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AS. 410.610 – Biochemistry –Fall 2019 Problem Set #1, Exams of Biochemistry

University of LeedsBiochemistry

Question 1 1. Which of the following statements is true about water? Select all answers that apply. A. it is non-polar B. it fully ionizes at neutral pH C. liquid-phase water is more dense than solid-phase water D. it is a weak acid E. it is a weak base 0.5 points Question 2 1. The Ka for a weak acid (HA) is known to be 1.9 x 10-5. What is the ratio of [A-] to [HA] at a pH of 6? Explain your answer & show all mathematical work. Answer: [A-] / [HA] is 19:1 at pH = 6. Reasoning: Using Henderson-Hasselbach Equation: pH = pkA + log10 [conjugate base] / [weak acid] pH = pKa + log ([A¯] / [HA]) pH = (pKa = -log Ka)+ log ([A¯] / [HA]) pH = 6 Ka = 1.9 x 10-5 (weak acid) 6 = (-log (1.9 x 10-5)) + log ([A¯] / [HA]) 6 = (4.721246399) + log ([A¯] / [HA]) 6 - (4.721246399) = + log ([A¯] / [HA]) 1.278753601 = + log ([A¯] / [HA]), cancel the log (right side) and base-log left

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oup being titrated around the point labele
ely a(n)__________________(basic or aci
12.5 at point (C), then this must be a
AS. 410.610 – Biochemistry –Fall 2019
Problem Set #1
-Answers: red.
-Correct answers: Green (i.e., 100% credit)
Question 1
1. Which of the following statements is true about water? Select all answers that apply.
A. it is non-polar
B. it fully ionizes at neutral pH
C. liquid-phase water is more dense than solid-phase water
D. it is a weak acid
E. it is a weak base
0.5 points
Question 2
1. The Ka for a weak acid (HA) is known to be 1.9 x 10-5. What is the ratio of [A-] to [HA] at a pH of 6?
Explain your answer & show all mathematical work.
Answer: [A-] / [HA] is 19:1 at pH = 6.
Reasoning:
Using Henderson-Hasselbach Equation:
pH = pkA + log10 [conjugate base] / [weak acid]
pH = pKa + log ([A¯] / [HA])
pH = (pKa = -log Ka)+ log ([A¯] / [HA])
pH = 6
Ka = 1.9 x 10-5 (weak acid)
6 = (-log (1.9 x 10-5)) + log ([A¯] / [HA])
6 = (4.721246399) + log ([A¯] / [HA])
6 - (4.721246399) = + log ([A¯] / [HA])
1.278753601 = + log ([A¯] / [HA]), cancel the log (right side) and base-log left
101.278753601 = ([A¯] / [HA])
19.0000000021 or
Concentration ratio: At pH = 6, the ratio of [A-] / [HA] is 19:1.
1 points
Question 3
1. For questions 3-6, consider the titration curve below. The chemical species being titrated by a
strong base must be a , since it is capable of donating proton(s).
Reasoning: Three steps on the graph, lent 3 H+’s, and the pH is high, so triprotic acid.
A. weak base; no
B. strong base; three
C. monoprotic acid; one
D. diprotic acid; two
E. triprotic acid; three
0.5 points
Question 5
0.5 points
Question 4
1. The chemical gr d (C) in the
graph is most lik dic) group.
Blank 1
Reasoning: (c) pKa is ~
asic group
(like
an amino grp).
basic
pf3
pf4
pf5
pf8
pf9
pfa

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/ oup being titrated around the point labele ely a(n)__________________(basic or aci 12.5 at point (C), then this must be a AS. 410.610 – Biochemistry –Fall 2019 Problem Set #

