Biological and Biochemical Foundations of Living Systems

Given the locations of the MlyI cleavage site and recognition sequence, MlyI is NOT a: A. Type I restriction enzyme. B. Type II restriction enzyme. C. Type III restriction enzyme. D. Type IV restriction enzyme

A is correct. A restriction enzyme is an enzyme that cuts DNA at or near specific recognition nucleotide sequences, known as restriction sites. Type I enzymes cleave at sites remote from the recognition site; they require both ATP and S-adenosyl-L-methionine to function. Since the passage shows a cleavage site very near a recognition site, MlyI must not be a Type I enzyme. <--- based on the passage that I missed when reading! ________________ Level 3/Correct by GUESSING!!!!, lucky me ________________ B: Type II enzymes cleave within or at short specific distances from their recognition sites and often require magnesium. C: Type III enzymes cleave at sites a short distance from their recognition sites and require ATP (but do not hydrolyze it). S-adenosyl-L-methionine stimulates this reaction, but is not required. D: Type IV enzymes target modified (e.g. methylated, hydroxymethylated) DNA.

Given passage information, Giardia most likely transports which of the following proteins external to the organism itself? A. cwp3 B. Histone H4 C. β-(1-3)-N-acetyl-D-galactosamine D. ESV

A is correct. According to paragraph 2, the CWM is "composed of at least 3 cyst wall proteins (cwp1-3)," one of which is cwp3, choice A. The CWM is eventually attached to the intestinal cell membrane that the parasite inhabits. This suggests that there must be a mechanism for transporting this protein from the interior of the parasitic cell to the host cell membrane _____ Level 2 /Correct/ I read the passage thoroughly and got the answer from the pasage itself. BRAVO JAZZ! ______ incorrect: B: According to the passage, histone H4 is associated with the cwpgenes in the nucleus. C: The passage states that β-(1-3)-N-acetyl-D-galactosamine is a component of the cyst wall. D: According to the passage, ESV is used to modify and arrange cell wall components inside the parasitic cyst. The question asks for a component that is transported outside of the parasite.

The symptoms of botulinum took 12-18 hours to be observed in the patient. How is this best explained? A. Growing bacteria produce the toxin. B. Botulinum toxin is immediately filtered by the kidneys and excreted. C. Toxins can be absorbed through mucous membranes or respiratory tract. D. Most of the toxins are polar and they take time to cross the nuclear membrane.

A is correct. If growing bacteria produce the toxin, then it will take time for the bacteria to generate sufficient quantities of the toxin to cause illness. ____________ Level 2/Incorrect (D)/guessed! ____________ B: If the toxin were immediately filtered and excreted by the kidneys, the symptoms would not appear. C: This is a true statement, but it does not explain why the toxin's initial effect takes so long to manifest. D: The toxin has its effect on the neuromuscular junction and does not cross the nuclear membrane.

The catalyst for TGFβ production most likely plays a direct role in: A. gene silencing. B. ribosomal inactivation. C. transfer of an acetyl group from acetyl-CoA. D. gene activation.

A is correct. The passage states that HDAC1 catalyzes the production of TGFβ and acts to remove acetyl groups from lysine residues of histones. This is important to gene regulation because DNA is wrapped around histone proteins, and DNA expression is regulated by the addition and removal of acetyl groups. ***Deacetylation attracts DNA to histones more tightly, inhibiting transcription. In summary, histone acetylation generally serves to increase gene expression, while deacetylation, along with DNA methylation, acts to silence gene expression. (VERY IMPORTANT TO REMEBER AND UNDERSTAND THIS!) _____________________________________________ Level 3 1st attempt: Incorrect ( C) _____________________________________________ Incorrect explanations: B: Nothing in the passage suggests that HDAC1 has any effect on ribosomes. This is also seems unlikely, since HDAC1 must be found in the nucleus (where it acts on histones), but ribosomes are found outside the nucleus. C: When HDAC removes the acetyl group from lysine residues, it is transferring that group to CoA. D: Histone acetylation, not deacetylation, would allow for increased gene expression.

In vitro, free oligonucleotides, such as mRNA transcripts, have a short half-life in the cytosol. What mechanism is used to stabilize these molecules? A. Polyadenylation B. Phosphorylation C. Enzymatic digestion D. Active transport across the nuclear membrane

A is correct. The poly-A tail is important for the nuclear export, translation, and stability of mRNA. The tail is shortened over time, and, when it is short enough, the mRNA is enzymatically degraded. ____________________________ Level 2/Incorrect/ I guess the answer because my brain was on fire while reading the passage ____________________________ B: By definition, nucleotides are already phosphorylated. When the PO43- group is removed, the molecules are called nucleosides. C: This is the opposite of what the question is asking, as degradation would lead to a faster breakdown. D: Active transport across the nuclear membrane would not stabilize an mRNA transcript.

Which molecule could be used to detect toxin D in the cultures described in the passage? A. An antibody B. A phospholipid C. A radiolabeled thymine D. An antigen

A is correct. To recognize the antigens on the botulinum protein, an antibody could be used. Antibodies are built to recognize and bind specific proteins. ____ Level 2/Incorrect(B)/ I guessed! ____ Incorrect explanations: B: Phospholipids are largely hydrophobic molecules that contain long alkyl chains. They would not interact or bind with the target proteins. C: Radiolabeled thymine would be useful in identifying nucleotides, but not for identifying proteins. D: Antigens are surface proteins that serve as identifying markers on cells and molecules. They interact with antibodies and would not be able to bind or isolate the toxin. Do not confuse antibodies with antigens!

Polarity

A lack of electrical symmetry in a molecule. Charge differences on opposite ends of a structure. lone pair presents

Distillation

A process that separates the substances in a solution based on their boiling points If all are liquids, one may be able to utilize distillation, which aims to separate liquids by utilizing the difference between their boiling points. The liquids are initially held in the same round-bottom flask, termed the distilling flask. This flask is positioned above a heat source, typically a Bunsen burner. The top of the flask is connected to a column, which leads to a downward-sloping glass condenser over a receiving flask. The condenser is held within a glass casing through which cold water is pumped. As the round-bottom flask is heated, the liquid with the lower boiling point will begin to vaporize, and its vapor will travel up the column and re-condense to fall into the receiving flask. The eventual result is a receiving flask that is rich in the lower-BP liquid, while the distilling flask will still contain the liquid(s) with the higher BP. If boiling points are very high, a vacuum may also be used to lower atmospheric pressure, which lowers the boiling points of all substances involved.

