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Brews & Views Bulletin Board Service * Brews and Views Archive 2006 * Archive through February 15, 2006 * Counting yeast cells for pitching < Previous Next >

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Milan Bartolec
Member
Username: Littlebro

Post Number: 130
Registered: 03-2003
Posted From: 24.22.175.129
Posted on Wednesday, January 25, 2006 - 07:20 am:   Edit Post Delete Post View Post/Check IP    Ban Poster IP (Moderator/Admin only)

I've made a starter for my lager. Starter volume is 1.250 Liters (10x the XL smack pack). I want to take a sample and count the # of cells to determine what my pitching rate is going to be. Got a microscope and hemocytomer tried it out and no problem counting the cells.

My question is this: Where do you pull the sample from?
1. The wort itself,
2. The slurry at the bottom,
3. Stir the whole thing up and then pull a sample?

TIA
 

Fredrik
Senior Member
Username: Fredrik

Post Number: 2877
Registered: 03-2003
Posted From: 62.20.8.114
Posted on Wednesday, January 25, 2006 - 08:04 am:   Edit Post Delete Post View Post/Check IP    Ban Poster IP (Moderator/Admin only)

In order to make an accurate count you need a representative sample. Sometimes you need to make dilutions.

To take a sample but pulling a pipett into a thick sediment and put on the slide is IMO no good idea. The flow into a narrow pipett can be changed, also you would get a too high cell density in on the slide so it would be hard to count.

I'd stir it all up, to make it homogenus. Then take a sample from the suspension quickly before is sediments.

Then check it. If there cells are so dense that it's hard to count - take a new sample and make a 1:10 diltion and investigate that. Then you know the wort has the cell density of x10 of what you find under the microscope.

If you have time it's always a good idea to take more than one sample and make new slides of each. If there is huge sample to sample variation something is probably wrong.

My guess is that you have some 200-300 billion cells in that culture.

/Fredrik
 

Bill Pierce
Moderator
Username: Billpierce

Post Number: 4519
Registered: 01-2002
Posted From: 24.57.229.8
Posted on Wednesday, January 25, 2006 - 02:50 pm:   Edit Post Delete Post View Post/Check IP    Ban Poster IP (Moderator/Admin only)

I'll echo what Fredrik said about diluting the sample (sometimes even 100:1) in order to make it easier to count the cells. Then you multiply the count by the dilution factor. Use a pipette and measure as accurately as you can. I'm assuming you know the procedure for counting cells that are touching the hemacytometer grid lines on two of the sides but ignoring those that are touching the other two sides.

If you have access to methylene blue stain, it's best to use it to identify the healthy cells. You ignore counting the ones that aren't healthy.
 

Milan Bartolec
Member
Username: Littlebro

Post Number: 131
Registered: 03-2003
Posted From: 24.22.175.129
Posted on Wednesday, January 25, 2006 - 05:17 pm:   Edit Post Delete Post View Post/Check IP    Ban Poster IP (Moderator/Admin only)

thanks guys, I'll use that proceedure. Here are some notes from some trial runs I did taking yeast after my primary was done.

Washed Yeast.

I did originally take a sample right from the slurry, and it was way too dense to count. So I dilluted it 10x, still too dense. I was able to dillute it to 50x and count (Bill, I've read up on how to count the cells that touch the lines).

I counted 250 cells [(250*5)*(50)*(1x10^4)] = 6.25*10^ what is that?- 6.25 billion cells/ml

Then I took a sample from the wort of my starter (completely unrelated beer&yeast) - (at low kraussen), counted 123 cells in 5 of the medium squares X pattern. Dillution 10x
[(123*5)*(10)*(1x10^4)] = 6.15 x10^7 = 61.5 million cells/ml.

Is my math OK here?

I'll pitching this into a lager wort of 5 gallons @ 11.86 Deg. Plato. This should be fine, but I've read that you want 12-15 million cells/ml for each Deg. plato. In which case I would need 144 - 180 million cells/ml. I'm not worried about it, but I just would like to confirm my grasp of the concepts I've been reading.

Also, I've got the methylene blue powder, but there are no directions on how to mix it. I've read that 1 gram/L should be fine, but I don't want to mix that much at one time, and don't have anything that will really measure 1 gram that accurately. Any suggestions using tsp. 1/4 tsp. or even color of final mixed product (dark blue, light blue)?

Thanks again
 

Bill Pierce
Moderator
Username: Billpierce

Post Number: 4525
Registered: 01-2002
Posted From: 24.57.229.8
Posted on Wednesday, January 25, 2006 - 05:32 pm:   Edit Post Delete Post View Post/Check IP    Ban Poster IP (Moderator/Admin only)

For what it's worth, 1 tsp. is 4.93 milliliters. From what I remember, the stain should be a medium to dark blue.
 

