dttom

Possibly the denitrification releases CO2, but as the plant die, if it is decomposed, majority of the carbon will be oxidized to CO2.

So think about these questions: What's your unit for the growth rate? Are you looking for overall growth rate, or the instantaneous growth rate?

I think an increase in blood CO2 makes one more alert, mediated by chemosensitive cells in nervous system.

Here you are performing an artificial selection to push the population size to the bigger size. I would not say being bigger would be due to a single gene, especially if you are dealing with mammals like dogs. But even without this assumption I could still claim that, the selection force, imposed by you, to favor bigger individuals, provided with enough generations, would act on genetic level that, any allele, or even allele combination, that results in a bigger phenotype would have their frequency increased in the gene pool. To address you question on how a pair of medium parents could give rise to a bigger progeny, variation is constantly emerging and disappearing in the population, and sometimes they may be hidden, so that their phenotype could only be expressed under certain conditions, plus there is always mutation acting.

To say an allele is dominant, you need a reference context, or what allele are you comparing it with. Then a dominant allele would be one that is able to cause a phenotype by single copy.

My routine practice is, after preparing the molten medium, I flame the neck of the bottle before removing the lid, then I pour the medium into plates with the plate-lid partially open, and I pour plates stack by stack. However, I am taking a lab course now, and the lab manual tells me to pour one plate each time, and open the lid completely while pouring in medium. In this case, one additional step comes in, that is, after pouring all the medium, the medium poured in the plate is flamed before letting them set. The tutor asked for two reasons to account for that. Because the lid is completely removed and plates are poured one by one, the whole process is performed rather slow and allows more chances for contamination, I could think of one reason that it helps kill comtaminants that get in in the prcoess, but got no idea of another one, could anyone help? Thanks.

First, I'm new to statistics, so I only know the very basics like the SD, mean, mode .etc. I've conducted an experiment so that I have two sets of data in hand, I want to know if the difference observed in the experiment set and the control set is significant or not. If I have the sample size, mean and SD of each group, could I calculate the confidence level. Someone suggested T-test and p-value, I surfed the internet for info, but I just cannot grasp the idea clear, so could someone explain them to me? Any help would be appreciate a lot, thanks.

Ok, I see the reason. By the way, as mentioned, I could not see any band even I conducted a gradient from 37c to 55c. To check if it is due to genomic DNA defect, I may include a positive control. But in the PCR mix (forward and reverse primers, dNTPs, DNA Pol (I use Taq), buffer, genomic DNA and water (please tell if I miss any ingradient)), another possible source of error might originate from the pair of primers, I would also like to know if there is any method that could ressolve the error source to one kind of primer (whether the problem is based on the forward or reverse primer).

I am trying to amplify a portion of DNA from a genome (and get no band), I also included a positive control while I was doing a temperature gradient from 37c to 55c. For the (+)control, I got one band (with its size fit) at 55c but none at 37c. If I got no band with a higher temperature I can deduce it is due to the inability of primer to bind onto genome template, but now it is the lower temperature which showed no band. With a lower temperature, multiple bands due to multi-target makes sense, but I do not know how such a band pattern is result. Hope get any help, thanks.

If I remember right, we share merely 65% with bacteria, and this number does not need to bother with HGT nor LGT, consider all organism back converge to single origin, it is not diffcult to predict such a major similarity. And great morphological difference could be resulted not simply from informational blueprint but differential utilisation. Yes, sterile infants also share the percentage, as a species. We should share 99% with chimps... Not sure about what your question asking is ..about.

This is not a homework question, that I work on integration of normal distribution and found some difficulties. The difficulty could be summarised in integrating this: f(x) = e^(x^2), I am not familiar to integrating exponential function to power x. Hope could get some help, thanks.

Yes, you got what I mean, the two bands are both higher than target. And CCC DNA, occurring as supercoil, should be faster than linear forms. But if considering them as open circular form, so that they ran slower, the hypothesis of (V/I/V) could explain that, just adding another structure of (V/V) to account for the band near to 10k (two bands exceeding 10k, one near band to and one far band from 10k), upon single enzyme cut it is released to 10k band. I suspect that the problem is I used too much vector or the concentration of reactant DNA is too high to encourage intermolecular ligation rather than intramolecular one.

The best explanation I could think of is the insert (I) and vector (V) align in an arrangement of (V/I/V|circular), V/V junction ligated in tail-to-tail alignment, so that the enzyme which cut at the very beginning of the vector release an insert (should be 2.5k) in 2k band and vector dimer (should be 9k) in 10k band.

