August 31st - In class and homework

I collected your evaluations of the 3 web sites we did in class Friday. In addition, we briefly looked at the levels of organizations in life shown in our textbook pages 4 & 5; as well as the chart on page 27 (Eleven themes that unify Biology). At this point, we talked about the expectations about the basics of chemistry (background necessary for this class), and I handed out Learning objectives, as well as a number of review questions to test your mastery of these basics. The answers will be shown below. The homework is to complete these questions (if necessary), and to brush up on whatever aspects of chemistry that you are not comfortable with.

Answers to Chemistry questions
Part A -1-b, 2-e, 3-d, 4-d, 5-b, 6-b, 7-a, 8-d, 9-a, 10-b, 11-c
Part B - 1-B, 2-A, 3-B, 4-C, 5-A, 6-A, 7-D, 8-C, 9-D, 10-B
Part C - 1-B, 2-B, 3-A, 4-C, 5-B, 6-E, 7-C, 8-B, 9-C, 10-D
Part D - 1-D, 2-A, 3-B, 4-B, 5-C, 6-b, 7-B, 8-A, 9-B, 10-B
Part E - 1-D, 2-D, 3-E, 4-B, 5-A, 6-B, 7-C, 8-A, 9-E, 10-E, 11-E, 12 - B, 13-C, 14- E, 15-A, 16-B, 17-E, 18-A

August 28th Class activities and Homework

Today in class we looked over a number of Biology Animations from the following web site: http://nhscience.lonestar.edu/biol/bio1int.htm . Our in class assignment was to briefly watch a number of sites found here, and then to evaluate 3 of them. These sites have to do with material we will cover for at least the 1st 9 weeks of class Our homework is to finish this assignment (if necessary).

August 27th - Activities in Class and Homework

In class today, we completed a brief investigation brainstorming about the relationship between catalase (found in chicken liver) and hydrogen peroxide. We came up with a hypothesis addressing conditions necessary for optimal catalase activity. We finished the class with a 10 question pretest on Chapter 1 material, and briefly going over the answers. Our homework is to read sections 1.4 (pgs. 15 - 19) and 1.6 (pgs. 26 & 27).

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Now, please begin reading pages 2 - 14

Most recent test (Evolution Unit)

A couple of comments concerning our last test. The average for both classes (together) was almost 81% - nice job! There was only one question I reevaluated a bit - question #2 dealing with the attempt to "evolve" winter wheat. The correct answer should have been "C" (Lamarck) as the point was the scientist was trying to cause changes in the wheat during its life - and these changes would be passed to the next generation (inheritance of acquired characteristics). However, I also allowed "D" (Darwin) NOT because his theory included this idea, but because the question did not take into account whether or not some of the variations of wheat did not survive (perhaps the scientist only selected those that survived and reproduced). Therefore I felt C or D would be OK. The only other question missed by more than 1/2 the students was number 40 - I think the diagram threw a lot of you - but I was hopeful you would understand that discrete ancestors can NOT merge into one taxon.

Now, for the Hardy-Weinberg problem(s)

Version "A"
480 out of 4000 show the recessive phenotype (aa): aa= 480/4000 = .12 or 12% therefore a= the square root of (.12) =.35 or 35%
A= 1-(.35) = .65 or 65% AA=(.65) squared =.43 or 43% Aa = 2(.65)(.35) =.455 or 46%

Version "B"
1280 out of 5000 show the recessive phenotype (aa): aa= 1280/5000 = .256 or 26% therefore a= the square root of (.256) =.506 or 51%
A= 1-(.51) = .49 or 49% AA=(.49) squared =.24 or 24% Aa = 2(.49)(.51) =.450 or 50%

Answers to Chapter 24 & 25 Sample Test Questions

Chapter 24

3.d 4.e 5.c 6.a 8.c 9.b 11.b 12.a 14.c 16.e


Chapter 25
6.b 7.d 8.e 9.e T/F- 1.T 2.F 3.T 5.T

Answers to chapter 23 "Structure your Knowledge" and Chapters 22 and 23 "Test your Knowledge"

Chapter 23 "Structure Your Knowledge"

1. a. The Hardy-Weinburg theorem states that allele frequencies within a population will remain constant from one generation to the next as log as only Mendelian segregation and sexual recombination of alleles are involved. This state requires five conditions: A large population, mating is random, mutation and migration are negligible and no selection pressure operates.

b. The main equation is P squared + 2 pq + Q squared = 1. In the Hardy-Weinburg equation, p and q refer to the frequencies of 2 alleles in the gene pool. The frequency of homozygous offspring are p x p or p squared, and q x q or q squared. Heterozygous individuals can be formed in two ways, depending on whether the sperm of the ovum carries the p or q allele, so their frequency is equal to 2pq. Also, since there are only two forms of the gene (in this equation), p + q= 1.

2. Genetic variation is retained within a population due to the presence of diploidy (2 copies of an allele per gene) and balancing selection. Diploidy masks recessive alleles from selection when they occur in the heterozygote. Thus, less adaptive or even harmful alleles are maintained in the gene pool, and are available should selection pressures change. Balanced selection maintains several alleles at a gene locus in a population and leads to balanced polymorphism. In situations where there is heterozygote advantage, the two alleles will be retained in stable frequencies within the gene pool. Frequency-dependent selection, in which varieties present in larger quantities are selected against by predators or other factors, is another cause of balanced polymorphism.


Answers to Chapter 22 "Test Your Knowledge:

1.b 2.c 3.e 4.a 5.c 6.a 7.e 8. d
9.d 10.d 11.c 12.e 13.b 14.(d) 15.c 16.c


Answers to Chapter 23 "Test Your Knowledge"
1.e 2.a 3.c 4.c 5.d 6.d 7.e 8.d
9.b 10.e 11.e 12.c 13.d 14.a 15.b 16.b
17.c 18.b 19.b 20.(c) 21.a 22 e 23 d 24.(c)
25.e