Science desk
< December 19	<< Nov \| December \| Jan >>	December 21 >

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

December 20

Do fish drink?

Well, do they? Is it accurate to speak of a fish drinking? If they don't drink, what do they do, exactly? FreeKnowledgeCreator (talk) 03:55, 20 December 2018 (UTC)[reply]

Googling "do fish drink" yields many items. This one from Quora, for example, says saltwater fish drink and freshwater fish don't.[1] ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:57, 20 December 2018 (UTC)[reply]

All aquatic organisms that are too large to meet their oxygen requirement by body absorption arrange water flow through their gills by action one cannot entirely call drinking or breathing. DroneB (talk) 18:51, 20 December 2018 (UTC)[reply]

This is kind of tangential to the question. Yes, all fish with gills need water flow over them to respirate. But, as stated in the link above, saltwater fish actually lose water from doing this through osmosis, because the seawater is saltier than their bodies. So, they drink water, just like you and me. Freshwater fish pull water into their bodies through their gills, because their situation is the reverse of saltwater fish. The excess water has to be excreted through urination. Since they take in plenty of water through their gills, they don't drink. --47.146.63.87 (talk) 21:55, 20 December 2018 (UTC)[reply]

Hmm, that generally makes sense, but not completely, and the sourcing is not exactly rock-solid. The part that doesn't make sense is that even freshwater fish presumably take water into their gut along with their food, and it's not clear how you distinguish that from "drinking". I could easily believe that they don't drink except incidentally to eating. --Trovatore (talk) 22:24, 20 December 2018 (UTC)[reply]

"Saltwater fish actually lose water from doing this through osmosis, because the seawater is saltier than their bodies. So, they drink water, just like you and me." -- But what are they drinking? They are drinking the same seawater, so they are actually getting into their system even more salt, so what is the point of that? --2001:16B8:118B:3800:903B:C589:11CC:913B (talk) 07:39, 21 December 2018 (UTC)[reply]

I was at an aquarium a few months ago, where a vet explained that in the wild, marine mammals get non-salt water from the foods they eat. I did not have a chance to ask if they had any sort of active transport for salt excretion. I would be interested to see a comparison of average body salt-content going up the food chain, on the assumption that a predator could trivially be more saline by exchange with the ocean but would be harder to lose salt. DMacks (talk) 08:35, 21 December 2018 (UTC)[reply]

A slightly better reference is here [2]. It starts to get into technical detail like osmoconformers and osmoregulators. For example sharks have high concentrations of urea and TMAO so they are as "salty" (osmolar) inside as out, so they don't have to drink for water balance. Freshwater fish don't want to drink (though they might anyway - they don't want to eat fishhooks either) because like us the more they drink the more they have to pee. With very watery urine they can set things to rights when they do drink, or absorb water through their gills, but I don't know if their pee can be more watery than water, or if this implies they have to take in ions from food, sediment, etc. to compensate... Even this reference, which is far more complete, doesn't seem to directly answer the question of how "complete" the reabsorption of Na+ and Cl- is, or just how "dilute" the urine really is. Meanwhile the salt water fish actually do drink because they lose water through their gills and have to make up for that. So they take in salty water to their digestive tract, then pee a lesser amount very salty urine for a net intake of water minus salt. Wnt (talk) 15:11, 22 December 2018 (UTC)[reply]

Mercury and gallium

Mercury has a melting point of -39°C while the boiling point is 357°C; gallium has a melting point of 30°C and boiling point 2203°C. Is Mercury's low melting and boiling points are due to the same reasons? If so, how the boiling point of gallium so extremely high relative to its melting point? Planet Star 05:51, 20 December 2018 (UTC)[reply]

Gallium and mercury are not in the same column of the periodic table. 62.49.80.34 (talk) 09:49, 20 December 2018 (UTC)[reply]

More to the point than simply not being in the same column is what that means. Transition metals with unpaired electrons tend to have higher boiling points (there is some simple discussion of that here. Gallium has one unpaired electron in its 4p orbital, while mercury has zero unpaired electrons. These unpaired electrons allow for the formation of stronger interatomic forces with neighbors, leading to a higher boiling point for gallium. Certainly this also has an impact in the melting point differences as well, but other factors may also come into play there (crystal structure, perhaps?). --OuroborosCobra (talk) 14:33, 20 December 2018 (UTC)[reply]

