AAMC chemistry and physics

labeling theory

theory that society creates deviance by identifying particular members as deviant

work function equation

Work Function = h x threshold frequency

nearsightedness focal length f=r/2

-In nearsightedness, the lens focuses an image not on the retina but in front of the retina. Therefore it order to properly place the focused image a diverging, or concave lens must be used. -In farsightedness, the image is projected BEHIND the retina, and for correction, a converging or convex lens must be used The radius of curvature is kind of like imagining the the lens was a full circle and then taking that circle's radius. Therefore, if you want to move an image forward (myopia) you increase the radius, whereas if you want to move the image backwards you decrease the radius (hyperopia). You can view this better by looking at a concave/convex mirror and completing the full circle if you want to visualize the radius a bit better. What is the effect produced by the PRK technique designed to correct nearsightedness? Jack Westin Advanced Solution: This is a passage-related question that involves going back to the passage for key details. After the author explains and shows us Figure 1 with the schematic of the laser device, we get into the function of excimer lasers. We're told, "Excimer lasers are used in the photorefractive keratectomy (PRK) technique designed to correct vision impairment." However, we want to make the distinction between the PRK procedure used to correct farsightedness versus the PRK procedure used to correct nearsightedness. The author tells us "To correct nearsightedness, the laser beam is directed onto the central part of the cornea, resulting in a flattening of the cornea (Figure 2B)." Why would this work? Because a patient with nearsightedness has too curved of a cornea and rays from objects are forming images in front of the retina instead of on the retina. A normal eye is not as elongated, and images focus on the retina instead of in front of the retina like we have in nearsightedness. By flattening the cornea, we're adjusting for this elongation and increasing the radius of curvature so the image properly focuses on the retina instead of in front of the retina.We want an answer choice consistent with flattening of the cornea. A.The density of the cornea is increased.AAMC Standard Solution:The density of the cornea is unchanged due to the PRK technique because it is an intrinsic property of the tissue. Jack Westin Advanced Solution: The density of the cornea is increased. The density of the cornea does not change. Flattening the cornea and removing corneal material will change the mass and volume of the cornea in a relatively equal amount. While mass and volume will change individually, the ratio remains the same. Density stays the same. B.The radius of curvature of the cornea is increased.The correct answer is B.AAMC Standard Solution:According to the passage, to correct nearsightedness, the laser beam is directed onto the central part of the cornea, resulting in a flattening of the cornea. This means that the radius of curvature of the cornea is increased. Jack Westin Advanced Solution: The radius of curvature of the cornea is increased. This answer choice is consistent with our breakdown. We have to visualize what happens in nearsightedness. A patient with nearsightedness has too curved of a cornea and rays from objects are forming images in front of the retina instead of on the retina. We want objects to form images on the retina. By flattening the cornea, we're adjusting for this elongation and increasing the radius of curvature so the image properly focuses on the retina instead of in front of the retina. This is a strong answer choice. C.The index of refraction of the cornea is increased.AAMC Standard Solution:The refraction index of the cornea is unchanged due to the PRK technique because it is an intrinsic property of the tissue. Jack Westin Advanced Solution: The index of refraction of the cornea is increased. Reasoning here is going to be similar to answer choice A. The index of refraction is not going to change. Answer choice B remains the best option. D.The thickness of the cornea at the apex is increased.AAMC Standard Solution:The thickness of the cornea at the apex is decreased because the laser removes layers of tissue from the cornea at the apex. Jack Westin Advanced Solution: The thickness of the cornea at the apex is increased. This is the opposite of what happens in this process. We can reference Figure 2B and what the author tells us in the passage. The laser is removing corneal matter and flattening the cornea. Only answer choice B correctly lists the effect produced by the PRK technique.

Rules for oxidation numbers

1. Ionic compound: each element takes its charge. 2. Covalent compound: more electronegative takes negative charge as if ionic. 3. Sum of all oxidisation numbers in compound=0 4. Sum of all oxidisation numbers in ion=charge on ion 5. Elements have oxidation number=0 e.g. Br₂, P₄, He, Zn 6. Oxygen has oxidation number=-2 exept as peroxide when ox no=-1 7. Hydrogen has oxidation number=+1

Adenine structure

6-amino purine

Based on the passage, as compared to a wild-type mouse, an Acc2-/-/Cpt1-/- double knockout mouse is most likely to exhibit: A. increased insulin secretion. B. decreased fatty acid oxidation. C. decreased triglyceride synthesis. D. increased malonyl-CoA production.

Based on the passage, as compared to a wild-type mouse, an Acc2-/-/Cpt1-/- double knockout mouse is most likely to exhibit: Jack Westin Advanced Solution: This question has a similar setup to our previous question because we will be referencing the passage for some key information, but we'll ultimately rely on our general knowledge to come up with our final answer.We can start with Acc2 -/-. We're told in the passage ACC2 regulates fatty acid oxidation through the conversion of acetyl CoA to malonyl-CoA (shown in Reaction 1), so the absence of ACC2 can adversely affect fatty acid oxidation.Next, we move on to Cpt1 -/-. Carnitine palmitoyltransferase 1 is an integral part of the carnitine shuttle which functions to transport fatty acids that impermeable to the mitochondrial membranes into the mitochondrial matrix. Once in the matrix, β-oxidation takes place. However, the absence of Cpt1 means there are no fatty acids in the mitochondria, so oxidation is not happening. A. increased insulin secretion. Jack Westin Advanced Solution: increased insulin secretion. This answer choice is tangentially related to what we would see, but we likely wouldn't see increased insulin secretion. We do anticipate seeing increased glucagon as a result of the decreased β-oxidation, but not necessarily insulin. When blood glucose levels decrease, glucagon is released to promote beta-oxidation, not insulin. B. decreased fatty acid oxidation. Jack Westin Advanced Solution: decreased fatty acid oxidation. This answer choice is consistent with our breakdown of the question. We can focus our attention on CPT1 which we mentioned is an integral part of the carnitine shuttle that functions to transport fatty acids into the matrix. When we don't have this shuttle, we don't have the necessary fatty acids in the mitochondria and oxidation is not happening. C. decreased triglyceride synthesis. Jack Westin Advanced Solution: decreased triglyceride synthesis. This gets into the effect of the absence of ACC2. The author presents Reaction 1 in the passage which we can pull up here:Absence of ACC2 means more Acetyl-CoA available for fatty acid synthesis (catalyzed by ACC1). Increased fatty acid synthesis would mean the opposite of this answer choice. We expect increased triglyceride synthesis. D. increased malonyl-CoA production. Jack Westin Advanced Solution: increased malonyl-CoA production. This ties into our reasoning for answer choice B. The absence of CPT1 means no carnitine shuttle transporting fatty acids into the mitochondria and no β-oxidation or formation of acetyl CoA. We will likely see decreased malonyl-CoA production. How do you know you are truly ready? That's where the Jack Westin MCAT Diagnostic Tool comes in. The solution is B. Insulin secretion is stimulated by blood glucose levels. CPT1 is an essential enzyme in the carnitine shuttle, transporting fatty acids into the mitochondria for oxidization. The absence of CPT1 would prevent the transport of fatty acids into the mitochondria for oxidization, irrespective of the presence or absence of ACC2. In the absence of the conversion of acetyl-CoA to malonyl-CoA by ACC2, more acetyl-CoA would be available for the reaction catalyzed by ACC1, which provides the malonyl-CoA needed for fatty acid synthesis. Fatty acids are then used to synthesize triglycerides. The absence of CPT1 would prevent the transport of fatty acids into the mitochondria, where fatty acids are oxidized to produce acetyl-CoA.

Based on the passage, what can be determined about the composition of the mRNA that encodes the protein from which ghrelin is cleaved? It is composed of: A. more than 28 amino acids. B. exactly 84 nucleotides. C. exactly 87 nucleotides. D. more than 87 nucleotides. Ghrelin is a 28-amino-acid peptide hormone that is cleaved from a longer protein produced by the stomach and small intestine.

Based on the passage, what can be determined about the composition of the mRNA that encodes the protein from which ghrelin is cleaved? It is composed of: Jack Westin Advanced Solution: This question is similar to the previous question in this set in that we're ultimately going to rely on our general knowledge to answer, but we still need to pick out some key information from the passage to make sure we're answering this specific question correctly. Note, the test-maker says "Based on the passage" which doesn't always mean we rely exclusively on the passage, but we likely pick out some key details.First thing we want to note is the author talks about the makeup of ghrelin in Paragraph 4. We're told ghrelin is a 28-amino-acid peptide hormone that's actually cleaved from a longer protein. However, we're focused on mRNA, which means we're talking nucleotides. Each amino acid of the 28 is coded by 3 nucleotides. However, we can't forget one major detail: the STOP codon! That means at minimum we're looking at:(28 amino acids x 3 nucleotides/amino acid) + (3 nucleotides STOP codon) = 87 nucleotides.Why do I say minimum? Because the author explicitly tells us ghrelin is cleaved from a longer peptide chain. 87 is going to be the very least number of nucleotides, when in reality that number is going to be higher. Let's glance at our answer choice: A. more than 28 amino acids. Jack Westin Advanced Solution: more than 28 amino acids. The test-maker mentions we're talking about the composition of mRNA so we're dealing in nucleotides, not full amino acids. B. exactly 84 nucleotides. Jack Westin Advanced Solution: exactly 84 nucleotides. This answer choice incorrectly forgets to include the 3 nucleotides from the STOP codon, and also disregards the fact the ghrelin is cleaved from a longer peptide chain. C. exactly 87 nucleotides. Jack Westin Advanced Solution: exactly 87 nucleotides. At first glance this may sound like a viable answer, but we remember that key detail from the passage: ghrelin is cleaved from a longer peptide chain. We expect the actual number to be higher than 87, not exactly 87. D. more than 87 nucleotides. Jack Westin Advanced Solution: more than 87 nucleotides. This answer choice is consistent with our breakdown. We know the number of nucleotides has to include the STOP codon, so we're looking at 87 nucleotides just from the cleaved portion, but ghrelin is cleaved from a longer peptide chain meaning we have more than 87 nucleotides. Answer choice D is our best answer choice. Want more MCAT Prep? We have an MCAT Podcast! The solution is D. The mRNA is made of nucleotides, not amino acids. Exactly 84 nucleotides account for the 28 amino acids in ghrelin but omits the STOP codon. Additionally, the passage indicates that ghrelin is cleaved from a longer peptide chain. The minimum number of nucleotides for the 28-amino-acid peptide ghrelin is 87, including the STOP codon. However, the passage indicates that ghrelin is cleaved from a longer peptide chain. Each amino acid is coded by 3 nucleotides; thus, 84 nucleotides are needed to code for the 28-amino-acid peptide ghrelin. To this number, the triplet that codes for the STOP codon needs to be added. Therefore, 87 is the minimum number of nucleotides for the 28-amino-acid peptide ghrelin. However, the passage indicates that ghrelin is cleaved from a longer peptide chain. For this reason, the total number of nucleotides is higher than 87.

Capacitance equation

C=Q/V

Nersnt Equation function

E = Eo - (RT/nF) log (Q) At 298 K and in base 10 log form, E = Eo - (0.06/n) log (Q)

Electric field strength in a capacitor

E = V/d =Q/EoA

Magnetic force on a moving point charge

F = qvBsin(angle) RHR: thumb velocity, fingers: mag field, palm: force vector

uniform electric field between two plates

Electric field strength is equal in magnitude and has the same direction at all points in the region. The answer to this question is D because the intensity of a uniform electric field E is related to the voltage V and distance d over which it is applied as E = V/d = 4.5 kV/(0.5 m) = 9 kV/m.

In phosphorylation of tyrosine which atom is exchanged for Phosphate

H of the hydroxyl groups

Difference between condensation to produce ester and Fischer esterification

Fischer Esterification is an organic reaction which is employed to convert carboxylic acids in the presence of excess alcohol and a strong acid catalyst to give an ester as the final product

GABA receptors

GABAa (ionotropic) and GABAb (metabotropic) hyper polarize by opening cl channels and cl enter the cells

proteolytic cleavage

Hydrolysis of a protein by a proteolytic enzyme, eg. trypsin, chymotrypsin

Hydroquinone to Quinone

Na2Cr2O7, H2SO4

Oligomers

Proteins with more than one subunit. Ex. Dimer, trimer, tetramer, etc. Each unique subunit is identified by a different Greek letter.

Equlibrium expressions doesn't include

Pure liquids and solids

Ammeter, voltmeter, ohmmeter

Series Parallel around small resistance Large resistance small resistance

Calorimeter phase

all phases

phosphatides

any of a class of compounds that are fatty acid esters of glycerol phosphate with a nitrogen base linked to the phosphate group.

nucleoside

base + sugar

Properties of water

cohesion, adhesion, excellent solvent, solid is less dense than liquid, specific heat high specific heat high bp, fp, surface tension, polarity

STP (standard temperature and pressure)

conditions of 0.00°C (273 K) and 1 atm pressure and 22.4 L

Damping

decrease in amplitude

carboxylation of hydroquinone

done by treating a substrate with carbon dioxide

Status Groups (Weber)

groups of people with similar kinds of attributes or identities such as those based on religion, ethnicity, or race

Graham's Law of Diffusion and Effusion

heavier gas molecules diffuse and effuse slower, lighter gas molecules diffuse and effuse faster; rate1/rate2=(MM2/MM1)^1/2 ; MM=molar mass

sphinogolipids

lipids with an attached amino alcohol. Cell membrane protection and cell signaling. abundant in the nervous system

Prosthetic Groups are

nonprotein components of an enzyme REQUIRED for enzyme activity

homodimer in sds page

one band

P450

oxidizing enzymes that detoxify things. This improves solubility which allows them to enter the kidneys and be released as waste

phosphonic acids

phosphates with OR, OH group

how to do logs without calc

so if -log(1.6X10-4) -(-4+0.16)= 4-0.16 if log(1.6X10-4) -4+0.16 if -log(1.6X104) -(4+0.16)

Polarity Ranking of the Functional Groups

polarity ranking: COOH > Amide > OH > Amine > Aldehyde and ketone > ester > ether > alkane

somatic nervous system

the division of the peripheral nervous system that controls the body's skeletal muscles

What is the average power consumed by a 64-year-old woman during the ascent of the 15-cm-high steps, if her mass is 54 kg?

90 W. Make sure if you are using time for 30 steps you are accounting for 30 steps and if doing 1 step then time for 1 step

Analyte

A substance that is being identified or measured in a laboratory test.

Gap junctions (communicating junctions)

Animal cell junctions that function similarly to plasmodesmata in plant cells - they are membrane-lined channels that join a cell to an adjacent cell - Allows for ions, sugars, amino acids and other small molecules to pass between cells

Guanine structure

2-amino-6-oxy purine

DNA polymerase I, II, III

Can elongate existing DNA strand (primer) Cannot initiate DNA synthesis

In designing the experiment, the researchers used which type of 32P labeled ATP?

because the phosphoryl transfer from kinases comes from the γ-phosphate of ATP. Therefore, the experiment should require γ32P-ATP. The gamma phosphate is the farthest and will be transferred first

peptide bond forms

by the removal of a water molecule, by a dehydration reaction, and between two amino acids

ester + water

carboxylic acid + alcohol

lactose structure

galactose-beta-1,4-glucose reducing

Myelination

increases the speed of axonal conduction decreases the capacitance

When a question asks about "always" or "never",

it has to be something that is very common with that reaction and is always bound to or not bound to happen

amino acids

occurs naturally as L and most have S configuration except cysteine

fatty acid oxidation (beta oxidation)

saturated fatty acids- dehydrogenase to create double bond (produce FADH2), then produce NADH to create ketone, breaks off acetyl-CoA, repeat unsaturated fatty acids- isomerase to move double bond, then produce NADH to create ketone, break off acetyl-CoA, repeat in mitochondrial matrix needs *2 ATP* to initially activate fatty acid need 1 FAD, 1 NAD+ for each 2 C removed produces *1 FADH2, 1 NADH*

factors affecting solubility of solids and gases in water

solids= temp, inc temp inc solubility and vice versa gases= temp inversely propotional inc temp dec solubility in water and vice versa external pressure: direct prop

Adding an electronegative group on Carboxylic acid (with drawing) benzene, halogens etc

stabilize the conjugate base and decreases the pKa of the acid making dissociation more likely

response rate

the number or percentage of surveys completed by respondents and returned to researchers

electromagnetic spectrum

the range of wavelengths or frequencies over which electromagnetic radiation extends.

titrant vs titrand

titrant is known concentration and the titrand is unknown concentration

Power units are

​​​​​​​​​1W = 1 J/s= 1 kgm^2s^-3= 1ftlbs^-1

Bascity

EDG increases bascitity, EWG decrease Bascitity measure of how well a base shares its lone pairs of electrons with a proton

Michael addition reaction

We're *forming a C-C bond*, and we're *breaking a C-C Pi bond*. After we do the addition, we form an enolate - so adding acid gives us the neutral product

Sign conventions for mirrors

absolute value of m< 1 = image is reduced and >1 image is enlarged For single lens and mirror UV upright images arevirtual IR inverted images are real NO image is formed when an object is a focal lengtha way

elastic potential energy equation

U = 1/2 k x^2

Gamma decay releases

photon / energy

Reynolds number

predicts whether flow is laminar or turbulent

The most likely function of the ebulliator is to:

prevent superheating of the liquid to be distilled *The passage mentions that the ebulliator was added to introduce small air bubbles into the system. This is the same function provided by a boiling chip at atmospheric pressure. The air bubbles break the surface tension of the liquid being heated and prevent superheating and bumping.

Resistance and temperature

resistance changes with temperature, resistance of metals increases as temperature increases

coordinate covalent bond

forms when one atom donates both of the electrons to be shared with an atom or ion that needs two electrons to form a stable electron arrangement with lower potential energy No nonmetal to nonmetal is a covalent bond. Metal to nonmetal is ionic and nonmetal to transition metals are coordinate covalent because the nonmetal is donating both electrons to make the bond, the transition metal does not give an electron to make the bond. A coordinate bond (also called a dative covalent bond) is a covalent bond (a shared pair of electrons) in which both electrons come from the same atom. A covalent bond is formed by two atoms sharing a pair of electrons. The atoms are held together because the electron pair is attracted by both of the nuclei. In the formation of a simple covalent bond, each atom supplies one electron to the bond - but that does not have to be the case.

Four organic compounds: 2-butanone, n-pentane, propanoic acid, and n-butanol, present as a mixture, are separated by column chromatography using silica gel with benzene as the eluent. What is the expected order of elution of these four organic compounds from first to last?

n-Pentane → 2-butanone → n-butanol → propanoic acid

Avagadro's Law

V1/n1 = V2/n2 Equal volumes of gases at the same temperature and pressure contain equal number of molecules

Squalene

a cholesterol precursor found in whale liver and plants

Cylinderical Lens

focus the light into a line which doesn't change the location at which the image is formed

test cross

Testing a suspected heterozygote by crossing it with a known homozygous recessive

steroid hormones list

- Glucocorticoids: affect glucose metabolism, cortisol - Mineralocorticoids: maintain fluid and electrolyte balance, aldosterone - Androgens: secondary sex characteristics

Hill's Coefficient (cooperativity)

- value indicated nature of binding by the molecule - if > 1, positively cooperative binding is occuring (binding of one will increase the binding of others) - if < 1, negatively cooperative binding is occuring (binding of one will decrease the binding of others) - if = 1, enzyme does not exhbit cooperative binding Yeah as I understand it you have it correct. Hills coefficient > 1 is positive cooperativity (shift from T —> R, increasing subunits affinity) and < 1 is negative cooperativity (shift from R —> T, decreasing subunits affinity)

FLAT PEG (anterior pituitary)

-FSH -LH -ACTH -TSH -Prolactin -ENdorphins -GH (growth hormone)

Heterogenous catalyst definition

A catalyst in a different state to the reactants

What is the electronic configuration of the Co(II) center found in vitamin B12?

Ar 3d7 Co(II) is a dication and is formed from the atomic element by the loss of two 4s electrons. As a consequence, only seven 3d electrons remain in the valence shell Electrons are taken away from higher energy levels first

Two friends accept internships with a city council member even though they do not agree with many of the council member's policies. Which is most likely to happen if they are in a state of dissonance? The students will:

Cognitive dissonance has 3 solutions; 1) change your attitude 2) change your behaviour 3) neither In this scenario, changing their attitude is in line with the first solution. Option 4 really doesn't make any sense, as it has nothing to do with cognitive dissonance. Quitting the internship after a period of time, option 2 also doesn't make any sense. If they don't agree with the council member, why would they work there and immediately quit? That brings us to option 1 & 3. When it comes to cognitive dissonance, in order to solve it we have to either align ourselves with it or leave entirely. In option 1, this satisfies that criteria. In order to get past cognitive dissonance, they will change their attitude to be in line w/ the council member. Hence, why it's the better answer. Option 3 is also an option, but they're still faced with the dissonance. On the MCAT, cognitive dissonance is meant to be solved so that you don't feel it anymore. If you retain your original belief, you'll still be in that state of dissonance. However, if you change your attitude then you'll solve the issue and be past it. It's one of those things AAMC expects you to figure out as to how they want to define their teams and their solutions, and then apply them. Jack Westin Advanced Solution: In psychology, "cognitive dissonance" describes the mental stress or discomfort experienced by an individual who holds two or more contradictory beliefs, ideas, or values at the same time, or is confronted by new information that conflicts with existing beliefs, ideas, or values. When we experience cognitive dissonance, we are motivated to decrease it because it is psychologically, physically, and mentally uncomfortable. We can reduce cognitive dissonance by bringing our cognitions, attitudes, and behaviors in line—that is, making them harmonious. We can change the discrepant behavior, change our cognitions through rationalization or denial, or add a new cognition. A adapt their attitudes to be more in line with the council member. Jack Westin Advanced Solution: While the two friends may not like to hear it, this is going to be the most likely. We reduce dissonance by bringing our cognitions, attitude, and behaviors in line. We expect they will change their behaviors and beliefs to match their behavior while working for the city council member. B quit the internships after a period of time. Jack Westin Advanced Solution: It's possible this happens, but it's more likely the friends will change their attitude and beliefs to be more in line with those of the council member. C continue in the internships but retain their original beliefs. Jack Westin Advanced Solution: This seems very unlikely. While the friends may try to hold on to their beliefs, it's more likely they will change their beliefs as they continue to work for the city council member. We would need more information about the specific beliefs to make this claim. D do as little work as possible so as not to advance the council member's agenda. Jack Westin Advanced Solution: This is a change in behavior, but it's easier for the friends to change their beliefs rather than risk being fired for doing very little work. Changing their beliefs would be the best way to decrease cognitive dissonance. Answer choice A is our best answer. Need extra help breaking down AAMC logic? Find out how you can meet with one of our expert MCAT tutors! Solution: The correct answer is A. This is a Psychology question that falls under the content category "Individual influences on behavior." The answer to this question is A because research on cognitive dissonance has found that people tend to change their attitudes to match their behaviors, rather than change their behavior. It is a Knowledge of Scientific Concepts and Principles question because you must demonstrate your knowledge of cognitive dissonance theory. This may be an ultra late response , but... I just got this question wrong too and upon reading up on KA notes.. It appears that A was the correct answer because according to the cognitive dissonance theory, the feelings of discomfort experienced from cog. dissonance can lead to alterations in our beliefs/behaviours. Choice A also seems like one of the 4 things someone could do to reduce that discomfort of feeling cognitive dissonance (modify our cognitions). Choices B, C, and D do not seem to fit any one of the 4 things we could do to reduce the discomfort of feeling cognitive dissonance.

Indicators

Litmus: Acid = red; Base = blue; Neutral = purple Phenolphthalein: pH < 8.2 = colorless; pH > 8.2 = purple Methyl Orange: pH < 3.1 = red; pH > 4.4 = yellowBromophenol Blue: pH < 6 = yellow; pH > 8 = blue

charge on a capacitor formula

Q= CV Q = charge (units: coulomb) C = capacitance (units: farads F) V = voltage differential (units: volts )

vapor pressure

The ability of the molecules to move to the gas phase. If strong intermolecular forces then low vapor pressure high boiling point and vice versa High atmospheric pressure high bp When a liquid is heated, its molecules obtain sufficient kinetic energy to overcome the forces holding them in the liquid and they escape into the gaseous phase. By doing so, they generate a population of molecules in the vapor phase above the liquid that produces a pressure—the vapor pressure of the liquid.

basic

The cations of the weak bases are acidic cations

Eating suppresses ghrelin secretion. However, this suppression does not appear to be caused by the presence of food in the stomach or duodenum, nor by stomach distension. Given this, which of the following physiological conditions is likely to suppress ghrelin secretion following a meal?

The solution is C. The levels of circulating glucagon will decrease after a meal due to an increase in blood glucose levels. The osmolarity of the ileum will increase, not decrease, after a meal. The levels of circulating insulin will increase after a meal due to an increase in blood glucose levels. The levels of circulating insulin will increase, not decrease, after a meal.

If sounds produced by the human vocal cords are approximated as waves on a string fixed at both ends, and the average length of a vocal cord is 15 mm, what is the fundamental frequency of the sound? (Note: Use 3 m/s for the speed of sound through the vocal cord.) _

This question requires you to apply knowledge of sound production by vibrating cords and strings, and to identify the physical properties of such cords and their relationship with sound characteristics: frequency, propagation speed, and amplitude. In particular, the relationship between the sound speed and frequency for a string fixed at both ends must be used. The fundamental frequency of a vibrating string is given by f = v/(2L) = (3 m/s)/(30 × 10-3 m) = 1/10-2 Hz = 100 Hz.

homogenous catalyst

a catalyst that is in the same phase as all the reactants and products in a reaction system

galvanic cell (voltaic cell)

a cell in which energy released from a spontaneous chemical reaction is used to generate electricity anode - cathode + spontaneous electrons move from negative to positive electrons move from anode to cathode

phosphate ester

a compound formed by reaction of an alcohol with phosphoric acid

phase diagram

a graph showing the conditions at which a substance exists as a solid, liquid, or vapor

inductive effects on acidity

electronegative elements near acidic protons increase acid strength because they weaken the bond by drawing electrons

benzoin structure

used to clean the skin before applying an adhesive dressing

Speed of a wave on a string

v = √(tension / mass per unit length) linear density

Imine Formation

- Ammonia (NH3) is added to the carbonyl, resulting in the elimination of water, and generating an imine. - Imine can undergo tautomerization and form enamine - Example of condensation reaction since a small molecule is lost during the formation of a bond between two molecules.

Cytosine structure

-Pyrimidine w/ 3 N (2 in ring, 1 branched) and 1 O. -2 double bonds.

Which process is most likely responsible for the failure of a gamete to receive a copy of a particular chromosome? A. Recombination B. Linkage C. X inactivation D. Non-disjunction

Which process is most likely responsible for the failure of a gamete to receive a copy of a particular chromosome? Jack Westin Advanced Solution: This is a standalone question that relies on our general knowledge of genetics. Our correct answer choice is going to explain the likely reason for the failure of a gamete to receive a copy of a particular chromosome. Best way to answer this question is to define all four of our answer choices first. We can find the one that best answers the specific question being asked. A. Recombination Jack Westin Advanced Solution: Recombination. Recombination involves the formation of genetic combinations in offspring that are not present in the parents. That happens from independent assortment and crossing over and introduces genetic diversity into gametes during meiosis. This does not explain the failure of a gamete to receive a copy of a particular chromosome. B. Linkage Jack Westin Advanced Solution: Linkage. Linkage is the property of genes of being inherited together. The segregation of alleles into gametes can be influenced by linkage, but is not responsible for the failure of a gamete to receive a copy of a particular chromosome. C. X inactivation Jack Westin Advanced Solution: X inactivation. X inactivation has to do with inactivation of a copy of an X chromosome. The inactive X chromosome is packaged into heterochromatin. This inactivation does not explain the failure of a gamete to receive a copy of a particular chromosome. D. Non-disjunction Jack Westin Advanced Solution: Non-disjunction. Nondisjunction occurs when chromosomes fail to segregate during meiosis; when this happens, gametes with an abnormal number of chromosomes are produced. This can occur in Anaphase 1 and Anaphase 2 and can lead to gametes with too many or too few chromosomes. This is the exact answer we're looking for and is most likely responsible for the failure of a gamete to receive a copy of a particular chromosome. Want more MCAT Prep? We have an MCAT Podcast! The solution is D. Recombination occurs when genetic material is broken and recombined. Linkage occurs when genes tend to be inherited together. X inactivation occurs when one X chromosome in a female becomes almost completely inactivated during embryonic development. Non-disjunction occurs when sister chromatids fail to separate during cell division.

Which statement accurately describes the properties of maltose? A It is a reducing disaccharide in which only one anomeric carbon is involved in the glycosidic bond. B It is a reducing disaccharide in which both anomeric carbons are involved in the glycosidic bond. C It is a nonreducing disaccharide in which only one anomeric carbon is involved in the glycosidic bond. D It is a nonreducing disaccharide in which both anomeric carbons are involved in the glycosidic bond.

Which statement accurately describes the properties of maltose? Jack Westin Advanced Solution: We can answer this question using external information, or general knowledge. It's a very broad question, so quick glance at our answer choices shows we have to determine if it's reducing or nonreducing, and which anomeric carbons are involved in the glycosidic bond.Maltose is a disaccharide that is composed of two linked glucose units and can be a product in the breakdown of starch. The two glucose units that make up maltose are joined by an alpha 1-4' linkage.In the cyclic form of a sugar, the anomeric carbon is the carbon that was part of the carbonyl group in the straight-chain structure. Maltose's two glucose molecules are linked a way to leave one anomeric carbon that can open to form an aldehyde group. C1 is the anomeric carbon, so only one anomeric carbon is involved in the glycosidic bond. The anomeric carbon of our left glucose here. The anomeric carbon on the right side can open up to form an aldehyde group, so we can classify maltose as a reducing sugar. A reducing sugar is capable of acting as a reducing agent because it has a free aldehyde group or a free ketone group. All monosaccharides are reducing sugars, along with some disaccharides. A It is a reducing disaccharide in which only one anomeric carbon is involved in the glycosidic bond. Jack Westin Advanced Solution: This sounds like the breakdown we just went through. One anomeric carbon involved in the glycosidic bond, the other can open up to form an aldehyde group, so we classify maltose as a reducing disaccharide. B It is a reducing disaccharide in which both anomeric carbons are involved in the glycosidic bond. Jack Westin Advanced Solution: Both anomeric carbons aren't involved in the glycosidic bond, we can eliminate this answer choice for contradicting our breakdown of the question. C It is a nonreducing disaccharide in which only one anomeric carbon is involved in the glycosidic bond. Jack Westin Advanced Solution: Answer choices C and D both say maltose is a nonreducing disaccharide, which we determined isn't true. D It is a nonreducing disaccharide in which both anomeric carbons are involved in the glycosidic bond. Jack Westin Advanced Solution: Answer choices C and D both say maltose is a nonreducing disaccharide, which we determined isn't true during our breakdown of the question. We're left with our correct answer, answer choice A. Want more MCAT Prep? We have an MCAT Podcast! Solution: The correct answer is A. This is correct. This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is A because maltose contains a (1→4) glycosidic bond, which means that the carbons involved are the C1 and C4 of the respective monosaccharides. The C1 is the anomeric carbon and a disaccharide that has a C1 carbon that is not involved in a glycosidic bond is said to have a hemiacetal end. This is the requirement to be a reducing sugar. It is a Knowledge of Scientific Concepts and Principles question because you must recall the properties of maltose.

Can enzymes produce chiral products?

Yes

electrolytic cell

an electrochemical cell used to cause a chemical change through the application of electrical energy requires battery nonspontaneous E is neg both electrodes are inert but then oxidation reduction happens the anode is + and the cathode is - electrons move from anode to cathode but since nonspontaneous could say that electrons are moved to neg charge even though they don't want to

A team of researchers at a pharmaceutical company tests a new cancer drug. The researchers have concluded that the drug is effective, but other scientists CANNOT replicate the findings. Which is the most likely explanation for the lack of replicability of the original results? Jack Westin Advanced Solution: What a frustrating situation! We're told researchers are testing a new cancer drug that they deem effective, but other scientists cannot replicate the findings. We want to explain a possible reason for this unfortunate lack of replicability of the original results. We can go through the four explanations listed as answer choices, define each one, and decide which is the most likely explanation for the lack of replicability of the original results. A. Base rate fallacy Jack Westin Advanced Solution: Base rate fallacy. The base rate fallacy has to do with incorrectly judging a situation like not considering all the information at a researchers' disposal. Researchers might erroneously ignore base rates in favor of individuating information. That is not what happens in our question stem B. Hindsight bias Jack Westin Advanced Solution: Hindsight bias is the belief that the event just experienced was predictable, even though it really wasn't. This is not the reason for the lack of replicability of the original results. C. Observer bias Jack Westin Advanced Solution: Observer bias occurs when there is bias on the part of the specific researcher - this could occur because of the researcher's prior knowledge of the study, for example. This is a strong possibility in this case and the initial results may not have been obtained properly. That bias is what caused the initial conclusion that the drug is effective, despite the results not being replicable. This is a strong answer choice. D. Public verifiability Jack Westin Advanced Solution: Public verifiability is likely a distractor. This is not listed on AAMC's content outline and not a term commonly used when talking about research results. This answer choice is instead referring to the reason the subsequent scientists are testing the initial drug. We can eliminate this answer choice. Answer choice C remains our best option. How do you know you are truly ready? That's where the Jack Westin MCAT Diagnostic Tool comes in. The solution is C. The base rate fallacy refers to the error people make when they ignore the base rates (i.e., prior probabilities) when evaluating the probabilities (or frequencies) of events. The hindsight bias (i.e., the "I knew it all along" effect) refers to the tendency for a person to overestimate how well he or she could have successfully predicted a known outcome (i.e., a "forecast," given before the outcome was known). Any bias on the part of the observers recording the data could have contaminated the original results. In this case, it is possible that proper precautions (for example, ensuring that observers were "blind" with respect to which patients received the drug and which patients received the placebo) were not taken. Public verifiability is the reason other scientists are attempting to replicate the original findings.

another name for it is experimentar bias or observer-expectancy effect. google: Observer bias (also called experimenter bias or research bias) is the tendency to see what we expect to see, or what we want to see. When a researcher studies a certain group, they usually come to an experiment with prior knowledge and subjective feelings about the group being studied. in relation to this question, the answer is observer bias because the researchers were able to attribute their own knowledge to the study to determine that the drug is effective. the study can't be replicated because the study didn't properly blind the observers/patients which causes an observer bias to be present. also never seen public verifiability before but usually when I come across obscure terms like this, I usually mark them as wrong just on the fact that I've never seen it before and don't know what it is.

