Chem - Section bank 1

proteolytic cleavage is a type of

hydrolysis reaction

what equation would you use for the magnitude of the electric field

E = V/d

what equation would you use for electromagnetic energy

energy equation. E = P x t

phenols, thiophenols, and alkylbenzenes all share

aromatic rings

what must happen in order to break a bond

photon energy must be larger than the bond energy; high frequency

primary alcohol

Amine+Anhydride→

Amide+Carboxylic acid

anhydrides produce _____ (during nucleophilic acyl substitution reactions)

carboxylic acid byproducts

amides are ____ so they go into the _____

neutral, ether layer/organic layer

Charged molecules go to

aqueous layer

phosphatid (lipid)

pyrophosphate

phosphonic acid

how could liposomes be able to fluoresce during size-exclusion chromatography?

fluorescent dye could be trapped inside

Liposomes can be difficult to detect during experiments like size-exclusion chromatography since they

do not absorb visible light

If products do NOT rearrange after forming, what should you think?; ie stable to mixing

Kinetic control.

If products CAN rearrange into more stable forms, what should you think?; ie form new products after mixture

Thermodynamic control.

what equation would you use for this:

Assume the hydrolysis of ATP proceeds with ΔG′° = –30 kJ/mol.

ATP + H₂O → ADP + P_i Which expression gives the ratio of ADP to ATP at equilibrium, if the [P_i] = 1.0 M? (Note: Use RT = 2.5 kJ/mol.)

ΔG′° = -RT ln(k)

mol=

M x V or grams/molar mass

1.0 mL=

1.0×10^-3L

Which cation is most likely to be found in place of Fe(II) in the square planar binding domain of hemoglobin?

Co 2+

_______ catalyze the transfer phosphate groups from ATP to target proteins and are classified as _______

kinases, transferase

K_M is the concentration of substrate at which the kinetics experiment reaches

half maximum velocity

What happens when a larger amino acid is replaced by a smaller amino acid?

steric hinderance decreases

Why is alanine commonly used in mutation studies?

Because it is small and minimally disruptive.

What kind of side chains increase steric hindrance?

Large bulky side chains,

Ex. tryptophan, phenylalanine, isoleucine

What kind of side chains decrease steric hindrance?

small amino acids, like glycine and alanine

Deprotonation of the reacting water will make it

more nucelophilic (more acidic)

what would be a good reason to use alanine as the replacement residue for each of the single site variants used in a study?

reduces the interaction of the side chain with other active site components.

k_cat is related to

catalysis

what type of mutation is E147D

a conservative mutation - both amino acids have similar properties and charges

what type of bond is broken by this enzyme:

“Lysozyme is an enzyme that kills gram-positive bacteria by cleaving N-acetylglucosamine oligosaccharides”

glycosidic - oligosaccharides are chains of sugars, glycosidic bonds link sugars

this is expressed as y = mx + b

When performing experiments to measure the k_cat of an enzyme, the substrate concentration should be:

saturating

Why do enzymes sometimes protonate substrates during catalysis?

To improve leaving group ability or stabilize intermediates.

“enzyme was prepared by diluting 0.100 mL of the commercial preparation to 25.0 mL in the buffer solution”

buffer equation - 25.0/.100

Dnmt3a is active as a homotetramer - it showed a single band at 35 kDa what is its molecular weight

140

After substrate exposure, the reactions were initiated by addition of ³H-labeled methyl-donating substrate, which undergoes β^– decay with a half life of 4500 days

What fraction of the starting amount of tritium remains after 13,500 days?

A. 1/8

B. 1/3

C. 1/2
D. 3/4

A. 1/8

purified with nickel-nitrilotriacetic acid-agarose chromatography and histine tagging is an example of

affinity chromatography

Avidin: strongly binds biotin

So what happens? Biotinylated substrate (has biotin attached) binds tightly to: avidin plate

So during washing:

it stays attached

if a protein as no disulfide bonds what happens in Nonreducing SDS-PAGE

things run as monomers

if a protein has disulfide bonds what Happens in Nonreducing SDS-PAGE

everything stays linked together

on an SDS page what has the highest electrophoretic mobility

the smallest molecular weight (kDa)

in a mixture of 2:1 hexane/water what layer would steroid hormone, estrogen be in

the hexane layer

in a mixture of 2:1 hexane/water what layer would the peptide hormone, insulin be in

the aqueous layer

a protein with a low pI would be _______ at pH 7

negatively charged

Which amino acid in the substrate binding pocket is most likely to interact with the substrate through its backbone as opposed to its side chain?

glycine

What is the most likely reason for the decreased K_M values observed in the D45G variant compared to the other two versions of GalK with respect to each substrate?

The substrate binding pocket is a better fit.

