Introduction: Definitions:

During embryoic development, cells specialize in carrying out particular functios. In all animals except sponges, the process is irreversible: once a cell deifferentiates to serve a function, it and its descendants can never serve any other. A sponge cell that had specialized to serve one function (such as lining the cavity where feeding occurs) can lose the special attributes that serve that funciton and change to serve another function (such as being a gamete). Thus a sponge cell can dedifferentiate and redifferentiate. Cells of all other animals are organized into tissues, each of which is characterized by cells of particular morphology and capability. The fact that all other animals differnetiate irreversibly suggests that organisms with bodies contining cells specialized to serve particular functions may have an advantage over those with cells that potentially have multiple functions.

In the process of embryonic development, the cells of mot types of animals organize into three layers (called germ layers): an otuer extoderm, an inner endoderm and an intermediate mesoderm.

Blastula: Shortly after fertilization, an animal zygote first undergoes a series of mitotic divisions, called cleavage, which produces a ball of cells, the blastula.

Blastopore: In most naimals, the blastula fold inward at one point to form a hollow sac with an opening at one end called the blastopore.

Endothermy: Mammals are endothermic, an adaptation that has allowed them to be active at any time of the day or night. Also, more efficient blood circulation provided by the four chambered heard and more efficient respiration provided by the diaphragm make possible the higher metaboix rrate on which endothermy depedns.

Gastrula: An embryo at this stage is called a gastrula. The subsequent growth adn movement of the cells of the gastrula differ form one group of animals to another, relfecting the evolutionary history of the group.

Larva: Embryos of most kinds of animals develop into a larva, which looks unlike the adult of the species, lives in a different habitate, and eats different sorts of food. In most groups it is very small. A larva undergoes metamorphosis, a reorganization, to transform into the adult body form.

Mammals: Most mammals are herbivores,e ating mostly or only plants. Cellulose forms the builk of a plant’s body and is a major source of food for mammalian herbivores. Mammals do not have the necessary enzymes, however, for breaking the links between glucose molecuels in cellulose. Herbivorous mammals rely on a mutualistic partinership with bacteria in their digestive tracts that have the necessary cellulose splitting enzyems. Mammls such as cows have huge, four chambered fermentation vats derived form esophagus and stomach. The first chamber is the largest and holds a dense population of cellulose digesting bacteria. Rodents, horses, rabbits and elephants, by contrast, have relatively small sotmachs and instead difgest plant material in their alrge intestine, like a temrite. Even with these complex adaptations for digesting cellulose, a mouthful of plant is less nutritious than a mouthful of meat. Herbivores must consume alrge amounts of plant material to gain sufficient nutrition.

Placenta: In most mammla species, females carry their developing young internally in a uterus, nourishing them through the placenta. The placenta is a specialized organ that brings the bloodstream of the fetus into close contact with the bloodstream of the mother. Food, water and oxygen can pass across and wastes can pass over to the mother’s blood and be carried away.

Primates: are the mammalian group that gave rise to human species. In general, humans and chimpanzees exhibit a level of genetic similartiy normally found between closely related species of the same genus. In fact, there is a 99% sequence similarity between chimpanzees and humans.

Tissues: The cells of all animals except sponges are organized into structural and funcitonal units called tissues –collections of cells that are specialized to perform specific tasks.

Development of Particular Types of Animals:

Sponges:

As is true of many marine invertebrate animals, larval sponges are free swimming. After a sponge larva attaches to an appropriate surface, it metamorphoses into an adult and remains attached to that surface for teh rest of its life.

A unique feature of sponge cells is theri ability to differentiate form one type to anotehr, and to dedifferentiate from a specialized state to an unspecialized one. If a sponge is put thorugh a fine sieve or coarse cloth so that the cells are separated, they will seek each other out and reasemble the entire sponge -a phenomenon that does not occur in any other animal.

Flatworms:

Like sponges, a flatworm lacks a circulatory system for teh transport of oxygen and food molecules. The thin boyd of a flatworm allows gas to diffuse between its cells and the surrounding enviornment (oxygen diffuses in and carbon dioxide diffuses out).

The reproductive systems of flatworms are coplex. Most are hermaphroditic, each individual containing both male and female sexual strcuture. In most freshwater faltworms, fertilized eggs are laid in cocoons strung in ribbons and hatch into miniature adults. In constast, some marine species pass thorugh a larval stage, the fertized egg undergoing spiral cleavage, and the embryo giving rise to a larva that swims or drifts until metamorphosing into an adult, at which point it settles in an appropriate habitate.

Flatworms are known for their regenreative capacity: when a single individual of some species is divided into two or more parts, an entirely new flatworm can regrow what is missing from each bit.

