Uworld 2

How will the extent of van der Waals interactions be affected in the hydrophobic region IV in Figure 2 if a protein variant is made that replaces valine with alanine in this region? A.The van der Waals interactions will decrease because there will be fewer hydrocarbon units in the side chain. B.The van der Waals interactions will decrease because alanine has a dipole moment. C.The van der Waals interactions will increase because there will be more hydrogen bonds. D.The van der Waals interactions will remain the same because both valine and alanine are nonpolar.

A.The van der Waals interactions will decrease because there will be fewer hydrocarbon units in the side chain. helps if u see difference in chart

Which of the following procedures best describes the use of diffractive phenomena in separating the individual components of an electromagnetic wavefront? A.Single-slit diffraction of monochromatic light B.Double-slit diffraction of polychromatic light C.Grated diffraction of polychromatic light D.Grated diffraction of monochromatic light

C.Grated diffraction of polychromatic light

cofactors act like coenzymes

So therefore stabilize the transition state since this is what enzymes do

pitch and frequency

The frequency of a sound is associated with its perceived pitch (or tone). A sound with a high frequency is perceived to have a high pitch, and a sound with a low frequency (such as those near the apex) is perceived to have a low pitch.

condesation reactions

actions in which molecules release water during bond formation are known as condensation reactions.

All amino acids are glucogenic except

leucine and lysine

degenerate orbitals

orbitals that have the same energy

Photorefraction with a camera at a distance of 50 cm away from a child produces a completely dark image of the pupil. If the child's retina is 2.5 cm from the eye lens, what is the lens strength of the eye? (Note: Use the thin lens formula, S = 1/o + 1/i.) A.38 D B.42 D C.48 D D.52 D

42 D Lens strength is a measure of a lens's refractive power; the greater the lens strength, the more light will bend. The inverse of a lens's focal length is equal to its lens strength (S = 1/f); the closer the image is focused to the lens, the greater the lens strength. The given thin lens equation relates the strength of a lens (S) to the distance to the object (o) and to the created image (i): S=1o+1i According to the passage, a completely dark photographed pupil is produced from an eye lens that does not require correction because the image is properly focused at the retina. Therefore, the image distance is equal to the lens-retina distance (i = 2.5 cm). Because the child is looking at the camera during photorefraction, the object distance is equal to the distance to the camera (o = 50 cm). In calculations involving optics, image distance i and object distance o are both positive for converging (convex) lenses, like the eye lens. Lens strength is measured in units of diopters (D), which are equal to inverse meters (m−1). Therefore, these lengths are converted from cm to meters: o = 0.5 m and i = 0.025 m. Using these values in the above equation, S=10.5 m+ 10.025 m S=2 m−1+ 40 m−1=42 D (Choice A) The image distance would be negative for a diverging lens. Incorrectly using i = −2.5 cm would result in a lens strength of −38 D. (Choice D) If the inverses of i and o are not used in the thin lens equation, the incorrectly calculated lens strength would be 52 cm. Educational objective: Lens strength (refractive power) is the inverse of the lens's focal length and is measured in units of diopters (D). A diopter is equal to an inverse meter (m−1). Lens strength can be determined from object distance and image distance using the thin lens equation: S = 1/o + 1/i.

The C=S double bond in carbon disulfide (CS2) consists of a combination of one σ bond and one π bond. Compared to the bond dissociation energy (bond strength) of the π bond, the bond dissociation energy of the σ bond portion of the C=S bond is: A.greater. B.less. C.the same. D.proportional.

A However, if the strengths of the σ bond and π bond contributors are considered separately, π bonds are weaker than σ bonds. The end-to-end orbital overlap in σ bonds is more efficient than the side-to-side orbital overlap in π bonds. This causes σ bonds to exist in a more stable, lower energy state. As a result, breaking a σ bond requires more added energy than breaking a π bond (ie, a σ bond has a greater dissociation energy).

Each of the following molecules can be converted to oxaloacetate by a single enzymatic reaction EXCEPT: A.pyruvate. B.malate. C.alanine. D.aspartate.

A metabolic pathway is a series of enzymatic steps that interconvert metabolic intermediates to produce energy (catabolism) or structural molecules (anabolism). Many metabolic pathways intersect and have an intermediate in common. This intermediate may be synthesized from multiple precursors, depending on the pathway used in their synthesis. Oxaloacetate is an intermediate in several metabolic pathways, including gluconeogenesis, the citric acid cycle, and amino acid degradation. Each uses a different precursor and a different enzyme to catalyze the synthesis of oxaloacetate. In the first step of gluconeogenesis, the enzyme pyruvate carboxylase catalyzes the carboxylation of pyruvate to form oxaloacetate (Choice A). In this reaction, CO2 is added to pyruvate, and ATP is hydrolyzed to provide the necessary energy. In the final step of the citric acid cycle, oxaloacetate is synthesized by the oxidation of malate, a reaction catalyzed by malate dehydrogenase (Choice B). This reaction requires the simultaneous reduction of NAD+ to NADH. NADH can then enter the electron transport chain and provide electrons necessary for the production of ATP. In amino acid degradation, amino acids are converted to α-keto acids by transamination, in which an amino group is transferred from an amino acid to α-ketoglutarate to synthesize glutamate (Choice D). The α-keto acid derivative of aspartate is oxaloacetate. Alanine can be deaminated to yield pyruvate, which can then be converted to oxaloacetate by pyruvate carboxylase. This process requires two enzymatic steps instead of one.

The reaction between the toilet bowl cleaner and the aluminum foil forms aqueous AlCl3 in addition to the gas that is produced. Which of the following best explains this outcome of the reaction? A.Aluminum undergoes a double replacement reaction with the aqueous HCl in the cleaner. B.Aluminum participates in a single replacement reaction with the aqueous HCl in the cleaner. C.The AlCl3 forms by a combination reaction between Al and the aqueous HCl in the cleaner. D.The AlCl3 forms by a decomposition reaction between Al and the aqueous HCl in the cleaner.

A single replacement reaction involves the replacement of one type of atom with another in a reaction between a compound and a neutral element in its stable form. A neutral element that is sufficiently reactive will lose or gain electrons and take the place of one of the atoms in the compound. The atom that it replaces will then lose or gain electrons to become a neutral element. In the reaction discussed in this question, elemental aluminum Al(s) replaces hydrogen in HCl to form AlCl3. In turn, the hydrogen atoms combine to form elemental hydrogen gas (as indicated in Table 2), which exists as H2 because elemental hydrogen is most stable as a neutral diatomic molecule. (Choice A) The reaction is not a double replacement reaction because two ions are not exchanging their respective counterions. (Choice C) The reaction is not a combination reaction because two elements are not combining to form one new stable compound. (Choice D) The reaction is not a decomposition reaction because a compound is not breaking down into two or more compounds or elements.

