N/S FL 5

Power formula

(P = Fv).

Chromaffin cells (adrenal medulla)

- modified ganglionic sympathetic neurons - secrete hormones epinephrine and norepinephrine - enhance "fight-or-flight" response which increases heart rate and blood pressure and dilates bronchioles

Thus, the three-dimensional structure of globular proteins in aqueous solution will minimize interactions between hydrophobic amino acid residues and water by incorporating them into the interior of the protein.

. A similar outcome occurs in the formation of micelles or bilayer membranes by amphipathic phospholipids, with the hydrophobic alkyl tails clustered together and the polar phosphate heads facing the aqueous solution.

Operons are relatively simple and mechanistic systems that allow a bacterium to respond to changes in its environment by increasing or decreasing the expression of certain genes as appropriate. Operons can be under positive or negative control. In negative control, a repressor prevents transcription by binding to the operator (a sequence upstream of the first protein-coding region), while in positive control, an activator stimulates transcription.

. The classic examples for the MCAT, the lac operon and the trp operon, both involve negative control, but differ in that the lac operon is inducible and the trp operon is repressible. In a negative inducible operon, the repressor is normally present and the genes are not expressed except under specific conditions. In a negative repressible operon, the genes are usually transcribed, but transcription can be halted by binding of the repressor in appropriate conditions.

NMDA receptor

A receptor site on the hippocampus that influences the flow of information between neurons by controlling the initiation of long-term potentiation

epimers

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

open system

A system in which matter can enter from or escape to the surroundings.

closed system

A system in which no matter is allowed to enter or leave

isolated system

A system that can exchange neither energy nor matter with its surroundings.

Reasoned well Question 40 What is the primary process responsible for the loss of latent heat and entropy from the ocean at the air-sea interface in Table 1?

A. Precipitation Show Explanation B. Condensation Condensation should result in a positive latent heat for the ocean as it is a process that releases energy. C. Evaporation A negative latent heat implies that the phase change that is occurring is consuming energy. Therefore, the phase change that is occurring is melting, evaporation or sublimation. The fact that this process is occurring at the air sea interface means that the phase change must involve the gas phase. That fact rules out melting. Since sublimation is not an option, will go with evaporation. It makes sense that with evaporation, the ocean is losing entropy at the expense of the atmosphere which is gaining gas particles. D. Melting Show Explanation

After addition of 25 mL of NaOH, the pH of the solution was greater than 5. If 50 mL of NaOH represents the volume of titrant required to reach the equivalence point, then 25 mL is the volume added at the half equivalence point. At this point, one half of the original tolbutamide present will have been converted to its conjugate base, and their concentrations will be equal.

According to the Henderson-Hasselbach equation, when the values of protonated acid and conjugate base are equal, pH = pKa + log 1 = pKa + 0 = pKa. At the half-equivalence point, pH of solution equals the pKa of the analyte. According to paragraph 3, tolbutamide's pKa is 5.3.

faculative anaerobes

An organism that makes ATP by aerobic respiration if oxygen is present but that switches to fermentation under anaerobic conditions.

If the neomycin resistance gene on the pC27-53 and pC27-53X plasmids was abnormally susceptible to mutation, the most likely effect on the results from experiment 1 would be that: A. the number of colonies observed only for CRC157 and CRC184 transfected with pC27-53 would decrease. Show Explanation B. there would be a decrease in the number of colonies in all trials. If the neomycin resistance gene was unusually susceptible to mutation, then there would be an increased likelihood of a mutation rendering the resistance gene non-functional. Without that resistance, fewer colonies would be able to grow in the gentamicin solution. Thus there would be fewer colonies in all trials. C. the number of colonies observed only for CRC169 transfected with pC27-53 and pC27-53X would increase. Show Explanation

Another important context of antibiotic use is in genetic engineering techniques in which recombinant DNA is added to plasmids and introduced into bacteria (very commonly E. coli). Not all bacteria in a sample will take up the plasmids, so researchers must isolate those that have done so. For this purpose, plasmids used for genetic engineering generally contain antibiotic resistance genes, such that treatment with an antibiotic can kill off E. coli cells that did not take up any plasmids and select for those that did.

glycine is the only achiral amino acid, meaning that a solution of glycine does not rotate plane-polarized light. This stems from glycine's alpha carbon, which is attached to two hydrogen atoms as well as the amino and carboxylic acid termini. Since this does not constitute four distinct substituents, glycine is achiral; all 19 other standard amino acids are chiral.

Another property displayed by only one standard amino acid is the ability to form special bonds between sulfur atoms known as disulfide linkages. The amino acid that forms these bonds is cysteine, which has a thiol (-SH) group in its side chain. Disulfide bonds arise when one cysteine's sulfur atom connects to another, losing the attached hydrogen atoms in the process. These bonds are a key part of protein tertiary structure.

Titration is the process of finding the concentration of an unknown solution (the analyte) by reacting it with a solution of known concentration (the titrant). The analyte is generally placed in an Erlenmeyer flask, while the titrant is placed in a burette so that the volume of solution added can be monitored. The titrant is added to the analyte until the endpoint is reached. Calculations are then performed to find the unknown concentration of the analyte. Titrations are typically performed for acid/base reactions but are not limited to them.

At equivalence points during the titration, the number of acid or base groups added to the solution is equivalent to the number of base/acid groups in the original unknown solution. We can calculate our unknown concentration or volume using the formula NaVa = NbVb, where N and V are the normality (mol/L) and volume of the acidic and basic solutions, respectively. It is important to convert from molarity (M) to normality (N) for polyprotic acids and polyvalent bases. The flat regions of titration curves represent buffering solutions (a roughly equal mix of an acid/base and its conjugate), while the steep, near-vertical sections of the curve contain equivalence point(s), which indicate that enough of the titrant has been added to completely remove one equivalent (acid or base group) from each of the original molecules in the unknown solution. Species with multiple acid or base groups (e.g. H3PO4 or Ca(OH)2) will have multiple equivalence points during the titration. The final key point of any titration is the endpoint. To be successful, there must be some method for observing the endpoint of the reaction. The type of titration reaction that is being used will determine the method used for observing the endpoint. For example, in an acid-base titration, a specific pH value will be the endpoint (monitored by color-changing indicators), while for precipitation reactions, the endpoint is realized by the appearance of a precipitate. Regardless of the details of the reaction involved, the goal of titrations is always to use known volumes/concentrations to determine unknown volumes/concentrations.

