Midterm 2 BIOC 202 Flashcards

Question 1

Q

rate of reaction

Answer

A

Given A -> P
We can measure the rate of reaction (i.e. velocity) as either the disappearance of substrate A over time or the appearance of product over time
I.e. reaction rate = V = -△A/△t = △p/△t
Note V is usually expressed in moles/unit time or M/unit time

Question 2

Q

V = k[A]

Answer

A

If we measure the disappearance of A, the rate is directly related to [A]
V = k[A]
Where k is a rate constant (NOT an equilibrium constant)
Rate constant: proportionality constant that relates rate to [S] at a specific temperature
This is known as a first order reaction and usually has units of S-1

Question 3

Q

second order reactions

Answer

A

2A -> P
Where V = k[A][A] = k[A]2
Or, A + B -> P
Where V = [A][B]
Second order reactions usually express k as units of m-1 sec-1

Question 4

Q

Michealis-Menten curve and equation

Answer

A

In a series of separate tubes, take a fixed amount of enzymes and with different amounts of substrate
Then, measure the amount of product formed vs. time
Velocity is the slope of this curve
We see that there is a threshold where [P] no longer changes with time -> the curve plateaus
This is due to product reaching equilibrium (mixture of [S] and [P]) with the substrate

Question 5

Q

How can we relate reaction rate (velocity) to [S]?

To make the math easier, assume to obey Michaelis Menten kinetics:

Answer

A

We only examine early times in the reaction where [P] is low.
Therefore, we can ignore the reverse reaction
[S]»[E], thus we can assume [ES] doesn’t alter [S] -> accurate for biological situations
A steady state exists such that the rate of [ES] formation = rate of [ES] consumption

Question 6

Q

calculate an initial velocity (Vo) from Michealis-Menten curve and equation

Answer

A

For each [S] in fig 6-11, as the slope of the curve at the early times/beginning of reaction (when it is linear)
Note the units of Vo would be the change in the concentration of product/change in time, in this case um/min
This can be graphed as Vo vs. substrate concentration [S] (Fig. 6-12)
Vo is the fastest given curve on Fig 6-11
MUST HAVE STEADY STATE- > for michaelis menten curve
When [S] is low, the curve is linear
As [S] gets higher and higher, the curve begins to plateau -> saturating the enzyme

Question 7

Q

Michealis-Menten curve observation

Answer

A

We observe a hyperbole curve such that low [S], Vo is linear [S], but as [S] increases, Vo levels off

Question 8

Q

Michaelis-menten equation

Answer

A

Vo = Vmax([S]/([S] + Km))
Where:
[S] : substrate concentration
Vo: initial velocity (rate of reaction)
Vmax: the maximum velocity of the rection
I.e. when all of the active sites are saturated with substrate
Km: Michaelis constant
the substrate concentration at which the enzyme catalysed reaction proceeds at ½ its maximum rate
Also a rate constant equal to: (K-1 + K2)/ K1

Question 9

Q

Michealis-Menten curve and equation

When [S] &laquo_space;Km

Answer

A

when [S] &laquo_space;Km, we can say Vo = Vmax( [S] / Km )
Or, Vo = Vmax/Km ([S])
Both Vmax and Km are constants
(note that Vmax does change for different enzyme concentrations)
Vo is directly proportional to [S] when [S] is very low

Question 10

Q

Michealis-Menten curve and equation

When [S] = Km

Answer

A

When [S] = Km -> Vo = Vmax(½)

Vo = Vmax (Km/(Km + Km))

Question 11

Q

Michealis-Menten curve and equation

When [S]&raquo_space; Km

Answer

A

Vo = Vmax ([S]/[S])

I.e. the enzyme is substrate and the velocity is independent of [S] when [S] is very high

Question 12

Q

Km

Answer

A

Km is an important characteristic of an enzyme catalysed reaction
It provides a measure of [S] required for effective catalysis to occur
The higher the Km, the less sensitive the enzyme is to the substrate -> need more of the substrate reach Km
The lower the Km, the less [S] needed for enzyme activity -> better in most cases
It is a rate constant and doesn’t change with enzyme concentration
Putting it another way: it tells us how sensitive the enzyme is to its substrate
(i.e. the higher the Km, the less sensitive the enzyme is to a substrate)
Km is dependent on the enzyme, the type of substrate, pH, temperature, and ionic strength (how much salts there are)

Question 13

Q

Vmax

Answer

A

the maximum rate of enzyme catalysed reaction when all active sites are saturated with substrate
It does change with enzyme concentration
I.e. as [E] increases so does Vmax
We need a way to standardise Vmax with respect to [E], so we can compare different enzymes
This is where Kcat comes in

Question 14

Q

Kcat

Answer

A

Kcat: the maximum amount of substrate an enzyme can convert into product in a given time (maximum rate of enzyme)
Technically measured in min-1 or sec-1
But could say molecules of substrate converted [S] to product [P] per minute per molecule of enzyme (active site of enzyme)
Note: Kcat is the maximum rate per enzyme active site when the active site is saturated
Kcat: turnover number = K2 (rate constant)
Kcat = Vmax/[E]T
Where [E]T = number of active sites
Standardising Vmax to the [E] -> Vmax becomes independent of [E] by dividing by [E]T
Why not just [E]? - because some enzymes contain more than one active site

Question 15

Q

How can we calculate Km and Vmax (and thus Kcat)?

