part a cells are molecular - university of pennsylvania
TRANSCRIPT
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Part A
Cells are Molecular
http://www.cdgs.com/_about.html
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Dusty Carroll
Helical Wheel Project
"Helix Diagram" Project
This project is to be the first page in your Course Book. It involves working out and drawing a helical wheel, as in textbook Figure 5.21, for part of the helical region of a fragment of the muscle protein myosin. It will allow us to investigate the properties of an extended helical region of myosin.
There is actually a PDB X-ray structure for the heavy meromyosin fragment. It has extended helical regions, including one that is 41 residues long, which extends from residues 861 through 901. It is file 1I84 (one-eye-84). That part of the sequence is as follows:
861 AKDEELQRTK ERQQKAEAEL KELEQKHTQL CEEKNLLQEK L 901
The question to investigate is whether there is any pattern of hydrophobic and polar amino acid side chains in this sequence which would contribute to the formation of the coiled-coil structure that is found for fully intact myosin molecules. This is just a fragment of the whole chain, with most of the coiled-coil residues chopped off, and so it is of interest to find out if this part of the coiled-coil structure, the part closest to the molecule’s globular head group, has a significant pattern along the helix that would provide some of the driving force for the formation of the coiled-coil structure present in the entire, intact myosin molecule.
Let's explore along the chain by having each person do a part along the sequence. We should start with residue 861 and each make a helical wheel diagram for 14 residues. Each diagram needs to be labeled with circles numbered with the actual numbers of your residues in the part of the sequence you are drawing. So your numbers will follow a sequence starting with number like 861 and higher. Also, each amino acid needs to be labeled with its 3-letter abbreviation (not just the one-letter abbreviations given in the sequence above). Then, in addition, each residue name should be labeled in parentheses after it with a P for Polar or an H for Hydrophobic. For example, Alanine would be labeled as Ala(H), etc.
The idea is to draw your helical wheel and get it labeled the way you want it and then redraw it so it is clean and clear for your Course Book.
Finally, a brief comment is needed stating and explaining whether or not you found any pattern of hydrophobic and polar residues in your part of the helix that might contribute to stabilization of the coiled-coil structure.
The helical wheel diagram should be drawn so that the lowest-numbered amino acid residue in your part of the sequence is on top, that is, so that you are looking along the chain from lower to higher numbers, with the lowest number in the largest circle, etc. The sequences we will examine overlap, but they will all give different diagrams, and they encompass the entire helix. So then we will have our answer!
14-Residue Sequences
861 through 874 863 through 876 864 through 877 865 through 878 867 through 880 869 through 882 870 through 883 871 through 884 873 through 886
874 through 887 875 through 888 877 through 890 879 through 892 880 through 893 882 through 895 884 through 897 886 through 899 888 through 901
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Helical wheel
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Dusty Carroll
Lesson Plan: The acid-base properties of a peptide
Target Audience This lesson is created for an AP chemistry class during the unit on “Acids and Bases”. They will have
already completed the unit on basic organic chemistry and will be familiar with the amine and
carboxylic acid functional groups. Within the acid-base unit, they will know the difference between
weak and strong acids and bases; how to calculate pH; and the meaning of pKa. They will have
completed a lab on titration curves of weak acids. They will have been introduced to the concept of
buffers on the day before this lesson.
Objectives 1. Use previous knowledge of molecular bonding to correctly draw structural diagrams for amino
acids and a dipeptide.
2. Contrast the acid/base character of a free amino acid with that of an amino acid in the middle of a
peptide or protein chain.
3. Match pKa values for an amino acid to the shape of a titration curve.
4. Explain why certain amino acids in a protein may act as a buffer in the body.
Introduction
The following will be written on the board as an opening activity:
Discussion:
• This is called an amino acid because of the functional groups
• Recall amino acids from biology studies (join to form proteins)
• Recall Lewis definition of acids and bases
o Identify acidic/basic groups on the amino acid
o Draw the +1, -1, and zwitterion forms to recall the accepting and donating of a proton
o Note that at physiological pH, aa is in zwitterion form
• This is the simplest amino acid (glycine) and is amphoteric
• Some amino acids have buffering abilities
o BUT…most of the amino acids are found in proteins or in shorter chains called
dipeptides, tripeptides, etc.
o They are joined by peptide bonds
� The N of one aa will form a bond with the carboxylic acid C of another aa with a
loss of water
Correct the Lewis structure by adding lone pairs or double bonds wherever necessary. Then circle
and identify the functional groups.
H O C
O
C
H
H
N
H
H
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Try This:
Analyze:
• Are the structures the same if you put alanine on the left or on the right of glycine? (no)
o Protein properties are based on the specific arrangement of amino acids
• Since amino acids are found in long chains of proteins, will an amino acid in the middle of the
chain still have its acidic and basic functional groups?
o No. N can still act as a proton acceptor, but side chains will be more accessible, and
some of those are acidic or basic.
o The side chains of amino acids determine their buffering ability
Activity:
See attached worksheet
Summary:
• Glycine and alanine have pKa values that are representative of the amino groups and carboxyl
groups on all amino acids.
o In general, the –COOH have values lower than 3 and the –NH2 have values higher than
9.
• In order to act as a buffer, a molecule must have a pKa very similar to the pH of the
environment it is working in.
o pH of the blood is 7.4
o remember that the amino acid functional groups will not contribute to the acidity or
basicity when they are in a protein because they have become the peptide bond
o side chains are the only hope for a pKa near the pH of 7.4.
o histidine had a pKa of 6.0.
� this was the free amino acid form
� when bound in a protein, the pKa increases to above 7, making it a potential
buffer for blood
� Actually, it is a major component of the protein, albumin
• Albumin is a major acid-base buffer for blood plasma
End with albumin structure
Go to http://www.rcsb.org/pdb/explore.do?structureId=2BXI
Under Display Options, click “WebMol”
Change “AllAt” to “MainCh”
Uncheck “HetAt”
Click and drag to show 3-D aspect
Click “Select”, choose “His”, click Apply. Note # of histamines in albumin. Click “close”.
To change mouse from rotate to zoom, click “control” and check “zoom”
Show the structure that is formed when glycine forms a peptide bond with alanine. First, correct
the structure of alanine:
H O C
O
C
H
CH3
N
H
H
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Homework
Kotz & Treichel, page 870-871 #36, 37, 41, 44, 60
References
Garrett, R. H., and Grisham, C.M., Principles of Biochemistry With a Human Focus. Brooks/Cole &
Thomson Learning, 2002.
Kotz, J.C. and Treichel, Jr., P., Chemistry and Chemical Reactivity, 4th ed.. Saunders College
Publishing, 1999.
http://cti.itc.virginia.edu/~cmg/Demo/titr.html
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Name
Exploring amino acid properties using titration curves
The following activity will help you identify some amino acids that may contribute to the acidity or
basicity of the proteins of which they are a part.
Draw the various structures of glycine in the boxes as described. Make sure to write in any necessary
formal charges:
At low pH, an amino acid
will exist in its most
protonated form. (Lewis
acids will not have ionized
yet and Lewis bases will
have accepted a proton.)
Draw the structure of glycine
at low pH.
At mid-range pH, an amino
acid will exist in its neutral
state, but with two opposite
formal charges. The
carboxylic acid proton will
have dissociated from the
molecule. Draw glycine at
middle pH.
At high pH, an amino acid
will have dissociated all of
its ionizable protons. Draw
glycine at high pH.
Go to http://cti.itc.virginia.edu/~cmg/Demo/markPka/markPkaApplet.html
Click “Begin the plot”
The titration curves will be similar to those you created in class when you determined the Ka of acetic
acid. When read from the left, the amino acid titration begins at low pH. Base is added to increase the
pH. At the right side of the graph, any ionizable protons have been removed as the pH of the solution
is now very high. The difference on these graphs is that the x-axis is not the amount of base added.
Instead, it is numbered to show the number of protons dissociated from the molecule.
1. Choose “glycine” then click the “graph” button, then click “show pKa’s”.
a. Why are the pKa values shown at 0.5 and 1.5 protons?
b. Write in the pKa values for the glycine equilibria in the box above.
→ →
← ←
pKa=____ pKa=____
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c. There are other hydrogens on the glycine molecule. Why are these not ionizable?
d. Explain how the titration curve proves there are only two ionizable hydrogens in glycine.
