2. Amino Acids and Proteins

Question 1

What are the non-polar, hydrophobic amino acids? For each explain their shape, and their single letter name

Answer 1

GAVLIMP

glycine - no R 
alanine - CH3
valine - isopropyl
leucine - methyl-isopropyl 
isoleucine - secondary-butyl
methionine - C-C-S-C
proline - attaches to itself

Question 2

T or false, phenylalanine and tryptophan are also hydrophobic.

Answer 2

True,

they are not a part of GAVLIMP b/c they are aromatic

Question 3

What are the polar, hydrophilic amino acids? For each explain their shape, briefly.

Answer 3

STNC-Q

Serine - S - OH group 
Threonine - T - ethanol 
Asparagine - N - amino group (shorter)
Cysteine - C - SH 
Glutamine - Q - amino group (longer) 

tyrosine (Y) is also polar, but placed with the aromatic compounds. Polar amino acids form H-bonds with water.

Question 4

What are the basic amino acids?

Answer 4

Lysine - K - Long tail to an NH3
Arginine - R - shorter tail to an NHNH2=NH
Histidine - H - ring containing two Ns

Question 5

What are the acidic amino acids?

Answer 5

Aspartic Acid - D - shorter COOH

Glutamic Acid - E - longer COOH

Question 6

How can we differentiate E from D and Q from N?

Answer 6

Q and N are polar. Q (glutamine) is longer
E and D are acidic. E (glutamate) is longer

glut = longer.

Question 7

Explain the protonation status of D, E, K, R, and H ar physiological pH (7.4)

Answer 7

D and E are acidic (low pKa values). These are deprotonated (anionic) at physiological pH

K and R are basic (higher pKa values). These are protonated (cationic) at physiological pH.

H (pKa = 6.5) exists equally in protonated and deprotonated forms at 7.4. (His goes both ways)

Question 8

What are the aromatic amino acids?

Answer 8

tyrosine - Y - phenol group - polar
Tryrptophan - W - two aromatic rings - nonpolar
phenylalanine - F - benzene ring - nonpolar

Question 9

What amino acids are phosphorylated by kinases?

Answer 9

hydroxyl-containing amino acids: tyrosine, serine, and threonine.

Question 10

What amino acids contain sulfur?

Answer 10

Cysteine - C - SH

methionine - M - CCSC

Question 11

What is an essential amino acid?

Answer 11

An AA that we cannot synthesize and must be obtained by our diets. Most non-polar AAs are essential.

Question 12

What are the pKa values for the carboxyl group and amino group of the amino acid backbone?

Answer 12

COOH - pKa = 2

NH2 - pKa = 9/10

Question 13

What is the Henderson-Hasselbalch equation?

Answer 13

pH = pKa + log ([base] / [acid])

Question 14

Henderson-Hasselbalch equation: What happens if pH = pKa?

Answer 14

pH = pKa + log ([base] / [acid])

if pH = pKa, then log ([base] / [acid]) = 0. This means that [base] = [acid] since log(1) = 0

Question 15

Explain the relationship of pH and pKa with the protonation status of an R group.

Answer 15

pH < pKa; the acidic group will be mostly protonated

pH > pKa; the acidic group will be mostly deprotonated

e.g. at physiological pH (7.3), carboxylic acid groups of the AA backbone (pKa = 2) are almost entirely deprotonated.

Question 16

What is an isoelectronic point? What is a zwitterion?

Answer 16

isoelectronic point (pI) - the pH at which a molecule has a neutral net charge.

An amino acid that is at its isoelectronic point (glycine at 7.4 has a + and a - charge) is called a zwitterion.

Question 17

How can we calculate the isoelectronic point for an amino acid with two functional groups?

Answer 17

We just take the average of the pKa values. For glycine, we take

(9 + 2) / 2 = 5.5

Question 18

during polypeptide synthesis, what end is made first?

Answer 18

The amino end is made first and the COOH is made last.

Read N –> C

Question 19

In the dipeptide, is ALA - GLY the same molecule as GLY - ALA?

Answer 19

NO, since the amino acid terminals are different

H3N - ALA - GLY - COOH

H3N - GLY - ALA - COOH

connectivity is different.

Question 20

Why does peptide bond formation require a dehydration reaction?

Answer 20

It requires energy input = dehydration. Hydrolysis releases energy

NH2 attacks carbonyl oxygen

Question 21

T or f, cystine formation is an oxidation reaction.

Answer 21

True, since they reactants lost hydrogen

oxidation = lose electrons, lose hydrogens, or gain oxygen

Question 22

t or f, the inside of the cell is called a reducing environment because it contains many anti-oxidants.

Answer 22

True!

Question 23

What is the implication of the cells reducing environment on disulfide bridges?

Answer 23

The formation of a disulfide bond (cystine) is an oxidation reaction. However, because the inside of the cell is a reducing environment, the cell will reduce the bond back to two thiol groups.

Thus, Disulfide bridges are only found in extracellular proteins!!!

Question 24

What is capable of protein denaturation (protein unfolding)?

Answer 24

urea, extreme temperature, pH, or salt concentrations.

peptide bonds are NOT broken, only protein shape

Question 25

What holds together protein secondary structure?

Answer 25

Amino acid backbone hydrogen bonding The alpha-carboxyl oxygen of one amino acid, hydrogen bonds with the alpha-amino proton of an amino acid three residues away.

Question 26

Why are proline residues not found within an alpha-helix (n=2)?

