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Why do raw prawns turn red after sitting in vinegar?
Yesterday morning I put some raw prawns (shell removed) into vinegar (more specific this one), and put them in fridge. When I came back at night all of them turned red, which looks cooked, except those not covered by vinegar.
Why do the raw prawns change colour?

Heat and Acids denature proteins.

To me, this should be more suitable for Chemisty-SE.

The way I understood lobsters was that their shells have molecules of many different colors, giving them a mottled green-brown appearance. When cooked, the non-red molecules are destroyed, leaving only the red color. Vinegar may do something similar to the non-red molecules by being acidic, maybe the pigments are like pH indicators. I know shrimp aren't lobsters, but are probably close enough.

The prawn has proteins (meat, muscles).
Beef is red because the muscle proteins in it have a molecule that binds oxygen called myoglobin (whereas our red blood cells have hemoglobin that allows them to carry oxygen better).
In fish you know the meat is white because there is no myoglobin present (in the muscle proteins), but another molecule that still binds oxygen but doesn't look red.
When you add heat or acids (vinegar) to the meat, the proteins get denatured, change their shape and their colour too. You get red shrimps by heating the proteins in them.
Hope this helps

@Him , please let me know if you understood my answer and if you found it helpful please mark the question as checked


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Do human somatic cells have 46 chromosomes all the time?
As far as I know, a human somatic cell has the cell cycle:
Interphase and Replication (mitosis). In interphase, the cell has 2 growth phases (G1 + G2) and a synthesis one (S phase being in between G1 and G2). After G2 ends, the cell will start entering mitosis.
Until the S phase, the cell has only 23 chromosomes right? And only after the DNA has duplicated the number of chromosomes doubles (to 46).
Please correct me if I am wrong? And will the cell in the G0 phase (after duplication and before G1) have 23 or 46 chromosomes?

Is this question also considering chromosome aberrations like Trisomy 21? https://en.wikipedia.org/wiki/Trisomy#Human_trisomy        Because that would break the question wouldn't it?

No, it refers to a general somatic cell division, assuming no chromosome aberrations.

Most of the time DNA is not present as chromosome as  such.

It is also interesting to note that cancer cells often have more than 46 chromosomes due to mutations

Humans have 46 chromosomes, 23 come from the mother and 23 come from the father. I think The easiest way to remember this is to think of the sex chromosomes--if you are male, you have one X chromosome from your mother and one Y chromosome from your father. During S-phase, the cell will make another copy of the X chromosome and the Y chromosome, so you will (technically) have 4 sex chromosomes in the cell before it divides, although you only have 2 UNIQUE sex chromosomes. After division, each daughter cell will have 2 sex chromosomes (one X and one Y). The same holds for all the other chromosomes.
So, to answer your question, yes, in G2 phase, you have double the amount of chromosomes in your cell (46*2 = 92), but you only have 46 UNIQUE chromosomes in your cell.
EDIT: As many people have pointed out, there are examples that do not conform to this rule. For example, both skeletal and liver cells can maintain a second copy of their chromosomes (making them 2n). This, however, is the exception. In my opinion, it's more valuable to know the basic rules that govern chromosomal number.
As an exercise I'd suggest to think about what it means that a cell has '2n' chromosomes? What happens if a germ cell obtains 2n chromosomes and passes that to a viable offspring? Is that offspring now '2n', or do they just have double the number of chromosomes as the parent? When does a duplicated chromosome become an it's own chromosome and not just a "duplicate copy"?

So let me see if I got this right. In interphase(till S-pase you have 46 chromosmes not 23) and after you duplicate the number you will have in G2 phase a bigger cell nucleus with 92 chromosmes not 46. And in mitosis each daughter cell will have 46 chroosmes.

yes, minus some semantic issues, I think you have got it. I doubt you'll ever find someone agreeing that you "have 92 chromosomes". You merely have 2 copies of 46 unique chromosomes. BTW, if this answer is satisfactory, please accept my answer by clicking the checkmark next to my post

Liver cells can be polyploid...

Thank you @Istbl; but saying that you have two pairs of 46 chromosomes at one time is similar to having 96 chromosomes, still in interphase but after S-phase . @biogirl : Liver cells can have multiple chromosomes because they have more than one nuclei or in one of the nuclei they have more than 46 chromosomes ?

As @biogirl said there are somatic cells in the body of a human that do not conform to the single nucleus, 2n genomic DNA. Skeletal muscle cells are polynuclear, so they have multiple nuclei with their own set of chromosomes.

@Vladimir In one nuclei, they have more than 46 chromosomes. See http://www.nature.com/nature/journal/v467/n7316/edsumm/e101007-09.html

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What concentration of sugar is optimal for baker's yeast fermentation?
I'd like to do an experiment on what concentration of sugar would be best for fastest yeast fermentation. I'm thinking to put some active dry yeast into different concentrations of sugar water and measure the fermentation activity after a certain amount of time (like 15 minutes or so). I know at a certain concentration the yeast will effectively die as the sugar solution becomes hypertonic, but I don't really know what range would be most effective right now. Any papers I've found are on yeast for alcohol.

When glucose and fructose are fermented separately, the uptake profiles indicate that both sugars are utilized at similar rates. However, when fermentations are conducted in media containing an equal concentration of glucose and fructose, glucose is utilized at approximately twice the rate of fructose. The preferential uptake of glucose also occurred when sucrose, which was first rapidly hydrolyzed into glucose and fructose by the action of the enzyme invertase, was employed as a substrate. Similar results were observed in the fermentation of brewer's wort and wort containing 30% sucrose and 30% glucose as adjuncts. In addition, the high levels of glucose in the wort exerted severe catabolite repression on maltose utilization in the Saccharo~ myces uvarum (carlsbergensis) brewing strain. Kinetic analysis of glucose and fructose uptake in Saccharomyces cerevisiae revealed a Km of 1.6 mM for glucose and 20 mM for fructose. Thus, the yeast strain has a higher affinity for glucose than fructose. Growth on glucose or fructose had no repressible effect on the uptake of either sugar. In addition, glucose inhibited fructose uptake by 60% and likewise fructose inhibited glucose uptake by 40%. These results indicate that glucose and fructose share the same membrane transport components

References
D'Amore, Tony, Inge Russell, and Graham G. Stewart. "Sugar utilization by yeast during fermentation." Journal of industrial microbiology 4.4 (1989): 315-323.
Thirwell, A.J. and R.L. Busby. 1982. Liquid sugar as a brewing adjunct. Inst. Brew. Proc. Conv. (Aust. and N.Z. Sect.) 17:77 84.
, http://www.123HelpMe.com/view.asp?id=121963

you can also add this citation because the other two cannot be found (but they seem valid because they are cited)

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How many ATP molecules are formed?
What is the number of ATP molecules formed during the photosynthetic processes which consume 8 molecules of $\text{H}_2\text{O}$ due to noncyclic electron transport and subsequent photophosphorylation?
Assume that quinone cycle facilitates the transfer 4H+ to the lumen of thylakoid membrane for every two electrons passing through electron transport system and one ATP is formed for every 3H+ moving down the proton gradient by the mediation of F0F1-ATPase
MY ATTEMPT: In the lumen, due to hydrolysis 16 H+ are formed. Due to quinone cycle 32 H+ are taken from stroma and transferred to the lumen and NADP+ takes 48 H+ so in stroma there is deficiency of 80 and there are only 48 H+ in the lumen. If I assume that if there were initially hundred H+ on both sides then after non cyclic transport inside the lumen there are 148 H+ and in stroma there are 20 H+ so to have equilibrium then there should be 84 H+ on both side so total 64 H+ should travel then I am getting 64/3 ATP which is wrong. Where I am going wrong?
Screenshot of question paper

I had attempted but iam getting 64/3 ATP as answer which is wrong.

I have no knowledge in photophosphorylation, but isn't ATP synthase creates 3 ATP for every 10 H+? and so maybe it's 64/3.3 ATP

Sorry for the poor quality of the image, but its just as a reference for my answer.
You have made the question too complex. From the figure, we find that 1 H2O gives 2 H+ and 2 e-. 2 e-, through quinone cycle, provide 4 H+ i.e. total 6 H+ which form 6/3 = 2 ATP i.e.
1 H2O => 6 H+ => 2 ATP
Multiply this equation by 8 and you get:
8 H2O => 48 H+ => 16 ATP
So, 16 ATP should be the answer.
Reference:
Pradeep's: A textbook of Biology for Class XI, 2015, Vol. II, page IV/106

Let us continue this discussion in chat.

I threw an answer up, but according to my calculation, and ignoring ATP from NADPH, 16 ATP should be correct. I upvoted this.

@JM97 is it okay now?

@aayush Srivastav in your answer you are not considering the fact that reduction of NADP to NADPH is causing a deficiency of protons which effects the proton gradient

@JM97 I don't think that considering H+ gradient from NADP+ reduction is important because your question does not say to consider it (like the 4 H+ from Q cycle). Also, considering it is not only out of the scope of question (as said by @Kendall) but also it will always give a value in fraction.

So if we do not consider the role of NADP then only 40 protons must travel so only 40/3 ATP ARE GENERATED

I don't know whether its important to maintain H+ equilibrium inside & outside thylakoid. I have only studied that each proton delivered inside thylakoid builds up a proton gradient, so each of those protons should be removed from inside.

Think about this one: you have 100 H+ inside & outside. By splitting 2 H2O, you get 12 H+ more inside and 10 H+ less outside i.e. 112 inside and 90 outside. So, to have equilibrium there should be 101 H+ on each side i.e. 11 H+ should travel out, giving 11/3 ATP, which is obviously wrong as 4 ATP are actually produced.

That is the real question which I was really asking, WHY DONT WE APPLY PRICIPLES OF CHEMISTRY IN THIS PROCESS?

Now if that is your real question, then I think you should edit your question or ask another question which clearly states your problem, and maybe you will get better answers in chemistry SE (I'm just guessing).

