Final Exam

Question: Please describe how RGD works to inhibit integrin function.

 Ans.: The RGD peptide out-competes the normal RGD binding site for integrin. The normal site is a large wall protein that allows the integrin to be firmly anchored and to develop tension and compression forces when the gravity vector changes. But when integrin is bound to the RGD peptide it is not firmly anchored to the ECM, and it cannot develop tension and compression forces when the orientation of the cell changes.

 

Question: Please clarify the significance of the gravity-directed calcium current across Ceratopteris spores.

Ans.: There is a calcium current across the Ceratopteris spore that moves in the direction opposite of the gravity vector, and this current appears to help direct the rhizoid to grow out at the site of Ca+2 entrance (i.e., at the botoom of the cell). When the spore cell is flipped upside down,  the Ca+2 current is also flipped upside down, so that after the reorientation of the cell the Ca+2 current still moves in the direction opposite to gravity (i.e., it enters the bottom and exits out the top),  and the rhizoid still grows out on the bottom of the spore cell. The reorientation of the calcium current is very fast (less than 45 sec), which means that the activities of the pumps and channels that produce the current must also change very quickly. The most likely way this change could occur so fast would be if the pumps and channels would be post-translationally modified, probably by phosphorylation (mediated by protein kinases) and/or de-phosphorylation (mediated by protein phosphatases). There is no direct evidence yet that this actually happens, nor is there any way to predict a priori whether, if phosphorylation indeed is what changes the activity of the pumps and channels, the changes would be phosphorylation or de-phosphorylation or both.

 

Question: In class you drew a picture with a root section in between agar containing 3H labeled auxin , and auxin flow downwards. You show that flipping the root reverses the auxin flow direction, but wouldn't the flow still go downwards (because of aux1 and pin3 relocalization)?

Ans.:  The experiment you refer to used de-capped root sections. The PIN3 relocalization appears to be restricted mainly to the columella cells in the root cap, and appears to direct mainly lateral movement of auxin to the root flanks. In the more long-distance flow downward toward the cap in protophloem cells, it seems as though PIN1 may play a bigger role, and there is no evidence yet that this PIN would change its localization upon gravity stimulation.  So polar auxin transport toward the tip in de-capped root sections would not be likely to change readily if the root were turned upside down.

 

Quest.: Are PIN proteins always destroyed frequently or only during orientation changes with respect to gravity? Isn't it wasteful to constantly turnover PIN proteins?

Answer: You raise a good point, and I do not know the answer. Possibly in the absence of an orientation change the turnover rate is much slower.....

 

Quest.: If PIN3 is the transporter, there should be asymmetry in AUX1 when the root is turned on its side, there are two strong inference possibilities: One is, transporters need to be asymmetrically distributed or activated. What is a second alternative?

Ans.: I probably said that AUX1 might be asymmetrically distributed or activated, or that it could be evenly distributed, and the differential distribution of auxin after gravitstimulation could be due entirely to the differential activity of one of the PIN proteins (e.g., PIN3). Current data would seem to favor the latter hypothesis.

 

Quest.:   I do not understand what columella cells have to do with gravity response, when talking about PIN and also the function of Brefelden?

Ans.: The columella cells are the cells in the root cap that are enriched in amyloplasts and they appear to be the root cells most important for gravity sensing. Auxin transport must pass through these cells, so, since PIN controls the exit of auxin from cells, PIN function in the columella cells is very important. Brefeldin is a drug that blocks the process of secretion through the Golgi, and it blocks the redistribution of PIN in root cap cells. This means that the secretory system is critical for PIN distribution in the root cap.

 

Quest.: What is the significance that secretory activity is required for redistribution of PIN3?

Ans.:  This observation means that the asymmetry of PIN3 distribution before and after gravistimulation is maintained by differential secretion, and it renders less likely the hypothesis that the asymmetry of auxin distribution is due to differential activation of stable transporters in the membrane.

