TISSUE CULTURE MATHEMATICS Cell and Media C o n c e n t r a t i o n s and V o l u m e s
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Submitted by
A A R O N E. F R E E M A N T h e Children's Hospital o f A k r o n B u c h t e l A v e n u e at B o w e r y Street A k r o n , Ohio 4 4 3 0 8
I.
INTRODUCTION M a n y p r o b l e m s in tissue culture a r i t h m e t i c involve cell and m e d i a c o n c e n t r a t i o n s and volumes. M a n y p r o b l e m s o f this t y p e can be u n d e r s t o o d and solved with t h e following simple guidelines and formulas.
II. PROCEDURE Solution: CV = C' V'
A. Universal f o r m u l a for calculating all concent r a t i o n , v o l u m e p r o b l e m s : CV = C' V'. C = C o n c e n t r a t i o n of desired solution. V = V o l u m e desired. C' -- C o n c e n t r a t i o n o f s t o c k solution. V' = V o l u m e o f s t o c k solution.
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4 0 0 X 103 = 4 . 3 X 10 s X 4X l0 s = 4.3X l0 sX X =
P r o b l e m : W h a t v o l u m e o f 200 m M glutamine is required to m a k e 4 liters o f m e d i u m containing 2 mM glutamine?
4 - 4.3
= 0.93 ml
qs 0.93 ml o f cell harvest t o 400 m l
Solution: CV = C' V' B. 2 X 40O0 = 2 0 0 X 8000 = 2 0 0 X X = 40 ml
Preparation o f a w o r k i n g solution f r o m a s t o c k solution. Make = T a k e ; S t o c k c o n c e n t r a t i o n = V o l u m e 1.
qs 40 ml to 4 0 0 0 ml Proof:
P r o b l e m : Make a 6% solution f r o m an 84% solution. Solution: T a k e 6 ml of the 84% solution and dilute to 84 ml.
40 m l X 200 p m o l e s / m l = 8000 p m o l e s
Proof: 8000 u m o l e s 4000 ml
6 / 8 4 X 84% = 6%
= 2 # moles/ml 2.
P r o b l e m : Make a 3mM solution f r o m a s t o c k 6 7 m M solution.
= 2mM 2. P r o b l e m : F r o m a cell harvest containing 4.3 X 10 s cells/ml, p r e p a r e 400 m l o f m e d i u m containing 103 cells/ml.
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Solution: T a k e 3 ml of t h e 6 7 m M s t o c k and dilute t o 67 ml. Proof:
3/67 X 6 7 m M = 3mM
C. Preparation of a working solution f r o m two stock solutions (one weaker, one stronger). H -- D = B H = High stock concentrate. L = L o w stock concentrate. D -- L = A D = Desired stock concentrate. A = V o l u m e o f H to use. B = V o l u m e o f L t o use.
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D. The h e m a c y t o m e t e r 1. Each c h a m b e r of a Neubauer t y p e hemac y t o m e t e r is divided into 9 large squares, each side o f which is 1 m m long. The coverglass is s u p p o r t e d 0.1 m m over the hemac y t o m e t e r . Thus, when properly filled, the fluid v o l u m e in a n y one large square is 1 m m • l m m • 0.1 m m o r 0 . 1 m m ~. Since 1000 m m ~ = 1 cm ~ , 0 . 1 m m ~ = 10 ~ cm 3. T h u s :
Set up y o u r f o r m u l a as follows: A
H D
B
L
Total cell c o u n t • dilution factor • 10 ~ =N
Cross subtract:
N u m b e r o f squares c o u n t e d N = N u m b e r of cells in sample per ml
2. The counts can be misleading if the sample is n o t a true aliquot or if the cells were n o t r a n d o m l y distributed in the hemac y t o m e t e r . To help avoid this, make y o u r dilutions so that there are no m o r e t h a n 40 cells per square. On one side c o u n t all 9 squares or 200 cells, whichever e o m e s first. If the c o u n t s b e t w e e n sides differ b y m o r e t h a n ~^~zu~o,c o u n t a second sample.
Problem: Make a 22% solution f r o m an 85% s t o c k and from a 10% stock. Solution: 85.
