Unit 1 - Lecture Quizzes

Question 1

what is the enthalpy ot a chemical reaction that is spontaneous at 3/°C and has an entropy change of 35 J^-1mol^-1?

a) 5.6 kJ/mol
b) 10.8 kJ/mol
c) 35 kJ/mol
d) 62.4 kJ/mol
e) Not enough information is given to answer this question.

Answer 1

e) Not enough information is given to answer this question.

Question 2

For a given reaction, 18.6 kJ × mol^-1 of heat is absorbed and the change in entropy is 62 J × K^-1 × mol^-1. What is the lowest temperature above which the reaction will be spontaneous?

a) 43 k
b) 294 K
c) 300 K
d) 3K
e) 308 k

Answer 2

c) 300 K

Question 3

Which of the following statements is false?
a) Hydrogen bonding strength depends on the orientation of the atoms involved
b) lonic interactions are stronger in non-polar environments than in polar
environments.
c) Ionic interactions do not depend on the relative orientation of the ions
d) Hydrogen bonds are weaker than covalent bonds
e) Van der Waals interactions are too weak to be a significant attractive force.

Answer 3

e) Van der Waals interactions are too weak to be a significant attractive force.

Question 4

Which one of the following is NOT an example of a hydrogen bond?
a) -C-H—–O
b) -O-H—–N
c) -N-H—– O
d) -N-H—–N
e) -O-H——O

Answer 4

a) -C-H—–O

Question 5

Which molecule would you expect to be non-polar?

A. H2O
B. CO2
C. NH3
D. CHCl3
E. These molecules have equal polarity

Answer 5

B. CO2

Question 6

For a chemical reaction, if the change in enthalpy is negative and the change in entropy is positive, the reaction will be:

A. Never spontaneous
B. Always spontaneous
C. Sometimes spontaneous
D. Not enough information is given

Answer 6

B. Always spontaneous

Question 7

Which compound is the most acidic?

A. Acetic acid (pKa = 4.76)
B. Trifluoroacetic acid ( pKa = 0.18)
C. Phosphate (pKa = 6.82)
D. Boric acid (pKa=9.24)
E. Sodium hydroxide

Answer 7

B. Trifluoroacetic acid ( pKa = 0.18)

Question 8

If your kidneys are unable to function properly, you may expect to have:

A. Decreased levels of HCO 3- in the blood, resulting in a decrease of blood pH
B. Increased levels of HCO 3- in the blood, resulting in a decrease of blood pH
C. Decreased levels of HCO 3- in the blood, resulting in an increase of blood pH
D. Decreased levels of HCO 3- in the blood, resulting in an increase of CO 2 concentration
E. None of the above

Answer 8

A. Decreased levels of HCO 3- in the blood, resulting in a decrease of blood pH

Note:
- Equilibrium shifts left due to insufficient HCO 3- (renal problems) or high CO 2 (poor lung function)

Question 9

A solution contains a typical protein containing many acidic and basic amino acid residues. If the pH of the solution is less than the pI of the protein:

A. The protein will tend to be deprotonated, causing it to have a net positive charge
B. The protein will tend to be protonated, causing it to have a net positive charge
C. The protein will tend to be deprotonated, causing it to have a net negative charge
D. The protein will tend to be protonated, causing it to have a net negative charge
E. The protein will have a net charge of 0.

Answer 9

B. The protein will tend to be protonated, causing it to have a net positive charge

Question 10

What is the approximate isoelectric point of the following hexapeptide. Consult your notes to obtain relevant pKa values.

Asp-Leu-Ile-Glu-Lys-Gly?

Question 11

Which of these statements about protein folding is
false?

A. The native state is stabilized by the high entropy of water (hydrophobic effect)
B. The native state is destabilized by the low entropy of protein
C. Certain residue orientations (phi/psi angles) are more favourable than others
D. When folding, a protein samples all possible conformations to find the native state
E. The native state is stabilized by favourable enthalpic interactions, such as hydrogen bonds

Answer 11

D. When folding, a protein samples all possible conformations to find the native state

..false, proteins don’t randomly test of states they are guided through various conformations that are unstable until it finds stable structure..

Question 12

Which DNA sequence would you expect to have the
highest melting temperature?

A. GCCGCGGCCCGG
B. ATCAATACTTAT
C. GCAACTACTGCA
D. AACCTTGGAAAG
E. Not enough information is given

Answer 12

A. GCCGCGGCCCGG

Question 13

Which of the following mutations cannot result in a
frameshift?

