Information

Can you freeze an ATP assay sample for later use?


I am going to use adult Drosophila to do an ATP assay (ATP determination kit A22066). The most convenient for me is to freeze the samples and perform the experiment later.

However, the protocol provided by the kit producer keeps saying the assay is extremely sensitive, so I'm not sure if it's okay to keep the homogenized samples under -80°C for a short period of time.


Pure ATP can be stable when stored frozen but I would be worried, perhaps unjustifiably, about storing a homogenized sample. You should try dividing a homogenized sample in half and assaying one half right away and the other half after freezing to see how the signals compare.


Adenosine 5′-triphosphate–Agarose

If available for a given product, the recommended re-test date or the expiration date can be found on the Certificate of Analysis.

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Transportation information can be found in Section 14 of the product's (M)SDS.To access the shipping information for this material, use the link on the product detail page for the product.

How can I hydrate Product A2767, Adenosine 5´-triphosphate-Agarose?

The resin can be hydrated by placing it in excess water, approximately 50 mL per g of resin, for at least 30 minutes. Remove the lactose stabilizer by washing the resin on a Buchner funnel with gentle vacuum, using approximately 100 mL of water per g of resin. Do not allow the resin to dry. Resuspend the resin in excess water or starting buffer to pack the column bed.

How can I elute my proteins from Product A2767, Adenosine 5´-triphosphate-Agarose?

Specifically bound proteins can be eluted with 10-100 mM ATP or ADP. Nonspecifically bound proteins can be eluted with 2 M NaCl or KCl in water or 7 M urea.

How can I regenerate Product A2767, Adenosine 5´-triphosphate-Agarose?

Since N6-[N-(6-aminohexyl)-carbamyl]-ADP is a substrate for acetate kinase and pyruvate kinase, these enzymes can be used to regenerate ATP from ADP on the resin under conditions favoring the reverse reaction. Wash the resin with 20 volumes of 50 mM Tris HCl buffer (pH 8.2), with 0.2 mM EDTA and 100 mM KCl. Incubate the resin overnight at 2-8 ° C in an ATP-regenerating mixture containing 0.2 mM phosphoenol pyruvate, pyruvate kinase (10 units per mL of resin), 5 mM MgCl2, 0.2 mM EDTA and 100 mM KCl in 50 mM Tris HCl buffer (pH 8.2). Wash the resin with 25 column volumes of 2 mM ATP in 2 M KCl and reequilibrate the resin with 25 column volumes of sample buffer. Alternatively, the ATP regenerating mixture can contain 20 mM acetyl phosphate, acetate kinase (6.8 units per mL of resin), and 3 mM MgCl2 in 100 mM Tris HCl (pH 7.6). If acetate kinase is used, prewash the resin with 25 column volumes of 100 mM Tris HCl (pH 7.6), but use the same final wash and reequilibration steps as given for pyruvate kinase.

Can I store Product A2767, Adenosine 5´-triphosphate-Agarose, after I have hydrated it?

For the short term, it is possible to store the hydrated resin in 20% ethanol and dilute buffer at 2-8 ° C. In general, the hydrated resin can be stored refrigerated in water or buffer containing a bacteriostat, such as 0.02% sodium azide or thimerosal. Do not autoclave or freeze the hydrated resin. The resin can be used several times without loss of effectiveness. However, the ATP will slowly hydrolyze to ADP over time, and under certain conditions this process may be accelerated during usage. The resin can be regenerated please see the regeneration FAQ for this product.


REVIEW article

Olivier Braissant 1* , Monika Astasov-Frauenhoffer 2 , Tuomas Waltimo 2 and Gernot Bonkat 3
  • 1 Department of Biomedical Engineering, Faculty of Medicine, University of Basel, Allschwil, Switzerland
  • 2 Department Research, University Center for Dental Medicine, University of Basel, Basel, Switzerland
  • 3 Alta-Uro AG, Basel, Switzerland

Viability and metabolic assays are commonly used as proxies to assess the overall metabolism of microorganisms. The variety of these assays combined with little information provided by some assay kits or online protocols often leads to mistakes or poor interpretation of the results. In addition, the use of some of these assays is restricted to simple systems (mostly pure cultures), and care must be taken in their application to environmental samples. In this review, the necessary data are compiled to understand the reactions or measurements performed in many of the assays commonly used in various aspects of microbiology. Also, their relationships to each other, as metabolism links many of these assays, resulting in correlations between measured values and parameters, are discussed. Finally, the limitations of these assays are discussed.


Abstract

Since mica is a substitute for glass in the in vitro actin motility assay, I examined the structure of heavy meromyosin (HMM) crossbridges supporting actin filaments by quick-freeze deep-etch replica electron microscopy. This method was capable of resolving the inter-domain cleft of the monomeric actin molecule. HMM heads that are not bound to actin, when observed by this technique, were straight and elongated in the absence of ATP but strongly kinked upon addition of ATP or ADP·inorganic vanadate to produce the putative long-lived analog of HMM-ADP·inorganic phosphate. The low-magnification image of the ATP-containing acto-HMM preparation showed features characteristic of sliding actin filaments on glass coverslips. At high magnification, all the HMM molecules were found attached to actin by one head with the majority projecting perpendicular to the filament axis, whereas in the absence of ATP, HMM exhibited two-head binding with a preponderance of molecules tilted at 45°. Detailed examination of the shape of HMM heads involved in sliding showed a rounded, and flat appearance of the tip and comparatively thin neck portion as if the heads grasp actin filament, in contrast to rigor crossbridges which have a pear-shaped configuration with more gradual taper. Such configurations of HMM heads were essentially the same as I observed previously on acto-myosin subfragment-1 (S1) by the same technique, except for the presence of an additional neck portion of HMM which makes interpretaion of the images easier. Interestingly, under actively sliding conditions, very few heads were tilted in the rigor configuration. At first glance, the addition of ADP to the rigor-complex gave images rather like those obtained with ATP, but they turned out to be different. The contribution of the structural change of crossbridges to the force development is discussed.


Detection of Mycoplasma in cell cultures

Mycoplasma is a prokaryotic organism that is a frequent and occult contaminant of cell cultures. This organism can modify many aspects of cell physiology, rendering experiments that are conducted with contaminated cells worthless. Because of their small size, Mycoplasmas can pass through filters used to prevent bacterial and fungal contamination and potentially spread to all the cultures in a laboratory. It is essential that all new cell cultures entering a laboratory and all cell banks are tested for the presence of Mycoplasma. It is recommended that two techniques be used, selected from a PCR-based method, indirect staining and an agar and broth culture. This protocol describes these three tests for detecting Mycoplasma, which take from 1 d to 3–4 weeks, and such tests should be an obligatory component of quality control in every tissue culture laboratory.


Diffusion, Membranes, and Cryogenics: Can We Freeze Cells?

I cut a beet into identical weight sections (14 g), and placed then in jars with varying salt concentrations: 0%, 10%, 20%, 30%, 40%, and 50%. I set up 3 sets of these, and placed 1 set in a freezer, 1 set in the refrigerator, and the final set at room temperature. After 24 hours I measured cell damage by measuring the concentration of beet pigment in the water the rupturing of cells would release the pigment into the water. I used a spectrophotometer to measure the percent transmission of light as a measure of pigment concentration in the water.

There was little cell damage in the refrigerator and none in the room temperature jars, but the freezing caused considerable cell damage. However, when cells were preserved with a salt concentration greater than 20%, there was considerably less damage, and little cell damage at a concentration of 40%.

The significance of these findings is that there is a way to save cells and preserve them. This also indicates that someday, people may be able to freeze animals or humans and bring them back to life.

Problem

What will pass through a cell membrane?

What will pass through a membrane? Starch and sugar are being tested in this experiment to see if they will go through a membrane. The starch molecule is much bigger than the sugar molecule, being that the sugar I used is a simple sugar (glucose). The sugar should be able to go straight through the membrane, however, the starch will be too big to.

I hope to accomplish being able to make a "cell" and really be able to see how a membrane guards the cell. An experiment on cells last year in school prompted my research and sparked my curiosity for chemistry altogether.

Is it possible to preserve cells during freezing?

Being that it is almost impossible to freeze a cell with no protection and still have it live, this project may show a way. Is it possible? It will be (easily) possible to freeze a cell and have it live and function properly.

I hope to accomplish the preservation of cells in freezing conditions. An idea from my science teacher prompted my research. I was very anxious to see how to preserve cells even after freezing them.

Background Information

What Will Pass Through a Membrane?

A membrane is a thin layer of lipids (fats) and proteins around a cell. These decide what can come into or come out of a cell. This movement of going into or out of a cell is called osmosis. Naturally particles go from a place of high concentration to a place of low concentration (diffusion). (The particles might not do this if they are forced to move.)

The two groups within carbohydrates are called starches and sugars. The general formula for carbohydrates is (CH2O)n. ("n" is the number of carbons in the backbone) One carbohydrate is called a simple sugar or a monosaccaride. Glucose, like galactose and fructose, is an isomer. An isomer is a compound which differs in structure but not in molecular composition. Disaccharides are two monosaccharides combined in a condensed reaction. Polysaccharides are simple sugar building blocks bonded together to form chains.

