BIS102 - Objectives Study Guide

Interpret data on cellular components to hypothesize what major biomolecules are present in e. coli.

1. H20 (70% of weight)

2. Ions = Na, K, SO4, etc. (1% of weight)

3. Carbohydrates (3% of weight)

4. Amino Acids (0.4% of weight)

5. Nucleotides (0.4% of weight)

6. Lipids (2% of weight)

7. Proteins (15% of weight)

8. RNA (6% of weight)

9. DNA (1% of weight)

Connect the versatility of carbon bonding and free rotation of single c--c bonds to diversity in molecular structures.

Identify and name the biologically important functional groups presented in the lecture slides.

Main Functional Groups to Remember

1. Methyl (-CH3)

2. Ethyl (-CH2-CH3)

3. Phenyl (-CCHCHCHCHCH, ring)

4. Carbonyl (Aldehyde, -CH=O)

5. Carbonyl (Ketone, -CR=O)

6. Carboxylate/Carboxyl (-COO-)

7. Hydroxyl (-OH)

8. Amino (protonated, -NH3+)

9. Phosphoryl (-OPOHOO-)

10. Sulfhydryl (-SH)

11. Disulfide (-SSR)

12. Thioester (-C=OSR)

Name the four major classes of large biomolecules in cells.

1. Proteins

2. Polysaccharides

3. Nucleic Acids

4. Lipids (Not necessarily a macromolecule, not a polymer, but very important)

Compare the properties of water and other common solvents and describe the molecular reason behind the differences.

In comparison to other solvents:

- Water has a higher melting point

- Water has a higher boiling point

- Water has a higher heat of vaporization

This is a result of the Hydrogen Bonding between water molecules, as a greater amount of heat/energy is required to break these bonds.

Compare and contrast H bonds between water molecules and covalent bonds within water molecules.

Hydrogen Bonds Between Water Molecules

- Electronegativity of Oxygen atom induces a dipole

- Intermolecular attraction is due to electronegative structure of water

- Longer & Weaker than Covalent Bonds (take ~20KJ/mol to break, whereas CBs take ~470KJ/mol to break)

Discuss how the structure of water contributes to the cohesion of liquid water.

Attraction Between Adjacent Water Molecules = Great Internal Cohesion

- Due to electronegative structure of water

- Water is Both an electron ACCEPTOR/DONOR

In Liquid Water:

- H20 molecules are disorganized and are in constant motion

- "Flickering Clusters" = Short lived groups of H20 molecules interlinked by H-bonds

Diagram a water molecule's interactions with its neighbors.

Compare and contrast interactions between water molecules in liquid vs ice.

Water Molecules in Ice:

- Each H20 molecule is fixed in space

- Regular Crystal Lattice Structure (Why ice is less dense than liquid water...ice can float in liquid water)

Water Molecules in Liquid:

- One H20 molecule interacts with 3.4 other molecules (on AVG)

- H20 molecules are disorganized and are in constant motion

- "Flickering Clusters" = Short lived groups of H20 molecules interlinked by H bonds in water

Describe the atoms that can interact in hydrogen bonding and in what context they exist in.

Hydrogen Bonds are NOT unique to H20

Can Occur Between:

- The Hydroxyl group of an alcohol and water

- The Carbonyl group of a ketone and water

- Peptide groups in polypeptides

- Complementary bases of DNA

- Molecules with an existing/induced dipole moment

Can NOT occur Between:

- Two H atoms (both partially positive)

- Between molecules that do not have an existing/induced dipole (ex: C-H, no dipole)

Compare and contrast hydrophobic, hydrophilic, and amphipathic molecules.

Hydrophilic = Compounds that dissolve easily in water; Contains C-H & -OH groups (Polar)

Hydrophobic = Compounds that do NOT dissolve easily in water; Mostly C-H Chains (Nonpolar)

Amphipathic = Contains BOTH polar and non polar groups (hydrophilic & hydrophobic components of compound)

Describe how water can so easily disrupt the bond between Na and Cl in table salt when these ions are so hard to separate in air.

Water Breaks the Bond Btwn. Na and Cl Because:

- It shields the ions charges so they can separate

- It interacts electrostatically with charged solutesO

- Partially negative oxygen atom interacts with the Na+ (non-random orientation of water molecules)

- Partially positive hydrogen atom interacts with the Cl- (non-random orientation of water molecules)

Explain why biologically important gases like oxygen and carbon dioxide are not soluble in water and discuss how we transport these gases in our water-based bodies.

