Class 4: Biologically Important Molecules

Peptide Bond Characteristics

Due to resonance, the peptide bond acquires the following characteristics (resonance creates alternative of peptide bond): - partial double bond character - cannot rotate; makes protein very rigid - planar (sp2 hybridized) - amide H slightly acidic due to carbonyl

If R group amino acid has 4 carbons (isoleucine) how many carbons must be in aldehyde for Strecker Synthesis?

5 Since isoleucine has four carbons in the side chain, the starting aldehyde must have five carbons - the side chain plus the α carbon

pI for acidic amino acids

< 6

The molecular formula of an extracted triglyceride is C45H86O6. What is the molecular formula of the carboxylate ion obtained when the triglyceride is saponified?

= C14H27O2 When the triglyceride is saponified, it forms glycerol and three equivalents of the carboxylate anion. With the loss of glycerol, the fat loses three carbons and five hydrogens. The three equivalents of carboxylate ion have a combined molecular formula of C42H81O6. Divide that molecular formula by three and that gives C14H27O2.

ATP

Adenosine triphosphate (ATP) is a nucleic acid molecule that remains a single nucleotide. Unlike a DNA or RNA nucleotide, the ATP nucleotide has three phosphate groups attached to its ribose sugar.

ATP

adenosine triphosphate

D-Glucose

aldohexose = 6C aldehyde sugar - will form 6 membered ring = pyranose

Alpha- amino acid

an amino acid in which the amino group is bonded to the carbon atom next to the -COOH group - An α-amino acid (like the twenty naturally occurring amino acids) contains an amine and a carboxylic acid functional group bound to the same carbon atom

Epimers

are stereoisomers that differ at exactly one stereocenter. (think alpha-anomer and beta-anomer of carbohydrate cyclization; one OH on the anomeric C will be up and the other down because the nucleophilic addition can occur either on top or on bottom)

pI for neutral amino acids

around 6 2+10 / 2 =6

four interactions that are responsible for the tertiary structure of a protein

disulfide bonds (the only covalent bond), hydrogen bonds, hydrophobic interactions, and electrostatic interactions.

Fructose is a: A. pentose and aldose, and forms a furanose ring. B. pentose and ketose, and forms a furanose ring. C. hexose and ketose, and forms a furanose ring. D. hexose and ketose, and forms a pyranose ring.

hexose and ketose, and forms a furanose ring 6C = C6H12O6, a ketose, with the carbonyl group at C-2. Because of this, fructose usually forms a furanose (or five-membered) ring in solution, due to nucleophilic attack of the hydroxyl on C-5 onto the ketone. Hexose monosaccharides that are also aldoses form pyranose rings.

In order to prevent disulfide bond formation in a protein with multiple cysteine residues, the protein should be stored:

in a reducing environment - Disulfide bond formation is an example of an oxidation reaction. In order to prevent disulfide formation via oxidation, the protein should be stored in a reducing environment

Isoelectric point

is the pH at which an amino acid has no net charge (pH = pI) = where pKa1 and pKa2 are on either side of the zwitterionic form = pI is the pH at which 100% of the an aa is in the zwitterionic form - shifting pH to higher number (more than 6) = charge will be more (-) - shifting pH to lower number (less than 6) = charge will be more (+) pI > pH (protonated form, +) pI < pH (deprotonated form, -) pI for neutral AA (where carboxylic acid group is deprotonated (COO-) and amino group is protonated (NH3+) = roughly 6

D-Fructose

ketohexose = 6C ketone sugar - will form 5-membered ring = furanose "Fructose forms 5 membered furanoses"

Only alpha amino acid to have an R configuration

- cysteine - has a -CH2SH side chain of higher priority than the carboxyl group, and is therefore R in configuration.

A salt bridge is a type of electrostatic interaction that can play an important role in protein structure. Under physiological conditions (pH = 7.4), which of the following amino acid pairs is capable of forming a salt bridge?`

(not asking about a disulfide bridge) - want an ionic bond (+, -) - the amino acid pair aspartic acid and arginine is capable of forming a salt bridge. - A salt bridge must form between two ionized amino acids. At physiological pH, acidic side chains are deprotonated and negatively charged, while basic side chains are protonated and positively charged. The only answer choice that contains one acidic and one basic amino acid is "aspartic acid and arginine."

