Photosynthesis In Higher Plants

Step 1: Understanding the Formation of Oxaloacetic Acid in C Pathway
- In C plants, Phosphoenol pyruvate (PEP) carboxylase fixes CO to form oxaloacetic acid (OAA) in mesophyll cells.
- This reaction helps plants survive in high-temperature conditions by minimizing photorespiration.

Step 1: Understanding the Calvin Cycle
- The Calvin cycle fixes 6 CO molecules to synthesize 1 molecule of glucose. - The process consumes 18 ATP molecules and 12 NADPH molecules per glucose molecule produced.

Step 1: Functions of Eukaryotic RNA Polymerases:
In eukaryotes, there are three distinct RNA polymerases in the nucleus with specific functions:

• RNA Polymerase I: Transcribes rRNAs (28S, 18S, and 5.8S). So, A matches with II.
• RNA Polymerase II: Transcribes precursor of mRNA (hnRNA). So, B matches with III.
• RNA Polymerase III: Transcribes tRNA, 5S rRNA, and snRNAs. So, C matches with I.

Step 2: Matching with Options:
The sequence is A-II, B-III, C-I. This corresponds to Option (C).

Step 1: Understanding the Photosynthesis Equation:
The simplified chemical equation for photosynthesis is often written as: Energy values ( ) associated with the formation of 1 mole of glucose are approximately 686 K.cal (or 677.2 K.cal as per specific textbook data used here).
Step 2: Calculating Molecular Masses:

• Glucose ( ): (Given).
• Oxygen ( ): (Given approx 193 g).
• Carbon Dioxide ( ): .
• Water ( ): .

Step 3: Matching Values to X, Y, Z:

• X ( absorbed): 264 g.
• Y ( absorbed): Based on the simplified equation stoichiometry matching X, it corresponds to 6 moles, i.e., 108 g. (Note: The complete equation involves 216g input, but the options only support the simplified 108g value).
• Z (Energy): 677.2 K.cal (matches Option C).

Step 4: Final Answer:
The correct set is X = 264 g, Y = 108 g, Z = 677.2 K.cal.

DNA polymerase synthesizes DNA, not polypeptides, so statement I is wrong. Tailing occurs at the 3' end, not 5', so statement IV is wrong. Splicing and capping statements (II and III) are correct. Therefore, option 3 is correct.

1. Length of DNA = 510 Å. One base pair length ≈ 3.4 Å.
2. Total base pairs = 510 / 3.4 ≈ 150 base pairs.
3. Total nucleotides = 150 × 2 = 300 nucleotides.
4. DNA contains 20% 6-aminopurines (A+G). In B-DNA, A pairs with T (2 H-bonds), G pairs with C (3 H-bonds).
5. Approximate total hydrogen bonds: A-T pairs = 0.2×150 = 30 pairs × 2 H = 60 H-bonds,
G-C pairs = 120 pairs × 3 H = 360 H-bonds.
6. Total H-bonds = 60 + 360 = 390. Hence correct answer is (1) 300 nucleotides, 390 Hydrogen bonds.

1. In the light reaction of photosynthesis, electrons are passed through Photosystem II and Photosystem I.
2. Ferredoxin is an intermediate electron carrier.
3. NADP+ is the final acceptor of electrons in Photosystem I to form NADPH.
4. Phacophytin acts as primary electron acceptor in PSII, not final.
5. Hence, the final electron acceptor is NADP+. Correct answer is (4) NADP+.

Equation A shows breakdown of glucose to pyruvate with net 2 ATP and NADH — this is Glycolysis. Equation B shows oxidation of pyruvate to CO2 and reducing equivalents with ATP generation — this is Kreb's (TCA) cycle. Equation C shows conversion of pyruvate to ethanol and CO2 via NADH — this is Alcoholic Fermentation. Matching each process with its correct equation confirms option 4.

1. Length of DNA = 510 Å. One base pair length ≈ 3.4 Å.
2. Total base pairs = 510 / 3.4 ≈ 150 base pairs.
3. Total nucleotides = 150 × 2 = 300 nucleotides.
4. DNA contains 20% 6-aminopurines (A+G). In B-DNA, A pairs with T (2 H-bonds), G pairs with C (3 H-bonds).
5. Approximate total hydrogen bonds: A-T pairs = 0.2×150 = 30 pairs × 2 H = 60 H-bonds,
G-C pairs = 120 pairs × 3 H = 360 H-bonds.
6. Total H-bonds = 60 + 360 = 390. Hence correct answer is (1) 300 nucleotides, 390 Hydrogen bonds.

1. In the light reaction of photosynthesis, electrons are passed through Photosystem II and Photosystem I.
2. Ferredoxin is an intermediate electron carrier.
3. NADP+ is the final acceptor of electrons in Photosystem I to form NADPH.
4. Phacophytin acts as primary electron acceptor in PSII, not final.
5. Hence, the final electron acceptor is NADP+. Correct answer is (4) NADP+.