Molecule 3g03 Test 3

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  • 1/25 Questions

    A portion of the nucleotide sequence from the DNA coding strand of the chick ovalbumin gene is shown here.      CTCAGAGTTCACCATGGGCTCCATCGGTGCAG-      CAAGCATGGAA-(149bp)-ATTCTTTGGCAGATGT-      GTTTCCCCTTAAAAAGAA What is the partial amino acid sequence of the encoded protein?

    • MGSIGAASME-(50 a.a.)-FFGRCVSP
    • MSGIAAGSME-(50 a.a.)-EEGRCVSP
    • MAASMEGSIG-(50 a.a.)-FFGRCVSP
    • MGSIGAASME-(50 a.a.)-RCVSFFGP
    • MAASMEGSIG-(50 a.a.)-FCVSFGRP
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Molecule Quizzes & Trivia
About This Quiz

The 'molecule 3G03 Test 3' assesses knowledge in molecular biology, focusing on DNA to protein translation, isoelectric points of amino acids, and buffer preparation. It is designed for advanced learners aiming to enhance their understanding of genetic coding and biochemical applications.

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  • 2. 

    The isoelectric point of Tyr is

    • 5.7

    • 9.83

    • 6.33

    • 5.0

    • 10.0

    Correct Answer
    A. 5.7
    Explanation
    The isoelectric point of an amino acid is the pH at which the molecule has no net charge. At a pH below the isoelectric point, the amino acid will have a net positive charge due to the presence of excess protons. At a pH above the isoelectric point, the amino acid will have a net negative charge due to the loss of protons. Therefore, in this case, the isoelectric point of Tyr is 5.7, which means that at a pH of 5.7, Tyr will have no net charge.

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  • 3. 

    The isoelectric point of His-Tyr is

    • 7.68

    • 4.12

    • 5.46

    • 9.89

    • 5.7

    Correct Answer
    A. 7.68
    Explanation
    The isoelectric point of a molecule is the pH at which it carries no net electrical charge. In the case of His-Tyr, the isoelectric point is 7.68. This means that at a pH of 7.68, the His-Tyr molecule will have an equal number of positive and negative charges, resulting in a neutral overall charge. At pH values below 7.68, the molecule will carry a net positive charge, while at pH values above 7.68, it will carry a net negative charge.

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  • 4. 

    The isoelectric point of Ala-His-Asp-Arg-Val-Gln is

    • 7.95

    • 11.17

    • 4.97

    • 7.0

    • 3.03

    Correct Answer
    A. 7.95
    Explanation
    The isoelectric point (pI) of a peptide or protein is the pH at which it carries no net electrical charge. It can be calculated by averaging the pKa values of its ionizable groups. Each amino acid in the sequence has a specific pKa value for its side chain. In this case, Ala, His, Asp, Arg, Val, and Gln all have ionizable groups with different pKa values. By calculating the average pKa values and determining the pH at which the net charge is zero, the isoelectric point is found to be 7.95.

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  • 5. 

    To make a phosphate buffer at pH 12.38 starting with one liter of 10 mM phosphoric acid (H3PO4; pKs are of 2.15, 6.82, and 12.38), you could add

    • 5 millimoles of KOH

    • 15 millimoles of KOH

    • 25 millimoles of KOH

    • 20 millimoles of KOH

    • You can't make a buffer by adding HCL or KOH

    Correct Answer
    A. 25 millimoles of KOH
    Explanation
    To make a phosphate buffer at pH 12.38, you need to add a base to the phosphoric acid to increase the pH. The pK values of phosphoric acid indicate that at pH 12.38, the majority of the acid will be in its conjugate base form. Therefore, you need to add a base such as KOH to neutralize the acid and create the buffer. The correct answer of 25 millimoles of KOH ensures that there is enough base to react with the acid and achieve the desired pH of 12.38. Adding a lower amount of KOH would result in a lower pH, while adding a higher amount would result in a higher pH.

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  • 6. 

