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March 4

patenting life forms

is it true that life form ave been made available for patent after shell manage to ave an oil spill bacteria patented and that now only humans arent viable for patent? -clockwork fromage —The preceding unsigned comment was added by 216.113.96.176 (talk) 03:41, 4 March 2007 (UTC).[reply]

I don't know the precise answer to your question, but in the USA one can apply for a plant patent. -- mattb @ 2007-03-04T05:28Z

It's certainly false that human genes are not patentable. That's a very hot field right now. See here, for example. They mention the breast-cancer-related gene BRCA1, which I believe is under patent. It's madness. Also, original poster: please sign your posts with ~~~~, and have your h key looked at. :) --TotoBaggins 18:01, 4 March 2007 (UTC)

Your first bit about the oil spill bacteria is a reference to Diamond v. Chakrabarty, which ruled that you could patent new forms of life. As for humans, you can patent various aspects about the genes (how to detect them, the specifics of the coding sequence, how to use them, etc.) though in truth you aren't patenting the genes themselves. People often get confused on this point (not helped by misleading editorials by Michael Crichton). The practical effects of such a patent is that any therapy or detection method in relation to the gene would require royalties, jacking up the price for something we are not accustomed to paying for in such a way (though in a sense it is not too different than patents on medicines). In any case humans themselves cannot be a form of intellectual property in the USA (it would violate the anti-slavery elements of the Constitution) though all sorts of things relating to their genes and organs can be patented, so it is a very fuzzy line. --24.147.86.187 18:29, 4 March 2007 (UTC)[reply]

colloid osmotic pressure

—The preceding unsigned comment was added by 202.83.86.67 (talk) 06:01, 4 March 2007 (UTC).[reply]

colloid osmotic pressure?--VectorPotential^Talk 15:17, 4 March 2007 (UTC)[reply]

Larva

Does anyone know what type of larva these pictures depict? http://user.it.uu.se/~svens/larverna/normal.html 08:39, 4 March 2007 (UTC)

Ewwwww, that's disgusting. You'd better not fall asleep leaning against a tree around there ! I would guess they are some type of tent caterpillar. You might want to go to the Language Ref Desk and ask to translate the title, or translate it for us, if you already know what it means. Giving us the location, if known, would also help. The way they drop on a vertical line might be critical to identification. StuRat 11:33, 4 March 2007 (UTC)[reply]

... Wow. --Jeffrey O. Gustafson - Shazaam! - <*> 12:17, 4 March 2007 (UTC)[reply]

Looks like the Oak Processionary to me. They've become quite a plague here in the Netherlands. Not only do they consume entire trees. Their hairs have rather adverse affects on humans as well. PvT 13:27, 4 March 2007 (UTC)[reply]

Um, if you follow the link at the top of that page which says "More information about the caterpillars" the species is given as Yponomeuta evonymella. And, er, holy crap! I usually like caterpillars but that's a bit much. --YFB ¿ 22:27, 4 March 2007 (UTC)[reply]

Here in Texas we have 'bagworms' (which are catapillars - not worms) that do produce large webs containing bunches and bunches of catapillars - that look a lot like what those photos show - but nothing remotely on the scale of that infestation! The webs form in the trees where two branches meet. I've seen them two or three feet across - but nothing like as big as those photos - WOW! Our article on Bagworm moth doesn't describe what I've seen here in Texas - and I'm not really sure whether the species we get here are the same as the one described. I'm not a native Texan and it wouldn't surprise me to find that the locals are using the wrong names for whatever beasts we have. SteveBaker 03:48, 5 March 2007 (UTC)[reply]

Yeah - it looks like the species that local Texans are calling bagworms are really Eastern tent caterpillars - which fits with what StuRat believes. The article on the Eastern tent catapillars mentions that they are often mis-named as bagworms. SteveBaker 04:01, 5 March 2007 (UTC)[reply]

Lunar Eclipse

Whilst watching the lunar eclipse on 3rd march 2007 in England the moon, once 'covered', had a red tint to it. The only explanation for that i can imagine is that the light coming through the earths atmosphere is refracted and then carries onto the moon, how does that sound for an hypothesis? Thanks in advance for any answers, please bear in mind that i am not an astrophysicist (nor a astrochemist or astrobiologist)Dave 12:13, 4 March 2007 (UTC)[reply]

Your hypothesis is correct. This phenomena is known as Earthshine. It was studied by André Danjon and can be measured with the Danjon scale. Gandalf61 12:40, 4 March 2007 (UTC)[reply]

