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Q2 – Week 10 – 1/24 – 1/28

                                                                                                                                  Jump to:  Tuesday,   Wednesday,  Thursday,  Friday                                                                                                                                   ___________________________________________________________

1/24 – Monday  –   A Day – 

Main focus –                                                                                                                                                        
a)  To determine the spontaneity of a reaction by using the values of ΔS, ΔG, and ΔH.
b)  To Introduce the concept of equilibrium and relate it to ΔG.

 

1.  Let’s Start with a review of the Gibbs Free Energy Formula:      ΔG =  ΔH – TΔS

I will write the 4 possible conditions including the Always Rule, Never Rule, and the Sometimes Rule (entropy, enthalpy driven)

2.  Let’s Relate this to the “Cool Uplifting Demo” with the diagram below:

3.  Solidification Demo – 

Today’s Notes on the board:

      

                                                                       

Hand warmer demo:  Is this spontaneous? Why must heat be given off in order to for this solution to become a solid?  When has this physical process reached equilibrium?  

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1/24 – Monday’s Homework: –

1.  Make another submission to last nights Form -after today’s lesson!                                                                                                    
2Read me!   Look at the 2 reactions below.  The first reaction is the phosphorylation of glucose which is required in the 1st step to get Free Energy out of the glucose. As you can see the first reaction is not spontaneous (ΔG is not negative!) and it is the driven by the second reaction that is spontaneous (ΔG is negative). In this coupled process the the overall ΔG is -17 to show that the spontaneous reaction of ATP was spontaneous enough to DRIVE the first reaction. You may have learned that ATP is the energy currency of the body AND THE reason it is needed by all life sustaining chemical reactions is to COUPLE the endergonic reactions of life that require a steady supply of Free Energy. We will learn that we get ATP from the breakdown of glucose in our body (cellular respiration).
                             
3.  Complete ATP worksheet with textbook (Chapter 8).

 

4. Review with the key – There are mistakes in my key. Question 1 b should have 2 phosphates NOT 3 and 7b should be Exergonic NOT Endergonic! I feel bad and I am over it.

ATP Cycle worksheet .pdf
View Download

ATP Cycle worksheet Key.pdf
View Download


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1/25 – Tuesday –  B  Day – 

Main focus –                                                                                                                                                          
a)  To identify an ATP molecule and how it channels Free energy from the break down of nutrients (catabolic) to endergonic chemical reactions (anabolic). 
   
b) To introduce the concept of equilibrium and its implications with the flow of free energy into a cell.                                                                                                                                                                                                            
                                                                                                                                                                                     1.  To review the homework packet and collect.                                                                                                                                                                                               

2. Equilibrium and ΔG =

Equilibrium .pdf
View Download

Equilibrium and ΔG = 

Equilibrium = the death of a spontaneous process = ΔG = ZERO!!!!!
Equilibrium = the rate of the forward = the rate of the reverse reaction
Equilibrium = the amount of the reactants and products stay constant
Enzymes are biological catalysts that decrease the activation energy for both the forward and the reverse reaction.  They could help reversible reactions reach equilibrium faster.
 
Given the following reaction:    NO2  + NO2   —>   N2O4
Equilibrium – Equilibrium intro demo’s: Tanks, Film canisters, solidification, Oscillating Reaction.
     
              A) REVIEWED PARTS OF THE Energy diagram of chemical reactions. 
              activated complex, change of free energy, activation energy, what enzymes do   
 
3: Enzyme discussion with videos:  

                               

 
Enzyme videos:

 

   

                                         

Metabolism, Thermo and enzyme Presentation:

Equilibrium Tank Demo:    

   _____________________

1/25 – Tuesday’s Homework: –

     

1. Please complete the following form below on chapter 8.
    
 

 
1:  ATP/Enzyme Form 

End of Tuesday!

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1/26 – Wednesday –  B Day 

 
Main focus –                                                                                                                                                          
a) To perform the Catalase Lab                                                                                                                       

Catalase Lab –                              2H2O2     —->    2H2O    +    O2

We will use a yeast suspension (12g packet into 250 ml or warm water) to produce the catalase.
 
We will use store bought 3% hydrogen peroxide as the substrate.
 
independent variables:
 
Amount of Catalase
Temperature (cold and warm)
 Amount of Hydrogen Peroxide
Boiled enzyme
pH (high and low) 
 
Catalase Lab Grodski.pdf
View Download

   _____________________

1/26 – Wednesday’s Homework: –

1. Please complete the form below on the catalase lab.  You will need the “second” video for question 5 that I will repost below:  
 
Enzyme videos:
   
                                                                                                  
 
1:  Catalase Lab Form 

End of Wednesday..

