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(if you scroll down - you will see the info!)
Monday, November 10, 2008
Saturday, November 8, 2008
Workshop #8
A. Oxygen is a waste product of non-cyclic photophosphorylation because of the breakdown of H2O during the process. H2O breaks down into O2 + H+ + e- and the e- is needed to power the proton pump and carry out the rest of the cycle.
B. The lifespan of red blood cells from athletes is less than that from a sedentary human because athletes breathe more toxic O2 which breaks down biomolecules.
C. The master must have made the soy sauce in an O2 free environment, as it is rich in lactic acid and ethanol. My soy sauce tasted "vinegary" because I made it in an environment where O2 is present and acetic acid was produced.
D. With freely isolated mitochondria, pyruvate, ADP and Pi in solution, ATP is produced. This is because the pyruvate is used to produce acetic acid for the citric acid cycle to produce NADH which powers ETS which produces ATP.
E. An experiment similar to D is conducted but without the presence of pyruvate and ATP is still detected. Why? There are still proton gradients that will still allow the process to continue for some time, but will eventually stop making ATP when the gradients run out of energy.
F. If the proton gradient is unrestricted, the proton concentration will go to Keq and there will no longer be a gradient. If there is no gradient then no ATP will be produced. Mitrochondria in these cells contain this extra complex as another way for electrical energy to generate heat and insulate an animal.
G. The molecules that were radioactive in the student's urine is the H2O. The radioactive O2 was used in a coupled reaction to form H2O in ETS.
H. NEED: ATP/sec + glucose/sec
100,000 PL/1,000 sec x 5 ATP/1 PL = 500 ATP/sec needed to support the generation of a new membrane.
1 glucose transporter/cell. So,
500 ATP/sec x 1 glucose/2 ATP = 250 glucose/sec needed to support that metabolic rate.
B. The lifespan of red blood cells from athletes is less than that from a sedentary human because athletes breathe more toxic O2 which breaks down biomolecules.
C. The master must have made the soy sauce in an O2 free environment, as it is rich in lactic acid and ethanol. My soy sauce tasted "vinegary" because I made it in an environment where O2 is present and acetic acid was produced.
D. With freely isolated mitochondria, pyruvate, ADP and Pi in solution, ATP is produced. This is because the pyruvate is used to produce acetic acid for the citric acid cycle to produce NADH which powers ETS which produces ATP.
E. An experiment similar to D is conducted but without the presence of pyruvate and ATP is still detected. Why? There are still proton gradients that will still allow the process to continue for some time, but will eventually stop making ATP when the gradients run out of energy.
F. If the proton gradient is unrestricted, the proton concentration will go to Keq and there will no longer be a gradient. If there is no gradient then no ATP will be produced. Mitrochondria in these cells contain this extra complex as another way for electrical energy to generate heat and insulate an animal.
G. The molecules that were radioactive in the student's urine is the H2O. The radioactive O2 was used in a coupled reaction to form H2O in ETS.
H. NEED: ATP/sec + glucose/sec
100,000 PL/1,000 sec x 5 ATP/1 PL = 500 ATP/sec needed to support the generation of a new membrane.
1 glucose transporter/cell. So,
500 ATP/sec x 1 glucose/2 ATP = 250 glucose/sec needed to support that metabolic rate.
Wednesday, October 22, 2008
Workshop #6
6B)
- diffuse ions out of the cell
- uni-directional transporter wants equilibrium in order to maintain cell's volume
- phosphoenolpyruvate has the highest - delta G value
- an enzyme with binding sites for ADP and Pe-P
- radio label phosphate and filtrate
- enzymes work both ways depending on concentration
- add radio labeled glucose into the cell without ATP
- wait
- filtrate and see if glucse is present:
-if gluscose is present, passive
-if glucose is not present, active
6F)
- Change polymerase binding sites from RNA to DNA
6G)
- Double stranded DNA needs more energy
6H)
- heating breaks H-bonds between bases
- cooling reforms the double strands
6I)
- its easier to find the codes
Sunday, October 12, 2008
Workshop #5, w/solutions
Here is workshop # 5 and its guide. Use the guide as a reference, DO NOT study from it. Use your notes and olek's slides - they will be much more useful and informative.
Workshop #4 w/ solutions
Hey guys, here is the workshop, and Caitlin's solutions. Thanks Cait!