Which molecule is likely to have a higher zero-point energy than 206Pb? 204Pb 208Pb 238U 235U

A According to the passage, heavier isotopes have lower zero-point energy. Conversely, lighter isotopes must have higher zero-point energies—which is what the stem asks about. Answer A is the only one of the four options that is a lighter isotope than lead-206. Lead-208 is heavier, and both Uranium isotopes are considerably more massive than Lead-208.

Hyperventilation (rapid breathing) occurs in response to respiratory acidosis because hyperventilation: increases blood pH by decreasing CO2 concentrations. increases blood pH by decreasing HCO3- concentrations. decreases blood pH by decreasing CO2 concentrations. decreases blood pH by decreasing HCO3- concentrations.

A dumb mistake

Magnesium sulfate

"Addition of magnesium sulfate" - Is the purpose of this to just get rid of excess water from liquid organic solutions post aqueous layer seperation (using a sep funnel)? This is not correct because it doesn't "freeze dry" the protein.

glut 2 and glut 4

*GLUT 2*: low affinity transporter in hepatocytes and pancreatic cells. Captures the excess glucose after a meal primarily for storage. Km is high (15 mM), so the liver picks up excess glucose and stores it preferentially after a meal when blood glucose levels are high. *GLUT 4*: in adipose tissue and muscle and responds to glucose concentration in peripheral blood. Rate of transport increased by insulin, which stimulates movement of additional GLUT 4 transporters to the membrane. Km close to glucose levels in blood (~5 mM). Glut 2 is in pancreatic and hepatic tissues and glut 4 is in muscle and adipose tissue. Glut 4 is 0 order bc very small levels of glucose uptake gets it to vmax very quickly. So adding more glucose (substrate) doesn't affect the rate of rxn, just like a traditional 0 order rxn. The way to get more glucose to cross the membrane in muscle and adipose cells is to get more glut4 receptors to the membrane with open sites.

Ideal gas vs. real gas (definition)

- Deviations occur when temperature is extremely low or pressure is extremely high - Effects represented by Van der Waals equation - In summary, real gases occupy more volume than ideal gases but they have less pressure A. Increased intermolecular attractions decreases pressure in real gas B. Increased molecular volume increases volume in real gas as shown in the Vanderwaals equation, but its not required in the MCAT.

Aldol Condensation Reaction

- The aldol condensation involves two reaction series, the aldol addition reaction and the condensation reaction. Dehydration occurs through an elimination (technically, E1cB) mechanism to form an α, β-unsaturated aldehyde (enal) or ketone (enone), not the aldol. * The first step of aldol condensation involves a strong base like hydroxide abstracting a proton from the α-carbon (not the β-carbon) of a carbonyl compound (aldehyde or ketone) to form the enolate. * The second step of aldol condensation involves the enolate attacking the aldehyde or ketone through a nucleophilic acyl addition mechanism (not substitution). Only carboxylic acid and its derivative can undergo nucleophilic acyl substitution. - Examples are dehydration, nucleophilic addition , and aldol reaction.

base peak (mass spec)

- The mass of the base peak is the mass of the most stable carbocation fragment you can produce when a C-C bond from the molecule you're studying is broken. C-C bonds are broken because the mass spectrometer fires electrons like bullets at the molecule, and in doing so, often knocks one of the bonding electrons away, breaking the bond.

Function of DNA polymerase

1.) attach free floating nucleotides to the old strand of DNA and then bonds the nucleotides together using a strong covalent bond 2.) proof reads to see if there are any mistakes between the bonded nucleotides

For all rules

B= magnetic field = finger v or I = thumb F= palm for F= ILB sin/ F=qvB sin For magnetic field due to current carrying wire thumb = I Fingers= force

Based on reactions 1 and 2, researchers interested in increasing the efficiency of the chromium(VI) removal process should add which compound to the untreated water? NaOH HCl HClO4 KMnO4

Choice B is the best answer. Answer A should be dismissed. If NaOH were added to the untreated water, it would increase the pH of the reaction mixture, which—according to reactions 1 and 2—requires acidic conditions. Answers C and D are both strong oxidizers and would therefore be entirely counterproductive in a process intended to reduce chromium(VI). Choice B, adding HCl to the untreated water, would increase acidity and provide additional hydrogen ions for reactions 1 and 2, making B the correct choice.

transcription factors

Collection of proteins that mediate the binding of RNA polymerase and the initiation of transcription.

The presence of which type of intercellular connections between endothelial cells of brain capillaries results in the blood-brain barrier? A.Desmosomes. B.Gap junctions C.Intercalated discs D.Tight junctionsThe correct answer is

In the capillaries that form the blood-brain barrier, endothelial cells are wedged extremely close to each other, forming so-called tight junctions. The tight gap allows only small molecules, fat-soluble molecules, and some gases to pass freely through the capillary wall and into brain tissue. The presence of which type of intercellular connections between endothelial cells of brain capillaries results in the blood-brain barrier? Jack Westin Advanced Solution: This is a pseudo-standalone question that is tangentially related to what we covered in the passage. We're focused on different types of intercellular junctions and connections:Best way to answer this question is to go through the four different options listed as answer choices and define each one. We want to pick the best answer that represents the intercellular connections we find in the blood-brain barrier. The blood-brain barrier is highly selective and molecules are not able to cross into the CNS. While vital nutrients are able to cross, toxins and pathogens are usually not able. We want an answer choice that is consistent with this. A.DesmosomesAAMC Standard Solution:Desmosomes are intercellular junctions that function as anchors to form strong sheets of cells. Jack Westin Advanced Solution: Desmosomes. Animal cells may contain junctions called desmosomes which act like spot welds between adjacent epithelial cells. A desmosome involves a complex of proteins and some of these proteins extend across the membrane, while others anchor the junction within the cell. B.Gap junctionsAAMC Standard Solution:Gap junctions are intercellular junctions that provide cytoplasmic channels between adjacent cells. Jack Westin Advanced Solution: Gap junctions. Gap junctions are channels between neighboring cells that allow for the transport of ions, water, and other substances between cells. This is not what we're looking for in our answer choice. C.Intercalated discsAAMC Standard Solution:Intercalated discs are specialized intercellular junctions between cardiac muscle cells that provide direct electrical coupling among cells. Jack Westin Advanced Solution: Intercalated discs. Intercalated discs are synonymous with cardiac muscle, not with the blood-brain barrier. This answer choice is out of scope. D.Tight junctionsThe correct answer is D.AAMC Standard Solution:Tight junctions are intercellular junctions that prevent the movement of solutes within the space between adjacent cells. In blood capillaries, neighboring endothelial cells form tight junctions with one another to restrict the diffusion of harmful substances and large molecules into the interstitial fluid surrounding the brain. Jack Westin Advanced Solution: Tight junctions. Tight junctions create a watertight seal between adjacent cells. At the site of a tight junction, cells are held tightly against each other by many individual groups of tight junction proteins called claudins. These claudins interact with a partner group on the opposite cell membrane. The groups are arranged into strands that form a branching network, with larger numbers of strands making for a tighter seal. That's what we have at the blood-brain barrier. The purpose of these tight junctions in the CNS is to keep liquid from escaping between cells and preventing toxins and pathogens from entering. This allows a layer of cells to act as an impermeable barrier. Answer choice D is our best option.

Which statement best applies Rogers's concept of incongruence to SDT's suggestion for how healthcare professionals can promote autonomous motivation? Jack Westin Advanced Solution: At first glance, this is like a hybrid question in that we'll have to reference some key information from the passage, but ultimately, we'll use our external information to answer the question. The author tells us in Paragraph 4, "Healthcare professionals can increase autonomous motivation by supporting their patients' initiatives and offering them different choices for treatment." We want to find a statement that best applies Roger's concept of incongruence to this suggestion. Carl Rogers believed that people strive to become self-actualized—the "best version" of themselves or "ideal self". The gap between the "ideal" self and the real self can cause discomfort, unpleasant feelings, and lead to defensive behaviors - this gap is called incongruence.

Incongruence refers to the gap between a person's actual self and ideal self.

Nucleophilicity trends

Increases with more negative charge Increases down a group Increases to the left of a period

Based on the results of the kinetics studies and the observed Tm, what is the best conclusion regarding the role of the metal centers in catalysis and enzyme conformational stability?

Jack Westin Advanced Solution: We need to explain the role of the metal centers in the catalytic center, so our two Zinc ions. Their role in catalysis (so we'll use our two constants in table 1) and in conformation stability (so we'll use melting point). Each metal was given a distinct letter in the figure in the passage, so we'll find the role of each. We're going to use the table and information in the passage, and external knowledge to see how the different substitutions in Table 1 affected each of our Zinc ions.Here we have Table 1 and our Figure. Quick recap of the information in our Table that we looked at in our previous questions. What is our wild type? Wild type (WT) refers to the prototypical form of enzyme, as it occurs in nature. What do we see in the experimental data? We touched on this a bit already. Each variant that differs from the wild type has a decreased melting temperature, Tm, meaning both ions provide conformational stability that corresponds with a higher melting temperature. Next, the catalytic constant, Kcat, measures the number of molecules of substrate turned over into product per unit time. Kcat is positively proportional to efficiency. So, the higher your kcat is, the more reactions happen per unit time when you have enough substrate. The fact that kcat is so much higher for the wild type compared to the other variants means both ions are involved in catalysis also. Final answer is going to involve: both ZnA and ZnB provide confirmation stability and are involved with catalysis. A ZnA is responsible for catalysis, and ZnB provides conformational stability. Jack Westin Advanced Solution: These are both true statements, but not necessarily exclusively true. Zn-B is also responsible for catalysis, and Zn-A is also responsible for conformational stability. B ZnB is responsible for catalysis, and ZnA provides conformational stability. Jack Westin Advanced Solution: This is the same answer as answer choice A. Both statements are true again, but the answers are only partially complete. C Both ions are involved in catalysis, and both provide conformational stability. Jack Westin Advanced Solution: This matches our breakdown exactly. Answer choices A and B both said essentially the same thing, so answer choice C is our best answer so far. D Both ions are involved in catalysis, but neither provides conformational stability. Jack Westin Advanced Solution: The second part of the answer choice contradicts our breakdown. We said the melting temperature proved both ions provided conformational stability as well. We can eliminate answer choice D and we're left with our correct answer, answer choice C. How do you know you are truly ready? That's where the Jack Westin MCAT Diagnostic Tool comes in. Solution: The correct answer is C. This is a Biochemistry question that falls under the content category "Principles of chemical thermodynamics and kinetics." The answer to this question is C because the data suggest that replacement of any of the residues has a significant effect on conformational stability (as evidenced by the lowering of Tm) as well as the catalytic rate (as evidenced by the lower values for kcat). This strongly suggests a significant role for both Zn ions in the catalytic sequence. It is a Data-based and Statistical Reasoning question because you are asked to analyze and interpret data presented in Table 1 to arrive at the conclusion.

Is power related to frequency and wavelength

NO

Addition of which disaccharide to a solution of Ag2O in NH3(aq) will NOT result in the deposition of shiny silver mirror on the walls of the reaction vessel?

So basically the reducing sugars can reduce Ag and be oxidized so looking for carbon next to O anomeric carbon to be directly attached to OH. In this case sucrose is not reducing so it will not deposit silver.

Michael mentes

So in summary (correct me if I'm wrong): So a higher a higher Ka means a higher affinity A higher Kd means a lower affinity (more likely to dissociate) Higher Km means lower affinity And Kt is like Km in that it is the substrate concentration at which half the maximum transport capacity is reached. An enzyme's Km describes the substrate concentration at which half the enzyme's active sites are occupied by substrate. A high Km means a lot of substrate must be present to saturate the enzyme, meaning the enzyme has low affinity for the substrate.

Function of salt bridge

Source of unreactive/inert cations and ions which flow back-and-forth into the anode and cathode chambers to counterbalance charges as they build up so cations in sb goes to cathode and anions in sb goes to anode

What pair of compounds found in Table 1 can form extensive networks of intermolecular hydrogen bonds with both participating?

The formation of a hydrogen bond requires both a hydrogen bond donor: a molecule with a hydrogen bonded to a highly electronegative atom such as oxygen or nitrogen, and a hydrogen bond acceptor: an electronegative atom (such as O or N) with an available lone pair of electrons. Of the pairs listed, only glycine and methanol are able to act both as hydrogen bond donors and acceptors. Choice C is the best answer. carbon dioxide as an option but not because it can't donate hydrogens Carbon dioxide is a symmetrical, linear molecule, so despite having two C=O bonds, both dipoles cancel each other out and the molecule is non-polar overall. The question also asked for both molecules participating in hydrogen bonding, and CO2 has no hydrogens to be a hydrogen bond donor anyway.

The side chain of which amino acid is used by transaminases to bind to pyridoxal phosphate in the enzyme's resting state?

The side chain depicted in Figure 2, is -(CH2)4NH2 which corresponds to lysine, or Lys. The resting state is shown as Compound 7 in Figure 2.

What mass of Compound 1 (MW = 800 g/mol) is contained in the solution used to prepare liposomes that elute at 20 mL by size-exclusion chromatography?

The solution concentration was 0.10 mM for the liposomes that elute at 20 mL. Since the total volume of solution used in their preparation was 1 mL, the mass of lipid can be calculated as: (0.10 × 10-3 mol/L) × (1 × 10-3 L) × (800 g/mol) = 8 × 10-5 g = 80 µg. It is a Scientific Reasoning and Problem Solving question because you are asked to determine and use a formula to solve a scientific problem. Elution volume is the amount of solvent needed to elute it from the column. This lets you look at the chromatography experiment to see that the liposomes eluted after 20 mL of solvent were made from 0.10M. They ask you to calculate the mass used to prepare that 0.10M sample. They were synthesized with 1mL of Compound 1 (you have that highlighted).

What does the behavior of liposomes prepared from compounds 1 and 2 upon mixing indicate about the energetics of their transformations? Liposomes prepared from: A both Compound 1 and Compound 2 are under kinetic control. B Compound 1 are under kinetic control, but those prepared from Compound 2 are under thermodynamic control. C Compound 1 are under thermodynamic control, but those prepared from Compound 2 are under kinetic control. D both Compound 1 and Compound 2 are under thermodynamic control.

What does the behavior of liposomes prepared from compounds 1 and 2 upon mixing indicate about the energetics of their transformations? Liposomes prepared from: Jack Westin Advanced Solution: In the passage we found there were differences when we mixed the compounds. Compound 1 was stable to mixing, but Compound 2 formed new liposomes. We want to know why exactly this difference existed. That means we're going to reference the experiment in the passage, but then we'll use our general knowledge to explain the behavioral differences. Quick glance at our answer choices shows we'll have to distinguish between kinetic and thermodynamic control.The passage says "Liposomes formed from Compound 1 were stable to mixing, but mixing those from Compound 2 formed new liposomes with an average size expected for the effective final lipid concentration." Compound 1 doesn't change; Compound 2 quickly results in the formation of new liposomes.We have to distinguish between kinetic control and thermodynamic control, so let's do some quick background. Short reaction times favor kinetic control, whereas longer reaction times favor thermodynamic reaction control.Example: product A forms faster than product B because the activation energy for product A is lower than that for product B, but product B is more stable. In such a case A is the kinetic product and is favored under kinetic control and B is the thermodynamic product and is favored under thermodynamic control. The pathway that's favored depends on the condition of the reaction.If a reaction is under kinetic control, it is said that the preferred product/major output product of the reaction is the one that is formed most quickly and has the lowest activation energy barrier. It's usually associated with irreversible reactions. It explains why Compound 1, after forming the liposomes, doesn't form more energetically stable liposomes upon mixing.Thermodynamic control, is preference for a more stable product (lower free energy). No matter which product forms first, if there is a lower energy state product that can be formed, then it will be formed. This is associated with reversibility. That explains why Compound 2 is able to reverse the liposome forming reaction, and form more energetically stable liposomes after mixing. Our answer is kinetic control for Compound 1 and thermodynamic control for Compound 2. A both Compound 1 and Compound 2 are under kinetic control. Jack Westin Advanced Solution: Remember, we said only compound 1 is under kinetic control. Compound 2 is under thermodynamic control. This contradicts our breakdown. B Compound 1 are under kinetic control, but those prepared from Compound 2 are under thermodynamic control. Jack Westin Advanced Solution: This answer is saying exactly what I said in our breakdown of the question. C Compound 1 are under thermodynamic control, but those prepared from Compound 2 are under kinetic control. Jack Westin Advanced Solution: This answer choice is the opposite of our breakdown, so answer choice B remains the best option. D both Compound 1 and Compound 2 are under thermodynamic control. Jack Westin Advanced Solution: Only Compound 2 is under thermodynamic control, not Compound 1. Compound 1 is under kinetic control. We can stick with our correct answer choice, answer choice B. Need extra help breaking down AAMC logic? Find out how you can meet with one of our expert MCAT tutors! Solution: The correct answer is B. This is a General Chemistry question that falls under the content category "Principles of chemical thermodynamics and kinetics." The answer to this question is B. The mixing experiments demonstrate that liposomes formed from Compound 1 cannot attain their thermodynamically preferred state and are therefore under kinetic control. Mixing liposomes of different sizes of Compound 2, on the other hand, results in the formation of new liposomes which are of an intermediate size indicating that they rapidly attain their thermodynamically preferred state. It is a Scientific Reasoning and Problem Solving problem because you are asked to bring together theory, evidence, and observations to draw a conclusion. I got this question wrong but afer reading a post someone made on another thread I actually agree with the AAMC logic. From the passage: "The researchers mixed liposomes of different sizes and observed that those formed from Compound 1 were stable to mixing, but mixing those from Compound 2 formed new liposomes with an average size expected for the effective final lipid concentration." Compound 1 never formed new liposomes because it was limited to kinetic conditions. It stayed at its less stable liposome form. On the other hand, compound 2 was able to rearrange into a more stable product because it was subjected to thermodynamic conditions and thus able to rearrange into a more stable product. If a molecule CAN change into its more stable product given the conditions it's in it will. Compound 1 didn't because the conditions didn't favor the thermodynamic product. Compound 2 has more "wiggle room" to form the more stable thermodynamic product. Compound 1 doesn't. Also makes sense why thermodynamic products are related to reversible reactions and kinetic products are related to irreversible reactions. Reversible meaning if the thermodynamic product can be formed it will. Kinetic products are essentially stuck where they are. Even if they wanted to form the more stable product they just can't given the conditions. 32 ReplyShare level 1Impossible_Phase_409·1 yr. ago Late to the party on this but figured I'd answer since this is the top result on google for this Q. So I looked up the actual paper they took this passage from and it said this: "Therefore, single-chain lipids can jump between membranes and easily arrange in new ways giving rise to different aggregate structures in response to changing solution conditions. In other words, single-chain lipids behave more like a system under thermodynamic control that constantly seeks more energetically favorable structures as the environment changes. Conversely, double-chain lipids, such as diacyl phospholipids, of similar hydrocarbon length are not able to escape into solution from a membrane environment due to the high energetic cost of exposing their large hydrophobic surface area to water. The result is that, once double-chain phospholipids are arranged into bilayer membrane vesicles, they are kinetically trapped in that state and, thus, are less responsive to solution conditions. " So based on this, I reasoned that because compound 2 formed new liposomes with a size that was EXPECTED it is thermodynamic/favorable and compound 1 did not change with the conditions so it was the less favorable & kinetic.

acetyl group

The two-carbon molecule made from pyruvate during pyruvic acid oxidation (the transition reaction).

An experimental setup designed to measure the resistance of an unknown resistor R using two known resistors R1 and R2, the variable resistor R3, a voltage source, and a voltmeter is shown. Which relationship gives the value of R when R3 is adjusted so that the voltmeter reading is zero?

This is a Physics question that falls under the content category "Electrochemistry and electrical circuits and their elements." The answer to this question is A, because when the voltmeter reading is zero, the voltage across R is equal to the voltage across R1 and from Ohm's law, IR = I1R1, where I and I1 are the currents through the resistors. Moreover, IR3 = I1R2. By taking the ratio of these two equations, it follows that R/R3 = R1/R2, which is equivalent to R = R3 × R1/R2. It is a Reasoning about the Design and Execution of Research question because you must reason about the features of this experimental setup and the association between the variables that enable the use of this particular design for its stated purpose.

What expression gives the amount of light energy (in J per photon) that is converted to other forms between the fluorescence excitation and emission events?

This is asking about the energy dissipated which is basically the difference between the energy for excitation and emission deltaE=hdeltaf=(hc/w1)-(hc/w2)=hc(1/w1-1/w2)

A protein with which properties will most likely have the largest negative net charge at pH 7? _

This is correct. This Biochemistry question falls under the content category "Separation and purification methods." The answer is A because a protein with a low pI would be negatively charged at pH 7. This protein, being anionic, would bind to an anion exchange column (eliminates choices C and D). Largest negative net charge implies the presence of a large quantity of negatively charged amino acids, allowing the protein to bind tightly to the column. A high concentration of NaCl would be required for elution (eliminates choice B). This is a Reasoning about the Design and Execution of Research question because you must understand how the design of ion-exchange chromatography experiments affects the separation of proteins with various charges.

melting temperature (Tm)

To answer this question I believe you look at the Y axis and go to the .5 or 50% point as Tm is the point where 50% of DNA or protein is folded/unfolded. I understand where your logic was in looking at the midpoint because that's exactly what I did 2 ReplyShare level 1arp325_·4 yr. ago I believe the definition for melting temperature is the point where there is 50% denaturation. I guess it's just another fact to remember.

Muscle rigidity or stiffness

Tone is increased, adversely affecting movement. People with PD may demonstrate cogwheel motions, which are jerky, sometimes painful movements with joint mobility, most commonly in the upper extremities. Fatigue becomes a barrier to occupational performance because of the increased effort needed to produce voluntary movement.

What can be inferred from the results of Scheme A about the relative thermodynamic stabilities of Compounds 4a and 4b? A. They are equally stable. B. Compound 4a is less stable than Compound 4b. Compound 4a is more stable than Compound 4b. D. Nothing can be inferred about relative thermodynamic stabilities.

What can be inferred from the results of Scheme A about the relative thermodynamic stabilities of Compounds 4a and 4b? Jack Westin Advanced Solution: The passage states that upon dehydration of the cyclized product, a thermodynamic mixture of products is formed. Recall that there are two major factors influencing which products of a reaction are preferentially formed: thermodynamics and kinetics. Thermodynamic control will favor energetically stable products even if they require a higher activation energy and more time to produce. Kinetic control will favor products that form more quickly; these products will often be less energetically stable but are formed faster due to lower activation energies.According to Scheme A, 60% of the product is compound 4a and 40% is 4b. A. They are equally stable. Jack Westin Advanced Solution: If both products were equally stable, thermodynamic control would produce roughly equal amounts of 4a and 4b. There is more 4a produced than 4b, so 4a should be more stable. B. Compound 4a is less stable than Compound 4b. Jack Westin Advanced Solution: For this to be true, the amount of 4a produced would need to be less than the amount of 4b produced. This is the opposite of what we see in Scheme A. C. Compound 4a is more stable than Compound 4b. Jack Westin Advanced Solution: This is true; compound 4a must be more stable than 4b because more 4a is produced (60% > 40%) and we are told that a thermodynamic mixture is produced indicating thermodynamic control. Again, thermodynamic control of a reaction will favor the energetically stable product. D. Nothing can be inferred about relative thermodynamic stabilities. Jack Westin Advanced Solution: This is incorrect. As noted above, relative stability can be inferred because we are told in the passage that the products produced are a thermodynamic mixture. Answer choice C is the best answer. This is an Organic Chemistry question that falls under the content category "Nature of molecules and intermolecular interactions." The answer to this question is C because there is a higher percentage of Compound 4a than Compound 4b at equilibrium (a thermodynamic mixture). For a thermodynamic mixture of isomeric products, the relative mole ratio of products is directly related to the relative stability of these products. It is a Scientific Reasoning and Problem Solving question because the concept of thermodynamic stability of isomers must be correlated to percent ratio of products.

dehydrogenase enzymes

What enzymes transfer electrons to NAD+ are involved in oxidation and reduction

What is the general structure of a hydroxamic acid? Jack Westin Advanced Solution: I don't know about you, but I certainly don't know the structure of hydroxamic acid off the top of my head! Luckily, there is a very brief description in the passage that notes the acetylation of hydroxylamine nitrogen. The key description is the acetylation (carbonyl with a methyl group attached) of the nitrogen in the hydroxylamine (alcohol attached to an amine).

What is the general structure of a hydroxamic acid? Jack Westin Advanced Solution: I don't know about you, but I certainly don't know the structure of hydroxamic acid off the top of my head! Luckily, there is a very brief description in the passage that notes the acetylation of the hydroxylamine nitrogen. The key description is the acetylation (carbonyl with a methyl group attached) of the nitrogen in the hydroxylamine (alcohol attached to an amine). A. Jack Westin Advanced Solution: A. The nitrogen on the left side is lacking an alcohol attachment so it is missing the hydroxyl- part of the hydroxylamine. B. Jack Westin Advanced Solution: B. There should not be an oxygen atom present between the carbonyl and the nitrogen; the nitrogen needs to have two distinct functional groups attached, the acetyl group and the alcohol. C. Jack Westin Advanced Solution: C. This is the correct answer. There is a nitrogen with an alcohol attached as well as a carbonyl with an R group attached. D. Jack Westin Advanced Solution: D. While there is indeed an alcohol attached to an amine, there is an extra carbon that does not belong between the nitrogen and the carbonyl.

What was the most likely purpose of adding bovine serum albumin to the kinetics experiments in the passage? Bovine serum albumin: A acts as a co-catalyst for the reaction. B prevents the esterase from adhering to the walls of the vessel. C helps buffer the solution from changes in pH. D is a non-specific target for protease contaminants.

What was the most likely purpose of adding bovine serum albumin to the kinetics experiments in the passage? Bovine serum albumin: Jack Westin Advanced Solution: The students used 0.1 mg/ml bovine serum albumin in their setup. This question wants to know the purpose of using this BSA. The passage tells us BSA mobilizes proteins and lipids, but doesn't add any additional information. BSA is used to stabilize some enzymes during the digestion of DNA, and to prevent adhesion of the enzyme to reaction tubes, pipette tips, and other vessels-hence the mobilization of proteins and lipids mentioned in the passage. A acts as a co-catalyst for the reaction. Jack Westin Advanced Solution: BSA does not bring about the catalysis in the reaction. Co-catalysts would do that, while BSA only mobilizes proteins and lipids. This contradicts information given in the passage. B prevents the esterase from adhering to the walls of the vessel. Jack Westin Advanced Solution: Again, we said BSA mobilizes proteins and lipids. That mobilization can prevent the esterase from adhering to the walls of the vessel. C helps buffer the solution from changes in pH. Jack Westin Advanced Solution: There was no mention of BSA buffering the solution from changes in pH. This contradicts information given in the passage. D is a non-specific target for protease contaminants. Jack Westin Advanced Solution: Again, not a purpose of this experimental setup. This contradicts my initial breakdown of the question. We can stick with answer choice B-BSA mobilizes proteins and lipids, and it prevents the esterase from adhering to the walls of the vessel. Need extra help breaking down AAMC logic? Find out how you can meet with one of our expert MCAT tutors! Solution: The correct answer is B. This is a Biochemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to this question is B since the passage specifically states that albumin is a protein that mobilizes proteins and lipids in serum. In the context of a kinetics experiment it is perfectly logical to assume that albumin is added to maintain homogeneity and prevent the enzyme from adhering to walls and other surfaces which would inhibit its activity. It is a Reasoning about the Design and Execution of Research question because you are asked to reason about the features of a research study.

When the stopcock is opened, how far up the tube will the water rise? A. 1 m B. 5 m C. 10 m D. 20 m

When the stopcock is opened, how far up the tube will the water rise? A. 1 m Jack Westin Advanced Solution: A. Close, but not quite. Double check the arithmetic on your exponents, this answer is off by a factor of just 10. B. 5 m Jack Westin Advanced Solution: B. We only worked with orders of 10 in our calculations so our answer should be an order of 10. C. 10 m Jack Westin Advanced Solution: C.This is the correct answer. When the stopcock is opened, the fluid will rush into and up the tube until the pressure inside the tube, contributed by the fluid, is equal to the atmospheric pressure outside of the tube. This will occur when the fluid reaches a height of 10 m. D. 20 m Jack Westin Advanced Solution: D. As with answer choice B, we only worked with orders of 10 in our calculations so our answer should be an order of 10. The correct answer is 10 m.

When two amino acids are joined via a peptide bond, what is the mass of the byproduct of this reaction? (Note: Assume that the amino acids were not modified by protecting groups.) A. 17 amu B. 18 amu C. 32 amu D. 44 amu

When two amino acids are joined via a peptide bond, what is the mass of the byproduct of this reaction? (Note: Assume that the amino acids were not modified by protecting groups.) Jack Westin Advanced Solution: Peptide bonds are formed via a dehydration reaction. The product of a dehydration reaction is water. What is the atomic mass of water?Atomic mass H2O = 2(mass of H) + mass of O = 2(1)+16=18 amu A. 17 amu Jack Westin Advanced Solution: A. This answer is incorrect and fails to account for the mass of the second hydrogen in water. B. 18 amu Jack Westin Advanced Solution: B. This is correct; the atomic mass of water is 18 amu. Atomic mass H2O = 2(mass of H) + mass of O = 2(1)+16=18 amu C. 32 amu Jack Westin Advanced Solution: C. This is more aligned with the atomic mass of O2, not the atomic mass of H2O. D. 44 amu Jack Westin Advanced Solution: D. This answer is also incorrect and much too large. B is the best answer

Which experimental condition is NOT necessary to achieve reliable data for Michaelis-Menten enzyme kinetics? A. Initial velocity is measured under steady state conditions. B. Solution pH remains constant at all substrate concentrations. C. The concentration of enzyme is lower than that of substrate. D. The reaction is allowed to reach equilibrium before measurements are taken.

Which experimental condition is NOT necessary to achieve reliable data for Michaelis-Menten enzyme kinetics? A. Initial velocity is measured under steady state conditions. Jack Westin Advanced Solution: A. This is an important assumption for Michaelis-Menten kinetics. Measuring under steady state conditions ensures that the substrate concentration is not influenced by the formation of the enzyme-substrate complex and is instead true to the measurement. B. Solution pH remains constant at all substrate concentrations. Jack Westin Advanced Solution: B. If the pH were to change, the enzyme may not be able to function. Remember that enzymes are proteins and susceptible to pH changes just the same. A constant pH that optimizes enzyme function is a key experimental condition. C. The concentration of enzyme is lower than that of substrate. Jack Westin Advanced Solution: C. This is necessary and therefore not the correct answer. If the enzyme concentration were greater than that of the substrate, the effect of changing the substrate concentration would be lost to poor experimental design. Not all of the enzymes would be catalyzing reactions and the change in substrate concentration would not be noticed. D. The reaction is allowed to reach equilibrium before measurements are taken. Jack Westin Advanced Solution: D. Not only would this not be necessary, but this would be bad when obtaining Michaelis-Menten kinetics data. Once the reaction reaches equilibrium, changing the substrate concentration will not affect the rate of the reaction. Also, the researchers would not be able to obtain the initial rate/velocity because the reaction is past that point. The data obtained would essentially be useless for Michaelis-Menten kinetics. The question is a NOT question making this the best answer. Enzymes will help a reaction reach equilibrium faster, once its at equilibrium the enzyme doesnt really do much. Enzyme will only speed up a reaction if deltaG is permitting, at equilbrium deltaG is 0.

Which of the following best describes the functional relationship between distance d and time t in the figure? A. t d -1 B. t d -2 C. d t D. d t2

Which of the following best describes the functional relationship between distance d and time t in the figure? A. t d -1 Jack Westin Advanced Solution: A. The question stem asks for the functional relationship between distance and time. The way this answer choice is written suggests that time is proportional to or responding to distance, specifically the inverse of distance. Time is on the x-axis and distance changes in response to time; the distance does not determine the time. Also, the inverse suggests a negative relationship, whereas here the relationship is positive. B. t d -2 Jack Westin Advanced Solution: B. Like the answer choice above, this answer choice is a poor fit because it suggests the time changing in response to the distance. Also, it is inversely proportional to the second power whereas the relationship shown above is positively proportional to the second power. C. d t Jack Westin Advanced Solution: C. This represents a linear relationship; as we can see above, the relationship between distance and time is exponential and not linear. D. d t2 Jack Westin Advanced Solution: D. This is the best answer. The distance is roughly equal to the time squared. For instance, at 6s, the distance is about 62 or 36, at 8s just around 60 (82=64) and at 10s about 100 distance units.