Binary fission

Binary fission can take place very quickly, and it is easy for bacteria to exhaust the resources available to them in a given setting. The bacterial growth curve describes this process. First, when bacteria are introduced to a new environment, they adapt to it during the lag phase, which takes place before appreciable growth. After adapting to the environment, they embark on an exponential growth process, known as the exponential or log phase. Eventually, the environment stops being able to sustain exponential growth, and growth ceases in the stationary phase. Finally, the resources in the environment are exhausted completely and the bacteria die in the death phase.

Sister chromatids During synapsis, interaction occurs between chromatids from homologous chromosomes, not between sister chromatids. You can remember this because sister chromatids are genetically identical, so crossing over between them would have no effect. . Chromosomal centromeres Centromeres are chromosomal regions linking sister chromatids, not homologous chromosomes. C. Homologous chromosomes Synapsis is the pairing of homologous chromosomes during prophase I of meiosis. According to paragraph 2, the synaptonemal complex is "a structure which facilitates interaction between synapsing chromatids."

C. Homologous chromosomes Synapsis is the pairing of homologous chromosomes during prophase I of meiosis. According to paragraph 2, the synaptonemal complex is "a structure which facilitates interaction between synapsing chromatids." This protein structure bridges together synapsing chromatids from homologous chromosomes during synapsis (and possible recombination). D.

The most common function of proteins in the body is not enzymatic, but structural. Structural proteins are fibrous proteins that have an elongated shape and provide structural support for cells and organ tissues. T he first type of fibrous proteins are keratins, which form the skin, hair, and nails. Keratins are classified as soft or hard according to their sulfur content (i.e. the relative number of cysteine residues in their polypeptide chains). The low-sulfur keratins of the skin are much more flexible than the high-S, hard keratins. The second type of fibrous proteins you must know are the actin and myosin proteins of muscle tissue. Actin and myosin interact to form cross-linkages that allow the sliding of the filaments over each other in muscle contraction, which takes place through the contraction and relaxation of the sarcomere, the fundamental unit of all muscle fibers. When muscle contracts, the actin and myosin filaments slide over each other and the H-zone (myosin-only region), Z-lines (sarcomere boundaries), and I-band (actin-only region) all shrink, while the A-band (the entire myosin region) remains the same size. The opposite occurs upon muscle relaxation. A third type of structural protein you should know for test day is collagen, which is found in tendons, forms connective ligaments within the body, and gives extra support to the skin. Collagen is a triple helix formed by three proteins that wrap around one another. Many collagen molecules are cross-linked together in the extracellular space to form collagen fibrils to provide structural support for the cell. Elastin polypeptide chains are cross-linked together to form flexible, elastic fibers that give stretched tissues flexibility and the ability to recoil spontaneously as soon as the stretching force is relaxed.

Collagen is a triple helix formed by three proteins that wrap around one another. Many collagen molecules are cross-linked together in the extracellular space to form collagen fibrils to provide structural support for the cell. Elastin polypeptide chains are cross-linked together to form flexible, elastic fibers that give stretched tissues flexibility and the ability to recoil spontaneously as soon as the stretching force is relaxed.

isomer

Compounds with the same formula but different structures.

Unlike UV-Vis spectroscopy, which use larger energy absorbance from electronic transitions, IR spectroscopy relies on the much smaller energy absorbance that occurs between various vibrational and rotational states. Only molecules that undergo a net change in the dipole moment during vibrational and rotational motion can absorb IR radiation.

Diatomic molecules (e.g. O2, N2, and Br2) do not return IR signals because no net change in the dipole moment occurs. Molecules respond to the influx of energy by either stretching or bending. Stretching is a result of changing distances in a bond between two atoms on the molecule. Bending is any change in the angle between two bonds on the molecule. The various types of vibrations and rotations absorb at different frequencies within the infrared region, resulting in unique spectral properties for different molecular species.

Transformation refers to the ability of some bacteria to absorb genetic material directly from the environment.

Early experiments in bacterial genetics revealed that harmless strains of Streptococcus pneumoniae could be made virulent by exposing them to virulent bacteria that had been lysed with heat.

complementary DNA strands. Of the three main types of intermolecular attractions, hydrogen bonds are the strongest, because of the strength of the O-H, N-H, or F-H dipole. Dipole-dipole interactions are those that occur between polar molecules that do not contain O-H, N-H, or F-H bonds. These attractive forces draw the negative end of one dipole closer to the positive end of another. Finally, London dispersion forces are the weakest of the three forms of intermolecular attraction. All molecules experience these attractions, which can be thought of as synonymous to van der Waals forces for the sake of the MCAT. London dispersion forces stem from instantaneous dipoles, which are transient dipoles created by the random movement of electrons within a molecule

Even nonpolar molecules display these instantaneous dipoles, so even they will experience London dispersion forces. However, polar molecules that display hydrogen bonding or dipole-dipole attraction will not be affected by London dispersion forces to a significant extent, since those other attractions are much stronger, which is why London dispersion forces are often associated with nonpolar molecules.

Fatty acids are - charged at ph 7.4

Fatty acids play three crucial roles in the body: they are the constituents of complex membrane lipids, stored fat in the form of triglycerides (esters of 3 fatty acids joined together with glycerol, a 3-C triol), and the precursors for the synthesis of bioactive lipids (e.g. diacylglycerol). At physiological pH (pH 7.4), the carboxyl group is ionized, rendering fatty acids negatively charged.

C. θincident ≥ θcritical The question describes total internal reflection, when a light ray bounces inside a medium with a higher refractive index than the surrounding medium. For a light ray to totally internally reflect, rather than exit and refract (bend), the light ray must strike the edge of the glass tube at an angle equal or greater than the critical angle.

For a light ray to totally internally reflect, rather than exit and refract (bend), the light ray must strike the edge of the glass tube at an angle equal or greater than the critical angle.

It is often convienient to describe the relative positions of the resonances in an NMR spectrum.

For example, a peak at a chemical shift, δ, of 10 ppm is said to be downfield or deshielded with respect to a peak at 5 ppm, or if you prefer, the peak at 5 ppm is upfield or shielded with respect to the peak at 10 ppm.

In a phenomenon known as splitting, each signal is affected by protons on atoms adjacent to the carbon to which the proton is attached. Splitting patterns are predicted the n + 1 rule, which states that any peak will be split into a number of smaller peaks equal to the number of adjacent hydrogen atoms plus one.