  ______________________________________________________________________________________________________________________

1/27 – Thursday –  A Day 

Main focus –                                                                                                                                                          
a)  To complete the Catalase Lab
                                                                                                                                              
b)  To identify acidic and basic pH ranges through concentrations at equilibrium.                                                                                                                                                                                          

1. All lab groups need to complete the pH and boiled enzyme tests.

 
2.  The pH lesson  –  Classwork –       
 

pH worksheet.pdf
View Download

 
ph worksheet key p.pdf
View Download
          

pH NOTES:

  
auto – ionization of water:  
  
This called auto-ionization of water.
 
                                                      H2(l)  →    H+ (aq)   +   OH– (aq)             Kw =  1 x 10-14
                                                             Proton               Hydroxide
 
                                                                                         or
 
                        H2(l)  +  H2(l)  →    H3O+ (aq)   +   OH (aq)        Kw =  1 x 10-14
                                                             Hydronium         Hydroxide
                               
From 2 reactions above which are really the same
                                                                          you can see ONE H+ or H3O+ for every ONE OH.
 
                    Acids will increase the Hor the  H3O+ concentration above normal* conditions.
 
                               Bases will increase the OH– concentration above normal* conditions.
 
*Normal conditions refer to water that is NEUTRAL : (  Hor the  H3O=   OH– )  = pH = 7
 
                                                                 H2(l)  →    H(aq)        +   OH– (aq)                                                                   
                                H2(l)  +  H2(l)  →    H3O(aq)   +   OH (aq)   
 
                                        Keq =   [Products]   =  Kw = [H+ H3O+][OH]
                                                     [Reactants]
 
*We can use H3Oand H+ interchangeably.  I will write Hgoing forward but remember I can use H3Oat any time and it would mean the same thing.  (Whether you have a free proton or a a proton held by water does not matter)
Now the value of Kw (equilibrium constant for the auto-ionization of water) equals =  Kw = 1 x 10-14
                                                                                      
                                                                      Kw = [ H+ ]    x     [ OH]                                                                                                                                 
                                           1 x 10-14   = [ H+]     x    [ OH]
 
If one Hdissociates for every ONE OH–  H2(l)  →    H(aq)        +   OH– (aq)    then:                                
                                           1 x 10-14   =  [ H+]     x    [ OH]                                                                                                                                                                             
                                          1 x 10-14    =  [1 x 10-7]     x    [1 x 10-7 ]                                           
                
As you can see the concentration (molarity, M) of Hand  OH–  ARE THE SAME IN PURE WATER! 
                                                       
They must be the same concentration to equal 1 x 10-14 !
Remember when you multiply exponents you add!
 
*THIS is WATER AT NORMAL CONDITIONS.                           
 
 Now we use a convention to measure this balance or unbalance of ions in water.  It is called pH.
 
                                     pH = – log (base 10) of the Hor  H3Oconcentration.
 
                            Pure water at 4 degrees Celsius the [H+] or  [H3O+]  = 1 x 10-7
                            if you put  -log (1 x 10-7) in your calculator you will get :  pH  = 7
 
So basically pH is a numerical system that measures “stuffs” the entire concentration (small value) into a exponent (base 10) AND MAKES IT positive for Ease  
 
                                                                            pH = – log  [H+ 
 
a)What if the [H+] = 1 x 10-5  ?  What is the pH?   Is this an acid, basic or neutral solution?
 
b)What if the [H+] = 0.001  ?  What is the pH?  Is this an acid, basic or neutral solution?
 
c)What if we have a [HCl] = 0.1 M   What is the pH?  Is this an acid, basic or neutral solution?
 
d)What if we have a [HCl] = 5.6 x 10-6   What is the pH?  Is this an acid, basic or neutral solution?
 
e)What if you have [NaOH] = 0.001    What is the pH?  Is this an acid, basic or neutral solution?                                                                 
e)What if you have [NaOH] = 3.34 x 10-5   What is the pH?  Is this an acid, basic or neutral solution?
 
*Notice the above problems are strong Acids and Base, which means complete dissociation.
 
 answers on bottom of today’s homework post..   
4. Complete the Form Below: You will only have 1 submission to the form tonight.
 
Please look over the data from Sintia and Will to answer questions regarding Catalase activity for some questions.
 
