Chapter 4 - Amino acids and Proteins

Question 1

What are the Acidic amino acids? Give all three names . At physiological pH, what state are they in?

Answer 1

The only acidic amino acids are aspartic acid and glutamic acid

aspartate = asp = D 
glutamate = glu = E

they are both deprotonated at physiological pH

Question 2

What are the Basic amino acids? Give all three names . At physiological pH, what state are they in?

Answer 2

Lysine, Arginine, and Histidine

Lysine = lys = K 
Arginine = Arg = R 
Histidine = His = H 

at physiological pH lysine and arginine are protonated but Histidine is variable

“His goes both ways”

Question 3

What are the Hydrophobic i.e. non-polar amino acids? Give all three names .

include the two aromatic non-polar AA’s

Answer 3

Glycine - gly - G 
Alanine - Ala - A
Valine - val - V
Leucine - leu - L
Isoleucine - ile - i
Methionine - met - M 

Think GAVMIL

Phenylalanine - phe - F
Tryptophan - trp - W

Question 4

What are the Polar amino acids? Give all three names. Note these amino acids have a polar enough group to make hydrogen bonds but enough to be basic or acidic.

Answer 4

Serine - Ser - S
Threonin -Thr -T
Tyrosine - tyr - Y 
Asparginine - Asn - N
Glutamine - gln - Q
Cysteine - cys - C

Think STINKY-Q (STNCY-Q)

Question 5

What is the polar and non-polar sulpher containing amino acids?

Answer 5

polar - cysteine - cys - C

non-polar - methionine - met - M

Question 6

what is the most rigidly shaped amino acid which is non-polar

Answer 6

proline - pro - p

Question 7

Will an acidic group be protonated or deprotonated when

pKa > pH
pKa < pH

Answer 7

when pKa > pH the acidic group will be mostly protonated

when pKa < pH the acidic group will be mostly deprotonated

this can be visualised by looking at the henderson - hasselbalch equation

Question 8

what is an amino acids isoelectric point? how do you calculate it?

Answer 8

the pH value at which that amino acid has no net charge. This is calculated by adding the pKa values that straddle a charge of 0 and dividing them by 2.

Question 9

when counting the number of acidic / basic groups of a polypeptide, do we count R groups and terminals? or just R groups.

Answer 9

R groups and terminals

phe - glu - gly - ser - ala

there are 3 basic / acidic groups
the two terminal amino and carboxyl groups + the glutamic acid group.

Question 10

what is proteolysis?

Answer 10

proteolysis aka proteolytic cleavage is a way in which peptide bonds are cleaved. This is done by proteins called proteases or peptidases.

these cleaving enzymes often only cleave at specific AA residues.

Question 11

what is a cystine molecule?

Answer 11

two cysteine molecules which have formed a disulphide bond. Note: the cystine molecule is more oxidized than cysteine since hydrogen content has been removed

Question 12

what are antioxidants?

Answer 12

molecules that prevent oxidation reactions

Question 13

the inside of the cell is a common place for anti-oxidants. outside of the cell is said to be the oxidizing environment for this reason. With this knowledge, proteins with disulphide bonds would be located where?

Answer 13

outside the cell since the formation of this bond is an oxidation reaction which is hindered inside the cell

Question 14

true or false, protein denaturation disrupts peptide bonds

Answer 14

false, when a protein becomes denatured it is normally due to loss of shape.

Question 15

what 4 ways can you denature a protein?

Answer 15

1. change pH
2. change temperature to an extreme
3. alter the salt concentration (tonicity)
4. urea (fucks with H bonds)

Question 16

what forces hold together protein secondary structure?

Answer 16

hydrogen bonds which form alpha helices and beta sheets.

note: a lot of secondary H bonds come from the “backbone” of the peptide (the amino and carboxyl H bonds)

Question 17

What forces hold together tertiary structure of a protien

Answer 17

Tertiary structure primarily focuses on R group interactions which includes, H bonds, disulphide bonds, ionic interactions, and hydrophobic interactions

Question 18

what is quaternary structure of a protein?

Answer 18

this is when 2 or more separate poly-peptides (i.e. proteins) interact to form a complex.

same forces as tertiary expect for intrinsic peptide bonds.

Question 19

Enzymes: what are hydrolazes, isomerazes, and ligases

Answer 19

Hydrolase –> hydrolyzes chemical bonds (break with water)
isomerase –> rearranges chemical bonds
ligases –> link things via chemical bonds (DNA ligase)

Question 20

Enzymes: what are lyase, kinase, polymerase, phosphatase.

Answer 20

lyase –> breaks chemical bonds not through hydrolysis
kinase –> phosphorylates chemicals
polymerase –> polymerization reactions
phosphatase –> removes phopshate groups

Question 21

what 3 amino acids are subject to phosphorylation?

Answer 21

tyrosine, serine, and threonine due to their hydroxy groups. Note these are all polar AA’s

Question 22

what are cofactors and co-enzymes?

Answer 22

cofactors = typically inorganic molecules which help the activity of an enzyme. Most vitamins and metal ions are cofactors

coenzymes = organic cofactors

Question 23

what is the main difference, if any, between kinases and phosphorylase enzymes?