Which of the following processes would be directly impaired in von Gierke disease? Glycogenolysis Gluconeogenesis Pentose phosphate pathway A.I and II only B.I and III only C.II and III only D.I, II, and III

A.I and II only Maintaining blood glucose levels is critical for proper body function, so several biological processes exist to maintain glucose levels between meals. They take place predominantly in the liver, and all produce glucose 6-phosphate (G6P). G6P must be dephosphorylated by G6Pase to be released as glucose into the bloodstream, so all of these processes will be directly impaired by the inactivation of G6Pase. They include the following: Glycogenolysis is the breakdown of glycogen into glucose. It is the first source of glucose during fasting and requires G6Pase in its final step (Number I). Gluconeogenesis is the synthesis of glucose from pyruvate and other precursors. It is essentially glycolysis in reverse with a few differences, including the conversion of pyruvate to oxaloacetate and the use of phosphatases such as G6Pase instead of kinases to remove phosphate groups (Number II). Gluconeogenesis and glycolysis are connected by the Cori cycle in which lactate, produced in the muscle during anaerobic glycolysis, is sent to the liver. There it is converted to pyruvate and undergoes gluconeogenesis. The newly synthesized glucose is then returned to the muscles through the bloodstream. (Number III) The pentose phosphate pathway is required for nucleotide and NADPH synthesis. Although the pathway includes G6P as a substrate, it does not require G6Pase activity, so it will not be directly impaired by von Gierke disease. Educational objective: Several biological processes, including gluconeogenesis, glycogenolysis, and the Cori cycle, help buffer blood glucose levels. All of these processes require glucose 6-phosphatase (G6Pase) to catalyze the final step to release free glucose. The pentose phosphate pathway uses glucose 6-phosphate as a substrate, but does not produce free glucose or require dephosphorylation and is unaffected by changes in G6Pase activity.

All of the following could confirm apoptosis due to increased oxidative stress in cells with an L156P mutation in Complex V EXCEPT: A.inhibition of caspase activity. B.increased cytosolic cytochrome C. C.activation of proteolysis. D.increased mitochondrial superoxide concentration.

A.inhibition of caspase activity. Apoptosis is the controlled, programmed death of cells. It occurs in response to a variety of circumstances, including oxidative stress as well as DNA damage and certain developmental events. Apoptosis is initiated when the mitochondrial membrane is permeabilized in response to these environmental cues. This allows the release of cytochrome C from the mitochondria, resulting in increased cytosolic cytochrome C (Choice B). Cytochrome C induces proteolysis (Choice C) and other degradative pathways by activating caspase proteases (inactive caspase is found in healthy cells). Educational objective: Apoptosis (programmed cell death) can be caused by certain developmental events, DNA damage, and reactive oxygen species. It is induced when cytochrome C is allowed to leave the mitochondria and enter the cytosol, where it activates caspase. Caspase in turn activates several degradative pathways such as proteolysis.

Piezoelectric crystals likely create sound by: A.rapidly expanding and contracting. B.emitting high-energy photons. C.generating heat waves. D.sending pulses of electricity. "The piezoelectric crystals generate sound when an alternating voltage is applied"

A.rapidly expanding and contracting. Compressions (areas of high pressure) and rarefactions (areas of low pressure) are formed through the back-and-forth vibrations of the molecules of the medium. The passage states that piezoelectric crystals produce sound when an alternating voltage is applied to them. It is reasonable to conclude that the expansion and contraction of the crystals will cause molecules in the surrounding medium to vibrate, creating sound waves. (Choice B) A photon is a packet of energy that makes up electromagnetic radiation (light). The emission of photons would generate light, not sound. (Choices C and D) Sound energy is propagated in the form of pressure waves, not heat waves or electric waves. Educational objective: Sound is propagated in the form of pressure waves by the vibrations of the molecules in a medium. The rapid expansion and contraction of crystals creates pressure waves (sound) by vibrating nearby particles.

The image of an object projected onto a fixed screen through a lens using red light is clear and focused. Using the same set-up, the image is slightly unfocused when violet light is used. This is because violet light: A.refracts more than red light. B.is more polarized than red light. C.is unaffected by spherical aberrations. D.has greater phase shifts in glass.

A.refracts more than red light. Dispersion refers to the phenomenon of different frequencies of light having slightly different refractive indices. Light with higher frequency (such as violet light compared to red light) has a higher index of refraction in a lens and therefore refracts more. The formation of blurry images due to dispersion is known as chromatic aberration. Lens aberrations are real-world departures from ideal lens behavior that can result in blurry images. Chromatic aberration refers to the formation of blurry images due to the effects of dispersion through a lens. The refractive index of light increases with its frequency, and violet light is higher in frequency compared to red. As a result, violet light will refract more and converge before the screen, resulting in a blurry image. (Choice D) The phase shift of a wave refers to its displacement from another wave. Phase shifts (or lack thereof) in light do not affect its refractive properties.

enzymes at a higher concentration in solution will exhibit a _______ Vmax than enzymes with the same kcat value present at a lower concentration.

An enzyme's turnover number kcat represents the number of substrate molecules converted to product per second by each enzyme in solution under saturating conditions. The maximum velocity Vmax of an enzyme-catalyzed reaction is kcat multiplied by the enzyme concentration [E]. Therefore, enzymes at a higher concentration in solution will exhibit a higher Vmax than enzymes with the same kcat value present at a lower concentration.

Which of the following statements is true regarding the bonds in the tastant molecule L-methionine? The pi bond in C=O has a smaller dissociation energy than the sigma bond. The C=O bond is stronger than the C-O bond. The pi bond in C=O is more stable than the sigma bond. A.I only B.I and II only C.II and III only D.I, II, and III

B.I and II only Sigma bonds, which are generally the first bond formed between atoms, are created when the electron bonding pair is between the two atoms such that their atomic orbitals overlap end-to-end along the internuclear axis. Because of the efficient overlap of the atomic orbitals, sigma bonds are low in energy and are very stable. Every additional bond formed between two atoms after a sigma bond is a pi bond. Pi bonds are created by the sideways overlap of p orbitals along a plane perpendicular to (ie, above and below) the internuclear axis. Because the overlap in electron density is not as efficient as for sigma bonds, pi bonds exist in a higher energy state and are not as stable as sigma bonds. As a result, they require less energy to be broken than sigma bonds (ie, they have a smaller dissociation energy) (Number I). Although individual pi bonds are weaker than sigma bonds, a double bond is composed of both a sigma and a pi bond, and therefore is stronger than a single bond (Number II). (Number III) Because the pi bond in C=O is higher in energy than the sigma bond, it is less stable.

Myopia can be corrected with a: A.diverging lens, which creates real and inverted images. B.diverging lens, which creates virtual and upright images. C.converging lens, which creates real and inverted images. D.converging lens, which creates virtual and upright images.

B.diverging lens, which creates virtual and upright images.

