Eeb 162 Final

Front 1

Explain C4 photosynthesis using the following terms: HCO3-, PEP carboxylase, mesophyll, Kranz Anatomy, plasmodesmata, malate, separation of pathways in space, and water use efficiency (WUE).

Back 1

C4 photosynthesis: monocots, grasses, sedges, eudicot shrubs, high light and open environments (i.e. deserts); can keep stomata more closed than C3, no photorespiration.

- CO2 converted to HCO3-

- HCO3- fixed to 3-carbon PEP by enzyme PEP carboxylase in the cytoplasm of cell

- a 4-C compound (malate) is produced

- malate diffuses passively through plasmodesmata to bundle sheath cells (cell 2)

--> bundle sheath cells contain many chloroplasts that run Calvin Cycle

- malate decarboxylated down to 3-C pyruvate and CO2

--> pyruvate diffuses back to mesophyll cells

--> CO2 released to bundle sheath chloroplasts

-----> CO2 enters full C3 Calvin Cycle reaction in bundle sheaths

Kranz Anatomy: C4 plants have this K. Anatomy--enlarged bundle sheath, lots of chloroplasts

- no mesophyll cell is more than 2-3 cells away from the bundle sheath

- mesophyll cells are very close to bundle sheath cells

- many plasmodesmata connect the two

Separation pathways in space: some plants without K. Anatomy can operate C4 photosynthesis within single cells; separate PEP-carboxylase reactions from Calvin Cycle reactions within the cell

Water Use Efficiency: C4 plants can open stomata less for the same amount of CO2 fixation under high radiance (saving water)

Front 2

Why does C4 photosynthesis have a cost? Describe direct and indirect costs.

Back 2

Yes, C4 photosynthesis has a cost: regeneration of PEP consumes 2 ATP.

If O2 was low and photorespiration did not occur, C4 plants require more quanta of light per CO2 than C3 plants for the same CO2 fixation.

Front 3

Under what conditions would C4 plants have an advantage over C3 plants? Why?

Back 3

C4 plants operate efficiently in drier soil,high light, and open environments (deserts) than C3 plants.

Temperatures higher --> C3 plants increase their photorespiration because rubisco reacts more quickly with O2. (Also, O2 becomes more soluble at higher temperatures, making C4 plants more advantageous.)

C4 plants can lose less water, so keeping their leaf water potential higher during transpiration and and operate better on drier soil.

Front 4

Under what conditions would C3 plants have an advantage over C4 plants? Why?

Back 4

As CO2 levels increase, C3 species may possibly be favored and replace C4 grasses in the world's arid zones. The more CO2 rpesent, the less likely O2 will be fixed with RuBP (oxygenase reaction).

Front 5

What 3 possible environmental conditions might promote the evolution of the C4 photosynthetic pathway? Why?

Back 5

C4 plants tend to originate in arid regions.

1. heat

2. drought

3. salinity

... are important conditions that promote evolution of C4 photosynthetic pathway.

Front 6

Explain CAM photosynthesis using the following terms: stomata, rubisco, malate, PEP carboxylase, water use efficiency (WUE).

Back 6

CAM photosynthesis: found in cacti, euphorbias, bromeliads, orchids, and agaves

Stomata are closed during the day. CAM plants have succulent stems.

Similar to C4 in that PEP carboxylase fixes HCO3-...

- CO2 is released to normal C3 Calvin Cycle

- CAM allows increased CO2

--> fixes HCO3- with PEP carboxylase at night* when stomata are open

- malate is stored in vacuole

--> PEP is generated by breakdown of starch from chloroplast

During the day, malate is released from the vauole and breaks down to pyruvate and CO2 in the chloroplast light reactions and C3 Calvin Cycle run.

Water Use Efficiency: 10x higher than C3 plants because stomata are only open at night, and CAM plants have very water-resistant cuticles (waxy).

