Photosyntheis and the Oxygen Electrode

How does the oxygen electrode work

1. O2 generated by photosysnthesi

2. treavels to a platinum electrode

3. undergoes a reduction reaction

4. generates H2O2

5. Causes a current to be generated

6. Proportional to the amount of O2 in the system

For oxygen to be generated

• must be a flow of electrons through the photosynthetic pathway

• electron acceptors and donors must be present!

→

It is possible to substriute with artificial electron acceptors and donors

• to measure the in vitro activity of photosynthetic machinery

Examples of artifical e- acceptors

1. Ferricyanide→ only accept electrons from PSI

2. Pheylquinone→ only accepts electrons from plastoguinone of PSII

   • not from PSI !

Preparation of lettuce chloroplasts

1. remove central veins

2. chilled mortar and homegice with ice cold sucrose medium

3. pour over a wet muslin

4. and squeeze through 4 layers into beaker

5. immediately into centrifuge tube

6. centrifuge for 3 mins

7. chloroplasts with sediment→ pellet isn’t pure but is good enough

8. pour off supernantent

9. add more sucrose medium and sqirl to suspend pellet

Why such a high sucrose concentration in the buffer?

• This is the approximate concentration of solutes in the chloroplasts, making this an isotonic buffer.

• This means the osmotic potential of the buffer matches that of the biological sample,

• so there is no danger of osmolysis of the organelles.

Measuring chlorophyll content

1. acetone and chloroplasts into centrifuge tube

2. cetifuge

3. zero spectrometer and put in

4. calculate amount of chlrophyll in chloroplast smaple→ can use beer’s law but limitations!

5. replace chlroplasts on ice in the dark

Why kept on ice?

• When you grind up the tissue you will disrupt the lysosomes and vacuoles.

• This will release protease enzymes which will degrade your sample, therefore keeping the extracts on ice will reduce the activity of these enzymes and protect your sample.

Why use acetone to extract chlorophyll?

• Chlorophyll is a lipid-soluble molecule,

• and is found bound to chlorophyll-binding proteins in the thylakoid membrane

• Acetone dissolves the membrane and releases the chlorophyll.

Why can’t really use Beer’s law

• got a micture of chlorophyll a and b which has different absorption peak range thingy

Calibrating the oxygen electrodes

1. few crystals of sodium dithionite→ reducing agent

2. will remove all dissolved O2

3. do not put lid on

4. reading should fall to 0 in 30 seconds

5. wash reaction vessel with distilled water to remove sodium dithionite

6. shake RO water to aerate and saturate with oxygen

7. press + to get reading close to 1000 mV

8. Reading can now be converted to O2 concentration

   • since solubility of O2 in air-saturated water is 0.255 micromol O2ml-1

   • and the volume you are placing is 1.5mml

Determining if herbicides affect photosynthesis

1. Remove any liquid in current chamber

2. Add to O2 electrode

   • buffer (with KCN)

   • ferricyanice

3. Add N2 into liquid until reading is 200mV

4. allow to stabilise

5. add chloroplasts to liquid under the surface of the liquid

6. put the lid on

7. cover with a black pastic bag

8. start timer and take readings every 20 seonds

9. repeat more without plastic bag

10. Remove and add herbicide

11. Take readings again

Why use KCN in the buffer?

• KCN is toxic because it is an inhibitor of cytochrome c oxidase, which is one of the components of the respiratory electron chain.

• Use to inhibit mitcohondria respiration so only measure oxygen related to photosynthesis

Determining the target of the herbicide

1. repeat again but add phylquinone too

2. Also add the herbicide

3. Only accepts electrons from PSII

Results

1. Herbicide 1: must target PSII becaue when PhQ is used→ oxygen evolution is inhibited

2. Herbicide 2: reduction with ferricyanide but no reduction with PhQ→ does not target PSII→ do not know if it targets PSI

3. Herbicide 3: no reeduction in O2 with ferricyanide→ herbicide does not target photosynthesis e.g linezoid 1 inhibits protein translation in the chlroplast but not in nuclei

Plot the electrode readings from Table 1 against time on the graph paper provided.

By this means you can obtain a value for the rate of photosynthetic oxygen evolution, expressed as µmolO2 min-1 mgChl-1 .