Glycogen, or animal starch, is a place to store energy in animals. Glycogen is made up of glucose molecules strung together in a highly branched chain for animals. In plants, however, glucose in the form of a polysaccharide and a monosaccharide is also called a starch. Starches are metabolic reserves which are manufactured by green plants through photosynthesis. (Starches occur in the form of grains.) There are two basic forms of starches, unbranched chains that coil, and branched chains which are similar to glycogen.

Feezing Cells

If the liquid in a cell is not reduced when the cell freezes, then when the water in the cell freezes and expands, the membrane of the cell will break. (A cell membrane is the outermost part of the cell which lets the objects come into or go out of the cell through diffusion.) The inside of the cell will then leak out and the cell dies. (There are some cells in certain animals such as the wood frog which are protected from freezing due to inner body liquids.)

Some water may be taken out of cells that are not protected, through, by using a hypertonic solution. ( Through there are many types of hypertonic solutions this one would probably contain either sugar or salt in it.) This hypertonic solution would also probably also have to totally surround the cell so that there will be diffusion taking place going out of the cell. Diffusion is the movement going into or out of a cell from a place of high concentration to a place of lower concentration. Not all things, however can diffuse into a cell. The object must be small enough to pass through the cell membrane. (The cell membrane is made of mostly lipids which have very tiny holes in them.)

Hypothesis

What Will Pass Through a Membrane?

I expect to see the glucose go through the membrane and the starch not being able to because a starch molecule is too big. The glucose molecule is a simple sugar and a rather small molecule. It should be able to fit through the membrane easily.

Freezing Cells

I expect to see the salt preserving the beets when they are frozen because the salt will help the water in the beet cell (some of it) to diffuse out. This will leave enough room for the water in the cell to expand and still fit in the cell.

Materials

*Benedict's solution with dropper

*Iodine with dropper

*about 30 test tubes

*beakers (6 or 7)

*heating surface

*rubber bands and/or tape

*graduated cylinder (10ml.)

*test tube rack

*Beets (1 or 2 small ones)

*sharp knife

*salt (large container)

*18 baby food jars (or at least jars around that size)

*spectrophotometer (colorimeter)

*a scale that weighs grams

Procedure

  1. Make a scale of different colors made when .1%, .09%, .08%, .07%, .06%, .05%, .04%, .03%, .02%, of glucose is added to Benedict's Solution and put in a hot water bath for 5 minutes.
  2. Label the jars (test tubes) and take pictures.
  3. Place 5 ml. of .1% glucose into two test tubes.
  4. Place 10 ml. of .1% glucose into two different test tubes.
  5. Place a piece of membrane over the top of each test tube and make sure to "seal" the sides of the membrane with tape and/or rubber bands.
  6. Place each test tube into a beaker (upside down) filled with 50 ml. of water.
  7. Place 5 ml. of water into a test tube with a tablespoon of starch. (Do this twice to ensure accuracy.)
  8. Place a piece of membrane over the top of each test tube and make sure to seal the sides of the membrane with tape and/or rubber bands.
  9. Place each test tube into a beaker (upside down) filled with 50 ml. of water.

10. Wait overnight and put 5 drops of Benedict's Solution in each test tube that the glucose is in.

11. Place each in a hot water bath and compare how much extra water diffused up with the pictures taken.

12. Put 5 drops of Iodine in each starch beaker, then place them in a hot water bath for 5 minutes. If the color changes darker, then the starch passed through the membrane.

  1. Use the ruler to cut out 12- 1 cm 3 sized beet sections.
  2. Weigh the beets, and make sure that all of them weight the same size. (14 grams.)
  3. Fill 3 jars with plain tap water.
  4. Fill another three jars with 10% salt content in 50 ml. of water.
  5. Fill another three jars with 20% salt content in 50 ml. of water.
  6. Fill another three jars with 30% salt content in 50 ml. of water.
  7. Fill another three jars with 40% salt content in 50 ml. of water.
  8. Fill the last three jars with 50% salt content in 50 ml. of water.
  9. Split the jars into three sets each having 0%, 10%, 20%, 30%, 40%, and 50% salt content in 50 ml. of water.

10. Place a beet cube in each of the jars of 2 of the sets. Leave the third set empty!

11. Place one of the sets with beets in it in the freezer. Place the other two sets in the refrigerator.

12. After a day or two, take all of the beet jars out.

13. Use the spectrophotometer to measure the Absorption and Transmittance percentage for each of the beet jars. Use the light setting 530 nm. Use the set of jars with no beets in them as the blank samples (control).

Feezing Cells: Effect of Salt Concentration on Cell Damage

Discussion

What Will Pass Through a Membrane?

The data in this project tells us where a rough boundary line is of how big a molecule can be and still be able to diffuse through a membrane. This could help us understand why digestion is so important and takes so long to actually happen. This could also tell us what kinds of things you could actually have in your cells.

Freezing Cells

The data for this project tells us how much salt is needed for the preservation of cells after freezing them. Damaged cells release their pigment into the water, thus decreasing light transmission. 0% salt solution after freezing resulted in 0% light transmission, showing almost complete cell damage. In a 20% salt solution, the preservation rate increased dramatically, indicated by an increase in light transmission. At 40% salt content, most cells seem to have been prevented from rupturing. The refrigerator data acts as a control and shows little cell damage, indicted by high light transmission. There was a drop in light transmission at 40% salt and above. I do not have an explanation for this.

The data also tells us that it might be possible to freeze cells and bring them back to life afterwards. This is one of the first steps in the preservation of animals.

Conclusion

What Will Pass Through a Membrane?

My hypothesis that a starch molecule would be too large to go through a membrane was correct. I was also correct that glucose was able to diffuse. This project proves that even though both starches and sugars are carbohydrates, they are different in size. Starch molecules are too complex to diffuse through a membrane. The project proves that not all molecules can pass through a membrane.

This information helps us understand the process of digestion. The food molecules have to be digested to a certain size for the cell to "accept" it.

Freezing Cells

My hypothesis, that the salt would preserve frozen beet cells, was correct. The jars with the most salt in the freezer were only affected by the freezing a little bit, while the jars with little salt were bright purple from dead cells.

This project proves that there is a way to save cells and preserve them. This also proves that one day people might be able to freeze animals and then bring them back to life. The significance of these findings is that if later in the history of the world, people could be frozen, then brought back to life so many years later, they would be able to tell much about history.

My recommendations for future study is to use a variety of cells and try preservation on as many as you possibly can it may help history! Another recommendation would be to do further tests to see if the frozen cells actually survive after thawing. Also one could use different types of preservation to see what works the best.

Bibliography

Starr, C., Biology: Concepts and Applications, Belmont, CA, 1991, pp. 27 - 29, 40 - 43.

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Use materials and procedures

Culture of diatoms

Xenic cultures of Thalassiosira weissflogii were grown in an f/2 + medium (Guillard 1962) in 34 psu GF/F-filtered artificial seawater (Instant Ocean) at 66 μmol m −2 s −1 continuous illumination. For the first batch culture experiment (i.e., the nutrient limitation time course), diatoms were cultured in 6 × 2 L Erlenmeyer flasks, covered with aluminum foil, and filled with 1 L of f/2 + medium. Then, 1 mL of an exponentially growing starter culture was transferred to each flask as inoculum, and the experiment was conducted at room temperature (approximately 20°C). For the second set of experiments, diatoms were grown in 2-liter flasks containing 1 liter of f/2 + medium with loose plastic screw caps at the same illumination. Two flasks for each temperature treatment were immersed in water baths at 15°C, 20°C, and 25°C with temperature-controlled baths. Separate smaller batch cultures of the same alga grown in an f/2 + medium at room temperature and the same light conditions were used to test extraction and analytical procedures.

Extraction of ATP

A variety of extraction procedures were tested however, most resulted in a strong inhibition of the luciferin-luciferase reaction. Inhibitory effects were observed with detergents such as Triton X-100 and Dodecyltrimethylammonium bromide (DTAB), as well as with digestion with acid and bases that needed to be neutralized in a separate step, and thus, had a high salt content. These chemicals interfered with the bioluminescence assay in a similar way to the chemical extraction procedure discussed below. We found that a simple extraction procedure using boiling-hot ultrapure water was highly compatible with the ATP assay, corroborating the findings by Yang et al. ( 2002 ). We also explored a chemical extraction procedure using P-BAC, a mixture of benzalkonium chloride (1% w/w fin. conc.) with phosphoric acid (5% w/w fin. conc.) in a 25 mM Tricine buffer (Welschmeyer and Kuo 2016 ). Note that the density of 85% phosphoric acid is 1.6845 kg L −1 , and concentrations are expressed in weight percentages.

