Midterm 1 Chee 370

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Midterm Quizzes & Trivia

Questions and Answers
  • 1. 

    At the end of glycolysis, the original carbons of the glucose molecule form:

    • A.

      Six molecules of carbon dioxide

    • B.

      Two molecules of pyruvate

    • C.

      Two molecules of citric acid

    • D.

      Two molecules of fructose

    • E.

      Two molecules of NADH

    Correct Answer
    B. Two molecules of pyruvate
    Explanation
    At the end of glycolysis, the original carbons of the glucose molecule form two molecules of pyruvate. Glycolysis is the process in which glucose is broken down into two molecules of pyruvate. This occurs in the cytoplasm of the cell and is the first step in cellular respiration. Each glucose molecule is converted into two molecules of pyruvate, which can then enter the next stage of respiration, the citric acid cycle, to produce more energy. Therefore, the correct answer is two molecules of pyruvate.

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  • 2. 

    The fat substitute Olestra contains a sucrose backbone with six to eight fatty acids attached. How is this different from a naturally occurring fat?

    • A.

      It isn't; Olestra and natural fats have the same structure, just different tastes

    • B.

      Naturally occurring fats contain a glycerol and three fatty acids

    • C.

      Naturally occurring fats contain a glycerol, two fatty acids, and a phosphate group

    • D.

      Naturally occurring fats contain a sucrose backbone and three fatty acid chains

    Correct Answer
    B. Naturally occurring fats contain a glycerol and three fatty acids
  • 3. 

    The small, circular loops of DNA in prokaryotic cells that are separate from the main chromosome and may harbor genes are called

    • A.

      Cristate

    • B.

      Ribosomes

    • C.

      Plastids

    • D.

      Pili

    • E.

      Plasmids

    Correct Answer
    E. Plasmids
    Explanation
    Plasmids are small, circular loops of DNA that are separate from the main chromosome in prokaryotic cells. They are capable of replicating independently and can be transferred between cells. Plasmids often carry genes that provide advantages to the cell, such as antibiotic resistance or the ability to produce certain enzymes. Therefore, plasmids play a crucial role in horizontal gene transfer and contribute to the genetic diversity of prokaryotes.

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  • 4. 

    Termination of RNA synthesis is ultimately determined by

    • A.

      Exhaustion of RNA polymerase activity

    • B.

      Special protein factors

    • C.

      Specific nucleotide sequences on the template strand

    • D.

      CG-rich sequences followed by AT-rich sequences

    Correct Answer
    C. Specific nucleotide sequences on the template strand
    Explanation
    Termination of RNA synthesis is ultimately determined by specific nucleotide sequences on the template strand. These sequences act as signals for the RNA polymerase to stop transcription. When the RNA polymerase reaches these termination sequences, it releases the newly synthesized RNA molecule and detaches from the DNA template. This process ensures that RNA synthesis ends at the appropriate location and prevents the production of unnecessary or faulty RNA molecules. Exhaustion of RNA polymerase activity, special protein factors, and CG-rich sequences followed by AT-rich sequences may play roles in other aspects of transcription regulation but are not directly involved in termination.

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  • 5. 

    Which of the following is NOT a common energy carrier in the cell?

    • A.

      Electron carriers

    • B.

      ADP

    • C.

      Phosphate

    • D.

      ATP

    Correct Answer
    B. ADP
    Explanation
    ADP, or adenosine diphosphate, is a common energy carrier in the cell. It is converted to ATP (adenosine triphosphate) during cellular respiration, where energy is stored in the phosphate bonds. ATP is then used as a source of energy for various cellular processes. Therefore, the correct answer is ADP.

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  • 6. 

    The "backbone" of a nucleic acid molecule is made of

    • A.

      A sugar and phosphate groups

    • B.

      Nitrogenous bases

    • C.

      Amino acids

    • D.

      NAD+ and FAD

    • E.

