Photosynthesis 20 NEET Level Questions PDF | Chapter 13 Class 11 Biology | NEET 2027

*Chapter 13: PHOTOSYNTHESIS IN HIGHER PLANTS* 🌿  

*20 NEET LEVEL QUESTIONS - "NOT ON GOOGLE"* 🔥  

_With Detailed Solutions - 1-2Q concepts can come in NEET 2027_

---

### *SECTION A: CONCEPTUAL + NCERT DEEP* `10Q`

*Q1.* In C4 plants, Rubisco is active in bundle sheath cells but not in mesophyll. Reason?

A. PEPcase is absent in mesophyll

B. O2 concentration is low in bundle sheath = Less photorespiration

C. Calvin cycle does not occur in mesophyll

D. Bundle sheath chloroplasts are agranal

Ans: B + D 

Detail: In mesophyll, PEPcase fixes CO2 → OAA → Malate. Malate goes to bundle sheath and releases CO2. In bundle sheath, CO2 high + O2 low = Rubisco carboxylase activity increases, oxygenase decreases. Plus bundle sheath chloroplasts are agranal = Less PS-II = No O2 formed = Photorespiration negligible. This is why productivity is high in Maize, Sugarcane.

NEET Trap: Option C is wrong because mesophyll has CO2 fixation even if Calvin is absent.

*Q2.* When does cyclic photophosphorylation occur in light reaction?

A. CO2 high + NADP+ high

B. CO2 low + NADP+ low + ATP demand high

C. Water splitting high

D. Only at night

Ans: B

Detail: When Calvin cycle slows due to low CO2, NADPH is not used → NADP+ not regenerated. Electrons get jammed in PS-I. Then cyclic path starts: PS-I → FD → PQ → Cyt b6f → PC → PS-I. Only ATP is formed, no NADPH. ATP is used in regeneration step of Calvin. In C4 plants, bundle sheath also has more cyclic due to less PS-II.

*Q3.* What is the main advantage of Kranz anatomy?

A. Reduce water loss

B. CO2 concentration mechanism = Minimize photorespiration

C. Increase light absorption

D. Starch storage

Ans: B

Detail: Kranz = Wreath = Ring of bundle sheath cells around mesophyll. C4 acid forms in mesophyll, CO2 is pumped in bundle sheath. Result: CO2 near Rubisco = 10-20x atmospheric. So it doesn't compete with O2 = Photorespiration ~0. Productivity 2x of C3. Efficient even at 30-40°C.

*Q4.* What does Emerson enhancement effect prove?

A. Existence of PS-I and PS-II

B. CO2 fixation occurs in dark

C. Chlorophyll a is main pigment

D. Water is source of O2

Ans: A

Detail: Emerson observed: 680nm light + 700nm light together gives photosynthesis rate > sum of individual. Because 680nm = PS-II excited, 700nm = PS-I excited. Both together complete electron flow = Non-cyclic. This proved 2 photosystems exist. If only one existed, no enhancement.

*Q5.* When do stomata open in CAM plants and why?

A. Day, to take CO2

B. Night, to save water loss

C. Open 24 hours

D. Only during rain

Ans: B

Detail: CAM = Crassulacean Acid Metabolism = Desert plants = Opuntia, Pineapple. Day temp high = stomata close = reduce transpiration loss. Night cool = stomata open = take CO2 + PEPcase makes OAA → Malic acid = Store in vacuole. Day stomata closed, malic acid releases CO2 = Calvin cycle. So RQ is 0 in day because no CO2 goes out.

*Q6.* How many ATP and NADPH are needed to fix 6 CO2 in Calvin cycle?

A. 12 ATP + 12 NADPH

B. 18 ATP + 12 NADPH

C. 6 ATP + 6 NADPH

D. 30 ATP + 18 NADPH

Ans: B

Detail: 1 CO2 fix = 3 ATP + 2 NADPH used. 3 ATP = 2 in reduction step + 1 in regeneration. 2 NADPH = only in reduction. So 6 CO2 = 6×3 = 18 ATP, 6×2 = 12 NADPH. Total to make 1 Glucose. Light reaction gives ATP:NADPH = 3:2 ratio. Asked in NEET 2019.

*Q7.* Main difference between Chlorophyll a and b?