Content Foundations: Bicarbonate Buffering

Acid-base buffers confer resistance to a change in the pH of a solution when hydrogen ions or hydroxide ions are added or removed to solution. An acid-base buffer typically consists of a weak acid and its conjugate base. The most important buffer to know for the MCAT is the bicarbonate buffer system, which is shown below. H2O (aq) + CO2 (g) ⇌ H2CO3 (aq) ⇌ H+ (aq) + HCO3− (aq) Carbonic acid (H2CO3) has the conjugate base of HCO3−. Buffers work because the concentrations of the weak acid and its salt are large compared to the number of protons or hydroxide ions added or removed. When protons are added to the solution from an external source, some of the bicarbonate in the buffer is converted to carbonic acid, using up the protons added; when hydroxide ions are added to the solution, protons are dissociated from some of the carbonic acid in the buffer, converting it to bicarbonate and replacing the protons lost. Buffers resist pH changes best when the pH values are at or near the pKa value for the acid/base used, because that is when the conjugate acid and base have equal concentrations. Optimal buffering occurs when the pH is within approximately 1 pH unit from the pKa value of the system. The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log [conjugate base] / [acid]. On Test Day, you may see the bicarbonate buffer concept tested from a chemical, biochemical, or biological perspective. For example, hyperventilation (rapid deep breathing) results in excess CO2 being expelled from the blood, causing the pH to rise. In response, the buffer needs to release more H+ to lower the pH back to physiological norms. An additional fact to be aware of is that other mechanisms in the body are also used to regulate pH, since carbonic acid works best at a pH below physiological conditions, because its pKa1 (pKa1 = 6.3, pKa2 = 10.3) is much lower than the normal pH of blood (7.4). Adapted from Next Step MCAT Chemistry and Organic Chemistry Ch.7, where more information on this topic can be found.

Which molecule is least likely to be able to travel via the macula communicans pathway? A. Insulin B. Cl- C. Na+ D. Lysine

Ais correct. In paragraph 3, the passage states that this system is devoted to the movement of ions and fluids. Insulin, a peptide hormone, would be too large to use this pathway. _________________ Level 1/correct/ I answered by common sense only w/o gettignt he evidence fromt he passage itself, which is WRONG! read carefully jaz! _________________ B, C: Cl- and Na+ are ions and will be able to move through the channels described in the passage. D: Amino acids, like lysine, are small molecules that will be able to move through the channels described in the passage.

Content Foundations: Amino Acid Structure and Properties

Amino acids include both an amino (-NH2) and a carboxylic acid (-COOH) functional group. This can refer to a wide range of structures, but the MCAT focuses on the 20 amino acids that are coded for in human cells to build proteins. Chemically, these amino acids are known as α-amino acids because they have a single carbon at the α position, adjacent to the carbonyl carbon in the -COOH group. The central α-carbon has four substituents: -NH2, - COOH, -H, and -R. The -R group, or side chain, is the only part that differs between amino acids and determines their individual properties. Amino acid properties include polarity, acid-base behavior, chirality, and more specific chemical behaviors. Regarding polarity, all 20 amino acids can be categorized as polar (hydrophilic) or nonpolar (hydrophobic). Polar amino acids tend to be located on the exterior of globular proteins, facing the watery environment. In contrast, nonpolar amino acids are generally buried in the interior of these proteins, protected from the aqueous cytosol or extracellular fluid. Certain polar amino acids have side chains that can act as acids or bases. This behavior results from the presence of certain functional groups: carboxylic acids in the two key acidic amino acids and basic nitrogen-containing groups in the three basic AAs. Other properties are specific to a small number of amino acids. For example, 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. Nonetheless, it is important to keep in mind that amino acid structure, properties, and chemical behavior encompass a very large amount of MCAT science content. To study these topics in greater depth, consult your biochemistry book or other MCAT sources.

Content Foundations: Cell Junctions

Amino acids include both an amino (-NH2) and a carboxylic acid (-COOH) functional group. This can refer to a wide range of structures, but the MCAT focuses on the 20 amino acids that are coded for in human cells to build proteins. Chemically, these amino acids are known as α-amino acids because they have a single carbon at the α position, adjacent to the carbonyl carbon in the -COOH group. The central α-carbon has four substituents: -NH2, - COOH, -H, and -R. The -R group, or side chain, is the only part that differs between amino acids and determines their individual properties. Amino acid properties include polarity, acid-base behavior, chirality, and more specific chemical behaviors. Regarding polarity, all 20 amino acids can be categorized as polar (hydrophilic) or nonpolar (hydrophobic). Polar amino acids tend to be located on the exterior of globular proteins, facing the watery environment. In contrast, nonpolar amino acids are generally buried in the interior of these proteins, protected from the aqueous cytosol or extracellular fluid. Certain polar amino acids have side chains that can act as acids or bases. This behavior results from the presence of certain functional groups: carboxylic acids in the two key acidic amino acids and basic nitrogen-containing groups in the three basic AAs. Other properties are specific to a small number of amino acids. For example, 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. Nonetheless, it is important to keep in mind that amino acid structure, properties, and chemical behavior encompass a very large amount of MCAT science content. To study these topics in greater depth, consult your biochemistry book or other MCAT sources.

Content Foundations: Vitamins, Minerals, Cofactors, and Coenzymes

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. 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.

Anchoring junctions

Anchoring junctions include adherens junctions, which are associated with cadherins. The essential idea behind anchoring junctions is that they connect cytoskeletal components of the cell with other cells and/or the extracellular matrix, thereby contributing to the overall structural stability of tissues. Adherens junctions involve cadherin-mediated connections between actin filaments and other cells and the extracellular matrix. Desmosomes also involve cadherin, but in this case cadherin connects intermediate filaments to other cells. Hemidesmosomes are junctions in which integrins connect the intermediate filaments of cells to the extracellular matrix

Content Foundations: Antibodies and Antigens

Antibodies let the body know when it needs to mobilize the immune response. To do this, they must recognize substances/cells that need to be eliminated and be recognized by other components of the immune system. The term "antigen" is used to refer to what antibodies recognize. There is no specific structural property that defines an antigen, although they often happen to be macromolecules (especially proteins) expressed on the surface of a cell or a viral envelope/capsule. However, external substances like pollen can also serve as antigens, causing pollen allergies. The structure of an antibody provides a bridge between these two functions. Antibodies have a Y-shaped structure consisting of two heavy chains and two light chains that are linked by disulfide bonds. Five classes of antibodies exist, classified according to the details of their heavy chains: immunoglobulin (Ig) A, IgD, IgE, IgG, and IgM. The "top" ends of the Y-shaped structure (that is, the part with both the heavy and the light chains) have a hypervariable antigen-recognizing area, and the rest of the antibody structure can be recognized by other cells of the immune system. The specific site on an antigen that is recognized by an antibody is known as the epitope. Extensive random recombination of the antigen-recognizing area of the antibody (also known as the paratope) allows the generation of antibodies that recognize potentially infinitely many types of antigens. Antibodies are used in the adaptive immune system, most notably by B cells. Antigen-antibody interactions are also used in many biotechnology-related applications, most notably western blotting, in which antibodies are used to visualize proteins of interest after gel electrophoresis.