Reducing Sugars

*hemiacetal formation always starts with an aldehyde or ketone - OH attacks carbonyl under mildly acidic conditions (H+) If a cyclic sugar is an hemiacetal (labile): - it is in equilibrium with the linear (carbonyl) form - mutarotation occurs - it is a reducing sugar If a cyclic sugar is an acetal (react with a second equivalent of alcohol (OH); could be another sugar or a methanol: - it is NOT in equilibrium with the linear form (stable NOT labile) - does NOT mutarotate - it is a NON-reducing sugar

Benedict's test was performed on a sample of lactose and a sample of fructose. Which of the following would be observed?

- Both lactose (Gal-β-1,4-Glu) and fructose yield a brownish-red precipitate. - All monosaccharides (such as fructose) give a positive Benedict's test, due to the free hemiacetal functional group

Carbohydrate names are based on...

- D or L designation (remember R goes with D and L goes with S) - type of carbonyl: aldose (aldehyde) or ketose (ketone) - number of carbons: pentose (5C) or hexose (6C) - ring size: furanose (5-membered) or pyranose (6-membered)

Each of the following contains a correct matching of molecule to macromolecule class EXCEPT: A. antibodies are proteins with quaternary structure, and testosterone is a hydrophobic steroid derivative. B. genomic DNA, plasmid DNA, tRNA, rRNA and mRNA are all examples of nucleic acids. C. starch and cellulose are polysaccharides with predominantly β-glycosidic linkages. Correct Answer D. triacylglycerides are lipids, and are more reduced than dietary proteins such as actin, myosin and albumin.

- Each of these statements is true except choice C. Starch is used for energy storage in plants, and can be digested by humans. It therefore contains α-linkages, not β-linkages. Note that lipids are more reduced than both carbohydrates and proteins; they can be oxidized more in our bodies and therefore yield more energy.

Met (M)

- Met contains a thioether (like an ether that has an S atom instead of an O atom)= fairly NP

Strecker Amino Acid Synthesis (A > I > RM)

- Nonstereoselective synthesis of an amino acid from an aldehyde (aldehyde is starting material) - only able to make primary amino acids (not proline) - the aldehyde carries the amino acid R group = starting material contains R group - synthesis beings with imine (N=C) formation (Nu- addition) = aldehyde + (KCN or NH4Cl) = imine intermediate with R group + (KCN = nuc) = alpha-aminonitrile + (acidic conditions, H30+, heat) = hydrolyzes nitrile into carboxylic acid and protonates amine - produces racemic mixtures and amino acids (i.e. L, D-Valine) 1ST STEP: Primary amine makes an imine

Summary of Lipids: FAs and Phospholipids

- Saponification is the base-mediated hydrolysis of a triglyceride (or esters more generally) to form amphipathic soaps - Fatty acids and their carboxylate anions form micelles in aqueous solution - phospholipids form bilayers in aqueous solution - the saturation level of a FA determines the physical properties of the compound, most notably its phase at RT and adjusts membrane fluidity

Amphoteric

- amino acid can act as either an acid or a base

Strecker Synthesis preview - what does it use?

- ammonium and cyanide salts to transform aldehydes into alpha-amino acids - NH4Cl, NaCN (under acidic, H+ H30+ conditions)

Gabriel-Malonic Ester Synthesis (P > E > SN2 > NaOEt, R-Br)

