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Hello! I would appreciate aid in all the questions for thissituational exercise, thank you very much !A community experienced an outbreak of a bacterial infectionaf...

Question

Hello! I would appreciate aid in all the questions for thissituational exercise, thank you very much !A community experienced an outbreak of a bacterial infectionaffecting the gastrointestinal system. Among children,symptoms were found to be abdominal pain and life-threateningdiarrhea. The commonality between patients was observed to be withthe place they visited - the playground, specifically theswings. You are a medical practitioner in the area and youare tasked with identifying the bacterial

Hello! I would appreciate aid in all the questions for this situational exercise, thank you very much ! A community experienced an outbreak of a bacterial infection affecting the gastrointestinal system. Among children, symptoms were found to be abdominal pain and life-threatening diarrhea. The commonality between patients was observed to be with the place they visited - the playground, specifically the swings. You are a medical practitioner in the area and you are tasked with identifying the bacterial species responsible for the outbreak, through DNA profiling of the 16S rRNA gene. 1. Describe the culturing method (streak-plate, pour-plate etc.) and media (selective, differential or enriched) you will use to sample bacteria from the suspected origin. Explain. 2. How can you confirm whether the bacteria you obtained from the origin is the same as the one infecting the children? Which sample would you get? Describe the culturing method you will use and explain. 3. You have successfully diluted your culture into a few colonies of bacteria. However, you suspect two closely related species that give the same symptoms similar to the ones the children experienced. How can you elucidate between the two without DNA extraction? 4. To finally confirm the bacterial species, you decided to undergo DNA extraction. From the samples from question 1 and question 2, provide a flowchart for a reasonable DNA extraction method for both samples given the following reagents and apparatuses. Note: Other than pH, exact quantities are not needed. Meat tenderizer Ammonium sulfate Lysozyme Phenol-Chloroform Mortar and pestle 5. One of your colleagues decided to do parallel studies of the same samples. However, he forgot to add the meat tenderizer in the process of DNA extraction. The relevant absorbances of two samples are tabulated below: Sample A260 A280 A 0.780 0.610 B 0.779 0.408 Between sample A and sample B, which one most likely belongs to your colleague? Why?



Answers

The proof that DNA, not protein, is the carrier of genetic information involved a number of historical experiments, including transformation or horizontal gene transfer (HGT), which is the uptake and expression of extracellular DNA.

A. As described in Figure 14.3, transformation or HGT was first reported by Griffith in 1928 in an experiment in which the following occurred:

1. heat-treated, pathogenic bacteria recovered their pathogenicity when incubated with non pathogenic bacteria
2. plasmids were transferred to nonpathogenic bacteria from pathogenic bacteria through conjugation
3. non pathogenic bacteria acquired pathogenicity when incubated in a broth containing heat treated, pathogenic bacteria
4. polysaccharide cell capsules from pathogenic bacteria were transferred to non pathogenic bacteria

B. Griffith’s experiment, however, left undetermined the identity of the cellular component that encoded genetic information. The identity of DNA as the carrier of genetic information was resolved through the experiments by Martha Chase and Alfred Hershey because they observed the following:

1. injections with a serum containing chemically isolated polysaccharides and non pathogenic bacteria were not lethal
2. pathogenic bacterial DNA that was radioactively labeled using a phosphorus isotope was not present in mice that died
3. bacteriophages from a bacterial culture grown in a nutrient-containing medium and radioactively labeled using a sulfur isotope transferred the label to bacteria incubated in an unlabeled nutrient-containing medium
4. bacteriophages from a bacterial culture grown in a nutrient-containing medium and radioactively labeled using a sulfur isotope did not transfer the label to bacteria incubated in an unlabeled nutrient-containing medium

C. Transformation and transduction increase variation within populations of bacteria and archaebacteria by the following:

a. transferring DNA among different species
b. transferring free DNA across the cell membrane without energy expenditure
c. transferring DNA between different strains of the same species of bacteria
d. phagocytosis of bacteriophages

The evolution of antibiotic resistance via HGT poses a challenge to medical technology. On the other hand, transformation is often assayed by incorporating an antibiotic-resistance gene in the plasmid to be transferred into the host organism. In natural environments, bacterial and archaebacterial cells become competent (able to transport DNA through the cytoplasmic membrane) in response to stress such as UV radiation, high population density, or heat shock. Such conditions are often difficult
to model in the laboratory, where competence can be induced by high concentrations of divalent cations, $Ca^{+2}$ or $Mg^{+2},$
or electrical shock. In either setting, extracellular DNA can be transported into the cell, and (to a good approximation) uptake is proportional to the concentration of extracellular DNA.