Which of the following is a second period element that is a covalent network solid in its standard state? A. Carbon B. Phosphorous C. Oxygen D. Iodine

Which of the following is a second period element that is a covalent network solid in its standard state? Jack Westin Advanced Solution: This question is a mouthful; let's break it down. The first requirement is that the element is in the second period, or the second row on the periodic table. The second requirement is that it should be a solid made from covalent bonds when in its standard state. Putting it all together, the question is asking for an element in the second row of the periodic table that is a solid with covalent bonds. A. Carbon Jack Westin Advanced Solution: A. Carbon is indeed in the second period and also happens to be a solid in its standard form, also known as graphite. We're off to a great start on this last question. B. Phosphorous Jack Westin Advanced Solution: B. Phosphorus is in the third period, not the second. C. Oxygen Jack Westin Advanced Solution: C. While oxygen is in the second period, it is a gas and not a solid in its standard state. D. Iodine Jack Westin Advanced Solution: D. Iodine is in the fifth period, not the second period. Only carbon meets the question stem requirements.

Which pathway depicts the sequence of cellular compartments traversed by newly synthesized GABA-receptor subunits as they move to the cell surface? A. Rough endoplasmic reticulum, endosome, and Golgi complex B. Rough endoplasmic reticulum, Golgi complex, and secretory transport vesicle C. Golgi complex, endosome, and lysosome D. Golgi complex, rough endoplasmic reticulum, and secretory transport vesicle .

Which pathway depicts the sequence of cellular compartments traversed by newly synthesized GABA-receptor subunits as they move to the cell surface? Jack Westin Advanced Solution: This is another pseudo-standalone question. We're told we have newly synthesized GABA-receptor subunits move to the cell surface. Glancing at our answer choices, we want an answer choice in terms of the rough ER, endosomes, lysosomes, vesicles, or Golgi complex in some combination of three. We recall GABA receptors (proteins) are located in the cell membrane and are synthesized in the rough ER; proteins can also be modified here. Vesicles will transport proteins into the lumen of the Golgi apparatus. As the proteins and lipids travel through the Golgi, they undergo further modifications that allow them to be sorted. These newly-modified proteins and lipids are then tagged with phosphate groups or other small molecules so that they can be routed to their proper destinations. The modified and tagged proteins are packaged into secretory vesicles that bud from the trans face of the Golgi. While some of these vesicles deposit their contents into other parts of the cell where they will be used, other secretory vesicles fuse with the plasma membrane and release their contents outside the cell. A. Rough endoplasmic reticulum, endosome, and Golgi complex Jack Westin Advanced Solution: Rough endoplasmic reticulum, endosome, and Golgi complex. An endosome is a membrane-bound compartment inside a eukaryotic cell that functions to bring materials from outside the cell into the cell. We're focused on the movement of receptor subunits to the cell surface, not on material coming into the cell. B. Rough endoplasmic reticulum, Golgi complex, and secretory transport vesicle Jack Westin Advanced Solution: Rough endoplasmic reticulum, Golgi complex, and secretory transport vesicle. This answer choice is consistent with the breakdown we came up with. This is a superior answer to option A. C. Golgi complex, endosome, and lysosome Jack Westin Advanced Solution: Golgi complex, endosome, and lysosome. Reasoning here is similar to answer choice A. An endosome is a membrane-bound compartment inside a eukaryotic cell that functions to bring materials from outside the cell into the cell. We're focused on the movement of receptor subunits to the cell surface, not on material coming into the cell. Answer choice B remains superior. D. Golgi complex, rough endoplasmic reticulum, and secretory transport vesicle Jack Westin Advanced Solution: Golgi complex, rough endoplasmic reticulum, and secretory transport vesicle. While the different organelles here are correct, they are out of order. The receptor subunits are synthesized in the RER before making their way to the Golgi complex. Answer choice B is going to remain our correct answer.

Which solution component will have the lowest concentration at the end of the kinetic assay described in the passage? A Lactate B ADP C ATP D NAD+ n order to study the kinetics of GalK catalysis, researchers coupled the phosphorylation of galactose with two other enzymatically catalyzed reactions. The coupled reactions were the pyruvate kinase catalyzed conversion of phosphoenolpyruvate to pyruvate and the lactate dehydrogenase catalyzed conversion of pyruvate to lactate. The reaction was monitored by the decrease in the concentration of NADH. The resulting kinetic data with respect to both ATP and galactose are shown in Table 1. The experiments were conducted using wild-type (WT) GalK and two variant forms of GalK where Asp45 was substituted with either Ala or Gly.

Which solution component will have the lowest concentration at the end of the kinetic assay described in the passage? Jack Westin Advanced Solution: To answer this question, we can reference the passage for the specific aspects of our reactions. But we're going to need general biochemistry knowledge to identify the full reactions taking place.we have an excerpt from our passage here. Galactokinase transfers a phosphate group from ATP to galactose, producing ADP. This ADP can be converted back to ATP by pyruvate kinase. PEP + ADP in the presence of pyruvate kinase yields pyruvate and ATP.Pyruvate is converted to lactate by lactate dehydrogenase and NADH is converted to NAD+ in the process. Considering the four answer choices: lactate, NAD+, and ATP are all end products. ADP is a reactant that is part of the conversion from PEP to pyruvate. ADP is going to have the lowest concentration of the answer choices listed. That means our correct answer choice is answer choice B and we can eliminate answer choices A, C, and D. A Lactate B ADP C ATP D NAD+ Want more MCAT Prep? We have an MCAT Podcast! Solution: The correct answer is B. This is correct. This Biochemistry question falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is B because the GalK reaction produces ADP, but this ADP is immediately converted back to ATP by pyruvate kinase. Lactate and NAD+ are also both end products of the assay. It is a Reasoning about the Design and Execution of Research question because you must understand the experimental setup to understand how the concentrations of the solution components will change.

Which technique separates proteins independently of their charge? Which technique separates proteins independently of their charge? A. Native PAGE B. Gel filtration chromatography C. Ion exchange chromatography D. Isoelectric focusing

Which technique separates proteins independently of their charge? Jack Westin Advanced Solution: This is a standalone question that relies on our knowledge of different biochemical techniques. The best way to answer this question is to define each of the four answer choices and identify the answer choice that separates proteins independently of their charge. A. Native PAGE Jack Westin Advanced Solution: Native PAGE. Native PAGE is a technique for separating proteins based on both size and conformation. The mobility of proteins in Native PAGE is reliant on both size and intrinsic charge. B. Gel filtration chromatography Jack Westin Advanced Solution: Gel filtration chromatography. In gel filtration chromatography, proteins as separated based on size. As the solution travels down the column some smaller particles get stuck in the pores. The larger molecules simply pass by the pores because they are too large to enter the pores. This is exactly what we're looking for to answer this question. C. Ion exchange chromatography Jack Westin Advanced Solution: Ion exchange chromatography. Ion exchange chromatography is a technique used to separate proteins according to their charge. This is the opposite of what we're looking for in this question. This method is based on the attraction between oppositely charged ions, so a cationic stationary phase is used to separate anions in cation-exchange chromatography, and an anionic stationary phase is used to separate cations in anion exchange chromatography. D. Isoelectric focusing Jack Westin Advanced Solution: Isoelectric focusing. Isoelectric focusing is a laboratory technique in which a protein migrates along a membrane until it reaches its pI. a pH gradient is created on top of a membrane. One end of the membrane is connected to a negatively charged cathode, while the other end of the membrane is connected to a positively charged anode. When proteins are dolloped onto the membrane, their baseline charge will determine whether they will be attracted to the cathode or to the anode. We can stick with answer choice B as the technique that separates proteins independently of their charge. Want more MCAT Prep? We have an MCAT Podcast! The solution is B. Native PAGE is used to separate molecules based on their electrophoretic mobility, relying on length, conformation, and charge. Gel filtration chromatography separates protein only on the basis of their size. Ion exchange chromatography allows separation of the molecules based on their charge. Isoelectric focusing allows separation of the molecules based on their isoelectric point.

Which two classes of enzymes are needed in the two-step conversion of cytosine to 5hmC? A. Transferase and oxidoreductase B. Hydrolase and ligase . C. Oxidoreductase and hydrolase D. Transferase and ligase

Which two classes of enzymes are needed in the two-step conversion of cytosine to 5hmC? Jack Westin Advanced Solution: According to the passage, the conversion of cytosine to 5hmC requires methylation and hydroxymethylation. The enzyme for the first step needs to be able to add a methyl group and the enzyme of the second step needs to be able to hydroxylate that methyl group. A. Transferase and oxidoreductase Jack Westin Advanced Solution: A. The first step requires a methylation, and a transferase is able to transfer the functional group (such as a methyl group) from one substrate to another which would account for the methylation enzyme. A hydroxylation, the addition of an alcohol into another compound, is a type of reduction-oxidation reaction. An oxidoreductase catalyzes reduction and oxidation reactions. This is the best answer. B. Hydrolase and ligase Jack Westin Advanced Solution: B. Neither of the two steps mentioned above require cleavage of a bond using water which is the function of a hydrolase, so this answer is incorrect. C. Oxidoreductase and hydrolase Jack Westin Advanced Solution: C. For the same reason noted in answer choice B, this answer is also incorrect. Neither a methylation nor the hydroxylation of the methyl group require water-mediated bond cleavage. D. Transferase and ligase Jack Westin Advanced Solution: D. Transferase and ligase - As in answer choice A, a transferase would be a great way to ensure that a methyl group is transferred from one molecule to the molecule of interest. However, a ligase catalyzes the bonding of two larger molecules, such as a DNA ligase joining two fragments of DNA. However, a hydroxylation is a redox reaction that adds an alcohol to the molecule. Adding an alcohol is not the same as joining two molecules together making A the best answer.

Why did the researchers choose "A" as the replacement residue for each of the single site variants used in the study? Use of "A": A reduces the net charge on the bimetallic center. B increases strength of substrate binding to the active site. C increases the conformational rigidity of the active site and the enzyme. D reduces the interaction of the side chain with other active site components.

Why did the researchers choose "A" as the replacement residue for each of the single site variants used in the study? Use of "A": Jack Westin Advanced Solution: The researchers performed several amino acid substitutions in their study. This question wants to know why the substituted amino acid was amino acid "A", or alanine, for every site variant. To answer this question, we'll go back to the passage to reference the substitutions taking place. We'll use our external knowledge about amino acids to explain the reasoning for the alanine substitution and the effect it has.We have our table from the passage that shows all of our variants. Like we mentioned, there is an alanine substituted in every variant. H80A means the 80th residue was histidine, but has now been replaced by alanine. D113A means the 113th residue was aspartate, but has now been replaced by alanine. And amino acid E is glutamate and that's replaced by alanine.So that's 6 substitutions from amino acids to alanine. We can pull up the amino acid chart here for reference:What do we know about alanine? Alanine is classified as hydrophobic (It's the 2nd smallest nonpolar amino acid. And it only has an uncharged CH3 as its side group). The previous esterases are all electrically charged. The acidic asparate and glutamate both have negative charges on their side groups. Basic histidine has a positive charge on its side group.I've listed quite a few parts to our breakdown so far, but biggest change here is going from larger, charged side groups to an uncharged and much smaller side group. Alanine is more unique for being a smaller side group. There are quite a few other amino acids that are uncharged. Let's look at our answer choices and see if our breakdown can help us narrow them down or pick one. A reduces the net charge on the bimetallic center. Jack Westin Advanced Solution: Like we just mentioned, the researchers could have used a variety of different, uncharged amino acids. Using alanine specifically isn't consistent with answer choice A. B increases strength of substrate binding to the active site. Jack Westin Advanced Solution: Again, not the case here. The Michaelis constant decreases with the substitution. This answer choice contradicts what's happening in the passage; we're expecting weaker binding. C increases the conformational rigidity of the active site and the enzyme. Jack Westin Advanced Solution: This is not true; the substitution should do the opposite of this. D reduces the interaction of the side chain with other active site components. Jack Westin Advanced Solution: This sounds plausible. We have one of the smallest side chains of any amino acid on alanine. The side chain is also uncharged like we mentioned, so that would decrease the interactions even more. Need extra help breaking down AAMC logic? Find out how you can meet with one of our expert MCAT tutors! Solution: The correct answer is D. This is a Biochemistry question that falls under the content category "Principles of chemical thermodynamics and kinetics." The answer to this question is D. The researchers wanted to reduce or eliminate the interactions of each side chain with the active site and the best way to accomplish this while having a minimal impact on the overall structure of the enzyme is to use alanine. Alanine possesses a small hydrophobic side chain (methyl group) but participates in α helices and β sheets extensively. It is a Reasoning about the Design and Execution of Research question because you are asked to reason about the features of a research study that suggests relationships between the variables.

What is the balanced equation for the nonproductive reaction when lysine is the substrate? Jack Westin Advanced Solution: This question requires careful examination of the second main paragraph: the first step has two parts. In the first part of the first step, NADPH and FAD react to make NADP+and FADH-. The second part of the first step involves FADH- reacting with O2 and H+ to make FADH-OOH. In the nonproductive reaction, the passage tells us that FADH-OOH decomposes into H2O2. This means that the products of the nonproductive reaction are NADP+ and H2O2. A. H+ + NADPH + O2 → NADP+ + H2O2 Jack Westin Advanced Solution: A. This answer choice has the correct products, NADP+ from the first step and H2O2 from the decomposition of FADH-OOH at the end when lysine is the substrate. B. 2H+ + O2 → H2O2 Jack Westin Advanced Solution: B. While H2O2 is indeed one of the products, this answer fails to account for the other product of the nonproductive reaction, NADP+. C. FAD + NADPH + H+ → FADH2 + NADP+ Jack Westin Advanced Solution: C. This answer is incorrect because FADH2 is not one of the net products listed above. D. 2H2O + O2 → 2H2O2 Jack Westin Advanced Solution: D. Like answer choice B, this answer does not take NADP+ into account and is therefore incorrect. The two key products of the nonproductive reaction. Answer choice A is the correct answer. Researchers studied the kinetics of SidA-catalyzed Reaction 1. They discovered that the reaction proceeds in two steps. In the first step, NADPH reacts with FAD to form NADP+ and FADH-. FADH- quickly reacts with O2 and H+ to form FADH-OOH. This step is pH-independent and occurs at the same rate regardless of which, if any, substance occupies the active site. At this point, FADH-OOH either hydroxylates Compound 1 or decomposes nonproductively to form H2O2 if any other substance (such as L-lysine) occupies the active site.

Write out all the reactions that are mentioned in the passage: NADPH + FAD —> NADP + FADH FADH + O2 + H+ —> FADH-OOH FADH-OOH —> H2O2 + FAD Net reaction (I.e cross out everything that shows up on opposite sides of the equation): NADPH + O2 + H+ —> H2O2 + NADP+

Higher melting point and protein

a higher melting temperature is indicative of a more stable protein, as more energy is needed to unfold the protein. L203A has an approximate Tm of 50°C, therefore, it is the most stable and even more stable than the WT protein.

ideal gases

a hypothetical gas whose molecules occupy negligible space and have no interactions, and that consequently obeys the gas laws exactly.

Phospholipids

a lipid consisting of a glycerol bound to two fatty acids and a phosphate group.

Linkage

a phenomenon in which alleles that are located in close proximity to each other on the same chromosome are more likely to be inherited together

tight junctions

a specialized connection of two adjacent animal cell membranes such that the space usually lying between them is absent.

Disproportionation reactions

a type of redox reaction in which one element is both oxidized and reduced, forming at least two molecules containing the element with different oxidation states

Which of the four DNA bases contains the largest number of hydrogen bond acceptors when involved in a Watson-Crick base pair? A A B C C G D T J.

ack Westin Advanced Solution: Like most standalone questions, this question relies completely on knowing your content. To answer this question, we can consider DNA base pairing: A A Jack Westin Advanced Solution: Adenine contains 1 hydrogen bond acceptor B C Jack Westin Advanced Solution: Cytosine contains 2 hydrogen bond acceptors, so it's going to be our best answer choice. C G Jack Westin Advanced Solution: Guanine contains 1 hydrogen bond acceptor. D T Jack Westin Advanced Solution: Adenine contains 1 hydrogen bond acceptor. We can eliminate answer choices A, C, and D. Answer choice B, cytosine, has N and O hydrogen bond acceptors, so cytosine is our correct answer. Want more MCAT Prep? We have an MCAT Podcast! Solution: The correct answer is B. This is correct. This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is B because the hydrogen bond acceptors are N and O. Adenine contains 1 donor and 1 acceptor, thymine contains 1 donor and 1 acceptor, guanine contains 2 donors and 1 acceptor, and cytosine contains 1 donor and 2 acceptors. This is a Knowledge of Scientific Concepts and Principles question because you must recall structure of a Watson-Crick base pair.

Solubility equilibrium constant general convention

always solid to product Q < Ksp. The solution is unsaturated, and more of the ionic solid, if available, will dissolve. left to right Q = Ksp. The solution is saturated and at equilibrium. Q > Ksp. The solution is supersaturated, and ionic solid will precipitate. right to left

carbamate

amino termini structures that stabilize the T state

supercritical fluid

any substance at a temperature and pressure above its critical point, where distinct liquid and gas phases do not exist So when pressure is high enough, the molecules in a gas become so pressed together that the material basically becomes more like a liquid than a gas. On that same note, when you heat a liquid to a high enough temperature, the molecules in the liquid are moving so fast that they are breaking the intermolecular forces so quickly that they are acting like gasses more than liquids. With both high temp and high pressure, it basically becomes a super critical fluid because the material is not really acting the way either gas or liquid should Also at this point, the pressure is so great that the density of liquid phase equals that of the gaseous phase; when this happens gas and liquid are indistinguishable

A gas that occupies 10 L at 1 atm and 25oC will occupy what volume at 500 atm and 25oC? _exactly 0.020 ,somewhat more than beccause the volume is not neg somehwat more because of repulsions

because of the space occupied by the individual gas molecules Ex: The ideal gas law makes the assumption that molecules have no volume. This assumption is adequate when the gas is at 1 atm, but when the pressure is increased to 500 atm the volume of the gas molecules is no longer negligible. Ideal gas - consists of a large number of particles (atoms or molecules) in constant motion - collisions between particles and the walls of a container are perfectly elastic (kinetic energy is conserved) - attractive forces between particles do not exist - intermolecular forces are 0 expect during instantaneous collisions - the total volume of particles is very small compared with the volume of the container So basically the ideal gas law is based on a couple of assumptions that allow us to predict behavior of gases. These assumptions are: there is NO intermolecular forces between the molecules there is NO molecular volume (There are more assumptions that you can look up but these are the important ones for this). So the way I like to think of it is that we are basically thinking of a gas as empty space. So it's a direct function of the pressure and volume of its container! If you compress this container super small, you will get to the point where there is 0 volume... but we know, that real gases have actual molecules... so it can never actually be 0 volume. The same thing is true for pressure. If we neglect the size of individual gas particles (as we do to make ideal gas calculations) and increase the pressure ALOT then we'd eventually get to a point where our prediction is much smaller than what it should be. The two exceptions to the ideal gas law that the MCAT likes to test a lot are: at MODERATELY high pressures and/or temperatures, a gas will occupy less volume than predicted due to intermolecular attractive forces bt molecules (they are attracted so they get closer together than predicted) OR at EXTREMELY high temperatures and/or pressures a gas will occupy more volume than predicted due to the volume occupied by individual gas particles!

Which of the following compounds has the same geometry as methane? H2S XeF4 SiCl4 CO2

c draw the molecule and count the total number of electrons based on periodic table

When a weak acid (HA) is titrated with sodium hydroxide in the presence of an indicator (HIn), the pH at which a color change is observed depends on the: final conc of HA initial conc of HA final conc of HIn pka of HIn

d The indicator will change color over a specific pH range. The range at which the color change takes place depends on the point at which HIn is converted to In-, and this depends on the pKa of the indicator which is answer D.

CPT1

essential enzyme in the carnitine shuttle, transporting fatty acids into the mitochondria for oxidization rate-limiting enzyme of beta-oxidation

Spherical lens

focus the light into a point which don't change the location at which the image is formed

sucrose linkage

glucose-a-1,2-fructose non-reducing

maltose structure

glucose-alpha-1,4-glucose reducing sugar

Phospholipids structure

glycerol, 2 fatty acids, and a phosphate group Two fatty acids ester-linked to a single glycerol plus a charged head group phosphate is the charged group

reducing agents examples

good reducing agents have a lot of H's. LiAlH4--> strong reducing agent (reduces any functional group to an alcohol)... reduces carboxylic acid to alcohol...+ 3 to +1 oxidation state NaBaH4--> weak reducing agent (reduces ketones or aldehydes to alcohol) +2 to +1 oxidation state LiAlH4, NaBH4, H2, metals, NaDH high oxidation potential

Which change in the protonation state of Glu487 is most likely responsible for the change in MP activity at either low or high pH? The change in activity at: A low pH is due to the protonation of Glu487. B low pH is due to the deprotonation of Glu487. C high pH is due to the protonation of Glu487. D high pH is due to the deprotonation of Glu487.

his guys right. Both of those answers are true. Glu is protonated at low pH, and its deprotonated at high pH. However, the passage states that catalytic activity is driven by Glu protonating the substrate in the active site. So, while Glu is protonated at low pH, that wouldn't affect the catalytic activity. If its deprotonated (at high pH) then it wont be able to protonate the substrate. You gotta combine your knowledge of amino acids with info from the passage on this one. The wording is a little tricky because generally you don't thing of Glu being deprotonated at "high" pH. It gets deprotonated at neutral pH and even mildly acidic pH. But "high" is a relative term here. Which change in the protonation state of Glu487 is most likely responsible for the change in MP activity at either low or high pH? The change in activity at: Jack Westin Advanced Solution: To answer this question, we want to know the change in state, and the pH at which this happens. We're going to need to revisit Figure 2 from the passage, and we're combining that with our external knowledge of amino acids. If you need to, make sure to review the amino acid chart I added to question 88.Figure 2 shows that at very high and very low pH, relative activity is decreased. In terms of protonation, at low pH, functional groups are more likely to be protonated. Low pH means acidic. At high pH, functional groups are more likely to be deprotonated. High pH means more basic.The passage points out that catalytic activity is driven by Glutamate protonating the substrate in the active site. Glutamate itself is an acidic amino acid. If glutamate is protonated at low pH, that wouldn't affect the catalytic activity. If its de-protonated (at high pH), then it won't be able to protonate the substrate and we'd have a decrease in catalytic activity. A low pH is due to the protonation of Glu487. Jack Westin Advanced Solution: We said at low pH we would be expecting protonation, so that checks out. But we also said protonating glutamate at low pH wouldn't affect catalytic activity. B low pH is due to the deprotonation of Glu487. Jack Westin Advanced Solution: This answer not only contradicts our breakdown, but it also contradicts itself. At low pH we expect protonation, not deprotonation. We can automatically eliminate answer choice B. C high pH is due to the protonation of Glu487. Jack Westin Advanced Solution: Another answer choice that contradicts our breakdown. At high pH we expect deprotonation. D high pH is due to the deprotonation of Glu487. Jack Westin Advanced Solution: This is consistent with our breakdown. Deprotonating glutamate would mean it can no longer act as a general acid during catalysis. We can stick with our correct answer choice, answer D. Need extra help breaking down AAMC logic? Find out how you can meet with one of our expert MCAT tutors! Solution: The correct answer is D. This is correct. This Biochemistry question falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is D because the pH-dependence of MP activity shows that activity decreases at both low pH and high pH. At low pH, protonation of functional groups is the most likely cause of decreased activity and at high pH deprotonation of functional groups is the most likely cause of decreased activity. The passage states that Glu487 protonates the substrate prior to nucleophilic attack. This means that Glu487 must be able to act as a Brønsted acid during catalysis, which implies that the pKa of this side chain must be significantly higher in the enzyme active site than it is in solution, otherwise it would exist in the deprotonated form in the active site. If the pH is raised sufficiently to cause deprotonation of Glu487, it will no longer be able to participate in catalysis. This is a Data-based and Statistical Reasoning question because you must interpret patterns in data to make a prediction about the change in enzyme activity.

Numbering of nitrogenous bases

https://www.reddit.com/r/Mcat/comments/tn6kjw/can_someone_explain_the_numbering_on_nitrogenous/

In its lowest-energy electron configuration, zinc has a: Jack Westin Advanced Solution: According to the periodic table, zinc has an atomic number of 30, is in the fourth period, is the last element in its respective d-block and has 12 valence electrons. A. filled 3d energy level and a filled 4s energy level. B. half-filled 3d energy level and a filled 4s energy level. C. filled 3d energy level and a half-filled 4s energy level. D. half-filled 3d energy level and a half-filled 4s energy level.

i think the question would have explicitly said "zinc ion" if it was looking for the configuration for Zn2+. because it says "zinc," i would say it's safe to assume that it means the uncharged metal form. and yes, you are right that Zn2+ would lose electrons from the 4s subshell before 3d. In its lowest-energy electron configuration, zinc has a: Jack Westin Advanced Solution: According to the periodic table, zinc has an atomic number of 30, is in the fourth period, is the last element in its respective d-block and has 12 valence electrons. A. filled 3d energy level and a filled 4s energy level. Jack Westin Advanced Solution: A. The 3d orbital follows the 4s orbital according to subshell filling convention. The s orbital holds two electrons and the d orbital holds ten electrons. Zinc has 12 valence electrons which should completely fill both the 4s and 3d orbitals making this the correct answer. B. half-filled 3d energy level and a filled 4s energy level. Jack Westin Advanced Solution: B. This could apply if zinc had 7 valence electrons instead of 12 valence electrons. C. filled 3d energy level and a half-filled 4s energy level. Jack Westin Advanced Solution: C. This could apply if zinc had 11 valence electrons instead of 12 valence electrons. D. half-filled 3d energy level and a half-filled 4s energy level. Jack Westin Advanced Solution: D. This could apply if zinc had 6 valence electrons instead of 12 valence electrons. Only answer choice A accounts for all 12 valence electrons.

When asked whether a person who is afraid of spiders would be diagnosed as having a psychological disorder, a psychologist replies, "It depends on whether or not this fear interferes with the person's life." The psychologist appears to rely most heavily on which criterion of abnormality? Jack Westin Advanced Solution: This is a classic behavioral standalone question in that it relies on our external knowledge, specifically vocabulary. In order to classify behavior as abnormal, psychologists generally look at 4 criteria: violation of social norms, statistical rarity in the population, personal distress and maladaptiveness. We want to define each of our four answer choices and find the one that is consistent with psychological disorders interfering with one's life. A. Distress Jack Westin Advanced Solution: Distress. Distress has to do with negative feelings (like stress) or abnormal behavior that can be debilitating if it is continuous. This is inconsistent with what the psychologist replied to the person in the question stem. B. Maladaptiveness Jack Westin Advanced Solution: Maladaptiveness. Maladaptiveness means the behavior interferes with the person's normal life, which is exactly what the psychologist in the question stem used to define a psychological disorder. This is a strong answer choice. C. Statistical deviancy Jack Westin Advanced Solution: Statistical deviancy. Data tend to fall within a certain level, but statistical deviancy, like the name suggests, is when we see deviation from this expected level. That is not consistent with what the psychologist replied and is not the best criterion of abnormality. Answer choice B remains the best option. D. Violation of social norms Jack Westin Advanced Solution: Violation of social norms. Social norms are the explicit or implicit rules specifying what behaviors are acceptable within a society or group and include sanctions, folkways, mores, taboos, and anomie. Being afraid of spiders does not violate social norms and is inconsistent with the reply from the psychologist; this is not the appropriate criterion of abnormality within the context of this question. Answer choice B is going to be our best answer.

if my friend dies, i am distressed (very sad and distraught), and to cope, i cry a lot and write in my journal. this is not abnormal, it's a reasonable way to react to the situation. if my friend dies and i start stealing and doing drugs and beating people up in the street as my coping mechanism, i am displaying maladaptive behavior (behavior that is dangerous to myself and/or others). this is abnormal, you should not respond this way. similarly, if i'm distressed by spiders i might see one and be like "ewww someone kill it!" (not greatly interfering w my life). but if i have maladaptive behavior regarding spiders, i might have an immobilizing panic attack or i might refuse for the rest of my life to ever go back to the spot where i saw the spider even though it was in my bedroom and now i have nowhere to sleep (greatly interfering w my life). 7 ReplyShare level 2Pensive_NightsOP·2 yr. ago Thank you!!! This makes so much more sense. 2 ReplyShare level 1HabeasCormeum·2 yr. ago[524] The "interferes with the person's life" portion is the definition of maladaptive. Distress is mental pain, which is perfectly normal in a lot of situations. 3 ReplyShare level 1[deleted]·2 yr. ago I guess maladaptiveness is debilitating where as distress just causes anxiety and worries 1 ReplyShare

Discriminating stimulus

in operant conditioning, a stimulus that elicits a response after association with reinforcement (in contrast to related stimuli not associated with reinforcement) C. discriminating stimuli. Jack Westin Advanced Solution: discriminating stimuli. When an organism learns to respond differently to various stimuli that are similar, it is called stimulus discrimination. The numbers represent discriminating stimuli. The participants were presented with numbers, but they had to respond correctly in order to receive the monetary incentive. Participants had to discriminate between stimuli so this is a strong answer choice.

Astrology vs. Astronomy

is a pseudoscientific system of mysticism that correlates personality traits and compatibility with the star sign/major constellation you were born under via temporal displacement. Astronomy is the scientific study of the celestial bodies, stars, planets, moons, and observable space phenomena.

Epimers of glucose

mannose (C2) and galactose (C4)

Alcohols more polar than ketones and aldehydes Bec

of hydrogen bonding

native gel

one without SDS or another compound that would denature the proteins being separated Smaller molecules and more compact molecules will move farther in gel electrophoresis. Compact means higher density essentially. So like the same number of nucleotides, but clumped together more. (think euchromatin vs heterochromatin). Smaller size, but same weight and number of nucleotides. 3 ReplyShare level 1NirvaNaeNaeOP·7 yr. ago·edited 7 yr. ago Does more compact mean smaller size or less heavier but same number of nucleotides? 1 ReplyShare level 2Dako-Man·7 yr. ago Smaller size, same mass, same number of nucleotides Under denaturing conditions, gel electrophoresis will show you the size of the molecules. Because each one migrates the same distance on the denaturing, you know they all have the same size/amount of bases. Native on the other hand, depends on how the structure is actually folded, because it depends on friction (does not depend on charge with RNA/DNA because the charge is exactly proportional to the mass, unlike with protein). Thus a structure with more surface area (less compact) will not move as far down the gel as a more compact structure.

Quinone to hydroquinone

opposite way (reduction)

Molecular ion peak

parent ion peak--peak furthest to the RIGHT-represents the initial molecular cation=molecular weight of the original molecule.

Na+/K+ ATPase pump

primary active-transport protein that hydrolyzes ATP and releases energy used to transport sodium ions out of cell and potassium ions in both against the concentration gradient

CCK

produced by the small intestine when you eat and tells brain you're full, suppressing appetite

Erythromycin interferes with protein synthesis by binding to which ribosomal subunit? Jack Westin Advanced Solution: This is a passage-based question, but we'll also use our general knowledge and what we know about ribosomes. Ribosomes are macromolecular structures composed of rRNA and polypeptide chains. They are formed of two subunits (in bacteria and archaea, 30S and 50S; in eukaryotes, 40S and 60S), that bring together mRNA and tRNAs to catalyze protein synthesis. The author tells us in the passage that treatment of pertussis includes erythromycin which we're focused on in this question. Erythromycin inhibits bacterial protein synthesis by binding to the 23S rRNa component of the large subunit of the bacterial ribosome. The large subunit is the 50S subunit in bacteria.

prokaryotes= 50 and 30 70 eukaryotes= 60 and 40 80

Based on the information in the passage, which description of an enzyme-substrate covalent intermediate is most likely correct? The substrate is covalently attached to: Jack Westin Advanced Solution: First thing we'll want to do as we go through this question is to pick out the necessary information we need from the passage. We're given some background information about G3PPs in Paragraph 1. The author says, "The active site of G3PP contains Asp14 and Asp16, which act as a nucleophile and general acid, respectively, during catalysis." Nucleophiles are typically electron-rich and can donate electrons to form covalent bonds. What this tells us is that the substrate is covalently attached to a nucleophile, or Asp14 in this situation. We know it's unlikely to be Asp16 which is the general acid Quick glance at our answer choices shows that we were looking to pick between Asp14 and Asp16 (which we've done at this point), and also decide between the phosphorus and oxygen atom of the phosphate group. The nucleophile attacks an atom on the substrate-either the isotopically labeled oxygen or phosphorus.The oxygen that we see here is going to be the one that is doing the attacking, but it's not going to attack another oxygen. Rather, we expect it will attack a phosphorus.The substrate is covalently attached to Asp14 through the phosphorus atom of the phosphate group. Answer choice A is our best option. A.Asp14 through the phosphorus atom of the phosphate group.The correct answer is A.AAMC Standard Solution:The covalent intermediate will occur through the nucleophilic substitution by the side chain carboxyl of Asp14 at the electrophilic phosphorus atom in the substrate, displacing a leaving group. B.Asp14 through the oxygen atom of the phosphate group.AAMC Standard Solution:Oxygen in the substrate is not an electrophile because it has a partially negative charge and therefore cannot accept the nucleophilic carboxylate from Asp14. C.Asp16 through the phosphorus atom of the phosphate group.AAMC Standard Solution:Incorrect. There is no P-O bond formation between Asp16 and phosphate. The side chain carboxylic acid of Asp16 acts as a general acid, not a nucleophile. D.Asp16 through the oxygen atom of the phosphate group.AAMC Standard Solution:There is no P-O bond formation between Asp16 and phosphate. The side chain carboxylic acid of Asp16 acts as a general acid, not a nucleophile. Additionally, the oxygen in the substrate is not an electrophile because it has a partially negative charge.

so they are asking about how the substrate, glycerol 3-phosphate, covalently attaches to the active site of enzyme G3PP. In the hydrolysis reaction, we know that the active site Asp that will covalently attach to the substrate is the one that's a nucleophile because the nucleophile is the one that is capable of bond formation, not a general acid. We are told that Asp14 is the nucleophile, so we can immediately eliminate options C and D. Next, they are asking what atom on the substrate the Asp14 attacks. We know that Asp has a carboxyl group in the side chain, which means the oxygen on the Asp is the one that attacks. We know the nucleophile in the reaction. The oxygen (from Asp14) is not going to attack another oxygen, it's going to attack a phosphorus, and hence we can deduce that the electrophilic atom on the substrate is the phosphorus, not an oxygen. Thus, the answer is A. Hope that helps!

