NEET Biology Plant Kingdom Morphology and Physiology MCQs Set 02

Answer: (c) increase in height of the body

Explanation:
1. Body mass can fall again (loss of fat or water), the thymus shrinks with age, and a bulging belly can disappear with fat loss, so these changes are reversible.
2. Increase in height is a growth in the skeleton that cannot be undone, so it is irreversible.

Teacher's Note:
1. True growth involves permanent increase in structural parts.
2. Changes in weight due to fat or water are not true growth.

Answer: (d) 8, 6

Explanation:
1. Each binomial is a different species, so there are 8 species.
2. Genera are Pisum, Allium, Saraca, Mangifera, Cleome and Solanum (three species), which gives 6 genera.

Teacher's Note:
1. In a binomial the first word is the genus and the second the species.
2. Different species of the same genus share the first word.

Answer: (a) 1

Explanation:
1. (a) is true, since porins are channel proteins in the outer membrane of gram negative bacteria that allow passage of small molecules.
2. (b) is false, because mesosomes are infoldings of the plasma membrane and do not contain porins.
3. (c) is false, since peptidoglycan is in the cell wall.
4. Only one statement is true.

Teacher's Note:
1. Porins are present in the outer membrane of gram negative bacteria, mitochondria and chloroplasts.
2. Mesosomes help in respiration and DNA replication.

Answer: (c) saprophytic, plasmodium, spores dispersed by air currents

Explanation:
1. Slime moulds are saprophytic protists that move on decaying leaves and logs.
2. Under suitable conditions they form an aggregation called a plasmodium, and during unfavourable conditions it forms fruiting bodies bearing spores that are dispersed by air currents.

Teacher's Note:
1. Slime moulds are placed in kingdom Protista.
2. The spores have true walls and are very resistant.

Answer: (b) A virus causes potato spindle tuber disease

Explanation:
1. Potato spindle tuber disease is caused by a viroid, which is a free RNA without a protein coat, and not by a virus.
2. The other three statements are correct.

Teacher's Note:
1. Viroids are infectious RNA and prions are infectious proteins.
2. Lichens do not grow in polluted air as they are sensitive to \(\mathrm{SO_2}\).

Answer: (b) Phaeophyceae

Explanation:
1. Green algae (Chlorophyceae) and blue green algae (Cyanophyceae) are common in fresh water.
2. Brown algae (Phaeophyceae) are found almost exclusively in marine habitats.

Teacher's Note:
1. Examples of Phaeophyceae are Laminaria, Fucus and Sargassum.
2. Red algae are also mostly marine.

Answer: (d) green algae

Explanation:
1. Chlorophyll a and b are found in green algae and all land plants.
2. The embryo evolved after the green algal stage, leading to liverworts, then to leafy gametophytes (mosses) and later groups.
3. The group with chlorophyll b but no embryo is therefore the green algae.

Teacher's Note:
1. Land plants evolved from green algae, which share chlorophyll a and b, starch and cellulose walls.
2. Embryos are retained in all land plants (embryophytes).

Answer: (b) Gymnosperms

Explanation:
1. Seed ferns (Pteridospermales) are extinct plants with fern-like leaves that bore seeds.
2. Because they produced naked seeds, they are classified as primitive gymnosperms.

Teacher's Note:
1. Seed ferns are considered ancestors of cycads and other gymnosperms.
2. Ferns (pteridophytes) do not produce seeds.

Answer: (d) Parallel venation: Smilax

Explanation:
1. Neem has pinnately compound leaves and silk cotton (Bombax) has palmately compound leaves.
2. Ziziphus is a dicot with reticulate venation.
3. Smilax is a monocot but unusual in showing reticulate venation, so "parallel venation: Smilax" is incorrect.

Teacher's Note:
1. Most monocots have parallel venation, but Smilax, Dioscorea and Colocasia are exceptions.
2. Parallel venation is typical of grasses.

Answer: (c) China rose

Explanation:
1. In China rose (Hibiscus) the flower is hypogynous, with the ovary placed above the other floral parts, so the ovary is superior.
2. Cucumber and guava have an inferior ovary (epigynous flowers), and the rose is perigynous with the ovary lying at the middle level (half inferior).

Teacher's Note:
1. Hypogynous: superior ovary, as in mustard, brinjal and China rose.
2. Epigynous: inferior ovary, as in guava, cucumber and apple.