Some plants are facultative CAM, meaning they shift from C3 to CAM depending on environmental conditions:

- more heat, water or other stress --> plants express CAM genes

- normal conditions --> plants carry out normal C3 photosynthesis

CAM is very expensive due to additional metabolic costs and very slow photosynthetic rates.

- not competitive under high resource conditions with temperate climate.

Front 7

5 typical parameters of the photosynthetic light response curve?

Back 7

1. saturation irradiance

2. light-saturated assimilation rate (Amax)

3. dark respiration rate (y-intercept)

4. quantum yield (slope of left hand portion)

5. light compensation point (x-intercept) where photosynthesis = dark respiration --> so net photosynthesis = 0

Front 8

Difference in Sun vs. Shade plants for light response curve? (3)

Back 8

Sun plants...

1. higher saturation irradiance

2. higher dark respiration rates

3. higher light compensation points

Front 9

2 ways leaves are specialized for CO2 uptake?

Back 9

1. major part of leaf is airspace --> internal resistances reduced

2. large internal cell surface areas for CO2 uptake, keeping diffusion distance between stomata and chloroplast small

Front 10

Predict what a cross-section of a sun leaf would look like.

Back 10

Sun leaves are thick with more palisade mesophyll layers for direct light penetration.

Front 11

Predict what a cross-section of a shade leaf would look like.

Back 11

Shade leaves are thinner, with more spongy mesophyll to capture diffused light.

Front 12

What is photoinhibition?

Back 12

Occurs when reaction center PSII shuts down (inactive) due to excess light and plants experience stress. Can be dynamic/chronic.

Front 13

2 ways to reduce photoinhibition?

Back 13

1. higher nitrogen and rubisco concentration

2. large pool of xanthophyll cycle components (dissipate excess light energy)

Front 14

Rate at which sap travels through phloem?

Back 14

1 meter/hour

Front 15

What is the driving force of sap in phloem sieve tubes?

Back 15

bulk flow

Front 16

4 facts that support pressure-flow transport?

Back 16

1. sieve pores (connection between cells) are open

2. phloem transport in a single sieve tube occurs in one direction at a time

3. flow does not require energy

4. lack of energy requirement proven by fact that phloem transport still maintained in chilled leaf petioles (in which respiration and ATP production would be suppressed)

Front 17

Explain phloem structure and function, including the following terms: sieve elements, sieve cells, companion cells, girdling, translocation, sieve areas, P-proteins, and non-reducing sugars.

Back 17

- sieve elements make up the phloem and are living cells specialized for the translocation of fluids received from mesophyll --> companion cells --> sieve elements in minor veins of the leaf

- sieve elements consist of sieve cells which are found in gymnosperms* and contain sugars and organic molecules in sap

- companion cells are associated with sieve tube elements; involved in rapid exchange of solutes with dive tube elements via numerous plasmodesmata; also bring ATP to the sieve elements and play critical metabolic roles, including protein synthesis

- when bark is remove in a circle around a tree through girdling, sugar transport between roots and leaves ceases and tree dies

- sieve elements have sieve areas in their cell walls, where pores connect to other sieve elements, allowing transport between cells

- sieve tube elements in angiosperms* contain P-proteins which seal damaged sieve elements by plugging off the sieve plate pores when cut/punctured

- sugars in phloem are non-reducing (e.g. sucrose, raffinose) and are less likely to react with other chemicals in the plant during transport

Front 18

Which types of water transport, and which driving forces, are involved in the pressure-flow model of phloem transport?

Back 18

- bulk flow, driven by a pressure gradient, moves phloem sap

- pressure gradient established via loading/unloading of sugars into sieve tube cells, which leads to osmosis of water in/out of sieve tube cells, generating pressure

Front 19

Why do phloem loading and unloading require water to flow into the phloem from the xylem?

Back 19

Water osmosis in/out of sieve tubes establishes the pressure gradient needed to move the sucrose solution. The accumulation of sugars in the sieve elements from phloem loading generates a low/negative solute potential (psi s), causing a low water potential (psi w). Water diffuses into the sieve elements, increasing pressure potential (psi p), which pushes the sucrose solution downwards, which then diffuses into sink cells for plant storage/use.

