Biochemistry Exam: MCQ Quiz!
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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.
- 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 ± 1Explanation
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.Rate this question:
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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. KeratinExplanation
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.Rate this question:
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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 proteinsExplanation
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.Rate this question:
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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 acidExplanation
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.Rate this question:
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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/decreasesExplanation
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.Rate this question:
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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 correctExplanation
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.Rate this question:
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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 correctExplanation
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.Rate this question:
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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 sitesExplanation
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".Rate this question:
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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 translationExplanation
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.Rate this question:
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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.Rate this question:
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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, 1Explanation
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.Rate this question:
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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 treatmentExplanation
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.Rate this question:
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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 dehydrogenaseExplanation
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.Rate this question:
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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%.Rate this question:
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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; phosphorolysisExplanation
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.Rate this question:
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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 - geneticExplanation
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.Rate this question:
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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 aboveExplanation
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.Rate this question:
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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 kinaseExplanation
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.Rate this question:
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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 reducedExplanation
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.Rate this question:
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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.Rate this question:
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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. CitrateExplanation
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.Rate this question:
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Mar 22, 2023Quiz Edited by
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Apr 26, 2010Quiz Created by
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