    To make an acetate buffer at pH 4.76 (pK = 4.76) starting with 250 mL of 0.2 M acetic acid (pK = 4.76), you could add:

    • 0.05 moles of HCL

    • 0.025 moles of NaOH

    • You can't make a buffer by adding HCL or NaOH

    • 0.05 moles of NaOH

    • 0.2 moles of HCL

    Correct Answer
    A. 0.025 moles of NaOH
    Explanation
    Adding 0.025 moles of NaOH to the solution of acetic acid will create an acetate buffer at pH 4.76. The pK value of acetic acid is 4.76, which means that at this pH, the concentration of acetic acid and acetate ions will be equal. By adding NaOH, it will react with acetic acid to form sodium acetate and water. This reaction will increase the concentration of acetate ions, while the concentration of acetic acid will decrease. This shift in concentrations will create a buffer solution with a pH close to the pK value.

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  • 7. 

    Given the Henderson-Hasselbalch equation as pH = pK + log {[A-] /[HA]}, what is the probability of finding the form of histidine shown in Figure 2 at pH 4 assuming the various pK's of histidine are:      pK amino = 10.0; pK carboxyl = 2.0; pK imidazole = 6.0

    • (1/1000001)x(1/101)x(1/101)

    • (1/100001)x(1/101)x(1/101)

    • (1/10001)x(1/1001)x(1/1001)

    • (1/1001)x(1/10001)x(1/10001)

    • (1/101)x(1/100001)x(1/100001)

    Correct Answer
    A. (1/1000001)x(1/101)x(1/101)
  • 8. 

    O2 binds to the heme group of myoglobin such that binding is _____ when the oxygen concentration is equal to the dissociation constant

    • 1/4 - maximal

    • 1/8 - maximal

    • 1/2 - maximal

    • 1/16 maximal

    • None of the rest

    Correct Answer
    A. 1/2 - maximal
    Explanation
    When the oxygen concentration is equal to the dissociation constant, the binding of O2 to the heme group of myoglobin is at its maximum. This means that half of the available binding sites on the myoglobin molecule are occupied by oxygen molecules.

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  • 9. 

    The _______, which are identical in all members of a protein family, are essential for the structure and/or function of the proteins and cannot be replaced by other residues.

    • Homologous proteins

    • Invariant residues

    • Variable residues

    • Hemoglobin

    • All of the rest

    Correct Answer
    A. Invariant residues
    Explanation
    Invariant residues are the correct answer because they refer to the amino acid residues that remain the same in all members of a protein family. These residues play a crucial role in maintaining the structure and function of the proteins and cannot be substituted by other amino acids. Therefore, they are essential for the proper functioning of the proteins in the family.

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  • 10. 

    BPG binds to deoxyhemoglobin but not to oxyhemoglobin.  It therefore ______ hemoglobin's oxygen-binding affinity by stabilizing the deoxyhemoglobin conformation.

    • Increases

    • Reduces

    • Improves

    • Completely abolish

    • None of the rest

    Correct Answer
    A. Reduces
    Explanation
    BPG binds to deoxyhemoglobin, which is the form of hemoglobin that does not have oxygen bound to it. By binding to deoxyhemoglobin, BPG stabilizes its conformation. This stabilization reduces hemoglobin's oxygen-binding affinity, meaning that it makes it less likely for oxygen to bind to hemoglobin. Therefore, the correct answer is "reduces."

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  • 11. 

    The tubulin dimer has binding sites that bind

    • ATP and ADP

    • GTP and ATP

    • GDP and ADP

    • GTP and ADP

    • GTP and GDP

    Correct Answer
    A. GTP and GDP
    Explanation
    Tubulin is a protein that forms microtubules, which are crucial for cell structure and movement. The tubulin dimer refers to two tubulin proteins that come together to form a functional unit. These dimers have binding sites that can bind guanosine triphosphate (GTP) and guanosine diphosphate (GDP). GTP is typically bound to tubulin when it is incorporated into the growing end of a microtubule, providing energy for polymerization. As the microtubule grows, GTP is hydrolyzed to GDP. Therefore, the correct answer is GTP and GDP.

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  • 12. 