Look at aluminum and boron above gallium, they have higher melting points despite each having one unpaired p electron. The melting points decrease from boron to gallium, but after that going further down the group the melting points increase through indium, thallium, and man-made element nihonium, but boiling points decrease continuously all the way down. How is it that melting points decrease over these first three elements then increase the last three but boiling points decrease all the way from top to bottom? And what makes gallium to have so low of a melting point that it melts on a person's hand but other elements in that group don't despite all having only one p electron? Could it be that presence of filled electrons in the 3d orbital influenced the melting point of gallium? Planet Star 02:58, 21 December 2018 (UTC)[reply]

According to the above cited gallium wikipedia article, the low melting point is due to Ga's tendency to form dimers, though it is not immediately clear why the other elements in its group do not form dimers.--Wikimedes (talk) 04:51, 21 December 2018 (UTC)[reply]

Quora has the answer about why gallium has the lowest melting point of the group despite it is in the middle of the periodic table column. Planet Star 04:34, 22 December 2018 (UTC)[reply]

Quora's answers don't look exceptionally convincing to me, though I'm not sure I'll do better. But it should be clear that this region of the periodic table has a lot of low-melting compounds: the components of galinstan, which (debatably) melts more readily than water, include indium and tin which are relatively easy to melt (hence tin soldiers, which also involve lead, just below tin). Cadmium and zinc and thallium also melt fairly readily compared to most metals. So we more or less have two boundaries - one, the edge of the transition metals where mercury (and Cd and Zn) has (according to our article) a "pseudo-noble-gas configuration", and the other being the diagonal line where things like aluminum and germanium start to act less like metals, beyond which we get into covalent chemistry that is either hard as diamond or airy as oxygen depending on how the numbers work out. Our article on thallium sort of references the inert pair effect, a relativistic effect on the lower rows of the periodic table - I have no idea if this has the impact of sort of shoving thallium to the left on the periodic table, i.e. with higher melting point, due to the status of the s electrons, while perhaps pulling low-melting lead and bismuth into this general category. But it should be clear that the post-transition metals have a general trend to melt at low temperatures; that article gives more discussion and a lovely figure I've placed at right. Wnt (talk) 15:44, 22 December 2018 (UTC)[reply]

How many pregnancies at one time?

At any one time, roughly what percentage of women in the world aged between 18 and 50 are pregnant? -- Jack of Oz ^{[pleasantries]} 07:48, 20 December 2018 (UTC)[reply]

Googling that subject, the following US information turns up, which isn't a full answer, but it's interesting.[3] ←Baseball Bugs ^{What's up, Doc?} carrots→ 11:52, 20 December 2018 (UTC)[reply]

Current rate of fertility (average number of children per woman) is current about 2.35 - which would mean each woman being pregnant for 21.15 months - which would probably need to be increased to allow for those pregnancies which do not result in a live birth. Over the 32 year span suggested (which may be too short, as in many parts of the world women start having children much younger than 18) that would mean the average woman spending just under 18% of the period pregnant. Logically, that would also mean that at any given time about 18% of them should be pregnant - perhaps 20% is you allow for the miscarriages, still births and abortions. Wymspen (talk) 13:01, 20 December 2018 (UTC)[reply]

I can't really see where the 18% comes from. 21.15/(32 x 12) = 5.5 %. Incidentally, I used a different method and found almost the exact same result (i.e. 5.5% excluding miscarriages and abortions). - Lindert (talk) 13:24, 20 December 2018 (UTC)[reply]

I didn't specify pregnancies that go full-term. -- Jack of Oz ^{[pleasantries]} 16:21, 20 December 2018 (UTC)[reply]

I know, that part is just easier to calculate. As far as miscarriages go, [this page] (I was not able to post the url, as the site is blacklisted, however it is a well researched article, titled "Making Sense of Miscarriage Statistics" and reviewed by an OB/GYN, with links to scientific studies) has some interesting statistics: 1) up to 75% of conceptions may end in misconception, although this also counts cases where implantation fails to take place, and many argue that pregnancy does not start until that point. After confirmed implantation, this number drops to 31% 2) After a woman discovers she's pregnant, the probability of a misconception is about 15%. Nearly all misconceptions occur in the first 12 weeks, children who survive this period have a 96-97% chance of surviving the pregnancy (naturally).