For example, if a hydrogen atom is positioned on a terminal carbon adjacent to a carbon bound to one additional hydrogen atom, the peak that represents the first hydrogen atom will be split into a doublet (1 adjacent hydrogen atom + 1 = 2).

Fatty acids can be saturated (no C=C pi bonds) or unsaturated (1 or more C=C pi bonds). As with all pi bonds, the presence of the rigid sp2 geometry affects the molecular geometry of fatty acids. The site of unsaturation in a fatty acid is indicated by the symbol ∆ and the number of the first carbon of the double bond relative to the carboxylic acid group (-COOH), which is the highest priority carbon (i.e. #1).

For example, oleic acid is a 16-carbon fatty acid with one degree of unsaturation between C9 and C10, and is labeled as 16:1∆9. Another form of nomenclature uses the number of carbon atoms, followed by the number of sites of unsaturation (e.g. palmitic acid is a 16-carbon fatty acid with no unsaturation and is labeled 16:0).

The three main subcategories of structural isomers are chain isomers, functional isomers, and positional isomers. Chain isomers have different arrangements of the carbon 'skeleton.' Positional isomers have a given functional group in different locations (e.g., 1-pentanol vs. 2-pentanol).

Functional isomers are isomers where the molecular formula remains the same, but the type of functional group in the atom is changed. For example, a compound with an oxygen atom in addition to several carbon atoms and the corresponding number of hydrogens could be an alcohol with an -OH group, or an ether with a C-O-C group. Positional isomers have a given functional group in different locations (e.g., 1-pentanol vs. 2-pentanol).

W =

F⋅d⋅cos(θ),

Restriction enzymes, also known as restriction endonucleases, occur in nature in prokaryotes and archaea. In these organisms, they act as a defense system against invading viruses by cleaving foreign DNA. This activity is utilized in the laboratory to cleave target DNA, fragments of which can then be ligated together in a process termed genetic recombination. Restriction enzymes cleave DNA at very specific locations, or recognition sites, which vary from enzyme to enzyme. These sites correspond to sequences of 4 to 8 bases. Recognition sites usually contain some degree of symmetry, often in the form of palindromic sequences. In a palindromic sequence, the sequence of bases when read from 5′ to 3′ on one strand is the same as the sequence of the other strand when it is read from 5′ to 3′. When a restriction enzyme cleaves a DNA sequence vertically across the recognition site, the resulting fragments have "blunt" ends, whereas "sticky" ends result from restriction enzymes that cleave a DNA sequence in a zig-zag fashion. For example, the double-stranded sequence below represents the restriction site of the enzyme EcoRI. This site is cleaved after the guanine on both strands, resulting in sticky ends: GAATTC CTTAAG In contrast, the restriction enzyme SmaI cleaves between the central cytosine and guanine nucleotides on both strands, resulting in blunt ends: CCCGGG CCCGGG

GAATTC CTTAAG In contrast, the restriction enzyme SmaI cleaves between the central cytosine and guanine nucleotides on both strands, resulting in blunt ends: CCCGGG CCCGGG Sticky ends are particularly desirable in a laboratory setting because they ensure that the DNA fragments are ligated, or connected together, in the proper orientation. For example, consider the EcoRI restriction site shown above. Cleavage by this restriction enzyme produces a 4-base overhang, meaning that the cleaved fragment must be ligated with another fragment that has a similar, albeit complementary, overhang. This ligation is performed with DNA ligase, and fragments may come from different locations within a genome or even different species, as long as their cleavage produces these complementary overhanging sequences.

Oncogenes can arise from the mutation of other genes, termed proto-oncogenes.

If not mutated, proto-oncogenes do not promote cancer, but certain mutations or inappropriately elevated gene expression can effectively turn them into oncogenes.

To understand this principle, let us carefully compare what would happen if hydrophobic residues within a protein face an aqueous solution versus what would happen if hydrophilic residues face an aqueous solution, keeping in mind the polarity and excellent hydrogen-bonding properties of water . If hydrophilic—that is, polar—residues are facing the aqueous solution, then water will be able to hydrogen-bond freely with those residues, meaning that it will have relatively high entropy, which is energetically favorable. In contrast, water molecules will not be able to hydrogen-bond effectively with nonpolar residues, and as a result will form a highly-ordered solvation shell to minimize interactions with those residues. This highly-ordered shell represents a decrease in entropy, which is energetically unfavorable. Thus, the three-dimensional structure of globular proteins in aqueous solution will minimize interactions between hydrophobic amino acid residues and water by incorporating them into the interior of the protein. A similar outcome occurs in the formation of micelles or bilayer membranes by amphipathic phospholipids, with the hydrophobic alkyl tails clustered together and the polar phosphate heads facing the aqueous solution.

If water is able to bond on hydrophilic side.... high entropy Opposite= low entropy

. has a genome where nearly all material codes for protein. Nearly 95% of the human genome does not code for proteins or RNA. In contrast, the genomes of both prokaryotes and unicellular eukaryotes largely lack introns. In these organisms, most genetic material does code for protein products.

In contrast, the genomes of both prokaryotes and unicellular eukaryotes largely lack introns. In these organisms, most genetic material does code for protein products. Prokaryotes code for protein products

Broadly speaking, aerobic respiration requires oxygen to produce energy, while anaerobic respiration produces energy in the absence of oxygen. These methods differ in the amount of energy produced, processes involved, and organisms that utilize them. In general, aerobic respiration produces more energy (in the form of ATP) than anaerobic respiration. As we might expect, humans must use aerobic respiration for long-term survival, but human cells can function anaerobically for short periods. In human metabolism, the electron transport chain (ETC) and oxidative phosphorylation are aerobic processes, since oxygen is the final electron acceptor in the ETC. Therefore, oxygen is required to produce the proton gradient that leads to the formation of ATP via ATP synthase. The Krebs cycle (also known as the citric acid cycle) is also considered aerobic, even though it does not use oxygen directly, since this cycle requires the reduction of NAD+ to NADH. The opposite process (NADH to NAD+) happens in the ETC. When the ETC ceases to function in the absence of oxygen, then, NADH builds up and NAD+ becomes depleted, and the Krebs cycle eventually stops. Finally, glycolysis is anaerobic, since it does not require oxygen.

In fact, cells use glycolysis to function in low-oxygen conditions, since it produces ATP - albeit less than the cell would create aerobically. Like the Krebs cycle, glycolysis requires NAD+, so a cytosolic process is required to regenerate it from NADH under anaerobic conditions. This process is fermentation. In yeast cells, ethanol fermentation converts pyruvate into ethanol and carbon dioxide. In human cells and some bacteria, lactic acid fermentation uses the lactate dehydrogenase enzyme to convert pyruvate into lactate. In both cases, NADH is converted to NAD+, allowing glycolysis to continue.