      ATP molecules

    Correct Answer
    A. A sugar and phosphate groups
    Explanation
    The backbone of a nucleic acid molecule is made of a sugar and phosphate groups. This is because nucleic acids, such as DNA and RNA, are composed of repeating units called nucleotides. Each nucleotide consists of a sugar molecule (deoxyribose in DNA and ribose in RNA) and a phosphate group. These sugar-phosphate units form a strong and stable backbone that provides structural support to the nucleic acid molecule. The nitrogenous bases, such as adenine, guanine, cytosine, thymine (in DNA), and uracil (in RNA), are attached to the sugar molecules and project inward from the backbone, forming the genetic code.

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  • 7. 

    Which cluster of terms accurately reflects the nature of DNA replication in prokaryotes? 

    • A.

      Random point of initiation, bidirectional, semiconservative

    • B.

      Fixed point of initiation, bidirectional, conservative

    • C.

      Random point of initiation, unidirectional, semiconservative

    • D.

      Fixed point of initiation, unidirectional, conservative

    • E.

      Fixed point of initiation, bidirectional, semiconservative

    Correct Answer
    E. Fixed point of initiation, bidirectional, semiconservative
    Explanation
    In prokaryotes, DNA replication starts at a fixed point of initiation, where the replication machinery binds to the DNA molecule. The replication process then proceeds bidirectionally, meaning that replication occurs in both directions from the initiation point. Additionally, DNA replication in prokaryotes is semiconservative, meaning that each newly synthesized DNA molecule consists of one original strand and one newly synthesized complementary strand. Therefore, the cluster of terms "fixed point of initiation, bidirectional, semiconservative" accurately reflects the nature of DNA replication in prokaryotes.

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  • 8. 

    Complex, three-dimensional, teritiary structures of proteins are characterized by

    • A.

      An absence of hydrophilic amino acids

    • B.

      A lack of cysteines in the amino acid sequence

    • C.

      A helical shape

    • D.

      Disulfide bonds

    Correct Answer
    D. Disulfide bonds
    Explanation
    The correct answer is disulfide bonds. Complex, three-dimensional, tertiary structures of proteins are characterized by the presence of disulfide bonds. Disulfide bonds are covalent bonds formed between two cysteine amino acids, where the sulfur atoms are linked together. These bonds play a crucial role in stabilizing the protein structure and are responsible for maintaining the protein's shape and stability. They are particularly important in extracellular proteins, where they provide resistance to degradation and proteolysis. Therefore, the presence of disulfide bonds is a key characteristic of complex protein structures.

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  • 9. 

    During which step of aerobic respiration is oxygen used?

    • A.

      Conversion of pyruvate to acetyl CoA

    • B.

      Glycolysis

    • C.

      Fermentation

    • D.

      Krebs cycle

    • E.

      Electron transport system

    Correct Answer
    E. Electron transport system
    Explanation
    The electron transport system is the step during aerobic respiration where oxygen is used. This process occurs in the inner mitochondrial membrane and involves the transfer of electrons from electron carriers to oxygen. Oxygen acts as the final electron acceptor, allowing the production of ATP through oxidative phosphorylation.

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  • 10. 

    If the DNA of a certain organism has guanine as 30% of its bases, then what percentage of its bases are adenine?

    • A.

      10%

    • B.

      0%

    • C.

      20%

    • D.

      40%

    • E.

      30%

    Correct Answer
    C. 20%
    Explanation
    If the DNA of a certain organism has guanine as 30% of its bases, then the percentage of adenine can be determined using Chargaff's rule. According to this rule, in DNA, the percentage of adenine is equal to the percentage of thymine, and the percentage of guanine is equal to the percentage of cytosine. Since guanine is 30%, cytosine must also be 30%. Therefore, the percentage of adenine is equal to the percentage of thymine, which is 20%.

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  • 11. 

    A peptide bond forms between which of these groups?

    • A.

      Phosphate and hydroxyl

    • B.

      Carboxyl and aldehyde

    • C.