According to the article, Earthshine is something rather different. Lunar eclipse doesn't seem to give a name for this phenomenon, though it does indicate it is caused by refraction. Algebraist 13:55, 4 March 2007 (UTC)[reply]

When I read about this in the newspaper, they said it was caused by particles in the earth's atmosphere. - Mgm|^(talk) 15:17, 4 March 2007 (UTC)[reply]

Yes, it's refraction of light through the Earth's atmosphere. The Sun's light is refracted into the Earth's shadow, where it strikes the Moon, otherwise it would be almost black. As to why it looks red, (and this is my educated guess; something I should probably know but don't) I believe it's because that's the (optical) wavelength that's scattered least as it passes through the atmosphere, and so is the strongest component of the refracted light. (See Rayleigh scattering, Mie theory.) One way of looking at it would be to realise that the light that 'skims' the Earth and reaches the Moon is passing through all the places on Earth where it's near sunrise or sunset. The other wavelengths get scattered more in our atmosphere, turning the sky that we see - and the light that passes through our atmosphere and out the other side - predominantly red. I guess you could say the Earth's atmosphere acts as a giant red optical filter, although the physics are rather different, so I wouldn't want to carry the analogy very far.

Similarly, I imagine dust is relevant because it controls the level of scattering. The density (and size) of dust particles in our atmosphere controls the balance of scattering between different wavelengths. The dustier the atmosphere, the more everything else gets scattered (proportionally to the red light), and the more significant will be the red colouring (the other colours are 'missing'). After Krakatoa blew, for instance, there were supposedly many beautiful, deep red sunsets all around the world for many months. If that happened before an eclipse, the moon would have appeared a much deeper red, rather than the coppery or orange colour we often get. Spiral Wave 17:59, 4 March 2007 (UTC)[reply]

Certainly that's the easiest way to understand it. When we stand here on earth at sunrise or sunset, the sky goes bright red for a while around sunrise and sunset. If you imagine yourself standing on the moon looking towards the sun with the earth covering it - you'd see a sunrise/sunset all around the atmosphere of the earth - that's the only light you'd see - so the moon is being lit with red light. So then you're down to explaining the red sky around sunrise/sunset - which is Rayleigh scattering. SteveBaker 01:02, 5 March 2007 (UTC)[reply]

special realtivity factor and v(∞)

we know from special relativity that the transformation factor ∑=sqr{1-(v^2/c^2}. now, consider there is an observer with ∑1 notice an event with ∑2 in away that ∑1=(∑2)^2,abviously we can say that for n-number of different observers notice n-number of ∑ ,we write

∑(n)= {1-(v(n)^2/c^2)}^n.

as v→0 , n→∞, we can write ∑(n)= ∑(v∞)={1-(v(∞)^2/c^2}= lim {1-(1/(n)}^n.as n→∞,

and we get v(∞)=0.79c. the question is, what does v(∞) mean if it exists???? 80.255.40.168 13:13, 4 March 2007 (UTC)MATH FREAK[reply]

It will help a lot if you standardize your notation a bit. The usual statements are made in terms of

\gamma :={\frac {1}{\sqrt {1-\left({\frac {v}{c}}\right)^{2}}}}

, with

\beta :={\frac {v}{c}}

commonly defined for simplicity. You seem to be saying that you have a series of frames with a geometric progression of γ values. Since

\gamma \geq 1

, any such series with at least two distinct values will tend to positive infinity, so we can't deduce a "starting value", which is what it appears your

v(\infty )

is meant to be. I don't understand where you got

\lim _{n\rightarrow \infty }\left(1-{\frac {1}{n}}\right)^{n}={\frac {1}{e}}

at all (if you even meant that; your parentheses don't match up and the numerical value is wrong). --Tardis 15:09, 5 March 2007 (UTC)[reply]

WHAT CAN SPECIAL RELATIVITY INDICATE?

from the special relativity we have the transformation of lengh , time and mass, r=∑R , t=T/∑ and m=M/∑ , now for the same event with (r,t,m) of same ∑ we have, mr=constant (1), m=constant.t (2), rt=constant(3).

for example of we take d(3)= rdt+tdr=0 we will get (v=-dr/dt)(classical form) THE QUESTION IS, WHAT KIND OF SIGNS THIS DISCUSSION HOLDS? HOW CAN WE USE THE OBOVE RELATIONS? WHAT THEY INDICATES? 80.255.40.168 13:42, 4 March 2007 (UTC)MATH FREAK[reply]

I'm not sure I understand the question. --Tbeatty 05:29, 5 March 2007 (UTC)[reply]

