NEET Biology Plant Kingdom Morphology and Physiology MCQs Set 01

Answer: (a) carbon dioxide

Explanation:
1. The overall reaction is \( \ce{6CO2 + 12H2O -> C6H12O6 + 6O2 + 6H2O} \).
2. The oxygen released as \( \ce{O2} \) comes from the splitting of water, while the oxygen atoms of glucose come from carbon dioxide (also forming some of the water produced).

Teacher's Note:
1. Isotope studies with \( \ce{^18O} \) showed that the released oxygen comes from water.
2. Carbon and oxygen of glucose are derived from \( \ce{CO2} \).

Answer: (c) 18 molecules of ATP are synthesised

Explanation:
1. The Calvin cycle uses ATP; it does not synthesise it. For the synthesis of one glucose from 6 \( \ce{CO2} \), 18 ATP and 12 NADPH are consumed.
2. The other three statements are correct: RuBisCO fixes \( \ce{CO2} \) to RuBP, the first stable product is 3-phosphoglycerate (PGA), and NADPH reduces PGA.

Teacher's Note:
1. The Calvin cycle is the dark (synthetic) phase driven by ATP and NADPH from the light reaction.
2. 3 ATP and 2 NADPH are used per \( \ce{CO2} \) fixed.

Answer: (b) the amount of \( \ce{CO2} \) released is more than the \( \ce{O2} \) consumed

Explanation:
1. Organic acids are richer in oxygen than carbohydrates, so less oxygen is needed to oxidise them.
2. For example, malic acid: \( \ce{C4H6O5 + 3O2 -> 4CO2 + 3H2O} \), giving \( RQ = \frac{4}{3} > 1 \).
3. So more \( \ce{CO2} \) is released than \( \ce{O2} \) consumed.

Teacher's Note:
1. RQ of carbohydrate is 1, fats less than 1 and organic acids more than 1.
2. RQ is the ratio of \( \ce{CO2} \) released to \( \ce{O2} \) consumed.

Answer: (c) (i) biological nitrogen fixation; (ii) \( \ce{O2} \)

Explanation:
1. Reduction of \( \ce{N2} \) to \( \ce{NH3} \) by nitrogenase in microbes is biological nitrogen fixation: \( \ce{N2 + 8e- + 8H+ + 16ATP -> 2NH3 + H2 + 16ADP + 16Pi} \).
2. Nitrogenase is a Mo-Fe protein that is highly sensitive to molecular oxygen, so the process needs anaerobic conditions. In root nodules leghaemoglobin acts as an oxygen scavenger.

Teacher's Note:
1. Leghaemoglobin protects nitrogenase by binding oxygen.
2. Electrical and industrial fixation (Haber process) do not use enzymes.

Answer: (c) protein

Explanation:
1. Cellulose, starch and glycogen are homopolymers of glucose; chitin is a polymer of N-acetylglucosamine.
2. Proteins are heteropolymers made of different amino acids.

Teacher's Note:
1. Homopolymer: only one type of monomer. Heteropolymer: different monomers.
2. Proteins and nucleic acids are heteropolymers.

Answer: (a) To protect the tails from the aqueous environment

Explanation:
1. Phospholipids are amphipathic: the head is polar (hydrophilic) and the tails are non-polar (hydrophobic).
2. In the bilayer the heads face the aqueous surroundings (outside and cytoplasm) while the tails are in the inner side, shielded from water.

Teacher's Note:
1. Singer and Nicolson's fluid mosaic model describes the lipid bilayer with embedded proteins.
2. Hydrophobic parts avoid water.

Answer: (c) in diffusion, substances move independently of each other

Explanation:
1. In diffusion each substance moves independently along its own concentration gradient.
2. In mass flow, substances move together in bulk as a result of a pressure gradient.
3. Diffusion is slow and suitable only for short distances.

Teacher's Note:
1. Mass flow is the movement of substances in bulk, e.g. in xylem and phloem.
2. Diffusion is passive and independent of pressure.

Answer: (c) a Mo-Fe protein

Explanation:
1. Nitrogenase consists of two components: the Fe protein and the Mo-Fe protein.
2. So it is a molybdenum-iron protein, which is why molybdenum is an essential micronutrient for nitrogen fixation.