A. Insertion
B. Deletion
C. Substitution
D. Dinucleotide repeat expansion
E. All of these mutations can results in a frameshift

Answer 13

C. Substitution

Question 14

What would be the resulting pH if one drop (0.05 ml) of 1.0 M HCI were added to one liter of pure water (assume initial pH 7.0?

a) 2.7
b) 4.3
c) 5.0
d) 9.7
e) 7.0 (there would be no significant change)

Answer 14

b) 4.3

Question 15

What is the resulting pH if 1 mL of 10 M HCI is added to 1 liter of 0.1 M sodium phosphate buffer (pH 7.00). The pKa of sodium phosphate is 6.82.

a) 6.82
b) 6.98
c) 7.01
d) 7.19
e) Cannot be determined

Answer 15

a) 6.82

Question 16

Which of the below statements is false?

a) Bicarbonate is an important extracellular buffer.
b) A buffer is the least effective when the solution pH is the same as the buffer’s pKa.
c) Phosphate is an important intracellular buffer.
d) Many biomolecules can behave as buffers.
e) Le Chatelier’s principle is important for pH regulation in the body.

Answer 16

b) A buffer is the least effective when the solution pH is the same as the buffer’s pKa.

…false, that’s when its most effective..

Question 17

Which of the below statements is false?

a) A buffer is the most effective when the solution pH is the same as the buffer’s pKa.
b) In aqueous solutions the concentration of OH and H* is always equal to 10-7 M.
c) Carbon, nitrogen, oxygen, and hydrogen are some of the most abundant elements in living things.

Answer 17

b) In aqueous solutions the concentration of OH and H* is always equal to 10-7 M.

..Its actually equal to 10^14M…

Question 18

What would be the resulting pH if one drop (0.05 ml) of 1.0 M HCI were added to one liter of pure water (assume initial pH 7.0)?

a) 2.7
b) 4.3
c) 5.0
d) 9.7
e) 7.0 (there would be no significant change)

Answer 18

b) 4.3

Why?
- As HCI is a strong acid, the final concentration of (H*] would be the same as the concentration of HCI added, i.e. 1.0 M × (0.05 ml)(1000 ml) = 5x10-S M. Taking the negative log, the pH would be 4.3.

Question 19

What is the resulting pH if 1 mL of 10 M HCI is added to 1 liter of 0.1 M sodium phosphate buffer (pH 7.00). The pKa of sodium phosphate is 6.82.

a) 6.82
b) 6.98
c) 7.01
d) 7.19
e) Cannot be determined

Answer 19

a) 6.82

Why?
- Initially, 7.00 = 6.82 + log(HPO42-1/H-PO4 )), thus [HPO47-11H2P04] = 1.51.
- Since the total buffer concentration (HPO43-1+(H,PO4] = 0.1 M, then solving for (HPO-j: 0.1 M = (H-PO41+
(1.51)(HPO42-J, thus [HyPO4] = 0.IM/2.51 = 0.04 M, or 0.04 mol in 1 litre. Thus, [HPO42-J= 0.1 - 0.04 = 0.06 M
(0.06 mol in 1 litre).
- Adding 10 mmol (0.01 mol) of HCI will titrate that amount of (HO 2- to [HyPO4 1, so the final ratio will be (0.06-0.01)/(0.04+0.01) = 0.05/0.05 = 1.
- Since log(1) = 0, the final pH will be 6.82 (which happens to be equal to the pK).

Question 20

The hydrophobic effect is largely responsible for correct protein folding and the association of lipids into bilayers and other structures. The basis of the hydrophobic effect is:

a) Increased dielectric constant of water relative to non-polar solvents.
b) Increased water entropy due to minimized contact of water with non-polar groups.
c) Increased hydrogen bonding in water molecules surrounding non-polar molecules.
d) Vibration of water molecules adjacent to non-polar molecules.
e) Attraction of closely packed non-polar groups due to van der Waals interactions.

Answer 20

b) Increased water entropy due to minimized contact of water with non-polar groups.

Question 21

The side chain of aspartic acid normally has a pK of 3.9. At which pH value would you expect about 90% of the Asp side chains to be in the negatively charged (COO) form?

a) рн 2.9
b) рн 3.9
c) pH 4.9
d) рн 5.9
e) pH 8.6

Answer 21

c) pH 4.9

Why?
- According to the Henderson-Hasselbalch equation, the value of log[A*J/ [HA] would be +1 when the ratio of [A”/[HA] is 10, or about 90% of Asp side chains in the carboxylate form, corresponding to a pH value one unit above the pK.

