Kathryn is leaving for the airport in five minutes. If you haven't sent her your problem #2 answers yet, she asks that you cut and paste them into the body of your e-mail. She'll be able to check e-mail when she's in Antanarivo Sunday and Monday, but Internet connections are very slow, and an attachment may not be downloadable. It might be a good idea to send me a copy of your answers too, just to make sure that one of us has a copy.
September 2008 Archives
I've heard from a couple of people who are having difficulty getting started on Problem #2. Here's the hint I gave them, and I thought I'd pass it along to everyone.
To answer the second question you're going to need to think of a way to compare fitnesses. You don't have to use WinBUGS to answer that question. You could do a chi-squared contingency test, rather than using WinBUGS. But to get full credit for your answer on this question, you'll need to do one or the other.
- Remember the kind of selection we've been focusing on, namely viability selection.
- Remember the definition of viability selection, namely the probability of zygote to adult.
- Remember the assumptions in the problem statement, namely that 90%
of those diagnosed with severe malaria dia and that anyone who is going
to get severe malaria gets it before they have a chance to reproduce.
To answer the second question you're going to need to think of a way to compare fitnesses. You don't have to use WinBUGS to answer that question. You could do a chi-squared contingency test, rather than using WinBUGS. But to get full credit for your answer on this question, you'll need to do one or the other.
If you've already finished Project #1 and are looking for something to do, you'll find a link to Problem #2 on the Detecting viability selection lecture page. There isn't a link to a dataset, because the data you'll need are included in the problem statement.
I've just posted the spreadsheet with consolidated results from the structure analysis. You can follow the highlighted link in this post, or you can find the link along with all of the other project materials on the lecture detail page for 10 September, when Project #1 was assigned.
Hi All,
I am still waiting on the last few people to send me their Structure data. I teach from 2:30-6:30 this afternoon, so if everyone gets me data, the spreadsheet will go out this afternoon. Otherwise I will check my email at 6:30 and add whatever data is left. I will be gone the rest of the evening, so if you don't have your data to me by 6:30, it will be considered late.
Cheers
Kathryn
I am still waiting on the last few people to send me their Structure data. I teach from 2:30-6:30 this afternoon, so if everyone gets me data, the spreadsheet will go out this afternoon. Otherwise I will check my email at 6:30 and add whatever data is left. I will be gone the rest of the evening, so if you don't have your data to me by 6:30, it will be considered late.
Cheers
Kathryn
I've heard from a couple of people that they're getting an error message when they try to run Structure: "Could not create Java virtual machine." I'm not sure what's causing the problem, but here's one thing to try if you're running into that problem.
- Reinstall Structure, and when you get to the "Install Type" screen, check the radio button for "Custom".
- Click "Next", and make sure all of the check boxes are checked, especially the one labeled "Jre".
There appears to be a bug in the theta=0 model in Hickory. I'll see if I can find a workaround, but unless you hear otherwise, plan to do only the full model and f=0 model in Hickory.
Sorry about that.
We went over how to run Structure pretty quickly on Friday, and I've heard from a couple of people who are having some problems. Here's a list of instructions so it's all in the same place.
Since we need results from this analysis by Tuesday morning so that we can post the results, it would be a very good idea to see if you can get the analysis running today so that we can go over any problems tomorrow, even if you don't finish the analysis tonight.
- Double click on the Structure icon and start the program
- From the "File" menu, select "New project".You'll get a "Project Wizard" box labeled "Step 1 of 4".
- Give your project a name, e.g., "humans" (without the double quotes)
- Select the directory where you want the results to be stored, probably the directory where you downloaded the file humans.stru.
- Select the humans.stru file and click "Next >>". You'll get a "Project Wizard" box labeled "Step 2 of 4"
- There are 263 individuals in the data set, the ploidy is 2 (diploid), the number of loci is 100, and the missing data value is -9. Enter those values in the respective boxes and click "Next >>". You'll get a "Project Wizard" box labeled "Step 3 of 4".
- Check the box labeled "Row of marker names", and click "Next >>". You'll get a "Project Wizard" box labeled "Step 4 of 4".
- Check the box labeled "Putative population origin for each individual", check the box "Other extra columns", enter 2 in the box labeled, "Number of extra columns",and click "Next >>". You'll get a "Confirmation" box. Click "Proceed" and you sho uld have the data appearing as a spreadsheet in Structure.
