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17 Cards in this Set
- Front
- Back
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The bacterium E. coli requires simple organic molecules for growth and energy—it is therefore a: A) chemoautotroph. B) lithotroph. C) lithograph. D) photoautotroph. E) phototraph. F) chemoheterotroph. |
F) chemoheterotroph. |
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The three-dimensional structure of macromolecules is formed and maintained primarily through noncovalent interactions. Which one of the following is not considered a noncovalent interaction? A) hydrogen bonds B) hydrophobic interactions C) ionic interactions D) van der Waals interactions E) carbon-carbon bonds |
E) carbon-carbon bonds |
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The enzyme fumarase catalyzes the reversible hydration of fumaric acid to l-malate, but it will not catalyze the hydration of maleic acid, the cis isomer of fumaric acid. This is an example of: A) biological activity. B) chiral activity. C) racemization. D) stereospecificity. E) stereoisomerization. |
D) stereospecificity. |
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The pH of a sample of blood is 4.7, while gastric juice is pH 7.7. The blood sample has: A) 1,000 times lower [H+] than the gastric juice. B) 1,000 times higher [H+] than the gastric juice. C) 0.006 times lower [H+] than the gastric juice. D) 1.2333 times lower [H+] than the gastric juice. E) 0.1898 times the [H+] as the gastric juice. |
B) 1,000 times higher [H+] than the gastric juice. |
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Which of the following statements about buffers is true? A) A buffer composed of a weak acid of pKa = 5 is stronger at pH 4 than at pH 6. B) At pH values lower than the pKa, the salt concentration is higher than that of the acid. C) The strongest buffers are those composed of strong acids and strong bases. D) When pH = pKa, the weak acid and salt concentrations in a buffer are equal. E) The pH of a buffered solution remains constant no matter how much acid or base is added to the solution. |
D) When pH = pKa, the weak acid and salt concentrations in a buffer are equal.
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The Henderson-Hasselbalch equation: A) allows the graphic determination of the molecular weight of a weak acid from its pH alone. B) relates the pH of a solution to the pKa and the concentrations of acid and conjugate base. C) employs the same value for pKa for all weak acids. D) is equally useful with solutions of acetic acid and of hydrochloric acid. E) does not explain the behavior of di- or tri-basic weak acids |
B) relates the pH of a solution to the pKa and the concentrations of acid and conjugate base. |
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Consider an acetate buffer, initially at the same pH as its pKa (4.76). When hydrochloric acid (HCl) is mixed with this buffer, the: A) pH remains constant. B) pH rises more than if an equal amount of NaOH is added to an acetate buffer initially at pH 6.76. C) pH rises more than if an equal amount of NaOH is added to unbuffered water at pH 4.76. D) sodium acetate formed precipitates because it is less soluble than acetic acid. E) ratio of sodium acetate to acetic acid in the buffer falls. F) ratio of acetic acid to sodium acetate in the buffer falls. |
E) ratio of sodium acetate to acetic acid in the buffer falls. |
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Which of the following statements about aromatic amino acids is correct? A) All are strongly hydrophilic. B) Histidine’s ring structure results in its being categorized as aromatic or basic, depending on pH. C) On a molar basis, tryptophan absorbs less ultraviolet light than tyrosine. D) The major contribution to the characteristic absorption of light at 280 nm by proteins is the tryptophan R group. E) The presence of a ring structure in its R group determines whether or not an amino acid is aromatic. |
D) The major contribution to the characteristic absorption of light at 280 nm by proteins is the tryptophan R group. |
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Which of the following tripeptides would be expected to be the most hydrophobic: A) KYG B) DYA C) GYR D) WYA E) WYI |
E) WYI |
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At pH 9.5, arginine (pKs are alpha-carboxylate 1.82, alpha-amino 8.99, guanidino 12.48) would be charged as follows: A) 0 alpha-carboxylate, 0 alpha-amino, +1 guanidino, +1 net charge B) +1 alpha-carboxylate, 0 alpha-amino, -1 guanidino, 0 net charge C) -1 alpha-carboxylate, 0 alpha-amino, +1 guanidino, 0 net charge D) -1 alpha-carboxylate, +1 alpha-amino, +1 guanidino, +1 net charge E) -1 alpha-carboxylate, -1 alpha-amino, +1 guanidino, 0 net charge |
C) -1 alpha-carboxylate, 0 alpha-amino, +1 guanidino, 0 net charge |
