Tuesday, March 20, 2012

Procedure for submitting essays for regrading

When grading your essays, the TAs and I strive to make the score that you receive on your essay accurately indicate how well you have answered the question. At times it is possible that the score that you received on your essay does not correlate with how well you answer the question so this essay resubmission policy is designed to correct those rare mistakes.

Interpreting your essay scores.

Scores of 18 - 20 indicate that your grader thought you gave an excellent answer to the question. Score of 20/20 are reserved for answers that are exceptionally excellent.

Scores of 16 - 17 indicate that your grader thought you gave a "good" answer to the question. "Good" answers differ from "excellent" answers because they were not as well organized, lacked relevant details, or included some confusing or incorrect thoughts.

Score of 14 - 15 indicate that your grader thought you gave an "adequate" answer to the question. Students receiving these scores knew much of the relevant information, but were not able organize it clearly, left out critical information, or contained incorrect information

Scores of 12 -13 indicate that your grader thought that you gave a "poor" answer that did not answer answer the question well.

Scores of 11 and below indicate that your grader thought that you failed to answer the question correctly.

Guidlines for Submitting Essays

1. Answers written in pencil can not be submitted for regrading.

2. Do not turn in essays until you have carefully examined the answer keys. Take a look at the answer key and carefully compare your answer to the key. Ask the following questions. Did your answer actually answer the question that was asked (or did you simply tell me all of the information that you knew about the topic)? Did your answer include all of the details shown on the answer key? Was your answer as well organized as the answer key? Was your answer written in clear and concise English?

3. I only want to look at tests where you feel there is a severe difference between the score you earned and how well you answered the question (that means 3 or more points). I do NOT want to see any 19/20s!

4. To resubmit an essay please type out a brief paragraph describing why you think that your answer deserved more points that it was awarded. Staple the essay to this sheet (make sure you put your name on it) and leave the sheet in a box that is located outside of my office in McClellan Hall (room 215) by 5:00 PM on Monday March 26.

Answer Key- Midterm #2 Essays

(20 points) you may use both sides of this page to answer this question.

In a temperate rainforest in Washington State, a moss is found growing on an oak tree (Angiosperm). Discuss the characteristics that make the oak evolutionarily advanced relative to the moss. Be sure to discuss any selective advantages associated with the advanced traits.

Mosses, the first terrestrial plants, are primitive compared to an oak tree. Important evolutionary advances that have occurred between bryophytes and angiosperms include the move from haploid to diploid dominance, the development of vascular tissues, heterospory, the production of seeds, and the production of flowers.

The diploid dominant stage of the oak is advantageous relative to the haploid dominant stage of the moss because deleterious mutations can be masked in the heterozygote in the diploid stage.

The production of vascular tissues, xylem and phloem, allow plants to move water, nutrients, and sugar around the plant. Coupled with the development of true leaves, stems, and roots, the production of a vascular system allows plants to no longer be limited to inhabiting moist environments and allows them to increase in size. Thus, the plants possessing vascular tissues can exploit more ecological niches and achieve greater reproductive success.

The production of heterospores allows plants to produce microgametophyte and megagametophyte generations. In angiosperms, the microgametophyte develops into the pollen grain and the female gametophyte is an 8 celled structure located in the ovule. Because the sperm are able to move to the ovule in the pollen tube via cellular processes, heterosporous plants no longer require swimming sperm so they are not limited to living in environments that are wet during the reproductive season. Thus, heterosporous plants can exploit more ecological niches.

Reproducing by seeds is a large advantage compared to reproducing by spores. Inside an angiosperm seed, the embryonic plant is protected from desiccation by the seed coat and following germination, the developing embryo is provided resources held by the endosperm cells. Thus, an individual seed has a much higher chance of survival than an individual spore.

Reproducing via flowers is advantageous over the reproductive style of the moss. First, flowers can aid in the dispersal of pollen from one plant to another. Second, in mosses, the zygote is formed in the archegonia. In flowering plants, the zygote is produced inside of the ovule which is held inside of the ovary of the flower. The ovary provides protection to the developing zygote. Ultimately, the ovary develops into the fruit which can help to disperse the seeds either by animals or the wind.

(20 points)
Plants convert electromagnetic energy in light to potential energy stored in glucose by the process of phtotosyntheis. Discuss, in detail, the energy conversions that occur during this process.

The process of photosynthesis converts electromagnetic energy in light into potential energy stored in glucose. Important energy conversions occur in the following processes, (1) in a photosystem, (2) in the light dependent reactions, and (3) in the light independent reactions.

In a photosystem light energy is converted into potential energy held in an excited electron. Photosystems are clusters of antennae pigments, a reaction center, and a primary electron acceptor that are located in thylakoid membranes. When antennae pigments absorb a photon of light energy and electron is excited to a higher energy level. When the electron returns to resting stage, the energy released is used to excite an electron in an adjacent pigment, a process known as resonance. Resonance funnels energy to the reaction center chlorophyll a where an electron is excited to a higher energy level and then that electron, and its potential energy, is passed to the primary electron acceptor.