To compare the hot-water with the P-BAC extraction protocols directly, we designed an experiment in which the concentration of ATP is the same for the final step of the luminescence assay (Fig. 1). For these experiments, 5 mL each of six samples was filtered onto GF/F in the same filter rack (6-positions of 25 mm glass filtration manifold fitted with a stainless-steel grid [Millipore]) and using the same vacuum. The time required from transferring the filters from the filter rack to the first extraction step was also the same in the liquid-nitrogen hot-water and P-BAC procedures. For the hot-water extraction procedure, the filters were placed in 15 mL polypropylene centrifuge tubes (Falcon) and shock frozen in liquid nitrogen. The centrifuge tubes can be stored at −80°C or used after a few minutes. However, boiling-hot ultrapure water must be poured directly on the still frozen filters to inactivate ATPases and not be thawed slowly. The centrifuge tubes were then vortexed for 10 s each and immersed in a beaker with boiling-hot water for

15 min (Fig. 1A). After cooling to room temperature, ultrapure water was added to the 5 mL mark in the centrifuge tubes and vortexed again for a few seconds to mix the sample homogenously. For P-BAC extraction, samples were soaked at room temperature for

30 min in cryovials filled with 1 mL P-BAC (Welschmeyer and Kuo 2016 Fig. 1B). Samples can be stored frozen at this stage or analyzed after the extraction interval. For the final analysis using the firefly reagent, 50 μL from the hot water or 10 μL from the P-BAC extractant were used. This approach kept the ATP concentrations the same in the final analytical step for both extraction methods (i.e., 50 μL from 5 mL extractant = 10 μL from 1 mL extractant). To a subset of samples, a 50 μL of standard (ranging in concentrations from 3.24 to 162 nM) depending on the type of sample, was added, chased with 3 mL of ultrapure water for dilution (see the Discussion section for the requirement of a dilution step in the P-BAC procedure). Finally, 50 μL of firefly extract was added (CellTiter-Glo 2.0). For test samples, we used cultures of T. weissflogii grown in an f/2 + medium, collected surface water from the Elizabeth River at the Old Dominion University Sailing Center (16 September 2020, Latitude: 36.885 N, Longitude: 76.308 W, salinity: 16.4 psu, temperature: 24.0°C), and mesopelagic water offshore at 300 m (06 October 2020, latitude: 36.734 N, longitude: 74.629 W, salinity: 34.4 psu, temperature: 12.7°C).

Growth rate experiments with T. weissflogii

In all experiments, sampling occurred daily, beginning at the day of inoculation of the f/2 + medium. After agitating the flasks to resuspend and evenly mix settled cells, 25 mL were withdrawn from each flask using a sterile disposable pipette. Then, 5 mL were immediately filtered through a 25 mm diameter, 0.2 μm polycarbonate filter (Isopore type GTTP) at a vacuum of 200 mbar in a filter station preloaded with six filters. The filtrate was captured in 15 mL polypropylene centrifuge tubes (Falcon) placed below the filter funnel inside the vacuum flasks. The filters were quickly transferred into 15 mL polypropylene centrifuge tubes, and

4.5 mL boiling-hot ultrapure water (18.2 MΩ) was added. The centrifuge tubes were vortexed for a few seconds and then immersed in a beaker containing hot water (> 90°C). The hot ultrapure water breaks up cells and inactivates ATPases (Yang et al. 2002 ). For unfiltered whole water, triplicate samples of 0.5 mL were added to 15 mL centrifuge tubes chased by

4 mL boiling-hot water and vortexed. Of the filtrate, 0.5 mL was transferred to the centrifuge tubes and treated in the same fashion as the whole water samples. All tubes were kept in hot-water baths for approximately 15 min. The samples were subsequently cooled to room temperature and analyzed the same day. Since ATPases are largely inactivated at this stage, samples can also be frozen at −20°C or −80°C for later analysis if desired. Using a muffled Pasteur pipette, ultrapure water was added dropwise to exactly the 5 mL mark (markings on the 15 mL Falcon centrifuge tubes are sufficiently accurate). For analysis, 50 μL of sample, 3 mL of ultrapure water, and 50 μL of the ATP working stock (for the internal standard only, see below), and 50 μL of firefly extract (CellTiter-Glo 2.0, Promega Corp.) were combined in 6 mL plastic scintillation vials (Pico Prias, Perkin Elmer) and briefly vortexed.

We programmed the scintillation counter to run samples in sequence repeatedly up to 10 times, and only used the data of the second cycle. The other cycles were used to observe the decaying luminescence signal (Supplementary Fig. 1). Repeated runs are typically unnecessary, and if samples are run only once, we recommend providing a reaction time of approximately 30 min before the first vial is counted to count over the more linear time trajectory. We always alternated samples and corresponding internal standards.

Standards

10 μM using ultrapure water. The stock solution was then divided into individual 15-mL polypropylene centrifuge vials and stored frozen at −20°C. The exact concentration of the

10 μM standard was determined using a spectrophotometer (Shimadzu UV-2401PC) at a wavelength of 259 nm, a 1 cm cuvette and a molar absorptivity coefficient of 15.4 × 10 3 M −1 cm −1 (Karl 1993 ) in the equation

(1)

Calculations

(2) where [ATP] is the concentration of the internal standard (i.e., 16.4 nM), CPMsample the photon counts per minute for the sample, the average value of 4 to 6 blanks (50 μL of firefly reagent added to 3 mL ultrapure water only), the average counts per minute for the standard vials, the average value of replicate samples, Vstd the volume of the standard added to the scintillation vial in μL, R the ratio between volume of the extract (numerator) and volume of sample filtered (denominator), and Vextr the volume of the extract added to the scintillation vial (μL). If many samples are counted, as in our case, the most accurate procedure is to calculate the standard values by using a regression equation with time stamp of counting as explanatory variable and then applying this regression to every sample based on its time stamp in the counting protocol. This procedure, rather than using the values from the adjacent vials, reduces variance caused by standard-to-standard variability. However, if ATP concentrations are expected to be very different between samples, we recommend subtracting the values of the individual vials from those containing the internal standards instead. To arrive at the final ATP values in the results, the values were further multiplied by two to correct for the extraction efficiency difference between material collected on filters and whole water samples (Fig. 2). This factor was highly consistent between the two main experiments and in other experiments with T. weissflogii.

Cell counts

T. weissflogii was enumerated in a Z2 Coulter Counter Multisizer with a 100 μm orifice tube. Seven measurements each were made for particle sizes larger than 10 μm, the first values discarded, and the next six measurements recorded (the first measurement may still contain air bubbles).

(3)

Then, 5 mL of each sample was preserved in 2% (fin. conc.) 0.2 μm syringe-filtered formaldehyde stabilized with 10% methanol for bacterial counts. The following day, 1 mL of the formaldehyde-preserved sample was filtered onto 25 mm diameter 0.2 μm pore-size black polycarbonate filters (Isopore type GTBP). The filters were transferred onto microscope slides. One or two drops of a mounting medium containing DAPI (Vectashield, Vector Laboratories) was added, covered with a cover slip, and stored at −20°C until counting. Bacteria were enumerated under an epifluorescence microscope (100 × oil immersion objective lens, 2 × loupe, 10 × ocular magnification, equipped with a DAPI filter cube) by counting 60 fields (0.025 mm 2 each) per slide (30 vertical, 30 horizontal across the filter). Field sizes were determined using a stage micrometer with 0.01 mm increments.

T. weissflogii carbon was calculated using the equation provided by Strathmann ( 1967 ) as 110 pg cell −1 based on a diameter of 15 μm. This per-cell carbon value matched well the raw data on T. weissflogii (formerly Thalassiosira fluviatilis) in Strathmann ( 1967 ) and was very similar to the values calculated using equations reported in Menden-Deuer and Lessard ( 2000 average between regressions of small and large diatoms). For prokaryotes, 20 fg carbon cell −1 was used. This number is suitable for bacteria grown in culture but likely lower for prokaryotes found in the wild (Fukuda et al. 1998).

Literature comparisons

To convert ATP per dry weight from various literature sources (e.g., tables in Karl 1980 ) into molar concentrations (moles of ATP per-cell volume), we needed to make assumptions regarding the dry mass density per-cell volume. We used a universal conversion of 0.2 g cm −3 (pg μm −3 ) dry-matter content with the acknowledgment that the dry-weight to volume ratio may change from 0.1 to 0.57 g cm −3 (Bratbak and Dundas 1984 Norland et al. 1987 Simon and Azam 1989 ). A factor of 0.2 g cm −3 is also a suitable approximation for bacteria given a similar wet-weight to dry-weight relationship (Bakken and Olsen 1983 ). To test the conversion factor on algae, we calculated the dry-weight to volume ratio of a wide range of phytoplankton from small to large, from Thalassiosira pseudonana to Lingulodinium (previously Gonyaulax) polyedra using the allometric volume to carbon conversions in Menden-Deuer and Lessard ( 2000 ). Using a 50% carbon to dry-weight conversion, we arrived at 0.26 g cm −3 for T. pseudonana (the smallest alga in Menden-Deuer and Lessard 2000 ) and 0.13 g cm −3 for L. polyedra (the largest alga in Menden-Deuer and Lessard 2000 ), which brackets the conversion factor we used. For Skeletonema costatum, we used the mean of ATP (0.041 pg cell −1 Sakshaug 1977 ) and the range of cell volumes for S. costatum as reported in Strathmann ( 1967 ) to calculate molar concentrations of ATP. In the absence of other information for ciliates, we used the lorica volume of tintinnid ciliates in Verity and Lagdon ( 1984 ) to calculate molar concentrations of ATP from pg per cell in that taxonomic group.

Statistical analysis

Residuals were tested for normality using the Komolgorov–Smirnov test (Lilliefors 1967 ). Since, in several cases, residuals did not conform to the normality assumption of parametric tests, p-values were based on 10,000 randomizations of the data and custom F-distributions (Manly 2007 ).


Can you freeze an ATP assay sample for later use? - Biology

The presence of antibodies to Newcastle disease virus in chickens is detected by serological testing. The results of these tests are used for three purposes.

