Biochemistry Exam: MCQ Quiz!

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Biochemistry Exam: MCQ Quiz! - Quiz

How do you feel about biochemistry? Biochemistry is the study of biological experiences at the molecular level. The aim is to understand the fundamental chemical principles that govern complex biological systems. Biochemistry is utilized to learn about the biological activities which occur in cells and organisms. Biochemistry may be used to study the properties of biological molecules for several different purposes. Take this brilliant quiz and become an impressive student of biochemistry.


Questions and Answers
  • 1. 

    A weak acid can act as a buffer......

    • A.

      At pH values=pK±0.1

    • B.

      At pH values = pK ± 1

    • C.

      At pH values = pK ± 2

    • D.

      Only at body temperatures

    • E.

      None of the above

    Correct Answer
    B. At pH values = pK ± 1
    Explanation
    A weak acid can act as a buffer at pH values equal to the pK (the negative logarithm of the acid dissociation constant) plus or minus 1. This is because at these pH values, the weak acid is able to donate or accept protons, helping to maintain the pH of a solution relatively stable. Buffering capacity decreases as the pH moves further away from the pK value. Therefore, pH values equal to pK ± 1 provide the optimal range for a weak acid to act as a buffer.

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  • 2. 

    Which of the following is considered to be a fibrous protein?

    • A.

      Hemoglobin

    • B.

      Immunoglobulin

    • C.

      Keratin

    • D.

      Myoglobin

    • E.

      None of the above

    Correct Answer
    C. Keratin
    Explanation
    Keratin is considered to be a fibrous protein because it is a structural protein found in the hair, nails, and skin. Fibrous proteins are characterized by their long, elongated shape and their ability to provide strength and support to tissues. Hemoglobin and immunoglobulin are globular proteins involved in transportation and immune response, respectively. Myoglobin is a globular protein involved in oxygen storage. Therefore, the correct answer is Keratin.

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  • 3. 

    Disulfide bonds most often stabilize the native structure of:

    • A.

      Dimeric proteins

    • B.

      Extracellular proteins

    • C.

      Hydrophobic proteins

    • D.

      Intracellular proteins

    • E.

      All of the above

    Correct Answer
    B. Extracellular proteins
    Explanation
    Disulfide bonds are covalent bonds formed between two cysteine residues in a protein. These bonds play a crucial role in stabilizing the native structure of extracellular proteins. The extracellular environment, which is typically oxidizing, provides the necessary conditions for disulfide bond formation. These bonds contribute to the structural integrity and stability of extracellular proteins, allowing them to withstand harsh conditions and maintain their functional conformation. Therefore, the correct answer is extracellular proteins.

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  • 4. 

    In sickle cell anemia, the basis of the malfunction of the hemoglobin molecule is:

    • A.

      Faulty binding of the heme groups

    • B.

      Incorrect secondary structure

    • C.

      Reduced affinity for oxygen

    • D.

      Substitution of a single amino acid

    • E.

      None of the above

    Correct Answer
    D. Substitution of a single amino acid
    Explanation
    In sickle cell anemia, the basis of the malfunction of the hemoglobin molecule is the substitution of a single amino acid. This substitution involves the replacement of a glutamic acid with a valine in the beta chain of the hemoglobin protein. This change alters the shape of the red blood cells, causing them to become rigid and assuming a sickle shape. This abnormal shape leads to various complications, including the blockage of blood vessels and reduced oxygen-carrying capacity of the red blood cells.

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  • 5. 

    When BPG (2,3-Bisphosphoglycerate) binds to ______ site(s) on hemoglobin ________ its affinity for O2

    • A.

      Four/decreases

    • B.

      Four/increases

    • C.

      One/decreases

    • D.

      One/increases

    • E.

      None of the above

    Correct Answer
    C. One/decreases
    Explanation
    When BPG (2,3-Bisphosphoglycerate) binds to one site on hemoglobin, its affinity for O2 decreases. This is because BPG binds to a specific site on hemoglobin and stabilizes the T-state conformation, which has a lower affinity for oxygen. This allows for the release of oxygen in tissues where oxygen concentration is low, such as during exercise or at high altitudes. Therefore, the binding of BPG to one site on hemoglobin decreases its affinity for oxygen.

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  • 6. 

    The complementarity-determining regions (CDRs)

    • A.

      Are found at the antigen-binding sites

    • B.