I think we can look at it like this

Your noncyclic pathway liberates 4H+ from two water molecules. We're doing that because to get one O2, you've got to split 2H2O. This happens in photosystem II. So in the next step (you know, really broadly) the electrons are passed to cytochrome b6-f complex, where each electron allows it to pump 2H+ into the thylakoid space. Since we got 4e- from splitting 2H2O, from that we can get 8H+. So net H+ = 8 + 4 or 12H+/O2, or we can rewrite it as 4ATP/O2.
For 2H2O you also got 2NADPH, which can also be used to get ATP. I think it's 3ATP/NADPH, so you get  (6+4)ATP/2H2O and repeat four times:
8H2O = 40ATP
Ignoring NADPH oxidation:
8H2O = 16ATP

please see the z scheme given at Wikipedia yours is different , according to Wikipedia only 2e are needed for reduction of NADP to NADPH but according to yours 4e are reducing 2NADP, SO WHICH ONE IS CORRECT?

The reduction of NADP to NADPH requires two electrons per NADPH produced, diagram. So, either representation is correct, but I utilize 4e > 2NADPH because the light reactions don't just release radical oxygen, they release O2, hence I always begin with 2H2O. @JM97

why don't you consider the fact that reduction of NADP to NADPH also contributes to proton gradient

Because you specified noncyclic electron transport.

reconsidering it again from starting then 8 water molecules gives 16 protons and 16e then first inside thylakoid lumen there is an increase of 16H+ second quinone cycle facilitates transfer of 32H+ from stroma to lumen so there is decrease of proton of stroma and increase of protons in lumen. Third 16e will reduce only 8NADP to 8NADPH so there is decrease of only 8e on stroma. Overall:inside lumen +48 , in stroma -40 so to get equilibrium as per chemistry principles only 44H+ should pass therefore we get 44/3 ATP.

do you want to say that in non cyclic electron transport there is no role of reduction of NADP to NADPH in creating proton gradient? If I follow your assumption then also 48 H+ all will not be transferred only  40H+ will transport so that in equilibrium on both sides so that there will be 8H+ on both sides

The NADP does play a role in the proton gradient, but it's outside the scope of the textbook question. To be specific, though, the concentration of protons in the lumen is higher than the stroma. So you would expect more H+ in the lumen than the stroma.

As we know that for 4 H²O molecules one non cyclic electron transport occur which gives us 4H+ ions ..this 4H+ ions gives 2 ATP so 8H+ ions will gives us 4 ATP....so
Ans= 4ATP molecules /8H+ ions

ACCORDING TO ME THE ANSWER IS 24 ATP THAT IS BECAUSE
8H+ IS PRODUCE DE TO PHOTOLYSIS.
16 H+ WILL BE TRANSFERRED INTO LUMEN.
SO 8 ATP WILL BE PRODUCE BY ATPASE {TOTAL 24 H+]
4 NADPH2 WILL PRODUCE 12 ATP
4 ATP WILL BE RELEASED FROM ELECTRON WHILE MOVING IN PS1.
SO TOTAL 8+12+4 = 24 ATP.

First: Please don't write in all caps as this is seen as yelling. And then it would be nice to show some references.


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Constant or variable number of chiasmata during recombination?
During recombination, is the number of chiasmata consistent for each gamete and are the chiasmata regions consistent within a single organism?

Welcome to Biology.SE. What do you mean by [..] are the chiasmata regions consistent within a single organism?? Do you mean "Are there regions of high recombination rate and other regions of low recombination rate?"?

Maybe I should rephrase the question, how many recombination events occur in a single germ cell? Is recombination really random or is recombination deterministic? If not, what influences randomness in recombination. Does the cell nucleus secrete some enzyme the randomly glues homologous chromosomes at random sites?

There is a large amount of variation in the distribution and number of chiasmata or crossing overs (COs). The total number of chiasmata per cell can vary within the same organism.
Where chiasmata occur along the chromosome is also not consistent cell to cell. Hotspots are 1-2kb regions that experience elevated recombination compared to neighboring genomic regions. A way to think of the distribution of COs is that ~80% of recombination happens in hotspots which comprise ~10% of the genome.
Recombination depends on the chromatin context, so chiasmata are rare near the centromere and positive interference results in chiasmata being spaced out along the chromosome. A good prediction of total chiasmata number per cell is at least 1 per chromosome. Some researchers argue that one CO per chromosome arm is more accurate.
Below is a link to a good article that expands on some of the ways recombination varies in humans.
http://www.nature.com/nrg/journal/v8/n1/abs/nrg1947.html

Thanks!!! Your answer along with the article turned out to be very helpful!!!


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What factors limit a shape and size of an organ
We all know that anatomy deals with structure of human system. I am interested in knowing the factors that determine the shape and size of each organs in the body. For example, we know that kidneys are bean shaped, stomach is a C- shaped organ.
During the developmental stage, like mitosis, and other cellular pathways, what factors sets a boundary for cell division, so that an organ takes its respective size and shape.

Externally applied forces, contact signaling and intracellular checkpoint systems work in concert and underly tissue development, but as far as I know it's very poorly understood! Interesting, related article

To be precise, size and shape of an organ involves mechanobiology. Thanks Kendall for the supportive article.

Every organ is going to be way to broad, each organ has its own pathways.


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Do any animals prepare food?
I was just wondering if any animals (are smart enough to) enhance the flavor of their food by, for example, mixing ingredients or seasoning or letting it cure in some ways. In the first episode of Life (on Netflix), they show a monkey who dries some oily nuts for a week but that's just to make it easier to break the shell. But do any animals prepare their food in any way similar to human cooking?

You should accurately define what you mean by any way similar to human cooking. For the moment it is unclear whether you example from episode of Life would be a good answer or not. Maybe you are interested in the use of heat in order to pre-digesting food before consumption.

Any form of changing it from the raw state

Then, the answer is in the question! Yes, monkeys for example dry nuts until the shell is easier to break. I am voting to close as unclear.

Though I know how this sounds, it's true: my dog would steal duck eggs and bury them here and there (I sometimes came across them in my garden), and when, according to her own mysterious tastes, the buried duck eggs had reached their peak of "fermentation", she'd unearth them and eat them in the garage (thereby offending our olfactory sensibilities.) That was an animal capable of shopping (there were other kinds of eggs) and preparing food. Short of busting a fire, that's not so different from cooking. I'm sure other animals do similar things.

Your question consists numerous questions: 1. Are animals smart enough to prepare foods? 2. Do animals prepare food? And by prepare, I think you're asking about A. cooking, B. seasoning, and/or C. modifying (e.g., like the monkey you describe).

My horse will "prepare" hay by dunking a mouthfull in her water trough, swishing it around a bit, and then eating it.

Bees make honey and then eat it.

And she's now taught the other horse to do the same thing - or he's copied her.

Some animals do like;
Ants prepare their meat not by heating but by marinating it with digestive enzymes to create a glistening protein slurry. With their hourglass figures, adult ants have such tiny waists that solid food can't pass through to their abdomens. Biologists already knew that the blob-shaped larvae predigest meat. Some scientists had suggested that the adults feed meat to the larvae and return later for some regurgitated protein slurry.
some animal chew and regurgitate for their young, monkeys have been observed washing potatoes, and other animals bury and allow food to rot before consumption.
But cooking is certainly unique to humans. There’s no other species that does it. There’s obvious reasons for that because we’re the only ones that can make fire which is a pre-requisite. In a away fire comes first and cooking becomes a process after it. It’s becoming clear that really cooking provides quite a number of advantages. Richard Wrangham from Harvard has been doing a lot of experiments looking at how cooking can change the nutritional value of food. What it seems is that the process of warming food up: in a sense denaturing it has a number of effects. One is that food is much more tender. That we know. If you eat a cooked carrot instead of a raw carrot it’s much easier. We can spend less time chewing, we can swallow it faster and we can digest it faster. It seems that it’s an extension of things we see in other animals. Animals who use techniques often in their stomachs to tenderise food seem to try and make it more easily absorbable. If we turn to the other question of when all this happened the real question is when do we first find evidence of fire? That seems to be about half a million years ago or so. We don’t find direct evidence of cooking then but we do see over the next 100,000 years or so the beginnings of things like burnt stone which suggests that meat is cooked. It’s probably goes quite a long way back in our evolutionary history and some people would argue it’s really a very major change in the way we are able to live and survive.
The second possible advantage for cooking is that it improves digestion. We’ve done a model study here particularly with Zhongquan Sui when she was here – she’s now at the University of Perdue. She found that yes, cooking does to a certain extent improve digestion. You only need cook something for a fraction of the time that you actually do in order for it to be digested properly. The cooking times that people adopt when they normally say this is cooked seem to reflect strong mechanical changes in the food. In other words, these are things that affect you perception of food texture and allow you actually to eat it very much easier, very much shorter eating times than you would do if they’re raw. That I think I would give as the essential answer at the moment.

References:

http://www.sciencenews.org/view/generic/id/6910/title/Ant_Iron_Chefs_Larvae_fix_dinner_but_dont_sneak_snacks
http://www.thenakedscientists.com/

It would be good if you cite an article from a scientific journal rather than a popular science site. These sites would have cited to the original study, if not then the blog cannot be considered trustworthy.


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ChIP-seq for histone modification not in agreement with RNA-seq for expression
I have ChIP-seq for H3K79me2 and H3K36me3 and RNA-seq data for treated and untreated samples. Those two histones mark active genes. Lets say, hypothetically, a peak caller finds differential sites at gene A for both of those histone modifications. However, when I run RNA-seq analysis tools (like edgeR or DESeq2) this gene is not marked as differentially expressed (well it has a FDR value of >0.05).
There might be several technical reasons for RNA-seq analysis methods not to find this gene as differentially expressed because

it is really not differentially expressed
or it is not differentially expressed enough for the tools to detect it

However, I am more interested in the biological aspect. If two histones mark gene A as active in treated samples, one would expect that it would be expressed in RNA-seq. What mechanism could lead to the result that however the genes is marked as active by histones, it will not be differentially expressed in RNA-seq?

Your statement that those two post-translational modifications of histones mark active genes is not supported by your experimental evidence (in every case). Perhaps a more accurate statement would be something like "often mark active genes" or "are typically associated with active genes?"

