MCAT Biochemistry Flashcards
Why are triacylglycerols used in the human body for energy storage?
A) They are highly hydrated, and therefore can store lots of energy.
B) They always have short fatty acid chains, for easy access by metabolic enzymes.
C) The carbon atoms of the fatty acid chains are highly reduced, and therefore yield more energy upon oxidation.
D) Polysaccharides, which would actually be a better energy storage form, would dissolve in the body.
C
Triacylglycerols are highly hydrophobic and therefore not highly hydrated (which would add extra weight from the water of hydration, taking away from the energy density of these molecules). The fatty acid chains produce twice as much energy as polysaccharides during oxidation because they are highly reduced. The fatty acid chains vary in length and saturation.
After a brief period of intense exercise, the activity of muscle pyruvate dehydrogenase is greatly increased. This increased activity is most likely due to:
A) Decreased ADP
B) Increased acetyl-CoA
C) Increased NADH/NAD+ ratio
D) Increased pyruvate concentration
D
In most biochemical pathways, only a few enzymatic reactions are under regulatory control. These often occur either at the beginning of pathways or at pathway branch points. The pyruvate dehydrogenase (PDH) complex controls the link between glycolysis and the citric acid cycle, and decarboxylates pyruvate (the end product of glycolysis) with production of NADH and acetyl-CoA (the substrate for the citric acid cycle). After intense exercise, one would expect PDH to be highly active to generate ATP. ADP levels should be high because ATP was just burned by the muscle. Acetyl-CoA is an inhibitor of PDH, causing a shift of pyruvate into the gluconeogenesis pathway. A high NADH/NAD+ ratio would imply that the cell is already energetically satisfied and not in need of energy, which would not be expected in intensely exercising muscle.
Enzyme X creates peptide bonds between amino acids leucine and valine, as well as peptide bonds between leucine molecules and between valine molecules. If enzyme X catalyzed the production of leu2val2, how many different linear tetrapeptides would be possible?
A) 4
B) 6
C) 8
D) 16
B
Keeping in mind that a tetrapeptide (i.e. an oligopeptide or even if it were a protein) has an amino end and a carboxyl end, it means that LLVV would be different to VVLL where L = leucine and V = valine. Thus there are 6 possibilities:
LLVV VLLV VVLL LVLV LVVL VLVL Basic stats is only required for the 2015 MCAT. If that is what you are considering then you should describe this problem as 4 choose 2 or: 4!/(2!)(2!) = 4*3*2*1/2*2 = 24/4 = 6.
Enzymes are one of the key proteins in cells and facilitate chemical reactions. How does a typical enzyme catalyze a chemical reaction?
A) It eliminates the activation energy of a chemical reaction
B) It reduces the activation energy of a chemical reaction
C) It provides additional energy to the chemical reaction
D) It contains a covalently bound substrate, which is needed for the chemical reaction
B
Enzymes reduce the activation energy of chemical reactions by binding substrate(s) into energetically favorable orientations. They greatly reduce the activation energy, but do not eliminate it completely. Enzymes are completely reusable—they do not participate in the reaction as a substrate. Even when cofactors are needed for the successful action of the enzyme and though they may be tightly bound, they are not covalently bound to the enzyme. Enzymes do not add additional energy to chemical reactions as heat would, for example.
What is the most correct definition of competitive inhibition?
A) Another enzyme completes the same reaction and uses the available reactant
B) An inhibitor binds directly to the active site and prevents the reactant from binding
C) A metal cofactor prevents the binding of the reactant by binding at the active site
D) An inhibitor binds to the enzyme at a binding site, and prevents the enzyme from catalyzing the reaction
B
The correct answer is B. Metal cofactors and ATP are often used to help boost enzyme activity, and are not competitive inhibitors. Non-competitive inhibition (D) occurs when an inhibitor is able to prevent the enzyme from binding with the reactant by binding to the enzyme at a site away from the active site, and change the enzyme’s conformation so it cannot bind to the reactant. Competitive inhibition occurs when the inhibitor competes directly with the reactant at the active site, and this substrate takes the place of the reactant and prevents the reaction from occurring.
At what enzymatic step in the citric acid cycle (Krebs cycle) does substrate level phosphorylation occur?
A) Pyruvate to acetyl coenzyme A
B) Succinyl coenzyme A to succinate
C) Malate to oxaloacetate
D) Pyruvate to oxaloacetate
B
There is only one step in the citric acid cycle that causes substrate level phosphorylation: the conversion of succinyl coenzyme A to succinate. GDP and inorganic phosphate combine to form GTP, which is roughly equivalent to the formation of one molecule of ATP. All others do not produce ATP. Oxaloacetate may be formed from pyruvate though it requires the dephosphorylation of one molecule of ATP. Moreover, this reaction is not considered one of the classic reactions in the citric acid cycle. In the conversion of malate to oxaloacetate, NAD+ loses a proton to form NADH.
The conversion of threonine to isoleucine is a five-step enzymatic pathway. The end product, isoleucine, fits into the allosteric site of the enzyme at step 1, preventing its normal function. This is an example of:
A) Enzyme specificity
B) Competitive inhibition
C) Enzyme enhancement
D) Feedback inhibition
D
When the product of an enzymatic pathway inhibits any of the previous steps in the pathway, this is referred to as feedback inhibition. Enzyme specificity refers to the conditions appropriate for an enzyme to work. Competitive inhibition occurs when a substance occupies the active site that normally binds to the substrate involved in the reaction. The question states that isoleucine fits into the allosteric site.
In a neutral solution, most amino acids exist as:
A) Positively charged compounds
B) Zwitterions
C) Negatively charged compounds
D) Hydrophobic molecules
B
Most amino acids (except for acidic and basic amino acids) have two sites for protonation; the carboxylic acid and the amine. At neutral pH, the carboxylic acid will be deprotonated (-COOH^-) and the amine will remain protonated (-NH3^+). This dipolar ion is a zwitterion.
At pH 7, the charge on a glutamic acid molecule is:
A) –2
B) –1
C) 0
D) 1
B
Glutamic acid is an acidic amino acid because it has an extra carboxyl group. At neutral pH, both carboxyl groups are deprotonated and thus negatively charged. The amino group has a positive charge because it remains protonated at pH 7. Overall, therefore, glutamic acid has a net charge of -1. Notice that you do not even need to know the pI values to solve this equation; as an acidic amino acid, glutamic acid must have a pI below 7.
Which of the following statements is most likely to be true of nonpolar R groups in aqueous solution?