      Hydroxyl and carboxyl

    • D.

      Carboxyl and amino

    • E.

      Amino and aldehyde

    Correct Answer
    D. Carboxyl and amino
    Explanation
    A peptide bond forms between the carboxyl group of one amino acid and the amino group of another amino acid. This bond is formed through a dehydration synthesis reaction, where a water molecule is removed. This bond is crucial in the formation of proteins, as it links the amino acids together to create a polypeptide chain.

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  • 12. 

    Which of the following are among the major components of prokaryotic ribosomes?

    • A.

      12S rRNA, 5.8S rRNA, and proteins

    • B.

      15S rRNA, 5.8S rRNA, and 28S rRNA

    • C.

      18S rRNA, 5.8S rRNA, and proteins

    • D.

      16S rRNA, 5S rRNA, and 23S rRNA

    • E.

      Lipids and carbohydrates

    Correct Answer
    D. 16S rRNA, 5S rRNA, and 23S rRNA
    Explanation
    The major components of prokaryotic ribosomes are 16S rRNA, 5S rRNA, and 23S rRNA. These ribosomal RNAs (rRNAs) form the structural backbone of the ribosome and are essential for its function in protein synthesis. The 16S rRNA is involved in the decoding of mRNA, while the 23S rRNA catalyzes the formation of peptide bonds. The 5S rRNA helps stabilize the overall structure of the ribosome. Proteins are also present in ribosomes, but they are not the major components. Lipids and carbohydrates are not part of the ribosomal structure.

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  • 13. 

    A denatured protein differs from a normal protein because it

    • A.

      Is composed of nucleotides

    • B.

      Contains many disulfide bonds

    • C.

      Does not contain amino acids

    • D.

      Has lost its usual secondary and tertiary structures

    Correct Answer
    D. Has lost its usual secondary and tertiary structures
    Explanation
    A denatured protein differs from a normal protein because it has lost its usual secondary and tertiary structures. This means that the denatured protein has undergone a structural change that disrupts its folded shape. Secondary structure refers to the local folding patterns, such as alpha helices and beta sheets, while tertiary structure refers to the overall 3D arrangement of the protein. The loss of these structures can be caused by various factors such as heat, pH changes, or exposure to certain chemicals.

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  • 14. 

    Select three posttranscriptional modifications often seen in the maturation of mRNA in eukaryotes

    • A.

      3'-capping, 5'-poly(A) tail addition, splicing

    • B.

      Heteroduplex formation, base modifiction, capping

    • C.

      5'-capping, 3'-poly(A)tail addition, splicing

    • D.

      5'-poly(A) tail addition, insertion of introns, capping

    • E.

      Removal of exons, insertion of introns, capping

    Correct Answer
    C. 5'-capping, 3'-poly(A)tail addition, splicing
    Explanation
    The correct answer is 5'-capping, 3'-poly(A) tail addition, splicing. These are three common posttranscriptional modifications that occur during the maturation of mRNA in eukaryotes. 5'-capping involves the addition of a modified guanine nucleotide to the 5' end of the mRNA, which protects it from degradation and helps in the initiation of translation. 3'-poly(A) tail addition involves the addition of a string of adenine nucleotides to the 3' end of the mRNA, which also aids in stability and translation. Splicing is the process of removing introns, non-coding regions, from the mRNA and joining together the remaining exons, which are the coding regions. This ensures that the final mRNA molecule is ready for translation.

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  • 15. 

    Which statement is generally true regarding protein synthesis?

    • A.

      The 23S rRNA plays a role in elongation; the 16S rRNA plays a role in translocation

    • B.

      The 23S rRNA plays a role in termination; the 16S rRNA plays a role in elongation

    • C.

      The 23S rRNA plays a role in translocation; the 16S rRNA plays a role in elongation

    • D.