Workshop #4 Notes
4B) Free Nucleotides + Polynucleotide => Free nucleotides (monophosphates- AMP,GMP, UMP) link together by expending ATP (or any other tri-phosphate) to form a phosphodiester covalent bond. Weak hydrogen bonds form between the base pairs (3 H bonds between A and U or T, 2 btwn C and G) of the polynucleotide and the newly-forming polynucleotide. Template-directed RNA would help to stabilize the reaction and make it occur faster.
4C) A "cell-free" system = E.Coli without membranes in a test tube
How do we get polypeptides?
Amino Acids bind to tRNA with the expended energy of ATP or another high-energy intermediate
tRNA's anti-codon (3 nucleotide bases) bind to their specific codon (corresponding base pairs) on an mRNA sequence
This process re-occurs with another tRNA binding to mRNA, and the amino acids (attached at the top of the tRNA) are attached by a peptide bond, generating a polypeptide
4D) Codons on mRNA, Anti-codons on aa-tRNA/tRNA
Also, know that you should associate the amino acid with the correct CODON on mRNA.
4E) AUGUUUGGGCCCUUU => Met- Phe- Gly- Pro- Phe
Sequence of mutant mRNA giving rise to the mutant amino acid sequence (shown below)
AUGUUUAGGCCCUUU=> Met- Phe- Arg-Pro- Phe - There is an A nucleotide base instead of a G.
Mechanism of Mutation:
-Know that transcription is template RNA (DNA) to mRNA and translation is mRNA that codes for a protein.
One possible cause for mutation: Transcription error (different base pair codes for a different amino acid)
Another cause for mutation: One of the base pairs is not included in transcription, and the entire protein sequence changes.
4F) -Precursors for the formation of Phospholipids: phosphoglycerol and fatty acids
-As soon as you make a phospholipid, it is integrated into the protocell boundary membrane.
-Lack of an enzyme or inhibited enzyme that separates the phospholipids will limit the growth of the membrane.
4E) For this to occur, : rate of RNA replication > rate of phospholipid formation
- There must be enough RNA produced before it splits, so that there's enough for both daughter cells
Friday, October 10, 2008
Friday, September 26, 2008
Workshop #3 Solutions
3B. Looking at the diagram on the back, one would assume that glutamic acid would be more soluble in water than alanine (because of polarity or the fact that it says "hydrophillic/phobic" in the tables. Note: Olek says those are words to avoid...). After careful consideration, however, we can see that alanine is more soluble in water for a rather simple reason- it's size. Glutamic acid is bigger, so it would have to break way more water-water bonds to dissolve in water than alanine would. This means dissolving alanine in water is more favorable in terms of deltaG because it takes more energy to break the water bonds for glutamic acid.
3C. 1) There are MANY configurations/conformations/tertiary structures (all the same thing). But only the most stable of the structures act as an enzyme. (Derek aside: what are other random things that can affect the structures? temperature, charges, pH).
2) We would have to know BDEs and relative number of conformations in order to predict the largest ratio.
3D. Sucrose (models on BIO 110 site) {but is a glucose attached to a fructose}.
3E. There are many different conformations for the enzyme, sucrase. The enzymes that are the most stable tend to fit nicely around a sucrose molecule. The sucrase molecule sneaks up on the sucrose molecule and wraps itself around it. How? There are charges on both molecules that attracts them together. Sucrase essentially "pulls" on the attached glucose and fructose molecule, stretching out the bond in between them. This makes the bond weaker so a "less crazy water molecule" can hit it and possibly break it. Now, not only can the "crazier" water molecules that were able to break sucrose before, break the sucrose, but so can the "less crazy." This, in effect, increases the reaction rate of sucrose hydrolysis forming glucose and fructose.
3F. Filter! You could test different enzymes on molecules that you know the size of (and the size of what they would break up into). This way, you could filter out the broken up molecules from the not-yet-broken-up molecules. You can compare the TIME that it takes for each enzyme to break these molecules.
3G. What you would need to know about these polypeptide chains is that they have charges going along the chain. When they roll up and configure the way that they do, the + and - charges on the chain tend to line up together (because opposites attract). These + and - charges are present on H+s and OH-s. Knowing this, you could engineer the enzyme to effectively wrap around a molecule.