Yes the two bands are 2k and 10k. But I'm not sure if it is due to ligated construct, as I also ran the plasmid extract before enzyme cut, there were also two bands, both exceed 10k (maybe owing to circular alignment). The enzyme I used cut only at the very beginning of the vector, no site available in insert, vector I used is ~4.5k, if it is due to ligated plasmid there should be a band of so, but there was none.

I am contructing a plasmid by inserting a PCR product into a vector. After I did the PCR, I checked the product by running a gel, no problem, the product is then enzymatically cut and gel-ran to remove small fragment, no problem was revealed by the gel. Then I enzymatically cut open the vector, remove the small fragment by gel-run (and check the size, no problem), extract the vector backbone. Next I ligated the PCR product (after enzyme treatment) to the linearised vector by one-end ligation, made blunt the one-end ligated product, ligated the blunt end with ligase to close up the plasmid. After this I transformed the plasmid by electroporation into E.coli. My vector has an ampicillin resistant gene and I spread the transformed bacteria onto an Amp plate; incubated overnight, I picked a single colony and did a liquid culture to grow up the bacteria (supposed to be containing the desired plasmid) followed by plasmid extraction by commercial kits. To those plasmid extract, I performed a single enzyme cut, and gel-run it to look for desired band which is ~7kbp. But I got two bands of 2k and 10k. I'm regretful for not checking the one-end ligated product by running a gel, so it is not promising to have the insert attached to the vector. Even taking this into consideration, empty vector alone should give a band of 4.5k but was not observed in my gel. And I grew it on Amp plate, the competent cells should not have Amp resistance, the option of being contaminated by another bacteria is not so probable... Hope someone could help indicate what problem may have occurred. Thanks.

DNA sets itself to mutate... not so correct. mutation will occur, but the virus has trace mechanism to repair, so they appear to mutate. In other organisms (so called more stable ones), again mutations do occur, but be repair to a larger extent, they just don't appear to be so much mutated... So mutation is not a result of an active mechanism, but from passive ones like error in replication, environmental mutagens.etc.

Females must have their W from her mother, as her mother is of wild type, there is not such possibility assuming there is no mutation.

Doesn't it just mutate, but without much concern on repair so the observed mutation rate is high?

current attempts use viral vector system, but I'm concerned with the resident provirus, RNAi has to be continually effecting for its not killing the cell.

No, because N2 is derived from B female, so it could only be Z'Z (received affected Z from B female). So if you breed it with ZW, amongst the female you should get half of them as Z'W.

Oh, sorry, you can't do that, then what about breeding them with normal female? Will this get 50% female B trait? Because the information about crossing of B trait male and B female is based on single breed, it is not confident enough to be included in consideration. The following is based on your first two statements: 1) Your mutation is sex-linked. 2) If a female carries an affected W or an affected Z, it is a B female. 3) A male is a B male if it is homozygous for the mutation on Z. 4) Mutation on Z should not be allele to that on W (otherwise it is not sex-linked), so you may have two mutation, one on Z, another on W for B trait. Just my personal opinion...

your yeast should be autoactive to express BD-X, so when you fused your yeast with AD-Y plasmid can they interact (potentially) to activate (if they do interact) a third set of reporter.

You said B male crossed with normal female got B female and normal male, so just label this B female as B1. And B female crossed with normal male got B female and normal male, so this B female is B2, normal male as N2. First, when N2 is inbred, will you have 25% B male in progeny? Second, when you talked about B male crossing B female, you are using B1 or B2? If you are using B1 I may have an idea... 1) When a female got an affected W' or Z' or both, she expresses B trait. 2) B male occurs as Z'Z'.

What if there are variants of your ERV so that it doesn't kill its host? Wouldn't it confer the variant an advantage over others? So host is the vector how the virus propagates, if their burst kills all of their host, they're doomed to extinction. Yet if they didn't affect their host to help their propagation (like most of the symptoms of sneezing, coughing.etc), they got weaker propagation capacity and would be overrided by other intraspecific competents. There hence always an ideal equilibrium position (ratio) between virulent forms and milder forms. Even in your presumed example, the affected species would not go extinct.

maybe it occurs in a gradual way, possible. if you consider a sudden shift from ATP producing to non-producing and remains at advantages, there is a problem (error!) of saltationism.