      The 23S rRNA plays a role in translocation; the 16S rRNA plays a role in initiation

    Correct Answer
    D. The 23S rRNA plays a role in translocation; the 16S rRNA plays a role in initiation
    Explanation
    The 23S rRNA plays a role in translocation, which refers to the movement of the ribosome along the mRNA during protein synthesis. On the other hand, the 16S rRNA plays a role in initiation, which is the process of starting protein synthesis by binding to the mRNA and the start codon.

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  • 16. 

    Fibers of the cytoskeleton are composed of primarily

    • A.

      Lipids

    • B.

      ER

    • C.

      Nucleic acids

    • D.

      Proteins

    • E.

      Polysaccharides

    Correct Answer
    D. Proteins
    Explanation
    The fibers of the cytoskeleton are primarily composed of proteins. The cytoskeleton is a network of protein filaments that provide structural support and shape to the cell. These proteins include microtubules, microfilaments, and intermediate filaments, which are involved in various cellular processes such as cell division, cell movement, and maintaining cell shape. Lipids, ER, nucleic acids, and polysaccharides are not the main components of the cytoskeleton.

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  • 17. 

    What is the significance of the conversion of pyruvate to lactate during fermentation?

    • A.

      The citric acid cycle is initiated

    • B.

      The oxidation of pyruvate becomes possible

    • C.

      ATP is produced

    • D.

      Pyruvate becomes available to enter mitochondrial matrix reactions

    • E.

      NAD+ is regenerated for use in glycolysis

    Correct Answer
    E. NAD+ is regenerated for use in glycolysis
    Explanation
    During fermentation, pyruvate is converted to lactate in order to regenerate NAD+. NAD+ is an important coenzyme in glycolysis, which is the initial step of cellular respiration. Glycolysis produces ATP and NADH, but NADH cannot be used directly in glycolysis. By converting pyruvate to lactate, NAD+ is regenerated, allowing it to be used again in glycolysis. This is crucial for sustaining the production of ATP in cells when oxygen is limited, as in anaerobic conditions.

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  • 18. 

    What primarily determines the shape of animal cells, which lack cell walls?

    • A.

      Cytoplasm

    • B.

      Ribosomes

    • C.

      Endoplasmic reticulum

    • D.

      Nucleus

    • E.

      Cytoskeleton

    Correct Answer
    E. Cytoskeleton
    Explanation
    The cytoskeleton primarily determines the shape of animal cells. Unlike plant cells, animal cells lack cell walls, so they rely on the cytoskeleton for structural support and shape. The cytoskeleton is a network of protein filaments that extends throughout the cell and provides structural integrity. It helps maintain the cell's shape, enables cellular movement, and supports various cellular processes. Therefore, the cytoskeleton is the main determinant of the shape of animal cells.

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  • 19. 

    Covalent bonds joining two nucleotides within a single strand of DNA form between

    • A.

      Deoxyribose and a phosphate group

    • B.

      A phosphate group and adenine

    • C.

      Ribose and a base

    • D.

      The phosphate groups of both

    • E.

      Adenine and thymine

    Correct Answer
    A. Deoxyribose and a phosphate group
    Explanation
    Covalent bonds joining two nucleotides within a single strand of DNA form between deoxyribose and a phosphate group. This is because DNA is composed of nucleotides, which consist of a deoxyribose sugar molecule, a phosphate group, and a nitrogenous base. The covalent bond forms between the 3' carbon of one deoxyribose and the 5' carbon of the adjacent deoxyribose, with the phosphate group connecting the two sugars. This bond is known as a phosphodiester bond and it forms the backbone of the DNA strand.

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  • 20. 

    DNA polymerase III adds nucleotides

    • A.

      To internal sites in the DNA template

    • B.

      In the place of the primer RNA after it is removed

    • C.

      To both ends of the RNA primer

    • D.

      To the 3' end of the RNA primer

    • E.

