Chemical Bonding Test 2

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Chemical Bonding Test 2 - Quiz

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Questions and Answers
  • 1. 

    A nucleoside is a:

    • A.

      Sugar + base

    • B.

      Sugar + base + phosphate group

    • C.

      Sugar + acid

    • D.

      DNA + base

    • E.

      RNA + sugar

    Correct Answer
    A. Sugar + base
    Explanation
    A nucleoside is formed by the combination of a sugar molecule and a base. This combination does not include a phosphate group, acid, DNA, or RNA. Therefore, the correct answer is "Sugar + base."

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  • 2. 

    DNA rich in which of the following base pairs has the highest melting temperature (TM)?

    • A.

      A:U

    • B.

      A:T

    • C.

      G:C

    • D.

      G:A

    • E.

      G:T

    Correct Answer
    C. G:C
    Explanation
    DNA rich in G:C base pairs has the highest melting temperature (TM) because G:C base pairs form three hydrogen bonds, which are stronger than the two hydrogen bonds formed by A:T base pairs. The higher number of hydrogen bonds in G:C base pairs leads to stronger stability and a higher melting temperature.

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  • 3. 

    In the peptide backbone, there is no rotation around ______ due to its partial double bond character

    • A.

      N - N bond

    • B.

      C - C bond

    • C.

      C - N bond

    • D.

      C - H bond

    • E.

      N - H bond

    Correct Answer
    C. C - N bond
    Explanation
    The peptide backbone consists of alternating amino acids connected by C-N bonds. These bonds have partial double bond character due to resonance, which restricts rotation around the C-N bond. This lack of rotation is important for maintaining the secondary structure of proteins, such as alpha helices and beta sheets. Therefore, the correct answer is C - N bond.

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  • 4. 

    In the peptide backbone, rotations around the N - C alpha and C alpha - C bonds are restricted by

    • A.

      Ionic bonding

    • B.

      Van der Waals interaction

    • C.

      H-bonding

    • D.

      Steric interactions

    • E.

      None of the rest

    Correct Answer
    D. Steric interactions
    Explanation
    Steric interactions refer to the repulsive forces between atoms or groups of atoms that are in close proximity to each other. In the peptide backbone, the N - C alpha and C alpha - C bonds are rigid and allow limited rotation due to steric hindrance. The bulky side chains of amino acids can clash with each other, limiting the rotation around these bonds. Therefore, steric interactions play a crucial role in restricting the rotations around the N - C alpha and C alpha - C bonds in the peptide backbone.

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  • 5. 

    You have a short piece of synthetic RNA that you want to use as a probe to identify a gene in a sample of DNA. The RNA probe has a tendency to hybridize with sequences that are only weakly complementary. Which of the following statement is correct?

    • A.

      You should increase the temperature to improve your chances of tagging the correct sequence

    • B.

      You should decrease the temperature to improve your chances of tagging the correct sequence

    • C.

      Decreasing the temperature will melt out imperfect matches between the probe and the DNA

    • D.

      Increase the temperature will increase the chance of tagging the wrong sequence

    • E.

      None of the rest

    Correct Answer
    A. You should increase the temperature to improve your chances of tagging the correct sequence
    Explanation
    Increasing the temperature will improve the chances of tagging the correct sequence because the RNA probe has a tendency to hybridize with weakly complementary sequences. By increasing the temperature, the binding between the probe and the DNA will be more stringent, allowing for better discrimination of the correct sequence from imperfect matches.

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  • 6. 

    A portion of the nucleotide sequence from the DNA coding strand of the chick ovalbumin gene is shown here. CTCAGAGTTCACC ATG GGC TCC ATC GGT GCA G- CA AGC ATG GAA -(149bp)-A TTC TTT GGC AGA TGT- GTTTCCCCTTAAAAAGAA What is the partial amino acid sequence of the encoded protein?

    • A.

      MGSIGAASME-(50 a.a.)-FFGRCVSP

    • B.