Answer

A

Provided in tables - look it up

2. Use the Lineweaver Burk plot (aka double reciprocal plot)

Question 16

Q

Lineweaver Burk plot (aka double reciprocal plot)

Answer

A

This is the inverse of the Michaelis-Menten equation -> slope is Km/Vmax
1/Vo = 1/Vmax + (Km/Vmax)(1/S)
Y = b + m(x) -> linear equation => Lineweaver Burk Plot
we can calculate: 1/Vmax (y-intercept) and -1/Km (x-intercept)
Not all enzymes obey Michaelis-menten kinetics

Question 17

Q

Irreversible Inhibition

Answer

A

(usually covalent bonds or so many weak interactions) inhibitor that stays bound to an enzyme for long periods of time
Usually binds to and blocks the active site - should resemble substrate
E.g. penicillin (antibiotic) - structure of penicillin resembles/mimics the substrate of glycopeptide transpeptidase (GPT) D-ala-d-ala-peptide
GPT catalyses the final step in the synthesis of bacterial peptidoglycan cell walls
Penicillin binds in the active site and forms a covalent bond with a key serine residue in the active site -> blocks original substrate from binding

Question 18

Q

Reversible inhibition

and its 3 classes

Answer

A

the inhibitor associates and disassociates with the enzyme rapidly 
Grouped as 3 classes:
1. Competitive inhibitors 
2. Noncompetitive inhibitor
3. Uncompetitive inhibition

Question 19

Q

Competitive inhibitors

Answer

A

the inhibitor binds to the active site and competes with the substrate for it
Only one can bind the active site at a time
most common
Catalysis is slowed as there is less ES
Adding the inhibitor increases Km (more [S] to reach Vmax) but Vmax remains unchanged
Competitive inhibitors can be overcome by adding more and more substrate as Vmax remains the same
The inhibitor resembles the substrate
E.g. methotrexate- anticancer drug and makes DNA base T
Methotrexate binds 1000x better than Dihydrofolate by competitively inhibiting dihydrofolate reductase (needed to synthesise nucleotides)
Competes with substrate (dihydrofolate) -> give cancer patient as methotrexate almost to the point of killing the patient
Methotrexate and Dihydrofolate looks very similar

Question 20

Q

Noncompetitive inhibitor

Answer

A

binds to the enzyme at the allosteric site (does not block active site) and alters the enzyme
The substrate can bind but the products are not formed because the noncompetitive inhibitor alters the shape of active site
Cannot compete by adding more substrate
The substrate is no longer perfectly complementary to the active site
Km does not change but Vmax decreases
In effect you are lowering the number of functional enzymes
E.g. low dose doxycycline can non-competitively inhibit collagenase
The remaining enzymes still have the same Km

Question 21

Q

Uncompetitive inhibition

Answer

A

(rare) inhibitor binds to the ES (enzyme-substrate) forming an ESI, preventing formation of product
Like noncompetitive inhibition, this results in a decrease in Vmax (i.e. lowering number of function enzymes)
Km is also decreased because as ES -> ESI, [ES] decreases -> Enzyme appears to be more sensitive to the substrate -> promotes increased substrate binding to the enzyme (ES formation), lowering Km

Question 22

Q

Chymotrypsin functions

Answer

A

a serine protease in the small intestine
Proteins need to be cleaved rapidly and specifically
1. Protein turnover - repairing/replacing damaged protein
2. Digestion - must break proteins down into amino acids when we eat
3. Activation or inactivation of enzymes
4. Formation of multiple polypeptides from a single one

Question 23

Q

Proteases cleave peptide bonds by _____ BUT peptide bonds are stable due to ____

Answer

A

hydrolysis; resonance

Question 24

Q

the carbonyl carbon is less electrophilic (i.e. less δ+) and not as reactive as a regular carbonyl so you need a…

Answer

A

strong nucleophile -> slow cleaving reaction

Question 25

Q

what is Chymotrypsin and what does it do?

Answer

A

(a serine protease in the small intestine) cleaves peptide bonds on the C-terminal side of large hydrophobic aa residue
Such as Phe, Trp, Tyr, Met (similar Km but each will have its own unique Km)
Does not recognize small aa because they do not get enough interaction to stay long enough in the active site -> Km for small hydrophobic aa would be much higher compared to bigger aa

Question 26

Q

Chymotrypsin aa residue numbering (primary structure) reflects the numbering as ..

Answer

A

a single polypeptide (even though the enzyme has been cut into 3 pieces) -> keeps original numbering

Question 27

Q

Chymotrypsin is composed of

Answer

A

It is composed of 3 chains formed by the cleavage of a single polypeptide (tertiary structure)
With a highly reactive serine (S195) in the active site that can act as a strong nucleophile
Chymotrypsin amino acid residue numbering (primary structure) reflects the numbering as a single polypeptide even though it has been cut into 3 pieces)

Question 28

Q

brief mechanism of chymotrypsin

Answer

A

O^δ- on the serine is thought to attach the carbonyl, cleaving the peptide bond and forming an acyl enzyme intermediate (one of the products is attached to the enzyme via covalent bond)
The intermediate is cleaved by water (i.e. water is acting as a nucleophile) to generate a new carboxyl group and the free enzyme

Question 29

Q

chymotrypsin

Tertiary structure shows that S195 can H-bond to the..

Answer

A

imidazole ring of His57 (good buffer - good at being a proton donor and acceptor)
This positions S195 and polarises the hydroxyl group (it has alkoxide ion character -> becomes more negative)
The His is acting as a base catalyst because it is pulling the proton away from S195

Question 30

Q

Asp102 can also H-bond with

Answer

A

His57, positioning His57 and making it a better proton acceptor

Question 31

Q

the catalytic triad

Answer

A

Asp102, His57, and Ser195

Question 32

Q

overview of chymotrypsin mechanism (7 steps)

Answer

A

Substrate binds the active site
The alkoxide ion (O-) on S195 does a nucleophilic attack on the carbonyl carbon of the substrate
The tetrahedral intermediate rearranges to the acyl intermediate
A water molecule enters the active site and His57 H-bonds with it, making the water hydroxide-like (reactive)
Water’s oxygen (O^δ-) acts as a nucleophile, attacking the carbonyl carbon
The tetrahedral anion intermediate rearranges, reforming the carbonyl and separating the product from serine195
The new carboxyl peptide leaves and the enzyme is ready to catalyse another reaction

Question 33

Q

chymotrypsin mechanism

step 2

Answer

A

The alkoxide ion (O-) on S195 does a nucleophilic attack on the carbonyl carbon of the substrate
This causes the carbon to transiently adopt a tetrahedral geometry, forming an oxyanion tetrahedral intermediate
Note: the His57 acted as a base catalyst to facilitate this
This oxyanion is stabilised by H-bonds from the enzyme backbone (specifically Gly 193 and Ser 195) in the oxyanion hole

Question 34

Q

chymotrypsin mechanism

step 3

Answer

A

The tetrahedral intermediate rearranges to the acyl intermediate
The carbonyl reforms and the peptide bond is cleaved
His57 acts as an acid catalyst donating the proton to the amine group to promote this (the collapse of the tetrahedral intermediate)
The new amino peptide fragment leaves the active site
The cleavage reaction of chymotrypsin does not stop here because One product is still bound to serine195
Next series of steps are almost identical to the previous steps except different nucleophile and bond to be cleaved