 

Question.:  When you change orientation of the root from the vertical to the horizontal position, this causes a slight bias of auxin movement from the left to the right side. This was observed when GFP gene was attached to  DR5 promotor that is only activated by high [auxin]. Since auxin now moves more to the right side of the root instead of moving to the base of the root (in the vertical position), the binding of auxin to the auxin receptor on the right side of the root will cause a signal transduction pathway that leads to the polyubiquitination of repressors of auxin-regulated genes and ALSO (?) polyubiquinates PIN proteins that are not on the right side of the root. This allows for a redistribution of the PIN proteins in response to the movement of auxin towards the right side of the root when the root is turned to a horizontal orientation. Am I thinking this whole process correctly, please correct any concepts that I am confused on?? If the specific movement of auxin enhances the redistribution of the PIN proteins in the root, then when does the redirection of the Golgi vesicles to the right side of the root occur?

Answer.: In general your understanding is correct. The only change I would make would be to say that when the root is moved to the horizontal position, the PIN proteins do not move to the right side, they move to what is now the new BOTTOM of the cells. Thus after the horizontal positioning of the root, the PIN proteins accumulate along the new BOTTOM of the cells, and this allows auxin to accumulate along the bottom side of the root, where it inhibits the extension growth of the cells on that side. This makes the root bend downward. In order for the PIN proteins to be distributed along the bottom of root cells in horizontally positioned roots, Golgi vesicles have to be directed to that side. So the redirection of the Golgi vesicles is what allows the PIN protein distribution to change.

 

Quest.: What was the main purpose of the experiment that involved a GFP hooked to a promoter called DR 5? In my notes I don't the outcome of the experiment.

Ans.: GFP is a reporter gene, and DR5 is a promoter that responds to auxin. So in plants transformed with the  DR5-GFP construct one can follow where auxin is by visualizing the distribution of GFP. The recent paper reporting this experiment was important in that it showed that gravitropic root curvature correlates with asymmetric auxin flux through the Lateral cells of the root.

 

Quest.: How do transgenics help get the dsDNA get into plant cells for RNAi?

Ans.: The transgene encodes an mRNA that includes sequences for both the sense and antisense strands of a given mRNA, so that this mRNA ends up being expressed as a dsRNA

 

Question: Regarding Figures 3 C. and Figure 4 C of the paper on Molecular dissection of the GA/ABA pathways, I thought that ABA repressed GA in the signal transduction pathway....If this is correct, then why is the normalized HvA1-GUS and HvA22-GUS activity so high when ABA is expressed? What is HvA1-GUS and HvA22-GUS?

Answer: ABA turns on the promoters for the HvA1 and HvA22 genes. These promoters are placed in front of the reporter GUS gene in the HvA1-GUS and HvA22-GUS constructs. So you would expect that in the presence of ABA, the expression of HvA1-GUS and HvA22-GUS would be high. You are correct that ABA represses the ability of GA to turn on gene expression, and it does so by blocking the ability of GA to turn on GAMyb.

 

Question: Ethylene suppresses ABA, which suppresses GA, which allows growth...is this right?

Answer: Yes, this signaling sequence appears to be the best description of how flooding induces rapid growth in deepwater rice.

 

Question:  In the experiment on the HVA 1 and HVA 22, it showed that RNA i for GAMyb and the presence of ABA gave a high expression of HvA1-gus, but what does that prove or mean?  Or in other words what happens when high levels of HvA 1 are expressed or what does HvA1 do?

Ans.: HVA1 and HVA22 are promoters that are activated by ABA. When they are hooked up with GUS, the GUS expression means that these promoters were turned on. The results with GAMyb(RNAi) show that suppressing the expression of the mRNA for GaMyb using RNAi has no effect on the ability of ABA to turn on an ABA-regulated gene. Thus GAMyb, which in Fig. 3B was shown to be critical for GA to turn on Amy-GUS, is not critical for turning on ABA-regulated genes.

 

Question: How is slender involved in switching aleurone cells to their secretory mode? Is the inhibition of the slender gene expression by GA the switch for the cells to change their morphology to start producing vesicles?

Answer:     Yes, and the new, GA-induced cell morphology is referred to as the vacuolated form.

 

Question: Please explain to me the different techniques used to knock out specific genes in Arabidopsis versus barley and the significance of this? I remember you said that in barley, RNAi can be used but then I got lost from there.