In the light dependent reactions, potential energy in an electron excited in a photosystem is converted to chemical energy in ATP and NADPH by the process of electron flow. Cyclic electron flow converts potential energy in an excited electron to chemical energy in ATP. The excited electron from PS II moves down an electron transport chain. Energy released each time the electron moves from one molecule to the next is used to actively transport hydrogen ions from the stroma inside the thylakoid space. The resulting hydrogen ion concentration gradient powers the process of chemiosmosis that results in the production of ATP. In non-cyclic electron flow, ATP is produced by a similar mechanism as the excited electron moves from PSII to PSI. When the electron is re-excited in PSI it undergoes a different pattern of electron flow that results in the production of NADPH.

The light independent reactions convert the potential energy stored in the short term energy storage molecules ATP and NADPH into potential energy in glucose that can be stored for a longer period of time. The Calvin Cycle, which takes place in the stroma, uses the ATP and NADPH produced in the light dependent reactions to covert carbon dioxide into glucose. ATP and NADPH are required to convert PGA into glyceraldehyde phosphate which can either be used to created glucose or can be converted into RuBP, a reaction that requires the addition of energy from ATP.

Wednesday, March 7, 2012

BAC Results for Second Midterm

The BAC met this afternoon to discuss questions from the second midterm. The BAC representatives came to talk to me about two questions.

A botanist discovers a new species of land plants with a dominant sporophyte, chlorophylls a and b, and cell walls made of cellulose. Which of the following structures, if present, would be least useful in assigning the plant to the correct group?
(a) endosperm
(b) seeds
(c) sperm that lack flagella
(d) flowers
(e) spores

The correct answer to this question was listed as (e). A land plant with a sporophyte dominant phenotype which means that it must be a fern, gymnosperm, or angiosperm. Endosperm would eliminate F and G, seeds would eliminate G and A, sperm that lacks flagella would eliminate F, flowers would eliminate A, whereas all three groups produce spores. Thus, (e) remains the sole correct answer.

A flask containing photosynthetic green algae and a control flask containing water with no algae are both placed under a bank of lights, which are set to cycle between 12 hours of light and 12 hours of dark. The dissolved oxygen concentrations in both flasks are monitored. Predict what the relative dissolved oxygen concentrations will be in the flask with algae compared to the control flask.
(a) the dissolved oxygen in the flask with algae will always be higher
(b) the dissolved oxygen in the flask with algae will always be lower
(c) the dissolved oxygen in the flask with algae will be higher in the light, but the same in the dark
(d) the dissolved oxygen in the flask with algae will be higher in the light, but lower in the dark
(e) the dissolved oxygen in the flask with algae will not be different from the control

This is a question from the textbook's test bank, so I don't know exactly what the question writer was thinking. When I read the question I thought "oxygen added by photosynthesis during the light and oxygen removed by cellular respiration both during the light and the dark. Thus, I agreed with the textbook's answer of (d).

However, several of you were thinking more deeply about the question than the question write and I were so you realized that it was not possible to determine the relative rate of oxygen production and use given the information provided. Thus, it was possible that answers (a), (c), or (d) were correct. Thus, this question will be omitted from the exam and everyone will receive credit for this question.

Thanks again to the members of the BAC and the BAC reps for the time and effort that they have put into the BAC so far this semester.

MC Answer Key for Second Midterm

Here are the answers for Midterm #2. If you have concerns about any of the problems then please write them on a sheet of paper and leave them under the "BAC" sign on the ledge across from Dr. Dini's office before 3:30 this afternoon. The BAC will meet today at four to review any questions related to this midterm and I will report the results of the BAC.

Answers Form 1

1. b
2. e
3. b
4. b
5. d
6. b
7. c
8. a
9. c
10. e
11. a
12. b
13. c
14. c
15. c
16. d
17. e
18. c
19. a
20. c
21. e
22. d
23. b
24. b
25. a
26. b
27. d
28. d
29. d
30. e

Answers Form 2

1. e
2. a
3. b
4. c
5. b
6. c
7. c
8. b
9. e
10. c
11. b
12. b
13. d
14. c
15. a
16. e
17. d
18. b
19. b
20. a
21. d
22. e
23. c
24. a
25. c
26. b
27. d
28. d
29. d
30. e

Friday, March 2, 2012

Review Sessions for Midterm #2

The SI Leaders will hold their Marathon Review Sessions for the second midterm as follows.

Suzanne- Sunday 7 - 10 PM. Holden Hall room 104
Jeffrey- Monday 7 - 10 PM. Holden Hall room 104

I will hold my final Group Office Hours (tear) on Monday from 6 - ?? in Biology Rm.21

If you have questions over the weekend feel free to send me an email.