D. Identify a factor that might affect transformation or HGT. Then, design a plan to evaluate the dependence of transformational efficiency (defined as the number of transformations per gram of extracellular DNA) of plasmids that transfer antibiotic resistance to a particular strain of Escherichia coli that is not resistant on that factor.

We're looking at a toxin from Clostridium before engines, and the C terminus of this toxin binds to cloud in four but not cloud and one and the N terminus will then insert into a cell membrane and former hole that kills herself. So if we remove the n terminus and we're just looking at the C terminus, we're incubating an epithelial sheet that expresses cloud in one and four with this C terminus, and we find that when we do so, clouds and four disappears. But the cells aren't dying. The the tight junctions, the number of strands in them, is decreasing and they're less highly branched. But if we wash away the toxin, they recover back to the wild type. So we have some questions about this part. One is how is it that there are too tight junction strands remaining, even though all of cloud and four has disappeared. So remember that we're not just expressing climbing before also applied in one which the toxin doesn't mind too. So this has to be because of cloud in one so cloud and one must be unaffected e clouding four, including one aunt forming diners, or, if they are somehow only the clowning for is affected, and can one can somehow still do its job? Most likely? Most likely, the toxin is binding. Cloud inform on Imus, and that's stopping them from working without affecting cloud and one B. We noticed that the toxin will work if added to be both lateral side of the sheet, but not be a pickle. How can this be? Well, if it's only working on the base lateral side, that must be where its target is. So the binding sites of sites of plowed in four must be on the basal lateral sides. So maybe clouding four completely is on the basic lateral side. Maybe all of the climate for monomers are delivered to that side. Maybe there's an orientation going on where all of the clouds and four all face one direction, and that would mean that all their binding sites would be on the same side. So that's two possible ways this could happen. But this this is the reason it only works if we soak the basal lateral side. And, of course, um, it's not penetrating through the toxic VM. It's not penetrating through the tight junctions to get people said. Because they are tight junctions, they prevent the movement of molecules through them, so they're staying on the basal lateral side.