When used in place of spHM, which peptide would be most likely to achieve the same experimental results?

spHM and not sHM has phosphorylated threonine so negatively charged and so replacing T with E ie Glutamic acid would give that negative charge instead of Q Glutamine

Fatty acid synthesis

starts with acetyl-CoA and malonyl-CoA (from acetyl-CoA, using bicarbonate), activated to acetyl-ACP and malonyl-ACP acetyl-ACP to acetyl-FAS (with fatty acid synthase attached) fatty acid synthase helps combine malonyl-ACP with acetyl (release CO2), NADPH to remove ketone, then NADPH to remove double bond in cytosol need *2 NADPH* for each 2 carbons added

equivalence point

the point in a titration where the number of moles of hydrogen ions equals the number of moles of hydroxide ions

Assumptions of Michaelis-Menten

•[S] much >>> [E] -percentage of total substrate bound by the enzyme at any one time is small. •[ES] does NOT change with time. -Steady state assumption -synthesis of [ES] = degradation of [ES] •Only initial velocities (Vo) are used in the analysis of enzyme reactions. 1- the substrate concentration is greater than the enzyme concentration.(% of total substrate bound by the enzyme is small) 2-the initial velocity (V0) of the reaction is used to analyze enzymatic reactions (concentration of P is very small) Due to the divergent features of enzymes a general standardization of enzyme assays is not possible, rather special rules can be given as follows: pH: Preferentially the pH of the pH optimum of the respective enzyme is chosen, as far as possible at or near the physiological pH (~7.5). 2. Buffers and ionic strength: To stabilize the pH, buffers are used, and their pKa value should correspond to the pH optimum of the enzyme assay. Buffer concentrations of about 0.1 M are suitable for most enzyme assays, some (halo- and thermophilic) enzymes require a considerably higher ionic strength. 3. Temperature: One of three favoured temperatures should be chosen: • 25 °C, the most frequently used one, easy to maintain, but giving relatively low enzyme activities. • 30 °C, a compromise between 25 °C and the physiological temperature, especially for temperature sensitive enzymes. • 37 °C, the physiological temperature, relatively high enzyme activity, but more difficult to maintain. Different temperatures are needed for special cases (e.g. thermophilic enzymes). 4. Concentrations of substrates, and cofactors: should be saturating, as far as possible 100Km, but at least 10Km. 5. Concentration of the enzyme: as low as possible, but enough to observe the progressing reaction. 6. Concentrations of additives: (stabilizers, antioxidants, thiol reagents, protease inhibitors, complexing reagents) as required for efficiency. Generally all assay components must be compatible with one another, increase of ionic strength and influence on the pH of the assay must be taken into account. 7. Conditions of the particular enzyme assay must accurately be specified in the protocol.

A common column material used in size-exclusion chromatography is dextran, a polysaccharide of glucose. Which type of interaction most likely occurs between proteins and the dextran column material? A. Aromatic B. Hydrophobic C. Salt bridge D. Hydrogen bonding

A common column material used in size-exclusion chromatography is dextran, a polysaccharide of glucose. Which type of interaction most likely occurs between proteins and the dextran column material? A. Aromatic Jack Westin Advanced Solution: A. Glucose is not aromatic so we have no reason to believe that a polysaccharide made up of glucose will have aromatic properties. B. Hydrophobic Jack Westin Advanced Solution: B. Glucose, and therefore the polysaccharide dextran, is hydrophilic and not hydrophobic. It has a lot of hydroxyl groups or alcohols which make it polar rather than nonpolar. C. Salt bridge Jack Westin Advanced Solution: C. Salt bridges are used in electrochemistry to prevent the buildup of charges and are a combination of hydrogen and ionic bonding. This does not apply to glucose or dextran. D. Hydrogen bonding Jack Westin Advanced Solution: D. The oxygens in the hydroxyl functional groups in glucose are capable of hydrogen bonding to the side chains that may be exposed on different proteins. It is the only answer choice with interactions that are applicable to glucose and therefore dextran, making it the correct answer.

Approximately how many moles of Al3+ are reduced when 0.1 faraday of charge passes through a cell during the production of Al? (Note: Assume there is excess Al3+ available and that Al3+ is reduced to Al metal only.)

A faraday is equal to one mole of electric charge. Because each aluminum ion gains 3 electrons, 0.1 faraday of charge will reduce 0.1/3 moles of aluminum, or 0.033 moles of aluminum.

Prostaglandins

A group of bioactive, hormone-like chemicals derived from fatty acids that have a wide variety of biological effects including roles in inflammation, platelet aggregation, vascular smooth muscle dilation and constriction, cell growth, protection of from acid in the stomach, and many more.

Sanger sequencing is used to determine the nucleotide sequence of DNA. It requires the addition of dideoxyribose, which prevents additional growth of the nucleotide polymer. Which molecule is most likely used in Sanger sequencing?

A is correct; Sanger sequencing works by using a DNA nucleotide without a 3' hydroxyl to 'poison' the reaction. Because the normal DNA sugar is referred to as "deoxyribose," this sugar without either a 2' or a 3' hydroxyl group is called "dideoxyribose." Answer B is not correct since this is a normal ribose ring. Answer C is not correct since this is a deoxyribose ring, but not a dideoxyribose, which would still allow growth of the nucleotide chain. Answer D is not correct since there is not a phosphate to allow attachment to the nucleotide chain, and because it is not dideoxyribose.

Lorentz Force Law

A law stating that magnetic force increases as the magnitude of an electric charge or magnetic field increases

Dithiothreitol (DTT)

A strong reducing agent which breaks disulfide bonds in proteins. Disulfide reducing agents include tris (2-carboxyethyl) phosphine hydrochloride (TCEP), beta-mercaptoethanol (BME), and dithiothreitol (DTT).

Epimers

A subtype of diastereomers that differ in absolute configuration at exactly one chiral carbon

Desmosomes (anchoring junctions)

A type of cell junction in animal cells that functions like rivets, fastening cells together into strong sheets. Intermediate filaments made of sturdy keratin proteins anchor them in the cytoplasm. They attach muscle cells to each other in a muscle.

A vDNA sequence encoding a protein is inserted into a host genome by IN. A 5-GGCAACUGACUA-3 B 5-TAGTCAGTTGCC-3 C 5-CCGTTGACTGAT-3 D 5-UAGUCAGUUGCC-3

A vDNA sequence encoding a protein is inserted into a host genome by IN. The protein is translated from the hypothetical mRNA sequence shown. 5-GGCAACUGACUA-3 Based on the passage, the segment of the original viral genome that encoded this protein had what nucleotide sequence? Jack Westin Advanced Solution: The protein is translated from the hypothetical mRNA sequence shown.5'-GGCAACUGACUA-3' Based on the passage, the segment of the original viral genome that encoded this protein had what nucleotide sequence? To answer this question, we're finding the segment of the original viral genome that encoded this protein. We have to be very careful with verbiage and make sure we have the proper segment's nucleotide sequence. Note, the question stem says "based on the passage," but that doesn't always mean we automatically go back to the passage. This is almost like a standalone question, so we're going to reason this out slowly.We have the hypothetical mRNA sequence, so let's work backwards to get to the RNA of the initial HIV: the retrovirus. mRNA is synthesized from DNA in the nucleus. It's synthesized from a DNA template, but it's not an exact copy, we have a complementary strand. So that DNA is going to be CCGTTGACTGAT, or in other words, complementary bases to that hypothetical mRNA sequence. But we want the original viral genome. RNA is converted to DNA through reverse transcriptase, but again, we don't have an exact copy. We have complementary RNA. We're going to have uracil again instead of thymine, but what that means is, we're going to have the same nucleotide sequence as the hypothetical mRNA. A 5-GGCAACUGACUA-3 Jack Westin Advanced Solution: This matches our prediction, and the sequence in the question stem. We like this answer, but we still want to compare with the remaining options. B 5-TAGTCAGTTGCC-3 Jack Westin Advanced Solution: We have thymine here, not uracil, which we know we'd have in in our initial RNA. This answer choice is the DNA strand complement to our correct answer. That's not what we're looking for, so we can eliminate this answer choice. C 5-CCGTTGACTGAT-3 Jack Westin Advanced Solution: This answer choice also has thymine instead of uracil. Again, this could be correct if we were asking about the viral DNA. We wanted the sequence of the initial retrovirus RNA. We can eliminate this answer choice D 5-UAGUCAGUUGCC-3 Jack Westin Advanced Solution: This is the same as answer choice B, only we replaced thymine with uracil. We can eliminate this answer choice and we're left with our correct answer, answer choice A. Want more MCAT Prep? We have an MCAT Podcast! Solution: The correct answer is A. This is a Biology question that falls under the content category "The structure, growth, physiology, and genetics of prokaryotes and viruses." The answer to this question is A because according to the passage, viral DNA integrated into a host cell genome by integrase would originate from a retrovirus. mRNA transcribed from retroviral DNA is either used to synthesize viral proteins, or used as the RNA genome for progeny viruses. Thus, the sequence of the nucleotide in the original viral genome will be the same as that of the transcribed mRNA. It is a Scientific Reasoning and Problem Solving question because you are asked to recall that HIV is a retrovirus with an RNA genome and use your understanding of reverse transcriptase to use a given transcribed mRNA sequence to determine the original viral genome sequence.

What are the relative potentials for force and power generation by a slow-twitch muscle fiber and a fast-twitch muscle fiber of the same diameter? A. The slow-twitch and fast-twitch fibers possess the same potentials for force and power generation. Jack Westin Advanced Solution: A. Based on the passage, it is correct that the two fibers in the question stem with similar diameters will produce similar forces. However, recall that power is the force produced multiplied by the velocity at which that force is generated. As the name implies, fast-twitch fibers contract at faster velocities than slow-twitch fibers. If the force is the same and the velocity is greater for fast-twitch fibers, then the fast-twitch fibers generate more power. B. The slow-twitch fiber is capable of generating more force than the fast-twitch fiber, while the potential for power generation is the same. Jack Westin Advanced Solution: B. This contradicts the passage which says similar muscle diameters produce similar strengths and strength is the production of force. C. The fast-twitch fiber is capable of generating more power than the slow-twitch fiber, while the potential for force generation is the same. Jack Westin Advanced Solution: C. As noted in the explanation for answer choice A, the power produced by fast-twitch fibers is greater than the power produced by slow-twitch fibers because the two will produce the same force when they are the same diameter, but the fast-twitch fibers do so with a greater velocity. This is the correct answer. D. The fast-twitch fiber is capable of generating more force than the slow-twitch fiber, while the potential for power generation is the same. Jack Westin Advanced Solution: D. Like answer choice B, this directly contradicts the passage. Muscles of similar diameter should produce similar strengths and forces. Answer C remains the best answer.

A. The slow-twitch and fast-twitch fibers possess the same potentials for force and power generation. Jack Westin Advanced Solution: A. Based on the passage, it is correct that the two fibers in the question stem with similar diameters will produce similar forces. However, recall that power is the force produced multiplied by the velocity at which that force is generated. As the name implies, fast-twitch fibers contract at faster velocities than slow-twitch fibers. If the force is the same and the velocity is greater for fast-twitch fibers, then the fast-twitch fibers generate more power. B. The slow-twitch fiber is capable of generating more force than the fast-twitch fiber, while the potential for power generation is the same. Jack Westin Advanced Solution: B. This contradicts the passage which says similar muscle diameters produce similar strengths and strength is the production of force. C. The fast-twitch fiber is capable of generating more power than the slow-twitch fiber, while the potential for force generation is the same. Jack Westin Advanced Solution: C. As noted in the explanation for answer choice A, the power produced by fast-twitch fibers is greater than the power produced by slow-twitch fibers because the two will produce the same force when they are the same diameter, but the fast-twitch fibers do so with a greater velocity. This is the correct answer. D. The fast-twitch fiber is capable of generating more force than the slow-twitch fiber, while the potential for power generation is the same. Jack Westin Advanced Solution: D. Like answer choice B, this directly contradicts the passage. Muscles of similar diameter should produce similar strengths and forces. Answer C remains the best answer.

What percentage of NS1-2 amino acids are disordered based on the disorder predictions in Figure 1? 39% 49% 60% 97%

Answer A is correct; 39% of the NS1-2 protein is disordered. As stated in the Figure 1 legend, a DP value greater than 0.5 suggests that this region is disordered. Therefore, the regions at the N-terminus of the protein are predicted to be disordered because DP > 0.5. There is also a hint in Figure 2, where a region of the NS1-2 full protein is labeled as "Disordered, a.a. 1-142." The x-axis of Figure 1 represents the number of amino acids in the protein and can be used to estimate both the number of disordered amino acids (142) and the total number of amino acids (350). 142/350 = 40.5%. There is a small region among the disordered section that may be ordered, and a small section in the ordered section that may be disordered, but these appear to cancel one another out. Answer B is not correct; this would be the percentage if there were ~200 amino acids with DP > 0.5. Answer C is not correct; this value may be obtained if the student erroneously calculates amino acids for which DP < 0.5. Answer D is not correct; Figure 1 demonstrates that nowhere near 100% of NS1-2 is disordered. These answer choices illustrate the fair manner in which the AAMC usually presents numerical values. Most numerical answer choices differ by a significant margin. For example, in this case the student had to estimate the answer as 39%. We did NOT require you to pick between 30%, 35%, 40% and 45%. This is a prevalent problem among many questions found in status quo test prep resources--the answers are too close to justify given a "no calculators allowed" exam that requires estimation.

What do the mass spectral fragmentation data for the muconaldehyde isomers reveal about the relative stability of geometric isomers? Under electron impact ionization: a trans,trans geometry is more stable than a cis,cis geometry. a trans,trans geometry results in a higher boiling point than a cis,cisgeometry. cleavage of the C1-C2 bond is more favored by a cis,cis geometry than a trans, trans geometry. cleavage of the C2=C3 bond is favored equally in both geometric isomers.

Answer A is true. Because the molecular ion peak is the most abundant for the trans,trans isomer (note that the relative abundances are shown in parentheses, 100% for trans-trans and only 47% for cis-cis), it can be concluded that it is the most stable of the two isomers under electron impact ionization. Choice B is a true statement, but doesn't address molecular stability towards fragmentation. Yes, the AAMC does use--in fact, they frequently use--correct or true statements as distractors. Just because a statement is true generally does not mean it appropriately answers the question stem or applies to the current situation. We include answer choices such as this for the specific purpose of exposing you to this fact now so that you can establish good habits and avoid missing points on test day. The stem specifically asks about what we learn from Table 1 about stability under electron impact ionization...it does not follow to answer the question with a statement about boiling point. Consistently using all of the Altius strategies will help prevent such mistakes. If you are "Restating the Question Stem" and "Considering each answer choice as 'probably' or 'probably not'..." you are less likely to fall for a distractor of this type. If your brain hears the stem read as a complete statement, with the "mismatched boiling point" answer read as the second half of that statement, it is far more likely to recognize that something is amiss. Answer C is not supported by the data. The 81 m/z peak is due to a cleavage at the C1-C2 position, but both isomers have a peak at 81 m/z of similar magnitude. Choice D is not supported by the data because no fragments consistent with the C2=C3 bond cleavage appear in the mass spectrum.

gas X has a density of 1.44 g/L and gas Y has a density of 1.54 g/L. Which gas diffuses faster? gas X bec it has lower molar mass

Answer Key: A Avogadro's law states that equal volumes of different gases at the same temperature and pressure contain the same number of molecules. According to the density values given, 1 liter of gas X weighs 1.44 g and 1 liter of gas Y weighs 1.54 g. Since each sample has the same number of molecules, gas X must have a lower molar mass than gas Y. A lighter molecule diffuses faster than does a heavier one, so gas X diffuses faster than gas Y. Answer choice A is the best

As blood passes through actively contracting skeletal muscle tissue, the affinity of hemoglobin for oxygen in the muscle tissue: A. increases as a result of an increase in plasma temperature. B. increases as a result of an increase in plasma PO2. C. decreases as a result of a decrease in plasma pH. D. decreases as a result of a decrease in plasma PCO2.

As blood passes through actively contracting skeletal muscle tissue, the affinity of hemoglobin for oxygen in the muscle tissue: Jack Westin Advanced Solution: This is a standalone question that asks about actively contracting skeletal muscle tissue. We can think about what this contraction entails. Skeletal muscle mainly attaches to the skeletal system via tendons to maintain posture and control movement. For example, contraction of the biceps muscle, attached to the scapula and radius, will raise the forearm. Think of going to the gym and lifting weights-that involves actively contracting skeletal muscle tissue. The first set of an exercise might be easier, but muscle use can quickly overwhelm the ability of the body to deliver oxygen. Muscle fibers must switch to anaerobic metabolism and produce lactic acid, at which point the muscle begins to fatigue. Anaerobic respiration therefore causes decreased pH and reduced affinity of hemoglobin for oxygen. Additionally, that increased temperature, such as from increased activity of the skeletal muscle, causes the affinity of hemoglobin for oxygen to be reduced. We want an answer choice consistent with this breakdown of actively contracting skeletal muscle tissue. A. increases as a result of an increase in plasma temperature. Jack Westin Advanced Solution: increases as a result of an increase in plasma temperature. While it is true we expect increased temperature, we said we expect a decrease in the affinity of hemoglobin for oxygen in the muscle tissue. B. increases as a result of an increase in plasma PO2. Jack Westin Advanced Solution: increases as a result of an increase in plasma PO2. This answer choice is the opposite of our breakdown. We expect affinity to decrease, and we actually have to use anaerobic respiration. This is going to be our worst answer choice thus far. C. decreases as a result of a decrease in plasma pH. Jack Westin Advanced Solution: decreases as a result of a decrease in plasma pH. This answer choice is consistent with our breakdown, and is factually correct, unlike answer choices A and B. Note, answer choice A mentioned that increase in temperature, but incorrectly stated that affinity would increase. We always want to pick the best answer, and an answer that is factually correct. Answer choice C is going to be our best answer choice. D. decreases as a result of a decrease in plasma PCO2. Jack Westin Advanced Solution: decreases as a result of a decrease in plasma PCO2. First part of this answer choice is consistent with our breakdown of the question. We do expect affinity will decrease, but not because of a decrease in plasma PCO2. This effect would be seen in the opposite carbon dioxide environment. As the level of carbon dioxide in the blood increases, more H+ is produced, and the pH decreases. The increase in carbon dioxide and subsequent decrease in pH reduce the affinity of hemoglobin for oxygen. Answer choice C remains our best option. Want more MCAT Prep? We have an MCAT Podcast! The solution is C. Affinity would decrease with an increase in plasma temperature. Affinity would increase when PO2 increases. However, PO2 in muscle cells decreases with exercise. Affinity would decrease with a decrease in plasma pH, and during prolonged exercise, anaerobic respiration would decrease the plasma pH. Affinity would increase with a decrease in plasma PCO2.

What is the best description of the catalytic mechanism of GK? Catalysis occurs through: A an ordered mechanism in which a ternary complex is formed. B an ordered mechanism in which no ternary complex is formed. C a random order mechanism in which a ternary complex is formed. D a random order mechanism in which no ternary complex is formed.

As you said, ternary means 3 things are bound at once -- in the case of that question and many others that we're likely to see, the ternary complex is enzyme + ATP + other substrate! A non-ternary complex would be one in which the enzyme does not require ATP to bind as well, such as many of the isomerases etc. What is the best description of the catalytic mechanism of GK? Catalysis occurs through: Jack Westin Advanced Solution: This is a very open-ended question, but glancing at our answer choices we see that we need to distinguish between an ordered and random order mechanism. We also have to establish if a ternary complex is formed. We can revisit the passage to get specific details about the catalysis mechanism.The passage says "ATP and glycerol both occupy the catalytic cleft prior to catalysis, and it has been determined that glycerol binds first." The glycerol always binding first means it's an ordered mechanism, not random.We have both ATP and glycerol occupying the catalytic cleft. A ternary complex is composed of three parts, so we can confirm a ternary complex is formed. A an ordered mechanism in which a ternary complex is formed. Jack Westin Advanced Solution: This is exactly the answer we're looking for based on my breakdown. Let's go through our additional answers, just to be thorough. B an ordered mechanism in which no ternary complex is formed. Jack Westin Advanced Solution: First part of this answer is correct, but a ternary complex is formed. Answer choice A remains superior. C a random order mechanism in which a ternary complex is formed. Jack Westin Advanced Solution: This answer choice mentions a random order mechanism, but we know that's not the case. Glycerol always binds first, so it's ordered. D a random order mechanism in which no ternary complex is formed. Jack Westin Advanced Solution: This answer choice also mentions a random order mechanism. Glycerol always binds first, so it's ordered. Best answer choice here is going to remain answer choice A. Need extra help breaking down AAMC logic? Find out how you can meet with one of our expert MCAT tutors! Solution: The correct answer is A. This is correct. This Biochemistry question falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is A because the passage states that both ATP and glycerol occupy the catalytic cleft, which means that a ternary complex is formed. Since glycerol binds first, it is an ordered mechanism as opposed to a random order mechanism in which either substrate could be first to bind. This is a Knowledge of Scientific Concepts and Principles question because you must recall the standard categories of reaction mechanisms for bisubstrate reactions. Correct Your Answer:A Correct Answer:A Content & Skills

Assume that in the study with the rat tissues, fluid flows at a speed of 0.30 mm/s through a typical capillary opening caused by a burst microbubble. Given this, which of the following is closest to the volume flow rate of fluid passing through the opening? A. 4.5 × 106 μm3/s B. 7.5 × 106 μm3/s C. 1.2 × 107 μm3/s D. 4.5 × 107 μm3/s

Assume that in the study with the rat tissues, fluid flows at a speed of 0.30 mm/s through a typical capillary opening caused by a burst microbubble. Given this, which of the following is closest to the volume flow rate of fluid passing through the opening? Jack Westin Advanced Solution: The final sentence of the passage gives the size of the opening as ~2.5 × 104 μm2. The question stem asks for the volume flow rate, so we need to multiply the microtubule opening by the speed at which the fluid flows.Volume flow rate = area * velocity = 2.5 × 104 μm2 * 0.30 mm/s =2.5 × 104 μm2 * 0.30× 103 μm/s = 0.75 × 107 μm3/s = 7.5 × 106 μm3/s A. 4.5 × 106 μm3/s Jack Westin Advanced Solution: A. This answer choice is lower than the correct answer; 2.5*3=7.5, not 4.5. B. 7.5 × 106 μm3/s Jack Westin Advanced Solution: B. This is the correct answer as calculated above. Volume flow rate = area * velocity = 2.5 × 104 μm2 * 0.30 mm/s =2.5 × 104 μm2 * 0.30× 103 μm/s = 0.75 × 107 μm3/s = 7.5 × 106 μm3/s C. 1.2 × 107 μm3/s Jack Westin Advanced Solution: C. This answer choice incorrectly divides the 0.3 by 2.5 but the two need to be multiplied. D. 4.5 × 107 μm3/s you can't multiply mm units by um^2 units. You have to convert 0.30 mm/s to um/s, so it becomes 300 um/s. Now you can multiply 300um/s by 2.5*10^4 um^2. This will give you 750*10^4 um^3/s, which is 7.5*10^6um*3/s. AAMC's explanation has the wrong units because the answer choices are in terms of um^3/s but the explanation is in mm^3/s without converting the actual number.

Based on the data in Table 1, what is the purification yield from the culture supernatant after the final ion-exchange chromatography step? A 0.1% B 20% C 40% D 60%

Based on the data in Table 1, what is the purification yield from the culture supernatant after the final ion-exchange chromatography step? Jack Westin Advanced Solution: We're going to answer this question as a percent of our initial supernatant. We're going to need our data from the passage, and we're going to use dimensional analysis to calculate the number of units in each step, and ultimately the percentage purification yield.Supernatant step: 3000 mg x 0.1 units/mg = 300 unitsAmmonium sulfate precipitation: 1000 mg x 0.2 units/mg = 200 unitsIon-exchange chromatography: 30 mg x 4.0 units/mg = 120 unitsSize-exclusion chromatography: 8 mg x 10.0 units/mg = 80 unitsIon-exchange chromatography: 3 mg x 20.0 units/mg = 60 unitsWe started with 300 units in our supernatant and are left with 60 units in our final step. We divide:60 units final/300 units starting = 20% purification yieldThis was a math problem with no rounding, so we can use our calculated value to compare all of our answer choices at once. Answer choice B is going to be our correct value, and we can eliminate answer choices A, C, and D. A 0.1% B 20% C 40% D 60% Need extra help breaking down AAMC logic? Find out how you can meet with one of our expert MCAT tutors! Solution: The correct answer is B. This is correct. This Biochemistry question falls under the content category "Separation and purification methods." The answer is B because, while the specific activity represents a measure of solution purity, the activity units themselves provide the best measurement of yield. The total number of activity units in the initial culture was 3000 mg × 0.1 units/mg = 300 units. After the final step, the activity units that remain are 3 mg × 20 units/mg = 60 units. This represents a 60/300 = 0.2 or 20% yield. This is a Reasoning about the Design and Execution of Research question because you must reason about the effectiveness of an experimental process.

Based on the results shown in Figure 2, what effect does mutation have on M1 mRNA? A The folded structures of wild-type and MBmutant M1 mRNA are similar, but the structure of HPmutant M1 mRNA is less compact. J B The folded structures of wild-type and MBmutant M1 mRNA are similar, but the structure of HPmutant M1 mRNA is more compact. C Wild-type and MBmutant M1 mRNA have the same number of base pairs, but HPmutant M1 mRNA has fewer base pairs. D Wild-type and MBmutant M1 mRNA have the same number of base pairs, but HPmutant M1 mRNA has more base pairs.

Based on the results shown in Figure 2, what effect does mutation have on M1 mRNA? Jack Westin Advanced Solution: To answer this question, we're going to need Figure 2 in the passage to see the experimental results. Then we're using general knowledge to actually interpret the results.Gel electrophoresis can be used to separate macromolecules based on size and charge-smaller molecules will migrate across the gel more quickly than larger molecules.The native gel shows the HPmutant band further down the gel than the wild type, or the MBmutant bands. This means the folded structure of HPmutant is more compact than the other two.When all three samples are denatured, they all migrate the same distance along the denaturing gel, meaning the number of nucleotides is identical. If you are unsure about gel electrophoresis, make sure to reference the following figure: A The folded structures of wild-type and MBmutant M1 mRNA are similar, but the structure of HPmutant M1 mRNA is less compact. Jack Westin Advanced Solution: The first part of the answer choice is correct, and we saw this from the native gel. But the second part of the answer choice says HPmutant M1 is less compact. The HPmutant band migrated more quickly, meaning it is more compact than the wild type and MBmutant. B The folded structures of wild-type and MBmutant M1 mRNA are similar, but the structure of HPmutant M1 mRNA is more compact. Jack Westin Advanced Solution: Again, first part of the answer choice is correct. Second part correctly says the structure of HPMutant M1 mRNA is more compact, just like we said. Answer choice B is our best answer choice so far. C Wild-type and MBmutant M1 mRNA have the same number of base pairs, but HPmutant M1 mRNA has fewer base pairs. Jack Westin Advanced Solution: We expect all 3 to have the same number of base pairs based on the results of the denaturing gel electrophoresis, so answer choice B remains superior. D Wild-type and MBmutant M1 mRNA have the same number of base pairs, but HPmutant M1 mRNA has more base pairs. Jack Westin Advanced Solution: This is the same issue as answer choice C. All 3 variants have the same number of base pairs; our correct answer is answer choice B. How do you know you are truly ready? That's where the Jack Westin MCAT Diagnostic Tool comes in. Solution: The correct answer is B. This is correct. This Biochemistry question falls under the content category "Separation and purification methods." The answer is B because the native gel shows that the size and shape of wild-type and MBmutant are similar, but HPmutant is more compact because it migrates further in the gel. The denaturing gel confirms that the number of nucleotides in each mRNA is the same. It is a Data-based and Statistical Reasoning question because you must use the results of the electrophoresis experiments to compare the folded structures of the biomolecules.

blood-brain barrier

Blood vessels (capillaries) that selectively let certain substances enter the brain tissue and keep other substances out Blood brain barrier, like all cell membranes, are composed of hydrophobic, non polar groups. Small non polar things can diffuse across (CO2, small non polar. O2, small non polar.) A FATTY ACID has a polar head group (Hydroxyl, -OH) attached to an alkyl chain. There are exceptions to this rule though, for example Testersterone has some polar groups, but it is planar (because of the benzene rings), but is mostly non polar. A fatty acid is polar, however, due to the carboxylic acid (hence, fatty "acid") 4 ReplyShare The blood brain barrier is highly lipophilic. Any molecule that is lipophilic will easily cross the BBB. That is why certain allergy medications are labelled as drowsy and non-drowsy, the drowsy anti-histamines enter the BBB and cause sedation. R3MD had a point in not memorizing specific functional groups, just know the general idea that molecules that are small, non-charged, and lipophilic will easily cross the BBB. Lipids can pass the BBB. Also glucose and amino acids enter the BBB through special transport proteins found on the membrane, they do not readily diffuse like other lipophilic molecules.

unique or define characteristics

By a defining characteristic, I think they mean something that is unique to transcription factors. TFs do need to be in the nucleus to exert their effects, but there are other proteins in the nucleus that are not TFs. But if a protein has a DNA binding domain, it is very likely that it's a TF. So even though both can be true, C is correct because it's more specific to TFs

Human speech is generated in the vocal cords as the lungs push air past them. What property of the vocal cords is changed so that the frequency of sound can be altered? volume density tension number

C vocal cords are like guitar strings. when you push on them you are altering the tension. wave frequency depends on the source that's generating the wave. The strings properties rather than the medium in which the wave is travelling. Process of elimination: A) are you making the strings longer, shorter, bigger smaller when you play the guitar? NO so strike VOLUME. realistically you cannot change the volume of your vocal cord unless there's a pathology which is beyond the scope of the MCAT. B) Density? The only time I heard about density and sound is when it comes so the speed of sound moving from one medium to another. Are you changing the medium here? NO. Strike. C) Tension sounds strange here..the frequency of the source depends on the source that's generating the wave i.e. the vocal cords when I play the guitar I push on the strings... hmmmm tension looks good if they are lose the sound will be different vs when they are tight ,,, may be this one. D) Number???? I am not increasing the number of my vocal cords or the number of the strings of the guitar. STRIKE... The air pushes through the vocal cords causing them to vibrate and make sound. Therefore it's sound of a string which involves tension and a transverse wave.

CD4 and CD8

CD stands for cluster of differentiation, it's a glycoprotein on the outside of T-lymphocytes used in recognition of a major histocompatibility complex. A helper T-lymphocyte has the CD4 glycoprotein on its surface. CD4 binds to major histocompatibility complex type 2 (MHC-II), which is often expressed on such cells as B-lymphocytes. A cytotoxic T-lymphocyte has the CD8 glycoprotein on its surface. CD8 binds to MHC-I, which is expressed on the surface of every cell in the body.

During glycolysis, pyruvate CH3C(=O)CO2- is reduced to lactate CH3CH(OH)CO2- by nicotinamide adenine dinucleotide (NADH). What is the balanced reaction for this conversion?

CH3C(=O)CO2- + NADH + H+ → CH3CH(OH)CO2- + NAD+ Solution See if the atoms are equal on each side, the see if charges are equal and opt for equal neutral charges then for equal non-neutral charges

surface tension of alveolar fluid

Causes alveoli to assume smallest possible diameter Accounts for 2/3 of lung elastic recoil Prevents collapse of alveoli at exhalation surface tension causes the water molecules or fluids cohesiveness to increase and there is no upward force acting on the molecules so they all bind together more closely and form a film. they only have forces from the bottom drawing the im so low surface area such as water droplet net inward force

When illuminated, the solar cell is analogous to a battery. In this context, which portion of the solar cell acts as the positive terminal? Pt/F:SnO2 electrode F:SnO2 electrode Both electrodes Nanowires

Choice A is the correct answer. As described in the passage, the electrons leave the bottom electrode and flow upwards through the load. Therefore (because electrons carry negative charge) the current (being the opposite of electron flow) is flowing downward through the load. Since current gets "pushed out" of the positive terminal of a battery, the top electrode, Pt:F:SnO2, must be the positive terminal, and choice A is correct. Although choice D may be tempting, because the electrons do leave the nanowires in a fashion similar to an electrode, it is incorrect because the nanowires are not serving as electrodes; the two electrodes for the solar device were defined in the passage.