2. How would you expect the titration curve for alanine to compare to that of glycine? Explain.
a. On the website, click “clear”, then choose alanine. How did the graph compare to your
prediction?
3. Think about the groups that make up an amino acid. What would have to be true about the rest of
the structure of a given amino acid in order for the titration curve to yield 3 “bumps” or 3 pKa values?
4. Go to http://cti.itc.virginia.edu/~cmg/Demo/analyzeAA/histidine/histidine.html
a. Does this structure satisfy the requirement you listed in #3? How?
b. Which group will be the 2nd one to dissociate a hydrogen?
c. What is the pKa of this group?
Make sure your answers to all questions are complete and that you understand them. These will
be the basis for the discussion we will have in order to wrap up this lesson!
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Name Answer Key
Exploring amino acid properties using titration curves
The following activity will help you identify some amino acids that may contribute to the acidity or
basicity of the proteins of which they are a part.
Draw the various structures of glycine in the boxes as described. Make sure to write in any necessary
formal charges:
At low pH, an amino acid
will exist in its most
protonated form. (Lewis
acids will not have ionized
yet and Lewis bases will
have accepted a proton.)
Draw the structure of glycine
at low pH.
At mid-range pH, an amino
acid will exist in its neutral
state, but with two opposite
formal charges. The
carboxylic acid proton will
have dissociated from the
molecule. Draw glycine at
middle pH.
At high pH, an amino acid
will have dissociated all of
its ionizable protons. Draw
glycine at high pH.
Go to http://cti.itc.virginia.edu/~cmg/Demo/markPka/markPkaInstr.html
Click “Begin the plot”
The titration curves will be similar to those you created in class when you determined the Ka of acetic
acid. When read from the left, the amino acid titration begins at low pH. Base is added to increase the
pH. At the right side of the graph, any ionizable protons have been removed as the pH of the solution
is now very high. The difference on these graphs is that the x-axis is not the amount of base added.
Instead, it is numbered to show the number of protons dissociated from the molecule.
1. Choose “glycine” then click the “graph” button, then click “show pKa’s”.
a. Why are the pKa values shown at 0.5 and 1.5 protons?
The pKa is equal to the pH of the solution when it is half neutralized. Since the x-axis is the number of
protons dissociated, the pKa falls at the point where a proton is half-way dissociated. There are two
ionizable protons, so there are two pKa values.
b. Write in the pKa values for the glycine equilibria in the box above.
→ →
← ←
pKa= 2.3 pKa= 9.6
N CH C
H
O
OH
H
H
H
N CH C
H
O
OH
H
H
N CH C
H
O
OH
H
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c. There are other hydrogens on the glycine molecule. Why are these not ionizable?
Protons are only ionizable if the dipole of the solvent is strong enough to pull the proton away from the
atom it is bonded to. The other hydrogens in the molecule are either attached to carbon (nonpolar, so
no attraction for water) or nitrogen (polar bond, but strongly held).
d. Explain how the titration curve proves there are only two ionizable hydrogens in glycine.
The titration curve has two regions which flatten (buffering zone) then increase sharply. Each of these
regions corresponds to one proton being dissociated.
2. How would you expect the titration curve for alanine to compare to that of glycine? Explain.
Should be very similar because their structures only differ by one methyl group on the side chain. The
amino and carboxylic acid functional groups are the same.
a. On the website, click “clear”, then choose alanine. How did the graph compare to your
prediction?
The shape of the graph was very similar. The pKa values were only very slightly different.
3. Think about the groups that make up an amino acid. What would have to be true about the rest of
the structure of a given amino acid in order for the titration curve to yield 3 “bumps” or 3 pKa values?
The side chain would have to have an acidic or basic group on it.
4. Go to http://cti.itc.virginia.edu/~cmg/Demo/analyzeAA/histidine/histidine.html
a. Does this structure satisfy the requirement you listed in #3? How?
Yes, there is a nitrogen in the ring of the side chain that has an ionizable hydrogen when protonated.
This is a basic side group, but is acidic at low pH.
b. Which group will be the 2nd one to dissociate a hydrogen?
The N of the imidazole ring (the side chain).
c. What is the pKa of this group?
The pKa of this group is 6.0.
Make sure your answers to all questions are complete and that you understand them. These will
be the basis for the discussion we will have in order to wrap up this lesson!
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Dusty Carroll
Lesson Plan 2 – Antibodies
Target Audience and Background Junior level college-prep chemistry class. This lesson will be taught during the unit on bonding and
intermolecular forces. Students will be familiar with molecular structures and will have learned about
both intra- and intermolecular forces of attraction. They will have completed a previous lab with a
conclusion of “like dissolves like” in terms of polarity/nonpolarity.
Objectives 1. To correctly draw chemical structures based on models of “antigens” built by teacher using a
model kit.
2. To create the binding site that might be found on an antibody specific to the fictional antigens.
(binding site will be created using structures of real amino acids)
Lesson Length 89 minutes (lab period)
Introduction The following will be written on the board as an opening activity:
Discussion Points
• These are examples of some amino acids that make up the proteins in your body
• Highlight the parts that are the same
• Show that the side chains are the only thing distinguishing one aa from another
• Show how the peptide bond forms and tell students they will need to form these peptide bonds
for several amino acids as part of the activity
Transition
• Since this unit will be taught in the winter, someone is likely to have a cold. Use this as a
starting point for the idea of germs (antigens) and the antibodies in the body.
• Antibodies are proteins
• Antibodies are specific to the antigen they are fighting
• Probe student knowledge about antibodies and how they work
o Shape is important (see, molecular structure DOES mean something!)
o But shape is not enough- a molecule could be the right shape, but could bounce right off
if there are areas on each that repel.
• What makes the antigen “stick” to the antibody so it can be destroyed?
o Our favorite… intermolecular forces!
Correct the Lewis structures by adding lone pairs or double bonds wherever necessary.
H O C
O
C
H
H
N
H
H
H O C
O
C
H
CH3
N
H
H
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• Which ones?
o You name it, it can work.
• Show this picture:
• from
http://www.schoolscience.co.uk/content/5/chemistry/proteins/Protch7pg2.html
o Red are the heavy chains and white are the light chains
o Highlight the binding region
• Explain that the antigen would be attracted to, and would fit into this region. Then show this
picture:
• from
http://www.accessexcellence.org/RC/VL/GG/antiBD_mol.html
o The antibody is a complex structure which is only symbolized here by a long strand
o The part we will work on today is only the binding site
o Realize the binding site is not a separate piece, but is a region of the chain
Activity Review
• Review the use of model kits with students
Introduce activity
• Explain that they will use their understanding of all of the forces to create the binding site for
several different antigens created by the teacher.
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• Give the handout for the activity, explain directions and arrange groups.
• Make up several fictional structures with some model kits. As long as the bonding is valid,
don’t worry that the molecule does not actually exist. Give a variety of polar and nonpolar
regions. Throw in some ionic regions if possible. Suggestions for antigen structures:
o Ring structures with polar and nonpolar substituents
o Long chain structures with some twisting due to either hydrogen bonding or
hydrophobic interactions
o Short chain structures with branching
o Possibly a globular structure if enough model kits are available
• Label each structure with letters A, B, … then draw the Lewis structure as a reference when
checking student work.
Summary/Assessment
• Student groups will submit their worksheets
• Discussion of whether separate groups would have created similar binding sites or not for a
given antigen
o There may be several arrangements that could work for an antigen, but there are a finite
number of antibodies in the body.
o They can be created to match an antigen, so we can assume that there is a “best”
arrangement.
• Worksheets will be graded based on accurately matching side chains of amino acids to the
antigen in such a way that the forces should act to attract and trap the antigen.
References
Websites as listed with pictures
Garrett, R. H., and Grisham, C.M., Principles of Biochemistry With a Human Focus. Brooks/Cole &
Thomson Learning, 2002.
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Worksheet: Creating an antibody! Names
In this activity, you will use your knowledge of the various attractive forces you have learned to create
a string of amino acids that will bind to a fictional antigen in the body. You will do this for two
antigens.