Which of the following is NOT a function of the sympathetic nervous system? A. Increased heart rate B. Pupillary constriction C. Vasodilation D. Vasoconstriction

B is correct. Activation of the sympathetic nervous system triggers a fight-or-flight response. This includes several physiological responses: pupil dilation, increased rate and force of heart contraction, blood vessel dilation in skeletal muscle and constriction in gastrointestinal organs, and inhibition of peristalsis by the digestive tract. The parasympathetic system has the opposite effect. _______________ level 2/ Correct/ yes! _______________ Incorrect explanations: A: Increased heart rate is a valid response caused by sympathetic nervous system activation. C, D: The SNS causes vasodilation in skeletal muscle to ready the body for "fight or flight" and promotes vasoconstriction to the viscera, slowing down the digestive tract.

Chitin's flexibility and strength make it favorable as a biodegradable surgical thread. All of the following describe the portion of a chitin molecule shown below EXCEPT: A. carbohydrate. B. polypeptide. C. disaccharide. D. polymer.

B is correct. Chitin (C8H13O5N)n is a long-chain polymer of a N-acetylglucosamine, a derivative of glucose. The molecule shown comprises two of these rings attached via a β-1,4 linkage. As such, it can be classified as a carbohydrate, a polymer, and a disaccharide. It does not contain amino acid monomers joined via amide bonds, so it cannot be classified as a polypeptide. ________ Level 1/ correct/ I guessed and lucky!

Which of the following is an element of humoral immunity? A. Phagocytes B. Immunoglobulins C. T cells D. MHC I

B is correct. Humoral immunity is part of the body's adaptive immune response. It is provided by B cell activity, which promotes an antibody, or immunoglobulin, response. Antibodies can recognize polysaccharide, phospholipid, and nucleic acid antigens to help the body fight extracellular bacteria, viruses, and toxins. A, C, D: These are elements of cell-mediated immunity. __________ level 2/correct by guessing with much thinking! you are lucky but need to review more! ____________

Which of the following is most likely to use a protein channel to cross the eukaryotic cell membrane? A. Aldosterone B. Ca2+ C. O2 D. CO2

B is correct. The eukaryotic cell membrane is comprised of a phospholipid bilayer, cholesterol, and various transmembrane protein pores, channels, and gates. The largely hydrophobic nature of the lipid bilayer prevents ions and polar solutes from diffusing across the membrane, while allowing for the free movement of hydrophobic molecules. Calcium ions would be unlikely to pass directly through the cell membrane and thus require an ion channel ___________________ Level2/correct/ yes! ---------------- A: Aldosterone is a steroid hormone built from nonpolar cholesterol, which allows it to freely cross the membrane. C, D: Oxygen and carbon dioxide are small, nonpolar molecules, which allows them to freely cross the membrane.

In which of the following cell processes does thymidine kinase most likely play a role? A. Osmosis B. Mitosis C. Translation D. Conjugation

B is correct. The passage states that thymidine kinase is used to predict cancer activity. Cancer is a disease categorized by rapid, unregulated cell growth and replication. The name "thymidine" (a nucleotide base) also indicates that the enzyme has a key role in DNA synthesis and therefore in cell division. You should also know that the "kinase" part of the name indicates that the enzyme catalyzes the addition of a phosphate group to its substrate. ________ Level 2/incorrect/ answer by guessed! but if you read carefully! im sure you will get the answer, especially about cancer! _________________________ A: Osmosis refers to the movement of particles, usually water, across a membrane. This process is not associated with thymidine kinase. C: "Thymidine" resembles the word "thymine," a base typically only present in DNA. Translation does not directly involve DNA; instead, it is the production of protein from an mRNA template. D: Conjugation is the transfer of genetic material (plasmids) between bacterial cells by direct cell-to-cell contact or via a bridge-like connection between two cells. This process does not relate to thymidine kinase.

The structure of lysine is shown below. This molecule is best described as: A. aromatic. B. polar and basic. C. nonpolar and not aromatic. D. hydrophobic.

B is correct. The structure of lysine is dominated by a single carboxylic acid group (R(CO)OH) and two basic amine (NH2) groups. Therefore, it is best described as polar and basic. ____ Level 1 Incorrect (C) _____ A: Aromaticity requires a ring structure with a conjugated sigma- and pi-bond electron system. Lysine lacks a ring structure entirely. C, D: Lysine has several polar and ionizable funct

With which of the following do the arginine side chains found on histone proteins most likely interact A. Thymine groups on DNAB. Adenine groups on DNAC. Phosphate groups on DNAD. Oxyribose on RNA

C is correct. Arginine is a basic amino acid, meaning that it is positively charged at physiological pH. Positive species interact best with negative species, and of the answers listed, only phosphate groups are negatively charged. ________ Level 1 /Incorrect/ I was just guessing when answering the qs because it's actually an easy qs to answer by not reading the passsage. the main thing is re-read the fundamental concept of DNA.Histone. though I dont know the AAs yet. from my understanding, Phoshate groups on DNA has negative charges that attracted to positive charge in histones. they both attracted very strongly. __________

What is a likely neurological symptom caused by the toxin's effect on acetylcholine release? A. Tetanus B. Muscle spasms C. Flaccid paralysis D. Nausea

C is correct. By inhibiting acetylcholine release, the toxin interferes with nerve impulses and causes flaccid (sagging) paralysis of muscles. The presence of acetylcholine at the neuromuscular junction is required for skeletal muscle contraction. Flaccid paralysis is a neurological condition characterized by weakness or paralysisand reduced muscle tone without other obvious cause (e.g., trauma). This abnormal condition may be caused by disease or by trauma affecting the nerves associated with the involved muscles. ______________________________________ Level2/Incorrect/ we can answer the qs by not looking at the passage but I was just guessing when it comes to this qs. _______________________________________ Incorrect explanations: A: Inhibiting acetylcholine release would not cause the spastic paralysis seen in tetanus. B: Inhibiting acetylcholine release would not cause repeated depolarization/activation of the muscle. D: Inhibiting acetylcholine release would have no effect on nausea levels, which are controlled by the medulla.

In patients with a certain medical condition, the intercellular path described in the passage is damaged in the large intestine, causing lower levels of transport. What change would be expected? A. Increased Cl- in the lumen of the colon B. Decreased Cl- in the lumen of the duodenum C. Decreased waste fluidity in the colon D. Increased waste fluidity in the colon

C is correct. Paragraph 3 states that this pathway can be used to transport ions and fluids into the lumen of the large intestine. The addition of fluid softens the waste material and allows it to move. Without this mechanism, fewer ions will be secreted into the lumen and less water will follow. This will results in a dehydrated, more solid waste in the colon. _________________ (Level 2/ correct based on reading the passage in p#3/ good job jaz!!! keep it up!) ______ A: Lower levels of transport will result in lowered Cl- levels in the lumen of the colon. B: The question stem referenced only the large intestine. The duodenum is the region of the small intestine that is most proximal to the stomach. D: This would be caused by increased, not decreased, chloride ion transport.

