Plant Physiology (Bot 328), Test #1 NAME: KEY Spring 1995 Please provide succinct answers in the space provided under each question. Unless otherwise noted in the margin the value of each question is 3 points. 1. (a) What is the third rule of Strong Inference (i.e., third step of inductive inference) described by John Platt? Ans.: Carry out the experiment so as to get a credible result. (b) Give an example of an instance in which an apparent failure to follow the third rule led to proposals that a major hypothesis in plant physiology might not be correct. Ans.: In 1994 pressure probe measurements of xylem pressure suggested that there was not a negative pressure on the fluid in the xylem, counter to the hypothesis that predicts) is the main force driving transpiration. In 1995 other, more rigorous methods showed the pressure probe measurements were incorrect. 2. Regarding studies on pollen tube growth, (a) Scientists radioactively labelled an RNase to investigate a major question on mechanisms of self-incompatibility. What was the question and what was the answer? Ans.: Q: Is the RNase encoded by the S1 gene taken up by the pollen tube? A: Yes. (b) What are TTS (transmitting-tissue specific) proteins and what dual role do they play in pollen tube growth? Ans.: These are proteins secreted by the transmitting tissue of the style of flowers, and they both promote the growth of the pollen tube and serve as a kind of chemoattractant to guide tube growth. (c) Scientists have noted that TTS proteins show different levels of glycosylation depending on their location in transmitting tissue. What is thought to be the functional significance of these differences? Ans.: When the TTS proteins bind to the pollen tube they are de-glycoylated, and the sugars released may play a role in promoting the growth of the pollen tube or in guiding it. 3. (a) How many sperm cells are typically present in a pollen tube as it nears the entrance to the megaspore mother cell, and what are the different functions of each of these sperm cells? Ans.: Two, one to fertilize the egg and one to fuse with two central cells to form a triploid endosperm nucleus. (b) Following fertilization, there is an initial cell division. What is predictable about the plane of this cell division (i.e., where the new wall will be laid down), and what are the different fates of the two cells produced by this division? Ans.: The plane of cell division is at right angle to axis of growth; one of the two cells produced develops into the body of the embryo and the other forms the suspensor that anchors the embryo in the endosperm. (c) Following the maturation of the embryo, it typically undergoes a dramatic physiological change that makes it more capable of surviving extremes of environmental conditions. What is the specific physiological change that occurs and what term is used to designate this phase of seed development? Ans.: The physiological change is dehydration, and this phase is called dormancy. 4. (a) What are expansins, what is their function, and what experimental evidence was obtained to infer what their function was? Ans: Expansins are wall proteins that appear to increase the extensibility of plant cell walls. The evidence for this function is that when expansins are added back to walls that have lost their extensibility by being boiled in water, their extensibility is partially restored. (b) Indicate two ways that have been used experimentally to inhibit expansin function, and indicate how results of these studies were used to argue for the physiological significance of expansins in plant growth physiology. Ans.: Boiling in water, treatment with heavy metals, treatment with neutral pH. All of these treatments also inhibit the extensibility of isolated plant cell walls, so these results suggest that expansins may be the wall components that are altered when these treatments inhibit wall extensibility. (c) What question is being addressed by the data in Figure 1 (page 5)? What probable answer to this question is provided by the data? Ans.: Q.: Do expansins promote growth by breaking bonds in (hydrolyzing) cellulose? A.: No. 5. (a) What experimental evidence favors the conclusion that wall peroxidases could be involved in the regulation of growth? Ans.: There is an inverse correlation between wall peroxidase activity and growth: the more activity the lower the growth rate. (b) Wall peroxidases have also been implicated in plant defense responses, and for this function, peroxidases are thought to modify another protein in the wall. What is this other wall protein, and how could peroxidase modification of this protein contribute to plant defense responses? Ans.: Wall extensin. Peroxidase could cross link wall extensins to each other by forming intermolecular iso-dityrosine bonds, and this cross-linking could "toughen" the wall and make it less easily penetrable by pathogens. (c) Antibodies to glucanases block hormone-induced growth, and these results have been interpreted to mean that glucanases play an important role in mediating hormone-induced growth. Propose an alternative hypothesis to explain why antibodies to glucanases could block growth, and describe an experiment using anti-peroxidase antibodies that has been done to test this alternate explanation. Ans.: Alternate hypothesis is that the antibody effect was non-specific because the binding of any antibody to any wall component could cause steric hindrances that would inhibit growth. Test experiment was to study the effect of anti-peroxidase antibody on growth. Since peroxidase activity is thought to inhibit growth, then antibodies to peroxidase should promote growth, but test shows that anti-peroxidase antibodies inhibit growth just like anti-glucanase antibodies. 