It is difficult to draw any conclusion without further experimentation. There may be many other factors that prevent the expression of the gene including factors like post-transcriptional regulators. Some histone modifications like the ones you are mentioning are also a bit dicey and there can be bivalent modifications too. However, if there is a strong reason to speculate that these marks are associated with repression then you can try these:

You can verify how many genes are there whose expression is not in accordance with the histone marks. If this for a small number of genes then you can investigate them separately. Use a statistical test to verify the significance of your correlation between histone mark and expression. If the discordance is widespread, then I would suggest you to repeat the RNAseq (easier and cheaper than repeating ChIP-seq). BTW, how were the read qualities? Depending on that you can decide which experiment needs repetition. Repetition will help you to know if the observation is indeed correct (a biological replicate); possibly you may end also up concluding that these marks are actually dicey.
If a few genes show discordance, then just do Real-Time PCR for these genes. RT-PCR is more sensitive and another technique of quantification is anyways good. Use a few concordant genes also, in the experiment.

Which statistical test are you talking about? I have about 40 genes that are not supported by RNAseq, also I have genes that are supported by RNAseq but not by those histones. Sp, I have king of venn diagram of number of genes RNAseq vs active histone modification.

@Tonja 40 genes discordant: how many agree with the RNAseq? I am not sure about which statistical test to use here. Basically you have to see what is the probability of RNAseq not agreeing with Histone modification.

ChIP-seq isn't perfect. Even between technical replicates, you get a fair amount of variation, especially for broad marks like those you're using. It's rather uncommon to see people use H3K79me2 and H3K36me3 to determine if a gene's expressed or not. Using H3K4me3 and H3K27ac or H3ac is a more common method of marking promoters of transcribed genes.
40 genes discordant wouldn't worry me much unless you really have low reads for either set of sequencing experiments. @WYSIWYG's suggestion to use RT-PCR to verify your RNA-seq data for a few genes is a good one.

I checked the promoter histones as well and I have also some genes that are not supported by RNA-seq. I know I could validate the results but that is encode data. So, I would like just to know the theoretical reason for that.

What is encode data? The RNA-seq?

@JaredAndrews I think Tonja is referring to the TFBS ChIP-seqs..

@WYSIWYG If that's the case, I'd suggest looking at different marks, as they certainly have them if they also have H3K79me2 and H3K36me3. The epigenome is complex, flexible, and variant. As such, you should look at the pieces as a whole whenever possible.


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What type of Interspecific relationship does this graph depict?
The other day in class, our AP Biology teacher presented us with the following graph and asked us to determine which of the following interspecific relationships it represents:

(A) commensalism
(B) predation
(C) mutualism
(D) competition
(E) parasitism.

She explained to us that since there is not enough information in the graph to explain why species A suddenly drops off after time "x" and species B suddenly rises after time "x", the best answer out of the choices is (A) commensalism.
But to me, that seems to be a faulty line of thinking. After all, my teacher arrived at that answer only under the assumption that the graph/question itself is good to begin with. I tried finding graphs of commensalism using Google Images, and I actually found the original source of the graph that our teacher presented us.
Now, the question on Regentsprep actually gives us a point of reference and states that we are considering two herbivores in a grassland environment, so in my opinion it's reasonable to conclude that the relationship between species A and B is competition.
The problem is that my teacher never gave us a frame of reference: she just told us to identify the relationship in this graph without any "backstory". So what are your thoughts: is it possible to conclude that "A" is the best answer choice if we are not given any information aside from the graph?

I agree that the question as your teacher presented it is very unclear, and even the question as it is presented in the original source is still not a great  question. I would say that commensalism is tied for the worst answer with mutualism, as with commensalism you would expect one population to go up while the other remains the same.

Competition makes the most sense, IMO.
For every other type of relationship, A and B should be dependent and roughly proportional: If parasites kill their host, they die in proportion. If predators eat all their prey, they drop in numbers due to food scarcity until food repopulates, the graph would almost look like a double helix. Mutual/commensals also share a dependent and proportional graph. The inverse proportionality of the sample chart, to me, gives away competition: one organism is flat out winning against another through some method.

Yeah, I agree with why it's competition. That explanation makes so much more sense to me than my teacher's reasoning. EDIT: also, I looked online for competition graphs just now and found a reputable source with competition graphs (bottom of the page) that are very similar in behavior to the graph my teacher showed us: both are steady up until time "x", after which one takes over the other.


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In cell division, are daughter cells identical?
I understand that after a cell replicates, there will be two daughter cells instead of one. But wouldn't one of them be the old cell that created the second one?
The old cell having gone through G0, interphase, and mitosis and after dividing wouldn't it be somewhat different from the new cell (the old cell's telomeres having shortened, having already undergone the cell cycle and so on)?

There is one large cell with the duplicated chromosomes that then is split down the middle to make the two daughter cells. Its like taking a slice of bread and ripping it in two, to a degree.

Here is a picture to expand on my comment better. From wikipedia:

In the above diagram, the red chromosome represents one homolog while the blue chromosome represents the other homolog in the pair. After replication in Interphase, you have two homologs, each consisting of duplicated sister chromatids. You can see how we have the one enlarged cell in Mitosis be cleaved into two daughter(diploid) cells.

Actually, the answer is not obvious. @RoSiv gives the textbook case of symmetric cell division, where the two new cells can indeed be considered identical, and this is valid in many cases. But there are also cases of asymmetric cell division, where the "mother" and "daughter" cell are clearly different.
In asymmetric cell division, the parent cell is polarized already before division by gathering certain proteins at one end of the plasma membrane. Therefore, during replication one of the the two new cells obtains certain proteins while the other does not, which makes them distinct. See this article for details.
Asymmetric cell division occurs among other things during embryo development, where it is important to establish body plan asymmetry. It is also the basis for the stem cell concept: stem cells are thought to divide asymmetrically so that the "mother cell" remains a stem cell (stays immortal) while the "daughter cell" becomes mortal and differentiates.

I see. Thank you for further explanations and detailing.


text:
Simple Mendelian sex-linked chromosome fruit fly question
From my molecular biology textbook:

When red-eyed flies (dominant) were mated with white-eyed flies (recessive), most, but not all, of the F1 progeny were red eyed. Furthermore, when the red-eyed males of the F1 generation were mated with their red-eyed sisters, they produced about one-quarter white-eyed males, but no white-eyed females. In other words, the eye color phenotype was sex-linked.

First, wouldn't recombination scramble the eye color alleles among the sex chromosomes?
Secondly, even without recombination, the F1 generation will be heterozygous and all F1 flies will have white and red alleles. Some F1 males will have a white X and some F1 females will have a white X, in which case it should be possibly to have an F2 white fly.
Can someone explain this?

I think the text is slightly misleading. But before going into the detail genetics of eye color let's answer your first question.
First, wouldn't recombination scramble the eye color alleles among the sex chromosomes?
Two X chromosomes do recombine but one X and one Y chromosome do not recombine (to the exception of the so-called pseudo'autosomal region).
I am not sure why recombination would change anything here. I think the eye color of flies in determined by a single locus.
How is this possible?
Let's go through the text slowly because I think the text is a little misleading. I am suggesting the only scenario I could think of that would explain the observed pattern. Note that this scenario also depends on the interpretation of the text and I would need more info on the experience to make sure my interpretation is right.

When he mated red-eyed flies (dominant) with white-eyed flies (recessive), most, but not all, of the F1 progeny were red eyed.

Actually, only males are sometimes white eyed here. And to be more accurate, if sex was uncorrelated form eye color in the parental lines, then half of the males should have white-eyed.
What was happening here is that red is dominant, white is recessive and the locus for eye color exists only on the X chromosome. As a consequence, all F1 females are now heterozygous (and display the dominant red-eyed phenotype) and all males are hemizygous, half of them having the red-eyed phenotype while the other half have the white-eyed phenotype.

Furthermore, when Morgan mated the red-eyed males of the F1 generation with their red-eyed sisters, [..]

Let's forget about the white-eyed males from above, we will just cross the red-eyed males with their sisters.

they produced about one-quarter white-eyed males, but no white-eyed females.

By they produced about one-quarter white-eyed males what is meant is that a quarter of all individuals were white-eyed males. That is half of the males were white-eyed. The other three quarters of the individuals were red-eyed.
The males received their X chromosome from their mother only (as they can only receive the Y chromosome from their father). Half of the males have therefore received the white-eyed allele while the other half have recevied the red-eyed allele. For the females, they all received from their mother either a white-eyed allele from their mother or a red-eyed allele. As we selected only father that had red eyes in the previous step, all father carry exclusively the red-eyed allele. As a consequence, all females receive a red-eyed allele from their father. Therefore half the females are homozygous red-eyed or heterozygous. In any case, all females are red-eyed.

Wow, that was really clear and easy to understand! Just one question: doesn't crossing over occur between X-chromosomes in females?

Yes, crossing over occurs between X-Chromosomes in females.

In the first cross, if we talk of a single mating pair of flies, one parent is red eyed, one is white eyed. The ambiguity about sex leaves two options, 1) red eyed male crossed to white eyed female, 2) white eyed male cross to red eyed female. If it were the former (red eye male) then the mother must have two X chromosomes both of which have do not carry the dominant red eye allele, and all daughters would be red eyed and all sons white eyed (their X always comes from the mother).
If it were the latter (white eye male) that would leave two further possibilities, mothers are homo- or heterozygous for the dominant red eye allele. If the mother is homozygous for the red allele then all offspring should have red eyes, all will receive an X that carries the dominant red eye allele.  If the mother is heterozygous then half of the offspring will be red eyed and half white eyed, because both of the fathers sex chromosomes do not have the red allele (the X and the Y) and only one of the mothers carries it.
In the exercise, you have some white eyed flies in the first offspring, but not all, and most are red eyed. Given the wording of the question (that red eyed flies were mated to white eyed flies, and not a red eyed fly was mated to a white eyed fly) I would suggest that the first generation was produced by crossing multiple red eyed females to multiple white eyed males, and that the females were a mixture of hetero- and homozygotes. Therefore some of the offspring were white eyed. The point being, with the first cross it is impossible to tell whether the females used were each hetero- or homozygotes, you can't deduce their genotype.