Parameters
• You will have a handout showing the 20 amino acids that you may use to build your antibody
• The amino acids must be connected by the peptide bond shown during class
• You must use a variety of amino acids and cannot have more than two in a row that are the
same
• You may twist the antibody chain in any way allowed by the model kit
• Hydrogen bonding must be linear (I’ll explain this) when built with your model kit
• Polar groups must be correctly oriented such that they either attract or repel each other as
appropriate
o These polar groups are not freely moving, so it is not enough to just be polar, the
orientation in your molecule must also be correct
• Your chain must contain at least 15 amino acids and must loop around on itself at least twice
o The loops must stay together because of intermolecular forces
o The amino acid side chains must be able to “fit” in the orientation you have for them
Structures from http://dl.clackamas.cc.or.us/ch106-08/alphabet.htm
alanine
arginine
asparagine
aspartic acid
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cysteine
glutamic acid
glutamine
glycine
histidine
Isoleucine
leucine
Lysine
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methionine
Phenylalanine
proline
Serine
threonine
Tryptophan
tyrosine
Valine
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Select an antigen
Draw the Lewis structure for this antigen, then draw a schematic showing the overall shape of the
antigen.
Build your antibody using the above parameters. Write the 3-letter codes for your chain in the order
you have selected [ex: Arg-Lys-Tyr-Ile-…].
Draw a schematic of the shape of your antibody with the antigen bound to it. Number the structural
features that allow the antigen to stick to the antibody, then, list the specifics next to the picture.
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Select an antigen
Draw the Lewis structure for this antigen, then draw a schematic showing the overall shape of the
antigen.
Build your antibody using the above parameters. Write the 3-letter codes for your chain in the order
you have selected [ex: Arg-Lys-Tyr-Ile-…].
Draw a schematic of the shape of your antibody with the antigen bound to it. Number the structural
features that allow the antigen to stick to the antibody, then, list the specifics next to the picture.
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Dusty Carroll
Lesson Plan 3 – Hemoglobin
Background This lesson is for an AP chemistry class toward the end of the year. It will use hemoglobin as a model
to answer questions from a variety of chemistry units. Students are assumed to have a good
understanding of the basic principles involved in the AP chemistry curriculum and will be asked to
apply their knowledge to problems regarding hemoglobin. Students will also have studied hemoglobin
from a biology standpoint in the previous year and should be familiar with its general purpose. The
structure of hemoglobin and of the heme group will be given to them. The internet and their textbook
will be available to them for research.
Objectives 1. Recall various chemical principles as they apply to a new problem.
2. Recognize that biological processes involve chemistry
Lesson This will be a PIM activity for an 89 minute lab period. Below is the PIM diagram created by the
University of Pennsylvania. After the diagram, I have listed my expectation for what students will
know and what they should consider as part of the activity. Following, there is a worksheet that
students may use to take notes throughout the process.
Review of Hemoglobin
• Found in red blood cells
• A big protein made of several chains wrapped around each other in a specific structure (see
picture in textbook-noted below)
• Purpose is to carry oxygen to the body from the lungs
• The oxygen molecule binds to an iron atom in the hemoglobin molecule
• The iron atom is coordinated inside a heme group and the heme group is coordinated to the
protein through the iron (this leaves one coordination site available for binding oxygen) – see
picture in textbook noted below)
• There are more than one heme group inside a hemoglobin (students will determine that there
are four when they look at the picture)
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Initial Question
Posed on the accompanying worksheet as a series of questions.
Existing Information
• Students will have a reference for viewing the chemical structure of hemoglobin and the heme
group. The internet will also be available if they choose to use it. (pictures are below)
• Students should recall the structure of CO and of O2 in order to determine how one might
substitute for the other.
• Students should recall LeChatelier’s Principle in order to decide where stresses might affect the
equilibrium of the hemoglobin complex with oxygen.
• Students should infer from the structure that each heme group can bind one oxygen molecule,
for a total of four oxygen molecules per hemoglobin.
• Students should recall what a coordination complex is and should describe the octahedral
shape.
Hemoglobin Structure (Students will be referred to this picture in their textbook)
from
http://www.daviddarling.info/images/hemoglobin.jpg (just like the book picture)
Heme Group Structure (Students will be referred to this structure in their book)
from
http://www.daviddarling.info/images/heme.gif (just like the book picture)
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Reflect and Organize
• Students should use existing information to formulate reasonable explanations for each of the
questions.
• Students should first attempt to answer individually, then should communicate within their
group to attempt a more complete and accurate explanation.
Results
• Groups should form a preliminary answer to all questions
• All group members should agree and understand
Peer Review
• Pairs of students will find a pair from another group to work with. The pairs will exchange
information and ideas and will make changes or additions as appropriate.
New Information Needed
• If new information is needed, original groups or peer review groups may use internet, textbook,
or other students as a resource.
Back to original groups
• Original groups will reconvene and discuss any new or modified information.
• Once agreed upon, each student will write the accepted answers in their notes.
Community Knowledge
• Each group will be called upon to explain fully one part of the original question. That group
will write on the board any pertinent chemistry. After the group has finished presenting their
answer, other groups may add, comment or modify information as necessary.
• If there are any remaining misconceptions or omissions, teacher will fill in the gaps as
appropriate.
What I’m looking for (besides good and logical explanations):
Why is it said that you should not sit in a closed garage with the car running?
CO is a product of vehicle exhaust. CO can bind to the heme group better than oxygen can, maybe
because it has resonance structures with a negative charge on C, making C a good Lewis base for
complexing to Fe2+. [Resonance structures for CO should be drawn during their description) It is also
a similar size and shape. If CO is bound to the heme group, there is no room for oxygen. Then oxygen
will not be transported to the brain and vital organs and the body will die from lack of oxygen.
I’ll then add to their knowledge:
-CO has 25,000 times greater affinity than O2 for the heme group, but because of structural effects, its
binding affinity is reduced to only 250 times that of oxygen. (Explain the idea of sterics briefly)
-There is some CO in blood, but such low concentration that this is not a problem.
If iron is so good at binding oxygen, then how does it release the oxygen outside of the lungs?
Hb + O2 ⇔ HbO2
There is an equilibrium factor here. In the lungs, oxygen is abundant and the equilibrium shifts to the
right (hemoglobin complexes with oxygen). In the body tissues, oxygen is not abundant and the
equilibrium shifts to the left (releasing oxygen from the complex).
I’ll then add to their knowledge:
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-pH levels can also shift the equilibrium (free protons in solution can bind to hemoglobin, therefore
releasing the oxygen molecule)
-CO2 levels can also shift the equilibrium, party because of the protons it produces when dissolved in
water.
What is the stoichiometry involved?
Each hemoglobin has 4 heme groups. Each heme group can bind one oxygen molecule. So each
hemoglobin can bind 4 oxygen molecules
I’ll also mention:
Once it binds one oxygen molecule, the shape is changed in such a way that the heme groups flatten
out and that makes it easier to bind the other three oxygen molecules.
If iron has an octahedral coordination, why does oxygen bind to the complex at an angle instead of
along the coordination axis?
Because the lone pair on the oxygen is at an sp2 angle (not linear with the axis of the oxygen
molecule). The lone pair is what coordinates to the iron in the heme group, so this tips the molecule at
a 60o angle.
I’ll also mention:
The CO would come in along the axis, but the sterics prevent this, therefore reducing the binding
affinity.
References Garrett, R. H., and Grisham, C.M., Principles of Biochemistry With a Human Focus. Brooks/Cole &
Thomson Learning, 2002.
Zumdahl, S. S., Chemistry, 4th ed. Houghton Mifflin Company, NY, 1997
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An Inquiry on the chemistry of hemoglobin Name
Introduction Hemoglobin is the molecule responsible for transporting the oxygen you breathe from your lungs to
the rest of your body. Hemoglobin is a large protein in your blood. It contains heme groups which are
responsible for complexing with the oxygen. Look at the structure of hemoglobin in your book
(Zumdahl) page 976. You will notice that hemoglobin is composed of long chains of amino acids.
The heme groups are represented in this diagram by a disc shape. The structure of the heme groups
can be found on page 975.
Initial Question How does chemistry help us to understand how hemoglobin works in the body?
-Why is it said that you should not sit in a closed garage with the car running?
-If iron is good at binding oxygen, then how does it release it outside of the lungs?
-What is the stoichiometry involved?
-If iron has an octahedral coordination, why does oxygen bind to the complex at an angle
instead of straight along the coordination axis?
Notes Write down everything you know that might help to answer these questions. Once you are satisfied,
work with your group to develop an answer to the questions. Your teacher will give you instructions
about when you might talk with other groups of students. At the end of the process, make sure to write
down the final version of your answer in your notes.