Reptiles: Many lizards, including anoles, skinks and geckos, ahve the ability to lose their tails and then regenerate a new one. This ability allows these lizards to exape from predators.

Birds: Modern birds lack teeth and have only vestigial tials, but they still retain many reptilian characteristics. For instance, birds lay aminiotic eggs. Also, reptilian scales are present on the feet and lower legs of birds. Feathers are modified reptilian scales made of keratin, jsut like hair and scales. Feathers provide lift for flight and conserve heat. Feathers develop from tiny pits in the skin called follicals.

The revved-up metabolism need to power flight requires efficient blood circulation, so taht the oxygen captured by the lungs can be delivered to the flight muslces quickly. In the heart of most living reptiles, oxygen rich blood coming from the lungs mixes with oxygen poor blood returning from the body becasue the wall dividing the ventricle into tow chambers is not compelte. In birds, the wall dividing the ventricle is complete, and the two blood cirulations do not mixe. So flight muscles receive fully oxygenetaed flood.

Mammals: There are about 4000 living species of mammals which is less than the number of fishes, amphibians, reptiles, or birds. Most large, land swelling veterbrates are mammals. Mammals are distinguished form other classes of verterates by hair and mammary glands.

–Bats: are the only mammals capable of powered flight. Like the wings fo birds, bat wings are modified forelimbs, but bats have developed wings by modifying theirforearms in different ways.

Circulatory System:

Open circulatory system:

In an open circulatory system, the blood passes from vessels into sinuses (open areas within the body), mixes with body fluid that bathes the cells of tissues, then reenters vessels in another location.

Closed circulatory system:

In a closed circulatory system, the blood flows entirely within blood vessels, so it is physically separated for other body fluids. Blood moves thorugh a closed circulatory system faster and more efficiently than it does thorugh an open system.

Development of Plants:

Meristems: Meristems are clusters of small cells with dense cytoplams and proportinately large nuelei taht act as stem cells. One cell divides to give rise to two cells, of which one reamins meristematic, while the other udnergoes differentiation and contributes to the plant body. In this way, the population of meristem cells is continually renewed. Molecular genetic evidence supports the hypothesis that animal stem cells and plant meristem cells may share some common pathways of gene expression.

Human Cell Atlas (maps every cell of the human body)

See also types of cells

See also fertilization and embryonic development

Introduction:

Cell development starts as a fertilized egg (Zygote) that must undergo cell division to produce the new individual. Thus extensive cell divisions is required. However, in many cases it does not incuded much growth as the egg cell itself is quite large.

As cells divide, precise changes in gene expression result in differences between cells that ultimately result in cell specialization. In differentiated cells, certain genes are expressed at particular times, but other genes may not be expressed at all.

Pattern formation involves cells abilities to detect positional information that guides their ultimate fate. In animal embryos, the timing and number of cell divisions are species specific, and this period of rapid cell division following fertilization is called cleavage. During cleavage, the enormous mass of the zygote is subdivided into a larger and larger number of smaller and smaller cells, called blastomeres. Cleavage is not done by an increase in the overall sie of the embryo. The G1 and G2 phases of the cell cycle, during which a cell increases its mass and size, are greatly shortened or climinated alltogether during cleavage. This is controlled by cyclins and clin-dependent kinases. For exaple, zebrafish blastomeres divide once every 15 minutes during cleavage to create an embryo with a thousand cells in just under 3 hours. As a comparison, cycling adult human intestinal epithelial cells divide on average once every 19 hr.

As development continues the form of the body, its organs and anatomical features is generated. Morphogenesis may invovle cell death, cell division, cell migration, changes in cell shape and differentiation.

Cell Differentiation:

A human contains about 300 ytes of differentiated cells. They can be distinguised based on the particular porteins they synthesize, their morphologies and their specific functions. During development, cells become different from one another because of the differential expression of subsets of genes, both at different time and in different locations of the devleoping embryo.

Before differentiation takes place, cells make a molecular decision to become a particular cell type. The standard procedure to test whether a cell or group of cells is determined is to move the donor cell(s) to a different location in a host embryo. If the cells of the transplant develop into the same type of cell that they would have if left undisturbed, they they are considered to be determined. For example, a cell in the prospective brain region of an amphibian embryo at the early gastrula stage has not yet been determined if transplanted elsewhere in the embryo. It will develop according to the site of transplant. By the late gastrula stage, however, additional cell interactions have occurred and determination has taken place. The cell will develop as neural tissue o matter where it is transplanted.