6. (a) (i) Compare and contrast the structure and function of phytochelatins and metallotheioneins. (ii) What induces the production of these agents in plants? Ans.: (i) Metallothioneins are proteins; phytochelatins are peptides formed by linking units through the gamma carboxylic acid group of glutamic acid with the alpha amino group of a cystine. (ii) Both agents chelate heavy metals, and the production of both is stimulated by an increased presence of heavy metals in the responding cells. (b) Until recently metallotheioneins were thought to occur only in animals. What is the experimental evidence that a gene encoding a metallotheionein-like protein in plants actually codes for a protein that functions as a metallotheionein. Ans: The gene was used to transform a yeast mutant that could not make metallotheioneins and could not grow in the presence of heavy metals. This transformation with the plant gene complemented the genetic defect in the yeast and allowed it to grow in the presence of heavy metals. 7. (a) How do the reaction center chlorophylls in PSII and PSI differ from one another? How did this difference aid the original experiments that led to the conclusion that there were two different photosystems in photosynthesis? Ans: The reaction center chl in PSII absorbs at a lower wavelength (680 nm) than the reaction center chl in PSI (700 nm). The original experiments showed that far-red light (near 700 nm) was synergistic with red light (near 680 nm) in promoting the light reactions of photosynthesis. (b) Light harvesting complexes funnel light energy into the reaction center chlorophylls of PSII and PSI. If PSII is more active than PSI, so that electron flow from PSII to PSI is backed up, what mechanism corrects this problem? Ans.: When PSII is more active this results in a higher ratio of reduced plastoquinone to the oxidized form. This activates a protein kinase which phosphorylates proteins in Light Harvesting complex of PSII, inducing it to move to PS I, thus relatively increasing the light harvesting ability of PSI and restoring balance between the two photosystems. (c) Point out three distinctions between cyclic photophosphorylation and non-cyclic photophosphorylation. Ans.: (1) cyclic p. uses only PSI; non-cyclic uses both PSI & PSII; (2) cyclic p. does not produce NADPH directly, non-cyclic does; (3) cyclic p. does not result in the photolysis of water; non-cyclic does. 8. (a) What phenomenon accounts for results shown in Fig. 2? Why is this phenomenon beneficial to CAM plants? Ans.: In CAM plants stomates are open only at night, so their uptake of CO2 and their rate of water loss are both higher at night. This helps the CAM plants avoid water loss during the day, which is particularly advantageous in a desert environment, where many CAM plants grow. (b) How does the distribution of PEP carboxylase in C4 plants and in CAM plants differ? Ans.: In C4 plants PEP carboxylase is restricted to mesophyll cells, which do not have RUBISCO; in CAM plants it is in the same cells as RUBISCO, so some mechanism has to assure that the PEP carboxylase will be much more active than RUBISCO during the day. (c) The decarboxylation of malate in CAM plants is handled in part by a mitochondrial NAD-Malic enzyme (NAD-ME). In this regard what is the significance of the results shown in Fig. 3? Your answer should explain why NAD-ME has to be controlled differentially in the light and dark periods. Ans.: The decarboxylation of malate should not occur during the night when PEP carboxylase is active, otherwise this would result in a futile cycle: the malate produced by carbon fixation through PEPC would be decarboxylated as fast as it was produced. Figure 3 shows that the activity of NAD-ME is lower during the night, so by this mechanism the above noted futile cycle could be avoided. 9. (a) Indicate three lines of evidence that a proton pump is involved in phloem loading in plants. Ans.: (i) alkalinization of the external medium inhibits sugar uptake by leaf discs; (ii) the uptake of sugar by leaf discs is accompanied by an uptake of protons from the medium; (iii) proton pump distribution is highest in transfer cells responsible for taking up sucrose against a gradient. (b) Would you expect there to be a high concentration of a proton-pumping ATPase along the plasma membrane of sieve element cells in sugar beet? Explain your answer. Ans.: No, because the cells responsible for the active uptake of sucrose from the apoplasm (and thus the cells with the proton-pumping ATPase) are the companion and transfer cells, not the sieve tube cells themselves. 10. (a) What is the driving force for water movement in plants? Does the force alone account for sustaining the continuity of the water column throughout the plant? Explain. Ans.: The water potential gradient between the soil and the air, culminating in the evaporation of water from leaf cells into the surrounding air (which has the lowest water potential), is the driving force. This force by itself does not account for sustaining the water column. The cohesion of water molecules to each other and the adhesion of water molecules to the xylem cell walls also contribute. (b) What structure in a leaf controls the rate of transpiration out of the leaf? By what general mechanism would water stress affect whether the transpiration rate was faster or slower? Ans.: The leaf stomates control the rate of transpiration. At high water stress, the water potential of the guard cells (which regulate the relative pore size of the stomates) rises because salt is pumped out of them. The decreased water potential causes water to leave the guard cells, they become flaccid, and this shape change closes the pore of the stomates.