A lot of useful information can be found on Wikipedia article Immunohistochemistry
.
Fluorescent tags on secondary antibody allow for higher signal and labeling multiple targets using orthogonal pairs of antibodies. For example, we can label at the same time tubulin and actin.
Secondary antibodies, though, are bulkier and can decrease resolution, especially in super-resolution microscopy. Using multiple antibodies also can increase chance of off-target labeling (when antibody binds elsewhere rather than it's antigen).
Showing posts with label methods. Show all posts
Showing posts with label methods. Show all posts
Sunday, August 25, 2019
The Cell 9-8
The Cell (Alberts) 9-8 question
Antibodies that bind to specific proteins are important tools for defining the locations of molecules in cells. The sensitivity of primary antibody, the antibody that reacts with the target molecule, is often enhanced by using labeled secondary antibodies that bind to it. What are the advantages and disadvantages of using secondary antibodies that carry fluorescent tags versus those that carry bound enzymes?
Show answer
Sunday, November 1, 2009
#11 Amino acids identification
GRE-ST questions #162 and 163 are about amino acids: Arg, His, Cys, Pro and Trp.
A student was given a task of identifying the contents of five bottles of amino acids from which the labels had fallen off. Each of the original bottles contained one of the following: arginine, histidine, cysteine, proline, and tryptophan.Show answer
162. Which of the following methods could be most
readily employed to identify tryptophan?
(A) Electrophoresis
(B) Ultraviolet spectroscopy
(C) Gel filtration
(D) Analytical ultracentrifugation
(E) Optical rotation
163. Each amino acid was subjected to paper electrophoresis in a pH 9.5 buffer, and then the amino acids were visualized by spraying the paper with ninhydrin. Which amino acid is at the point labeled (1) in the figure above?
(A) Arginine
(B) Histidine
(C) Cysteine
(D) Proline
(E) Tryptophan
First one question is rather easy. Tryptophan have one very useful feature that called fluorescence. That is emission of light after illumination. Trp is one of 20 amino acids and widely occur in proteins and its fluorescence provide useful data about protein's folding and structure. Trp's fluorescence is one of the most intense among all fluorescent amino acids (others are Phe and Tyr). Trp excitation wave length is 280 nm and emitted light has wave length 300-350 nm. Ultraviolet is part of spectrum from 10 to 400 nm. So right answer is B, because other amino acids from the list are not fluorescent.
Next part of question refers mainly to amino acids' behavior in different pH. So it is necessary to investigate polarity of substances in solutions with different acidity.
But what does pH actually means? It is defined as pH=-Log[H+], i.e. minus decimal logarithm of H+ (protons) concentration in solution. Neutral pH is 7 because in water there are 10^-7 M/liter of H+ ions due to H20 decay into H+ and OH-.
So, briefly, high pH means very low concentration of H+, and low pH's mean high concentration of H+, or low concentration of OH-. Actually, pH and pOH (defined in the same manner) are closely connected with each other. For pure water pH=pOH, of course.
Now lets go further. Practically all amino acids can be distinguished using pKa – pH level at which molecule is not charged. That very simple experiment described in the question. If there is any pKa and if acid (actually, amino acid residue) have NH2 or OH groups, then at pH higher than pKa protons will be taken away from acid so -OH will become -O- (oxygen with negative charge) and at lower that pKa pH acid's NH2 will become NH3+ (one positive charge). Alberts provide several examples, that make everything clear:
Using pKa values is quite hard to solve this problem, really. Best picture can be granted by pI value. That is isoelectric point, pH level, at which whole molecule of acid is electroneutral. There is a nice table of pI values on the web. If pH higher than pI, then protons will be taken away and charge will become negative. Otherwise at pH lower than pI molecule will be protonated. If you check linked page, you will find, that all molecules from questions have pI lower than 9.5 and only arginine have higher value. Hence, Pro, His, Cys and Trp will have negative charge and only Arg – positive. That is why four molecules will move on one side from point of sample application, and only one – on the other. So it will be Arg, the answer is A. Note, that picture is electrophoresis results, so + and - are signs of CAthode and ANode, which attracts CAtions and ANions respectively.
Additional information
OK, we figured out – using something called pI – right answer. But why does it happens to be right? And why – what is more important – why argynine differs so much from other amino acids?
