Module 1 Flashcards

1A - Amino Acids - Carbohydrates and Lipids - Membrane Receptor & Transport; Material Transport - Enzymes 1B ECM, DNA & Protein Synthesis, Gene Expression and Mutations, Bioenergetics, Biological Oxidation & Oxidative Phosphorylation

1
Q
  1. Which of the following amino acids, when deficient, will cause a deficit of catecholamines as part of the sympathetic nervous system response?
    a. Tryptophan
    b. Histidine
    c. Tyrosine
    d. Glutamate
A

c. Tyrosine

Catecholamines (norephinephrine, epinephrine) are derived from tyrosine. The protein amino acids phenylalanine and tyrosine serve as precursors of epinephrine, norepinephrine, and DOPA (dihydroxyphenylalanine). (p19 Harpers)

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2
Q
  1. Selenium deficiency in the diet, leading to Keshan disease, a condition characterized by heart failure, also impairs the body’s ability to clear free radicals. The amino acid that requires selenium is?
    a. Synthesized in the nucleus
    b. Produced during translation through modification of cysteine
    c. Synthesized in the ribosomes
    d. Produced during translation through modification of serine
A

d. Produced during translation through modification of serine.

Selenocysteine is structurally similar to cysteine; both derived from serine. It has selenium residue instead of sulfur. Impairments in human selenoproteins have been implicated in tumorigenesis and atherosclerosis, and are associated with selenium deficiency cardiomyopathy (Keshan disease). (p286 Harpers)

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3
Q
  1. Which of the following amino acids will most likely cause akink in a helical peptide?
    a. Serine
    b. Alanine
    c. Proline
    d. Cysteine
A

c. Proline

Proline, owing to its uncanny structure (imino acid), causes it to disrupt helices by introducing folds and kinks.

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4
Q
  1. Glutathione is a tripeptide composed of the following, EXCEPT?
    a. Cysteine
    b. Glutamate
    c. Methionine
    d. Glycine
A

c. Methionine

Glutathione is a tripeptide that is composed of (Glu-Cys-Gly) Glutamate, Cysteine, and Glycine that participates in the metabolism of xenobiotics and the reduction of disulfide bonds, is linked to cysteine by a non-α peptide bond. (Harper p23)

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5
Q
  1. In sickle cell anemia, the altered hemoglobin structure has which of the following changes to the amino acids involved?
    a. Polar to non-polar
    b. Basic to acidic
    c. Acidic to basic
    d. Non-polar to polar
A

a. Polar to non-polar

Sickle cell is a disease that results from a substitution of a polar amino acid known as glutamate with a nonpolar one valine at position six of the beta polypeptide unit of hemoglobin. The substitution happens as a result of a change in one of the bases in the beta-globin gene from adenine to thymine. (Harper p417)

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6
Q
  1. A patient is diagnosed with familial hypercholesterolemia where there is a mutation in the receptors needed for lipoprotein transport and docking into the liver. Which of the following is a most likely outcome?
    a. Poor wound healing
    b. Hemolytic anemia
    c. Decreased plasma lipoproteins
    d. Early onset cardiovascular disease
A

d. Early Onset Cardiovascular Disease

Inherited defects in lipoprotein metabolism lead to the primary condition of either hypo- or hyperlipoproteinemia. For example, familial hypercholesterolemia (FH), causes severe hypercholesterolemia and is also associated with premature atherosclerosis. (Harpers p275)

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7
Q
  1. Which of the following stabilizes the secondary structure of peptides?
    a. Salt bridges
    b. Hydrophobic interactions
    c. Peptide bonds
    d. Hydrogen bonds
A

d. Hydrogen Bonds

Secondary structures are usually the general conformation or how the peptide is folded. Alpha helices, which is one of the abundant secondary structures for globular proteins, are stabilized by hydrogen bonding between the amide nitrogens and carbonyl carbons of peptide bonds. While for the Beta Sheets, which can be found more prominent in fibrillar proteins, are also being stabilized by hydrogen bonds for folding and alignment.

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8
Q
  1. A patient with osteogenesis imperfecta has a mutation in the gene encoding for collagen leading to poor bone and soft tissue maturation and development. The resulting collagen protein most likely has an altered?
    a. Primary protein structure
    b. Secondary protein structure
    c. Primary, Secondary, Tertiary, and Quarternary protein structures
    d. Tertiary protein structure
A

c. Primary, Secondary, Tertiary, and Quarternary protein structures

Osteogenesis Imperfecta (brittle bones) is characterized by abnormal fragility of bones. -Exome sequencing has emerged as an alternative to whole genome sequencing as a means for diagnosing rare or cryptic genetic diseases like the osteogenesis imperfecta. -Since it affects the sequencing, it has already affected the protein structure starting from the primary.

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9
Q
  1. Which of the following amino acids is neutrally charged at physiologic pH?
    a. Histidine [Considered correct after revisions]
    b. Aspartate
    c. Arginine
    d. Leucine
A

a. Histidine (Considered Correct after Revisions) AND d. Leucine

Histidine plays unique roles in enzymatic catalysis. The pKa (6.0) of its imidazole proton permits histidine to function at neutral pH as either a base or an acid catalyst without the need for any environmentally induced shift.

Leucine is an amino acid with aliphatic side chains that is branched and it is a non-polar which makes it a neutrally charged at physiologic pH

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10
Q
  1. A mutation in the peptide portion of a glycoprotein will most likely impair?
    a. Cell-to-cell communication
    b. Stability of connective tissue
    c. Activation of thyroid hormones
    d. Intercellular transport of lipids
A

a. Cell-to-cell communication

Glycoproteins are covalently associated with carbohydrates and it is functionally active in cell to cell interactions.

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11
Q
  1. An ion transporter containing 5 sub-units uses which level/s of protein structure?
    a. Quaternary only
    b. Primary, Secondary, Tertiary, and Quarternary
    c. Primary, Secondary and Tertiary only
    d. Primary and Secondary only
A

b. Primary, Secondary, Tertiary, and Quarternary

Many proteins consist of more than one polypeptide chain in their functional states. Each polypeptide chain in such a protein is called a subunit. -All levels of protein structure are used in 5 sub-units ion transporter since polypeptide chain will not be produced when it does not start with the primary protein structure which is responsible for the sequence of the chain. In secondary structure, it is responsible for the folding of the chain. In tertiary structure, it is responsible for the 3-D folding pattern. Then, quaternary structure is the arrangement of subunits and the nature of their interactions.

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12
Q
  1. A procedure that obliterates disulfide bonds between proteins, as in rebonding of hair, alters which level of protein structure?
    a. Primary
    b. Secondary
    c. Tertiary
    d. Both primary and secondary
A

c. Tertiary

Tertiary protein structures are stabilized by noncovalent and covalent interactions; and some of these are hydrophobic interactions, hydrogen bonds, salt bridges, and disulfide bonds – where the sulfhydryl groups of cysteinyl residues in a tertiary protein are bonded. Any alterations affecting these interactions in particular, like the disulfide bond, will lead to the denaturation of a tertiary protein.

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13
Q
  1. Which of the following amino acids actively participates in N-linked glycosylation?
    a. Asparagine
    b. Tyrosine
    c. Histidine
    d. Serine
A

a. Asparagine

Glycoproteins are proteins that contain oligosaccharide chains (glycans) covalently bound to amino acids. This bond is formed from a process called glycosylation, which is the enzymatic attachment of sugars, and the linkage can either be O-linked or N-linked.

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14
Q
  1. Ganglioside GM1 serves as the receptor for the cholera toxin, which is responsible for causing voluminous diarrhea in cholera. To gain entry into intestinal cells, the cholera toxin specifically binds to the
    a. Alcohol head group
    b. Monosaccharide (glucose or galactose) moiety
    c. Oligosaccharide moiety
    d. Phosphate group
A

c. Oligosaccharide

Moiety Gangliosides are glycosphingolipids, which contain sphingosine and sugar residues as well as a fatty acid. GM1, a complex ganglioside, is known to be the receptor in the human intestine for cholera toxin. Merritt et al. (1994) states that cholera toxin (CT) links to the pentasaccharide moiety of ganglioside GM1.

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15
Q

Please refer to the image. Technically, D-mannose and Dgalactose are _____ with respect to each other.

a. Diastereoisomers
b. Enantiomers
c. None of the choices
d. Constitutional Isomers

A

a. Diastereoisomers

Diastereoisomers are a type of stereoisomer that are not mirror images of each other. This is as opposed to enantiomers, which are non-superimposable mirror images, and constitutional isomers, which differ in the way the components of the molecule are attached and arranged. (Tymoczko, Berg & Stryer, 2015, pg. 168)

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16
Q
  1. When glycoproteins lose these saccharide residues from their oligosaccharide chains, they are endocytosed by the liver and thus cleared from circulation.
    a. Galactose
    b. Fucose
    c. Sialic acid
    d. Mannose e. N-acetylgalactosamine
A

c. Sialic Acid

“Loss of terminal sialic acid residues accelerates clearance of plasma glycoproteins from the circulation.”