Question 22

Only the amino acid can spontaneously form covalent bonds between two different polypeptide chains, while
is often used to monitor protein purification because of its ability to absorb ultraviolet light.

a) proline; tyrosine
b) cysteine, tryptophan
c) methionine; cysteine
d) glutamic acid; valine
e) methionine, histidine

Answer 22

b) cysteine, tryptophan

Why?
- When oxidized, two cysteine residues can form a disulfide bond between (or within) polypeptide chains, while tryptophan has the highest UV absorbance of all the amino acids (at 280 m).

Question 23

What is the approximate net charge of the pentapeptide Ala-Arg-Ser-Glu-Asn at pH 7?

a) -2
b) -1
c) 0
d) +1
e) +2

Answer 23

c) 0

Why?
- At this pH, the charged groups will consist of the N-terminal amino group (+), the guanidinium side chain of Arg (+), the Glu carboxyl side chain (-), and the C-terminal carboxy group (-). Note that peptide bonds are not charged. Thus, the net charge of the peptide is zero, as these all cancel out. N.B. while you are not expected to memorize exact pK values, you should know the amino acid side chain structures and thus which are negative, neutral, and positive at pH 7.

Question 24

What is the approximate isoelectric point of the following hexapeptide. Consult your notes to obtain relevant pKa values.

Asp-Leu-lle-Glu-Lys-Gly?

a) 3.7
b) 4.0
c) 6.55
d) 9.75

Answer 24

b) 4.0

Why?
- There are 5 protons that can be dissociated from this functional group (the side chains of Asp, Glu, and Lys, as we as the N- and C- terminus).
- The isoelectric point of this peptide is enclosed by two pa values (3.9 and 4.1). The average of these two values is 4.0, which is the isoelectric point.

Question 25

If you wanted to best visualize the potential role of a specific amino acid residue in the interaction of a protein with a small molecule, you would likely _____________? a) carry out gel filtration chromatography on the protein b) examine a ribbon model of an X-ray crystal structure of the protein c) examine a wireframe model of an X-ray crystal structure of the protein d) calculate the amount of beta-sheet secondary structure of the enzyme e) determine the amino acid sequence of the protein

Answer 25

c) examine a wireframe model of an X-ray crystal structure of the protein Why? - The wireframe representation of all atoms (except hydrogens) provides the most detailed view to examine potential chemical and binding processes.

Question 26

The formation of an alpha helix by a protein always involves? a) Arrangement of multiple peptide chains to form the alpha helix. b) Peptide bond isomerization c) Arrangement of the peptide chain so that R groups are pointed away from the helix d) Phosphorylation of side chains e) Hydrogen bonding between neighbouring amino acid residues

Answer 26

c) Arrangement of the peptide chain so that R groups are pointed away from the helix Why? - An alpha helix consists of one peptide chain arranged in a coil with R groups pointed away from the helix axis and with hydrogen bonding between amino acid residues that are 4 residues apart. Phosphorylation or peptide bond isomerization is not required for an alpha helix to form.

Question 27

Each of the following statements describes a difference between RNA and DNA, EXCEPT: a) DNA contains deoxyribose while RNA contains ribose. b) Thymine base in DNA is replaced by uracil in RNA. c) DNA typically occurs in a double helix conformation while RNA does not. d) RNA can catalyze chemical reactions while DNA does not. e) DNA contains only purines while RNA contains only pyrimidines.

Answer 27

e) DNA contains only purines while RNA contains only pyrimidines. Why? - Both DNA and RNA contain both purines (G, C) and pyrimidines (A, T or U). All other statements are correct.

Question 28

Yarmouth County of Nova Scotia has the highest prevalence in the world of Niemann-Pick C1 (NPC1) disease, a fatal inherited disorder of intracellular cholesterol transport. Most Nova Scotian NPC1 patients are of Acadian descent and are homozygous for a point mutation that results in replacement of glycine with tryptophan at position 992 of the amino acid sequence (G992W). This would be an example of what type of mutation? a) Nonsense b) Silent c) Missense d) Frameshift e) Splice variant

Answer 28

c) Missense Why? - This is a missense mutation, i.e. resulting in substitution with a different amino acid. Nonsense and silent mutations introduce stop codons or do not result in amino acid substitutions, respectively. Frameshift mutations change the translational reading frame, while splice variants normally result in incorrect exon joining.