Data for Catalase Lab.pdf
View Download

Connections FORM!

pH screencast lecture:    

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1/27 -Thursday’s Homework: –

   

1.  Catalase Lab – DUE Friday

 
        On the back of the lab please
 
        a)  list each independent variable that was tested
        b)  Explain the outcomes based on your knowledge of enzymes, collision theory, or  
             energy why each variable caused each outcome.
 
 
2 . Please read the connection notes below
 
3.  Complete the connections form below ( you need to view a short video on cellular respiration) to complete the last three questions.

Connection Notes!

Acidic solutions (pH that is lower than 7) and basic (pH that is higher than 7) solutions
can potentially be harmful to enzymes due to the fact that these tertiary proteins have R groups that will change according to pH of the solution.
 
Acidic R groups have -COOH endings that will become -COO–  (have a negative charge) because the H+ will be taken by the base in a basic solution.  So proteins that have amino acid that have R groups that are acidic will become negatively charged and they will attract other R groups that are positively charged. If the R groups change their attractive partners then THEIR SHAPE and thus THEIR FUNCTION changes. (Basic R – groups will also become charged (+) in the higher pH solutions.)
 
So R – groups that have ionic attractions will be be affected by slight changes in pH because there will more ions in the solution that these R could attract too.  This means that changes in the pH will cause changes in the amount of H+ and OH- in the cell which will cause the R groups of amino acids to attract the H+ (if the R group is negative) and attracts the OH- (if the R group is positive) AND thus change the shape of the tertiary folded protein. Changing the shape is called denaturing and denaturing changes the function of the protein when the active site is changed.
Form and Function!

 

So Arginine, Histidine, Lysine have positive R groups that will be affected by OH (Bases)
 
Aspartic Acid and Glutamic Acid have negative R groups that will be affected by H+ (acids)
 
Proteins (enzymes) that have less of these amino acids with these R groups will be more resistant to pH changes and will denature at lower or higher pH’s as compared to enzymes that have more of these amino acids.
 
Remember the 4 types of attractive forces that occur in folded tertiary proteins:
 
Remember that tertiary proteins are long chain polypeptides (amino acid linked together) that take an active form (shape) by the R groups attracting each other.  There are 4 types:

    a) Ionic – positives attract negatives

            –H+ and OH compete with these!

    b) H – bonding – electron deficient Hydrogens

            – Strongest attractive force    

    c) Disulphide Bridges – Sulfur bonding to Sulfur

– Strongest force (chemical bond)
that keeps R groups together

    d) Hydrophobic – LDF’s – (non-polar attractions)

– weakest attraction

 

A permanent change in the shape of the protein is called DENATURING.  A denatured protein or enzyme will have no enzymatic properties as the active site will no longer have the shape to interact with the substrate.
 
Enzymes can become denatured if the temperature is too high as this high energy state will cause R group attractions to break and unfold the protein. “How did the Boiled enzyme do in our lab?
Remember the Taq polymerase in our PCR – Taster Lab? Taq polymerase is an enzyme that has a higher temperature tolerance that most enzymes do not have .  It is an example of the heat tolerant bacteria (Thermostable)that is in the graph below. They have a lot more disulphide bridges R group connections which are actually chemical bonds, and much less hydrophobic R – attractions in our folded proteins. 
 
All enzymes have optimum pH ranges and optimum temperatures that they work most effectively:
 
SO DNA does not just code for the amino acids that make certain shapes they also code for certain amino acid combinations that can make them more or less stable in certain environment.
 
Connections!!!!!       

 

Answers for question in notes above.

a) pH = 5   Acidic solution (below 7)
b) pH = 3   Acidic solution (below 7)
c) pH = 1    Acidic solution (below 7)
d) pH = 5.25  Acidic solution (below 7)
e) pOH = 3  thus  pH = 14 – 3 =  pH = 11  Basic solution (above 7 – more OH than H+)
f) pOH = 4.48  thus pH = 14 – 4.48 = 9.52  Basic solution (above 7 – more OH than H+)

Cellular Respiration – Bioflix video:    

 
3:  Connections Form 

End of Thursday..

– -___________________________________________________________________________________________________________________

1/28 – Friday – B Day

Main focus –                                                                                                                                                          
 
a)  To identify the 3 major parts of Cell Respiration. 
b)  To identify the reduction and oxidation half reactions.

 

 

1. pH review with pH probe demo
 
2.  Complete the catalase lab in terms of the graphs!  
 
The reaction from the lab :  H2O2  –> O2  +  H2O + Heat was moving toward Equilibrium!
Also because group placed 3 ml of hydrogen peroxide with 7 ml of catalase there was an initial very fast reaction rate but it decreased quick as most of the substrate was reacted and used up.  The forward reaction was becoming less spontaneous and the reverse was becoming more spontaneous!
 