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17
Q
  1. The following statements regarding glycosaminoglycans are correct, EXCEPT

A. They extend perpendicularly from core proteins of proteoglycans in a bottlebrush-like structure.

B. As a group, they are considered as the heteropolymeric type of polysaccharides.

C. Genetic inability of the body to break them down leads to development of mucopolysaccharidoses.

D. An example would be a repeating disaccharide unit containing neuraminic acid and galactosamine.

A

d. An example would be a repeating disaccharide unit containing neuraminic acid and galactosamine.

This choice is incorrect, as the structure of a glycosaminoglycan consists of a repeating disaccharide unit containing an amino sugar and uronic acid. Reference: Module 1A.5 Carbohydrates of Physiological Significance PDF, Page 11

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18
Q
  1. Medications such as Ibuprofen (Advil) and Mefenamic acid (Dolfenal) work by inhibiting or preventing the production of prostaglandins, eicosanoids responsible for mediating pain and inflammation. Which of the following enzymes is the most likely target of these drugs for inhibition?
    a. Cyclooxygenase
    b. Phosphodiesterase
    c. Nucleotide cyclase
    d. Lipoxygenase
A

a. Cyclooxygenase

Indomethacin and ibuprofen inhibit cyclooxygenases by competing with arachidonate. (Rodwell, Bender, Botham & Kennelly, 2018, pg. 226) Reference: Rodwell, V. W., Bender, D. A., Botham, K. M, Kennelly, P. J. & Weil, P. A. (2018). Harper’s illustrated biochemistry [31st ed.]. McGraw-Hill Education.

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19
Q
  1. Which of the following is the most appropriate nomenclature for alpha-linoleic acid? Examine the image carefully!
    a. Omega-9 PUFA
    b. 18:3 (Δ 9,12,15)all-cis
    c. 18:3 (Δ 9,12,15)all-trans
    d. Omega-3 MUFA
    e. 20:4 (Δ 1,9,12,15)all-cis
A

b. 18:3 (Δ 9,12,15)all-cis

Carbon atoms are counted from the carboxyl carbon (C-1). Succeeding carbons C-2, C-3, and C-4 are also known as ⍺, 𝛽, and 𝛾 carbons. DΔN denotes the number of double bonds (D) and position (N) from the carboxyl carbon (C-1). Trans isomers have hydrogen on opposite side; cis isomers have hydrogen bonds on the same side. Structure shows an 18-carbon fatty acid with 3 double bonds located at the 9th, 12th, and 15th carbon from C-1 all in the cis formation (18:3Δ9,12,15)-all cis. Reference: Dr. Van Haute - Lecture on Carbohydrates and Lipids of Physiologic Significance (Part 3)

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20
Q
  1. The straight-chain form of a monosaccharide derivative will be unable to cyclize if
    a. At least one of the parent monosaccharide’s hydroxyl groups is replaced by a hydrogen.
    b. When a hydroxyl group of a parent monosaccharide has been replaced with an amino group.
    c. The carbonyl group of the parent monosaccharide’s aldose or ketose is reduced to an alcohol.
    d. When there is one carboxyl group on each end of the straight-chain monosaccharide.
A

c. The carbonyl group of the parent monosaccharide’s aldose or ketose is reduced to an alcohol.

Cyclic structures are formed by the reaction between the aldehyde or ketone group with an alcohol. Loss of aldehyde or ketone by reduction prevents cyclization. (Tymoczko, Berg, and Stryer, 2015, pg. 169-170) Reference: Tymoczko, J. L., Berg, J. M., & Stryer, L. (2015). Biochemistry: A short course (p. 169-170). New York: W.H. Freeman & Company.

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21
Q
  1. Which of the following occurs during the cyclization of a linear fructose molecule?
    a. A ketal is formed following a nucleophilic attack on the hemiketal group by a hydroxyl group from another monosaccharide.
    b. Two forms that are non-superimposable mirror images of each other emerge in solution at equilibrium.
    c. A hemiacetal is formed following a nucleophilic attack on the ketone group by a hydroxyl group on the same molecule.
    d. A hemiketal is formed following a nucleophilic attack on the aldehyde group by a hydroxyl group on the same molecule.
    e. A new asymmetric carbon is designated, with two anomeric forms of the molecule emerging at equilibrium.
A

e. A new asymmetric carbon is designated, with two anomeric forms of the molecule emerging at equilibrium.

Anomers are isomers that differ at a new asymmetric carbon atom formed on ring closure. Furanose-ring of fructose has anomeric forms ⍺ and 𝛽 at the anomeric carbon C2 (see image below). These two structures are present at equilibrium. (Tymoczko, Berg, and Stryer, 2015, pg. 170) Reference: Tymoczko, J. L., Berg, J. M., & Stryer, L. (2015). Biochemistry: A short course (p. 170). New York: W.H. Freeman & Company.

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22
Q
  1. Which of the following carbohydrates would not be able to lose its electrons to oxidize another chemical species?
  • Sucrose
  • Lactose
  • Fructose
  • Galactose
  • Glucose
A

a. Sucrose

Reducing agents are capable of reducing its electrons to oxidize another chemical species. Since sucrose is a non-reducing sugar, it may not perform such function. Reference: Handout 1 Carbohydrate and Lipid Chemistry, Module 1A, Dr. Van Haute’s Lecture

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23
Q
  1. Compared with palmitic acid (16:0), oleic acid (18:1 Δ 9 cis) has a lower melting point because oleic acid has
  • a kink produced by the double bond.
  • a longer hydrocarbon chain.
  • a trans-double bond at carbon 9.
  • a more saturated chain.
A

a. a kink produced by the double bond.

Oleic acid contains a double bond, hence making it an unsaturated fatty acid. Melting point of fatty acids decreases/lowers because of this double bond, which produces a kink in the chain. Reference: Rodwell, V., Bender, D., Botham, K., Kennelly, P., & Weil, A. P. (2015b). Harper’s Illustrated Biochemistry (30th ed., p. 213). McGraw-Hill Education / Medical.

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24
Q
  1. A medication called Montelukast (brand name, Singulair) antagonizes or blocks the leukotrienes responsible for bronchoconstriction (narrowing of the airways) during asthmatic attacks. Leukotrienes are eicosanoids derived specifically from which of the following? Choose the best answer.
  • Eicosapolyenoic acid
  • Arachidonic acid
  • Ceramide
  • Phosphatidylinositol
A

b. Arachidonic Acid

Leukotrienes are derived from arachidonic acid through the lipoxygenase pathway. Reference: Rodwell, V., Bender, D., Botham, K., Kennelly, P., & Weil, A. P. (2015b). Harper’s Illustrated Biochemistry (30th ed., p. 240). McGraw-Hill Education / Medical.

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25
Q
  1. Antiphospholipid antibody syndrome (APLAS), a disorder strongly associated with recurrent miscarriages, frequently manifests as having detectable antibodies in the blood directed against
  • phosphatidylinositol.
  • phosphatidylserine.
  • phosphatidylcholine.
  • phosphatidylethanolamine.
  • diphosphatidylglycerol.
A

e. Diphosphatidylglycerol

Antibodies against diphosphatidylglycerol (cardiolipin) often develop in antiphospholipid antibody syndrome (APLAS). Reference: 2023 Biochem Ratio

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26
Q
  1. Plasmalogen and platelet activating factor specifically have which of the following distinctive components in their structures?
  • A mono- or oligosaccharide attached to sn-3 of glycerol via a glycosidic bond
  • An amino alcohol head group attached to sn-3 of glycerol via a phosphodiester bond
  • A hydrocarbon chain attached to sn-1 of glycerol via an ether bond
  • A fatty acid attached to sn-2 of glycerol via an ester bond
A

c. A hydrocarbon chain attached to sn-1 of glycerol via an ether bond

The fatty acid at the sn-1 carbon of a glycerophospholipid is replaced by an unsaturated alkyl group attached by an ether (rather than an ester) linkage to the core glycerol molecule References: Rodwell, V. W., Bender, D. A., Botham, K. M., Kennelly, P. J., & Weil, P. A. (2018). Harpers illustrated biochemistry. New York: McGraw-Hill Medical. Ferrier, D. R. (2017). Lippincott illustrated reviews: Biochemistry. Philadelphia: Wolters Kluwer.

