Showing posts with label Class discussions. Show all posts
Showing posts with label Class discussions. Show all posts

Wednesday, November 2, 2011

Cellular respiration review

Follow this link to watch my review of cellular respiration. It may take a while to load because it is fairly long, but give it some time.

Or, you can watch it right here.

Tuesday, November 1, 2011

Cellular Respiration Overview

As I was looking over the work from today when I was out I saw this on the back sheet of one of the packets.


This outline is great, and would be a great thing for you all to know. I will elaborate on this a bit in class tomorrow, but this is a great basic overview.




Saturday, October 15, 2011

Chapter 36 Packet Answers


Here are the answers for the Chapter 36 review packet. Sorry if it is hard to read, I am trying out my iPad camera as a document scanner. Let me know in the comments if there are any answers you are unsure of.

Friday, October 14, 2011

Kidney animations

Ok, here is an attempt to get the kidney animations on here. Looks like the Flash player didn't embed correctly, at least on my MacBook. But you can still download the files and view them.



Friday, September 23, 2011

Chapter 5 Review Packet Answers

Once again, any questions regarding the quiz on Monday, post them in the comments.


Interactive Questions


5.1  a. hydroxyl
b. carbonyl
c. aldose
d. ketose
e. ring


5.2


















5.3 a. monosaccharide
b. CH2O
c. fuel
d. carbon source
e. glycosidic linkage
f. disaccharides
g. polysaccharides
h. glycogen
i. animals
j. starch
k. cellulose
l. chiton


5.4




















Monday, September 19, 2011

Chapter 3 Review Answers

Here are the answers for the chapter 3 review packet that I promised. If you have questions, post them in the comments, and I will answer them as quickly as I can.

Interactive Questions
3.1

3.2
a. water molecules
b. absorbed
c. released
d. specific heat
e. heat of vaporization
f. evaporative cooling
g. heat
h. snow
i. ice

3.3
a. hydrophobic - nonpolar
b. hydrophilic - polar
c. hydrophilic - ionic
d. hydrophibic - nonpolar


3.4
a. 45 g
0.5 mol/1 L x 90 g/1 mol = 45 g/L

b. 228 g
2 mol/1 L x 57 g/1 mol x 2 L = 228 g

3.5

10e-3
              10e-6     8   basic
10e-7                   7   neutral
10e-1    10e-13         acidic


3.6
Donor - H2CO3
Acceptor - HCO3-


a. Left
b. right


Wednesday, March 2, 2011

Monophyletic, Paraphyletic, Polyphyletic

Hope this post clears up some of the confusion about these terms.



Here is a link to the amniote family tree, along with discussion about the different proposed groupings.

Thursday, January 13, 2011

Reproduction and Development Powerpoints

Here are the big PowerPoints for the Animal Reproduction (Chapter 46) and Development (Chapter 47) lectures. For those of you that need to take the test, remember that the end of the Marking Period is arriving fast. Tests MUST be made up by the 27th, or they will go into the gradebook as a zero.



Monday, January 3, 2011

Course Evaluation Results

I thought this was kinda cool so I wanted to share it. This is a word cloud showing the most commonly appearing words in your responses to the course evaluation survey.

Question 1: Three words to describe Dr. Himmelheber as a teacher are:




created at TagCrowd.com












Question 2: If I were to change one thing about the way Dr. H teaches it would be:




created at TagCrowd.com




Looking at the results, obviously the PowerPoints and notes are an issue, along with the exam, but at least I am interesting. It is a little hard to tell since it is out of context, but I will attempt to address these issues.

Tuesday, December 14, 2010

Immune System via Zelda and FFXII (or XIII...I really don't know)

This is pretty much self-explanatory. And awesome.



As soon as I get some more time to poke around the Google results, I will post a few more...science-y ones. Till then, enjoy!

Friday, December 3, 2010

Brain Myths and Other Stuff

Some quick Google searching turned up a bunch of sites with the title "Top X Brain Myths Debunked", with X being anywhere between 10 - 12. Here are links to a couple that were brought up in class.

New wrinkles as you learn something

You only use 10% of your brain

If there are more that we talked about in class that you remember, you can link to them in the comments, or if you can't find anything, let me know and I will look them up.

Here is a link to a game where you can replicate the studies done in split brain patients. Remember, these are patients whose corpus callosum, the thick band of nerves that connects the left and right hemispheres of the cerebrum, has been surgically split to treat epilepsy.

Thursday, October 21, 2010

Entropy and Time

Just saw this, and thought I would pass it along. Pretty good review of the second law of thermodynamics.

Favorite quote:
After all, you are a very highly ordered person, I'm assuming. If you are somehow a disordered homogeneous cloud of gas, then please accept my apologies.