Phloem unloading raises (psi s) and (psi w). Water flows out, causing (psi p) to fall.

Front 20

What are P-proteins and callose? When are each made specifically?

(NOT COMPLETE)

Back 20

- both contained in sieve tubes of ANGIOSPERMS (not gymnosperms!)

- P-proteins can be tubular, fibrillar, granular or crystalline in form; they seal damaged sieve elements by plugging off the sieve plate pores; they are under positive pressure, so when cut/punctured, the contents surge toward the cut end; during surging, the P-protein is trapped on the dive plate pores, helping to seal the sieve element

callose:

- over the long term, callose is synthesized in the sieve element and deposited in sieve tube plate pores = wound callose. Callose can be deposited and degraded in response to transient stresses like high temperature or mechanical stimulation.

- long term equivalents of P-proteins

- i.e. scar tissue

Front 21

What are differences between apoplastic loading and symplastic loading? Describe differences in pathway, and in sugar molecules involved.

Back 21

Phloem loading refers to the loading of sugars into companion cells and sieve elements. Occurs in 1 of 2 pathways:

1. apoplastic loading: occurs in plants with phloem parenchyma cells. Requires energy. H+-ATPase transports H+ into the apoplast, and a sucrose-H+ symporter transports sucrose from the apoplast into the sieve element-companion cell complex. No plasmodesmata involved connecting the dive elements and companion cells to surrounding cells.

2. symplastic loading: occurs in plants with intermediary cells. No energy required. Sucrose synthesized in mesophyll diffuses from the bundle sheath cells into the intermediary cells to sieve elements via small plasmodesmata. In the intermediary cells, raffinaose and stachyose synthesized. These sugars can diffuse to the sieve element through larger plasmodesmata. Involves abundant plasmodesmata connecting sieve elements and companion cells to surrounding cells.

Front 22

Which types of companion cells are involved in apoplectic loading?

Back 22

1. ordinary companion cells: relatively few plasmodesmatal connections to cells except its own sieve element; symplastically isolated from surrounding cells (involved in apoplectic loading).

2. transfer cells: similar to ordinary companion cells, but has finger-like walls ingrowths of cell walls facing away from the sieve tube element, increasing surface area for solute transfer from the apoplast (involved in apoplectic loading)

3. intermediary cells: numerous plasmodesmatal connections with bundle sheath cells, well-suited for taking up solutes via cytoplasmic connections (involved in symplastic loading)

Front 23

Which 4 factors affect the specific sugar transport pathways from source to sink leaves?

Back 23

1. proximity of organs: how close the source is to the sink

2. development: root and shoot apices are key sinks during vegetative growth, seeds and fruits during reproductive growth

3. vascular connections: leaves preferentially supply sinks to which they are directly connected (orthostichous)

4. modification of pathways: wounding/pruning can change the translocation pathways

Front 24

Why does apoplastic loading require energy?

Back 24

It involves using the H+-ATPase protein which requires ATP to move H+ into the apoplast to create a proton motive force (later used to symport sucrose molecules into the cell)

Front 25

Why do plants respire?

Back 25

Respiration releases energy stored in C compounds for cellular use and makes C precursors for biosynthesis, free energy is released and transiently stored as ATP, which can be readily utilized for the maintenance and development of the plant, including active transport.

Front 26

What is the overall equation of respiration?

Back 26

C12H22O11 + 12O2 --> 12CO2 + 11H2O

(oxidation of sucrose)

H2O reduced to O2 (60 ADP + Pi converted to 60 ATP + H2O)

Front 27

What are the glycolysis reactions?