First, what is pI actually? It is mean of pK(amino group) and pK(carboxyl group) for molecule, which have only one of those. pI=(pK(NH2)+pK(-OH))/2. When pH is rising from pK of -OH more and more groups undergo deprotonating. Similarly, when pH decrease from pK of NH2, more and more groups undergo protonating, so there is point of equilibrium, where number of protonated groups equal number of deprotonated groups, thus net charge is zero. We assume that this equilibrium point is average of pK's.
But there are several amino acid residues that have their own -NH2 or -OH groups and even not one as Arg (2 -NH2), His (1) and so on. In this case pI is counted as sum of pK-s of all -OH and -NH2 groups divided by two. That is why Arg has pI much greater than any amino acid's that lack -NH2 groups at all.
Next part of question refers mainly to amino acids' behavior in different pH. So it is necessary to investigate polarity of substances in solutions with different acidity.
But what does pH actually means? It is defined as pH=-Log[H+], i.e. minus decimal logarithm of H+ (protons) concentration in solution. Neutral pH is 7 because in water there are 10^-7 M/liter of H+ ions due to H20 decay into H+ and OH-.
So, briefly, high pH means very low concentration of H+, and low pH's mean high concentration of H+, or low concentration of OH-. Actually, pH and pOH (defined in the same manner) are closely connected with each other. For pure water pH=pOH, of course.
Now lets go further. Practically all amino acids can be distinguished using pKa – pH level at which molecule is not charged. That very simple experiment described in the question. If there is any pKa and if acid (actually, amino acid residue) have NH2 or OH groups, then at pH higher than pKa protons will be taken away from acid so -OH will become -O- (oxygen with negative charge) and at lower that pKa pH acid's NH2 will become NH3+ (one positive charge). Alberts provide several examples, that make everything clear:
Using pKa values is quite hard to solve this problem, really. Best picture can be granted by pI value. That is isoelectric point, pH level, at which whole molecule of acid is electroneutral. There is a nice table of pI values on the web. If pH higher than pI, then protons will be taken away and charge will become negative. Otherwise at pH lower than pI molecule will be protonated. If you check linked page, you will find, that all molecules from questions have pI lower than 9.5 and only arginine have higher value. Hence, Pro, His, Cys and Trp will have negative charge and only Arg – positive. That is why four molecules will move on one side from point of sample application, and only one – on the other. So it will be Arg, the answer is A. Note, that picture is electrophoresis results, so + and - are signs of CAthode and ANode, which attracts CAtions and ANions respectively.
Additional information
OK, we figured out – using something called pI – right answer. But why does it happens to be right? And why – what is more important – why argynine differs so much from other amino acids?
First, what is pI actually? It is mean of pK(amino group) and pK(carboxyl group) for molecule, which have only one of those. pI=(pK(NH2)+pK(-OH))/2. When pH is rising from pK of -OH more and more groups undergo deprotonating. Similarly, when pH decrease from pK of NH2, more and more groups undergo protonating, so there is point of equilibrium, where number of protonated groups equal number of deprotonated groups, thus net charge is zero. We assume that this equilibrium point is average of pK's.
But there are several amino acid residues that have their own -NH2 or -OH groups and even not one as Arg (2 -NH2), His (1) and so on. In this case pI is counted as sum of pK-s of all -OH and -NH2 groups divided by two. That is why Arg has pI much greater than any amino acid's that lack -NH2 groups at all.
Tuesday, October 27, 2009
#7 Bacteria and mammalian protein in vitro synthesis
GRE-ST question #29.
When bacteria produce mammalian proteins,Show answer
cDNA is used rather than genomic DNA. Which of the following is the best explanation?
(A) It is easier to clone cDNA than genomic DNA of comparable size.
(B) It is easier to clone RNA than DNA.
(C) It is not possible to clone the entire coding region of the gene.
(D) Most eukaryotic genes have introns that cannot be removed in bacteria.
(E) Most eukaryotic gene promoters do not function in bacteria.
cDNA – complementary DNA, synthesized from mature mRNA (matrix RNA, protein-coding RNA, which is translated by ribosomes).
Intron – non-coding DNA region, that is transcribed into precursor RNA, but removed during splicing while RNA maturing.
Promoters – DNA regions, that signals begin of protein-coding sequences.
This question is uncovered by Alberts et al. in a tiny piece of text about cDNA libraries. Shortly, they remind that bacteria or yeast can't delete non-sense sequences (cos' their DNA have none of those) which are removed during splicing in mammalian. Hence right answer is D.
Intron – non-coding DNA region, that is transcribed into precursor RNA, but removed during splicing while RNA maturing.