3H. Yes, it is important! Some proteins are more soluble in water than others because of their size, polarity, etc... You can make a protein relatively insoluble in water but more soluble in a lipid by make it nonpolar. Lipids are nonpolar, so they'd go together great.
Wednesday, September 17, 2008
Workshop # 2 Answers
A) Check out Workshop #1 Post
B) The model of ATP can be found on Blackboard (Make sure you have three phosphate groups).
C) Calculate the delta G (AKA Free Energy) for ATP + H2O ---> ADP + Pi given the following starting P/R ratio values in 4 different muscle cells (MC):
Note that delta G0' (AKA G naught) is given as -30 kJ/mole and the formula for delta G is:
delta G = delta G0' + [(6 kJ/mole) log10(P/R)]
MC #1) P/R=1.0 x 10^-5
delta G = -30 kJ/mole + [(6 kJ/mole) log10(1.0 x 10^-5)] = -60 kJ/mole
MC #2) P/R=1.0 x 10^2
delta G = -30 kJ/mole + [(6 kJ/mole) log10(1.0 x 10^2)] = -18 kJ/mole
MC #3) P/R=1.0 x 10^5
delta G = -30 kJ/mole + [(6 kJ/mole) log10(1.0 x 10^5)] = 0 kJ/mole
MC #4) P/R=1.0 x 10^7
delta G = -30 kJ/mole + [(6 kJ/mole) log10(1.0 x 10^7)] = 12 kJ/mole
1) The muscle cells that are able to do any work at all are MC #1 and MC #2 because their delta G is negative.
2) The strongest muscle cell is MC #1 because it has the largest negative value, thus it has more free energy to expend.
D) The structure of glycine and di-glycine can be found on Blackboard in the "memorize" section.
If the mass of O =16, N = 14, C = 12, and H = 1, the total mass of a molecule of glycine would be 75.
Note that glycine is a free amino acid, and amino acids make up proteins. So, di-glycine is the simplest polypeptide, which means that it's the simplest protein.
The structure of a di-nucleotide can be found with all the other structures that need to be memorized. Remember that you add the second nucleotide on to the C3'. The phosphodiester link should look like this : C3' - O - P - O - C5. Also, don't forget that the bases are attached to the C1.
E) The Data can be found in the Workshop #2 handout. Without an enzyme, the reaction is very slow and we notice no change in the first few minutes. With an enzyme added to the solution, we see that the mass of the molecule starts to change fairly quickly. This is because the enzyme is helping to break the bonds of the polypeptide. The half-life of the polypeptide with the enzyme present is about 60 seconds becuase we see that the original polypeptide has only about half of its original mass at this point.
F) In this case, we are given a high concentration of ATP and AMP in a solution. We need to figure out how 4 AMPs will link together to create a polynucleotide. In order for bonds to form between two AMPs, we need energy. ATP is going to provide us with this needed energy. When a water molecule with enough kinetic energy (KE) collides with a molecule of ATP, it will break off a phosphate group and release energy. This release of energy is then used to create the bonds to connect two AMPs together. This process must be repeated until we are left with four AMPs (nucleotides) stuck together.
Since we are only given ATP and AMP to work with at this point in our experiment, the only possible sequence of bases we can have in any given nucleotide is: A - A - A - A.
If GTP and GMP were added to the solution as well, we would be capable of having any combination of As and Gs in our nucleotide. Ex: A - G - G - A
The paragraph from the text is trying to explain that ATP and GTP are energy-rich molecules, and that when their bonds are broken, energy is released into the system.
G) In order to understand this part, it is important to first understand the diagram showing how much KE water molecules have. Some water molecules have a low KE and some have a high KE. The amount of KE required to break the bonds of most biomolecules is fairly high, so only a handful of water molecules will have enough KE to break these bonds. The average amount of KE that water molecules have (1 x 10^-20 J/molecule) is much lower than the KE required to break bonds.
If the newly discovered biomolecule really had an activation energy (G*) of 1 x 10^-20 J/molecule (which is the average for water) then its bonds would be constantly breaking. This is ridiculous becuase we would never be able to use this biomolecule in our bodies because we're mostly made up of water. So, whatever you do, don't believe this report!
H) 500 A proteins and 500 B proteins are put into a solution. After 10 minutes, a lot of the A and B proteins have combined together to form complexes. 450 A-B complexes are found as well as 50 free A molecules and 50 free B molecules. The ratio does not change anymore.