Back 27

Glycolysis involves reactions carried out by soluble enzymes in the cytoplasm and in the plastid. It occurs in all living organisms and in plants. Substrate is sucrose, malate, and pyruvate. Sucrose is split into glucose, fructose, and UDP-glucose. Products are converted to glucose-6-phosphate and fructose-6-phosphate. Products are split into two 3-C triosphosphates: glyceraldehyde-3-phosphate and dihydroxyacetone phosphate. These steps require energy, and the next steps release energy, yielding a small amount of ATP and NADH. The triosphosphates are oxidized to produce two 3-C organic acid molecules (PEP), which can be converted to pyruvate yielding ATP or oxaloacetate, then malate. The pyruvate and malate can enter the mitochondrion to be used in the Krebs/Citric Acid Cycle.

Front 28

When does anaerobic respiration occurs? What are its end-products?

Back 28

Anaerobic respiration occurs when oxygen is unavailable (e.g. when roots are flooded). The fermentation pathway reduces pyruvate to convert NADH (from glycolysis) to NAD+ converting pyruvate to ethanol + CO2, or lactic acid. This allows glycolysis to continue in the absence of Krebs/Citric Acid Cycle and oxidative phosphorylation.

Fermentation (4 ATP per sucrose) vs. Krebs/Citric Acid Cycle (60 ATP)

Front 29

Where is the pentose phosphate pathway? Why can't it run in the chloroplast?

Back 29

The pentose phosphate pathway is located in the cytosol and plastids. The pathway is inhibited by the accumulation of fructose-6-phosphate and glyceraldehyde-3-phosphate (formed in the chloroplast by the Calvin Cycle), and therefore, cannot run in the chloroplast.

Front 30

Describe the Citric Acid Cycle, indicating what are the inputs and outputs. Use the following terms: mitochondrion, pyruvate dehydrogenase, succinate dehydrogenase.

Back 30

The Citric Acid Cycle occurs in the matrix of the mitochondrion to generate energy for plant use. Pyruvate from glycolysis is oxidized completely to CO2, generating major accounts of reducing power in the forms of NADH and FADH2.

The input is pyruvate, which is decarboxylated by pyruvate dehydrogenase to acetyl-CoA, releasing CO2 and NADH. The acetyl-CoA enters the cycle and additional oxidative decarboxylations occur, yielding NADH, FADH2, and ATP as the outputs. Reactions are catalyzed by enzymes in the matrix of the mitochondrion except for succinate dehydrogenase, which operates in the inner mitochondrial membrane.

Front 31

a. What is oxidative phosphorylation?

b. Where does it occur?

c. what are its inputs and outputs?

d. What is the role of the chemiosmotic gradient in oxidative phosphorylation? Where does it exist?

Back 31

a. a metabolic pathway consisting of electron transport proteins bound to the inner of the 2 mitochondrial membranes that converts NADH and FADH2 generated in glycolysis and the Citric Acid/Krebs Cycle to ATP via e- transfer along an e- transport chain. This transfer releases a large amount of free energy, which is used by ATP synthase to generate ATP.

b. inner mitochondrial membrane

c. inputs: NADH, FADH2, ADP, Pi, O2

outputs: 60 ATP per sucrose, NAD+, FAD, H2O

d. The chemiosmotic gradient of H+ (in the inter-membrane space vs. the matrix) drives the generation of ATP from ADP + Pi by the F0-F1-ATP synthase. H+ is being pumped into the inter-membrane space. The gradient also plays a role in the movement of ATP and organic acids of the Citric Acid/Krebs Cycle, out of the mitochondria.

Front 32

How does plant respiration change with

a. flooding?

b. increased temperature?

Back 32

a. decreases (unless the plant has special adaptations)

b. increases substantially

Front 33

How do plants adapt to tolerate flooding? Name 2 ways.

Back 33

1. aerenchyma: ducts in the shoot conducting air to the roots

2. pneumatophores: root outgrowths that protrude out of the water

Front 34

What is the driving force for solute transport?

Back 34

chemical potential difference x membrane permeability

Front 35

What does the Nernst equation determine? What does the Goldman equation determine?

Back 35

Nernst: gives the concentration gradient expected across a membrane at equilibrium if the voltage across the membrane is known

Goldman: modified version of the Nernst equation, which predicts the diffusion potential from Na+, K+, and Cl- (the ions with the highest cellular concentrations)

Front 36

What are the 3 kinds of solute transporters? How do they differ?

