
#1
Feb2610, 09:55 PM

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There are 14 bottles of wine and one is tainted such that,
a single sip of the tainted bottle is lethal. It takes up to 24 hours for full effect of the toxicity. If you have only 4 lab animals to test all 14 bottles of wine; what experimental design might you use, to determine which bottle was lethal? You need to determine the answer after 24 hours. For those who heard this question on a popular radio program, (or have heard this question before) please let some of the others give this a try. I am curious of your approach, if you come from a biology academic background, versus those with a crossdisciplinary education. (hint: some maths may be used) 



#2
Feb2810, 06:30 PM

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<nudge> So getting to the solution(s) of this question is not hard. After I worked out a design, I passed this by a h.s. senior and he figured out the approach with not much difficulty.




#3
Mar110, 08:26 AM

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I can think of an approach, but maybe I should let other people take a guess?




#4
Mar110, 11:52 AM

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Experimental Design  test your skills
I'm not sure I can do it if a control is needed.




#5
Mar110, 12:55 PM

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Actually I've read up on the technique for my work, so this riddle is pretty straightforward to me. Only I'm not out to identify the tainted bottle of wine by examining the lethality in rabbits, but I'm out to identify protein interactions by examining growth of yeast on selective plates.
The question would be something on the line of: you have 6144 proteins, one of those has an interaction with your favorite bait protein (resulting of growth of the yeast on the selective plates). You only have money to examine three 384well plates. How would you positively identify the one gene out of the 6144 total that interacts with your bait protein? 



#6
Mar110, 09:25 PM

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It appears that space is constrained to three 384well plates, but not time. Can we assume, you have time to incubate more than one plate? In the o.p., lets make an assumption that only a small dose is necessary, to test positive for lethality (e.g. 1 ul) of wine. The line "You need to determine the answer after 24 hours" is to be interpreted that, you need to find out the answer at the conclusion of 24 hours. So time is a constraint. You only have time to run one experiment. 



#7
Mar210, 02:46 AM

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The O.P. is a tricky question, because it says that a sip is lethal after about 24 hours. This does not mean that you can do a single experiment in those 24 hours, because toxicity is in the dose. You could give the rabbits a higher dose and see a result quicker, but I am not sure how the rabbits would react to a higher dose of alcohol. That might be lethal to them as well. 



#8
Mar210, 10:15 AM

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#9
Mar210, 10:26 AM

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The approach I made for your question is to divide the total number of proteins (6144) by the number we may add per well and see how many wells it would require. For example, if we put 2 proteins/well, we would need 6144/2 = 3072 wells. We are constained by having only 1152 wells, so I continued finding common divisors. The first divisor that requires less than or equal to 1152 wells, happens to be 6 proteins/well. It requires 6144/6 = 1024 wells. If one of those 1024 wells indicates a positive result, you will have narrowed the field down to 6 proteins. (ie. the 6 proteins in the well that tested positive). If we have enough wells available for a 2nd experiment, we could then test those 6 proteins singly per well and see which it is. So total wells needed would be 1024+6 = 1030. This is within the total wells available 1152 and satisfies the given physical criteria However you also want it to be a costeffective. I take that to mean it should take the least amount of labor (effort to set up), materials (number of plates used and incubator space), and time. So I continued looking for common divisors of 6144, until I found one that used the smallest number of wells (require the least amount of labor to set up). I found 64 proteins per well, only requires 6144/64= 96 wells. After the first experiment, one of those 96 wells would test positive and you have narrowed the field down to 64 proteins. The 2nd experiment (testing one of those 64 proteins per well) would tell you which one it is. So total number of wells needed is 96+64 =160. If you could use the same plate for more than one experiment, you are within 384 wells/plate and you could find the unique protein in two experiments using a single plate. If you cannot use the plate again, after the first incubation, then at most you would need 2 plates. The next common divisor is 96 proteins/well initially uses 6144/96 = 64 wells, total wells required 96+64 =160. This has the same cost effectiveness as the last one (using 64 proteins/well). If you continue this process, the total number of required wells become larger than 160. (ie 128 proteins/well requires total 176 total wells, 256 proteins/well requires total 280 total wells, etc.). Based on this selection process, I would choose 64 proteins/well using a total of 160 wells. It is well within the physical constraints given and good costeffectiveness. 