- another way to produce alpha-amino acids - nonstereoselective synthesis of an amino acid from N-phthalimidomalonic ester - N-phthalimidomalonic ester must first be synthesized - first, remove acidic alpha-H from phthalimide via strong base (KOH) - then (-)N attacks malonic ester (sigma -> sigma bond breaking = SN2); malonic ester is a diester with a LG - phthalidomide is a strong nuc N- attacks malonic ester between 2 carbonyls and LG leaves - gain a temporary ester (COOEt), acidic alpha-H (N-CH), protected carboxylic acid and protected amine (N-CH) *the protected carboxylic acid and protected amine = will be apart of final AA - then treat again with base =NaOEt removes acidic alpha H, this carbon now become an enolate (charged) and attacks PhH2C-Br (THIS IS THE AMINO ACID R GROUP) attacks - now must get rid of protecting groups like the ring the amine is wrapped up in - acidic conditions, H3O+, causes N to leave ring structure = IMIDE (almost have AA) = hydrolysis of esters and imide. Ester groups will be converted into carboxylic acids = ester hydrolysis reaction - heat causes decarboxylation (-CO2) final product: AA with (L,D) because either of the 2 carboxylic acids could have left - both because non-stereoselective reaction

Biomolecule Reflect

- biopolymers (trains) are composed of monomers (train cars) - understanding structure and reactivity in the monomer explains the polymer

Biomolecule Preview

- consider taking a train from NYC to Chicago - many cars make up train and each serve a purpose and contributes to overall train - rules: engine goes in front, caboose in the back - remaining cars randomized and length can be short/long - cars must connect end-to-end (must be single entity) - so even though some cars must come first or last, the remaining cars can be connected in many ways, making a different train every time - the train does NOT spontaneously connect Biomolecules link to together in a similar fashion: - enzymes - assemble monomers (train cars) into polymers (a train) in biosynthesis reactions - polymerization reactions are typically uni-directional and highly specific

Weak acids

- do not dissociate completely ex: RCOOH (carboxylic acid) RNH3+ (ammonium ion) - equilibrium constant Ka < 1 because the equilibria will favor left hand side of reaction so we will have more reactants than products at equilibrium - pKa = -log(Ka) so pKa > 0 = weak acid *the higher the pKa the weaker the acid

The formation of a peptide bond is not thermodynamically favorable, How are peptides formed and maintained inside cells?

- during protein synthesis, stored energy is used to force peptide bonds to form - once the bond is formed, even though its destruction is thermodynamically favorable, it remains stable because the activation energy for the hydrolysis reaction is so high - hydrolysis is thermodynamically favorable but kinetically low hydrolysis = any reaction in which water is inserted in a bond to cleave it *ex: hydrolysis of an ester under both basic and acidic conditions, and of peptide bond (amide bond) to form a free amine and a carboxylic acid is thermodynamically favorable f

Micelles

- fatty acids are amphipathic (contain both hydrophobic/NP and hydrophilic/P portions) = soap - in aqueous environment, fatty acids aggregate into micelles - hydrophilic charged "head" groups are on outside - hydrophobic area with nonpolar tails is on inside = they have LDF (van der Waals forces) and this is what holds the micelle together *in an aqueous environment, fatty acids aggregate into micelles **grease molecule (cleaning your pan) will be associated with hydrophobic tail of micelle/FA and globule will be washed away with water because water can interact with polar charged head groups of micelle

Fatty Acids & Saponification

- fatty acids are stored in the form of triacylglycerols and can be liberated by saponification **Saponification (aka base-promoted ester hydrolysis = break ester bonds RO-Carbonyl) liberates free fatty acids - uses base to break ester (3 equivalents for triacyglycerol) - elimination of LG (OEt(-)) - protonation of OEt(-) -> EtOH - form carboxylate anion (charged, blood soluble) (forms carboxyl salts in the case of FAs)

Glucose and galactose are: A. diastereomers. B. anomers. C. meso compounds. D. conformational isomers.

- glucose and galactose are both hexoses and aldoses. They differ only in their absolute configuration at C-4, and so are diastereomers. - Note that they are also epimers (a subclass of diastereomers that differ at exactly one chiral center). - Anomers are a subtype of epimers but are due to ring closure, and this is the not the case for glucose and galactose (choice B is incorrect). - Meso compounds are optically inactive molecules that have an even number of chiral centers and an internal plane of symmetry (choice C is incorrect). - Conformational isomers differ in sigma bond rotation (choice D is incorrect).