Which of the four DNA bases contains the largest number of hydrogen bond acceptors when involved in a Watson–Crick base pair?

C

Which of the four DNA bases contains the largest number of hydrogen bond donors when involved in a Watson–Crick base pair?

G

The side chain of which amino acid can form a bond that is similar to a peptide bond?

lysine

Substituting residues in a peptide with which type of amino acida will most likely result in a peptide with an increased pI?

basic amino acids

What is the primary means of stabilization of phosphate bound to MP based on the binding site depiction shown in Figure 1?

hydrogen bonds

Which statement accurately describes the properties of maltose?

A. It is a reducing disaccharide in which only one anomeric carbon is involved in the glycosidic bond.

B. It is a reducing disaccharide in which both anomeric carbons are involved in the glycosidic bond.

C. It is a nonreducing disaccharide in which only one anomeric carbon is involved in the glycosidic bond.

D. It is a nonreducing disaccharide in which both anomeric carbons are involved in the glycosidic bond.

A. It is a reducing disaccharide in which only one anomeric carbon is involved in the glycosidic bond.

Both anomeric carbons bonded is this reducing

no

glycerol - 3 - phosphate

if something is a homodimer, then the disruption of the quaternary structure during SDS would result in a

single gel band

if something is a heterodimer, then the disruption of the quaternary structure during SDS would result in

two bands

the passage states that both ATP and glycerol occupy the catalytic cleft, which means that a

ternary complex was formed.

ATP and glycerol both occupy the catalytic cleft prior to catalysis, and it has been determined that glycerol binds first. Glycerol is stabilized in the cleft through interactions with Arg¹⁸, Arg⁸⁴, Glu⁸⁵, and Asp²⁴⁶.

What is the best description of the catalytic mechanism of GK? Catalysis occurs through:

A. an ordered mechanism in which a ternary complex is formed.

B. an ordered mechanism in which no ternary complex is formed.

C. a random order mechanism in which a ternary complex is formed.

D. a random order mechanism in which no ternary complex is formed.

A. an ordered mechanism in which a ternary complex is formed.

To determine a protein’s thermodynamic stability, chemical denaturation studies can be performed. Assuming that only the native and unfolded states can be observed under experimentally available conditions, what is the most likely shape of the curve for the dependence of the fraction of folded protein upon denaturant concentration?

sigmoidal

Purpose:

• Turn large biomolecules into ions without breaking them apart.

Examples:

• Proteins

• Peptides

• Lipids

• Carbohydrates

All ions are accelerated by the same voltage, so they have the same kinetic energy

maldi - tof

Why use a matrix in maldi tof

To absorb laser energy and prevent fragmentation of the analyte.

Conjugation and electron delocalization =

greater stability = easier detection = absorption at longer wavelengths.

• Resonance

• Conjugation

• Delocalization

• Aromatic stabilization

• Extended π system

• Lower transition energy

• Longer wavelength absorption

"This molecule is probably more stable and more likely to be observed experimentally."

More conjugation → smaller

HOMO-LUMO gap → longer λ absorbed

The energy of the photon must equal the

energy gap

Requires:

• High-energy light

• Short wavelength

Examples:

• Molecules with little conjugation

Large HOMO-LUMO gap

Requires:

• Low-energy light

• Long wavelength

Examples:

• Molecules with lots of conjugation

Small HOMO-LUMO gap

allows electrons to spread out (delocalize).

conjugation

Why does Compound 2b absorb at a longer wavelength?

Compound 2b has greater conjugation, resulting in a smaller HOMO-LUMO energy gap.

A molecule has extensive electron delocalization. What is expected?

• Smaller HOMO-LUMO gap

• Lower transition energy

• Longer wavelength absorption

Which ion arrives first? in maldi tof?

Ion A:
m/z = 50

Ion B:
m/z = 100

ion A

Why is MALDI useful?

Because it ionizes large biomolecules without extensive fragmentation.

The instructor prepared a solution of NaOH(aq) by dissolving 8 g of NaOH(s) (MM = 40.00) in 2 L of H₂O, as shown in Equation 1. The temperature of the solution rose during the mixing process.:

Enthalpy decreased, entropy increased 

in ester hydrolysis what bond is broken:

RCOOR′+H2​O

Acyl-oxygen bond

CH₃​Br + OH− → CH3₃ OH+Br−

How many bonds are broken and formed?

Broken: C-Br (1)

Formed: C-O (1)

1 broken, 1 formed

Peptide hydrolysis

RCONHR

Broken: C-N bond

Formed: C-O bond

Which Carbon is the Reacting Carbon?

C=O because it is electrophilic

carbonyl carbon - most electrophilic

What geometry change occurs when a nucleophile attacks a carbonyl?