Question 35

Q

chymotrypsin mechanism

step 5

Answer

A

Water’s oxygen (O^δ-) acts as a nucleophile, attacking the carbonyl carbon
His57 acts as a base catalyst stealing the proton from water, forming another short-lived tetrahedral oxyanion

Question 36

Q

chymotrypsin mechanism

step 4

Answer

A

A water molecule enters the active site and His57 H-bonds with it, making the water hydroxide-like (reactive)

Question 37

Q

chymotrypsin mechanism

step 6

Answer

A

The tetrahedral anion intermediate rearranges, reforming the carbonyl and separating the product from serine195
His57 acts as an acid catalyst, donating the proton back to serine 195
A “new” carboxyl peptide and free Ser 195 OH are generated

Question 38

Q

chymotrypsin mechanism

step 7

Answer

A

The new carboxyl peptide leaves and the enzyme is ready to catalyse another reaction
Note: His57 acts as both an acid and a base catalyst

Question 39

Q

chymotrypsin mechanism

This is how the peptide bond is cleaved but how is the enzyme specific?

Answer

A

The active site contains a deep hydrophobic pocket (Aka S1 Pocket) -> lined with 2 hydrophobic residues: Trp, Met
Large hydrophobic groups can bind to the pocket
The catalytic triad mechanism for cleaving peptide bonds can be found in other proteases
Some are similar in structure to chymotrypsin
E.g. trypsin and elastase have very similar structures to chymotrypsin -> come up with new enzymes by changing the pocket

Question 40

Q

The active site contains a deep hydrophobic pocket (Aka S1 Pocket) -> lined with 2 hydrophobic residues: Trp, Met

Answer

A

Trp’s Nitrogen is pointing away from the pocket so the pocket is a deep hydrophobic pocket
Target peptide must be partly unwound -> pulled through active site if hydrophobic residue comes and gets into the pocket, it stops long enough to do the mechanism cleavage

Question 41

Q

their substrates vary due to changes in the S1 pocket

- trypsin

Answer

A

Trypsin has an Asp189 at the bottom of the pocket, adding a negative charge and binds positively charged amino acid residues such as Arg and Lys
Trypsin does not cut C-terminals of large hydrophobics, Trypsin cuts C-terminal to large positively charged amino residues

Question 42

Q

their substrates vary due to changes in the S1 pocket

- elastase

Answer

A

Elastase has 2 valines in the sides of the S1 pocket, which narrows the pocket and only SMALL hydrophobic residues can bind (e.g. alanine, valine)

Question 43

Q

Other serine proteases do not have structures similar to chymotrypsin but use the same catalytic mechanism
e.g. protease

Answer

A

3 key aa in active site: Ser, His, Asp (catalytic triad)

Question 44

Q

There are 3 other groups of proteases:

Answer

A

Cysteine proteases
Aspartyl proteases
Metalloproteases

Question 45

Q

Metalloproteases

Answer

A

They use a metal to make water a better nucleophile 
Matrix metalloproteases (involved in disease characterised by gum tissue breakdown)

Question 46

Q

Aspartyl proteases

Answer

A

Uses 2 aspartyl groups to make WATER a strong nucleophile
The 2 aspartyl suspend the water and kind of play tug-a-war
E.g. HIV’s protease is an aspartyl protease

Question 47

Q

Cysteine proteases

Answer

A

Uses cysteine as a nucleophile

E.g. COVID-19 main protease (SARS COV-2) - drug tries to inhibit this protease

Question 48

Q

why do we need Hb?

Answer

A

O2 is not soluble enough to supply enough in the blood to supply all tissue
So red blood cells (RBCs) use Hb to carry O2 from the lungs (increase [O2], decrease [CO2], pH 7.4) to a working tissue (decrease [O2], increase [CO2], pH 7.2)
Production of CO2 will lower pH
Hb is perfectly suited to transport O2

Question 49

Q

Hemoglobin structure

Answer

A

a heterotetramer consisting of 2𝛂 2𝜷 subunits (i.e. 𝛂-globin & 𝜷globin -> dimer-dimer)
Alpha subunit is 144 aa 7 𝛂-helices
Beta subunits is 146aa 8 𝛂-helices
Share similar structure and sequences -> may have evolved from one another
Arranged in 2 𝛂𝜷 subunits

Question 50

Q

Hemoglobin

Arranged in 2 𝛂𝜷 subunits

Answer

A

𝛂1𝜷1 interface -> interfaces are very stable with many weak interactions
𝛂2𝜷2 interface
𝛂1 and 𝛂2 are identical
𝛂1𝜷1 - 𝛂2𝜷2 -> interface has few weak interaction - more labile
Most of the interactions that allows Hb to function as an O2 transporter are happening in this interface
A lot of what haemoglobin does occurs in that interface

Question 51

Q

heme group

Answer

A

Each Hb subunit contains 1 heme group -> This is where the O2 binds
1 hemoglobin molecule can bind 4 oxygen molecules
Heme is made up of a protoporphyrin ring with a Fe2+ in the ring centre
Fe2+ has 6 coordination sites
The four are occupied by nitrogens from the ring
The 5th is occupied by a nitrogen below the ring
The 6th is above the ring and is where the O2 binds

Question 52

Q

When the 6th site of the heme group is empty

Answer

A

the iron sits a little outside the ring (too big to fit)

Question 53

Q

When it binds O2 to Hb

Answer

A

When it binds O2, the electron density changes and the iron “pops” into the ring -> pulls the histidine towards the ring
The histidine is attached to the alpha-helix, so in turn the alpha helix moves, disrupting and forming interactions in the a1B1 - a2B2 interface
Histidine is part of an alpha-helix -> when histidine moves, the alpha helix moves with it
This small conformational change increases the other hemes ability to bind O2 by shifting the equilibrium towards the high affinity state (R state)

Question 54

Q

Hb binds O2 cooperatively

Answer

A

Hb with 3 O2 bound has a 20x greater affinity for O2 than when no O2 is bound
aka allosteric interaction: change in shape or activity of a protein (enzyme) that results from the binding of a molecule at a point other than the active site
As Hb binds O2, its affinity for O2 increases
The O2 bound active site of heme groups are acting as an allosteric site because they are altering the other heme groups ability to bind O2

Question 55

Q

As more and more O2 binds to Hb….

