Rubric 2

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Rubric 2

Use of Accurate Scientific Language

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Reflection 1.

Chem504. Kinetics and regulations.

Enduring Understanding:

The study of the rates and mechanisms of chemical reactions and the factors on which they depend.

To understand how most biochemical reactions are catalyzed by enzymes? Enzymes are protein catalysts. Reactivity and selectivity controlled by enzyme shape the functional groups binding and specificity.

The Michaelis-Menten model describes the kinetics of many enzymes.

K_eq(With a capital K) is the equilibrium constant which is essential for understanding many natural processes we covered in biochemistry class such as oxygen transport by hemoglobin in blood and acid base homeostasis in our body.

k (small k) is the reaction rate coefficient with units of concentration/time if and only if the reaction is 0- order.

K_M a term I learned in biochemistry and its relevance to K_eq and k.

1. Baseline Reflection

As I studied enzymes catalyzed reactions in biochemistry class it was important to know that at equilibrium an enzyme (E) binds a substrate (S) to form an enzyme-substrate complex (E-S). The E-S complex can either dissociate or irreversibly convert the substrate to a product (P). The Michaelis-Menten equation (see later evidence) describes the relationship between the rate of substrate conversion by an enzyme and the concentration of the substrate. In this equation, V is the rate of conversion, V_max is the maximum rate of conversion, [S] is the substrate concentration, and K_M is the Michaelis constant. The Michaelis constant is equivalent to the substrate concentration at which the rate of conversion is half of V_max.

The evidence below is a power point slide shown in Chem. 504 class. It was the first time I was introduced to the new term K_M. It is also called the kinetic activator constant. The slide lists facts about K_M, but I still could not relate it to neither K_eq nor k.

Chem 504 Power point slide

2. Later Reflection.

Here is my understanding to this symbol:

The Michaelis Constant, K_M determins enzyme-substrate interaction. This value is often dependent on pH, temperature, and ionic strength. The K_Mis able to detect two factors: One is the concentration of substrate when the reaction velocity is half that of the maximal velocity; thus, the Michaelis constant measures the concentration of substrate required for a significant catalysis to take place. From the graph below one can determine the value of K_M. It is found at the substrate concentration when the reaction rate is half of its maximum value (V_max/2).

Chem. 504. K_M at V_max/2.

Secondly, it is able to detect the strength of binding of the enzyme-substrate complex (ES). Its value includes the affinity of substrate for enzyme, also the rate at which the substrate bound to the enzyme is converted to product. Using the below equation, only if k₂ is much smaller than k_₁ will K_M equal a binding affinity (see equation 1 and Michaelies constant equation below).

Michaelis Constant equation

So for enzymatic reactions which exhibit simple Michaelis–Menten kinetics, the enzymatic reaction is assumed to be irreversible, and the product does not bind to the enzyme. The equation for the Michaelis Constant is

K_M= (k_-1 + k₂)/k₁ sometimes it can be seen as [ES] = [E][S]/K_M. using this equation, a high K_M indicates weak binding and low K_M indicates strong binding.

The later evidence I present here is quiz five, July 30, 2009 Chem 504. Question eight askes if k₂(the rate of the reaction to form product) is small what does this say about K_M?

From the relation between k₂and K_M I explained that if k₂ is small then it is the rate determining step. K_M would be low because it is directly related to k₂ which means that the binding between the enzyme and the substrate at V_max/2 is strong.

In question six for the same quiz, the graph shows the complete understanding of how the Michaelies-Menton equation is a combination of the first and zero order. It is first order when the enzyme is still binding with the substate and a zero order at equilibrium where the break down equals the formation of the enzyme-substrate complex ( [ES]_break = [ES]_formation ).

Chem 504. Quiz five

Question nine below is another evidence of understanding which enzyme substrate reaction has a better K_M. I chose [S]₁ because as shown in the diagram, the lower K_M the better binding between the substrate and the enzyme at V_max/2.

Chem 504. Quiz five

Furthermore, the evidence below shows the post activity skill exercises question one. Containing a table for different enzymes with different K_M. The question askes which enzyme has the highest affinity for substrate? using the provided table , C has the lowest K_M so it has the greatest affinity for the substrate.

Chem 504. Post activity. SKILL EXERCISES

The last evidence shown below (Chem.504. Post activity) is continuing to question one skill activity where technology is used to find K_M. Given the substrate concentration and V the rate of conversion for the three substrates A, B, and C, we can plot the double recipricols of both V and [S] for each substrate to find the linear equations for the three substrates using excel as shown in the figure below. From the linear equations we can determine K_M.

We can also find the value of K_M using the following equation: 1/V = K_M/ V_max( 1/[S]) + 1/V_max..

To conclude, we can determine K_M from both; the graph and the equation.

Chem 504. Post activity. SKILL EXERCISES

Reflection 2.

CHEM 505 Environment Chemistry

Enduring Understanding:

Statistical analysis of data: the general linear model is most often used for converting qualitative environmental observations into quantitative expressions.