Question 33 A patient presents in the emergency department having ingested a large quantity of tolbutamide. Intravenous administration of which of the following compounds is most likely to increase the rate of urinary excretion of the drug? A. KCl KCL is a salt that is very close to neutral (neither acidic nor basic). Because Cl- is the conjugate base of a strong acid (HCl), it will only be able to act as a base when pH is extremely low. B. NaHCO3

In order to increase the percentage of drug excreted in the urine, it is necessary to decrease the fraction of tolbutamide capable of reabsorption, or diffusion out of the lumen of the nephron. Paragraph 3 says that for weakly acidic drugs, the uncharged state is capable of diffusion through membranes much more than the charged form. Thus, to prevent reabsorption, we must maximize the charged form of tolbutamide. Wecan do this to a weak acid by deprotonating the drug (rendering it negatively charged) via administration of a base. This will increase blood and urinary pH (i.e. decreasing [H+]) and increase the fraction of ionized drug present. Urinary alkalinization can be accomplished by administration of a basic salt, like NaHCO3 (the salt of the conjugate base of carb Answered well

All the enthalpic (ΔH) changes presented in Table 1 are positive, indicating that all three mechanisms are endothermic for all three compounds in both solvents.

In other words, all three mechanisms must overcome an enthalpic barrier to move forward. The lower the enthalpic barrier, the more thermodynamically favorable the reaction is. Thus, lower ΔH values are associated with higher antioxidant activity. Compound 1 generally shows lower ΔH values than compound 2, and the mechanisms have lower ΔH values in water than in benzene, making choice D correct.

us, increasing the reactant concentration or decreasing the product concentration will shift the reaction towards the product side, and vice versa.

In reactions with gases, increasing the volume (decreasing the pressure) will shift the equilibrium to the side with more moles of gas (and vice versa). For a reaction where ∆H > 0, increasing the temperature will shift the reaction toward the products, while decreasing it will shift the reaction toward the reactants; the opposite pattern is found if ∆H < 0.

Antibiotic resistance is a major problem in modern medicine, and the genes responsible for antibiotic resistance are often found on plasmids (non-chromosomal bits of circular prokaryotic DNA) and spread through conjugation, which can be thought of as the bacterial equivalent of sexual reproduction. (However, conjugation is actually not a method of reproduction at all because it does not produce new bacterial cells.'

Instead, it is a form of gene transfer.) Conjugation involves the transfer of a plasmid through a bridge that is created when a sex pilus on one bacterium (often known as F+, which refers to the presence of the fertility factor, or as male) attaches to another bacterium (generally known as F−). During this process, the fertility factor itself is duplicated and transferred, converting the F− cell into an F+ cell.

tautomers

Isomers that can interconvert by changing the location of a proton

The relative preponderance of products versus reactants is described by the equilibrium constant (Keq), which is the concentration of products at equilibrium (raised to the power of their stoichiometric coefficients) divided by the concentration of reactants (raised to the power of their stoichiometric coefficients). For a sample reaction X (g) + 3 Y (g) ⇌ 2 Z (g), Keq = [Z]2/[X][Y]3. An important caveat is that solids and pure liquids should not be included in the equilibrium expression, and it is also important to note that Keq varies with temperature. A reaction that favors the products—in other words, that has a numerator greater than its denominator— will have a large Keq, while a reaction that favors the reactants will have a small Keq.

Le Châtelier's principle is an MCAT-critical concept related to equilibrium. This principle states that if an equilibrium mixture is disrupted, it will shift to favor the direction of the reaction that best facilitates a return to equilibrium. A simpler version is that a reaction will shift in the direction that relieves the stress put on the system, where "stress" can refer to a change in reactant or product concentration, temperature, pressure, or volume. Thus, increasing the reactant concentration or decreasing the product concentration will shift the reaction towards the product side, and vice versa. In reactions with gases, increasing the volume (decreasing the pressure) will shift the equilibrium to the side with more moles of gas (and vice versa). For a reaction where ∆H > 0, increasing the temperature will shift the reaction toward the products, while decreasing it will shift the reaction toward the reactants; the opposite pattern is found if ∆H < 0.

An enzyme (a biological catalyst) may require another chemical compound to be present in order for it to carry out its biological functionality. In general, such "helper" molecules are known as cofactors. Cofactors can be either inorganic (with some common examples including metal ions such as Mg2+, Zn2+, and Cu+) or organic, and organic cofactors are sometimes known as coenzymes. Many coenzymes are vitamins or derivatives of vitamins, and they often contribute to the function of enzymes by carrying certain functional groups from one place to another in a reaction. Perhaps the most well-known example of this is coenzyme A, which transfers acyl groups from one place to another. Coenzymes that are tightly (or even covalently) bonded to their enzyme are known as prosthetic groups. A famous example of an organometallic prosthetic group is heme, which contains an iron ion in the center of a porphyrin ring, and is attached to oxygen-transport proteins such as hemoglobin and myoglobin.

Many coenzymes are vitamins or derivatives of vitamins, and they often contribute to the function of enzymes by carrying certain functional groups from one place to another in a reaction. Perhaps the most well-known example of this is coenzyme A, which transfers acyl groups from one place to another. Coenzymes that are tightly (or even covalently) bonded to their enzyme are known as prosthetic groups. A famous example of an organometallic prosthetic group is heme, which contains an iron ion in the center of a porphyrin ring, and is attached to oxygen-transport proteins such as hemoglobin and myoglobin.

Although there are many chromatography techniques, they share the principle that the molecules in a mixture are applied onto a stationary phase (usually a solid), while a fluid known as the mobile phase (generally a solvent chosen to match the target molecules, e.g. polar or nonpolar) containing the molecules of interest travels through the stationary phase. Molecules of interest in the mobile phase will interact with the stationary phase with different levels of intensity.

Molecules that interact more strongly with the stationary phase will take longer to pass through it, whereas molecules that interact more weakly with the stationary phase will pass through it more quickly. Common factors that shape these interactions include molecular characteristics related to adsorption, polarity- or charge-based affinity for the stationary/mobile phase, and differences in molecular weight. A chiral stationary phase can also be used to separate stereoisomers based on the principle that the various enantiomers of a compound may interact differently with such a stationary phase.