A) They are hydrophilic and found buried within proteins
B) They are hydrophilic and found on protein surfaces
C) They are hydrophobic and found buried within proteins
D) They are hydrophobic and found on protein surfaces
C
Nonpolar groups are not capable of forming dipoles or hydrogen bonds; this makes them hydrophobic. Burying hydrophobic R groups inside proteins means they don’t have to interact with water, which is polar. Choices A and B are incorrect because nonpolar molecules are hydrophobic; not hydrophilic. Choice D is incorrect because they are not generally found on protein surfaces; rather they are found within proteins.
Which of these statements concerning peptide bonds is FALSE?
A) Their formation involves a reaction between an amino group and a carboxyl group
B) They are the primary bonds that hold amino acids together
C) They have partial double bond character
D) Their formation involves hydration reactions
D
Peptide bonds are the primary covalent bond between the amino acids that make up proteins. They involve a condensation reaction between the amino group of one amino acid and the carboxyl group of an adjacent amino acid. The peptide bond has a partial double bond character because the double bond can resonate between C=O and C=N. Thus, the peptide bond has a partial double bond character and exhibits limited rotation. Formation of the peptide bond is a condensation reaction - specifically a dehydration reaction involving the loss of water - not a hydration reaction involving the addition of water.
How many distinct tripeptides can be formed from one valine molecule, one alanine molecule, and one leucine molecule?
A) 1
B) 3
C) 6
D) 27
C
There are three choices for the first amino acid, leaving two choices for the second, and one choice for the third. Multiplying those numbers gives us a total of 3 x 2 x 1 = 6 distinct tripeptides. (Using the one-letter code for valine (V), alanine (A), and leucine (L), those six tripeptides are VAL, VLA, ALV, AVL, LVA, and LAV.)
Which of these is most likely to be preserved when a protein is denatured?
A) Primary structure
B) Secondary structure
C) Tertiary structure
D) Quaternary structure
A
Denaturing a protein results in the loss of three-dimensional structure and function. Because the denaturation process does not normally result in breaking the peptide chain, the primary structure should be conserved. All of the other levels of structure can be disrupted.
An α-helix is most likely to be held together by:
A) Disulfide bonds
B) Hydrophobic effects
C) Hydrogen bonds
D) Ionic attractions between side chains
C
The ɑ-helix is held together primarily by hydrogen bonds between the carboxyl groups and amino groups of amino acids. Disulfide bridges and hydrophobic effects are primarily involved in tertiary structures, not secondary. Even if they were charged, the side chains of amino acids are too far apart to participate in strong interactions in secondary structure.
Which of the following is least likely to cause denaturation of proteins?
A) Heating the protein to 100°C
B) Adding 8 M urea
C) Moving it to a more hypotonic environment
D) Adding a detergent such as sodium dodecyl sulfate
C
High salt concentrations and detergents can denature a protein, as can high temperatures. But moving a protein to a hypotonic environment - that is, a lower solute concentration - should not lead to denaturation.
A particular α-helix is known to cross the cell membrane. Which of these amino acids is most likely to be found in the transmembrane portion of the helix?
A) Glutamate
B) Lysine
C) Phenylalanine
D) Aspartate
C
An amino acid likely to be found in a transmembrane portion of an ɑ-helix will be exposed to a hydrophobic environment, so we need an amino acid with a hydrophobic side chain. The only choice that has a hydrophobic side chain is phenylalanine. The other choices are all polar or charged.
Which of these amino acids has a chiral carbon in its side chain?
I. Serine
II. Threonine
III. Isoleucine
A) I only
B) II only
C) II and III only
D) I, II, and III
C
Every amino acid except glycine has a chiral ɑ-carbon, but only two of the 20 amino acids - threonine and isoleucine - also have a chiral carbon in their side chains as well. Just as only one configuration is normally seen at the ɑ-carbon, only one configuration is seen in the side chain chiral carbon.
Adding concentrated strong base to a solution containing an enzyme often reduces enzyme activity to zero. In addition to causing protein denaturation, which of the following is another plausible reason for the loss of enzyme activity?
A) Enzyme activity, once lost, cannot be recovered
B) The base can cleave peptide residues
C) Adding a base catalyzes protein polymerization
D) Adding a base tends to deprotonate amino acids on the surface of proteins
B
Bases can catalyze peptide bond hydrolysis. Enzyme activity can be recovered in a least some cases. Choice D is a true statement, but fails to explain the loss of enzyme activity.
Which of these amino acids has a side chain that can become ionized in cells?
A) Histidine
B) Leucine
C) Proline
D) Threonine
A
Histidine has an ionizable side chain; its imidazole ring has a nitrogen atom that can be protonated. None of the remaining answers have ionizable atoms in their side chains.
In lysine, the pKa of the side chain is about 10.5. Assuming that the pKa of the carboxyl and amino groups are 2 and 9, respectively, the pI of lysine is closest to:
A) 5.5
B) 6.2
C) 7.4
D) 9.8
D
Because lysine has a basic side chain, we ignore the pKa of the carboxyl group, and average the pKa of the side chain and the amino group; the average of 9 and 10.5 is 9.75 or 9.8.
Which of the following is a reason for conjugating proteins?
I. To direct their delivery to a particular organelle
II. To direct their delivery to the cell membrane
III. To add a cofactor needed for their activity
A) I only
B) II only
C) II and III only
D) I, II, and III
D
Conjugated proteins can have lipid or carbohydrate “tags” added to them. These tags can indicate that these proteins should be directed to the cell membrane (especially lipid tags) or to specific organelles (such as the lysosome). They can also provide the activity of the protein; for example, the heme group in hemoglobin is needed for it to bind oxygen.
Collagen consists of three helices with carbon backbones that are tightly wrapped around one another in a “triple helix.” Which of these amino acids is most likely to be found in the highest concentration in collagen?
A) Proline
B) Glycine
C) Threonine
D) Cysteine
B
Because collagen has a triple helix, the carbon backbones are very close together. Thus, steric hinderance is a potential problem. To reduce that hinderance, we need small side chains; glycine has the smallest side chain of all: a hydrogen atom.
Consider a biochemical reaction A → B, which is catalyzed by A–B dehydrogenase. Which of the following statements is true?