What are the formal charges of the oxygen atoms in the acetate ion of the Cu(C2H3O2)2 used during the reaction if the effects from resonance are excluded? A.0 and 0 B.−2 and 0 C.0 and −1 D.−1 and −1

C.0 and −1 In the passage, the name of Cu(C2H3O2)2 is stated to be copper(II) acetate. The Roman numeral (II) indicates that the copper has a charge of +2. This means that each formula unit of the acetate anion must have a net charge of −1, as in C2H3O2−. This also indicates that acetate possesses one additional electron beyond those contributed by the valence of each atom. With this information, a Lewis structure obeying the octet rule can be drawn for the acetate anion. This is an individual resonance contributor that excludes effects from delocalization. Assessing the formal charge for each oxygen atom shows that the O atom in the C=O bond has a formal charge of 0 whereas the O atom in the C-O bond has a formal charge of −1.

When oxalate anions, C2O42−, are added to a solution of [Fe(H2O)6]3+, the formation of [Fe(C2O4)3]3− occurs in a series of steps that together result in the following overall reaction: [Fe(H2O)6]3+(aq) + 3 C2O42−(aq) → [Fe(C2O4)3]3−(aq) + 6 H2O(l) During the first step of the reaction, two new coordinate bonds are formed to produce an intermediate ion of: A.[Fe(H2O)4(C2O4)]+(aq). B.[Fe(H2O)2(C2O4)2]−(aq). C.[Fe(C2O4)3]3−(aq). D.[Fe(H2O)5(C2O4)]2−(aq).

Complex ions (coordination complexes) consist of a central metal ion surrounded by one or more ions or molecules called ligands that are bound to the metal center by coordinate bonds. The coordinate bonds form because the ligands surrounding the metal center act as Lewis bases and donate a lone pair of electrons to the metal center, which acts as a Lewis acid. These Lewis acid-base coordinations hold the complex together, but stronger Lewis bases can displace weaker Lewis bases as ligands within the complex. In the [Fe(H2O)6]3+ complex ion, an uncharged water molecule serving as a ligand is a weak Lewis base and can be displaced by the oxalate (C2O42−) anion, a stronger Lewis base. Each C2O42− anion displaces two water molecules and decreases the overall charge of the resulting complex ion by −2. Displacement of water within the complex occurs in a stepwise sequence. In the first step, two water molecules are displaced from [Fe(H2O)6]3+ by a single C2O42− anion to form an intermediate complex ion: [Fe(H2O)6]3+(aq) + C2O42−(aq) → [Fe(H2O)4(C2O4)]+(aq) + 2 H2O(l) In the second and third steps, additional C2O42− anions displace the remaining water ligands to form the final product. (Choice B) This is the intermediate ion formed during the second step of the reaction, not the first step. (Choice C) This is not an intermediate ion; it is the end product formed during the final third step of the overall net reaction. (Choice D) Both the ion charge and this formula are incorrect. The first step adds one C2O42− ion to the complex, but each oxalate ion displaces two water molecules and decreases the overall charge of the complex by −2. Educational objective: When a reaction occurs in a stepwise sequence, the species formed as products in earlier steps and then subsequently consumed as reactants in later steps are intermediate species. Intermediates do not appear in the overall net reaction.

coordinate bonds

Coordinate covalent bonds tend to maximize geometric space around a central atom. Transition metals generally form complexes with two, four, or six coordinate bonds. Therefore, the geometry of the coordinate covalent bonds will typically be linear (two bonds), tetrahedral or square planar (four bonds), or octahedral (six bonds).

Oxalate anions, C2O42−, added to a solution of [Fe(H2O)6]3+ displace water molecules within the complex ion to form [Fe(C2O4)3]3−, as shown in the reaction below: [Fe(H2O)6]3+(aq) + 3 C2O42−(aq) → [Fe(C2O4)3]3−(aq) + 6 H2O(l) Which of the following does NOT explain why C2O42− displaces H2O in the formation of [Fe(C2O4)3]3− ? A.C2O42− has fewer available lone pairs of electrons per oxygen atom than H2O. B.C2O42− is a stronger Lewis base than H2O. C.C2O42− forms coordinate bonds more readily than does H2O. D.C2O42− is negatively charged whereas H2O is uncharged.

Coordination complexes consist of a central metal ion surrounded by one or more ions or molecules called ligands that are bound to the metal center by coordinate bonds. The ligands act as a Lewis base, and the metal center acts as a Lewis acid. Stronger Lewis bases can displace weaker Lewis bases as ligands within a complex. Stronger Lewis bases tend to be those that have lone pair electrons on atoms with a net charge and/or a lower electronegativity. Charged species tend to be less stable (more reactive) than uncharged species because forming a bond produces a more stable, lower energy state. Likewise, atoms with a lower electronegativity tend to be better Lewis bases because the electrons are held more loosely and any net charge is less stabilized. Both C2O42− and H2O have available lone pair electrons on an oxygen atom (same electronegativity aspect), but C2O42− is a stronger Lewis base because it also has a negative charge on each of two of the oxygen atoms. This charge comes from the presence of an extra lone pair of electrons on each. Therefore, C2O42− does not have fewer lone pair electrons per oxygen atom than H2O, and this choice does NOT explain why C2O42− displaces H2O. (Choice B) This statement explains the formation of [Fe(C2O4)3]3− because weaker Lewis bases tend to form weaker coordinate bonds. C2O42− is a stronger Lewis base, which can displace H2O and form a stronger coordinate bond than H2O. (Choices C and D) These choices explain the reaction because charged species tend to be stronger Lewis bases and more reactive than uncharged species. The two additional lone pairs of electrons on two oxygen atoms in C2O42− give C2O42− its net charge and a greater tendency to donate electrons for coordinate bond formation. Educational objective: Stronger Lewis bases (electron pair donors) can displace weaker Lewis bases as ligands within a coordination complex. Lone pair electrons on atoms with a lower electronegativity tend be stronger Lewis bases than those with a higher electronegativity, and charged atoms with lone pair electrons tend to be stronger Lewis bases than comparable uncharged atoms.

he chemical structures of sulfur compounds H2S, SF6, S2Cl2, and S4N4 each contain only single (sigma) bonds. The compound that contains the longest bond between sulfur and an atom of another element is: A.H2S B.SF6 C.S2Cl2 D.S4N4

Covalent bonds between atoms are formed by sharing electrons, which is made possible through the overlap of the atomic orbitals from each atom. Covalent bonds made by the end-to-end overlap of atomic orbitals are called sigma (σ) bonds, whereas covalent bonds made by the side-to-side overlap of p orbitals are called pi (π) bonds. A single covalent bond consists only of a σ bond, a double bond consists of one σ bond and one π bond, and a triple bond consists of one σ bond and two π bonds. Because a σ bond involves end-to-end overlap of two orbitals, the length of a σ bond can be estimated as the sum of the atomic radii of the bonded atoms. Bonding between atoms with larger atomic radii positions the atomic nuclei farther apart and results in a longer σ bond (and vice versa). On the periodic table, atomic radii tend to decrease across a row (due to an increasing effective nuclear charge) and increase down a column (due to having an additional electron shell). In this question, the given compounds all have σ bonds to a sulfur atom. As a result, the longest bond between sulfur and another element will be formed with the element that has the largest atomic radius. Because Cl has a larger atomic radius than H, N, or F, the S-Cl bond in S2Cl2 will the longest bond with sulfur among the given compounds.