A. Chl a = C55H72O5N4Mg, Chl b = C55H70O6N4Mg

B. Chl a = Reaction centre, Chl b = Accessory

C. Chl a = Blue-green, Chl b = Yellow-green

D. All correct

Ans: D

Detail: Chl b has -CHO group, Chl a has -CH3. So Chl b is slightly more polar. Chl a = P680, P700 = reaction centre = direct e- release. Chl b = in LHC = light harvest + transfer to Chl a. In paper chromatography Chl b is lower = more adsorbed. Both have Mg-porphyrin head + Phytol tail.

*Q8.* Compare CO2 compensation point of C3 and C4.

A. C3 = 25-100 ppm, C4 = 0-10 ppm

B. C3 = 0 ppm, C4 = 50 ppm

C. Both same = 50 ppm

D. C4 has no compensation point

Ans: A

Detail: Compensation point = CO2 conc where photosynthesis = respiration. In C3, Rubisco has O2 affinity too = photorespiration high = CO2 released. So net uptake needs 25-100 ppm CO2. In C4, CO2 pump = high CO2 in bundle sheath = no photorespiration = +ve photosynthesis even at 0-10 ppm. So C4 survives in drought.

*Q9.* In bacterial photosynthesis, H2O is not used. What is the electron source?

A. CO2

B. H2S, Thiosulphate, H2

C. Glucose

D. O2

Ans: B

Detail: Purple sulphur bacteria = H2S → S + 2H+ + 2e-. Green sulphur bacteria same. So no O2 released = Anoxygenic. Only PS-I type, no PS-II. This experiment proved O2 source in plants is H2O, not CO2 = Van Niel hypothesis. Purple non-sulphur use organic acids.

*Q10.* How does electron move from P680 to P700 in Z-scheme?

A. Direct jump

B. PQ → Cyt b6f → PC → P700

C. FD → NADP+

D. Water → P700

Ans: B

Detail: Z-scheme = Non-cyclic. P680* → Pheophytin → QA → QB → PQ → Cyt b6f complex = H+ pump → PC = Plastocyanin → P700+. P700* → A0 → A1 → FeS → FD → NADP+ reductase → NADPH. Z shape forms in redox potential graph. 4H+ pump ≈ 1 ATP. 2H2O → O2 + 4H+ + 4e- gives 4e-.

---

### *SECTION B: NUMERICAL + APPLICATION* `5Q`

*Q11.* One chloroplast releases 10 O2 molecules in 1 sec. How many H2O split and minimum photons needed?

A. 20 H2O, 80 photons

B. 10 H2O, 40 photons

C. 20 H2O, 40 photons

D. 5 H2O, 20 photons

Ans: A

Detail: 2H2O → O2 + 4H+ + 4e-. So 1 O2 = 2 H2O split. 10 O2 = 20 H2O. Quantum requirement = 8 photon per O2 = 10×8 = 80 photons. Quantum yield = 1/8 = 12.5%. This calculation comes in NEET.

*Q12.* Total ATP needed to make 1 Glucose in C4 plant?

A. 18 ATP

B. 24 ATP

C. 30 ATP

D. 38 ATP

Ans: C

Detail: C3 Calvin = 18 ATP. C4 needs extra 2 ATP per CO2 for PEP regeneration: Pyruvate → PEP by Pyruvate-dikinase. 6 CO2 = 6×2 = 12 extra ATP. Total = 18 + 12 = 30 ATP. NADPH same = 12. So C4 uses more energy but efficient at high temp because no photorespiration loss.

*Q13.* At 25°C, C3 plant photosynthesis rate is 20 mg CO2/dm²/hr. What happens to C4 rate at 40°C?

A. Decrease to 10

B. Same 20

C. Increase to 35-40

D. Zero

Ans: C

Detail: C3 optimum 20-25°C, at 40°C Rubisco oxygenase increases = photorespiration high = net PS less. C4 optimum 30-40°C, CO2 pump + no photorespiration = rate increases. Maize rate at 40°C ≈ 2x of C3. So tropical crops are C4.

*Q14.* If radioactive 14CO2 is given to C4 plant, where will radioactivity be found in 5 sec?

A. 3-PGA

B. Malic acid / Aspartic acid

C. RuBP

D. Glucose

Ans: B

Detail: Hatch-Slack pathway: In mesophyll, PEP + 14CO2 → OAA 4C → Malate/Aspartate 4C. This is first stable product. 3-PGA forms later in bundle sheath via Calvin. Hatch-Slack proved this in Sugarcane in 1966. In C3, 3-PGA is first.

*Q15.* A leaf shows RQ = 0.7. What can be the substrate?