Based on the above, it's probably reasonable to say that the average lenght of a misconceived pregnancy is at most around 6 weeks. That would mean these pregnancies cumulatively account for +/- (31/69 x 6/40 =) 0.07 times the pregnancy duration of all succesful pregnancies. This would change 5.5% to (1.07 x 5.5%=) 5.9%

Stillbirths account for about 1% of pregnancies, so this is hardly statistically significant.

When it comes to abortion, statistics are undoubtedly incomplete, especially in countries where abortion is not legal/tolerated. That said, most abortion also take place in the first trimester. If 50% of pregnancies are aborted, at an average of 8 weeks, that would add (2 x 8/40 x 5.9 =) 1.2%, but it's probably less than that. This brings the total to 7.1 % (a rough estimate). - Lindert (talk) 17:26, 20 December 2018 (UTC)[reply]

Thanks, that's most interesting. -- Jack of Oz ^{[pleasantries]} 21:51, 21 December 2018 (UTC)[reply]

Besides not carrying to full term, you would also have to decrease the result by half a percent because of twin pregnancies. – b_jonas 23:26, 21 December 2018 (UTC)[reply]

Tuned response function.

I have interest in the equation that gives the amplitude of the current (or the voltage across the resistor) in a voltage exicted series RLC circuit but using only the parameters: Q and omega0,(Where Q is the quality factor of the complete circuit and omega0 is the undamped resonant radian frequency of the circuit). I need the equation to be a function of omega (the driving frequency),omega0 and Q only please. I have searched all over the web and all the books i can find. Yet I cannot find the formula for the amplitude against radian frequency. This is not homework but a personal interest. Its too hard for any homework and doesnt seem to be covered anywhere. Help!!!213.205.242.154 (talk) 23:45, 19 December 2018 (UTC)

Can someone ask that sparky spinner to look at this, He.she seems to be very clever.213.205.242.154 (talk) 23:47, 19 December 2018 (UTC) --213.205.242.154 (talk) 12:49, 20 December 2018 (UTC)[reply]

See RLC_circuit#Series RLC circuit that gives a 2nd order differential equation whose solution is a complex number because the current is not in phase with the applied voltage. At the resonant frequency V(t) = sin (W₀t) and I_R = V(t) Q / (L W₀).

At higher frequency than resonance W > W₀, I_R is smaller and lags the phase of V(t). At lower frequency than resonance W < W₀, I_R is smaller and leads the phase of V(t).

I_R is zero in these 3 cases:

V is not time-varying i.e zero or d.c.
V alternates at infinite frequency
R is infinite

RLC filter describes applications of this second-order tuned circuit as an electrical filter. DroneB (talk) 18:25, 20 De6cember 2018 (UTC)

thanks but I need an equation telling me the amplitude of the response at any frequency only in terms of the circuit wo and its Q.

I get this formula for amplitude of impedance of circuit as a function of input frequency ω. Assume a sinusoidal input. It also depends on resistance R. ${\sqrt {R^{2}+\left({\frac {\omega QR}{\omega _{0}}}-{\frac {\omega _{0}QR}{\omega }}\right)^{2}}}$ This ignores phase shift. Graeme Bartlett (talk) 11:15, 21 December 2018 (UTC)[reply]

Ah yes that's more useful even though it still contains R, i can get rid of that by changing to equation that gives the voltage across the R. In that case, R disappears and om left with an expression contain g only omega, omega0 and Q which was my goal. Thanks for helping me toward my goal.80.2.20.132 (talk) 22:24, 21 December 2018 (UTC)[reply]