Oxygen & CO2 = Nonpolar Gases (do not easily dissolve in water, only in tiny amounts)

How We Move O2 & CO2 in our Aqueous System

- Molecular Rearrangment

Ex: CO2 can change into a form that is H20 Soluble CO2 --> H2CO3 --> HCO3- (very soluble bicarbonate ion)

- H20 Soluble Carrier

Ex: Hemoglobin

Describe four weak interactions.

1. Hydrogen Bonding = Occurs only when a dipole moment exists between molecules

2. Hydrophobic Interactions = Repulsion between hydrophobic molecules and water; NOT an attraction

3. Ionic Interactions = Between charged atoms or groups; Attraction (primarily) or Repulsion

Ex: -NH3+ ||| -OOC-

- Strength depends on the environment of the ions and the distance between them

- "Salt Bridges" in proteins

4. Van Der Waals Interactions = When 2 uncharged atoms are very close

- As nuclei get closer, they get repelled by their e- clouds

- Random variations in which the e- can create a transient dipole --> Transient dipole induces an opposite transient dipole in a nearby atom --> Weak Interaction

- Net Attraction = Maximal contact at a certain point --> VDW contact

Explain the molecular basis of hydrophobic interactions.

When a Hydrophobic Molecule Enters Water:

- Highly order H20 molecules form "cages" around the hydrophobic alkyl chains

- To reduce order: Nonpolar regions cluster together to present the smallest area to the aqueous environment (To be in line with the 2nd Law of Thermodynamics)

- Results in Greatest Thermodynamic Ability.

State the Gibbs free energy equation and describe its components.

G = Change in Free Energy; Amount of energy available to do work; Rxn at constant T & P

- Units = J/mol

H = Change in enthalpy; Reflects the kinds/#'s of chemical bonds & non-covalent interactions broken/formed; Heat content of reacting system

- Units = J/mol

T = Temperature

- Units = Kelvin

S.= Entropy Change; Change in randomness/disorder

- Units = J/mol k

What is a spontaneous reaction and how do the components of the Gibbs free energy equation contribute to spontaneity?

Spontaneous Reaction Happens:

- w/o NET input of energy

- when energy is released

- in the forward direction

- exergonic (not necessarily exothermic)

- favorably

- not necessarily quickly

- (-)deltaG = rxn happens for sure, but no information on the pathway, mechanism, or speed/rate

- could be irreversible if deltaG is large enough

Gibbs Free Energy Equation:

- (-)deltaG = Exergonic

- (+)deltaG = Endergonic

- (+)deltaH = Endothermic

- (-)deltaH = Exothermic

- (+)deltaS = Entropy is Increasing (Rxn increases randomness/disorder)

Compare and contrast the following terms: endergonic, exergonic, endothermic, exothermic.

Endergonic: Requires free energy - {(+)deltaG}

Exergonic: Releases free energy - {(-)deltaG}

Endothermic: Requires enthalpy; The reacting system takes up heat from the surroundings - {(+)deltaH}

Exothermic: Releases enthalpy; Reaction releases heat; Products have less heat than reactants - {(-)deltaH}

Draw curves for an exergonic and an endergonic reaction on a graph where the x-axis is the reaction coordinate and the y-axis is free energy.

If two reactions in a cell are coupled, describe how one would calculate the total free energy change of the two reactions.

Endergonic and Exergonic Rxns are Couple

- Sum of the free energy change

Ex: Three Reactions (R1, R2, R3)

R1 = deltaG1, R2 = deltaG2, R3 = deltaG3

R1 +R2 --> R3, SO...

delaG3 = deltaG1 + deltaG2 (additive)

Calculate a reaction's Keq given concentrations of the products and reactants at equilibrium.

Be able to determine if a set of concentrations for products and reactants are representative of equilibrium conditions given Keq for the reaction.

Keq = [products]eq/[reactants]eq

Ex: ATP --> ADP + Pi

Measured concentrations are:

[ATP] = 5mM

[ADP] = 0.5 mM

[Pi] = 5 mM

Keq (given) = 2 x 10^5 M

Keq (calculated) = [ADP][P]/[ATP] = 0.5mM(5mM)/5mM = 0.5mM = 5 x 10^-4 M < Keq (given), SO far from equilibrium

Compare standard conditions and biochemical standard conditions for reactions. Why is there a difference?

Standard Conditions:

- 25 degrees C

- 1 ATM

- Initial concentration of reactants and products are 1M

VS

Biochemical Standard Conditions:

- [H+] = 10^-7 M

- [H2O] = 55.5 M

- All other standard conditions

Calculate delta G given delta G standard and concentrations of products and reactants (given the values of R and T).

deltaG = deltaG' + RTln[products]/[reactants]

- Ratio of P/R @ equilibrium = Keq

- Ration of P/R not @ equilibrium = Q (mass action ratio)

- deltaG' = Can look up value

- R = 8.315 J/mol K

- T = Kelvin (~298 K, 25C)

At Equilibrium, deltaG = 0:

deltaG' = -RTlnKeq

Do practice problems given for practice calculations.