Determination usually takes places in stages, with a cell first becoming partially committed. In a chicken embryo, for example, tissue at the base of the leg bud normally gives rise to the thigh. I this tissue is transplanted to the tip of the identical looking wing bud, which would normally give rise to the wing tip, the transplanted tissue will develop into a toe rather than a thigh. The tissue has already been determened, but it is not yet committed to being a particular part of the leg. Thus, it can be influenced by the positional signaling at the time of the wing bud to form a tip but in this case, a tip of leg.

Cells initiate development by using transcription factors to change patterns of gene expression. When genes encoding these transcription factors are activated, one of their effects is to reinforces their own activation. This reinforcement makes the developmental switch deterministic, initiating a chain of events that leads down a particular developmental pathway.

Cells in which a set of regulatory genes have been activated may not actually undergo differentiation until some time later, when other factors interact with the regulatory proteins and cause it to activate still other genes. Nevertheless, once the initial “switch” is thrown, the cell is fully committed to its future developmental path.

Cells become committed to follow a particular developmental pathway by differential inheritance of cytoplasmic determinants, which are maternally produced and deposited into the egg during oogenesis or via cell-cell interaations.

Plant Development:

A major difference between animals and plants is that most animals are mobile. Plants, in contrast, are anchored in position and most therefore endure whatever environment they experience.

Instead of creating a body in which every part is specified to have a fixed size and location, a plant assembles its body throughout its life span from a few types of modules, such as leaves, roots, branch nocdes and flowers. Each module has a rigidly controlled structure and organization, but how the modules are utilized is flexible; it can be adjusted to environmental condtiions.

Plasnts develop by building their bodies outward, creating new parts from groups of stem cells that are contained in structures called meritems. As meristomatic stem cells cotinually divide, they produce cells that can differentiate into teh tissues of the plant. The plant cell cycle is regualted by the same mechanisms as animal cells and yeast by cyclins and cyclin-dependent kianses. For example, overexpression of a Cdk inhibitor can result in strong inhibition of cell division in leaft meristems, leading to significant changes in leaf size and shape.

Introduction:

The long-term preservation of suspension of cells by freezing and stroage at cryogenic temperatures is a well-established technique. A wide variety of single cell types in addition to multicellular structures and organisms may be kept indefinitely through suspension in a cryopreservation fluid with subsequent freezing at a controlled rate of temeprature reduction. The cryogenic storgae vessels in which the cell suspensions are frozen are typically stored for extended preriods in a vacuum-insulated tank that is refigerated by liquid nitrogen, although mechnical refrigeration systems may also be applied. Cell suspensions that are preserved by such methods may also be transported to remote locations while maintaining cryogenic temperature conditions of -196 to -150c by means of vacuum-insulated flasks wherein liquid nitrogen, as a refigerant to maintain temperature, has been introduced into an adsorbent material. Such a container is often refered to as a “dry-shipped”. Alternatively, if shortterm tempperature shifts to a range of -80C to 50C are not exceedingy deleterious to cell viability, an insulated container using dry-ice as a refigeerant may be applied. Upon arrival at the destination the frozen cell suspension may be transferred to a local extended-term storage system, or may be recovered from the refigerated shipping system and used directoy. Prior to use, frozen cell suspension must be tahwed and retrieved fromt eh storage vessel. (WO 2019/2900145). 

Types of Cryogenic Storage Devices/Vessels:

Feezing Bag:

A common type of cyrogenic storage vessel is a flexible-wall storage bag also referred to as a “feezing bag”, a “cryopreservation storage bag” or a “cryostorage bag”. Common procedrues for cryostorage bag usage include the filling of the bag to a selected volume, after which, the tubing extension is often heat sealed at a location enar to the ports and the remainder of the tubing assembly is severed and disposed. In some cases, a suer will elave a short extension of tubing attached to the port area with intermitten seals such that, post thawing, segments of cell solutions may be isolated for varous testing pruposes. The edge of the bag opposite to the port edge may comprise an extension of the sealed region in which may be located a slot by which the bag may be suspended on a hanging apparatus.(WO 2019/2900145)

As a result of the flexible properties and fragile design of cryostorage bags, various procedures are generally followed and limits applied when preparing and feezing these types of crysorage vessels. For example, to control the thickness of the cryostorage bag vessel, the bags are typically limited in fill volume, and frozen in an orientation such that the major planar surfaces of the bag are perpendicular to the gravitions force vector (also referred to as a “falt” orientation). In some instances, the filled bags may be frozen while being stored wihtin a rigid cassette in order to limit and control the thickness of the frozen cryostorage bag and contents. However, despite efforts taken during the feezing process, the natural expansion of the aqueous solution in the flexible cryostorage bag resutls in uneven surface variations, dimensions, and solution thickness. (WO 2019/2900145).