1. To assess the efficacy of Newcastle disease vaccine in laboratory and field trials.

2. To assess the level of Newcastle disease virus antibodies in the field.

3. Serum known to contain antibodies to Newcastle disease virus is used to confirm the presence of Newcastle disease virus in a test sample of allantoic fluid. Such a sample would be obtained during the isolation of virulent Newcastle disease virus. See Section 15 .

There are two assays commonly used to carry out serological testing for Newcastle disease virus antibodies.

1. Haemagglutination inhibition (HI) test. The HI test is a convenient and commonly used assay that requires cheap reagents and is read by eye.

2. ELISA (Enzyme linked immunosorbent assay). This is a colourimetric assay and requires the use of a sophisticated instrument to read the optical density of the reactions. ELISA kits for Newcastle disease virus antibody detection are prepared and sold commercially. Detailed instructions are supplied with the kits. They are usually quite expensive.

In this manual a protocol for the HI test based on the test described by Allan and Gough will be used for serological testing. (Allan and Gough, 1974 a.)

Serum samples are collected for testing for the presence of antibodies to Newcastle disease virus. Blood is collected as described in Section 7. The blood forms a clot in the syringe in a few minutes. Once the blood clots, the syringes of blood can be kept with the needle upright to prevent serum filling the needle cap. The serum will separate from the clot within a few hours at room temperature or in approximately 2 hours at 37°C. Storage at 37°C will help the serum separate from the clot.

  • Syringes with blood samples.
  • Sterile glass Pasteur pipettes.
  • Microfuge tubes.
  • Storage tubes. (1.8 mL Nunc cryotubes are ideal but microfuge tubes are a cheaper alternative.)
  • Sharps container and discard bag for biological waste.

1. Remove the plunger from the syringe. Transfer the serum to a microfuge tube by pouring or using a glass Pasteur pipette.

2. Dispose of needles, syringes, clots and pipettes in appropriate containers.

3. Often the serum will contain red blood cells. Centrifuge for 30 seconds in a microfuge centrifuge or allow to settle under gravity overnight at 4°C to pellet the cells. Do not freeze the samples at this step. Freezing will lyse the red blood cells.

4. Transfer clear serum to a second tube.

5. Remember to label each tube after the transfer of the serum. This ensures the serology results can be applied to individual birds and groups.

Store ampoules of serum at -20°C.

Storage at 4°C is acceptable for a short period, up to 2 weeks.

Samples collected in the field may end up at room temperature overnight and often do not separate well.

It has been noted at the John Francis Virology Laboratory that in some samples the serum does not separate from the clot. In these cases, the whole clot is centrifuged. This usually results in a small amount of serum separating.

It is important that the serum samples are clear. Pink samples contain lysed red blood cells. When pink samples are tested observe both test and control samples carefully to determine effect of the colour on the results of the test.

Pink coloured samples can affect results of an ELISA, which is a colourimetric assay.

An overview of the Haemagglutination Inhibition (HI) test

Antibody response to the haemagglutinin protein in the Newcastle disease virus envelope can be measured by the HI test.

When serum containing these antibodies is mixed with Newcastle disease virus, the antibodies bind to the haemagglutinin protein in the envelope of the virus. This blocks the haemagglutinin protein from binding with the receptor site on chicken red blood cells.

Thus the haemagglutination reaction between the virus and the red blood cells is inhibited.

By performing two-fold serial dilutions on the serum prior to testing, the concentration of the serum antibodies can be expressed as an HI titre to the log base 2.

Standardization of the HI test

It is important that there is correlation between the results of tests carried out by different technicians and in different laboratories. For this reason HI tests should be standardized both within a laboratory and between laboratories.

Standardization is achieved by following a standard protocol. This will include:

Using a standard 4 HA units of Newcastle disease virus antigen.

Using standard positive anti-serum and negative serum.

Including a serum control for each test serum to detect the presence of non-specific agglutinins.

Using a standard 1 percent dilution of red blood cells.

The use of control serum in the HI test

Positive and negative control sera are tested to avoid errors in the interpretation of the results of the HI test. Inconsistencies in the results of the HI test may be caused by variations in reagents and procedures.

Examples of possible variations include:

The source and exact percentage by volume of the red blood cells.

The exact amount of antigen used in the HI test.

Accuracy of dilutions.

Temperature.

Time allowed for the antibody/antigen reactions to occur.

Quality of test serum samples (haemolysed serum samples may be responsible for non-specific reactions)

Two standard sera are used.

1. A standard negative control serum known to contain no antibodies to the Newcastle disease virus. It has no HI titre and does not agglutinate chicken red blood cells.

2. A standard positive control serum also known as a standard anti-serum. The HI titre of the anti-serum will have been established by repeated titration.

Variability in HI titres of standard positive serum

It is sometimes observed that the standard HI titre of the standard anti-serum tested on the same day with the same antigen preparation and protocol may vary. A difference in HI titre of one dilution that is one log base 2 (2 1 ) can be regarded as due to random errors and an inherent variability in biological responses. However if the titre of the standard positive serum differs by more than one dilution or log base 2, then the test is invalidated. In this case, fresh antigen must be prepared and tested and the HI test repeated.

Three categories of standard sera

1. International standard reference serum. Certain laboratories prepare reference serum to a very high standard. Standard reference Anti-Newcastle disease serum is available for purchase. It is used as a reference serum in the preparation of national and laboratory standard sera.

The National Institute for Biological Standards and Control (NIBSC) in the United Kingdom can supply an international standard reference serum. The potency of the international reference serum has been determined by the supplier and is expressed in International Units (IU) per ampoule of freeze-dried serum.

Email [email protected]
Tel 44 (0) 1707 654753
Fax 44 (0) 1707 646730
Website http://www.nibsc.ac.uk

2. National standard serum is prepared by a national laboratory and distributed to collaborating laboratories as required. This serum is prepared by mixing sera with known HI titres. A series of HI tests are carried out to establish the HI titre of the national standard. This is compared with the HI titre of the international standard reference serum if available. The national standard serum is then stored in multiple aliquots and distributed to collaborating laboratories within a country.

3. Laboratory standard serum is prepared by some laboratories in order to conserve supplies of national standard serum. The laboratory standard is prepared by mixing serum samples with known HI titres. Comparative HI testing of the laboratory standard and the national standard is used to establish the HI titre of the laboratory standard.

Preparation of national and laboratory standard sera

Each laboratory will require a supply of negative and positive control serum in order to carry out HI tests on serum samples.

Collection of HI negative serum

Collect serum from chickens that have had no exposure to Newcastle disease virus. The serum should show no inhibition of viral haemagglutination activity when tested by the HI test. It is often difficult to find serum without any HI activity. In this case use serum with low HI titres of 2 1 .

Collection of HI positive serum

There may be stored serum that is suitable for this purpose. If not, vaccinate several chickens with I-2 Newcastle disease vaccine. Two weeks after the primary vaccination, give a booster vaccination. Two weeks later, take a serum sample from each chicken and test the HI titre. Collect extra serum from chickens with a titre of 2 5 or above. The volume of blood collected from each chicken depends on the size of the chicken. Pool all the serum samples with HI titres of 2 5 or above and test the HI titre of the pooled serum.

Comparative HI testing of international reference and national standard sera

Information supplied by NIBSC in 2001 indicated their international standard reference contained 320 IU per ampoule.

The serum can be reconstituted in PBS. A suitable volume of PBS would be 2 mL, which would give 4 IU in the 25 µL test sample. Refer to instructions provided by the manufacturer.

Use the standard HI test described in this section to determine the titre of the international reference standard and the national standard.

Test both samples in triplicate on the same microwell plate.

Carry out a series of at least three tests on different days.

Once the HI titre of the international standard serum and the national standard serum have been established, prepare a series of two-fold dilutions of the serum that are spread around the endpoint used to establish the HI titres. Titrate the dilutions that range from complete inhibition to no inhibition as an additional check on the HI titres. The results of testing these dilutions will confirm the HI titre of the undiluted serum.

Example of confirming HI titre of standard samples by testing a series of dilutions of the sample.

Repeated testing of the pooled serum established the HI titre at 2 5 .

Prepare 1/8, 1/16, 1/32 and 1/64 dilutions of the pooled serum.

Test each dilution for HI titre.

Confirmatory results: The results tabled confirm that the HI titre of the serum used to prepare the dilutions had a titre of 2 5 .

Table 3: Results confirming HI titre of serum = 2 5

The test results will give a HI titre for both the International reference serum and the national laboratory serum being tested. The HI titre for 4 IU of the International reference serum can be used to determine the number of IU in the national reference serum.

The standard HI test of 25 µL of the international reference serum containing 4 IU and has a HI titre of 8 (2 3 ). The national serum tested HI titre of 64 (2 6 ).

How many IU does the national serum contain?

4 IU in a sample with a HI titre of 8.

How many (?) IU in a sample with a HI titre of 64.

Not all laboratories use the same protocol for testing HI titre and IU are independent of the test system used.

Serological results of samples expressed in IU should give equivalent results when tested by different systems.

By comparing the HI titre of the standard national serum with the HI titre of the international reference serum enables you to express the results of HI tests in IU. However it is not very often that you will be required to express the results of HI testing in IU.

Most publications of Newcastle disease serology results describe the assay used and express HI titres to log base 2.

