ArabidopsisProteomicsPart 2Part 3Name: ___________________________ Lab room, day, time: __________________TA name: _______________Arabidopsis1. Controls are important for determining the validity of experimental results.a. (1 points) For the genomic PCR, which reactions were used as negative controls? Positive controls? Defend your answers.All reactions using water instead of template were negative controls because they were expected (in all cases) to not produce a result. The one reaction that may be less obviously a negative control is the reaction using the T-DNA-specific insert primers against the wild type plant. This should not have produced a PCR band because there should not have been any T-DNA in the wild type plants. The PCR using genome-specific primers against the wild type plant was the only positive control because the primers should not have any interrupting T-DNA insertion between them and so should have generated a band.b. (0.5 points) There is one more reaction that could have been used as a positive control with the genomic DNA. Explain what this reaction should have.The “missing” positive control would be one for a T-DNA insert. This would need the T-DNA primers combined with a DNA template that definitively had an insert present.2. (1 point) What should the test PCR results be (band or no band) for the following situations? Assume that T-DNA inserts are disruptive to RNA expression.a. Genomic PCRs with gene-specific and T-DNA test primer pair sets on:i. A test plant lacking T-DNA insertsii. A test plant heterozygous for a T-DNA insertiii. A test plant homozygous for T-DNA insertsb. RT-PCR using a gene-specific primer pair on:i. A test plant lacking T-DNA insertsii. A test plant heterozygous for a T-DNA insertiii. A test plant homozygous for T-DNA insertsSummary chart:gDNA RT-PCRGene-specific primerpair T-DNA primer pairGene-specificprimer pairtest plant test plant test plantNo T-DNA + - +Heterozygote + + +Homozygote for T-DNA - + -3. (1 points) Ideally we want to identify a plant that is homozygous for a disruptive T-DNA insert. If that does not happen, we can self-cross a heterozygote and test the offspring (plants, such as Arabidopsis, that do not have mechanisms to preventit can fertilize themselves). What percent or fraction of those offspring would we anticipate might show an altered response to extracellular ATP? Briefly explain.This is just a roundabout way of asking for the genotype/phenotype outcomes of a monohybrid cross. We want a homozygote for the insert and we’re crossing a heterozygote. ¼ or 25% of the offspring should end up with a T-DNA insert in both chromosomes so that’s the same fraction we would anticipate to respond differently to ATP.ProteomicsYou will work with your own observations this week. Part 24. (1 point) Attach either a hand-written (on graph paper!) or printed copy of your standard curve or (with prior approval) e-mail an electronic copy to your TA.5. (0.5 point) Using your standard curve, fill in the following tables. Note that you may need fewer rows than provided. If you need more, write the extra informationto the side of the relevant table or add extra rows.Species “A”Distance Migrated (mm) Molecular Mass (kD)Species “B”Distance Migrated (mm) Molecular Mass (kD)Species “C”Distance Migrated (mm) Molecular Mass (kD)Species “D”Distance Migrated (mm) Molecular Mass (kD)Species “E”Distance Migrated (mm) Molecular Mass (kD)6. (0.5 point) Complete and turn in the table from p. 86 of the lab manual (or a copy of it). 7. (0.5 point) Fill out the following table based on the p. 87 table:Number of proteins in common between speciesSpecies A Species B Species C Species D Species ESpecies ASpecies BSpecies CSpecies DSpecies E8. (1 point) Following the guidelines given, construct a cladogram depicting the relationship between the 5 examined species. Defend your construction by explaining the arrangement you chose.Answer is officially going to depend upon the data they get. Their cladogram needs to be constructed based on the information they have. Highest shared proteins should be placed together, least common furthest away, and the remainder in the middle. Species A (shrimp) is most likely to have the least proteins in common, so should be the outlier/outgroup at the first branch at the base of the cladogram. Species B and C are have different names and phenotypic differences but are actually the same species (steelhead trout and rainbow trout are both Oncorhynchus mykiss), so hopefully those two samples end up similar!Note that cladograms can actually be “rotated” around each branch point, so this does (unfortunately) complicate evaluating the grading of them. In sample the picture below, all 4 cladograms are equivalent:9. (0.5 point) Consider all of the information that would normally contribute to the construction of a cladogram of these four species. Would you expect an “official” cladogram (i.e. one you might find in a textbook) to match the one you constructed from your protein data? Explain your answer.This one can be answered either way.If they think it should match, they should discuss the fact that generally all the other differences in morphology and biochemistry and such are also likely to be reflected in theprotein profiles of the samples looked at. Generally, closely related species will share mostly all features in common while distantly related species are going to have many features that are different, even if only slightly. If they think they should not match, this is because official cladograms are based on a synthesis of varied information types. A “real” cladogram involves morphology, anatomy, molecular data (DNA sequences, protein sequences), developmental data, fossils, etc. Their cladogram only has a single information type so is likely to be inaccurate (the specific reasons why will be discussed in the last lecture).Part 310. (1 points) Good procedures should have positive and negative controls. Did your Western blot have either? Explain for each (positive and negative) and identify what it was (if it was present).There is a positive control, the myosin and actin standards. We would expect that the standards should react, so that makes them a positive control.There isn’t clearly a negative control, however. One of the samples (shrimp) likely didn’t react, but we were testing that, not expecting it, so it is not a negative control. They