11.5 (mL) = 9983 (nmoles/min)

(11.5 total fraction volume obtained from table 1)

Total SDH Activity of homogenate (HF) = 9983 (nmoles/min)


 

Example of how to calculate % recovery of activity in contrast to homogenate

  1. % of recovery           100%
  2. % of MF = (Total SDH X MF mitochondria fraction) / (Total SDH Homogenate fraction)
  3. 2998.8 (nmoles/min) / 9983 (nmoles/min) X 100% = 30.01%

 

Example of how to calculate specific SHD activity for the homogenate fraction

Specific SDH activity refers to the activity of an enzyme per milligram of total protein as stated by ..

(Total SDH Activity nmoles/per min) / (Protein amount in fraction’s Total Volume mg)

  1. For HF = 9983(nmoles/min) / 2116.719 (mg) = 4.716 nmoles / min / mg  (Specific SDH Activity)
  2.  Convert to µmoles = To do this divide the amount of substance by 1000

= 4.716 X 10-3 µmoles / min / mg protein   

 

 

 

 

 

 

 

 

Discussion

Overall, the results obtained showed a relatively good separation had occurred. The evidence for this can be seen in table 3 column D, in the differing amounts of protein in each fractions total volume, as well as the differing values in specific SDH activity and total succinate dehydrogenase activity from table 6, both show that the organelles have been efficiently isolated from each other. Out of the three fractions obtained from the homogenate, the supernatant had the highest, Total SDH Activity this was unexpected, since SDH is only found in the mitochondria, the MF fraction should have had the highest value, which indicates some fraction contamination, as there is mitochondria activity detected in the supernatant fraction.

The reason for this could have been due to, the inaccuracies caused in the initial experiment (table 1) when decanting the supernatant fraction between each centrifugation, some of the pellet could have been poured in with the supernatant affecting all of the following results.

From looking at table 3, the percentage of protein recovery in relation to the initial homogenate was not 100%, instead there was a 35.266% loss. The reason for this could be as proteins are held by relatively weak forces such as hydrophobic, van der Waal and salt bridges according to … to create a biologically active structure. The bonds can be easily disrupted by temperature, as the tubes were constantly being taken out of ice and returned this could have also had a significant effect on the catalytic activity of SDH, as the room temperature was not cold. During homogenisation, cells like lysosomes can also burst releasing enzymes such as protease which are known to degrade proteins within the source material, to minimize the activity of protease the temperature needs to be maintained at 25oC according to …, therefore there is a high probability of protease being one of the causes for low recovery. However to reduce this effect BSA (bovine serum albumin) was added, which prevents proteolysis of the target protein according to…

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To improve the separation of the organelles, an alternative would be rate zonal centrifugation which should provide purer fractions. This method works on the basis of different masses of water soluble proteins, the proteins are separated by centrifugation through a solution of increasing density, usually sucrose is used. The heavier particles sediment the fastest followed by smaller and less dense ones, this creates discrete bands which can be collected as fractions and assays can be performed. This is especially useful in separation of other organelles from the mitochondria. A small amount of detergent is added, and absorbed by the cell through endocytosis, making the lysosomes less dense which gives a clean uncontaminated separation from the mitochondria. 

In, the experiment a hand-held homogenizer was used; this perhaps was not the most accurate way of releasing cell content as insufficient force was used so whole bits remained in the solution. To improve this, a rotor based or an ultrasonic homogeniser could have been used which would provide more efficiency in organelle separation.

The results obtained in table 3 column A (average absorbance for the fractions), for the homogenate and nuclear fraction were much higher than the absorbance values, taken for the BSA standard curve. This could have been due to inaccuracies in pipetting, and collecting the samples such as not shaking the tubes properly, or due to an error in centrifugation of the samples in the initial experiment.

The supernatant had a high protein concentration, reference table 3 row D, this is because a high concentration of soluble proteins, were left in the solution such as peroxisomes,  lysosomes and other large macromolecules, fragments of Endoplasmic reticulum, and microsomes. Organelles specific marker molecules could be used to extract these. For example cytochrome c protein is present only in mitochondria, so the presence of this protein in a fraction of lysosomes would indicate contamination. Also, catalase is present only in peroxisomes; ribosomes, only in the rough ER and acid phosphatase, only in lysosomes.

Conclusion

In conclusion the aim of this particular subcellular fractionation was achieved, a fraction enriched in nuclei and a fraction enriched in mitochondria were obtained. The specific SDH activity was the highest in the mitochondrial fraction as was predicted and the total SDH activity in the homogenate. However the percentage activity recovered was the highest in the supernatant fraction which indicates contamination of mitochondria in the fraction. Overall there has been some contamination, however majority of the results conclude what was intended, slight changes to laboratory equipment and a change in some of the methods used would mean less contamination and purer fraction could be obtained avoiding results such as finding SDH activity in the supernatant.

References:

 

      Lodish, H.F., 2016. Molecular cell biology Eighth edition..; Global., New York: W.H. Freeman.

      Hames, B.D., ebrary, Inc & Hooper, N. M, 2005. Biochemistry3rd ed., New York; Abingdon: Taylor & Francis.

      Bonner, P. (2018). Protein purification (2nd ed., Basics (Taylor & Francis)). Boca Raton: Taylor & Francis.

      Doumas, B.T.,1975.Standards for total serum protein assays-a collaborative study. Clinical biochemistry,21 (8), 1159-1166. Available at: https://www.ncbi.nlm.nih.gov/pubmed/1169135 [Accessed: Nov 8, 2018]

      Berg.J.M.,2002. Biochemistry.5th ed., New York;Basingtodistributor:W.H.Freeman and Co.

      Lodish, H.,Berk,A.,Zipursky,S.L., et al., 2000. Purification of cells and their parts. Molecular Cell Biology.4th ed, Available at: https://www.ncbi.nlm.nih.gov/books/NBK21492/ [Accessed: Nov 8, 2018].

 

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