Friday, May 23, 2014

STANDARDS

Okay, so here are where you can find my standards for this semester. I have not received my Unit 10 test results yet, therefore I do not know if there are standards that I could use from there, but here is what I was able to come up with.

SP3 - HERE

SP4 - I received a 1 on the Unit 8 test for this standard.

SP5 - (HERE) I received a 3 on this standard on the Unit 8 test, so I will show my understanding of my errors and explain the concepts.

#28 and #29 are both questions that have to do with Hardy-Weinberg and Mendelian Genetics.

28. In a population of squirrels. the allele that causes bushy tail (B) is dominant, while the allele that causes bald tail is recessive (b). If 64 of the squirrels have a bushy tail, what is the frequency of the dominant allele?

I said the frequency was 0.8, but I was completely wrong. The frequency is actually 0.4 and this can be proven using the Hardy-Weinberg concept. In Hardy-Weinberg problems, the equation  p² + 2pq + q² = 1. In this equation, p equals all of the alleles in individuals who are homozygous dominant (BB) and half of the alleles in people who are heterozygous (Bb) for this trait in a population, and q equals all of the alleles in individuals who are homozygous recessive (bb) and the other half of the alleles in people who are heterozygous (Bb). Because there are only two alleles in this question, the frequency of one plus the frequency of the other must equal 100%. In other words, p + q = 1. That means that p = 1-q. In this problem, we are looking for the dominant allele, which is p. So we know that 64 squirrels have bushy tails. And the frequency must equal 100%, so we subtract 64 from 100 to get the number of squirrels with bald tails, which ends up being 36. 36 is q squared, therefore we must square root 36 and we get 0.6, which is just q. Now we have q, so we can plug that into + q = 1 or p = 1-q, and we discover that p equals 0.4. Therefore, the frequency of the dominant allele is 0.4. 

29. A llamas coat color is controlled by a gene that exists in two allelic forms. If a homozygous yellow llama is crossed with a homozygous brown llama, the offspring have gray coats. If two gray-coated offspring were crossed, what percentage of their offspring would have brown coats?

Okay, for this one, I said that 0% would have brown coats, but I had originally chosen 25%. I guess I just second guessed myself. But anyway, the cross between the two gray offspring can be done on what is called a Punnett square. 

 Basically, this is what I first drew and then I second guessed myself and changed my answer. As you can see, 1/4 of the offspring would have brown coats, 25%.

This might also be able to go under SP9 (HERE).

SP6 - HERE

SP7 - HERE

SP8 - HERE

SP9 - (HERE) As I mentioned before, I believe I can use my explanation for standard 5 to get my   standard 9, as well

I also have HBs that I need standards for:

WebbHB5/2 - I can construct models that connect the movement of molecules across membranes with membrane structure and function.DIFFUSION (HERE)

I received a 4 for this standard on the Unit 5 test, so allow me to explain what I did wrong and show my understanding for this topic.

The questioned explained that hospital patients are often given intravenous gluids (IVs) to maintain proper levels of water and salts in the body. Great care is used in preparing these solutions. If a manufacturer accidentally prepared a batch of I V fluid that contained much more than the usual amount of salt, harm to the patient could result. The most likely effect on a patient if this incorrectly prepared IV fluid was used is that _________________.

I said that water would move into body cells and cause them to burst, but the actual answer is that water would move out of the body cells and cause them to dehydrate. Obviously, this is the correct answer because obviously salt can cause dehydration. I obviously did not read the question correctly. When the body consumes to much salt, excess salt absorbs the body's water content, causing dehydration. 

WebbHB7/1 - I can use a model to illustrate that cellular respiration is a chemical process whereby the bonds of food molecules and oxygen molecules are broken and the bonds in new compounds are formed resulting in a net transfer of energy. CELLULAR RESPIRATION (HERE)

I received a 3 for this standard on the Unit 7 test. Please allow me to explain what I did wrong and show my understanding for this topic.

Okay, so basically everything I got incorrect on this test involves cell respiration. Therefore, I am just going to explain the whole process of cell respiration.

The first stage is glycolysis, which occurs with oxygen. Basically, a six carbon sugar, glucose, is split into two molecules of a three carbon sugar. During this process, 2 molecules of ATP, 2 molecules of pyruvic acid and 2 NADH are produced. When there is no oxygen, glycolysis allows cells to make small amounts of ATP. This process is called fermentation. The second stage is the Krebs Cycle, which occurs only when there is oxygen; however, it does not use oxygen directly. During this stage, several compounds with the capability of storing electrons are produced, as are 2 ATP molecules. These compounds, known as NAD and FAD, are reduced in the process. These reduced forms carry the electrons to the next stage, which is the electron transport. This stage does require the use of oxygen directly. The electron transport "chain" is a chain of electron carriers in the membrane of the mitochondria in eukaryotic cells. Throughout this stage, the electrons are passed to oxygen. In the process, a gradient is formed, ultimately producing ATP. Yay!

I was able to illustrate cellular respiration through the Pineapple and Jello lab, which is on my website. Click on the "HERE" above. 

WebbHB8/2I can construct an explanation based on evidence for how the structure of DNA determines the structure of proteins which carry out the essential functions of life through systems of specialized cells.HORMONES (HERE)

I received a 2 on the Unit 8 test. Plus, I have my podcast on thyroid hormones under this standard on mys website.

WebbHB9/1I can evaluate the evidence for the role of group and individual behavior on individual and species’ chances to survive and reproduce. ANIMAL BEHAVIOR (HERE)

My pill bug lab is under this standard on my website. 


I have not received the results for my Unit 10 test; therefore, I do not know if I need any standards for unit 10. When I find out, I will add on to this post. 

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