A) The reaction will proceed until the enzyme concentration decreases
B) The reaction will be most favorable at 0°C
C) A component of the enzyme is transferred from A to B
D) The free energy change (ΔG) of the catalyzed reaction is the same as for the uncatalyzed reaction
D
Enzymes catalyze reactions by lowering their activation energy, and are not changed or consumed during the course of the reaction. While the activation energy is lowered, the free energy of the reaction, ∆G, remains unchanged in the presence of an enzyme. A reaction will continue to occur in the presence or absence of an enzyme; it simply runs slower without the enzyme. Most physiological reactions are optimized at body temperature, 37℃. Finally, dehydrogenases catalyze oxidation-reduction reactions, not transfer reactions.
Which of the following statements about enzyme kinetics is FALSE?
A) An increase in the substrate concentration (at constant enzyme concentration) leads to proportional increases in the rate of the reaction
B) Most enzymes operating in the human body work best at a temperature of 37°C
C) An enzyme–substrate complex can either form a product or dissociate back into the enzyme and substrate
D) Maximal activity of many human enzymes occurs around pH 7.4
A
Most enzymes in the human body operate at maximal activity around a temperature of 37℃ and a pH of 7.4, which is the pH of most body fluids. In addition, as characterized by the Michaelis-Menten equation, enzymes form an enzyme-substrate complex, which can either dissociate back into the enzyme and substrate or proceed to form a product. An increase in the substrate concentration, while maintaining a constant enzyme concentration, leads to a proportional increase in the rate of the reaction only initially. However, once most of the active sites are occupied, the reaction rate levels off, regardless of further increases in substrate concentration. At high concentrations of substrate, the reaction rate approaches its maximal velocity and is no longer changed by further increases in substrate concentration.
Some enzymes require the presence of a nonprotein molecule to behave catalytically. An enzyme devoid of this molecule is called a(n):
A) Holoenzyme
B) Apoenzyme
C) Coenzyme
D) Zymoenzyme
B
An enzyme devoid of its necessary cofactor is called an apoenzyme and is catalytically inactive.
Which of the following factors determine an enzyme’s specificity?
A) The three-dimensional shape of the active site
B) The Michaelis constant
C) The type of cofactor required for the enzyme to be active
D) The prosthetic group on the enzyme
A
An enzyme’s specificity is determined by the three dimensional shape of its active site. Regardless of which explanation for enzyme specificity we are discussing (lock and key or induced fit), the active site determines which substrate the enzyme will react with.
Enzymes increase the rate of a reaction by:
A) Decreasing the activation energy
B) Increasing the overall free energy change of the reaction
C) Increasing the activation energy
D) Decreasing the overall free energy change of the reaction
A
Enzymes increase the rate of reaction by decreasing the activation energy. They do not affect the overall free energy, ∆G, or the reaction.
In the equation below, substrate C is an allosteric inhibitor to enzyme 1. Which of the following is another mechanism necessarily caused by substrate C?
A —————-> B —————-> C
Enzyme 1 Enzyme 2
A) Competitive inhibition
B) Irreversible inhibition
C) Feedback enhancement
D) Negative feedback
D
By limiting the activity of enzyme 1, the rest of the pathway is slowed, which is the definition of negative feedback. Choice A is incorrect because there is no competition for the active site with allosteric interactions. While many products do indeed competitively inhibit an enzyme in the pathway that creates them, this is an example of an allosterically inhibited enzyme. There is not enough information for Choice B to be correct because we aren’t told whether inhibition is reversible. In general, allosteric interactions are temporary. Choice C is incorrect because it is the opposite of what occurs when enzyme 1 activity is reduced.
The activity of an enzyme is measured at several different substrate concentrations, and the data are shown in the table below.
[S] (mM) [v(mmol/sec)] 0.01 1 0.05 9.1 0.1 17 0.5 50 1 67 5 91 10 95 50 99 100 100
Km for this enzyme is approximately:
A) 0.5
B) 1
C) 10
D) 50
A
While the equations given in the text are useful, recognizing relationships is even more important. You can see that as substrate concentration increases significantly, there is only a small change in the rate. This occurs as we approach Vmax. Because the Vmax is near 100 mmol/min, Vmax/2 equals 50 mmol/min. The substrate concentration giving this rate is 0.5mM and corresponds to Km.
Consider a reaction catalyzed by enzyme A with a Km value of 5 × 10^–6 M and Vmax of 20 mmol/min.
At a concentration of 5 × 10^–6 M substrate, the rate of the reaction will be:
A) 10 mmol/min
B) 20 mmol/min
C) 30 mmol/min
D) 40 mmol/min
A
Relationships are important. At a concentration of 5 × 10^–6 M, enzyme A is working at one-half its Vmax because the concentration is equal to the Km of the enzyme. Therefore, one-half of 20 mmol/min is 10 mmol/min.
Consider a reaction catalyzed by enzyme A with a Km value of 5 × 10^–6 M and Vmax of 20 mmol/min.
At a concentration of 5 × 10^–4 M substrate, the rate of the reaction will be:
A) 10 mmol/min
B) 15 mmol/min
C) 20 mmol/min
D) 30 mmol/min
C
At a concentration of 5 × 10^–4 M, there is 100 times more substrate than present at half maximal velocity. At high values (significantly larger than the value of Km), the enzyme is at or near its Vmax, which is 20 mmol/min.
The conversion of ATP to cyclic AMP and inorganic phosphate is most likely catalyzed by which class of enzyme?
A) Ligase
B) Hydrolase
C) Lyase
D) Transferase
C
Lyases are responsible for the breakdown of a single molecule into two molecules without the addition of water or the transfer of electrons. Lyases often form cyclic compounds or double bonds in the products to accommodate this. Water was not a reactant, and no cofactor was mentioned; thus lyase remains the best answer choice.
Which of the following is NOT a method by which enzymes decrease the activation energy for biological reactions?
A) Modifying the local charge environment
B) Forming transient covalent bonds
C) Acting as electron donors or receptors
D) Breaking bonds in the enzyme to provide energy
D
Enzymes are not altered by the process of catalysis. A molecule that breaks intramolecular bonds to provide activation energy would not be able to be reused.
A certain cooperative enzyme has four subunits, two of which are bound to substrate. Which of the following statements can be made?