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27
Q
  1. Lung surfactant keeps alveoli from collapsing especially during expiration (exhalation). The major component of lung surfactant, dipalmitoylphosphatidylcholine (that is, phosphatidylcholine wherein the 2 fatty acid esters are palmitic acid; see structure in image) is a
  • Sphingophospholipid
  • Glycerophospholipid
  • Ether glycerolipid
  • Glycosphingolipid
A

b. Glycerophospholipid

A glycerophospholipid consists of a glycerol backbone and an ester. In the structure, the one encircled is a glycerol backbone connecting the 2 esters and phosphatidylcholine. In DPPC, positions 1 and 2 on the glycerol are occupied by palmitate. Reference: Ferrier, D. R. (2017). Lippincott illustrated reviews: Biochemistry. Philadelphia: Wolters Kluwer.

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28
Q
  1. When two carbohydrates are epimers,
  • One is an aldose and the other is a ketone
  • They rotate plane-polarized light in the same direction
  • One is a pyranose and the other is a furanose
  • They differ in the configuration around one carbon atom only
  • They differ in length by one carbon
A

d. They differ in the configuration around one carbon atom only

Isomers that differ in configuration of the -OH and -H on carbon atoms 2, 3, and 4 of glucose are defined as epimers of each other.

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29
Q
  1. This is a key molecule needed for protein degradation
    a. Clathrin
    b. Ubiquitin
    c. Heat shock protein
    d. COPI
A

b. Ubiquitin

Lysosomal protease, a major pathway for protein degradation, does not use ATP but involves ubiquitin. Ubiquitin is a small, highly conserved protein that tags various proteins for degradation in proteasomes. The ubiquitin pathway is associated with the disposal of misfolded proteins and regulatory enzymes that have short half-lives. Polyubiquitinated target proteins enter proteasomes in the cytosol, which is composed of 4 rings with the protease active sites and 1 or 2 caps that recognize the polyubiquitinated substrates and initiate degradation. Target proteins unfold via ATPase in the cap. Target proteins are degraded to small peptides in the core and further degraded by cytosolic peptidases once out of the proteasome. Reference: Harper’s Illustrated Biochemistry 31st edition. Chapt. 49

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30
Q
  1. Alternating charged and uncharged series
    a. Nuclear importation
    b. Nuclear exportation
    c. Mitochondrial importation
    d. Peroxisomal importation
    e. ER importation
A

b. Nuclear exportation

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31
Q
  1. N-terminal signal of 5 to 10 hydrophobic amino acids
    a. Peroxisomal importation
    b. ER importation
    c. Mitochondrial importation
    d. Nuclear importation
    e. Nuclear exportation
A

b. ER importation

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32
Q
  1. Positively charged amino acids series
    a. Nuclear importation
    b. Mitochondrial importation
    c. ER importation
    d. Peroxisomal importation
    e. Nuclear exportation
A

a. Nuclear importation

In nuclear import, proteins to be imported carry a nuclear localization signal (NLS). One example of an NLS is the amino acid sequence (Pro)2(Lys)3-Arg-Lys-Val, which is markedly rich in basic residues.

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33
Q
  1. After protein synthesis, what is the correct sequence of events for exocytosis?
    a. Nucleus to Golgi Complex to ER to Plasma Membrane
    b. ER to Mitochondria to Golgi Complex to Plasma Membrane
    c. Plasma Membrane to Golgi to ER to Nucleus
    d. ER to Golgi Complex to Vesicles to Plasma Membrane
A

d. ER to Golgi Complex to Vesicles to Plasma Membrane

Proteins synthesized and sorted in the RER branch include many destined for various membranes (eg, of the ER, Golgi apparatus [GA], plasma membrane [PM]) as well as lysosomal enzymes, and also those for export from the cell via exocytosis (secretion).These various proteins may thus reside in the membranes or lumen of the ER, or follow the major transport route of intracellular proteins to the GA. In the secretory or exocytotic pathway, proteins are transported from the ER → GA→ PM and then released into the external environment. Reference: Harper’s Illustrated Biochemistry 30th ed. Chap 49 pg. 608

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34
Q
  1. 3 amino acids at carboxy terminus
    a. Peroxisomal importation
    b. Nuclear importation
    c. Mitochondrial importation
    d. ER importation
    e. Nuclear exportation
A

a. Peroxisomal importation

Reference: Doc Raymond ppt

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35
Q
  1. Alternating positively charged amino acids with hydrophobic amino acids
    a. Mitochondrial importation
    b. ER importation
    c. Nuclear importation
    d. Nuclear exportation
    e. Peroxisomal importation
A

a. Mitochondrial importation

Reference: Doc Raymond ppt

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36
Q
  1. Which of the following vesicles is responsible for anterograde protein transport?
    a. V Snares
    b. COPI
    c. T Snares
    d. COPII
A

d. COPII

A number of proteins possess the amino acid sequence KDEL (Lys-Asp-Glu-Leu) at their carboxyl terminal KDEL-containing proteins first travel to the GA in vesicles coated with coat protein II (COPII). This process is known as anterograde vesicular transport. In the GA they interact with a specific KDEL receptor protein, which retains them transiently. They then return to the ER in vesicles coated with COPI (retrograde vesicular transport), where they dissociate from the receptor, and are thus retrieved. HDEL sequences (H = histidine) serve a similar purpose. Reference: Harper’s Illustrated Biochemistry 30th ed. Chap 49 pg. 621

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37
Q
  1. Sphingomyelin is a molecule identified in a plasma membrane. Which of the following describes its location in that membrane?
    a. Intercalated between phospholipids
    b. The outer membrane leaflet
    c. Extending into the environment
    d. In a transmembrane arrangement
A

b. The outer membrane leaflet

Sphingomyelin are found in the outer leaflet of the cell membrane and are particularly abundant in specialized areas of the plasma membrane known as lipid rafts. Reference: Harper’s Illustrated Biochemistry 31st ed. Chap 21 pg. 200

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38
Q
  1. A plasma membrane is observed to have caveolae. These structures are:
    a. Composed of disorderly phospholipids
    b. Cholesterol-enriched membrane invaginations
    c. Intercalated between cholesterol molecules
    d. Components of phospholipids
A

b. Cholesterol-enriched membrane invaginations

Lipid rafts are specialized areas of the outer leaflet of the lipid bilayer enriched in cholesterol, sphingolipids, and certain proteins. Caveolae may derive from said lipid rafts. Reference: Harper’s Illustrated Biochemistry 31st ed. Chap 40 pg. 466

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39
Q
  1. A ligand receptor is identified in a plasma membrane of a living cell. The arrangement of that ligand receptor within the membrane is best described as a/an:
    a. Lipid-anchored protein
    b. Glycolipid
    c. Integral membrane protein
    d. Peripheral protein
A

c. Integral membrane protein

Most membrane proteins fall in the integral class, meaning that they interact extensively with the phospholipids and require the use of detergents for their solubilization. Reference: Harper’s Illustrated Biochemistry 31st ed. Chap 40 pg. 464-465

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40
Q
  1. The degradation of improperly folded Endoplasmic Reticulum proteins occurs in the:
    a. Mitochondria
    b. Lysosomal compartment
    c. ER lumen
    d. Proteasome in the cytoplasm
A

d. Proteasome in the cytoplasm

Improperly folded ER proteins are exported and degraded in the cytosol. Dr. Raymond Cruz’ ppt. Slide 51

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41
Q
  1. Which of the following statements regarding membranes is FALSE?
    a. Proteins are the most abundant type of macromolecules within cell membranes
    b. The model that best describes the plasma membrane is the fluid mosaic model
    c. Membrane leaflets have an asymmetric distribution
    d. Plasma membranes are selectively permeable structures of eukaryotic cells
A

a. Proteins are the most abundant type of macromolecules within cell membranes

The basic structure of all membranes is the lipid bilayer. This bilayer is formed by two sheets of phospholipids in which the hydrophilic polar head groups are directed away from each other and are exposed to the aqueous environment on the outer and inner surfaces of the membrane. Reference: Harper’s Illustrated Biochemistry 30th ed. Chap 40 pg. 497

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42
Q
  1. Lactate dehydrogenase has five isoenzymes. The following statements regarding these isoenzymes of LDH are true EXCEPT:
    a. They produce different migration patterns during electrophoresis
    b. They are made up of different subunits
    c. They are characteristically produced by different types of tissues
    d. They bind to the same substrate but produce different products
A

d. They bind to the same substrate but produce different products

Lactate dehydrogenase is comprised of two domains, an N-terminal NAD+-binding domain and a C-terminal binding domain for the second substrate, pyruvate. Reference: Harper’s Illustrated Biochemistry 30th ed. Chap 5 pg. 40

