apricimo

No not michaelis menten. This is a general partition function.

Hello, In deltaG = deltaG' +RTlnK once at equilibrium deltaG = 0 so deltaG' = -RTlnK This is supposed to be at standard temp, ph and conentration but I see it often that people will find the Keq and sub it in and get deltaG' at whatever concentration and temp, and ph they are working with for their system. That is, they are not at standard conditions. Why is this so? Is just the matter of defining your own "standard" conditions? How then do you universally say this is deltaG for this enzyme and this enzyme and be sure that their conditions are the same?

Hello, I was wondering if anyone could expand on how taking the derivative dlnY/dlnx of a partition function that describes the different species ligating species, Y=1+kx, in this case a ligand, x, binding to protein with a single site creates the function for average ligation(k is the association constant). That is, after taking this derivative you get Y'=kx/(1+kx). My question is why does this work and where does this derivative come from? Is it somehow derived to get to this point or is it simply an observation?

I think I am too ignorant for this explanation. Can you expand on this in more detail. Thank you

thank you

Hello, Lets say I have an equation y=mx^2. If I plot this equation in excel and then set the x-axis to plot in log is the same thing as y=m*log(x^2)? Having difficulty in trying to discern what the log plot actually is doing.

So people who argue that evolution violates the second law of thermodynamics are________ Fill in the blank People who say those who say the second law of thermodynamics is violated in evolution are wrong because we are not talking about an isolated system. These people are______ fill in the blank

Can we consider our universe an isolated system?

Can someone explain what the difference is in taking the derivative with respect to P of PV=nRT vs V=nRT/P ? T is kept as a constant.

So to tentatively summarize: de Sitter binary star experiment showed that the speed of light is constant? I don't think that if you measure the speed of light on earth you can say you are in a different frame of reference when you measure the speed of light in June and then in January. The source of light that is measured is the sun so no matter what time of year the fram of reference is technically the same. Is that right?

Hello, I am trying to install Rosetta Commons on my Mac (Snow Leopard) and if anyone has done this please help me out if you will. Or if anyone has ever used the terminal to install any of these types of programs that require "Scons" and so forth please be in contact with me via private chat where we can exchange an email. Thank you

Does anyone know why one would do a slant vs. a petri dish type of media setup?

ah oh ok thank you

Hi I was wondering how come when they show the calculation of a second moment for a particular variable they don't also square the binomial function but only the term outside it. for example this is how they show it [math] \begin{array}{l} \overline {n_1 ^2 } = \sum\limits_{n = 0}^M {W(n_1 )} n_1 ^2 \\ \overline {n_1 ^2 } = \sum\limits_{n = 0}^M {\frac{{N!}}{{n_1 !(N - n_1 )!}}p^{n_1 } q^{N - n_1 } } n_1 ^2 \\ \end{array} [/math] why not also square the n inside the W function?

ok... got it

In a book I am reading they list these steps and I don't understand the simplification... BTW u is a random variable and bar over u is the average for notation clearification [math] \begin{array}{l} \Delta u = u - \overline u \\ \overline {\Delta u} = \overline {(u - \overline u )} = \overline u - \overline u = 0 \\ \overline { < \Delta u > ^2 } = \sum\limits_{i = 1}^M {P(u_i } )(u_i - \overline u )^2 > 0 \\ \overline {(u_i - \overline u )^2 } = \overline {(u^2 - 2u\overline u + \overline u ^2 )} = \overline {u^2 } - 2\overline {uu} + \overline u ^2 \\ \end{array} [/math] and then they simplify to [math] \overline {(u_i - \overline u )^2 } = \overline {u^2 } - \overline u ^2 [/math] where did the middle term go?

Ideally from mycelia but it seems that spores form throughout the growth. Maybe my prep sucks. I've isolated the gemomic DNA before, but have had a hard time since. I use a bactozol kit it works nicely if there are no spores. What medium do you use for your growth to get your species to produce natural products?

Hello, I am working with the mentioned species and I am trying to isolate the genomic DNA. This species makes spores after some time but overall spores form throughout the growth period. Does anyone know a good strategy for isolating genomic DNA from spores? Or have experience with working with sporulating streptomyces and isolating their DNA? Thank you.

The correct answer to taking the derivative [math] \frac{{d\ln y}}{{d\ln x}} [/math] of [math] y = \frac{{kx}}{{1 + x}} [/math] is [math] \frac{{d\ln y}}{{d\ln x}} = \frac{x}{y}\frac{{dy}}{{dx}} = \frac{x}{y}\left( {\frac{k}{{1 + x}} - \frac{{kx}}{{(1 + x)^2 }}} \right) = \frac{{(1 + x)x}}{{kx}}\left( {\frac{k}{{1 + x}} - \frac{{kx}}{{(1 + x)^2 }}} \right) = \frac{{1 + x}}{k}\left( {\frac{k}{{1 + x}} - \frac{{kx}}{{(1 + x)^2 }}} \right) = 1 - \frac{x}{{1 + x}} [/math]

Well, that's all fine, but it's also a bit of a stretch. Newtons laws worked for how long until they had to be modified. Thanks for the explanation about the prior experiment though.

It's my understanding that they set out to see whether aether existed or not. By splitting light into two beams they measured if there was a difference in the path length should it travel through aether. They did this by turning their platform in to different angles with respect to "aether" and found there was no difference. How from that does it follow that light is the same speed independent of the observer?

Yeah so that's what I've been trying to just satisfy within myself. Whether this assumption is based on calculations or real experiments. Anyone know this please reply.

Ok... Three observers... One is standing still and the other two are zooming past each other in opposite directions at the speed of light respectively. The standing still observer sees two ships traveling at at a combined speed of 2c. Two questions. How can the standing still observer see the ships and how can the observers iside the ships see each other? If a ship is already traveling at the speed of light then for you to see it a photon has to bounce off the ship and then into your eye. By the time the ship is seen it has already come and gone. For the two guys inside the ship... They have come and gone twice as fast... Is that really it? So the answer is you cannot see each other in "real time". There is some lag time by the time anyone sees anyone and that lag time will depend on the distance from each observer? Did I answer my own questions?

Yes Thank you I realize that. That's why I said forget the (Lorentz) transformations and just intuitively think about it. No need to reply if you don't add anything in the end and just want to show you know more but then you don't even share that. That's the part thats difficult to conceptualize. I'm just thinking in terms of seeing with my eyeballs. If I am in a spaceship and I am flying at the speed of light and the speed of light is 3 x 10^8 m/s then in order for me to SEE that spaceship the photons would have to be bouncing off the spaceship and then hitting my eyeball at 2c. Or will my physiology be affected? i.e. I literally won't see the other spaceship?

If two objects are zooming past each other in opposite directions at the speed of light are their combined speeds not 2 x the speed of light? I mean forget transformation of coordinates and all that. Just thinking physically about two objects travelling past each other.