In the use of statistics, it is important to note that corralation leads to the identification of a process that can be used to relate certain observations.

The T-test is a statistical tool used to evaluate hypotheses about group- level differences in outcomes and results of experiments. I learned to use this statistical tool first in Chem. Ed 536 when we were asked to evaluate a survey of attitudes towards astronomy. I learned that t-test is a tool to evaluate hypotheses about group-level differences in results and outcomes. Specifically, I learned how to use two different applications of the t-test (listed below) in evaluating two kinds of hypotheses:

1. The first sample t-test is one in which the level of outcomes for a group is compared to a known standard.

2. The second sample t- test is one where the outcome levels of two groups are compared to each other.

1. Baseline Reflection

I chose this particular subject to demonstrate my growth from the time I was introduced

to it, with zero "null" understanding, in Chem. Ed 536, where we were taught to calculate

it using our T1 calculator as shown in the baseline evidences below, to the heavy use of it

In Chem. 505 Environmental Chemistry as demonstrated in the later evidences. We then continued in applying this tool using

excel and extended it to find the P-value . These expressions were new to me and I think to all

the teachers in my cohort. I still remember Jenny Line raising her hand in Chem. 536 asking

“What is this T-test?” It was a new concept for all of us and a great introduction to Chem. 505

where we used it in depth.

In our first class Chem Ed 536. September 2007 we worked on a surveys of attitudes toward astronomy were questions designed to identify attitudes about research and research methods. As a group we categorized the questions into positive and negative attitudes which were both subdivided into three subdivisions:

a. Applications

b. Ability

c. Enjoyment.

We created a code for the survey worksheet along with the scale marking the responses using excel worksheet. Data used for this t-test are the personal answers vs. questions. -1, +1, 0 indicating positive, negative or neutral attitudes. The spreadsheet we used is demonstrated below.

Chem Ed 536. Data spreadsheet

Evidence below is a reflection I submitted in our October class Chem 536, 2007 about our first session we attended in September. The fifth paragraph, marked in red, is an evidence of our correct work as a group in creating the spradsheet with the average of both the answers for the questions and the average of answers of each of the seventeen participants.

Refliction 1. Chem Ed. 536

The survey was a great opportunity to practice calculating the t-test. The evidence below shows two of the t-test preformed for the same survey through my T1 84 SE Calculator, copied to my laptop then embedded as a screen shot into the survey that I submitted.

Chem Ed.536.using calculator to find t-test.

2. Later Reflection:

Evidences in this reflection demonstrate my understanding of the fundamentals of how and when this tool is important and useful, and whether the means are significantly different.

A central question of concern is “How are outcomes different for different groups?” The t-test assesses whether the means of two groups are statistically different from each other. In other words the question that t-test addresses is whether the means are statically different. The evidence below (Chem Ed. 536. Deferences between means) shows relatively little over lap between the two bell shaped curves which indicates most distinct differences between groups in the low variability at the bottom of the figure and high overlap of the bell shape distribution in the high variability case (second situation) indicating least difference between groups.

Chem Ed. 536. Deferences between means.

Another evidence piece I present from Chem. 505 is problem set # 1, where we stretched our understanding of t-test to the P-value. Here is how I understand the statistical significance; When a test is done in microsoft excel it gives a value which is a probability associated with t-test, this is the interpretation of the P-value. If this value is obtained through t-test and is lower than a chosen P value of 0.05, the results would be considered statistically significant. This later evidence below includes microsoft excel table that calculates P-value as a result of t-test. It shows the summary output of electrodes 1, 2, 3. The calculated P-value of 0.96 is higher than the chosen P-value of 0.05 which means the results are significant and the null hypothesis should not be rejected.

Chem. 505. problem set # 1

In the following later evidence Chem 505. continue problem set #1. I explained the acceptance of Ho (null hypothesis) by calculating the amount of overlap between the ranges of the given values. These values of regions of uncertainty show an overlap of 0.214. When the values are within one standared diviation of each other, then 0.214 being greater than 0.16 should explane the acceptance of the null hypotheses.

Chem 505. continue problem set #1

Later Evidence below (Chem 505 statistics set) shows complete understanding of the level of significance affer metacognition for question #1. It explaines the decision of the risk level tolarated in rejecting the null hypothesis. So if we have a 95% confidance level, the risk value is 0.05. Any value below this amount will lead to the rejection of the null hypothesis.

Chem 505. Metacognition. statistcs set #1

In conclusion I can summarize my growth in the following:

a. The t-test is used to compare the means of two independent groups.

b. H_o (null hypotheses) is when the means of the test groups are equal.

c. H_a (alternative hypotheses) is when the means of the two groups are not equal.

d. A low P-value (can be obtained from both microsoft excel when performing t-test or a T1

84 SE calculator) means there is an evidence to reject the null hypotheses in favor of the

alternative hypotheses, or, there is an evidence that the difference in the two means are

statistically significant. The P-value is a probability of obtaining a result at least as extreme

as the one that was actually observed assuming the null hypotheses is true.

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