Storage/Shipping cassette:

For some procedrues, a cryostorage bag is encloded within a storage or shipping cassette as part of a freezing process. eous solution in the flexible cryostorage bag resutls in uneven surface variations, dimensions, and solution thickness. (WO 2019/2900145

Biolife Solutions (US 17/046,502, published as US Patent Application No: 20210137787 and WO2019/200145) discloses a storage system which is configured to work with commercially available cryostorage bags which includes a protective interface cusion having an upper and lower half configured to receive a cryostorage bag. The upper and/or lower halves can include one or more recesses which achieve a desired shape or configuation of a solution in the cryostorage bag during and following a freezing process. The uper and/or lower halves may futher include one or more openings to provide access to the bag ports or bag port extension tubing. The upper and lower halves include one or more cutouts and/or recesses formed on an interior surface of the halves, wherein these features act as a mold to acheive a desired shape or configuation of a solution in the cryostorage bag during and folloiwng a feezing prcoess. Portions of the cryostrage bag expand into the one or more cutouts and/or recesses as the liquides of the cryostorage bag solidify during the freezing process. Upon subsequent removal of the protective interface cushion device, the resultant and desired shape or configuation of the cryostorage bag and its contents may assist in storing the cryostorage bag. In some instances, the resultant and desired shape or configuation of the cryostorage bag and its contents is utilized to process the contents of the cyrostroage bag, suchh as to provide optimized interaction with external thawing instrument equipment and/or components. Following the freezing prcoess, the one or more cutouts and/or recesses in the upper and lower halves of the protective interfact cusion device provide cushioning to the frozen contents and protect fragile elements of the cryostorage bag, such as the heat-sealed seams, as well as prevent undersirable movement or shifting of the cryostorage bag relative to the protective interface cusion device and other elements of the storage system. In some embodiments, the protective interface cusion device includes a thermoinsulative material configured to absorb impact, shock and acceleration forces at crogenic temepratures. The protective interface cusion device may include a fibrous synthetic polymer material, as at cryogenic temperatures, polymeric fibers may remain sufficiently flexible to sustain a compressible cusioning effect. In some embodiments, the storage system further includes a protective shell casette including a first and second half, each half having an interior surface for receiving and housing at least a portion of the protective interface cusion device and bag. In some embodiments, the protective interface cusion devide is provided without an overflow region, and is further configured to limit the total volume of fill volume of the cryostroage bag. In place of an overflow region the protective interface cusion device includes an expanded or enlarged exclusion region. The volume limitations and external pressrues provdied by the exclusion region displaces the entire liquid contents of the bag into the fill volume. Thus, liquid contents are not reliant on gravity for displacement into the fill volume. Accordingly the protective interface cusion device is compantbile for freezing in a variety of orientations. 

Sartorius Stedim North American Inc. (US 2012/0017609) disclsoes a system for use in feezing, storing and thawing biophamraceutical materials which incluedes a fexible sterile container for holding the biopharmacetucial material and a holder which is more rigid and fiedly connected to the container means. The flexible container 10 may be formed of a laminated film which includes a plurality of layers. The sterile, flexible container 10 may be configured (e.g., shaped and dimensioned) to be received in, and integrally connected to a supporting strcuture, such as a frame or holder. The container 10 may can include one or more ports or conduits to allow filling or draining of biopharmaceutical materials. A typical process is that one or mroe of the cotainers is integrally formed or fixedly connected to a holder. 

Common Cell Types: Polypeptides for use in pharmaceutical applications are mainly produced in mammalian cells such as CHO cells, NSO cells, Sp2/0 cells, COS cells, HEK cells, BHK cells, PER.C6® cells and the like (WO 2009/010269). Recombinant therapeutic proteins are commonly produced in several mammalian host cell lines including murine myeloma NSO and Chinese Hamster Ovary (CHO) cells (WO 2008/091740) and (Van Reis, “Bioprocess membrane technology, Science 297 (2007) 16-50) . 

Cell Growth mediums: 

A mammalian cell culture begins with a mammalian tissue. The tissue is dissociated either mechanically or enzymatically or by a combination of the two to yield a mixture of single cells and small clups of cells. The mixture is inoculated into an appropriate liquid growth medium that ordinarly includes salts, glucose, certain amino acids, vitamin and blood serum (about 20% of the medium). The serum is included as a means of providng components that have not all been identified but are needed if the cells are to live and grow.

Fetal calf serum (FCS) is considered the best available serum for this purpose.

Cohn Fractions:  MacLeod (EP0440509 A2) discloses growth medium components which when added to cell growth media can be used as a replacement for FCS. The medium is free from active virsues and essentially immunogloublin free. MacLeod also describes a process for producing the medium using Cohn fractions, adjusting the pH to 7-8, adding polyethylene glycol and then diluting the resultant liquid with water or a suitable buffered saline. Subsequently the resultant solution is pasteurised by heating.