- To do that select "Parameter Set -> New". You'll get a "New Parameter Set" box popping up.
- We want a burnin of 10,000 iterations and a sample of 50,000 iterations. Enter those numbers in the respective boxes and click "Ok".
- You'll get an "Input" box popping up asking you for the name of your parameter set. You can pick anything you want. I usually pick "default". Enter the name you want in the box and hit "Ok". The box will disappear, and you'll see in the list of things on the left side of the screen a new Parameter set with the name that you just provided.
- Select "Project -> Start a job", and a "Structure Scheduler" box will pop up.
- Highlight the parameter set you created, set K from 2 to 10 by entering the numbers in the appropriate boxes. Leave the number of iterations set at 1.
- Click "Start". The "Strucutre Scheduler" box should disappear, a "Structure Job Log" box should appear, and you'll start seeing numbers appearing in the black box at the bottom of the screen.
Since we need results from this analysis by Tuesday morning so that we can post the results, it would be a very good idea to see if you can get the analysis running today so that we can go over any problems tomorrow, even if you don't finish the analysis tonight.
I've just uploaded the spreadsheet for the Structure part of the assignment to the website. Here's the link. There will also be a link to it from the assignment page shortly.
Remember the t-shirt experiment I referred to? Well, a paper published today in PLoS Genetics provides evidence at the population level that mate choice in humans is influenced by the genotype at MHC loci. European American couples are significantly more dissimilar than would be expected if mating occurred at random.
I just uploaded a new version of the NEXUS file for GDA. None of the data have changed. So if you've already run analyses with GDA, you don't need to run them again. But the new file has an addition that you'll find helpful in answering the second question. I'll describe the addition in lecture today.
By the way, there may be a problem with the theta=0 analysis in Hickory. Kathryn and I are checking it out now and we hope to know for sure before lecture. If we don't know by then, I'll post a notice as soon as we know how to work around the problem.
I found a small error in the notes for today's genetic structure lecture. It's buried in the "Gory details" section, so you probably won't care. But if you do, and if you've printed out that page, you might want to print out a new copy.
Hi all,
For any of you who are thinking to track me down on Monday morning, I've got some bad news. I'm have a meeting and then a doctor's appointment so probably won't be around until about 10:30.
Kathryn
For any of you who are thinking to track me down on Monday morning, I've got some bad news. I'm have a meeting and then a doctor's appointment so probably won't be around until about 10:30.
Kathryn
Several people have asked for clarification on the bonus question. Let me see if I can state the question in a way that makes my intent clearer.
What frequency of homozygotes would you expect in the population if the allele frequencies you just estimated are characteristic of the population and genotypres are in Hardy-Weinberg proportions? Suppose that you didn't know there was a null allele, i.e., you assumed that those individuals who are homozygous null were bad samples that could be ignored. What frequency of "homozygotes", i.e., phenotypes you'd score as homozygous even if they really aren't, would you expect to find in your sample?
Let me know if that doesn't help, and I'll try again.
What frequency of homozygotes would you expect in the population if the allele frequencies you just estimated are characteristic of the population and genotypres are in Hardy-Weinberg proportions? Suppose that you didn't know there was a null allele, i.e., you assumed that those individuals who are homozygous null were bad samples that could be ignored. What frequency of "homozygotes", i.e., phenotypes you'd score as homozygous even if they really aren't, would you expect to find in your sample?
Let me know if that doesn't help, and I'll try again.
If you're having trouble with Problem #1, you're not alone. I've spoken with many of you and shared some hints, so I thought I'd write them down and pass them along.
First, if I told you there were five alleles segregating in this population (A0, A113, A119, A123, and Au) with frequencies p0, p113, p119, p123, and pu I'll bet you could figure out how to calculate the frequency of each offspring category in this table, right?
So do that and figure out how the phenotype categories in the problem correspond to the genotype frequencies in this table. (You'll have to do this for the other maternal genotype too, obviously.)
That's the hard part, and that's the part I'm really interested in. That's biology (or at least genetics). The rest is translating the formulas you have in that table into WinBUGS.
For that, stare at the code we used for estimating allele frequencies in the ABO blood group system a bit, try your hand at a solution, and if you get stuck, stop by or e-mail Kathryn or me for some additional hints.