Phospholipids

- have a structure similar to a triacylglycerol (but contain a phosphate group instead of a 3rd FA) - are a major component of cell membranes (lipid bilayers) - the saturated hydrophobic tail is why many things cannot get through the phospholipid bilayer

Henderson-Hasselbalch equation

- low pH means high H+ - lower pKa (same as higher Ka) describes a stronger acid that can donate a proton even when there are already excess protons (high[H+], low pH)

Fischer projection of amino acids

- most oxidized group on top (i.e carboxylic acid) - group on left and right are facing out of page at you - group top/bottom are facing into the page away from you

Summary of Gabriel Malonic synthesis

- most primary amino acids can be synthesized this method - the side chain is incorporated via and SN2 (enolate attacks R group and kicks out LG) - non-stereoselective so both D and L forms of the amino acid are formed

Essential AA

- must be taken in the diet in times of need (non-essential = can be synthesized in the body in times of need)

Nucleic Acids

- nucleotides are building blocks (monomers) of nucleic acids (DNA, RNA = polymers) Each nucleotide contains: - ribose (or deoxyribose) sugar - purine (AG) or pyrimidine (CUT) base joined to C1 of the ribose ring - 1, 2, or 3 phosphate units joined to C5 of ribose ring

Proline and alpha helix

- proline forces it to kink the polypeptide chain, hence proline residues never appear within the alpha-helix

What is the difference between a disulfide bridge involved in 4 structure and one in 3 structure?

- quaternary disulfides are bonds that form between chains that aren't linked by peptide bonds - tertiary disulfides are bonds that form between residues in the same polypeptide

Peptide Bond Hydrolysis

- requires enzymes - peptide bond hydrolysis is NON-SPONTANEOUS and non-selective peptide bond hydrolysis and requires acid, high temperatures, and long reaction times to accomplish (amide is a stable carboxylic acid derivative) - remember peptide bond hydrolysis is thermodynamically favorable but kinetically low since Ea is so high - proteases digest a protein by catalyzing hydrolysis of specific peptide bonds - the peptide backbone always contains a repeating sequence of N-C(alpha)-C ex: 4aa = 3 peptide bonds (n-1) **proteins or peptides can be non-selectively hydrolyzed using acid, or selectively hydrolyzed using enzymes (protease)

Gel Electrophoresis

- separates AAs based on differences in charge - remember that the pI of an AA determines its charge at a given pH apply concepts of: pH < pI = (+) charge pH > pI = (-) charge *Zwitterion (overall neutral) will not move in a electrophoresis experiment

Summary of Benedict's test

- sugars exist in equilibrium between their straight chain form (which contains an aldehyde or ketone) and their cyclic forms (which contains a hemiacetal) - sugars that contain acetals rather than hemiacetals at the anomeric carbon can no longer equilibrate with the straight chain form in solution - the Benedict's test indicates the presence of a reducing sugar (will gain O-bonds) through the formation of a red precipitate

Hydrophobic Interactions

- the fatty acids in micelles AND phospholipids in lipid bilayers interact by van der Waals forces (dipole and hydrophobic forces) between the fatty acid tails - saturated FAs pack better and form solids due its straight linear form (stacking); higher melting points - cis-Unsaturation increases fluidity of fatty acids (kinks) and the cell membrane (integral for health) - cis unsaturated fats form liquids *Unsaturation increases fluidity of fatty acids and the cell membrane - unsaturated fats have low melting points and are usually liquids

Which of the following is an accurate characterization of lipid molecules? A. Phospholipids and triacylglycerides are made of glycerol and 3 fatty acids, but steroids are not. B. Steroids such as testosterone and estrogen can simply diffuse across the phospholipid bilayer. C. Lipases hydrolyze triacylglycerides to liberate fatty acids from the steroid backbone. D. Steroids, triacylglycerides and phospholipids all have some amphipathic character.

-Steroids are hydrophobic and so can diffuse across the plasma membrane without the help of the protein transporter (choice B is correct).