  • Answers: red.
  • Correct answers: Green (i.e., 100% credit) Question 1
    1. Which of the following statements is true about water? Select all answers that apply. A. it is non-polar B. it fully ionizes at neutral pH C. liquid-phase water is more dense than solid-phase water D. it is a weak acid E. it is a weak base 0.5 points Question 2
    2. The Ka for a weak acid (HA) is known to be 1.9 x 10 -5. What is the ratio of [A-] to [HA] at a pH of 6? Explain your answer & show all mathematical work. Answer: [A-] / [HA] is 19:1 at pH = 6. Reasoning: Using Henderson-Hasselbach Equation: pH = pkA + log10 [conjugate base] / [weak acid] pH = pKa + log ([A¯] / [HA]) pH = (pKa = -log Ka)+ log ([A¯] / [HA]) pH = 6 Ka = 1.9 x 10 -5^ (weak acid) 6 = (-log (1.9^ x^ 10-5)) + log ([A¯] / [HA]) 6 = (4.721246399) + log ([A¯] / [HA]) 6 - (4.721246399) = + log ([A¯] / [HA]) 1.278753601 = + log ([A¯] / [HA]), cancel the log (right side) and base-log left 10 1.278753601^ = ([A¯] / [HA]) 19.0000000021 or Concentration ratio: At pH = 6, the ratio of [A-] / [HA] is 19:1. 1 points Question 3
    3. For questions 3-6, consider the titration curve below. The chemical species being titrated by a strong base must be a , since it is capable of donating proton(s). Reasoning: Three steps on the graph, lent 3 H+’s, and the pH is high, so triprotic acid. A. weak base; no B. strong base; three C. monoprotic acid; one D. diprotic acid; two E. triprotic acid; three 0.5 points Question 5 0.5 points Question 4
      1. The chemical gr d (C) in the graph is most lik dic) group. Blank 1 Reasoning: (c) pKa is ~ asic group (like an amino grp). basic
  1. The titration curve shown in Question #3 is that of a free amino acid in aqueous solution. Based on the graph, which particular amino acid is most likely represented? Simply name the amino acid. Blank 1 Reasoning: -Titration curve clearly =3 ionizable groups, -must be Tyr, Cys, Lys, His, Arg, Asp, or Glu (only free amino acids w/ionizable side chains). -(A) pKa = 2+ something (2.25 maybe), -(B) pKa = 9 -(C) pKa = 12. -Only amino acid within this pKa is arginine. 0.5 points Question 6
  2. Assume that the equilibrium represented around point (A) in the titration can generically be described as: H 3 A + OH-^ ---> H 2 A-^ + HOH. What is the pH at which the ratio of [HA2-] to [H 2 A-] is 25:1? Show all work & clearly explain your answer. Answer: pH at [HA2-] / [H 2 A-] is 25:1 = 10 Reasoning:
  3. Equilibrium (A) = [HA2-] to [H 2 A-] is 25:
  4. Equilibrium (B) = pKa = 9 (or 9.04 if it is really arginine)
  5. Triprotic acid H 2 A-^ + OH----> HA2-^ + HOH Per the H/H equation: pH = pKa + log([HA2-]/[H 2 A-]) -pKa at point (B) is approximately 9 (or precisely 9.04 if arginine), so: -pH = 9 + log(25/1) = 9 + 1.4 = 10. 1 points Question 7
  6. For Questions #7-13, consider the following peptide: His-Met-Asp-Tyr-Phe-Ser The code for this peptide, using one-letter symbols, is. 0.5 points Question 8
  7. How many distinct "ionizable" groups exist within this peptide? His-Met-Asp-Tyr-Phe-Ser = 5 +3 Ionizable side chain +1 Ionizable carboxul grp +1 Ionizable amino terminal grp A. 2 B. 3 C. 4 D. 5 E. 6 0.5 points Arginine HMDYFS
  1. Based on your answer for Question #11, would you categorize this peptide as "acidic" or "basic"? Explain. Answer: Acidic Reasoning: -pI is far below neutral pH (7). -Therefore, at a low pH, this peptide would lose > ½ of its dissociable protons, and must have acidic groups in its majority. 1 points Question 13
  2. At a pH of 10, would you expect this peptide to be retained for a longer time within an anion exchange column or a cation exchange column? Clearly explain your reasoning. Answer: Longer within an anion exchange column Reasoning: -pH = 10, -Most of this peptide's ionizable groups are negatively or neutrally charged. -The peptides negatively-charged groups should electrostatically bind to the positively-charged immobile phase of the column. -Therefore, the peptide would be expected to be retained longer in an anion exchange column 1 points Question 14
  3. The amino acid residues commonly found within a β turn are: A. Ala & Gly. B. two Cys. C. hydrophobic. D. Pro & Gly. E. those with ionized R groups. 0.5 points Question 15
  4. Cellular proteins are oftentimes post-translationally modified. Choose one of the following PTMs: N- linked glycosylation, phosphorylation, ubiquitination, or GPI-anchor. Clearly indicate your choice, then address the following: (a) How is the PTM attached to the protein of interest? At which amino acid residue(s)? What enzyme(s) is involved, if any? (b) Is the PTM relatively stable or highly dynamic? Explain. How does the PTM become detached from the protein of interest? What enzyme(s) is involved, if any? (c) What is the function of the PTM? Provide one specific example. Answer: GPI-anchor (a, b) The PTM (or posttranslational modification; a covalent enzymatic modification which occurs after biosynthesis) is attached to the GPI (or glycophosphatidylinositol; a glycerol-based phospholipid) by the end of an amino acids chain terminated by the amino acid residue (i.e. free carboxyl group) –COOH, named the C-terminus. This C-terminus was created from the N- terminus of the protein when translated from messenger RNA. The hydrophobic C-terminal sequence (attached to the ER membrane by its hydrophobic C terminus) is stable. But is useful when cleaved off and replaced by the GPI-anchor after being inserted in the ER membrane (i.e., PTM become detached from the protein of interest). Cleavage of the group by an enzyme that hydrolyses phospholipids into fatty acids (e.g., Phospholipase C; PLC) will result in controlled release of the protein from the membrane; proteins released from membranes in enzymatic assays are glypiated proteins. PLC will cause release of GPI-linked proteins: Thymus gland