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- - 1 0 = 12)
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T h e r e f o r e : Use 12 ml o f the 85% solution and 63 ml o f the 10% solution.
Example: Dilution f a c t o r = 1 : 8 Total c o u n t o n side one = 2 1 5 / 7 sq. Total c o u n t on side t w o = 2 3 6 / 8 sq. C o u n t in sample = 30 ~'5-1,.• 8 X 104
Proof: 12 ml • 85 gms/100 ml + 63 ml • 10 g m s / t 0 0 ml 75 ml
10.2 gms + 6.3 gms 75 ml 22 gms/100
= 240 • 104 = 2.4 • 106 cells/ml
C A U T I O N : Some o f the older h e m a c y t o m e t e r s are n o t 0.1 m m deep. If t h e y are 0.2 m m deep, each square will contain a volume o f 0.2 m m 3 and the f o r m u l a must be adjusted. Use 5 • 103 instead o f 104 in the formula.
16.5 gms 75 ml = 22%
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F. N o r m a l i t y When chemicals react, t h e y do so on an a t o m i c and n o t on a weight basis. Therefore, since a t o m s differ in size and weight, t h e use o f m o l a r i t y gives us a b e t t e r insight t h a n weight as to t h e activity o f a chemical. However, certain a t o m s have d i f f e r e n t valences and molecules dissociate. To d e t e r m i n e equivalent activities, t h e valency m u s t b e considered. N o r m a l i t y is n o t h i n g m o r e t h a n an expression o f equivalent reactivities based o n the h y d r o g e n ion unit.
E. Solution c o n c e n t r a t i o n A m o l e is t h e m o l e c u l a r weight o f t h e s u b s t a n c e in grams. A 1 m o l a r solution contains 1 m o l e per liter. T h u s the solution is 1 m o l a r regardless of t h e v o l u m e used. O n e ml is 1 m o l a r (1M) and so is 500 ml. H o w e v e r , one liter contains 1 m o l e , and 500 ml contains o n l y 0.5 m o l e . Usually we need to k n o w b o t h m o l a r i t y and actual weight o f the material in grams. T h e m o s t useful conversions are as follows: 1 Molar = m o l e c u l a r weight in grams/ liter or m g / m l . 1 m M o l a r = m o l e c u l a r weight in mg/liter or pg/ml. F r o m this we can see t h a t :
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HC1 = H + + CI" 1 Molar HC1 = 1 N o r m a l HC1 C a ( O H ) 2 = Ca ++ + 2(OH)-. Thus, it takes 2 m o l e s of HC1 to neutralize 1 m o l e o f Ca(OH) 2. 1M C a ( O H ) 2 = 2 N o r m a l Ca(OH)2.
X MW
M = molarity. X = number of mg/ml. MW = m o l e c u l a r weight in m g / m l .
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III.
1. P r o b l e m : What is the m o l a r i t y o f a solu t i o n o f glucose (MW = 180) containing 360 mg in 20 ml?
Solution: M
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X
. MW
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360/20 . 180
.
18 180
These e x a m p l e s are i n t e n d e d as aids for w o r k e r s w h o have already a t t a i n e d a general b a c k g r o u n d in tissue culture p r o c e d u r e s . For e x a m p l e : it is assumed t h a t t h e reader will be aware of h e m a c y t o m e t e r c o u n t i n g p r o c e d u r e s or will read t h e b r o c h u r e provided with each one. T h e r e is no s h o r t c u t f o r good training. H o w e v e r , every p r o c e d u r e should be carried o u t as quickly and efficiently as possible and these simple f o r m u l a s m a y help achieve t h a t goal. IV.
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2. P r o b l e m : If the m o l e c u l a r weight is 60, h o w m a n y grams are in 54 m l o f a 0.1mM solution? Solution: l m M = m o l e c u l a r weight in mg/liter. Since MW = 60, a l m M solution contains 60 mg/liter. Therefore 0.1mM solution contains 6 m g / l i t e r or 0.6 m g per 100 ml. 54
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DISCUSSION
=0.324mg
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REFERENCE Pierce, C o n w a y , and R. Nelson Smith. 1958. General Chemistry Workbook. W. H. Freem a n and Co., San Francisco and L o n d o n .
