Prepare A Nuclei And Mitochondrial Fraction Biology Essay

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The objective of this experiment is to prepare a nuclei and mitochondrial fraction using differential centrifugation, from a rat liver homogenate sample. The amount of activity of mitochondria in the fractions can be measured using succinate dehydrogenase (SDH) as a marker.

To measure the percentage recovery of the SDH of Mitochondrial, Nuclei and supernatant fractions in comparison to the Homogenate and to Calculate the specific and relative activity of SDH in each fraction.

Figure 1: Shows a typical animal cell with the individual organelle components.

Figure 2: Shows the typical features and functions of the organelles of interest in this report.

Figure 1 + 2 Created on Microsoft paint with reference to Essential Biology (2004)

Individual organelles differ in size but are all usually around 10nm in diameter. There is a small surface area and size/density depends on the organelle, the smaller organelles being lysosomes and ribosomes.

Mitochondria differs in cell type depending on the energy demand of that organ, the more ATP that is required in a particular organ the more mitochondria found. E.g. more mitochondria found in heart and liver cells than in a white blood cell like a lymphocyte. Smaller organelles include lysosomes and ribosomes.

Metabolism can be detected using various methods such as use of inhibitors. These can be both competitive and non-competitive, an example is seen with arsenic with inhibits pyruvate dehydrogenase. Another method is with the use of radioisotopes to measure activity aswell as histochemistry, immunocytochemistry and electromicroscopy.

Preparation of the homogenate occurs in various stages. Firstly the homogenisation of liver cells. This can be done using a Potter - Elvehjem homogeniser to extract organelles without damaging the actual cell. This is a simple and effective homogenisation method. A small gap is made within the cell wall which is then pressurised which forces the contents i.e organelles, cytoplasm etc. out of the cell. This occurs at a low temperature and mild pH, and to keep the isotonic solution a sucrose buffer is used, therefore since there is the same water potential inside the cell and outside the cell there is no net movement of water (osmosis) and thus the cell remains the same size. Homogenized cells also must be kept at low temperatures to prevent autolysis (the degradation of a cell by its enzymes). (www.bookrags.com).

Figure 3 shows a classic Potter - Elvehjem homogeniser Image taken from (umwcellbiology.org)

The second stage is fractionation of the homogenate sample. This process is called centrifugation and can be further split into either a differential centrifugation or a density gradient centrifugation. The differential centrifugation splits the impure fraction into separate compartments due to the size of the various organelles in question and there density. The centrifuge applies a gravitational force onto the sample to separate components. The rate of centrifugation is determined by the acceleration or speed applied to the homogenate and is usually measured in revolutions per minute (RPM) or g. Depending on the density of the organelles will determine their isolation at a given speed. The higher density organelles and the bigger organelles separate at a lower speed centrifugation. (K. Wilson 2005). The separation forms a pellet which is the precipitate proportion of the sample and the component of interest and a supernatant which is the liquid component. The supernatant readily decanted from the sample without removing the precipitate. Diferemces in centrifugation occur due to the techniques used, differential centrifugation is based upon the sedimentation rate of particles and thus the sedimentation rate separates them based on size and density. After initial sedimentation the largest particles separate first into pellet and supernatant (K. Wilson 2005). Density gradient centrifugation separates organelles using a media. Various media can be applied and depending on the particles will be best for certain types and may not work well for others. (K. Wilson 2005).

The 4 fractions we will obtain are nuclei, mitochondrial, supernatant and homogenate.

Various tests can be carried out to distinguish between fractions and to determine their actual purity, testing for specific enzymes can code for the activity occurring in the cell fractions therefore indicating the most abundant component.

Some tests include: Testing for DNA in both nuclei and mitochondrial fractions. This is because DNA is contained within the nucleus but also within the mitochondria. This is because relating to the endosymbiotic theory mitochondria was a separate aerobically respiring bacterial cell which was later engulfed by an early eukaryotic cell to merge into one aerobically respiring cell. Mitochondria is maternally inherited in the case of the majority of multicellular organisms, this is due to the higher number of mtDNA molecules in the ooecyte and much fewer in a sperm cell which are mostly degraded before fertilization takes place.