This question is making sure you understand restriction enzymes, how they cut and how they're sticky or blunt ends will react with different portions of DNA. Create a new hours double strand set of DNA. So part A is asking us. But a Nico are one enzyme cut would look like so in one of the tables in the textbook. It gives us the sequence for ICO, our wine. It's going to look for a G, A, A, T T C and its complementary strand of C T T A A. G. And here the ICO are one enzyme is going to cut along this pathway. It's going to separate the two ends and create two fragments of DNA along this point. So for part A, we're going to have a long strand of DNA mixed with It's a and its complementary par shin of the strand with a T. T. A sticky end, and this is going to be separated from the rest of the fragment of the single strand of DNA that we just cut you have. It's three prime and five prime, so these are the kind of things you'd see. Just remember, we have a longer strand of DNA continuing out along both sides. But this is the site we're looking at. That was just cut for B. It's just asking us to attach a nucleotide triphosphate to the sticky ends. So we're just going to attach the complementary bases for each of these fragments. So here on the three prime Strand, we have our t t a. So we're going to attach our complimentary A T T. And then on the second fragment we're going to attach are complementary pairs like ways. So we have a okay to to. So these are new fragments we've created with blunt ends. Now, since we've added our nuclear tied try phosphates to each of those for part C, it's just asking us to attach kind of these blunt ends together to create a single strand of DNA has just been reassembled. So here we'll have our G A T T fragment, and it's going to attach and, like, eight to our A a two TC and Mike wise on the bottom. That's May. So here we have our new DNA strand when are previously made and nucleotide triphosphate added strands have been connected. Just remember, we have our five, prime to three Prime and the complementary strand as well. For Part D, it's asking us to take our original sticky and fragments and to digest any single strand of DNA that's present. So here we're going tohave are sticky ends made with just an original eco r one cut and we're going to digest and remove our UNP aired nucleotides. So in the end, we're going to get very short fragments of just a G here, iguana zine and a cytosine and the other fragment with the site is seen on top and a quantity nonviolent. So here's our new strand we've created. If we're going to digest those UNP aired nucleotides from the sticky ends for party, it's going to ask what happens if we stick our original eco r one cut to what we've created here with just our single nuclear tides. So if you imagine we're going to take our sticky and pears Ah, and we're going to attach this three prime to five prime to our single nucleotide fragment of just the guanine from part D. So here we're just taking this single nucleotide fragment of the five prime and attaching it to the ico are one cut here. Likewise on top. It's just gonna be the complementary, but with this cytosine attached or with a guanine, because we're attaching our a a TTC to our single nuclear Tiggy from before. So for this instance, we are attaching to this five prime to three prime end here, and we are attaching our A, t, t c and G fragment. So this is what they're looking for, which is, ironically, just the ICO are one restriction site as well. Basically, what we've done is we've cut our original strand of DNA removed, are single nucleotides and added them back again for part F. There's actually a misprint in the book. Um, we're looking for the restriction kind of product if we're using a P V U two enzyme. Yeah, So if you notice in the book, the ICO are five. Restriction site is listed as the same as the PV You one PV YouTube PTU to actually has a different restriction site of C A, G, C T G, and its complementary G T c g A C. And it's going to create blunt ends by cutting straight through the center. Try that and read So for going for a PV you to cut, we're going to make the fragments C a g by prime three prime GTC and the fragments ctg g a c So this is what they're looking for for part G. We're going to attach this fragment we've made and add it to the blunt ends that we made in part B. So we're going to add our TV You to to this section here. So, Fergie, we take our five prime three prime initial strands C G G c, and we're going to add it to those blunt ends of our five prime to three Prime and our three prime five Prime. We're going to make this new strand here. So we've just attached are blunt ends together along this point, and this is our new strand with me for part h. They're asking us to apply this knowledge to kind of make us try and get a different plasmid structure where