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Concept Map
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Concept Map
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Part B
Life involves chemical reactions
and enzyme catalysis
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Thermodynamic Cycles direction sheet
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Thermodynamic Cycles Project
1. The energy difference between isomeric free radicals using a thermodynamic cycle:
(CH3)3C. + H
. � (CH3)2CHCH3 ∆Eo = -91 kcal/mol
CH3CH2CH2CH3 � CH3CH2CH2CH2. + H
. ∆Eo = 98 kcal/mol
(CH3)2CHCH3 � CH3CH2CH2CH3 ∆Eo = 2 kcal/mol
_________________________________________________________________
(CH3)3C. � CH3CH2CH2CH2
. ∆Eo = 9 kcal/mol
2. The free energy difference for binding of a transition state to an enzyme.
E + S � ES ∆Go = 2.8 kcal/mol
ES � Tc‡ ∆Go
‡ = 13.3 kcal/mol
Tu‡ + E � S + E ∆Go
‡ = -25.8 kcal/mol
Tu‡ + E � Tc
‡ (∆Go
‡)b = -9.7 kcal/mol
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Dusty Carroll
Lesson Plan 4: Getting to know Lactase
Background Information
Lactase is the enzyme responsible for breaking down the lactose in your body. Lactose is a
disaccharide that is found in milk. Human babies produce lactase in large quantities, but the
production of the enzyme decreases significantly into adulthood. Deficiency of the lactase enzyme
causes many people to have trouble digesting the lactose found in milk. This leads to the familiar
problem of lactose intolerance. (1)
Lactase can act as a catalyst for several different biological reactions. The lactase enzyme is the only
human enzyme that can cleave a β-glycosidic linkage like that found in lactose. The specific reaction
that is the focus of this lesson is the breakdown of lactose into the two monosaccharides, galactose and
glucose as seen in the reaction below:
OOH OH
OH
OH
CH2OH
O
OH OH
OH
OH
CH2OH
OOH
OH
OH
CH2OH
O
O OH
OH
OH
CH2OH
+ H2O
LACTASE
LACTOSE
GALACTOSE
GLUCOSE
Much research has been done on the mechanism of this reaction. In order to understand the
mechanism, however, a brief overview of the structure of the enzyme is necessary.
The lactase enzyme is found in humans and other organisms. In the human intestines, lactase is
combined with another enzyme called phlorizin hydrolase to form a transmembrane enzyme complex
called lactase- phlorizin hydrolase. It has been shown that the lactase portion of this enzyme complex
is the only portion active in the breakdown of lactose. (2) The mechanism under study for this lesson
comes from the lactase found in the Escherichia coli bacteria. It was here that molecular biologists
discovered the phenomenon of enzyme induction. Basically, the presence of lactose induced the
biosynthesis of an enzyme to split it. (3) The lactase-lactose system became the focus of much
research. Essentially, the lactase enzyme is genetically regulated. In other words, the action of the
enzyme results from its synthesis which is regulated by the genes that code for it.
The remainder of this lesson will concentrate on the independent lactase enzyme (not that complexed
with the phlorizin hydrolase). The structure of lactase is rather complex. Its crystal structure contains
four identical subunits. Each subunit contains a chain of 1023 amino acid residues. When this
structure was determined, it was the longest polypeptide for which an atomic structure had been
obtained. (3) It is a very large enzyme and scientists continue to query about the biological reasons for
such a large structure. Since each region of the enzyme seems to have a clear purpose, the common
belief is that portions of the molecule were useful in certain ways and it just sort of…happened. The
30
large size does not appear to have any reason other than that resulting from the combination of all of its
parts. (3,4)
Primary Structure (5)
Again, the lactase enzyme consists of four identical subunits or chains. Below is the sequence of
amino acids that make up just one of those chains. The sequence is shown using the one-letter
abbreviations for the amino acid residues.
GSHMLEDPVVLQRRDWENPGVTQLNRLAAHPPFASWRNSEEARTDRPSQQLRSLNGEWRFA
WFPAPEAVPESWLECDLPEADTVVVPSNWQMHGYDAPIYTNVTYPITVNPPFVPTENPTGCYS
LTFNVDESWLQEGQTRIIFDGVNSAFHLWCNGRWVGYGQDSRLPSEFDLSAFLRAGENRLAV
MVLRWSDGSYLEDQDMWRMSGIFRDVSLLHKPTTQISDFHVATRFNDDFSRAVLEAEVQMC
GELRDYLRVTVSLWQGETQVASGTAPFGGEIIDERGGYADRVTLRLNVENPKLWSAEIPNLYR
AVVELHTADGTLIEAEACDVGFREVRIENGLLLLNGKPLLIRGVNRHEHHPLHGQVMDEQTM
VQDILLMKQNNFNAVRCSHYPNHPLWYTLCDRYGLYVVDEANIETHGMVPMNRLTDDPRW
LPAMSERVTRMVQRDRNHPSVIIWSLGNESGHGANHDALYRWIKSVDPSRPVQYEGGGADTT
ATDIICPMYARVDEDQPFPAVPKWSIKKWLSLPGETRPLILCEYAHAMGNSLGGFAKYWQAF
RQYPRLQGGFVWDWVDQSLIKYDENGNPWSAYGGDFGDTPNDRQFCMNGLVFADRTPHPA
LTEAKHQQQFFQFRLSGQTIEVTSEYLFRHSDNELLHWMVALDGKPLASGEVPLDVAPQGKQ
LIELPELPQPESAGQLWLTVRVVQPNATAWSEAGHISAWQQWRLAENLSVTLPAASHAIPHLT
TSEMDFCIELGNKRWQFNRQSGFLSQMWIGDKKQLLTPLRDQFTRAPLDNDIGVSEATRIDPN
AWVERWKAAGHYQAEAALLQCTADTLADAVLITTAHAWQHQGKTLFISRKTYRIDGSGQM
AITVDVEVASDTPHPARIGLNCQLAQVAERVNWLGLGPQENYPDRLTAACFDRWDLPLSDM
YTPYVFPSENGLRCGTRELNYGPHQWRGDFQFNISRYSQQQLMETSHRHLLHAEEGTWLNID
GFHMGIGGDDSWSPSVSAEFQLSAGRYHYQLVWCQK
From this list it is easy to see why scientists are curious about the size of the enzyme. The four chains
together amount to 4092 amino acid residues!
Secondary Structure
Each of the four chains is analyzed as having five separate domains. Each domain serves a different
purpose in the enzyme. Some act to help the polypeptide chain attain its tertiary structure. Some act to
hold one chain to another to aid in the formation of the quaternary structure. Some join with a domain
on another chain to form the active site for the molecule. Below is a picture of the five domains found
on chain A of lactase. These images are taken from the CATH Protein Structure Classification
website. (6)
Note that each domain appears to have different secondary structures. The 3
rd domain is the one where
most of the helices appear. The 4th domain has no helices and consists of only beta sheets. The other
domains contain mostly beta sheets and a few sections of helices. The seemingly individual nature of
each domain may be necessary to facilitate the folding of the entire chain. (4)
31
Tertiary Structure
Together, these five domains form just one chain of the enzyme with a molecular weight of 116,570.8
D:
Jacobsen, Zhang, DuBose & Matthews, in their Nature article (4) show a similar structure of the one
chain with the domains labeled. The picture in the article is a stereo view of the chain.
This picture will not be shown on the posted version of this lesson. It was copied from the pdf file of the original Nature
paper. Readers will have access to the picture if they have access to the journal.
Quaternary Structure
Many pictures of the full enzyme are available. (3,4,5,7) The overall structure is a homotetramer
consisting of four identical chains. It is thought that the individual monomers form first, followed by
formation of dimers, then dimerization of the dimers to form the tetramer. (7) Each domain has a
hydrophobic core, consistent with the notion that the monomers formed individually before joining.
Some of the domains then interact with each other through polar networks on their exterior surfaces.
There are three major regions of interface formed from complementary regions of the monomers.
There are four active sites in each tetramer. Each site is formed from the interaction of two of the
monomers. Each site is also marked by two metal ligands, Na+ and Mg
2+. In the image from the
Juers, et.al. article, the domains on each monomer are colored such that each domain on a given
monomer is a different shade of the same color.
32
This picture will not be shown on the posted version of this lesson. It was copied from the pdf file of the original Protein
Science paper. Readers will have access to the picture if they have access to the journal.