The electrolysis of aqueous silver nitrate involves two separate half-reactions that occur at the anode (Reaction 1) and the cathode (Reaction 2). 2H2O(l) → O2(g) + 4H+(aq) + 4e− Reaction 1 Ag+(aq) + e− → Ag(s) Reaction 2 Which of the following net reactions occurs during the hydrolysis? A.2H2O(l) + Ag+(aq) → O2(g) + Ag(s) + 4H+(aq) B.2H2O(l) + 4Ag+(aq) → O2(g) + 4Ag(s) + 2H2(g) C.2H2O(l) + Ag(s) → O2(g) + Ag+(aq) + 4H+(aq) D.2H2O(l) + 4Ag+(aq) → O2(g) + 4Ag(s) + 4H+(aq)

D. 2H2O(l) + 4Ag+(aq) → O2(g) + 4Ag(s) + 4H+(aq) For an oxidation-reduction (redox) reaction to be balanced, the same number of each type of atom and the same net charge must be present on both sides of the reaction equation. Redox reactions can be expressed as two half-reactions (one for the oxidation and one for the reduction, which occur together). Adding the two half-reactions together returns the net reaction. Species present on both sides of the summation cancel out. In a balanced scenario, the total number of electrons released from one half-reaction will match the total number of electrons used by the second half-reaction, causing the electron terms to cancel completely. If not, multiples of one (or both) of the half-reactions may be used in the summation. The net reaction resulting from Reactions 1 and 2 can be found by adding these half-reactions together. In order for the electron terms to fully cancel out, a multiple of Reaction 2 (all terms multiplied by 4) must be used. Canceling the 4e− terms and summing the reactants side and the products side of the half-reactions yield a net reaction with both an atom balance and a net charge balance (+4 on each side).

Consider the trends in atomic radii on the periodic table. The bond in citrate that is the longest and weakest with the lowest bond dissociation energy is the: A.C-H bond. B.C=O bond. C.O-H bond. D.C-C bond.

D.C-C bond. Increasing bond length= decreasing bond dissociation energy The energy required to break a bond, known as the bond dissociation energy, is related to the bond length. Atoms with smaller radii (near the top of the periodic table) tend to form shorter, stronger bonds than atoms with larger radii. Same element usually gives you longer bond length since same size and electronegativity (carbon-Carbon) Bond dissociation energy is the energy required to break a chemical bond; it is related to the bond length, or the distance between the nuclei of two bonded atoms. Shorter bonds are stronger and require more energy to break. Atoms with small atomic radii can form short, strong bonds whereas atoms with larger radii form longer, weaker bonds. Each subsequent row on the periodic table indicates an additional electron shell, and therefore a larger atomic radius, so atoms in the first row will have smaller atomic radii than atoms in the second row, and so on. Double bonds require more energy to break than single bonds. Citrate includes bonds between carbon and carbon, carbon and oxygen, carbon and hydrogen, and oxygen and hydrogen. Of the choices given, the C-C bond is the only single bond that does not involve a first row atom (hydrogen). Therefore, the C-C bond will be the longest and require the least energy to break.

Which experimental procedure(s) must scientists use to determine Vmax and Km of an enzymatic reaction using the Michaelis-Menten model? They must ensure that: they only measure the initial reaction rate for each substrate concentration. total enzyme concentration is much greater than the Km of the reaction. each initial substrate concentration tested is much greater than enzyme concentration. A.I only B.III only C.I and II only D.I and III only

D.I and III only Michaelis-Menten kinetics are determined by measuring the rate of product formation in the presence of varying concentrations of substrate. The reaction rate (initial slope) at each substrate concentration is then plotted as a single point, with concentration on the x-axis and rate on the y-axis, yielding a hyperbolic curve (Michaelis-Menten plot) described by the Michaelis-Menten equation. The Michaelis-Menten equation relies on three assumptions: The free ligand approximation states that substrate concentration [S] is constant during the reaction. This approximation is only true during the initial phase of the reaction, before a significant amount of substrate is converted to product. Substrate can also be depleted when it binds the enzyme to form the enzyme-substrate complex (ES). To ensure that ES formation does not significantly impact [S], the total concentration of enzyme in solution should be much smaller than any substrate concentration tested (Number III). The steady state assumption states that the concentration of ES remains constant over the course of the reaction, allowing the rate of product formation to remain constant. Once [S] becomes significantly depleted, ES levels decrease and the reaction slows. The irreversibility assumption states that the reaction proceeds only in the forward direction, and product does not get converted back to substrate. Once enough product accumulates, the reverse reaction occurs at non-negligible levels and further slows the net rate of product formation. Each of these assumptions holds true only during the initial phase of the reaction, before substrate is depleted or product accumulates (Number I). (Number II) Km is the substrate concentration at which ½Vmax occurs. Because enzyme concentration must be substantially smaller than all substrate concentrations tested, enzyme concentration must be much smaller than Km, not larger. Educational objective: The Michaelis-Menten equation models the initial rates of a reaction at various substrate concentrations. Rates are measured as the slope of the initial, linear phase of the reaction before substrate is depleted and product accumulates. Enzyme concentration should be much lower than all substrate concentrations tested.

In the reaction to form the cis isomer shown in Figure 1, which of the following structural sites function(s) as a Lewis base? A.Cu only B.N only C.O only D.N and O only

D.N and O only Coordination complexes consist of a central metal ion surrounded by one or more ions or molecules called ligands, which are bound to the metal center by coordinate bonds. The ligands act as a Lewis base (electron pair donor), and the metal center acts as a Lewis acid (electron pair acceptor). Ligand structures that can form more than one coordinate bond by donating more than one pair of electrons have more than one site that can function as a Lewis base. During the formation of copper(II) glycinate (Figure 1), glycine displaces acetate in Cu(C2H3O2)2 because stronger Lewis bases can displace weaker Lewis bases as ligands within a complex. Stronger Lewis bases tend to be those that have lone pair electrons on atoms with a net charge and/or a lower electronegativity. As a ligand, glycine has an available pair of electrons to donate on both the oxygen atom in the carboxylic group and on the nitrogen atom of the amino group, and both sites in the structure of glycine function as Lewis bases (electron pair donor) in the reaction shown in Figure 1. This is indicated by the formation of the Cu−N and Cu−O coordinate bonds in the complex. (Choice A) In the reaction, copper is present in the form of Cu2+ ions that are supplied by the Cu(C2H3O2)2 solution. The Cu2+ ion is electron-deficient and acts as a Lewis acid (electron pair acceptor). (Choices B and C) The oxygen atom in the carboxylic acid group and the nitrogen atom in the amino group in the structure of glycine can both donate a lone pair of electrons. As such, both sites can function as Lewis bases. Educational objective: Within coordination complexes, a ligand acts as a Lewis base, and the metal center acts as a Lewis acid. Ligand structures that can form more than one coordinate bond by donating more than one pair of electrons have more than one site that can function as a Lewis base.