I will try to get some more posts up here, maybe not the once-a-week pace of the summer, but something more regular. And maybe there will be some extra credit possible for "comment of the week" or something.

Tuesday, July 13, 2010

Summer Assignment: 7.13.10

Hello and welcome to the official start of the AP Biology Summer Assignment. I know that I told most of you that the first post would be up last week, but a number of factors made it impossible for me to get anything up here. This will be the first post that you can earn points for, and from now until the end of the summer, they should be up once a week. Your comments will be moderated, which means that I will read them first before they show up on the blog, so don't worry if your comment does not show up immediately, it just means that I am not at the computer and have not read through all the comments yet.

Any questions or concerns can be emailed to me, or posted in the comments. On to the links for the week...

The proton shrinks in size

Not exactly biology-related, and it is only 0.00000000000003 millimeters, but that is a 4% difference, and any change in one of the three basic particles of an atom should be pretty big news.

Pohl and his team have a come up with a smaller number by using a cousin of the electron, known as the muon. Muons are about 200 times heavier than electrons, making them more sensitive to the proton's size. To measure the proton radius using the muon, Pohl and his colleagues fired muons from a particle accelerator at a cloud of hydrogen. Hydrogen nuclei each consist of a single proton, orbited by an electron. Sometimes a muon replaces an electron and orbits around a proton. Using lasers, the team measured relevant muonic energy levels with extremely high accuracy and found that the proton was around 4% smaller than previously thought.
Could be something to keep track of...though one researcher quoted in the story does seem to think that this new result could also be an error.


"Sinister Motion" may influence soccer referees

Or, just an excuse to post something about the World Cup. Once again, the third-place match ended up being much more entertaining than the final. The Dutch obviously knew they had to play physical to counter Spain's offensive capabilities. They seemed willing to take as many yellow cards as they needed to get the game to penalty kicks..,not sure it was the best strategy, and I'm still not sure why this was not given a straight red.



Why you should never arm wrestle a saber-tooth tiger.

To be honest, I only clicked through to this story because of the headline, but it is fairly interesting. Because of their oval-shaped cross section, saber-tooth cats had relatively weaker canine teeth than do modern cats, who have conical-shaped teeth. This was made up for by their stronger forelimbs.

Despite their vulnerable canines, prominent muscle attachment scars on sabertooth limb bones suggest the cat was powerfully built. Saber-toothed cats may have used their muscular arms to immobilize prey and protect their teeth from fracture, she explained.

Monday, December 7, 2009

Genetics Practice Problems Solutions

Here are the answers and explanations for the genetics review problems from class today. There will be five problems on the test.

1. Colorblindness is a sex-linked, recessive trait. Remember that sex-linked traits are on the X chromosome, which means that men only have one copy of the gene. In this example, we are told that the man has colored vision, so his genotype must be XCY. The woman also has colored vision, so she could be XCXC or XCXc. Since one of her sons is colorblind, she must be XCXc, since sons inherit their X chromosome from their mothers.

2. This is a relatively simple dihybrid cross. If we use B for brown eyes and b for blue eyes; and H for brown hair and h for blonde hair, the genotype of the man is BbHH, and the woman is bbhh. Doing the cross, their children have a 50% chance of being BbHh and a 50% chance of being bbHh. Answering the question posed on the sheet, there is 0 chance their children will have blue eyes and blonde hair.

3. This is an example of a dihybrid cross where one of the genes displays incomplete dominance - red and white flowers give pink. We are crossing two F1 plants, so the phenotype of both parents is TtRr. For this type of dihybrid cross with complete dominance, we would normally expect the 9:3:3:1 ratio. However, with incomplete dominance there is a new phenotypic class, since the heterozygous individuals are distinct from the homozygous dominant individuals. The expected phenotypic ratios are then 3 tall, red-flowered; 6 tall, pink-flowered; 3 tall, white-flowered; 1 dwarf, red-flowered; 2 dwarf, pink-flowered; and 1 dwarf, white-flowered.

4. A dihybrid cross with a lethal allele. If an individual is homozygous recessive for the l allele, they will not survive, and are not counted in the phenotypic ratios for the answer. The parental genotypes are LlBb and Llbb. After throwing out the individuals with the lethal gene combination, the phenotypic ratio in the offspring is 1 normal-legged, brown; 1 normal-legged,white; 2 deformed-legged, brown; and 2 deformed-legged, white.

5. Gene linkage. You absolutely must know how to analyze these types of data, and tell the difference between parental and recombinant phenotypes. The data that are presented are from a testcross on the F1 generation. The genotypes for this cross are CcShsh crossed with ccshsh (remember, a test cross is always performed with a homozygous recessive individual). The phenotypes of the parents are colored, full seeds and colorless, shrunken seeds.