lymphocyte

Which of the conformations illustrated above (a or b) is more likely PrPc^ (the normal cellular form of PrP) and which is more likely PrPSc^ (the pathogenic form of PrP)? Why? Answer: (a) is PrPc, and (b) is PrPSc^. Reasoning: -PrPc^ and PrPSc^ tertiary conformations differ by amounts of secondary structures (i.e., beta-sheet structures).

  • PrPc^ has very little beta-sheet while PrPSc^ is has high amounts of beta-sheet structure. -(a) is highly composed of alpha-helical, turns, and coil structures. -(b) has (at least; two strand in green, two strands in red) two beta sheet structures -Therefore, (a) is PrPc, and (b) is PrPSc^. 1 points Question 19
  1. The red arrows within the (b) conformation are best categorized as: A. alpha helices B. beta-turns C. tertiary structure D. parallel beta-strands E. antiparallel beta-strands 0.5 points Question 20
  2. Questions #20-23 refer to the figure shown below. Approach each question independently. Figure Legend: Tagging schematic. The figure shows various steps involved in the identification of interacting proteins by using an epitope-tagging strategy. The cDNA of interest is first cloned into a vector that provides an epitope tag. This is followed by transfection of the tagged "bait" into the cell of interest. The cells are then lysed and the lysates purified by affinity purification using an antibody against the epitope. Proteins bound specifically to the bait protein are eluted by competitive elution using a peptide that encodes the epitope. The proteins are then resolved by gel electrophoresis followed by mass spectrometric identification.

STEP 1: Epitope-tagged protein (encoded by cDNA) STEP 2: Transfect into cells STEP 3: Lyse cells STEP 4: Affinity-based purification using immobilized antibody against the epitope STEP 5: Competitive elution with peptide encoding the epitope STEP 6: Gel electrophoresis STEP 7: Excise bands and analyze by mass spectrometry Question #20: According to the gel (Step 6), which protein has the highest molecular weight? A. Protein A B. Protein B C. Protein C D. Protein D E. YFP 0.5 points Question 21

  1. Are the bands indicated in Step 6 (SDS-PAGE) likely visualized by western blot or by a general protein stain (such as Coomassie or silver)? Explain your answer. Answer: General protein stain Reasoning: -Can’t be a western blot: All the proteins (in the complex) appear on the gel. A WB would only show one of the bands because an antibody against only one protein is used. -Can be a ‘general protein stain’. 1 points Question 22
  2. Step 7 states "Excise bands and analyze by mass spectrometry." (A) Following excision of the bands (simply cutting each band out of the gel), briefly describe each step, in order, that must be taken prior to analysis in the mass spectrometer. (B) What's the big experimental question that the researchers hope to answer following Step 7?

0.5 points C. anion exchange chromatography. D. hydrophobic interaction chromatography. E. size exclusion chromatography. Question 26

  1. For the table shown in Question #24, the purification step labeled "CM-sepharose" is most likely a(n): A. salting in/out step. B. cation exchange chromatography. C. anion exchange chromatography. D. hydrophobic interaction chromatography. E. size exclusion chromatography. 0.5 points Question 27
  2. For the table shown in Question #24, the purification step labeled "Gel filtration" is most likely a(n): A. salting in/out step. B. cation exchange chromatography. C. anion exchange chromatography. D. hydrophobic interaction chromatography. E. size exclusion chromatography. 0.5 points