Promoters – DNA regions, that signals begin of protein-coding sequences.
This question is uncovered by Alberts et al. in a tiny piece of text about cDNA libraries. Shortly, they remind that bacteria or yeast can't delete non-sense sequences (cos' their DNA have none of those) which are removed during splicing in mammalian. Hence right answer is D.
Monday, October 26, 2009
#4 Membrane protein arrangement
Following problem is from my current MCB class at the Universuty.
Show answer (its quite long, with images, black-jack and hookers)
You would like to study some protein of red blood cell membrane. So you prepare ghosts of that cells and extract all proteins from the bilayer. Via SDS gel electrophoresis you select some protein with mass 106 kDa. Using some organism you get appropriate antibodies to that protein. They are polyclonal and monospecific.
After that you get new ghosts and reconstitute them into closed empty membranes. The experiment is designed in such way (it is its imperfection), that some ghosts will be right side out and the others will be inside out. Those different membranes are easily separated using chromatography because outer surface of natural red blood cells is covered with oligosaccharides and the inner is not (we will talk about it again in later posts about membrane proteins formation).
OK. You separate right out and inside out membranes into two different test tubes and add into them set of different proteases – enzymes that break down polypeptide chains (proteins) in different sites. Of course, some scraps will be washed into buffer and easily taken away.
To extract parts, that are still in the membrane, you use detergent (e.g. Triton X100) and sediment then your solution. Proteins will get to the bottom of the test tube after centrifugation.
After SDS gel electrophoresis of precipitate you got following sets of lines:
– right side out vesicles: 22.8, 19.2 and 38.4 (kDa)
– inside out vesicles: 14.4, 13.2, 10.8 (kDa)
which are visible because of reaction with antibodies you got earlier. You are able to see not all the protein's parts that were still in the membrane of course.
The question is what assumption can you make about that protein arrangement inside red cell membrane?
Show answer (its quite long, with images, black-jack and hookers)
That is really experimental exercise, an example of protein study. The idea here is that we know several types of protein-membrane interactions or how some protein can penetrate membrane.
As we know nothing about glycolipids links and another stuff, lets assume that protein snakes membrane in some way. Next, it is important to know what protein's composition is inside bilayer. It can be either alpha-helix (numbers 1, 2 and 4) or beta-barrel (number 3).
Alpha-helices that pass through the membrane are chains of 15 to 30 amino-acids residues. If one acid weight 100 to 200 Da, then one helix's molecular weight is 1.5 to 6 kDa.
If there is hydrophilic residue, it hides inside if the helix. Otherwise hydrophobic residues are exposed to the hydrophobic phospholipids' tails.
Barrels, made of several (8 to 22) beta-strands are something different. Strands form walls pore in the membrane often filled with water. As in helices, hydrophobic and hydrophilic residues' orientation persist: later are exposed to solute (water). But the difference is in linkage between strands and helices of multi-pass proteins. Barrel's "wall" is tightly fixed and rigid because of cross hydrogen bonds between residues when alpha-helices are much more mobile.
Barrel-type proteins are restricted to mitochondrial, chloroplast and bacterial outer membranes. vast majority of multipass transmembrane proteins in eucaryotic cells and in the bacterial plasma membrane are constructed from transmembrane alpha-helices, as Albers et al. wrote. This happens because of greater mobility of helices relative to each other that cause possibility of open-close switching of the channel.
But lets return to the problem. First, we will check if it is possible, that protein snakes through membrane as several helices.
Assume that antibodies will be produced to actually any helix and no parts will be unlabeled in the gel. We can do so because protein is rather large – 500 to 1000 amino acids residues. That means that it is impossible that protein snakes membrane, for example, 10 times, but we see only three chains (see image below as explanation; red are labeled antibodies and transparent regions are cleaved by protease so numbered regions are electrophoresis result).
Half-inserted helices as in picture 4 from first diagram are not possible because they are protected from washing away and can only act as additional parts that will not be shown in experiment with another side-out vesicles or there will be too much parts separated by protease.
So in "alpha-helices"-approach only possible answer is 5 transmembrane helices as on the picture. The question is only where N- and C-terminal amino acids are located. They cannot be on one side of the membrane, you can see in easily just drawing several snakes through bilayer. But it also cannot be determined during our data.
Also there is no need to distinguish exterior and interior any more. Lets just say, that from one side there is one stack of polypeptides and another from the opposite.
