Nick Crawford Evolution and more…

21Feb/090

Quiz 1

You're responsable for all HW assigned in Chapters 2-6. Good luck.

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16Feb/092

New measures of genetic differentiation

Lou Jost has pretty neat paper out describing new measures genetic diversity that accurately account for allelic differences.  I've written some PYTHON scripts to parse genepop files, measure allele frequencies, and generate [some of] these measures of diversity.  Enjoy.

http://www.ngcrawford.com/django/jost/

Jost L. 2008. GST and its relatives do not measure differentiation. Molecular Ecology,17(18), 4015 - 4026

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14Feb/090

Don’t work too hard this weekend

Enjoy this classic flatworm sex clip.

You need to a flashplayer enabled browser to view this YouTube video

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12Feb/090

Wordpress Update Issue (and fix)

Using the 'Automatic' update button I got the following error:

Fatal error: Cannot redeclare pclziputilpathreduction()...

To fix this I had to turn off the 'Wordpress Automatic Upgrade Plugin' that was formerly required for automatic updates.  More details here.

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12Feb/092

Tuesday is a Monday Schedule Next Week

Next week my Monday classes will be taught on Tuesday the 17th, thus my Tuesday class will have two homework assignments due on the 24th.  However, if Tuesday students want the first assignment graded earlier they can leave it in my box in BRB 507.  If you go this route just be sure to put your section number in an obvious place.

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7Feb/090

Chapter 5 Question 14(i): Complete linkage

Another question:

I understand the cross between AB/ab x AB/ab creates 2 phenotypes. 

But...  with the 1.Ab/2.aB x 3.Ab/4.aB cross, however, I get 3 different phenotypes, A-bb (1x3), A-B- (1x4) and aaB- (2x4). I do not see where I went wrong.

(I have numbered the allele sets to show how I crossed them)

The first cross is essentially a mono-hybrid cross because of the complete linkage causes A and B to act as one gene.

The second cross makes more sense if you remember that A and B are always linked and on the same chromosome.  So Ab/aB could be written as [AB]b/a[AB] where [AB] is essentially one gene.  This cross is really a mono-hybrid with two homozygous parents for AB (e.g. AB x AB).

update 2.9.09

The second portion could also be interpreted as two sets of linkages: A linked b and B linked a.  If this is the case then you would end up with three possible phenotypes in the F1 progeny.

  1. AbBa
  2. AbAb
  3. BaBa
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7Feb/090

Chapter 5 Question 14(g): Recessive Epistasis

From a studious student:
I was hoping you could help me with the homework. I am working on problem 14 and I am confused on 2 parts, g and i.    

g: I recognize that there is recessive epistasis, but I am a little confused as to what that means.  The answer key says that two of the classes of the F2 have the same phenotype.  Which ones?  When I do the cross, I get the typical 4 phenotypes, 1: A-B-, 2: aaB-, 3: A-bb and 4: aabb.
1: Both genes show
2: Only a shows
3: Only b shows
3: Perhaps both a and b show?

Recessive epistasis means that the recessive homozygote (aa or bb) is 'dominant' to any combination of dominant genes.  Thus you should see the following genotype to phenotype pairs:

  1. A-B- = unique phenotype x
  2. --bb = unique phenotype y
  3. --aa = unique phenotype z
  4. aabb = one of the phenotypes from 2 or 3 (e.g. y or z)
* take a look at page 60 and figure 3.14 in the text if you're still confused.
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3Feb/090

Homework 1

I've completed grading.  I'll have it in class next week or if you're desperate you can come by my office and pick it up. 50/50 chance I'll be there though.

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2Feb/092

Questions concerning HW 3

Here's a link  to the homework assignment: BI 206 HW3 Use the comments to ask me questions.

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2Feb/090

RSS

If you want to follow the course blog posts via RSS (= really simple syndication) here's the link:

feed://www.ngcrawford.com/category/courses/feed/

For feed aggregators I like Netvibes, but iGoogle is also nice.  In addition to iGoogle, google also provides Google Reader a standalone RSS reader similar to Gmail.

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