Question 29

Which of the following statements is false? a) The DNA double helix has a polar exterior. b) DNA can be used to create complex 3D objects. c) G-quadruplexes may have functional roles in the body. d) DNA sequences have low information density. e) Hydrogen bonding is important for determining DNA base complementarity.

Answer 29

d) DNA sequences have low information density. ..False, they have high info density...

Question 30

The DNA double helix : a) Is destabilized by hydrogen bonding b) is stabilized by the hydrophobic effect c) is a linear molecule with no three- dimensional shape d) does not associate with protein e) is stabilized by CI-

Answer 30

b) is stabilized by the hydrophobic effect Why? - The correct answer is B. Hydrogen bonding between complementary nucleotides stabilizes the DNA double helix (A is incorrect), and the double helix is stablizdb by stacking interactions and the hydrophobic effect (B is correct). The DNA double helix has a 3D shape to it (like all molecules), and associates with proteins such as histones and transcription factors (D is incorrect). Cations such as Mg2+ stabilize the DNA double helix (E is incorrect).

Question 31

During cation exchange chromatography: a) positive charges on the surface of proteins interact with negatively charged resin. b) negative charges on the surface of proteins interact with negatively charged resin. c) positive charges on the surface of proteins interact with positively charged resin. d) negative charges on the surface of proteins interact with positively charged resin. e) all of the above

Answer 31

a) positive charges on the surface of proteins interact with negatively charged resin.

Question 32

The 6x histidine affinity tag: a) causes proteins to enter a stationary phase and elute from a column more slowly. b) is composed of a mixture of histidine and arginine residues c) allows proteins to be purified by ion exchange chromatography d) binds tightly to cations such as Ni2+ e) permanently binds proteins to a stationary phase

Answer 32

d) binds tightly to cations such as Ni2+

Question 33

Mutagenic replacement of isoleucine (Ile) with glycine (Gly) in the interior of a globular enzyme would most likely result in a loss of function because of... a) altered packing due to the smaller Gly side chain b) loss of disulfide bond formation by Ile c) loss of the negative charge from Ile the increased d) hydrophobic nature of Gly versus Ile e) loss of hydrogen bonding from the Ile side chain

Answer 33

a) altered packing due to the smaller Gly side chain Why? - This would be an example of a non-conservative amino acid replacement (i.e. by an amino acid quite different in size, charge and/or chemical properties), and would not normally be tolerated in the interior of a tightly packed globular protein. Isoleucine is a fairly hydrophobic (non-polar) amino acid (thus D is incorrect) with a side chain that is not ionizable, lacks a sulfhydryl group, and is incapable of forming hydrogen bonds, and (C, B and E are incorrect).

Question 34

Which of the following is most critical for maintaining the native tertiary structure of a globular protein? a) Hydrogen bonds between regions of the polypeptide chain in the folded protein. b) The hydrophobic effect of burying non-polar residues in the protein interior. c) Disulfide bonds between Cys residues in the protein. d) Salt bridges (i.e. ionic bonds between opposite charged side chains). e) None of these choices.

Answer 34

b) The hydrophobic effect of burying non-polar residues in the protein interior. Reason: - All of these factors (plus optimized van der Waals forces in the packed protein interior) can contribute to stabilize the native folded protein structure, but the hydrophobic effect of burying non-polar amino acids is dominant for most proteins. The major factor opposing the native state is the increase in chain entropy (i.e. conformational possibilities) in the unfolded state.

Question 35

A researcher has isolated a protein that appears to be involved in synthesis of biofuels in algae. It elutes from a gel filtration column with an apparent native mass of 120 kDa (based on elution of comparable standard proteins), while SDS-PAGE analysis in the presence of mercaptoethanol reveals two protein bands of similar intensity corresponding to 25 and 35 kDa. Based on this information, this protein is most likely to be: a) a homotetramer b) a heterotetramer c) a homodimer d) a heterodimer e) cross-linked by disulphide bonds in its native form

Answer 35

b) a heterotetramer Why? - Since the protein appears to have equal amounts of 25 kDa and 35 kDa subunits, with a total MW of 120,000, it is most likely a heterotetramer composed of two 25 kDa and two 35 kDa subunits. SDS-PAGE in the presence of the reducing agent mercaptoethanol would reduce (i.e. break) any disulfide cross links between or within subunits, but the data do not allow you to conclude whether these are present in the native protein: many intracellular multi-subunit proteins, such as hemoglobin, do not require disulfide bonds to retain their quaternary structure.