Patients with excess fat are more likely to require larger therapeutic doses of which vitamin? A. Vitamin B1 B. Vitamin C C. Vitamin D D. Vitamin B3

C is correct. The lipid-soluble vitamins are A, D, E, and K. The water-soluble vitamins are B and C. Greater amounts of subcutaneous fat sequester more of the lipid-soluble vitamins and lower their release into the circulation. Thus, excess fat increases the initial dose of vitamin required to achieve a particular effect. **LIPID SOLUBLE VITAMINS= ADEK ________________________________________________________ Level 2/ Correct/ Yes! i learned from Dr. Gray's podcast before ______________________________________________________ A, B, D: Since the vitamins listed in these answer choices are water-soluble, their absorption is unlikely to be affected by excess fat.

Some proteins and antibodies are susceptible to pH damage. In order to prevent this damage, what would be the most efficient and effective additional step to perform at the end of the elution process? A. 0.1 M glycine•HCl, pH 3.0 B. 0.1 M NaCl, pH 7.0 C. 1 M Tris•HCl, pH 8.5 D. 0.7 M KOH, pH 13.85

C is correct. The passage describes the final step of the process as lowering the pH to elute the proteins off of the beads. To reduce the risk of damage, we need to neutralize any excess acid. 1 M Tris•HCl, pH 8.5 is the buffer that will most effectively dissociate most protein:protein and antibody:antigen binding interactions without permanently affecting protein structure. Eluted protein fractions are best neutralized immediately by addition of alkaline buffers. ______________ Level 2/Incorrect/ Answered by guessing because I thought 7 ph is the neutral number for basic and acid rxn ______________ A: Note that the question asks for an additional step to perform after the elution process has taken place. Acid was already added to elute the proteins from the beads, so the solution in choice A will only continue to lower the pH, potentially damaging the molecules. B: This solution has a neutral pH and will not be able to neutralize the acidity of the final eluate. D: This solution is far too strong; a very high pH can damage the molecules as much as an excessively low pH.

The muscle of the dual layered system described in the passage is: A. striated. B. multi-nucleated. C. regulated by the parasympathetic nervous system. D. innervated by the somatic nervous system.

C is correct. The passage states that the dual layered system is comprised of smooth muscle. Smooth muscle is uni-nucleated, non-striated, and under the control of the autonomic nervous system (ANS). Its activity is also regulated by parasympathetic and sympathetic branches of the ANS. ______________________________ Level 1/incorrect (B) and answered by guessing -----------------------

Content Foundations: Cell Junctions

Cell junctions describe ways in which neighboring cells interact with each other to form structures. There are three main types of cell junctions that you should be aware of: anchoring junctions, gap junctions, and tight junctions.

Chromatography

Chromatography is a broad set of separatory techniques based on relative affinity, or tendency for a compound to attract to a certain solvent or structure. Specifically, sample molecules vary in their affinities for a mobile (moving, typically solvent-based) phase versus a stationary (static) phase. In column chromatography, the stationary phase is a vertical column packed with an adsorbent with carefully chosen properties. This adsorbent can attract sample molecules based on charge, size, or affinity for specific ligands.

Content Foundations: Membrane Transport

Content Foundations: Membrane Transport The plasma membrane of eukaryotic cells is primarily composed of a lipid bilayer of amphipathic phospholipids with hydrophilic heads and hydrophobic tails. It also contains cholesterol and membrane proteins. Transmembrane proteins are membrane-spanning proteins with hydrophilic cytosolic and extracellular domains and a hydrophobic membrane-spanning domain. Additionally, peripheral proteins are only transiently attached to integral proteins or peripheral regions of the lipid bilayer, and lipid-anchored proteins are covalently bound to membrane lipids without actually contacting the membrane directly. Membrane transport is accomplished through several mechanisms. Some molecules, like small gases, can directly diffuse through the membrane. This is known as simple diffusion, and is an example of passive transport because no energy is necessary. Osmosis is a type of simple diffusion in which water moves in or out of the cell to attempt to equalize concentrations of solute. Facilitated diffusion is another form of passive transport where no energy is necessary because molecules diffuse down their concentration gradient, but a transmembrane channel is necessary because the molecule may be too large or polar for simple diffusion. Ions are often transported through facilitated diffusion, and aquaporins are facilitated diffusion channels for water that augment osmosis. In primary active transport, energy is used directly to move a solute against its gradient through a transmembrane channel. Secondary active transport is a more complicated system in which the energy stored in an electrochemical gradient established via primary active transport is used to facilitate the movement of a solute. An example is the sodium-calcium exchanger, which allows three Na+ ions to flow down their concentration gradient, which was previously established by a primary active transport mechanism, into the cell, while transporting one Ca2+ ion out. Endocytosis is used to ingest larger materials. It is divided into pinocytosis and phagocytosis. In pinocytosis, cells engulf liquid substances, while in phagocytosis, they engulf solid particles. The basic pathway of endocytosis involves recognition of a target molecule at the plasma membrane, followed by invagination and the formation of a vesicle on the inside of the cell. Exocytosis can be thought of as endocytosis in reverse. It is used to release hormones, neurotransmitters, membrane proteins and lipids, and other mate

Content Foundations: Neuromuscular Junction

Content Foundations: Neuromuscular Junction Muscles receive the signal to contract at the neuromuscular junction. An action potential propagates down a motor (efferent) neuron. The neurotransmitter acetylcholine is then released into the neuromuscular junction. Acetylcholine binds to receptors on the cell membrane, which is known as the sarcolemma in muscle cells, and the sarcolemma then depolarizes in response. This results in an action potential, and when the action potential reaches the sarcoplasmic reticulum, Ca2+ is released into the sarcoplasm (the term used for the cytoplasm in muscle cells). Once in the sarcoplasm, Ca2+ can bind to troponin, which allows contraction to take place.

The functions of the ESVs of Giardia most closely resemble what cell organelle? A. Lysosomes B. Golgi apparatus C. Nucleolus D. Centriole

Correct answer is B. Golgi apparatus because based ont he passed the function of ESV is to modify or partition cyst wall cargo during secretory transport. and this reflect a similarity function of Golgo apparatus. ____ Level 1 / Correct / I can answer this very easily during the test ___ Incorrect explanations: A: Lysosomes function as the digestive system of the cell, serving both to degrade material taken up from outside the cell and to digest obsolete components of the cell itself. C: The nucleolus is the site of cellular rRNA transcription and processing; it also serves as the location of ribosome assembly. D: The centriole, which helps organize microtubules, is a cylindrical structure consisting of nine triplets of microtubules in the centrosomes of most animal cells. **remember the PMAT process! centriole plats a central role during this process :) review the PMAT!