Answer

A

the a1B1 dimer rotates relative to the a2B2 dimer by approximately 15% (big twist/conformational change)
The crystal state of hemoglobin will shatter when they switch from T to R state due to the dramatic conformational change
As more and more O2, shifting equilibrium/increasing amount of time in R state

Question 56

Q

Hb has 2 major states:

Answer

A

T state: untwisted deoxyHb (with low affinity for O2)

R state: twisted oxyHb (with high affinity for O2)

Question 57

Q

How does Cooperative binding of O2 help with hb transport of O2?
Cooperative binding illustrated in the O2 binding curve below

Answer

A

X-axis: partial pressure of O2 in the blood = [O2] in blood
Lungs operate at ~100 torr
Working tissue that is actively metabolising is ~20 torr
Y-axis: fractional saturation of Hb = # of heme sites that have O2 bound/ total amount of heme sites available
1.0 = every heme site has O2 bound
If we want to measure how much O2 is transported from lungs to tissue => difference in saturation of O2 in lungs and tissue is the amount of O2 transported/released to tissue from the lungs

Question 58

Q

How does Cooperative binding of O2 help with hb transport of O2?
If Hb was only in the R state

Answer

A

would bind to O2 in the lungs but never release it in the tissue
Affinity too high for O2 -> acting like Myoglobin - bad O2 transportation molecule

Question 59

Q

How does Cooperative binding of O2 help with hb transport of O2?
If Hb was only in the T state

Answer

A

would only bind a small amount of O2 in the lungs and release into tissue

Question 60

Q

How does Cooperative binding of O2 help with hb transport of O2?
Because Hb binds to O2 cooperatively…

Answer

A

it can exist in the R state in the lungs and become saturated with O2
Then travel to the tissue where it would convert to the T state and release its O2
The concentration/partial pressure of O2 will affect how Hb ability to bind to O2

Question 61

Q

2,3 BPG (2,3-bisphosphoglycerate; (-) charge)

Answer

A

Pure hemoglobin still has a high affinity for the R state -> body solves this by adding molecule 2,3 BPG
2,3 BPG is shifting equilibrium to favour the T state
deoxyHb binds 2,3 BPG
This stabilises the T form and lowers the affinity for O2
Binds in the pocket formed by/between the 2B-subunits
The pocket has positively charged aa residues (His + Lys)
2,3BPG is negatively charged
Part of the transition from T -> R state involves the removal of 2,3 BPG
R state is too small to bind 2,3 BPG
The more 2,3-BPG, Hb can stay in T state for higher O2 concentration
2,3 BPG allows the cell to reposition the curve along the x-axis to the right, increasing O2 release in tissue
2,3 BPG shifters the O2 binding curve to the right
Allosteric interaction because the binding of 2,3-BPG on the Hb site other than the heme but it is altering the Heme’s ability to bind to O2

Question 62

Q

The concentration of 2,3 BPG stays ______in the RBC

Answer

A

constant ; does not change in the lungs or tissues

2,3-BPG gets pushed off of the hb but stays in the cytosol of the RBC

Question 63

Q

The Bohr effect

Answer

A

Higher [H+] and [CO2] promote the release of O2 from oxyHb
High H+ and CO2 concentration in a working tissue -> working tissue is consuming O2, producing CO2 and protons as a part of its metabolism
When CO2 dissolves in the blood, it produces protons as well
The harder the cell works, the more Hb wants to unload to O2 - further enhancing Hb’s ability to carry O2 to rapidly respiring/working tissues
Decrease in pH (i.e. increase in H+) shifts the equilibrium shifts to the right
Hb releases more O2 in the working tissue
Does not affect O2 binding in the lungs -> saturated with O2
High [CO2] shifts the curve to the right even more
High [CO2] produces H+
CO2 + H2O <=> H+ + HCO3-
By carbonic anhydrase
Main way to carry CO2 to the blood

Question 64

Q

The molecular basis of the Bohr effect

Answer

A

a) H+: the Beta subunit has His residues in the a1B1-a2B2 interface with pKas close to 7
In conditions of low pH (high H+), they become protonated and positively charged
They can then form new ionic interactions that stabilise the T state (shifts equilibrium to T state)-> lowers affinity for O2
b) CO2: CO2 can bind to free terminal amino groups (N-terminal) to form carbamate
The N terminal amino groups are located in the a1B1 - a2B2 interface
Carbamate ions can form new ionic interactions that stabilise the T state
Hb can carry H+ and CO2 back to the lungs (not the major transporter of CO2 though)
Most of the CO2 is transported back as bicarbonate

Answer 65

A

allosteric;
I.e. they bind at sites other than the heme but they alter the heme’s ability to bind O2
O2 is behaving as a substrate and an allosteric compound

Answer 66

A

high affinity; They behave like they are the R-state (not in R state because there is only one conformation when the subunits are by themselves)
Only get low affinity or possibility of having low affinity T state when the 2 dimers come together to form tetramer

Answer 67

A

You won’t get cooperative binding or the allosteric effects of CO2 and protons with just the heme (biggest reason) -> no R or T state for optimal O2 transport
The heme ion favours carbon monoxide (CO) and binds 25,000x better than O2
Hb lowers the affinity for CO
The iron in a heme can be oxidised to Fe3+ (not expected to know)
He-Fe3+-O-O-Fe3+-He
(forms very tightly and no longer exchanges O2)

Answer 68

A

High in O2
Pushing equilibrium to R -> Vast majority of Hb are in R state
O2 binds Hb and promotes R-state
2,3-BPG is excluded from the Beta pocket -> equilibrium favours the R-state and bind O2 with high affinity
Either the molecule snaps into R and pops out the 2,3-BPG or the 2,3-BPG unbinds for a moment and the heme has already bound to O2
less T state stabilisation as we lose the ionic interactions
Slightly higher pH (7.4)
His deprotonates as pH is above their pKas
Low in CO2
CO2 diffuses off -> losing carbamate