In the second cross, both parents have red eyes. What this tells you is that the fathers have an X chromosome with the dominant red eye allele, and the mothers have at least one X with the dominant red eye allele. They produce sons and daughters, among which, 1/4 are white eyed males, and none are white eyed females. The implied/logical further outcome is then that 1/4 of the offspring were red eyed males, and 1/2 (2/4) were red eyed females. In the second cross the mothers must be heterozygous (have just one red eye carrying X chromosome). The key difference here is that you can establish the genotype of the parental flies.
The cross below shows this. The sons, who inherit an X from their mother and Y from their father, are white eyed half of the time because there is a 50% chance of inheriting a red eye X from the mother and 0% chance of inheriting a red eye X from the father. Females are always red eyed because the X comes from the father, and he has only one X (which carries the red eyed allele). Half of the daughters will be heterozygous, and half homozygous, for the red eye allele, but it is dominant so all have the red eye phenotype.

I assume that recombination cannot occur in males (D. melanogaster has no recombination in males) and therefore has nothing to do with the solution, but that depends which species of fruit fly the example refers to - could you clarify? It appears this is this book, and therefore it is D. melanogaster.

Wow, thanks so much for the thorough and illustrated answer.

The book is Molecular Biology 5th Edition by Robert Weaver, ISBN 0073525324, chapter 1.1 page 3.

And yes, the book says on the same page that this experiment involved Drosophila melanogaster.

@rg255 On a side note: What programming language or software did you use to make these diagrams? +1 Thanks

Im ashamed to admit I just used PowerPoint! I could have done it with R but didn't bother - I had power pint open to work on a presentation I'm giving next week anyway @remi.b

Ha ha no shame. I was curious as I never made these kind of diagrams and would not really know how to do with R (except eventually with iGraph).


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What is the maximum altitude where humans can survive?
What is the highest altitude that a human can survive without being pressurized?  Let's assume that oxygen and heat/insulation are not the limiting factors?
Why I asked.

An interesting question. If you've done any of your own research into this could you share it. I wouldn't want to see this question get closed for "homework" reasons.

The sequence in which the surviving crewman in 2001: A Space Odyssey space-walks from the pod to the airlock of the spaceship, manually opening the outside door, entering, closing the door, and pressurizing the airlock raised the question: Is it possible?
The answer was: Yes, barely. A few more seconds and the character, if actually experiencing the transfer as acted, would have lost consciousness.

Air pressure drops by half for each three miles of altitude.

Short answer
Between 62,000 and 63,500 feet (18,900 and 19,350 meters) blood begins to boil at body temperature. This altitude, referred to as the Armstrong limit, is generally considered to be the absolute limit compatible with life. At this point, humans cannot survive without pressurization measures.
Background
Atmospheric pressure drops at higher altitudes. As you go higher into the atmosphere, gases start to expand due to the lower atmospheric pressure. This is what causes your ears to pop when driving up in the mountains or when an airjet lifts off. At altitudes higher than 10,000 feet (3.5 km) people unaccustomed to the reduced atmospheric pressure may become short of breath or experience dizziness. At altitudes above 20,000 ft (6.1 km) supplemental oxygen is required to support life and at 34,000 ft (10.4 km), 100% oxygen is necessary to equal the partial pressure of oxygen in sea level air. Above 40,000 ft (12.2 km), 100% oxygen must be administered pressurized (NASA).
Altitudes above 50,000 ft (15.2 km) are considered near-space and man requires a pressurized suit to be safe. At 55,000 ft (16.8 km), atmospheric pressure is so low that water vapor in the body starts to boil at ambient temperatures, causing the skin to inflate. At 63,000 ft (19.2 km) blood at normal body temperature (98 F / 37oC) starts to boil. At altitudes above 65,000 ft atmospheric pressure approaches that of space, i.e., a vacuum (NASA).
Between 62,000 and 63,500 ft (18.9 - 19.4 km), blood at body temperature (37oC) begins to boil. This is the point beyond which humans absolutely cannot survive in an unpressurized environment. This point is named the Armstrong limit, after Harry George Armstrong, who founded the U.S. Air Force’s Department of Space Medicine in 1947. On Earth, the Armstrong limit is somewhere between 62,000 and 63,500 feet (18.9 and 19,4 km), or about 12 miles.
As a reference, the highest point on earth is the peak of the Mt. Everest at 29,000 ft (8,8 km).
Above the Armstrong limit, water in the soft tissues of your body vaporizes, causing gross swelling, though the tight seal of your skin would prevent you from actually bursting apart. Your eyes, likewise, would refrain from exploding, but continued escape of gas and water vapor leads to rapid cooling of the mouth and airways (Goslin, 2008).
Water and dissolved gas in the blood forms bubbles in the major veins, which travel throughout the circulatory system and block blood flow. After about one minute circulation effectively stops. The lack of oxygen to the brain renders you unconscious in less than 15 seconds, eventually killing you. When the pressure gets very low there is just not enough oxygen, which is thought to be the first and most important concern (Goslin, 2008).
Note that under conditions of near-vacuum, death is not instantaneous. At one three-hundred-eightieth of atmospheric pressure at sea level, dogs were shown to survive up to 90 seconds. During their exposure they weren't in a great state (the dogs were bloated, ice covered their tongue, and many more horrible facts to be read in (Goslin, 2008)). Monkeys were even shown to cope for 3.5 minutes under these conditions.
Reference
- Goslin, Sci Am February 2008

Excellent answer, comprehensive and well-sourced.


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Why don't antibodies generally bind to food and drugs?
Are these excluded thru central tolerance?  What if you ingested something with a unique molecular structure that you hadn't ingested before?

Maybe my answer to your other question answers somehow this one?

B cells containing receptors against food's or drug's molecular structures may exists. However if the naive B cell binds the molecule without a second signal coming simultaneously from CD4 Tfh cells (linked recognition) the B cells recognizing this antigen will become anergic or will die by apoptosis. This occurs with antigens to which we are chronically exposed, like food.
Reference: Janeway's Immunobiology, 8th edition. Chapter 8 (The development and survival of lymphocytes). In particular section 8-26, p320-321.

I would add that many food allergy structures are recognized in hypersensitivity reactions because they oddly enough resemble helminthic peptide antigens. Furthermore, lipid/CHO-type antigens which aren't peptide can't be handled in the same way as peptide antigens, hence we have gamma-delta T cells and surface receptors like TLR-4 that have to do this job. Since antibodies get generated generally due to peptide antigen presentation by APCs, there's a lack of robust antibody production to non-peptide antigens.

I’m not an immunologist, but I think it fair to say that antibodies may well be able to bind to certain food or drugs. However there are several reasons that this may not cause their removal or trigger an immune response. These include:

Location. If the food is in the stomach I imagine that the digestive environment favours breakdown to sugars, amino acids etc and exposure to antibodies is minimal.
Chemistry. Certain substances such as fat, nucleic acids, are weakly antigenic.
Size. Small molecules are not immunogenic unless attached to a larger one.

Why are small molecules not immunogenic?
I think this is probably the key point here, so I'll expand on it a little in general terms.
If an antibody binds a foreign protein, this is large enough that other antibody molecules can bind it too. And the antibody is bivalent, so it may bind a second foreign protein. This leads to a cross-linked matrix that can be recognized by macrophages and engulfed. If a small molecule binds to an antibody this cannot happen and the molecule may well diffuse off again.
As well as binding those soluble antibodies (IgG) already present, antigens cause the antibody-producing cells (B-cells) to divide to produce more of the particular antibody. This happens because the B-cells have a membrane bound variant (IgM) with the same specificity on their surface. When a molecule binds to two of these membrane antibodies a signal is triggered to the cell. This can happen with a large molecule, but a small molecule can bind only one IgM on the surface and this expansion of antibody-producing cells will not happen.
This size requirement for immunogenicity can be understood in the context of the size of the bacteria and viruses which one presumes the immune system evolved to combat.
More Information
As I mentioned, I deliberately answering the question in general terms. If you wish to know more detail (and about cellular immunity) the following on-line book chapters (although a little out of date) may be of interest.

Berg et al. Ch. 33
Alberts et al. Ch. 24

hypersensitivity (allergy) is an autoimmune response, and it is caused by ingestion (peanuts) or inhalation (pollen)

@FoldedChromatin — You should critcize the question, not my answer, as you are criticizing its premise. And perhaps you would care to elaborate and explain how this occurs in molecular terms and why most people do not suffer from such allergies. As I said, I'm no immunologist, but your comments have contributed nothing to my understanding. If, as I assume, you have down-voted my question, you could point out to everyone which particular parts of my answer are incorrect.


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Is DNA transcription inactive during mitosis ie no proteins are made? How does the cell survive?
I've read that during mitosis, DNA exists as heterochromatin, a form that is unable to be accessed for transcription.
Does this mean no new proteins are made during what can be 20% of a cell's life?  How does the cell survive?

To make proteins you  just need mRNAs. And most of the synthesis (transcription) is done in the S-phase, including the mRNAs required for later stages.

Consider two values, the average half-life of an mRNA, and the average half-life of a protein.  A cell may not need to transcribe it's essential genes non-stop.


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How to search for a protein sequence of an specific taxa in NCBI?
My example: Sequence of HBB (Hemoglobin Beta) in Fishes.
How can I do a good search to obtain above sequences in NCBI? (I know hot to download them into a FASTA file)

Try going to http://www.ncbi.nlm.nih.gov/protein and searching for ' (hemoglobin beta[Protein Name] OR (hemoglobin[All Fields] AND beta[All Fields])) AND "Fish"[ORGN] '.
Search filters can be changed with the panel to the left of the search results.


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How can I find the most studied SNP of a gene?
How can I find the most studied SNP of a gene and diseases that the most studied SNP has been related to?
I search in dbSNP database but I can't find it.
What is the process?

I have selected the BRCA1 gene.

Can you provide the gene of interest?
You can usually search Entrez for the gene symbol in the dbSNP database, which will pull up a NCBI Entrez page for SNPs located in the gene and will also give an indication of their clinical significance (benign or otherwise).
Alternatively, if you have a specific disease in mind, you can check more specific databases for a SNP, such as COSMIC for cancer-related SNPs.
Lastly, merely Googling the gene of interest (for example, "EZH2 SNP") in Google Scholar will bring up articles that present SNPs of interest.
Some combination of those methods should yield what you want, but I might be able to provide a better answer with more info.