During the first 48 hours, the pH of the reaction mixture is sampled as it passes from P1 to P3. Which pattern is expected? pH will steadily increase pH will steadily decrease pH will equilibrate to 7.0 pH will remain constant

Choice A is the correct answer. Because reactions 1 and 2 both consume H+ions, the reaction mixture must increase in pH (i.e., become less acidic) over time (pH = -log[H+]). Figure 2 indicates that all of the chromium(VI) is being reduced to chromium(III) during the first 48 hours, so we know that hydrogen ions are being consumed at a high rate. In fact, because no chromium(IV) appears at P3 until after 48 hours, we know that between P1 and P3 100% of the chromium (IV) is being converted to chromium (III). This process will consume a large amount of H+ ions (per Reactions 1 and 2), decreasing the acidity and increasing the pH. Answer B states exactly the opposite, and is therefore incorrect. Answer C is not possible. Because reactions 1 and 2 require acidic conditions, they could not proceed to any appreciable degree in a neutral solution. For the reasons already stated, Answer D cannot be true. The pH cannot remain unchanged if hydrogen ions are being consumed.

The mitochondria of brown fat adipocytes contain a high concentration of the proton channel thermogenin in the inner mitochondrial membrane. Thermogenin generates heat via non-shivering thermogenesis. Which energy transformation best explains this process? Electrical potential, to kinetic, to chemical Electrical potential to thermal Chemical, to kinetic, to electrical potential Electrical to chemical

Choice B is the correct answer. When thermogenin is introduced into the inner mitochondrial membrane, the energy stored in the electrochemical gradient is decoupled from ATP production. Instead of being used to generate ATP, this energy is released as heat. This makes only Answer B plausible because it is the only option that includes thermal energy.

Sucrose is a disaccharide composed of glucose and fructose. Which of the following gives the correct structure of sucrose?

Choice C is the correct answer. Sucrose is a disaccharide composed of one glucose and one fructose unit. Glucose forms a six membered ring, while fructose forms a five membered ring, and thus the answer must contain one five- and one six-membered sugar ring. Answer A, B, and D all contain only six membered sugars. This question illustrates the fact that many MCAT questions are easier than they first appear. In this case, the student did not need to know the exact structure of both monosaccharides, or even the correct linkage. Knowledge that fructose forms a five-membered ring was sufficient to select the correct answer.

What is the most likely result of a mutation in chymotrypsin from Asp-102 to Glu-102? Total loss of protein catabolism Large increase in protein catabolism Small decrease in protein catabolism No change to protein catabolism

Choice C is the correct answer. The mutation from Asp to Glu is a complementary substitution. The only difference between Asp and Glu is a single additional carbon between the alpha carbon and the carboxylic acid functional group. This minor change, with no change to the functional group itself, is unlikely to significantly impact chemistry. The increased size of the amino acid may distort the active site slightly, but the difference being only one carbon, activity is likely to be retained. Answer A is incorrect since the enzyme will remain active. Answer B is incorrect because there is no reason to suspect that a complementary substitution will dramatically increase catabolism. However, any physical change to the enzyme is likely to have at least some impact on activity, making Answer D incorrect. Remember that enzyme active sites are usually very tight spaces with specific geometry related to their function. It may also be worth remembering that allosteric enzyme regulation is the result of a change (sometimes large, sometimes relatively small) in the conformation of the enzyme. All of these basic concepts suggest that a complementary substitution would have some effect, but would not dramatically alter enzyme function. Finally, notice that all three distractors use definitive language ("Total," "Large," and "No"). Answer C is the only option that allows for some nuance. While this is not always the case—sometimes definitive answer choices can be correct—keeping an eye out for over-stated logic can assist you in eliminating wrong answer choices.

The lens shown in Figure 1 creates an image of the rotating and static diffusers on the sensor. If the focal length of the lens is 25 mm, the image formed at the sensor will be: TRY THIS QUESTION AGAIN WITH KAPLAN BOOKS virtual and upright. virtual and inverted. real and upright. real and inverted.

Choice D is the correct answer. It can be seen from Figure 1 that this lens is a converging lens. It converges rather than diverges the light according to the lines used to represent the light passing through the lens. Further, the object (the static/rotating diffuser) is farther away than the focal distance of the lens (202.5 mm vs. 25 mm). Thus, a ray diagram drawn for this circumstance would be similar to the following: The rays truly meet on the right hand side of the lens; thus it is a real image. The arrow that is formed is upside down, so it is an inverted image. These facts make answer choice D correct. A virtual image, as in choices A and B, would be the answer if the rays did not meet up on the right hand side of the lens. This would occur if the object was much closer, closer to the lens than the focal length of the lens, which it is not. An upright image, as in choices A and C, would be the answer if the image arrow is produced right side up. To avoid drawing the ray diagram shown, a student can recall the PRI and NVUmnemonics taught in the Altius Student Study Manual. Positive images are Real and Inverted, and Negative images are Virtual and Upright. Therefore, one must only identify one of these three characteristic and then the other two are known. In this case, the lens drawn in Figure 1 is clearly a converging, or positive lens. For positive lenses, as long as the object distance is greater than the focal length, a positive lens will always create a positive image. Therefore, from the PRI mnemonic, we also know that the image is real and inverted.

Aqua running is a form of cardiovascular conditioning in which a person runs while partially submerged in water. If an aqua runner (mass 80 kg, weight 784 N) is exactly 50% submerged in water, what is the apparent weight of the runner? (Note: Assume the density of both the water and the runner is 1.0 g/cm3.) 157 N 196 N 261 N 392 N

Choice D is the correct answer. This question tests your knowledge of Archimedes' Principle, or buoyancy. Since the person and the water are the same density, if the person were fully submerged, the water would support all of the person's weight and the runner would be apparently weightless (in the sense of not pushing against the bottom of the pool). The person is only 50% submerged, though, so the water only supports half of the person's weight. Therefore, the apparent weight is the actual weight divided by 2: 784/2 = 392 N. For a more technical analysis, you could say that the actual weight is mpersong = (rpersonVperson)g. The buoyant force is upward and equal to rfluidVsubmergedg. The normal force, or what the problem called the "apparent weight" is equal to the difference between the two (draw a free body diagram if that is not obvious): N = rpersonVpersong - rfluidVsubmergedg = rVpersong - r(½ Vperson)g = ½ rVpersong = ½ mg. That is half of the person's given weight (mg, 784 N), or 392 N, or choice D. Answers A, B and C all give incorrect fractions of the aqua runner's original weight. For a person 50% submerged, the fluid would have to be more dense than the person in order for the apparent weight to be less than 50%.

Function of RNA polymerase in transcription

Creates a "bubble" around the DNA and separates the DNA strands - begins the process of transcription

One company sells a defibrillator for home use that uses a 9-volt DC battery. The battery is rated at 4.2 A•hr (amp•hour). Roughly how much charge can the battery deliver?

Current has units of Amperes (A) which is charge per unit time, or C/s. The question is asking for how much CHARGE is delivered, meaning we need the answer to be in Coulombs. They gave us 4.2 amp •hour, which has the units 4.2 C/s • hour. In order to convert that to just coulombs, we convert hour to seconds (1 hour = 3600 seconds), and if we multiply 4.2 C/s x 3600s, this leaves us with 15,120 C Because they tell you it is 4.2 amp/hour lol. The SI unit for current is the Ampere, they are giving you the extra information in the unit. Amp/Hour is not a SI unit and is given as extra info to solve the question. 1hr=3600sec*4= Choice C. 1 ReplyShare level 2Scientist-LocalOP·1 yr. ago But what if they said how much charge would be delivered in 30 minutes or two hours? It changes the answer. It assumes that we are considering only at 1 hour ? It's strange 1 ReplyShare Continue this thread level 1WAGUSTIN·1 yr. ago522 (132/130/130/130) an amp•hour is a unit of charge. They're quite literally telling you how much charge it can deliver. You just have to convert the way it's expressed into coulombs.

Which change in a property of pilin will be observed after modification by PptB?

Decrease in isoelectric point The correct answer is D. AAMC Standard Solution: Based on the passage, PptB is a phosphotransferase, which adds a phosphoglycerol to pilin subunits. The addition of this group (phosphoglycerol) adds a negative charge to pilin and thus decreases its isoelectric point. Jack Westin Advanced Solution: Decrease in isoelectric point. This is going to be a viable option. The transfer of a negatively charged phosphate group is going to decrease isoelectric point. Think about negatively charged amino acids. The presence of additional acidic groups is going to lower the isoelectric point. Answer choice D is our best option.

Dissociation vs. Dissolution

Dissassociation- Partner withdraws Dissolution- Partnership ends for Nacl lattice The breakdown of Nacl molecules from each other endo is dissolution and that of water from each other The breakdown of Nacl in to Na and Cl is dissociation Solvation is when ions solute is being surrounded by ions of solvent exo

Suppose researchers need a purified sample of ᴅ-limonene with minimal contamination. To accomplish this, ᴅ-limonene could be doubly distilled: under a nitrogen atmosphere. below atmospheric pressure. under UV light. using filtered, purified air.

Distilling under nitrogen excludes air, and therefore oxygen, from reacting with the limonene. Because the main premise of the passage is that D-Limonene undergoes automatic oxidation in the presence of oxygen, excluding oxygen is a logical requirement if one is to obtain a pure sample. Answers B, C, and D are not applicable because they do not exclude air. Answer A would provide a distilled sample that is oxygen-free. wasn't given any info on UV or purified air

formula for cell potential

E cell = Ecathode- EAnode delta G= -nFE delta G = -RTlnk G= delta G + RTlnQ IT=nF I=q/t F=q/n If you're calculating deposition of something at the anode/cathode, use this equation: Mol M = It/nF, where Mol M is "moles of metal," I = current, t = time, n = moles of electrons, and F = Faraday's Constant. Mnemonic: "Calculating Moles of Metal? It is Not Fun."

fluorescence

Final point is that fluorescence is not cause by *inter*molecular interactions, it is caused when an excited electron is emitted, so I am pretty sure that even if the question stem asked why the fluorescent dye itself exhibited fluorescence, option C would still not be correct since it is due to excitation and emission of energy, not interactions between molecules.

Experimental data suggest NS1-2 exists as a dimer. Which size-exclusion chromatography spectra is consistent with an NS1-2 mixture which may contain a folded monomer, an unfolded monomer, or a folded dimer?

First, the examinee should recall that in size-exclusion chromatography larger molecules migrate faster and smaller molecules migrate more slowly. This occurs because the smaller the molecule is, the more time it will spend migrating through tiny pores in the stationary phase (usually gel or agarose beads). Larger molecules can only migrate through the largest of such pores, or may be too big to pass through any of the pores, resulting in a faster migration to the bottom of the column. A dimer is larger, and therefore migrates faster, than a monomer, making Answer B the correct answer. Answer A is not correct since the dimer would migrate faster than the folded monomer, not vice versa. Answer C is incorrect since the unfolded monomer has a larger radius and would therefore migrate faster than the folded monomer. Answer D is not correct because the dimer will again migrate faster than a monomer—either folded or unfolded. If the student is not confident as to the relative size of folded vs. unfolded proteins, the data in Table 2 makes this trend clear: unfolded proteins have much larger radii than folded proteins. This question is an excellent example of how the AAMC will require you to demonstrate understanding of a basic science concept (the order of elution from a size-exclusion chromatography column based on size), but then go one step further by applying that information to a visual representation. Many students may understand the concept, but be unable to select the correct graph.

Rate order table

For normal rate vs concentration zero order= horizontal line If you double the concentration the reaction rate stays constant first order= linear increasing line If you double the concentration the reaction rate doubles second order= exponential If you double the concentration the reaction rate quadruples

To determine a protein's thermodynamic stability, chemical denaturation studies can be performed. Assuming that only the native and unfolded states can be observed under experimentally available conditions, what is the most likely shape of the curve for the dependence of the fraction of folded protein upon denaturant concentration? A Hyperbolic B Linear C Sigmoidal D Exponential

Good question, not sure if I have a perfect answer but here's how I thought of it. Think of a graph with % denaturation on the y-axis and [denaturant] on the x-axis. Upon addition of a small amount of denaturant, the enzyme concentration is much greater than the denaturant and is thus not very sensitive to it and the graph would have a small positive slope. Upon addition of a significant amount of denaturant, a large change in the % denaturation is expected which would account for a strong positive slope as seen in a sigmoidal curve. Once the enzyme has denatured completely, addition of more denaturant will not affect % denaturation and the graph will become horizontal, thus finishing the shape of a sigmoidal curve. 7 ReplyShare level 2****bitchesgetpolio·6 yr. ago Just to add to this. Proteins can be dense, with multiple subunits and tons of ionic, van der whaals and hydrogen bonds. Therefore they are somewhat resistant to denaturants at low concentrations. As your concentration slowly increases bonds will break and reform on the surface to remain energetically favourable. Get that concentration up and these bonds will rapidly break allowing the denaturant to work deeper into the protein, until all molecules are denatured. Key concept. Always remember protein folding and unfolding is cooperative so always sigmoidal graph. Same thing with DNA.

Which extraction procedure will completely separate an amide from the by- A Add 0.1 M NaOH(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the ether layer by evaporating the solvent. Jack Westin Advanced Solution: This is consistent with our breakdown of the question. Strong base will quench the unreacted anhydride. Meaning it's now unreactive. The strong base deprotonates the anhydride and we have a charge difference. We add diethyl ester, which is analogous to our organic solvent. The charge difference allows us to dissolve the amide, but the carboxylate won't dissolved. So, evaporating the solvent separates the amide and leaving only the carboxylate. B Add 0.1 M HCl(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the ether layer by evaporating the solvent. Jack Westin Advanced Solution: The first part of the process is consistent with our process. But using a strong acid instead of a strong base doesn't allow for the deprotonation of the anhydride and give us that charge difference. We can still maintain that A is our best option. C Add 0.1 M NaOH(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with HCl(aq). Jack Westin Advanced Solution: Our amide is in the organic layer and we have to filter it to get our amide. This contradicts what I said in the breakdown of the question. D Add 0.1 M HCl(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with NaOH(aq). Jack Westin Advanced Solution: This is similar to answer choice C, but also, we're using a strong acid instead of a strong base. This contradicts our breakdown as well. We're left with our correct answer, answer choice A.

Hey OP! You're over thinking it. When performing extractions we typically want the thing we are keeping (amide) to end up in the organic layer and the thing we are getting rid of (acid anhydride) in the aqueous layer. Something that really helped me with OChem is remembering Acid Base reactions happen more quickly than nucleophilic additions and eliminations. We want to protonate/ deprotonate the carboxylic acid anhydride so it ends up in the aqueous layer at the bottom of the sep funnel and can be decanted away. Because the carboxylic acid anhydride is acidic, we want to deprotonate it to give it a negative charge. Because Acid/Base reactions tend to happen first. for the unreacted carboxylic acid anhydride, it will be deprotonated first (though the AAMC is inappropriately using the term quench here lol). This gives the carboxylic acid anhydride a negative charge, allowing it to dissolve in the aqueous layer. The other species we know is in solution is the amide, which the base can't deprotonate so it stays in the organic layer made of diethyl ether. Because we want to use a base to deprotonate the acid anhydride, this eliminates B and D. Because the amide end up in the organic layer (diethyl ether) and the carboxylic acid anhydride ends up in the aqeous layer, this eliminates C, leaving A as the correct answer. There are two concepts being tested here. So first you'll have to know that the reaction between an amine and carboxylic anhydride gives you an amide and carboxylic acid. Since we are adding excess anhydride we can assume that all amine are turned into amide. Now we have to deal with the excess carboxylic anhydride, and quench just means converted to an unreactive form. Adding NaOH or HCl here doesn't make a difference - they serve as basic/acidic catalysts to convert anhydride to carboxylic acid. But the choice does matter for separation. Keep in mind that extraction takes advantage of a molecule's polarity/charge. We can deprotonate COOH to make COO- but at the same time not affecting the charge of amide. Finally, what can be dissolved in diethyl ether? I think you can get to the answer for there! for the HCL debate if i had added HCL then i would have lowered the pH so COOH would be COOH but Nh2 would be in the aqueous layer since it would have been NH3+. that answer choice would look like use HCL, Amide in aq layer, then evaporate alright so we got excess anhyride right, becuz all the answer start off with quench unreacted anhyride so we know that. So we have anhyride reaction with an amine to produce 2 things a amide and COOH + the unreacted anhydride. cool stay with me now an Anhdyride + NaOH will produce 2 COOH because you are breaking it down with a base. so now the reaction products have COOH original + NH2 amide + COOH broken from anhydride by base NaOH (alt you can use HCl becuz it wouldnt matter that step is just to get rid of excess anhdyrides just for this step but it matters later so keep it in mind- to break an anhydride if doest matter using hcl or naoh but it matters later for the question. now what we gotta do is we have to seperate an amide from COOH. We know that becuz its extraction. We have to seperate based on properties, since they are both pretty polar we need to seperate based on charge cuz that be easiest ya feel me. so the way to seperate by charge would be to remove proton from COOH to make it COO-, and NH2 keep it same. NOW remeber when i said the NaOH and HCl debate matters now. we need to increase the pH to deprotonate COOH to COO- and since NH2 has a high Pka(think nh2 end of AA how their pka on the backbone is 9) if i raise by pH by base to 7, all my COOH is now COO- BUT my NH2 still kept its H to its still NH2. NOW we basically have 2 things COO- and Nh2. so COO- is hella charged and Nh2 aint. Diethyl ether is an organic solvent, meaning it will dissolves things not charged. so you add diethyl ether, which says to COO- like "you cant sit with us" and tell Nh2 come here booboo you mine forever. So now you just have Nh2 in organic layer. Then you filter it out and evaporate to get Nh2

The electric field inside each of the conductors that forms the capacitor in the defibrillator is zero. Which of the following reasons best explains why this is true? Jack Westin Advanced Solution: This is a passage-based question that focuses on something we learned in Paragraph 2 of the passage where the author talks about defibrillators. The author tells us, "A portable defibrillator works by storing electric energy in a capacitor. Electrodes, sometimes called "paddles," are placed on a patient's chest, and current can be passed from one electrode to the other once a switch is flipped by a medical professional."All conductors contain electric charges. When exposed to potential difference, the positive charges in a conductor will migrate towards the negative end, the negative charges in the material will move towards the positive end of the potential difference. This flow of charge is electric current. We want to explain why the electric field inside the conductors that form the capacitor is zero. The effect of electric fields can be found by superimposing its E vectors, and in this case that effect ends up as a net zero. The charges are all functioning to essentially cancel out and make the electric field acting on them zero. That can happen once electrons have reached an equilibrium and the electric field produced in the conductor cancels with any external field. That's a property of conductors. A.All of the electrons in the conductor are bound to atoms, and thus there is no way for an external electric field to penetrate atoms with no net charge.AAMC Standard Solution:Conductors are characterized by the existence of free electrons that carry current. Jack Westin Advanced Solution: All of the electrons in the conductor are bound to atoms, and thus there is no way for an external electric field to penetrate atoms with no net charge. This answer choice is extreme, first of all. We typically stay away from answers that begin with "all." Furthermore, conductors also have free electrons that can travel and orient themselves on the surface of the conductor. B.Free electrons in the conductor arrange themselves on the surface so that the electric field they produce inside the conductor exactly cancels any external electric field.The correct answer is B.AAMC Standard Solution:Conductors contain both atom-bound electrons and free electrons. Free electrons arrange themselves on the surface of conductors, and their collective electric field produced inside the conductor cancels any external electric field. The resulting electric field inside the conductor is zero. Jack Westin Advanced Solution: Free electrons in the conductor arrange themselves on the surface so that the electric field they produce inside the conductor exactly cancels any external electric field. This answer choice is consistent with what happens with conductors. We can imagine free electrons arrange themselves on the surface like we mentioned in our breakdown of the question. The charges are all functioning to essentially cancel out and make the electric field acting on them zero. That can happen once electrons have reached an equilibrium and the electric field produced in the conductor cancels with any external field. This is going to be our best answer choice. C.Free electrons in the conductor arrange themselves on the surface and throughout the interior so that the electric field they produce inside the conductor exactly cancels any external electric field.AAMC Standard Solution:Free electrons arrange themselves only on the surface of the conductor. If they also arranged inside, the electric field inside the conductor would move the electrons even in the absence of a battery. Jack Westin Advanced Solution: Free electrons in the conductor arrange themselves on the surface and throughout the interior so that the electric field they produce inside the conductor exactly cancels any external electric field. Free electrons are present on the surface, but not throughout the interior. The reason we get the electric field to be zero is because of the free electrons on the surface and produce an electric field that cancels the external field. Adding another electric field changes that net zero and balance. D.All electrons in the conductor, both free and bound, arrange themselves on the surface so that the electric field they produce inside the conductor exactly cancels any external electric field.AAMC Standard Solution:Bound electrons cannot arrange themselves on the surface of the conductor due to the binding effects. Jack Westin Advanced Solution: All electrons in the conductor, both free and bound, arrange themselves on the surface so that the electric field they produce inside the conductor exactly cancels any external electric field. Reasoning here is similar to answer choice A. We avoid answer choices that begin with "all." Furthermore, bound electrons are not part of the rearrangement on the surface. Conductors have free electrons that can travel and orient themselves on the surface of the conductor. We can stick with answer choice B as our best answer choice.

Honestly I would just memorize that the electric field inside any conductor is zero. The electrons always arrange themselves in such a way that this is always true.

A person is struggling to achieve generativity, rather than stagnation. In terms of Erikson's theory, this person is a(n): Jack Westin Advanced Solution: This is another standalone question that relies on external knowledge, specifically vocabulary. Erikson's eight stages of psychosocial development include trust vs mistrust, autonomy vs shame/doubt, initiative vs guilt, industry vs inferiority, identity vs role confusion, intimacy vs isolation, generativity vs stagnation, and integrity vs despair.Generativity vs Stagnation: When people reach their 40s, they enter the time known as middle adulthood, which extends to the mid-60s. The social task of middle adulthood is generativity vs stagnation. Generativity involves finding your life's work and contributing to the development of others through activities such as volunteering, mentoring, and raising children. A. adolescent. Jack Westin Advanced Solution: adolescent. This corresponds to identity vs. role confusion. In adolescence (ages 12-18), children face the task of identity vs role confusion. According to Erikson, an adolescent's main task is developing a sense of self. Adolescents struggle with questions such as "Who am I?" and "What do I want to do with my life?" Along the way, most adolescents try on many different selves to see which ones fit; they explore various roles and ideas, set goals, and attempt to discover their "adult" selves. B. young adult. Jack Westin Advanced Solution: young adult. This corresponds to intimacy vs. isolation. Intimacy vs Isolation: People in early adulthood (20s through early 40s) are concerned with intimacy vs isolation. After we have developed a sense of self in adolescence, we are ready to share our life with others. C. middle-aged adult. Jack Westin Advanced Solution: middle-aged adult. A middle-aged adult would most likely be struggling to achieve generativity rather than stagnation. This is our correct answer. D. elderly adult. Jack Westin Advanced Solution: elderly adult. This corresponds to integrity vs despair. Integrity vs Despair: From the mid-60s to the end of life, we are in the period of development known as late adulthood. Erikson's task at this stage is called integrity vs despair. He said that people in late adulthood reflect on their lives and feel either a sense of satisfaction or a sense of failure. Answer choice C is the best answer. According to Erikson's theory, adolescents struggle with identity versus role confusion. According to Erikson's theory, young adults struggle with intimacy versus isolation. According to Erikson's theory, a middle-aged person struggles with generativity versus stagnation. According to Erikson's theory, an elderly person struggles with integrity versus despair.

I know that generativity vs stagnation is around 40-65. Can someone explain to me how I'm supposed to reason that these people are middle-aged vs. elderly. I guess 65+ is your stereotypical elder but what sort of LOGIC am I missing here? Or is this just a question that has no real logic to and you're just supposed to know? I'm assuming because it was a discrete it was just pure knowledge, so am I just an idiot?

Which statement best explains how the concept of external motivation (used by SDT) is different from extrinsic motivation? Compared to external motivation, extrinsic motivation: Jack Westin Advanced Solution: Our approach to this question is going to be similar to our approach to the previous question. This is a passage-based question that relies on us going back to the passage to pick out key information, but ultimately, we'll need to rely on knowing our vocabulary to answer this question correctly. We can keep Paragraph 4 in mind as we go through each term. Paragraph 4 is when the author explains self-determination theory (SDT). The author mentions controlled behaviors are due to pressure from social contacts (external motivation). Extrinsically motivated behaviors are performed in order to receive something from others—such as a promotion, praise, candy, money, or attention. Extrinsic motivation is therefore a broader term that can actually include external motivation. A. is a broader term that includes external motivation. Jack Westin Advanced Solution: is a broader term that includes external motivation. This answer choice matches our breakdown of the question. We said extrinsically motivated behaviors are performed in order to receive something from others. That can be in addition to appeasing social contacts. That pressure from social contacts falls under the umbrella of extrinsic motivators, but there are additional motivators. B. is a narrower term that refers to external reinforcers. Jack Westin Advanced Solution: is a narrower term that refers to external reinforcers. This is the opposite of our breakdown. We said extrinsically motivated behaviors are performed in order to receive something from others. That includes external motivation, so extrinsic motivation is a broader term, not a narrower term. C. excludes social punishers and reinforcers. Jack Westin Advanced Solution: excludes social punishers and reinforcers. Reinforcement and punishment are principles of operant conditioning that increase or decrease the likelihood of a behavior. This also falls under extrinsic motivation. D. excludes internal states that direct behavior. Jack Westin Advanced Solution: excludes internal states that direct behavior. This is not something that is different about external motivation versus extrinsic motivation. Internal states are excluded from both. We can stick with answer choice A as our best answer.

I'm going to break this down by wrong answers. D. Is wrong because external and extrinsic motivation have nothing to do with internal drive(which I'm sure you knew) C. Is wrong because punishments and reinforcements are both external forces and there is no distinction made as to whether external or extrinsic motivation motivate you to do or not do something but simply motivation in a particular direction B. Is wrong because like I mentioned for C. motivation can be a punishment or a reinforcer This leaves the ever so general A. which isn't a sexy answer by any means but it's not wrong and is therefore the correct answer

Movement of electrons and current

I'm not sure about reversing but the way I think of it, the flow of the current (AKA the flow of positive charge) is always opposite to the flow of electrons (the flow of negative charge). Currents naturally flow from high to low potentials (like electric field lines) so electrons must flow from low to high potential. Maybe this is going a little too in depth but, a battery does the work of separating the +/- charges and creating that potential difference in the terminals to allow electrons and a current to flow through the circuit. When a battery is "turned on", it releases the stored potential energy inside and the current spontaneously flows through from high to low or the positive terminal to the negative terminal, meaning that the electrons are naturally flowing from the negative to the positive terminal. I hope this makes sense!

If [E]T was the concentration of lactase in the kinetics trials, what expression gives the concentration of lactase in the commercial preparation of this enzyme? If [E]T was the concentration of lactase in the kinetics trials, what expression gives the concentration of lactase in the commercial preparation of this enzyme? A [E]T × 25 B [E]T × 50 C [E]T × 200 D [E]T × 500 The students prepared stock solutions of Compound 1 in four different concentrations in pH 7 buffer. A stock solution of the enzyme was prepared by diluting 0.100 mL of the commercial preparation to 25.0 mL in the buffer solution. Experiments were initiated by mixing 1.0 mL of each substrate solution with 1.0 mL of the enzyme solution. The initial rates Vo were measured for each trial. The students then plotted 1/Vo versus 1/[S] (Figure 1) to determine KM, Vmax, and [E]T for the four trials.

If [E]T was the concentration of lactase in the kinetics trials, what expression gives the concentration of lactase in the commercial preparation of this enzyme? Jack Westin Advanced Solution: We need to know about the concentrations of our two samples so we can write an expression that relates the two; we'll get that from our passage.As always, carefully look at the units and find the concentrations for each step. Stock solution was made by diluting 0.100 mL of the commercial preparation to 25.0 mL in the buffer solution. We can divide the two. This means the commercial preparation was diluted 1🡪250.Next, 1.0 mL of substrate solution was mixed with 1.0 mL of the enzyme solution to further dilute it. It's now half as concentrated so 1🡪2 here.Final solution means we multiply: 1/250 x ½ to get 1/500th of the commercial preparation. So, the concentration of lactase in the commercial preparation of the enzyme is [E]T x 500. This was a calculation problem and we can compare all four of our answer choices at once. Our correct answer is answer choice D. We can eliminate answer choices A-C for being incorrect values. A [E]T × 25 B [E]T × 50 C [E]T × 200 D [E]T × 500 Need extra help breaking down AAMC logic? Find out how you can meet with one of our expert MCAT tutors! Solution: The correct answer is D. This is a General Chemistry question that falls under the content category "Unique nature of water and its solutions." The answer to this question is D because the commercial preparation was first diluted by 1 → 250 to prepare the stock solution, and it was further diluted by 1 → 2 by mixing with the substrate stock solution to perform the kinetics experiments. It is a Scientific Reasoning and Problem Solving question because you are asked to bring together theory, evidence, and observations to draw a conclusion.

What quantity of NAG3 was required to reach the equivalence point in the titration? Jack Westin Advanced Solution: We're thinking back to the passage and HEW lysozyme was titrated with injections of NAG3. How much NAG3 was needed to reach the equivalence point, and our answer is given in different units of moles. We'll need to visit the figures in the passage and get specific details. But we'll also have to know about titrations, and specifically equivalence points from our general knowledge. Let's solve for our quantity in either nanomoles or micromolesThe number of mol of NAG3 must equal number of mol of HEW at the equivalence point.The information about NAG3(25 injections, 10 micro L and 2.5mM) in the passage is about the TOTAL amount of NAG3 added in the titration but we don't actually know what amount of NAG3 was needed to reach equivalence.The only thing the titration graphs tell you is that to reach equivalence you need a 1:1 mol ratio of HEW to NAG3. However, if you look back at the passage info about HEW, it tells us that there is 1 mL (1×10^-3 L) and .1mM (1×10^-4 Molar) of HEW. We can convert this to moles. Multiplying, we get 100nmol or 0.1 micromoles (1×10^-7 mol) of HEW.We said there are an equal number of moles. So, we'd also need 100nmol or 0.1 micromoles of NAG3 to reach the equivalence point.We can compare all four of our answer choices at once. Our correct answer is answer choice B. We can eliminate answer choices A, C, and D for being incorrect values. A 25 nmol B 100 nmol C 25 µmol D 100 µmol How do you know you are truly ready? That's where the Jack Westin MCAT Diagnostic Tool comes in. Solution: The correct answer is B. This is a General Chemistry question that falls under the content category "Unique nature of water and its solutions." The answer to this question is B because the graphs in Figure 1 both imply a 1:1 mole ratio of NAG3 was added at the equivalence point and the solution contained 1.0 mL of 0.10 mM = (1 × 10-3 L)(0.10 × 10-3 mol/L) = 1.0 × 10-7 mol = 100 nmol. It is a Data-based and Statistical Reasoning question because you are asked to analyze and interpret data presented in Figure 1 to arrive at the conclusion.

If you look at the titration curves you can see the point at which the slope is greatest which is the equivalence point. The molar ratio on the x axis shows this point to be around 1 so that's how you determine the molar ratio as 1:1. From this point you just need to determine the amount of moles added. You could do it either by calculating the moles of HEW in your sample or by determining how many titrations of NAG you added to reach the equivalence point (looks like around 4 or 5). You can calculate it either way but determining the moles of HEW is easier as you know the exact molarity and volume at the start. You can use that molar amount because we already know from the graph it's 1:1. Using NAG you would have to find the moles of each titration and then multiply that by the number by 4 or 5. Hope that helps 2 ReplyShare level 2Altruistic_Bug_1264OP·9 mo. ago Since the ratio is the same for HEW and NAG3 i can use the values of HEW to calculate NAG3? 1 ReplyShare Continue this thread level 1Tigster17·9 mo. ago Ok, so the equivalence point of a titration curve is where the # of moles substance A = # moles substance B (aka where the molar ratio is 1 because 1:1 is 1)... We will use HEW as substance A and NAG as substance B. So, at equivalence, the moles of HEW must equal moles of NAG. Well, that's great but we need to know how many moles there are of either of them. We don't have enough info to find the number of moles of NAG, but we do have the info for number of moles of HEW. Molarity = mol/L and we rearrange this formula for mol. So, mol = Molarity x L. We are given this info from HEW (0.1 mMol x 1 mL). This will give us the number of moles of HEW. And we know that the number of moles of HEW equals the number of moles of NAG. So, this calculation should give you the answer. Hope this helps. I SUCK at titration stuff.

Suppose that a blood vessel of cross-sectional area A carries microbubbles at a speed v into a capillary bed. If the capillary bed is made up of n capillaries, each with cross-sectional area a, with what speed will the blood flow in the capillary bed?