This is our model. Lets say, top is exterior and bottom is interior. Our model is essential for check if such composition ever possible. parts inside membrane are helices, and so, for example, part 1+2 can weight 10.8 kDa, because it is on the exterior of red blood cell and protected while protease cleave inside-out vesicle.
What's next? I suggest to write down hole system and solve it for each part (1, 2, 3 etc.). But there will be 6 equations (for number of conditions) and 11 unknown variables, that depends actually on helices weights. Vice versa weights of helices are functions of parts' weights (it is almost so).
My idea is to set different weights and check if our system will have positive and reliable solution on 1st, 3rd, 5th, 7th, 9th and 11th parts' weights.
Beta-barrels are not likely to be kind of answer because it is made of several, usually more that 5 strands, which ruins our model. But if we assume that there are 5 strands, everything will be OK if we just change even numbers in system for beta-barrel reliable. I hope it will fit, but check it up.
Phew! Get this tiny bonus from me.
As we know nothing about glycolipids links and another stuff, lets assume that protein snakes membrane in some way. Next, it is important to know what protein's composition is inside bilayer. It can be either alpha-helix (numbers 1, 2 and 4) or beta-barrel (number 3).
If there is hydrophilic residue, it hides inside if the helix. Otherwise hydrophobic residues are exposed to the hydrophobic phospholipids' tails.
Barrels, made of several (8 to 22) beta-strands are something different. Strands form walls pore in the membrane often filled with water. As in helices, hydrophobic and hydrophilic residues' orientation persist: later are exposed to solute (water). But the difference is in linkage between strands and helices of multi-pass proteins. Barrel's "wall" is tightly fixed and rigid because of cross hydrogen bonds between residues when alpha-helices are much more mobile.
Barrel-type proteins are restricted to mitochondrial, chloroplast and bacterial outer membranes. vast majority of multipass transmembrane proteins in eucaryotic cells and in the bacterial plasma membrane are constructed from transmembrane alpha-helices, as Albers et al. wrote. This happens because of greater mobility of helices relative to each other that cause possibility of open-close switching of the channel.
But lets return to the problem. First, we will check if it is possible, that protein snakes through membrane as several helices.
Assume that antibodies will be produced to actually any helix and no parts will be unlabeled in the gel. We can do so because protein is rather large – 500 to 1000 amino acids residues. That means that it is impossible that protein snakes membrane, for example, 10 times, but we see only three chains (see image below as explanation; red are labeled antibodies and transparent regions are cleaved by protease so numbered regions are electrophoresis result).
Half-inserted helices as in picture 4 from first diagram are not possible because they are protected from washing away and can only act as additional parts that will not be shown in experiment with another side-out vesicles or there will be too much parts separated by protease.
So in "alpha-helices"-approach only possible answer is 5 transmembrane helices as on the picture. The question is only where N- and C-terminal amino acids are located. They cannot be on one side of the membrane, you can see in easily just drawing several snakes through bilayer. But it also cannot be determined during our data.
Also there is no need to distinguish exterior and interior any more. Lets just say, that from one side there is one stack of polypeptides and another from the opposite.
This is our model. Lets say, top is exterior and bottom is interior. Our model is essential for check if such composition ever possible. parts inside membrane are helices, and so, for example, part 1+2 can weight 10.8 kDa, because it is on the exterior of red blood cell and protected while protease cleave inside-out vesicle.
My idea is to set different weights and check if our system will have positive and reliable solution on 1st, 3rd, 5th, 7th, 9th and 11th parts' weights.
1+2=10.8As you see, if we define all even weights (weights of helices actually) we will surprisingly fast achieve all pieces weights that are wanted just to be positive and reliable as polypeptides' weight in kDa. Thus if all helices weights (remember they are even numbers in system) are from 1.5 to 6 kDa, other pieces weights will vary from approximately 4 to 9 kDa (1st), 2 to 10 (5rd), 3 to 11 (9th), 26 to 35 (3rd), 10 to 19 (7th) and 13 to 17 (11th). Hence there will be from 10 to 175 amino acids residues average in each "tail" of protein that faces either exterior of the cell or cytoplasm. That is truly reliable (I loooove this word sooooo much, indeed!) because, for example, there is a protein called band 3 in the red cell membrane (no, question is not about it) with 12 alpha helices and about 930 amino acids in polypeptide. So if count a bit and draw picture of band-3 we'll see, that there are about 75 amino acids for each tail that is outside the bilayer. If we get average number of amino acids in "tails" from our exercise it will be (10+175)/2 = 92. What is quite close to band-3 result.