Mouse bone marrow/GM-CSF:

When suspensions of mouse bone marrow are cultured in the presence of GM-CSF, 3 types of myeloid cells expand in numbers (a) neutrophils predominate but do not adhere to the culture surface. (b) Macrophages are firmly adherent to the culture vessel. (c) dendritic cells arise from cellular aggregates that are attached to the marrow stroma. The aggregates become covered with sheet-like cell processes and eventually release typical single dendritic cells.

Production of Antibodies in cell culture (see antibody production)

Protocols Involving Dendritic Cells:

Infecting Dendritic cells with LP

In this experiment, we will infect dendritic cells which have been obtained from the bone marrow of mice. See on how to obtain such dendritic cells.

Prepare 5% RPMI without Antibiotic 

Collect Your DC cells: (They are in the medium)

reagents: –prepare 10% RPMI without Antibiotic  –pour some PBS into 50 ml tube and let warm up *in incubator for 10 minutes if want; needs to be room temp)

1) At this point your DCs have been in culture for 6-10 days. 

2) Your cells will be a mixture of macrophages which adhere to the plate and dendritic cells (most of which do not adhere). Using pipette, siphone off the liquid being careful not to scrape and stire up the macrophages which are adhered to the place. Place liquid into 50 ml tube. 

(3) Add about 1 ml (room temp) PBS to each well plate in order to wash the rest of DC. Room temp PBS will not lift the macrophages. Place liquid into your 50 ml tubes.

(4) spin at 10 C at 1100 RPM for 10 min.

(5) Remove supernatent

(6) resuspend cells and transfer to tube using 10 ml medium (use 10% RPMI but without antibiotic since antibiotic will inhibit the infectivity of Lp)

(7) count the cells as below. (always keep cells on ice while doing anything!)

Prepare LP:

reagents: –blank cuvette, sample cuvette, transfer pipette, 2×2 square of parafilm –also turn on machine so warms up

(5) Put saline (about 5 ml) into 15 ml tube. take a   [this is a tube which has been taken from -80, streaks out onto agar and incubated for 48 hours) which is on special agar (BCYE). Use swab end to collect and put into your tube. Repeat with cotton swab if necessary. You should just barely see letter in back of tube. (if you do not see you can add more saline)

(6) take 1 ml of LP-solution and transfer to disposable cuvette. (place parafilm around cuvette!)

(7) take OD reading. get blank in drawer. put blank in. Change wave lenght to 620. change parameter so can print out. Convert reading using graph into number CFU. (write this on the tube!)

<Turn on> (takes 5 min) 1st place blank in. <1><enter>”change parameters”  change the second alpha to “620” (the first stays the same at “800) <Y> autozero <start> (need to be at 0). Place sample <start> again (your reading should be close to 1 (e.g., .834 converts to 21×108 so write on tube 2.1 x 109) hit return and turn machine off

(8) ) take your DC cells out of centrifuge from (4) above and pour off supernatent into sink. Transfer pellets into 15 ml tube having about 10 ml RPMI (if not expecting a lot of cells can use 5 ml). (so you can pipette a 3 ml RPMI volume up and down in one 50 ml tube and do the same for the others to get that 10 ml volume. You will need to know this final volume for calculations below)

Count your DC cells:

(9) take 50 ul from your tube and add to 50 ul tryptan blue (you can do this in a well plate) Look under microscope low magnification 1st for grid. count 5 squares. (corners and middle)

# cell counted X 2 X 5 X 104 (for the cuvette) x 10 =  your total # cells per ml

[Let say your total number is 20 x 106 per ml. We want a final concentration of 1 X 106 per ml. What volume should we resuspend our cells in to get this final concentration?

20 x 106 / X = 1 X 106 / 1 ml

x= 20 ml.   So if our starting concentration was 10 ml we would need to add another 10 ml to get this final concentration. (use larger 50 ml tube to make this appropriate volume)

(10) Now your are ready to Transfect your cells.

Transfecting Cells:

(1) The infection ratio is 10:1. So figure out how much of your Lp you should use to infect your cells. Suppose you have 4 million cells. You would use 14 ul of LP. So spin your cells down (want to infect in about 1 ml), add 14 ul of LP and incubate at 37 C for 30 min.

Ex. you have 4 million cells and 2.8 x 109 cell/ml bacteria. How much LP should you add? 