Which of the following sulfur compounds from Table 1 has a chemical structure that contains both ionic and covalent bonds? H2S,SO2,SO3,H2SO4,Na2SO4,K2SO4,SF6,S2Cl2,S4N4 A.Hydrogen sulfide B.Sulfuric acid C.Sulfur trioxide D.Potassium sulfate

D.Potassium sulfate Ionic bonds are formed between two atoms when the valence electrons from one atom are transferred to another atom, resulting in the formation of charged ions held together by strong electrostatic attractions. In contrast, covalent bonds are made between two atoms by sharing electrons through the end-to-end overlap of atomic orbitals. The type of bond that forms between two atoms depends on the relative difference in electronegativity between the atoms. A large difference in electronegativity promotes ionic bond formation, but a small difference in electronegativity promotes covalent bond formation. As a result, ionic bonds generally form between a metal and a nonmetal whereas covalent bonds generally form between two nonmetals; however, some exceptions do exist because the actual ionic character of a bond exists on a continuum. Using the metal and nonmetal classifications of the elements as generalization for determining bonding types in the four compounds given in this question, only potassium sulfate (K2SO4) contains a metal. Therefore, potassium must form a metal-nonmetal (ionic) bond in the structure in addition to the nonmetal-nonmetal (covalent) bonds present between the nonmetal atoms in the structure.

Treatment with which of the following could help counteract hypoglycemia in patients with von Gierke disease? "It affects roughly 1 in 100,000 individuals, and the resulting loss of both hormonal sensitivity and enzymatic activity leads to hypoglycemia (low blood sugar)." A.Insulin B.Vitamin A C.Epinephrine D.Starch

D.Starch Hypoglycemia can normally be countered with hormones that upregulate gluconeogenesis and glycogenolysis. Normally, glucagon would help counteract hypoglycemia by upregulating gluconeogenesis and glycogenolysis in the liver. However, because von Gierke disease is insensitive to hormones, it inhibits both processes regardless of glucagon levels. Blood glucose levels must therefore be controlled by diet instead. Starch is a polysaccharide composed of several linked glucose molecules. It is slowly broken down to its constituent glucose molecules in the digestive tract by the enzyme amylase. As glucose molecules are released from starch, they enter the bloodstream and help counteract hypoglycemia.

Which of the following bond types are found in the calcium phosphate present in 35% of kidney stones? Ionic Polar covalent Nonpolar covalent A.I and II only B.I and III only C.II and III only D.I, II, and III

Different atoms have different affinities for electrons, expressed as electronegativity. The type of bond that forms between two atoms depends on the electronegativity difference between the atoms. When atoms have an electronegativity difference greater than about 1.7, electrons are fully transferred from the less electronegative atom to the more electronegative atom. This process forms charged particles called ions and creates an ionic bond. Metals such as the calcium in calcium phosphate have low electronegativity and almost always participate in ionic bonds (Number I). Atoms with an electronegativity difference between about 0.5 and 1.7 do not fully transfer electrons but instead share electrons unequally. The atom with the higher electronegativity gains a partial negative charge as it exerts a stronger pull on the electrons whereas the less electronegative atom gains a partial positive charge. This separation of charge forms an electric dipole, and bonds of this nature are called polar covalent bonds. The phosphate ion is composed of a phosphorus atom covalently bonded to several oxygen atoms. Oxygen is the second-most electronegative atom on the periodic table, and the covalent bonds that it forms are almost always polar, as in the case of phosphate (Number II). (Number III) Nonpolar covalent bonds form between atoms with a difference in electronegativity no greater than 0.5. These bonds involve equal sharing of electrons, and they frequently form between two atoms of the same type (eg, carbon-carbon bond). Carbon-hydrogen bonds are also considered nonpolar because both atoms have similar electronegativities. No such bonds exist in calcium phosphate.

Which cofactor is regenerated by lactate synthesis? A.NAD+ B.NADH C.NADP+ D.NADPH

During glycolysis, NAD+ is reduced to NADH as glyceraldehyde 3-phosphate is oxidized to 1,3-bisphosphoglycerate; therefore, NAD+ is required for glycolysis to continue. Under aerobic conditions, NAD+ can be regenerated by oxidation of NADH in the electron transport chain (ETC). Under anaerobic conditions, the ETC cannot regenerate enough NAD+, and another source is required. Under these conditions, lactate synthesis takes place and oxidizes NADH to NAD+ as pyruvate (the final product of glycolysis) is reduced to lactate. (Choice B) Lactate synthesis consumes NADH to regenerate NAD+. (Choices C and D) NADP+ and NADPH are used in the pentose phosphate pathway and anabolic pathways such as fatty acid synthesis, but not in glycolysis. They therefore do not need to be regenerated or consumed by lactate synthesis. Educational objective: Lactate synthesis is coupled to the synthesis of NAD+ from NADH. Under anaerobic or oxygen-poor conditions, this process provides the NAD+ necessary to continue glycolysis.

Entropy and Order for surrounding molecule of hydrophobic or hydrophilic proteins

Each water molecule must participate in hydrogen bonds, but the side chains of hydrophobic amino acids cannot hydrogen bond with water. Therefore, water molecules in the vicinity of hydrophobic residues must hydrogen bond with each other instead, and can only do so by forming a rigid, highly ordered network called a solvation layer around the hydrophobic molecule. Hydrophilic amino acids, on the other hand, readily hydrogen bond with water, so formation of a solvation layer around these residues is unnecessary. The water molecules surrounding hydrophilic residues are more disordered than those surrounding hydrophobic residues.

enzyme kinetics

Enzymatic reactions are typically zero-order or first-order reactions. In zero-order reactions the rate is only dependent on the rate constant because substrate concentrations exceed the Km. First-order reactions depend on substrate concentration and occur when the Km is greater than the substrate concentration.

Malonyl CoA

Formed from Acetyl-CoA and HCO3 via the Acetyl-CoA carboxylase (ACC). Serves as a regulator of FA catabolism and precursor in FA synthesis. (fatty acid precursor)

H2CO3----> Decomposition

H2O + CO2

Determining Trigonal Pyramidal vs Tetrahedral

If all the electron domains are from bonds, the molecule's geometry will be the same as its electron group geometry. If lone pairs are present, the molecule's geometry will be different from electron group geometry. In this case, the highlighted phosphorus and carbon atoms have only bond domains and will be tetrahedral and trigonal planar, respectively. Nitrogen has four domains with one lone pair and will be trigonal pyramidal. Just 3 electron groups is Trigonal PLANAR

resonance

In physics, resonance is the tendency of a system to vibrate with increasing amplitudes at some frequencies of excitation. These are known as the system's resonant frequencies (or resonance frequencies). The resonator may have a fundamental frequency and any number of harmonics.

effect of increasing enzyme concentration on a reaction..