    Specificity pocket in trypsin refers to a hole on the enzyme surface at the active site that accommodates the residues on the _____ side of the _____ peptide bond

    • C-terminal, catalytic triad

    • C-terminal, scissile

    • N-terminal, catalytic triad

    • N-terminal, serine protease

    • N-terminal, scissile

    Correct Answer
    A. N-terminal, scissile
    Explanation
    The specificity pocket in trypsin refers to a hole on the enzyme surface at the active site that accommodates the residues on the N-terminal side of the scissile peptide bond. This means that the pocket is specifically designed to bind and interact with the amino acids located on the N-terminal side of the peptide bond that will be cleaved by trypsin. The scissile bond refers to the peptide bond that will be cleaved by the enzyme.

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  • 13. 

    Hemoglobin is 50% saturated with oxygen when pO2 = 26 torr. If hemoglobin exhibited hyperbolic binding (as myoglobin does) with 50% saturation at 26 torr. Which of the following statements is correct? (pO2 =100 in the lungs and pO2 = 30 in the tissues; fractional saturation YO2  = pO2 /  {K  + pO2} and K=p50=pO2 when YO2=0.5)

    • The fractional saturation when pO2 = 30 torr is 0.24

    • The fractional saturation when pO2 = 100 torr is 0.59

    • Hemoglobin can only use 25% of its capacity to deliver O2 to the tissues

    • Hemoglobin is fully oxidized in the tissues

    • None of the rest

    Correct Answer
    A. Hemoglobin can only use 25% of its capacity to deliver O2 to the tissues
  • 14. 

    Given hexacoordinate Fe(II) in heme is bright red and pentacoordinate Fe(II) is blue. Which of the following statements is correct?

    • Oxidized hemoglobin Fe(III) is bright red

    • Iron in oxygenated hemoglobin is pentacoordinated Fe(II)

    • Hemoglobin in the venous blood is bluish in colour

    • Hemoglobin in the venous blood contains hexacoordinate Fe(II)

    • None of the rest

    Correct Answer
    A. Hemoglobin in the venous blood is bluish in colour
    Explanation
    The correct answer is "Hemoglobin in the venous blood is bluish in colour." Venous blood is blood that has been oxygenated by the tissues and is returning to the heart to be pumped to the lungs for reoxygenation. When hemoglobin is deoxygenated, it has a bluish color, which is why venous blood appears bluish.

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  • 15. 

    Which of the following statements is correct about hemoglobin?

    • Decrease in [H+] promotes the shift of hemoglobin from the oxy to the deoxy conformation

    • Increase in oxygen affinity improves oxygen delivery to the muscle where it is needed

    • Hemoglobin in the blood passing through highly active muscle has less oxygen affinity

    • Fetus hemoglobin has higher p50 than that of mother

    • None of the rest

    Correct Answer
    A. Hemoglobin in the blood passing through highly active muscle has less oxygen affinity
    Explanation
    Hemoglobin in the blood passing through highly active muscle has less oxygen affinity because increased metabolic activity in the muscle leads to a decrease in pH and an increase in carbon dioxide levels. This acidic and high carbon dioxide environment causes a decrease in oxygen affinity of hemoglobin, facilitating the release of oxygen to the muscle tissues where it is needed.

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  • 16. 

    Lampreys are among the world's most primitive vertebrates and are therefore interesting organisms to study. It has been shown that lamprey hemoglobin forms a tetramer when deoxygenated, but when the hemoglobin becomes oxygenated, the tetramers dissociate into monomers. The binding of oxygen to lamprey hemoglobin is influenced by pH; as in human hemoglobin, the deoxygenated form is favoured when the pH decreases. Glutamate residues on the surface of the monomers play an important role. Which of the following statements is incorrect?

    • Negatively charged glutamate side chains on the surface of the oxygenated lamprey monomer would resist association due to the charge-charge repulsion

    • When the pH decreases, excess protons would bind to the glutamate side chains, neutralizing them

    • When pH decreases, lamprey hemoglobins form monomer

    • Lamprey hemoglobins exist as tetramers in metabolically active tissues

    • Lamprey hemoglobins exist as monomers in lung tissues

    Correct Answer
    A. When pH decreases, lamprey hemoglobins form monomer
    Explanation
    The given statement is incorrect because when the pH decreases, lamprey hemoglobins do not form monomers. It has been mentioned in the passage that the deoxygenated form of lamprey hemoglobin is favored when the pH decreases, but it does not specify that it forms monomers. The passage only states that the tetramers dissociate into monomers when the hemoglobin becomes oxygenated. Therefore, the statement that lamprey hemoglobins form monomers when the pH decreases is incorrect.