A web site that used to give advice

Hello everybody ! I am an autodidact. I like that kind of site that advices those who want to become accomplished scientists by themselves, for instance http://www.phys.uu.nl/~thooft/theorist.html or Wikiversity. There used to be (or there is still I don't know ?) another one dealing with mathematics as far as I can remember. Can you help me to find it again ? Thank you so much. —The preceding unsigned comment was added by 85.68.79.152 (talk) 14:10, 4 March 2007 (UTC).[reply]

Is it just my warped sense of humor that sees the irony in an "autodidact" asking for instruction? --Tbeatty 17:05, 4 March 2007 (UTC)[reply]

Indeed. The article on Autodidacticism refers almost immediately to "...Jean Paul Sartre's Nausea [which] depicts an autodidact who is a self-deluding dilettante." Nocternal 23:38, 4 March 2007 (UTC)[reply]

Well, I'm pretty sure we can assume that the meaning "Self-educated person" should be assumed here. I have to confess that 'self-educated' scientists are the bane of many people's existance. 99% of the cranks and weirdo's claim to be self-educated and wear that as a badge of pride - when in fact they tend to have dived into the deep end of physics without attempting to work up gradually from the basics. This tends to result in people who have a superficial knowledge of stuff like quantum theory without the basic stuff needed to back it up. Most (if not all) of the nut jobs who claim to have built Water fuelled cars or anti-gravity rays claim to be self taught. Get a formal college education - trust me on this. SteveBaker 00:57, 5 March 2007 (UTC)[reply]

Perhaps MathWorld will help. The idea of self-teaching is great. However, you should not discount the importance of peer review in any subject. There are other smart people (perhaps smarter than you or I!) and they are able to help check your work for error. Nimur 01:16, 5 March 2007 (UTC)[reply]

Echoing the sentiment that you really should get some kind of formal education if your aim is to "become an accomplished scientist". Formal learning institutions constantly test you on the breadth of your knowledge in various subjects. While there's a lot to be said for teaching yourself, you really can't test whether you know as much about a self-taught subject as you think you do. What's more, nearly all employers looking for scientific talent will require formal education; how else can they be sure that you really know what's what? -- mattb @ 2007-03-05T01:45Z

An "accomplished" scientist must be a member of the community of scientists. An outsider will never really be able to communiate well, nor will they be in tune with the priorities of the scientific community. College education in engineering or pure science is the best way you can become a member of the scientific community. Nimur 04:56, 5 March 2007 (UTC)[reply]

Congenital blindness

I have a fictional character who I'm developing, and it is an integral part of her personality that she is blind. Looking on the Blindness page there are only two things (i.e. symptoms, diseases, conditions etc) that cause congenital blindness, as Nasrin is born without sight. Are there any more, and are any of these idiopathic or would not have given visible symptoms to her parents? Also, I draw her with her pupils not black; as if they are hollow and 'visibly' sightless. Would any of this actually happen? Lady BlahDeBlah 20:37, 4 March 2007 (UTC)[reply]

Everyone in Little Orphan Annie's world seems to be born without pupils or irises: [1]. StuRat 19:30, 5 March 2007 (UTC)[reply]

Don't underestimate the power of dramatic license if you are creating a fictional character. Nimur 21:56, 4 March 2007 (UTC)[reply]

Yea, I know...but I like to be very accurate and I'm stingy with details especially when it comes to medicine in my stories. Plus Nasrin already has a lot of poetic licence in her; a black woman with bright lime green eyes and she's a botanokinetic - she can manipulate plants with her mind. I'd like to find a real reason why she could be blind. Lady BlahDeBlah 22:03, 4 March 2007 (UTC)[reply]

I'm sorry... you're a stickler for medical fact, but you have a form of psychokinesis embedded into your character? This pains me greatly. Psychokinesis has been widely discredited (many times, as often as scientists are willing to waste their time formally studying it). If you are seeking medically or scientifically accurate literature, you should first focus on this blaring error. Alternatively, you can indulge in fiction, create an entertaining and interesting story, and try to be creative with your prose. Nimur 01:22, 5 March 2007 (UTC)[reply]

It might just be possible that she's making a fantasy world or story of some kind, and she wants things to be accurate up to that. I mean, I would find including something not obviously fiction, such as a made-up medical condition, in a story uncomfortable. Vitriol 03:10, 5 March 2007 (UTC)[reply]