Trigonal planar → tetrahedral

add more substrate to overcome

competitive inhibition

hydrolyzed

broken bond wih water

A glutamate residue (Glu-1) in the active site donates a proton to an oxygen atom. Another glutamate on the opposite axial side (Glu-2) acts as a nucleophile to liberate D-glucose and generate an α-D-galactopyranosyl-modified enzyme intermediate. Then, Glu-1 deprotonates water, and the resulting hydroxide ion acts as a nucleophile to liberate β-D-galactose and regenerate the enzyme.

Based on the description provided, if lactose was hydrolyzed under the action of lactase in O-18 labeled water, in which location(s) would the label appear?

A. Neither the galactose nor the glucose products

B. The glucose product only

C. The galactose product only

D. Both the galactose and the glucose products

C. The galactose product only

A glutamate residue (Glu-1) in the active site donates a proton to an oxygen atom. Another glutamate on the opposite axial side (Glu-2) acts as a nucleophile to liberate D-glucose and generate an α-D-galactopyranosyl-modified enzyme intermediate. Then, Glu-1 deprotonates water, and the resulting hydroxide ion acts as a nucleophile to liberate β-D-galactose and regenerate the enzyme.

For what mechanistic reason does G1 of lactase first act as a Brønsted acid during catalysis?

A. G1 becomes a better nucleophile.

B. G2 becomes a better nucleophile.

C. Glucose becomes a better leaving group.

D. Galactose becomes a better leaving group.

C. Glucose becomes a better leaving group.

What is the K_M of Compound 1 with respect to lactase?

A. 0.05 mM

B. 0.10 mM

C. 5 mM

D. 20 mM

D. 20 mM

The initial rates V_o were measured for each trial. The students then plotted 1/V_o versus 1/[S] (Figure 1) to determine K_M, V_max, and [E]_T for the four trials.

What method did the students use to calculate V_max?

A. V_max was calculated as K_M × k_cat.

B. V_max was determined as the inverse of the y-intercept of the graph shown in Figure 1.

C. V_max was calculated as [S] × k_cat for the trial with the highest substrate concentration.

D. V_max was determined as the fastest initial rate of Compound 2 formation obtained during the kinetics experiments from the passage.

B. V_max was determined as the inverse of the y-intercept of the graph shown in Figure 1.

The students prepared stock solutions of Compound 1 in four different concentrations in pH 7 buffer. A stock solution of the enzyme was prepared by diluting 0.100 mL of the commercial preparation to 25.0 mL in the buffer solution. Experiments were initiated by mixing 1.0 mL of each substrate solution with 1.0 mL of the enzyme solution

If [E]_T was the concentration of lactase in the kinetics trials, what expression gives the concentration of lactase in the commercial preparation of this enzyme?

A. [E]_T × 500

B. [E]_T × 50

C. [E]_T × 200

D. [E]_T × 25

A. [E]_T × 500

cytosine modified at C-5 position

is something migrates further in the gel.

that means it is more compacts

Lane 1 = control

Lane 2 = mutant

Band appears only in Lane 1.

Question:

What happened?

Mutant lacks detectable protein.

Lane 1 band is darker than Lane 2.

Question:

Interpretation?

Lane 1 has more protein.; band intensity = amount of proteins

After siRNA treatment, band nearly disappears.

Protein expression decreased.

Based on the results shown in Figure 2, what effect does mutation have on M1 mRNA?

A. The folded structures of wild-type and MBmutant M1 mRNA are similar, but the structure of HPmutant M1 mRNA is less compact.

B. The folded structures of wild-type and MBmutant M1 mRNA are similar, but the structure of HPmutant M1 mRNA is more compact.

C. Wild-type and MBmutant M1 mRNA have the same number of base pairs, but HPmutant M1 mRNA has fewer base pairs.

D. Wild-type and MBmutant M1 mRNA have the same number of base pairs, but HPmutant M1 mRNA has more base pairs

B. The folded structures of wild-type and MBmutant M1 mRNA are similar, but the structure of HPmutant M1 mRNA is more compact.

Example

Normal protein: 100 kDa

Mutant: 70 kDa

truncated protein

Researchers use:

Anti-phosphotyrosine antibody

Question:

What is detected?

Only phosphorylated proteins

Example

Protein X total amount unchanged.

Phosphorylated Protein X increases.

increased activation

Protein band doubles.

Actin band doubles too.

Interpretation?

More total sample loaded - Need normalization.

Example

Protein:

50 kDa monomer

100 kDa dimer

Reducing gel:

50 kDa

Nonreducing gel:

100 kDa

aubunits were linked by disulfide bonds - reducing agent breaks S-S bonds.

Example

SDS-PAGE shows: 25 kDa

Protein known to be homodimer.

Question:

Native protein size?

25×2=50 kDa