Name the element which is an activator of RuB is CO and PEP case.

Name the element which is an activator of RuB is CO and PEP case.
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Name the element which is an activator of RuB is CO and PEP case.
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The C4 plants are photosynthetically more efficient than C3 plants because (a) the CO2 efflux is not prevented (b) they have more chloroplasts (c) the CO2 compensation point is more (d) CO2 generated during photorespiration is trapped and recycled through PEP carboxylase.

Description : The C4 plants are photosynthetically more efficient than C3 plants because (a) the CO2 efflux is not prevented (b) they have more chloroplasts (c) the CO2 compensation point is more (d) CO2 generated during photorespiration is trapped and recycled through PEP carboxylase.

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Bundle sheath cells (a) are rich in PEP carboxylase (b) lack RuBisCO (c) lack both RuBisCO and PEP carboxylase (d) are rich in RuBisCO.

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The carbon dioxide acceptor in Calvin cycle/ C3-plants is (a) phosphoenol pyruvate (PEP) (b) ribulose 1, 5-diphosphate (RuDP) (c) phosphoglyceric acid (PGA) (d) ribulose monophosphate (RMP).

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The primary acceptor, during CO2 fixation in C3 plants, is (a) phosphoenolpyruvate (PEP) (b) ribulose 1, 5-diphosphate (RuDP) (c) phosphoglyceric acid (PGA) (d) ribulose monophosphate (RMP).

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Carbon dioxide acceptor in C3-plants is (a) PGA (b) PEP (c) RuDP (d) none of these.

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is the measure of whether a plant is now using more or less energy to manufacture its product than it did in the past. a) Total Dissolved Solids (TDS). c) Plant Energy Performance (PEP) b) Revolutions Per Minute (RPM) d) Option (a) and (b)

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Last Answer : Plant Energy Performance (PEP)

is the measure of whether a plant is now using more or less energy to manufacture its product than it did in the past. a) Total Dissolved Solids (TDS) b) Plant Energy Performance (PEP) c) Revolutions Per Minute (RPM) d) Option (a) and (b)

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PEP = (RYEE – current year energy)/RYEE = (23011 – 23834)/23011 = (-) 0.0369 ie (-) 3.7 % …………………….1.5 marks Since the PEP is negative, it implies that the energy conservation measure did not yield reduction in energy consumption, action to be taken to improve the plant performance.

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Last Answer : Production factor (PF) = 28750/34000 = 0.846 …………………….1 mark Ref year equivalent energy (RYEE) = Ref Year Energy Use (RYEU) x PF   = 27,200 x 0.846= 23011MWh

A manufacturing industry plans to improve its energy performance under PAT through implementation of an energy conservation scheme. After implementation, calculate the Plant Energy Performance (PEP) with 2015-16 as the reference year. What is your inference? Given that:  The current year (2016-17 ) Annual Production – 28,750 T ,  Current year (2016-17 ) Annual Energy Consumption– 23,834 MWh,  Reference year (2015-16 ) production - 34,000 T,  Reference year (2015-16 ) Energy consumption - 27,200 MWh.

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Last Answer : Production factor (PF) = 28750/34000 = 0.846 …………………….1 mark Ref year equivalent energy (RYEE) = Ref Year Energy Use (RYEU) x PF   = 27,200 x 0.846= 23011MWh

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