If you think about it , it is simple continuity equation so one blood vessel has Av and other has nav2 so v2 = Av/na or Q=Av and A= na v=Q/na v=Av/na

The ᴅ designation in ᴅ-limonene is an abbreviation for dextrorotatory. This indicates that ᴅ-limonene rotates polarized light: counterclockwise and could contain a chiral carbon with either the R or the S configuration. clockwise and could contain a chiral carbon with either the R or the S configuration. counterclockwise and must contain a chiral carbon with the S configuration. clockwise and must contain a chiral carbon with the R configuration.

In organic chemistry, ᴅ and ᴌ are used to indicate if an isomer is dextrorotatory or levorotatory, respectively. Dextrorotatory isomers rotate polarized light clockwise. These letters are not associated with any particular R or S absolute configuration, and thus answers C and D can be eliminated. It may be tempting to assign R/S configuration using the limonene structure shown in the passage--which includes the necessary stereochemistry to make that decision. However, the question stem only asks what the ᴅ designation indicates about ᴅ-limonene. That designation alone does not give any information about R/S. Answer A incorrectly uses counterclockwise for the rotation. Answer B is thus the best answer.

A reaction is designed to produce ammonia from the gas phase equilibrium of nitrogen and hydrogen. N2(g) + 3 H2(g) 2 NH3(g) Introducing a catalyst into the system will cause the amount of ammonia at equilibrium: inc remain the same decr to change in a manner that depends on the value of eq constant

In short, enzymes do not change the equilibrium state of a biochemical reaction. ΔG0 and Keq remain the same. Instead, the enzyme reduces the activation energy needed for the reaction to proceed, and thus increase the rate of reaction. Say for example, Keq = 1, and you start with reactants only. Under normal conditions it may take 100 years, but eventually the concentration of reactants will equal the concentration of products. Now start over, but add an enzyme. In this case we reach equilibrium in 100 seconds, but it still results in a 1:1 ratio of reactants to products. Why? Enzymes work by stabilizing reaction intermediates. And as such, they catalyze reactions in either direction! Enzymes enhance rates. They don't change free energy differences nor do the affect equilibrium constants. b ans The position of equilibrium in a reaction is determined solely by the difference in free energy between products and reactants. A catalyst speeds the reaction by lowering the activation energy of the reaction. It can do nothing to affect the energies of products and reactants and thus cannot affect the position of the equilibrium. It only allows the reaction to reach equilibrium more quickly. Answer choice B is correct.

If red litmus paper is dipped into the Na2CO3 solution, it will:

In water, carbonate will undergo the following reaction: CO32-(aq) + H2O(l) → HCO3-(aq) + OH-(aq). Red litmus paper will turn blue in a base. Thus, C is the best answer.

if the two compound has diff stability but same constituents around it

It can't have inductive effect

If the third codon in the coding region of the Rdl GABA receptor cDNA is replaced with an amber codon (for example, TAG) and the modified Rdl GABA receptor cDNA is expressed in frog oocytes, functional full-length receptors will: Jack Westin Advanced Solution: This is a pseudo-standalone question that asks us to identify the effects of introducing an amber codon. A stop codon is one of three codons (UAA, UAG, and UGA) that signals the end of translation. If there is a mutation which occurs and causes an early stop codon in an mRNA sequence this is called an amber codon. The example in our question stem says the third codon in the coding region is replaced with TAG. mRNA is going to be the same, but we replace T with U. We know UAG is one of three codons that signals the end of translation, and full-length receptors are not made. The only answer choice consistent with this is answer choice C which says functional full-length receptors will not be synthesized. A.accumulate in the Golgi complex.AAMC Standard Solution:An amber codon is a stop codon. If the third codon is replaced with a stop codon, the protein will not be synthesized, and for this reason, it cannot accumulate in the Golgi. B.accumulate in the rough endoplasmic reticulum.AAMC Standard Solution:An amber codon is a stop codon. If the third codon is replaced with a stop codon, the protein will not be synthesized, and for this reason, it cannot accumulate in the rough endoplasmic reticulum. C.not be synthesized.The correct answer is C.AAMC Standard Solution:An amber codon is a stop codon. If the third codon is replaced with a stop codon, the protein will not be synthesized. D.lose resistance to dieldrin.AAMC Standard Solution:An amber codon is a stop codon. If the third codon is replaced with a stop codon, the protein will not be synthesized, which will make the oocytes resistant to dieldrin simply because the agent dieldrin reacts with is absent.

It has been a while since I've done this, but if its TAG on the DNA wouldn't it be UAG on the coding tRNA? TAG->AUC->UAG Isn't cDNA complementary to the mRNA strand? So the mRNA would be AUC? 4 ReplyShare level 1blackninja20·3 yr. ago Retake 8/1: (130/127/129/129) 506 --> 515 The cDNA is the sense strand aka coding strand. This means that the mRNA will be the except same except the T will be replaced with U. cDNA: TAG mRNA: UAG Remember stop codons are UAG, UGA, and UAA (U Are Gone, U Go Away, and U Are Annoying)

A researcher measures the concentration of CO2(aq) in a solution at equilibrium. Which additional quantity must be measured in order to calculate the Henry's Law constant kH for the gas? The solution is C. The atmospheric pressure Patm is not a factor in the Henry's Law constant equation. The volume V of the solvent is not a factor in the Henry's Law constant equation. The Henry's Law constant kH relates the solubility of a gas S to the pressure of that gas Pg above the solution and is written as S = kH•Pg. Therefore, in addition to the concentration of CO2(aq) in a solution at equilibrium, the partial pressure of the gas Pg must be measured.​ The vapor pressure Pv of water is not a factor in the Henry's Law constant equation.

It says that at constant temperature, the partial pressure of a gas is directly related to its solubility in a liquid. Thus, the higher the partial pressure, the more soluble (I.e will dissolve in) it is in the liquid. You can apply this to the respiratory system. The more higher the partial pressure of oxygen is, the more it will dissolve in the blood. Likewise is true for carbon dioxide.

A description of the structures of four proteins is shown in the table. ProteinDescriptionProtein 132 kDa monomerProtein 2Disulfide-linked homodimer comprised of 19 kDa monomersProtein 3Homotrimer comprised of 25 kDa monomersProtein 4Homodimer comprised of 38 kDa monomers Which protein has the highest electrophoretic mobility in SDS-PAGE under non-reducing conditions? (Note: There are no disulfide interactions unless stated in the table.)

Jack Westin Advanced Solution: For this question, we need some background on SDS-PAGE gels. We have to know proteins can be chemically denatured before running them on a gel and proteins are separated exclusively by size. The chemical SDS denatures proteins, and so denaturing SDS-PAGE gels are used for these experiments to separate proteins by size. We're told there are no disulfide interactions unless we're explicitly told there are. Disulfide bonds between cysteine residues are not fully broken by SDS alone. The key here is finding the smallest subunits that will be broken down. The protein with the smallest subunits has the highest electrophoretic mobility in SDS-PAGE. Let's go through our 4 options and consider size. A Protein 1 Jack Westin Advanced Solution: Protein 1 is a monomer, and at first glance through our answer choices this would be the smallest protein of the 4 options listed. However, we don't just want the smallest protein, we want the smallest subunits. We'll keep the 32 kDa size in mind and keep comparing. B Protein 2 Jack Westin Advanced Solution: We want to make sure to not get tricked here. In the question stem, we're told there are no disulfide interactions unless we're explicitly told there are. Disulfide bonds between cysteine residues are not fully broken by SDS alone. That means the size here is going to be 38 kDa and Protein 1 is still our superior answer. C Protein 3 Jack Westin Advanced Solution: This homotrimer is comprised of 25 kDa monomers, meaning when it's broken down, we'll have 25 kDa subunits. Protein 3 will have the highest electrophoretic mobility of the first three answer choices we compared. D Protein 4 Jack Westin Advanced Solution: Protein 4 is similar to protein 3 in that we'll have subunits that are 38 kDa, despite the homodimer itself being larger. We're still going to stick with the smallest subunits in this situation: Protein 3. Want more MCAT Prep? We have an MCAT Podcast! Solution: The correct answer is C. This is correct. This Biochemistry question falls under the content category "Separation and purification methods." The answer is C because in an SDS-PAGE gel that is run under non-reducing conditions, proteins 1, 3, and 4 will run as monomers. Protein 2 will run as a dimer because the disulfide bonds between Cys residues are not reduced. The running masses will then be: A = 32, B = 38, C = 25, D = 38. As the smallest one, Protein 3 will have the greatest electrophoretic mobility. This is a Reasoning about the Design and Execution of Research question because you must understand the experimental conditions used in SDS-PAGE and how addition or lack of a reductant affects the result of the analysis.

When performing experiments to measure the kcat of an enzyme, the substrate concentration should be: limiting. equal to 1/2 Km. equal to KM. saturating.

Jack Westin Advanced Solution: Kcat, the catalytic constant, measures the number of molecules of substrate turned over into product per unit time (usually seconds). The higher your kcat is, the more reactions happen per unit time (i.e. Vmax). All of this is taking place when conditions are optimal, and you have enough substrate. So, Kcat is positively proportional to efficiency. Let's see if we can find an answer choice consistent with what we've laid out for Kcat. A limiting. Jack Westin Advanced Solution: I mentioned in the brief breakdown of the question and Kcat, we want conditions to be optimal, and we need sufficient substrate. Limiting substrate goes against what I said. B equal to 1/2 Km. Jack Westin Advanced Solution: The Michaelis constant, Km, can help compare the affinity of an enzyme for a substrate. A lower Km means a higher affinity of an enzyme for its substrate. The higher km is, the more concentration of substrate you need before you reach optimal conditions for Kcat, so km is inversely proportional. When measuring Kcat, substrate concentration does not have to be equal to ½ Km. C equal to KM. Jack Westin Advanced Solution: This is similar to answer choice B in that we're going to say that when measuring Kcat, substrate concentration does not have to be equal to Km. We're still hoping for a better answer choice. D saturating. Jack Westin Advanced Solution: This matches what we said in the breakdown of this question; we want conditions to be optimal, and we need sufficient substrate. This is the best of the four answer choices listed. Want more MCAT Prep? We have an MCAT Podcast! Solution: The correct answer is D. This is correct. This Biochemistry question falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is D because the kcat is used to describe the rate-limiting step of catalysis under saturating conditions of substrate. This is a Reasoning about the Design and Execution of Research question because you must understand the design of an experiment used to measure the maximum velocity of an enzymatic reaction. Correct Your Answer:D Correct Answer:D Content & Skills SIRS3CC5EFC5BCH

When experiments are performed on enzymes that display traditional Michaelis-Menten kinetics, what shape does the graph of V0 versus substrate concentration [S] have? _

Jack Westin Advanced Solution: Let's consider the following graph and put a name to the shape. We even have our equations up top, this is a hyperbolic relationship. A A hyperbolic dependence on [S] Jack Westin Advanced Solution: This matches exactly what we said in the breakdown of the question and what we see in the image above. This is likely going to be our best answer. Let's compare to be thorough. B A linear dependence on [S] Jack Westin Advanced Solution: Note the equations and the actual graph above. The relationship here is not linear. We can still stick with our best answer choice, answer choice A. C A sigmoidal dependence on [S] Jack Westin Advanced Solution: This is similar to answer choice B in that it's factually incorrect. A remains the best option. D A parabolic dependence on [S] Jack Westin Advanced Solution: Just like B and C, this does not describe what we're seeing in the above graph, nor is it consistent with our equations. We'll stick with our best answer here, answer choice A. Want more MCAT Prep? We have an MCAT Podcast! Solution: The correct answer is A. This is correct. This Biochemistry question falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is A because traditional Michaelis-Menten kinetics describes a hyperbolic dependence of V0 on substrate concentration. This is a Knowledge of Scientific Concepts and Principles question because you must recall the shape of a typical Michaelis-Menten plot. Correct Your Answer:A Correct Answer:

Which scenario is NOT an accurate representation of the McDonaldization of society? _ A A customer cleaning their table and disposing of waste after eating at a restaurant B A doctor seeing a patient at his or her residence rather than at a medical facility C A supermarket chain using self-scan machines at check-outs, in place of employees D A chain of coffee shops offering the same menu and décor across the United States

Jack Westin Advanced Solution: McDonaldization is a phenomenon that occurs when society, its institutions, and its organizations are adapted to have the same characteristics that are found in fast-food chains. These include efficiency (optimal method for accomplishing a task and minimizing time), calculability (quantifiable objectives-high quantity equals quality), predictability and standardization (same service at any McDonald's chain), and control (employees are standardized and use technology). Make sure to pay close attention to the question stem. We want a scenario that is NOT a representation of the McDonaldization of society. A A customer cleaning their table and disposing of waste after eating at a restaurant Jack Westin Advanced Solution: This would make the process of serving a customer more efficient. By having customers clear their own table, that is less time and resources the McDonald's has to spend on the customer. B A doctor seeing a patient at his or her residence rather than at a medical facility Jack Westin Advanced Solution: This is the opposite of answer choice A because this would decrease efficiency for the doctor. Because we have that decrease in efficiency, this is NOT a representation of the McDonaldization of society and is a good answer. The doctor would rather see patients quickly in their medical facility instead. C A supermarket chain using self-scan machines at check-outs, in place of employees Jack Westin Advanced Solution: This has to do with control. By using self-scan machines, the supermarket becomes more efficient and predictable. This is a representation of the McDonaldization of society so we can eliminate this answer choice. D A chain of coffee shops offering the same menu and décor across the United States Jack Westin Advanced Solution: This has to do with predictability and standardization. We expect the same service at any McDonald's chain, and the coffee shop offering the same menu and décor is achieving that same predictability and standardization. This is a representation of the McDonaldization of society so we can eliminate this answer choice. We can stick with answer choice B as the best answer choice. Want more MCAT Prep? We have an MCAT Podcast! Solution: The correct answer is B. This is a Sociology question that falls under the content category "Social interactions." The answer to this question is option B because it does not represent the elements of McDonaldization which include efficiency (option A), calculability, uniformity (option D), and technological control (option C). It is a Knowledge of Scientific Concepts and Principles question because you are asked to recognize the sociological term McDonaldization of society and identify its constituent elements.

What is the most accurate description of the natural substrate of GalK? A It is a five-carbon aldose. B It is a five-carbon ketose. C It is a six-carbon aldose. D It is a six-carbon ketose.

Jack Westin Advanced Solution: Note the natural substrate of GalK is galactose, not glucose. We need to describe galactose's structure. This is almost like a standalone, or discrete, question.Galactose is a monosaccharide and six-carbon aldose. Even if you did not remember the structure of galactose, recall it is an epimer of glucose. Galactose and glucose are both epimers at C-4. Epimers are stereoisomers that differ in their configuration at a single stereocenter. The position of the -OH group on carbon C-4 is in opposite directions in galactose and glucose. Answer choice C matches our breakdown exactly, galactose is a six-carbon aldose.

Thermal denaturation experiments can be used to follow the transition of double-stranded DNA into single-stranded DNA. Which of the following parameters affects the Tm of dsDNA in this experiment? pH of solution Ionic strength of solution Length of DNA strands

Jack Westin Advanced Solution: Something I want to point out here. If we are able to eliminate any of the three options listed, we can get to our correct answer. What do I mean by that? Every one of the three options is listed in three answer choices. For example, if we eliminate option I, that would allow us to eliminate answer choices A, B, and D, and pick C as the correct answer. Let's go through the three choices and see if they affect the Tm of dsDNA in the experiment.pH of solution. This could mean an increase or decrease in pH. In acidic solution, we would have protonation of bases, so that disrupts the usual base-pairing and leads to errors. This is a viable option.Ionic strength of solution. If we have a solution with a lot of positive ions, we have the phosphate groups stabilize. If we gave increased stability, we have a higher melting point and it's more difficult to denature. This is also a viable option.Length of DNA strands. The length of a DNA strand determines how many hydrogen bonds are in the strand. A longer DNA strand would be more stable because there are more hydrogen bonds. That means this is also a viable option.We went through all three choices here and determined each one was a viable option and could affect the Tm of dsDNA in the experiment. The only answer choice that's consistent with that is answer choice D: I, II, and III. A I and II only B I and III only C II and III only D I, II, and III This is correct. This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is D because all of these parameters would affect the thermodynamic stability of the DNA double helix. Significant drops in pH would result in protonation of hydrogen-bond acceptors, leading to a loss in base-pairing interactions. The presence of positive ions in solution (particularly Mg2+) leads to stabilization of the DNA fold via shielding of the repulsion between phosphate groups within the DNA backbone. The length of DNA strands would also play a role. Longer DNA strands are held together by more hydrogen bonds, meaning that more energy is required to denature the double-stranded DNA. This is a Reasoning about the Design and Execution of Research question because you must understand the consequences of changing variables on the results an experiment.

What is the reading of the energy meter in Figure 1 when an appropriate laser is used in PAC to dissociate a particular chemical bond? . A Em B ΔHu C ΔHnr D 0

Jack Westin Advanced Solution: That means we're using a specific laser to dissociate a specific chemical bond, and we want to know the energy meter reading. We can answer this question using content, but a quick glance at our answer choices shows that the answer could be in terms of the specific variables discussed in the passage.The passage say excess energy after the laser hits the sample is absorbed and temperature increases. But in this case, we're using an appropriate laser, for a particular chemical bond.Everything works in a particular way, and the energy meter doesn't pick up anything. It would only pick up if energy was to come off, and then you would know that there is an unexpected variable in your measurements.Looking at the diagram, you can see that the basic flow of energy goes from laser (photon energy) -> to the lens (releases heat, causing sound waves) -> microphone -> and is picked up by the energy meter.When the laser/photon energy is just right for a particular chemical bond, no excess heat is released from bond breakages inside the cell, and no sound waves are produced. The energy meter will read 0.EmΔHuΔHnr0Glancing at our answer choices, answer choice D matches our breakdown exactly. Answer choice B is the difference between the laser pulse energy, answer choice A, and the heat detected, answer choice C. Energy is conserved, as we're using an appropriate laser. We're going to eliminate answer choices A through C.

Which ideal solution exhibits the greatest osmotic pressure?

Jack Westin Advanced Solution: This is a standalone question so what I like to do is give a brief overview of the content being tested. In this case, we're dealing with osmotic pressure. Osmotic pressure is a measure of the tendency of water to move into one solution from another by osmosis. The higher the osmotic pressure of a solution the more water wants to go into the solution. We want to pick the solution with the greatest concentration of solute particles. Why is that? Because of the definition we just covered. There will be greater movement of water in that situation. Quick glance at our answer choices shows multiple molarities, but also compounds that can dissociate into ions. Much like these compounds can be broken down, let's break down each answer choice. A 0.1 M MgCl2 Jack Westin Advanced Solution: At first glance, one might incorrectly glance at this answer choice and see 0.1 M and move on. However, this solution can dissociate into multiple ions. MgCl2 can dissociate into three ions, which also means we have 0.3 M in solute particles. We know this question entails knowing more than just which number in the answer choices is the largest, so the fact that we multiplied this molarity by three ions makes this a good start. Not always the case, but we know AAMC typically has something up their sleeve, so we like answer choice A for now. B 0.2 M NaCl Jack Westin Advanced Solution: Similar to answer choice A, we're not taking this answer choice at face value. We know sodium chloride can dissociate into two ions. What does that do to our molarity? It can be multiplied by 2 to give us 0.4 M. This is now our best answer. C 0.2 M CaCl2 Jack Westin Advanced Solution: Similar methodology here, but we expect calcium chloride to dissociate into three ions. What does that do to molarity? We multiply by 3 to give us 0.6 M. This is our best answer so far. We can eliminate answer choices A and B. D 0.5 M Glucose Jack Westin Advanced Solution: This is tricky because glucose does not ionize into multiple molecules in solution like we expect from answer choices A-C. While this is our second-best answer, we still have a superior answer with answer choice C. Want more MCAT Prep? We have an MCAT Podcast! Solution: The correct answer is C. This is a General Chemistry question that falls under the content category "Assemblies of molecules, cells, and groups of cells within single cellular and multicellular organisms." The answer to this question is C because the solution listed has the greatest concentration of solute particles. CaCl2 dissociates into three ions, making the solution listed 0.6 M in solute particles. It is a Scientific Reasoning and Problem Solving question because you are asked to determine which formula applies to the stated scenario and apply it correctly to make a prediction.

Liposomes derived from Compound 2 are prepared at pH 8.5 from two different solution concentrations (0.10 mM and 0.20 mM) as described in the passage. What is the expected appearance of the size-exclusion chromatograph of the liposomal suspension that results after mixing equal volumes of these? but mixing those from Compound 2 formed new liposomes with an average size expected for the effective final lipid concentration.

Jack Westin Advanced Solution: We can solve this question by using information given in the passage, specifically Figure 1. Compound 2 is able to reverse the liposome forming reaction, and form more energetically stable liposomes after mixing. We'll analyze our figure with this in mind.Mixing equal volumes of the 0.10 mM and 0.2mM will yield a suspension equal to a concentration of 0.15mM.The combined concentration of 0.15mM, has an elution volume of about 16 or 17 mL. So that 16mL is the only place we would see the fluorescence bump.And just for comparison purposes: Individual elution volumes for 0.10 mM and 0.2mM are at about 20 mL and ~12 mL A Jack Westin Advanced Solution: Answer choice A has two bumps at roughly 12 ml and 20 ml. We just said these are individual elution volumes for 0.1 millimolar and 0.2 millimolar concentration. B Jack Westin Advanced Solution: Answer choice B has the same two bumps at 12 mL and 20 mL. We're getting the same bumps as answer choice A, so there's a chance we end up eliminating answer choice along with answer A for saying the same thing. C Jack Westin Advanced Solution: Answer choice C has a single bump at roughly 25 mL. This would be the case at a very low concentration, not the roughly 0.15 millimolar concentration in this question. D Jack Westin Advanced Solution: Answer choice D has a single bump at roughly 16 mL, which is consistent with our breakdown. We can pick our correct answer, answer choice D, and eliminate all our other answer choices. Want more MCAT Prep? We have an MCAT Podcast! Solution: The correct answer is D. This is a Biochemistry question that falls under the content category "Separation and purification methods." The answer to this question is D because mixing the two suspensions will result in the same liposomal suspension that is created when the lipid concentration is 0.15 mM. This is the average of the two concentrations, since the volumes were identical. It is a Scientific Reasoning and Problem Solving problem because you are asked to bring together theory, evidence, and observations to draw a conclusion.

Based on the data in Figure 3, what is the melting temperature Tm of wild-type M1 mRNA in a solution containing 100 mM KCl?

Jack Westin Advanced Solution: We're going to need Figure 3 from the passage, but we're also going to have to be able to understand the data and associate the graph to melting temperature.The question is asking about the solution containing 100 mM potassium chloride, or the most solid line in the figure above. That was the one that has the most unfolding at the lowest temperature.For nucleotides, Temperature of Melting (Tm) is defined as the temperature at which 50% of double-stranded DNA is changed to single-standard DNA. When does that happen? When the fraction unfolded is at 0.5, or 50% on the graph. We can estimate the temperature at that point to be slightly above 50 degrees Celsius, we can call it roughly 53 or 55 degrees.Looking at the answer choices, none of our options are within more than 8 degrees Celsius of each other, so our approximation should be good enough here. The only answer choice that matches our low 50s range is answer choice B. A 45°C B 53°C C 64°C D 80°C Need extra help breaking down AAMC logic? Find out how you can meet with one of our expert MCAT tutors! Solution: The correct answer is B. This is correct. This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is B because the Tm is the temperature at which 50% of the molecules are denatured or the fraction folded is 0.5. This occurs at approximately 53°C in 100 mM KCl. It is a Data-based and Statistical Reasoning question because you must interpret the data from the graph of thermal denaturation to determine melting temperature.

Which of the two products was detected during the experiment? Jack Westin Advanced Solution: What is this question referencing? It's asking about the experimental results specifically, so we can reference our passage.The passage mentions Compound 1 was utilized because it produced two compounds: Compound 2a and Compound 2b. One product is intensely colored, so we can say that compound was clearly detected if the students made this observation. We're looking for the colored compound, but we also want to know the reasoning behind the intense color. What causes the color to appear? Extensive conjugation causes color to appear in organic compounds. In transition metals, it's the d orbital that gives color. Let's look at our two compounds closely.Only B has a conjugated system, meaning Compound B will be the one that's colored, and it's the one that was detected. Having a "conjugated system" for our sake simply means consecutive alternating double bonds. Double bonds indicate there are overlapping pi orbitals and this delocalizes the electrons. Highly delocalized compounds (or highly conjugated compounds) are better at absorbing and emitting light. At its core, really all this question is asking is which product is emitting light, and why. We're looking for an answer like Compound 2b, because of the conjugation. A Compound 2a; it has three aromatic rings. B Compound 2a; it is a diphenol. C Compound 2b; it has extensive delocalization of electrons. D Compound 2b; it is a mixed phenol and quinone.

Jack Westin Advanced Solution: What is this question referencing? It's asking about the experimental results specifically, so we can reference our passage.The passage mentions Compound 1 was utilized because it produced two compounds: Compound 2a and Compound 2b. One product is intensely colored, so we can say that compound was clearly detected if the students made this observation. We're looking for the colored compound, but we also want to know the reasoning behind the intense color. What causes the color to appear? Extensive conjugation causes color to appear in organic compounds. In transition metals, it's the d orbital that gives color. Let's look at our two compounds closely.Only B has a conjugated system, meaning Compound B will be the one that's colored, and it's the one that was detected. Having a "conjugated system" for our sake simply means consecutive alternating double bonds. Double bonds indicate there are overlapping pi orbitals and this delocalizes the electrons. Highly delocalized compounds (or highly conjugated compounds) are better at absorbing and emitting light. At its core, really all this question is asking is which product is emitting light, and why. We're looking for an answer like Compound 2b, because of the conjugation. A Compound 2a; it has three aromatic rings. Jack Westin Advanced Solution: That goes against our prediction of Compound 2b first of all. Secondly, we want the greater number of conjugated double bonds. The fewer aromatic rings works against answer choice A. B Compound 2a; it is a diphenol. Jack Westin Advanced Solution: Does that affect our answer? It sounds like it could be a better choice than A, but still not the compound we predicted. C Compound 2b; it has extensive delocalization of electrons. Jack Westin Advanced Solution: This sounds like our breakdown, just in different words. Double bonds indicate there are overlapping pi orbitals and this delocalizes the electrons. We like answer choice C the best so far. D Compound 2b; it is a mixed phenol and quinone. Jack Westin Advanced Solution: Much like answer choice B, this doesn't exactly answer our question, even if it's technically a true statement. We're sticking with option C as the best answer. How do you know you are truly ready? That's where the Jack Westin MCAT Diagnostic Tool comes in. Solution: The correct answer is C. This is an Organic Chemistry question that falls under the content category "How light and sound interact with matter." The answer to this question is C because Compound 2b has a much larger set of conjugated double bonds than Compound 2a. In the passage it was stated that Compound 1 was chosen as a substrate since it gave rise to an equilibrium mixture of products, one of which was intensely colored. It is a Reasoning about the Design and Execution of Research question because you are asked to reason about the features of a research study.

An ITC experiment is conducted by injecting Compound 1 into a solution of Protein A giving large measured heats and a high affinity constant K1. A CITC experiment under identical conditions but in the presence of 10 mM Compound 2 gives the same heat curve and Kapp = K1. What change to the experimental conditions can result in measurable heat differences and Kapp < K1? Jack Westin Advanced Solution: We have the same heat curve and an apparent affinity constant equal to association constant in two different compounds. How can we change the conditions to have heat differences and have the association constant for Compound 1 be greater than the apparent affinity constant?The affinity constant (KNAG3) essentially tells us the affinity of an enzyme for substrate. A higher affinity constant means more of the enzyme substrate complex is formed from the starting concentrations of enzymes and substrate.Apparent affinity constant considers a second substrate. So in our passage we had enzume+NAG3, and our second substrate is NAG. So we now have a KAPP, not KNAG3. So our KNAG3 doesn't change with the 2nd substrate. Only the Kapp does.So in our graph, while we're increasing [NAG], KNAG3/Kapp is also increasing. But we said KNAG3 isn't changing. That means KAPP is decreasing.Adding more NAG will decrease KAPP to compensate and keep the slope the same. KNAG3 doesn't change. A Dilute all of the solutions. Jack Westin Advanced Solution: The net effect here would cancel out. B Increase the concentration of Compound 1 in both titrations. Jack Westin Advanced Solution: We're trying to decrease the apparent affinity constant between Protein A and Compound 1 here. C Increase the concentration of Protein A in both titrations. Jack Westin Advanced Solution: Same basic reasoning as answer choice B. We're ultimately trying to decrease apparent affinity between Protein A and Compound 1. D Increase the concentration of Compound 2 in the second titration. Jack Westin Advanced Solution: This is what we mentioned we're trying to do in our breakdown of the question. We increase concentration 2, our NAG, and this ultimately decreases Kapp. We're left with our correct answer, answer choice D: Increase the concentration of Compound 2 in the second titration. Want more MCAT Prep? We have an MCAT Podcast! Solution: The correct answer is D. This is a General Chemistry question that falls under the content category "Unique nature of water and its solutions." The answer to this question is D since the experiment depends on Compound 2 successfully competing for the binding sites available on Protein A and this is the only response that will improve the chances of this happening. It is a Reasoning about the Design and Execution of Research question because you are asked to reason about the features of a research study that suggests relationships between the variables.

Maybe it'll help to understand the difference between the different constants.K1=affinity constant higher K1 or Ka means higher affinity, hence the name.Kapp=affinity constant=relating to inhibition or competition for the enzyme.Kd=Km=dissociation rate constant (notice how this is the inverse) lower Kd or Km means higher affinityIn the Q-stem, it first introduces comp1 & protA, their affinity is K1. They stated that its pretty high.Then they introduce comp2, and from the experimental name, we know that it's a competitive inhibitor, this is where Kapp comes into play. (or you can think of this as K2)comp1-protA complex measures K1comp1-protA+comp2 measures K2 (we assumed that addition of an inhibitor, should lower K2/app, because this K is still the measurement of comp1-protA)Comp2 is an inhibitor for comp1, meaning it binds at the active site of protA to inhibit binding of comp1 to protA.The q-stem then says that upon addition of 10mM of comp2, the Kapp=K1. This means the addition of the comp2 didn't really do anything.The last sentence then asks what can we adjust to make Kapp<K1. Meaning how can we make Kapp smaller. Just think that you can't change K1 anymore. To decrease Kapp=to decrease the affinity between comp1-protA w/ inhibitor, we increase comp2.

Most cytochrome P450 enzymes alter the activity of drugs by: A. phosphorylating them. B. dephosphorylating them. C. oxidizing them. D. reducing them.

Most cytochrome P450 enzymes alter the activity of drugs by: Jack Westin Advanced Solution: This is a passage-related question that functions almost like a pseudo-discrete. The author introduces cytochrome P450 enzymes in Paragraph 3 and we get some background information:Cytochrome p450 enzymes appear a few times in AAMC's practice material but it's not explicitly listed on the content outline. They are membrane-bound proteins that enable oxidation reactions. We want an answer choice that deals with redox reactions. When the body detects a new drug/toxin or increased concentrations of a drug, the body increases the concentration of cytochrome P450 to metabolize the drug effectively. Cytochrome P450 alters the activity of drugs by acting as an oxidizing agent to allow drugs and toxins to be metabolized. A. phosphorylating them. Jack Westin Advanced Solution: phosphorylating them. We mentioned in our breakdown that we're looking for an answer choice that deals with redox reactions. Typically, kinases will be the ones phosphorylating proteins in the body. B. dephosphorylating them. Jack Westin Advanced Solution: dephosphorylating them. We mentioned in our breakdown that we're looking for an answer choice that deals with redox reactions. This answer choice is describing phosphatases which function to dephosphorylate. C. oxidizing them. Jack Westin Advanced Solution: oxidizing them. This answer choice matches our breakdown of the question and the cytochrome P450 enzyme background information we went over. Cytochrome P450 enzymes function as monooxygenases. When the body detects a new drug/toxin or increased concentrations of a drug, the body increases the concentration of cytochrome P450 to metabolize the drug effectively. Cytochrome P450 alters the activity of drugs by acting as an oxidizing agent to allow drugs and toxins to be metabolized. This is our best answer choice. D. reducing them. Jack Westin Advanced Solution: reducing them. While cytochrome P450 itself is reduced, the enzymes function to alter the activity of drugs by oxidizing them. This is going to be the opposite function of what we're looking for. Answer choice C is our best answer choice. Need extra help breaking down AAMC logic? Find out how you can meet with one of our expert MCAT tutors! The solution is C. Cytochromes are proteins that are involved in redox reactions. Specifically, cytochrome P450 acts as a monooxygenase, catalyzing an oxidation reaction. Phosphate groups are not added to the drug during this process. Cytochromes are proteins that are involved in redox reactions. Specifically, cytochrome P450 acts as a monooxygenase, catalyzing an oxidation reaction. Phosphate groups are not removed from the drug during this process. Cytochrome P450 acts as monooxygenases, where an oxygen atom is inserted into a substrate (the drug of interest), thereby resulting in the oxidation of the substrate. Cytochrome P450 acts as a monooxygenase, catalyzing an oxidation reaction. In this process, oxygen molecules are reduced, but the substrate molecules (the drug of interest) are oxidized.

increase muscle tone mean

Muscle tone is the amount of tension (or resistance to movement) in muscles. Our muscle tone helps us to hold our bodies upright when we are sitting and standing. Changes in muscle tone are what enable us to move. Muscle tone also contributes to the control, speed and amount of movement we can achieve.

trimethyllysine

Okay, so you need to know that the amino acids in the study are the ones listed in the legend of figure 1. You'll see that all the amino acids are lysines and an ornithine (Just lysine with 1 less carbon in the R group). Lysines are characterized as basic at physiological pH (7.4) because hey will have a positive charge after taking up an additional proton. The question says to ignore trimethyllysine since it cannot take up a proton since the N has four bonds to carbon already, so it's technically not basic. If you could recognize that the amino acids in question are basic (+ charge) it should follow that they're also hydrophilic. What was your logic on this question, just to see where it disconnected?