Preparing a laboratory standard serum

Each laboratory should receive a national standard serum from a central laboratory. The HI titre and activity in International units of the serum will have been determined by rigorous testing as described above. To conserve the supply of national standard serum some laboratories will decide to prepare a secondary laboratory standard serum.

Collect positive serum as described above. Do comparative testing of the laboratory and national standards. Test each sample in triplicate on the same microwell plate. Repeat several times and analyze HI titres for both samples. Record the HI titre for the laboratory standard.

Store 1 mL aliquots of the laboratory standard at -20°C. This standard is used every time the HI test is carried out and should always show the same HI titre when tested with the 4 HA units of antigen.

Preparing a national standard serum without an international reference serum

Many laboratories will not be able to obtain an international reference serum with which to compare their national standard serum. In this case, more rigorous HI testing of the pooled HI positive serum samples will be required to establish the HI titre of the standard serum. It is suggested that the serum is tested up to ten times. Use freshly prepared 4 HA units of antigen for each test and carry out the test on different days if possible. The results will be a range of HI titres. The most frequently occurring titre can be considered the HI titre of the standard serum. All future HI tests carried out on the standard serum should give this titre.

Storage of standard serum

Once the positive and negative standard sera have been tested, aliquots can be prepared, labeled and stored frozen. Storage at -70°C is optimal (but -20°C is adequate). Do not thaw and refreeze the samples frequently. Representative samples should be thawed and tested to confirm that there is no loss of titre in the storage process.

Preparation of Newcastle disease virus antigen for use in HI tests

Antigen is prepared by inoculating embryonated eggs with a sample of Newcastle disease virus and harvesting the allantoic fluid four days later. Part of the first batch of I-2 Newcastle disease vaccine prepared from the I-2 working seed can be set aside for storage as antigen. A volume of 50 mL to 100 mL is adequate for a large number of tests. Centrifuge the sample at 1 200 g to clarify and remove any contaminating red blood cells. Store the antigen in one mL aliquots at -20°C.

Preparation of 4HA units of Newcastle disease virus antigen

The standard amount of Newcastle disease virus used in the haemagglutination inhibition (HI) test is 4HA units. It is necessary to prepare and test a suspension of Newcastle disease virus containing 4HA units in order to carry out the HI test. This involves a series of following steps.

1. Titrate the stored suspension of virus to be used as the antigen in the HI test. See Section 10 Calculate the HA titre.

2. Calculate the dilution factor required to produce 4 HA units. A simple way is to divide the HA titre by 4.

3. Apply the dilution factor and dilute the original suspension of antigen in PBS to produce an adequate volume of 4HA antigen to carry out the HI test. Allow 2.5 mL for each microwell plate.

4. Titrate the diluted (4HA) suspension of virus. This is a back titration to check the diluted antigen contains 4 HA units.

5. Read HA titre. It should equal 4HA units. If not adjust the dilution and titrate again.

6. Use the 4HA unit dilution of antigen in an HI test to test the standard positive and negative serum. The HI titre of the laboratory standard positive serum should equal the predetermined titre.

The results of the back titration of the diluted antigen and the HI titre of the standard positive are both used to confirm the antigen has been diluted to a concentration equivalent to the standard 4 HA units.

If the HI titre of the positive control serum is less than the standard titre, the antigen is too concentrated. Prepare a new dilution and test again.

Conversely if the HI titre of the positive control serum is too high the antigen is too dilute. Prepare a new dilution and test again.

Preparation of 4HA units of antigen for HI test in 10 microwell plates:

The HA titre of the antigen was tested according to the protocol described above. The HA titre = 128

Calculation of dilution factor to prepare 4 HA units: 128/4 = 32

Calculation of volume of 4HA unit dilution of antigen required:

Allow 2.5 mL per plate total volume required = 10 × 2.5 mL = 25 mL

Apply dilution factor = 25 mL/32 = 0.781 mL = 781 µL

Preparation of the diluted antigen: mix 781 µL of original virus suspension with 24.219 mL of diluent. PBS is a suitable diluent.

Note: In this case it would be easiest to prepare 32 mL of the 4HA antigen. This would use 1 mL of the original suspension diluted in 31 mL of PBS.

  • Thawed serum samples in racks
  • V-bottom microwell plates and covers
  • PBS
  • 1 percent washed red blood cells
  • V-bottom reagent trough
  • 25 µL single and multichannel pipettes and tips
  • Microwell plate recording sheet.
  • Newcastle disease virus antigen diluted to 4 HA units per 25 µL
  • Standard positive and negative serum

1. Fill in recording sheets to record how samples will be dispensed into microwell plates.

2. Calculate the number of plates required and number each plate.

3. Dispense 25 µL of PBS into each well of the plates.

4. Shake each serum sample and dispense 25 µL into the first well and the last (control) well of a row of a microwell plate.

5. Use a multichannel pipette to make two-fold serial dilutions along the row until the second last well from the end. The last well is the serum control. Do not dilute this well. See Appendix 4 for instructions on carrying out two-fold serial dilutions.

6. Add 25 µL of the 4HA dilution of antigen to each well excluding the control wells in the last column. See Section 10 for preparation of 4HA units of antigen

7. Gently tap the sides of the microwell plates to mix the reagents. Cover plates with a lid. Allow to stand for 30 minutes at room temperature.

8. Add 25 µL of 1 percent washed red blood cells to each well including the control wells in the last column.

9. Gently tap the sides of the microwell plates to mix the reagents. Cover the plates with a lid. Allow to stand at room temperature for 45 minutes.

10. Read the settling patterns for each serum sample. Read the control serum well first then read the patterns in the other wells.

11. Record the pattern observed in each well on a microwell plate recording sheet. Determine the endpoint. This is the point where there is complete inhibition of haemagglutination.

12. Record the antibody level for each serum sample. This is expressed as a log base 2. For convenience, the titre is often recorded as just the log index. For example a titre of 2 6 would be recorded as 6.

Interpretation of results

In the wells where antibodies are present there will be haemagglutination inhibition. The red blood cells will settle as a button.

In the wells where antibodies are absent, the red blood cells will agglutinate.

The end point of the titration is the well that shows complete haemagglutination inhibition. Sometimes it is not easy to determine. Look at the size of the button as an indication of the degree of haemagglutination inhibition. Use the control well as a point of comparison. Be consistent in determining the endpoint.

The neuraminidase enzyme present in the virus particle will eventually break the bond between the virus and red blood cells. This process is called elution.

When elution occurs, the red blood cells are no longer agglutinated. They roll down the side of V-bottom microwell plates to resemble the negative settling pattern, a tight button.

Some Newcastle disease virus strains elute more rapidly and the test must be read before this occurs.

Usually elution takes longer than 45 minutes. A control well with virus and red blood cells is useful to determine elution time.

Some sera may contain substances other than antibodies that inhibit viral haemagglutinin. These substances are described as non-specific inhibitors and are rarely observed with chicken serum and Newcastle disease.

Some chicken sera contain substances that will agglutinate chicken red blood cells. The settling pattern of the agglutinated cells is similar to that produced by Newcastle disease virus. These natural agglutinins are present in low concentration. The control serum well will indicate the presence of natural agglutinins.

Adsorption of natural agglutinins

If natural agglutinins are present in a serum sample, they can be removed by adsorption with chicken red blood cells. The serum can then be retested in the HI test.

  • Suspension of 10 percent washed chicken red blood cells.
  • Microfuge (Eppendorf) tube
  • Micropipette, 200 mL to 1 000 mL and tips
  • Discard tray
  • Microfuge centrifuge
  • Serum samples

1. Place 200 mL of the 10 percent washed red blood cells into a microfuge tube.

2. Centrifuge for 15 seconds.

3. Remove the supernatant.

4. Shake the serum sample and remove 500 mL of serum. Some samples may contain less volume. Use a new tip for each sample.

5. Add the serum to the red blood cells in the microfuge tube. Mix gently.

6. Stand for not less than 30 minutes at 4°C.

7. Centrifuge for 15 seconds.

8. Remove the serum (supernatant) immediately and transfer to a clean microfuge tube.

9. Store the adsorbed sera at 4°C overnight or at -20°C for longer periods.


Freeze-Thaw Cycles and Why We Shouldn’t Do It

Freeze-thaw—you know it’s bad for your samples, don’t you? While working in the lab, you have most likely heard someone say ‘aliquot your protein/cells/DNA/RNA to avoid too many freeze-thaw cycles.’ But do you actually understand why?

You probably thought that avoiding freeze-thaw cycles had something to do with damaging cell structure as well as proteins or DNA/RNA—and you would be right. But you might be surprised to know that freeze-thaw cycles can damage your samples in several other ways, and we don’t quite understand how all of them work.

Damaging your samples during freeze-thaw cycles can cause problems with downstream processes. For example, multiple rounds of freezing and thawing can damage protein structures, which can interfere with study protein kinetics using surface plasmon resonance. Even minor DNA damage can result in uninterpretable data from PCR.

Your samples are not the only concern when it comes to freeze-thaw cycles. At the moment, a lot of research is going into the cryopreservation of embryos and gametes. Humans aren’t the only ones using assisted reproductive technology. Animal husbandry professionals, zoologists, and others are using assisted reproductive technology to increase farming production or aid in the preservation of endangered species. Many studies have shown that the freeze-thaw process can affect DNA integrity in sperm and also hinder embryo development.