Fredrik
Senior Member
Username: Fredrik

Post Number: 2881
Registered: 03-2003
Posted From: 213.114.44.246
Posted on Wednesday, January 25, 2006 - 06:20 pm:   Edit Post Delete Post View Post/Check IP    Ban Poster IP (Moderator/Admin only)

Milan, 1 g/Liter is perfect. In fact I'm using a methylene blue just diluted in plain distilled water to 0.05% (0.5g /liter) and it works fine. My solution is dark blue in the bottle.

Regarding the calculations, I belive the volume in squares may depend on the brand of your chamber. Not all are the same. Refer to your documentation.

Mine looks like this.



The cell density is

# counts per square / Square-volume * dilution

If you count several squares you just average.

I really don't know what yours are. But as an example. If your mid-square are like my "b" squares, then they are 1e-6 ml each.

If you have 123 cells in 5 squares? - then you have a total volume of 5 times square volume. In this example (one sqaure = 1e-6 ml) you have

123 cells / 5e-6 ml * 10 dilution = 246 million/ml

About the pitching rates, note that 1 million/ml/P refers to the wort after pitching ie. 5 gallons wort - not the cell density in your starter.

5*3785*12*1 = 227 billion cells

/Fredrik
 

Fredrik
Senior Member
Username: Fredrik

Post Number: 2882
Registered: 03-2003
Posted From: 213.114.44.246
Posted on Wednesday, January 25, 2006 - 06:36 pm:   Edit Post Delete Post View Post/Check IP    Ban Poster IP (Moderator/Admin only)

If you can't it make it exactly 1g/L don't worry, it will work fine anyhow. Just estimate. There is no reason whatsoever that it has to be 1.000 g/L.

/Fredrik
 

Milan Bartolec
Member
Username: Littlebro

Post Number: 132
Registered: 03-2003
Posted From: 24.22.175.129
Posted on Tuesday, January 31, 2006 - 11:49 pm:   Edit Post Delete Post View Post/Check IP    Ban Poster IP (Moderator/Admin only)

So Frederik,
Your second post above fills in some gaps for me. My hemocytomer has the same calibrations - no documentation came with it but it does have the same depth of 0.10 mm.

Let me make sure I understand completely what you are saying, because I also have a follow up question: When I counted the yeast cells in my starter, I counted 246 million cells/ml. The volume of the starter was 1.3 Liters or 1301 ml.
1301 ml * 246 million cells = 3.2 x 10^11 = 320 Billion cells. (That is how many yeast cells are in my starter)

The 227 Billion cells (from your post above) is the recommended # of yeast cells needed for a 5 gallon batch of 12P wort. Right?

Using our calculations, it appears that I am pitching an adequate amount of yeast - 320 Billion cells. Right? (320 billion > 227 billion)

If I got all that right, here is my follow-up question:
Q: How do I calculate what volume of a starter will yeild the correct # of cells to pitch? To put it another way, say I have an XL smack pack that contains 100 Billion cells in 125 ml of wort. I need to have 300 billion cells for my next beer.
Using the guidelines for stepping up no more than 10x the starter volume, I would step up to a wort volume of 1.25 L. Is there a guideline that tells me how much my yeast will multiply? Is it 10x?
Or is the rate of growth 3 to 5 times (Noonan) for lager yeast?
 

Fredrik
Senior Member
Username: Fredrik

Post Number: 2894
Registered: 03-2003
Posted From: 213.114.44.246
Posted on Wednesday, February 01, 2006 - 09:09 pm:   Edit Post Delete Post View Post/Check IP    Ban Poster IP (Moderator/Admin only)

Yes Milan, it sounds like you got it right!

To actually know the *exact* number you get is not so easy, and the reason is that the way you do it matters. A stirrer give you more cells, more oxygen and more nutrition also give more cells.

But assuming you have no stirrer, and aerate once, the standard assumption of 5% biomass yield gives approximately this ballpark formula.

Step up one XL pack in X [liter] 1.040 wort to get the cells needed for 1 million/ml/P to ferment V liter of OG wort:

X = (0.25*V*OG-100)/67

Volyme in liter, OG in "gravity points" (ie fr 1.060 you enter 60)

Example: 19 liter of 1.060 wort;

X = (0.25*19*69-100)/67 ~ 2.8 liter 1.040 starter

This is a simple ballpark formula only, but to be absolutely sure what you have, you really need to count the cells. **If you use a stirrer, or shake in oxygen alot you will get far more cells.**

Your numbers for a 1.3 stepupf of a XL pack indicates a high yield - did you use a stirplate or did ou shake it alot? If so, you need to find your own formula. The exact yield is too complex for a simple formula in such a case. Also 100 billion is the typical count or something. I'm sure there is a variation in the packs too. I suspect many packs may havem ore cells.

For practical purposes the formula I suggested should be like a pessimistic prediction, if you don't stir & aerate once. Anything beyonf that will give you more cells. And exactly how much more is not easy to tell, you can only guess. The problem is howto quantify aeration and howto quantify stirring.

/Fredrik
 

ELK
Senior Member
Username: Elkski

Post Number: 1392
Registered: 01-2003
Posted From: 67.177.25.240
Posted on Wednesday, February 01, 2006 - 11:29 pm:   Edit Post Delete Post View Post/Check IP    Ban Poster IP (Moderator/Admin only)

wrong room!
 