First, if I told you there were five alleles segregating in this population (A0, A113, A119, A123, and Au) with frequencies p0, p113, p119, p123, and pu I'll bet you could figure out how to calculate the frequency of each offspring category in this table, right?
| Maternal Genotype | Seed genotype | |||||||||
| A0A123 | A0A0 | A0A113 | A0A119 | A0A123 | A0Au | A123A0 | A123A113 | A123A119 | A123A123 | A123Au |
| ? | ? | ? | ? | ? | ? | ? | ? | ? | ? | |
That's the hard part, and that's the part I'm really interested in. That's biology (or at least genetics). The rest is translating the formulas you have in that table into WinBUGS.
For that, stare at the code we used for estimating allele frequencies in the ABO blood group system a bit, try your hand at a solution, and if you get stuck, stop by or e-mail Kathryn or me for some additional hints.
I just realized that there's a problem with the way I presented the data for the first mother in the problem set. This is what's in the problem set as posted:
Here's the problem. A0 is a null allele. That means that when we score the offspring we can't tell an A0A123 offspring from one that's A123A123, just as we can't tell whether a human being with blood type A is homozygous for the a allele or heterozygous for a and o. So this line of the table should really read
In other words the A123 phenotype among offspring of this mother includes two genotypes, A123A123 and A123A0.
| Maternal Genotype | Seed genotype | ||||
| A0A123 | A0A0 | A0Ax | A0A123 | A123Ax | A123A123 |
| 1 | 10 | 0 | 10 | 3 | |
Here's the problem. A0 is a null allele. That means that when we score the offspring we can't tell an A0A123 offspring from one that's A123A123, just as we can't tell whether a human being with blood type A is homozygous for the a allele or heterozygous for a and o. So this line of the table should really read
| Maternal Genotype | Seed phenotype | |||
| A0A123 | A0 | A0Ax | A123Ax | A123 |
| 1 | 10 | 10 | 3 | |
In other words the A123 phenotype among offspring of this mother includes two genotypes, A123A123 and A123A0.
Just so we're all on the same page, I thought it would be a good idea to let you know two things about what and when we're expecting to receive your answers to Problem #1.
- We're expecting you to turn in printed copies of your answers, including the WinBUGS code you used to solve the second problem.
- We'll ask that you turn in your answers at the beginning of class on Monday. Talk with Kathryn if you think you're going to have trouble making that deadline.
Continue reading Problem #1 due date clarification.
Hello everyone! If you would like me to take a look at your WinBUGS code (or another aspect of the homework you are having problems with), I am available on Thursday (Sept 4) from 9-12, Friday (Sept 5) from 2-4. I'm expecting that you will have already given the problem a try (or two) and that you have identified the area(s) that you need help with. I'll need to be able to see your code, and it will probably be very helpful to bring your computer along if at all possible.
My office is in the Pharmacy/Biology building, room 302. You can ignore any signs that are up regarding alarms. (The alarms are deactivated during the day.) My extension is 6-5731 if you get lost!!
Cheers,
Kathryn
My office is in the Pharmacy/Biology building, room 302. You can ignore any signs that are up regarding alarms. (The alarms are deactivated during the day.) My extension is 6-5731 if you get lost!!
Cheers,
Kathryn
If you're running Windows Vista, you are liable to run into permission problems when you try to install the Keys.ocf file that comes as part of your registration. To avoid this problem, I suggest installing WinBUGS in your Public directory (c:/Users/Public). It's possible to tweak the permissions, but sticking it somewhere else is easier.
If you take a look at the lecture schedule you'll notice that I didn't say anything at all about inbreeding on Friday. We'll talk about inbreeding tomorrow, and my plan is to cover both the basics of inbreeding and the basics of testing for departure from Hardy-Weinberg in lecture tomorrow. To accomplish that, I plan to skip the section on chi-squared tests in the testing Hardy-Weinberg notes, and I'll gloss over the details on self-fertilization, skipping directly to partial self-fertilization in the inbreeding notes.
If you haven't tried using WinBUGS yet, I encourage you to do so soon. It's not the easiest program to learn, and in addition to learning how to use it, you have a problem that you'll need to solve with it. That means you'll need to write your own WinBUGS code. I'll be amazed if you manage to solve the problem without asking for help, and I can't promise that Kathryn and I will be able to help much if you wait until next weekend to give it a shot.
Oh, you'll find some hints in the comments to the Problem #1 update that you may find helpful.
Oh, you'll find some hints in the comments to the Problem #1 update that you may find helpful.