I'm basically writing a superhero story. I know it's been discredited; I'm GLAD it's been discredited, because it gives me free rein over what powers I can use. That and I get to destroy superhero stereotype with the story I'm writing. Don't worry, it's complete fiction - god I know nobody can really control plants with their mind, hahaha!!! But Ajsh is right, apart from this creative deviation I'm wanting everything else to be correct, I've done lots of research into things like kidnapping laws, pilot licences, trauma units, Basque culture, New York University...all I want at the moment is a real, plausible, legitimate reason for Nasrin's blindness that is not poetic licence... Lady BlahDeBlah 12:30, 5 March 2007 (UTC)[reply]

I don't know about pupils — although you could always say that her eyes were physically damaged by, say, staring into the sun without knowing — but Leber's congenital amaurosis, linked from the blindness article, is recessive, so it could easily be the case that Nasrin's parents were carriers without anyone knowing. --Tardis 15:24, 5 March 2007 (UTC)[reply]

You don't know noone can as not everyone has tried, and someone might have just not told you. In my books if I'm not sure how to do something like that, I just avoid mentioning it too much and let people reading it make it up for themselves if they have to.

Pupils are black because they are a window into a dark, pigmented, cavity. Thus they are black because they are hollow. Lens and corneal opacity, though, would make the pupil less dark. ^tucker/_rekcut 18:07, 5 March 2007 (UTC)[reply]

You're very helpful, thank you ppl. You've given me enough info to name Nasrin's condition...p'raps I should remember this place for next time I can't find something. (though, I can kinda 'know' because it's my book = my rules...)Lady BlahDeBlah 18:44, 5 March 2007 (UTC)[reply]

I did something like that, with the plants, but probably not as well as the person in the book as I really exist, no really I do, but I can't prove it yet.

Question about Tennents Super alcoholic beverage

My dad drinks Tennents super, and is interested to know how many calories there are in each can. RSVP Ahadland 20:39, 4 March 2007 (UTC)[reply]

Does it not say on the side of the can? --Seans Potato Business 00:32, 5 March 2007 (UTC)[reply]

Most alcoholic drinks in the United States are not required to display nutritional information or ingredients. This, I think, is a terrible loophole in the regulation of product quality. Nimur 01:24, 5 March 2007 (UTC)[reply]

In the UK, lots of companies (for food, drink, in fact anything) put "customer care line" telephone numbers on the labels, and they would be perfect for this kind of question. However, I suspect that the can won't have that either. For a rough estimate, according to Cleaveland Police, UK, a pint of lager has about 170 calories, which is "a small sausage roll".[2] --h2g2bob 02:18, 5 March 2007 (UTC)[reply]

I wondered if there'd ever be a use for those customer care line thingies :P Vitriol 03:15, 5 March 2007 (UTC)[reply]

I think drinking Super, calories are the least of your worries

chemistry

Hello, do you know if exists a substance that heated become coloured? —The preceding unsigned comment was added by 83.184.84.180 (talk) 21:10, 4 March 2007 (UTC).[reply]

Yes, there are many substances which become colored when heated, the most obvious examples I can think of are transition metal complexes, specifically compounds that are halfway between high and low spin who are prone to changes in spin with slight changes in bond distance, which in turn leads to a change in color. --VectorPotential^Talk 21:11, 4 March 2007 (UTC)[reply]

I'm sorry, I've forgotten : a transparent substance—The preceding unsigned comment was added by 83.184.84.180 (talk) 21:21, 4 March 2007(UTC).

Yes, depending on their spin many transition metal complexes are either spectacularly colored, or simply colorless, which is why the change from uncolored to colored is so noticable --VectorPotential^Talk 21:23, 4 March 2007 (UTC)[reply]

I mean: before you can see through this substance , than you heat it and it become colored (you can't see through metals).—Preceding unsigned comment added by 83.184.84.180 (talk • contribs)

Just as an example, your blood is a highly colored mostly transparent solution, that takes its color from Heme, a complex between a tetrapyrrole ring, protoporphyrin IX, and Iron, a transition metal (: however heme is oxygen sensitive, not temperature sensative, so this isn't really an answer to your original question --VectorPotential^Talk 21:42, 4 March 2007 (UTC)[reply]

Copper(II) sulfate is a blue powder when in hydrated solid form. It turns white when heated (dehydrated). Nimur 22:00, 4 March 2007 (UTC)[reply]

Powered sulphur goes from a pretty yellow to a dark red when heated. SteveBaker 00:38, 5 March 2007 (UTC)[reply]

Just to clarify here, the OP is after a TRANSPARENT substance which reacts when heated to become NOT transparent. I'm not sure of the answer my self but my mum has a black mug that becomes white with a photo on it when you pour hot water into it, it's quite obviously a layer of something that becomes transparent so the opposite definitely exists, have no idea what it is called though. Vespine 04:15, 5 March 2007 (UTC)[reply]

Transition metal complexes are aqueous solutions of transition metals (not the solid form - obviously solid cobalt is never colorless!). ugen64 04:49, 5 March 2007 (UTC)[reply]

you can use the same metal many times or you have to change it every times?