Another image of the overall structure is found on the RSCB Protein Data Base (5) as seen below:
Here you can see how the interactions of the four monomers form into a single tetramer.
33
Kyte-Doolittle Hydropathy Plot (8)
This plot is calculated for a single 1023 residue chain of the lactase enzyme.
With a window size of 9, this plot shows many peaks below the midline. This corresponds to surface
regions of the globular protein. With so many peaks, it appears that there is a large surface area which
is hydrophilic. (Hydrophilic regions are given negative values.)
With a window size of 19, possible transmembrane regions show as peaks above 1.8. This lactase
chain shows no peaks and therefore is unlikely to have any transmembrane regions. The lactase-
phlorizin hydrolase enzyme noted in the introduction is a transmembrane enzyme. According to this
plot, the lactase portion of that enzyme would not be the portion spanning the interior and exterior of
the membrane.
Reaction Mechanism
The Enzyme Commission code for lactase is 3.2.1.23.
• 3 – Hydrolases
o 2 – Glycosylases
� 1 – Glycosidases (enzymes hydrolyzing O- and S-glycosyl compounds
• 23 – lactase or beta-galactosidase
34
In other words, lactase acts as a catalyst for the hydrolysis of the O-galactosidic bond in the sugar,
lactose. The exact mechanism of the reaction has been studied and Juers, Heightman, et.al. have
summarized previous work and given clarification to the proposed mechanism. (9) Lactase, aka β-
galactosidase, hydrolyzes its substrate (lactose) while allowing the constituent monosaccharides to
keep their stereochemistry. The reaction is a two step reaction. The first step is cleavage of the
glycosidic bond.
O
H
HO
H
HO
H
H
OHH
OH
O
OH
O
H
HO
H
H
OHHOH
OH
Glu537
C
O–O
Glu461
C
O O
H
O
H
HO
H
HO
H
H
OHH
OH
O
OH
O
H
HO
H
H
OHHOH
OH
Glu537
C
OO
Glu461
C
O O
H O
H
HO
H
HO
H
OHH
OH
O
OH
O
H
HO
H
H
OHHOH
OH
Glu537
C
OO
Glu461
C
O O
H
It is believed that this process takes place with a mechanism somewhere between that of an SN1 and
that of an SN2. The Glu537 from the active site of the enzyme acts as a nucleophile toward the
anomeric carbon of the galactosyl group. This forms an intermediate with enzyme Glu537 in the
alpha-glycosidic orientation. As seen in the above reaction, this is facilitated by a concerted
protonation of the glycosidic oxygen. This particular step is not well-proven, yet, but it is one
explanation. The acid responsible for protonation may be the Glu461 from the active site of the
enzyme.
The second step is the transfer of the galactosyl
product from the nucleophile of the enzyme
(Glu537) to an acceptor molecule. This step is
believed to occur in an SN1 release of the
nucleophile. During this process, the carbocation
(oxocarbenium ion) transition state is thought to
be stabilized by interactions between Glu537,
Tyr503 and the oxygen on the galactosyl ring.
Glu461 abstracts a proton from the acceptor
molecule, allowing the acceptor molecule to act
as a nucleophile toward the oxocarbenium ion.
There is some debate as to whether the metal
ligands in the active site play a role in the
catalysis. They appear to facilitate the catalysis,
but the exact mechanism for this is not yet
determined. The major metal ion ligands found
in the enzyme are magnesium and either sodium
or potassium ions. An alternate theory of
mechanism for the first step of this reaction
involves magnesium forming a complex by direct electrophilic attack on the glycosidic oxygen. This
reaction will not be further discussed, but it is important to realize that there are still several
possibilities for showing the actual reaction mechanism for this catalysis.
Functional Amino Acid Residues within Lactase
Glutamic acid: In its anionic state (like Glu537), this acts
as a nucleophile. In its neutral state (like Glu461), it acts
as an acid and donates a proton as described in the text.
H2N CH C
CH2
OH
O
CH2
C
OH
O
Tyrosine: The –OH group on the side chain participates
in hydrogen bonding in order to stabilize the transition
state.
H2N CH C
CH2
OH
O
OH
35
Catalytic Function
Catalytic efficiency values are altered by pH and also by the absence of magnesium for the lactase
enzyme. (9) Though the mechanism through which magnesium ion works is not clear, it appears that
its presence is necessary for optimal catalysis by lactase. This may be some sort of enzyme regulation
by the metal ions, but the exact parameters are still unknown. The efficiency values are listed by Juers,
Heightman, et.al. as:
• Kcat ≈ 60 s-1
• Km ≈ 1 mM or 1 x 10-3 M (This value is reported elsewhere as 4 mM) (10, 11)
These values are used in a ratio to determine the catalytic efficiency of the enzyme as follows (1):
111 000,601/60/ −−−
== MsmMsKk mcat
Enzymes whose ratios are in the range of 109 s-1M
-1 are considered to be the most efficient. This
shows that lactase has a pretty good efficiency.
Target Audience for Lesson This is an interdisciplinary lesson intended for a combination AP chemistry/AP biology class. The
school in which I teach puts great value on interdisciplinary lessons, so it is quite realistic to combine
these two classes. It will be necessary for the chemistry teacher and the biology teacher to collaborate
on this lesson. In addition to the background information in this lesson, both teachers should have a
working knowledge of basic chemical kinetics.
The AP biology students will have a strong background in the structure and function of enzymes. The
AP chemistry students will have a strong background in the kinetics of chemical reactions.
Objectives • Students will gain a basic overview of the structure and function of the lactase enzyme.
• Students will describe the differences between a unimolecular process and a bimolecular
process.
• Students will use given data to determine the kinetic aspects of a given reaction.
Lesson Using the classroom projector, a structure of lactase will be on the screen (from the Protein Data Base
website5). Students will be broken into groups of four with approximately equal numbers of chemistry
and biology students per group. Students will complete the “Introductory Worksheet” together for 5
minutes.
Discussion
• Review the correct answers to the worksheet
• Using the structure on the projector, point out the various aspects of structure that can be seen
o The four chains are colored differently
o Helices and beta sheets as symbolized
o Strands that are not helices or beta sheets
o Show the regions where the active sites are
o Discuss the meaning of “active site” (the specific portion of the enzyme that interacts
with the substrate)
36
• Connect the idea of an enzyme to that of a catalyst (small group discussions)
o Write the following on the board:
� Speeds up reaction
� May interact with substrate but is not used up
� Lowers activation energy for reaction
o Tell groups to discuss what this means. Chemistry and biology students should have a
different perspective to share with each other. Highlights should be:
� Speeds a reaction
• Necessary in the body because without enzymes, reactions would be too
slow
� May interact with substrate but is not used up in the reaction
• Substrate is the reactant that the enzyme works on
� Lowers the activation energy
• Provides a lower energy path for the reaction to begin, yet overall free
energy of the reaction is not affected
o Have groups write the overall reaction for the following:
� “lactose, a disaccharide, is broken into galactose and glucose by hydrolysis
when the enzyme lactase is present”
• Student reactions should match the reaction in the introduction. All of
these structures are in any AP biology textbook and several AP
chemistry textbooks. Students just need to make an equation out of it.
• “lactase” should be written above the arrow; water and lactose are the
reactants; galactose and glucose are the products; students should
recognize the 1:1 stoichiometry throughout the reaction.
� Ask whether the reaction mechanism is obvious from this reaction (no).
� Ask how a mechanism might be determined.
• Only by experimentation
• Determining the rate equations and rate constants
• The exact mechanism for this process is under debate.
o May be unimolecular or bimolecular
� Chemistry students: explain to the bio students how this relates to the reaction
order
• For elementary steps, molecularity is the same as order (uni = 1st order,
etc)
� Biology students explain to the chem. students how this relates to the enzyme
and substrate
• For a unimolecular reaction the rate depends only on the concentration of
the substrate
• For a bimolecular reaction the rate depends on the concentration of the
enzyme and of the substrate
• Summarize: More experimentation is necessary to determine the exact reaction mechanism for
the lactase enzyme. This may be accomplished through kinetic studies similar to the basic
kinetics we study at this level.
Follow-up Complete “Putting it all together” sheet as a group. Each student should have the answers. Turn in the
most legible copy.
37
AP Chem/Bio Enzyme Lesson
Introductory Worksheet
Try to answer the following questions without the help of those in your group.