You will need to include the following pieces of information: - The solubility of O2 in air-saturated water is 0.255 µmol O2 ml-1 at 20o C - The concentration of chlorophyll in your chloroplast suspension, after any volume adjustments you may have made - Remember that photosynthesis doesn’t occur in the dark!

1. Photosynthetic rate= rate in ligh- rate in dark Substrating background rates!)

2. Covert from mV→ O2 concnetration

   • e.g 1000mV = 0.255, so 1mV =2.55 ×10-4

3. O2 concentraion→ O2 amount

4. Determne amount of chlorophyl in the chamber chl c→ amount

5. Divide by amount of chlorophyll

In the practical we will use a buffer with the following composition to extract chloroplasts: 0.38M sucrose, 0.07 M K₂HPO₄, 0.01 M KCl, adjusted to pH 7.5. Why?

• It maintains an appropriate physiological pH

• It resembles the high concentration of sucrose in the normal cytosol

• It provides an isotonic solution which prevents the chloroplasts from bursting.

Which part of the molecule is responsible for the molecule being able to absorb light directly?

• Aromatic porphyrin ring

Which processes take place in the stroma of a chloroplast? 

• NADPH production

• Calvin cycle

• ATP production

What type of proteins are chlorophyll molecules associated with in the cell?

Integral membrane proteins in the thylakoid membrane

Artifical electron transort chain acceptors

• Phenl quinone

• Ferricyanide

Source of electrons for plastoquinone

• PSII

Source of electrons for Ferredoxin

• PSI

Source of electrons for Pheylquinone

PSII

Source of electrons for ferricyanide

PSII and PSI

Source of electrons for Plastocyanin

Cytochrome bgf

Role of sodium dithionite

• reducing agent

Role of H2O

• polar electron donor

Plastoquinone

• lipid soluble hydrogen carrier

Plastocyanin

• electron carrying protein including copper

NADP+

• Polar electron acceptor

IF PSII is stopped, what parts of photophosphylation are stopped

1. O2 prodution

2. NADPH production

3. Linear electron flow

A student set up a chloroplast extraction in an appropriate buffer within an oxygen electrode, as in the week 7 experiment.

She added PhQ as an electron acceptor and paraquat, a PSI inhibitor, to her chloroplast preparation after a short time and measured the oxygen electrode reading at regular intervals.

Which graph below best reflects her results?

A student sets up the oxygen electrode to measure photosynthesis using the instructions in the practical schedule. When they start recording, there is a large decrease in the O₂ concentration over time when the chloroplasts are in the dark. What is the most likely mistake they have made in setting up the electrode?

They forgot to add KCN

• without KCN present there will still be respiration occurring which will consume the oxygen present

Best buffer to use

• The buffer should have a pH of between 7 and 7.5 to match cytosolic pH,

• and the 0.38 M sucrose concentration makes the buffer isotonic to the chloroplasts to help maintain their structure.

An oxygen electrode was calibrated with sodium dithionite to define 0 mV, and air saturated water to define 1000 mV. Chloroplasts were prepared as per the IA cells practical, then 40 µL of chloroplasts were added to an electrode chamber containing 1410 μL phosphate buffer, 40 μL ferricyanide and 10 μL KCN. The following data were recorded:

1. Lightrate-dark rate

   • Light→ 558-42

   • Dark→ 359-352

   • Rate= 130mVmin-1

2. convert mV to O2 using the solubility of oxygen

   • convert o O2

   • 0.03315

3. Convert concentraion to amount

   • n=cxv because we have 1.5ml

   • n= 0.03315 ×1.5 = 0.0497

An experiment was set up as follows: 

• An oxygen electrode was calibrated such that 1000 mV corresponds to fully air-saturated water. 

• 6 g of rat liver was homogenised, and 3 mL of mitochondria were extracted. 

• 20 μL of mitochondria were added to 2.98 mL of an ADP containing buffer in an oxygen electrode chamber. 

Given that a decrease of 133 mV was recorded over 60 seconds, and assuming that the solubility of  O₂ in water under the experimental conditions is 0.255 μmole O₂ mL^-1, calculate the rate of respiration as μmole O₂ min^-1 g_liver^-1.

1. mV → O2 concentration

2. Concentration→ amount in chamber

3. Amount in chamber→ amount in total extact

4. divide by g of liver in tissue