4+5+6=13.2
8+9+10=14.4
2+3+4=38.4
6+7+8=22.8
10+11=19.2
Beta-barrels are not likely to be kind of answer because it is made of several, usually more that 5 strands, which ruins our model. But if we assume that there are 5 strands, everything will be OK if we just change even numbers in system for beta-barrel reliable. I hope it will fit, but check it up.
Phew! Get this tiny bonus from me.
Questionnaire #1
That is first series of short (with short description as well as answer) questions from GRE Subject Test.
#5
#13
#31
#34
#5
Most of the dry mass in the trunk of a tree was originally derived fromShow answer
(A) the soil
(B) light energy
(C) amino acids
(D) CO2
(E) glucose
As soon as dry mass of the trunk of a tree consists mainly of oligosaccharides its source should contain carbon and hydrogen. Soil is too much common answer for me. Light energy can't be right answer because energy is not converted into matter easily. Amino acids is strange idea also. So let's check two options: CO2 and glucose. As this question – obviously – touches photosynthesize process, which yields glucose and other carbohydrates, I pick up CO2 as an answer.
#13
Which of the following types of molecules is always found in virions?Show answer
(A) Lipid
(B) Protein
(C) Carbohydrate
(D) DNA
(E) RNA
First, we should define what virion is. Surprisingly, it is just one instance of object virus. So you talk about virus (e.g. fag T4) properties, but when you meet one in the shop, it is virion.
What do we know about virions? They are simple systems which are not really cells because of lack of membrane and other important compartments as well as functions and facilitates. They have got capsid – hard-breaking protein-lipid armor around internal stuff. Inside they have some proteins and nucleic acid: either DNA or RNA, either single- or double-strained molecules. So the answer is proteins as molecules, that are found in all viruses and thus virions.
What do we know about virions? They are simple systems which are not really cells because of lack of membrane and other important compartments as well as functions and facilitates. They have got capsid – hard-breaking protein-lipid armor around internal stuff. Inside they have some proteins and nucleic acid: either DNA or RNA, either single- or double-strained molecules. So the answer is proteins as molecules, that are found in all viruses and thus virions.
#31
The difference between the molecular weight of sucrose and that of the sum of the molecular weights of its components (glucose and fructose) isShow answer
(A) 0
(B) 1
(C) 16
(D) 18
(E) 180
If you remember scheme structure of sucrose as O_O (:-) you get the idea of question. The reaction of such polymerization looks like that:
Glu-OH + OH-Fru = Glu-O-Fru + HOH
Thus, difference is in one water molecule which is 2 hydrogen atoms and 1 oxygen. The answer is D: 18 = 2*1 + 16.
Glu-OH + OH-Fru = Glu-O-Fru + HOH
Thus, difference is in one water molecule which is 2 hydrogen atoms and 1 oxygen. The answer is D: 18 = 2*1 + 16.
#34
An alpha-helical conformation of a globular protein in solution is best determined by which of the following?Show answer
(A) Ultraviolet-visible absorbance spectroscopy
(B) Fluorescence spectroscopy
(C) Electron microscopy
(D) Analytical ultracentrifugation
(E) Circular dichroism
I really looooove fluorescence. So I hope it is the right answer, but lets be patient.
Electron microscopy is not good idea, because it can't show us structure of the protein as x-ray crystallography make it possible.
Ultracentrifugation, I suppose, will not help us because it will just sediment peptides and nothing more.
Circular dichroism is damn strange stuff I don't know much about. But google fetched me nice page about circular dichroism. I got the idea of that method: difference in absorbances of left- and right-polarized light is higher if peptide is structured. It is rather similar to fluorescence anisotropy. High anisotropy shows that fluorophore is fixed in space (and time :-) It can also predict stable structures as alpha-helices or beta-layers.
Electron microscopy is not good idea, because it can't show us structure of the protein as x-ray crystallography make it possible.
Ultracentrifugation, I suppose, will not help us because it will just sediment peptides and nothing more.
Circular dichroism is damn strange stuff I don't know much about. But google fetched me nice page about circular dichroism. I got the idea of that method: difference in absorbances of left- and right-polarized light is higher if peptide is structured. It is rather similar to fluorescence anisotropy. High anisotropy shows that fluorophore is fixed in space (and time :-) It can also predict stable structures as alpha-helices or beta-layers.
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