Answer:  A principle method used in Arabidopsis is by random insertion of T-DNA sequences, followed by an analysis to determine which gene was disrupted by the insertion.  There are now national repositories where scientists can acquire mutants with T-DNA insertions in any one of the vast majority of genes in Arabidopsis. This method is feasible in small genomes like Arabidopsis, but is problematic in large genomes like that of barley (ca. 10 times bigger than that of Arabidopsis).  So in barley it is more feasible to do targeted gene suppression by RNAi.

 

Question:   Regarding wounding reponses, I have in my notes that methyl jasmonate activates JA receptor that activates proteinase inhibitors.  I also have in the absence of proteinase, insects are not able to benefit from the plant's nutrition.  I think i may have wrote that wrong because I think it would be that in the presence of proteinase that happens?

Ans.: No, there is no misprint. It is true that in the absence of protease, insects are not able to benefit from the plant's nutrition, because they cannot digest the protein. Thus the induction of protease inhibitors (which decrease protease activity) by injury should be (and has been proven to be) a good defense mechanism.

 

Question.:  What is the compound that attracts parasitoid wasps, is it volicitin or a volicitin- induced plant chemical? Also, does volicitin repels moths from egg laying on those leaves...or does a plant chemical repel the moths?

Ans.: The compounds that attract parasitoid wasps are volatile compounds whose production is induced by volicitin.  These same volicitin-induced compounds also repel moths.

 

Quest.: You drew a picture describing how to measure xATP in plants without cuticle. Would the drop of water added dilute the actual xATP concentration?

Ans.: Yes, the drop of water would significantly dilute the actual xATP concentration. So the experiment with overexpressing MDR only showed that the relative level of xATP goes up as a function of MDR overexpression.  Absolute levels are technically very difficult to measure, although new methods are being designed.

 

Question: How do you know membrane bound ATP pumps don’t influence the apyrase-MDR activity?

Ans.: In class we described evidence that: (1) the MDR transporter transports out ATP as well as toxins, and (2) these two transport processes may be linked by a symport mechanism. As far as I know there are no publications to date that describe transporters that are specific for transporting only ATP out of cells.

 

 

 

Exam 3

1) Question:   I have a question about Figure 3 in the  the handout titled "Molecular Dissection of the Gibberellin/Abscisic Acid Signaling pathways by Transiently expressed RNA Interference in Barley Aleurone Cells."  In part B of Fig 3, there was the condition where GA3 is not present and Ubi1-GAMyb(RNAi) is not present either and the results show that there's barely any (~ 0%) expression of GUS. Does this result indicate that since GA3 is absent, there can't be activation of the transcription factors needed to turn on the genes encoding amylase?

Answer: Yes

2) Question: If amylase isn't present, does this mean that the promotor Amy-GUS can't be turned on so this results in no expression of the GUS gene that was attached to the Amy-GUS promotor?

Answer:  Amylase does not turn on the Amy-Gus promoter; GA does this. Amylase is the enzyme encoded by one of the genes turned on by GA.  Amy-GUS is the gene construct turned on by GA.

3) Question:  I also wanted to make sure, is Ubi1-GAMyb (RNAi) the RNAi for the transcription factors?

Answer: GAMyb is the transcription factor that is needed for GA-regulated genes to be turned on. The Ubi1-GAMyb (RNAi) construct is the one that constitutively produces the dsRNA for GAMyb, and plants expressing it are suppressed in their level of the GAMyb mRNA, and thus suppressed in their level of the GAMyb transcription factor.

 

4) Question:   In class you said that if the slender gene is knocked out then you won't need GA to activate the expression of the transcription factor gene because GA was only needed to induce the expression of the TF since slender was present to inhibit it. Does this mean that in normal conditions, GA's activity to induce TF expression can override Slender's inhibitory activity on the TF?

Answer: GA induces the production of GAMyb by suppressing the expression of slender. If slender is suppressed by RNAi, then no GA is needed for GAMyb (and the genes it turns on) to be expressed.

 

5) Question: Is the Amy-GUS promoter directly activated by GA or does it need GA present to activate the TF that will activate the amylase gene and thus activate the Amy-GUS promoter?

Answer: For the amylase promoter to be turned on by GA, GA first needs to turn on the transcription factor that will activate the amylase promoter and thus turn on the Amy-GUS gene.