C. Anaphase II

Nondisjunction is the failure of chromosomes to separate properly during anaphase I of meiosis or the failure of sister chromatids to separate properly during anaphase II of meiosis

Solvents have a notable effect on the rates of these reactions. Polar protic solvents (such as water and ethanol) tend to stabilize ions in solution. Since they can stabilize the carbocation, these protic solvents are best used for SN1 reactions. However, SN2 reactions rely not on a carbocation, but on a strong nucleophile displacing the leaving group. Protic solvents tend to stabilize (weaken) this nucleophile, so they should not be used for SN2 procedures. Instead, polar aprotic solvents, such as acetone, are a better choice.

Protic solvents tend to stabilize (weaken) this nucleophile, so they should not be used for SN2 procedures. Instead, polar aprotic solvents, such as acetone, are a better choice.

Question 44 The synaptonemal complex with which TEX11 interacts serves as a protein-DNA bridge between what nuclear bodies? A. Sister chromatids During synapsis, interaction occurs between chromatids from homologous chromosomes, not between sister chromatids. You can remember this because sister chromatids are genetically identical, so crossing over between them would have no effect. B. Chromosomal centromeres Centromeres are chromosomal regions linking sister chromatids, not homologous chromosomes. C. Homologous chromosomes Synapsis is the pairing of homologous chromosomes during prophase I of meiosis. According to paragraph 2, the synaptonemal complex is "a structure which facilitates interaction between synapsing chromatids." This protein structure bridges together synapsing chromatids from homologous chromosomes during synapsis (and possible recombination). D. Meiotic spindles In meiosis, the spindle apparatus attaches to chromosomes through the kinetochore, not through the synaptonemal complex. 33.07% A 7.49% B 56.07% C 3.36% D Your answer was incorrect. 56.07% of students answered this question correctly. Content Foundations: Mitosis and Meiosis In eukaryotes, the process of asexual cell division is known as mitosis. Mitosis takes place in four phases: prophase, metaphase, anaphase, and telophase. Prophase prepares the cell for mitosis: the DNA condenses such that distinct chromosomes become visible, as sister chromatids (or copies of a given chromosome) join at a region known as the centromere. The kinetochore assembles on the centromere, and is the site where microtubule fibers that extend from the centrosome and form the mitotic spindle attach to pull the sister chromatids apart in later stages of mitosis. Other microtubules known as asters extend from the centrosome to anchor it to the cell membrane. Additionally, the nuclear envelope and the nucleolus disappear, and the mitotic spindle forms. In metaphase, the chromosomes line up at the middle of the cell along an imaginary line that is known as the metaphase plate. In anaphase, the sister chromatids are separated and pulled to opposite sides of the cell by shortening of the microtubules attached to the kinetochores. Telophase can be thought of as the opposite of prophase, as a new nuclear envelope appears around each set of chromosomes and a nucleolus reappears within each of those nuclei. The process of mitosis is completed by cytokinesis.

Question 44 The synaptonemal complex with which TEX11 interacts serves as a protein-DNA bridge between what nuclear bodies? A. Sister chromatids During synapsis, interaction occurs between chromatids from homologous chromosomes, not between sister chromatids. You can remember this because sister chromatids are genetically identical, so crossing over between them would have no effect. B. Chromosomal centromeres Centromeres are chromosomal regions linking sister chromatids, not homologous chromosomes. C. Homologous chromosomes Synapsis is the pairing of homologous chromosomes during prophase I of meiosis. According to paragraph 2, the synaptonemal complex is "a structure which facilitates interaction between synapsing chromatids."

What is the most likely genotype of individual SG_459382 from Experiment 3 regarding eye color only? A. Gg or gg Show Explanation B. Gg or gG The only possible genotype is a heterozygous one. This is because the test cross in Figure 3 reveals that some offspring will have green eyes and some will have black eyes when one parent is homozygous recessive for eye color and hence has black eyes (genotype gg). The only way this can occur is if the other parent sometimes passes on a dominant G allele, causing the offspring to have a Gg genotype and green eyes, and at other times passes on a recessive g allele, causing the offspring to have a gg genotype and black eyes.

Review genetics

uestion 25 The researchers later used SDS-PAGE and size-exclusion chromatography to separate different mixtures containing both CP8 (a 76-kDa protein) and Zp_127 (a 40-kDa protein). CP8 would be expected to: A. travel farther during SDS-PAGE and elute more quickly during size-exclusion chromatography. Show Explanation B. travel farther during SDS-PAGE and elute more slowly during size-exclusion chromatography. This answer choice incorrectly presents how protein size would affect SDS-PAGE and size-exclusion chromatography. Traveling further in SDS-PAGE would be associated with a smaller protein, which would elute more slowly in size-exclusion chromatography. C. travel a smaller distance during SDS-PAGE and elute more quickly during size-exclusion chromatography. Kilodaltons (kDa) are units of size used for proteins. Since CP8 is larger than Zyg_27, it will move more slowly through the polyacrylamide gel that forms the matrix during SDS-PAGE. However, size-exclusion chromatography includes a stationary phase with many small pores, which smaller proteins become trapped in while passing through the column. For this reason, larger proteins (like CP8) travel more quickly during such chromatography procedures. D. travel a smaller distance during SDS-PAGE and elute more slowly during size-exclusion chromatography. This answer choice incorrectly presents how protein size would affect SDS-PAGE and size-exclusion chromatography. Traveling further in SDS-PAGE would be associated with a smaller protein, which would elute more slowly in size-exclusion chromatography. 5.1% A 6.25% B 76.53% C 12.12% D Your answer was incorrect. 76.53% of students answered this question

SDS PAGE - larger moves slower size exclusion- larger ones elute quickly because they do not get caught up in beads

In a technique known as sodium dodecyl sulfate-polyacrylamide gel electrophoresis (SDS-PAGE), the strong anionic detergent SDS is used to denature native proteins into their unfolded polypeptide states. SDS imparts most proteins with an even distribution of charge per unit mass, and denatures secondary and tertiary structure (except for disulfide bonds).

SDS-PAGE is a useful technique because it allows proteins to be separated only based on mass. More specifically, the electrophoretic mobility displayed by these proteins will be a linear function of the logarithms of their molecular weight.