Answer 23

kinases use ATP to add a phosphate group to a molecule

phosphorylase enzymes use a free inorganic phosphate to add to

phosphatase enzymes remove the phosphate group

Question 24

what is allosteric regulation?

Answer 24

allosteric regulation is when a molecule binds to a site on the enzyme other than the active site to increase or decrease that enzymes activity

Question 25

what is feedforward stimulation?

Answer 25

This mechanism competes with negative feedback to drive the forward process.

E.g. A –> B —> C —> D

one molecule of A can allosterically bind the enzyme that converts C to D to “activate” it.

Question 26

What unit is enzyme kinetics measured in?

Answer 26

Moles / second

Question 27

true or false both substrate and enzyme concentration affect reaction rate. What is saturation?

Answer 27

true, however we typically only consider substrate concentration since enzyme concentration is often quite fixed. If we assume enzyme conc. is fixed, when we add substrate the rxn rate changes proportionally

at the start of the rxn if we double substrate then rxn rate doubles (first order). However eventually the enzymes active site becomes too occupied. The curve slows down until eventually no amount of substrate added will increase rate. This is called saturation of enzymes.

Question 28

What is Vmax and Km?

Answer 28

Both of these values assume constant conc. of enzymes.

Vmax –> the fastest rate enzymes can catalyze a rxn. This is the point when the enzyme is saturated (i.e. no amount of increased substrate will increase rate)

Km –> this is the substrate concentration when the enzyme is at half of Vmax. This value explains substrate-enzyme affinity.

Question 29

If a substrate-enzyme complex has a low Km, what does that say about affinity?

Answer 29

recall Km is the amount of substrate at half full rate (half Vmax)
Therefore, a low Km means a small amount of substrate was required to get to that value, therefore the enzyme must have a fairly high affinity for the substrate.

Question 30

what is positive cooperative binding? What does the graph look like for a cooperative enzyme?

Answer 30

these are enzymes that contain more than one active site. When one molecule is bound the affinity increases to bind another (hemoglobin)

Cooperative enzyme graph (substrate conc. vs. reaction rate)
start = low affinity = gradual slope
middle = increasing affinity = steep slope
end = close to saturation = levels off

loos like an S

Question 31

Explain competitive inhibition. (action, Vmax, Km, graph, etc.)

Answer 31

competitive inhibition is when a molecule directly binds the active site of an enzyme to prevent substrate-enzyme interaction.
This type of inhibition can be overcome by increasing substrate levels.

Vmax - no effect (increase enough [S] will get you there)
Km - must increase [S] therefore Km increases

Graph would remain lower then normal the whole way but eventually match the height of the original graph.

Question 32

Explain Non-competitive inhibition. (action, Vmax, Km, graph, etc.)

Answer 32

A non-competitive inhibitor will bind to an allosteric site of the enzyme which will decrease the catalytic function of that enzyme. However, this binding typically still allows the enzyme to bind the substrate (at the active site) - however the catalytic function is no longer working.

cannot overcome this with increase [S]

Vmax - reduced
Km - unchanged (since binding still occurs)

graph - lower rate at every conc. of substrate

Question 33

Explain un-competitive inhibition. (action, Vmax, Km, graph, etc.)

Answer 33

an uncompetitive inhibitor can only bind the substrate-enzyme complex!! therefore it must wait for them to bind prior to its own binding.

Once bound, it inhibits the E-S complex from forming products by making them stay bound together (decreases Vmax). Since they remain bound, this increases their apparent affinity for each other (lowers Km)

Vmax - decrease
Km - decrease

graph - since the uncompetitive inhibitor must wait for the E-S complex, the reaction rate will seem very quick (steep slope) and then level off drastically

Question 34

explain mixed-type inhibition.

Answer 34

mixed type inhibition occurs when the inhibitor can bind the free enzyme or the E-S complex. In situations like this, Vmax is always lowered but Km is variable.

E.g. If the free enzyme has a higher affinity for the inhibitor then Km will be unchanged since binding occurs but no catalytic function occurs
but if the E-S complex has a higher affinity for the inhibitor than the Km decreases (since apparent E-S affinity has increased)

Question 35

Lineweaver Burke- Plot

What is the slope, x, and y intercepts of the graph.

Answer 35

The slope = Km / Vmax

X intercept = -1/Km

y intercept = 1 / Vmax

Question 36

As y values decrease (i.e. 5 , 4, 3 etc. ) what happens to the value of Vmax?

Answer 36

since the graph is reciprocal, decreasing values for y indicate an increase in V (rate)

Similarly, decreasing X values indicate an increase in substrate concentration.

Question 37

on a lineweaver burke plot (linear line of reciprocal functions) how would a non-competitive inhibitor effect the graph?

Answer 37

we know that a non-competitive inhibitor lowers Vmax but doesn’t alter Km.

Therefore the y intercept much change
The X intercept must remain the same.

since Vmax is lowered and the graph is reciprocal, we need to show this lowered Vmax by increasing Y values.

therefore the Y intercept must increase.