wed 10 19 05

Lecture 10/19
Test results are online. He looks at improvement to curve grades, so keep working hard. Problem set is posted and due next Wednesday.
• Chapter 6 Enzymes
• 3 ½ billion years- need enzymes catalyst to start protein building and life.
• Power of catalyst = we eat 20-40 pounds of ATP a day and enzymes can recycle this mass.
• Table 6-1 cofactors- metal or cations- help enzyme function. Bind reversible but usually sticks. Cofactors are not transformed by enzymes.
• Table 6-2 Coenzymes are also required for activity. It is a truly revisable interaction. Usually require more than just protein; dietary vitamins. NAD very much needed but usually not produced, need dietary precursor or nicotiniz acid (niacin) not a coenzyme. It has to be transferred into it by enzyme.
• Table 6-3 4,200 known genes of the E.Coili, 2,000 of then have been assigned to metabolic function. Given 3 numbers to for function.
• Figure 6-1 the substrate binds to an enzyme. The active site cleaves the internal peptide bond. (In the picture) The red spot is Amino R groups that are important to catalyst enzyme. The neighbor stick is called the active site. Most of the grey stuff is needed because if you mess with the structure the activity stops.
• Figure 6-2 Free energy chemical profile. Down hill is a chemically favored. Free energy barrier ΔG‡ S-> P
• Overhead Reaction velocity for unimolecule reaction:
Vcat = K{S} + Cuncat.
K is the rate constant. Cuncat is usually 0
K is large enough to ignore Cuncat
How do G‡ and K related? G‡ is the activation energy and the K is the rate.
From transition state theory
K inversely proportional to e^ΔG‡/RT
R is the gas constant, T is absolute K
• ΔG‡ p->S has a larger free energy. Can reach equilibrium
• Figure 6-3 Enzymes do not change chemically. Can accelerate catalyst to equilibrium. They lower the free energy barrier. Usually hills because of substrate interaction. The enzyme is unchanged.
• Table 6-5 Very effective in increasing the rates.
• Figure 6-5 (Fault to picture-creating 2 sticks forward should =backwards) A) energy is used to break stick. B.) Lock and key – poor –model- enzyme active site optimized binding to the stick, energy lowered first because of magnetic interaction, makes bigger a bigger energy to climb. C.) Use of transition state. Some energy for ES (enzyme substrate) complex, but the enzyme complements the shape, with catalyst bending, makes ‡ less dramatic, so it can happen faster.
• Figure 6-6 ΔG‡ for energy enzyme catalyst fashion does not mean it is the same pace in the backwards motion.
• Figure 6-7 There is entropy (chemical potential energy): linkage of bonds, entropy changes, changing the rates.
• Figure 6-8 Directed acid-base properties. In active site, give chemical properties. R groups can be changed by what is in their neighborhood. Can be helped by enzymes. To make stronger base, put close to acid. Make water by adding H, which makes it a stronger acid and a better leaving group. This guides the rate.

Midterm points are posted, and the ave is:

The average is: 56.7%
and the points are posted at:
http://bio.classes.ucsc.edu/bio100/E1scores.html
but I don't rem him ever saying whether or not he grades on a curve... if you remember him saying something about that, please post a comment below:)

High is 86%
Low is 28%

Here are the actual scores in decending order, without IDs...
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A different set of notes for sunday's review...

Sunday Review 10/16
(I am going to make this short, because I am a little on the tired side so I hope this helps.)
• Know amino acids: mainly L, know the structure and how they look, fisher projections, R is coming at you H is in the board, H matters where amino acids are located, understand importance of the R group position.
• Know the how to use acid base chemistry. pH=pKa + a/ha, understand titration curves and how to read them
• Understand bonds in the peptide
• Know 1 2 3 4 structure characteristics, the importance of H bonding, alpha helixes, beta sheets
• Hemoglobin! Inactive is T state, active is R state. Binds to O2, CO2, CO and BPG, understand fig. 5-10, fig 5-16 and fig 5-17
• Sugars, know how to change conformations, know how to make rings, (If it was LEFT UP to me, I would be DOWN RIGHT pissed that I had to learn this) remember this for positions of OH when going from chain to ring. If you look at the two points sticking of the ring (like 1 and 4) if they are in trans than it is beta, if they are cis than it is alpha. Know how linkage happens, pg 246, fig 7.16a.
• For nucleotides, know what he covered very well because it was very little, nucleotides are base, P and sugar; nucleosides is just base and sugar, know bases and basic structure and chemistry, fig 8.8, know Avery and the mouse experiment, know Hershey and Case and radioactive experiment, figure 8-33 thru 36
* I know this is short, I hope it helpful is some why. GOOD LUCK!!