Km is still the same..affinity would not change The maximum velocity (Vmax) of the enzyme-catalyzed reaction, which occurs when the enzyme is saturated with substrate, is equal to the product of the catalytic rate constant (kcat) and the concentration of enzyme in solution: Vmax=kcat [Etot] Therefore, increasing enzyme concentration would result in an increase in the maximum velocity of the reaction and a decrease in the y-intercept.

Lyases

Lyases break bonds by forming double bonds or rings (double-bond equivalents) elsewhere.

Group A mice, but not group B mice, could use which of the following dietary supplements to help maintain blood glucose levels after day 10? A.Cholesterol B.Alanine C.Acetyl-CoA D.Palmitic acid

Mice in group A were not given tetracycline at any point and so have functional G6Pase. They therefore will be able to use any metabolites that can enter gluconeogenesis to help sustain their blood glucose levels. Amino acids (except for leucine and lysine) can be converted into glucose through gluconeogenesis. For example, alanine can be converted to pyruvate by deamination. Mice in group A will be able to use alanine to make glucose in the liver, thereby maintaining blood glucose levels. Mice in group B were given tetracycline at day 10 and will be unable to carry out gluconeogenesis after that point. Therefore, glucogenic substrates such as alanine will not help maintain their blood glucose levels. (Choice A) Cholesterol is a lipid used as a structural molecule in cell membranes and in the synthesis of steroid hormones. It is not used in glucose synthesis. (Choice C) Although acetyl-CoA can be converted to the glucogenic molecule oxaloacetate through the citric acid cycle, it must first combine with oxaloacetate to do so. The two carbons gained by adding acetyl-CoA to oxaloacetate are effectively lost as CO2 during the cycle, so acetyl-CoA cannot provide additional carbon for glucose synthesis. (Choice D) Most fatty acids, including palmitic acid, are broken down to acetyl-CoA, which cannot be converted to glucose.

Light from an incandescent lightbulb passes through an ideal linear polarizer. The intensity of the light is 100 lm before passing through the filter, but 50 lm immediately after. Which of the following explains this discrepancy? A.Refraction occurs as light enters the polarizer. B.Light waves with magnetic fields oriented perpendicular to the axis of polarization are absorbed. C.Light waves with electric fields oriented perpendicular to the axis of polarization are absorbed. D.Reflection occurs before light enters the polarizer.

Polarization refers to the alignment of transverse wave oscillations in a particular orientation within the x,y,z-coordinate system. For example, a transverse wave traveling along the x-axis causes oscillations that may occur exclusively along the y- or z-axis, or at some angle between the y- and z-axes. Polarization is unique to transverse waves because only transverse waves cause oscillations perpendicular to the direction of propagation. Consequently, electromagnetic radiation can be polarized but sound cannot. A waveform is said to be linearly polarized if waveform oscillations take place within only one plane of the coordinate system. By standard convention, the overall orientation of electromagnetic radiation is the same as the orientation of the electric field component. Polarization filters are optical devices that reorient the polarization of electromagnetic radiation so that radiation exiting the filter is polarized in only one orientation. A linear polarization filter, for example, allows for the transmission of electromagnetic radiation oriented parallel to the axis of polarization but inhibits the passage of radiation oriented perpendicular to this axis. Most light sources emit unpolarized light comprising waveforms oriented in all directions. Even for unpolarized light, however, approximately half of the intensity of electromagnetic radiation is polarized along one of two perpendicular axes. Because only electromagnetic radiation with an electric field oriented parallel to the axis of polarization passes through, the total intensity of light will therefore decrease by 50% when nonpolarized light passes through a linear polarization filter.

The anion SO32− and sulfur trioxide, SO3, have the same elemental composition. These two chemical species are: A.the same compound, because both have the same elemental composition. B.the same compound, because both have the same number of electrons and bonding configuration. C.different compounds, because the atoms in SO32− are held together by ionic bonds. D.different compounds, because SO32− has two more electrons and a different bonding configuration.

Polyatomic ions are covalently bonded groups of two or more atoms that have a net formal charge. If the total number of electrons exceeds the total number of protons in the group, the polyatomic ion will be negatively charged (and vice versa). Each electron present in a polyatomic group provides one unit of negative charge. For example, in SO32− (a sulfite anion), the number of protons in the group is exceeded by two electrons. Some polyatomic ions have an elemental composition that is the same as a neutral (uncharged) covalent molecule, but the polyatomic ions and the neutral covalent compounds are chemically distinct. For example, the sulfite anion (SO32−) and the covalent compound sulfur trioxide (SO3) both consist of one S atom and three O atoms, but the two species are not the same compound. As seen by comparing the Lewis structures for each, both the number of electrons and the bonding configuration of SO32− are different from those of SO3. The additional pair of nonbonding electrons on the S atom of SO32− gives the S atom a formal charge of +1 and causes the structure of SO32− to adopt a trigonal pyramidal geometry with a net charge of −2. In contrast, SO3 lacks the nonbonding electrons on the S atom, which gives the S atom a formal charge of +2 and causes the molecule to adopt a trigonal planar geometry with no net charge. These characteristics give SO3 and SO32− different chemical and physical properties.

Proteases

Proteins can be degraded by protease enzymes. During peptide hydrolysis, a water molecule is used to cleave the C-N bond in the amide linkage using an acid (H+) or base (OH−). In contrast, the reverse of amide hydrolysis is amide bond condensation, which involves the formation of a water molecule from two amino acids to form a larger peptide.

Laser Doppler flowmetry uses an infrared laser to measure blood flow velocity. Compared to ultrasound, the observed frequency shift using this technique is: A.smaller. B.the same. C.larger. D.zero.

The Doppler effect can be approximated by the relative velocity between the source and the observer and the speed of the wave in the medium (∆ff=vc) . The magnitude of the frequency shift is inversely proportional to the wave's speed. The Doppler effect of light will result in smaller observed frequency shifts because the speed of light is much greater than that of sound.