Monocytes, in conjunction with epithelium-derived factors, can act to facilitate which biological process? A. Fatty acid oxidation B. Transfection C. Lipid synthesis D. Host immune response

D is correct. Monocytes are the largest type of white blood cells. Factors secreted by the epithelium help form a cascade that leads to accumulation of immune cells at the site of an injury. ___________________________________________ Level 1 1st attempt: Correct ___________________________________________ Incorrect explanations: A: While oxidized fatty acids have been shown to aid in monocyte activation, monocytes do not facilitate fatty acid oxidation. B: Transfection is the process of introducing nucleic acids to cells, usually by way of a non-viral vector. C: Synthesis of fatty acids takes place in the cytoplasm and involves 1) initiation via the formation of acetoacetyl carrier protein and 2) an elongation cycle where carbon units are successively added to the growing chain. Neither of these processes involve monocytes.

Which nucleophile was likely used to prepare the oligonucleotides used in the purification procedure? A. NaOH B. (CH3)3CCl C. HCN D. NaN3

D is correct. Paragraph 2 states that the oligonucleotides are azide-modified. Azide (N3-) is a powerful nucleophile which can readily add to available electrophiles, such as nucleotide molecules. ______ level 2/incorrect (c)/ answered by guessing!what i had in mind when reading the qs is HCN is "a strong molecule" ____ A, B, C: These reagents do not include N3-.

Which of the following will cause a blood pH of 8.2? A. Decreased O2 concentration B. Decreased tidal volume C. Increased H2O concentration D. Increased respiratory rate

D is correct. Physiological pH is 7.4; thus, 8.2 represents an increase in pH, or alkalosis. The bicarbonate buffer system of the blood is given by: H+ (aq) + HCO3- (aq) ↔ H2CO3 (aq) ↔ H2O (l) + CO2 (g) According to Le Châtelier's principle, a decrease in products will cause the equilibrium to shift to the right. An increased respiratory rate will cause excess CO2 (a product) to be exhaled. The resulting rightward shift will cause a drop in H+ concentration and, thus, an increase in pH. _________ Level2/Correct/ Yes - i have learned this before from dr.Gray podcast! ___________ Incorrect explanations: A: O2 concentration itself does not directly impact blood pH, but the hypoventilation that could cause decreased O2 concentration would cause blood pH to fall below 7.4. (ACIDOSIS) B: This will cause blood pH to drop below 7.4 due to decreased exhalation of CO2. (Normal tidal volume is approximately 500 mL, and includes the volume of air that fills the alveoli in the lungs, as well as the volumeof air that fills the airways.) decreased tidal volume means decreased exhalation CO2??? need to clarify this*** C: Adding more water to the blood would not cause the system to become more basic.

A molecule enters a cell and creates pores in the inner mitochondrial membrane. Will oxidative phosphorylation continue to generate ATP? A. Yes, because the cell membrane is still complete. B. Yes, because ATP is required for cell function. C. No, because there is no available oxygen. D. No, because the proton gradient will be dissipated.

D is correct. The energy released by oxidative phosphorylation and used to make ATP is generated from potential energy in the form of an H+ gradient, which creates an electrical potential across the mitochondrial membrane. If this membrane is disrupted, this gradient cannot be maintained. Without this gradient, oxidative phosphorylation cannot continue to produce ATP. _________ Level 1/incorrect (A)- towards the end, I am already tired and can't concentrated _________ A: Although the cell membrane is intact, oxidative phosphorylation requires intact mitochondial membranes. B: While it is true that ATP is required for cell function, oxidative phosphorylation cannot provide that ATP if the mitochondrial membrane has been disrupted. C: Nothing in the question suggests that there is a shortage of oxygen as a result of toxin invasion.

Studies show that macula communicans are common but not found in tissues such as the gall bladder and skeletal muscle. One possible explanation for this could be that these organs are: A. not under autonomic nervous control. B. not involved in digestion of materials. C. not involved in secretion of materials. D. not involved in absorption of materials.

D is correct. The passage states that gap junctions allow for the exchange of ions and fluids. One such reason for their absence would be that the organs in question have no need for absorptive mechanisms. _____ Level 2 /incorrect (2) because macula found in intestine which is part of the digestion tract (which is also wrong). ____________ A: The gall bladder is under autonomic control, while skeletal muscle is under somatic control. B, C: The gall bladder is directly involved in secreting bile for digestion, but neither the gall bladder nor skeletal muscle is involved in the absorption of nutrients.

What is the expected ploidy of a cyst that has completed an encystation? A. 2nB. 4nC. 8nD. 16n

D is correct. The protozoa will typically be diploid, and as stated in the second paragraph, cyst formation involves a round of DNA replication, followed by nuclear replication, followed by a second round of DNA replication. This process is illustrated below, and the presence of 4 4n nuclei yields a final total ploidy of 16n. _________ Level 3 Incorrect comment: I miss understood the 2nd passage and just quickly looking for an answer which is 4 that it stated on the passage that is very appealing for me to choose! READ CAREFULLY (Next step passage)

The easiest method to separate the two subunits of the botulinum protein for subsequent analytical purposes would be: A. gas chromatography.B. mass spectrometry.C. thin-layer chromatography.D. size-exclusion chromatography

D is correct. The second paragraph states that the two subunits are different in size (100 kDa vs. 50 kDa). Size-exclusion chromatography could therefore be used to separate the two proteins by size. ______ Level 2/correct/I guessed and lucky! review the separation system) _____ A: Gas chromatography separates molecules based on affinity and would thus be less useful than size-exclusion chromatography, which separates based on size alone. Gas chromatography also requires that the molecules be vaporized, and typical proteins are much too large to be easily vaporized. B: Mass spectrometry is used to measure the size of one molecule and would not be useful to carry out a separation of two molecules. Subjecting a sample to mass spec ionizes the molecule and breaks it into smaller ion fragments. After conducting such a procedure, you would not be left with two samples that could be subjected to further analysis. C: Thin-layer chromatography separates molecules based on affinity and would not be useful to separate molecules by size.