Notes from Ananta's review...

Notes from Ananta's Sunday review....

Remember to review sections 1 and 2
Know the Henderson Haselback eq.

You only need to know the L amino acids, but you better know the basic structure
(amine, cooh, H, R)
Each R group effects pka, and it matters where they are oriented in the chain.
pka is at half way between the two (equal parts)
Know which R groups are basic/acidic/hydrophobic/philic...
The zwitterion is the ion at pi (net neutral)

Structure:
1' amino acid sequence
2' alpha helixes, beta sheets, beta turns, etc
3' how the 2' comes together: alpha-beta-alpha
4' how monomers come together and attract each other: dimers, trimers, etc

2' is stabilized by H bonds (fig 4-4, 4-7)
alpha helix: polarity (dipole moment)
H bonding occurs between N-H----O=C

Beta sheets: parallel vs. antiparallel scheme... par is less stable/antipar is more stable because it is head on.
beta turn: fig 4-20 only possible for anit-par

Hemoglobin....
very like myoglobin
stucture is slightly different
binds O2, CO2, CO, BPG.
See fig 5-10
ph is analogous to how much CO2 in the signodal curve. (more CO2 at 7.2)
H+ protonates Histadine (fig 5-9) (see fig 5-16 too)
BPG at higher alts (fig 5-17) pg 172
At higher alts you bind less O2, but BPG helps you release more of it at a time so you end up with more in your bloodstream.
Hemoglobin forms 4' structures.
Read up on sickle cell anema.

Sugars:
It is an L sugar if the C4 is up, and R if it is down:
If it was LEFT UP to me, I would be DOWN RIGHT angry to have to learn this.

To know if a linkage is an alpha or a beta linkage:
Look at the one with the anameric carbon, and if C 1 and 4 are cis then it is alpha, and if it is trans, it is beta. (look at homework prob 2)

Nucleotides:
8.1, 8.2, and some of 8.3... KNOW IT WELL (fig 8.34)
a nucleotide has: a base, a sugar, and a phosphate group...
a nucleoside only has the first 2 'sides': a base and a sugar

Be sure to understand the logic of the experiments he talked about.

Puridines: A, G: 2 rings (2 H bonds)
Pyrimidines: C, T, U: 1 ring (3 H bonds) (more stable)
(the above may not be correct... so look it up, and make sure you know it)

DNA vs RNA
Make sure yo ukonw the chemical instability that results in RNA self-hydrolysis. (fig 8-8 and 8-4 and section 8.3)

Look at 8-33 thru 8-36 (mutations)

Homework 3 help/hints/links...

A great place to get images:
http://www.steve.gb.com/science/molecules.html


1)



2)



3)NeuAC(2->6)GalNAc
http://marukin-bio.com/en/products/substra/nana_e.htm
http://www.bme.jhu.edu/~kjyarema/monosaccharides/monosaccharides2.htm
(you will have to take the break out of the next 2 links...)
http://www.mercksource.com/pp/us/cns/cns_hl_dorlands.jspzQzpgzEzzSzppd
ocszSzuszSzcommonzSzdorlandszSzdorlandzSzdmd_a_07zPzhtm#913343

http://www.mercksource.com/pp/us/cns/cns_hl_dorlands.jspzQzpgzEzzSzppd
ocszSzuszSzcommonzSzdorlandszSzdorlandzSzdmd_s_11zPzhtm



4)
a & c : http://oregonstate.edu/instruction/bb451/winter2005/stryer/ch31/Slide2.jpg
b: if you don't know this answer you are seriously behind...
d:http://digestive.niddk.nih.gov/ddiseases/pubs/lactoseintolerance/
http://www.galactosemia.org/student_new.htm
http://www.galactosemia.org/galactosemia.htm
e:
f:
g:


5)

Not-so-much a review from Eric's review (sat)