The experiments you would use to generate the data would be radio labeling and then filtration.
The free energy (delta G) of complex formation must be negative because we have way more products than reactants when the solution has reached equilibrium. P/R > 1 thus delta G is negative.
I) Good luck on the exam everyone.
B) The model of ATP can be found on Blackboard (Make sure you have three phosphate groups).
C) Calculate the delta G (AKA Free Energy) for ATP + H2O ---> ADP + Pi given the following starting P/R ratio values in 4 different muscle cells (MC):
Note that delta G0' (AKA G naught) is given as -30 kJ/mole and the formula for delta G is:
delta G = delta G0' + [(6 kJ/mole) log10(P/R)]
MC #1) P/R=1.0 x 10^-5
delta G = -30 kJ/mole + [(6 kJ/mole) log10(1.0 x 10^-5)] = -60 kJ/mole
MC #2) P/R=1.0 x 10^2
delta G = -30 kJ/mole + [(6 kJ/mole) log10(1.0 x 10^2)] = -18 kJ/mole
MC #3) P/R=1.0 x 10^5
delta G = -30 kJ/mole + [(6 kJ/mole) log10(1.0 x 10^5)] = 0 kJ/mole
MC #4) P/R=1.0 x 10^7
delta G = -30 kJ/mole + [(6 kJ/mole) log10(1.0 x 10^7)] = 12 kJ/mole
1) The muscle cells that are able to do any work at all are MC #1 and MC #2 because their delta G is negative.
2) The strongest muscle cell is MC #1 because it has the largest negative value, thus it has more free energy to expend.
D) The structure of glycine and di-glycine can be found on Blackboard in the "memorize" section.
If the mass of O =16, N = 14, C = 12, and H = 1, the total mass of a molecule of glycine would be 75.
Note that glycine is a free amino acid, and amino acids make up proteins. So, di-glycine is the simplest polypeptide, which means that it's the simplest protein.
The structure of a di-nucleotide can be found with all the other structures that need to be memorized. Remember that you add the second nucleotide on to the C3'. The phosphodiester link should look like this : C3' - O - P - O - C5. Also, don't forget that the bases are attached to the C1.
E) The Data can be found in the Workshop #2 handout. Without an enzyme, the reaction is very slow and we notice no change in the first few minutes. With an enzyme added to the solution, we see that the mass of the molecule starts to change fairly quickly. This is because the enzyme is helping to break the bonds of the polypeptide. The half-life of the polypeptide with the enzyme present is about 60 seconds becuase we see that the original polypeptide has only about half of its original mass at this point.
F) In this case, we are given a high concentration of ATP and AMP in a solution. We need to figure out how 4 AMPs will link together to create a polynucleotide. In order for bonds to form between two AMPs, we need energy. ATP is going to provide us with this needed energy. When a water molecule with enough kinetic energy (KE) collides with a molecule of ATP, it will break off a phosphate group and release energy. This release of energy is then used to create the bonds to connect two AMPs together. This process must be repeated until we are left with four AMPs (nucleotides) stuck together.
Since we are only given ATP and AMP to work with at this point in our experiment, the only possible sequence of bases we can have in any given nucleotide is: A - A - A - A.
If GTP and GMP were added to the solution as well, we would be capable of having any combination of As and Gs in our nucleotide. Ex: A - G - G - A
The paragraph from the text is trying to explain that ATP and GTP are energy-rich molecules, and that when their bonds are broken, energy is released into the system.
G) In order to understand this part, it is important to first understand the diagram showing how much KE water molecules have. Some water molecules have a low KE and some have a high KE. The amount of KE required to break the bonds of most biomolecules is fairly high, so only a handful of water molecules will have enough KE to break these bonds. The average amount of KE that water molecules have (1 x 10^-20 J/molecule) is much lower than the KE required to break bonds.
If the newly discovered biomolecule really had an activation energy (G*) of 1 x 10^-20 J/molecule (which is the average for water) then its bonds would be constantly breaking. This is ridiculous becuase we would never be able to use this biomolecule in our bodies because we're mostly made up of water. So, whatever you do, don't believe this report!
H) 500 A proteins and 500 B proteins are put into a solution. After 10 minutes, a lot of the A and B proteins have combined together to form complexes. 450 A-B complexes are found as well as 50 free A molecules and 50 free B molecules. The ratio does not change anymore.