Back 36

1. channels: allow passive transport. Solutes diffuse through extremely rapidly. May have structures called gates that open/close the pore in response to external signals (e.g. light, voltage channels, hormone binding). Can be specialized to allow only inward/outward ion movement.

2. carriers: do not have pores that extend completely across the membrane. The substance binds to the carrier protein, and a conformational change in the protein deposits the substance on the other side of the membrane. Typically ions move through a carrier 106x slower than a channel. Carrier-mediated transport may be passive or active.

3. pumps: move ions against the eletrochemical gradient. Requires energy.

Front 37

How do non-leguminous plants take up nitrogen from soil to the leaf mesophyll and assimilate it into amino acids? Name key enzymes involved.

Back 37

Plant roots absorb NO3- (nitrate) from the soil solution via high affinity nitrate-proton co-transporters. The nitrate is transported via xylem to leaf mesophyll cells, and reduced to nitrate (NO2-) in the cytosol by nitrate reductase. Nitrate is transported to chloroplasts in the leaf cells, where it is converted to ammonium (NH4+) by nitrate reductase again. The NH4+ is assimilated into glutamine by glutamine synthetase, and glutamate is produced by glutamate synthetase. Glutamate is transported to the cytosol, and amino acid biosynthesis proceeds.

Assimilation of nitrogen requires huge energy expenditure. Plants need to use 12 ATP to convert NO3- into glutamine.

Front 38

Why is nodule formation by nitrogen fixing bacteria in legumes a case of "everyday endosymbiosis"?

Back 38

Biological nitrogen fixation by bacteria accounts for most of the conversion of atmospheric N2 into ammonium. The most common symbiosis occurs between the Legume family (forms nodules) with soil bacteria.

Nodule formation involves special development of root hairs to allow entry of bacteria into the infection thread, formed by plant cell vesicles, and the fusion of the infection thread with cell membrane.

Front 39

Define with equations and give typical units for the following:

a. relative growth rate

b. specific leaf area

c. leaf area ratio

d. root mass fraction

e. unit leaf rate

Back 39

a. relative growth rate

- RGR = ln (mass 2/mass 1) / (t2-t1) = [ln (mass 2) - ln (mass1)] / (t2-t1)

- RGR (g g-1 time-1) = leaf area ratio (m^2 g-1) x unit leaf rate (g m-2 t-1)

- indicates growth relative to mass

- tends to decline as plants increase in size

- RGR is determined both by morphology and by physiology

- ULR, LAR, SLA all contribute to a high RGR

b. specific leaf area

- SLA (m2 g-1) = leaf area / leaf mass

- SLA (m2 g-1) = [thickness (m) x density (g m-3)]-1

- SLA higher for leaves that are thinner or less dense

c. leaf area ratio

- LAR (m2 g-1) = leaf area / plant mass

- LAR (m2 g-1) = specific leaf area (m2 g-1) x leaf mass fraction (g g-1)

- index of morphology

d. root mass fraction

- root mass fraction (g/g) = root mass / plant mass

- RMF can be a factor in the plant's ability to take up water and nutrients

e. unit leaf area

- ULR (g m-2 t-1) = mass accumulated per leaf area per time

- unit leaf rate is like photosynthetic rate per leaf area, but integrated over growth time, and is also affected (diminished) by respiration of the whole plant, and by loss of tissues (e.g. leaf fall)

- longer leaf lifespan = higher ULR because of lower loss of mass

- index of physiology

Front 40

A grass has 4x the relative growth rate of a tree. But it only has double the photosynthetic rate. Explain why the grass has such a higher RGR.

Back 40

The grass grows faster relative to its dry mass, thus a higher RGR.

High (100x) specific leaf area (SLA) and leaf mass fraction (LMF) contribute to a high LAR and thus RGR for grass.

Front 41

Why might relative growth rate decline during early plant ontogeny? Give 3 possible reasons, related to classical growth analysis components.