Answer 69

A

Low O2
O2 comes off Hb since the concentration of O2 in environment is low
Losing R state stabilisation, which was stabilised before by high O2 binding
Opens way for 2,3-BPG to bind to Hb -> promotes the T-state and releasing of O2
High CO2 -> as tissue is consuming O2 and producing CO2
More T state stabilisation -> equilibrium shifts to T state
Slightly lower pH (pH 7.2)
His protonate
Carbonate formed from CO2 binding to terminal amino groups

Answer 70

A

a highly integrated network of chemical pathways that enable a cell to extract energy from the environment and use the energy for biosynthetic purposes

Answer 71

A

Catabolism: reactions that transform fuel into useful energy (ATP)
Anabolism: reactions that use energy to form complex structures from simple ones
Synthesising molecules (e.g. proteins)

Answer 72

A

Glucose + O2 -(oxidation)-> CO2 + H2O + energy (ATP)
Includes 10 steps each catalysed by enzymes

Overall: Glucose + 2NAD+ + 2ADP + 2Pi → 2 pyruvate + 2ATP + 2NADH +2H+ + 2H2O
Net ATP: +2 ATP (4 ATP generated from Payoff phase, consumed 2 ATP during the prep phase)

Only thing that keeps RBCs alive

Answer 73

A

The individual reactions must be specific for one set of products (Product of one reaction must be the substrate for the next reaction)
ALL reactions must be thermodynamically favourable/spontaneous
The △G must be negative

Answer 74

A

One way is to couple it to an exergonic reaction (evolution selected for this)
The overall free energy change for a chemically coupled series of reactions = the sum of the free energy changes of the individual steps
For cases where △G is positive (not thermodynamically favourable -> phosphorylation onto glucose - cell’s way of saying “you’re mine”)
Coupling -> e.g. use the very -△G of ATP hydrolysis to drive the +△G of phosphorylation
If 2 reactions are coupled -> can add up individual △G°’ to get the sum of that reaction

Answer 75

A

In the coupled reaction, is a phosphoryl transfer not ATP hydrolysis reaction not occurring
ATP is directly donating its phosphate to glucose = phosphoryl transfer
Glucose + ATP -> glucose-6-phosphate + ADP (-16.7 kJ/mol)

Answer 76

A

the standard free energy change
The knot “°” = standard state -> Where T: 298° K (25°)
all concentrations are at 1M
At pH 7, hence the prime (‘) -> Ignore proton concentration, since pH is buffered

Answer 77

A

Adenosine triphosphate (ATP) hydrolysis

Answer 78

A

ATP is considered the common energy currency of the cell
ATP has 3 major components:
Sugar: pentose called ribose
Purine base: adenine
Series of phosphate groups
closest to the sugar -> Alpha phosphate -> Beta phosphate -> Gamma phosphate -> farthest from the sugar

Answer 79

A

ATP + H2O ⇋ ADP + Pi ; △G°’ = -30.5 kJ/mol
Pi: orthophosphate (aka inorganic phosphate)
More common
ATP + H2O ⇋ AMP + PPi ; △G°’ = -46.6 kJ/mol
PPi: pyrophosphate (inorganic)

Answer 80

A

Ignore H+ product because pH is 7
Ignore water because it is the solvent and there is so much water and the concentration is constant
△G’ = △G°’ + RT(ln([ADP][Pi]/[ATP])
△G°’ = -30.5 kJ/mol
[ADP] = 3.0mM
[ADP] = 0.8 mM
[Pi] = 4mM
T = 310° K
R: 0.008315 kJ/mol K
Always convert [] to M
△G’ = -30.5 kJ/mol + (0.008315 kJ/mol K)(310K)*(ln(0.008315 x 4x10-3/3x10-3))
△G’ = -30.5 kJ/mol + 2.58 kJ/mol *(ln(1.07 x 10-3))
△G’ = -30.5 kJ/mol + 2.58 kJ/mol *(-6.84)
△G’ = -30.5 kJ/mol - 17.6 kJ/mol
= -48.1 kJ/mol -> more spontaneous in cell conditions
This is another way that cells can deal with a reaction that is non-spontaneous because they can alter the ratio of products to reactants

Answer 81

A

cell conditions; This is because concentration of products & reactants are different
Altering product/reactant ratio is another way cells can drive a reaction forward or even make a nonspontaneous reaction spontaneous
If concentration of substrate goes down or concentration of products goes up -> more spontaneous

Answer 82

A

(very negative △G°’)
1. When converting ATP to ADP, there is less charge repulsion from the negative charges
Relief of charge repulsion
At pH 7, ATP: -4 and ADP: -3
2. Resonance stabilisation of Pi
3. Ionisation of ADP at pH 7; ADP-2 ⇌ ADP-3 + H+ (products are more favourable)
4. Greater solvation of the products

Answer 83

A

When you first hydrolyze ATP, the product you get is ADP -2 with OH group (left over from water during attack)
The pKa of OH group is less than 7 -> so at pH 7 the OH group will lose proton -> ionise to form ADP-3 (immediate product of ATP hydrolysis)
Consume immediate product of hydrolysis -> equilibrium shifts towards the products

Answer 84

A

More waters can form H-bonds with the products (ADP and Pi) compared to substrate (ATP) which stabilises products more -> favours products
Although entropy of water is decreasing, the enthalpy of as a result of forming H-bonds is becoming more negative -> enthalpy wins

Answer 85

A

phosphoryl transfer from ATP

Answer 86

A

movement of a phosphate group from 1 molecule to another
This is a way to couple reactions
Looks like a one step mechanism -> but it has multiple steps
Glutamate + ammonia -(coupled with ATP hydrolysis)-> Glutamine

Answer 87

A

Other molecules can also transfer phosphate
Based on their △G°’ of hydrolysis (of the phosphate) we can compare the phosphoryl transfer potential (PTP)
The more negative the △G°’ of hydrolysis -> the greater the phosphoryl transfer potential (PTP)
Thus, something with a higher PTP can easily phosphorylate something with a lower PTP

Answer 88

A

(one of the ways cells make ATP): If you have a molecule whose △G°’ is more negative that ATP, you can use that molecule to phosphorylate ADP into ATP
e.g. Creatine phosphate + ADP ⇌ ATP + creatine