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At which end does polymerization of microtubules occur?
My book says that polymerization and depolumerization of microtubules occurs on the + end however, I've found a note that says that depolymerization occurs on the - end. I need help please :) thank you

From my Campbell's Biology textbook (Ninth Edition):

Because of the orientation of tubulin dimers, the two ends of a microtubule are slightly different. One end can accumulate or release tubulin dimers at a much higher rate than the other, thus growing and shrinking significantly during cellular activities. (This is called the "plus end", not because it can only add tubulin proteins but because it's the end where both "on" and "off" rates are much higher.)

So in summary: polymerization and depolymerization occur on both the + and - ends. The "+" and "-" merely designate the speed/efficiency of polymerization and depolymerization.

I think this is also a good reference that explains how proteins assist in MT nucleation and constrain the direction of MT polymerization.

Practically speaking, microtubule nucleation occurs at the plus end. This is well established in vivo and in vitro. Notably, the minus ends within most cells are "capped", so dynamics do not occur (think of the centrosome).
However, recently there is a report of MT polymerization at the minus end in a cell; this is far from the consensus among professionals at the moment, and requires much more study. Nonetheless, the rate of any growth and shrinkage at the minus end is significantly slower than the plus end, as stated above.
So for a test question, plus end is the site of polymerization.

Both polymerization and shrinkage of microtubules take place at plus end and minus end is anchored within MTOC's..hence is the dynamic end


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The Semi-Conservative Model of DNA Replication: Question
My Campbell's Biology textbook contains the following diagram related to the semi-conservative model of DNA replication proposed by Watson and Crick. I have highlighted where my confusion arises in red:

So, I understand what goes on in the first replication--that's pretty much straightforward. However, what I don't understand is why in the second replication, the light blue strand isn't paired with a dark blue strand. After all, aren't the light blue strand and dark blue strand complementary, per the results of the first replication?
After the light blue and dark blue DNA strands separate to serve as templates in Replication #2, why don't we see two dark+light double-helices?

A small note for the future. Ending the title by : question is really not informative as the post will obviously contain a question.

I believe the reason you are having trouble understanding the concept is due to a poor usage of colors in the diagram. Don't focus on the colors, but on the concept. It's the same for both replication events. Each strand of a double helix is used as a template to make a new complimentary strand, giving rise to two new DNA helices from the original. In each new double helix, one strand should match that of the double helix it came from before, as it was the template and is the same strand. While the other strand, the complimentary strand, has been newly synthesized to match the complimentary strand. In this diagram, every newly synthesized strand is shown in light blue.
Start by looking at the first double-helix of DNA (navy blue). During replication, the DNA is unwound and each navy strand is used as a template to create the newly synthesized (complimentary) strand, which is shown in light blue. In this first all navy double helix, you have two navy strands that are each used as a template strand and new complimentary strands (light blue) are synthesized to match . This produces the second two helices (navy and light blue). They each have one strand from the original helix (navy), and one new strand (light blue).
The process is exactly the same for the second replication event, only, a new color was not introduced to show a newly synthesized strand, so one cannot differentiate between the original (template) strand and the new complimentary strand. Both are pictured as light blue.
For the final replication event, only focus on the top navy and light blue helix first. Just as before, the two strands are separated from each other and are both used as templates to create a new strand. The navy blue strand is used as a template for a new light blue strand to be transcribed. This produces the top-most helix, where the original strand is shown as navy and the newly transcribed (complimentary) strand is light blue.
Just as the navy strand was a template, the light blue strand from the bicolored helix is used as a template strand as well to create the second all light blue helix. This light blue template strand is used to synthesize a new light blue strand, creating the all light blue helix. One of the strands is the original light blue one from the navy and light blue helix, while the other strand is the newly synthesized one.
It might've been easier if they had used a different color for the final round of replication, like orange. Then to show the newly synthesized strands in the final round, the 1st and 4th double helix would consist of a navy (original) strand and an orange (newly synthesized) strand. The second and third double helices would then have each had a light blue (original) strand and an orange (newly synthesized) strand. This would have made it easier to understand where each strand originally came from.

Thanks for your thorough response. But in the second round of replication, why isn't the light blue template strand's complementary strand just another navy blue strand? I would imagine the base pairs of the complementary strand would be the same as those of the navy blue strand. Does it have to do with 5' and 3' arrangements?

Ah, I see. This diagram is set up so that you can easily follow which strand was the original template strand that is from the original helix. The navy color is not signifying whether a strand is 5'-3' or 3'-5'. If you want to think about it like that then, yes, one of the light blue strands for the final helix, IS exactly the same as a navy strand, so it could be shown as navy.  But really, navy is used to signify the original strands of the first helix so you can follow the movement of original strands started out with. Try drawing your own replication event of a double helix to help. :)

Yep, I took your advice and drew a diagram using 5'___3' and the parental DNA does in fact end up in all of the "daughter" strands. Not sure why the book writers didn't use the navy after replication #2. EDIT: nevermind, I see what you mean. So they just represented the original parent strands with navy rather than using navy for any strand that matches the parental sequence, right?

Exactly!! In the EDIT, you have it! That's all they were trying to do.


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How was gene therapy able to cure diseases through the transformation of actively dividing cells?
I thought that gene therapy, when performed on target cells that regenerate themselves constantly, can be effective for a limited time only. I.e., the effect gradually wears off after a while, depending on how fast the target cells renew themselves. In other words, the patient needs to continue receiving gene therapy every so often to keep getting desired effect.
On the other hand, when targeting cells that do not divide/renew, such as an eye, gene therapy can and actually did cure blindness once and for all, and the patient no longer needs gene therapy injections.
So what about the other diseases that gene therapy allegedly cured? I think they include blood disorders like hemophilia and some serious immune system disorders? I thought they mainly involved cells that do regenerate, like bone marrow and liver?

Short answer
Both dividing and non-dividing cells can permanently be modified with gene therapy by using vectors that incorporate their payload in the host's chromosomal DNA
Background
Vector-based gene therapy can be applied by using vectors, such as modified retroviruses or adenoviruses, that incorporate their payload in the host's chromosome.
Whether the transformed cells are replicating yes or no, the parent as well as any possible daughter cells will carry the new DNA. For example, immunodeficiency has been treated successfully with gene therapy, but the first trials resulted in blood-bone cancers, because the stem cells started to proliferate uncontrollably, hence dividing cells definitely can be permanently transformed. In fact, when germ line cells are transformed in this way, even the offspring of the host will inherit it.
Sources
- American Medical Association
- Learn Genetics
- Utah University

I would also add that "gene therapy" is an increasingly broad term. Viral vectors are one way of altering patient cells, but there are other ways, depending on the cell type. Current gene therapies to treat soft tissue tumors like leukemia include removing patient stem cells, altering them and finally re-inserting them. So far, no cancers have resulted from that cure. Here's a description of the technique: http://www.discoverymedicine.com/Runzhe-Chen/2015/10/cd19-chimeric-antigen-receptor-t-cell-therapy-for-the-treatment-of-b-cell-lineage-acute-lymphoblastic-leukemia-3/

@Forest - the means to obtain the target cells differ, the final method to get the DNA into them is based on vectors afaik

Ah, I see now. Good point.


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Assays to determine competitive binding versus non-competitive
I'm looking for both simple and complex assays or technologies than can be used to determine if two competing molecules are competitive or noncompetitive.  I figure xray crystallography is a clear one, but say I can't get crystals to form.  What other techs or assays could I use to determine this?
It's a protein-protein interaction with a small molecule competing with one of the proteins.  This particular set has not been done before. All interactants are purified.

Can you please describe your setup in a little more detail? Are your ligands small molecules, proteins, or something else? Do you have the purified target protein? Is it a receptor or some other kind of enzyme? Has anyone worked with it in the literature before?

If you've got an enzyme and a good activity assay and you want to test if the molecule is an inhibitor, try building Lineweaver-Burke Plots. It can take a lot of work though, because each plotted point is a reaction rate, and determining each reaction rate can take significant effort. Because the plots are reciprocal measures, small errors are easily magnified, so you have to be very very careful to minimize any error whatsoever. But if done well you can determine competitive, non-competitive, or mixed inhibition.

http://www.ncbi.nlm.nih.gov/books/NBK92001/ has some good guidelines for mechanism of action studies.

Now that I see you are looking at protein-protein interaction, would fluorescence anisotrpy help? If you labeled one protein with a fluorophore, you should be able to observe a change in tumble rate as the other protein binds or falls off.

I'll have to edit the question, I'm looking for label free methods.  Sorry about that!


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What is the purpose of requiring two separate binding systems for the antibody response?
I've read that in most cases, B-cell activation requires helper T-cells.  This requires antigen binding by both antibodies and T-cell receptors, using two different antigen-binding proteins, recognizing two different epitopes.
Why is this so complex?  Would the overall antibody system still work if B-cells required only antibody/B-cell receptor binding?

What do you mean by "antibody mechanism"?