1. Next to each of the following descriptions or pictures, write one of the following:
Primary Structure
Secondary Structure
Tertiary Structure
Quaternary Structure
Several chains of
amino acids can come together to form a large 3-
dimensional structure which is made up of several
units held together by covalent or intermolecular
forces.
The flat arrows represent beta
sheets
The spirals represent alpha helices
This shows how sections of amino acids interact
with each other in long chains.
The sequence of amino acids that make up the
enzyme protein. Below are a few amino acids
that may be included in the sequence:
H2N CH C
CH2
OH
O
CH2
C
OH
O
H2N CH C
CH2
OH
O
OH
The alpha helices
and beta sheets can interact with each other to
form a larger 3-dimensional structure that has a
particular shape.
2. What is a catalyst?
3. What is an enzyme?
38
Putting It All Together
Work together in your chem/bio groups to answer the following questions. Make sure that EACH of
you knows how to answer the questions and is able to understand what you write down. This may NOT
be the last time you see these questions!!!
1. You have learned how the study of enzymes can depend on some chemistry principles. Using this
simpler example, answer the following questions.
From the 1999 AP Chemistry Test
2 NO(g) + Br2(g) � 2 NOBr(g)
A rate study of the reaction represented above was conducted at 25ºC. The data that were obtained
are shown in the table below.
Experimen
t
Initial
[NO]
(mol L–1)
Initial
[Br2]
(mol L-1)
Initial Rate of
Appearance of
NOBr (mol L–1 s–
1)
1 0.0160 0.0120 3.24x10–4
2 0.0160 0.0240 6.38x10–4
3 0.0320 0.0060 6.42x10–4
(a) Calculate the initial rate of disappearance of Br2(g) in experiment 1.
(b) Determine the order of the reaction with respect to each reactant, Br2(g)and NO(g). In each case,
explain your reasoning.
(c) For the reaction,
(i) write the rate law that is consistent with the data, and
(ii) calculate the value of the specific rate constant, k, and specify units.
(d) The following mechanism was proposed for the reaction:
Br2(g) + NO(g) � NOBr2(g) slow
NOBr2(g) + NO(g) � 2 NOBr(g) fast
Is this mechanism consistent with the given experimental observations? Justify your answer.
2. You have learned that the rate of enzyme-catalyzed reactions may depend on various factors. Think
about the conditions inside your body and list several other factors which may influence the rate of a
reaction inside your body.
3. It was mentioned that in the quaternary structure of an enzyme, the individual chains may be held
together by covalent or intermolecular forces. What are the intermolecular forces we have discussed
previously in this class?
39
AP Chem/Bio Enzyme Lesson
Introductory Worksheet Try to answer the following questions without the help of those in your group.
1. Next to each of the following descriptions or pictures, write one of the following:
Primary Structure
Secondary Structure
Tertiary Structure
Quaternary Structure
Several chains of
amino acids can come together to form a large 3-
dimensional structure which is made up of several
units held together by covalent or intermolecular
forces.
The flat arrows represent beta
sheets
The spirals represent alpha helices
This shows how sections of amino acids interact
with each other in long chains.
The sequence of amino acids that make up the
enzyme protein. Below are a few amino acids
that may be included in the sequence:
H2N CH C
CH2
OH
O
CH2
C
OH
O
H2N CH C
CH2
OH
O
OH
The alpha helices
and beta sheets can interact with each other to
form a larger 3-dimensional structure that has a
particular shape.
2. What is a catalyst? Something that speeds up a chemical reaction without being used up in
the process.
40
3. What is an enzyme? A biological catalyst. An enzyme is a protein in the body that helps
important biochemical reactions occur at an acceptable rate.
Putting It All Together
Work together in your chem/bio groups to answer the following questions. Make sure that EACH of
you knows how to answer the questions and is able to understand what you write down. This may NOT
be the last time you see these questions!!!
1. You have learned how the study of enzymes can depend on some chemistry principles. Using this
simpler example, answer the following questions.
From the 1999 AP Chemistry Test
2 NO(g) + Br2(g) � 2 NOBr(g)
A rate study of the reaction represented above was conducted at 25ºC. The data that were obtained
are shown in the table below.
Experimen
t
Initial
[NO]
(mol L–1)
Initial
[Br2]
(mol L-1)
Initial Rate of
Appearance of
NOBr (mol L–1 s–
1)
1 0.0160 0.0120 3.24x10–4
2 0.0160 0.0240 6.38x10–4
3 0.0320 0.0060 6.42x10–4
(a) Calculate the initial rate of disappearance of Br2(g) in experiment 1.
(b) Determine the order of the reaction with respect to each reactant, Br2(g)and NO(g). In each case,
explain your reasoning.
(c) For the reaction,
(i) write the rate law that is consistent with the data, and
(ii) calculate the value of the specific rate constant, k, and specify units.
(d) The following mechanism was proposed for the reaction:
Br2(g) + NO(g) � NOBr2(g) slow
NOBr2(g) + NO(g) � 2 NOBr(g) fast
Is this mechanism consistent with the given experimental observations? Justify your answer.
Answer
Note: Some of the equations show the infinity sign instead of the multiplication sign. My
apologies…I’ll have to reload my equation editor.
(a) Since the disappearance of 1 Br2 produces 2 NOBr, then the rate would be half as much or
rate = -1.62x10–4.
(b) With respect to Br2: rate = k [NO]m[Br2]n
41
In expt. 2 the [Br2] is twice the concentration in expt. 1, as well, the initial rate of expt. 2 is
twice the initial rate of expt. 1, while [NO] remains constant. Therefore, it is 1st order with
respect to [Br2], n = 1.
With respect to NO: rate = k[NO]m[Br2]1
expt. 2: 6.38x10–4 = k (0.0160)m(0.024)
k = (0.0160)m(0.0240)
6.38∞∞∞∞10–4
expt 3: 6.42x10–4 = k (0.0320)m(0.0060)
k = (0.0320)m(0.0060)
6.42∞∞∞∞10–4
(0.0160)m(0.0240)
6.38¥10–4 =
(0.0320)m(0.0060)
6.42∞∞∞∞10–4
solving: m = 1.97 or m = = = = 2
therefore, 2nd order with respect to [NO].
(c) (i) rate = k[NO]2[Br2]
(ii) k = rate
[NO]2[Br2]
= 3.24∞∞∞∞10–4 mol L–1 s–1
(0.0160 mol L–1)2(0.0120 mol L–1)
= 105 L2mol–2s–1
(d) No; since the rate determining step is the slowest step (and in this case, the first step), then
the rate for this proposed mechanism depends only on the cencentration of the reactants in
the first step and would be: rate = k[NO][Br2]
2. You have learned that the rate of enzyme-catalyzed reactions may depend on various factors. Think
about the conditions inside your body and list several other factors which may influence the rate of a
reaction inside your body.
Body temperature is fairly regular, but the reactions may be affected when the body has a fever
or changes temperature. pH values may affect reaction rates. Depending on where the reactions
are occurring, different concentrations of salts or other molecules from the diet may affect the
reaction rates.
3. It was mentioned that in the quaternary structure of an enzyme, the individual chains may be held
together by covalent or intermolecular forces. What are the intermolecular forces we have discussed
previously in this class?
Intermolecular forces (IMF) are the forces resulting from molecules interacting with each other.
*Hydrogen Bonds – the strongest of the IMF; the attraction for a very electropositive hydrogen
for the lone pairs of an electronegative atom. Results with H bonds to N, O, or F.
42
*Dipole-Dipole Forces – the attraction of polar molecules for each other (opposite ends, of
course)
*Dipole-Induced dipole – a polar molecule induces a dipole moment in a nonpolar molecule,
therefore causing an attraction
*London Disperson Forces – Temporary fluctuations in the electron clouds surrounding
molecules cause temporary dipole moments in molecules. These are then attracted.
(Those forces involving ions are studied under a separate classification.)
Literature Cited
1. Garrett, R.H. and Grisham, C.M., Principles of Biochemistry With a Human Focus, Brooks/Cole
and Thomson Learning, 2002.