Which step of the procedure corresponds to the release of DNA-coupled conjugates? A. B B. C C. E D. D

D is correct. We are looking for the step in which the DNA-coupled antibodies are released. Paragraph 2 outlines each step of the procedure shown in Figure 1. Beginning with step A, DNA oligonucleotides are hybridized to the oligonucleotides in the mixture. In step B, agarose beads are used to capture the biotinylated capture oligonucleotides, both in the form of conjugates and free oligonucleotides. In step C, the unconjugated antibodies are removed by washes. In step D, the MlyI enzyme is used to cleave the captured oligonucleotide hybrids, allowing both conjugates and oligonucleotides to be eluted from the solid support. Thus, step D is our answer. ****At that point, you should answer the question and move on. However, note that in step E, the eluate is then incubated with protein G beads, while free oligonucleotides are removed by washes. Finally, in step F, purified conjugates are eluted from the solid support. ________ Level 2/Correct by guessing! luck you jazz! _______

Content Foundations: Key Organelles

Eukaryotic cells contain several membrane-bound organelles. The three most commonly-tested organelles on the MCAT are the nucleus, mitochondria, and ribosomes. The nucleus contains DNA and a sub-organelle known as the nucleolus, which is responsible for ribosome assembly. Mitochondria are sometimes known as the powerhouse of the cell; they are self-replicating organelles with their own DNA that are the location of the citric acid cycle (kerb cycle) and oxidative phosphorylation in eukaryotes. Ribosomes themselves are not membrane-bound organelles, and they are not unique to eukaryotes (although prokaryotic ribosomes have a different structure). They play the crucial role of protein translation. Once proteins are synthesized, the Golgi apparatus modifies them and packages them into membrane-bound vesicles that are then sent to the ultimate destination of the proteins. In addition, the endoplasmic reticulum is a net-like organelle that extends out from the nuclear membrane. It is divided into the rough ER and the smooth ER. The rough ER is rough because it is covered with ribosomes, which are the site of protein synthesis. In contrast, the smooth ER does not have ribosomes, and is involved in lipid metabolism (both synthesis and breakdown), the production of steroid hormones, and detoxification. Lysosomes are the garbage disposal system of the cell; material from outside the cell enters the lysosomes through endocytosis, while material from inside the cells enters through autophagy. (digestive system) They contain hydrolases that operate best at acidic pH levels, and the lysosomes are therefore kept at a pH of 4.5-5.0. Finally, peroxisomes play a major role in the metabolism of very-long-chain lipids by breaking them down to medium-chain lipids that are transported to the mitochondria for further processing and play a role in detoxification of substances such as ethanol.

centrifugation

Finally, centrifugation utilizes a rapidly spinning apparatus to separate particles by density. More dense particles, such as cells, gravitate toward the bottom of the spun tube, while less dense substances remain at the top in a liquid termed the supernatant. This liquid can then be poured off, and further separation or analysis can be conducted.

Content Foundations: Small and Large Intestine

For the MCAT, it is important to understand the basic structures and functions of human organ systems, including the digestive system. A key component of the digestive tract is its two tubelike intestines: the small intestine and the large intestine. The small intestine, located adjacent to the stomach, is the first of these structures. Partially digested food - at that point termed chyme - enters the small intestine from the stomach through the pyloric sphincter. This intestinal system has three parts: the duodenum (closest to the stomach), the jejunum, and the ileum (DJE). As chyme enters the duodenum from the acidic stomach, it is first neutralized by bicarbonate ions secreted by intestinal cells. In addition, the small intestine also performs two key functions: it continues and completes the chemical digestion (or enzymatic breakdown) of dietary molecules, and it begins the absorption of the product molecules into the body. The first of these functions is accomplished by enzymes, including trypsin, pancreatic amylase, and disaccharidases, as well as bile, a non-enzymatic substance that facilitates the digestion of fats. The second is assisted by the small intestine's large surface area, or "brush border," which is specialized for nutrient absorption. In contrast, only absorption - not digestion - takes place in the large intestine. This structure is found between the small intestine and anus, and it serves as a "last resort" for the absorption of already-broken-down nutrients. The large intestine also absorbs a large amount of water, which helps transition watery chyme into solid feces. Finally, this intestine is host to a huge community of intestinal bacteria, known as gut microbiota. Some of these bacteria provide their human host with benefits, most notably the synthesis of vitamin K, which is necessary for human health.

Gap junctions

Gap junctions are formed by Connexin proteins, which connect cells in a way such that diffusion can take place between them, enabling communication, without involving direct contact between the cytoplasmic fluids of each cell. Gap junctions are relatively less common, but they play certain crucial roles within the body. The most important example for the MCAT is in cardiac muscle, where gap junctions allow cells to contract at the same time.

Content Foundations: Nucleophilicity

In organic chemistry, reactions happen when a bond is formed between an electrophile and a nucleophile. Electrophiles are species that 'want' more electrons. Often, electrophiles are positively charged or partially positively charged. In contrast, nucleophiles are species that have an excess of electrons, and therefore 'want' to use those electrons to form a bond with an electrophile. This is known as nucleophilic attack. Nucleophiles can often be recognized by the presence of at least one free pair of valence electrons, although nucleophilic attack can also take place using electrons from a pi or sigma bond. The factors that make a molecule a strong nucleophile include greater electron density—that is, more lone pairs and, if possible, more charge—and the lack of steric hindrance.

Content Foundations: Ploidy, Aneuploidy, and Nondisjunction

Ploidy refers to how many copies of each chromosome a cell has. In humans, the vast majority of cells are diploid (2n), meaning that they contain two copies of each chromosome (except for the sex chromosomes; females have two X chromosomes and males have an X and a Y chromosome). Such cells are known as somatic cells—that is, the cells of the body. In contrast, germ cells (i.e., ova and spermatozoa) are haploid (n), meaning that the only have a single copy of each chromosome. Aneuploidy results from having too many or too few copies of a given chromosome. This results from nondisjunction in anaphase during cell division. Having only one copy of a chromosome is known as monosomy, and having three copies is known as trisomy. Aneuploidy is commonly discussed as occurring in meiosis, and indeed, this is the only way for aneuploidy to be inheritable. For this reason, nondisjunction during meiosis is the cause of aneuploidies such as Down syndrome (trisomy 21) or Turner syndrome (monosomy X). However, nondisjunction during mitosis can also occur, and this is extremely common in cancer cells. *nondisjunction: the failure of one or more pairs of homologous chromosomes or sister chromatids to separate normally during nuclear division, usually resulting in an abnormal distribution of chromosomes in the daughter nuclei.

Content Foundations: Restriction Enzymes

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 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. Adapted from Next Step MCAT Biology Ch. 5, where more information on this topic can be found.

Content Foundations: Separations

Separation techniques are widely used in organic chemistry to prepare purified substances for analysis or reaction. The best technique for a given scenario often depends on the phases of the substances being separated.