Immunity can be a double edge sword. It is aimed to detect and destroy invading pathogens but it can also target the self, as exemplified in autoimmune diseases. To avoid this, many failsafe mechanisms exists.
For example, imagine an innocuous foreign antigen gets into your body and reaches the lymph node. There it may find a naive B cell that has a receptor that recognizes it. Without a double check mechanism this B cell will mature and begin to produce antibodies against this antigen, triggering an immune response that is not beneficial (e.g. in allergy) and is not necessary (the antigen is harmless).
Because of this, activation of the adaptive immune system requires usually two signals. In the context of B cells the other signal usually comes from effector CD4 T cells (of the type T follicular helper Tfh) and is called linked recognition, because both the naive B cell and the Tfh cell have to be specific for the same antigen.
So a possible scenario may look something like this: a pathogen's antigen is taken by DCs. If there is an ongoing inflammatory response this DCs will get fully activated (a process called licensing) and migrate to the lymph nodes. There they present the antigen to naive T cells. Naive T cells specific for the antigen mature and become effector T cells that may include CD8 and several classes of CD4 T cells, including CD4 Tfh cells. The antigen is also transported to the lymph nodes and presented to naive B cells. Naive B cells that recognize the antigen AND also are stimulated by Tfh specific for the same antigen then mature into plasma cells that secrete antibodies.
For harmless antigens, even though DCs may phagocyte them they will not get fully activated in the absence of an inflammatory response. So the activation chain is broken. This system fails somehow in the case of allergy.
Another point is that this mechanism is related to the generation of antibodies against microbial protein antigens. Animals lacking T cells cannot produce them and so they are called thymus-dependent (TD) antigens. Other microbial antigens like bacterial polysaccharides can induce antibody production without the help of T cells. This so called thymus-independent (TI) antigens provide the second signal by direct recognition of the antigen (e.g. through TLRs) or by inducing extensive crosslinking of B cell receptor (BCR). TI antigens include LPS and bacterial DNA. TI antigens appear earlier than TD antigens and although provide some protection they are less efficient than the TD ones because TI-specific B cells do not undergo affinity maturation and isotype class switching as TD antigens do.
Reference: Most of what I describe is found in Janeway's Immunobiology, 8th edition. Chapter 10 (The humoral immune response). In particular the section 10- "B-cell activation by helper T-cells".


text:
How to identify genes in Ralstonia that synthesize PHB and promote granule formation?
The compound polyhydroxybutyrate (PHB) is of considerable industrial interest as a biodegradable substitute for plastic. PHB is synthesized from glycerol by the bacterium Ralstonia eutropha. PHB accumulates to high levels in proteindelimited inclusion bodies called PHB granules that are thought to protect the compound from degradation. PHB stains with the blue dye polyphorin 3Z27, so Ralstonia colonies on polyphorin plates have a blue color when glycerol is present (and PHB is made), but a white color if glycerol is absent (and PHB is not made).
Since bacterium E. coli is not able to synthesize PHB, but it would be useful to engineer an E. coli that does so. What approaches can I use with E. coli to identify the Ralstonia genes that synthesize PHB and promote granule formation?

As far as I can tell the original results for this kind of question were presented back in the late 80s - early 90s: Poly-P-hydroxybutyrate (PHB) Biosynthesis
in Alcaligenes eutrophus H16, Cloning of the Alcaligenes eutrophus genes for synthesis of poly-beta-hydroxybutyric acid (PHB) and synthesis of PHB in Escherichia coli., Molecular basis for biosynthesis and accumulation of polyhydroxyalkanoic acids in bacteria.


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Relation between heterozygosity and allelic diversity in founder effects/bottleneck?
Can someone try to explain me why allelic diversity falls faster than heterozygosity, reminding you that we're talking about bottleneck or a founder effect? Look at this graphic:

It's clear to me that both are related, so both have to decrease.
It is clear that allelic diversity has a huge fall because there was lost of alleles (specially the rare and uncommon ones that, despite contributing less to allelle frequencies, each one has the same weight to allelic diversity, take a look at the other graphic to understand what I'm talking about).

Now, heterozygosity is specifically related to allele frequencies, and since the most rare/uncommon don't contribute much to this, the only alleles that will have a large impact on the decrease of heterozygosity are the more/most common.
Since that after a bottleneck, (probably) the alleles remaining are the most common, heterozigosity will decrease less than allelic diversity.
I'm aiming for some complementary ideas about this, specially when it comes to statistical models that describe heterozygosity. It's hard for me to understand the relation between allele frequencies and heterozygosity when it comes to this case - in what manner do allele frequencies influence H?

What is the source of the graphs?

Unfortunately, I don't know. It was provided to me without references.

So this is part of a homework/course work question?

Well, yes. Is there a problem?

No, not really. We just normally add a tag to homework questions (which I've now done). However, it is a bit unclear to me what your specific question is, since you seem to understand/explain why allelic diversity should decrease more than heterozygosity. Are you looking for a mathematical relationship between heterozygosity and allelic diversity?

Yes! I've an idea but I would like to complement it with a mathematical relationship.


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Why is exocrinous pancreatic cancer so lethal when compared to others?
I know it seems far fetched, but I would really like to understand why an interference with the regulation of production of pancreatic exocrine products (and which are they) is so lethal. I know that people having it generally live for one year or less (top is circa 5 years).
Why?

Sure. Go to [https://en.wikipedia.org/wiki/Pancreatic_cancer] and in the vastness there's the info.

I added it in the question. +1

One of the (but not the only) reasons for the high mortality rate of pancreatic cancer is the fact that it is so difficult to screen for. The symptoms (if any) are most often dismissed as general back or gut pain, and it can take quite a bit of testing before the true cause is found, by which time the cancer has already become metastatic. Unlike the digestive system, there is no good non-invasive way to image the pancreas (such as with a colonoscopy for colon/rectal cancer) except using radiological methods such as X-rays, so there is no safe way to provide for standard screenings.


text:
0.1 M sodium citrate in 10% ethanol and DNA solubility
I understand how DNA precipitation works in the presence of salt (such as 0.3 M sodium acetate or 0.2 M sodium chloride) and alcohol (30 ~ 50 % isopropanol or 60 ~ 75% ethanol)
However, in the Trizol kit manual, DNA could be washed and precipitated in 0.1 M sodium citrate in 10% ethanol
At only 10% ethanol and 0.1 M sodium ion, how can DNA be in a "compact/insoluble" form so it can be precipitated by a low speed centrifugation (2000 x g for 5 mins)?

The sodium ions will bind to and neutralize the (-) charge on the phospho backbone of DNA. This reduces hydrogen bonding sites for water, which then results in less solubility.
Edited for clarity and because I didn't answer the question the first time. Sorry about that.
The extraction of DNA is from the interphase layer, which is a very small fraction of the original sample. The extraction of the interphase layer is then mixed with 100% EtOH (300 ul per 1.0 Trizol used in that sample). This should actually be 60% or more EtOH based on my experience using this method.
That is the precipitation step.
The fact that the wash has only 10% EtOH and 0.1M sodium citrate and that is still sufficient to keep DNA precipitated revolves around 0.1M sodium citrate is usually 0.3M sodium ions (sodium citrate can be mono-, di- or tri-sodium, but unless specified..sodium citrate implies tri-). Remember for the wash, you don't need to do a quick precipitation, you just need to prevent it from solubilizing again.
I am removing my comments that contained the same information.

This doesn't address why only 0.1 M sodium citrate and 10% EtOH is used while most precipitation protocols require 0.3 M sodium acetate in 70% EtOH


text:
How does current delivered by a neural stimulator propagate in the brain?
I have been trying to find good literature on how a stimulus current propagates through a medium that contains both gray and white matter (like the brain). Is it safe to assume that a current will spread through grey matter more easily than white matter, because of all the insulation provided by the myelin? Is it then safe to assume that, if you stimulated tissue which has white matter on one side of the electrode and grey matter on the other, the current would preferentially flow through the grey matter?
As an example I tried to sketch something that would represent a deep brain stimulation electrode in the subthalamic nucleus (STN) sending pulses through its bottom electrode (let's assume the return electrode is the implant case and far away from the target structure). As the STN is close to the internal capsule, I'm wondering if any assumptions can be made about the tissue activation region. More specifically I'm wondering if the current would spread more easily through the STN and have more trouble getting past the internal capsule (exciting the fibres there if the current is strong enough but maybe not getting past those fibres because of all the myelin?). So could we imagine the activating region of the stimulus pulses to be biased toward the STN rather than the white matter tracts?
Hope this clears things up.

If an electrode is placed in the white matter and current flows, it goes per definition through the white because it has no option to do otherwise. I think it would be good to include a sketch of the anatomical situation you are referring to in question #2. +1 for the nice question.

Hi @Christiaan thanks for your help yet again! I added a little sketch of a hypothetical DBS scenario that started my questioning. But the question is much more generic, basically I'm wondering if some assumptions about tissue activation can be made when the general physiology surrounding stimulus electrodes are known or if overall it's all way too complex to make any such claims and think about it this way.


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Why does the open tree of life have 2600 species of rose?
The open tree of life has 2754 species and subspecies with the prefix Rosa.
Wiki says:

There is significant disagreement over the number of true rose
  species. Some species are so similar that they could easily be
  considered variations of a single species. Lists of rose species
  usually show between 100 and 150.

The thing is that a subspecies in the open tree taxonomy (OTT) should be written with a different latin name, and I can find at least 400 OTT rosa that are listed as individual species and not subspecies.
Is the OTT not very coherent in terms of accepted species?

OTT does not claim the taxonomy to be perfect, as it was generated by a synthesis of a lot of existing taxonomies. This might lead to an overrepresentation of some species/subspecies, because they have been classified differently in different taxonomies.


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At any given moment, how much energy is stored in the human body as ATP?
At any given moment, approximately how much energy is stored in the human body as ATP in the ADP-P-bond?
This of course depends on what type of cell it is and the activity of the individual in question. The calculation should be fairly simple, as we know that ATP hydrolysis release about 30 kJ/mol. Hence, the question could actually be re-phrased to $"$At any given moment, how much ATP does the human body have?$"$

how much googling have you done? I'd start with "how many ATP molecules in human body" and then multiply that estimate by 30kJ/mol

All I can find is how much ATP the human body uses per time unit. I'm not seeking that but rather how much present ATP there is at a given time.

that is generally expected that you'll add what you;ve found to your question. Is makes it much better

what do you think of following approximation: ATP is not a long-term storage hence every day you consume all ATP you produced from food/fat. That means that on average day-to-day you have X mol ATP/min * 24 hr? where X is the number of ATP/min body burns.

I'm not looking for that either. If you froze time and started to count all ATP molecules in a human body, that's the number.

then you will have to specify when exactly you freeze time: 1 hour before breakfast, right after dinner? Numbers will vary greatly, I suppose

ATP burned per minute is not a useful number because the turnover is so high. 2000 kcal/day is dozens of kilograms of ATP so obviously ATP is turned over more than once a day, but there's probably more than one molecule of ATP being passed around between all the ATP synthases.
This blog claims 250 grams. Taking the estimate of ATP concentrations(1-10 mM) from wikipedia and multiplying by a 60 kg person pops out 150 grams (for 5 mM). The ATP/ADP ratio is about 5 to 1 under physiological conditions, so I'm comfortable leaving out that source of error.
So: Probably a few hundred grams, depending on a lot of things. Which is about .2-.7 moles, for reference. 6-20 kJ is not a lot, in fact even at the upper end of estimates you're looking at maybe 8 dietary calories, and probably more like 3.