A) The affinity of the enzyme for the substrate has just increased
B) The affinity of the enzyme for the substrate has just decreased
C) The affinity of the enzyme for the substrate is at the average for this enzyme class
D) The affinity of the enzyme for the substrate is greater than with one substrate bound
D
Cooperative enzymes demonstrate a change in affinity for the substrate depending on how many substrate molecules are bound and whether the last change was accomplished because a substrate molecule was bound or left the active site of the enzyme. Because we cannot determine whether the most recent reaction was binding or dissociation, Choices A and B are incorrect. We can make absolute comparisons though. The unbound enzyme has the lowest affinity for substrate, and the enzyme with all but one subunit bound has the highest. The increase in affinity is not linear, so Choice C is not necessarily true. An enzyme with two subunits occupied must have a higher affinity for the substrate than the same enzyme with only one subunit occupied.
Which of the following is LEAST likely to be required for a series of metabolic reactions?
A) Triglyceride acting as a coenzyme
B) Oxidoreductase enzymes
C) Magnesium acting as a cofactor
D) Transferase enzymes
A
Triglycerides are unlikely to act as coenzymes for a few reasons, including their large size, neutral charge, and ubiquity in cells. Cofactors and coenzymes tend to be small in size, such as metal ions or small organic molecules. They can usually carry a charge by ionization, protonation, or deprotonation. Finally, they are usually in low, tightly regulated concentrations within cells. Metabolic pathways would be expected to include both oxidation-reduction reactions and movement of functional groups.
How does the ideal temperature for a reaction change with and without an enzyme catalyst?
A) The ideal temperature is generally higher with a catalyst than without
B) The ideal temperature is generally lower with a catalyst than without
C) The ideal temperature is characteristic of the reaction, not the enzyme
D) No conclusion can be made without knowing the enzyme type
B
The rate of reaction increases with temperature because of the increased kinetic energy of the reactants, but reaches a peak temperature because the enzyme denatures with the disruption of hydrogen bonds at excessively high temperatures. In the absence of enzyme, this peak temperature is generally much hotter. Heating a reaction provides molecules with an increased chance of achieving the activation energy, but the enzyme catalyst would typically reduce activation energy. Keep in mind that thermodynamics and kinetics are not interchangeable, so we are not considering the impact of heat on the equilibrium position.
In a single strand of a nucleic acid, nucleotides are linked by:
A) Hydrogen bonds
B) Phosphodiester bonds
C) Ionic bonds
D) Van der Waals forces
B
Nucleotides bond together to form polysaccharides. The 3’ hydroxyl group of one nucleotide’s sugar joins the 5’ hydroxyl group of the adjacent nucleotide’s sugar by a phosphodiester bond. Hydrogen bonding is important for holding complementary strands together, but does not play a role in the bonds formed between adjacent nucleotides on a single strand.
Which of the following statements regarding differences between DNA and RNA is FALSE?
A) DNA is double-stranded, whereas RNA is single-stranded
B) DNA uses the nitrogenous base thymine; RNA uses uracil
C) The sugar in DNA is deoxyribose; the sugar in RNA is ribose
D) DNA strands replicate in a 5′ to 3′ direction, whereas RNA is synthesized in a 3′ to 5′ direction
D
Because we are looking for the false statement, we have to read each choice to eliminate those that are true or find one that is overtly false. Let’s quickly review the main differences between DNA and RNA. In cells, DNA is double-stranded, with a deoxyribose sugar and the nitrogenous bases A, T, C, and G. RNA, on the other hand, is usually single-stranded, with a ribose sugar and the bases A, U, C, and G. Choice D is false because both DNA replication and RNA synthesis proceed in a 5’ to 3’ direction.
Which of the following DNA sequences would have the highest melting temperature?
A) CGCAACCATCCG
B) CGCAATAATACA
C) CGTAATAATACA
D) CATAACAAATCA
A
The melting temperature of DNA is the temperature at which a DNA double helix separates into two single strands (denatures). To do this, the hydrogen bonds linking the base pairs must be broken. Cytosine binds to guanine with three hydrogen bonds, whereas adenine binds to thymine with two hydrogen bonds. The amount of heat needed to disrupt the bonding is proportional to the number of bonds. Thus, the higher the GC-content in a DNA segment, the higher the melting point.
Which of the following biomolecules is LEAST likely to contain an aromatic ring?
A) Proteins
B) Purines
C) Carbohydrates
D) Pyrimidines
C
Aromatic rings must contain conjugated ⫪ electrons, which require alternating single and multiple bonds, or lone pairs. In carbohydrate ring structures, only single bonds are present, thus preventing aromaticity. Nucleic acids contain aromatic heterocycles, while proteins will generally contain at least one aromatic amino acid (tryptophan, phenylalanine, tyrosine).
For a compound to be aromatic, all of the following must be true EXCEPT:
A) The molecule is cyclic
B) The molecule contains 4n + 2⫪ electrons
C) The molecule contains alternating single and double bonds
D) The molecule is planar
C
For a compound to be aromatic, it must be cyclic, planar, conjugates, and contain 4n + 2⫪ electrons, where n is any integer. Conjugation requires that every atom in the ring have at least one unhybridized p-orbital. While most examples of aromatic compounds have alternating single and double bonds, compounds can be aromatic if they contain triple bonds as well; this would still permit at least one unhybridized p-orbital.
Which of the following enzymes is NOT involved in DNA replication?
A) Primase
B) DNA ligase
C) RNA polymerase I
D) Telomerase
C
During DNA replication, the strands are separated by DNA helicase. At the replication fork, primase creates a primer for the initiation of replication, which is followed by DNA polymerase. On the lagging strand, Okazaki fragments form and are joined by DNA ligase. After the chromosome has been processed, the ends, called telomeres, are replicated with the assistance of the enzyme telomerase. RNA polymerase I is located in the nucleolus and synthesizes rRNA.
How is cDNA best characterized?
A. cDNA results from a DNA transcript with noncoding regions removed.
B. cDNA results from the reverse transcription of processed mRNA.
C. cDNA is the abbreviation for deoxycytosine.
D. cDNA is the circular DNA molecule that forms the bacterial genome.
B
cDNA (complementary DNA) is formed from a processed mRNA strand by reverse transcription. cDNA is used in DNA libraries and contains only the exons of genes that are transcriptionally active in the sample tissue.
Which of the following statements regarding the polymerase chain reaction is FALSE?
A) Human DNA polymerase is used because it is the most accurate
B) A primer must be prepared with a complementary sequence to part of the DNA of interest
C) Repeated heating and cooling cycles allow the enzymes to act specifically and replaces helicase
D) Each cycle of the polymerase chain reaction doubles the amount of DNA of interest
A
The polymerase chain reaction is used to clone a sequence of DNA using a DNA sample, a primer, free nucleotides, and enzymes. The polymerase from Thermus aquaticus is used because the reaction is regulated by thermal cycling, which would denature human enzymes.