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43
Q
  1. Which enzyme would have the highest affinity for its substrate?
    a. Enzyme X, Km = 5.5
    b. Enzyme Y, Km = 2.25
    c. Enzyme W, Km = 3.7
    d. Enzyme Z, Km = 50
A

b. Enzyme Y, Km = 2.25

A numerically small (low) Km reflects a high affinity of the enzyme for substrate, because a low concentration of sub strate is needed to half saturate the enzyme—that is, to reach a velocity that is ½ Vmax. Reference: Lippincott’s Illustrated Reviews Biochemistry 6th ed. Chap 5 pg. 59

44
Q
  1. Aldolase, a glycolytic enzyme, catalyzes the cleavage of fructose 1,6 bisphosphate by initially forming a Schiff base from the reaction by a lysine residue in the enzyme and the carbonyl carbon of the substrate. Identify the mechanism of action.
    a. Desolvation effect
    b. Strain effect
    c. Proximity and orientation effect
    d. General acid/base catalysis
    e. Covalent catalysis
A

e. Covalent catalysis

The process of covalent catalysis involves the formation of a covalent bond between the enzyme and one or more substrates. Covalent catalysis is particularly common among enzymes that catalyze group transfer reactions. Residues on the enzyme that participate in covalent catalysis generally are cysteine or serine, and occasionally histidine Reference: Harper’s Illustrated Biochemistry 30th ed. Chap 7 pg. 63

45
Q
  1. Identify the type of feedback inhibition involved in this sequence if E acts as an inhibitor of the conversion of A to B: A → B → C → D → E
    a. Cooperative feedback inhibition
    b. Multivalent feedback inhibition
    c. Simple feedback inhibition
    d. Sequential inhibition
A

c. Simple feed-back inhibition

Types of Feed-back inhibition :

  1. Simple feed-back inhibition. The final product (E) inhibits the step from A to B.
  2. Co-operative feed-back inhibition. Both final products (D, E) inhibit the first step of their own synthesis together.
  3. Multivalent feed-back inhibition.
  4. Inhibition at a ramification of a biosynthesis pathway (sequential inhibition)
46
Q
  1. Medications such as Ibuprofen (Advil) and Mefenamic acid (Dolfenal) work by inhibiting or preventing the production of prostaglandins, eicosanoids responsible for mediating pain and inflammation. Which of the following enzymes is the most likely target of these drugs for inhibition?
    a. Lipooxygenase
    b. Cyclooxygenase
    c. Phosphodiesterase
    d. Nucleotide cyclase
A

b. Cyclooxygenase

Prostanoid synthesis involves the consumption of two molecules of O2 catalyzed by cyclooxygenase (COX) (also called prostaglandin H synthase), an enzyme that has two activities, a cyclooxygenase and peroxidase. COX is present as two isoenzymes, COX-1 and COX-2. The product, an endoperoxide (PGH), is converted to prostaglandins D and E as well as to a thromboxane (TXA2) and prostacyclin (PGI2). Each cell type produces only one type of prostanoid. The NSAID aspirin inhibits COX-1 and COX-2. Other NSAIDs include indomethacin and ibuprofen, and usually inhibit cyclooxygenases by competing with arachidonate. Since inhibition of COX-1 causes the stomach irritation often associated with taking NSAIDs, attempts have been made to develop drugs which selectively inhibit COX-2 (coxibs).

47
Q
  1. Identify the type of inhibitor
    a. Non-competitive
    b. All of the choices
    c. Competitive
    d. Uncompetitive
A

c. Competitive

Competitive - inhibitor has no effect on Vmax but raises K′m

48
Q
  1. Which of the following is true regarding the binding of a substrate to the enzyme’s active site prior to a reaction?
    a. There is an associated formation of weak attractive bonds between the enzyme and substrate
    b. The reaction results in a permanent change in the enzyme
    c. The enzyme is stabilized by the formation of peptide bonds
    d. There is a joining together of non-complementary regions between the active site and the allosteric site of the enzyme
A

a. There is an associated formation of weak attractive bonds between the enzyme and substrate

Enzymes lower the activation energy by utilizing the free energy called binding energy. Binding energy is the free energy released by the formation of a large number of weak interactions between a complementary enzyme and substrate. Reference: Biochemistry: A Short Course 3rd ed. Chap 6.4 pg. 105 Emil Fischer proposed that enzymes and their substrates interact to form an enzyme-substrate (ES) complex Reference: Harper’s Illustrated Biochemistry 30th ed. Chap 7 pg. 62

49
Q
  1. Which of the following is true of covalent modification of enzymes?
    a. It is a temperature-dependent process
    b. It usually uses small molecules such as CH3 for activation and inactivation
    c. It is irreversible
    d. Proteolytic cleavage is a common covalent modification
A

b.It usually uses small molecules such as CH3 for activation and inactivation

COVALENT MODIFICATION

  • The covalent attachment of another molecule can modify the activity of enzymes and many other proteins
  • In these instances, a donor molecule provides a functional moiety that modifies the properties of the enzyme
  • Most modifications are reversible, unlike proteolytic reactions
  • Phosphorylation and dephosphorylation are the most common methods of covalent modification (sulfation, acetylation, methylation are less common)
50
Q
  1. The removal of a hydrogen ion from water by one aspartic acid residue in the active site of HIV protease makes water more nucleophilic; the donation of another hydrogen ion by another aspartic acid residue facilitates the breakdown of the tetrahedral intermediate. Identify the mechanism of action.
    a. Covalent catalysis
    b. General acid/base catalysis
    c. Proximity and orientation effect
    d. Desolvation effect
    e. Strain effect
A

b. General acid/base catalysis

51
Q
  1. Identify the enzyme class.
    a. Oxidoreductase
    b. Isomerase
    c. Hydrolase
    d. Lyase
A

b. Isomerase

Isomerase, a class of enzymes that catalyze reactions involving a structural rearrangement of a molecule. This step in glycolysis is catalyzed by triose phosphate isomerase. Isomerases catalyze isomerization reactions (geometric/structural changes). This step completes the first phase of glycolysis. Reference: Cruz, R. (n.d.).Powerpoint presentation: Enzymes.

52
Q
  1. Which of the following statements regarding the allosteric regulation of enzymes is true?
    a. Allosteric modulators bind irreversibly at the active site and induce conformational changes in the enzyme
    b. Cellular response is faster with allosteric control than by controlling enzyme concentration in the cell
    c. Allosteric regulation is primarily dependent on an adequate supply of protons for the creation of covalent linkages
    d. It is usually the mode of regulation for the last step in reaction pathways since this step produces the final product
A

b. Cellular response is faster with allosteric control than by controlling enzyme concentration in the cell

ALLOSTERIC ENZYMES

  • Bind small, physiologically important molecules to modulate enzymes
  • Usually contain multiple subunits and frequently catalyze the committed step early in a pathway
  • The small regulatory molecules to which they bind are known as effectors
  • The hallmark of effectors is that when they bind to enzymes, they alter the catalytic properties of an enzyme’s active site
  • Those that increase catalytic activity are known as positive effectors (ACTIVATORS)
  • Effectors that reduce or inhibit catalytic activity are negative effectors

Allosteric modulators do not bind to the active site.

Allosteric regulation is dependent on the binding of allosteric activator/inhibitor on the enzyme which causes a conformational change of the active site.

Allosteric regulators usually regulate the first step in the reaction pathway and not the last step.

53
Q
  1. The binding of the substrate to the active site of the enzyme holds the reacting groups in close contact and in the proper alignment for reactions to occur. Identify the mechanism of action.
    a. Covalent catalysis
    b. Proximity and orientation effect
    c. Desolvation effect
    d. Strain effect
    e. General acid/base catalysis
A

b. Proximity and orientation effect

“Proximity: Reaction between bound molecules doesn’t require an improbable collision of 2 molecules. They’re already in “contact” (increases local concentration of reactants).”

“Orientation: Reactants are not only near each other on enzyme, they’re oriented in optimal position to react. The improbability of colliding in correct orientation is taken care of.” Reference: Cruz, R. (n.d.).Powerpoint presentation: Enzymes.

54
Q
  1. An enzyme increases the rate of a reaction by lowering:
    a. the activation energy needed to bring the reactants to the transition state
    b. All of the choices
    c. the initial energy of the reactants in the chemical reaction
    d. the standard free energy change between the reactants and the products
A

a. the activation energy needed to bring the reactants to the transition state

“When the activation energy is lower, more molecules have the energy required to reach the transition state and more product will be formed faster… The essence of catalysis is stabilization of the transition state.”

55
Q
  1. Which enzyme is used to assess liver function?
    a. Transmethylase b.

Transacetylase

c. Transketolase
d. Transaminase

A

d. Transaminase

Aminotransferases, also called transaminases, serve a diagnostic function. “The presence of alanine and aspartate aminotransferase in the blood indicates liver damage. Liver damage can occur for a number of reasons, including viral hepatitis, long-term excessive alcohol consumption and reaction to drugs such as acetaminophen. Under these conditions, liver cell membranes are damaged and some cellular proteins, including the aminotransferases, leak into the blood.”

