DOC PREVIEW
TAMU BICH 410 - Exam 1 Study Guide
Type Study Guide
Pages 6

This preview shows page 1-2 out of 6 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 6 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 6 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 6 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

BICH 410 1st EditionExam # 1 Study Guide Chapter: 1 – 4Chapter 1: The Facts of LifeI. H, O, C and N make up 99+% of atoms in the human body. Why is this?A. Covalent bond formation by electron pairing and the stability and versatility of these products B. What is bond energy?i. Energy required to break a bond. H-H, H-C, C-C and C-O bonds have relatively high bond energies meaning they are very stable bondsII. Weak forces maintain biological structure and determine biomolecular interactionsA. Weak forces are very important!! They may be “weak” but all of them working together can make for strong interactionsB. Weak forces restrict organisms to a narrow range of environmental conditionsi. Conditions such as pH, temperatureii. Very important for building proteins because they need to constantly be made and unmade which would be difficult with very strong bonds in placeIII. The last section of the chapter was about organelles in the cell which she did not stress. She said that there probably wouldn’t be a question from this section. Spend your time studying other things. Chapter 2: WaterI. Hydrophobic interactions are considered the “secret of life” what does hydrophobic mean?A. Water fearing…i. The way water interacts with things (with hydrogen bonds and hydrophobic interactions is what makes life the way it is)B. Nonpolar solutes interact differently with water by “organizing” iti. The H-bond network of water reorganizes to accommodate the nonpolar soluteii. This decreases entropy (disorder) because the system is becoming more orderediii. Amphiphilic molecules interact favorably with both polar and nonpolar environments. Amphipathic molecules contain both polar and nonpolar groups. These terms mean the same thing!! Example: fatty acids, alcoholsiv. Amphiphilic molecules in aqueous solution form MICELLESKeq= [H+][OH-]/[H20]pH=-log[H+] pH+pOH=14Ka= [H+][A-]/[HA]SUPER IMPORTANT: Henderson Hasselbalch equation:pH=pKa+log[A-]/[HA]C. Buffers are solutions that resist changes in pH as acid and base are addedi. Most buffers consist of a weak acid and its conjugate baseD. Why is water so special?i. Very good solvent for ionic and polar solutesii. Poor solvent for nonpolar substancesiii. Hydrophobic interactions=membranes form and the cellular nature of life is creatediv. High dielectric constant= forms ionsv. High heat capacity enables temperature regulationhttps://ecampusprod.tamu.edu/bbcswebdav/pid-1146621-dt-content-rid-7408414_1/courses/BICH.410.1511.M2/Water.pdfimportant link to understand water if you are still confusedChapter 3: ThermodynamicsI. TerminologyA. The system: the portion of the universe with which we areconcerned (eg.: the cell)-The surroundings: everything else- Isolated system cannot exchange matter or energy- Closed system can exchange energy- Open system can exchange either or bothB. Enthalpy,H,the total energyFree Energy,G,The energy available to do work.Entropy,S,Disorder-A system’s thermal energy that is unavailable for work.C. Standard state free energy change ΔG° is the value of delta G when temperature is 298 K, pressure at 1 atm, and concentrations at 1 M. i. Biochemical free energy change, ΔG°’, has to be at a pH of 7. Equations:ΔG= ΔH – TΔSΔG’ = ΔG°’ + RT ln[C][D]/[A][B]ΔGo’= -RT lnK’eqVery Important Concept:If ΔG = 0, the reaction is at equilibriumIf ΔG < 0, the reaction proceeds as writtenIf ΔG > 0, the reaction proceeds in the opposite directionII. Why are ATP and ADP such useful molecules??A. Large negative free energy change on hydrolysis is due to:• electrostatic repulsion (oxygens with negative charge)• stabilization of products by ionization and resonance• entropy factorsi. Hydrolysis relieves electrostatic repulsion and increases energy (favorable) Memorize this! Make sure you know which form is more stable, which is keto, which is enol and that the process is called tautomerizationHow can reactions in cells run against their thermodynamic potential? (formation of ATP)Reactions are driven in the thermodynamically unfavorable direction via coupling with highly favorable processesAlso a very important reaction!! Example of couplingChapter 4: Amino AcidsI. Structure and PropertiesA. All contain a central tetrahedral carbon atom i. There are 20 common amino acids each with a different R group or side chainii. They join to each other through peptide bonds via dehydration because water is releasediii. There are Non-polar amino acids, polar, uncharged amino acids, acidic amino acids, basic amino acids. There really isn’t a way around this, you have to memorize the 20 amino acids and their R groups.B. The alpha carboxyl always has a pKa of 2 and the alpha amino has a pKa of 9. i. Some amino acids have R groups with pKas that also need to be memorized. ii. Aspartic Acid, Asp, D: pKa (β-carboxyl) = 3.9iii. • Glutamic Acid, Glu, E: pKa (γ-carboxyl) = 4.3iv. • Histidine, His, H: pKa (imidazole) = 6.0v. • Cysteine, Cys, C: pKa (sulfhydryl) = 8.3vi. • Tyrosine, Tyr, Y: pKa (phenolic OH) = 10.1vii. • Lysine, Lys, K: pKa (ε-amino) = 10.5viii. • Arginine, Arg, R: pKa(guanidino group) = 12.5ix. • Serine, Ser, S: pKa (hydroxyl) = 13x. • Threonine, Thr, T: pKa (hydroxyl) = 13xi. Average 2 +9 and any pKas of the side groups to get pI xii. Example: (2+9+8.3+6)/4=6.325C. All amino acids are chiral but glycinei. L-amino acids predominate in natureii. R,S nomenclature is better esp for amino acids with 2 chiral centersD. The chapter talks about NMR and different technologies but she said that this wouldn’t be on the test.E. Peptide Bondi. Found in the trans conformationii. Has partial 40%double bond character==== important!!iii. Length of 0.133 nm which is longer than double bond but shorter than singleiv. the six atoms of the peptide bond group define a plane – the amide planeEach unit of an amino acid is called a residue.2 residues= dipeptide, 3 residues=tripeptide, etc.One polypeptide chain - a monomeric protein• More than one - multimeric protein• Homomultimer - one kind of chain• Heteromultimer - two or more different chainsF. Sequence of Amino Acidsi. Unique of every proteinii. Is a form of genetic informationiii. Is read amino to carboxyl terminusHighly recommend watching one or more of these links if you are confused about one or two topicshttps://ecampusprod.tamu.edu/bbcswebdav/pid-1174837-dt-content-rid-7463739_1/courses/BICH.410.1511.M2/KHAN%20Academy%20links.pdfTEST INFO:Ch 1-


View Full Document

TAMU BICH 410 - Exam 1 Study Guide

Type: Study Guide
Pages: 6
Documents in this Course
Load more
Download Exam 1 Study Guide
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Exam 1 Study Guide and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Exam 1 Study Guide 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?