-Another concept that you need to understand is that if you increase the substrate too much you could reach the limit of the enzyme. That means you could be reaching the substrate saturation point where the enzyme is at its fastest and having more of the substrate will not make the reaction go any faster.
 
Catalyzed reaction – (what biological enzymes would do)

 Catalyzed reaction – (what biological enzymes would do)
 
A catalase demo:
 
Todays Demo:             2H2O2     ->    2H2O    +    O2
 
Very small rate of reaction until a catalyst was added.
Just lowering the Ea (activation) but not increasing the ΔG .
Elephant’s Toothpaste:
 
 
 Free energy to drive Life’s Processes:                                                                                                                                         
 
3.  Reviewed Cell Respiration basicsProton motor force with videos of ATP -synthase 
Cellular Respiration Overview – Note taking started!!
 
Cellular Respiration Chart.pdf
View Download
 
Cell respiration Bio Notes pics 4 pages.pdf
View Download
 
Complete oxidation formula: when you combine all 3 parts of cellular combustion
 
Complete Cellular Respiration Formula:                         
 
                        C6H12O6 (s)       +       6O2       —–>       6CO2(g)        +       6H2O (l)
 
      glucose      –>      Electron Carriers (NAD+, FAD)          –>      Electron transport System  (Oxygen bonds with H+ to make H2O)
   High Energy                          oxidize glucose                                     high energy electrons used to pump H+ ions against gradient
       Electrons                                                                                                 these H+ rush out of membrane giving atp synthase energy    
       

   ____________________

1/28 -Weekend’s Homework: –

1. Please review your graded Genetics Test.   I will link the test to everyones personal email as some students are still finishing.   Use the key to understand what you did wrong. If you find an error in my grading or have a question please write it down an ask me Monday. 

Link to the Genetics Test: https://drive.google.com/file/d/1bH3jtLSyDGbs4p0M1_OAQ-sDGwDgrlTV/view?usp=sharing

Once you have reviewed your tests, which I will collect Monday, please use the link below to make a total of 6 posts in the Genetic Test Forum.

2 unique posts that you create –

3 posts that you respond to others in the class – 

LINK to the TEST 3 – Genetics Test 3 Forum

2. Please read the following:

Friday (today) we learned that the energy that we harness from glucose (from the food we eat) is due to high energy electrons that exist in C – H bonds.  These electrons are shared equally because the Carbon and the Hydrogen have almost the same attraction for electrons which allows electrons to have the most freedom.  Electrons with the most freedom have the greatest potential energy.  When they are held closer to one or another atom the electrons “feel” the nucleus more thus have less potential energy.  
 
The “slow” burn of glucose occurs as we take the high energy of the C-H electrons and convert them into lower energy electrons. This is accomplished by slowly taking these electrons away from the glucose in each stage of cellular respiration ( Glycolysis, Pyruvate oxidation, Citric Acid Cycle, Oxidative Phosphorylation) and we use these high electron carriers that get reduced WHEN THEY pull electrons away from glucose or glucose frags!  These high energy carries CAUSE THE OXIDATION of glucose!!  
REDUCTION: Gaining electrons from glucose
 
                                                  NAD+   +  H+  +    2e–    —->  NADH
 
                                      FAD    +   2e–   +   2H+   —>  FADH2
 
These electron carriers deliver the high energy electrons that come from glucose to the electron transport system AND they pump H+ ions into the inter-membrane space THAT LEADS to making ATP directly!   These electrons lose their energy and are pulled along the electron transport chain by oxygen who combines with it  make water. 
OXIDATION:  Losing electrons to the Electron Transport Chain
 
                                      NADH  —->   NAD+   +  H+  +    2e–    
                        
                                     FADH—>     FAD    +   2e–   +   2H+   
 
 In water the electrons are held more tightly or are pulled more to the oxygen (who loves to grab electrons) These electrons are held closer to one of the atoms in a bond and have less freedom leading to a release of energy which is used to make ATP.
 
Do not forget there is whole host of enzymes that catalyze the many steps needed to slowly grab the high energy electrons.  There are also co-enzymes (vitamins!) that help carry these electrons into electron transport system.  They include NADH (Vitamin – Niacin), FAD (Vitamin – Vitamin B2 – riboflavin). These are important vitamins!!!! 
 
Do some vitamins give us energy? No but they help get the free energy!!!                                                                                            

3. Complete the Form below: 

 
3:  Energy and Cellular Respiration Form:

End of 2nd quarter!!