You could just set R to 1. Then you can plot a graph. Using a logarithmic ω axis will make the plot symmetrical. Originally there were three parameters R L and C, but you only gave two (Q and ω₀) so there is still one needed. L and C can be written as a function of QR and ω₀. Then the formula I gave is just the impedance of resistor, inductor, and capacitor added, but since two are imaginary it is square root of the sum of squares. Graeme Bartlett (talk) 12:00, 22 December 2018 (UTC)[reply]

All that has been very helpful. Thank you.80.2.23.3 (talk) 00:55, 28 December 2018 (UTC)[reply]

Direct vision repair

What precisely is this in the context of heart surgery, and what should it contrasted to? Ericoides (talk) 18:00, 20 December 2018 (UTC)[reply]

See Cardiac_surgery#Types_of_cardiac_surgery, SURGERY BY DIRECT VISION IN THE OPEN HEART DURING HYPOTHERMIA and AORTIC STENOSIS—Surgical Treatment Under Direct Vision, Using the Heart-Lung Machine. DroneB (talk) 18:34, 20 December 2018 (UTC)[reply]

Ah, OK. The Hypothermia pdf contrasts it to "feel". Thanks, Ericoides (talk) 19:02, 20 December 2018 (UTC)[reply]

activity coefficient as a function of membrane potential

If my solution near a transmembrane protein is locally more depolarized or closer to 0 mV with respect to the outside of the cell, would ionic interactions between chains of a protein by stronger or weaker? I'm trying to figure out how to use the Debye–Hückel equation here. Any advice? In this case, I'm being asked to brainstorm some novel mechanistic bullshit about hERG (that's honestly the standards of this assignment) for a "literature review" (lol) -- essentially, stuff that would violate WP:SYN if it were on Wikipedia -- but I ideally want to brainstorm some evidence-based bullshit.

Anyway, in popular explanations of the lysine-rich and arginine-rich voltage-sensing domains of voltage-gated ion channels to undergrads, as the membrane potential depolarizes, the positive membrane depolarizes. As such he positive potential "repels" the lysine/arginine gates away from the cytosol, inducing conformational change that opens the channel.

Recently, someone pointed out to me that there are several issues with this explanation. The first is that the conformational change happens when the membrane potential is at less positive, but still negative potentials, e.g. -40 mV. The second is that the lysine/arginine residues in most of these channels (I am writing about the hERG channel, a cardiac potassium channel) are in a very hydrophobic region; they are already poorly stabilized. I pointed out that for the potassium channels, the positive voltage-sensing gates "retreat" into a negatively-charged pocket of acidic residues. Thus, in my original conception, negative membrane potential serves as a substitute for stabilization for the pocket and competes for the pocket. When the membrane potential gets closer to zero, retreating into the pocket is now the only energetically stable state left.

However, when I looked at some of the PDB structures of the hERG channel (5VA2), we can see that many of the basic residues are actually close to many acidic residues on a nearby neighboring domain (S5), implying tightly coupled electrostatic interactions. How would changing membrane potential (from -90 mV to 0 mV) affect electrostatic interactions in a hydrophobic transmembrane region of the cell membrane, especially in an ion channel protein? Yanping Nora Soong (talk) 20:34, 20 December 2018 (UTC)[reply]

Fascinating assignment. What the heck is this for, a graduate course in biophysics or something? Violating WP:SYN is always to be encouraged, except where you'll get in trouble for it.

I haven't done the necessary literature review to even start looking at this properly, and I assume there has to be some context of what resources should be familiar here. But arguing off the top of my head, then looking at the figure in the article for a number, I'm thinking that a phospholipid bilayer is, what, maybe 36 angstroms of insulator, going by our figure? (I assume a better figure can be researched, if relevant) And in that 36 angstroms we might blow through as much as -90 mV of potential difference. So, I mean, if you had that kind of field over a full centimeter, you'd have -9,000,000 volts difference between the electrodes. We can imagine then that these proteins feel a lot of force. A full positive or negative charge moving by a full angstrom would be comparable to electron volts: -90 meV * (1/36) = -2.5 x 10^-3 eV, if I didn't do something stupid, which is like 28 kelvin degrees' worth of energy or 0.25 kJ/mol. Not bond breaking, to be sure, but a significant factor yanking positive bits one way and negative bits the other. I can't begin to fathom a way to model the effect on the structure aside from, well, compiling some computer program to calculate protein secondary structure from scratch and then adding code pulling/pushing on the charges. I suppose someone must have done it though... Wnt (talk) 02:24, 22 December 2018 (UTC)[reply]