      MSGIAAGSME-(50 a.a.)-EEGRCVSP

    • C.

      MAASMEGSIG-(50 a.a.)-FFGRCVSP

    • D.

      MGSIGAASME-(50 a.a.)-RCVSFFGP

    • E.

      MAASMEGSIG-(50 a.a.)-FCVSFGRP

    Correct Answer
    A. MGSIGAASME-(50 a.a.)-FFGRCVSP
    Explanation
    The given nucleotide sequence is translated into the corresponding amino acid sequence using the genetic code. The correct answer, "MGSIGAASME-(50 a.a.)-FFGRCVSP," represents the partial amino acid sequence encoded by the DNA sequence.

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  • 7. 

    The genome of the bacterium Carsonella ruddii contains 159 kb of DNA with 182 ORFs. What can you conclude about the habitat or lifestyle of this bacterium?

    • A.

      C ruddii with such a small genome and only 182 genes, must be some sort of parasite rather than a free-living bacterium

    • B.

      C ruddii with such a small genome and only 182 genes, must be an aerobic organism rather than anaerobic organism

    • C.

      C ruddii with such a small genome and only 182 genes, must be a thermophilic organism

    • D.

      C ruddii with such a small genome and only 182 genes, must be a species of plant

    • E.

      None of the rest

    Correct Answer
    A. C ruddii with such a small genome and only 182 genes, must be some sort of parasite rather than a free-living bacterium
    Explanation
    The small genome size and limited number of genes suggest that Carsonella ruddii is likely a parasite rather than a free-living bacterium. Parasites tend to have smaller genomes and fewer genes compared to free-living organisms because they rely on their host for many essential functions. This adaptation allows them to streamline their genome and conserve energy by not carrying unnecessary genes.

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  • 8. 

    Which of the following statements is correct about PCR experiments using DNA polymerase from E. coli instead of from thermophilic bacteria?

    • A.

      DNA polymerase is destroyed by the high temperature required for separating the double stranded DNA

    • B.

      After separating the two DNA strands and allowing the primers to anneal with the template, more DNA polymerase has to be added

    • C.

      The best temperature for primer extension is between 37 - 47 degree celsius

    • D.

      Fresh DNA polymerase has to be added at each reaction cycle

    • E.

      All of the rest are correct

    Correct Answer
    E. All of the rest are correct
    Explanation
    The correct answer is that all of the rest of the statements are correct. This means that using DNA polymerase from E. coli instead of thermophilic bacteria does not destroy the polymerase at the high temperature required for separating the double stranded DNA. After separating the DNA strands and allowing the primers to anneal, more DNA polymerase does need to be added. The optimal temperature for primer extension is between 37 - 47 degrees Celsius. Finally, fresh DNA polymerase does need to be added at each reaction cycle.

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  • 9. 

    For amino acids His, Phe, Ser, Ile, Asp and Trp, the correct order of solubility from high to low in water at pH 6.5 is

    • A.

      Asp, His, Ser, Trp, Ile, Phe

    • B.

      Phe, Ile, Trp, Ser, His, Asp

    • C.

      Ser, Asp, His, Ile, Trp, Phe

    • D.

      Ser, Asp, His, Trp, Ile, Phe

    • E.

      His, Ser, His, Trp, Ile, Phe

    Correct Answer
    A. Asp, His, Ser, Trp, Ile, Phe
    Explanation
    The solubility of amino acids in water is influenced by their chemical properties. Asp (aspartic acid) is a polar amino acid with a negatively charged side chain, making it highly soluble in water. His (histidine) is also polar, but its solubility is slightly lower than Asp due to its positively charged side chain. Ser (serine) is a polar amino acid with a hydroxyl group, which enhances its solubility. Trp (tryptophan) is a nonpolar amino acid, making it less soluble in water. Ile (isoleucine) and Phe (phenylalanine) are both nonpolar amino acids, with Phe being less soluble than Ile due to its larger aromatic side chain. Therefore, the correct order of solubility from high to low is Asp, His, Ser, Trp, Ile, Phe.