Teacher's Note:
1. Molybdenum is a component of nitrogenase and nitrate reductase.
2. Both enzymes need Mo.

Answer: (a) telophase

Explanation:
1. During telophase the chromosomes reach the poles, decondense, and a nuclear envelope re-forms around each set.
2. The nucleolus, Golgi complex and ER reappear. Thus two daughter nuclei are formed in telophase (karyokinesis completes), followed by cytokinesis.

Teacher's Note:
1. Karyokinesis (nuclear division) ends at telophase; cytokinesis divides the cytoplasm.
2. Anaphase only separates the chromatids.

Answer: (d) Ascomycetes and Deuteromycetes

Explanation:
1. Asexual reproduction in fungi is by zoospores or aplanospores (Phycomycetes) and by conidia (Ascomycetes and Deuteromycetes).
2. Conidia are borne externally on conidiophores, as in Penicillium and Aspergillus.

Teacher's Note:
1. Penicillium forms chains of conidia.
2. Basidiomycetes usually reproduce asexually by fragmentation and rarely by conidia.

Answer: (b) Mostly pteridophytes are homosporous

Explanation:
1. Most pteridophytes produce spores of a single kind (homosporous); Selaginella and Salvinia are heterosporous exceptions.
2. All gymnosperms are heterosporous, but some (e.g. Cycas) do not bear two types of cones on the same plant; vascular plants have a diplontic or haplo-diplontic cycle depending on the group, and Chara is monoecious.

Teacher's Note:
1. Homosporous plants produce one type of spore; heterosporous plants produce micro- and megaspores.
2. Heterospory led to the seed habit.

Answer: (c) chemoautotrophic nutrition

Explanation:
1. Oxygenic photosynthesis occurs in cyanobacteria and in plants; flagellar movement occurs in many protists and others; carbon dioxide fixation occurs in plants and algae.
2. Chemosynthetic (chemoautotrophic) nutrition, obtaining energy by oxidising inorganic compounds, occurs only in some bacteria (Monera).

Teacher's Note:
1. Nitrifying bacteria such as Nitrosomonas are chemoautotrophs.
2. Among all five kingdoms, chemoautotrophy is exclusive to Monera.

Answer: (a) garlic

Explanation:
1. Ginger (rhizome), potato (tuber) and Colocasia (corm) store food in modified underground stems.
2. In garlic (a bulb) the food is stored in the fleshy scale leaves, not in the reduced stem.

Teacher's Note:
1. Underground stems: rhizome (ginger), tuber (potato), corm (Colocasia), bulb (onion, garlic).
2. In bulbs the stem is a small disc.

Answer: (d) Liliaceae family

Explanation:
1. A trimerous flower has floral parts in multiples of three. This is the monocot condition.
2. Liliaceae (a monocot family) has flowers with 3+3 perianth lobes, 3+3 stamens and 3 carpels. Solanaceae, Fabaceae and Brassicaceae are dicots with pentamerous or tetramerous flowers.

Teacher's Note:
1. Monocot flowers are trimerous; dicot flowers are tetramerous or pentamerous.
2. Floral formula of Liliaceae: \( \ce{P_{(3+3)}A_{3+3}G_{(3)}} \).

Answer: (d) both (a) and (c)

Explanation:
1. In Funaria (a moss) the young sporophyte is green and photosynthetic, though it depends on the gametophyte.
2. In Selaginella (a pteridophyte) the dominant sporophyte is a green, independent plant. Spirogyra is a haploid alga in which only the zygote is diploid.

Teacher's Note:
1. In bryophytes the gametophyte is dominant; in pteridophytes the sporophyte is dominant.
2. The sporophyte of Funaria remains attached to the gametophyte.

Answer: (b) consumer-decomposer protista

Explanation:
1. Slime moulds are saprophytic protists that live on decaying twigs and leaves.
2. They engulf organic material, so they are consumer-decomposers.

Teacher's Note:
1. Under suitable conditions they form an aggregation called plasmodium that can grow up to several feet.
2. Spores are dispersed by air currents.