Normal blood value for alanine and aspartate aminotransferase activity: 5-30 units/l and 40-125 units/l (value can reach 200-300 units/l depending on the extent of liver damage.

56
Q
  1. When a substrate binds to the hydrophobic environment of the active site of the enzyme, the shell of water molecules surrounding the substrate dissociates, rendering the reacting groups available. Identify the mechanism of action.
    a. Covalent catalysis
    b. Strain effect
    c. Desolvation effect
    d. General acid/base catalysis
    e. Proximity and orientation effect
A

c. Desolvation effect

“The close association between the active site and the substrate means that water is usually excluded from the active site unless it is a reactant. The nonpolar microenvironment of the cleft enhances the binding of substrates as well as catalysis.”

57
Q
  1. A newly discovered drug, called Super X, was formulated. It is said to have a similar structure as glucose, and it competes with glucose for the active site of hexokinase, a rate-limiting enzyme for glucose breakdown. Based on this, Super X probably exerts its effect via what mechanism?
    a. Proteolytic enzyme cleavage
    b. Active site inhibition
    c. Allosteric inhibition
    d. Gene regulation
A

b. Active site inhibition

Active site- recognition site for binding substrates; provides the environment wherein chemical transformation takes place.

Competitive inhibitors- binds to the active site thereby blocking access by the substrate

58
Q
  1. The binding of the carbohydrate chain to the active site of lysozyme changes the conformation of the sugar from a chair to a less stable boat conformation making it easier to cleave. Identify the mechanism of action.
    a. Desolvation effect
    b. Proximity and orientation effect
    c. Strain effect
    d. Covalent catalysis
    e. General acid/base catalysis
A

c. Strain effect

Enzymes that catalyze lytic reactions, chemical transformations that involve breaking a covalent bond, typically bind their substrates in a conformation that is somewhat unfavorable for the bond targeted for cleavage. This strained conformation mimics that of transition state intermediate. T_he resulting strain selectively distorts or stretches the targeted bond, weakening it and making it more vulnerable to cleavage_

59
Q
  1. Arrange the following events in their correct sequence to trigger synaptic transmission from a presynaptic neuron to a postsynaptic cell.
  2. Depolarization of the membrane
  3. Travelling action potential
  4. Neurotransmitter release
  5. Increased calcium influx

a. 4, 2, 1, 3
b. 2, 1, 4, 3
c. 1, 2, 3, 4
d. 2, 3, 1, 4

A

b. 2, 1, 4, 3

Action potential produces depolarization then the excitatory transmitter opens Ca2+ channels causing the neurotransmitter release Reference:Ganong’s Review of Medical Physiology 26th ed. Chap 6 pg. 124-126

60
Q
  1. Which of the following statements regarding collagen hydroxylation is correct?
    a. The hydroxylation of collagen occurs simultaneously with the translation of the mRNA template
    b. All of the choices
    c. All proline and lysine residues of collagen are hydroxylated
    d. Vitamin C is an essential cofactor in the hydroxylation process
A

d. Vitamin C is an essential cofactor in the hydroxylation process

Hydroxyproline is formed by the posttranslational hydroxylation of peptide-bound proline residues catalyzed by the enzyme prolyl hydroxylase, whose cofactors are ascorbic acid (vitamin C) and α-ketoglutarate. Reference: Harper’s Illustrated Biochemistry 31st Ed. page 593 / Dr. Cruz ppt

61
Q
  1. Which of the following is a correct enzyme: cofactor pair?
    a. Lysyl oxidase: Cuprous ion
    b. Prolyl hydroxylase: Pyridoxal phosphate
    c. Aspartate aminotransferase: Thiamine pyrophosphate
    d. Alanine aminotransferase: Biotin
A

a. Lysyl Oxidase: Cuprous Ion

  • Transaminase enzymes require pyridoxal phosphate
  • Prolyl hydroxylase requires Ascorbic Acid (Vitamin C), NOT pyridoxal phosphate
  • Lysyl Hydroxylase requires Ascorbic Acid (Vitamin C) and Cuprous Ion
62
Q
  1. Which of the following is not a manifestation of an abnormality in collagen biosynthesis?
    a. Poor wound healing
    b. Skin hyperextensibility
    c. Hematuria
    d. Hyperammonemia
A

d. Hyperammonemia

Abnormalities in Collagen Synthesis:

  • Ehlers-Danlos syndrome - hyperextensibilty of skin, abnormal tissue fragility, increased joint mobility
  • Alport syndrome - Presenting sign is hematuria and patient may develop End-stage renal disease
  • Scurvy-Bleeding gums, subcutaneous hemorrhage, poor wound healing
  • Hyperammonemia is not an abnormality in collagen biosynthesis.
63
Q
  1. A frameshift mutation
    a. is the result of an unequal crossing-over
    b. results from an insertion or deletion in a DNA sequence.
    c. substitutes a pyrimidine for a purine, or vice versa.
    d. substitutes a purine for a pyrimidine only e. leaves the amino acid sequence unchanged.
A

b. Results from an insertion or deletion in a DNA sequence

Frameshift Mutation - insertions and deletions that do not involve three nucleotides or multiples of three nucleotides change the translation of all the mRNA downstream of the mutation

64
Q
  1. All of the following statements are true of collagen EXCEPT:
    a. It has glycine residues at every third position of the alpha chain
    b. It has many different genetic types
    c. It undergoes extensive post translational modification
    d. It has intramolecular desmosine cross-links in its structure
A

d. It has intramolecular desmosine cross-links in its structure

65
Q
  1. Marfan syndrome is due to mutations in the gene for:
    a. Laminin
    b. Fibrillin
    c. Estrogen
    d. Collagen
A

b. Fibrillin

  • Most cases of Marfan’s syndrome (an autosomal dominant disease), are caused by mutations on chromosome 15 for Fibrillin
  • Marfan’s syndrome patients have lens dislocation (ectopia lentis), long digits (arachnodactylyl) with hyperextensible joint, and weakness of the aortic media leading to dilation of ascending aorta

Laminin- In muscular dystrophy, Laminin 2 is deficient; causing loss of function resulting in muscle weakness and atrophy

Estrogen- Inhibits Osteoclasts formation

Collagen:

  • Epidermolysis Bullosa → Collagen Type VII
  • Alport Syndrome → Collagen Type IV, found in basement membrane of renal glomeruli; Presenting sign: hematuria and may develop into end stage renal disease
  • Ehler’s Danlos → Collagen Type IV (most serious; spontaneous artery rupture on bowel, reflecting abnormal Collagen Type III); Type VI (due to deficiency in lysyl hydroxylase)
66
Q
  1. True of bone tissue:
    a. Osteoclasts synthesize most of the proteins found in bone
    b. The inorganic component is mainly hydroxyapatite
    c. The major protein is Type IV Collagen
    d. The sole cation that can be incorporated in bone tissue is calcium
A

b. The inorganic component is mainly hydroxyapatite

67
Q
  1. What is the most abundant amino acid seen in mature collagen fibers?
    a. Hydroxyproline
    b. Glycine
    c. Hydroxylysine
    d. Proline
A

b. Glycine

The side chain of glycine, an H atom, is the only one that can fit into the crowded center of a three-stranded helix. Hydrogen bonds linking the peptide bond NH of a glycine residue with a peptide carbonyl (C═O) group in an adjacent polypeptide help hold the three chains together

68
Q
  1. Which of the following statements is not true of glycosaminoglycans?
    a. They are long, branched homopolysaccharides attached to amino acids
    b. They are highly viscous molecules
    c. They contain an amino sugar and a uronic acid residue
    d. They are highly negatively charged molecules
A

a. They are long, branched, homopolysaccharides attached to Amino Acids

Glycosaminoglycans

  • Most abundant heteropolysaccharide in the body
  • Has different functions
  • Made of
    • Complex amino sugars and uronic acids
    • Sulfate groups (except for Hyaluronic Acid)
    • Unbranched polysaccharides made of disaccharide units contain either of the two modified sugars– N-acetylgalactosamine (GalNac) or N-acetylglucosamine (GlcNac)
    • Negatively charged
    • High viscosity to solution
    • Ideal for lubricating fluids, structural integrity to cells, and cell migration passageways
69
Q
  1. The strength of collagen is due to the presence of aldol cross links within the structure. Where are mature, cross-linked, collagen fibers synthesized?
    a. In the cis portion of the Golgi complex
    b. Inside the nucleus
    c. Extracellularly
    d. In the rough endoplasmic reticulum
A

c. Extracellularly

After processing and assembly of type I procollagen is completed, it is secreted into the extracellular space. During or following exocytosis, extracellular enzymes, the procollagen peptidases, remove the N-terminal and C-terminal propeptides.