Protein Structure (bonds/forces)

1 - amino acid sequence = peptide bonds (NCC) 2 - 3D folding of the straight chain peptide/backbone only = backbone H-bonds (N-C-C only; no R groups) between alpha helix and B-pleated sheet 3 - folding of the secondary peptide into most stable 3D conformations BUT BETWEEN A SINGLE POLYPEPTIDE ex: myoglobin (polar R groups on outside, NP R groups tucked away on inside; R groups dictate this level of folding and therefore structure/function) R group interactions: 1. H-bonds (in 2 structure are from backbone; 3 structure are from R groups in protein and R groups) 2. S-S bonds (covalent and strongest bond type) 3. van der Waals interactions (hydrophobic forces) 4. ionic bonds (salt bridges) 4 - association of multiple peptide chains (protein complex) = governed by same interactions as in tertiary structure (H-bonds, disulfide, van der Waals, ionic bonds); BUT BETWEEN DIFFERENT POLYPEPTIDES ex: Hemoglobin has 4 polypeptides that act as a single complex ex: DNA POL enzyme is multiple proteins

Palmitoleic acid

16:1 - Which naturally occurring biological molecule would be most likely to bind in the same site as ibuprofen? - palmitoleic acid, because fatty acids consist of a hydrocarbon tail and a carboxylic acid, which most closely resembles ibuprofen based on the structure shown in Figure 1

Pro (P) - what can it not do?

= R attached to alpha-amino group = secondary amino acid Proline cannot act as a hydrogen bond donor when it is in the middle of a polypeptide sequence. For most amino acids, the hydrogen attached to the amide nitrogen at the peptide bond may act as a hydrogen bond donor. However, due to the covalent bond between proline's amino group and its sidechain, proline lacks an amide hydrogen when it is part of a polypeptide chain, and therefore cannot act as a hydrogen bond donor. (G, A, L can all act as proton donors, etc) - disrupts H-bonds in proteins

Lys (K)

= accepts H+ pKa = 10

Arg (R)

= accepts H+ pKa =12-13

Gly (G)

= achiral (only achiral amino acid) - not and L and a D form; only an L

Glycosidic Linkage

= aka acetals, connect two monomeric sugars To name linkage: 1. indicate the anomeric form of the first (leftmost) sugar (alpha or beta) 2. then indicate the 2 Cs connected by the bridging oxygen (ex: alpha-1,4)

Basic amino acids

= all accept H+

Peptide Bond Formation

= also known as amide bond (carboxylic acid derivative) - peptide bonds form through an addition-elimination mechanism (pi bond breakage) - peptide bond formation is NON-SPONTANEOUS and requires enzymes to catalyze this process - ie ribosomes *this reaction requires protecting groups or biological machinery to achieve; you can't expect an acidic group (COOH) to react with a very basic group (NH2), rather than the nucleophilic addition that you want, the OH hydrogen will protonate the NH2 group (acid/base chemistry)

His (H)

= contains secondary nitrogen (what makes it basic) - accepts H+ - can also be acidic (due to its pKa being close to physiological pH so it can act to donate a proton = acid) - His is neutral a physiological pH (so it remains uncharged at higher pH's as well and positively charged at pH < 7) (His is deprotonated at physiological pH; pKa < pH -N with double bond and LP is where protonation +/deprotonation = neutral occurs pKa = 6

Denaturation of Proteins

= disrupts a protein's 3D structure without breaking peptide bonds (2,3,4 structure only) (primary structure can only be broken via enzymes = proteases!!) This is achieved through denaturing agents: - high temperature (think egg whites) - extreme pH - changing tonicity (salt concentration) - addition of urea (disrupts H-bonds, salt bridges can't form either) After denaturation, proteins adopt a random coil and lose function If you re-natured a protein it would re-fold depending on: - severity of denaturation (more denaturing agents) - complexity of original folding - chaperone proteins assist with protein folding and they are no longer in solution of denatured protein * removing the denaturing agent SOMETIMES allows protein renaturation

Glu (E)

= donate H+ due to CH2COOH (carboxylic acid side chain) (GLUE) (E & D are GLUE-D together in the alphabet) pKa = 4

Asp (D)

= donate H+ due to CH2COOH (carboxylic acid side chain) pKa = 4

Cys (C)

= forms S-S bonds = thiol (similar to OH group) Has R configuration - ALL OTHER AMINO ACIDS HAVE S