      To the 5' end of the RNA primer

    Correct Answer
    D. To the 3' end of the RNA primer
    Explanation
    DNA polymerase III adds nucleotides to the 3' end of the RNA primer. This is because DNA polymerase III can only add nucleotides in the 5' to 3' direction, meaning it can only add nucleotides to the end of the RNA primer that has a free 3' hydroxyl group. The RNA primer is initially synthesized in the 5' to 3' direction, so when it is removed, DNA polymerase III can extend the DNA strand by adding nucleotides to the 3' end of the RNA primer.

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  • 21. 

    The term peptidyltransferase relates to

    • A.

      Discontinuous strand replication

    • B.

      Elongation factor binding to the large ribosomal unit

    • C.

      5' capping of mRNA

    • D.

      Peptide bond formation during protein synthesis

    • E.

      Base additions during mRNA synthesis

    Correct Answer
    D. Peptide bond formation during protein synthesis
    Explanation
    Peptidyltransferase is an enzyme found in the ribosome that plays a crucial role in protein synthesis. It catalyzes the formation of peptide bonds between amino acids, which is a key step in the process of translating mRNA into a polypeptide chain. Therefore, the correct answer is "peptide bond formation during protein synthesis".

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  • 22. 

    A short segment of an mRNA molecule is shown below. The polypeptide it codes for is alson shown: 5'-AUGGUGCUGAAG : methionine-valine-leucine-lysine Assume that a mutation in the DNA occurs so that the fourth base (counting from the 5' end) of the messenger RNA now reads A rather than G. What sequence of amino acids will the mRNA now code for?

    • A.

      Methionine-leucine-leucine-lysine

    • B.

      Methionine-methionine-leucine-lysine

    • C.

      Methionine-valine-methionine-lysine

    • D.

      Methionine-lysin-leucine-lysine

    • E.

      Methionine-valine-leucine-lysine

    Correct Answer
    B. Methionine-methionine-leucine-lysine
    Explanation
    The given mRNA sequence is 5'-AUGGUGCUGAAG. The mutation occurs at the fourth base, which changes from G to A. This mutation leads to a change in the codon that the mRNA codes for. The original codon was GUG, which codes for the amino acid valine. However, due to the mutation, the codon becomes AUG, which codes for the amino acid methionine. Therefore, the correct answer is methionine-methionine-leucine-lysine.

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  • 23. 

    Triglycerides are

    • A.

      Always composed of carbon rings

    • B.

      Non-polar and hydrophobic

    • C.

      Made from glycerol and nucleic acids

    • D.

      Lacking carboxyl groups (-COOH)

    • E.

      Polymers of amino acids

    Correct Answer
    B. Non-polar and hydrophobic
    Explanation
    Triglycerides are non-polar and hydrophobic molecules because they are composed of three fatty acids attached to a glycerol backbone. The fatty acids consist of long hydrocarbon chains, which are non-polar and repel water. This property makes triglycerides insoluble in water and causes them to form droplets or separate from aqueous solutions. Additionally, the absence of carboxyl groups (-COOH) in triglycerides further contributes to their non-polar nature.

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  • 24. 

    An intron is a section of

    • A.

      Transfer RNA that binds to the anticodon

    • B.

      RNA that is removed during RNA processing

    • C.

      Carbohydrate that serves as a signal for RNA transport

    • D.

      DNA that is removed during DNA processing

    • E.

      Protein that is clipped out posttranslationally

    Correct Answer
    B. RNA that is removed during RNA processing
    Explanation
    During RNA processing, introns are sections of RNA that are removed. Introns are non-coding regions of RNA that do not contain instructions for protein synthesis. They are transcribed from DNA but are not involved in the final protein product. Introns are removed through a process called splicing, where they are cut out of the RNA molecule and the remaining exons are joined together to form the mature mRNA molecule. This allows for the production of a functional protein from the coding regions of the RNA molecule.

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  • 25. 

    GTP provides energy for

    • A.

      Transcription

    • B.

      Translation

    • C.

      Protein folding

    • D.