70
Q
  1. Glycosaminoglycans (GAG) are also known as mucopolysaccharides. What is the primary defect seen in the group of diseases generally referred to as mucopolysaccharidoses (MPS)?
    a. A defect in the Golgi complex
    b. Accumulation of GAG in the cell
    c. A mutation in the DNA for GAGs
    d. Impaired GAG synthesis
A

b. Accumulation of GAG in the cell

The mucopolysaccharidoses (MPSs) are a family of inherited disorders of GAG degradation. (Rodaks, p.481)

They are due to lysosomal storage abnormalities, which allow the proliferation and non-degradation of GAG in cells. There is no synthesis abnormality involved because the problem is in degradation. No DNA mutation or Gogli complex defect is involved.

71
Q
  1. All information in the DNA codes for protein
    a. True
    b. False
A

b. False

The entire haploid genome contains sufficient DNA to code for nearly 1.5 million average-sized protein coding genes (i.e., ~2200 bp of protein- coding DNA). However, early studies of mutation rates and of the complexities of the genomes of higher organisms suggested that humans have significantly fewer than 100,000 proteins encoded by the ~1% of the human genome that is composed of exonic DNA.

This implies that most genomic DNA is non protein coding—that is, its information is never translated into an amino acid sequence of a protein molecule.

72
Q
  1. A 40-year-old female presents with a low grade fever, nonproductive cough and was diagnosed with pneumonia. She was advised to take erythromycin. Which of the following is the mechanism of action of erythromycin?
    a. inhibits the initiation factor 1
    b. inhibits the 50s ribosomal subunit
    c. binds to the Shine Dalgarno sequence
    d. inhibits the incoming aminoacyl tRNA
A

b. Inhibits the 50s ribosomal subunit

73
Q
  1. There is only one kind of amino acyl tRNA synthetase
    a. True
    b. False
A

b. False

Each codon represents a particular amino acid, and each codon (20) is recognized by a specific tRNA

74
Q
  1. A 4 month old infant with rhinitis and coughing was tested positive for Bordella pertussis. He was given an antibiotic that inhibits translocation of peptidyl-tRNA on 70s ribosomes. This patient is most likely treated with
    a. erythromycin
    b. tetracyclin
    c. streptomycin
    d. lincomycin
A

a. Erythromycin

Erythromycin is an antibiotic that inhibits protein synthesis by inhibiting translocation during translation. It specifically acts on the 50s subunit of the ribosome.

Tetracyclin inhibits aminoacyl tRNA binding on the A-site

Streptomycin inhibits codon-anticodon initiation and elongation

Lincomycin inhibits peptidyl transferase

75
Q
  1. A 13-year-old teenager has been advised to start with tetracycline for acne treatment. Which of the following best describes the mechanism of action of this antibiotic?
    a. inhibits translocation
    b. inactivates peptidyl transferase
    c. inhibits prokaryotic peptidyl transferase
    d. inactivation of the aminoacyl tRNA binding site
A

d. inactivation of the aminoacyl tRNA binding site

Tetracycline is a broad spectrum polyketide antibiotic produced by the Streptomyces genus of Actinobacteria. It exerts a bacteriostatic effect on bacteria by binding reversible to the bacterial 30S ribosomal subunit and blocking incoming aminoacyl tRNA from binding to the ribosome acceptor site.

76
Q
  1. Which proofreading activity is critical in determining the accuracy of nuclear DNA replication and thus the base substitution rate in human chromosomes?
    a. 5’ to 3’ exonuclease activity of DNA Pol epsilon
    b. 3’ to 5’ endonuclease activity of DNA Pol epsilon
    c. 5’ to 3’ polymerase activity of DNA Pol delta
    d. 3’ to 5’ exonuclease activity of DNA Pol delta
A

d. 3’ to 5’ exonuclease activity of DNA Pol delta

5’ to 3’ Polymerase – ability to add nucleotides

3’ to 5’ Exonuclease – proofreading capacity

5’ to 3’ Exonuclease – removes primers

77
Q
  1. In Fragile X Syndrome, the FMR1 gene is silenced by methylation at the
    a. DnaA sequence
    b. CpG on the promoter site of the gene
    c. CpG located within the gene
    d. TATA box of the promoter sequence
    e. CCAAT sequence
A

c. CpG on the promoter site of the gene

FMR protein is needed for normal brain development. Full mutation causes the FMR1 gene to shut down or silence. Silencing of the FMR1 gene means no production of the FMR protein. Non production of the FMR protein is due to the methylation of the cytosine bases on the CGG repeat located on the CpG island proximal to the promoter region of the gene. The presence of numerous methyl groups on the CGG repeat will inhibit the formation of the transcriptional apparatus on the promoter site.

Reference: Prof. Torres handout re Fragile X Syndrome

78
Q
  1. A double stranded DNA genome isolated from a virus in the stool of a child with gastroenteritis was found to contain 20% thymine. What is the percentage of cytosine in this genome?
    a. 20
    b. 30
    c. 40
    d. 60
A

b. 30

In DNA molecules, the concentration of deoxyadenosine (A) nucleotides equals that of Thymidine (T) nucleotides (A=T), while the concentration of deoxyguanosine (G) nucleotides equals that of deoxycytidine (C)nucleotides (G=C) Chargaff’s Rule (A+G=C+T) In this case, 20% Tymine=20% Adenine → 40% 100%-40%= 60% (C-G) 60%/2= 30% 30% Cytosine 30% Guanine

Reference: 2023 Trans on Nucleic Acid (Dr. Sheila Torres)

79
Q
  1. Termination codon has no tRNA
    a. True
    b. False
A

a. True

Termination occurs when the A site reaches a stop codon. There are no anticodon complementary to stop codon present in tRNA, the ribosome ”pauses” until it “falls off” the mRNA and the polypeptide chain terminates. This process is facilitated by a release factor protein that binds into the ribosomal A site containing a stop codon to help with protein release.

80
Q
  1. An isolated primary RNA transcript has the following architecture. Which part contains the poly A tail?
    a. 3’UTR
    b. coding region
    c. intron
    d. 5’UTR
A

a. 3’UTR

Messenger RNA

  • Composed of the 5’UTR, 3’UTR, and coding region
  • The 5’UTR is composed of:
  • Shine Dalgarno sequence (prokaryotic)
  • Kozac Consensus (eukaryotic) - G cap (eukaryotic)
  • 3 U’TR contains the poly A tail

Reference: Prof. Shiela’s discussion on DNA and Protein synthesis

81
Q
  1. The extracellular domain of fibronectin bind to components of the extracellular matrix through recognition of this group of 3 amino acids:
    a. Lysine, Proline, Glutamic acid
    b. Valine, Leucine, Isoleucine
    c. Alanine, Glutamine, Asparagine
    d. Arginine, Glycine, Aspartic acid
A

d. Arginine, Glycine, Aspartic acid

Fibronectin contains an Arg-Gly-Asp (RGD) sequence that binds to the integrin receptor. This sequence is shared by a number of other proteins present in the ECM that bind to integrins present in cell plasma membranes, and its presence in synthetic peptides enables them to inhibit the binding of fibronectin to cells Reference: Harper’s Illustrated Biochemistry 31st Ed. page 597

82
Q
  1. A 25-year-old female presents with high grade fever, headache, loss of appetite and abdominal pain. She has been diagnosed with typhoid fever, for which she has been started with Chloramphenicol. This drug inhibits prokaryotic:
    a. aminoacyl tRNA synthetase
    b. elongation factor
    c. translocase
    d. peptidyl transferase
A

d. peptidyl transferase

The mode of action of chloramphenicol is in the ribosome specifically the prokaryotic peptidyl transferase. Chloramphenicol is an antibiotic. Chloramphenicol is effective against a wide variety of microorganisms, but due to serious side-effects (e.g., damage to the bone marrow, including aplastic anemia) in humans, it is usually reserved for the treatment of serious and life-threatening infections (e.g., typhoid fever). Chloramphenicol is bacteriostatic but may be bactericidal in high concentrations or when used against highly susceptible organisms.

83
Q
  1. A mutation that replaces pyrimidine with purine
    a. transversion
    b. frameshift
    c. transition
    d. silent
A

a. transversion

Single-base changes (point mutations) may be transitions or transversions. In the former, a given pyrimidine is changed to the other pyrimidine or a given purine is changed to the other purine. Transversions are changes from a purine to either of the two pyrimidines or the change of a pyrimidine into either of the two purines, as shown in Figure 37–3.

