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Photosynthesis worksheet.

1. During photosynthesis plants use ____________ energy of the sun and convert it into _____________ energy, which is stored in the leaves as ____________ molecules. Plants are called ________________ organisms because they make their own organic molecules from ___________ raw materials.

2. During chemiosmosis in chloroplasts the proton gradient across the membranes is as great as 1.0 to 2.0 pH units. Which side (internal or external) of the membrane has the lowest pH?

3 a. Using chemical reaction equations summarize what happens during the light and dark reaction of photosynthesis.

Light Reaction :

Dark Reaction :

b. Which contains more energy and why:

i) one molecule of NADP or one molecule of NADPH?

NADPH because that is the reduced form.

ii) Three molecules of CO₂ or one molecule of phosphoglyceraldehyde and why?

1 molecule of Phosphoglyceraldehyde.

4. How many molecules of CO₂ must enter the Calvin cycle in order for the plant to produce one molecule of a sugar containing 24 carbon atoms? (Hint : Use stoichiometry from 3a)

24 molecules of CO₂.

All the carbon in the sugar comes from CO₂. 6 molecules of CO₂ need to be fixed to form 1 molecule of glucose (a 6 carbon sugar). So 24 molecules of CO₂ should enter the calvin cycle to produce a sugar that has 24 carbon atoms.

5. During an experiment conducted in a lab it was found that Pisum sativum (pea plant) has a maximum rate of photosynthesis at 25^o C. The maximum photosynthesis rate of this green plant was found to be 10 mg glucose/dm²/hr.

Given, that the heat of combustion of glucose is 686 KJ/mole and the molecular weight of glucose is 180 grams, calculate the energy, in KJ, that could be theoretically made over a 18 hr lighted period, by a single leaf which has a surface area of 2 dm².

KEY : Amt of. Glucose produced by 1 leaf = 0.01 x 2 x 180 = 3.6 g

= 3.6 /180 moles = 0.02 moles

Energy produced by 2 moles of glucose = 686 x 0.02 = 13.72 KJ

If a cattle eats 40 leaves from the above plant, how much energy (in ATP) will be produced by the animal after complete oxidation? (Avagadro’s number = 6.023 x 10²³). Assume surface area and photosynthesis rate to be the same for all the leaves, which is given by 4a).

KEY : Moles of glucose produced by 1 leaf = 0.02 moles ( from part a)

Moles of glucose produced by 40 leaves = 40 x 0.02 = 0.8 moles

No. of glucose molecules in 80 moles = 0.8 x 6.023 x 10²³

= 4.8184 x 10²³

1 molecule of glucose on complete oxidation yields = 36 ATP (net yield is 38 – 2 = 36)

So total ATP yield = 4.8184 x 10²³ x 36 = 173.4624 x 10²³ATP

How many molecules of CO₂ must have entered the calvin cycle to produce the amount of ATP obtained in 5(b). (Hint: use the no of molecules of glucose produced from 5b and the stoichiometric relationships in Q.4 to calculate no of molecules of CO₂ needed)

KEY : To make 1 molecule of glucose we need 6 molecules of CO₂ ( since all the C in glucose comes from CO₂)

for 4.8184 x 10²³ molecules of glucose we need = 4.8184 x 10²³ x 6 CO₂

= 28.9104 x 10²³ CO₂ molecules

6. A student in the plant biology lab isolated chloroplasts from leaves and equilibrated them in an acidic solution at pH 4.0. After the chloroplast’s thylakoid compartments reached a pH of 4.0, the chloroplasts were collected and transferred to a basic solution at pH 8.0. This caused the chloroplasts to make ATP, even if placed in the dark. Explain?

KEY : Light reaction is required to create the proton gradient across the thylakoid membrane. In this case this requirement is overcome and a pH gradient is created by incubating the chloroplast in buffers with varying pH, pH 4 first and then in pH 8. During incubation at pH 4.0, a H+ gradient builds up across the membrane by diffusion. When this chloroplast is placed in a buffer with pH 8.0, the inside being more acidic, the H+ ions will want to migrate out which can be utilized by ATP synthase to generate ATP.

7. The 24 power plants in Baltimore metro area collectively emits about 40,300 tons of CO₂ per day (www.carma.org), thereby increasing the level of greenhouse gases in this region. To counter this, environmentalists plan to plant trees in the area. If on average 500 moles of glucose is synthesized by a tree per day, how many trees need to be planted to reabsorb all the CO₂ released into the atmosphere every day by the factory. (1 ton =1000kg, molecular weight of CO₂ is 44g)

KEY: g of CO₂ emited /day = 40,300 x 1000 = 40,300,000 kg = 40,300 x 10⁶ g

Moles of CO₂ emitted/day = 40,300 x 10⁶ g / 44 = 915.9 x 10⁶ moles/day

Moles glucose made/day by 1 tree = 500

6 moles of CO₂ must be fixed to make 1 mole of glucose (from question 4 !)

moles of CO₂fixed = 500 x 6 = 3000 moles of CO₂

No of trees needed = Moles of CO₂ emitted per day/ Moles of CO₂ fixed per day by 1 tree

= 9159 x 10⁵ / 3000 trees = 305, 300 trees !!!!!

b. If a tree on average occupies 6 x 10^-5 sq miles, what percent of the area of Baltimore city needs to be forested to counter the CO₂ emitted from power plants alone. (total area of Baltimore city is 92 sq miles).

KEY : Area occupied by 1 tree = 6 x 10^-5 sq miles

Area occupied by 305 300 trees = 6 x 10^-5 sq miles x 305300

= 18.318 sq miles

Area of Baltimore city = 92 sq miles

% of Baltimore city that needs to be forested = Area occupied by 305300 trees x 100

Area of Baltimore City

= 18.318 sq miles x 100

92 Sq miles

= 19.91 % (= ~ 20 % of Baltimore city !!! )

GO GREEN !!!!

LABORATORY EXERCISES ON PHOTOSYNTHESIS 1 PAPER CHROMATOGRAPHY OF PHOTOSYNTHETIC
NAME DATE PD CHROMATOGRAPHY PHOTOSYNTHESIS 1 WHAT
NAME DATE PERIOD CHAPTER 8 PHOTOSYNTHESIS 81 ENERGY AND

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