Benedict's Test

= identifies a reducing sugar (hemiacetal) via a redox reaction (reducing Cu2+ to Cu2O = precipitate) **a reducing sugar is one that will itself become oxidized - a postive test (red precipitate) indicates a reducing sugar is present - a negative test (no precipitate) indicates no reducing sugar is present *reducing sugars must contain an aldehyde or ketone (reducing sugar) if an acetal is present: stuck in its cyclic form If a hemiacetal is present: can open to straight chain and react with Benedict's reagent *only takes one hemiacetal to get a positive test

The conversion from an α-furanose to a β-furanose is an example of:

= mutaroation - In aqueous solution the α and β forms of glucopyranose are in dynamic equilibrium, constantly converting between one form and the other. As such, even if a pure sample of the α or β form were attained the process of ring opening and closing would begin to convert some of the sample to the other form. As such a pure sample, in solution, is unattainable. - you cannot use any of the separation techniques to isolate these isomers into pure compounds

Mechanism for linear to cyclic form of a sugar?

= nucleophilic addition reaction - SN reactions are substitution reactions. As the name implies, substitution indicates replacing one group for another, which is not the case here - Addition-elimination reactions add a group across a double bond while another group is eliminated to recreate the double bond - The formation of cyclic monosaccharides involves the nucleophilic attack of the C=O by an O—H group toward the end of the carbon chain. It is therefore adding the O—H group to the C=O.

Acidic and Basic side chains

= these donate or accept H+ - these are charged R groups

pI for basic amino acids

> 6

Under physiological conditions, which of the following amino acids is most likely to be found on the exterior of a body protein after folding is complete? A. E B. I C. M D. Y

A - E = acidic and charged at physiological pH (correct) B - I = NP C - M = NP D. - Y = P but bulky ring structure - The passage states that one of the major driving factors for protein folding is the hydrophobic effect, whereby hydrophobic (nonpolar) residues fold in such a way as to minimize their interaction with water. - As such, the majority of hydrophobic residues prefer to face toward the protein interior, away from water present at the protein surface. - In contrast, hydrophilic residues prefer to face toward the protein exterior and are therefore likely to be found on the protein surface. - At physiological pH, glutamate (E) will be in its deprotonated form with a full negative charge, making the side chain extremely polar and most likely to be found in the exterior of the protein (choice A is correct). - Both isoleucine (I) and methionine (M) are nonpolar, hydrophobic amino acid residues that are expected to be found on the protein interior (eliminate choices B and C). - While the side chain of tyrosine (Y) contains a hydroxyl group (OH) capable of forming hydrogen bonds, it also contains a large hydrophobic ring that makes it considerably less polar and thus less likely to be found in the protein exterior than glutamate (eliminate choice D; choice A is better).

glycerol

A three-carbon alcohol to which fatty acids are covalently bonded to make fats and oils.

If a mixture of glycine and lysine was dissolved in a pH = 1.5 solution and separated via gel electrophoresis, which of the following would occur? A. Both amino acids would move towards the positive electrode. B. Both amino acids would move towards the negative electrode. C. Glycine would move towards the positive electrode while lysine would migrate towards the negative electrode. D. Lysine would move towards the positive electrode while glycine would migrate towards the negative electrode.

B. Both amino acids would move towards the negative electrode. Lysine pKa1 NH3+ = 10 > pH (protonated, +) pKa2 COO- = 2 > pH (protonated, 0) pKa3 Lys = 10 > pH (protonated, +) Glycine pKa1 NH3+ = 10 > pH (protonated, +) pKa2 COO- = 2 > pH (protonated, 0) Both would be + and migrate to -

Carbohydrate Nomenclature

CnH2nOn (i.e C6H12O6); for every carbon molecule you have one water (think water hydrates carbs) - aldehydes and ketones are always up at the top (most oxidized C at the top of Fischer projection) - CH2OH will always be at the bottom - D = OH on right (not the CH2OH on bottom but of penultimate carbon) - L =OH on left **D form is the naturally occurring form (the form we use in the body)** LA = L, amino acids (L-amino group on left) DC = D, carbohydrates (D-OH group on right) If carbohydrate is not shown in Fischer projection - D- if the penultimate carbon is in the R configuration - L - if the penultimate carbon is in the S configuration * R/S applied to chiral centers ** (+)/(-), same as (d)/(l) = (CW)/(CCW) describes which direction a molecule rotates light; has NOTHING to do with D/L *** (+)/(-) is something you CANNOT figure out (but you can figure out R/S and D/L by looking at sugar structure)