      DNA replication

    Correct Answer
    B. Translation
    Explanation
    Translation is the process in which the genetic information stored in the mRNA molecule is decoded by ribosomes to synthesize proteins. During translation, the ribosomes read the mRNA sequence and assemble amino acids in the correct order to form a polypeptide chain, which eventually folds into a functional protein. Therefore, GTP (guanosine triphosphate) provides energy for translation, as it is hydrolyzed by the ribosomes to provide the energy needed for the formation of peptide bonds between the amino acids.

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  • 26. 

    What is the final electron acceptor in cellular respiration?

    • A.

      NADH

    • B.

      FADH2

    • C.

      ATP

    • D.

      Carbon dioxide

    • E.

      Oxygen

    Correct Answer
    E. Oxygen
    Explanation
    In cellular respiration, oxygen serves as the final electron acceptor. During the electron transport chain, electrons are passed through a series of protein complexes, ultimately ending with the transfer of electrons to oxygen. This process generates a proton gradient, which is used to produce ATP through chemiosmosis. Without oxygen as the final electron acceptor, the electron transport chain would not be able to function properly, leading to a decrease in ATP production and a disruption in cellular respiration.

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  • 27. 

    Where does the synthesis of pyruvate occur during glycolysis?

    • A.

      Ribosomes

    • B.

      Fluid portion of the cytoplasm

    • C.

      Intermembrane compartment

    • D.

      Inner membrane

    • E.

      Mitochondrial matrix

    Correct Answer
    B. Fluid portion of the cytoplasm
    Explanation
    During glycolysis, the synthesis of pyruvate occurs in the fluid portion of the cytoplasm. Glycolysis is the process by which glucose is broken down into pyruvate, and it takes place in the cytoplasm of the cell. The fluid portion of the cytoplasm, also known as the cytosol, is where the enzymes and molecules involved in glycolysis are located. It is within this fluid that the conversion of glucose to pyruvate occurs, producing energy in the form of ATP. Therefore, the correct answer is the fluid portion of the cytoplasm.

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  • 28. 

    If glucose is metabolized under completely anaerobic conditions, then pyruvate

    • A.

      Is converted to NADH

    • B.

      Is converted back to fructose until the concentration of oxygen increases

    • C.

      Is converted by fermentation to CO2 and ethanol or to lactate

    • D.

      Leaves the fluid portion of the cytoplasm and enters the mitochondrial matrix

    • E.

      Immediately enters the Krebs cycle

    Correct Answer
    C. Is converted by fermentation to CO2 and ethanol or to lactate
    Explanation
    When glucose is metabolized under completely anaerobic conditions, it undergoes fermentation. In fermentation, pyruvate is converted to either CO2 and ethanol or to lactate. This process allows for the regeneration of NAD+ from NADH, which is necessary for glycolysis to continue in the absence of oxygen. Therefore, the correct answer is that pyruvate is converted by fermentation to CO2 and ethanol or to lactate.

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  • 29. 

    What type of lipid is most important in biological membranes?

    • A.

      Oils

    • B.

      Trans fats

    • C.

      Phospholipids

    • D.

      Fats

    • E.

      Steroids

    Correct Answer
    C. Phospholipids
    Explanation
    Phospholipids are the most important type of lipid in biological membranes. They have a unique structure with a hydrophilic (water-loving) head and hydrophobic (water-hating) tails. This structure allows phospholipids to form a bilayer in cell membranes, with the hydrophilic heads facing the aqueous environment inside and outside the cell, and the hydrophobic tails facing each other in the middle. This arrangement provides a barrier that controls the movement of molecules in and out of the cell, making phospholipids crucial for maintaining cell integrity and regulating cellular processes.

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  • 30. 

    In cells, _____ of the chemical energy in a metabolized glucose molecule is used for ATP production and the rest is released as heat.

    • A.

      40%

    • B.

      25%

    • C.

      Less than 1%

    • D.