 

Exam 2

1. In the article SE and CC traffic control centers of phloem by oparka and turgeon, Fig 4 shows polymer trapping; it also says that raffinose and stachyose are "too samll to diffuse back to the mesophyll" this doesn't make sense, does it mean, "too big to diffuse back into the mesophyll"?

Ans.: I think this is a misprint. Stachyose is more than twice as big as sucrose.

2.In the Juergensen et all paper about the nematode induced syncitium, table 1 shows that Truernit el all 1996 found + and- Gus activity in the stem and leaf. how can this be? does the +/- mean there was localized activity?

Ans.: +/- simply means that the data were hard to interpret, so that the GUS signal could have been positive, but it was too low to be confident.

3. What is an apical hook? it is described in page 6 of the handout titled  "different photoreceptor pigments in plants are activated by different wavelenghts of solar radiation"as in, "5 minutes a day of dim red light  prevents maintenance of apical hook?

Ans.:  We mentioned earlier in class that when dicots are growing in darkness they take on a streamlined shape in which the hypocoty is bent in a hook shape and the cotyledons are folded in close to the stem. When these etiolated seedlings are exposed to light, the hook straightens out and the cotyledons open and expand.


4. What happens at the boundary between different directional flows (say up and down) in cytoplasmic streaming?

Ans. There is no boundary. The cytoplasm flows in a continuous movement around the periphery of the cell, which means it flows up on one side and down the other.


5. How many identical bases must a gene have relative to the small RNA fragments created by dicer to be destroyed?

Ans.: I know the identity must be more than 90% but can be less than 100%.

6. How is antisense and RNAi different? Is it that antisense introduces ssRNA
and RNAi introduces dsRNA to a cell?

Ans.: Antisense mRNA is expressed initially as a single stranded RNA, but then it presumably base-pairs with the sense mRNA, forming a dsRNA. After that the RNA destruction is thought to be essentially the same as what happens in RNAi.


 7.. Which cells in a plant express phy? cry?

Ans.: Phy and cry tend to be expressed most strongly in cells near the meristem that are not fully developed, and in other cells that are still in a growth phase.


8. What is the chromophore?

Ans.: The chromophore is the pigment structure that is bound to a pigment-protein holoprotein. The chromophore for cry and phototropin is a flavin; the chromophore for phy is an open chain tetrapyrrole.


9.  Why can't you label RDS to determine where and what the integrins it binds are?

Ans.: In principle this sounds like a good experiment.  I will check to see if it has been done in plants, and, if so, what the results were.


10. In figure 5 of Staves et all Gravity sensing in roots, don't the amyloplasts still lead to some gravity response at external medium density of 1016 kg/m3?

Ans.: Yes, the gravity response is depressed, but not zero.


11. Can you please tell us the relationship between water potential and solute concentration (i seem to have written down contradictory things from lecture).

Ans.: The water potential becomes more negative as the solute concentration increases.


12. How does low auxin concentration favor growth when high auxin concentration inhibits growth? (is this over some threshhold?) do the two hemispheres of a root change in auxin concentration during differential growth response (gravitropism)?

Ans.:  Part of the explanation of the bell-shaped dose-response curve for auxin effects on growth is that high auxin concentrations induce the production of ethylene, a growth inhibitor. Yes, the two hemisphers of a root change in auxin concentration during
differential growth response.


13. What is the difference between water potential and osmotic potential? Are they the same?

Ans.: They are not the same, but osmotic potential is a component of water potential.


14. Cavitation?

Ans.: This is an event in xylem that happens when tension overcomes the cohesive nature of water resulting in the breaking of the water column.

 15. I have a question about Figure 8 on the first page of the March 20handout ("CIP3, a New COP1 Target, Is a Nucleus-Localized Positive Regulator...") As I understand it, the graph shows hypocotyl length in Arabidopsis for wild type and several mutants under different light conditions and in the presence or absence of DEX (which controls whether or not the CIP4 gene is expressed).  Now, if RED light converts phytochrome to Pfr and thereby inhibits elongated hypocotyl growth and FAR RED light causes hypocotyl elongation (as under the forest canopy, when a plant senses far red radiation and then has a period of stem elongation to try to reach the sunlight), then why is it that in this graph, the hypocotyls of seedlings grown in red light are longer than the hypocotyls of seedlings exposed to far red light?