Content Foundations: Sympathetic and Parasympathetic Nervous Systems

The autonomic nervous system, which controls involuntary responses in the body, including things like sweating, blushing, and pupil dilation, is divided into the sympathetic and parasympathetic systems. The sympathetic nervous system stimulates the body in the classic "fight or flight" response, mediated by hormones such as epinephrine and norepinephrine. If the body needs to get ready for action, it will dilate the pupils, raise the heart rate, and increase blood flow to skeletal muscles to prepare for sudden action. By contrast, the parasympathetic nervous system is the "rest and digest" system that increases blood flow to the digestive system, slows the heart rate, constricts the pupils, and generally exerts actions opposite to those of the sympathetic nervous system. In turn, the autonomic nervous system is part of the peripheral nervous system, which describes the nervous system throughout the body and is distinguished from the central nervous system, which corresponds to the brain and spinal cord. The peripheral nervous system is also divided into the visceral nervous system, which modulates the digestive system, and the somatic nervous system, which connects to skeletal muscle to allow for voluntary movement.

Content Foundations: Gene Expression

The central dogma of molecular biology states that information is passed from DNA to RNA to protein. This means that when a cell needs more of a certain protein, it can increase the degree to which the gene corresponding to that protein is transcribed. Transcribing more or less of a gene in response to the cell's needs is known as gene expression. It plays a major role in the differentiation of organs in multicellular organisms, and can also vary on shorter time scales in response to changing environmental conditions. The details of how gene expression is regulated in eukaryotes are quite intricate, but you should be aware of some key concepts for the MCAT. Promoters are regions of DNA that lie upstream to a given gene and initiate transcription by binding specific transcription factors that contribute to the binding of RNA polymerase. Additionally, expression is upregulated by enhancers, which are DNA sequences that can be located further from the gene of interest, and work by binding transcription factors that twist DNA into a hairpin loop, bringing distant regions into close proximity for transcription to begin. Silencers are the opposite of enhancers in eukaryotic cells; they are regions of DNA to which transcription factors known as repressors bind. Additionally, the methylation of C and A residues can reduce transcription. Methylation is associated with epigenetics, which refers to inheritable phenotypic changes involving mechanisms other than the alteration of the genome itself. Gene expression can also be regulated on the level of nucleosomes (i.e. chromatin and histones). Acetylation promotes transcription by attaching acetyl groups to lysine residues on histones, making them less positively-charged and causing a looser wrapping pattern that allows transcription factors to access the genome more easily.( LOOSEN THE WRAPPER SURROUD THE HISTONE WHICH WILL INCREASE GENE EXPRESSION) Finally, non-coding RNA plays a role in gene expression. MicroRNA (miRNA) strands are single-nucleotide strands incorporated into an RNA structure with a characteristic hairpin loop, while small interfering RNA (siRNA) molecules are short and double-stranded. Both tend to be approximately 22 nucleotides in length, and silence genes by interrupting expression between transcription and translation.

Content Foundations: Histones and DNA Packaging

The human genome (The human genome is the complete set of nucleic acid sequences for humans, encoded as DNA within the 23 chromosome pairs in cell nuclei and in a small DNA molecule found within individual mitochondria.) contains approximately 3 billion base pairs. To fit into human nuclei, which are generally about 6 μm (6 × 10−6 m) in size, DNA must be compressed. Subdividing the genome into linear chromosomes accomplishes some of this task, but the rest of the job is done by histones and chromatin. Histones are proteins that act as spools for DNA to wind around. They are composed of various subunits known as H1, H2A, H2B, H3, and H4. The core of a histone contains two dimers of H2A and H2B and a tetramer of H3 and H4, while H1 serves as a linking unit. Approximately 200 base pairs of DNA can be wound around a histone, and the complex formed by DNA and a histone is known as a nucleosome. The phrase "beads on a string" is often associated with the appearance of nucleosomes under electron microscopy, and chromatin refers to the structure formed by many nucleosomes. Two distinct forms of chromatin exist: euchromatin and heterochromatin. Euchromatin is a loose configuration that is difficult to see under light microscopy and allows DNA to be readily transcribed. Throughout interphase (i.e., most of the cell cycle), DNA generally exists as euchromatin. Heterochromatin is the tightly coiled, dense form of chromatin that is visible during cell division and is present to a lesser extent even during interphase. On a biochemical level, charge plays a major role in the interactions between histones and DNA. Histones are highly alkaline and are positively charged at physiological pH, which facilitates their interaction with the highly negatively charged phosphate groups on the backbone of DNA. Modifications like acetylation of histones reduce that positive charge, making histones interact with DNA less closely, which in turn facilitates transcriptional activity.

Content Foundations: Immune System

The immune system refers to the complex set of mechanisms that the body uses to protect itself against foreign invaders and malfunctioning cells originating from the body itself. The highest-level distinction in the immune system is between the innate (or non-specific) immune system, which responds generally to threats but does not learn to recognize specific foreign bodies/molecules, and the adaptive immune system, which does. The non-cellular component of the innate immune system includes anatomical barriers and signaling molecules such as cytokines and complement proteins, while the cellular component includes a range of white blood cell types (leukocytes) that play various roles in responding to threats. White blood cells include neutrophils, lymphocytes, monocytes (which differentiate into macrophages or dendritic cells), eosinophils, basophils, and mast cells. The various components of the innate immune system can act independently or be coordinated in the process of inflammation. The adaptive immune system includes B cells and T cells, both of which are lymphocytes that are produced in the bone marrow and mature in the lymphatic system. B cells recognize antigens and secrete large amounts of antibodies in response. The human body utilizes five classes of antibodies: immunoglobulin (Ig) A, IgD, IgE, IgG, and IgM, which differ in the details of their heavy chains. This response is known as humoral immunity. In contrast, T cells correspond to the cell-mediated branch of the adaptive immune system. T cells, which mature in the thymus, recognize cells that were originally self, but have been damaged by viral infections or have malfunctioned in ways likely to turn them into cancer cells. Then, various subgroups of T cells either directly attack compromised/foreign cells or mobilize responses to them based on antigen fragments that are presented by major histocompatibility complex (MHC) class I and II.

Content Foundations: RNA Transcription

The process of going from DNA to RNA—more specifically, messenger RNA (mRNA)—is called transcription. Transcription takes place in the nucleus, and it results in the creation of an mRNA copy of a gene that can then be transported to the cytosol for translation into a protein. The DNA helix must be unzipped for transcription to take place, which means that some of the same machinery used for DNA replication has to be engaged, especially enzymes like helicase and topoisomerase. RNA polymerase is the enzyme responsible for RNA synthesis. In eukaryotes, it binds to a promoter region upstream of the start codon with the assistance of transcription factors, the most important of which is the TATA box. RNA polymerase travels along the template strand in the 3'-5' direction, synthesizing an antiparallel complement in the 5'-3' direction. The template strand is known as the antisense strand, and the opposite strand is known as the sense strand, because it corresponds to the codons on the mRNA that is eventually exported to the cytosol for translation. The immediate product of transcription in eukaryotes is not mRNA, but heterogeneous nuclear RNA (hnRNA). hnRNA must undergo a set of post-transcriptional modifications to become mRNA. Examples commonly tested on the MCAT include the 3' poly-A tail, the 5' cap, and splicing. The 3' poly-A tail is a string of approximately 250 adenine (A) nucleotides added to the 3' end of an hnRNA transcript to protect the eventual mRNA transcript against rapid degradation in the cytosol. The 5' cap refers to a 7-methylguanylate triphosphate cap placed on the 5' end of an hnRNA transcript. Similarly to the 3' poly-A tail, it helps prevent the transcript from being degraded too quickly in the cytosol, but it also prepares the RNA complex for export from the nucleus. In splicing, noncoding sequences (introns) are removed and coding sequences (exons) are ligated together. (Remember that exons are expressed). Each gene normally has multiple distinct exons that can be ligated in different combinations; that is, if a gene had a set of four exons named A, B, C, and D, possible alternate splicing combinations could include ABCD, ABC, ACD, ABD, BCD, and so on. This dramatically increases the amount of different, but related proteins that can be expressed from a single gene. Splicing explains why there are over 200,000 proteins in the human body, but only approximately 20,000 genes. Splicing is carried out by the spliceosome, a combination of small nuclear RNAs (snRNAs) and protein complexes.

