9. Titration of valine by a strong base, for example NaOH, reveals two pK's. The titration reaction occurring atpK2 (pK2 = 9.62) is:a) —СООН + Он -> —соо" + Н20.b) -COOH + -NH2 -> -COO" +-NH2*.c) -co0 +-NH2* ->-COOH +–NH2.d) –NH3* + OH" -> –NH2 + H2O.e) –NH2 + OH -> –NH + H2O.10.In a highly basic solution, pH = 13, the dominant form of glycine is:а) NH2—CH2—СООН.b) NH2-CH2-coO".c) NH2-CH3*-coo.d) NH3*-CH2-COOH.e) NH3*-CH2-coo".

Answer 9 - option D) -NH3+ + OH → ¬NH₂ + H₂0 Explanation: Valine has two acid-base groups: the…

Answered: 9. Titration of valine by a strong…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Titration of alanine by a strong base, for example NaOH, reveals two pk's. Which is the titration reaction occurring at pK1 (pK1 = %3D 2.34)? O-NH2 + COH →-NH + H2O O-COOH + OH →-CO + H20 -COOH + -NH2 → -COO" + -NH2+ -CoO +-NH2* → -COOH +-NH2 O -NH3* + OH →-NH2 + H2O
This amino acid is: (Note that the non-ionized form is shown; remember that backbone groups are irrelevant to classification) a) Polarized (+) charged b) Nonpolar c) Polar and (-) charged d) Polar uncharged
38) A nucleic acid sequence is read from the A) free 3' hydroxyl; free 5' phosphate B) free 3' phosphate; free 5' hydroxyl C) free 5' phosphate; free 3' hydroxyl D) free 5' hydroxyl; free 3' phosphate to the
Draw the L enantiomer in a Fischer projection for each amino acid; identify the amino acid as neutral, acidic, or basic. Give the three-letter sign [3], and the one-letter symbol [4].arginine a. arginine b. arginine c. a. glutamic acid b. glutamic acid c. glutamic acidtyrosine b. valine c. arginine
6) Proteins can be modified by phosphorylation, which adds a phosphate group to the hydroxyl group of serine, threonine, or tyrosine residues. The R-group for phosphoserine is shown at right. A) The image below is an Isoelectric focusing strip that shows the unphosphorylated protein-of-interest (in blue). To which side of the unphosphorylated protein would you expect to see the phosphorylated protein? (Draw an arrow to indicate direction). Briefly justify your answer. un-phosphorylatable: Low pH Justify: B) To study the effects of phosphorylation, researchers often mutate a Ser/Thr to appear as though it is always phosphorylated or never phosphorylated at a particular site. What amino acid substitution should you use to preserve similar dimensions as Ser (or Thr) but make the side chain appear to be: constitutively (always) phosphorylated: Justify: Ser or Thr → O O=P-O O I CH₂ Ser or Thr➜ Phosphoserine High pH
92.17
25) Below are two amino acids: cysteine and histidine. a) Draw the reaction mechanism of a dehydration synthesis reaction between cysteine and histidine below: HS H Product 1 N OH HN H b) Draw two different products of the dehydration synthesis reaction from part a). Prove that these are two different products by labelling the N and C termini, peptide bond, and R-groups of each product. Product 2 OH
You have discovered a new hemoglobin variant you are calling Hb-21. You have noticed the following from your studies of Hb-21: - The binding constant of Hb-21 is the same at pH 7.1 and 7.4.- The binding constant curves generated by Hb-21 and O2 are sigmoidal.- The Hill coefficient for Hb-21 is 3.-The binding constant for Hb-21 with O2 in the presence of 5mM 2,3-bisphosphglycerate is about 6-times lower instead of the 8-times lower that is observed under these conditions for the wild type enzyme. Use this information to answer questions: Based on this information which of these features of hemoglobin would you conclude are altered in Hb-21 as compared to the wild type hemoglobin? ____ the cooperativity.____ the amount of T versus R state in the tissues. ____ the Bohr effect. What is the major issue for an individual with the Hb-21 variant? Explain exactly what aspect of hemoglobin’s function is altered.
2. | Calculate the overall charge (pH 7) on the following three polypeptides and answer the questions below. Assume the following pKa values: N-terminal -NH3®, 7.0; all -COOH groups, 4.0; Arg, 12.5; Cys, 8.4; His, 6.0; Lys, 10.0; Tyr, 10.0. (A) Ser-Tyr-Ser-Met-Glu-His-Phe-Arg-Trp-Gly-Lys-Pro-Val-Gly-Lys-Lys-Arg-Arg-Pro-Val-Lys-Val-Tyr-Pro-Asp-Ala -Gly- Glu-Asp-Gln- Ser-Ala-Glu-Ala-Phe-Pro-Leu-Arg-Glu-Phe (B) Ser-Tyr-Ser-Met-Glu–His-Phe-Arg–Trp–Gly-Ala-Pro-Val-Gly-Glu-Glu-Cys-Asp-Pro-Val-Glu–Val-Tyr-Pro-Asp- Ala-Gly-Glu-Asp-Gln-Ser-Ala-Glu-Ala-Phe-Pro-Leu-Glu-Phe-Cys-Ser-Tyr-Ser-Met-Glu-His-Phe-Asp-Trp-Gly- Asp-Pro-Val-Gly-Pro-Asp-Ala-Gly-Asp-Gln-Pro-Val-Gly-Glu-Glu-Cys-Asp-Pro-Val-Glu-Val-Tyr-Pro-Asp-Ala | (C) Gly-Ser-Val-Arg-Asp-Pro-Val-Lys-Glu–Val-Tyr-Pro-Asp- Lys-Ala-Gly-Arg-Glu-Ser-Arg-Ala (a) Which of the three peptides would elute first from a gel filtration column? (b) Which of the three peptides would migrate the fastest on SDS-PAGE (c) Which of the three peptides could be…
5) In the below two step transformation, the first step AG is positive i.e. 1.7 kJ/mol. Yet this two step coupled reaction takes place to form Fructose-1.6-bisphosphate. Explain why?" C-H CH-OH CH,-0 H-C-OH AG"- 1.7 KJ/mol (+0.4 kcalmol) C=0 AG"=-14.2 k/mol(-3.4 kcalmol) HO-C-H HO-C-H HO-C-H H-C-OH H-C-OH H-C-OH H-C-OH H-C-OH ATP ADP H-C-OH CH-0-P. CH,-0-P-o- CH,-o- Glucose-6-phosphate Fructose-6-phosphate Fructose-1,6-bisphosphate
1. Amino Acid 1 = W + X = 1 Draw amino acid 1 at a low pH a) As the D Fisher Projection Name of amino acid 1: b) As the S isomer: Arginine
Consider a short peptide with the sequence M-C-Q-L-Y-P-E-D-K. List the ionizable groups in this peptide, and the net charge of the whole peptide at pH a) 1.5, b) 7.0, and c)12.0.

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