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  • 10. 

    Which of the following two amino acid side chains can interact with one another via ion pair at certain pH

    • A.

      K and E

    • B.

      K and D

    • C.

      R and E

    • D.

      H and D

    • E.

      All of the rest

    Correct Answer
    E. All of the rest
    Explanation
    All of the rest of the amino acid side chains can interact with one another via ion pair at certain pH. This includes the combinations K and E, K and D, R and E, and H and D. Ion pairs are formed when the side chains of these amino acids have opposite charges and can attract each other. At certain pH levels, these amino acids can become ionized, allowing for the formation of ion pairs.

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  • 11. 

    Glutathione transferase consists of a dimer of two identical subunits. The dimer is in equilibrium with its constituent monomers. Site-directed mutagenesis studies were carried out on this protein in which two arginine residues were mutated to glutamines and two aspartates were mutated to asparagines. These substitutions caused the equilibrium to shift in favour of the monomeric form of the enzyme. Which of the following statements is correct about these arginines and aspartates?

    • A.

      They are likely to be found on the surface of the monomer forming intra-monomer ion pairs

    • B.

      They are likely to be found on the surface of the monomer forming inter-monomer ion pairs

    • C.

      They are likely to be found buried at the core of the monomer

    • D.

      They are likely to be found on alpha helices

    • E.

      None of the rest

    Correct Answer
    B. They are likely to be found on the surface of the monomer forming inter-monomer ion pairs
    Explanation
    The substitutions of arginine residues to glutamines and aspartates to asparagines in the glutathione transferase protein caused a shift in the equilibrium towards the monomeric form of the enzyme. This suggests that the arginine and aspartate residues are likely to be found on the surface of the monomer, forming inter-monomer ion pairs. This means that these residues are involved in interactions between the two identical subunits of the dimer, stabilizing its structure.

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  • 12. 

    The sequence of crinia-angiotensin, an angiotensin II-like undecapeptide from the skin of the Australian frog is Ala-Pro-Gly-Asp-Arg-Ile-Tyr-Val-His-Pro-Phe. Treatment of the peptide with chymotrypsin released two fragments. Which of the following correspond to the sequence of one of these two fragments.

    • A.

      APGDRIY

    • B.

      VHPP

    • C.

      APGDR

    • D.

      IYVHPF

    • E.

      PGDRI

    Correct Answer
    A. APGDRIY
    Explanation
    The given sequence of crinia-angiotensin is Ala-Pro-Gly-Asp-Arg-Ile-Tyr-Val-His-Pro-Phe. When treated with chymotrypsin, it released two fragments. One of these fragments is APGDRIY, as it matches the sequence of the peptide.

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  • 13. 

    The isoelectric point of His is

    • A.

      7.68

    • B.

      5.56

    • C.

      3.82

    • D.

      7.0

    • E.

      6.0

    Correct Answer
    A. 7.68
    Explanation
    The isoelectric point of His is 7.68. This means that at a pH of 7.68, the amino acid histidine will have no net charge. At pH values below 7.68, histidine will have a positive charge, and at pH values above 7.68, it will have a negative charge. The isoelectric point is the pH at which an amino acid or molecule has no net charge.

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  • 14. 

    The isoelectric point of His-Asp is

    • A.

      4.97

    • B.

      2.94

    • C.

      7.68

    • D.

      4.01

    • E.

      6.61

    Correct Answer
    A. 4.97
    Explanation
    The isoelectric point of a molecule is the pH at which it carries no net electrical charge. His-Asp is a peptide consisting of the amino acids histidine and aspartic acid. Histidine has a pKa value of approximately 6.0, while aspartic acid has a pKa value of approximately 3.9. The isoelectric point of a peptide is determined by the average pKa values of its constituent amino acids. Since the pKa value of histidine is higher than that of aspartic acid, the isoelectric point of His-Asp is expected to be closer to the pKa value of histidine, which is around 6.0. Therefore, the correct answer is 4.97.

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  • 15. 