Transcription

There are 3 main stages: 1)Initiation 2)Elongation 3)Termination **refer this to your note in the binder mcat!

Content Foundations: Biomolecules

There are four main classes of biomolecules: amino acids/proteins, lipids, carbohydrates, and nucleic acids. All of these classes except for lipids can occur as polymers(A polymer is a large molecule, or macromolecule, composed of many repeated subunits.), as proteins can be seen as polymers of amino acids, DNA and RNA are polymers of nucleic acids, and compounds such as starch, cellulose, and glycogen are polymers of carbohydrates. Amino acids contain a central carbon to which -NH2, -COOH, -H, and -R groups are added, and they combine by peptide bonds to form dipeptides, tripeptides, oligopeptides, and proteins. The key functionality of an amino acid is determined by its side chain (-R). Proteins are the building blocks of the body, and are involved in structural and signaling roles. Lipids are nonpolar (or amphipathic, with both polar and nonpolar areas) molecules that play roles in signaling, structural functions, and energy storage. The main categories of lipids that you must know for the MCAT are fatty acids and the derivatives thereof (triacylglycerols, phospholipids, and sphingolipids), cholesterol and its derivatives (steroid hormones and vitamin D), prostaglandins, and terpenes and terpenoids. Carbohydrates are important for energy storage and are used in metabolism. They are made up of carbon, hydrogen, and oxygen, often in the ratio Cx(H2O)y. Important monosaccharides include glucose, fructose, and galactose. Key disaccharides are sucrose (glucose + fructose), lactose (glucose + galactose), and maltose (glucose + glucose). Polysaccharides (starch and glycogen) are polymers of glucose that are used for energy storage in plants and animals, respectively. Nucleic acids are involved in the storage and transmission of biological information. They are made up of nucleotides, which have three components: a nitrogenous base, a five-carbon sugar (ribose in RNA and deoxyribose in DNA), and at least one phosphate group. RNA contains uracil (U) instead of thymine (T) and is generally single-stranded, whereas DNA is generally double-stranded. DNA is used for the long-term storage of genetic information, while RNA is used for gene expression.

After injury-induced cataract formation has begun, which of the following are LEAST likely to be found in nearby monocytes? A. TGFβ transporters B. TGFβ receptors C. Tight junctions D. Cytokines

Tight junctions is the correct answer. Tight junctions are areas where the membranes of two closely-joined cells adhere tightly and form a near-impenetrable barrier. This is a hallmark of epithelial barriers and is not likely to be a property of nearby monocytes. In addition, the passage says that the epithelial cells are undergoing a transition into mesenchymal cells (EMT) during this process. Thus, the number of epithelial cells would be decreasing as they turn into mesenchymal cells, which do not have such tight junctions. __________________________ Level 2 1st attempt : Incorrect (B) __________________________ Incorrect explanations: A: The passage states that monocytes export their TGFβ once it is synthesized. The cells must have a mechanism to transport this TGFβ into the aqueous humor. B: 5- TGFβ receptors are involved in recognizing TGFβ molecules. According to the passage, monocytes are able to recognize and respond to the presence of TGFβ to aid in the inflammatory response. This means the cells must have functional receptors on their surface. D: Cytokines are small proteins that are important to cell signaling processes, like inflammation and immune response. Cytokines modulate the balance between humoral and cell-based immune responses, and they regulate the maturation, growth, and responsiveness of particular cell populations.

Content Foundations: Cancer

While knowledge about specific diseases is generally not required for MCAT success, there are notable exceptions to this rule. One such exception is cancer, which can be summarized as abnormal gene expression. To make sense of this statement, let's review the steps involved in cancer development, or oncogenesis. The term "tumor" describes any abnormal proliferation of cells. Benign tumors remain localized, whereas malignant tumors (which are what the term "cancer" properly refers to) can invade other organs and tissues in the body in a process called metastasis. The first step in oncogenesis, tumor initiation, involves changes that allow a single cell to proliferate abnormally. This means that the cell must develop the ability to bypass regulatory steps of the cell cycle that normally help to limit mitotic proliferation. Tumor progression occurs as a cell develops the ability to proliferate even more aggressively, such that its descendants are selected for and come to predominate the growing tumor. In addition, malignant cells often undergo mutations that promote their own growth and the development of blood vessels to feed them (angiogenesis). Oncogenesis is most often associated with mutations that occur by random chance (and elude the normal DNA repair machinery in the cell) or as a result of mutagenic compounds known as mutagens or carcinogens. (Examples of mutagens include ultraviolet light and certain chemicals, such as reactive oxygen species.) These mutations alter the functionality of crucial genes in the cell. However, oncogenesis is also associated with dysregulation of gene expression, as the abnormally elevated expression of genes involved in growth and proliferation can help contribute to the development of a tumor. The genes involved in oncogenesis can be divided into two groups: oncogenes and tumor suppressor genes. The basic difference between them is that oncogenes function to promote abnormal growth and proliferation, leading to cancer, while tumor suppressor genes function to prevent tumorigenic properties. 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.

Recrystallization

is used to purify a solid product that contains impurities. This process involves the dissolution of the solid in a solvent and subsequent heating. The solid then dissolves and is cooled, causing it to solidify (crystallize) again. As the lattice structures of solids tend to exclude impurities, each subsequent recrystallization results in a progressively purer compound.

Tight junctions

tight junctions are found in epithelial cells. As the name suggests, the cells in tight junctions are linked very closely to each other, preventing solutes from being able to move freely from one tissue into another. A classic example is the blood-brain barrier, where the epithelial cells in blood vessels in the brain form very tight junctions that allow the close regulation of which substances from the bloodstream can enter the central nervous system. Some types of epithelial tissue have relatively few tight junctions; these are known as leaky epithelia. Examples include some parts of the kidney.