# Experimental Design - test your skills

1. Feb 26, 2010

### Ouabache

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 cross-disciplinary education.

(hint: some maths may be used)

Last edited: Feb 27, 2010
2. Feb 28, 2010

### Ouabache

<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. Mar 1, 2010

### Monique

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

4. Mar 1, 2010

### Andy Resnick

I'm not sure I can do it if a control is needed.

5. Mar 1, 2010

### Monique

Staff Emeritus
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 384-well plates. How would you positively identify the one gene out of the 6144 total that interacts with your bait protein?

Last edited: Mar 1, 2010
6. Mar 1, 2010

### Ouabache

how were you planning to use a control?
In your experiment, are you allowed to put more than one protein per well?
It appears that space is constrained to three 384-well 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.

Last edited: Mar 1, 2010
7. Mar 2, 2010

### Monique

Staff Emeritus
The experiment would be constrained in that you can only test 1152 reactions (3x384). You can put more than one protein in a well, but you need to come up with the most cost-effective way to identify the protein that interacts.

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. Mar 2, 2010

### Ouabache

Actually I am the OP'er, so I get to clarify my question . Let's include the wording, 'you only have time to run a single experiment'. You're right about considering the concentration of alcohol dosage. That is why I left the next clue, only a very small dose (1 ul of wine) is necessary to test positive. Also I didn't specify what kind of lab animal to use. What I am indirectly saying is that you can test more than one bottle of wine on the same lab animal in a single experiment.

Last edited: Mar 2, 2010
9. Mar 2, 2010

### Ouabache

For your protein experiment, let me take a guess at a cost-effective design. Since you can add more than one protein in a well, let's assume (unless you tell me otherwise) that the number of proteins you can add per well is not a constraint; such that putting all 6144 proteins in a single well is physically possible and will not confound the results.

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 cost-effective. 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 cost-effectiveness.

10. Mar 2, 2010

### Monique

Staff Emeritus
You've not incorporated the condition that the interaction must be seen six times and how are you going to handle false-positive interactions?

11. Mar 2, 2010

### Ouabache

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.

Last edited: Mar 2, 2010
12. Mar 3, 2010

### Andy Resnick

I was thinking of giving one animal *all* the different samples, to determine if in fact, there was a tainted sample. I suppose another animal has to be given a 'sham' procedure- an equivalent dose of benign fluid, handled in the same way, housed with the same animals, etc. to ensure any lethality arises from the wine, and not an environmental factor.

13. Mar 3, 2010

### Monique

Staff Emeritus
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.

Last edited: Mar 3, 2010
14. Mar 3, 2010

### Andy Resnick

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. Mar 4, 2010

### minorwork

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.

0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111​

Now you've 16 different ways to get results from 4 rats. Use them wisely young Jedis.

16. Mar 4, 2010

### Monique

Staff Emeritus
And we have a winner!! (unless Ouabache has something else in mind) With 4 lab animals, the number of combinations you can test are: 24=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. Mar 4, 2010

### minorwork

Don't hand out the cookie yet. Things are just starting to get interesting.

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 1-14, but some other description of it. Think, Deal or No Deal.

Last edited: Mar 4, 2010
18. Mar 4, 2010

### Ouabache

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.

Here you propose giving animal #1 a dose from each bottle #1,3,5,7,9,11,13,14 (lets call this set#1) If animal#1 dies, you have seven bottles to test (in set#1). If it lives we assume the tainted bottle is one of 2,4,6,8,10,12 (six bottles) (call this set#2). We only have enough time for one experiment.

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.

Last edited: Mar 4, 2010
19. Mar 4, 2010

### minorwork

You are correct. I was warming up to the problem. I had discarded it by the time I'd ended the post. However, all the animals will be given a dosage at the same time.

My efforts would be concentrated on designing a specific mix for each mouse. I'm going to use mice now. Each mouse would receive a drop (assuming one drop of the tainted bottle would for sure kill a mouse) from each of a specific combination of the sample bottles. What would be the size of that sample? 14 or 13. Would there be anything to be gained by not including a bottle, say #14? Yes. If #14 was tainted and not consumed by any mice then all the mice would live. 1111, four mice alive would mean the untested bottle was the tainted one. So now our 4 combinations of the remaining 13 must be designed to pinpoint the tainted bottle if #14 is not the poisoned bottle.

Each mouse will take a combination of solutions at the same time. In 24 hours the single experiment is complete. The results will be available 24 hours later determined by the specific combination of the mice (or particular mouse) that kicks the bucket.

Now I am a bit out of my pay grade and the time I wish to allott to designing the specific combinations for each mouse. I can, and have picked a specific bottle of the 14 to be the tainted bottle. If a specific combination of bottles per each designated mouse is presented I can tell you which mice will die. You can tell me which bottle is tainted and I can confirm that conclusion.

I'll present, publicly, bottle #9 as the poisoned bottle. I'll also hide a number on a stickup here for any to test against for results. You give me which bottle or bottle combinations was drunk by which rat and I'll give the numbers of the rat that died. I will give it like this:
1101 would mean that, starting with the least significant digit as representing Mickey, next would be Minnie, then Tom and finally Jerry as the most significant digit. In the case of 1101 we would have the death signified by a 0 which would mean Minnie would die. 1011 would mean Tom died.

Thinking out loud and wishing for a program that would give me results. 13 bottles. Give Mickey a drop from, oh, let's just say for starters, bottles 1-7. Mickey will live since he didn't get poisoned. Any combination of mice deaths with Mickey alive will mean that the suspect bottles are 8-13. I am not going to include bottle #14 in any administered solution. So that all the mice alive point to #14 as the poisoned bottle. Thanks Howie Mandell.

Of course if the poisoned bottle is not known it could be that Mickey might be dead in 24 hours. That would mean the tainted bottle is among the numbers 1-7.

Let's see if we can narrow down Bottle 9.

Minnie, Tom, and Jerry get what? Hmm. Isn't there some math genius available to do this fun stuff for me?

Hey how about you setting a bottle as poison and let me tell the combinations of bottles that each gets, then you telling which ones croak?

My ears are smoking. Later

20. Mar 5, 2010

### Monique

Staff Emeritus
It's simple, you make four different pools and assign bar codes. You administer one pool to each animal and note the outcome after 24 hours.

animal
1234
0001 bottle 1
0010 bottle 2
0011 bottle 3
0100 bottle 4
0101 bottle 5
0110 bottle 6
0111 bottle 7
1000 bottle 8
1001 bottle 9
1010 bottle 10
1011 bottle 11
1100 bottle 12
1101 bottle 13
1110 bottle 14

If animals 2 and 3 don't make it, it's code 0110 and that uniquely corresponds to bottle 6.