Test for histones which indicate nuclei fraction as well as testing for various enzymes such as ATPase found in cytoplasmic (supernatant) and mitochondrial fractions and phosphotase kinase indicating microsomes and golgi apparatus are present. Some enzymes are exclusive to the citric acid cycle which occurs in the mitochondria, therefore testing for these enzymes indicates the presence of mitochondria in a fraction. The enzyme marker to test for mitochondria which we use is succinate dehydrogenase which is exclusive to the inner mitochondrial membrane. Succinate dehydrogenase is formed only during the citric acid cycle so is only given as an indication of mitochondria. However, since during the homogenisation process the mitochondria could potentially burst spilling their contents into the cytoplasm (supernatant fraction), this does not therefore give an accurate indication of mitochondria present in a fraction. Succinate dehydrogenase breaks down succinate into fumurate, therefore the measurement of formazan indicates presence of succinate dehydrogenase.

Measuring Succinate Dehydrogenase Activity

(Red Formazan assay)

This occurs in 2 reactions:

1: succinate + FAD è fumarate + FADH2

SDH breaks down succinate into fumarate. This is an oxidation reaction since the succinate loses 2 electrons, in addition a reduction of the enzyme flavin adenine dinucleotide occurs (FAD gains 2 electrons) (FAD + 2 electrons è FADH2)

Figure 4: Shows the redox reaction which occurs with succinate and FAD.

Image taken from natuurlijkerwijs.com

SDH activity is measured by the formation of formazan a deep red compound formed from the reduction on a tetrazolium salt. The reduced FADH2 reduces tetrazolium salt (INT).

2: FADH2 + INT è FAD + formazan

Centrifugation and calculating the relative centrifugal field. (K. Wilson 2005)

G = W2r = 4 II2 r (rPM)2 = 1.11x10-5r (rPM)2

3600

G= Relative centrifugal force (RFC)

r = Radical distance from axis of rotation

w = Angular velocity

rPM = Revolutions per minute.

T = 9 ɲ (In Rt/Rb)

2 W2rp2 (Pp -P)

ɲ = Viscosity of medium

rp = Radius of particle

Pp = Density of particle

P = Density of medium

Rt = Radius to top of centrifuge tube

Rb = Radius to bottom of centrifuge tube.

There are many differences in types of centrifuges available and results depend on the speed of the centrifugation and whether a vacuum is present and the type of rotor used. (K. Wilson 2005)

Analysis of marker enzymes in subfractions determines the recovery of subcellular organelles, with comparison to previous tests, quantative data can be used to assess contamination of fractions. Showing whether the subfractionation method has been successful or not.

These tests also hold health benefits and implications e.g. microsome C causes cell death and can be found in mitochondrial fractions, however in cancer patients no microsome c is present, indicating no cell death will occur a common feature of cancer cells.

Enzyme measurement in subcellular fractions however does hold some implications such as the solubility of the environment which may cause differences in enzyme function. Another implication is latency of enzymes, this refers to whether proteins are bound to the enzyme which in turn activates them once bound signalling enzyme function. There may also be low recovery of enzymes in the fractions due to poor recovery of the organelles which they come from, in particular if the enzyme is confined only to a specific region.

Over the 3 week period centrifugation will separate the fractions according to size/density and separating the sample into the pellet and supernatant fractions. The speed of the centrifuge determines whether the pellets will separate. A lower speed is needed to separate the nuclei fraction due to the higher density, whereas the higher speed is needed to separate the supernatant due to the smaller density remaining organelles. (K. Wilson 2005). The protein content is also measure for each fraction using the biuret assay, absorbance values are given which determine the protein content of each fraction. Finally succinate dehydrogenase is measured. This causes a redox reaction and causes e- ions are released, using formazan as an indicator this changes the colour of solution red, showing a redox reaction has taken place.

From this research I can predict that the mitochondrial fraction is expected to have the highest results in specific activity due to fewer proteins present in that fraction.

