Nick Crawford » Courses

Vertebrate Zoology: Bi302

Nick — Fri, 22 Jan 2010 14:21:11 +0000

Welcome to Vertebrate Zoology lab (Bi302 Lab). I'll be using this portion of my website to post notes/slides and to answer your questions. Please make use of the comments. More to come.

Thanks for a great term

Nick — Thu, 30 Apr 2009 21:09:22 +0000

Just wanted to thank my students, all ~44 of you, for putting up with my stammering explanations of genetics. I sure learned a ton and I hope you did too. Good luck with your finals.

Shmeat

Nick — Tue, 31 Mar 2009 10:58:03 +0000

Hey, one of my students (erica) suggested this video.

Quiz 2

Nick — Thu, 12 Mar 2009 16:17:21 +0000

No quiz for my Monday classes until March 23rd. My Tuesday class should be prepared for a quiz on chapters 7 and 8 on Tuesday the 17th.

Quiz 1

Nick — Sat, 21 Feb 2009 17:34:57 +0000

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

Don’t work too hard this weekend

Nick — Sun, 15 Feb 2009 02:31:56 +0000

Enjoy this classic flatworm sex clip.

Tuesday is a Monday Schedule Next Week

Nick — Thu, 12 Feb 2009 12:47:54 +0000

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.

Chapter 5 Question 14(i): Complete linkage

Nick — Sat, 07 Feb 2009 13:53:43 +0000

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.

AbBa
AbAb
BaBa

Chapter 5 Question 14(g): Recessive Epistasis

Nick — Sat, 07 Feb 2009 13:28:10 +0000

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:

A-B- = unique phenotype x
--bb = unique phenotype y
--aa = unique phenotype z
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.

Homework 1

Nick — Tue, 03 Feb 2009 21:39:04 +0000

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.