Answer 26

1. proline binds to its own backbone, which removes the only hydrogen on its amino group. Thus, proline residues cannot hydrogen bond (which holds together helixes) 2. proline causes a physical kink in the peptide chain

Question 27

How does an enzyme work?

Answer 27

Enzymes are biological catalysts: They increase the rate of a reaction by lowering the activation energy (which is done by stabilizing the transition state). They DO NOT effect the ΔG or the Eq. enzymes are kinetic not thermodynamic

Question 28

``` What does each enzyme do? Kinase Polymerase Phosphatase Lyase Hydrolase Ligase ```

Answer 28

Kinase - adds (transfers) phosphate groups Polymerase - extends nucleotide chains Phosphatase - removes phosphate groups Lyase - breaks chemical bonds by means other than REDOX or hydrolysis Hydrolase - hydrolyzes chemical bonds Ligase - forms chemical bonds (e.g. in DNA lagging strand)

Question 29

What is reaction coupling?

Answer 29

Enzymes facilitate ΔG (-) reactions. However, many reactions are ΔG (+). Therefore, the cell couples very negative reactions with positive reactions, which drive the positive reaction. ATP hydrolysis is the primary reaction coupler.

Question 30

T or f, enzymes selectively promote unfavorable reactions via reaction coupling.

Answer 30

True. Enzymes only have a kinetic role in the reaction itself. But since they are designed for specific reactions, they are selective.

Question 31

Amino acids and sugars. Which isomers are found in animals / humans (D/L)?

Answer 31

D-sugars (think Delicious) | L-amino acids

Question 32

What are co-factors and what are co-enzymes?

Answer 32

co-factor: A small molecule or metal ion that facilitates enzyme catalysis. Vitamins are co-factors. co-enzyme: An organic (Carbon-containing) co-factor

Question 33

What is allosteric regulation?

Answer 33

The modification of an enzymes active site via interactions that occur at a different site on the enzyme (called the allosteric site).

Question 34

The reaction rate (mols of product per second) of an enzyme catalyzed reaction is dependent on what?

Answer 34

the [enzyme] the [substrate] however, the enzyme concentration is usually fixed, and therefore, the substrate concentration is only considered.

Question 35

On a V (rate) vs [S] curve, explain how changing [S] effects V?

Answer 35

At low [S], they are proportional. If you double [S], you double V. However, as you consistently increase [S], eventually the enzymes all become saturated. At this point, any increase in [S] does not change V. This is Vmax.

Question 36

What is Km on a V (rate) vs [S] curve?

Answer 36

Km is the [S] at which we are at Vmax / 2 how much substrate is needed to get to half maximum velocity.

Question 37

What is the implication of a high / low km?

Answer 37

A low Km indicates that not much substrate is required to get to half Vmax. If this is the case, then the enzyme affinity for that substrate must be high. If the Km is high, then the affinity must be low.

Question 38

Certain enzymes do not follow normal Vmax, Km kinetics. Explain enzyme cooperativity.

Answer 38

Cooperativity is a special form of allosteric regulation. An enzyme or multienzyme complex that has two or more active sites is capable of this. Essentially, the binding of one active site regulates the substrate affinity (+ or -) of the other active site. Positive cooperation: Oxygen binding on hemoglobin

Question 39

What curve is made with positive cooperation?

Answer 39

A sigmoid curve. At the start [S] effects V a little. Then once one AS is bound, the other AS is in a relaxed state, so the curve shoots up. It then plateaus at saturation.

Question 40

Enzyme inhibition: Explain Km and Vmax. | Competitive inhibition

Answer 40

Competitive inhibition, the inhibitor directly binds the enzymes active site. Thus, we need to increase the [S] to compete with this inhibitor. Vmax is unchanged Km - increases (since you need more substrate to get to half Vmax)

Question 41

Enzyme inhibition: Explain Km and Vmax. | Non-Competitive inhibition

Answer 41

Non-competitive inhibition, the inhibitor binds an allosteric site that alters the active site. Adding more substrate will not displace the allosteric inhibitor and therefore does not change anything. Vmax - decreased while the inhibitor is bound Km - unchanged (since the affinity of the active site has not changed). The AS can still bind the substrate, but ithe reaction will proceed more slowly.

Question 42

Enzyme inhibition: Explain Km and Vmax. | Un-Competitive inhibition

Answer 42

Un-competitive inhibition, the inhibitor can only bind the Enzyme once the substrate has already bound the AS (i.e. the inhibitor binds the enzyme-substrate complex at an allosteric site). Vmax - decreased since the inhibitor cannot be displaced. Km - decreased (the apparent affinity for the E has increased - the E-S complex is favored).

Question 43

Enzyme inhibition: Explain Km and Vmax. | Mixed inhibition

Answer 43

Mixed inhibition is a combination of non-competitive and un-competitive inhibition. It can bind E alone or the E-S complex. Vmax - decreased Km - varies on inhibitor

Question 44

Explain the Lineweaver-Burke plot. Slope, y-axis, x-axis.

Answer 44

A Lineweaver-Burke plot is a double reciprocal plot that plots enzyme kinetics on a linear graph. Slope = Km / Vmax y-intercept = 1/Vmax x-intercept = -1/Km

Question 45

How is an increase in reaction rate or an increase in Km represented on a Lineweaver-Burke plot.

Answer 45

Since a Lineweaver-Burke plot is a double reciprocal, increases in Vmax are seen as a decrease in y-value. increases in Km are seen as a decrease in x-value. Vice versa, if Km decreases, x-values increase.

Question 46

What is the slope of a Lineweaver-Burke plot?

Answer 46

Km / Vmax