All naturally occurring carbohydrates have the ___ configuration

D - animal carbohydrates are chemically derived from D-glyceraldehyde, and are thus all D

Charged side chains

D, E, K, R, H - Charged compounds are hydrophilic - often form salt bridges (ionic bonds) - CHARGED IONS SPECIES CANNOT CROSS BIOLOGICAL MEMBRANES

An aqueous solution contains a mixture of α- and β-glucopyranose. Which of the following techniques could be used to separate the two isomers into pure compounds? A. Column chromatography B. Fractional distillation C. Gas chromatography D. Pure samples of α- and β-glucopyranose cannot be attained by any of these methods.

D. Pure samples of α- and β-glucopyranose cannot be attained by any of these methods. In aqueous solution the α and β forms of glucopyranose are in dynamic equilibrium, constantly converting between one form and the other. As such, even if a pure sample of the α or β form were attained the process of ring opening and closing would begin to convert some of the sample to the other form. As such a pure sample, in solution, is unattainable.

A pure sample of solid α-D-glucose is dissolved in water and its specific rotation is promptly measured to be +112.2˚. What is the best explanation for the fact that when the same sample is remeasured after an hour the polarimeter reads +52.7˚? A. α-D-Glucose completely converts to the more stable β-D-glucose, which has a much smaller specific rotation. B. Some water evaporated from the solution leaving it more concentrated and changing its optical activity. C. An equilibrium is established between the α-anomer and the straight-chain form of the sugar, lowering the optical activity of the solution. D. The α-anomer partially converts to the β-anomer, and the equilibrium mixture of anomers has an optical activity of +52.7˚.

D. the α-anomer partially converts to the β-anomer, and the equilibrium mixture of anomers has an optical activity of +52.7°. The phenomenon described in this question is mutarotation. While it's true that the straight-chain form of the sugar is in equilibrium with the α-anomer, it's also in equilibrium with the β-anomer. Very little of the acyclic sugar is present, so it's not the presence of the acyclic form that changes the optical activity. While the β-anomer is more stable than the α-anomer, because of the equilibria already mentioned, complete conversion from α to β cannot be correct. β-D-Glucose in its pure form actually has a specific rotation of +18.7°, so the observed rotation of +52.7° is a weighted average based on the ratio of isomers in solution. Finally, while the concentration of a solution does have an impact on its optical activity, if a solution becomes more concentrated it will increase the magnitude of the rotation of light, not decrease it.

Carbohydrate Cyclization (what can anomers not do?)

Going from linear form to chair conformation will form: - LPs on OH on penultimate carbon attack anomeric carbonyl - can attack from either side bc anomeric carbonyl is sp2 - straight chain form and two cyclic forms exist in equilibrium isomeric relationship: - stereoisomers of rings = epimers = "anomers" = nucleophilic addition (pi bond breaks) & hemiacetal formation (CH3-OH) mutarotation = interconversion of anomers (they cannot be separated) alpha anomer = OH down beta anomer = OH is up

Strong acids

HCl, HBr, HI, HNO3, H2SO4, HClO4 - strong acids dissociate completely - equilibrium constant Ka is large > 1 (tells us that reaction favors the products and will dissociate) - pKa = -log(Ka); so small pKa < 0 = strong acid *the lower the pKa the stronger the acid

trans pi bonds vs. cis pi bonds

In fatty acids: - trans pi bonds decreases membrane fluidity - trans pi bonds are like saturated fatty acid packing - cis pi bonds create kinks in membrane and increase fluidity

All animal amino acids have the ___ configuration

L - all animal amino acids are derived from L-glyceraldehyde (because they share the same basic structure at the penultimate carbon) - this is why they all have L-configuration