Answer 89

A

In muscles, we have creatine phosphate
When ADP levels build, it starts to phosphorylate ADP into ATP
Reaction catalysed by creatine kinase
At standard state, the △G°’ creatine phosphate hydrolysis; -43.1 kJ/mol (more native than ATP hydrolysis)
△G°’ ATP synthesis; +30.5 kJ.mol
-43.1 kJ/mol + +30.5 kJ.mol -12.6 kJ/mol
Only lasts a few seconds as creatine is used up -> body muscles create ATP with ADP
Only lasts a few seconds because there is a limited amount of creatine phosphate
Creatine phosphate is used to generate ATP in muscle during the first few seconds of heavy muscle contraction when [ATP] are low
Note: the above reaction may not always have a negative △G°’

Answer 90

A

aldehyde or ketone compound with multiple hydroxyl groups with the formula (C-H2O)
-ose = sugar
Sugars can be drawn as linear aldehydes (aldoses) or ketones (ketoses) depending on placement of the carbonyl

Answer 91

A

(trioses)
Glyceraldehyde; aldotriose
3 carbons = triose
Dihydroxyacetone; ketotriose

Answer 92

A

Sugars have chiral carbons -> there exists different enantiomers
Other than dihydroxyacetone
Almost all aa -> L
Almost all sugars -> D
We can use fischer projections to help view this
Designate D and L
When looking at a fischer projection of a sugar with the carbonyl at the top
D-sugar: if the chiral centre furthest from the carbonyl has an OH on the right
L-sugar: If the OH is on the left
Note: there is only one enantiomer for each sugar and it is a perfect mirror image
The enantiomer for D-glucose is a L-glucose

Answer 93

A

multiple chiral centres -> can exist as diastereomers and enantiomers
Diastereomers: isomers that are not mirror image
For (C-H2O)6 ; get 16 different aldoses & 8 different ketones

Answer 94

A

Sugars that differ at only one chiral carbon
E.g. D-glucose & D-galactose ; D-glucose & D-mannose
Body only needs to flip one chiral centre to process in the body

Answer 95

A

because aldehydes and ketones can react with alcohols to form hemiacetals or hemiketals respectively
This can happen when the C5OH of an aldohexose attacks the C1 carbonyl forming a 6 membered ring known as pyranose
OR when the C5OH of a ketohexose attacks the C2 carbonyl, forming a five membered ring known as a furanose
OR when the C6OH of a ketohexose attacks the C2 carbonyl, forming a pyranose

Answer 96

A

the new chiral carbon is formed where the carbonyl was previously
Glucose, this is C1
Fructose, this is C2

Answer 97

A

Below the ring (i.e. opposite the CH2OH) -> designate as alpha
Above the ring (i.e. same side as the CH2OH) -> designate as Beta
These are known as anomers (special type of diastereomer)

Answer 98

A

~⅓ ɑ-anomer, ⅔ 𝛃-anomer and ≤ 1% open chain

Answer 99

A

Conversion between ɑ-anomer & 𝛃-anomer is known as mutarotation
This is a spontaneous process for free glucose and fructose in solution

Answer 100

A

Not a problem because if we are consuming ɑ-anomer glucose constantly and glucose is mutarotating between ɑ-anomer & 𝛃-anomer, ɑ-anomer will be favoured and continued to be formed
As long as the glucose is free to mutarotate, it does not matter which form the enzyme recognizes

Answer 101

A

furanose
Convention is to draw it as a furanose
fructose has 4 different structures but is predominantly in one structure

Answer 102

A

Glucose and fructose are sister -> fructose is the ketose version of glucose (aldose)

Answer 103

A

glycosidic bonds -> i.e. can attach sugars
The anomeric carbon is especially reactive because it has 2 oxygens pulling on it
Can link glycosylation, other sugars, purines, and pyrimidines

Answer 104

A

sugars to proteins
By far the most common type of glycosylation with OH groups is with serine
Can also be done with Asp (can be Glu but most enzymes recognise Asp)
Enzymes are stereospecific and will only produce one form
Note: these sugars can no longer mutarate (no OH group -> replaced with OR) -> they are locked in the ring

Answer 105

A

Both the composition in type of sugars and the linkage is important
Common disaccharides: lactose (B form), sucrose, trehalose
Sucrose: glucose (𝞪1 ->𝜷2) fructose
Lactose: galactose (𝜷1 -> 4) glucose
glucose of lactose can mutarotate -> free C1 OH

Answer 106

A

the sequence of reactions that breaks down 1 molecule of glucose into 2 molecules of pyruvate with the net production of 2 ATP
Glyco- (sugar); lysis (to break)
Anaerobic process (no O2 required) -> pretty much to only catabolic pathway in the body that can function without O2

Answer 107

A

Gummy bears are mostly glucose -> reacting with potassium chloride (has lots of O2 in it) -> glucose is gummy bear is oxidised to CO2 and water -> sound/light/heat energy released

Answer 108

A

CO2 and H2O via acetyl CoA, the Krebs Cycle, and oxidative phosphorylation (when O2 is present)
OR it can be fermented into lactic acid or ethanol (yeast and bacteria)
Pretty much all of earth’s alcohol supply is from fermentation of glycolysis

Answer 109

A

In order to make ATP, you must spend ATP

Answer 110

A

Preparative: the trapping and destabilisation of glucose
Cleavage of phosphofructose
Payoff: ATP generation and pyruvate generation
Glycolysis occurs in the cytosol -> some cell types do not have mitochondria
We want to cleave glucose into 2 3-carbon identical pieces -> if not, we would need 2 separate processes for each piece

Answer 111

A

enzymes that transfer a phosphate group from ATP to another molecule (or more uncommonly, vice versa - can occasionally use another molecule to create ATP out of ADP)
They phosphorylate things
Glucose induces a conformational change in hexokinase such that the glucose becomes enclosed in the protein except for the C6 OH which sticks out into active site and receives the phosphate (induced fit)

Answer 112

A

glucose is phosphorylated using ATP by hexokinase on C6, yielding glucose-6-phosphate
The phosphate is not going to allow the glucose to leave the cell, and in turn, this reaction traps the glucose in the cell
Charged group - can’t move back across the membrane
Destabilizes glucose
This is an irreversible reaction (1 sided arrow) in a cell
Phosphate group on C6 because the phosphate will be used to make ATP later, so it must be positioned properly
By phosphorylating on C6, you have left the ability for glucose ring to open and glucose
Glucose C6 can mutarotate
To make 2 identical pieces, if phosphate is added on C6, phosphate must also be added on to C1
Hexokinase closes around glucose -> bury all of the OH’s of the glucose such that only the C6 OH in the active site is positioned right next to the ATP -> induced fit
Excludes water to prevent the water from hydrolysing the ATP, ensuring phosphate is transferred onto C6