2.8 X 109 cells/ml (X) = 4.0 X 107 X=4.0 X 107 / 2.8 X 109        =1.4 X 10-2 = .014 ml = 14 ul of LP

Ex. From your OD reading you determine you have 2.3 x 10e9 cells/ml Lp. You have counted 5 million cells/ml of DCs. You u want an infection fatio DC:Lp of 1:10 how much Lp should you add?

2.3 X 109 cells/ml (X) = 5.0 X 107 X=21.7 ul

Remove Excess LP (wash the cells)

(1) add Hanks solution (HBSS) to the tubes (fill up the tubes with hanks). Always pour Hanks from the large stock flask into a smaller flask and from there into your tubes. Centrifuge for 10 min

(2) remove supernatent (make sure this goes into bleach since it is infected and let it sit for an hour!)

(3) repeat by filing tube with hanks, spin, and then removing supernatent into bleach

Resuspend pellet in appropriate amounts medium 

(1) If, for example, you had 4 million cells in the pellet, you might  add 4 ml media to get 1 million cells/ml. This will all depend on your experiment.

Measuring Cell Concentrations

Question (Concentrations: cells): You want to split cells into a new flask so that the flask with have a 50 ml volume and final cell concentration of 4 * 105 . Your flask with cells has a volume of 50ml and a concentration of 25 * 104. What the volume that should be take out of this flask to end up with your desired final concentration?

(CI VI = C2 V2)     (25 * 104) VI = (50 ml) (4 * 105)

VI = 80 ml                    This volume is clearly more than what is in the flask. So we will instead have to spin the cells down and add media instead. What then should be the volume of the second flask to get the desired final concentration?

(CI VI = C2 V2)     (25 * 104) 50 ml= (V2) (4 * 105)

V2 = 31.25 ml      So spin down the flask having 50 ml and then resuspend the cells in about 32 ml.

Question (Concentrations: cells): You want 15 ml of cells at a concentration of 50,000 cells/50 ul. After doing a cell viability count, you figure that you have 280K cells/ml in a flask which has the cells suspended in about 75 ml. What volume would you need to take out of that flask to get your desired concentration?

(CI VI = C2 V2)     (280Kcells/ml) VI = (15 ml) (50 K cells/0.05 ml)

VI = 53.5 ml

Question (Concentrations: cells): After counting cells, you find that you should have 10 million in the pellet that you spun down (in this example, you came up with the # by mutiplying #cells X 2 (dilution factor) X 104 (volume of hemacytometer) X #ml (cell suspension). You want a concentration of 1 X 106. What volume should you resuspend your cells?

10ml

Question (Concentrations: cells): You have a [cells] or 1 X 106 and you would like a final [cell] of 2 X 105. What volume should you take from the 1 X 106 and place into your plate well for a final volume of 1 ml.

CI VI = C2 V2)     (1 x 106 cells/ml VI ) = (1 ml) (2 x 105)

V1= 0.2 ml or 200 ul    You should thus add 200 ul (this amount can never change!) to your 1 ml volume. So you might add 200 ul of cells + 800 ul media (this amount can change to make room for other chemicals).

Question (Concentrations: cells): You have a [cells] or 8 X 106 cells/ml and you would like a final [cell] of 1 X 106. From outlining your experiment, you determine that you need to use 13 wells in a 24 well plate (to do everything you want to accomplish). Is it possible to conduct your experiment at this concentration?

Yes. But you can not use 1 ml of you cells (brought up in 8 ml) because you will not have enough cells to distribute this way. What you must use is a smaller volume. If you use 0.5 ml instead of 1ml per well you can still keep a [final] of 1 X 106 . This would in fact allow you to use up to 16 wells. 

So now what volume should you take out of your 1 ml of cells to get this correct concentration?

CI VI = C2 V2 (1 x 106 cells/ml VI ) = (1 x 106)(0.5 ml)

V1= 62.5 ul (this can not change!)  (Add this to your LPS + MDP up to a total volume of 0.5 ml)

Question (Concentrations: dendritic cells & LP infection): You determine that you have 16 million dendritic cells/ml and after your OD reading that you have 2.9 X 109 bacterial cells/ml. You would like to infect your dendritic cells with the bacteria at a ratio of 10:1. What volume of LP should we take to get a 10:1 infection?

One way to do this is to multiply the dendritic cells by a factor of 10 and then divide by the # bacterial cells. So

16 X 107 dendritic cells/ ml / 2.9 X 109 bacterial cells/ml 

= 5.5 x 10-2 ml  or 0.055ml or 55 ml

Question (Concentration in multi-well plates): After counting cells, you find that you have 8 million cells suspended in 5 ml. You want a concentration of 1 million per 200 ul. What should you do?