84
Q
  1. Which of the following regulators are said to act in “trans”?
    a. the lac operator and the mammalian transcription factors
    b. the lac repressor and the mammalian enhancers
    c. the lac transcription factors and the mammalian activators
    d. the tryp operon co-repressor and the lac operator
A

c. the lac transcription factors and the mammalian activators

Regulatory elements Classified into two types, based on their “location”

  • Cis-acting elements - DNA sequences that exert their effect only on nearby genes Example: TATA box, enhancers and silencers ν
  • Trans-acting elements - Regulatory proteins that bind to such DNA sequences Example: Reference: Gene Expression Compendium by Sheila Maria P. Torres M.S.
85
Q
  1. The chemiosmotic theory proposes a mechanism for the tight coupling of electron transport via the respiratory chain to the process of oxidative phosphorylation. Which of the following options is NOT predicted by the theory?
    a. The electrochemical potential difference across the inner mitochondrial membrane caused by electron transport is positive on the matrix side.
    b. Protons are pumped across the inner mitochondrial membrane as electrons pass down the respiratory chain.
    c. ATP synthesis occurs when the electrochemical potential difference across the membrane is discharged by translocation of protons back across the inner mitochondrial membrane through an ATP synthase enzyme.
    d. A proton gradient across the inner mitochondrial membrane generated by electron transport drives ATP synthesis.
    e. An increase in the permeability of the inner mitochondrial membrane to protons uncouples the processes of electron transport and oxidative phosphorylation.
A

a. The electrochemical potential difference across the inner mitochondrial membrane caused by electron transport is positive on the matrix side.

The chemiosmotic theory, proposed by Peter Mitchell in 1961, postulates that the two processes are coupled by a proton gradient across the inner mitochondrial membrane so that the proton motive force caused by the electrochemical potential difference (negative on the matrix side) drives the mechanism of ATP synthesis. Reference: Rodwell, V. et. al. Harper’s Illustrated Biochemistry. (2018) Section III, Chapter 13, p. 121.

86
Q
  1. Which of the following statements describes how glucagon stimulates transcription of PEPCK gene
    a. binding of the glucagon by cell surface receptors initiates a chain of events that results in the phosphorylation of CREB protein
    b. glucagon forms a complex with adenylate cyclase resulting in the synthesis of cAMP and activation of protein kinase
    c. glucagon forms a complex with glucocorticoid response elements associated with PEPCK
    d. binding of glucagon to the CRE that activates PEPCK gene
A

a. binding of the glucagon by cell surface receptors initiates a chain of events that results in the phosphorylation of CREB protein

87
Q
  1. Complex IV of the electron transport chain catalyzes the transfer of reducing equivalents from cytochrome c to molecular oxygen, resulting in the formation of water. Complex IV is thus a/an
    a. oxygenase
    b. oxidase
    c. hydroperoxidase
    d. dehydrogenase
A

b. oxidase.

Oxidase removes hydrogen from a substrate and transfers it to oxygen to form water or hydrogen peroxide Reference: Page 4/8 of the handout on Biological Oxidation provided by Dr. Van Haute

88
Q
  1. What is the role of allolactose to Lac Operon
    a. cleaves glucose
    b. inducer
    c. activates catabolite repressor protein
    d. repressor
A

b. inducer

The allolactose serves as the inducer and binds to the receptor protein which in turn cannot bind to the O site due to conformational change in its structure

Reference:. Doc Shiela’s Compendium on Gene Regulation and Mutation

89
Q
  1. All of the following elements are trans acting regulators except one. Which one is the exception?
    a. Initiator Element (Inr)
    b. TATA Binding Protein (TBP)
    c. Transcription Factor II D gene (TFIID)
    d. Transcription factor II F (TFIIF)
A

a. Initiator Element (Inr)

Trans-acting regulators are regulatory proteins that bind to such DNA sequences Ex: Transcription factors and other initiating factors (TBP)

Cis-acting elements are DNA sequences that exert their effect only on nearby genes Ex: TATA box, enhancers, silencers Reference: Doc Shiela’s Compendium on Gene Expression

90
Q
  1. Consider a biochemical reaction that started with equal amounts of substrate and product, and was allowed to reach equilibrium. If the value of the equilibrium constant (Keq) is 5.1, this means that
    a. ΔG0’ is positive
    b. ΔG0’ is zero
    c. –RT(ln Keq) is negative
    d. ln Keq is negative
A

c. –RT(ln Keq) is negative

Reference: Doc Van Haute’s Handout on Bioenergetics

91
Q
  1. The proliferation of cytotoxic T-cells is impaired upon infection with HIV-V. The defect has been traced to the expression of the viral-encoded enzyme that inactivates a host cell nuclear protein required for DNA replication. Which protein is a potential substrate for the viral enzyme?
    a. Catabolite activator protein
    b. TATA binding protein
    c. Single Strand Binding Protein
    d. Cap Binding Protein
A

c. Single Strand Binding Protein

Explanation# 1:

  • Prokaryotic Enzymes of Replication The single-stranded DNA that result from the helicase action is coated with single-stranded DNA-binding protein that prevents reannealing of parental strands and prevent hairpins and other super secondary structures from forming in the sing stranded DNA

Explanation# 2:

  • SSBs bind ssDNA with high affinity and in a sequence-independent manner and, in doing so, SSbs help to form central nucleoprotein complex substrates for DNA replication, recombination, and repair processes.
92
Q
  1. Which of the following types of mutation describes sickle cell anemia
    a. transversion, nonsense
    b. transition, missense
    c. transversion, missense
    d. transition, nonsense
A

c. Transversion, Missense

Purine- Adenine, Guanine (A & G)

Pyrimidines- Cytosine, Thymine (C & T)

Point Mutation Types:

  1. Transitions- interchanges of purines (A G) or pyrimidines (C T).
  2. Transversions- interchanges of purine for pyrimidine bases or vice versa.
    • Ex. (A C), (A T), (G C), (G T)

Substitution Mutation Types:

  1. Silent mutations have no effect on the protein.
  2. Missense mutations result in a single amino acid change in the translated sequence.
  3. Nonsense mutations result in an amino acid codon being replaced by a “stop” codon. Nonsense mutations end translation prematurely and result in a truncated protein.
93
Q
  1. MELAS Syndrome (Mitochondrial Encephalopathy, Lactic Acidosis, and Stroke-like episodes) is an inherited condition caused by mitochondrial DNA or tRNA mutations that result in electron transport chain complex deficiency. One form of such disease results in mitochondrial Complex IV deficiency. For patients with this type of the disease, what would be the net ATP produced for one molecule of NADH as an electron carrier?
    a. 0
    b. 1
    c. 4
    d. 2
A

d. 2

Complex 4 transfers 2 hydrogen ions to the interstitial space. Since it is inhibited and then instead of 10 hydrogen ions only 8 hydrogen ions will be formed. Since 4 hydrogen ions are required for 1 ATP synthesis. Therefore 2 ATP will be produced Reference: handout in Oxidative phosphorylation provided by Dr. Van Haute

94
Q
  1. A worker in a farm was tasked to sweep up a spill of what looked like white powder in one of their storage sheds. He was later found to be in respiratory distress, with eventual seizures occurring. On analysis, the powder was identified as rotenone. Given this information, which of the following is not expected to happen?
    a. Coenzyme Q would still be able to transfer electrons, albeit via an alternate route.

b .ATP synthase, though left uninhibited, will be producing lesser amounts of ATP.

c. Only NADH is rendered useless, but FADH2 can still work as an electron carrier.
d. Both NADH and FADH2 would not be rendered useless as electron carriers.

A

d. Both NADH and FADH2 would not be rendered useless as electron carriers.

With rotenone intoxication, the ETC is expected to:

  • Coenzyme Q would still be able to transfer electrons, albeit via an alternate route.
  • ATP synthase, though left uninhibited, will be producing lesser amounts of ATP.
  • Only NADH is rendered useless, but FADH2 can still work as an electron carrier.

Except: Both NADH and FADH2 would not be rendered useless as electron carriers – Since Complex I is inhibited by Rotenone by inhibiting electron transfer of NADH from Complex I Fe-S center to CoQ, then NADH will be rendered useless but not FADH2 that conducts its electron transfer to Complex II. FADH is still capable to transfer its electron, but with less amount since ATP produced through Complex II (1.5 ATP) is less than ATP produced through Complex I (2.5 ATP) Reference: MODULE 1B. 15.) SAMPLE CONCEPT MAP: THE ROTENONE POISONING CASE

95
Q
  1. A university student takes some tablets she is offered at a club, and without asking what they are, she swallows them. A short time later, she started to hyperventilate and feel hot all over. She was brought to the ER, where her temperature was found to be 41ºC. What is the most likely action of the tablets she took?
    a. Stimulation of mitochondrial ATP synthesis
    b. Uncoupling of mitochondrial electron transport and oxidative phosphorylation
    c. Stimulation of ATP transport out of the mitochondria into the cytosol
    d. Inhibition of mitochondrial electron transport
    e. Inhibition of ADP transport into mitochondria to be phosphorylated
A

b. Uncoupling of mitochondrial electron transport and oxidative phosphorylation

96
Q
  1. In redox reactions, these 2 groups of coenzymes react directly with substrates to obtain their reducing equivalents.
    a. NADH and FADH2
    b. H2O and CO2
    c. NAD+ and FAD
    d. NADPH and ATP
A

c. NAD+ and FAD

“NAD+ and FAD reacts directly with H+ to form NADH+ and FADH2 respectively.”