      More than 90%

    Correct Answer
    A. 40%
    Explanation
    In cells, approximately 40% of the chemical energy in a metabolized glucose molecule is used for ATP production and the rest is released as heat. This means that only a fraction of the energy obtained from glucose is converted into a usable form of energy (ATP), while the majority is dissipated as heat. This phenomenon is known as the efficiency of cellular respiration, where a significant amount of energy is lost in the process.

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  • 31. 

    How many CO2 molecules are generated from each pyruvate that enters the mitochondria?

    • A.

      Four

    • B.

      Three

    • C.

      Five

    • D.

      Two

    • E.

      One

    Correct Answer
    B. Three
    Explanation
    During the process of cellular respiration, each pyruvate molecule is converted into one molecule of acetyl-CoA, which enters the citric acid cycle. In the citric acid cycle, acetyl-CoA is further broken down, resulting in the release of three molecules of carbon dioxide (CO2) per pyruvate. Therefore, from each pyruvate that enters the mitochondria, three CO2 molecules are generated.

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  • 32. 

    Which of the following is NOT involved in the DNA replication process?

    • A.

      DNA replicase

    • B.

      DNA helicase

    • C.

      DNA ligase

    • D.

      DNA polymerase

    Correct Answer
    A. DNA replicase
    Explanation
    DNA replicase is not involved in the DNA replication process. DNA replicase is a fictional term that does not exist in biology. The actual enzyme involved in DNA replication is DNA polymerase, which is responsible for synthesizing new DNA strands by adding nucleotides to the growing chain. DNA helicase unwinds the double helix structure of DNA, DNA ligase joins the Okazaki fragments on the lagging strand, and DNA polymerase carries out the actual replication process.

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  • 33. 

    ATP is an energy carrier. Where is the energy actually located?

    • A.

      Inside the phosphate

    • B.

      Attached to the nucleotide

    • C.

      Attached to the phosphate group

    • D.

      In the bonds between phosphate groups

    • E.

      Between the sugar and the phosphate

    Correct Answer
    D. In the bonds between phosphate groups
    Explanation
    The energy in ATP is actually located in the bonds between phosphate groups. When one of these phosphate bonds is broken, a high-energy phosphate group is released, which can be used to fuel cellular processes. The energy is stored in these bonds due to the negative charges of the phosphate groups, which repel each other and require energy to keep them together. When the bond is broken, the released energy can be used by cells to perform work.

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  • 34. 

    DNA polymerase I is thought to add nucleotides

    • A.

      On single-stranded templates without need for an RNA primer

    • B.

      To the 5' end of the primer

    • C.

      In a 5' to 5' direction

    • D.

      In the place of the primer RNA after it is removed

    • E.

      To the 3' end of the primer

    Correct Answer
    D. In the place of the primer RNA after it is removed
    Explanation
    DNA polymerase I is able to add nucleotides in the place of the primer RNA after it is removed. This is because DNA polymerase I has the ability to remove RNA primers and replace them with DNA nucleotides during DNA replication. It has both exonuclease and polymerase activities, allowing it to remove the RNA primer in a 5' to 3' direction and simultaneously add DNA nucleotides in its place. This process is known as primer removal and synthesis, and it is an important step in DNA replication.

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  • 35. 

    The four polypeptides that are joined together to make hemoglobin represent which level of protein organization?

    • A.

      Tertiary structure

    • B.

      Primary structure

    • C.

      Secondary structure

    • D.

      Quaternary structure

    Correct Answer
    D. Quaternary structure
    Explanation
    The four polypeptides that make up hemoglobin represent the quaternary structure of protein organization. The quaternary structure refers to the arrangement of multiple polypeptide chains or subunits to form a functional protein. In the case of hemoglobin, it consists of two alpha and two beta subunits. The interaction and assembly of these subunits result in the functional hemoglobin molecule, which is essential for oxygen transport in the blood.

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  • Mar 22, 2023
    Quiz Edited by
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  • Dec 03, 2013
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