Answer: (d) inferior ovary

Explanation:
1. Liliaceae has perigonium of six tepals in two whorls of three, actinomorphic trimerous flowers, and endospermous seeds.
2. The ovary is superior, so an inferior ovary is not shown.

Teacher's Note:
1. Liliaceae is a monocot family, with examples like Allium, Aloe and Lilium.
2. Its flowers are bisexual, trimerous and hypogynous.

Answer: (b) monocot root

Explanation:
1. The dicot root has two to four (di- to tetrarch) xylem bundles.
2. The monocot root has more than six (polyarch).
3. Stems have endarch xylem and open or scattered vascular bundles.

Teacher's Note:
1. Roots show exarch xylem, in which protoxylem is towards the periphery.
2. Dicot stems have endarch xylem.

Answer: (b) phelloderm

Explanation:
1. Dedifferentiation is the process in which permanent cells regain the power of division (for example, cork cambium).
2. Redifferentiation is when the cells produced by these meristems mature and lose the power of division again.
3. Phelloderm is cut off by the phellogen and matures into permanent tissue, so it is redifferentiated. Phellogen and wound phellogen are dedifferentiated, and the primary cortex is a primary tissue.

Teacher's Note:
1. Dedifferentiation then redifferentiation occurs during secondary growth.
2. Secondary xylem and phloem are also redifferentiated tissues.

Answer: (d) It modifies and packages proteins

Explanation:
1. The Golgi apparatus is found in all eukaryotic cells, animal and plant, and is absent in prokaryotes.
2. It receives proteins from the ER, modifies them (for example by glycosylation), and packages them into vesicles.
3. ATP is produced in mitochondria.

Teacher's Note:
1. Cis face receives and trans face dispatches vesicles.
2. In plant cells the Golgi forms the cell plate during cytokinesis.

Answer: (b) ribosome with mRNA

Explanation:
1. Several ribosomes attached to a single mRNA strand form a polyribosome or polysome, which is also called an ergosome.
2. They allow many copies of a protein to be made from one mRNA at the same time.

Teacher's Note:
1. Polysomes make translation faster.
2. A nucleosome chain is chromatin.

Answer: (d) all of the above

Explanation:
1. Chromatin is made of DNA, basic histone proteins, non-histone proteins, and some RNA.
2. Therefore all the listed components are present.

Teacher's Note:
1. Chromatin condenses into chromosomes at cell division.
2. Histone proteins are rich in lysine and arginine.

Answer: (d) telophase

Explanation:
1. In telophase the chromosomes reach the poles and uncoil to form diffuse chromatin.
2. The nuclear envelope reforms around them, so the individual chromosomes can no longer be seen.

Teacher's Note:
1. Chromosomes are best seen at metaphase.
2. Telophase is the reverse of prophase.

Answer: (b) E \(\rightarrow\) F \(\rightarrow\) D \(\rightarrow\) C \(\rightarrow\) A \(\rightarrow\) B

Explanation:
1. Bivalents form during zygotene (E), and recombinase acts in pachytene during crossing over (F).
2. The synaptonemal complex dissolves in diplotene (D), and chiasmata terminalise in diakinesis (C).
3. Homologous chromosomes disjoin in anaphase I (A) and centromeres divide in anaphase II (B).
4. The order is E, F, D, C, A, B.

Teacher's Note:
1. Prophase I stages: leptotene, zygotene, pachytene, diplotene, diakinesis.
2. Centromere division occurs only in meiosis II.

Answer: (b) fatty acids

Explanation:
1. A triglyceride is formed when the three OH groups of glycerol esterify with three fatty acid molecules.
2. The \(\mathrm{R-CO-}\) units are fatty acid residues, and R is the hydrocarbon chain.

Teacher's Note:
1. Fats are triglycerides with saturated chains and oils with unsaturated chains.
2. Lecithin is a phospholipid with a phosphate group in place of one fatty acid.

Answer: (b) 3 carbon atoms

Explanation:
1. The simplest carbohydrates are trioses such as glyceraldehyde and dihydroxyacetone.
2. They have 3 carbon atoms with the formula \(\mathrm{C_3H_6O_3}\).

Teacher's Note:
1. Monosaccharides range from trioses to heptoses.
2. Glucose is a hexose.

Answer: (b) lysine

Explanation:
1. Lysine has two amino groups and one carboxyl group, so it is a basic amino acid.
2. Glutamic acid has two carboxyl groups, and alanine and valine each have one amino and one carboxyl group.