Through this research and years of experimentation in the lab, we’ve learned that a variety of factors are responsible for damage caused by freeze-thaw cycles.
Different Mechanisms Cause Instability During Freeze-Thaw Cycles
Ice Crystals

Ice crystals that are formed during the freeze-thaw process can cause cell membranes to rupture. Rapid freezing results in ice crystal formation in the outer parts of cells, which causes the interior of the cells to expand, pushing against the plasma membrane until the cell bursts. While slow cooling allows water to leach out and reduce ice crystal formation, slow cooling still leads to cell rupture due to an imbalance in osmotic pressure. If you are freezing live cells or microorganisms, both of these processes can greatly decrease viability.
Freeze Concentration

In addition to mechanically damaging cells, ice crystals can also cause the salts and proteins in the buffer to become concentrated. This problem is known as freeze concentration and can cause significant stress on the stability of proteins. Although the exact mechanism of ice-induced protein denaturation is not fully understood we do know that changes in the physical environment of the protein lead to stresses that can impact stability. For example, freeze concentration has been shown to cause protein unfolding at the ice:aqueous interface for several proteins, including, azurin, liver alcohol dehydrogenase and alkaline phosphatase.
Oxidative stress

Another common problem seen as a result of multiple freeze-thaw cycles is oxidative stress, which may be generated through different mechanisms. Ice crystal-induced damage to organelle structures could lead to activation of rescue systems that are associated with energy generation. This results in a subsequent increase in oxidative stress and production of reactive oxygen species (ROS, free radicals produced as by-products of reduction-oxidation, or redox, reactions). When the balance between ROS and antioxidants is lost, oxidative stress results in molecular damage to DNA, proteins, and lipids in the cell. Some studies have shown that thawed cells contain an increase in phosphorylated H2AX,a marker of double strand breaks in DNA.
Tips to Minimize the Damage Caused by Freeze-Thaw Cycles

There are two main ways to avoid the changes seen after freeze-thaw cycles:

Don’t do it. The easiest and most obvious solution is to prevent freeze-thaw cycles. As I said at the very beginning, one of the best ways to avoid multiple freeze-thaw cycles is to aliquot everything – your samples, your antibodies, your cells, and anything else you can think of.
Use cryoprotectants. In addition to aliquoting, add a cryprotectant to your cells/samples/etc. to help prevent the stresses caused by freezing.

Cryoprotectants, which are an important addition to samples on their freezing journey, were first discovered in the UK by Christopher Polge in 1949. He inadvertently supplemented an experimental freezing solution with glycerol, resulting in the unexpected survival of his experimentally frozen cells. I’m sure you’ve all used something in the lab that has had glycerol added for this purpose (e.g., antibodies or RNase). Even the antifreeze for your car has glycerol added.

There are two main classes of cryoprotectants:

Intracellular agents.These agents penetrate the cell to prevent the formation of ice crystals and, thereby, membrane rupture. There are several common reagents, including dimethylsulfoxide (DMSO), glycerol, and ethylene glycol. The most common agent used in the lab is DMSO, which provides a high rate of cell survival. However, some groups have shown that it can promote stem cell differentiation in neuronal cells. Also, DMSO can be cytotoxic at room temperature. Even though DMSO may have some drawbacks, it works well enough for most applications.
Extracellular agents. These agents do not penetrate the cell membrane but act by reducing the hyperosmotic effect in the freezing procedure. Common extracellular agents include sucrose, dextrose, and polyvinylpyrrolidone. Cells preserved in extracellular agents (e.g., sucrose) tend to have a lower viability after thawing than DMSO-preserved cells. This may be because extracellular agents don’t prevent ice crystal formation.

Whatever method you use, you should always be mindful of the changes you are causing in your samples during freezing and thawing. Especially those that cannot be seen—they could affect your results.


24.3 Homeostasis

In this section, you will explore the following questions:

  • What is homeostasis?
  • What factors affect homeostasis?
  • What are differences between negative and positive feedback mechanisms used in homeostasis?
  • What are differences between thermoregulation mechanisms in endothermic and ectothermic animals?

Connection for AP ® Courses

Animals must be able to maintain homeostasis—the ability to maintain dynamic equilibrium around a set point—while also being able to respond to changing conditions. For example, as an endotherm, your body temperature remains fairly constant around 37◦C or 98.6◦F. If your temperature climbs above the set point, you sweat to cool off if your temperature drops below the set point, you shiver to warm up. Your blood glucose levels also remain fairly constant because the liver removes glucose from the blood and converts it to glycogen when the body cells require glucose, glycogen is broken down. (You can probably hypothesize how your liver will respond if you eat a dozen jelly donuts!) The failure to maintain homeostasis can be detrimental and can even cause death. Consequently, negative and/or positive feedback loops regulate homeostasis.

Negative feedback mechanisms result in slight fluctuations above and below the set point. For example, if you were to consume a dozen jelly donuts, your blood sugar level would rise, and your pancreas would release insulin, a hormone involved in the conversion of glucose to glycogen, thus returning your blood glucose level to its appropriate set point. By comparison, positive feedback amplifies responses in the same direction, with the variable initiating the response moving the system even further away from the set point. There are fewer examples of positive feedback, but one is the onset of labor in childbirth when uterine contractions increase in strength with the secretion of oxytocin, another hormone. However, the loss of internal equilibrium due to positive feedback can be detrimental for example, a small area of damaged heart tissue can precipitate a heart attack which, in turn, damages even more cardiac muscle.

Information presented and the examples highlighted in the section support concepts outlined in Big Idea 2 of the AP Biology® Curriculum Framework. The AP ® Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP ® Biology course, an inquiry-based laboratory experience, instructional activities, and AP ® exam questions. A learning objective merges required content with one or more of the seven Science Practices.

Big Idea 2 Biological systems utilize free energy and molecular building blocks to grow, to reproduce, and to maintain dynamic homeostasis.
Enduring Understanding 2.C Organisms use feedback mechanisms to regulate growth and reproduction, and to maintain dynamic homeostasis.
Essential Knowledge 2.C.1 Organisms use feedback mechanisms to maintain their internal environments and respond to external environmental changes.
Science Practice 7.2 The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas.
Learning Objective 2.16 The student is able to connect how organisms use negative feedback to maintain their internal environments.
Essential Knowledge 2.C.1 Organisms use feedback mechanisms to maintain their internal environments and respond to external environmental changes.
Science Practice 5.3 The student can evaluate the evidence provided by data sets in relation to a particular scientific question.
Learning Objective 2.17 The student is able to evaluate data that show the effect(s) of changes in concentrations of key molecules on negative feedback mechanisms.
Essential Knowledge 2.C.1 Organisms use feedback mechanisms to maintain their internal environments and respond to external environmental changes.
Science Practice 6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models.
Learning Objective 2.18 The student is able to make predictions about how organisms use negative feedback mechanisms to maintain their international environments.
Essential Knowledge 2.C.1 Organisms use feedback mechanisms to maintain their internal environments and respond to external environmental changes.
Science Practice 6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models.
Learning Objective 2.19 The student is able to make predictions about how positive feedback mechanisms amplify activities and processes in organisms based on scientific theories and models.
Essential Knowledge 2.C.1 Organisms use feedback mechanisms to maintain their internal environments and respond to external environmental changes.
Science Practice 6.1 The student can justify claims with evidence.
Learning Objective 2.20 The student is able to justify that positive feedback mechanisms amplify responses in organisms.

Animal organs and organ systems constantly adjust to internal and external changes through a process called homeostasis (“steady state”). These changes might be in the level of glucose or calcium in blood or in external temperatures. Homeostasis means to maintain dynamic equilibrium in the body. It is dynamic because it is constantly adjusting to the changes that the body’s systems encounter. It is equilibrium because body functions are kept within specific ranges. Even an animal that is apparently inactive is maintaining this homeostatic equilibrium.

Homeostatic Process

The goal of homeostasis is the maintenance of equilibrium around a point or value called a set point. While there are normal fluctuations from the set point, the body’s systems will usually attempt to go back to this point. A change in the internal or external environment is called a stimulus and is detected by a receptor the response of the system is to adjust the deviation parameter toward the set point. For instance, if the body becomes too warm, adjustments are made to cool the animal. If the blood’s glucose rises after a meal, adjustments are made to lower the blood glucose level by getting the nutrient into tissues that need it or to store it for later use.

Control of Homeostasis

When a change occurs in an animal’s environment, an adjustment must be made. The receptor senses the change in the environment, then sends a signal to the control center (in most cases, the brain) which in turn generates a response that is signaled to an effector. The effector is a muscle (that contracts or relaxes) or a gland that secretes. Homeostatsis is maintained by negative feedback loops. Positive feedback loops actually push the organism further out of homeostasis, but may be necessary for life to occur. Homeostasis is controlled by the nervous and endocrine system of mammals.