Results:

Calculations:

Formazan's molar extinction coefficient (E490nm) = 20,100 M-1 cm-1

The specific activity and relative activity of the fractions can be determined by measuring the concentration using Beer- Lamberts Law: (www.chemguide.co.uk)

A = ε x l x C

A = Absorbance (no units)

ε = Epsilon. The adsorbtion coefficient M -1 cm -1

l= Cuvette's light path length, this is the length of solution a light passes through (always 1 cm)

C= Concentration of substance in M (moles in 1 litre)

Rearrange to give concentration:

C = A / ε x l

Units: M-1 x cm-1 = 1 / M x cm

C = A / ε x l Gives units: ( 1/ (1/M x cm) x cm). This can be simplified to give 1/ (1/M)

And further simplified to give units: M (moles per litre or dcm -3)

Know values:

ε = the formazan adsorption coefficient is 20,100 M -1 cm -1

A = refers to the absorbance at 490nm values for each fraction are found in the mean-control table section.

Using the equation: C = A / ε x l

We can work out the concentration of formazan formed in the reaction.

The concentration value is for 1 litre, therefore we must calculate the actual concentration from the actual assay volume used.

Concentration = amount/volume rearranged to give A = C x V

The final assay volume from week 3 is 6 ml* due to the addition of ethyl acetate.

* Note by mistake 6ml of ethyl acetate was added instead of 4 ml giving a different final volume to the other groups.

Converting 6ml into its litre value and x by the concentration gives the accurate mole product of formazan produced.

Reaction time needs to be included to give the accurate units. Activity units can be determined using the following equation.

Activity = Moles of formazan/reaction time (12 minutes)

This gives the activity in M -1

Calculating total activity and specific activity of the fractions.

Table 1: Total volumes from each cellular fraction.

Fraction

Total Volume (ml)

Homogenate

12

Nuclei Fraction

12

Mitochondrial Fraction

12

Supernatant Fraction

26

To do this we need to take into the account:

The total volume

The total protein of the fraction.

Dilution factor

The total volume values for each fraction can be found in table 1.

The sample of each fraction used was 0.2ml, therefore the amount of moles of formazan calculated is in 0.2ml.

(0.2 / total volume) x moles of formazan in 0.2ml

X by the dilution factor of each fraction to give the total activity for each fraction, the values are given in table 4.

To determine the specific activity we must consider the total protein of the fraction. Values are given in table 3.

Specific activity = Total activity of fraction/ total protein of fraction

Table 2: Bovine serum albumin (BSA) solution concentrations

 

Volume (ml)

BSA (10mg/ml BSA

0

0.2

0.4

0.6

0.8

1.2

1.6

0.1m NaOH

2.0 (blank)

1.8.

1.6

1.4

1.2

0.8

0.4

Table 3: Values for BSA standard curve.

See Graph 1 for the results from the corresponding fraction absorbance.

Protein Amount (mg)

0 (blank)

2

4

6

8

12

16

Absorbance at 550nm

0

0.105

0.184

0.275

0.354

0.511

0.531

Table 4: Protein amount in homogenate and subcellular fractions.

 

Homogenate 0.05ml

Nuclei 0.2ml

Mitochondria 0.2ml

Supernatant 0.2ml

Average Absorbance (550nm)

0.169

0.054

0.174

0.199

Protein amount in sample's aliquot (mg)

3.6

1.18

3.8

4.15

Protein concentration in fraction (mg/ml)

72

5.9

19

20.75

Protein amount in fraction's total volume (mg)

864

70.8

228

539.5

Graph 2: Shows the difference in protein amount amongst cellular fractions.

Table 5: Actual concentration of fraction after dilution.

 

Dilution Factor

Actual concentration (mg/ml)

Homogenate

20

3.6

Nuclei

3

2

Mitochondrial

20

0.95

Supernatant

1

20.75

Table 6: Formazan content absorbance at 490nm.