#10
Mar210, 11:45 AM

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You've not incorporated the condition that the interaction must be seen six times and how are you going to handle falsepositive interactions?




#11
Mar210, 09:13 PM

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I have more thoughts about the protein identification example, but elect to reserve comment until some solutions start coming for the original post.
For those out there who are diligently pondering how to find the tainted bottle of wine, let me add that it is not complicated. It is very simple compared to Monique's example. 



#12
Mar310, 07:44 AM

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#13
Mar310, 09:41 AM

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You can definitely test all the samples, there is also room for two negative controls
By knowing the number of lab animals, you can quite easily calculate how many samples can be tested. 



#14
Mar310, 02:01 PM

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Wait...what? 4 animals, 14 bottles to test (all at once, given the time constraints) and I have enough animals for 2 negative controls? I don't see how I can uniquely identify the poisoned bottle with 2 animals and a single test.
I can uniquely identify the bottle if I have 4 animals and no negative control. 



#15
Mar410, 12:36 PM

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I be a retired coal miner. This problem is similar to finding the odd weighted of 12 balls and whether it is light or heavy by weighing various combinations but three times only on a balance. This can be done. I used my wife to choose a ball and assign it heavy or light and as I put combinations in a right and left pan of an imaginary scale she would identify which side went up. Three weighings is all that it takes if you do it right.
Thinking out loud now. Applied to 14 bottles and 4 rats then. All the feedings of the solutions must be kept track of carefully. All feedings will be accomplished simultaneously. The observed dead rat at the end of 24 hours will determine the bottle containing the poison. Let's eliminate the odd number bottles. Prepare for rat 1 a solution containing a drop from each odd numbered bottle, and maybe a drop from bottle 14. If that rat lives then the poison will be in bottles 2, 4, 6, 8, 10, or 12. Only 6 to figure out. If it dies then the poison is in one of the other 7. This may not be the correct combination of drops for rat 1, but the conclusion of which of 4 rats dies will determine, in a well defined mix of solutions imbibed by each of 4 rodents, if it is possible, which bottle is tainted. 4 rats with two states. Dead or alive. A binary set of possibilities that may include that more than one rat die. Or, information is obtained even if no rats die in a type of scheme like Howie Mandell's "Deal or No Deal" game show where all the previous guess's leave only the untested bottle as the tainted one. Thus there are these possibilities of results for our fortunate rats. 0000Now you've 16 different ways to get results from 4 rats. Use them wisely young Jedis. 



#16
Mar410, 04:09 PM

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And we have a winner!! (unless Ouabache has something else in mind) With 4 lab animals, the number of combinations you can test are: 2^{4}=16. The number of samples you can put into a pool are: 16/2=8 (look at the above binary columns, each column has 8 samples). This means that when your number of samples increases exponentially, the number of pools to be tested increases linearly.




#17
Mar410, 04:25 PM

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What bottle is indicated by none of the rats dying at the end of 24 hours? HINT: I am not asking for the number of the bottle 114, but some other description of it. Think, Deal or No Deal. 



#18
Mar410, 07:08 PM

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I see some interesting ideas, but as minorwork has noted, we need to hold off handing out cookies until a complete design, is proposed.
Minorwork, on your odd & even thoughts, although a viable concept, it does not meet the constraints of this question. I elaborate below. We don't know whether the tainted bottle is in set#1 or set#2. We don't have enough animals left (three animals left) to test both sets simultaneously with the combined dosage you gave animal#1 (we do have enough animals to test 'either' set 1 or set 2. So this particular stategy will not work with our given conditions. 


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