Negative Feedback Mechanisms

Any homeostatic process that changes the direction of the stimulus is a negative feedback loop. It may either increase or decrease the stimulus, but the stimulus is not allowed to continue as it did before the receptor sensed it. In other words, if a level is too high, the body does something to bring it down, and conversely, if a level is too low, the body does something to make it go up. Hence the term negative feedback. An example is animal maintenance of blood glucose levels. When an animal has eaten, blood glucose levels rise. This is sensed by the nervous system. Specialized cells in the pancreas sense this, and the hormone insulin is released by the endocrine system. Insulin causes blood glucose levels to decrease, as would be expected in a negative feedback system, as illustrated in Figure 24.20. However, if an animal has not eaten and blood glucose levels decrease, this is sensed in another group of cells in the pancreas, and the hormone glucagon is released causing glucose levels to increase. This is still a negative feedback loop, but not in the direction expected by the use of the term “negative.” Another example of an increase as a result of the feedback loop is the control of blood calcium. If calcium levels decrease, specialized cells in the parathyroid gland sense this and release parathyroid hormone (PTH), causing an increased absorption of calcium through the intestines and kidneys and, possibly, the breakdown of bone in order to liberate calcium. The effects of PTH are to raise blood levels of the element. Negative feedback loops are the predominant mechanism used in homeostasis.

Positive Feedback Loop

A positive feedback loop maintains the direction of the stimulus, possibly accelerating it. Few examples of positive feedback loops exist in animal bodies, but one is found in the cascade of chemical reactions that result in blood clotting, or coagulation. As one clotting factor is activated, it activates the next factor in sequence until a fibrin clot is achieved. The direction is maintained, not changed, so this is positive feedback. Another example of positive feedback is uterine contractions during childbirth, as illustrated in Figure 24.21. The hormone oxytocin, made by the endocrine system, stimulates the contraction of the uterus. This produces pain sensed by the nervous system. Instead of lowering the oxytocin and causing the pain to subside, more oxytocin is produced until the contractions are powerful enough to produce childbirth.

Visual Connection

State whether each of the following processes is regulated by a positive or negative feedback loop.

a. A person feels satiated after eating a large meal.

b. The blood has plenty of red blood cells. As a result, erythropoietin, a hormone that stimulates the production of new red blood cells, is no longer released from the kidney.

a. This is regulated by a positive feedback loop as the stimulus (hunger) has changed direction in response to a signal (fullness).

b. This is regulated by a positive feedback loop as the stimulus (red blood cell release) has changed direction in response to a signal (presence of enough red blood cells).

a. This is regulated by a negative feedback loop as the stimulus (hunger) has changed direction in response to a signal (fullness).

b. This is regulated by a positive feedback loop as the direction of the stimulus has been maintained.

a. This is regulated by a positive feedback loop as the stimulus (hunger) has changed direction in response to a signal (fullness).

b. This is regulated by a negative feedback loop as the stimulus (red blood cell release) has changed direction in response to a signal (presence of enough red blood cells).

a. This is regulated by a negative feedback loop as the stimulus (hunger) changed direction in response to a signal (fullness).

b. This is regulated by a negative feedback loop as the stimulus (red blood cell release) changed direction in response to a signal (presence of enough red blood cells).

Set Point

It is possible to adjust a system’s set point. When this happens, the feedback loop works to maintain the new setting. An example of this is blood pressure: over time, the normal or set point for blood pressure can increase as a result of continued increases in blood pressure. The body no longer recognizes the elevation as abnormal and no attempt is made to return to the lower set point. The result is the maintenance of an elevated blood pressure that can have harmful effects on the body. Medication can lower blood pressure and lower the set point in the system to a more healthy level. This is called a process of alteration of the set point in a feedback loop.

Changes can be made in a group of body organ systems in order to maintain a set point in another system. This is called acclimatization. This occurs, for instance, when an animal migrates to a higher altitude than it is accustomed to. In order to adjust to the lower oxygen levels at the new altitude, the body increases the number of red blood cells circulating in the blood to ensure adequate oxygen delivery to the tissues. Another example of acclimatization is animals that have seasonal changes in their coats: a heavier coat in the winter ensures adequate heat retention, and a light coat in summer assists in keeping body temperature from rising to harmful levels.

Link to Learning

Feedback mechanisms can be understood in terms of driving a race car along a track: watch a short video lesson on positive and negative feedback loops.

Voltage-gated sodium channels occur in the cell membranes of nerve cells. They open in response to sodium entering the cell. This allows more sodium to enter the cell.

A scientist claims this is a positive feedback loop. What reasoning can be used to justify this claim?

  1. The voltage-gated sodium channels open in response to sodium influx. When a change happens in response to a change in conditions, it makes a positive feedback loop.
  2. The voltage-gated sodium channels close when there is enough sodium in the cell. This self regulation means this is an example of a positive feedback loop.
  3. The voltage-gated sodium channels open due to sodium and this causes more sodium to go through. The response reinforces the feedback, making this a positive feedback loop.
  4. The voltage-gated sodium channels are on the cell membrane. All channels through cell membranes are examples of positive feedback loops. This, this is an examples of a positive feedback loop.

Science Practice Connection for AP® Courses

Think About It

How are negative feedback loops used to regulate body homeostasis? How is a condition such as diabetes a good example of the failure of a set point in humans? Hypothesize and draw a diagram that shows what you think is the feedback failure for a person with diabetes.

Teacher Support

Negative feedback loops maintain the levels of some variable near a set point. In diabetes, a rise in blood glucose does not signal the production of insulin, which would normally lower blood glucose back to the set point. The Think About It question is an application of AP ® Learning Objective 2.16 and Science Practice 7.2 and Learning Objective 2.17 and Science Practice 5.3 because students are connecting negative feedback to the regulation of homeostasis and then, using blood sugar levels in humans as an example, explaining how a change in a negative feedback mechanism can have a deleterious effect.

Homeostasis: Thermoregulation

Body temperature affects body activities. Generally, as body temperature rises, enzyme activity rises as well. For every ten degree centigrade rise in temperature, enzyme activity doubles, up to a point. Body proteins, including enzymes, begin to denature and lose their function with high heat (around 50 o C for mammals). Enzyme activity will decrease by half for every ten degree centigrade drop in temperature, to the point of freezing, with a few exceptions. Some fish can withstand freezing solid and return to normal with thawing.

Link to Learning

Watch this Discovery Channel video on thermoregulation to see illustrations of this process in a variety of animals.

  1. Loose skin is thicker, which allows the excess heat to dissipate quickly through the skin.
  2. Loose skin brings more heat and blood to the body surface, facilitating heat loss.
  3. Loose skin contains greater skin area, which allows excess heat to dissipate as heat loss occurs through the skin.
  4. Loose skin has smaller skin area, which allows excess heat to dissipate as heat loss occurs through the skin.

Endotherms and Ectotherms

Animals can be divided into two groups: some maintain a constant body temperature in the face of differing environmental temperatures, while others have a body temperature that is the same as their environment and thus varies with the environment. Animals that rely on external temperatures to set their body temperature are ectotherms. This group has been called cold-blooded, but the term may not apply to an animal in the desert with a very warm body temperature. In contrast to ectotherms, poikilotherms are animals with constantly varying internal temperatures. An animal that maintains a constant body temperature in the face of environmental changes is called a homeotherm. Endotherms are animals that rely on internal sources for maintenance of relatively constant body temperature in varying environmental temperatures. These animals are able to maintain a level of metabolic activity at cooler temperature, which an ectotherm cannot due to differing enzyme levels of activity. It is worth mentioning that some ectotherms and poikilotherms have relatively constant body temperatures due to the constant environmental temperatures in their habitats. These animals are so-called ectothermic homeotherms, like some deep sea fish species.

Everyday Connection for AP® Courses

  1. increase vasodilation
  2. sweat
  3. move into shade
  4. increase metabolic rate

Heat Conservation and Dissipation

Animals conserve or dissipate heat in a variety of ways. In certain climates, endothermic animals have some form of insulation, such as fur, fat, feathers, or some combination thereof. Animals with thick fur or feathers create an insulating layer of air between their skin and internal organs. Polar bears and seals live and swim in a subfreezing environment and yet maintain a constant, warm, body temperature. The arctic fox, for example, uses its fluffy tail as extra insulation when it curls up to sleep in cold weather. Mammals have a residual effect from shivering and increased muscle activity: arrector pili muscles cause “goose bumps,” causing small hairs to stand up when the individual is cold this has the intended effect of increasing body temperature. Mammals use layers of fat to achieve the same end. Loss of significant amounts of body fat will compromise an individual’s ability to conserve heat.

Endotherms use their circulatory systems to help maintain body temperature. Vasodilation brings more blood and heat to the body surface, facilitating radiation and evaporative heat loss, which helps to cool the body. Vasoconstriction reduces blood flow in peripheral blood vessels, forcing blood toward the core and the vital organs found there, and conserving heat. Some animals have adaptations to their circulatory system that enable them to transfer heat from arteries to veins, warming blood returning to the heart. This is called a countercurrent heat exchange it prevents the cold venous blood from cooling the heart and other internal organs. This adaptation can be shut down in some animals to prevent overheating the internal organs. The countercurrent adaptation is found in many animals, including dolphins, sharks, bony fish, bees, and hummingbirds. In contrast, similar adaptations can help cool endotherms when needed, such as dolphin flukes and elephant ears.

Some ectothermic animals use changes in their behavior to help regulate body temperature. For example, a desert ectothermic animal may simply seek cooler areas during the hottest part of the day in the desert to keep from getting too warm. The same animals may climb onto rocks to capture heat during a cold desert night. Some animals seek water to aid evaporation in cooling them, as seen with reptiles. Other ectotherms use group activity such as the activity of bees to warm a hive to survive winter.