Just one example, glass is transparent, but becomes bright red if you heat it to a high enough temp. StuRat 19:20, 5 March 2007 (UTC)[reply]

Photochromic coatings on the lenses of eyeglasses can switch back and forth between opaque and translucent as many times as you need in response to light levels - I would imagine that something like that that would react to infrared light might do what you want.

everybody made me examples but nobody told me the exact name of the substance!

Symbol for a dynamo?

Equivalent circuit, from the Electrical generator article.

Does anyone know what the electrical symbol for a dynamo is? Thanks very much, Bioarchie1234 21:30, 4 March 2007 (UTC)[reply]

I so rarely encounter dynamos on my circuit diagrams or schematics, I wouldn't know it off the top of my head. I would prefer that it be labeled as a dynamo in plain text, regardless of the symbol. Anyway, it might be modeled as an AC source with some internal resistance. See the attached image from the electrical generator article (dynamo redirects there). Nimur 21:54, 4 March 2007 (UTC)[reply]

Thanks Nimur!Bioarchie1234 07:52, 5 March 2007 (UTC)[reply]

I always use this one:

+-----+
| O O |
|  O  |
| O O |
+-----+
| O   |
|  O  |
|   O |
+-----+

Oh wait, you said dynamo, didn't you ? :-) StuRat 19:17, 5 March 2007 (UTC)[reply]

That is a completely off-topic remark, and I... am laughing out loud at it! Very good. —Steve Summit (talk) 01:24, 6 March 2007 (UTC)[reply]

I have seen several symbols. They differ by industry, by country, and by era. There are special diagrams to distinguise series, shun, and compound machines. A given school, textbook, manufacturer, or industry should have a symbol library. Some use a circle with "G" in it to indicate a generator. Elsewhere that means a galvanometer. It is important to know what symbol linrary is to be useed on a project. If you are used to a resistor being a zigzag line, you may be surprised to find industries where it is shown as a rectangle. There is no universally accepted standard symbol set. Edison 22:41, 5 March 2007 (UTC)[reply]

Thanks for all your help. Especially the ~~dynamo~~ domino, StuRat! Bioarchie1234 18:33, 6 March 2007 (UTC)[reply]

Synapse behavior

Does anybody know how does the synapse in the brain perform summation? —The preceding unsigned comment was added by 213.217.212.29 (talk) 21:37, 4 March 2007 (UTC).[reply]

Our article, chemical synapse has this to say:

Integration of synaptic inputs

Generally, if an excitatory synapse is strong, an action potential in the pre-synaptic neuron will trigger another in the post-synaptic cell; whereas at a weak synapse the excitatory post-synaptic potential ("EPSP") will not reach the threshold for action potential initiation. In the brain, however, each neuron typically forms synapses with many others, and likewise each receives synaptic inputs from many others. When action potentials fire simultaneously in several neurons that weakly synapse on a single cell, they may initiate an impulse in that cell even though the synapses are weak. This process is known as summation. On the other hand, a pre-synaptic neuron releasing an inhibitory neurotransmitter such as GABA can cause inhibitory postsynaptic potential in the post-synaptic neuron, decreasing its excitability and therefore decreasing the neuron's likelihood to fire an action potential. In this way the output of a neuron may depend on the input of many others, each of which may have a different degree of influence, depending on the strength of its synapse with that neuron. John Carew Eccles performed some of the important early experiments on synaptic integration, for which he received the Nobel Prize for Physiology or Medicine in 1963. Complex input/output relationships form the basis of transistor-based computations in computers, and are thought to figure similarly in neural circuits. --Seans Potato Business 00:29, 5 March 2007 (UTC)[reply]

Basically, summation means that al the inputs to the neuron add up to produce a single response. There are three things that go into summation: total stimulus, stimulus timing, and stimulus location. More stimuli in a short period of time near the effector parts of the neuron have a greater effect. An analogy is toppling over your friend by pushing him. Pushing harder (total stimulus) will be more likely to topple him. If one small push doesn't topple him, but puts him off balance, then another small push, quickly following the first, is more likely to topple him over (temporal summation). Also, pushing his legs out will be more likely to topple him than pushing on his chest (spacial summation). ^tucker/_rekcut 17:57, 5 March 2007 (UTC)[reply]