Lewis acids vs Lewis Bases

The electron-deficient character of metal cations causes them to be attracted to electron-rich ions or molecules and act as Lewis acids. The strength of the attraction increases as the ionic charge increases or as the ionic radius decreases. As a result, metal cations with a smaller ionic radius and a higher positive charge such as Al3+ and Ti4+ are stronger Lewis acids than metal cations with a larger ionic radius and lower charge such as Ca2+ and Mg2+. Lewis Bases= donate electrons

During prolonged fasting, which of the following liver enzymes has upregulated activity? A.Glycogen synthase B.Pyruvate carboxylase C.Glucokinase D.Phosphofructokinase

The first step in gluconeogenesis is carboxylation of pyruvate to form oxaloacetate. This step bypasses the irreversible pyruvate kinase reaction and is catalyzed by the enzyme pyruvate carboxylase. In the well-fed state, pyruvate carboxylase is downregulated so that pyruvate can be used in other pathways, but during prolonged fasting, pyruvate carboxylase activity is upregulated in the liver to increase the rate of gluconeogenesis. (Choice A) Glycogen synthase consumes glucose to produce glycogen. During fasting, the liver must synthesize glucose rather than consume it, so glycogen synthase activity is downregulated. (Choices C and D) Glucokinase and phosphofructokinase are tightly regulated enzymes of glycolysis that phosphorylate glucose and fructose 6-phosphate, respectively. During prolonged fasting, the activities of these enzymes are downregulated in the liver, whereas the gluconeogenesis enzymes—glucose 6-phosphatase and fructose 1,6-bisphosphatase—are upregulated to bypass these irreversible steps and increase gluconeogenesis. Educational objective: During periods of fasting, the liver helps maintain blood glucose levels. The liver initially synthesizes glucose by degrading glycogen. However, after glycogen stores are depleted in prolonged fasting, gluconeogenesis is upregulated, increasing the synthesis of glucose from precursors such as pyruvate. In the fasting state, the liver downregulates key enzymes of glycolysis and upregulates enzymes of gluconeogenesis.

Ozone (O3) in the atmosphere protects against harmful UV radiation. Its formation proceeds in two steps and is initiated when molecular oxygen (O2) splits into two oxygen atoms upon absorption of UV light. Which of the following would be the second step if ozone formation goes to completion? A.O + O2 → O3 B.2O + 2O2 → 2O3 C.3O2 → 2O3 D.O + 2O2 → O3

The law of conservation of mass says that in a chemical reaction, atoms are neither created nor destroyed: they can only be rearranged. Therefore, in any chemical reaction the same number of each type of atom must be present on the left and right sides of the reaction arrow. Reactions may proceed in multiple steps, producing intermediates along the way. If the reaction goes to completion (the maximum possible amount of product is formed), all intermediate species must be consumed in subsequent steps to form the final products. In the formation of ozone (O3), first, one molecule of oxygen (O2) is split into two oxygen atoms as intermediates (O2 → 2O). The oxygen atoms are then consumed as each reacts with an oxygen molecule to form ozone. For the reaction to go to completion, the two oxygen atoms must combine with two O2 molecules to form two ozone molecules (2O + 2O2 → 2O3) in the second step. This yields the net reaction 3O2 → 2O3. (Choice A) O + O2 → O3 correctly shows the reaction of one oxygen atom with one O2 molecule to form one ozone molecule. However, if this were the second step it would not allow the reaction to go to completion because the resulting net reaction would be 2O2 → O3 + O. (Choice C) 3O2 → 2O3 shows the net balanced reaction for the conversion of O2 to O3, not the second step of the reaction. The second step must use the oxygen atoms formed in the first step. (Choice D) O + 2O2 → O3 is not a balanced equation. It has five oxygen atoms on the left side and only three on the right, leaving two oxygen atoms unaccounted for. Educational objective: According to the law of conservation of mass, atoms are neither created nor destroyed in a chemical reaction. The same number of each type of atom must be present on both the left and right sides of a reaction arrow. Multistep reactions that go to completion must consume all intermediates produced before or during the final step.

In addition to calcium oxalate stones, researchers might hypothesize that potassium citrate (a weak base) could reduce the risk of uric acid stones for which of the following reasons? A.Potassium reduces the hydroxide concentration in urine. B.Potassium increases the hydroxide concentration in urine. C.Citrate reduces the hydronium concentration in urine. D.Citrate increases the hydronium concentration in urine.

The pH of a solution is defined as the negative logarithm of the hydronium concentration, often expressed in a simplified manner as −log [H+]. pH is related to the negative logarithm of hydroxide concentration (pOH) by the equation pH = 14 − pOH. Therefore, pH can be increased by adding hydroxide ions or by removing hydronium ions. The passage states that citrate alkalinizes (increases the pH of) the urine, and Table 1 indicates that uric acid becomes more soluble with increasing pH. The more soluble a compound is, the less likely it is to precipitate and form kidney stones. Citrate is a weak base and therefore a proton acceptor. It can alkalinize a solution by removing protons from hydronium ions, converting those ions to water and reducing the total hydronium concentration. Therefore, citrate may increase the solubility of uric acid by reducing the hydronium concentration and may decrease the risk of uric acid kidney stones. (Choice A) Reducing the hydroxide concentration would decrease the pH. (Choice B) Increasing the hydroxide concentration requires deprotonation of water to form hydroxide. Potassium is positively charged and would repel positively charged protons, and so be unable to deprotonate water. (Choice D) Citrate decreases the hydronium concentration. Increasing the hydronium concentration would lower the pH and decrease uric acid solubility.

A kinetics experiment reveals that the enzyme fumarase catalyzes the dehydration of malate with a kcat of 900 s−1. If the substrate concentration is equal to the Km of the reaction, what is the reaction velocity when the enzyme concentration is 200 nM? A.3.6 × 102 µM/s B.9.0 × 101 µM/s C.4.5 × 10−3 µM/s D.2.2 × 10−4 µM/s

The rate V0 of an enzymatically catalyzed reaction is determined by the enzyme concentration [E], the substrate concentration [S], the Michaelis constant Km, and the turnover number (reactions per second per enzyme, known as kcat). This relationship is given in the Michaelis-Menten equation: V0=kcat[E][S]Km+[S] The product of kcat and [E] equals the maximum velocity Vmax of the reaction, which simplifies the equation to: V0=Vmax[S]Km+[S] The question states that the concentration of malate [S] is equal to Km. Therefore, Km can be substituted into the equation to yield V0=VmaxKmKm+Km=VmaxKm2Km Finally, the Km terms in the numerator and the denominator cancel each other to yield V0=Vmax2 Therefore, when the substrate concentration equals the Km of the catalyzed reaction, the reaction will proceed at half the maximum possible velocity, or ½Vmax. The question says the concentration of fumarase [E] is 200 nM, and the kcat is 900 s−1. Multiplying these terms yields a Vmax of 180,000 nM/s, which is equivalent to 180 µM/s (1,000 nM = 1 µM). Per the derived equation, dividing Vmax by 2 yields a reaction rate V0 of 90 µM/s, which is equal to 9.0 × 101 µM/s. (Choice A) 3.6 × 102 µM/s is double the maximum velocity Vmax of the reaction at the given enzyme concentration. The question states that the concentration of malate is equal to the Km of fumarase, so the reaction will proceed at ½Vmax. (Choice C) 4.5 × 10−3 µM/s results from incorrectly calculating Vmax by dividing kcat by the fumarase concentration [E] instead of multiplying. (Choice D) 2.2 × 10−4 µM/s results from incorrectly calculating Vmax by dividing the fumarase concentration [E] by kcat instead of multiplying. Educational objective: The rate of an enzymatic reaction at any given substrate concentration is described by the Michaelis-Menten equation. The maximum velocity Vmax equals the product of kcat and enzyme concentration [E]. When the substrate concentration equals the Michaelis constant Km, the reaction proceeds at ½Vmax.

fundamental frequency

The resonant frequency of an object is the frequency that causes relatively large oscillations compared to other frequencies. The fundamental frequency is the lowest resonant frequency, and each subsequent harmonic frequency is a multiple of the fundamental frequency.