On the day of the experiment, the subjects drank about 1 L of water on average and excreted about 400 mL of urine. The most likely explanation for the difference between water intake and urine excretion is that: Jack Westin Advanced Solution: This question is tangentially-related to the passage, but we should be able to answer this question using our general knowledge of osmoregulation. Osmoregulation is the active regulation of the osmotic pressure of bodily fluids to maintain the homeostasis of the body's water content. Although the kidneys are the major osmoregulatory organ, the skin and lungs also play a role in the process. That means that while a large amount of water is excreted through urine, we also will excrete water through the skin and lungs. A.the extra water was stored as blood.AAMC Standard Solution:The extra water cannot be stored as blood because this would increase the blood volume. Jack Westin Advanced Solution: the extra water was stored as blood. This is not what we expect from the water not excreted as urine. We can actually see in Figure 1, that blood volume does not fluctuate much with water consumption. Extra water being stored as blood would mean blood volume increases and blood gets diluted. B.water was consumed with the food that was eaten.AAMC Standard Solution:If the water was consumed with the food that was eaten, then it would be absorbed and accumulate in circulation, increasing the volume of blood. Jack Westin Advanced Solution: water was consumed with the food that was eaten. This answer choice has more to do with the timing of the water intake. We still do not have a significant change in blood volume or osmolarity according to our graphs. Instead, we expect water was excreted through the skin and lungs like we mentioned in our breakdown of the question. C.the extra water was excreted by the intestine.AAMC Standard Solution:Extra water could be excreted through the intestine if the subjects were having diarrhea. As the passage does not indicate this condition, this answer is not likely. Jack Westin Advanced Solution: the extra water was excreted by the intestine. Typically, water is absorbed in the intestines and not excreted by the intestine. Excretion by the intestine would mean diarrhea or watery stool, which the author did not mention. D.water was excreted via the skin and the lungs.The correct answer is

On the day of the experiment, the subjects drank about 1 L of water on average and excreted about 400 mL of urine. The most likely explanation for the difference between water intake and urine excretion is that: Jack Westin Advanced Solution: This question is tangentially-related to the passage, but we should be able to answer this question using our general knowledge of osmoregulation. Osmoregulation is the active regulation of the osmotic pressure of bodily fluids to maintain the homeostasis of the body's water content. Although the kidneys are the major osmoregulatory organ, the skin and lungs also play a role in the process. That means that while a large amount of water is excreted through urine, we also will excrete water through the skin and lungs. A.the extra water was stored as blood.AAMC Standard Solution:The extra water cannot be stored as blood because this would increase the blood volume. Jack Westin Advanced Solution: the extra water was stored as blood. This is not what we expect from the water not excreted as urine. We can actually see in Figure 1, that blood volume does not fluctuate much with water consumption. Extra water being stored as blood would mean blood volume increases and blood gets diluted. B.water was consumed with the food that was eaten.AAMC Standard Solution:If the water was consumed with the food that was eaten, then it would be absorbed and accumulate in circulation, increasing the volume of blood. Jack Westin Advanced Solution: water was consumed with the food that was eaten. This answer choice has more to do with the timing of the water intake. We still do not have a significant change in blood volume or osmolarity according to our graphs. Instead, we expect water was excreted through the skin and lungs like we mentioned in our breakdown of the question. C.the extra water was excreted by the intestine.AAMC Standard Solution:Extra water could be excreted through the intestine if the subjects were having diarrhea. As the passage does not indicate this condition, this answer is not likely. Jack Westin Advanced Solution: the extra water was excreted by the intestine. Typically, water is absorbed in the intestines and not excreted by the intestine. Excretion by the intestine would mean diarrhea or watery stool, which the author did not mention. D.water was excreted via the skin and the lungs.The correct answer is D.AAMC Standard Solution:Extra water is normally excreted through skin and lungs. The skin excretes water through the process of transpiration, and the lungs use water to humidify the air that enters the body. Jack Westin Advanced Solution: water was excreted via the skin and the lungs. This answer choice is consistent with osmoregulation. Although the kidneys are the major osmoregulatory organ, the skin and lungs also play a role in the process. We expect the difference between water intake and urine excretion is due to water being excreted via the skin and lungs.

If no braking occurs, a total of how much power would be required to keep the railcar moving at 40 m/s?

P=Force*velocity=Work/timeIn the absence of braking, the only force opposing motion is the one mentioned in the last sentence; there is a continuous decelerating force that is present any time the railcar is moving. This decelerating force is 1,000N.P=Force*velocity= 1,000N * 40m/s = 40,000 W = 40kW

Oxidizing agents examples

Peroxides, ozone, and peracetic acid O2, H2O2, CrO4-, Cr2O72-, NAD+

Critical point on phase diagram

Point on the boiling/condensing line that marks where a substance will only exist as a gas if it continues

Ca influx happens in the pre synaptic axon which triggers NT release like you said but the actual action potential begins when the voltage gated sodium channels open allowing depolarization to occur. It's a little technical difference but that's AAMC for you While you are right the Calcium influx causes neurotransmitter release, these neurotransmitters themselves do not individually causes APs. Between having neurotransmitters than are excitatory and inhibitory while also having concepts such as temporal and spatial summation, it would be too broad to say the Calcium influx always directly leads to an action potential. What does directly lead to an action potential is the summation of all NTs from the presynaptic neurons at the axon hillock.

Question 58 Of the events listed, which occurs first during action potential generation? Jack Westin Advanced Solution: This is a standalone question that relies on our knowledge of action potential generation. I always recommend you review action potentials any chance you get as you review:This is a comparison question so we can go through each of our four answer choices one-by-one. We're looking for the event that occurs first. A.Voltage-gated sodium channels open at the axon hillock.The correct answer is A.AAMC Standard Solution:When threshold is met at the axon hillock, voltage-gated sodium channels open, generating an action potential. Jack Westin Advanced Solution: Voltage-gated sodium channels open at the axon hillock. If the threshold of excitation is reached at the axon hillock, voltage-gated Na+ channels open, and the membrane depolarizes. We can see this in our visual above, but we can't comment just yet on if this event happens first until we go through our additional answer choices. B.Hyperpolarization stimulates the opening of ligand-gated potassium channels.AAMC Standard Solution:Voltage-gated (not ligand-gated) potassium channels open in response to depolarization, not hyperpolarization. Jack Westin Advanced Solution: Hyperpolarization stimulates the opening of ligand-gated potassium channels. At the peak action potential, K+ channels open and K+ begins to leave the cell. This is in response to depolarization. The membrane becomes hyperpolarized as K+ ions continue to leave the cell. The hyperpolarized membrane is in a refractory period and cannot fire. Potassium channels are voltage-gated. This answer choice is factually incorrect, so we can stick with answer choice A as our best option for now. C.Graded potentials propagate along the axon.AAMC Standard Solution:Graded potentials occur in the cell body and dendrites, not the axon. Jack Westin Advanced Solution: Graded potentials propagate along the axon. We don't want to get tricked here as location is key. Answer choice A mentioned voltage-gated sodium channels opening at the axon hillock. This choice mentions propagating along the axon which cannot happen until answer choice A happens. D.Calcium influx stimulates vesicle fusion and release of neurotransmitter.AAMC Standard Solution:Calcium influx and release of neurotransmitter does not occur until action potentials arrive at axon terminals. Jack Westin Advanced Solution: Calcium influx stimulates vesicle fusion and release of neurotransmitter. Once the threshold potential is reached, the neuron completely depolarizes. Depolarization causes the opening of voltage-gated calcium channels, and the influx of calcium into the cell triggers neurotransmitter release. This happens after action potentials arrive at axon terminals or, in other words, later than answer choice A. Answer choice A is our best answer.

intercalated discs of cardiac muscle

Represent specialized cell junctions between cardiac muscle cells. Interdigitating folds of sarcolemma at the end of adjacent cells, linking them structurally and functionally

vacuum distillation vs fractional vs simple

Simple = below 150 degrees and if the substances you're trying to distill are at least 25 degrees or greater apart in terms of boiling point Fractional = used for less than 25 degrees apart, better for substances that are close in bp Vacuum= when the boiling point of a substance is greater than 150, what you can do is actually lower the atmospheric pressure, and when atmospheric pressure = vapor pressure = bp, so this is the only one where you can actually change the bp of the substance

With which of the above metals can copper form a galvanic cell in which copper is reduced? _

Simplest way to look at it: What type of cell are we looking to create, and what are we trying to reduce? Since we are looking for a galvanic cell trying to reduce copper, we need to end up with a higher reduction potential than the given reduction potential for copper; this is because galvanic cells are spontaneous. (in contrast, electrolytic cells require external force or energy, which is usually voltage) The reduction potential of a cell equation is: Ecell = Ecathode - Eanode. Since Copper is getting reduced in this case, it will be at the cathode (Red cat). So, since we know that a positive number (copper) minus a negative number (lead and/or zinc) will always result in a positive number (Ecell) that is higher than what we started with, the answer can only be C and D. The other user gave a really good explanation but I'll just add on another thing. Generally, the more positive the E for each half-cell, the more likely it wants to be reduced. We can see in this reaction that since Cu's E is more positive than Pb and Zn, Cu is more likely to be reduced in a reaction with those two molecules. If combining reduction potential and they all are reducing don't change sign compare as it is If combining reduction and oxidation and using formula then change sign so we have one oxidation and one reduction

Which procedure would be the best negative control for endogenous GABA-receptor function in frog oocytes? A.Addition of GABA to mock-transfected frog oocytesThe correct answer is B.Transfection of wild-type GABA receptor into insect oocytes C.Addition of excess GABA to Rdl-transfected oocytes D.Transfection of wild-type GABA receptor into Rdl-expressing oocytes When frog oocytes transfected with cDNA encoding wild-type or Rdl forms of this gene for a GABA-receptor subunit are exposed to GABA with or without dieldrin, the changes in cell membrane potential shown in Figure 1 are seen.

So in this experiment, they're transfecting oocytes to express either a WT or Rdl GABA receptor. The problem is, these cells might also be expressing their own (endogenous) GABA receptors, and we don't know what effect that's having on the results. Thus, a good control would be to see what effect GABA has on a mock-transfected oocyte, so you can see what how exactly the endogenous receptors contribute to the results. Answer choice C talks about adding more GABA to the Rdl GABA receptor-transfected oocytes. This isn't great since the level of your independent variable is now different across the two groups, and you wouldn't be able to directly compare the activity of the WT to the Rdl receptor. Which procedure would be the best negative control for endogenous GABA-receptor function in frog oocytes? Jack Westin Advanced Solution: While this is another passage-based question, the key to answering this question is understanding the role of a negative control. A negative control is included in an experiment, but is not expected to change because of the experiment itself or influence the results of the experiment. It functions like a baseline and is not subject to the treatment of the experiment. By using a negative control, researchers can look for confounding variables and get more clarity about positive results. For example, if we're experimenting with a treatment for patients with a specific illness, the negative control would be used to see what happens if no treatment is administered. Once we have that baseline, we can note any changes that occur because of the positive control, or any independent variables. A.Addition of GABA to mock-transfected frog oocytesThe correct answer is A.AAMC Standard Solution:Transfection of GABA to mock-transfected frog oocytes would be the best negative control for endogenous GABA-receptor function. This procedure would ensure that the process of transfection per se does not generate a response to GABA. Jack Westin Advanced Solution: Addition of GABA to mock-transfected frog oocytes. A negative control functions to not generate a response, but rather give us a baseline. This method simply shows that transfection process itself does not affect any results and we have a good sense of the baseline if we have the addition of GABA. This is a strong answer choice. B.Transfection of wild-type GABA receptor into insect oocytesAAMC Standard Solution:The insect oocytes already have the GABA receptor, so transfection of more GABA receptors would not generate a negative control. On the contrary, it might increase the sensitivity to GABA. Jack Westin Advanced Solution: Transfection of wild-type GABA receptor into insect oocytes. This also does not help us as much because we want to focus on frog oocytes, not adding additional receptors into insect oocytes. A negative control provides us with a baseline of sorts so transfection into insect oocytes doesn't help us here. C.Addition of excess GABA to Rdl-transfected oocytesAAMC Standard Solution:Addition of excess GABA to Rdl-transfected oocytes would not be a good negative control on the function of GABA-receptors because Rdl is not responsive to GABA. In this case, no reaction is possible. Jack Westin Advanced Solution: Addition of excess GABA to Rdl-transfected oocytes. The author tells us in the passage Rdl is the mutant form of the GABA receptor gene. This answer choice involves adding GABA to these Rdl-transfected oocytes but Rdl is not responsibe to GABA. Answer choice A remains superior. D.Transfection of wild-type GABA receptor into Rdl-expressing oocytesAAMC Standard Solution:Transfection of wild-type GABA receptor into Rdl-expressing oocytes would restore sensitivity to GABA. This method is not a negative control but a rescue experiment. Jack Westin Advanced Solution: Transfection of wild-type GABA receptor into Rdl-expressing oocytes. This would essentially introduce the wild-type GABA receptor into an organism that has a mutant form of the GABA receptor gene. Once again, this is not a negative control. Answer choice A is our best option.

Based on the passage, which subunit of PTx is LEAST suitable for generation of vaccine against B. pertussis? A. S1 B. S2 C. S4 D. S5

Sorry, this is a little late, but one can infer that S1 is toxic because the passage says that it inhibits the heterotrimeric Gi protein. The toxin is composed of an A (S1) and B (S2-4) components. Since the B component is only involved in docking, not in any sort of inhibition, it follows that the A component is responsible for the toxicity by inhibiting Gi, which "paralyzes the cilia and interferes with the clearing of pulmonary secretions." So in essence, since it's the only part of the toxic protein (PTx) interfering with cell processes, we can assume that it is the only toxic subunit. I know it's a little late for this, but hopefully it helps someone else! It's the idea that you put the antigen inside the patient and allow them to build up natural immunity. That being said, you want to include the section of the protein that will be LEAST likely to cause harm/sickness in the patient when being injected. Therefore, all subunits EXCEPT the subunit with toxicity (1) will be best to avoid this. Based on the passage, which subunit of PTx is LEAST suitable for generation of vaccine against B. pertussis? Jack Westin Advanced Solution: This is a passage-based question and the test-maker explicitly asks "based on the passage." A few things we want to notice right away is we're looking for a subunit that is LEAST suitable for generation of a vaccine. Three of our answer choices will be suitable, while our correct answer is not suitable for generation of a vaccine. Additionally, we grouped the subunits into an A component and B component. Three of our answer choices are associated with the B component, and only one answer choice is associated with the A component. While this isn't an automatic reason to pick answer choice A, we do want to note that is the one answer choice that is not part of the B component. What we want to note is which of the subunits would produce an immune response (without an immune response we don't have an effective vaccine), but also be safe for use in humans. We're told the S1 subunit is actually inserted into the cytoplasm, while the B component simply binds to cell surfaces. The ADP ribosyl transferase (S1 subunit) is toxic, while the B component (contains S2, S3, S4, and S5 subunits) subunits would be safe for use. This confirms our initial instincts that we'll end up picking the option that is not grouped in with the others. We can pick the only answer choice that is not part of the B component, answer choice A: S1. A. S1 B. S2 C. S4 D. S5 Need extra help breaking down AAMC logic? Find out how you can meet with one of our expert MCAT tutors! The solution is A. When choosing an antigen for vaccine production, there are two aspects to consider: immunogenicity and toxicity. S1 has the ADP ribosyl transferase activity responsible for toxicity of PTx and therefore is least suitable for vaccine production. The S2 subunit is part of the B component of PTx that binds to the cell surface and has no toxic activity. The S4 subunit is part of the B component of PTx that binds to the cell surface and has no toxic activity. The S5 subunit is part of the B component of PTx that binds to cell surface and has no toxic activity.

specific gravity of water

Specific gravity of water 1g/cm^3 , 1000 kg/m^3 for values of SG less than or equal to 1 SG= displaced fluid volume/object volume

Squalene

Squalene is a precursor of the fused four-ring system common to cholesterol and all of the steroid hormones.

isomers

Starting from the basics: STRUCTURAL isomers have the same molecular formula, but a different structure. (Think - C4H8O could be an alcohol, an aldehyde, a ketone, whatever else) STEREOISOMERS, have the same atomic connectivity (so basically it's pretty much the same molecule), but a different arrangement in SPACE. A subset of stereoisomers are CONFORMATIONAL isomers - these differ by rotation around a single bond. This is where Newman Projections come in - staggered, eclipsed etc. Another subset of stereoisomers are CONFIGURATIONAL isomers -unlike conformational isomers, you actually have to make/break bonds to interchange between isomers. If the isomerism in configurational isomers is around a double bond, then these are GEOMETRIC isomers - these are cis/trans isomers. If however, they are non-superimposable mirror images of one another, then they are OPTICAL isomers (think handedness; when you look at your hands they're mirror images of one another, but if you place them on top of each other, then they're not superimposable). This is a type of configurational isomer (see the flow chart). Okay, this is where enantiomers and diastereoisomers come in. Enantiomers are non-superimposable mirror images of one another, which have opposite stereochemistry at EVERY chiral center. If you look at the flow chart, you can see that enantiomers are MIRROR images - they have the same properties as one another except rotation of plane polarised light; if one enantiomer is (+) or 'd', then the other is (-) or 'l' If you have equal concentrations of two enantiomers, they are not optically active (just think of the rotations being cancelled out) Diastereoisomers are NON-MIRROR images. They differ at some but NOT ALL chiral centers. Again, look at the flow chart. Meso compounds are molecules with chiral centers (meaning it is an asymmetric carbon compound, attached to 4 different groups), but they have an internal plane of symmetry. Therefore, they do not have optical activity. Google meso-tartaric acid for an example. Flow chart: http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/isomers.gif

Blue litmus paper

Stays blue for base, turns red for acids

Red litmus paper

Stays red for acids, turns blue for bases

Bascity vs. Nucleopohilicity

Steric Hindrance favors bascitity over Nucleophilicity - Nucleophiles must have VERY little hindrance, -primary nucleophiles most common -secondary atoms can often as as bases or nucleophiles -tertirary atoms will only act as bases Reactivity (low stability) favors bascicity over nucleophilicity - if an atom has a full negative charge it will almost always act as a base (halides are notable exceptions)

Cofactors, Coenzymes and Prosthetic Groups

Substances that bind to enzyme active sites to provide greater catalytic diversity -Chemically modified during catalysis -Cofactors e.g. metal ions bind reversible to apoenzymes to form holoenzymes -Prosthetic groups are extremely tightly bound e.g. heme -Coenzymes- formed from vitamins and often contain adenine for anchoring, some bound and released as cosubstrates e.g. NAD+, FAD Yes cofactors are typically metal ions that assist function of enzymes, they usually cannot function without them. Think Mg++ in Phosphofructokinase and Phosphoglycerate Kinase in Glycolysis. Think more of a stabilizing role than actually helping at the active site. Coenzymes are organic based molecules that bind to enzymes and aid in catalysis of the substrate. Usually they combine with an apoenzyme (inactive) to form the full holoenzyme. They tend to be loosely bound and have a higher Kd (easier to dissociate) Cosubstrates are just additional substrates of enzymes. They can interact with the enzyme on their own (as secondary substates of sorts) or in tandem with the primary substrate Prosthetic groups have a low Kd and a high Ka, good affinity for the enzyme/ protein. They can be organic or non organic and just facilitate whatever function is required. Prosthetic groups are permanent groups and attachments to the enzyme - i.e. the enzyme cannot function without them. These are typically metal ions that are core to the function. Cofactors are defined as being non-protein molecules that are needed by an enzyme to perform its function as a catalyst. Think of cofactors as being an umbrella term to describe ALL "helper molecules" that assist enzymes. An inactive enzyme without the cofactor is called an apoenzyme, while the complete enzyme with cofactor is called a holoenzyme. Cofactors can be divided into two categories: metal ions and coenzymes. Metal ions (inorganic) help enzymes by assisting in catalyzing oxidation/reduction reactions. Coenzymes are organic molecules that assist enzymes. Coenzymes can be further divided into two categories: prosthetic groups and cosubstrates. A prosthetic group is a coenzyme that is tightly attached to the enzyme (eg. the heme in hemoglobin). A cosubstrate is a coenzyme that is bound and released after a reaction (eg. NADH/NAD+). COFACTORS are non-protein helper molecules, and they can be organic or inorganic. If it is organic, then we can call it a COENZYME. If a coenzyme is tightly bound or covalently bound, then it can be called a PROSTHETIC GROUP. So think of it as just sub-levels, getting more and more specific. Cofactor is just the most generic, all-encompassing term. Examples of inorganic cofactors: Mg2+ ions in DNA polymerase help stabilise the negative charges on the phosphates Examples of organic coenzymes: NADH/NADPH, which serve as electron carriers and are derived from Vitamin B3. Although these are organic just like prosthetic groups they are free to move around and shuttle electrons. Also Acetyl-coA, whose full name is "Acetyl coenzyme A" and it carries around an acetyl group. Examples of prosthetic groups: FAD/FADH2 which is covalently attached to Complex 2 of the ETC and helps the enzyme succinate dehydrogenase take electrons from succinate. Succinate dehydrogenase is considered a "flavoprotein" because it is bound to FAD/FADH2. Prosthetic groups are just coenzymes that are bound in place. Hope that helps! I tried to find the most common examples of enzymes they're used with to help it "click". EDIT: my KA notes say that for the most part vitamins become organic cofactors/coenzymes. Minerals become inorganic cofactors only. Kaplan differentiates it by saying prosthetic groups are covalently bonded non protein molecules that are part of an enzyme's active site. Cofactors may be organic or inorganic, and aid in the enzyme's function without directly interacting with the enzyme

Increasing the enzyme concentation would increase the rate of production of H2O2 (thereby increasing absorbance), but wouldn't change the reading of glucose concentration to make it more precise, it would just increase it above the range further (choice A is out) Increasing the oxygen pressure would also shift the reaction to the right, producing more H2O2, and thereby increasing the absorbance and glucose concentration reading further above the range (choice B is out) Decreasing the content of the oxygen acceptor (which is glucose), this would yield an excess of unreacted H2O2, oxidized chromogen wouldn't be produced, so it's unlikely we'd have a reading at all (choice C is out) Dilute the sample with additional solvent, this would increase the ratio of oxygen acceptor solvent to glucose oxidase, decreasing the fraction of glucose oxidase and thereby decreasing the amount of H2O2 present that needs to react with an oxygen acceptor (glucose) to yield an absorbance rating 7 ReplyShare level 2Flutegal10·1 yr. agoTesting 1/15/22 I'm finishing up my FL4 review today and noticed your post - and your testing date! Gl with your MCAT tomorrow and thanks for the explanation! 4 ReplyShare Continue this thread level 2Rare_Thanks_9699·1 yr. ago this makes sense ... thank you! 2 ReplyShare level 1samyeeta·3 yr. ago Learned this in biochem lab in regards to concentration vs absorbance, but the idea is that you can't extrapolate data when your unknown concentration gives u an Absorbance value that is past the range of your standard curve (which relates known values of concentration and absorbance), so to make it a value within the standard curve you can dilute the concentration of the unknown and get an accurate data point for the absorbance on the standard curve to figure out the concentration of the dilution and then just multiply that value by how much you diluted it

Suppose a blood sample tested above the range (6.0 mg/dL) of the standards used in the experiment. What modification will provide a more precise reading by data interpolation as opposed to extrapolation using the same standards? Jack Westin Advanced Solution: This is a fairly common question type when it comes to experimental passages. We're asked about modifications to the research we read about in the passage. By doing so, the test-maker ensures that we have grasped the big picture of the experiment and passage. What do I mean by that? If we don't understand the initial experiment taking place in the passage, it's very unlikely that we will be able to explain the effects of different modifications.Both interpolation and extrapolation involve predicting values related to a data set. Interpolation is used to predict values between known data points, while extrapolation is used to predict values outside the range of data. Note the blood sample in the question stem tested above the range of the standards used in the experiment. We want to get that glucose concentration within the range of the standards used in the experiment so the researchers can use data interpolation as opposed to extrapolation. A.Increase the enzyme concentration.AAMC Standard Solution:Increasing the enzyme concentration will increase the quantity of absorber (chromogen) in solution, according to the protocol described in the passage. This will make the resulting absorbance value even further from the values in Table 1 and require additional extrapolation. Jack Westin Advanced Solution: Increase the enzyme concentration. This is the opposite of what we're looking for. We can recall from Paragraphs 1 & 2, that by adding more glucose to Reaction 1, we know we have additional chromogen in solution and optical absorbance is affected by this chromogen. This exacerbates our problem in this situation. We already have a blood sample that tested above the range of the standards used in the experiment, and answer choice A would increase this value. B.Increase the oxygen pressure.AAMC Standard Solution:Increasing the oxygen pressure will increase the quantity of absorber (chromogen) in solution, according to the protocol described in the passage. This will make the resulting absorbance value even further from the values in Table 1 and require additional extrapolation. Jack Westin Advanced Solution: Increase the oxygen pressure. This answer choice is going to be similar to answer choice A. We can look at Reaction 1 and consider the effects of increased oxygen pressure. The reaction will shift to the right so more reactants are produced. Once again, we know we have additional chromogen in solution and optical absorbance is affected by this chromogen. This once again exacerbates our problem in this situation. C.Decrease the content of oxygen acceptor.AAMC Standard Solution:The oxygen acceptor is glucose. Removing glucose from the solution is not feasible and defies the logic of the experiment, which involves quantifying the amount of glucose present. One would need to know exactly how much glucose was removed, and this would require a separate measurement. Jack Westin Advanced Solution: Decrease the content of oxygen acceptor. This answer choice is unreasonable. Removing glucose, the oxygen acceptor, would make it so we don't get any valuable information from the research. Removing glucose makes quantifying the amount of glucose present essentially pointless. D.Dilute the sample with additional solvent.The correct answer is D.AAMC Standard Solution:By adding solvent, the concentration of glucose will be lowered, and the resulting absorbance will fall within the range of the standards. This is easily accomplished, and the resulting calculations that account for the dilution are not difficult. Jack Westin Advanced Solution: Dilute the sample with additional solvent. This answer choice is consistent with our breakdown of the question. We said the blood sample in the question stem tested above the range of the standards used in the experiment. We want to get that glucose concentration within the range of the standards used in the experiment so the researchers can use data interpolation as opposed to extrapolation. How can we get that glucose concentration within that range? By diluting the sample with additional solvent and decreasing the concentration of glucose. The absorbance will then fall within the range of the standards and we can use data interpolation. Answer choice D is our best answer choice.

This is very late but I think I got it. First of all they give us the mass in grams when calculations for force are supposed to be done in kilograms. 20g x (kg/1000g)= 0.02 kg. which is mass in kg. average acceleration is final velocity - initial velocity divided by time: (Vf-Vi)/time. Here that is (35cm/s-25cm/s)=10 cm/s for velocity, but remember that velocity is in cm when it is supposed to be METERS!!!!!! so yet again we have to do another dimensional analysis calculation. 10cm x (1meters/100cm) = 10/100=0.1 m/s= velocity they gave you time which is 0.1 seconds acceleration is velocity/time so plug in (0.1 m/s)/(0.1 sec)= 1 m/s^2 Force=mass*acceleration Force=(0.02kg)(1m/s^2)=0.02 Newtons The explanation they gave honest to god made no sense to me, but doing it this way the math works out, which is fine I guess. But there were several steps here which is kinda unfortunate to do in a time constraint. If Anyone has any quicker way to solve this problem pls, PLS reply

Suppose a defibrillator successfully returns a baby's heart to normal beating. Suppose further that 20 g of blood enters the heart at 25 cm/s and leaves 0.10 s later at 35 cm/s. What is the estimated average force on the 20 g of blood as it moves through the baby's heart? Jack Westin Advanced Solution: This is going to be similar to Question 54 in that is a pseudo-standalone question. Force is equal to mass x acceleration. We already know mass of the blood is 20 grams or 0.02 kg.This 0.02 kg of blood enters the heart at 25 cm/s and leaves at 35 cm/s. That is a difference in 10 m/s in 0.1 seconds. The change in velocity over 0.1 seconds allows us to solve for acceleration:(10 m/s) / 0.1 seconds = 1 m/s2Multiplying force x acceleration gives us:0.02 kg x 1 m/s2 = 0.02 N.This was a math problem where we did minimal rounding or approximating. We solved for an exact force which is equal to answer choice A: 0.02 N. A.0.020 NThe correct answer is A.AAMC Standard Solution:According to Newton's second law, the average force is equal to the mass of blood multiplied by the average acceleration of the blood. The average acceleration is (35 cm/s - 25 cm/s)/0.10 s = 100 cm/s2 = 1 m/s2. The average force is 20 g × 1 m/s2 = 0.020 kg × 1 m/s2 = 0.020 N. B.0.20 NAAMC Standard Solution:Either the mass is incorrectly used as 0.20 kg, or the average acceleration is incorrectly computed as 10 m/s2. C.20 NAAMC Standard Solution:Either the mass is incorrectly used as 2.0 kg, or the average acceleration is incorrectly computed as 100 m/s2. D.2000 NAAMC Standard Solution:Either the mass is incorrectly used as 20 kg, or the average acceleration is incorrectly computed as 1000 m/s2.

Suppose that the railcar passes by a horn that is emitting a sound with frequency f. Which of the following describes the frequency f that the person on the railcar hears? A. f > f before passing the horn, f < f after passing it B. f < f before passing the horn, f > f after passing it C. f = f before passing the horn, f = f after passing it D. f > f before passing the horn, f > f after passing it

Suppose that the railcar passes by a horn that is emitting a sound with frequency f. Which of the following describes the frequency f that the person on the railcar hears? Jack Westin Advanced Solution: To answer this question, we'll need to understand the Doppler effect and the effect of relative motion on perceived frequency. Most of us can conceptualize this by imagining an ambulance as the source of the sound/horn. A. f > f before passing the horn, f < f after passing it Jack Westin Advanced Solution: As you get closer to the horn/the ambulance, the frequency you hear is higher than the actual frequency, and once you're moving away from the horn/the ambulance, the frequency you hear is lower than the true frequency. This answer choice corresponds to what we actually experience and is therefore the correct answer B. f < f before passing the horn, f > f after passing it Jack Westin Advanced Solution: This is the opposite of what we see above and is incorrect. As we approach the stationary source of the sound, perceived frequency increases and is greater than the actual frequency. As we move away from the stationary source of the sound, perceived frequency drops below the true frequency. C. f = f before passing the horn, f = f after passing it Jack Westin Advanced Solution: The perceived frequency is equal to the actual frequency when there is no relative motion. D. f > f before passing the horn, f > f after passing it Jack Westin Advanced Solution: While f 'should be greater than f before passing the horn, it should not remain greater after passing the horn. Answer choice A is the most appropriate application of the Doppler effect.

The complement to the WT sequence does NOT contain which nucleoside? A. Deoxyadenosine B. Deoxycytidine C. Deoxyguanosine D. Deoxythymidine .

The complement to the WT sequence does NOT contain which nucleoside? Jack Westin Advanced Solution: Let's take a quick peek at the wild type sequence: 5′-TTCCCTACCCTCCCCACCCTAA-3′. Well, that sequence doesn't have any "G"s meaning it does not contain deoxyguanosine. According to the base pairing rules, if G is not present on the original strand, its C complement should not be present on the complement sequence. No G to start, no C to pair. A. Deoxyadenosine Jack Westin Advanced Solution: A. Deoxythymidine is present in the wild type sequence so deoxyadenosine should be present in the complement. B. Deoxycytidine Jack Westin Advanced Solution: B. Deoxyguanosine is not present in the wild type sequence so deoxycytidine should not be present in the complement sequence making this answer choice correct. C. Deoxyguanosine Jack Westin Advanced Solution: C. Deoxyguanosine is not present in the wild type sequence but should be present in the complementary sequence because its complementary base pair deoxycytidine is present in the WT sequence. D. Deoxythymidine Jack Westin Advanced Solution: D. Like its base pair answer choice A, deoxythymidine should indeed be present in the complementary sequence because its deoxyadenosine is in the WT sequence. Only deoxycytidine is not present in the complementary sequence.

Enzyme activation energy

The difference in free energy between the ground state of a reacting substance and the transition state.

The energy of electromagnetic radiation is directly proportional to the number of photons,

The energy of electromagnetic radiation is directly proportional to the number of photons, and the intensity of electromagnetic radiation is defined as energy emitted per unit time. Thus, intensity is directly proportional to the number of photons emitted.

The glucose meter measures the current produced during Reaction 2. If 0.67 μmol of electrons were measured, what mass of glucose was present in the sample? (Note: The molar mass of glucose is 180 g/mol = 180 μg/μmol.) 20 μg B. 60 μg C. 90 μg Jack Westin Advanced Solution: D. 270 μg he blood glucose meter uses test strips loaded with an enzyme and additives, and is calibrated to take a fixed volume of solution (usually blood). It measures the amount of Reaction 2, a two-electron oxidation, that takes place. Based on the total amount of Reaction 2 and the volume of the sample, the concentration of glucose can be measured and displayed.