Fraction

Control

Test 1

Test 2

Mean-Control

Homogenate

0.132

0.58

0.52

0.42

Nuclei

0.21

0.352

0.326

0.13

Mitochondrial

0.057

0.391

0.265

0.27

Supernatant

0.132

0.52

0.33

0.29

Results for Homogenate:

From table 5, we have the absorbance of homogenate as 0.42 this divided by the adsorption co-efficient gives:

0.42/20,100 = 2.1 x 10 -5 M

The units for concentration are left as moles per litre (M). To get this into moles in the actual volume used (6ml not 1 litre) 2.1 x 10 -5 M x 0.006 lite = 1.3 x 10-7 M

Include the reaction time of 12 minutes to give moles per minute.

1.3 x 10-7 M /12mins = 1.0x10-8 M -1

To determine total activity and specific activity.

The total volume from table 1: for the homogenate is 12ml, however the sample used was only 0.2ml we therefore divide actual volume / used volume x concentration of H x dilution factor (20 in the case of the homogenate from table 5 values)

Total activity = (12/0.2) x1.0x10-8 M -1 x 20 = 1.2 x 10 -5 M -1

specific activity = 1.2 x 10 -5 M -1/ total amount protein in homogenate from table 4

1.2 x 10 -5 M -1/864= 1.3 x 10-8 M min-1

Results for nuclei fraction:

0.13/20,100 M-1 cm-1 = 6.5 x 10-6

In 0.006 litre : 6.5 x 10-6 x 0.006 = 3.9 x 10-8 M

3.9 x 10-8 M / 12 = 3.2 x 10-9 M min-1

Total activity = 3.2 x 10-9 M min-1 x (12/0.2) x 3 = 5.8 x 10 - 7M min-1

Specific activity = 5.8 x 10 - 7/ 70.8 = 8.2 x 10 -9 M min-1

Results for mitochondria:

C = 0.27/20,100 m-1 cm -1 = 1.3 x 10-5‑M

1.3 x 10-5‑M x 0.006 = 7.8 x 10-8 M

7.8 x 10-8 M / 12 = 6.5 x 10-9 M min-1

Total activity = 6.5 x 10-9 M min-1 x (12/0.2) x 20 = 7.8 x 10-6 M min-1

Specific activity = 7.8 x 10-6 M min-1/228 = 3.4 x 10-8 M min -1

Results for supernatant:

C = 0.29/20,100 m-1 cm -1 = 1.4 x 10-5‑M

1.4 x 10-5‑M x 0.006 = 8.7 x 10-8 M

= 8.7 x 10-8 M / 12 = = 7.3 x 10-9 M min-1

Total activity = 7.3 x 10-9 M min-1 x (26/0.2) = 9.4 x 10-7 M min-1

Specific activity = 9.4 x 10-7 M min-1/539.5 = 1.7 x 10-9 M min -1

Percentage recovery of Succinate Dehydrogenase for the fractions

This is done by dividing the amount of Succinate dehydrogenase in the individual fractions by the original homogenate and then multiplied by 100 to give a percentage.

Table 7: Shows the total activity for each of the fractions.

Fraction

Total Activity

Homogenate

1.2 x 10 -5 M -1

Nuclei

5.8 x 10 - 7M min-1

Mitochondria

7.8 x 10-6 M min-1

Supernatant

9.4 x 10-7 M min-1

Nuclei fraction:

(5.8X10-7/1.2x10-5 ) x 100 = 4.8%

Mitochondria fraction

(7.8x10-6/1.2x10-5 ) x 100 = 65%

Supernatant fraction

(9.4x10-7/1.2x10-5) x 100 =7.8%

Relative Specific Activity of Succinate Dehydrogenase

This is found by dividing the specific activity of the fractions (found above) by the specific activity of the homogenate (found above).

Table 8 shows the specific activity for each of the fractions:

Fraction

Specific Activity

Homogenate

1.4 x 10-8 M min-1

Nuclei

8.2 x 10 -9 M min-1

Mitochondria

3.4 x 10-8 M min -1

Supernatant

1.7 x 10-9 M min -1

Nuclei fraction

8.2 x 10 -9 M min-1 /1.4 x 10-8 M min-1 = 0.586

Mitochondrial fraction

3.4 x 10 -8 M min-1 /1.4 x 10-8 M min-1 = 2.429

Supernatant fraction

1.7 x 10 -9 M min-1 /1.4 x 10-8 M min-1 = 0.121

Discussion:

Note: There was very little protein found in the nuclei fractions total volume, this is abnormally low since we would expect this to be higher.