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

    • A.

      7.95

    • B.

      11.17

    • C.

      4.97

    • D.

      7.0

    • E.

      3.03

    Correct Answer
    A. 7.95
    Explanation
    The isoelectric point (pI) is the pH at which a molecule has no net charge. It can be calculated by averaging the pKa values of the ionizable groups in the molecule. In the case of Ala-His-Asp-Arg-Val-Gln, there are several ionizable groups: the amino group of Ala, the imidazole group of His, the carboxyl group of Asp, the guanidinium group of Arg, and the side chain carboxyl group of Gln. The pKa values of these groups are approximately 2.34, 6.04, 3.65, 12.48, and 2.17, respectively. By averaging these values, we can estimate the pI of Ala-His-Asp-Arg-Val-Gln to be around 7.95.

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  • 16. 

    To make a phosphate buffer at pH 9.6 starting with one liter of 10 mM phosphoric acid (H3PO4; pKs are of 2.15, 6.82, and 12.38), you could add

    • A.

      5 millimoles of HCL

    • B.

      15 millimoles of KOH

    • C.

      25 millimoles of HCL

    • D.

      20 millimoles of KOH

    • E.

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

    Correct Answer
    D. 20 millimoles of KOH
    Explanation
    To make a phosphate buffer at pH 9.6, you need to add a base to the phosphoric acid to raise the pH. The pKa values of phosphoric acid suggest that it can act as a weak acid and donate protons at different pH levels. In this case, to achieve a pH of 9.6, you need a base with a pKa higher than 9.6. Among the given options, KOH is the only base, and adding 20 millimoles of KOH will help neutralize the acid and create a phosphate buffer at the desired pH. HCL is an acid and cannot be used to create a buffer, so the other options are not suitable.

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  • 17. 

    To make an acetate buffer at pH 4.76 (pK = 4.76) starting with 500 mL of 0.2 M acetic acid (pK = 4.76), you could add:

    • A.

      0.05 moles of HCL

    • B.

      0.025 moles of NaOH

    • C.

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

    • D.

      0.05 moles of NaOH

    • E.

      0.2 moles of HCL

    Correct Answer
    D. 0.05 moles of NaOH
    Explanation
    To make an acetate buffer at pH 4.76, you need to add a weak acid (acetic acid) and its conjugate base (sodium acetate) in the correct ratio. Since acetic acid is a weak acid, it can react with a strong base like NaOH to form sodium acetate. By adding 0.05 moles of NaOH, you can neutralize 0.05 moles of acetic acid, resulting in the formation of sodium acetate. This will create a buffer solution with the desired pH of 4.76. Adding HCl or NaOH alone will not create a buffer because they are strong acids and bases respectively, and will completely dissociate in solution.

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  • 18. 

    A soccer coach keeps a couple of instant cold packs in her bag in case one of her players suffers a muscle injury. Instant cold packs are composed of a plastic bag containing a smaller water bag and solid ammonium nitrate. In order to activate the fold pack, the bag is kneaded until the smaller water bag breaks, which allows the released water to dissolve the ammonium nitrate. Which of the following statements about the dissolution of ammonium nitrate in water is correct?

    • A.

      Delta G > 0

    • B.

      Delta S > 0

    • C.

      Delta H < 0

    • D.

      Delta T > 0

    • E.

      Delta H < Delta S

    Correct Answer
    B. Delta S > 0
    Explanation
    The dissolution of ammonium nitrate in water is an endothermic process, meaning it requires energy input. When the solid ammonium nitrate dissolves in water, it breaks apart into individual ions, which increases the randomness or entropy of the system. This increase in entropy indicates that Delta S, the change in entropy, is greater than zero.

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  • 19. 

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

    • A.

      (1/1000001)x(1/101)x(1/101)

    • B.

      (1/100001)x(1/101)x(1/101)

    • C.

      (1/10001)x(1/1001)x(1/1001)

    • D.

      (1/1001)x(1/10001)x(1/10001)

    • E.