The glucose meter measures the current produced during Reaction 2. If 0.67 μmol of electrons were measured, what mass of glucose was present in the sample? (Note: The molar mass of glucose is 180 g/mol = 180 μg/μmol.) Jack Westin Advanced Solution: This question can be pretty tough at first glance, but it's actually not too bad. Hear me out: the question stem wants us to find the mass of glucose in the sample and gives us the mole-mass conversion factor, also known as the molar mass. If we can find out the moles of glucose, we can easily find the mass. Now the passage stated that the reaction is a two-electron oxidation which means that whatever moles of electrons were used is twice the amount of glucose in the sample. By keeping track of the variables we're given, the conversion factors and the units of whatever variable(s) the question is asking for, we can easily tackle MCAT questions. A. 20 μg Jack Westin Advanced Solution: A. This mass, 20μg, is too small; did you by chance multiply the denominators of the mole fractions to get 1/9 of 180μg? See below for the correct calculation. B. 60 μg Jack Westin Advanced Solution: B. This is the correct mass C. 90 μg Jack Westin Advanced Solution: C. 90 μg - This answer choice is too large and is likely the result of using ½ μmol. Remember that per the passage, the electron-to-glucose ratio is 2:1 and the number of moles of electrons was ⅔ μmol, making the moles of glucose ⅓ μmol. D. 270 μg Jack Westin Advanced Solution: D. 270 μg - This answer choice assumes that more than one μmol of glucose is present, when in fact there is only about ⅓ of a μmol of glucose present.

metabolic rates

The measure of a person's energy use. Depends on speed and efficiency of different enzymes. Changes according to activity levels. I think it's a common misconception that leaner individuals have a higher metabolic rate than heavier individuals. Think about it--if you have more tissue, you're going to need more energy to fuel all those extra cells. Side note: your assumption that leaner individuals having higher metabolism would be correct IF we are comparing individuals of the SAME weight but of a different body fat %. This is because a pound of muscle burns more energy than a pound of adipose tissue. Tissue involved in movement is going to use a lot more energy than tissue that's just used for storage. Also don't get tripped up by the wording. Suppressed in this case just means that ghrelin levels are lower than in normal weight/lean individuals--it does NOT mean that levels are zero. So keeping in mind that you'd expect obese individuals to have HIGHER metabolic needs than leaner individuals, you can arrive at B through PoE. - A & C: wrong because ghrelin is not secreted by the hypothalamus. A is also wrong because lean individuals have lower energy requirements. - D: wrong because lean individuals have lower energy requirements. 3 ReplyShare

What is the correct expression for the ΔG′° for the transition observed in the experiments described in the passage? A. ΔG′° = -RTe([native]/[unfolded]) B. ΔG′° = -RTe([unfolded]/[native]) C. ΔG′° = -RTln([native]/[unfolded]) . D. ΔG′° = -RTln([unfolded]/[native])

The passage states "The pH-dependent unfolding of each oligonucleotide solution was monitored using circular dichroism (CD) spectroscopy." This implies the experiment studied the process of DNA unfolding. Thus, the reaction is: Folded --> Unfolded The equilibrium constant, being products over reactants, is unfolded/native What is the correct expression for the ΔG′° for the transition observed in the experiments described in the passage? A. ΔG′° = -RTe([native]/[unfolded]) Jack Westin Advanced Solution: A. The Gibbs free energy equation includes the natural log "ln," not its inverse "e." B. ΔG′° = -RTe([unfolded]/[native]) Jack Westin Advanced Solution: B. The Gibbs free energy equation includes the natural log "ln," not its inverse "e." C. ΔG′° = -RTln([native]/[unfolded]) Jack Westin Advanced Solution: C. The passage describes a "denaturation" so the unfolded DNA is the product, and the folded DNA is the reactant. The equilibrium constant Keq is the product relative to the reactant, so equal to [unfolded]/[folded]. Finally, the Gibbs free energy equation is ΔG′° = -RTlnKeq = -RTln([unfolded]/[folded]). This answer choice is incorrect because the Keq is the inverse of what it should be. D. ΔG′° = -RTln([unfolded]/[native]) Jack Westin Advanced Solution: D. This is correct. As described above, Keq for the denaturation described in the passage is equal to [unfolded]/[folded] and when plugged into the Gibbs free energy equation gives: ΔG′° = -RTlnKeq = -RTln([unfolded]/[folded]).

Magnesium reacts with hydrochloric acid in a sealed container, as shown. The initial pressure inside the container is 1.20 atm and its volume is 100 mL. What is the pressure inside the container after 243 mg of magnesium has reacted? (Note: Assume the reaction is isothermal and the gas behaves ideally.) Mg(s) + 2HCl(aq) → MgCl2(s) + H2(g) 1.4 atm 2.4 atm 3.6 atm 25 atm

The pressure added to the system from the reaction can be calculated by determining the number of moles of H2 gas that form. The reaction given is: Mg(s) + 2HCl(aq) → MgCl2(s) + H2(g). The moles of H2 gas formed are therefore equal to the moles of Mg that reacted, and can be calculated from: n = 0.243 g/(24.3 g/mol) = 0.01 moles (converting 243 mg to 0.243 g). The added pressure is calculated by P = nRT/V using approximations: 0.08 L atm/molK for R (0.0821 rounded), 300K for T (303.15 before rounding) and 0.10 L for V. Solving gives pressure = 2.4 atm. Finally, this number must be added to the initial pressure: 2.4 + 1.2 = 3.6 atm (Answer C). Answer A results from using 30oC (forgetting to convert to Kelvin). Answer B results from not adding the initial pressure of 1.2 atm to 2.4 atm. Answer D results from incorrect conversion of 100 mL to 0.010 L instead of the correct 0.100 L. Remember to double-check your math. In this case, as in most calculations on the MCAT, the wrong answers will be based on common calculation errors.

Given that antibiotics like erythromycin are metabolized by estrogen-sensitive P450 enzymes within the liver, which graph best predicts the expected relative half-life of erythromycin in adult males versus adult females?

The solution is A. This is the only graph that correctly illustrates a quicker metabolism—or shorter half-life—of erythromycin in females. Given the continuous presence of GH in females, the cytochrome P450 enzymes are up-regulated, resulting in more rapid metabolism of certain drugs (like erythromycin). As a result, the half-life of erythromycin is expected to be lower in females (in other words, it would take less time to metabolize the drug due to the greater presence of P450 enzymes). In this graph, both the males and females have the same relative half-life. The passage indicates that in the continuous presence of GH in females, the cytochrome P450 enzymes are continuously up-regulated. Thus, there is a difference of half-life between the males and females. In this graph, both the males and females have the same relative half-life. The passage indicates that in the continuous presence of GH in females, the cytochrome P450 enzymes are continuously up-regulated. Thus, there is a difference of half-life between the males and females. The passage indicates that in the continuous presence of GH in females, the cytochrome P450 enzymes are continuously up-regulated. Thus, there is a difference of half-life between the males and females: there will be reduced, not increased, erythromycin half-life in females.

Which of the following hypotheses best accounts for the differences in the level of circulating ghrelin in individuals who are obese with no underlying condition and this level in individuals who are lean?

The solution is B. Ghrelin is secreted by the stomach and small intestine, not the hypothalamus. Figure 1 shows that the levels of circulating ghrelin are higher in lean individuals than in obese individuals with no underlying conditions. Additionally, lean individuals have lower, not greater, metabolic needs than obese individuals. Obese individuals have more energy stores than lean individual. For this reason, the levels of ghrelin, which increases appetite, are lower in obese individuals.​ Although it is true that lean individuals have lower metabolic needs than obese individuals, it is the stomach and small intestine that secrete ghrelin, not the hypothalamus. Ghrelin secretion is suppressed in obese individuals. However, obese individuals have more, not less, energy stores than lean individuals.

triacylglycerol

A lipid consisting of three fatty acids linked to one glycerol molecule; also called a fat or triglyceride.

The side chain of tryptophan will give rise to the largest CD signal in the near UV region when:

Asymmetry resulting from tertiary structural features causes the largest increase in CD signal intensity in the near UV region (250-290 nm) of peptides." The Q asks for "near UV region," -> scan for that phrase in the passage ->see the above sentence -> know that tertiary structure refers to 3D folding. alpha-helix and beta-sheet are examples of secondary structure, homie. free amino acids can't have tertiary structure. the answer is D. Asymmetry resulting from tertiary structural features causes the largest increase in CD signal intensity in the near UV region (250-290 nm) of peptides. The side chains of aromatic amino acid residues absorb in this region. Answer part of a fully folded protein means it has the tertiary structure

At STP, the volume of N2(g) produced by the complete decomposition of 1 mole of nitroglycerin would be closest to which of the following?

Remember at STP molar volume of an ideal gas 22.4 L/mol. so from equation we know that 1.5 moles moles of N2 are produced and so 1.5X22.4

acids and bases can dissolve each other

Bases can be either water soluble or insoluble. Insoluble bases react with acids, directly dissolving in the acid as they react. Soluble bases form hydroxide ions in solution, that then react with the acid as described in the above section Acids, bases and water. The latter substance, Ni(OH)2, is apparently insoluble in neutral water, but will dissolve if the solution is acidic. This behavior is typical of substances that feature basic anions.

Suppose that, under the experimental conditions employed, Compound 1 is "saturated" with cations. What graph depicts the expected change in rate of K+ transport from the "IN" to the "OUT" phases at pH 2 (Trial 1 conditions) as a function of increasing K+ ion concentration in the "IN" phase?

Compound 1 acts like a catalyst for cation transport across the organic phase. If Compound 1 is saturated with substrate, increasing the substrate concentration will not increase the rate. In a plot of velocity V versus [S] (classic Michaelis-Menten kinetic treatment) the rate of cation transport is at the limiting value, called Vmax. horizontal line grapgh on vmax

If the solar cell is illuminated by monochromatic light in the visible spectrum, which color would produce the most electrical energy per photon? Red Green Violet Ultraviolet

Choice B is the correct answer. The question is asking about which color is most efficient. The efficiency plot is given in Figure 3. The peak can be estimated by visual inspection to be at approximately 530 nm. Answer D can be eliminated because ultraviolet light is outside the visible spectrum, which is required by the question stem. The visible light spectrum ranges from 390 - 700 nm, with 390 nm being violet and 700 nm being red. Therefore, B must be the correct answer. Even if the student does not recall the wavelength associated with green, it is the only possible answer that could be somewhere between 390 nm and 700 nm. Students should know that red and violet define the two extremes of the visible spectrum and it is obvious that the peak in Figure 3 is nowhere close to either 390 nm or 700nm.

Which change in solution composition would cause a protein to elute from a hydrophobic interaction column? Decreasing pH Increasing pH Decreasing salt concentration Increasing salt concentration

Choice C is the correct answer. Hydrophobic interaction chromatography (HIC) relies on high salt concentrations to enhance or strengthen hydrophobic interactions. Therefore, decreasing the salt concentration weakens these interactions, causing the protein to dissociate from the column. There is not a significant role for pH in HIC, making A and B incorrect. There could be a small effect if charges were to be neutralized but this is not the primary effect. Answer D is not correct since this would strengthen interactions with the column, not weaken them. high salt high hydrophobic interaction low salt low hydrophobic interaction

The two parallel plate electrodes give the solar cell an inherent capacitance. If the distance between the plates in Figure 1 were increased from 20 µm to 40 µm, the capacitance would: increase by a factor of 2. increase by a factor of 4. decrease by a factor of 2. decrease by a factor of 4.

Choice C is the correct answer. The capacitance of a parallel plate capacitor is given by C = ε0A/d, where ε0 is a constant, A is the area of the plates, and d is the separation of the plates. From this formula you can see that if the separation is doubled, C goes down by a factor of 2, which is Choice C. Choices A, B and D do not reflect the relationship shown in this formula.

Electrons flow through the electron transport chain from electron donors to electron acceptors. Which member of the electron transport chain has the greatest electron affinity? Cytochrome C NAD O2 Coenzyme Q

Choice C is the correct answer. The electrons in the electron transport chain of the mitochondria are always passed to carriers with greater electron affinity. O2is the final electron acceptor and therefore must have the greatest electron affinity. The electrons are donated from NADH or FADH2 originating from the Kreb's cycle. Then, in a multi-step pathway, those electrons are transferred to ubiquinone, later to cytochrome c, and eventually to oxygen. The ranking of the electron affinities of the other answer choices is in the order of electron transfer: NADH < ubiquinone < cytochrome c.

The average height of an American adult is 1.70 m. When an individual of this height is standing, by how much will the blood pressure at the bottom of the feet differ from the blood pressure at the top of the head? (Densityblood = 1060 kg/m3) 2000 Pa 6000 Pa 10000 Pa 18000 Pa

Choice D is correct. The pressure in a static fluid varies with depth h according to P = Pair + ρgh, where the ρgh term arises from the weight of the fluid. In this case, the pressure at the top isn't from the air, but the same principle applies: Pfeet = Phead + ρgh. The pressure difference is thus ρgh = 1060×9.8×1.7 = 17660 Pa, which rounds to 18000 Pa. Since the answers are not particularly close to each other, the multiplication can easily be approximated as ρgh ≈ 1000×10×1.7 = 17000 Pa, which is closest to choice D. Choices A, B, and C are various options that could be obtained via multiplication errors or mistakes made in choosing which quantities to multiply, and are all incorrect.

The light from an OLED device is produced by the: excitation of an electron to a higher energy level. absorption of a high-energy photon. emission of a positron associated with the relaxation of an electron. emission of a photon associated with the relaxation of an electron.

D was easy dumb

Soaps are chemically modified natural products that can be derived from all of the following EXCEPT: phospholipids fatty acids tryacylglycerol cholesterol

D because it has four fused rings while others have long carbon chains given in passage

Prostaglandin

Prostoglandins are derived from arachidonic acid and have 20 carbon atoms and one 5-carbon ring.

Absorption of ultraviolet light by organic molecules always results in what process?

Different effects dominate at different frequencies. With IR, there isn't enough energy in the photon to excite electrons, so the energy goes into vibrations. With UV, there is always enough energy in the photon to cause an excitation of the electron, and that's where the energy goes, rather than vibration.

Dissolution is different from dissociation.

Dissolution is having each molecule being solvated by the solvent. Dissociation refers to coming apart, or in this case, breaking down into a proton and the conjugate base.

What modification of Compound 1 will favor transport of Na+ relative to K+?

For the first one, just know that Na+ is a smaller ion than K+ (atomic radius increases going down and to the right), so a smaller channel would favor Na+ over K+ because K+ would now have a harder time fitting through. Removing a "ring" as specified in the question would accomplish this. This is a question you see a lot in practice material.

electric field lines

Go from positive to negative Always at a right angle to the surface The closer together the lines are, the stronger the field is It is what would happen to a positive test charge if the source is positive or negative. By definition, electric field lines exit from positive and enter negative charges. The axon is negatively charged and the extracellular solution has a zero net potential, thus electric field lines have the direction shown in this image.

Weakest acid

HPO42- has a high negative charge and so dissociation of it will occur to the least extent. Options B-D include H3PO4 and its various conjugate bases. When you're working with a polyprotic acid like H3PO4, the conjugate acid with the fewest H's remaining (HPO4-2) is the weakest acid.

What happens to the pH of a soapy solution as a result of the introduction of hardness ions?

Hardness ions essentially bind to and precipitate the carboxylate ions in solution (weak bases). Weak bases in solution could react with water, forming hydroxide ions in the process. Therefore, if you remove a weak base from the solution, you generate fewer hydroxide ions and lower the pH. Is there more to this question? 1 is false because an increasing [H] LOWERS pH. 2. Positive ions are Lewis acids. So, there has to be a reaction. 3. If you're adding a lewis acid, this ph decreases, so [H] increases and [OH] decreased. 4. A Lewis acid is a lewis acid. I'm not 100% sure, but this is my thought process. Paragraph 2 says common soaps contain sodium salts of organic acids with long carbon chains (weak acids). These salts are weak conjugate bases.

Why must the person either lean forward or slide their feet under the chair in order to stand up?

I'll refer to "you" here to simplify, so picture yourself in the situation. Torque is applied because as soon as your butt gets off the chair (before you were in static equilibrium), you have your center of mass which is not aligned with your only point of contact (through your feet), meaning you have a force at a distance from your center of mass. Any force at a distance from the center of mass that cannot be traced straight through that center of mass produces torque (imagine extending the vector in both directions infinitely). Now, because torque is applied, you need to move your body forward to counteract it such that your center of mass aligns with the force vector that pushes into the ground, nullifying the torque. Thus, the whole movement should keep the body in rotational equilibrium. I don't like how the question doesn't specify what type of equilibrium, because it is true that you are moving all throughout. I've seen center of mass come up here and there and it's a good concept to have in your back pocket for physics and more specifically, statics questions. Hope this helps!

What is the mechanical work done by the cantilever when the extension increases from 10 nm to 15 nm?

If you're given a force vs displacement graph, the area under that graph is equal to Work. I might be way off base but if W = area under the curve then it is just the area of a triangle which is 1/2 base x height. Therefore -> 1/2 (5x10^-9)(1x10^-10) = 2.5x10^-19 J n the first question, you are finding mechanical work given a graph. You can find work by the area of the curve, as seen in the Kaplan books. This is actually one of the 3 ways to calculate work, if you need a refresher on these 3 I recommend revisiting the Kaplan physics book under the chapter called "work".This is just a fact, work = area. Thus for the first one you find the area of the second triangle because it is the one between 10-15nm. Well if u used W=Fd what force would you use? Force isn't constant... If they give you a F vs d graph and ask for work, it will most likely always ask for area under curve. Think of W=Fd as when the force is constant (unless u know the average force)... if we then took F x d, (which would look like a horizontal line on the graph) we would still be finding the area

If no braking occurs, a total of how much power would be required to keep the railcar moving at 40 m/s?

Jack Westin Advanced Solution: P=Force*velocity=Work/timeIn the absence of braking, the only force opposing motion is the one mentioned in the last sentence; there is a continuous decelerating force that is present any time the railcar is moving. This decelerating force is 1,000N.P=Force*velocity= 1,000N * 40m/s = 40,000 W = 40kW

Michael mentes

Ka is the association constant, so measures the ratio of enzyme-ligand to free enzymes and free ligands. This measures only binding affinity. Kd is the dissociation constant and measures the inverse. This also only measures binding affinity. Km takes into account that enzyme-ligand complexes can react to form products, and so adjusts for enzyme binding AND reacting. You know it as the amount of substrate needed for enzymes to be reacting at half max velocity. Kt should be analogous to Km in this regard, but for transport~~/channel~~ proteins.

When finding the net charges of amino acid?

Look for the pH given and give approx pKa and decide which one will be protonated and which one will be deprotonated and then add charges

How much work did an 83-year-old female do while stretching the rubber band to the limit of her strength?

My question is, why can't we use the formula F = -kx? It asks for work, and Work = Force * distance. So couldn't I have done Work = -kx*x? However, when I do that it doesn't give me the right answer (it gives me 8 J). The right answer comes from the formula Work = 1/2kx2, which makes sense, but why couldn't the other method work? Work and energy are interchangeable but work and force are not. I did the same thing as you when I did that problem because I forgot the other equation. I may be mistaken here, but force of an elastic band/spring is not interchangeable with the work done because the force changes as the distance of the band changes. The work is only due to the distance that the spring/band travels That formula W=Fd is a special case for a constant force .W = F*d gives you work only when the force is uniform over the distance. Because a band becomes harder to stretch as d increases, F is NOT uniform for this system, so you cannot use W = F*d Work represents a change in energy. The change in energy for the rubber band is 1/2(kx^2). K is 200. x is .2. Plug in the numbers and you get the answer The force is non constant as the elastic band is stretched hence the work is also non uniform

Nh3 basic or water

Nh3 because it is less electronegative and would be more willing to give up share electrons that O in water so if HCl is added in a mixture of NH3 and H2O it would react with NH3 because it is the more basic So HCl is a strong acid. NH3 is more basic than H2O (this is because the Nitrogen is less electronegative than Oxygen based on periodic table trend). Remember that a base is defined as a proton acceptor (Bronsted-Lowry Definition) and an electron donor (Lewis Acid definition). So what will happen is the Nitrogen in NH3 has a lone pair on it. It will donate that lone pair of electrons and take the H+ proton from HCl. Thus, it will become NH4+. So what happened was after we added the HCl we have less NH3 because that NH3 turned into NH4+. So, the equation will shift to the left in order to try to replenish that NH3. This means that the concentration of Cu(NH3)4 does NOT increase since it shifts left so actually decreases. NH3 is a stronger base than H2O so it reacts with HCl more favourably than it reacts with H2O. the reaction of NH3 + HCl uses up some of the reactants which is why we have a left shift (according to le chateliers principle).

A glass rod is rubbed with a silk scarf producing a charge of +3.2 × 10-9 C on the rod. (Recall that the magnitude of the proton and electron charges is 1.6 × 10-19 C.) The glass rod has:

Now, before rubbing, both the glass rod and the silk cloth are neutral. When the rod is rubbed with the cloth, due to the higher work function of glass as compared to that of silk, some free electrons from the glass get transferred to the cloth. Hence, the rod gets positively charged(due to the loss of electrons), while the cloth gets negatively charged (due to the gain of electrons). However, the total charge of the system remains the same. :The formula for charge is q = ne q is the amount of charge (in Coulombs) n is the number of electrons e is the elementary charge (1.6 x 10-19 C) you are looking for the total number of electrons that were ripped away from it (since you can't move protons). therefore you would divide the total charge 3.2 x 10-9 by 1.6 x 10-19. it ultimately comes down to realizing that removing protons is kind of a big deal (it's like a nuclear reaction) and probably wouldn't be able to happen just by rubbing a rod. however, electrons, especially valance electrons, can easily be removed from stuff. think about ionization on the periodic table - usually an element loses some electrons to become an ion. if it loses protons (in the nucleus) it would become an entirely new element (ie radioactive decay) or dimensional analysis so 1proton/electron = (1.6 x 10-19 C) When you remove the element becomes positive and when you add the element becomes negative.

Compared to micellular Compound 1, Compound 2 is structurally more rigid as a result of what type of interaction? A. Intermolecular hydrogen bonding B. Intermolecular covalent bonding C. Intramolecular hydrogen bonding D. Intramolecular covalent bonding

So what you know is that Compound 1 gets oxidized to become Compound 2--which is rigid. You know that Compound 1 has cysteine residues (by looking at the structure of Figure 1) and that cysteine residues get oxidized to form rigid disulfide bonds between molecules. While the compound also has hydrogen bonds, the existence of these bonds would not explain why the oxidation of Compound 1 resulted in a more rigid structure, as oxidation does not necessarily increase the strength of hydrogen bonding. Watch out though, always double check the environment is not reducing - or else disulfide bond formation is impossible. Seen a few trick questions like that! Disulfide bond/bridge is an actual covalent bond, whereas hydrogen bond is an intermolecular bonding interaction (electrostatic attraction) between a hydrogen atom bonded to a very electronegative atom and an adjacent atom with a long pair of electrons. The strength of a H bond (intermolecular interaction) is not homologous to that of a disulfide bond (covalent bond). iodine oxidation can result in disulfide bond

Based on Reaction 1 and the fact that the product of oxidation absorbs strongly in the visible region, the most likely structure of the product is the highly conjugated quinone that is depicted in structure D.

Spectroscopy involves light whose wavelengths are on the scale of what you're trying to investigate. For our purposes in chemistry, where bond lengths are measured in nanometers, we will be using UV and visible light. We focus on bond lengths because chemistry is essentially the study of electrons and their excitability. Relate these ideas in your head: UV spectroscopy involves shooting photons of UV light at conjugated structures which bounce about from one conjugated double bond to the next while 'venting' in the process and becoming a lower energy photon of visible light with a longer wavelength. As for oxidation, you will generally see it do things like take -OH groups and turn them into carbonyls - basically making double bonds for those UV photons to bounce around in Conjugation (usually) leads to color. It's not listed in the passage, but it's a known phenomenon because of the HOMO-LUMO energy difference.

steriospecificity

Stereospecific refers to the fact that it only produces one isomer. In the passage, it states that this reaction produces maleic acid and its isomer fumaric acid hence it isn't stereospecific since it creates two different isomers. The key here is to know that the reaction occurs through a carbocation intermediate. Upon losing that fourth bond and forming a carbocation you will lose all stereochemistry, so the reaction couldn't be stereospecific even if it wanted to. Like /u/eshado said, when they're talking about stereochemistry they don't mean E/Z, they mean S/R.

Students added HCl dropwise to a solution of phospholipids with an initial pH of 7.4. Upon reaching a pH of 4.0, what changes will have occurred to the [COO-] and [COOH] present on the phospholipid head? Increased [COO-], because water will abstract more protons. Increased [COOH], because water will donate more protons. Increased [COO-], because the [COOH] increased. Increased [COOH], because the [COO-] increased.

The correct answer is choice B. This problem is solved by understanding how Le Chatelier's principle applies to acid base chemistry. The dissociation for the carboxylic acid is RCOOH + H2O → RCOO- + H3O+. In this situation pH is DECREASING, which means the [H3O+] will be increasing. This will shift the dissociation equilibrium to the left, or reactant side, and increase the ratio of the protonated carboxylic acid. This proton could come from water or hydronium. Answers A and C are incorrect because the [COO-] will decrease not increase. Answer D is incorrect because both the acid and the base cannot increase simultaneously. Answer C makes this same claim, in reverse order, which provides a second reason to eliminate Answer C.

What thermodynamic and chemical changes (if any) occur during aspartate transamination (Figure 1)?

The free energy of the reaction as written in either direction is the same since Keq = 1, and ΔG = -RTln(Keq) and therefore ΔG = 0. When a reaction is thermoneutral ATP is neither produced, nor required for the reaction to proceed. ln 1 is 0 ln 0 is 1

What is the ratio of the minimum sound intensities heard by a 64-year-old male and a 74-year-old female?

The way I think of decibels is the number before the zero (i.e. 2 in 20) is how many "zeroes" there are to measure intensity. For example, 20 decibels = 2 zeros = 100 (10^2). 30 decibels = 3 zeroes = 1000 (10^3). And so on. So if you are doing a ratio of 20 dB to 40 dB, it would be 10^2/10^4. I don't really use the formula, this is the way it clicked for me. I realize it may be kind of confusing to think of it this way so maybe someone will have a better method for you. There are 2 equations, the one you wrote & this one: Bf = Bi + 10 log (If / Ii) So basically, If is the intensity that corresponds to the decibel Bf. And Ii is the intensity that corresponds to the decibel Bi. Here's how I did it: Bf = Bi + 10 log (If / Ii) The B will always be decibels, and the I will always be intensity. So it would be: 40 dB = 20 dB + 10 log (If / Ii) ---- subtract 20 ------- 20 dB = 10 log (If / Ii) ------ divide by 10 ------- 2 = log (If / Ii) ------ take 10 to the power on both sides so log gets cancelled out -------- 10^2 = If / Ii very 10 db the intensity increases by a factor of 10 so since you're going from 20 to 40 that would be 102 which is 100. Long form: 40 db = 10 log i/10-12 I = 10-8 20 db = 10 log i/10-12 I = 10-10 10-8/10-10 = 102 On the actual mcat don't do the math, it's a waste of time. Just remember every 10 db you increase the intensity will increase by a factor of 10. Another handy short cut is that doubling the intensity yields an increase of about 3.1 dB. DON"T cancel 20/40 and make it 1/2 just do it in terms of 10

Metal ions present in plasma are expected to:

Think of it this way: Ka is the association constant (w/ the metal) K0 is association constant (w/o the metal) If there was no difference in binding affinities, we would get 1. But we see that when metal ions are added, this ratio is less than 1. This means that Ka is lower than Ko and that it decreases the amount of CPFX bound to BSA (B).

Based on the data presented in figures 2 and 3, what is the most likely role of Y229 in protein stability and cAMP activation?

This may be a little late but I'm going over the problem so I'll explain it (hopefully this is useful for future test takers). So in the passage, it is hypothesized that L203, Y229, I204 and R241 are not involved in cAMP activation. The proteins are mutated (transformed to a different protein) and their WT being removed. For example, Leucine is in the 203 position of the polypeptide. But when it is mutated to alanine (L203A), it becomes alanine, therefore L203 was removed (I can explain this further if you still don't understand). Mutation = WT removed. So the graphs are showing that the researchers are testing how the mutation affects cAMP activation/protein stability. In the question stem (read carefully), it is asking about Y229 (wild type). Because Y229 was mutated (Y229A) in the experiment, we see that it does play a role in protein stability. Y229A (mutated form) decreased the protein stability in Figure 3. If you look at the graph, we see that the mutated protein (Y229A) unfolds quicker at lower temperature in comparison to the WT line. The mutation causes low protein stability and it indicates that Y229 (wild type) is important for protein stability. For figure 2, the legend (pay close attention to legends when analyzing graphs) indicates that Y229A has the same activation as WT and was not shown. Therefore, it's stating that even if Y229 is mutated, it doesn't matter because it doesn't change the cAMP activation anyways. So it's does play a role in cAMP activation (confirming the hypothesis stated in the passage).

The pressure and volume changes that occur during a cycle of breathing are illustrated graphically in the figure shown.

This question requires you to interpret graphically a thermodynamic process and the area it encloses when represented in pressure-volume coordinates. The area under a curve that represents graphically the relationship of two physical variables (one independent and the other dependent) has the physical dimension of the product of the measurement units of the variables. In this case, a thermodynamic process is shown in PV coordinates, hence the area under the curve has the dimensions of [N/m2] · [m3] = [N·m] = [J]. The unit of joule (J) is used for energy and work.

eye lens

This question requires you to know that the lens is convex, therefore it is converging. Converging lenses always have positive focal lengths. Some lenses converge the light rays whereas some diverge the light rays. The light entering our eyes must be converged at a single point. Therefore, the lens in our eyes must be a converging lens and a convex lens is a converging lens. Therefore, the lens present in a human eye is a convex lens.

Flurophores

Used as stains in fluorescence microscopy. Fluorescence, absorb, a relatively short wavelength of light (blue), become excited, and decays to the ground state by emitting a slightly longer wavelength of light. A fluorophore is a fluorescent chemical compound that can re-emit light upon light excitation. Fluorophores typically contain several combined aromatic groups or planar or cyclic molecules with several π bonds

The slope of a Hill plot is used to quantify enzyme cooperativity. Given the graph below, what is the Hill coefficient for GCK? 2 -1.1 0 1.6

Using a Hill Plot (as shown) we can determine the Hill Coefficient via the slope of the line. A student could estimate the slope of the line to be approximately 1.6 to 1.7, but this is not necessary. The slope of the line is obviously positive, and choice D is the only positive value. Answers A and B are incorrect as these are negative numbers, representing negative slopes. Answer C is incorrect as a line with a slope of 0 would be horizontal.

Which single bond present in nitroglycerin is most likely the shortest?

Yes. As was mentioned, the greater the bond enthalpy (aka energy required to break the bond) the stronger the bond is. Stronger bonds will be shorter than weaker bonds because the atoms are held more "tightly" together. The bond dissociation energy increases as the difference in the electronegativities of the bonded atoms increases. Thus a larger difference in electronegativity between the two bonded atoms will have a stronger and shorter bond. The passage says that the C-H bond has the highest energy, therefore it is the shortest. Remember that bond energy is inversely proportional to length. This question is just testing your understanding that bond lengths will be shorter if a small element is involved. Hydrogen is so small that it forms shorter bonds with carbon vs two similarly sized elements.

The speed of sound in a medium depends on

bulk modulus/ density

coordination number

coordination number, the number of atoms, ions, or molecules that a central atom or ion holds as its nearest neighbours in a complex or coordination compound or in a crystal. Thus the metal atom has coordination number 8 in the coordination complexes [Mo(CN)8]4- and [Sr(H2O)8]2+; 7 in the complex [ZrF7]3-; 4 in the complexes [Zn(CN)4]2-, [Cu(CN)4]3-, and [Ni(CN)4]4-; 2 in the complexes [Ag(NH3)2]+, [AuCl2]-, and [HgCl2]. Coordination numbers from 2 to 9 have been observed in complexes; higher coordination numbers occur, although rarely I think they say that because like if NH3 had a -1 charge for some reason (just imagine), then the charge in NH3 would be 4- total right, so that would mean the other ion would need 4+ charge, and then now the number "4" could represent both an oxidation number AND a coordination number. Whereas since NH3 is neutral, then it won't represent another ions oxidation number. 4 is only a coordination number. Does that make sense? I know it's kinda muddled logic

Adding an electron donating group like alkyl on carboxylic acid

destabilize the conjugate base , increases the pKa and making dissociation less likely

Spontaneous

positive E reduction potential

Transition state analog

resembles the transition-state structure of the normal enzyme-substrate complex Also just want to add that both agonists and antagonists NEED to be transition state analogs-they need to resemble the transition state so they have affinity for the enzyme, to either elicit the response (agonist) or reduce the response (antagonist).

Which amino acids can be phosphorylated?

serine, threonine, tyrosine, histidine Serine CAN be phosphorylated along with lysine, threonine, tyrosine, histidine and arginine. An easy way to remember is any sidechain with a hydroxyl group or basic side chain". So which is where my confusion came in because I knew 'His' could be in some pathways but I never came across the others anywhere else