CI VI = C2 V2 (8 x 106 cells/ml VI ) = (1 x 106)(0.2 ml)

V1= 125 ul 

Question (Concentration in multi-well plates): Suppose in question above that you want to set up a dilution of the cells much like an ELISA. you want to start your highest concentration of cells at 4 million.

4 million is 4x 25  so you would add 100 ul of your stock    to 100 ul media

2 million, add 50 ul        to 150 ul media

1 million, add your 25 ul    to 175 ul media

.5 million 12.5 ul    to 187.5 ul media

.25 million 6.25 ul    to 193.75 ul media

.125 million   3.125 ul to 196.875 ul media

Unthawing Cells:

(1) Obtain 2 50 ml tubes; place 10% C RPMI media in to one tube as working stock

(2) before cells are completely thawed, use 2 ml pipette to suck up cells (slowly mix them up and down a couple of times) and then transfer to empty 50 ml tube (let cells slide down the side of tube slowly as you rotate the tube)

(3) Using 2 ml pipette, transfer slowly 2 ml of RPMI media from working stock into tube with cells (do this slowly or cells will burst!). Repeat 4 more times until you have 10 ml total volume.

(4) Using 25 ml pipette transfer more media from working stock to tube with cells for final volume 50 ml

(5) spin 15 min at 1100 RPM (select RPMx1000) (can use big centrifuge outside room)

(6) aspirate with vacuum tube supernatent

(7) transfer 8 ml new media into tube (resuspend pellet by pipetting up and down)

(8) transfer contents of tube to small flask

Changing Media:

Molt-4 cells should be changed every 2-3 days. They grow quicker. maximum [cell] is 2×106 or they will die

(1) transfer cells in flask to tube and spin down

(2) remove supernatent

(3) Wash cells by adding about 50 ml hanks (HBSS) and resuspend pellet (you can add say 5 ml first and then take a 50 ul sample to count cells while spinning cells down)

(4) spin

(5) remove supernatent

(6) resuspend cells in 10 ml of 10% RPMI

Cell Mediums

RPMI + 1X P/S + 1X L-glu without serum (RO)  take the RPMI and dilute 1:100 with streptamycin and glutamine

Thawing Cells

Reagents: RPMI media (Roswell Park Memorial Institute) (Cellgro makes this). This is a special media for mammalian cells which have necessary vitaminnes, syrum and growth factors. 

(1) Look in log book (ex. “rack 10 box 2”) for location of cells (ex. MT-2 cells which are a cell line that express CD4+ receptor on their surface and are very susceptible to HIV infection). Mammalian cells will ideally be stored at 186 c whereas bacterial cells lines are typically stored at -70 or -80. Mammalian cells are frozen in presence of a 10% DMSO (dimethy sulfoxide). This means that mammalian cells will need to be stored in nitrogen machine. Take 2 cell tubes out. If you are not going to be working with the cells right away, put them at -80.

(2) Transfer cell tubes to ice bucket and then quickly thaw cell tubes in 37 c water bath. Do not overthaw since the presence of the DMSO can kill the cells!  You want to thaw just to the point the cells have thawed. 

(3) transfer cells using (2 ml pipette) into conical sterile tube (spray the tube and cap with 70% ETOH to help with sterility) which has 50 ml of RPMI media. 

(4) label the tube appropriately. (ex. MT-2, LN98074 F 1/19/99 T 6/16/03 JL 6/16/03)

Counting Cells

1) Look at the cells under a microscope. They will be clumped together.

2) Take a portion (~5 ml) from the flask of cells and transfer to sterile tube. Using the pipette, mix and break the cells by pipetting up and down about 10 times in the conical tube. This should be done under the hood.

3) Clean off with 70%ETOH a hemacytometer slide and cover slip with klenex. Transfer about 50 ul of the cells into a pipette from the conical tube and add 50 ul of Trypan Blue (can be purchased from Sigma) so that you end up with a 1:1 dilution.. Release the cells slowly into the groove of the hemacytometer so as to fill up the groove.

4) using a counter, count the number of live cells (they will have a shine about them) and dead cells (they will be stained by the trypan blue) in the 4×4 grid of the slide. The total volume of the grid is 0.1 ul. The total viability of cells can be determined by taking # of live cells/total # of cells * 100. For example, if you count 18 live cells out of a total of 34, this would be 18/34 * 100 = ~53 viability. In general, you want the viability up above 80%.

Question: 0.1 ul times what is equal to 1ml?

Answer: 0.1 ul x 1ml/1000ul = 0.0001 ml   So 0.0001 ml would need to be multiplied by 104 or 10, 000 to get 1 ml. This means that the total cells which we count in 0.1 ul under the microscope would have to be multiplied by 10,000 to get the total # of cells in ml of solution. So if you count 18 live cells, this means that you have 180K cells/ml. But since your are diluting in Trypan Blue at 1:1 you need to multiple this number by 2. So you would actually have 180K X 2.