it would no longer be digested by an eco r one site, but it will have a bam h one site. So from what we've learned in the previous questions, what we could dio is we could take our initial Iko are one G A, T T C and its complementary strand Cut it with the ICO are one. And we can either remove this single strand DNA using a different enzyme, or we can fill in here using our nuclear tied complementary pairs. And then we can just add additional nuclear tights on here That would give us a bam H one site as long as we don't begin with C and G along this next nuclear tied we add on. If we add in this CG pair equal are one would still be able to cut it. But if we add, say A and A and A T here equal are one will no longer recognize this sequence and it won't be able to cut now it takes a lot of time and scientific error. Um, restriction enzyme aren't perfect, and they don't give you 100% outcome and yield. So a better way to do this is to create our own DNA strand. But they have listed is we're going to make this artificial strand here and its complement so feel notice. What we're now including in here is our a t T See are these editions to this nucleotide sequence. So basically, this sequence is made by cutting it away with a Nico are one enzyme on both ends and you'll notice the Bam H one restriction enzyme of G A, T, C C, and its complement is located right here in the center. So by artificially creating this sequence of DNA, we have an opportunity to attach it to a Nico are one enzyme site, and it automatically has a bam H one enzyme site included. So, after we've introduced equal art one toe wherever this is attached, we can now make a new cut down the center of this sequence with a bam H one cut. So if you look in this previous example, if we're going to add DNT piece here, we create a blunt end, and blunt ends aren't very good at connecting up ends of DNA. Um, it's easier with sticky ends because complementary pairs like to stick and fix along to make double stranded DNA. Rather than bear up to blunt ends, there's just more free energy available and sticky ends make the reaction more spontaneous and more likely to occur and give a stronger hold rather than just attaching across the DNA rather than across the compliments. So for H, this is our better option. Two years, due to its advantages of sticky ends and its utility of both ends for a Nico are one enzyme and the inclusion of the Bam H one site straight through the center finally per part II. They're asking us to create different strands that are compatible with Iko are one and or PST one. So just as a reminder, all right, the equal are one cut site and the PST one cut site. So for equal are one It's the G A, T T c. And for PST wine, we have CTG, C, A G, and our method of cutting creating sticky ends. So what we want to do is create a sequence where you have hey is both being c N t. Do you is neither for being able to do an ICO are one site and for C, just a pst one site. So, for a what we can do is we can make an eco r one site along the far ends of the strand of DNA, and we can make the PST one sites closer into the center. So if we start with this sequence of DNA a long way, we're cutting. So here we'll have a a a T T. C bound to its G here would be the ICO, our wine cut site in the center. We could use whatever nucleotide sequence we want. It doesn't matter, because we're not going to cut along these lines. And along the right side, we'll have our PST one cut site. So here is our c g C. A sequence. If we wanna have unequal are one on Lee, What we're going to do is we can just change this sequence. This base pair here MPs Taiwan will no longer recognize our strand so we can have a a t t c whatever we want in the center. And then we could just add in a guanine here instead on create are complementary sequence in the middle. So here we stole of our eco r one cut site. However, on this and this sequence is not recognized by any of the enzymes. So for part c, we do just the opposite and change this base pair here and leave this original water alone. So here we'll put in a guanine instead of that side is seen. Keep our cytosine on this strand and drawn are complementary. So here we have our PST one cut site and no cuts site over here. Lastly, for our neither we can change both of these sites so that neither iko are one or P s t one will recognize them. We have a a t t. G. So there's no cut and G T g c A. So there's no cut, so you'll notice neither this strand or this strand is recognized by either equal are one or P s t. One. So there will be no cut along these points. So this sums up how we've located where to cut along Our enzyme restriction site had to attach blunt and sticky ends and how to form different pairs so that were either removing the restriction cut site or adding it into it. And this is what this question is looking to address