      Reside in the VL and VH regions of the IgG molecule

    • C.

      Reside in the CL and CH regions of the IgG molecule

    • D.

      A and B are correct

    • E.

      A and C are correct

    Correct Answer
    D. A and B are correct
    Explanation
    The complementarity-determining regions (CDRs) are regions within the variable domains of immunoglobulin (IgG) molecules that are responsible for binding to antigens. These CDRs are found at the antigen-binding sites, which are located in both the variable light chain (VL) and variable heavy chain (VH) regions of the IgG molecule. Therefore, both statements A and B are correct.

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  • 7. 

    Antibodies of the IgG class

    • A.

      Consist of four subunits

    • B.

      Are glycoproteins

    • C.

      Have inter- and intra-chain disulfide crosslinks

    • D.

      All three choices are correct

    • E.

      None of the choices are correct

    Correct Answer
    D. All three choices are correct
    Explanation
    The correct answer is that all three choices are correct. Antibodies of the IgG class consist of four subunits, are glycoproteins, and have inter- and intra-chain disulfide crosslinks. This means that they are composed of four protein subunits, contain sugar molecules, and have strong chemical bonds that connect different parts of the antibody structure.

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  • 8. 

    The two exonuclease activities of DNA polymerase I

    • A.

      Are coupled during the formation of "gapped DNA"

    • B.

      Degrade DNA in a 3' --> 5' direction

    • C.

      Degrade DNA in a 5' --> 3' direction

    • D.

      Occur at two different active sites

    • E.

      None of the above

    Correct Answer
    D. Occur at two different active sites
    Explanation
    The two exonuclease activities of DNA polymerase I occur at two different active sites. This means that the enzyme has separate regions or sites where each exonuclease activity takes place. One site is responsible for degrading DNA in a 3' --> 5' direction, while the other site degrades DNA in a 5' --> 3' direction. This allows DNA polymerase I to efficiently remove nucleotides from both ends of the DNA strand during the formation of "gapped DNA".

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  • 9. 

    The RNA primers used to initiate replication in E. coli

    • A.

      Are joined together by DNA ligase

    • B.

      Are removed by helicase + ATP

    • C.

      Are removed by Pol I "nick translation

    • D.

      Result in Okazaki fragments on the leading strand

    • E.

      None of the above

    Correct Answer
    C. Are removed by Pol I "nick translation
    Explanation
    The correct answer is that the RNA primers used to initiate replication in E. coli are removed by Pol I "nick translation". Pol I is an enzyme that is responsible for removing the RNA primers during DNA replication in E. coli. It replaces the RNA primers with DNA nucleotides, allowing for the synthesis of a continuous DNA strand. This process is known as nick translation, where the RNA primer is nicked and replaced by DNA synthesis.

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  • 10. 

    On the ribosome, mRNA binds ____; the peptidyl transferase reaction occurs ____

    • A.

      Between the subunits; on the large subunit

    • B.

      Between the subunits; on the small subunit

    • C.

      To the large subunit; on the small subunit

    • D.

      To the small subunit; on the large subunit

    • E.

      None of the above

    Correct Answer
    D. To the small subunit; on the large subunit
  • 11. 

    The pKa of an ionizable side chain, such as imidazole in histidine, can be determined with NMR because

    • A.

      Protonated imidazole has twice the number of protons

    • B.

      The chemical shift of His differs from that of His+

    • C.

      The pKa is near neutrality

    • D.

      Unprotonated imidazole is invisible in the spectrum

    • E.

      None of the above

    Correct Answer
    B. The chemical shift of His differs from that of His+
    Explanation
    The correct answer is that the chemical shift of His differs from that of His+. In NMR spectroscopy, the chemical shift is a measure of the electronic environment of a nucleus. When a molecule is protonated, the electronic environment changes, resulting in a different chemical shift compared to when the molecule is not protonated. Therefore, by comparing the chemical shifts of His and His+, it is possible to determine the pKa of the ionizable side chain.

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  • 12. 

    During the synthesis of a 100 residue protein, the number of GTP molecules used during initiation, elongation, and termination are, respectively:

    • A.

      1, 100, 1

    • B.

      1, 198, 1

    • C.

      1, 200, 1

    • D.

      2, 198, 2

    • E.