I would have expected that seedlings grown under far red light would have longer hypocotyls than seedlings grown under red light (I would have expected the "far red" bars to be about the same as the "dark"
bars).  Am I just getting red and far red effects mixed up, or is there a bigger concept that I am missing here?

Answer: In class we did not have time to provide the explanation, and this won't be on the test, but here is the answer. In darkness, 100% of the phy is in the Pr form. Pfr absorbs far-red light very efficiently, but, as it turns out, Pr also absorbs some far-red light, albeit very inefficiently. As a result when plants go from darkness into far-red light, the % Pr goes from 100% to about 97% Pr and about 3 % Pfr. For some phy responses, 3% Pfr is enough to trigger a response, especially when this % Pfr is maintained by the continuous application of far-red light. Thus continuous far-red light is known to ultimately suppress hypocotyl elongation.

 

 

 

Exam 1

Question: What does it mean to say that in gametophytic self-incompatibility the male factor (which is not known) is what actually prevents RNAase from destroying the RNA. Does the male factor actually bind to the receptors on the transmitting tract and if the male factor is a non-self, the male factor somehow stops the release of RNAase from the transmitting tract cells

Answer:  The RNase binding factor appears to be expressed generally in pollen, but it does not have a hypervariable region, so it cannot account for the specificity of the self-incompatible reaction. This suggests that the normal state of things is that the RNase secreted by the transmitting tract gets into all pollen, both self- and and non-self pollen.  When the pollen is non-self, the RNase-binding factor binds to it (induces its proteolysis?) and prevents it from hydrolyzing the pollen tube RNA. When the pollen is self, the RNase-binding factor does not bind to the RNase, and the enzyme is free to destroy the pollen tube RNA. The male self-incompatibility factor is not known, but the expectation is that it would somehow determine whether the RNase-binding factor blocks the activity of the SI RNase.

           

 Question: I still don't understand how the sucking up of minerals by the cell wall can lead to the transportation through the cortex and finally to the casparian strip? How can cell walls acting like a sponge transport minerals at far distances? (through the plasmodesmata)

 Answer: The movement through the wall is strictly by passive diffusion, driven by the concentration gradient in which the highest concentration of the minerals would be in the soil and the lowest would be near the Casparian strip (where active transport across membranes would deplete the mineral concentration). Plasmodesmata promote SYMPLASTIC transport only, not apoplastic transport.

 

Question: Do the experiments with ASK and ORE-9 verify that ORE-9 will interact with ASK to carry out ubiquitination of proteins during sensecence?

Answer: The experiments with ASK and ORE-9 only verify that these two proteins physically interact. Because ASK seems to function in other contexts primarily in the ubiquitination process, the expectation is that its interaction with ORE-9 is also related to the process of ubiquitination, but this expectation has not directly been demonstrated to be true.

 

Question: I don't understand Figure 5.4 on Growth vs. concentration of nutrient in tissue.

Answer: The main point of this Figure is that for every nutrient there is an optimal concentration for growth,  below and above which growth is impaired.

 

Question: I know that ORE9 destroy the repressor of SAG gene -the senescence activated gene. If we knock out ORE9, the repressor will still bind to SAG gene and we will delay senescence. What about those hormones ABA, MeJA, and ethylene? I know SAG genes are turned on by these hormones but what is the correlation between those hormones and ORE9?

Answer:  You are correct that knocking out ORE9 allows the repressor to continue to function and delay the onset of senescence. The main finding of the hormone studies is that knocking out ORE9 delays the senescence induced by all three hormones. The conclusion from this finding is that these three hormones share a common signaling pathway leading to senescence: i.e., all three pathways involve the participation of ORE9.

 

Question: Please clarify definition of methods plants use to overcome heavy metal toxicity. 

Answer: The definitions you gave were essentially correct but I modified them to make them more complete and precise. The modified definitions are given below:

Mycorrhizas: are root-associated fungal symbionts that can take up heavy metals and sequester them so that they do not enter the roots.

Exudates: are chemicals secreted by roots that can chelate heavy metals external to the plant to prevent their entry.

Phytochelatins: are small peptides produced by phytochelatin synthase that chelate heavy metals

Metallothioneins: are proteins that chelate heavy metals

Vacuolar compartmentation: is the process in which transporters on the vacuolar membrane package heavy metals into the vacuole thus removing them from the cytoplasm.