2. Panzer, P., Preuss, U., Joberty, G., and Naim, H. J. Biol. Chem. 1998, 273, 13861-13869.
3. Ullmann, A., “Escherichia coli Lactose Operon”, Encyclopedia of Life Sciences, Nature Publishing
Group, 2001, www.els.net
4. Jacobson, R.H., Zhang, X-J, DuBose, R.F., and Matthews, B.W., Nature, 1994, 369, 761-766.
5. RSCB Protein Data Bank http://www.rcsb.org/pdb/Welcome.do ID# 1DP0
6. http://cathwww.biochem.ucl.ac.uk/cgi-bin/cath/SearchPdb.pl?type=PDB&query=1DP0
7. Juers, D.H., Jacobson, R.H., et.al., Protein Sci. 2000, 9, 1685-1699.
8. Plot rendered from http://wrpsun3.bioch.virginia.edu/fasta_www/grease.htm Kyte, J. and Doolittle,
R. F., J. Mol. Biol. 1982 157, 105-132.
9. Juers, D.H., Heightman, T.D., et.al., Biochemistry, 2001, 40, 14781-14794.
10. http://employees.csbsju.edu/hjakubowski/classes/ch331/transkinetics/kmkcatvalues.htm
11. http://www.ncbi.nlm.nih.gov/books/bv.fcgi?rid=stryer.table.1055
43
Enzyme Catalysis and Rates Project
The Excel Spreadsheet below allows you to enter substrate consentrations and initial rate data for an enzyme reaction whereupon the sheet will automatically plot a Lineweaver-Burk plot and calculate the enzyme kinetic parameters KM and Vmax.
What is needed is just to type in your set of values of substrate concentration [S] and rate v, and the data will be plotted and analyzed by the speadsheet. So, you just type in pairs of [S] and v values in columns A and B, starting with A2 and B2 and proceeding down as far as the number of data points that you have.
The present project is this:
Some representative data are given below, and what we want to do is to change the numbers slightly and see what the resulting values of the enzyme kinetic parameters KM and Vmax turn out to be. So your project is to take each number given and change it by an amount that might represent a typical magnitude of experimental error, and then put those new numbers into the Excel document. Then, print out and hand in a copy of the resulting Excel document with the graph and the values of KM and Vmax that are calculated from your new numbers. The results will be at least a little different for each one of you, and they will then be collected and a report on the differences found will be posted for everyone's edification.
Assume a plus-or-minus 1% error in the concentration values and a plus-or-minus 5% error in the initial rate values. Then, choose numbers that differ in what you think of as a random way from the values given below, and use those (and not the numbers below) in your spreadsheet. For example, if the value for [S] is 0.001, the number you choose should be between 0.00099 and 0.00101, which are 1% below and above the original value of 0.00100. If the value for v is 65, the number you choose should be between 61.75 and 68.25, which are 5% below and above 65.00 (since 5% 0f 65.00 = 3.25).
The idea here is to assume rather random errors in each measured value, so your values should be chosen more-or-less randomly to be some greater than and some less than the numbers given below. You just choose what you would think might be a representative set of differences from the data below, based on this idea of possible random experimental errors. The hypothesis will be that your choice might really represent the kinds of experimental errors that might be encountered if you repeated the experiment below and got a new set of values for yourself.
Here are the 10 sets of substrate concentration [S] and rate v, given as [S] (in M, i.e., mol/L) first, followed after the comma by v (in micromoles/min):
Table 1: My Values
[S], M v, µmol/min 1/[S], L/mol 1/v, min/µmol
1.0090E-04 66.0000 9.9108E+03 1.5152E-02
4.9900E-04 62.5000 2.0040E+03 1.6000E-02
9.9500E-05 51.8000 1.0050E+04 1.9305E-02
4.9900E-05 42.0000 2.0040E+04 2.3810E-02
3.0080E-05 33.9000 3.3245E+04 2.9499E-02
2.0100E-05 27.2000 4.9751E+04 3.6765E-02
9.9900E-06 16.9900 1.0010E+05 5.8858E-02
5.0010E-06 9.5050 1.9996E+05 1.0521E-01
9.9900E-07 2.2009 1.0010E+06 4.5436E-01
5.0020E-07 1.1000 1.9992E+06 9.0909E-01
0.001, 65 0.0005, 63 0.0001, 51 0.00005, 42 0.00003, 33 0.00002, 27 0.00001, 17 0.000005, 9.5 0.000001, 2.2 0.0000005, 1.1
44
Lineweaver-Burk Ploty = 4.4629E-07x + 1.3937E-02
0.0000E+00
1.0000E-01
2.0000E-01
3.0000E-01
4.0000E-01
5.0000E-01
6.0000E-01
7.0000E-01
8.0000E-01
9.0000E-01
1.0000E+00
0.0000E+00 5.0000E+05 1.0000E+06 1.5000E+06 2.0000E+06 2.5000E+06
1/[S ], L/mol
1/v, min/mmol
B )Linear (B
Intercept at 1/[S] = 0, equals 1/Vmax 1.3937E-02 Vmax = 7.1752E+01
Intercept at 1/v = 0, equals -1/KM -3.1196E+04 KM = 3.2056E-05
Class Average:
45
Dusty Carroll
Lesson Plan 5: Metabolism
Target Audience The lesson is created for an AP chemistry class during their unit on electrochemistry. They will have a
basic understanding of redox reactions, standard reduction potentials, the relationship between energy
and electric potential, equilibrium, and Gibbs Free energy. This is a single 42-minute class period.
Objectives • Describe the basic structure and function of the mitochondria
• Provide electrochemical explanations for the energy used to drive the synthesis of ATP
• Use a variation of the Nernst equation to calculate Gibb’s Free Energy
Introduction I. Use Overhead Transparency: Structure of the Mitochondria
Point out the following aspects of the structure:
• Approximately 1-10 µm long
• Consist of two membranes
o Outer is mainly protein and acts as a filter (filters out large molecules)
o Inner is much more selective, allowing only essential molecules inside
� Folded into “cristae” which increases the surface area of the membrane
• Higher surface area allows many proteins/enzymes to be embedded,
allowing for transport of essential molecules
o Inner membrane “protects” the matrix which contains most of the enzymes necessary to
generate ATP
• Between the membranes is the “intermembrane space”
o Several enzymes that use ATP are found here
II. Brainstorm: What is ATP?
Expected answers:
• Adenosine triphosphate
• Where energy is stored
• What gives us the energy we need to live
Teacher discussion:
• Structure of ATP is shown at right
• ATP transports energy within cells
• Release of one phosphate group = free energy
release of -30.5 kJ/mol
• ATP is generated from ADP by an oxidative
process to add a phosphate
• Since the reverse is exergonic, this process must
be endergonic (require energy)
• Where does that energy come from??
III. Use Transparency: Mitochondrial Electron Transport Chain
• Recognize the reactions shown in the diagram are redox reactions
• Enzyme complexes in the inner membrane catalyze a series of reactions
46
• This series of reactions causes the proton concentration in the matrix to decrease greatly, while
the proton concentration increases greatly in the intermembrane space
• This causes a proton gradient and an electrical gradient across the membrane
• The free energy related to this situation can be calculated using the Nernst equation (Kotz &
Treichel, page 973) and adding in a factor for the potential across the membrane (G&G page
537).
o pH is a convenient way to express proton concentration, so the equation becomes:
ψ∆ℑ+−−=∆ )(303.2 inout pHpHRTG
For the transport of protons from the matrix to the intermembrane space.
Where molVkJ */485.96=ℑ (Faraday’s constant) and ∆ψ = the potential difference across the
membrane (created by the + protons on one side and the – charge left on the other side).
IV. Summarize
• Which would be more acidic, the matrix or the intercellular space?
o The intercellular space because there are more protons
• Which would have a higher pH?
o The matrix (high pH = less acidic)
• Which way would protons flow, if they could, across the membrane?
o From the intercellular space through the inner membrane and into the matrix
• The flow of protons is restricted, so they are “stuck” in the intercellular space. The ATP
Synthase enzyme in the membrane can allow protons to flow into the matrix. Let’s say the
difference in pH is equal to one pH unit. A typical membrane potential, ∆ψ = 0.18 V. What is
the free energy change for the inward flow of protons? (Body Temp = 37oC)
molkJV
molVJK
molKJG 23)18.0)(996485)1)(310)(314.8(203.2 =+−−=∆
• Notice the ∆pH is negative because the pH inside the matrix is higher than that outside in the
intermembrane space.
• The 23 kJ/mol is unfavorable for the transport of protons out of the matrix. This means that the
free energy for the reverse direction is favorable. This energy is used to drive the synthesis of
ATP from ADP.