Km can be an approximate measure for binding affinity.

Therefore lower Kd and also lower Km equate to higher affinity.

transferase

Transferases move a chemical group from one molecule to another. Kinases are a type of transferase that either move a phosphate group from ATP to a substrate to form ADP and phosphorylated substrate, or from a substrate to ADP to form ATP.

uncompetitive inhibitor decrease quantity of Km and Vmax must be....otherwise its not uncompetitive.

Uncompetitive inhibitors bind only enzyme-substrate complex and decrease both the Vmax and Km of an enzymatic reaction by the same factor. SO MUST BE A PROPORTIONAL DECREASE, NOT JUST A DECREASE!!!!! SAME RATIO

Which of the following statements best explains why hemoglobin is red when it binds to O2? A.O2 interacts with iron's d orbitals. B.O2 is a nonpolar ligand. C.O2 has lone pairs of electrons. D.O2 changes the surrounding protein structure.

When placed within a coordination sphere, the atomic d orbitals of a metal are no longer degenerate, so they will have different energies. This occurs because some orbitals are pointing toward the ligands, which are electron donors, and other d orbitals are pointing away from or between the ligands. The orbitals that point toward the ligands are higher in energy than the orbitals pointing away from or between them because electrons repel each other. The energy of the orbitals determines which wavelengths of light can be absorbed. Some ligands cause greater differences in the energy of the d orbitals than other ligands, resulting in the absorption of different wavelengths of light. These wavelengths are usually in the visible spectrum. When heme binds O2, the nature of the O2 ligand changes the energy of iron's d orbitals. This energy change causes heme to absorb blue-green light and reflect red light. (Choice B) Oxygen's polarity does not affect color. Color is due to electron transitions. (Choice C) Oxygen's lone pairs are a factor in the difference in energy in iron's d orbitals, but the red color is due to electron transitions between the d orbitals, not the lone pair of electrons on O2. (Choice D) Although hemoglobin's protein conformation does change upon binding O2, this change in conformation is not the cause of the red color in blood. Educational objective: The nature of the ligands in a coordination sphere causes the metal's d orbitals to have different energies. This energy difference determines the wavelength of light absorbed by the bonds. The wavelength of light that is reflected is often in the visible region of the electromagnetic spectrum.

When sound passes from air to the tympanic membrane, what changes occur to the sound's intensity and velocity? A.Both increase B.Both decrease C.Intensity increases; velocity decreases D.Intensity decreases; velocity increases

When sound crosses from one medium to another, a portion of the wave's energy is reflected. Therefore, sound waves lose energy (are attenuated) and their intensity decreases when passing from air to a solid structure, such as the tympanic membrane (Choices A and C). Wave velocity v is the product of wavelength λ and frequency f: v = λf When a wave moves from one medium to another, its frequency does not change. Changes in wave velocity are only due to changes in wavelength. The propagation velocity of sound waves depends on the properties of the medium: Velocity increases with temperature. Velocity is slowest in gases, faster in liquids, and fastest in solids.

London dispersion forces vs dipole dipole

While all molecules are attracted to each other, some attractions are stronger than others. Non-polar molecules are attracted through a London dispersion attraction; polar molecules are attracted through both the London dispersion force and the stronger dipole-dipole attraction.

combination reaction

a chemical change in which two or more substances react to form a single new substance

thin film interference

a phenomenon in which a spectrum of colors is produced due to the constructive and destructive interference of light waves reflected in a thin film Thin-film interference describes the multicolored array generated when polychromatic light is incident on an interface formed by two semitransparent media. Localized discrepancies within the thickness of a thin film (top media) contribute to variable interference patterns that produce the colorful array.

Lyases

add groups to or remove groups from double-bonded substrates Lyases either form two molecules by breaking a bond within one molecule or they combine two molecules into one.

dehydrogenase is what type of enzyme

an oxidoreductase

single-displacement reaction

chemical reaction in which one element replaces another element in a compound

30.) Given the unbalanced equation (Reaction 1) and the molecular weight of calcium citrate (498.5 g/mol), if 15 nmol of calcium oxalate is mixed with 15 nmol of potassium citrate, what is the approximate theoretical yield of calcium citrate? 3 CaC2O4 + 2 K3C6H5O7 → Ca3(C6H5O7)2 + 3K2C2O4 A.1,250 ng B.2,500 ng C.3,750 ng D.7,500 ng

heoretical yield is the maximum amount of product that can form during a reaction based on the limiting reactant, which is the reactant that will be entirely consumed if the reaction goes to completion. The limiting reactant can be determined from the balanced chemical equation and the number of moles of each reactant. The passage gives the unbalanced equation for the reaction between calcium oxalate and potassium citrate. The equation must be balanced to determine the limiting reactant. The balanced equation is: 3 CaC2O4 + 2 K3C6H5O7 → Ca3(C6H5O7)2 + 3 K2C2O4 The equation shows that the formation of 1 mole of calcium citrate requires 3 moles of calcium oxalate but only 2 moles of potassium citrate. The question states that 15 nmol of each reactant were mixed together. As such, a quantity of 15 nmol of calcium oxalate can form 5 nmol of calcium citrate: 15 nmol CaC2O4×1 nmol Ca3(C6H5O7)23 nmol CaC2O4=5 nmol Ca3(C6H5O7)2 The limiting reactant is the one that yields the least product. In this case, calcium oxalate is the limiting reactant. The amount of calcium citrate possible from the reaction can be calculated by converting nmol to ng using the molecular weight of calcium citrate (≈ 500 g/mol) evaluated on the nanoscale: Therefore, the theoretical yield is 2,500 ng of calcium citrate.

coordinate bonds

shared electron pair originates from the same atom

coordination sphere

the central metal plus the ligands bound to it

cordination number

the number of ions of opposite charge that surround each ion in a crystal -# of coordinate bonds Such coordinate bonds are often formed between electron-poor metal ions and molecules called ligands that contain one or more electron-rich atoms with available lone-pair electrons. The coordinately bonded metal and its ligands are called a complex.

enzymes do not alter

thermodynamics, free energy, Keq, etc.