Calculate delta G given Keq and vice versa (given the values of R and T) and describe how one value affects the other.

deltaG = deltaG' + RTln[products]/[reactants]

- Ratio of P/R @ equilibrium = Keq

- Ration of P/R not @ equilibrium = Q (mass action ratio)

- deltaG' = Can look up value

- R = 8.315 J/mol K

- T = Kelvin (~298 K, 25C)

At Equilibrium, deltaG = 0:

deltaG' = -RTlnKeq

Do practice problems given for practice calculations.

Describe how the deltaG for a reaction in one direction differs from the deltaG for the same reaction in the opposite direction?

The deltaG for the forward and reverse directions of a reaction are the same value, just opposite signs (+/-).

Ex: A -> B deltaG = 8kj/mol; B -> A deltaG = -8kj/mol

Describe the ionization of water and the balance that exists between the concentration of hydrogen and hydroxide ions in aqueous solutions.

H2O is Not Inert ---> Undergoes Ionization

H2O + H2O ->/<- H3O+ + OH-

H20 ->/<- H+ + OH-

Only 2 of every 10^9 molecules in H2O are dissociated at any given time

Pure H2O at 25 C = Neutral pH, so [H+]= 10^-7 M and [OH-] = 10^-7 M

Calculate the pH or pOH of a solution of strong acid or base.

Q: What is the pH of 0.1 M HCl? Note: If all HCl dissociates, then [H+] = 0.1 M...

A: pH = -log[H+] = -log[0.1] = 1 M

Q: What is the pH of 0.1 M NaOH?

A: pOH = -log[0.1] = 1 M

pH + pOH = 14, so pH = 13 M

Calculate pH, pOH, [H+], or [OH-] given sufficient data and explain the relationship between these variables.

pH = -log[H+]

pOH = -log[OH-]

There's a Balance...

[H+][OH-] = 10^-14 M^2

pH + pOH = 14

A solution of pH = 6 is 10x more acidic than a solution at pH = 7.

pH is a property of a solution, NOT a compound

Rank a list of acids in order of increasing strength given their Ka values or pKa values and explain how Ka or pKa values are derived.

pKa = -log[Ka]

Lowest pKa to Largest pKa:

- Phosphoric Acid (pKa = 2.14)

- Glycine, Carboxyl (pKa = 2.34)

- Carbonic Acid (pKa = 3.77)

- Acetic Acid (pKa = 4.76)

- Dihydrogen Phosphate (pKa = 6.86)

- Ammonium Ion (pKa = 9.25)

- Glycine, Amino (pKa = 9.60)

- Bicarbonate (pKa = 10.2)

- Monohydrogen Phosphate (pKa = 12.4)

Low pKa = STRONG acid, High pKa = WEAKER acid

Ka = [H+][A-]/[HA]

The stronger the acid, the greater its tendency to lose its proton (greater dissociation).

STRONG acid has a HIGH Ka value, WEAKER acid has a LOWER Ka value

Calculate pKa given Ka and vice versa.

Ka = [H+][A-]/[HA] -- pKa = -log[Ka]

Ex: CH3Cooh ->/<- H+ + CH3COO-

Ka = 1.74 x 10^-5 M = [H+][CH3COO-]/[CH3COOH]

pKa = -log[1.74 x 10^-5 M] = 4.76

To Find Ka from pKa:

- Ka = 10^-(pKa) - SO using example above...

Ka = 10^(-4.76) = 1.74 x 10^-5 M

Calculate the pH of a solution of weak acid given its pKa or Ka and its molarity.

Example:

Q: What is the pH of 0.1 M Acetic Acid? Ka = 1.74 x 10^-5 M

CH3COOH --> H+ + CH3COO-

(0.1 - x) (x) (x)

Ka = [products]/[reactants]

1.74 x 10^-5 = [H+][CH3COO-]/[CH3COOH]

1.74 x 10^-5 = [x][x]/[0.1 -x]

If 'x' is small enough, compared to 0.1, you can eliminate the 'x' from the denominator.

- Look at Ka

- [HA] initial > 100xKa

- Also, if less than 10% of HA

So, Ignoring 'x' in the Denominator:

x^2 = 0.1(1.74 x 10^-5)

x = 1.32 x 10^-3 = H+

pH = -log[H+] = -log[1.32 x 10^-3] = 2.88

Check if < 10%

1.32 x 10^-3 = x(0.1)/100

H+ = 1.32% of HA, so Yes!

Interpret a titration curve and be able to identity which species are predominantly present at different points of the curve.