Each of us burns approximately twice our body weight in ATP every day. So double your body weight and divide that number by the number of seconds in one day; that will give you the average mass of ATP. N.B. a person with a higher mass will have a higher answer.

Is there a reference for that?

My memory may not be what it used to be, but I thought I read a quotation by Paul D. Boyer (Nobel Prize recipient in 1997 for his work on ATP synthesis) where he cited that figure, but I may be off by a factor of 2 according to this slightly more well-known source which says "equivalent to your body weight": https://en.m.wikipedia.org/wiki/Adenosine_triphosphate


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Action of tetracycline on 30s ribosomal subunits
My question concerns the mode of action of 30s ribosomal antibiotc inhibitors — tetracyclines. According to some online resources, the antibiotics enter the A-site, which prevents other tRNA from entering the ribosome. Others say that the antibiotic prevents the ribosome from moving along the mRNA. Which one is correct?

Adding references and citations to the "resources" and also naming the particular drug you were interested in would help your chances of getting this question answered considerably.

I am giving the class of antibiotics.  30s ribosomal Inhibitors. Some examples: Aminoglycosides Tetracyclines.

Firstly, as AMR said, you should add the online resources  which you referred to. Secondly, different antibiotics can have different effects. In fact there can be different effects based on concentration. Moreover, it is now established that aminoglycosides are bactericidal because they not just prevent translation but also cause mistranslation leading to toxic products.

Could you expand on "how they prevent" translation.

@RadekMartinez can you first please add references and focus your question on one class of antibiotics such as aminogycosides.

The crystal structure of tetracycline bound to the 30S ribosomal subunit of Thermus thermophilus has been determined by Pioletti et al. and the major binding site (Tet-1) interferes with the location where the A-site tRNA binds as shown in the stereo pair, below:

They write:

“Thus tetracycline can physically prevent the binding of the tRNA to the A-site. This mode of interaction is consistent with the classical model of tetracycline as an inhibitor of A-site occupation, and hence offers a clear explanation of the bacteriostatic effects of tetracycline.”

So the proximal effect would appear to be the first alternative in the question, although the second may be an indirect consequence.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Skull size of Homo naledi
I want to 3D print the skull of the newly discovered Homo naledi.
The printing data can be found here: http://morphosource.org/index.php/Detail/MediaDetail/Show/media_id/7300
Where can I find the exact size of the skull so that I get a 1:1 copy? Google was of no help.

To 3-d print it, you need the volume right , not the area cause I know the volume

@Tusky: I think this would be helpful!

From the link:a cranial volume of 560 cm3 (34 cu in) for the males and 465 cm3 (28.4 cu in) for females, approximately half the volume of modern human skulls; average Homo erectus skulls are 900 cm3 (55 cu in). You could use these data?

Based on Figure 11 in the original publication (scale bar = 10 cm):

The anteroposterior length of the skull should be about 16.3 cm. I just printed out the figure and did the conversion.
So check that your 3D printing software estimates a little over 16 cm for the length from brow ridge to occipital. The full reconstructed skull will be longer of course, but this should give you a concrete measurement to compare to.

Yowza! That's quite the brow! Guess no hats for H. naledi!

From my research, I can deduce that the cranial volume for males is

560 cm3 (34 cu in)

and

465 cm3 (28.4 cu in) for females

Note: These skulls are approximately half the volume of modern human skulls average Homo erectus skulls are

900 cm3 (55 cu in)

I don't think you would need the area for 3-D printing, but you can use the volume!

From your research? It's in my comment :)

@Christiaan It was in the wiki link. I did some reading thats all


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Is HSV-vector-mediated miRNA expression in dorsal root ganglia stable?
My question is on the following article:
"Reduction of voltage gated sodium channel protein in DRG by vector mediated miRNA reduces pain in rats with painful diabetic neuropathy"
My question is, do you think the rats continued to have decreased levels of voltage-gated Na+ channels after this experiment, or did they likely gain them back up eventually, because dorsal root ganglion neurons regenerate themselves? I couldn't find this info in the article.

The vector used in this study is based on the herpes simplex virus genome (HSV).
Wildtype HSV mainly infects sensory neurons. After infection, it resides in a latent state in the nervous systems of the host for a lifetime. The viral genome persists in the cell, without integrating into the host genome (Marconi et al., 2000).
The article cited is devoid of background on the vector and its characteristics and I understand your issues. I think that the expression of the RNAi is quite stable in your linked article, because the authors infected neurons in the dorsal root ganglia. The dorsal root ganglia contains sensory neurons. It is an immuno-privileged region, because the blood-brain-barrier (BBB) limits the access of immune cells. Indeed, protein expression through HSV-transfection in the brain is stable, at least up to a year (Suzuki et al., 2008) and possibly longer. Peripherally, outside the BBB, the expression of HSV-plasmids may be compromised within weeks through cellular immunological responses targeting the expressing cells (Suzuki et al., 2008).
Regarding regeneration - The DRG contain mature neurons and is part of the peripheral nervous system. These neurons are known to regenerate readily after injuries, but these injuries are mostly investigated through physical damage (Muratori et al., 2015). Latent HSV infection does not kill neurons, and even an active infection seems to leave neurons relatively unscathed (Rautema, 2002).
References
- Marconi et al., In: Madame Curie Bioscience Database. Austin (TX): Landes Bioscience (2000)
- Muratori et al., (2015); 2015, 860546
- Rautemaa et al., Immunology (2002); 106(3): 404–11
- Suzuki et al., Molecular Therapy (2008); 16(10): 1727–36)


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"Prime" structure of DNA Double Helix: Confusion
In this video on DNA replication, the diagram shows the unwound DNA as still being anti-parallel, but the first diagram in this post on Biology SE shows that the individual strands are 5'____5' and 3'_____3'. So my question is the following:
Does a single, unwound strand of DNA run as either 5'______3' or 3'_______5', or does it run either 5'______5' or 3'________3'?

DNA strands always have one 3' end and one 5' end (since each nucleotide has one of each and a strand is formed by connecting the 3' side of one nucleotide to the 5' side of another nucleotide).  In a double helix DNA molecule, the two strands run in opposite directions.
The Pearson Education diagram in the post you referenced has one pair of the strand ends mislabelled.  The lefthand-most labels 5' above the 3' should have been 3' above 5'.

In the SE post that OP mentions, the diagram is immediately preceded by a link to the source page of the diagram (via the word "schema"), and on that page there is a "Homework:" paragraph noting that the diagram is mislabeled.

Oh, didn't see that!

A single, unwound strand of DNA runs 5' to 3', the video is correct. When wound properly, the strands run in reverse directions, 5' to 3' for the first, 3' to 5' for the other. That diagram is not helpful, though the post itself is quite good.


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Numbers in the names of tumor markers
In tumor markers such as CA 125, CA 19-9 and many other, CA stands for Carcinoma antigen, but what about the number?

There are dozens of cell marker genes. The number refers to antigen protein from one of those specific genes. The number itself can be a result of any of a variety of things, from the order in which the protein was discovered to the corresponding antibody being the 125th antibody produced against the cell line or organism. These are often used as biomarkers of various cancers. CA-125, for example, is sometimes found at high levels on the surface of blood cells in women with ovarian cancer.

So there is no systematic way for numbering ?

No, not as far as I can tell. CA-125 was named so because the antibody to it was the 125th antibody produced against the ovarian cell line the group was using. I'm not sure how CA 19-9 was named though.


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Are there any animals that gradually approaches the predator to search for food?
I was wondering if there is an animal (or insect, bird, etc.) that eventually comes close to its predator to search for food, but only does so if it couldn't find food in the environment it is in. So it first searches part of the environment farthest away from the predator, but gradually comes closer and closer to the predator as it searches for new food. That is, over time it dares more to come closer to a, lets say ambushing predator, since it has ran out of food (or perhaps for other reasons). Similar behavior is ok, but note that the key behavior that I want in the animal is that it "gradually" comes closer and closer to the predator.
Thank you.

Also see: Turkeys Circle Dead Cat :p

The behaviour that you describe is common in most animal species, as part of the natural trade-off between access to food, minimizing risk, habituation and hunger. Animals usually choose to forage in high-quality habitats that has a low risk of predation, but if food sources are depleted (or competition and/or territoriality is high) they will move to other areas with lower food quality or higher risk (since they are desperate for food).
In general, the effect of predators are both direct (killing of prey) and indirect (changes in prey behavior to avoid predation). The indirect effects means that prey will often choose to feed in habitats that are suboptimal from the perspective of forage quality, which will have negative effects on their foraging efficiency. These types of indirect effects of predators are often labelled as 'The landscape of fear' or the 'ecology of fear' (see e.g. Brown et al, 1999). However, as food resources run out this will trade-off against the fear of predators (along with possible habituation effects), which will lead to the behaviour you are asking for. You might also be interested in Gamelius et al (2014), as an example of how prey behaviour can be influenced by the presence of predators (here, roe deer and lynx), but there is a large number of studies looking into these issues. The review "Fear in animals: a meta-analysis and review of risk assessment" by Stankowich & Blumstein (2005) will also give you a good background and many examples.
The use of scarecrows is a simple way to understand the 'landscape of fear'. By placing a scarecrow in a field you strengthen the "landscape of fear" (by adding an artificial 'predator'), which will modify the behaviour of perceived pest species (prey), so that they avoid the areas around the scarecrow. However, if the animals you want to scare away ("pests") run out of food they will take larger risks and venture closer to the scarecrow to forage. Over time, they will also become habituated to the scarecrow, and will discover that it doesn't seem to pose a big threat, which will decrease the effectiveness of the scarecrow.
References

Brown et al. 1999. The Ecology of Fear: Optimal Foraging, Game Theory, and Trophic Interactions. Journal of Mammalogy
Gamelius et al. 2014. Habitat Selection and Risk of Predation: Re-colonization by Lynx had Limited Impact on Habitat Selection by Roe Deer. Plos One
Stankowich & Blumstein. 2005. Fear in animals: a meta-analysis and review of risk assessment. Proceedings B


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what is the principle behind Allele Frequency Net Database
I just wanted to know what is the basic principle of Allele*Frequency Net Database. How are the frequencies present in them calculated??