 

Question: I know that LHC is a portion of the PSII antenna complex and that it acts to distribute light energy between PSI and PSII. However, I don't know where it fits in Figure 10.15 "A tentative model for the organization of the thylakoid membrane". I know it is activated when Pq becomes predominantly reduced. When LHC is Phosphorylated, it moves to PSI, does that mean it moves pass Pq and Pc to get to PSI??? So then what is the function of Pq and Pc in term of electron transfer? My guess is Pq and Pc transfer e- from PSII to PSI and LHC is just a portion of PSII that decreases light absorption in PSII by decreasing the antenna associated with PSII.  Is this correct?

Answer: Your understanding is essentially correct. Regarding the question of whether the LHC moves past Pq and Pc to get to PSI, I am not sure a satisfactory experimental answer to this question has been obtained. Fig. 10.15 gives a 2-dimensional picture of the organization of the thylakoid membrane. If you consider the thylakoid membrane more realistically as a three-dimensional sheet, you could imagine a fraction of LHCII moving to PSI without disturbing the positioning of Pq or Pc.

 

Question: In the experiment of the tomato seed, the blotter/ membrane was exposed to LeEXP8 and LeEXP10 antisense RNA probes so that their can be hybridization between the RNA probes and the mRNAs of the endosperm and embryo. So does the specific (embryo / endosperm) region that shows hybridization indicate that there was expression of the EXP8 or EXP10 in that region since the mRNAs for those specific genes were made, thus allowinghybridization with their specific antisense RNA probes?

Answer: You are correct, hybridization indicates expression.  

 

Question:  Please explain the yeast two-hybrid experiment better.

 Answer: The ore-9 is expressed within a yeast cell and this cell can combine with any one of millions of other yeast cells each one expressing a different protein in a library. If the yeast cell expressing the ore-9 protein combines with a yeast cell from the library that is expressing a protein that can physically bind to ore-9 then when the ore-9 protein binds to that protein the combination of the 2 proteins will cause the expression of another protein. In the yeast two-hybrid system the bait protein is combined with one part of a transcription activation complex and the proteins in the library are combined with another part of a transcription activation complex, so that only if the bait protein combines with a partner from the library will the transcription activation complex be reconstituted in an active form. The transcription that is activated allows that yeast cell to express a protein that allows it to grow in a selection medium. So yeast cells that grow in the selection medium have the bait protein (e.g., ore-9) bound to a partner protein (e.g., ASK). By studying the cells that grow in the selection medium, scientists can find the gene(s) that express proteins that bind to the bait protein.



Question:  In the co-immuno precipitate system, the ASK protein is being tagged by GST and so ASK-GST is treated with antibodies. If ASK is partnered with ore-9, then treatment of ASK-GST with the antibody will bring down the ore-9 precipitate. The precipitate can be analyzed on a gel by putting a film over it to see the radioactivity of ore-9. My question on this is where is the radioactivity coming from?

Answer:  The ore-9 protein is labeled with radioactive amino acids during its synthesis in an in vitro translation system. Thus it is radioactive when it binds to ASK-GST, and its position on the
SDS-PAGE gel can be easily located by autoradiography. They concluded from both of these experiments that ASK is a partner of ore-9 because they verified that it binds to the F box of ore-9. So they think that ore-9 partners with ASK in ubiquination of proteins that repress the expression of SAG genes, thus delaying senescence. Please correct me if I have the wrong picture or the wrong facts! Thank you so much!!

Question: Could you please explain the SCR and actin gel runs we talked about in class today?

Answer: The SCR and actin gel is a technique for measuring how much mRNA for SCR and actin are expressed in different tissues. The technique involves four steps:             

1) Total mRNA is isolated from the tissue described in the legend to Fig. 3 (on the handout given in class) and electrophoresed on a gel to separate the many hundreds of different mRNAs represented. The different mRNAs are loaded into different lanes at the top of the gel, and when they migrate down the lane (the negatively charged mRNA migrates toward the positive pole), they separate into narrow bands according to their size.             

2) The bands of mRNA are transferred to a sheet of paper, where they stick in the same order they were on the gel.             

