A major obstacle to obtaining useful energy from a nuclear fusion reactor is
containment of the fuel at the very high temperatures required for fusion. The reason
such high temperatures are required is to: - ANSWER--A MAKES NO SENSE
WHY WOULD YOU USE HIGH TEMPS TO ELIMINATE THE NULEAR
FORCE
charge is always conserved no matter the carge
C- the higher density the material, the closer the nucleic are to each oter
Enantiomers can exhibit a difference in which chemical or physical property? -
ANSWER--- Enantiomers retain most properties (density, boiling point, IR spect)
- The strong nuclear force is what holds a nucleus together, fusion would want a
strong nuclear force to hold the. protons together
Which classification of amino acids applies to the Trp residues after photochemical
modification by CCl3CO2H? - ANSWER--Carboxylic is an acid
it is not asking about trp in general but rather the residues
Based on the passage, what most likely causes W96 to be accessible to CHCl3 at
75°C?
A.
Peptide bonds are broken, releasing W96. B.
Reduction of disulfide bonds occurs. C.
The protein unfolds and exposes W96 to the buffer. D.
CHCl3 extracts W96 from the protein interior. - ANSWER--Passage says:
temperature which fully denatures carbonic anhydrase but leaves its primary
structure intact
primary structure= amino acids Peptide bonds are not broke
,What is the name of the ionic compound used to make Buffer B? A.
Ammonium formate
B.
Ammonium carbonate
C.
Ammonium bicarbonate
D.
Ammonium acetate - ANSWER--formate: hco2 carbonate: Hco3-
bicarbonate: hco3- acetate: ch3cooh
Which chromatographic technique would most likely separate a mixture of native
carbonic anhydrase from carbonic anhydrase photochemically modified by
CCl3CO2H? - ANSWER--Passage says: Native carbonic anhydrase also has a net
charge of -2.9 at pH 8.0.
The photochemical adds carboxylic acids which are negatively charged
Pv=nrt - ANSWER--Helps you look at the relationship between everything
4 mols of CO react to give you 2 mol of product
n is halved that mean pressure is halved bc direct relationship 0.5*1
In which phase(s) will the MCS precursor be predominantly found after the
extraction step? - ANSWER--Passage MCS is a lipophillic precursor (it loves lipids)
so it is not water soluble
When the covalent attachment to alliinase is broken, PLP is still held rigidly in the
active site by a salt bridge and a π-stacking interaction. These interactions are most
likely provided by the side chains of which amino acids? - ANSWER--Can eliminate
B and D because they do not include pi stacking (tyrosine pi stacks)
, Now between Asp and Arg. Use passage info
Alliinase is a pyridoxal phosphate (PLP, vitamin B6) dependent enzyme. PLP is a
planar molecule with a 6-membered ring, a hydroxyl group, an aldehyde group, and
a phosphate. PLP is covalently attached to alliinase by forming an aldimine with the
side chain of the amino acid at position 251.
So negative needs a positive to stabilize (phosphate in the group when deprotonated
would be negative)
As the pH of a solution of this amino acid is raised, which group deprotonates first? -
ANSWER--Between I and III
I would be more acidic because the negative charged is stabilized through inductive
effect of the neighboring electronegative chlorines
In the case of III the negative charge would be unstable because of the close charge
seaparation from the amino group
units for capacitance - ANSWER--farad
If both the capacitor and the power supply in Figure 1 are adjustable, which of the
following changes would result in an increase in the charge on the capacitor?
Jack Westin Advanced Solution will be added here.
A.
Decreasing the area of the parallel plates
B.
Decreasing the separation between the parallel plates
C.
Removing the dielectric from the capacitor
D.
Decreasing the voltage of the power supply - ANSWER--Q=CV and also C=AE0/D
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