A. Glucose

B. Tripalmitin = Fat

C. Oxalic acid

D. Protein

Ans: B

Detail: RQ = CO2/O2. Fats have more C, less O = more O2 needed for oxidation = RQ <1. Tripalmitin C51H98O6 + 72.5O2 → 51CO2 + 49H2O. RQ = 51/72.5 = 0.7. Carb = 1, Organic acid = >1, Anaerobic = ∞. Germinating castor seed has RQ 0.7.

---

### *SECTION C: ASSERTION-REASON + MATCH + DIAGRAM* `5Q`

*Q16.* Assertion: No O2 is formed in cyclic photophosphorylation.  

Reason: PS-II is not involved in cyclic.

A. Both T, R correct explanation

B. Both T, R wrong

C. A T, R F

D. A F, R T

Ans: A

Detail: Cyclic = PS-I → FD → PQ → Cyt b6f → PC → PS-I. PS-II + water splitting absent = no O2. Only ATP from chemiosmosis. Occurs when NADP+ is low or ATP demand is high.

*Q17.* Match:

Column I           Column II

i. PEPcase        p. Bundle sheath, Calvin

ii. RuBisCO       q. Mesophyll, C4 acid form

iii. P680         r. PS-II reaction centre

iv. P700          s. PS-I reaction centre

A. i-q, ii-p, iii-r, iv-s

B. i-p, ii-q, iii-s, iv-r

C. i-r, ii-s, iii-p, iv-q

D. i-s, ii-r, iii-q, iv-p

Ans: A

Detail: PEPcase = Phosphoenolpyruvate carboxylase = Mesophyll cytosol = forms OAA. RuBisCO = Bundle sheath chloroplast = Calvin entry. P680 = PS-II, P700 = PS-I. This exact match came in NEET 2021.

*Q18.* Assertion: Double carboxylation occurs in C4 plants.  

Reason: First PEPcase, then RuBisCO fixes CO2.

A. Both T, R correct explanation

B. Both T, R wrong

C. A T, R F

D. A F, R T

Ans: A

Detail: 1st = In mesophyll, PEP + CO2 → OAA = C4 acid. 2nd = In bundle sheath, C4 acid releases CO2 + RuBP + CO2 → 3-PGA. So C4 = Dicarboxylic acid pathway. Increases efficiency.

*Q19.* According to Blackman's Law of Limiting Factors, high light + low CO2, which is limiting?

A. Light

B. CO2

C. Temperature

D. Water

Ans: B

Detail: Blackman: When multiple factors exist, rate is decided by factor in minimum. Light is high but CO2 low = CO2 limiting. Increasing CO2 increases rate, increasing light won't. Graph shows plateau.

*Q20.* Diagram: In chloroplast, what happens in granum and stroma?

A. Granum = Light reaction, Stroma = Dark reaction

B. Granum = Dark, Stroma = Light

C. Both Light

D. Both Dark

Ans: A

Detail: Granum = Thylakoid stacks = Chlorophyll + ETS + ATP synthase = Light reaction = ATP + NADPH + O2. Stroma = Enzymes of Calvin cycle = RuBisCO, etc = Dark reaction = Sugar synthesis. Stroma also has DNA + 70S ribosome = Semi-autonomous.

---

### *BONUS: DIAGRAM LABELS FOR NEET BOARD* 📝

*1. Chloroplast*

Outer membrane, Inner membrane, Stroma, Granum, Thylakoid, Lumen, Stroma lamella, DNA, 70S Ribosome, Starch grain, Lipid droplet

*2. Z-Scheme*

H2O → P680 → Pheo → QA → QB → PQ → Cyt b6f → PC → P700 → A0 → A1 → FeS → FD → NADP+ reductase → NADPH

Label: 4H+ pump, ATP synthase, O2 release, Photolysis

*3. C4 Pathway*

Mesophyll: PEP + CO2 → OAA → Malate/Aspartate

Bundle sheath: Malate → Pyruvate + CO2, CO2 + RuBP → Calvin → Sugar

Pyruvate → PEP = Back to mesophyll = 2 ATP used

*4. Calvin Cycle*

Carboxylation: RuBP + CO2 → 2×3-PGA = RuBisCO

Reduction: 3-PGA + ATP + NADPH → G3P

Regeneration: G3P → RuBP = 5/6 G3P used, 1/6 Glucose

*SAVE THIS* 📌  

2-3Q concepts from these 20Q can come in NEET 2027. Focus on C4, Cyclic, Compensation Point, Calvin ATP count.