Answer 113

A

Glucose-6-phosphate is converted into fructose-6-phosphate by phosphoglucose isomerase (aka phosphoglucoisomerase aka glucose-6-phosphate isomerase)
Isomerases move bonds around (double bonds) -> they do not move atoms other than Hydrogens around
Converted into fructose because fructose is more identical and we need as identical 2 pieces as possible
If glucose was cleaved, it would be a 4-carbon piece and a 2-carbon piece, whereas cleaving a fructose would give 2 3-carbon pieces (more symmetrical)
Converting an aldose into a ketose; positions of OH groups are the same between glucose and fructose
Multi Step process

Answer 114

A

Phosphofructokinase (PFK1) phosphorylates fructose-6-phosphate to fructose-1,6-bisphosphate (bis = at different part of chain; di = 2 connected together) using ATP
Irreversible reaction
The committed step -> if you cross this reaction, you are doing glycolysis
Key regulatory enzyme of glycolysis -> trying to control speed of glycolysis

Answer 115

A

Aldolase cleaves fructose-1,6-bisphosphate into dihydroxyacetone phosphate and glyceraldehyde-3-phosphate
Now have 2,3 carbon fragments -> The products are the simplest sugars -> trioses
Only difference in the triose -> one is aldose and the other is the ketose
Aldolase works on the linear form of fructose -> because the anomeric carbon for fructose is on C2, that sugar is mutarotate
Enzyme is promoting the ring to open -> only requires one cut
Carbon 1,2, 3 become dihydroxyacetone phosphate -> convert into aldehyde form by using triose sugar phosphate isomerase to move the double bond -> Reaction 5
converted to use the same enzyme for both 3-carbon pieces
Carbon 4,5,6 become glyceraldehyde-3-phosphate -> we only need aldose for rest of glycolysis
Reversible reaction

Answer 116

A

Fate of glucose carbons after cleavage by aldolase
We only the aldehyde (Glyceraldehyde 3-phosphate)
Use triose phosphate isomerase to move the double bond on Dihydroxyacetone phosphate
Triose phosphate isomerase converts dihydroxyacetone phosphate (DHAP) to glyceraldehyde-3-phosphate (GAP)
Reversible reaction
Works through an enediol intermediate
-There are 2 glyceraldehyde-3-phosphate
Everything going through the pathway from now on is x2 because there is 2 identical 3 carbon sugars

Answer 117

A

convert glucose-6-phosphate to open chain -> convert it to fructose-6-phosphate by repositioning the double bond -> then form the cyclic furanose
If you cannot mutarotate, then you cannot open the ring -> reaction 1 phosphorylates C6 instead of C1 (glucose-1-phosphate)
△G°’ = 1.7kJ/mol -> only a little positive
Increase glucose-6-phosphate or decrease fructose-6-phosphate
Reversible reaction

Answer 118

A

Arguably the most important reaction in glycolysis -> if this step does not occur, cannot generate ATP from glycolysis
GAP is oxidised and phosphorylated to 1,3-bisphosphoglycerate by glyceraldehyde-3-phosphate dehydrogenase
requires NAD+ and generates NADH
Dehydrogenase: remove hydrogen, and more importantly -> redox reaction (oxidise things)
1,3 bisphosphoglycerate (1,3-BPG) has a high phosphoryl transfer potential -> specifically in C1
Previously, we were consuming ATP to phosphorylate things -> now, we are attaching inorganic phosphate by coupling to drive phosphorylation

Answer 119

A

phosphoryl transfer reaction from 1,3BPG to ADP (phosphoglycerate kinase)
1,3-bisphosphoglycerate phosphorylates ADP forming ATP and 3-phosphoglycerate -> done by phosphoglycerate kinase
Example of substrate level phosphorylation
The formation of ATP from ADP in which the phosphate donor is a substrate (ATP) with higher phosphoryl transfer potential
Kinase runs in reverse
Because there were 2GAPs, generate 2 ATP -> net ATP in glycolysis is 0 at this step

Answer 120

A

Reaction 8: Intramolecular shift of the phosphate group from C3 to C2 (phosphoglycerate mutase)
3-phosphoglycerate has a lower phosphoryl transfer potential than ATP -> must convert 3-phosphoglycerate into a molecule with high phosphoryl transfer potential
Phosphoglycerate mutase converts 3-phosphoglycerate to 2 phosphoglycerate
Done because we need to control where carbonyl will appear and set up for next reactions
Mutase: an enzyme that causes an intramolecular shift of a group -> i.e. takes atoms off and puts it somewhere else

Answer 121

A

Enolase removes water from 2-phosphoglycerate forming an enol : phosphoglycerate
Redox dehydration -> removing water and inducing double blood (oxidation)
Electrons leaving with the water
Enolase makes enol
Enol is more unstable than keto -> molecule wants to be keto, but phosphate is in the way (enol form is trapped by the phosphate)
enol has high phosphoryl transfer potential

Answer 122

A

Pyruvate kinase transfers (named for reversed reaction- Substrate pyruvate is phosphorylated) a phosphate from phosphoenolpyruvate to ADP, forming ATP and pyruvate
In biological systems, this is an irreversible reaction
ATP gained here is for profit

Answer 123

A

Enz-His(in active site)-(P) + 3-phosphoglycerate(substrate) ⇌ [Enz-His + 2,3-bisphosphoglycerate]intermediate ⇌ Enz-His-(P) + 2-phosphoglycerate(product)

Answer 124

A

Prime the enzyme with a small amount of 2,3 bisphosphoglycerate (2,3BPG)

Answer 125

A

Oxidation: the loss of electrons (e-)
Reduction: the gain of electrons (e-)
In biological systems, usually accompanied by the addition or removal of pairs of H ions

Answer 126

A

oxygen; I.e. 2H+ + 2e- + ½O2 → H2O

Answer 127

A

carriers are used to transfer e- (and H+) to the electron transport chain or to biosynthetic reactions

