Questions: Photosynthesis in Higher Plants

Q1. Write the overall equation of photosynthesis and name the source of the oxygen released.

Answer (English): The overall equation is 6CO2 + 12H2O gives C6H12O6 + 6O2 + 6H2O in the presence of light and chlorophyll. The oxygen released comes from water, because water is split in the thylakoid lumen by the oxygen evolving complex of PS II. Carbon dioxide provides the carbon for sugar and does not supply the released oxygen.

Hinglish: Equation mein 6CO2 aur 12H2O se glucose, 6 oxygen aur 6 paani banta hai. Chhutne wali oxygen paani se aati hai kyunki PS II par paani split hota hai, CO2 se nahi.

Q2. What is a light harvesting complex and why is it needed?

Answer (English): A light harvesting complex is a group of a few hundred pigment molecules bound to proteins around one reaction centre. Its antenna pigments absorb photons over a wide area and pass the energy inward by resonance transfer. A single chlorophyll molecule captures very little light, so grouping many pigments around one reaction centre gives the leaf an efficient way to collect light energy and deliver it to a small number of centres.

Hinglish: Light harvesting complex mein kai pigment molecules ek reaction centre ke aas paas hote hain. Antenna pigments photon pakad kar resonance transfer se energy andar bhejte hain, jisse chhote area se aayi light bhi reaction centre tak pahunch jaati hai.

Q3. Why is PS II placed before PS I in the electron path?

Answer (English): The names were fixed by the wavelengths at which the two reaction centres absorb, with P680 of PS II absorbing at a shorter wavelength than P700 of PS I. In the working chain, however, PS II starts the sequence because it is the photosystem that splits water and supplies electrons. The electron travels from PS II through the carriers to PS I and finally to NADP+, so the functional order is PS II then PS I.

Hinglish: Naam absorption wavelength se aaye hain, isliye P680 wala PS II pehle number par hai. Kaam ke hisaab se PS II pehle chalta hai kyunki wahi paani split karke electron deta hai, phir electron PS I hokar NADP+ tak jaata hai.

Q4. How is a proton gradient built across the thylakoid membrane?

Answer (English): Three processes push protons into the thylakoid lumen. Water splitting at PS II releases protons directly into the lumen. Electron transport through the cytochrome complex pumps protons from the stroma into the lumen. Reduction of NADP+ on the stroma side removes protons from the stroma. Together they make the lumen acidic and the stroma relatively basic, and this gradient drives ATP synthase as protons return to the stroma.

Hinglish: Proton gradient teen tarah se banta hai: paani splitting se, cytochrome complex ke pumping se, aur NADP+ reduction se stroma ke protons hatne se. Lumen acidic ho jata hai aur wahi gradient ATP synthase ko chalata hai.

Q5. What is the primary acceptor of CO2 in C3 plants, and what is the first stable product?

Answer (English): The primary acceptor is ribulose bisphosphate, RuBP, a five-carbon sugar phosphate. RuBisCO joins CO2 to RuBP and the product splits into two molecules of 3-phosphoglycerate, the first stable product with three carbon atoms. This is why the pathway is called C3.

Hinglish: C3 mein primary acceptor RuBP hai aur enzyme RuBisCO; pehla stable product 3-phosphoglycerate banta hai jisme teen carbon hote hain.

Q6. Why is the Calvin cycle called light independent and not dark?

Answer (English): The Calvin cycle does not use photons directly, which is why it is light independent. It still depends on the light reaction for ATP and NADPH, so it cannot keep running in darkness. In a dark chamber the supply of ATP and NADPH soon stops and the cycle stalls, which makes the older name dark reaction a misleading one.

Hinglish: Calvin cycle light ko directly use nahi karti, isliye light independent hai, lekin ATP aur NADPH light reaction se aati hai. Andhere mein yeh ruk jaati hai, isliye ise dark reaction kehna galat hai.

Q7. How many turns of the Calvin cycle make one glucose, and what is the energy cost?

Answer (English): Each turn fixes one CO2 and needs three ATP and two NADPH. A glucose molecule has six carbon atoms, so six turns are required. The total cost for one glucose is 18 ATP and 12 NADPH. If the question asks for one CO2 fixed, the answer is three ATP and two NADPH.

Hinglish: Ek turn mein ek CO2 fix hoti hai aur 3 ATP, 2 NADPH lagte hain. Glucose mein 6 carbon hain, isliye 6 turns chahiye aur kul 18 ATP aur 12 NADPH kharch hote hain.

Q8. What is Kranz anatomy and why does it help a C4 plant?

Answer (English): Kranz anatomy is the wreath-like arrangement of large, thick walled bundle sheath cells around the vascular bundles, with a thin mesophyll layer outside and no intercellular spaces between the sheath cells. The mesophyll fixes CO2 with PEP carboxylase and the sheath runs the Calvin cycle. The tight sheath keeps the CO2 released inside from leaking out, so RuBisCO sees a rich supply and photorespiration stays low.

Hinglish: Kranz anatomy mein bundle sheath cells ki ring vascular bundle ke aas paas hoti hai. Mesophyll CO2 fix karke sheath ko deta hai, aur moti deewarein CO2 ko leak hone se rokti hain, isliye photorespiration kam hoti hai.

Q9. Why is photorespiration a waste for the plant?

Answer (English): When RuBisCO acts as an oxygenase it produces one molecule of 3-phosphoglycerate and one of phosphoglycolate. The phosphoglycolate has to be recovered through a long route that consumes ATP and releases CO2, and the final result is no sugar and no usable energy. Carbon that could have been fixed is lost, so the plant gains nothing and spends energy in the bargain.

Hinglish: RuBisCO jab oxygenase banta hai to phosphoglycolate banta hai jise recover karne mein ATP lagta hai aur CO2 nikal jaati hai. Koi sugar nahi banti, isliye yeh process plant ke liye waste hai.

Q10. State Blackman law of limiting factors with one example.

Answer (English): The law says that when several factors affect a process, the rate is controlled by the factor that is nearest its minimum value. On a bright day with normal air, light is plentiful while CO2 is at 0.04 per cent, so CO2 becomes the limiting factor and raising it raises the rate. On a dull day, light is nearest its minimum and adding CO2 has little effect until the light improves.

Hinglish: Blackman ka law kehta hai ki rate us factor se tay hoti hai jo minimum ke sabse kareeb ho. Dhoop wale din CO2 limiting hoti hai, aur kam roshni wale din light limiting hoti hai, chahe doosra factor kitna bhi ho.

Quiz pointer. The chapter quiz follows this page with five multiple choice questions. Attempt it without scrolling back, then read the explanation for any answer you get wrong, because the quiz retests the same enzyme, product and factor names used in these answers.