      None of the above

    Correct Answer
    B. 1, 198, 1
    Explanation
    During protein synthesis, GTP molecules are used in various stages. During initiation, one GTP molecule is used to start the process. During elongation, 198 GTP molecules are used as each amino acid is added to the growing protein chain. Finally, during termination, one GTP molecule is used to stop the synthesis. Therefore, the correct answer is 1, 198, 1.

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  • 13. 

    Peripheral membrane proteins

    • A.

      Are not accommodated by the fluid mosaic model

    • B.

      Are only of peripheral interest to membrane biologists

    • C.

      Can be dissociated from membranes relatively easily, e.g., high salt treatment

    • D.

      Can undergo transverse membrane motion, i.e. "flip-flop"

    • E.

      None of the above

    Correct Answer
    C. Can be dissociated from membranes relatively easily, e.g., high salt treatment
    Explanation
    Peripheral membrane proteins can be dissociated from membranes relatively easily, for example, through high salt treatment. This means that they are not strongly bound to the lipid bilayer and can be removed from the membrane without disrupting the overall structure. This characteristic distinguishes them from integral membrane proteins, which are firmly embedded within the lipid bilayer and require more disruptive methods for extraction. This property of peripheral membrane proteins is in line with the fluid mosaic model, which describes the dynamic nature of the cell membrane and the ability of proteins to move within it.

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  • 14. 

    During glycolysis the following reaction requires NAD+:

    • A.

      Alcohol dehydrogenase

    • B.

      Glyceraldehyde-3-phosphate dehydrogenase

    • C.

      Lactate dehydrogenase

    • D.

      Pyruvate dehydrogenase

    • E.

      All of the above

    Correct Answer
    B. Glyceraldehyde-3-phosphate dehydrogenase
    Explanation
    Glyceraldehyde-3-phosphate dehydrogenase requires NAD+ during glycolysis. This enzyme catalyzes the oxidation of glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate, and in the process, NAD+ is reduced to NADH. NAD+ is a coenzyme that acts as an electron carrier, accepting electrons during the oxidation reactions in glycolysis. This reduction of NAD+ to NADH allows for the production of ATP through subsequent steps in glycolysis.

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  • 15. 

    Since the energy released by all the reactions of glycolysis is 185 kJ/mol glucose, and ____ mol ATP is/are produced per mol glucose, the overall efficiency of glycolysis is about ____

    • A.

      1/ 16%

    • B.

      1 /33%

    • C.

      2 /33%

    • D.

      2 /50%

    • E.

      None of the above

    Correct Answer
    C. 2 /33%
    Explanation
    The correct answer is 2 /33%. This is because glycolysis produces a net of 2 molecules of ATP per molecule of glucose. The overall efficiency of glycolysis can be calculated by dividing the energy released by the reactions (185 kJ/mol) by the energy stored in the ATP molecules produced (2 x 30.5 kJ/mol). This gives us an efficiency of approximately 33%.

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  • 16. 

    Glycogen digestion in the gut begins with a ________ reaction; intracellular glycogen breakdown begins with a ________ reaction.

    • A.

      Hydrolysis; hydrolysis

    • B.

      Hydrolysis; phosphorolysis

    • C.

      Hydrolysis; phosphorylation

    • D.

      Phosphorolysis; hydrolysis

    • E.

      None of the above

    Correct Answer
    B. Hydrolysis; phosphorolysis
    Explanation
    Glycogen digestion in the gut starts with a hydrolysis reaction, which involves the addition of water to break down glycogen into its individual glucose molecules. On the other hand, intracellular glycogen breakdown begins with a phosphorolysis reaction, which involves the cleavage of glycogen by the addition of a phosphate group. Therefore, the correct answer is hydrolysis; phosphorolysis.

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  • 17. 

    The correct order, from fastest-acting to slowest, for the time scale of these regulatory mechanisms is:

    • A.

      Allosteric - covalent - genetic

    • B.

      Allosteric - genetic - covalent

    • C.

      Covalent - genetic - allosteric

    • D.

      Genetic - allosteric - covalent

    • E.

      None of the above

    Correct Answer
    A. Allosteric - covalent - genetic
    Explanation
    Allosteric regulation refers to the binding of a molecule to a site on an enzyme, causing a change in its activity. This mechanism acts quickly because it involves a direct interaction with the enzyme. Covalent regulation involves the addition or removal of a chemical group to an enzyme, which can alter its activity. This process takes longer than allosteric regulation as it requires enzymatic reactions. Genetic regulation involves changes in gene expression, which ultimately leads to the production of enzymes with altered activity. This mechanism is the slowest as it requires transcription and translation processes. Therefore, the correct order, from fastest-acting to slowest, is allosteric - covalent - genetic.