Back 41

1. self shading increases as the plant grows

- ULR decreases for larger plants due to self shading since few plants get ideal sunlight

2. LMF decreases as the plant grows

- as plants age, they begin to put more investment in stem than leaves, decreasing LMF

3. SLA decreases as leaves develop

- initially, the plant will have high SLA (broad, thin leaves) and this will lead to high amounts of photosynthesis and thus accumulation of products. This build-up causes a thickening of leaves reducing SLA and ultimately RGR.

Front 42

Define:

a. plasticity

b. adaptation

Back 42

plasticity: changes in plant form and physiology when plants are grown in different conditions. Plastic changes arise from interaction between environment and genetics.

adaptation: evolution of differences in plant form and physiology. Adaptive changes are totally genetic. Adaptive differences between plants would be observed if plants are grown in a common garden (where there would be no plasticity).

Front 43

How do plants tend to differ in form:

a. When grown in shade vs. sun?

b. When adapted to shade vs. sun?

c. When grown in high vs. low N supply?

Back 43

a. shade grown: increased SLA, increased RMF

b. shade adapted: decreased SLA, increased RMF

c. high N supply, increased SLA, decreases RMF

Front 44

Explain how shade-adapted plants tend to differ from sun-adapted plants using the terms: specific leaf area, dark respiration, root mass fraction, leaf lifespan.

Back 44

Plants adapted to shade typically have a lower SLA and higher RMF.

Shade-adapted plants have a lower specific leaf area and a higher root mass fraction because they are adapted to having a low demand for energy and a high tolerance of other stresses. Lower SLA corresponds to a longer leaf lifespan and thicker leaves. They are adapted to dark respiration and other stresses. Higher RMF allows shade plants to better compete with the large root systems of the overstorey trees, and to take occasional droughts that occur during long lifetimes in the shade.

Front 45

Define:

a. axial patterning

b. radial patterning

c. stem cells

d. indeterminate growth

e. shoot apical meristem

f. leaf primordia

g. root cap

Back 45

a. axial patterning: arraying of tissues and organs along a linear or polarized axis. (e.g. stem and root; cotyledons at top)

b. radial patterning: tissues arranged in a pattern extending from outside to center (e.g. concentric rings of tissues in roots)

c. stem cells: undifferendiated cells

d. indeterminate growth: showing no predetermined limit. Shoot apical meristem is indeterminate, producing phytomeres.

e. shoot apical meristem: generates the stem, leaves, and lateral buds

f. leaf primordia: young leaves formed by shoot apical meristem at the shoot tip

g. root cap: the meristematic zone that regenerates the primary root

Front 46

Describe one Arabidopsis developmental mutant. What did it reveal about plant development?

Back 46

Arabidopsis mutants indicate the gene expression that underlies development:

1. GNOM gene: seedlings with GNOM mutant lack roots and cotyledons. The GNOM gene is required for axial patterning.

2. Monopteros gene: seedlings with Monopteros (MP) mutant lack both a hypocotyl and a root, though they do form an apical region. However, these seedlings will form adventitious roots. Thus, MP gene is required for primary root formation, but no adventitious root formation.

Front 47

What is meant by the phrase: "Cell fate is determined by the position, not by clonal history?"

Back 47

In English ivy, a mutation can affect chloroplast differentiation. In variegated cultivars, the L2 layer contains the albino mutation, and L1 and L3 layers contain wild-type copy of the gene. The L1 layer forms the epidermis, which is colorless. Mesophyll tissue is typically derived from the L2 layer, so the leaves should be white.

But most leaves are variegated. The green tissue was derived from L1 or L3 layers, and the colorless from the L2 layers. So the green cells derive from occasionally differentiation of L1 and L3 layers cells into green mesophyll cells. Thus, these cells form mesophyll cells based on their location, not their clonal origin.

Front 48

a. What is phytochrome?

b. What forms can it have?

c. What wavelengths of light do they preferentially absorb?

d. Which form is physiologically active?

