Plant Development and Morphogenesis Questions and Answers

Answer: B

The quiescent centre is present in the root apical meristem.

Described by F. A. L. Clowes, the quiescent centre is a small group of cells at the tip of the root meristem, just behind the root cap, that divide very rarely. It keeps the surrounding stem cells (initials) undifferentiated and acts as a reserve that can repopulate the meristem after damage.

The shoot apical meristem is organised instead into a central zone, peripheral zone and rib zone, with the WUSCHEL-expressing organising centre beneath the stem cells; it has no quiescent centre. Leaf meristems and floral meristems are determinate and lack one too.

Answer: B

The mutated genes are the Class B gene(s).

In the ABC model, A alone specifies sepals (whorl 1), A with B specifies petals (whorl 2), B with C specifies stamens (whorl 3) and C alone specifies carpels (whorl 4). A normal flower is sepal-petal-stamen-carpel.

If B function is lost, whorl 2 has only A and becomes sepals, and whorl 3 has only C and becomes carpels, giving sepal-sepal-carpel-carpel, exactly as described. This is the phenotype of apetala3 and pistillata mutants.

Losing A lets C spread outward, giving carpel-stamen-stamen-carpel. Losing C lets A spread inward, giving sepal-petal-petal-sepal. A double mutant for A and C has no C activity, so it could not form carpels at all.

Answer: A

The expanded cell is (A), the tall, vertically elongated block.

Cellulose microfibrils are stiff in tension along their length. When they are wrapped around the cell horizontally, like hoops on a barrel, they resist any widening of the cell, but the wall can still yield parallel to the cell's axis, between the hoops. Turgor-driven expansion is therefore channelled into elongation along the axis, perpendicular to the microfibrils, which is how hypocotyl cells grow long.

(B), a widened cell, would need the hoops to stretch. (C), an unchanged cube, would mean no expansion at all. (D), a sphere, is the shape expected when microfibrils are laid down randomly, so the wall yields equally in every direction.

Answer: C

The best conclusion is that FT protein moves from leaves to the SAM and promotes flowering.

Observation (i) shows FT is made only in leaves, yet (ii) shows it is needed for the apex to flower, so something must travel from leaf to apex. (iii) shows that trapping the protein in leaf nuclei (with a nuclear localisation signal) stops it working, so the protein must leave the leaf cells. (iv) shows that destroying any FT transcript in the meristem has no effect, so the mRNA is not what acts there.

Together these point to the FT protein as the mobile florigen. A leaf-resident protein is ruled out by (iii), a mobile transcript by (iv), and a requirement for both in the SAM is ruled out by (iv) as well.

Answer: B

The correct match is P-iii, Q-i, R-ii.

In the ABC model, A alone gives sepals, A + B petals, B + C stamens and C alone carpels, and A and C repress each other.

• A mutants: C spreads to all whorls, giving carpel, stamen, stamen, carpel.
• B mutants: petals become sepals and stamens become carpels, giving sepal, sepal, carpel, carpel.
• C mutants: A spreads inwards, giving sepal, petal, petal, sepal.

Arrangement (iv), sepal, sepal, petal, petal, does not arise from a single class of mutant, and the other options swap these patterns.

Answer: B

The most likely result is many tetraploid cells in the epidermis.

The shoot apical meristem is layered. Because cells of the epidermal L1 layer divide only anticlinally (with new walls perpendicular to the surface), their daughters stay in the same layer. A cell in the central zone is a stem cell: it keeps dividing throughout the plant's life, and its descendants are displaced outward to form new tissue.

So a central-zone L1 cell that becomes tetraploid founds a clone that remains in the epidermis and grows with every division, giving a sector of many tetraploid epidermal cells by the adult stage. This is the logic of periclinal chimeras used to map meristem layers.

A single tetraploid cell would imply it never divided, which a stem cell does. The clone cannot spread to the whole SAM, because anticlinal division keeps it out of the inner L2 and L3 layers, and an all-diploid meristem would mean the tetraploid lineage was lost.

Answer: B

The pathway is A gene suppresses B, which promotes flowering transition.

• The b mutant flowers late (60 days against 30), so B promotes flowering.
• The a mutant flowers early (15 days) and has increased B transcript, so A normally represses B; without A, more B speeds flowering.
• The ab double mutant flowers as late as b alone (60 days), so B is epistatic to A: B acts downstream, and the early flowering of a depends entirely on B.

'A activates B' contradicts the rise in B transcript when A is lost. The options in which B regulates A contradict the b mutant's normal A transcript level and the epistasis of b.

Answer: C

The correct match is P-3; Q-4; R-2; S-1.

• CLAVATA3 encodes a small peptide that limits WUSCHEL expression and so keeps the shoot meristem at a constant size; clv3 mutants have enlarged meristems.
• CONSTANS responds to day length and activates FT, controlling the photoperiodic transition to flowering.
• SCARECROW directs the asymmetric division that separates endodermis and cortex, a cell-type specification in the root meristem.
• AGAMOUS is the floral C-class gene that specifies stamen and carpel organ identity.

Every other option swaps at least two of these well-defined roles, for instance giving SCARECROW floral organ identity or AGAMOUS root patterning.

Answer: C

Deleting class B and class C genes gives sepals in every whorl.

In the ABC model, A alone specifies sepals (whorl 1), A with B petals (whorl 2), B with C stamens (whorl 3) and C alone carpels (whorl 4). A and C repress each other, so when one is lost the other spreads into all four whorls.

Without B and C, A is expressed throughout and is the only function left, so every whorl forms sepals.

Losing A alone lets C spread: carpel, stamen, stamen, carpel. Losing A and B leaves C alone everywhere: carpels throughout. Losing B alone gives sepal, sepal, carpel, carpel.

Answer: A

The pathway is X suppresses Y and Y promotes root development.

Test it on each mutant. A gain-of-function X over-suppresses Y, so Y activity is low and roots fail to form: rootless. A loss-of-function Y removes the promoter of root development directly: rootless. In the double mutant Y is already absent, so the extra X makes no difference: rootless. All three observations fit.

If X promoted Y and Y promoted roots, gain-of-function X would give more roots, not none. If Y suppressed roots, losing Y would give roots, which rules out both pathways in which Y is a suppressor. The double mutant also tells us X acts through Y, because it looks like the Y mutant.