For this question we're looking at the code on and how this is interpreted to make different amino acids or parts of a protein. So the first part of this question, we're looking at different experiments by Crick and his associates, as well as experiments by Nirenberg and Matty. So the first experience by Crick and his associates were dealing with adding mutations, two different DNA sequences inside a virus. So you can imagine a virus could have a a DNA sequence such as this, of a t a g c a per se. What they did is they are going to add in a different number of mutations to see what affects it would have on the virus and its capabilities. So they noticed that of course the um mutated virus had a normal function. Two different viruses with either the addition of one or 2 nucleotides showed no correct function. And the virus that picked up three nucleotide editions also had a normal function. And this was used to prove the importance of three nucleotides to make up. One code on this is because in the addition of either one or two nucleotides, it's going to cause a frame shift mutation. And this is because three nucleotides are required to make the protein. So when one nucleotide is added, it's going to pull two nucleotides from the next three pair of nucleotides and that's going to alter all the next upcoming proteins and it can severely alter the capability for the virus to function. This is the same for the edition of two nucleotides because it's still going to grab the first nucleotide of every next amino acid grouping or code on coming up. However, the last option appeared to be normal because with the addition of three nucleotides, this would only correspond to one mutated protein inside the virus. And typically one mutation of a protein is not enough to completely alter and destroy the function of the cell so the rest of the amino acid groups, or Cody johNS located later in the DNA sequence, would still be normal, allowing for the virus to function normally. As for the experiments by Nirenberg and maturity, they discovered that with the addition of ribosomes and say a certain amino acid, if they only had say use or Azour gs inside of a solution, it would only create one type of amino acid. Whereas the addition of different nucleotides and amino acids would correspond to a completely different type of amino acid. So that just helped prove the specificity of the coding sequence where one type of code on Is going to produce one type of amino acid as long as the are in a. Zor ribosomes are present inside the solution as well. Moving on to part B of the question, they are asking us to calculate ratios are the percent chance of different combinations of code ons. So we have three different types. We have the first scenario where we have two guan means To one side a zine. We have one wanting to to cida zines. And we have the last scenario where we have only cida zines And of course the ratio of GS two season here is 5- six. So five wanting things for everyone cytosine. Now you can add these together to get the whole of the probability chance. So this would be equal to six parts to the whole when calculating this inside the problem. They make it a little complicated but it's just calculating simple chances and probabilities for each of the amino acid sequences. So we'll say we're calculating the ratio of this code on G. C. To the percent chance of randomly coding for all guan means. So the chance in this solution of coding just for a guangming is going to be five out of six because Pulling six times from the solution one time you're likely to get a cytosine, whereas five times you're likely to pull the guarani. So again for the second quantity you have a five out of six chance, whereas for the last nucleotide, the Cytisine, you have a one out of six chance or probability. And then they are just dividing this by the chance of getting all Gs in this other code on. So again the chance of getting a quantity is five out of six. So this is all they are doing in the problem and then they are just simplifying it and rewriting it As say 5/6 Squared times, say 1/6 divided by 5/6 Cube. So this is all they're doing in the problem. So you should be able to calculate the probabilities of this out between all of these where the ratio or the percent chance of getting one G 21 C. So you have a G c C sequence to a g g G. The chance of Or the probability of coding for this exact code on would be 5/6 Times 1/6 times 1/6 Divided by of course that 5/6 Cube. And for the last problem, the chance of getting all these side Busines is 1/6 times 1/6 times 1/6 divided by 5/6. Cute! You just plug that into a calculator to see your chance for getting each of these. And of course the other possible code ons such as G c G will have this same ratio because instead of this 5/6 here you have a 1/6. And instead of the 1/6 you would have the 5/6 and you should still get the same probability or the same chance of this occurring for part C. They give us an amino acid table for this and they are saying we have an equal proportion of cida zines and guan means and they are asking us the percentages for getting each of the amino acids in these groups. So if you've never read from one of these tables before you start on the left to check for the first amino acid. So say we have a C C c sequence that we are looking to code and amino acid for. So you'll start with C. You'll move on to the second position up on top. So you move over from the table until you get to the calm with the next see. And from there you move on to the 3rd position of the table where you pull from the left. So if you do that for any amino acid you should get the amino acid that is coded for by the code on. So for example C C C would code for the amino acid of pro lean. So keeping that in mind And the 1-1 ratio of wanting to cytosine. We should consider the different possibilities we have. We have a code on where we'll have three CS. 20 Gs. We'll have code ons with two Cida Zines, 21 guanine will have a one cytosine to guanine means and will have zero citizens to three Guangdong's and you can just right in the different combinations of these. So these should be some of the code ons and DNA sequences that you get. And then from there you can just use the table to figure out what amino acids these would code for. And you should get prowling for the C. C. C. Which I demonstrated above. You should also get pro lean here, should get arginine, Ellen, E, argentine, Alan E clay scene and placing and then you can just total these up to get a ratio. So we'll have to pro lean to to Argentine 2 to Alan E. Too too classy. That can be coded for in this scenario. Of course the percent chance of getting each of these is going to be equal because the chance of getting either G or C in each of these positions is one out of two. So no matter what happens, you have a one out of eight chance of getting any of these nucleotide or code on sequences. So you can just total these up And you should find that there are eight parts and because these are all equal and they have to, each of these amino acids has a one out of four or 25% chance of occurring or being coded for when the G to the C ratio is equivalent for the last part were being asked about our different stop sequences of say you A G. We also have you A and you G. A. When you look on the table and you code for each of these, you should see the stop inside the table rather than a three letter word for an amino acid or form a thin. The role of these is they do not code for any amino acid. So when we are creating our polymers of proteins and amino acids instead of adding an amino acid to the chain, it's going to add nothing. So it's basically going to break off and end the protein sequence. So these are responsible for terminating the elongation of the protein and for finalizing it. So these are going to allow the amino acid chain and protein to disassociate from the ribosomes and to enter the cytoplasm in their complete form. Now, with those for it should help you understand the purpose of the code on With its reliability and it's three nucleotide sequence.


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