Back 48

a. Phytochrome is a protein pigment that absorbs red and far-red light most strongly, and also blue light.

b. Phytochrome can interconvert between forms. In dark=grown or etiolated plants, phytochrome is present in a red light-absorbing form, Pfr. This switching of forms is known as photoreversibility.

c. Red light (650-680 nm) causes antagonistic responses

d. Pfr is the physiologically active form. Action spectra of processes best correlates with Pfr.

Front 49

a. Why do plants respond to red:far-red ratio as seeds?

b. Why do they respond as growing juvenile plants?

Back 49

a. Plants respond to red:far-red ratio as seeds because this signals that they have landed into a clearing which is ideal for growing.

b. etiolation: when plants grow thin and pale due to lack of light. Plants allocate energy to make thin long stems to escape shade. Plants under canopies have lower ratios because the shady leaves above it have already absorbed the red light so they choose to grow to obtain a higher ratio (more red light).

- they respond as growing juvenile plants in order to induce faster cell elongation to grow out of shade environments into open areas for competitive advantage

Front 50

Name the 3 types of genes responsible for regulating floral organ identity. What are their roles?

Back 50

1. meristem identity genes: encode transcription factors that are necessary for the initial induction of organ identity. Meristem identity genes must be active for the immature primordial at the shoot or inflorescence apical meristem to become a floral meristem.

2. floral organ identity genes: encode transcription factors that directly control floral identity. These are homeotic genes that act as major developmental switches that activate the genetic program of a particular structure.

3. cadastral genes: encode transcription factors that act as spatial regulators of the floral organ identity genes by setting boundaries for their expression.

Front 51

A mutant has a flower with only stamens and carpels. What type of activity is lost? how does the ABC Model predict this?

Back 51

Loss of type A activity results in the formation of carpels instead of sepals in the first whorl, and of stamens instead of petals in the second whorl.

ABC Model: each whorl is determined by a unique combination of 3 organ identity genes activities

- Type A activity alone specifies sepals

- Type A + Type B specifies petals

- Type B + Type C specifies stamens

- Type C activity alone specifies carpels

- Type A activity controls organ identity in the first and second whorls

- Type B activity controls organ identity in the second and third whorls

- Type C activity controls organ identity in the third and fourth whorls

Front 52

What happens when A, B, and C activity are all lost?

Back 52

Loss of A, B, and C activity results in the formation of a pseudoflower, in which all floral organs are replaced with green leaf-like structures. This shows that flowers are indeed modified leaves, as predicted in the 18th Century by Goethe.

Front 53

What is photoperiodism? Why do plants show photoperiodic flowering responses?

(INCOMPLETE?)

Back 53

Photoperiodism is the ability of an organism to detect day length, to allow an event to occur at a particular time of year. At the equator, day and night lengths are equal and constant throughout the year. As one moves toward the poles, the days become longer in summer and shorter in winter. Plants have evolved to sense these seasonal changes in day length according to their latitude.

Front 54

What experiment can you carry out to show that plants monitor day length by measuring the length of the night?

Back 54

Plants monitor day length by measuring the length of the night. Giving a short flash of light (i.e. a "night break") during a long night cancels the effect of the long night, ad disrupts flowering in SDPs, and triggers flowering in LDPs.

Phytochrome is the primary photoreceptor in periodism. If a night break is given as red light, it has the effect of disrupting SDP flowering, and triggering LDP flowering; this can be reversed by far-red light.

Front 55

What is "florigen"?

Back 55

The evidence in support of florigen comes from grating experiments showing that noninduced receptor plants were stimulated to flower by being joined to a leaf or shoot from a photoperiodically induced donor plant.

Recent work implicatees macromolecules, and especially specific mRNAs for given genes, which would travel in the phloem and into the shoot apical meristem via plasmodesmata.

Front 56

What are the differences between PGRs and animal hormones?

Back 56

- few PGRs relative to animal hormones

- PGRs can be made in most tissues

- PGR can act in the same tissue where it is produced

- each hormone elicits a variety of responses, and works together with PGRs. The same PGR can elicit different responses in different tissues, or even in the same tissue at different times of development.