Restriction endonucleases are used for which of the following?
I. Gene therapy
II. Southern blotting
III. DNA repair
A) I only
B) II only
C) II and III only
D) I, II, and III
D
Endonucleases are enzymes that cut DNA. They are used by the cell for DNA repair. They are also used by scientists during DNA analysis, as restriction enzymes are endonucleases. Restriction enzymes are used to cleave DNA before electrophoresis and Southern blotting, and to introduce a gene of interest into a viral vector for gene therapy.
How does prokaryotic DNA differ from eukaryotic DNA?
I. Prokaryotic DNA lacks nucleosomes.
II. Eukaryotic DNA has telomeres.
III. Prokaryotic DNA is replicated by a different DNA polymerase.
IV. Eukaryotic DNA is circular when unbound by centromeres.
A) I only
B) IV only
C) II and III only
D) I, II, and III only
D
Prokaryotic DNA is circular and lacks histone proteins, and thus does not form nucleosomes. Both prokaryotic and eukaryotic DNA are replicated by DNA polymerases, although these polymerases differ in identity. Eukaryotic DNA is organized into chromatin, which can condense to form linear chromosomes; only prokaryotes have circular chromosomes. Only eukaryotic DNA has telomeres.
Why might uracil be excluded from DNA but NOT RNA?
A) Uracil is much more difficult to synthesize than thymine
B) Uracil binds adenine too strongly for replication
C) Cytosine degradation results in uracil
D) Uracil is used as a DNA synthesis activator
C
One common DNA mutation is the transition from cytosine to uracil in the presence of heat. DNA repair enzymes recognize uracil and correct this error by excising the base and inserting cytosine. RNA exists only transiently in the cell, such that cytosine degradation is insignificant. Were uracil to be used in DNA under normal circumstances, it would be impossible to tell if a base SHOULD be uracil or if it is a damaged cytosine nucleotide.
Tumor suppressor genes are most likely to result in cancer through:
A) Loss of function mutations
B) Gain of function mutations
C) Overexpression
D) Proto-oncogene formation
A
Oncogenes are most likely to result in cancer through activation, while tumor suppressor genes are most likely to result in cancer through inactivation.
Which of the following is an ethical concern of gene sequencing?
A) Gene sequencing is invasive, thus the potential health risks must be thoroughly explained
B) Gene sequencing impacts relatives, thus privacy concerns may be raised
C) Gene sequencing is very inaccurate, which increases anxiety related to findings
D) Gene sequencing can provide false-negative results, giving a false sense of security
B
One of the primary ethical concerns related to gene sequencing is the issue of consent and privacy. Because genetic screening provides information on direct relatives, there are potential violations of privacy in communicating this information to family members who may be at risk. There are not significant physical risks and gene sequencing is fairly accurate.
Which of the following is NOT a difference between heterochromatin and euchromatin?
A) Euchromatin has areas that can be transcribed, whereas heterochromatin is silent
B) Heterochromatin is tightly packed, whereas euchromatin is less dense
C) Heterochromatin stains darkly, whereas euchromatin stains lightly
D) Heterochromatin is found in the nucleus, whereas euchromatin is in the cytoplasm
D
Euchromatin has a classic “beads on a string” appearance that stains lightly, while heterochromatin is tightly packed and stains darkly. Heterochromatin is primarily composed of inactive genes or untranslated regions, while euchromatin is able to be expressed. All chromatin is found in the nucleus, not the cytoplasm.
During which phase of the cell cycle are DNA repair mechanisms least active?
A) G1
B) S
C) G2
D) M
D
Mismatch repair mechanisms are active during S phase (proofreading) and G2 phase (MSH2 and MLH1). Nucleotide and base excision repair mechanisms are most active during the G1 and G2 phases. These mechanisms exist during interphase because they are aimed at PREVENTING propagation of the error into daughter cells during the M phase (mitosis).
What role does peptidyl transferase play in protein synthesis?
A) It transports the initiator aminoacyl-tRNA complex
B) It helps the ribosome to advance three nucleotides along the mRNA in the 5′ to 3′ direction
C) It holds the protein in its tertiary structure
D) It catalyzes the formation of a peptide bond
D
Peptidyl transferase is an enzyme that catalyzes the formation of a peptide bond between the incoming amino acid in the A site and the growing polypeptide chain in the P site. Initiation and elongation factors help transport charged tRNA molecules into the ribosome and advance the ribosome down the mRNA transcript. Chaperones maintain a protein’s three-dimensional shape as it is formed.
Which stage of protein synthesis does NOT require energy?
A) Initiation
B) Elongation
C) Termination
D) All stages of protein synthesis require energy
D
All three stages of protein synthesis (initiation, elongation, and termination) require large amounts of energy.
Topoisomerases are enzymes involved in:
A) DNA replication and transcription
B) Posttranscriptional processing
C) RNA synthesis and translation
D) Posttranslational processing
A
Topoisomerases, such as prokaryotic DNA gyrase, are involved in DNA replication and mRNA synthesis (transcription). DNA gyrase is a type of topoisomerase that enhances the action of helicase enzymes by the introduction of negative supercoils into the DNA molecule. These negative supercoils facilitate DNA replication by keeping the strands separated and untangled.
Enhancers are transcriptional regulatory sequences that function by enhancing the activity of:
A) RNA polymerase at a single promoter site
B) RNA polymerase at multiple promoter sites
C) Spliceosomes and lariat formation in the ribosome
D) Transcription factors that bind to the promoter but not to RNA polymerase
A
Specific transcription factors bind to a specific DNA sequence, such as an enhancer, and to RNA polymerase at a single promoter sequence. They enable the RNA polymerase to transcribe the specific gene for that enhancer more efficiently.
In the genetic code of human nuclear DNA, one of the codons specifying the amino acid tyrosine is UAC. If one nucleotide is changed, and the codon is mutated to UAG, what type of mutation will occur?
A) Silent mutation
B) Missense mutation
C) Nonsense mutation
D) Frameshift mutation
C
UAG is one of the three known stop codons, so changing tyrosine to a stop codon must be a nonsense (or truncation) mutation.
A double-stranded RNA genome isolated from a virus was found to contain 15% uracil. What percentage of guanine should exist in this virus’s genome?
A) 15%
B) 35%
C) 70%
D) 85%
B
The percentage of uracil must equal that of adenine due to base-pairing because the genome is double-stranded. This accounts for 30% of the genome. The remaining 70% must be evenly split between guanine and cytosine, so they each account for 35% of the genome.