Teacher's Note:
1. Basic amino acids: lysine, arginine and histidine.
2. Acidic amino acids: aspartic and glutamic acid.

Answer: (a) formation of \(6\times10^5\) molecules of \(\mathrm{H_2CO_3}\) every second

Explanation:
1. Carbonic anhydrase catalyses the reaction at about \(6\times10^5\) molecules per second.
2. This is about \(10^7\) times faster than the uncatalysed rate.
3. (b) is the same rate as (a) written per minute with a wrong power: \(6\times10^5\) per second is \(36\times10^6\) per minute, which does not match (b).

Teacher's Note:
1. Carbonic anhydrase is one of the fastest enzymes.
2. It is a zinc-containing enzyme found in red blood cells.

Answer: (b) It is sensitive to inhibitors which react with the lipid side chains

Explanation:
1. Facilitated diffusion uses carrier or channel proteins and moves molecules down their concentration gradient, so (a) and (d) are correct.
2. Water channels are made of 8 different types of aquaporins, so (c) is correct.
3. The transport is sensitive to inhibitors that react with the protein side chains of the carriers, not with lipids, so (b) is incorrect.

Teacher's Note:
1. Facilitated diffusion is passive, specific, saturable and inhibitable.
2. Active transport requires ATP and works against the gradient.

Answer: (d) P, N, K and Mg

Explanation:
1. Macronutrients are required in amounts above 10 mmol per kg of dry matter: C, H, O, N, P, S, K, Ca and Mg.
2. Fe, Mn, Co and Zn are micronutrients (trace elements).
3. The set P, N, K and Mg has only macronutrients.

Teacher's Note:
1. Remember the micronutrients: Fe, Mn, Cu, Zn, B, Mo, Cl, Ni.
2. Co is needed in very small amounts by legumes and certain microbes.

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

Explanation:
1. The OEC splits water into protons, electrons and oxygen.
2. It is associated with PS II and is located on the lumen (inner) side of the thylakoid membrane, so protons accumulate in the lumen.

Teacher's Note:
1. Water is the source of electrons for PS II.
2. Photolysis of water also releases protons that make the lumen acidic.

Answer: (b) 0

Explanation:
1. Carboxylation is the fixation of \(\mathrm{CO_2}\) to RuBP by RuBisCO to form two molecules of 3-PGA.
2. This step needs no ATP.
3. ATP is used later in the reduction and regeneration phases.

Teacher's Note:
1. The Calvin cycle needs 3 ATP and 2 NADPH for each \(\mathrm{CO_2}\) fixed.
2. RuBisCO is the most abundant enzyme on earth.

Answer: (d) A = \(\mathrm{H_2O}\), B = ADP, C = pyruvic acid

Explanation:
1. Enolase removes a molecule of water from 2-phosphoglycerate to give PEP, so A is \(\mathrm{H_2O}\).
2. Pyruvate kinase transfers a phosphate from PEP to ADP, forming ATP and pyruvic acid.
3. So B is ADP and C is pyruvic acid.

Teacher's Note:
1. This last step is a substrate level phosphorylation.
2. Two molecules of ATP are made per glucose in this step.

Answer: (d) \(\alpha\)-ketoglutaric acid

Explanation:
1. \(\alpha\)-Ketoglutaric acid has 5 carbon atoms.
2. Oxaloacetic acid has 4 carbon atoms, PGA has 3, and acetyl CoA has the 2-carbon acetyl group.

Teacher's Note:
1. In the Krebs cycle: citrate 6C, \(\alpha\)-ketoglutarate 5C, succinate 4C, oxaloacetate 4C.
2. RuBP is the 5C compound of the Calvin cycle.

Answer: (b) ripening fruits

Explanation:
1. Ethylene is a gaseous hormone produced in large amounts by tissues undergoing senescence and by ripening fruits.
2. It promotes fruit ripening.

Teacher's Note:
1. Ethylene is used commercially to ripen fruits such as mango and banana.
2. It is also used to break seed and bud dormancy in some plants.

Answer: (a) self pollinated

Explanation:
1. Cleistogamous flowers never open, so anthers and stigma lie close together in a closed flower.
2. Pollen germinates in the anther and the pollen tubes reach the stigma, so these flowers are always self pollinated.

Teacher's Note:
1. Cleistogamy assures seed set even without pollinators.
2. Examples: Viola (common pansy), Oxalis and Commelina.