- PGR may drive positive responses over a narrow range of concentrations and drive negative resonses at high concentrations.

PGR can stimulate certain physiological responses and inhibit others.

- PGRs are typically neutral or negatively charged, small particles.

Front 57

How can you prove that a given substance causes a given response? Describe experiments using a PESIGS scheme (6).

Back 57

Traditional approach was to apply a substance to a plant or excised part, and test for response. it was assumed that the substance was a hormone involved response.

1. parallelism: a change in the chemical must parallel the response

2. excision: removal of the chemical or its source (e.g. producing tissue) removes the response

3. substitution: addition of the chemical should substitute for the stimulus in eliciting the response

4. isolation: the chemical response relationship should be maintained in experimental systems isolated from the organism

5. generality: the same chemical response system should be found in several species

6. specificity: one chemical should elicit one response, which is only elicited by that chemical (criterion not typically applied)

Front 58

What is polarized auxin transport? Why does it require energy? Why is it essential for the plant?

Back 58

polarized auxin transport: the axial polarity of roots and shoots depends on the polarity of auxin transport. Polarized auxin transport is the principal cause of an auxin gradient extending from the shoot tip to the root tip.

Polarized auxin transport proceeds in a cell-to-cell fashion rather than via the symplast. Auxin exists the cell through the plasma membrane, diffuses across the cell wall, and enters the next cell through its plasma membrane. This process requires metabolic energy. The velocity of polarized auxin transport is faster than diffusion but slower than phloem translocation.

Polarized auxin transport is essential to the development of the basic shoot-root polarity. Treatment of embryos with auxin transport inhibitors results in severe developmental abnormalities and loss of polar growth at root and shoot apices.

Front 59

How does auxin stimulate cell growth?

Back 59

Auxin stimulates cell growth by increasing the extensibility of the cell wall. The acid growth hypothesis holds that auxin stimulates expression of an H+-ATPase, which increases the wall extensibility by stimulating expansins, proteins in the cell walls that loosen cell walls under acidic pH by weaking the hydrogen bonds between polysaccharides in the wall.

Front 60

What are the main functional roles of auxin? (5)

Back 60

1. transport

2. cell elongation

3. phototropism: growth towards light

4. gravitropism

5. developmental effects

Front 61

What are the main functional roles of gibberellins? (6)

Back 61

1. stem growth and root growth

2. transition from juvenile to adult phase

3. influence on floral initiation and sex determination

4. pollen development and tube growth

5. promotion of seed set and parthemocarpy

6. seed development and germination

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What are the main functional roles of cytokinins? (6)

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1. regulation of cell division and expansion

2. modify apical dominance and promote lateral

3. induce bud formation in moss

4. delay leaf senescence

5. promote movement of nutrients into leaves

6. promote chloroplast development

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What are the main functional roles of ethylene? (9)

Back 63

1. promotes fruit ripening

2. leaf epinasty

3. the "triple response" (reduced stem elongation, increased lateral growth (swelling) and abnormal horizontal growth

5. hook formation in etiolated seedlings

6. breaking seed and bud dormancy

7. induces formation of roots and root hairs

8. induces flowering in the pineapple family

9. enhances the rate of leaf senescence and promotes abscission

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What are the main functional roles of abscisic acid? (5)

Back 64

1. regulation of seed maturation, promotion of desiccation tolerance

2. promotes root growth and inhibits shoot growth during water stress

3. induces stomatal closure during water stress

4. promotes leaf senescence

5. promotes bud dormancy

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Name 3 additional PGRs.

Back 65

1. brassinosteroids: steroid PGR, recently discerned in plants; induce cell division and elongation

2. jasmonate: gaseous PGR, stimulated by damage or herbivory, and may communicate to other plants and induce defense responses

3. salicylic acid: PGR involved in defense responses, which stimulates thermogenesis (heating) and high respiration rates. Linked with resistance to pathogens.