3) The paper (called a "blot") is soaked in a solution containing radioactively labeled nucleotide probes that will recognize only one kind of mRNA. The two probes used for the results shown in Fig. 3 were probes that recognized only the mRNA for the SCR gene and the mRNA for the actin gene. The blot is washed so that the radioactive probes remain only on the bands where they bound.             

4) The band where the radiolabeled probes stick is visualized by exposing the blot to an x-ray film. The film turns black wherever there are labeled bands, in proportion to the amount of radioactive probe stuck to it.             

Different preparations of mRNA were run in different lanes, but mRNA for actin is present in every lane because it is "constitutively expressed". That is, all cells appear to make mRNA for actin, so the amount of actin mRNA in a lane can be a measure of how evenly the various lanes were loaded. This is important because you cannot compare the intensities of the labeled bands from lane to lane unless you load the same amount of total mRNA in each lane. If you compare the amount of actin mRNA loaded in each lane (by the amount of radioactive label stuck to it), you see that the loading was fairly even, although the amount in lane 5 is a little low and the amounts in lanes 7, 9, and 10 are a bit high. As regard the amount of mRNA for SCR, in lane 1, mRNA from stigmas were run. In that lane there was no SCR mRNA, indicating either that the gene for SCR is not turned on in stigmas (is not being transcribed into mRNA), or that if it is being made it is being destroyed faster than it is being made, so that it does not accumulate. The amount of labeled SCR band in lanes 2, 3, and 4 indicates that the anthers of the parent plant of the pollen were expressing the SCR mRNA during the time when the pollen was being made. In lane 5, the lack of labeling indicates that once the pollen grains (= microspores) were mature (and, presumably, the SCR was already packaged into the pollen coat), the anthers stopped expressing the mRNA for SCR. Lane 6 shows that leaves do no express SCR mRNA. Lane 7 shows that the mRNA for SCR is expressed in the pollen grains. Lines 8 and 9 show that pollen from a plant with the S13 allele makes an S13 type SCR. Lane 10 shows that a mutant plant called m1600, which has lost the self-incompatible characteristic by mutation, has lost the ability to make SCR. Lanes 11 -14 are different lines of S2S2 plants that have been transformed with the SCR6 gene. Three of these lines are successfully expressing SCR6 and one (in lane 12) is not.

Question: Could you please explain how figure 3B and C (pollination response...) relate to this?

Answer: Figures 3 B and C are what you would see through a microscope if you were trying to count the number of pollen grains that successfully germinated on a stigma and grew down a style. Fig. 3B shows what scientists saw when they placed pollen from transformed plants that were expressing the SCR6 gene on a stigma that was expressing the female version of the S6 incompatibility factor. That is, no growth of the pollen tubes. Fig. 3 C shows good growth of the pollen tube when pollen from a transformed plant that failed to express SCR6 (same as plant assayed in lane 12 in Fig. 3 A).

Question: Could you explain "what if SCR was expressed in the stigma?"

Answer: If the SCR mRNA was expressed in the stigma and was translated into an active SCR, then it would bind up the SRK receptor for SCR in the stigma, and then all the receptor sites for the SRK receptor would not be occupied, rendering the receptor unable to respond to SCR coming from pollen. This, of course, would render the self-incompatible reaction no longer functional. However, theoretically, the stigma could be making the mRNA for SCR, but not translating it into an active SCR. In this case the SRK receptor could still be available to respond to SCR from pollen, and so could successfully carry out the self-incompatible reaction.

Question: When DCMU is applied to the electron transport chain is the cytochrome complex inhibited from producing ATP or is it still able to function? I assumed that it stopped ATP production ...Can you clarify this for me please?

Answer: You are correct. Because DCMU prevents the transfer of electrons from QA to QB, and thus stops electron flow at that point, it prevents the delivery of electrons to the Cytb6f complex. If there is no electron flow through this complex, the translocation of protons into the thylakoid lumen and build-up of a proton gradient is suppressed. Without the proton gradient, ATP cannot be synthesized.

Question: In phloem transport, why exactly is it that the lower leaves serve as a source for the roots?

Answer: Empirically it is observed that sinks (like roots) typically tap the nearest sources (such as lower leaves) for their sugar supply. In principle, it would be possible for a lower leaf to serve as a source also for an apical meristem.