Answer 128

A

oxygen; I.e. 2H+ + 2e- + ½O2 → H2O

- O2 is garbage dump to get rid of spent electrons

Answer 129

A

NAD: nicotinamide adenine dinucleotide
NAD+/NADH has an H on ribose C2
NADP: nicotinamide adenine dinucleotide phosphate
NADP+/NADPH has a phosphate on ribose C2
Derived from ATP and niacin (vitamin B3)
NAD+ + 2e- + 2H+ ⇌ NADH + H+
Generally, NAD+/NADH is used in catabolic reactions
NAD is free to move around the cell but cannot cross membranes (too big and bulky)
NADP+/NADPH is used in anabolic reactions
Be able to identify the oxidised and reduced form of these molecules (do not need to know how to draw)

Answer 130

A

FAD: FLavin adenine dinucleotide
FMN: Flavin mononucleotide
Note: FMN is FAD minus AMP
Derived form ATP and riboflavin (vitamine B2)
FAD + 2H+ + 2e- ⇌ FADH2
- Be able to identify the oxidised and reduced form of these molecules (do not need to know how to draw)

Answer 131

A

NAD: nicotinamide adenine dinucleotide
NAD+/NADH has an H on ribose C2
NADP: nicotinamide adenine dinucleotide phosphate
NADP+/NADPH has a phosphate on ribose C2
Derived from ATP and niacin (vitamin B3)
NAD+ + 2e- + 2H+ ⇌ NADH + H+
Usually, When doing an oxidation reaction, 2 H+ come off
Binding 2 electrons and a proton = hydride H-

Generally, NAD+/NADH is used in catabolic reactions
NAD is free to move around the cell but cannot cross membranes (too big and bulky)
NADP+/NADPH is used in anabolic reactions
Be able to identify the oxidised and reduced form of these molecules (do not need to know how to draw)

Answer 132

A

FAD: FLavin adenine dinucleotide
FMN: Flavin mononucleotide
Note: FMN is FAD minus AMP
Derived form ATP and riboflavin (vitamine B2)
FAD + 2H+ + 2e- ⇌ FADH2
FAD and FMN will be bound to different enzymes - to pass electrons around within enzyme
Not free to move around cell
- Be able to identify the oxidised and reduced form of these molecules (do not need to know how to draw)

Answer 133

A

The pool of NAD+/NADH is small -> NAD+ must be regenerated in order for glycolysis to continue otherwise glycolysis will stop
I.e. something must be done with the high energy electrons that were generated
There are 3 fates of pyruvate

Answer 134

A

NAD+ is regenerated from NADH by the e- transport chain
O2 is the final e- acceptor, water (as O2 transforms into water) is the final e- carrier
Pyruvate is converted into acetyl CoA and oxidized further in the Krebs cycle

Answer 135

A

In mammals, lactate dehydrogenase reduces pyruvate to lactate in the process, converting NADH to NAD+ by fermentation
Pyruvate is the oxidized form of lactate
Lactate is the reduced form of pyruvate
Allows glycolysis to function under anaerobic conditions
RBCs do this as they do not have mitochondria and must ferment
Flight or fight mode: push muscles into fermentation mode
The final e- acceptor is pyruvate
The final e- carrier is lactate

Answer 136

A

In yeast and ecoli: pyruvate is first to decarboxylated by pyruvate decarboxylase (releasing CO2) to turn into acetaldehyde and then reduced to ethanol by alcohol dehydrogenase
Ethanol fermentation produces CO2 from carboxylic acid -> the CO2 is why there are bubbles in alcoholic beverages
Yeast will try to do this reaction even when there is oxygen -> try to eat up all the glucose, pump out ethanol into surrounding medium to kill everything around and aerobically break down ethanol
Notice that NADH is converted back to NAD+ in the second step
Acetaldehyde is the final e- acceptor
Ethanol is the final e- carrier

Answer 137

A

Phosphofructokinase (PFK)
Pyruvate kinase (PK)
*hexokinase
Different cell types will regulate glycolysis differently and may have different rates of glycolysis

Answer 138

A

High levels of [ATP] allosterically inhibit PFK
high levels of [AMP] activates PFK because high [AMP] = low [ATP]
High [fructose-2,6-bisphosphate] activates PFK -> only occurring in the liver
Hormonally induced cell signalling produces this molecule
This is how glucagon and insulin will affect the liver’s glycolytic rate
High [citrate] inhibit PFK
Citrate is an intermediate of the citric acid cycle (the Krebs Cycle) -> allows cross talk between Krebs cycle and glycolysis
Citate is also a biosynthetic precursor
High [H+] inhibit PFK
Normally not a problem due to buffering system and the pumps that pump out H+
When the cell gets too acidic -> stop making PFK
But if the cell needs ATP, the process will still occur (if the cell does not have ATP, the cell dies)
Glycolysis produces some protons

Answer 139

A

small changes in [ATP] result in larger % changes in AMP because concentration is kept low (more sensitive and allows cells to make much smaller changes)
ADP + ADP ⇌ ATP + AMP

Answer 140

A

High [ATP] inhibit PK in liver
High [alanine] inhibit PK -> alanine and pyruvate are sisters in terms of structure (are interchangeable)
Pyruvate and alanine levels are linked as alanine is made from pyruvate
High [alanine] => high [pyruvate]
High [fructose-1,6-bisphosphate] activate PK
The product of phosphofructokinase catalysed reaction is activating pyruvate kinase; The top of glycolysis ordering/stimulating the bottom of glycolysis to speed up -> feedforward stimulation
Upper fortune of glycolysis stimulating the lower portion
Pyruvate kinase is also activated by hormonal signalling

Answer 141

A

This regulation does not occur in the liver
Many cells will have High [glucose-6-phosphate] that will inhibit hexokinase allosterically
Prevents excessive trapping of glucose - hexokinase catalyzes the trapping of glucose

Answer 142

A

Catabolism of proteins, fats and proteins in the 3 phases of cellular respiration
Glycolysis: turned glucose into pyruvate and gotten 2 ATP and high energy e-
We want to use the products of glycolysis to put into Kreb’s Cycle -> need to be in the form of acetyl coA
There is bridging reaction catalysed by pyruvate dehydrogenase complex -> connects glycolysis and the Krebs cycle by converting pyruvate to acetyl coA

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Midterm 2 BIOC 202 Flashcards

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