The experiments you would use to generate the data would be radio labeling and then filtration.
The free energy (delta G) of complex formation must be negative because we have way more products than reactants when the solution has reached equilibrium. P/R > 1 thus delta G is negative.
I) Good luck on the exam everyone.
Thursday, September 11, 2008
Workshop 1 solutions/notes
Part A: Structures can be found under "Memorize" on blackboard.
Part B:
1) Radio-labelling: When the reaction has reached equilibrium, count radio-labelled molecules after filtering. Measured amounts are directly proportional to the amount of reactants/products in solution.
For PNP reaction, wait at least 48 hours for reaction to reach equilibrium, unless an enzyme is present, in which case wait appx. 30 seconds-1 minute.
2) If Keq>1, then there are more products than reactants (when reaction has reached equilibrium) -- Keq= [P]/[R] = [ADP][Pi]/[ATP]
Keq= 1.0x10^5 >> 1, so there are much more products than reactants, and delta G (dG) will be negative for this reacton.
Work (max) = delta G
dG=(-5.75 kJ/mol)log(Keq)
Part B:
1) Radio-labelling: When the reaction has reached equilibrium, count radio-labelled molecules after filtering. Measured amounts are directly proportional to the amount of reactants/products in solution.
For PNP reaction, wait at least 48 hours for reaction to reach equilibrium, unless an enzyme is present, in which case wait appx. 30 seconds-1 minute.
2) If Keq>1, then there are more products than reactants (when reaction has reached equilibrium) -- Keq= [P]/[R] = [ADP][Pi]/[ATP]
Keq= 1.0x10^5 >> 1, so there are much more products than reactants, and delta G (dG) will be negative for this reacton.
Work (max) = delta G
dG=(-5.75 kJ/mol)log(Keq)
dG=(-5.75)log(1.0x10^5)
dG=-28.75 kJ/mol x 1 mol ATP= -28.75 kJ = W(max)
Part C:
1) Given W=10 J to lift weight once **Coupling: Rxn able to do work (release kJ/mol)
W(max) = dG = -50 kJ/mol --> -dG rxn does POSITIVE work
(5.0x10^4) J / 1 mol =10 J / x mol
x = 2.0x10^-4 mol ATP
2)(2.0x10^-4 mol ATP) x (6.02x10^23 molecules)/1 mol = 1.204x10^20 molecules ATP
Part D:
1 cal = 4.19 J, 10 J of work done to lift once
1 cal/4.19 J x 10 J/1 lift = 2.3 cal/lift
2000 Cal/1 day x 1 lift/2.3 cal x 1000 cal/1 Cal = 8.7 x 10^5 lifts/day!
Part E: Atoms in bonds in wood are holding the table together enough to exert Normal force and oppose gravity. Measure bond strength (in Newtons) by adding more mass until table breaks.
dG=-28.75 kJ/mol x 1 mol ATP= -28.75 kJ = W(max)
Part C:
1) Given W=10 J to lift weight once **Coupling: Rxn able to do work (release kJ/mol)
W(max) = dG = -50 kJ/mol --> -dG rxn does POSITIVE work
(5.0x10^4) J / 1 mol =10 J / x mol
x = 2.0x10^-4 mol ATP
2)(2.0x10^-4 mol ATP) x (6.02x10^23 molecules)/1 mol = 1.204x10^20 molecules ATP
Part D:
1 cal = 4.19 J, 10 J of work done to lift once
1 cal/4.19 J x 10 J/1 lift = 2.3 cal/lift
2000 Cal/1 day x 1 lift/2.3 cal x 1000 cal/1 Cal = 8.7 x 10^5 lifts/day!
Part E: Atoms in bonds in wood are holding the table together enough to exert Normal force and oppose gravity. Measure bond strength (in Newtons) by adding more mass until table breaks.
Wednesday, September 10, 2008
But this is...
Welcome to BIO 110! This page is for you, as is workshop...so abuse it! Use this as a space to write all your questions, answer other people's questions, vent and freak out the night before the test...really its for whatever you need it for. It is intended to be an exam study helper, as the notes for ALL the workshops will be here. I am also hoping to post the actual workshops, so as you are studying you can refer to them directly, and then see the solutions on the blog. If you have any feedback or ideas for this blog PLEASE let me know - i am very big into making this your site. Good luck!
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