When trypsin converts chymotrypsinogen to chymotrypsin, some molecules of chymotrypsin bind to a repressor, which in turn binds to the operator and prevents further transcription of trypsin. This is most similar to which of the following operons?
A) Trp operon during lack of tryptophan
B) Trp operon during abundance of tryptophan
C) Lac operon during lack of lactose
D) Lac operon during abundance of lactose
B
The example given is a sample of repression due to the abundance of a corepressor. In other words, this is a repressible system that is currently blocking transcription. For the trp operon, an abundance of tryptophan in the environment allows for the repressor to bind tryptophan and then to the operator site. This blocks transcription of the genes required to synthesize tryptophan within the cell. The system described is a repressible system; the lac operon is an inducible system, in which an inducer binds to the repressor, thus permitting transcription.
Which of the following RNA molecules or proteins is NOT found in the spliceosome during intron excision?
A) snRNA
B) hnRNA
C) shRNA
D) snRNPs
C
shRNA (short hairpin RNA) is a useful biotechnology tool used in RNA interference. It is not, however, produced in the nucleus for use in the spliceosome. It targets mRNA to be degraded in the cytoplasm; it is not utilized in splicing of the hnRNA (heterogenous nuclear RNA). snRNA (small nuclear RNA) and snRNPs (small nuclear ribonucleoproteins), however, do bind to the hnRNA to induce splicing.
A 4-year old toddler with cystic fibrosis (CF) is seen by his physician for an upper respiratory infection. Prior genetic testing has shown that there has been a deletion of three base pairs in exon 10 of the CFTR gene that affects codons 507 and 508. The nucleotide sequence in this region for normal and mutant alleles is shown below (X denotes the missing nucleotide):
Codon number 506 507 508 509 510 511
Normal (coding strand) ATC ATC TTT GGT GTT TCC
Mutant (coding strand) ATC ATX XXT GGT GTT TCC
What effect will this mutation have on the amino acid sequence of the protein encoded by the CFTR gene?
A) Deletion of a phenylalanine residue with no change in the C-terminus sequence
B) Deletion of a leucine residue with no change in the C-terminus sequence
C) Deletion of a phenylalanine residue with a change in the C-terminus sequence
D) Deletion of a leucine residue with a change in the C-terminus sequence
A
In this table, we are given the sequence of the sense (coding) DNA strand. This will be identical to the mRNA transcript, except all thymine nucleotides will be replaced with uracil. With the deletion of these three bases, codon 507 changes from AUC to AUU in the transcript; these both code for isoleucine due to wobble. However, codon 508 (UUU in the transcript) has been lost. UUU codes for phenylalanine. The C-terminus sequence will remain unchanged because the deletion of three bases (exactly one codon) will not throw off the reading frame. For reference, the mutant reading frames would be as follows:
AUC AUU GGU GUU UCC
(5 codons instead of 6 in the original normal transcript)
A gene encodes a protein with 150 amino acids. There is one intron of 1000 base pairs (bp), a 5′-untranslated region of 100 bp, and a 3′-untranslated region of 200 bp. In the final mRNA, about how many bases lie between the start AUG codon and final termination codon?
A) 150
B) 450
C) 650
D) 1750
B
The intron will not be a part of the final, processed mRNA, and the untranslated regions of the mRNA will not be turned into amino acids. Translation will begin with codon 1 (which would be AUG). Because there are 150 amino acids, we can surmise that there will be 151 codons. Each codon will use 3 nucleotides, so 150 x 3 = 450 because codon 151 will be the stop codon.
Peptidyl transferase connects the carboxylate group of the one amino acid to the amino group of an incoming amino acid. What type of linkage is created in this peptide bond?
A) Ester
B) Amide
C) Anhydride
D) Ether
B
Peptidyl transferase connects the incoming amino terminal to the previous carboxyl terminal; the only functional group listed here with a carbonyl and amino group is the amide. Peptide bonds are thus amide linkages.
A eukaryotic cell has been found to exhibit a truncation mutation that creates an inactive RNA polymerase I enzyme. Which type of RNA will be affected by this inactivation?
A) rRNA
B) tRNA
C) snRNA
D) hnRNA
A
RNA polymerase I in eukaryotes is found in the nucleolus and is in charge of transcribing most of the rRNA for use during ribosomal creation. RNA polymerase II is responsible for hnRNA and snRNA. RNA polymerase III is responsible for tRNA and the 5S rRNA.
You have just sequenced a piece of DNA that reads as follows:
5′—TCTTTGAGACATCC—3′
What would the base sequence of the mRNA transcribed from this DNA be?
A. 5′—AGAAACUCUGUAGG—3′
B. 5′—GGAUGUCUCAAAGA—3′
C. 5′—AGAAACTCTGTAGG—3′
D. 5′—GGATCTCTCAAAGA—3′
B
To answer this question correctly, we must remember that mRNA will be antiparallel to DNA. Our answer should be 5’ to 3’ mRNA, with the 5’ end complementary to the 3’ end of the DNA that is being transcribed. Thus, the mRNA transcribed from this strand will be:
5’ - GGAUGUCUCAAAGA - 3’
mRNA contains uracil, rather than thymine.
Double-stranded RNA cannot be translated by the ribosome and is marked for degradation in the cell. Which of the following strands of RNA would prevent mature mRNA in the cytoplasm from being transcribed?
A) Identical mRNA to the one produced
B) Antisense mRNA to the one produced
C) mRNA with thymine substituted for uracil
D) Sense mRNA to the one produced
B
The mRNA produced has the same structure as the sense strand of DNA (with uracils instead of thymines). Because bonding of nucleic acids is always complementary but antiparallel, the antisense strand of mRNA would be the one that binds to the produced mRNA, creating double-stranded RNA that is then degraded once found in the cytoplasm.
A student is trying to determine the type of membrane transport occurring in a cell. She finds that the molecule to be transported is very large and polar, and when transported across the membrane, no energy is required. Which of the following is the most likely mechanism of transport?
A) Active transport
B) Simple diffusion
C) Facilitated diffusion
D) Exocytosis
C
We are asked to identify the type of transport that would allow a large, polar molecule to cross the membrane without any energy expenditure. This scenario describes facilitated diffusion, which uses a transport protein (or channel) to facilitate the movement of large, polar molecules across the nonpolar, hydrophobic membrane. Facilitated diffusion, like simple diffusion, does not require energy.