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At pH 3, aspartic acid (pKs are alpha-carboxylate 1.99, alpha-amino 9.9, beta-carboxylate 3.9 ) would be charged as follows: A) -1 alpha-carboxylate, +1 alpha-amino, -1 beta-carboxylate, -1 net charge B) 0 alpha-carboxylate, -1 alpha-amino, 0 beta-carboxylate, -1 net charge C) 0 alpha-carboxylate, +1 alpha-amino, +1 beta-carboxylate, +2 net charge D) +1 alpha-carboxylate, +1 alpha-amino, +1 beta-carboxylate, +3 net charge E) -1 alpha-carboxylate, +1 alpha-amino, 0 beta-carboxylate, 0 net charge |
E) -1 alpha-carboxylate, +1 alpha-amino, 0 beta-carboxylate, 0 net charge |
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The positive charge on proteins in electrospray ionization mass spectrometry is the result of: A) protons fired at the gas-phase protein molecules. B) protonated Asp and Glu residues. C) a low pH. D) protonated Arg and Lys residues. E) a high pH |
C) a low pH. |
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The average molecular weight of the 20 standard amino acids is 138, but biochemists use 110 when estimating the number of amino acids in a protein of known molecular weight. Why? A) The number 110 is based on the fact that the average molecular weight of a protein is 110,000 with an average of 1,000 amino acids. B) The number 110 reflects the number of amino acids found in the typical small protein, and only small proteins have their molecular weight estimated this way. C) The number 138 represents the molecular weight of conjugated amino acids. D) The number 110 reflects the higher proportion of small amino acids in proteins, as well as the loss of water when the peptide bond forms. E) The number 110 takes into account the relatively small size of nonstandard amino acids. |
D) The number 110 reflects the higher proportion of small amino acids in proteins, as well as the loss of water when the peptide bond forms |
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By adding SDS (sodium dodecyl sulfate) during the electrophoresis of proteins, it is possible to: A) determine a protein’s isoelectric point. B) determine an enzyme’s specific activity. C) determine the amino acid composition of the protein. D) separate proteins exclusively on the basis of molecular weight. E) preserve a protein’s native structure and biological activity. |
D) separate proteins exclusively on the basis of molecular weight |
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The first step in two-dimensional gel electrophoresis generates a series of protein bands by isoelectric focusing. In a second step, a strip of this gel is turned 90 degrees, placed on another gel containing SDS, and electric current is again applied. In this second step: A) the individual bands become stained so that the isoelectric focus pattern can be visualized. B) the individual bands become visualized by interacting with protein-specific antibodies in the second gel. C) the individual bands undergo a second, more intense isoelectric focusing. D) the proteins in the bands separate more completely because the second electric current is in the opposite polarity to the first current. E) proteins with similar isoelectric points become further separated according to their molecular weights. |
E) proteins with similar isoelectric points become further separated according to their molecular weights. |
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(A) A nonapeptide was determined to have the following amino acid composition: (Phe) 2, Lys, Arg, Gly, Trp, His, Leu, Met. The native peptide was incubated with 1-fluoro-2,4-dinitrobenzene (FDNB) and then hydrolyzed; 2,4-dinitrophenylhistidine was identified by HPLC. When the native peptide was exposed to cyanogen bromide (CNBr), no cleavage happened. Complete digestion of the native peptide with trypsin gave two tetrapeptides and a free amino acid. Chymotrypsin digestion produced two peptides. Determine the sequence of the peptide. A) Gly–Phe–Arg–Lys–Trp–Leu–Met–Phe–His. B) His–Leu–Gly–Lys–Arg–Trp–Phe–Leu–Met. C) His–Phe–Leu–Trp–Lys–Arg–Phe–Met–Gly. D) His–Leu–Phe–Lys–Phe–Trp–Gly–Arg- Met E) His–Leu–Phe–Trp–Arg–Lys–Phe–Met–Gly. |
B) His–Leu–Gly–Lys–Arg–Trp–Phe–Leu–Met. |
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(B) A nonapeptide was determined to have the following amino acid composition: (Lys)2, (Gly) 2, (Phe) 2, His, Leu, Met. The native peptide was incubated with 1-fluoro-2,4-dinitrobenzene (FDNB) and then hydrolyzed; 2,4-dinitrophenylmethionine was identified by HPLC. When the native peptide was exposed to cyanogen bromide (CNBr), an octapeptide and a blocked methionine were recovered. Incubation of the native peptide with LysC gave two tetrapeptides and free His. One of the tetrapeptides contained equimolar amounts of Phe, Lys, Leu and Met. Chymotrypsin digestion produced a dipeptide, a tripeptide and a tetrapeptide. The tetrapeptide contained a Lys, two Gly and a His. The native sequence was determined to be: (1) G-F-K-K-G-L-M-F-H (2) H-L-G-K-K-F-F-G-M (3) M-L-F-G-K-K-F-H-G (4) H-F-L-G-K-K-F-M-G (5) M-L-F-K-F-G-G-K-H |
(5) M-L-F-K-F-G-G-K-H |