Review Saturday 15th
So the study group was only about an hour long and it was just the TA kind of guiding our studying focus. He did not go over any problems or concepts, just what to focus on. Hope this helps!
• Know the 20 amino acids and be aware of their properties. (You do not need to memorize Table 3-1 but be aware of how it works) Have a general idea of how and where they loss their H’s.
• Understand the backbone and how the resident form makes the bond between C and N almost a double bond, which restricts flexibility.
• Be aware of measurements, like 5.4A a helix turn and 3 residues a turn.
• Know fundamental stuff, not really odd-ball stuff, like proteins that do not fit the rules.
• Know pH, pKa, pI, and all that comes with that
• Understand disulfide bonds.
• Know parallel and anitparallel and how they are different. (H bonds)
• Understand the helix and how the amino terminus is positive.
• Understand the two types of beta turns(figure 4-8) and the stable folding patterns (figure 4-20)
• Know hemoglobin.
• Understand the basics of carbohydrates: ones to focus on: Glyceraldehyde, ribose, glucose, mannose, galactose.
• Understand disaccharides and how the form.
• Be aware of the difference between starch, glycogen, and dextrans.
• Know structures nucleotides and nucleic acids.
• Know bases names, not structure.
• Be aware of the measurements in the helix.
• Be aware of names and experiments.


* I just wanted to add that this was all off the top of his head. He really had no idea what to tell us to study. He was just looking through the chapters and pulling stuff out. So, while this is a guide, I personally believe the test will be a little harder than he was thinking; so don’t base all your studying off of this if you want to be safe. Good luck!

friday 10 14 05

10/14 Lecture
Midterm
The midterm will be multiple choice. 50 questions. You can use a calculator. Cover everything though up to chapter 8. Will give materials to take the test so you do not need to bring a scantron. There are two study sessions: Saturday 5pm at Earth and Marine B206 and Sunday 5pm Thimm 1.

• Figure 8-7 Phosphodiester linkage. Two active hydroxyl groups attack. The P is tri-functional, one is deprotenated. Nucleic Acids are polyanions and there is a negative charge in the chain under physical neutrality. They are very soluble so they have cations to help stabilize.
• Have polarity, 1 is 5’ the other is 3’, which helps place enzymes. Depending on how you view it: if from the P that the 3’ is up and the 5’ is down, from the sugar the 5’ is up and the 3’ is down. Scientists look from sugar to denote polarity.
• Read from 5’ to 3’. Important for the information that DNA contains.
• Figure 8-8 In RNA, the O attacks the P group and leaves a cyclic P group, breaking the chain. Can happen at a high pH and can break the nucleotides. Does not happen in DNA because it does not have that OH, so you need different conditions and pressures. This protects selected the information in DNA and makes is a better “hard disk”. RNA is important to carry information. The P ring is short lived and is attacked by water, breaking the ring.
• Figure 8-9 Bases are hydrophic molecule. They are planer and are pack together like lifesavers. Have multiple residents forms, as shown in the picture. They are pH dependent.
• Figure 8-10 absorbed UV light.
• Figure 8-34 Because they stack on top of each other, they are very close together. When Thymine is adjacent to thymine, UV light will cause it to form a bond structure. DNA has police enzymes running up and down and will call for repair enzyme to break the cyclobutane thymine ring and removes it entirely. DNA polymers will fill it with thymine but if it does not happen fast enough and a gap is left for replication, you get mutations.
• Figure 8-11 A classic picture of Watson and Crick. (Notice the nose?)
• Figure 8-12 isolated bacteria to make plates of growth. Showed that you could grow deadly bacteria then killed it and inject it and the mouse would live. If you took heat killed bacteria and put it with bacteria that was not deadly the mouse would die. They tested to see if it was protein or DNA. They did not know; our professor used a peal necklace as an example. They thought the pearls, or proteins were the main factor, but it was the string, or nucleotides, that were important. Protein clipping enzyme did not destroy the transformability but with enzymes that attacked nucleic acids the transformability was lost. This really established that DNA carried out heritability not proteins.
• Figure 8-13 Used radioactive labeling, P for the DNA and S for the protein. Used a blender to take off the shells and centrifuged the solution. Found empty protein shells outside, and label DNA in the bacteria.
• Figure 8-14 Use of x ray diffraction by Franklin helped (more like stealing her work) Watson and Crick find the double helix model.
• Figure 8-15 Paring of A to T and G to C held by hydrogen bonds (they guessed the bond) help them set up a double helix. They were very lucky.
• Figure 8-16 G to C is about 50% stronger because of the H bonds (shown on Figure 8-11). This gives a major and minor angle making major or minor grooves in the helix.
• Chains run in the opposite directions.
• Figure 8-17 With the structure understood, they could explain replication. The helix splits and the daughter cells form complementary stands.
• Figure 8-11 One period in the helix is about 36 A, longer in water. The diameter is about 20 A.
• Figure 9-18 a Shows the flexibility of the coil, especially on bond 4.
• Figure 8-19 B form is the one Watson and Crick found. There is an A form and Z form. Under less polar conditions, you remove water and the coil gets tighter, which is the A form. This is more important to RNA.
• Figure 8-25 RNA is single stranded and can fold in to 2nd structure.
• Figure 8-27 These folds are more complex; the strands are in A form.