Amino Acid (AA) nomenclature

L- amino acids: amino group on left (naturally occurring, utilize this form the most) D - amino acids: amino group on right AA are AMPHOTERIC (able to act as an acid and a base) - One amino acid has both an L and D form so will have an enantiomeric pair

amide

NH2 - attached to carbonyl carbon

Which of the following hexapeptides will most readily dissolve in 0.1 M NaHCO3? A. EASVIR B. DGEPDI C. LIFMGA D. VLWAKR

NaHCO3 deprotonates carboxylic acids - determine which hexapeptide has the greatest number of carboxylic acid functional groups, one can determine which will most easily dissolve. All of the hexapeptides feature a terminal carboxylic acid group. The sequence DGEPDI contains three acidic residues (two D and one E residue), far more than the other choices, and will most readily dissolve.

Can you metabolize carbohydrates that evolved from L-carbohydrates?

No, enzyme activity depends on 3-D shape, and all animal digestive enzymes have active sites specific for substrate carbohydrates with the D configuration

Which of the following is true about polar amino acids? A. They are hydrophobic. B. Their R group contains only a hydrocarbon group. C. Their side chains are on the exterior of protein molecules. D. They have uncharged R groups.

Polar amino acids have their side chains on the exterior of protein molecules. These polar side chains are on the exterior of protein molecules so that they can interact with the polar solvent (H2O).

Polar side chains

S, T, N, Q, Y can all act as both hydrogen donors and acceptors

Sugars

Structure: - natural form is D (OH on right of penultimate carbon) - many structural/steroisomers - notation: D/L, aldose/ketose, pyranose/furanose, alpha/beta anomers Reactivity: - Addition gives cyclic and polymeric sugars (acetal chemistry) - mutarotation interconverts open-chain and cyclic forms - Benedict's test (test for mutarotation = hemiacetal present)

Amino Acids

Structure: - natural form is L - R groups (p, np, a, b) - synthesis via Strecker and Gabriel racemize alpha-carbon Reactivity: - pH vs pI determines charge on AA - Addition-elimination forms/hydrolyzes peptide bond (amide chemistry)

Lipids

Structure: - saturated vs. unsaturated (different intermolecular forces) - micelles form in aqueous solution Reactivity: - Addition-elimination forms/hydrolyzes triacylglycerol (ester chemistry) - saponification = basic hydrolysis

Aromatic side chains

Tyr (Y) - H bonds (polar) Phe (F) - NP; 1 ring Trp (W) - NP; 2 rings

Melting point of fatty acids

Unsaturated fatty acids have a lower melting point (liquids) than saturated fatty acids (solids), even though their molecular weights are notably different, - The compound with the higher molecular weight (when comparing either the pair of saturated or unsaturated fatty acids) will have the higher melting point.

Nonpolar solvents

dissolve nonpolar solutes/molecules - toluene, benzene, and n-heptane

phospholipids are like

detergents, substances that efficiently solubilize oils while remaining highly water soluble - detergents are like soaps but much stronger

AA protonation and charge states depend on pH

pH < pKa = HA (protonated) pH > pKa = A- (deprotonated) pH < pI = (+) charge pH > pI = (-) charge

Carboxylic acid group pKa (COO-)

pKa = 2 at a higher pH where there is more hydroxide in solution pH > pKa and will be deprotonated; carboxy H is the first proton removed (COO-) bc it is the most acidic group Keep making solution more basic ie. pH >> pKa next most acidic H will be removed (NH2 or R-group depending)

Histidine pKa

pKa = 6.5 (you would expect it to be more basic at 10) at pH = 7.4 pH > pKa - will be in its deprotonated form (neutral/basic form; doesn't have proton) *at pH's higher than physiological pH, His is deprotonated and neutral at pH < < 7 pH < pKa - will be in its protonated form (acidic form; has proton to give +-charge)

Amino group pKa (NH3+)

pKa = 9-10

soaps are

sodium salts of FAs (RCOO- Na+) - amphipathic - soap helps remove grease from your hands since grease is hydrophobic, it doesn't wash off easily in water because it is not soluble in water. So scrubbing your hands with soap causes micelles to form around the grease particles