A researcher treats a solution containing animal cells with ouabain, a poisonous substance that interferes with the sodium–potassium ATPase embedded in the cell membrane, and the cell lyses as a result. Which of the following statements best describes ouabain’s effects?
A) Treatment with ouabain results in high levels of extracellular calcium
B) Treatment with ouabain results in high levels of extracellular potassium and sodium
C) Treatment with ouabain increases intracellular concentrations of sodium
D) Treatment with ouabain decreases intracellular concentrations of sodium
C
This question requires an understanding of osmosis and the action of the sodium-potassium pump. When a cell is placed in a hypertonic solution (a solution having a higher solute concentration than the cell), fluid will diffuse out of the cell and result in cell shrinkage. When a cell is placed in a hypotonic solution (a solution having a lower solute concentration than the cell), fluid will diffuse from the solution into the cell, causing the cell to expand and possibly lyse. The sodium-potassium pump moves three sodium ions out of the cell for every two potassium ions it lets into the cell. Therefore, inhibition of the sodium-potassium pump by ouabain will cause a net increase in the sodium concentration inside the cell and water will diffuse in, causing the cell to swell and then lyse.
Resting membrane potential depends on:
I. The differential distribution of ions across the membrane.
II. Active transport processes.
III. Selective permeability of the phospholipid bilayer.
A) I only
B) I and III only
C) II and III only
D) I, II, and III
D
The polarization of the membrane at rest is the result of an uneven distribution of ions between the inside and outside of the cell. This difference is achieved through active pumping of ions (predominantly sodium and potassium) into and out of the cell and the selective permeability of the membrane, which allows only certain ions to cross.
Which of the following is NOT a function of the cell membrane?
A) Cytoskeletal attachment
B) Protein synthesis
C) Transport regulation
D) Second messenger reservoir
B
Ribosomes are the site of protein synthesis within a cell and are not coupled to the cell membrane. The cell membrane functions as a site for cytoskeletal attachment through proteins and lipid rafts. Transport regulation is accomplished through channels, transporters, and selective permeability, while the phospholipids act as a reagent for second messenger formation.
The dynamic properties of molecules in the cell membrane are most rapid in:
A) Phospholipids moving within the plane of the membrane
B) Phospholipids moving between the planes of the membrane
C) Proteins moving within the plane of the membrane
D) Proteins exiting the cell through exocytosis
A
Movement of individual molecules in the cell membrane will be affected by size and polarity, just as with diffusion. Lipids are much smaller than proteins in the plasma membrane and will move more quickly. Lipids will move fastest within the plane of the cell membrane because the polar head group does not need to pass through the hydrophobic tail region in the same way that it would if it were moving between the membrane layers.
Which lipid type is LEAST likely to contribute to membrane fluidity?
A) Unsaturated glycerophospholipids
B) Trans glycerophospholipids
C) Cholesterol
D) Unsaturated sphingolipids
B
Compounds that contribute to membrane fluidity will lower the melting point or disrupt the crystal structure. Cholesterol and unsaturated lipids are known for these functions. Trans glycerophospholipids tend to increase the melting point of the membrane and therefore decrease membrane fluidity.
A membrane receptor is most likely to be a(n):
A) Embedded protein with catalytic activity
B) Transmembrane protein with sequestration activity
C) Membrane-associated protein with sequestration activity
D) Transmembrane protein with catalytic activity
D
Membrane receptors must have both an extracellular and intracellular domain; therefore, they are considered transmembrane proteins. In order to initiate a second messenger cascade, they typically display enzymatic activity, although some may act strictly as channels.
Which of the following is NOT a cell–cell junction in animals?
A) Desmosomes
B) Gap junctions
C) Plasmodesmata
D) Tight junctions
C
Plasmodesmata are cell-cell junctions that are found in plants, not animals. Gap junctions, tight junctions, desmosomes, and hemidesmosomes are all found in animals, particularly in epithelia.
Which of the following is true of diffusion and osmosis?
A) Diffusion and osmosis rely on the electrochemical gradient of only the compound of interest
B) Diffusion and osmosis rely on the electrochemical gradient of all compounds in a cell
C) Diffusion and osmosis will proceed in the same direction if there is only one solute
D) Diffusion and osmosis cannot occur simultaneously
A
The movement of any solute or water by diffusion or osmosis is dependent only on the concentration gradient of that molecule and on membrane permeability.
The bulk movement of liquid into a cell through vesicular infoldings is known as:
A) Phagocytosis
B) Pinocytosis
C) Exocytosis
D) Drinking
B
The endocytosis (bulk uptake through vesicle formation) of fluid is known as pinocytosis. Phagocytosis is the endocytotic intake of solids, while exocytosis is a method of releasing vesicular contents. Drinking does not apply on a cellular level.
Which of the following is LEAST likely to be the resting membrane potential of a cell?
A) –70 mV
B) –55 mV
C) 0 mV
D) +35 mV
C
Cell membranes are most likely to have a resting membrane potential that is nonzero because the resting membrane potential creates a state that is capable of responding to stimuli. Signaling molecules and channels would not be as useful with a membrane potential of zero. The values given in the answer choices correspond to different stages of the action potential, but the key information is that a resting potential of 0 mV does not maintain gradients for later activity.
How does the inner mitochondrial membrane differ from the outer mitochondrial membrane?
A) The inner mitochondrial membrane is more permeable and lacks cholesterol
B) The inner mitochondrial membrane is less permeable and lacks cholesterol
C) The inner mitochondrial membrane is more permeable and has cholesterol
D) The inner mitochondrial membrane is less permeable and has cholesterol
B
The outer mitochondrial membrane is very permeable while the inner mitochondrial membrane is highly impermeable. The inner mitochondrial membrane is unique within the cell because it lacks cholesterol.
Which of the following statements conflicts with the fluid mosaic model?
A) The cell membrane is static in structure
B) Membrane components can be derived from multiple biomolecules
C) Hydrophobic interactions stabilize the lipid bilayer
D) Proteins are asymmetrically distributed within the cell membrane
A
The fluid mosaic model accounts for a dynamic membrane. In this model, membrane components contain both fatty and carbohydrate-derived components. Further, the membrane is stabilized by the hydrophobic interactions of both fatty acids tails and membrane proteins, which may be found on the cytosolic or extracellular side of the membrane, or may run directly through the membrane.
Which of the following is a sphingolipid?
A) Lecithin
B) Phosphatidylinositol
C) Cholesterol
D) Gangliosides
D
Gangliosides, along with ceramide, sphingomyelin, and cerebrosides, are sphingolipids.
