chlorophyll & DCPIP

Question 1

Investigation into the effect of a named factor on the rate of dehydrogenase activity in extracts of chloroplasts

Question 2

what is dehydrogenase?

Answer 2

-an enzyme found in plant chloroplasts
-it is needed in the LDS of photosynthesis
-dehydrogenase catalyses the reaction where NADP is reduced (gains electrons)

Question 3

how are electrons released?

Answer 3

-photoionisation of chlorophyll
-photolysis of water

(during the LDS)

Question 4

where is the dehydrogenase enzyme found?

Answer 4

in chloroplasts

Question 5

what is DCPIP?

Answer 5

-a redox indicator which is blue and turns colourless when reduced
-it can pick up the electrons from the LDR instead of NADP

Question 6

chemical equation of reduced DCPIP:

Answer 6

blue DCPIP
→ (electrons)
colourless DCPIP

Question 7

what can ammonium hydroxide do?

Answer 7

-an alkaline
-it could denature the dehydrogenase enzyme
-ammonium hydroxide can also accept electrons, so could pick up electrons instead of DCPIP and NADP

Question 8

hypothesis:

Answer 8

rate of reaction will decrease with ammonium hydroxide

Question 9

if a weedkiller is absorbing all the electrons
released from photoionised chlorophyll, what is the impact of this?

Answer 9

-stop/slow light-dependent reaction
-less production of reduced NADP
-light-independent reaction slowed/stopped
-photosynthesis slower/not completed

Question 10

why would weed killer which contains ammonium hydroxide be effective?

Answer 10

1) stops/slows the light-dependent reaction
2) stops/reduces production of NADPH
3) light-independent reaction slowed/stopped
4) overall photosynthesis is slower/stopped
5) will prevent growth/respiration – kills weed

Question 11

test tube A:

Answer 11

-DCPIP solution
-water
-chloroplast suspension
-wrapped in aluminium foil to exclude light

Question 12

test tube B:

Answer 12

-DCPIP solution
-water
-isolation medium

Question 13

what are test tube A & B?

Answer 13

-control tubes
-should be left until the end of your investigation

Question 14

test tube C:

Answer 14

-water
-chloroplast suspension

Question 15

what is the purpose of test tube C?

Answer 15

to use as a standard to determine when any colour change is complete (when the reaction has taken place)

Question 16

test tube X:

Answer 16

-DCPIP solution
-water
-chloroplast suspension

Question 17

what should be done with test tube X?

Answer 17

-mix the contents and start the timer
-record in seconds how long it takes for the contents of tube X to change colour from blue-green to green
-use tube C to help you determine when the colour change is complete

(repeat at least twice)

Question 18

test tube Y:

Answer 18

-DCPIP solution
-ammonium hydroxide
-chloroplast suspension

Question 19

what should be done with test tube Y?

Answer 19

-quickly mix the contents and start the timer
-record in seconds how long it takes for the contents to change colour from blue-green to green
-use tube C to help you determine when the colour change is complete
-if this has not taken place within 300 seconds record the colour at this point

(repeat at least twice)

Question 20

how to calculate rate of reaction

Answer 20

amount of change / time taken

Question 21

results: test tube C vs test tube X

Answer 21

test tube X has become the same colour as test tube C, this means that the DCPIP has become colourless & the reaction has occurred

Question 22

results: test tube C vs test tube Y

Answer 22

-test tube 5 has retained its blue green colour, and doesn’t look the same as test tube 3
-no reaction has taken place

Question 23

conclusion of the practical:

Answer 23

ammonium hydroxide did decrease the rate of dehydrogenase enzyme activity

Question 24

limitations of the practical:

Answer 24

end point is subjective, not everyone will agree that the greens are the same → use a colorimeter for a more subjective view
(cuvettes)

unequal distributions of light around the test tubes could affect the rate of reaction → set up four lamp from each direction to ensure equal light intensity

foil isn’t blocking out the light from all directions → fully cover the test tubes

Question 25

how to prepare the chloroplasts suspension (steps)

Answer 25

1) put about 50 cm² of isolation medium into a beaker 2) tear 8 spinach leaves into small pieces and put the pieces into the isolation medium in the beaker 3) do not put pieces of the midrib or the leaf stalk into the beaker 4) half fill a large beaker with ice and place a small beaker on top of the ice 4) put 3 layers of muslin over the top of the filter funnel and wet it with the isolation medium. rest the filter funnel in the small beaker on the ice. 5) pour the spinach and isolation medium into the blender and blend for about 15 seconds. pour the blended mixture back into the beaker 6) pour a little of your blended mixture through the muslin in the filter funnel. carefully fold and squeeze the muslin to assist the filtering process. repeat until most of the blended mixture has been filtered 7) label this filtrate which is in the small beaker on ice as chloroplast suspension 8) label five test tubes A, B, C, X and Y. stand these five tubes in the ice in the large beaker

Question 26

how far away should the lamp be?

Answer 26

around 10cm

Question 27

why must all solutions be ice cold?

Answer 27

to slow activity of the enzymes which could damage the chloroplasts that were released from blending the spinach leaves

Question 28

why must the isolation medium be an isotonic solution?

Answer 28

Isotonic means the solution has the same water potential as the chloroplasts and therefore will prevent the chloroplast bursting or shrivelling

Question 29

why were the spinach leaves blended?

Answer 29

to release the chloroplasts by breaking open the cells

Question 30

why did you filter the blended spinach?

Answer 30

to remove large pieces of cell debris and other organelles

Question 31

you were told not to put the leaf midrib or leaf stalk in the blender. suggest why.

Answer 31

few/no chloroplasts present / tough material so will not be cut up (by blender)

Question 32

what was the purpose of using a blender?

Answer 32

break open cells & release chloroplasts

Question 33

there are large pieces of tissue and other organelles in your chloroplast suspension. describe how you could isolate the chloroplasts from the other components in the suspension.

Answer 33

centrifuge at a low speed then centrifuge at a higher speed collect the pellet

Question 34

you were told to wrap tube A in aluminium foil. explain why.

Answer 34

-keep in the dark / exclude light -slow release of electrons from chlorophyll -stop photosynthesis

Question 35

tube B was a control experiment. explain how the results in this tube acted as a control experiment.

Answer 35

shows that the colour change in x isn’t due to isolation medium and is due to chloroplasts

Question 36

In your investigation, the colour change in DCPIP was caused by the transfer of electrons from chlorophyll. name the part of a chloroplast where this transfer of electrons takes place.

Answer 36

thylakoids

Question 37

explain how the cuvette kept in the dark and the cuvette containing no leaf extract acted as controls in this experiment.

Answer 37

dark: to see if DCPIP decolourised in the absence of light no leaf extract: to see it DCPIP would decolourise without chloroplasts

Question 38

the rate of photosynthesis in intact leaves can be limited by several factors including light, temperature and carbon dioxide. which of these factors will have little effect on the reducing capacity of the leaf extract?

Answer 38

CO2 as it only affects only the LIS

Question 39

describe how you might alter this practical to investigate the effect of light intensity on the light-dependent reactions of photosynthesis.

Answer 39

repeat with different distances of the lamp