Milan Bartolec
Member
Username: Littlebro

Post Number: 133
Registered: 03-2003
Posted From: 24.22.175.129
Posted on Thursday, February 02, 2006 - 02:49 am:   Edit Post Delete Post View Post/Check IP    Ban Poster IP (Moderator/Admin only)

Thanks Fredrik.

I pretty much figured that I would need to count to get the actual numbers. I did oxygenate the starter a couple of times in the first day, and swirled randomly during the next couple of days. Thanks for helping me better understand what I've been reading.

ELK "wrong room!" ??? I'm not sure I understand your post. Am I asking questions that are not appropriate for this board?

- Milan
 

Fredrik
Senior Member
Username: Fredrik

Post Number: 2895
Registered: 03-2003
Posted From: 213.114.44.246
Posted on Thursday, February 02, 2006 - 06:19 am:   Edit Post Delete Post View Post/Check IP    Ban Poster IP (Moderator/Admin only)

Maybe I can expand the logic in how to estimate the # of cells you get a bit.

Suppose you have 1300 liter of 10P wort. This means you have 1300*1.04*0.1 = 135 g extract.

Assume 65% fermentability, this means you have 135*0.65 ~ 87.8g fermentable sugar.

Now the real question is, given that the yeast consumes 87.8g sugar, how much does it grow on this given amount of food?

Like I mentioned before one of the standard assumptions is that during a normal beer fermentation (ie inital aeration and no stirring) you assume that you get new biomass(cells) of about 5% of the mass consumed sugar.

But this is a variable, and it's not constant. During optimal conditions when yeast is grown during controlled respiration this can be up to 50%.

This means that assuming 5% biomassyield you get 0.05*87.8 = 4.4g (dry) new yeast.

So, how much is 4.4 g(dr) yeast? Even this is a bit complex, as it depends on the strain, but I usually estimate with maybe 20 billion/gram. But I guess it could vary from 20-30 or so. This is also why it's hard to predict.

Assume 20 billion/g, then we have 4.4*20 = 88 billion cells.

Add that to the original cells of ~ 100 (XL pack) and we have now 188 billion.

320 billion indicates, given that we have 100 billion originally, and that your starter was 1.040 and not more, an apparent biomas yield of 12.5% in this simple formula.

Also from the wyeast website it seems the counts in their packs can vary +/- 36%, and they have previously expressed that their philosophy is to rather err on the high side than the low side, so you can guess that more often than not, maybe you even have a little more than 100 billlion cells.

Also the step is so short when you step it up less than a generation, that the formula almost breaks down. This could also explaine the apparent large yield.

It is best to see the calculations as a guideline only.

Eventually I'll try to improve that formula but I'm not there yet.

/Fredrik
 

Doug J
Member
Username: Doug_j

Post Number: 169
Registered: 06-2004
Posted From: 207.250.116.150
Posted on Thursday, February 02, 2006 - 09:35 pm:   Edit Post Delete Post View Post/Check IP    Ban Poster IP (Moderator/Admin only)

Sorry I lost track of this thread.

Where did you read that you want 12-15 million cells/ml per P?

That is ridiculously overpitching, 1 mc/ml/P is the standard
If there is something here, it's not really blank
 

Fredrik
Senior Member
Username: Fredrik

Post Number: 2896
Registered: 03-2003
Posted From: 213.114.44.246
Posted on Thursday, February 02, 2006 - 10:17 pm:   Edit Post Delete Post View Post/Check IP    Ban Poster IP (Moderator/Admin only)

I agree with Doug. I missed that.

Sometimes in brewing texts they write like

"12-15 million/ml"

wich is the same as 1 million/ml/P assuming your OG is 1.050-1.060 (which is normally the case).

I suspect that's what you read though.

/Fredrik
 

Milan Bartolec
Member
Username: Littlebro

Post Number: 134
Registered: 03-2003
Posted From: 24.22.175.129
Posted on Thursday, February 02, 2006 - 11:35 pm:   Edit Post Delete Post View Post/Check IP    Ban Poster IP (Moderator/Admin only)

I'm reading Noonan's "New Brewing Lager Beer" p 166 - Fermentation chapter.

"Approximately .5 to 6. Fluid oz of (10 to 14 grams) of pasty, thick yeast is needed for each gallon of wort to be pitched to give 12 to 15 x 10^6 cells per millileter of wort.

So you are probably right Fredrik, because a paragraph later, Noonan references a pitching rate ... "Conservative pitching rates (.8to 1 x 10^6 cells/milliliter per each dPlato of the wort)"

So in the first instance he does not reference the gravity of the wort, and in the second he does.

Its a wonder I have a difficult time comprehending what I read in the more advanced books - which is why I bought the microscope, hemocytomer, etc. - I'm a hands on learner.

I feel a RDWHAHB coming on. Which I am. Right now.

Thanks all for working through this with me - I have learned a ton.

- Milan