Wednesday 10 12 05

10/12 Lecture
• Figure 7-11 (he does not like these forms because they do not show position, he likes chair conformations) Sugars can polymerize into long chains. They share a glycosidic bond. The active alcohol attacks the hemiacetal in a specific way to create a bridge. Water is produced and water can be added to break the bond; either case you need an enzyme catalyst.
• In the text, maltose should be writing as alpha-d-glucopyranosyl-(alpha1-4)-d-glucpyranose. The attack happens at carbon 1 and it is an axial attack from the equatorial beginning.
• Correction to figure 7-11. In a-D-glucose, the OH severs as the leaving group so it needs to be in the beta position. In the alpha position, it blocks the alpha attack.
• Maltose is a disaccharide and has a polarity because of the active hemiacetal, which is where a ring carbon is joined to two oxygen’s, one in the ring and one on the OH. The active hemiacetal act as a substrate to continue chains and is also called the reducing sugar.
• Figure 7-12 Lactose is a Gal(beta 1 -> 4)Glc. Sucrose has an attack form beta to an alpha giving us Glc(alph 1 -> beta2)Fru. Sucrose is not an active reducer because the hemiacetal is involved in the bridging. This makes sucrose stable disaccharides so that it can act as a transport sugar. Trehalose is an alpha alpha attack.
• Figure 7-11 If we were to form a beta bond, instead of an alpha, it would lead to different polymers that would be used as structural proteins like cellulous.
• Figure 7-13 can get different polymers, heterpolysaccharides and homopolysaccharides. They can be unbranched or have more then one bond which creates a branch.
• Figure 7-14 A lovely pictures of real life starch and glycogen, showing that polymers can have different forms and functions.
• Figure 7-15b You can get a branch point at alpha1->4 and alpha1->6 but alpha1->6 is very rare.
• Figure 7-15c showing the structure. As the chain extends, water excludes itself because it can interact better with itself, which cause the chains to adapt by making 2 and 3 structures with the help of internal H bonds. Water can form H bonds with the chains but it is not strong enough to keep water happy so it stays with its own kind.
• Figure 7-16 shows the importance of equatorial and axial position. One of the sugars is flipped 180 degrees in relation to its neighbor to help bonding/folding.
• Figure 7-19 Bonds can be twisted and coil up as long chains.
• Figure 7-21 They coil up and form 2 strain coils, which look like ribbons. This allows the maximum amount of H bonds to form, making dense forms of polysaccharides. It drives higher level structure.
• Figure 7-22 Mixed polymers with proteins and lipids, usually found in bacteria. On the green rings attached to the proteins, there is an active carboxyl group that makes what looks like a peptide bond. Can be used in bacteria to protect themselves from their own defense mechanisms.

Chapter 8: Nucleotides and Nucleic Acids
• Figure 8-1 Primary function of nucleotides is in the formation of them. They can be broken into three parts, the sugar which is the anchor, the phosphate, and the base.
• Figure 8-3a The OH is all on one side, which creates a lot of hindrance, so attack comes from carbon four, makes a 5 member ring that has less hindrance.
• Figure 8-1 In a chain, the link is between the phosphate and active oxygen.
• Active carbon 1 bond to the Nitrogen from the base. Enzyme catalyst reaction. Very stable.
• Figure 8-2 Pyrimidines are single, 6 member rings: 2 nitrogen (2 atoms apart). Planer molecule. Nitrogen 1 does the attacking. Purines are two rings: 4 nitrogen. Planer. Nitrogen 9 does the attacking. Both attacks occur as beta.
• Figure 8-3b Sugar ring not planer. 4 of the 5 are almost coplanar, the odd one is out of the plane, either 2 or 3 carbon and in either exo or endo form.
• Figure 8-4 You do not need to know the structure. These are the different nucleic acid for DNA and RNA.
• Figure 8-5 Nature throws little curveballs. These are used to self i.d. Helps forging DNA/RNA in cells, because they use this curveballs to mark their DNA so that they know when forging DNA enter their cell. It is a security check for them.
• Figure 8-7 5’->3’phosphodiester linkage. Needs a enzyme catalyst. Strung together very fast, 6,000 per min. Has polarity. Difference between DNA and RNA is the 2’ carbon, either H (DNA) or OH (RNA).