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  • 18. 

    Mitochondrial ATP synthesis requires:

    • A.

      A [H+] gradient

    • B.

      A membrane potential

    • C.

      An intact inner mitochondrial membrane

    • D.

      All of the above

    • E.

      None of the above

    Correct Answer
    D. All of the above
    Explanation
    Mitochondrial ATP synthesis requires all of the above because a [H+] gradient, a membrane potential, and an intact inner mitochondrial membrane are all essential for the process. The [H+] gradient is necessary for the synthesis of ATP through the action of ATP synthase, which uses the energy from the gradient to convert ADP to ATP. The membrane potential, created by the movement of ions across the inner mitochondrial membrane, is also required for ATP synthesis. Additionally, the inner mitochondrial membrane must be intact to maintain the necessary conditions for ATP synthesis to occur.

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  • 19. 

    Which of the following enzymes does not catalyze a physiologically irreversible step in glycolysis?

    • A.

      Hexokinase

    • B.

      Phosphofructokinase

    • C.

      Phosphoglycerate kinase

    • D.

      Pyruvate kinase

    • E.

      None of the above

    Correct Answer
    C. Phosphoglycerate kinase
    Explanation
    Phosphoglycerate kinase catalyzes the conversion of 1,3-bisphosphoglycerate to 3-phosphoglycerate, which is a reversible step in glycolysis. This means that the reaction can proceed in both directions depending on the concentration of substrates and products. In contrast, hexokinase catalyzes the irreversible conversion of glucose to glucose-6-phosphate, phosphofructokinase catalyzes the irreversible conversion of fructose-6-phosphate to fructose-1,6-bisphosphate, and pyruvate kinase catalyzes the irreversible conversion of phosphoenolpyruvate to pyruvate. Therefore, the correct answer is Phosphoglycerate kinase.

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  • 20. 

    Following the addition of a mitochondrial respiratory inhibitor, components of the electron transport chain.

    • A.

      All components will be oxidized

    • B.

      Downstream of the blockage will be reduced

    • C.

      Upstream of the blockage will be oxidized

    • D.

      Upstream of the blockage will be reduced

    • E.

      None of the above

    Correct Answer
    D. Upstream of the blockage will be reduced
    Explanation
    When a mitochondrial respiratory inhibitor is added, it blocks the flow of electrons in the electron transport chain. This means that the components upstream of the blockage, which are responsible for donating electrons, will not be able to pass their electrons down the chain. As a result, these upstream components will be reduced, meaning they will gain electrons. This reduction occurs because these components are unable to pass their electrons to downstream components. Therefore, the correct answer is that upstream of the blockage will be reduced.

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  • 21. 

    For the following reaction: [FAD] + 2 Cyt c (Fe2+) + 2H+ -> [FADH2] + 2 Cyt c (Fe3+) the electron donor is _______; the reduced product is ________

    • A.

      2 Cyt c (Fe2+); [FADH2]

    • B.

      [FAD]; [FADH2]

    • C.

      2 Cyt c (Fe2+); 2 Cyt c (Fe3+)

    • D.

      [FAD]; 2 Cyt c (Fe3+)

    • E.

      None of the above

    Correct Answer
    A. 2 Cyt c (Fe2+); [FADH2]
    Explanation
    The electron donor in the reaction is 2 Cyt c (Fe2+) because it donates two electrons to the reaction. The reduced product is [FADH2] because it gains the two electrons from the electron donor, resulting in its reduction.

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  • 22. 

    Acetyl groups (e.g. for lipid biosynthesis) are transported out of the mitochondria indirectly using:

    • A.

      Acetyl carnitine

    • B.

      Acetyl phosphate

    • C.

      Citrate

    • D.

      Malate

    • E.

      All of the above

    Correct Answer
    C. Citrate
    Explanation
    Acetyl groups are transported out of the mitochondria indirectly using citrate. Citrate is formed in the mitochondria through the condensation of acetyl-CoA with oxaloacetate. Citrate can then exit the mitochondria and be converted back into acetyl-CoA in the cytoplasm, where it can be used for lipid biosynthesis. Acetyl carnitine and acetyl phosphate are not involved in the transport of acetyl groups out of the mitochondria.

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