Spermatogenesis is the process through which male gametes (spermatozoa) are generated. Like oogenesis, it involves meiosis, but the two processes have several important differences. Spermatogenesis is initiated during puberty and continues throughout the lifespan. The process of spermatogenesis takes approximately 3 months, and approximately 100 million viable sperm are produced daily. Spermatogenesis begins with spermatogonial stem cells and ends with mature spermatozoa, but has several intermediate stages. First, spermatogonial stem cells can either divide into descendent spermatogonial stem cells (thereby maintaining the supply) or differentiate into spermatogonia. Spermatogonia divide through mitosis into two primary spermatocytes. Primary spermatocytes go through meiosis I and divide into two secondary spermatocytes. This is where the transition from diploid (2n) to haploid (n) happens. Secondary spermatocytes then go through meiosis II, forming spermatids. A total of four genetically unique spermatids are formed from each primary spermatocyte.

Spermatids initially lack some of the most important features of the mature sperm cells that are released during ejaculation, and they gain those features in a process known as spermiogenesis. The main events of spermiogenesis are as follows: (1) formation of the acrosomal cap, which facilitates the ability of a sperm to fertilize an egg; (2) formation of a tail; and (3) loss of excess cytoplasm. Spermiogenesis results in non-mature spermatozoa that are incapable of independent movement, and are transferred to the epididymis to undergo maturation. Mature sperm cells are very compact, and contain roughly the bare minimum of structures necessary for their functionality. A mature sperm cell has a head, a mid-piece, and a tail. The head contains the cell's DNA and is surrounded by the acrosomal cap. The mid-piece contains abundant mitochondria, which are necessary because sperm cells require quite a bit of energy throughout their life cycle, and the tail provides motility.

Tautomers are structural isomers that interconvert with each other and exist in equilibrium. The most commonly encountered is keto-enol isomerism, in which a ketone interconverts with an enol (a structure with an C-OH group and an adjacent C=C bond, instead of a C=O bond and an adjacent C-C bond, as in a ketone). At room temperature, the keto form is favored, but the enol form contributes significantly to some reaction mechanisms, and the deprotonated intermediate (known as an enolate ion) is also important for some reactions.

Tautomerization also takes place between enamines and imines (the second most important example), lactams and lactims, and amides and imidic acids. It is important to understand that tautomers and resonance structures are not the same thing. Resonance is a phenomenon in which electrons are delocalized, and resonance structures are a somewhat crude way of representing a single underlying, delocalized structure of the molecule. In contrast, tautomers are two different structures that interconvert via the breaking and re-formation of bonds.

The most important law when looking at enthalpy is Hess's law: ΔHrxn = Σ∆Hproducts - ΣΔHreactants. This equation illustrates that enthalpy, like entropy, is a state function. This means that the ∆H accompanying a chemical reaction is independent of the mechanism by which the reaction occurs.

That is, when reactants are converted into products, the overall enthalpy change is the same whether it is done as one step or multiple steps.

Rules for Doppler Effect

The Doppler effect is used to analyze moving objects or fluids. A Doppler shift will be registered only if the fluid is moving relative to the source of the sound (the device). More specifically, at least some component of the fluid's velocity must exist in the same directional plane as the wave's velocity (otherwise, the device will register the fluid as not moving at all), and this component must be different from the velocity of the sound source.

When an electron drops from a higher-energy orbital back to its ground state, it releases energy as a photon (itself a form of electromagnetic radiation).

The amount of energy required for an electron to transition between orbitals is specific to each orbital in each element. Thus, substances will absorb or emit light of specific frequencies (corresponding to specific energy levels through the equation E = hf). The resulting atomic emission and absorption spectra are unique to each

A lower Km corresponds to higher affinity. However, Km is not the only way to measure enzyme affinity.

The association constant (Ka) can also be defined, using the mathematical formalism of equilibrium constants, as [ES]/[E][S], where [ES] is the concentration of the enzyme-substrate complex, [E] is the concentration of the enzyme, and [S] is the concentration of the substrate. The dissociation constant (Kd) is then the inverse of Ka, and can be defined as [E][S]/[ES].

When a wave-emitting detection device is used, its accuracy is optimized if the waves emitted by the device travel directly parallel to the waves that are being measured (θ = 0).

The deviation caused by non-parallel angles is known as the cosine effect, because the measured velocity will equal the true velocity × cos(θ).

Reaction coordinate diagrams are used to visualize the energy changes associated with a reaction over time. Typically, the reactants begin with a certain amount of energy. For the reaction to take place, an energy barrier termed the activation energy must be overcome. Overcoming this barrier produces a transient species termed the transition state, which is so unstable and high-energy that it is only present instantaneously. From there, the products are formed. =

The difference in energy between the reactants and products is the ∆G, or change in Gibbs free energy, of the reaction. If the products are lower-energy than the reactants, then the reaction is thermodynamically favorable and spontaneous. If the products are higher-energy than the reactants, then the reaction is nonspontaneous. An important concept to note is that spontaneity does not relate to the rate of the reaction, or the speed at which it progresses. Rate is a kinetic parameter and is determined largely by the activation energy, while spontaneity is a thermodynamic parameter and relates to ∆G. A lower level on the y-axis in a reaction coordinate diagram, corresponding to a lower-energy state, indicates greater thermodynamic stability.

try Reduction and oxidation (redox) reactions are often studied in the context of electrochemistry, generally focusing on reactions involving ionic compounds. However, reduction is defined as a decrease in the oxidation state of an atom, while oxidation is defined as an increase in the oxidation state of an atom, and these definitions can also be applied to reactions containing organic compounds. The following features are associated with reduction in organic chemistry: (1) gain of an electron, (2) decreased oxidation state, (3) formation of a C-H bond (e.g. alkene → alkane), and (4) loss of a C-O or C-N bond (or any bond between carbon and an electronegative atom). Conversely, oxidation is associated with (1) loss of an electron, (2) increased oxidation state, (3) loss of a C-H bond (e.g. alkane → alkene), and (4) gain of a C-O or C-N bond (or any bond between carbon and a highly electronegative atom).

