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VCU Biochem Exam 2 Questions And Answers
is binding of ligands reversible? - ANS-yes, it is essential for it to be reversible
cooperativity - ANS-In multisubunit proteins, conformational changes in one subunit can affect the others
functions of globular proteins - ANS-1. storage of ions and molecules (myoglobin and ferritin)
- transport of ions and molecules (hemoglobin, serotonin transporter)
- defense against pathogens (antibodies, cytokines)
- muscle contraction (actin, myosin)
- biological catalysis (chymotrypsin, lysozyme)
ligand - ANS-a molecule that binds to a protein
(typically a small molecule)
binding site - ANS-region in the protein where the ligand binds
(ligand binds with noncovalent forces)
what does the fraction of bound sites depend on? - ANS-1. the free ligand concentration (known)
- Kd (determined graphically or least-squares regression)
what is binding expressed in terms of when ligand is a gas - ANS-partial pressure
Kd (dissociation constant) - ANS-[P][L]/[PL]
Theta (fraction of occupied binding sites) - ANS-[L]/ ([L] + [Kd])
what does L = when theta is 0.5? - ANS-Kd
theta for oxygen binding to myoglobin - ANS-pO2/ (p50 + pO2)
Gibb's free energy equation relating enthalpy to entropy - ANS-Go = Ho -TS
Gibb's free energy equation related to Kd - ANS-Go = -RTlnKd
if the Kd is low (less than 10nM) - ANS-strong binding and doesn't want to separate
if the Kd is high (greater than 10uM) - ANS-weak binding and will readily let go
lock and key model - ANS-- high specificity
- complementary in size, charge, shape hydrophilic.hydrophobic character (preformed)
- If lock and key was right you would never get out of transition state to get to product, so its not right
induced fit - ANS-- conformational changes occur upon ligand binding
- allows for tighter binding of the ligand
- allows for high affinity for different ligands
- You don't want it to fit perfectly at first. You want it to fit perfectly later in the reaction during the transition state to promote the continuation of the rxn
main oxygen storage protein - ANS-myoglobin
circulating oxygen-binding protein - ANS-hemoglobin
Heterotropic - ANS-Different ligand affects binding of the normal ligand
hemoglobin structure - ANS-tetramer: 2 alpha subunits and 2 beta subunits
(each subunit is similar to myoglobin)
for hemoglobin, where does the strongest subunit interactions occur? - ANS-between unlike subunits (ex. large change at alpha1beta2 contact)
tense state - ANS-- More interactions, more stable
- Lower affinity for O
- tissues
R = Relaxed state - ANS-- Fewer Interactions, more flexible
- Higher affinity for O
- lungs
O2 binding triggers a ______ conformational change - ANS-T -> R
(breaking ion pairs between the α1- B2 interface)
The ____ difference between lungs and metabolic tissues increases efficiency of the O2 transport - ANS- pH
bohr effect - ANS-- shifts curve to the right
- lower affinity for oxygen (greater release to tissues)
- C
-2,3 DPG
carbamate - ANS-- yields a proton which contributed to bohr effect
- forms salt bridges stabilizing the t state
how is CO2 exported? - ANS-1. carbamate on amino terminal residues of each of the polypeptide subunits
- as dissolved bicarbonate
-Formed by carbonic anhydrase, and also producing a proton
2,3 BPG - ANS-- negative heterotropic regulator of hemoglobin function (sends signal to hemoglobin to release oxygen because body will need oxygen in ETC)
-stabilizes the t state
- Small negatively charged molecule,
binds to the positively charged
central cavity of Hb
Deoxyhemoglobin (in venous blood) appears what color? - ANS-purple
oxyhemoglobin (in arterial blood) appears what color? - ANS-red
The heme group is a strong chromophore that absorbs both in _____ and _____ - ANS-ultraviolet and visible range
transition state theory - ANS-- in the [S] there is a distribution of energetic forms and one such form is referred to as the transition state between S and P.
- The reaction only proceeds through this form
- rate of the reaction really is -dS/dt = k[S] where S is the transition state
formula for K* - ANS-K= [S]/[S]
K* is the equilibrium constant between the transition state and the ground state
the magnitude of the ____ determines the rate of the reaction - ANS-∆G*
The lower this value is the faster the reaction goes
as Keq increases, what happens to ∆G? - ANS-∆G decreases
how enzymes catalyze reactions - ANS-1. General acid/general base
- Electrostatic
- Change the reaction mechanism
- Proximity and orientation effects
- Entropy considerations
assay - ANS-- Monitor the rate of the reaction
- Can measure the disappearance of substrate as a function of time
- Can measure the appearance of product as a function of time
- Rate of the reaction = -d[S]/dt or d[P]/dt
1 unit of activity - ANS-amount of enzyme that catalyzes the conversion of 1 micromole of substrate to product in 1 min
specific activity - ANS-the number of units of activity per milligram of protein
kinds of assays - ANS-1. Fixed point assays
- Continuous assays
- Spectrophotometric assays
- Coupled assays
essential features of assays - ANS-- must be linear with time
- must be proportional to enzyme concentration
formula for Vo - ANS-Vo = Vmax[S]/ Km
slope for lineweaver burk graph - ANS-Km/Vmax
competitive inhibitor - ANS-- Km increases
noncompetitive inhibitor - ANS-- Km unchanged
uncompetitive inhibitor - ANS--Km decreases