Wednesday, February 29, 2012

Answer Key- Retest Essays

(20 points) Answer questions a & b on in the space provided on the front of this page. Answer questions c on the back of this page

a) Draw the graph that shows how the population growth rate varies over time in logistic growth when the initial population size is much smaller than the carrying capacity. 5 points

b) describe, in words, all of the information held in the graph you drew in part a. 5 points

Over time the population growth rate increases until the population growth rate reaches its maximum then the population growth rate decreases until the population growth rate equals zero where it remains.

c) Diagram the demographic transition and explain why patterns of population growth are so different between developed and developing countries. 10 points

The patterns of human population growth are quite different. Population sizes in develop nations have stabilized so there is no further growth. However, population sizes are still increasing rapidly in most developing nations. If human populations were regulated by the same factors that regulate other populations, then we might expect population growth rates to be lower in developing than developed countries because medical and agricultural facilities are not as well developed in developing countries. What explains the difference in population growth rates between the two types of nations? The demographic transition, a predictable pattern of changes in per capita birth and death rates that occurs as nations become more developed might help to explain this difference.

In pre-industrial societies, such as the Native Americans living on the Llano Estacado 2000 years ago, the difficulty of obtaining food and the lack of medical care made the per capita death rates quite high. Not surprisingly, people in pre-industrial societies responded to these high death rates by having lots of children (high per capita birth rates). Because the high per capita birth rates equaled the high per capita death rates, there was very little change in population sizes in pre-industrial societies.

As societies begin the transition to becoming fully developed, the initial improvement in food production, public health, and medical facilities causes a decline in the per capita death rates. As societies develop, death rates continue to decline until they stabilize at a minimal level. However, people continue to have high birth rates. Thus, over time in the transitional stage the difference between b and d increases which results in per capita growth rates increasing over time. As population sizes grow and the per capita growth rate increases, population growth rates increase dramatically.

Eventually during the transitional period the per capita birth rates begin to decline until they they reach a minimum value at the end of the transitional period. The reasons for the decline in birth rates is not well understood, but demographers have observed that birth rates decline with increasing education, economic opportunity, and access to family planning. In fully industrial nations, the low per capita birth rates equal the low per capita death rates so the population growth rate in developed countries is equal to zero.

Thus, the patterns of population growth are different between developed and developing countries arises because they are at different points along the demographic transition.

(20 points) You may use both sides of this sheet.

Why, at first glance, is the production of altruistic acts difficult to understand? Discuss three possible hypotheses to explain the production of altruistic traits.

Because natural selection causes genes that code for traits that make organisms better at surviving and reproducing in an environment to become more common (selfish traits), at first glance it is difficult to understand the presence of altruistic traits. Altruistic traits decrease the fitness of the actor, so it is difficult to see how those traits can become more common in a population over time. Although altruistic acts decrease the fitness of the actor, they increase the fitness of recipients of the act. Here are three hypotheses that might explain why we see altruistic acts.

Typically, organisms pass on their genes by reproducing. However, close relatives share some genes in common because of descent from a common ancestor. The kin selection theory suggests that organisms might be able to pass on their genes by helping their relatives to reproduce more than they would have wit out their help. Whether or not an organisms should be altruistic depends on the number of relatives helped, the relationship to the helped individuals, the cost of the act, and the benefit of the act. W. D. Hamilton developed Hamilton’s rule and predicted that animals should be altruistic whenever (add equation).

Sometime altruistic acts occur between non-related individuals so kin selection could not produce these behaviors. Reciprocal altruism, the notion that “I will be altruistic towards you now if you are altruistic towards me later on”, can help us to understand altruism among non-relatives. An individual might be willing to pay a cost now in hope of receiving a larger benefit sometime in the future. Reciprocal altruism requires long-term associations among individuals, the ability to recognize individuals, and the ability to remember who owes you and who doesn’t. Thus, reciprocal altruism should be limited to long lived, “smart” animals. Not surprisingly, reciprocal altruism is common in humans.

Some human altruistic acts actually results in the death of the altruistic (e.g., a soldier falling the grenade or a police officer dying to prevent a robbery). Because these acts are not limited to relatives and the act cannot be reciprocated (you can’t receive a benefit if you are dead) the neither kin selection or reciprocal altruism can explain these behaviors. Cultural selection has been proposed as a mechanism to explain these types of altruistic behaviors.

Gene have the ability to replicate themselves; is there anything else that is capable of self -replication. Ideas (memes) can replicate themselves and they differ in their survival and transmission ability. Thus, the differential survival and transmission of ideas can form an analogous process to natural selection. Cultural selection may explain self-sacrificing behavior in humans as well as other behaviors that apparently do not maximize the transmission of genes.

Sunday, February 26, 2012

Mastering Biology Homework Assignments

The final two homework assignments for my portion of the course are now online.

Due Dates

Homework #6. Cellular Respiration- Friday March 2nd, 2012 11:59 PM

Homework #7. Plant Diversity- Monday March 5th, 2012 11:59 PM