      (1/101)x(1/100001)x(1/100001)

    Correct Answer
    A. (1/1000001)x(1/101)x(1/101)
  • 20. 

    Which of the bases in Figure 5 is not found on DNA

    • A.

      5a

    • B.

      5b

    • C.

      5c

    • D.

      5d

    • E.

      5e

    Correct Answer
    C. 5c
    Explanation
    In Figure 5, the bases 5a, 5b, 5d, and 5e are commonly found on DNA. However, base 5c is not found on DNA. DNA consists of four bases: adenine (A), cytosine (C), guanine (G), and thymine (T). Base 5c does not correspond to any of these four bases, indicating that it is not found on DNA.

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  • 21. 

    Which of the molecules in Figure 6 is used in DNA sequencing reactions

    • A.

      6a

    • B.

      6b

    • C.

      6c

    • D.

      6d

    • E.

      6e

    Correct Answer
    B. 6b
    Explanation
    In Figure 6, molecule 6b is used in DNA sequencing reactions.

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  • 22. 

    In your laboratory, you plan to use CM (Figure 3) ion exchange chromatography to separate a peptide from a mixture of different peptides (shown below) at pH 7.0. Which of the following peptides would have the strongest binding affinity to the CM groups.

    • A.

      GLEKSLVRLGDVQPSLGKESRAKKFQRQ

    • B.

      GLEESLVDLGEVQPSLGAESRAKKFQDQ

    • C.

      GLKESLVKLGDVEPSDGLESFARVFQDQ

    • D.

      QDQFVRAFSELGDSPEVDGLKVLSEKLG

    • E.

      QRQFDEAHSEQGLSPQVDGLYVLSTELG

    Correct Answer
    A. GLEKSLVRLGDVQPSLGKESRAKKFQRQ
    Explanation
    The peptide GLEKSLVRLGDVQPSLGKESRAKKFQRQ would have the strongest binding affinity to the CM groups because it contains a higher number of positively charged amino acids, such as lysine (K) and arginine (R), which can interact with the negatively charged CM groups through ion exchange interactions at pH 7.0. These interactions result in stronger binding and better retention of the peptide on the CM ion exchange resin.

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  • 23. 

    Biochemists sometimes link a recombinant protein to a protein known as green fluorescent protein (GFP), which was first purified from bioluminescent jellyfish. The fluorophore in GFP (shown in Figure 2) is a derivative of three consecutive amino acids that undergo cyclization of the polypeptide chain and an oxidation. Identify the three residues that form the fluorophore.

    • A.

      SYG

    • B.

      TYG

    • C.

      STG

    • D.

      GTS

    • E.

      GYS

    Correct Answer
    A. SYG
    Explanation
    The three residues that form the fluorophore in GFP are SYG.

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  • 24. 

    According to Table 4, which of the following statements is incorrect.

    • A.

      MspI and AsuI generate sticky ends

    • B.

      EcoRI and PstI generate sticky ends

    • C.

      SauI and NotI generate sticky ends

    • D.

      AluI and EcoRV generate blunt ends

    • E.

      AsuI and SauI cleave palindromic DNA sequences

    Correct Answer
    E. AsuI and SauI cleave palindromic DNA sequences
    Explanation
    The given statement is incorrect because AsuI and SauI do not cleave palindromic DNA sequences.

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  • 25. 

    The sequence of the template DNA for the sequencing gel shown in Figure 1 is

    • A.

      3'AGTTFCTACC5'

    • B.

      3'GGTAGCAACT5'

    • C.

      3'CCATCGTTGA5'

    • D.

      3'TCAACAGTTG5'

    Correct Answer
    B. 3'GGTAGCAACT5'
    Explanation
    The correct answer is 3'GGTAGCAACT5'. This is because the template DNA sequence is read in the 3' to 5' direction, and the given answer matches the sequence provided in the question.

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  • Current Version
  • Mar 19, 2023
    Quiz Edited by
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  • Dec 11, 2012
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    ISci2A18
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