Question 36

The binding of one O2 to a molecule of hemoglobin normally results in..? a) the release of any other O2 that may have bound earlier b) a decrease in hemoglobin's ability to bind a second O2 c) an increased affinity for O2 in the remaining subunits d) dissociation of the heme prosthetic group e) dissociation of the hemoglobin subunits

Answer 36

c) an increased affinity for O2 in the remaining subunits Why? - The sequential binding of O2 to each subunit in hemoglobin tetramer shifts the Tense (I) - relaxed (R) state equilibrium towards the R state (i.e. more like myoglobin), increasing the affinity for subsequent O, binding to the remaining subunits. This is one example of a cooperative phenomenon

Question 37

You have discovered a new form of myoglobin from cockroaches that has a P50 of 40 torr. Whale myoglobin has a P50 of 3 torr. Which of the following statements is true? a) Cockroach myoglobin must have a sigmoidal oxygen binding curve. b) At a pO2 of 3 torr, cockroach myoglobin has very little oxygen bound. c) At a pO2 of 40 torr, cockroach myoglobin is nearly saturated with oxygen. d) Whale myoglobin is always less saturated with oxygen than cockroach myoglobin, no matter what the partial pressure of oxygen (pO2) is. e) At a pO2 of 500 torr, cockroach myoglobin is saturated with oxygen, but whale myoglobin is not.

Answer 37

b) At a pO2 of 3 torr, cockroach myoglobin has very little oxygen bound.

Question 38

Gel filtration chromatography estimates the molecular weight of a protein to be 100 kDa. However, when your co-worker analyzes the protein by reducing SDS-PAGE it is observed to only have a size of 25 kDa. Which statement below explains this result? A. Your co-worker is incompetent B. The protein has degraded prior to gel filtration chromatography C. The protein exists in solution as a homotetramer D. The protein aggregated during SDS-PAGE analysis E. All of the above

Answer 38

C. The protein exists in solution as a homotetramer

Question 39

A new type of myoglobin gene has been discovered in pokémon that has a P50 of 10 mmHg. This is significantly higher than human myoglobin, which has a P50 of 2.5 mmHg. This means that: A.Pokémon myoglobin has a higher affinity for O2 than human myoglobin. B. Pokémon myoglobin can only bind more oxygen than human myoglobin at high oxygen pressures. C. Pokémon myoglobin binds more oxygen than human myoglobin at low oxygen pressures. D.Human myoglobin binds more oxygen than pokémon myoglobin at all oxygen pressures. E. None of the above

Answer 39

D.Human myoglobin binds more oxygen than pokémon myoglobin at all oxygen pressures.

Question 40

Mutation of hydrophobic amino acids to serine would likely destabilize keratin by: A. Stabilizing beta sheet structure in keratin. B. Inducing kinks into the alpha helices of keratin. C. Inducing crosslinks between keratin peptide chains. D. Preventing the association of the keratin coiled-coil. E. None of the above.

Answer 40

D. Preventing the association of the keratin coiled-coil.

Question 41

Actomyosin _____ a) is composed of a thin actin filament and thick myosin filament. b) requires ATP to drive conformational change. c) is responsible for muscle contraction. d) All of the above. e) None of the above

Answer 41

d) All of the above.

Question 42

Which of the following statements is false? a) Hydroxylation of proline occurs post-translationally. b) Intermediate filaments such as collagen and elastin are often crosslinked. c) Microtubules are hollow and composed of tubulin. d) Kinesins transport cargo along microfilaments. e) Actin fibers are able to grow from both the (+) and (–) ends.

Answer 42

d) Kinesins transport cargo along microfilaments. mircotubules**

Question 43

In keratin ______ a) glycine is required at every third residue because it is the smallest amino acid. b) proline is not important. c) α-helices coil together to form a coiled-coil. d) cysteine provides crosslinks between fibers. e) aromatic amino acids hold the fiber together.

Answer 43

c) α-helices coil together to form a coiled-coil.

Question 44

Antibodies _________ a) Cannot be fused to (conjugated with) other molecules such as enzymes. b) contain an antigen binding region that recognizes a specific epitope. c) Degrade when reacted with fluorescent dyes. d) can be subdivided into a heavy, light, and intermediate chain. e) Have little use in the biotechnology industry.

Answer 44

b) contain an antigen binding region that recognizes a specific epitope.