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  • 17. 

    Since proteins are limited in their abilities to catalyze oxidation-reduction reactions, enzymes often employ ______ to assist with catalysis.

    • Leaving group

    • Leaving group

    • Coenzymes

    • Acid

    • Base

    Correct Answer
    A. Coenzymes
    Explanation
    Enzymes often employ coenzymes to assist with catalysis because proteins alone have limited abilities to catalyze oxidation-reduction reactions. Coenzymes are small organic molecules that work together with enzymes to facilitate and enhance their catalytic activity. They can act as carriers of specific functional groups or electrons, allowing enzymes to perform a wider range of chemical reactions. By binding to the active site of an enzyme, coenzymes can participate in the reaction and help to lower the activation energy, making the reaction more efficient.

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  • 18. 

    An uncatalyzed reaction has a rate of 4.1 x 10^-7.sec-1. When an enzyme is added the rate is 8.2 x 10^4.sec-1. Calculate the rate enhancement caused by the enzyme.

    • 3.2 x 10^4

    • 1.34 x 10^-2

    • 2.0 x 10^11

    • 7.4 x 10^-3

    • None of the rest

    Correct Answer
    A. 2.0 x 10^11
    Explanation
    The rate enhancement caused by the enzyme can be calculated by taking the ratio of the rate with the enzyme to the rate without the enzyme. In this case, the rate with the enzyme is 8.2 x 10^4.sec-1 and the rate without the enzyme is 4.1 x 10^-7.sec-1. Dividing these two values gives a rate enhancement of approximately 2.0 x 10^11. This means that the enzyme increases the rate of the reaction by a factor of 2.0 x 10^11.

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  • 19. 

    The imidazole side chain of histidine can function as either a general acid catalyst or a general base catalyst because:

    • In the physiological pH range both H+ and OH- are present at high concentrations

    • The imidazole group is a strong reducing agent at physiological pH

    • One guanidine group is protonated and the other is deprotonated at physiological pH

    • In the physiological pH range, the nitrogen in the ring can be easily protonated/deprotonated

    • The sulfur atoms in the ring can either gain or lose a proton at physiological pH

    Correct Answer
    A. In the physiological pH range, the nitrogen in the ring can be easily protonated/deprotonated
    Explanation
    The imidazole side chain of histidine can function as either a general acid catalyst or a general base catalyst because in the physiological pH range, the nitrogen in the ring can be easily protonated/deprotonated. This means that the imidazole group can readily donate or accept a proton, allowing it to act as both an acid and a base catalyst depending on the reaction conditions. This versatility makes histidine an important residue in enzyme catalysis, as it can participate in a wide range of chemical reactions.

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  • 20. 

    Which of the following statements does not apply to collagen?

    • It contains hydroxylated amino acids.

    • The polypeptide forms a left-handed helix.

    • Hydrogen bonds between the -OH groups of Hyp residues stabilize the helix.

    • There is a requirement for glycine at every third position.

    • The triple helical structure twists in the right-handed direction.

    Correct Answer
    A. Hydrogen bonds between the -OH groups of Hyp residues stabilize the helix.
    Explanation
    The statement "Hydrogen bonds between the -OH groups of Hyp residues stabilize the helix" does not apply to collagen.

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  • 21. 

    Why is the decreased affinity of fetal hemoglobin for BPG advantageous?

    • With fewer BPG molecules bound there are more heme residues available for O2 binding.

    • Decreased BPG binding biases the fetal hemoglobin toward the R state.

    • More free BPG is available to bind to adult hemoglobin, resulting in a shift of adult hemoglobin to the R state.

    • BPG is available to bind to fetal myoglobin, helping to release O2 in fetal muscle tissue.

    • None of the rest is correct.

    Correct Answer
    A. Decreased BPG binding biases the fetal hemoglobin toward the R state.
    Explanation
    The decreased affinity of fetal hemoglobin for BPG is advantageous because it biases the hemoglobin towards the R state. The R state is the relaxed state of hemoglobin, which has a higher affinity for oxygen. This means that with decreased BPG binding, more oxygen can bind to the heme residues of fetal hemoglobin, allowing for efficient oxygen transport to the developing fetus.