The following features are associated with reduction in organic chemistry: (1) gain of an electron, (2) decreased oxidation state, (3) formation of a C-H bond (e.g. alkene → alkane), and (4) loss of a C-O or C-N bond (or any bond between carbon and an electronegative atom). Conversely, oxidation is associated with (1) loss of an electron, (2) increased oxidation state, (3) loss of a C-H bond (e.g. alkane → alkene), and (4) gain of a C-O or C-N bond (or any bond between carbon and a highly electronegative atom). Oxygen-containing organic compounds exist on a spectrum of oxidation from alcohols (most reduced/least oxidized) to aldehydes/ketones (intermediate reduction/oxidation) to carboxylic acids (least reduced/most oxidized). A primary alcohol can be oxidized to an aldehyde by a mild oxidizing agent (such as PCC) or to a carboxylic acid by a strong oxidizing agent like NaCr2O7. A secondary alcohol will be oxidized to a ketone by either a mild or a strong oxidizing agent. A strong oxidizing agent will likewise oxidize an aldehyde to a carboxylic acid. A strong reducing agent, such as LiAlH4, can reduce a carboxylic acid directly to an alcohol, while weak reducing agents such as NaBH4 will not reduce carboxylic acids at all. A special agent, DIBAL, can reduce a carboxylic acid to an aldehyde when applied at a precise 1:1 ratio. Both mild and strong reducing agents can reduce aldehydes and ketones to primary and secondary alcohols, respectively.

The melting point of fatty acids increases as the length of the carbon tail increases. In addition, unsaturated fatty acids will have lower melting points than saturated fatty acids with the same number of carbon atoms. As a general rule, unsaturated fatty acids are liquids at room temperature while more saturated fats are solids. Fatty acids also demonstrate cis/trans isomerism. In cis fatty acids, the acyl groups are on the same side of the double bond, while in a trans fatty acid, the acyl groups are on opposite sides. Cis fats generally have higher boiling points, while trans fats have higher melting points. The lipid synthesis mechanism of the body can theoretically supply the body with near-sufficient fatty acid structures. However, the fatty acids linoleic acid and α-linolenic acid cannot be synthesized from precursors in the body, and are thus considered the essential fatty acids.

The lipid synthesis mechanism of the body can theoretically supply the body with near-sufficient fatty acid structures. However, the fatty acids linoleic acid and α-linolenic acid cannot be synthesized from precursors in the body, and are thus considered the essential fatty acids.

conjugation. Bacterial conjugation is the transfer of genetic material, ordinarily in the form of a plasmid, between bacterial cells by direct cell-to-cell contact or via a bridge-like connection between two cells.

The prototypical conjugative plasmid is the F plasmid, which is transferred from donor to recipient cell via a bridge-like structure known as a pilus.

The passage states that the correlation between BAC and extent of memory loss has already been established; this is not what the study is investigating.

The researchers are interested in whether the rate at which alcohol is consumed affects the extent of memory loss, so they measured BAC to ensure that it was relatively similar in all groups. This increases the validity of the data because it shows that differences in BAC cannot explain the differences in memory loss between groups.

Ka = [H+][A-]/[HA] The square brackets indicate the concentration of the respective aqueous species. Ka expresses how easily an acid releases a proton (i.e. its strength). In addition, this equation shows how the dissociation state of weak acids vary according to the [H+] level in the solution. A commonly tested family of acids on the MCAT are carboxylic acids (those containing -COOH), such as lactic acid and amino acids, which normally have a Ka of approximately 10-3 to 10-6. As this shows, expressing acidity in terms of the Ka constant alone involves inconvenient numbers that are not very intuitive.

Therefore, pKa was introduced as an index to express the acidity of weak acids, where pKa = -logKa. For example, the Ka values for lactic acid (HC3H5O3) and nitrous acid (HNO2) are 8.3 × 10-4 and 4.1 × 10-4, respectively. The pKa values for these acids are 3.1 and 3.4, respectively, which are simpler expressions that are easier to understand and compare. The smaller the pKa value, the stronger the acid (since as X increases, pX decreases). Therefore, the pKa values above tell us that lactic acid is a stronger acid than nitrous acid.

Unlike DNA, RNA tends to exist in single-stranded form, rather than as a double-stranded double helix. Additionally, the sugar in RNA is ribose, which contains a hydroxyl (-OH) group on its 2′ carbon. (DNA contains deoxyribose, which lacks this 2′ hydroxyl group.) In addition to mRNA, which codes for protein production, several forms of non-coding RNA exist.

These include transfer RNA (tRNA), which assists in translation, and small interfering RNA (siRNA) and microRNA (miRNA). siRNA and miRNA differ in their structure: miRNA strands are single-nucleotide strands incorporated into an RNA structure with a characteristic hairpin loop, while siRNA molecules are short and double-stranded. Both tend to be approximately 22 nucleotides in length, and both silence genes by interrupting expression between transcription and translation.

In most cases, the solubility of ionic substances in water increases with temperature, while the opposite pattern is observed for gases. .

This is because higher temperatures provide gases with more kinetic energy that they can use to escape the solution. Additionally, pressure favors the solubility of gases.

The Ka and pKa values of acids are inversely related to each other (since pKaa = -log [Ka], and to corresponding Kb and pKb values of their conjugate bases. Thus, the strongest conjugate base will be produced by deprotonating the weakest acid. As pKa values decrease, acid strength goes up.

Thus, the weakest acid present, acetazolamide has the greatest pKa value. You can also use the shortcut that at standard temperatures (which are present unless otherwise stated on the MCAT), pKa + pKb = 14. Thus as pKa increases, pKb decreases, which indicates a stronger base.

kinematic equations

V = Vo + at X = Xo + Vot + 1/2at^2 V^2 = Vo^2 + 2a(X - Xo)

Vitamins are non-macronutrient compounds that are vital for healthy functioning and cannot be synthesized in adequate quantities by the body, meaning that they must be obtained from external sources. The B vitamins and vitamin C are coenzymes used for various important reactions, and are water-soluble. Vitamins A, D, E, and K are lipid-soluble.

Vitamin A plays a key role in vision, vitamin D in calcium and phosphate absorption from the gastrointestinal tract, vitamin E is an antioxidant, and vitamin K promotes coagulation. Like vitamins, minerals contribute to essential roles in the body and must be obtained from the diet, but unlike vitamins, minerals are inorganic. The most important minerals in the body are calcium, phosphorus, magnesium, sodium, and potassium.

In the lac operon, when glucose is present but lactose is not, there is no need for the bacterium to express the genes needed for lactose metabolism. Therefore, the repressor is bound to the operator, transcription does not happen, and no lactose metabolism takes place.

When lactose is present, its metabolite allolactose releases the repressor from the operator, allowing transcription to happen and some degree of lactose metabolism to take place. However, intense lactose metabolism does not take place unless lactose is present in the relative absence of glucose, meaning that the bacterium really needs to metabolize lactose. In that case, cAMP causes the catabolite activator protein (CAP) to bind to the upstream CAP binding sequence. CAP promotes more intense expression of the genes on the lac operon.