I measured inter-residue distances in pymol and used the idea of the field inducing torque on charges linked to levers and pulleys on several domains (since glycines and prolines break alpha helices and would serve as fulcra -- thanks Ramachandran!) So rotating the S4 domain say, 15 or so degrees, results in yes thousandths of eV, but breaking an entire ionic bond between arginine and aspartate is on the order of eV. If I conceive of the ionic bonds as dipoles which the electric field then does work on, I get less unfavorable energetics with free energy changes on fractions of kJ and maybe a whole kJ if I'm lucky in how I frame my moment arm, but mutations in certain regions discussed in the literature (ie changing the lysines or arginines to aspartates) discuss open or closed states becoming more favorable by factors of up to 12 kJ/mol. I feel like I'm missing something. I mean the relative dielectric constant of the cell membrane being 5 would strengthen ionic bonds as well as the external electric field -- or am I wrong? I'm going to look at the literature a little more. Yanping Nora Soong (talk) 22:58, 22 December 2018 (UTC)[reply]

I could be totally wrong about this, but my thought is that dielectric constant affects capacitance -- the higher the constant, the more charge can be put on the plates of the capacitor, i.e., the aqueous surfaces of the membrane. Here I don't really care how many charges are on either side of the membrane, and certainly don't know, so I'm not measuring Coulomb force and not caring how much it is shielded by local dipoles. I only know the potential difference and the charge, and want an answer in electron volts or some interconvertible unit.

I'm not sure how you got to thousands of eV. My thought is that overall the protein won't have a lot of total positive or negative charges on it (once you cancel out the + and - on a helix or domain), and it's only being pulled by mV, so I don't see how you crack an eV of energy to be had. Maybe if there's some sequence like KRKKRKKRKK in there... but my feeling is you should recheck you didn't drop a "milli" somewhere. Wnt (talk) 23:10, 22 December 2018 (UTC)[reply]

Thousandths. (Also my MCAT score is 515 so I'm not completely clueless about science.) Yanping Nora Soong (talk) 14:58, 23 December 2018 (UTC)[reply]

Oh, crap... and that's actually what you had there, I'm the one who's careless. And if your model is that precise, you probably meant the relative dielectric constant, i.e. an extra abundance of charge around the edges of the transmembrane domain. The problem is ... if you are being this careful, without obvious errors, anyone trying to answer the question would have to be as well read on the topic as you are, and then spend as much time on the problem as you have, to really be able to compete to answer it. I mean, I can search and find relevant papers like [4] but you've probably read the same or better, and taken a lot more time to think about the ideas involved, and assuming you are doing graduate work, you're vastly more likely to be in a position to actually run one of those lovely little simulations if you can think up a good hypothesis to test. Wnt (talk) 19:12, 23 December 2018 (UTC)[reply]

Okay, so here's the rub: the dielectric constant is 25-30 on the outside of the surface of proteins, ~15 for polar species at the edge of the cell membrane, and 2 in the bulk hydrophobic region. How do I calculate the local electric field within the protein due to polarization of an external electric field across a symmetric (for simplicity) cell membrane? Is there a huge difference in behavior of the field if there's symmetry across the interlayer axis? Yanping Nora Soong (talk) 14:31, 24 December 2018 (UTC)[reply]

I certainly don't know. My guess is it sounds like at this point you are not just doing a back-of-the-envelope computation but simulating the effect of the field on the protein in a way that I'd guess would need significant programming time and computer resources to work out properly. Once you get to that level I'd think there is no dielectric constant, just nuclei and clouds of electrons responding to and altering the electric field specific to their circumstances. Looking at [5] apparently the protein can respond to the field by changes in protonation as well as shape... which I suppose also feeds back on the secondary structure issues you're interested in? Wnt (talk) 19:00, 24 December 2018 (UTC)[reply]