Below pKa = More HA than A-

At pKa = 50% HA; 50% A- (1/2 ionization)

Above pKa = More A- than HA

When 100% titrated, solution is 100% conjugate base (acid completely neutralized).

Every point on graph = Equilibrium point

Review graph in notes (L6) for more details

Describe the experimental procedure of performing a titration of an acidic solution.

General Steps

1. Obtain a measured volume of acid.

2. Add a small amount of strong base (ex: NaOH)

3. Measure pH & plot

4. Repeat until all acid is consume/neutralized

Describe how a buffer works to minimize pH changes.

Buffer = Aqueous system that resists pH changes when small amounts of acid (H+) or base (OH-) are added

- Consists of a weak acid and its conjugate base

Use the H-H equation to calculate variables pH or pKa when a quantity of strong base is added to a weakly acidic solution.

pH = pKa + log[A-]/[HA]

Example

Q: What is the pH when 0.02 moles of solid NaOH are added to 1L of a 0.1 M solution of CH3COOH (pH = 2.88)? pKa = 4.76

Every unit of OH added removes one unit of acid & creates one unit of base

CH3COOH + OH- --> CH3COO- + H2O

(0.1 mol/1L) (0.02 mol)

[CH3COO-] = 0.02 mol

[CH3COOH] = 0.1 - 0.02 = 0.08 mol

pH = pKa+ log[A-]/[HA]

pH = 4.76 + log[0.02]/[0.08]

pH = 4.16

Calculate the pH of a buffer solution with known concentrations of acid and conjugate base (given the pKa of the acid).

Q: What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4 in 1 liter? pKa = 6.86

H2PO4 ->/<- HPO4^(2-) + H+

pH = pKa + log [A-]/[HA]

pH = 6.86 + log [0.058]/[0.042]

pH = 6.86 + 0.13

pH = 7.0

Quantify the change in pH when a strong base is added to a buffer solution or pure water.

Q: If 1 mL of 10M NaOH is added to a liter of the buffer you just prepared, how much will the pH change? Strong Base to Buffer Solution

H2PO4- + OH- ->/<- HPO4^(2-) + H2O

0.042 mol 0.058 mol

10 mol/1000 mL = x mol/1 mL -> x = 0.01 mol

(HA) H2PO4- = 0.042 mol - 0.01 mol = 0.032 mol

(A-) HPO4^(2-) = 0.058 mol + 0.01 mol = 0.068 mol

pH = pKa + log [A-]/[HA]

pH = 6.86 + log[0.068]/[0.032]

= 6.86 + 0.33

pH = 7.19

Q: What is the pH if the same amount of NaOH is added to 1 liter of pure H2O? Strong base to H2O

NaOH -> Na+ + OH-

0.01 mole

[OH-] = 0.01 mol/L

pOH = -log[0.01 mol/L]

= 2

pH + pOH = 14, SO pH = 12

More significant increase in pH when strong base is added to H2O vs Buffer solution

Calculate the number of moles of acid and conjugate base needed to make a certain concentration of buffer at a given pH (given the pKa of the acid).

Q: How many moles of NaH2PO4 + Na2HPO4 would you combine to make 1 liter of 0.1M phosphate buffer @ pH = 7? pKa = 6.86

H2PO4 ->/<- HPO4^(2-) + H+

pH = pKa + log [A-]/[HA]

7 = 6.86 + log [A-]/[HA]

0.14/1 = log[HPO4^(2-)]/[H2PO4-] Cross Multiply

1.4[H2PO4-] = [HPO4^(2-)]

We Know... [HPO4^(2-)] + [H2PO4-] = 0.1 M

So Substitute... [H2PO4-] + 1.4[H2PO4-] = 0.1 M

2.4 [H2PO4-] = 0.1M

[H2PO4-] = 0.042 M/L = 0.042 moles

0.1 moles - 0.042 moles = 0.058 moles of HPO4^(2-)

Define what a polyprotic acid is and provide an example.

Polyprotic Acids

- Have more than one ionizable group

- Have more than one hydrogen to donate

Example: Amino Acids - (Histidine shown)

NH2+ --CH--COOH (pKa = 1.8)

(pKa = 9.2) |

CH2

|

CHCNHCHNH+ (forms ring, pKa = 6)

Calculate or determine conceptually if a ionizable group on a polyprotic acid will be protonated at a certain pH (given the pKa of that group).

Q: What % of Histidine in a 1M solution will have a protonated carboxylic acid group @ pH = 7.3?

pH A- HA

1.8 - When pH = pKa - 50% 50%

2.8 - 1 unit above pKa - 91% 9%

3.8 - 2 units above pKa - 99% 1%

@ 7.3, all of Histidine will have COO-