• Closing: ATP is a very efficient energy transporter, but the Law of Conservation of Energy
reminds us that the energy had to come from somewhere! (PS-it essentially came from the
food we ate.)
References
Garrett, R.H. and Grisham, C.M., Principles of Biochemistry With a Human Focus. Brooks/Cole &
Thomson Learning, 2002.
Kotz, J.C. and Treichel, Jr., P., Chemistry and Chemical Reactivity, 4th ed.. Saunders College
Publishing, 1999.
Image references are listed with the images.
47
Overhead Transparency
Structure of the Mitochondria
Image from
http://micro.magnet.fsu.edu/cells/mitochondria/mitochondria.html
48
Overhead Transparency
Mitochondrial Electron Transport Chain
http://en.wikipedia.org/wiki/Image:Etc2.png
49
Part C
Heredity is Molecular
http://www.answers.com/topic/evolution
50
DNA Sequence
51
DNA Sequence
52
DNA Structure
53
DNA Structure
54
DNA Structure
55
DNA Structure
56
Dusty Carroll
Lesson Plan 6: DNA to RNA How Protein Synthesis Works
Target Audience
AP chemistry class. I’ll be assuming they have the appropriate background knowledge of the structure
of DNA and RNA and proteins from their previous biology course. This lesson will focus on the more
chemical aspects of the protein synthesis process.
Objectives
• Recall the basic mechanisms of protein synthesis
• Explain the chemistry of the translation mechanism
Introduction to the Lesson
Use the “Reference for Nucleic Acids” handout
• Review the basic structure of nucleic acids
• Recall that DNA and RNA are similar in structure, but differ in one of their bases
• Review the basic structure of nucleotides
• Show the phosphodiester link
• Point out that continued polymerization causes a backbone of alternating sugars and phosphates
with bases as side chains
Use the worksheet “Overview of Protein Synthesis”
• Review structure of DNA
• Recall the 5 bases from previous handout
• Students draw proposed orientation of the bases for optimum hydrogen bonding
• Show overhead with correct orientation
• Review basic steps of protein synthesis
o Note that the complementary bases are such because of the chemical structure
Use the transparency “Translation: Chemical Aspects I”
• Show how the DNA stays in the nucleus and the mRNA takes the information out of the
nucleus to the ribosome.
• Notice how the mRNA has bases exposed (single strand, not double like DNA)
• The ribosome moves over the mRNA
• Each time this happens, a new amino acid is added to the peptide chain
• These amino acids are brought to the mRNA by tRNA
• The slide says, “Every amino acid is coded by a sequence of three bases” What does this
mean?
Use the transparency “Translation: Chemical Aspects II”
• A “codon” is a set of three nucleotide bases.
• Just as DNA has complementary bases, the three bases in the mRNA codon will have a
complementary codon that can interact with them. This is called the anticodon.
• The anticodon is found on tRNA
o tRNA binds to specific amino acids on one end and contains an anticodon on the other
end
o Show transparency “tRNA structure” – note the anticodon and the site of the amino acid
57
• When tRNA brings a new amino acid, an enzyme helps to form a peptide bond, attaching it to
the polypeptide chain
Use the transparency “Translation: Chemical Aspects III”
• This process is more clearly seen diagrammatically in this transparency
• Explain the process again and answer any questions that arise
Use the transparency “Translation: Chemical Aspects IV”
• This transparency shows the same process, but the diagram has some chemical structure to give
students an idea of how the amino acids continue to add to the polypeptide chain.
• Note that the bases are still drawn in cartoon form here, but remind students what they look like
(from handout and worksheet) and that they are complementary because of the hydrogen
bonding seen earlier
Summarize and Close
• Students previously accepted the explanation that the two strands of DNA stuck together
because of paired bases.
• They also accepted the explanation that mRNA magically copies the DNA information and
takes it away to make a protein.
• They now have a better understanding of the chemistry behind the magic!
References
Garrett, R. H., and Grisham, C.M., Principles of Biochemistry With a Human Focus. Brooks/Cole &
Thomson Learning, 2002.
McMurry, J. and Castellion, M.E., Fundamentals of General, Organic, and Biological Chemistry, 4th
ed. Pearson Education, Inc., 2003.
Image references noted within
58
Reference for Nucleic Acids
Remember that nucleic acids are polynucleotides.
A nucleotide is composed of three parts:
A sugar (5-membered monosaccharide)
A phosphate group
A nitrogen-containing cyclic compound that is a base
The 5 bases in DNA and RNA
N
NNH
N
NH2
Adenine
NH
NNH
N
O
NH2
Guanine
N
NH
NH2
O
Cytosine
NH
NH
O
O
Thymine
NH
NH
O
O
Uracil
The nucleotides polymerize in a specific way
Two nucleotides connected through a phosphodiester link
Long chains of nucleotides form DNA and RNA
Thymine is only found in DNA; Uracil is only found in RNA
The other three bases are found in both
O
OH
CH2OH
OH
OH
O
OH
CH2OH
OH
OH
ribose deoxyribose
or
P
O
O-
O-
N
NN
N
NH2
O
HO
HH
HH
PO
O-
O-
N
NH2
ON
O
HO
HH
HH
PO
O
HO
O-
59
Overview of Protein Synthesis
DNA exists as a double helix. Each strand contains a backbone of alternating sugars and phosphates.
The bases of the nucleotides stick out from the backbone. In DNA, the bases on opposite strands are
complementary because they form hydrogen bonds which hold the strand together.
Image from http://www.biologycorner.com/bio1/DNA.html
The complementary bases are oriented in such a way as to maximize the
hydrogen bonding capabilities. Using the structures in the “Reference for
Nucleic Acids” handout, determine the orientation that each pair of bases
must assume in order to maximize hydrogen bonding. Use the following
parameters to assist you:
• Guanine and Cytosine pair to form 3 hydrogen bonds
• Thymine and Adenine pair to form 2 hydrogen bonds
Draw your structures in the space below:
Guanine and Cytosine Thymine and Adenine
Steps for protein synthesis
• An enzyme breaks the hydrogen bonds between some base pairs in DNA in order to separate
the two strands
• Messenger RNA (mRNA) is assembled by enzymes to carry the DNA information out of the
nucleus and into the ribosome where protein synthesis can occur
• mRNA and transfer RNA (tRNA) work together along with enzymes to translate the DNA
information into a chain of amino acids (protein)
60
Overview of Protein Synthesis – Answer Key
DNA exists as a double helix. Each strand contains a backbone of alternating sugars and phosphates.
The bases of the nucleotides stick out from the backbone. In DNA, the bases on opposite strands are
complementary because they form hydrogen bonds which hold the strand together.
Image from http://www.biologycorner.com/bio1/DNA.html
The complementary bases are oriented in such a way as to maximize the
hydrogen bonding capabilities. Using the structures in the “Reference for
Nucleic Acids” handout, determine the orientation that each pair of bases
must assume in order to maximize hydrogen bonding. Use the following
parameters to assist you:
• Guanine and Cytosine pair to form 3 hydrogen bonds
• Thymine and Adenine pair to form 2 hydrogen bonds
Draw your structures in the space below:
Guanine and Cytosine Thymine and Adenine
N
N
N
N O
HN
N
N
NH
OH
H
H
sugar
sugar
N
N
N
NNH
N
N
O
O
H
H
sugar
sugar
Steps for protein synthesis
• An enzyme breaks the hydrogen bonds between some base pairs in DNA in order to separate
the two strands
• Messenger RNA (mRNA) is assembled by enzymes to carry the DNA information out of the
nucleus and into the ribosome where protein synthesis can occur
• mRNA and transfer RNA (tRNA) work together along with enzymes to translate the DNA
information into a chain of amino acids (protein)
61
Transparency
Translation: Chemical Aspects I
http://www.genomeeducation.ca/GEcurious/crashCourse/proteinsSynthesis.asp?l=e
62
Transparency
Translation: Chemical Aspects II
http://www.anselm.edu/homepage/jpitocch/genbio/translat.JPG
63
Transparency
tRNA
http://www.med.uottawa.ca/patho/devel/
64
Transparency
Translation: Chemical Aspects III
http://www.med.uottawa.ca/patho/devel/
65
Transparency
Translation: Chemical Aspects IV
http://www.med.uottawa.ca/patho/devel/