Which of the following factors would be most likely to cause acetic acid to completely dissociate in aqueous solution? A. Higher temperatures, which increase the pKa of the acid Increasing the pKa of a species implies that it becomes a weaker acid. Weak acids do not completely dissociate in solution. B. Enzymes that catalyze the forward reaction Enzymes allow a reaction to proceed faster, but they do not affect equilibrium concentrations. Also, enzymes by definition catalyze both the forward and reverse reaction of a particular process. C. Continuous addition of acetic acid to the solution This would force the equilibrium toward the product side of the reaction, causing more H+ to be present in the solution. However, it would not cause the acetic acid to completely dissociate.

Wrong answers

lac operon

a gene system whose operator gene and three structural genes control lactose metabolism in E. coli

zwitterion

a molecule or ion having separate positively and negatively charged groups.

decibel scale

a nonlinear scale of loudness based on the ratio of the intensity level of a sound to the intensity at the threshold of hearing

aprotic solvent

a solvent that has no acidic protons; a solvent with no O-H or N-H groups

velocity

accelaration * time

dissociation constant

an equilibrium constant between a ligand and a protein, such as a receptor or an enzyme Select kinetic properties of the γ and β enzymes are shown below. Which statement best describes how the two enzymes differ from each other? A. Enzyme β exhibits a lower degree of cooperativity than enzyme γ. Show Explanation B. Enzymes β and γ act on separate substrates. Show Explanation C. The maximal velocity of the reaction catalyzed by enzyme β is higher than that reached by the reaction catalyzed by enzyme γ. kcat represents the catalytic rate constant, often called the turnover number. It is a measure of how many bound substrate molecules turn to product in 1 second. We can calculate the Vmax of a catalyzed reaction using the equation Vmax = kcat [ET], where ET represents the total concentration of enzyme. Since enzyme β has a lower kcat value, its reaction must have a lower maximal velocity than the reaction catalyzed by enzyme γ. D. Enzyme γ has a greater binding affinity for its substrate than enzyme β. In biochemistry, the dissociation constant (Kd) is useful to quickly compare the binding affinities of enzymes to their substrates. The key is the word "dissociation." A low Kd value means that the ES complex does not readily dissociate into enzyme and substrate (in other words, it is relatively stable). The lower the Kd, the greater the enzyme-substrate binding affinity. The table shows us that enzyme γ has a lower Kd value than enzyme β.

Prokaryotes (including bacteria and Archaea) are defined by the absence of a nucleus and membrane-bound organelles.

are defined by the absence of a nucleus and membrane-bound organelles.

structural isomers

differ in the covalent arrangements of their atoms

Doppler equation

doppler shift = 2 x velocity of blood x transducer frequency x cos0 / propagation speed

non-conservative forces

forces that do work in such a manner that thermal energy is involved and mechanical energy is not conserved. examples are friction and fluid resistance along with horizontal pushing or pulling forces

Mendel's Law of Independent Assortment

genes found on different chromosomes are sorted into sex cells independently of one another

transmembrane domain

hydrophobic region of a transmembrane protein that anchors it in the membrane

biallelic expression

maternal and paternal inherited genes contribute to offspring phenotype

Eukaryotic cilia and flagella are composed of bundles of microtubules.

microfilaments. Microfilaments are composed of actin and are found in the cytoplasm of eukaryotic cells, as well as in muscle. Intermediate filaments are less dynamic than actin or microtubular filaments. They are not involved in ciliary or flagellar structure. Myosin is present in muscle and aids in the process of contraction.

lowest delta h

most dynamically favorable That is, when reactants are converted into products, the overall enthalpy change is the same whether it is done as one step or multiple steps.

conservative forces

path independent and do not dissipate the mechanical energy of a system

Leydig cells

produce testosterone

protic solvent

solvents with protons in solution, e.g. water or alcohol. large atoms tend to be better nucleophiles in here because they can shed the solvating protons around them and are more polarizable

sertoli cells

support spermatogenesis

For a light ray to totally internally reflect, rather than exit and refract (bend),

the light ray must strike the edge of the glass tube at an angle equal or greater than the critical angle.

granulosa cells

the majority of the cells surrouding an oocyte in a follicle. Granulosa cells secrete estrogen during the follicular phase of the ovarian cycle (before ovulation).

Where bacteria grow in a culture setting may depend on their metabolism. A classic example involves bacteria in a culture medium contained in a test tube with a loosely fitting cap. Oxygen would be expected to be present at the top of the tube and absent at the bottom.

therefore, obligate aerobes would only be found at the top, and obligate anaerobes only at the bottom. Facultative aerobes would be found throughout the tube, but with a greater distribution at the top due to the increased efficiency of aerobic metabolism. Aerotolerant anaerobes would be spread evenly throughout the tube.

Fluid regulation is crucial for the maintenance of life. There are two basic scenarios that the body can encounter: too little fluid and too much fluid. Having too little fluid in one's system manifests in three important ways: reduced blood volume (because relatively little water is present in the blood plasma), reduced blood pressure (a consequence of reduced blood volume—less liquid is present to exert pressure against the walls of the blood vessels), and increased blood osmolarity (the same solutes are present, but less solvent is available). Correspondingly, increases in the amount of fluid present in the system manifest as increased blood volume (because more water is present in the blood plasma), increased blood pressure (more blood volume means more pressure against the walls of the blood vessels), and decreased blood osmolarity (the same solutes are present, but more solvent is available).

wo major hormones respond to low fluid levels by increasing fluid retention: aldosterone (the main example of a class of steroid hormones known as mineralocorticoids) and anti-diuretic hormone (ADH), a peptide hormone that is also known as vasopressin. However, these two hormones have different mechanisms. Aldosterone works by increasing sodium absorption in the distal convoluted tubule and collecting duct of the nephron, which drives water absorption. Aldosterone also increases excretion of potassium and hydrogen ions in the urine. In contrast, ADH increases the permeability of the collecting duct to water, thereby increasing water absorption. Thus, ADH acts to reduce the osmolarity of blood by increasing the amount of water present without changing the solute levels, whereas aldosterone does not affect osmolarity because sodium reabsorption drives water absorption. Atrial natriuretic peptide (ANP; also known as atrial natriuretic factor [ANF]) is a hormone that the endocrine system uses to deal with the problem of excess blood volume. Essentially, it is the opposite of aldosterone. It is released in response to high blood volume and decreases sodium reabsorption in the distal convoluted tubule and the collecting duct, as well as increasing the glomerular filtration rate and inhibiting aldosterone release.