If we assume that these genes are going to follow the Mendelian laws of inheritance, we predict that the offspring would have equal numbers of the four possible phenotypic classes: colored, full; colored, shrunken; colorless, full; and colorless, shrunken. HOWEVER, that is NOT what the data show. Two of the phenotypic classes, colored, full seeds and colorless, shrunken seeds are MUCH more common than the other two. These two common phenotypes are called the PARENTAL phenotypes, since they resemble the parents of the cross. The other two phenotypes, which are much less common, are called RECOMBINANT phenotypes, since these gene combinations do not exist in the parental generation.

To calculate the map distance, we need to calculate the recombination frequency, which is simply the percentage of offspring that show recombinant phenotypes. For this problem, the answer is (515 + 489) / 8368 = 12%. This means that the genes are 12 map units apart.

6. This is an example of epistasis - one gene is influencing the expression of a second gene at a second location. In this case, dogs that are homozygous recessive for the e gene will be yellow, regardless of what alleles are at the location that determines pigment color (B for black and b for chocolate). The phenotypes of the parents are BbEe. Doing the cross results in a 9 black to 3 chocolate to 4 yellow labs.

7.

...........7..........3...................15..................5..........
-----/-----------/----------/---------------------------/-------------/---
.....b..........d...........a...........................c.............e

8. This is a simple incomplete dominance cross. The heterozygous individuals have green flowers. A cross of two green flowers gives results of 1 blue, 2 green and 1 yellow.

9. The genotype of the woman must be ii, since that is the only possibility for type O blood. Her baby, with type A blood, must have at least one i allele from the mother. Therefore, the babies genotype must be IAi. The IA allele must come from the father. The only man with an IA allele to contribute is man #2.

10. Pedigree A is an autosomal recessive trait. Pedigree B is a sex-linked trait because many more males exhibit the trait than females (7 vs 2). Pedigree C is a dominant autosomal trait. To differentiate recessive and dominant traits, there are a few things to look for. First is that recessive traits tend to skip generations. Look at generations I and III in pedigree A. For a dominant trait, at least one parent must exhibit the trait in order for it to be passed on to the offspring.

11. I treated this example like a regular dihybrid cross, but in this case the dominant allele will change based on the sex of the individual. The two parental genotypes are BbXX and BbXY. For female offspring, there will be 3 with hair for every 1 bald; and for males there will be 1 with hair for every 3 bald.

Wednesday, September 2, 2009

Summer Assignment: 9.2

Since this is the last post for the summer assignment, I would like to take the time to thank everyone for doing a great job. The comments were very thoughtful, and there seems to be some very interesting differences of opinion among you all; hopefully this will lead to some good discussions/debates in class.

There is no reading assignment this week. I want to know some of the topics you are interested in learning more about this year in class. It can be something that was covered over the summer, such as evolution, or anything else that you find interesting. We will do our best to cover as many of the topics as we can, but it may not happen till after the AP test.

Sunday, February 15, 2009

Chapter 35 and 36 review packet answers

Answers for chapter 35 and 36 review packets

Chapter 35

MATCHING

1...H .6...A
2...F .7...C
3...B .8...D
4...E .9...J
5...I .10...G

MULTIPLE CHOICE
1...A .9...E
2...E .10...D
3...B .11...E
4...C .12...B
5...B .13...C
6...A .14...C
7...E .15...E
8...C .16...D


CHAPTER 36

1...B .11...A
2...E .12...C
3...D .13...E
4...A .14...B
5...C .15...C
6...B .16...D
7...D .17...D
8...D .18...B
9...A .19...D
10...C .20...C

Thursday, January 15, 2009

Darwinism Test and PowerPoint

I was a bit disappointed by the level of surprise in class today about the test on Friday. I hand out the monthly syllabus for a reason: For you to know what is happening in class. I expect you to use the schedule, and not rely on me to tell you when things are happening. There should be no surprises when it comes to tests, quizzes, labs, or any other class activity.

I want you to use the syllabus for a few reasons. One, it makes it easier for me to plan. It should also make it easier for you to study, since you have a few week's notice of tests and quizzes. If you are absent from class, you can look at the schedule and see what we are going over that day. One of the most important reasons I do this is that this is what happens in a college course. You will get a syllabus at the beginning of the semester that lists all the tests. You will be expected to know when they are. The professor/instructor may not remind you that they are coming.

Please use the syllabus. There is one posted in the classroom, and I can get you another copy if you lost yours. There are two more quizzes on the syllabus for this month. Make sure you know when they are. I may not remind you in class about them.

Anyway, here is the complete PowerPoint for Chapter 22. I will be moderating comments till 10.30 tonight.