Adding heat to a closed biological system will do all of the following EXCEPT:
A) Increase the internal energy of the system
B) Increase the average of the vibrational, rotational, and translational energies
C) Cause the system to do work to maintain a fixed internal energy
D) Increase the enthalpy of the system
C
In a closed biological system, enthalpy, heat, and internal energy are all directly related because there is no change in pressure or volume. Because pressure and volume are fixed, work cannot be done.
Which of the following statements is true about the hydrolysis of ATP?
A) The free energy of ATP hydrolysis is independent of pH
B) One mole of creatine phosphate can phosphorylate two moles of ADP
C) The free energy of hydrolysis of ATP is nearly the same as for ADP
D) ATP yields cyclic AMP after two hydrolysis reactions
C
The hydrolysis of ATP is energetically favorable because there are repulsive negative charges that are relieved when hydrolyzed, and the new compounds are stabilized by resonance. This is true of both ATP and ADP. Some of the other answer choices are tempting, though. In Choice A, ATP hydrolysis relies on pH because a protonated ATP molecule contains less negative charge and therefore experiences less repulsive force. Choice B, the energy released by one mole of creatine phosphate upon hydrolysis is not sufficient to phosphorylate two moles of ADP; creatine phosphate donates one phosphate group to a molecule of ADP, so one mole of creatine phosphate will phosphorylate one mole of ADP. Choice D, the removal of two phosphate groups from ATP yields AMP, not cyclic AMP.
The reduction half-reaction in the last step of the electron transport chain is:
A. O2 + 4 e– + 4 H+ → 2 H2O
B. NADPH → NADP+ + e– + H+
C. NADP+ + e– + H+ → NADPH
D. Ubiquinone (Q) → Ubiquinol (QH2)
A
Reduction is a gain of electrons and choice B is a oxidation reaction. NADPH is a product of the pentose phosphate pathway. Ubiquinone transfers electrons during the course of the electron transport chain, but is not the final electron acceptor. This title belongs to oxygen.
The ability to exist in both an oxidized and a reduced state is characteristic of:
A) Adenosine triphosphate (ATP)
B) Electron carriers
C) Regulatory enzymes
D) Peptide hormones
B
In order to transport electrons, electron carriers like flavoproteins must be able to exist in a stable oxidized state and a stable reduced form. ATP can be dephosphorylated but is generally not oxidized or reduced. Regulatory enzymes may also be phosphorylated or dephosphorylated but are not generally oxidized or reduced.
With prolonged fasting, the brain can turn to which alternative fuel for energy?
A) Fructose
B) Lactate
C) Ketone bodies
D) Fatty acids
C
The brain is almost exclusively dependent on glucose for energy; however, in a prolonged fast, ketone bodies can be used for up to 2/3 of the brain’s energy requirement.
How do hormonal controls of glycogen metabolism differ from allosteric controls?
A) Hormonal control is systemic and covalent
B) Hormonal control is local and covalent
C) Hormonal control is systemic and noncovalent
D) Hormonal control is local and noncovalent
A
Hormonal controls are coordinated to regulate the metabolic activity of the entire organism, while allosteric controls can be local or systemic. The modification of the enzymes of glycogen metabolism by insulin and glucagon is either through phosphorylation or dephosphorylation, both of which modify covalent bonds.
Which of the following tissues is most dependent on insulin?
A) Active skeletal muscle
B) Resting skeletal muscle
C) Cardiac muscle
D) Smooth muscle
B
Adipose tissue and resting skeletal muscle require insulin for glucose uptake. Active skeletal muscle uses creatine phosphate and glycogen (regulated by epinephrine and AMP) to maintain its energy requirements.
Glucocorticoids have been implicated in stress-related weight gain because:
A) They increase appetite and decrease satiety signals
B) They increase the activity of catabolic hormones
C) They increase glucose levels, which causes insulin secretion
D) They interfere with activity of the leptin receptor
C
Short-term glucocorticoid exposure causes a release of glucose and the hydrolysis of fats from adipocytes. However, if this glucose is not used for metabolism, it causes an increase in glucose level which promotes fat storage. The net result is the release of glucose from the liver to be converted into lipids in the adipose tissue under insulin stimulation.
In the absence of oxygen, which tissue will experience damage most rapidly?
A) Skin
B) Brain
C) Red blood cells
D) Liver
B
The brain uses aerobic metabolism of glucose exclusively and therefore is very sensitive to oxygen levels. The extremely high oxygen requirement of the brain (20% of the body’s oxygen content) relative to its size (2% of total body weight) implies that the brain is the most sensitive organ to oxygen deprivation.
A respiratory quotient approaching 0.7 indicates metabolism primarily of which macromolecule?
A) Carbohydrates
B) Lipids
C) Nucleic acids
D) Amino acids
B
The respiratory quotient (RQ) gives an indication of the primary fuel being utilized. An RQ around 0.7 indicates lipid metabolism, 0.8 - 0.9 indicates amino acid metabolism and 1.0 indicates carbohydrate metabolism. Nucleic acids do no contribute significantly to the respiratory quotient.
Which of the following side effects would be anticipated in someone taking leptin to promote weight loss?
A) Drowsiness
B) Increased appetite
C) Irritability
D) Fever
A
Leptin acts to decrease appetite by inhibiting the production of orexin. Orexin is also associated with alertness, so decreasing the level of orexin in the body is expected to cause drowsiness. Even without this information, the answer should be apparent because the body tends to maintain an energy balance. If consumption decreases, energy expenditures are expected to decrease as well.
Which of the following statements is FALSE?
A) Growth hormone participates in glucose counter-regulation
B) T4 acts more slowly than T3
C) ATP stores are turned over more than 10,000 times daily
D) Catecholamines stimulate the sympathetic nervous system
C
ATP stores are turned over about 1,000 times per day, not 10,000.
Which process is expected to begin earliest in a prolonged fast?
A) Ketone bodies are used by the brain
B) Glycogen storage is halted
C) Proteins are broken down
D) Enzyme phosphorylation and dephosphorylation
D
A prolonged fast is characterized by an increase in glucagon, which accomplishes its cellular activity by phosphorylating and dephosphorylating metabolic enzymes. Glycogen storage is then halted, but this requires enzyme regulation by glucagon to occur. Later in the postabsorptive state, protein breakdown begins. Eventually, in starvation, ketone bodies are used by the brain for its main energy source.