Many animals, especially mammals, use metabolic waste heat as a heat source. When muscles are contracted, most of the energy from the ATP used in muscle actions is wasted energy that translates into heat. Severe cold elicits a shivering reflex that generates heat for the body. Many species also have a type of adipose tissue called brown fat that specializes in generating heat.

Neural Control of Thermoregulation

The nervous system is important to thermoregulation, as illustrated in Figure 24.22. The processes of homeostasis and temperature control are centered in the hypothalamus of the advanced animal brain.

Visual Connection

  1. Pyrogens circulate to the hypothalamus to reset the body’s “thermostat,” causing a rise in temperature.
  2. Pyrogens circulate to the thalamus to reset the body’s “thermostat,” causing a rise in temperature.
  3. Pyrogens cause an increase in the activity of the animal’s enzymes, which results in the temperature rise.
  4. Pyrogens entering the blood release some lipid substances, which ultimately cause the rise in temperature.

The hypothalamus maintains the set point for body temperature through reflexes that cause vasodilation and sweating when the body is too warm, or vasoconstriction and shivering when the body is too cold. It responds to chemicals from the body. When a bacterium is destroyed by phagocytic leukocytes, chemicals called endogenous pyrogens are released into the blood. These pyrogens circulate to the hypothalamus and reset the thermostat. This allows the body’s temperature to increase in what is commonly called a fever. An increase in body temperature causes iron to be conserved, which reduces a nutrient needed by bacteria. An increase in body heat also increases the activity of the animal’s enzymes and protective cells while inhibiting the enzymes and activity of the invading microorganisms. Finally, heat itself may also kill the pathogen. A fever that was once thought to be a complication of an infection is now understood to be a normal defense mechanism.

When faced with a sudden drop in environmental temperature, an endothermic animal will:

Which is an example of negative feedback?

Which method of heat exchange occurs during direct contact between the source and animal?

The body’s thermostat is located in the ________.

Why are negative feedback loops used to control body homeostasis?

An adjustment to a change in the internal or external environment requires a change in the direction of the stimulus. A negative feedback loop accomplishes this, while a positive feedback loop would continue the stimulus and result in harm to the animal.

Why is a fever a “good thing” during a bacterial infection?

Mammalian enzymes increase activity to the point of denaturation, increasing the chemical activity of the cells involved. Bacterial enzymes have a specific temperature for their most efficient activity and are inhibited at either higher or lower temperatures. Fever results in an increase in the destruction of the invading bacteria by increasing the effectiveness of body defenses and an inhibiting bacterial metabolism.

How is a condition such as diabetes a good example of the failure of a set point in humans?

Diabetes is often associated with a lack in production of insulin. Without insulin, blood glucose levels go up after a meal, but never go back down to normal levels.

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    DISCUSSION

    Receptor-mediated endocytosis via CCV is a primary mechanism for the efficient internalization of extracellular solutes. One of the reasons for the high efficiency is that the cellular components used in this process are efficiently recycled for multiple rounds of endocytosis. For example, clathrin is released from coated vesicles by the action of hsc70, and the resulting soluble clathrin is used in the formation of new coated pits. Although the mechanism for clathrin release and recycling is well characterized, very little is known about the mechanism and requirements for AP release and recycling. We developed a novel and quantitative assay to study the release of APs from CCV. Using this assay we have demonstrated that AP release requires hsc70 in addition to an unidentified cytosolic factor(s) and ATP.

    AP release is directly dependent on cytosol. Here we demonstrate that hsc70 is a rate-limiting cytosolic component for AP release. Its addition to crude cytosol significantly reduces the cytosol dependence and enhances the rate of the AP release reaction. Given that hsc70 is an abundant cytosolic component, it is paradoxical that its concentration is limiting. These limitations are likely explained by the numerous functions that hsc70 plays in the cell, e.g., maintenance of the native folded structure of cytosolic proteins (Hartl et al., 1994). Importantly, as previously reported (Schlossmanet al., 1984 Heuser and Keen, 1988 Greene and Eisenberg, 1990 Buxbaum and Woodman, 1995), hsc70, while necessary, is not sufficient to support the release of APs from CCV. Several lines of evidence suggest that, in addition to hsc70, other cytosolic factor(s) are required for AP release. First, neither hsc70 nor hsc70-depleted cytosol can support AP release, whereas together these distinct fractions support efficient AP release. Second, maximal AP release requires twofold more ATP than is required for hsc70- mediated clathrin release. Thus, the additional factor may directly bind ATP. Alternatively, this factor may interact with hsc70 in such a way that it alters the affinity of hsc70 for ATP. Previous studies (Gross and Hessefort, 1996) have identified a 66-kDa protein that does alter the affinity of hsc70 for ATP and is required to promote the recycling of hsc70. In addition to the differing ATP requirements, ADPβS inhibits clathrin release but not AP release. Third, cytosol-dependent AP release can be uncoupled from hsc70-dependent clathrin release. Finally, AP-releasing activity can be resolved from clathrin-releasing activity and total protein by gel filtration chromatography.

    Immunodepletion of hsc70 from cytosol indicates that it is required for AP release. One function of hsc70 is to release clathrin. However, our data suggest that hsc70 plays a dual role in the release of both coat constituents from CCV. Using an antibody (3C5) that recognizes the clathrin-binding site of hsc70 as a specific inhibitor, we demonstrated that complete clathrin release was not a prerequisite to AP release. We cannot rule out that rearrangements of the clathrin lattice mediated by hsc70 in the presence of 3C5 reflect its requirement for AP release. However, our findings that AP release can occur without release of clathrin are consistent with in vitro coat assembly assays that have established that APs can be integrated into the coat after clathrin assembly (Ahle and Ungewickell, 1989 Keenet al., 1991 Prasad and Keen, 1991). Furthermore, APs can be selectively removed from coated vesicles by limited proteolysis without dissociating the clathrin coat (Matsui and Kirchhausen, 1990Schroder and Ungewickell, 1991 Traub et al., 1995). Thus, our data suggest that, in addition to the well-documented function of hsc70 in releasing clathrin from CCV, hsc70 also plays a direct role in the release of APs.

    How can the 3C5 antibody selectively inhibit hsc70-mediated clathrin release while not interfering with the putative second role in AP release? One possibility is that AP release may utilize a domain of hsc70 that is distinct from the clathrin-binding site. Several proteins that interact with hsc70 at a site distinct from the peptide/clathrin-binding site have been identified these include DnaJ homologues, other heat shock proteins, and kinases (Perdew and Whitelaw, 1991 Cyr et al., 1994 Czar et al., 1994 Pratt and Toft, 1997). Clathrin release by hsc70, a mammalian DnaK homologue, requires the presence of the mammalian DnaJ family member auxilin (Ahle and Ungewickell, 1990 Ungewickell et al., 1995). DnaJ domain-containing proteins have been postulated to regulate the functional diversity of DnaK homologues by targeting them to different substrates (Cyr et al., 1994). It is possible that hsc70 functions to recruit or to interact with these other factors (possibly a novel DnaJ homologue) to affect AP release. Consistent with this, hsc70-depleted cytosol has a significantly reduced AP-releasing activity. This may be the result of the selective codepletion of an additional AP-releasing factor together with hsc70 due to their interaction in the cytosol.

    The dual role of hsc70 in the uncoating reaction not only suggests that AP and clathrin release may be coordinated but that they may also occur by a common mechanism. The kinetics of clathrin and AP release are similar and biphasic. There is an initial and rapid release of both APs and clathrin, which is complete within 3–5 min. This initial “burst” is followed by a much slower steady-state rate of release (Greene and Eisenberg, 1990). Addition of hsc70 to cytosol enhances the rate of this initial burst of AP release. These kinetics, together with the potential association of the cytosolic AP-releasing factor(s) and hsc70, suggest that AP release may occur in concert with clathrin release.

    Clathrin is released in a stable complex with hsc70, with up to three hsc70 molecules bound to the vertex of the triskelion (Schlossmanet al., 1984 Heuser and Steer, 1989). Thus, it is possible that APs are also released in a stable complex with the uncoating factor(s). Our preliminary efforts to detect such complexes or proteins associating with APs have proven unsuccessful to date. This may reflect inappropriate reaction conditions for the formation of a stable complex, dissociation of the complex after AP release through the subsequent action of other cytosolic proteins, or suboptimal gradient conditions. Hence, we cannot eliminate this mechanism for the release of APs until we have examined the process using purified components.

    Phosphorylation of APs is another possible mechanism for the release of APs. This is especially attractive since cytosolic APs have been shown to be preferentially phosphorylated (Wilde and Brodsky, 1996). In addition, hsc70 has been shown to associate with a kinase activity (reviewed in Pratt and Toft, 1997). This interaction would also explain the selective codepletion of the AP-releasing activity with hsc70. Consistent with the possibility that the AP-releasing factor is a kinase, ATPγS, which can support kinase activity, also supports AP release, whereas AMP-PNP does not. Efforts to detect phosphorylated APs after their cytosol-mediated release from CCV were unsuccessful. This result, however, does not eliminate the possibility that AP release is phosphorylation dependent because these reactions were performed in the presence of crude cytosol, and therefore the APs were accessible to cytosolic phosphatases. A more complete understanding of the mechanism and requirements for AP release awaits the purification of the cytosolic AP-releasing factor(s) responsible.