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  • 22. 

    Which of the following is correct about amino acid side chain?

    • The nucleophilicity of a side chain is inversely related to its acidity.

    • The less acidic a group is, the more nucleophilic it is.

    • The more basic a group is, the more likely it is to react with an electrophilic group.

    • The more electron-rich a group is, the more likely it is to react with an electrophilic group.

    • All of the rest are correct.

    Correct Answer
    A. All of the rest are correct.
    Explanation
    All of the statements mentioned in the answer are correct. The nucleophilicity of a side chain is inversely related to its acidity, meaning that the less acidic a group is, the more nucleophilic it is. Additionally, a more basic group is more likely to react with an electrophilic group, and a more electron-rich group is also more likely to react with an electrophilic group. Therefore, the correct answer is that all of the rest of the statements are correct.

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  • 23. 

    Lysozyme catalyzes the hydrolysis of a polysaccharide component of bacterial cell walls. The damaged bacteria subsequently lyse (rupture). Part of lysozyme's mechanism is shown in Figure 1. The enzyme catalyzes cleavage of a bond between two sugar residues (represented by hexagons). Catalysis involves the side chains of Glu 35 and Asp 52. One of the residues has a pK of 4.5; the other has a pK of 5.9. Which of the following statements is correct?

    • Glu 35 has a pK of 5.9 and Asp52 has a pK of 4.5

    • Glu 35 has a pK of 4.5 and Asp52 has a pK of 5.9

    • Lysozyme is inactive at pH 2.0 because both the Glu and the Asp are deprotonated

    • Lysozyme is inactive at pH 8.0 because the Asp would be unable to donate a hydrogen to cleave the bond between the sugar residues.

    • All of the rest are correct

    Correct Answer
    A. Glu 35 has a pK of 5.9 and Asp52 has a pK of 4.5
    Explanation
    The correct answer is Glu 35 has a pK of 5.9 and Asp52 has a pK of 4.5. This is because the question states that one of the residues has a pK of 4.5 and the other has a pK of 5.9. Therefore, Glu 35 must have a pK of 5.9 and Asp52 must have a pK of 4.5.

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  • 24. 

    Hydrophobic residues usually appear at the first and fourth positions in the seven-residue repeats of polypeptides that form coiled coils. The first and fourth side chains are buried in the coiled coil, but the remaining side chains are exposed to the solvent and therefore tend to be polar or charged. Which of the following sequences is more likely to appear in a coiled coil?

    • Ile-Gln-Glu-Val-Glu-Arg-Asp

    • Trp-Gln-Glu-Tyr-Glu-Arg-Asp

    • Ile-Gln-Glu-Val-Glu-Ala-Asp

    • Trp-Gln-Glu-Tyr-Glu-Ala-Asp

    • Gly-Pro-Hyp-Gly-Pro-Hyp-Gly (Hyp = hydroxyproline)

    Correct Answer
    A. Ile-Gln-Glu-Val-Glu-Arg-Asp
    Explanation
    In coiled coils, hydrophobic residues are typically found at the first and fourth positions of the seven-residue repeats. The first and fourth side chains are buried within the coiled coil structure, while the remaining side chains are exposed to the solvent. Therefore, the more likely sequence to appear in a coiled coil would be one that has hydrophobic residues at the first and fourth positions, and polar or charged residues at the remaining positions. Among the given sequences, "Ile-Gln-Glu-Val-Glu-Arg-Asp" fits this pattern, making it the more likely sequence to appear in a coiled coil.

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  • 25. 

    According to Figure 3, which of the following is not likely to be a serine protease zymogen?

    • Factor XI

    • Factor IX

    • Factor X

    • Prothrombin

    • Fibrinogen

    Correct Answer
    A. Fibrinogen
    Explanation
    Fibrinogen is not likely to be a serine protease zymogen because it is a soluble protein that plays a crucial role in blood clotting by converting to fibrin during the coagulation process. Serine protease zymogens are inactive forms of enzymes that require activation to become active proteases. Factors XI, IX, and X are all examples of serine protease zymogens involved in the coagulation cascade.

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  • Mar 21, 2023
    Quiz Edited by
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  • Dec 11, 2012
    Quiz Created by
    ISci2A18

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