Monday 10 10 05

10/10/05 Lecture
Now we are starting chap 7, which is Carbohydrates, and because of time constraints we are only doing 7.1 and 7.2
HW is dues next Monday.
Next Monday is the midterm, more information about the test to come during the week.
• Carbohydrates are the most abundant biomolecules on earth. They are made by photosynthesis, in abundance, so much that 10^18 of carbohydrates are fixed every year.
• Carbohydrates provide energy, which can be stored for use later. They can also be used for cell walls.
• All carbohydrates share similar chemistry
• Figure 7-1a Simple sugars start at 3 carbons. Characterized by CH2O, number of carbons equal oxygen. Every carbon but one is in the form of an active hydroxyl group. One carbon is a carbonyl group. This has special chemical properties: when it is a terminal carbon it is called an Aldoses. When it is not it is called a Ketoses.
• Figure 7.1 b You can add carbons and make longer chains.
• Figure 7.1.c Used for RNA and DNA. The D means sugars are optically actively.
• Figure 7.2 The stereoisomers are mirror images of each other. The L form is usually not relevant biologically, a few expectations. D rotates clockwise to the light source. The two forms are nonsuperimposable.
• Figure 7.3a Active carbon with OH is on the left on the next to last carbon, or penultimate carbon, gives a D sugar. D family is relative common, L is rare. In the figure, the ones with their names in boxes are the common aldoses.
• Figure 7.3b These are the ketoses
• Figure 7-4 If you look at these three, glucose is what the others are compared to; the other two are called epimers. They vary by optical rotation on one carbon. *All are nonsuperimposable which can be a trick question on a midterm because it looks like you can on paper, don’t forget these molecules are in 3D. To make this rotation, it needs an enzyme called epimerace.
• Figure 7-5 The condensation of aldehyde or ketone. Alcohol is the nucleophile. E- is taken by the Oxygen, which makes hemiacetal / hemiketal. This is reversible, which can happen fast without catalysts. It is the carbon bonded to 2 oxygen that remains active so that another alcohol can attack making a stable acetal / ketal and water. The second half is slower so it needs more help for a catalysts then the first step.
• Figure 7-6 Rings can form among chains; it can produce alpha and beta isomers. It can attack its self because it the of the single bonds. If Carbon 4 or 6 attacks it creates rings with too much strain. If carbon 5 attacks, it gives a 6 member ring, 5 carbons, 1 oxygen. The alpha form has the OH down and H up while the beta form has the OH up and the H down. All three forms are in rapid equilibrium, called mutarotation. It is important because it will create slightly skewed amounts so you do not have 50/50 of the two forms. Enzymes are picky and will only react with 1 or the other, their docking and the placement of the H bond is dependent on which conformation it is.
• Figure 7-7 This ring forming makes pyran or furan.
• Figure 7-8 resulting rings are puckered makes chair conformations. Two possible chair forms; there is an equatorial and axial position. Axial is up and down, equatorial is out to the side. When OH is in the equatorial position, it is happier and the H likes to be in the axial position.
• Figure 7-9 You can usually play games with the 2 and 3 carbons with out affecting the function. Be aware of all the different members of the glucose family, and what makes them different.
• Figure 7-10 Sugars can be oxidized by 2 electrons can form ester by eliminating H2O. You need to have an enzyme.
• Figure 7-11 Stringing sugars together by glycosidic bonds. It is a condensation reaction. Alcohol is it the nucleophile again and attacks the hemiacetal (the carbon bonded to two oxygen.) Once again, there is the dilemma between the alcohol wanted to be basic and the hemiacetal wanted to be acidic, so you need an enzyme to help make a boat confirmation.