Subcloning Cells

1) Cells will eventually use up their media and should be transferred to new media to maintain viability. One way to know when it is time to transfer the cells into new media is when the media becomes turbid (slight yellow color). So take a new flask and add about 35 ml of new media. To the old flash, add about 25 ml of media to provide those cells with new media. Label the flasks appropriately. Ex. MT-2, LN90741 F: 1/19/99 T 6/17/03 split 1:1 6/17/03.

Enriching Cells

(1) Count the total # of cells that you want to enrich. (ex. if you counted 24 lives cells above, you would multiply that by 10K and by 2 (for dilution factor) to get 480,000 cells/ml. Suppose your flask volume is 80ml you would multiply by 80 to get 38,400,000 cells total. Suppose you do this for a 2nd flask and get 59,200,00 cells total for 80ml in that flask. So you would add the total cells from both flasks to get 97,600,000 cells. We will want to have a final concentration of 1 * 108 cells/2ml.

97,600,000cells/(x)ml = 1 * 108 cells/2ml

x=1.952     So we are going to need to resuspend our cells in about 2 ml

(2) Prepare RPMI-5. (so if you have 20% you could add 10 ml R-20 and 40 ml R-0)

(3) Distribute cells into 50 ml tubes (yellow cap). So we could use 4 tubes with 40 ml each. 

(4) centrifuge 15 min, 800 RCF (or ~2160 RPM) (Accel=fast) (brake=slow)

(5) decant and resuspend pellet with 1 ml of solution(2) then transfer this volume to 2nd tube, and so forth until last tube where you bring the volume up to 2ml with solution(2) [cell concentration could be off so can also bring volume up to 2 ml with solution(2)]

(6) Add an equal volume of Ficol-Paque using glass pipette. Slowly release the Fico-Paque at the bottom of the tube being careful not to dispense any while moving the tube down or out. (create a balance tube having and equal volume of Ficol and RPIM-5]

(7) centrifuge again

(8) Using glass pipette suck out the fluffy layer in tube and transfer to new tube. Add  5 ml RPMI-5. 

(9) Centrifuge 10 min at 1080 RPM

(10) pour off supernatent and resuspend pellet with 10 ml of RPMI-5 

(11) resuspend pellet in 10 ml RPI-5. Check cell viability 1:5 by taking 10 mul of cells to 40 ul trypton blue.

(12) Divide cells in tube into flasks with desired volume (ex. 5 ml RPMI R-20 + 35 ml R-10. ex. 50 ml R-10)

Freezing Cells

Suppose that after counting cells (above) you count 53 live cells and 0 dead. So you would have 10,000 * 53 = 530,000 cells per 1 ml. You would actually have twice this amount or 1,060,000 cells/ ml in the flask since we diluted 1:1 with tryptan blue. 

Question: We want to take out 20 ml of our cells and resuspend those cells after centrifusion in what volume if we  want final [cell] to be 2 * 106 cells/ml? 

1,060,000cells/ml * 20 ml * = 21,200,000 cells total

21,200,000 cells/x ml = 2,000,000 cells/ 1 ml

21,200,000 / 2,000,000 = x

10.6 ml = x

(1) Obtain and label specialized cell tubes. 

(2) we want to get tubes cold to enhance freezing so place them at -20.

(3) prepare freezing media. We want a 20 ml total volume freezing media (DMSO which is dimethly sulfoxide)  . So get conical tube (yellow cap) and add 

–18 ml of FBS (since we 90% FBS (20 * .90 = 18) +

–2ml DMSA (since we want 10% DMSO)

Mix tubes by inverting. Label (ex. “FM 90% FBS + 10% DMSA JL”) and place at -20c

(4) Obtain the flask with cells that contains 100% viability and add 20 ml to a conical tube.

(5) centrifuge tube at 1100 RMS for 10 min (make sure tubes are balanced and swinging buckets in tact)

(6) pour off liquid (cells will be at bottom)

(7) Transfer freezing media (3) above to tube (6). Pipette solution up slowly with 25 ml pipette to dissolve the cells. Use 5 ml pipette to transfer 1.3 ml into 14 vials (so will be using 18.2 ml total). 

(8) Place cell tubes into freezing container which has isopropyl alcohol poured into the bottom. Place this container at -70 for 4 hours (we want to freeze cells slowly which is the exact opposite from thawing). Label the 14 tubes “B16H78 2 * 106 6/28/03 JL”.

Macrophages/dendrites/neutrophils: CD14, CD1a and CD64