Allele frequency is simply the frequency at which a specific allele is found in members of a population. Let's take the following example from the site you referenced:

The first column merely has the name of the allele, A*01. The second column is the population from which individuals were genotyped. Skipping ahead, the fifth column is the number of people genotyped from the population, 187 in this instance. Of those 187 people, 14.8% of them have at least one copy of the allele as stated in the third column. It's important to remember that each individual could potentially have two copies of the allele.
The fourth column gives the actual allele frequency, calculated by the number of A*01 alleles/number of total alleles (374, assuming no serious mutations). Of those 374 total alleles, 7.4% (0.074) of them are the A*01 allele, meaning 27-28 of the 187 total people had the allele. Since 14.8% of people had the allele, but the frequency of the allele was only 7.4%, we know that each person who has the allele only has one copy of it.


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Which of the three is true for insulin receptors?
I have seen the following question in a Cell Biology exam:
Which of the following is true"

Insulin has an hydrophobic Signal Peptide and the insulin receptor does not have an hydrophobic Signal Peptide

The insulin is coded by two genes, the insulin receptor has an hydrophobic sequence in the middle of the protein.

Receptors in the ER bound the SRP that is bounded to the insulin receptor, SRP is made of RNA and proteins.

I think that (1) is the right answer. However, people I talked with thought (3) is the right one.
I think 1 is right because I have never seen anywhere that insulin receptor has a signal peptide so I assumed it does not have.
I also think that SRP is a part of the insulin receptor, which is located on the ER (and not bounded to the ER by receptors), this is why I thought (3) is not the right answer.
Am I right, or did I miss anything?

Eeeeh, if you meant insulin receptor has a signal peptide which is removed during receptor maturation, then you are correct. http://www.uniprot.org/uniprot/P06213

Signal recognition particle (SRP) receptor, also called docking protein, is a dimer composed of 2 different subunits that are associated exclusively with the rough ER in mammalian cells. The surface of the rough endoplasmic reticulum (often abbreviated RER or Rough ER) (also called granular endoplasmic reticulum) is studded with protein-manufacturing ribosomes giving it a "rough" appearance. so i don't think (3) is the right statement.
the insulin receptors is made up of two hormones, insulin and glucagon, are at the center of this signaling system. there is no such things as hydrophobic sequence in the middle of the protein i believe.
also, Insulin is a peptide hormone produced by beta cells of the pancreatic islets and and insulin receptors doesn't have an hydrophobic signal peptide. so i think (1) is the correct answer.
Edit :
https://en.wikipedia.org/wiki/Signal_recognition_particle_receptor
https://www.healthline.com/health/type-2-diabetes/insulin#Function1
www.news-medical.net/health/What-is-Insulin.aspx


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How difficult is it to make a shRNA/miRNA/siRNA to silence/knockdown NaV1.7 voltage gated sodium channels in humans?
There have been various research projects that experimented with shRNA/miRNA/siRNA to specifically silence/knockdown NaV1.7 voltage gated sodium channels in small animals like rats & guinea pigs. (Example; my previous question: Is HSV-vector-mediated miRNA expression in dorsal root ganglia stable?
But I can't find any that's been done for humans... So I think it means scientists have not gotten to that stage (if they are interested in doing so), and I was wondering,

technically, how difficult do you think it is to make a shRNA/miRNA/siRNA that can silence/knockdown NaV1.7 in humans, given that they already know how to make it for NaV1.7 in small animals?
financially, how much money is needed to invent such a thing (and how hard to get a grant for it?)
if technically not too difficult, and not very expensive to invent just a sample, is this something not necessarily a professional researcher, but a student could try making as their research project?

Why should any ethics comittee approve such experiments in humans?

Because especially when it is a partial silencing/knockdown, people who are suffering from pain after failed pain therapies or people with pain disorders can benefit from it, I think

What about potential negative side effects? Life is usually more complex than people think.

That's true, there could be side effects, although no serious ones seem to have been reported so far (but only in animals...).

Knockdown experiments in humans are almost always carried out on cell lines. In fact, I am doing a few such experiments myself at the moment (using U6 promoter expressed shRNAs introduced by lentiviruses). Doing it on humans would be extremely problematic for ethics committees due to the host of possible side effects.

The major hurdle for novel treatments to make it into humans is the assessment of its merits and added value over existing, proven treatments. And more importantly, the safety for humans has to be validated to a sufficient extent. The notorious from bench to bedside (Goldblatt & Lee, 2010) bottle neck. To answer your questions:

Technically, going from animals to humans may be a small step. Just increase the dose according to body weight. However, ethically it is a giant step. In fundamental research, lab animals are typically investigated only for the positive effects (e.g., amelioration of symptoms). In preclinical research, the animals have to be tested on side effects. Systemic injections, and even localized administration of potentially harmful virus vectors carries considerable risk. What if it returns to an infectious state? What are the side effects? Further, while animals can be treated systemically via iv injections, humans don't like needles. The FDA will put every possible danger and every possible discomfort to the patient into the spotlight.
Getting grants is hard. Going from bench to bedside is even harder, as the FDA, institutional review boards and the scientific community at large will scrutinize the plans.
To undertake clinical trials with humans is a team effort involving medical professionals, review boards, patients or healthy volunteers, an independent monitoring committee and so forth. It's not something you'll be doing in a few months. It can take years or decades to bring novel treatments to patients. It doesn't take one grant - it takes many, and development costs easily run into the tens of millions.

Reference
- Goldblatt & Lee, Am J Transl Res (2010); 2(1): 1–18

@pewiwe - If you appreciate the answer, an upvote can help to show others that it's a useful answer. Thank-you comments are great, but SE works with a voting system. Thanks!

Aw sorry i can't seem to upvote it because I registered as a guest :(

@pewiwe - perhaps consider converting to a real account?


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Can a cell start a mitosis if it lacks energy or molecules to complete it?
I'm wondering whether a cell can start a mitosis if it lack molecules or energy to fully complete it. From what I'm reading on wikipedia the cell passes most of its time in the interphase in which the DNA is replicated and then there's the G2 checkpoint to make sure mitosis can start.
However I'm not sure all of this is enough to ensure that the cell has sufficient energy to fully complete the mitosis (is the cell "fueled" in energy and molecules during the mitosis or does it has everything needed before starting it?).
Edit: I'm considering this question for humans in particular, or animals in general if the answer is the same.
When I speak about "molecules" I mean any atom or molecule. In other words, whether the cell needs some material intake from the external environment during the mitosis or not. If not, I consider that it has all the necessary atoms/molecules to fully complete the mitosis without any external need.

When all checkpoints are working, then the most critical is the G1-S. The reason is that once the cell passes this checkpoint, it is committed to either replicate and divide or it will trigger apoptosis. The reason for this is that once DNA replication starts, there is no way for the cell to undo this, so it has to have everything it needs before it commits to DNA replication. There are other checkpoints, however these are to make sure that replication has proceeded properly, or that the daughter cells will receive the exact copies of the genome. AleksandrH's answer provides the basics.

Which organism are you considering? Free-living organisms such as yeast have a variety of mechanisms for sensing nutrients, and regulate proliferation depending on nutrient availability. In animal cells it's very different, as growth factors are the dominant driving forces, and here the role of nutrients as signals is more uncertain / controversial.

I've edited my question to make it clearer (humans basically).

The short answer: no. First, let's get an understanding of the cell cycle control system, as there are some important molecules involved in this system that regulate mitosis. Think of the control system as a series of stoplights: as you mention, there is one stoplight at the G2 phase. There are two additional checkpoints: one at G1 and one in the M (mitotic) phase, just before Anaphase. Why before anaphase? At this stage in mitosis, the cell checks to determine if all of the kinetochores of the duplicated sister chromatids have microtubules attached to them. This is to ensure proper distribution of chromatids in anaphase.
Ok, so back to your question: what are some of the molecules involved, or what are some molecules that the cell might need before it proceeds?

Cyclins - these are proteins that are produced in the late S-G2 phases of Interphase. They activate cyclin dependent kinases (Cdks)
Cyclin Dependent Kinases - as the name suggests, these proteins (kinases) are dependent on cyclin in order to operate. The best known Cdk is called MPF, short for maturation promoting factor.

Throughout the cell cycle, these Cdks, together with their bound cyclins, are able to phosphorylate various proteins in the cell as it undergoes mitosis. For instance, the MPF would phosphorylate lamins, which are proteins on the nuclear envelope that can initiate degradation of the laminar matrix. This is essential in Prophase, when the nucleus is disassembling. There are many other things it can phosphorylate. For a more complete list, look under "Targets of MPF" in the above link for MPF. Take a look at this diagram, taken from Campbell's Biology (9th Ed):

There is one other molecule worth mentioning:

Growth factors - these are proteins specifically released by neighboring cells (relative to the cell that is preparing to undergo mitosis). These are crucial for mitosis to even initiate. A great example of a growth factor is PDGF, which stands for "Platelet Derived Growth Factor". These PDGFs are released by platelets at sites of injury, such as a cut on your knee. The PDGFs stimulate cell division and help your skin to reheal!

There are many other essential nutrients that cells require before they can initiate mitosis; not all have been listed here. Also, I know I didn't necessarily answer the "energy" part of your question, but I hope this helps for your first part.

This isn't the whole story, but it is representative of what would be taught in introductory college level biology.

This answer doesn't really concern energy or nutrients. Cyclins, CDKs and growth factors are all signalling proteins, they merely relay information.

@Roland But the OP didn't ask about nutrients, they asked about energy and molecules, and I covered the molecule part (I mention near the end that I don't know the answer to the "energy" part).

Okay, I guess "molecules" is ambiguous, I would guess in this context it means material / building blocks, not signalling molecules. Perhaps the OP can clarify?

Sure, I wasn't too clear myself because molecules is indeed a very broad term.

I've edited my question, please let me know whether it's clearer now.