From the results we can determine that the this supports our prediction that

The mitochondrial fraction is expected to have the highest results in specific activity due to fewer proteins present in that fraction. Organelles have been isolated from each other as seen with the differing proportions of protein found in each fraction as well as the differing values for specific and total activity calculated.

However the homogenate is expected to have the highest total activity due to the higher amount of protein since all fractions are present.

However since protein was found in the cytoplasm or supernatant fraction, this indicates that there was an error in the separation of the fractions as SDH is present where it usually isn't found. Succinate dehydrogenase works by transferring 2 electrons from succinate which transfers it to fumerate, which blocks the rest of the reaction when it binds to FAD, from the measurement of formazan gives the value of activity.

Results show that the relative specific activity is highest in the mitochondrial fraction, as well as the percentage recovery of the fractions. Therefore demonstrating that the fractions were purified and that the homogenisation and centrifugation has been relatively successful in separating fractions.

However there were some inaccuracies from the results, this includes the very low protein amount found with in nuclei fraction, this was however predicted to contain a higher amount of protein due to the nature of the organelle and the enzymes contained within it. Another inaccuracy in this experiment is that SDH was found within the supernatant. This is primarily a marker for mitochondria so would not usually be found within the cytoplasm, however due to mitochondria bursting and releasing its contents into the cytoplasm during the homogenisation stage and centrifugation the enzyme succinate dehydrogenase was present. Since the test was carried out under the same conditions in a neutral pH buffer we can conclude that this was a fair test, however is it often found that the more molecules present in a

The separating of the homogenate could be improved by using another method of homogenisation, in this experiment we used a Potter - Lethem homogeniser which is a glass and plastic hand homogeniser. This perhaps isn't the most accurate at pressurising cells with the force needed to accurately release cell content. Alternative homogenisers include ultrasonic and rotor based homogenisers which may provide more accurate. (www.proscientific.com)

A different centrifugation method used. During this experiment differential centrifugation was used, however density gradients may provide more accurate at purifying a sample (www.coleparmer.co.uk). This method works by placing various layers after layer of gradient media such as sucrose in a tube with the heaviest layer at the bottom and the lightest at the top. The cell fraction to be separated is placed on top of the layer and centrifuged. Density gradient separation can be classified into two categories. Rate-tonal (size) separation. Isopycnic (density) in which organelles separate until their density matches the surroundings of the media in which they are. A very good medium for separating organelles is an iodinised media. (www.coleparmer.co.uk).

Accuracy of the absorbance and accuracy of obtaining the protein amount. Results are slightly low indicating inaccuracy in both collecting the samples and also measuring the absorbance, this could be due to error in homogenisation and centrifugation techniques but could also be due to error in the reading of absorbance using the Spectrophotometric assay since U.V wavelength has different absorbance levels if either oxidised or reduced enzymes absorb light therefore giving innacurate indication to enzyme present (www.millipore.com) . This may affect the absorbance levels in the fractions if specific enzymes are affected thus giving an altered absorbance level and therefore undermined protein amount. Another method to measure enzyme assay could be to use a caliometric method which measures heat radiance given off instead of the absorbance levels.

Some of the organelles which remain in the supernatant fraction are the smaller and less dense proportions of the cell such as ribosomes and lysosomes. Further centrifugation at a higher speed can be used to separate these smaller less dense organelles into pellets. This can also be used to further purify bacteria.

In conclusion we see that as predicted, the specific activity is highest in the mitochondrial fraction and the total activity is highest in the homogenate. The % recovery of each fraction and the relative specific activity for each fraction calculated shows a higher proportion in the mitochondrial fraction also. Overall the results indicate accurate laboratory skills and results conclude what was intended, however some slight changes to laboratory equipment would mean that some of the results such as SDH found in the supernatant may not come about in a future test.

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