Redox Reactions (Oxidation-reduction reactions)

  • Chemical reaction that involves a transfer of electrons between two chemical species (atom, molecule, or ion)
  • What is Reduced is the one accepting electrons. (the fuel in this example is carbohydrate)
97
Q
  1. Which of the following examples show negative transcriptional control?
    a. any situation that inactivates repressor
    b. inhibition of EF2
    c. binding of transcription factor to the inducer
    d. protein blocks the access of RNA Polymerase to the gene’s promoter region
A

d. protein blocks the access of RNA Polymerase to the gene’s promoter region

Types of Gene Regulation:

  • Positive Regulation (Positive Transcriptional Control) – Expression of the genetic information is increased by the presence of specific regulatory element (activator/inducer)
  • Negative Regulation (Negative Transcriptional Control) – Expression of the genetic information is diminished by specific regulatory element (repressor).

The repressor binds to the Operator to block the RNA polymerase from “going” to the structural gene in order to prevent it from transcribing unnecessary/unneeded protein. Reference: Doc Shiela’s Compendium on “Gene Regulation and Mutation”

98
Q
  1. A biochemical reaction occurs spontaneously if
    a. Both ΔE and ΔS are positive, with ΔE having a much greater magnitude
    b. The ΔE is negative, while the ΔS is positive
    c. Both ΔE and ΔS are negative, with ΔS having a much greater magnitude
    d. The ΔE is positive, while the ΔS is negative
A

b.The ΔE is negative, while the ΔS is positive

99
Q
  1. A mutation that will always give to zero activity of the gene product?
    a. gain mutation
    b. nonsense mutation
    c. same sense mutation
    d. gene amplification
A

b.nonsense mutation

Nonsense mutation - result in an amino acid codon being replaced by a STOP CODON. It ends translation prematurely and result in a truncated protein.

Silent mutation - replacement of a single nucleotide with another, but have no effect on the protein (new codon can code for the same amino acid)

Missense mutation - replacement of a single nucleotide with another, and results in a single amino acid change in the translated region

100
Q
  1. In one type of hemoglobin structure, base A of a codon was changed to base U. What possible results can happen?
    a. newly formed codon may dictate the same amino acid or a different amino acid
    b. the new hemoglobin is a mutant
    c. all of the choices
    d. oxygen carrying capacity of hemoglobin may change
A

c.all of the choices

This type of point mutation has a change in a single nucleotide of the organism’s DNA which affects RBC shape. Hence, there is a change in O2 carrying capacity of Hgb. The classic example is Sickle Cell Disease (SCD), caused by mutation of a single base, a T-A DNA Substitution, which in turns result to A-U change in mRNA corresponding to 6th codon of beta globin gene. The altered codon specifies a different amino acid, Valine rather than Glutamic Acid causing structural abnormality of the beta globin molecule leading to hemoglobin aggregation and RBC SICKLING.

101
Q
  1. An 8-year-old boy is seen by an ophthalmologist for slow eye movements, difficulty in seeing in all visual fields, and increased retinal pigmentation. Kearns-Sayre syndrome, a disorder due to a mutation in Complex II of the electron transport chain, is suspected. Transfer of electrons from which substrate would be impaired?
    a. Pyruvate
    b. Isocitrate

c .Succinate

d. Malate

A

c.Succinate

Complex II is also known as Succinate Coenzyme Q Reductase. From the name itself, Succinate dehydrogenase (SDH; succinate: ubiquinone oxidoreductase; mitochondrial complex II) plays a central role in mitochondrial metabolism, catalyzing the oxidation of succinate to fumarate and the reduction of ubiquinone (UQ) to ubiquinol (UQH2). Source: Harpers Illustrated Biochemistry (31st Edition); Rodwell et.al; page 120.

102
Q
  1. Consider a biochemical reaction that started with equal amounts of substrate and product, and was allowed to reach equilibrium. If the value of the equilibrium constant (Keq) is 0.25, this means that
    a. The direction of the reaction will proceed spontaneously from left to right.
    b. There are roughly 25 moles of substrate for every 100 moles of product.
    c. There are roughly 25 moles of product for every 100 moles of substrate.
    d. There was no net movement in either direction of the reaction from the beginning.
A

c. There are roughly 25 moles of product for every 100 moles of substrate.

Keq = [product]/[substrate]

25 mole of product divided by 100 moles of substrate, the resulting Keq is 0.25. Thus, “there are roughly 25 moles of product for every 100 moles of substrate.” Is the correct answer.

Reference: 2023 ratio

103
Q
  1. What is the role of tryptophan in Tryp Operon?
    a. inducer
    b. co-repressor
    c. repressor
    d. co-inducer
A

b. co-repressor

Tryptophan Operon

  • genes for the enzymes of tryptophan biosynthetic pathway
  • repressible genes=Genes whose expression is turned off by the presence of a repressor
  • The regulatory gene produces repressor but is inactive (apo-repressor)
  • The 5 structural genes produce proteins essential for tryptophan synthesis
  • In the absence of tryptophan, the operon is on
  • Inactive repressor cannot bind the operator site
  • Excess tryptophan represses transcription
  • Tryptophan acts as co-repressor that activates the apo-repressor
  • Binding of the activated repressor blocks RNA Pol transcription of the structural genes source:
104
Q
  1. Proteins that bind between the promoter region and the section of a gene which is transcribed thus prevents gene expression
    a. silencer
    b. repressor
    c. inducer
    d. operator
A

b. repressor

Binding of the repressor to the operator region does not allow the polymerase to bind to the promoter, hence stops gene expression.

105
Q
  1. Examine the following statements. Which statement best describes the direction a biochemical reaction will take?
    a. Under standard conditions, a reaction will proceed in a forward direction if the free energy ΔG0’ is positive.
    b. The direction of a reaction is independent of the initial substrate and product concentrations, because the direction is determined by the change in free energy.
    c. One can easily determine the rate at which a reaction’s equilibrium can be reached by looking at the ΔG0’.
    d. A reaction with +ΔG0’ proceeds in a forward direction if the substrate concentration is raised high enough, and its product is rapidly cleared by a subsequent reaction.
A

d. A reaction with +ΔG0’ proceeds in a forward direction if the substrate concentration is raised high enough, and its product is rapidly cleared by a subsequent reaction.

Recall that metabolic pathways must have an overall negative △ G°(provided by ATP hydrolysis) to favor the formation of products over the formation of reactants. In that same vein, it is possible for reactions in the pathway with positive △ G°’ to proceed in the forward direction if:

  • The concentration of that reaction’s substrate is raised to a high enough level. This increase in substrate concentration is provided by the reaction preceding the + △ G°’ reaction in question along the metabolic pathway.
  • The concentration of that reaction’s product is decreased to a very low level. This occurs when the subsequent reaction (which uses the + △ G°’ reaction’s product as substrate) clears the product.

These 2 processes cause disequilibrium as far as the + △ G°’ reaction is concerned because of more substrate and less product than usual compared to the levels present at equilibrium. In order to obtain equilibrium again, the + △ G°reaction has to move forward. This reiterates the fact that the direction if a reaction is dependent on the initial substrate and product concentration as reflected in Keq. While the general principle is “larger and more negative G°-> spontaneous remains true, △ G°, however is not an indicator of the velocity of the reaction or the rate at which equilibrium can be reached. This actually depends on the efficiency and amount of enzyme available to catalyze the reaction.

106
Q
  1. Examine the following statements regarding molecular oxygen, and choose the one which is the most accurate.
    a. It has a high positive redox potential, thus easily reduces other molecules.
    b. It has a negative redox potential, thus easily oxidizes other molecules.
    c. It has a negative redox potential, thus easily reduces other molecules.
    d. It has a high positive redox potential, thus easily oxidizes other molecules
A

d. It has a high positive redox potential, thus easily oxidizes other molecules

Redox Potential (E0’)

  • A measure (in volts) of the energy change when that compound accepts e–
    • Expression of the willingness of a compound to accept electrons.
  • The more positive the Redox Potential, the more willing it is to accept electrons.
    • (More positive = Better as Oxidizing agent) That’s why it can easily oxidize other molecules.

In contrast:

  • The more negative the Redox Potential, the more willing it is to give off electrons.
    • (More negative = Better as Reducing agent)