Solubility, Exam Questions (PDF)

Solubility is the max amount of solute that can dissolve in a given amount of solvent at a specified temperature and pressure to form a homogenous solution. Here we are converting g/L→mol/L (molarity) Molar mass of PbSO4=303 g/mol Soluble: Ammonium Alkali metals Nitrates Acetates Halides (except Ag,...

Solubility is the max amount of solute that can dissolve in a given amount of solvent at a specified temperature and pressure to form a homogenous solution. Here we are converting g/L→mol/L (molarity) Molar mass of PbSO4=303 g/mol Soluble: Ammonium Alkali metals Nitrates Acetates Halides (except Ag, Pb, Hg) Sulfates (except Ca, Sr, Ba, Pb) Nitrites Insoluble: Carbonates Sulfites Nitrides Sulfides Phosphates Hydroxides (except Ca, Sr, Ba) Oxides (except Ca, Sr, Ba) If we have ammonium or alkali metals with anything on the insoluble side, it will be soluble. Option A: We know that sulfates with Ba are not soluble. Option B: We know that Section: Biological and Biochemical Processes 1. An overview of the action potential is needed to understand this: 321 NOKIA 3 Na Out 2 K+ In 1 atp used Position 2: greatest # of voltage-gated Na channels open, which is depolarization, which is the opening of voltage gated Na+ channels. Position 4: is the membrane potential just after hyperpolarization. Here, voltage gated sodium channels will be closed, and the Na potassium pumps will be restoring the membrane to resting potential. Position 3: is the membrane being repolarized, which is when the voltage-gated potassium channels are open, but voltage gated sodium channels are closed. Phase Resting Membrane Potential Depolarization Threshold Rising Phase Peak Falling Phase Undershoot (Hyperpolarization) Refractory Period Description Neurons maintain a resting membrane potential around -70 millivolts (mV) inside the cell relative to the outside, primarily due to the unequal distribution of ions across the membrane. Stimulus causes sodium (Na+) channels to open, allowing Na+ ions to rush into the cell, making the interior less negative. If depolarization reaches a certain threshold (around -55 mV), voltage-gated sodium channels rapidly open, triggering an action potential. Voltage-gated sodium channels open further, causing a rapid influx of Na+ ions, resulting in a sharp increase in membrane potential toward positive values. Membrane potential reaches its peak value as sodium channels begin to close. Voltage-gated potassium (K+) channels open, allowing K+ ions to flow out of the cell, leading to repolarization as the membrane potential returns toward its resting value. Membrane potential briefly becomes more negative than the resting potential due to prolonged opening of K+ channels. Neuron enters a refractory period, temporarily unable to generate another action potential, ensuring one-directional propagation and regulating neural signaling frequency. 2. Equation 1 from the passage is: potential=0.06*log(K+extracellular/K+ intracellular. We should know that the resting membrane potential established by the sodium potassium pumps, which pumps Na+ out of the cell and K+ into the cell. we know that the extracellular concentration of Na is greater than the intracellular concentration, therefore the log of the ratio will be positive. If we were solving for the K+, we know that K+ extracellular is less that K+ intracellular which will give us an answer less than one and therefore the log of that would be negative. 8. Autosomal recessive is a pattern of inheritance for a trait or genetic disorder located on one of the autosomal chromosomes(chromosomes that are not sex chromosomes). In an autosomal recessive inheritance pattern: a. An individual must inherit 2 copies of the recessive allele, which would be one froe ach parent, to express the trait or genetic disorder. Term Definition DNA sequences located at the beginning of a gene that serve as binding sites for Promoter RNA polymerase and transcription factors. They initiate the process of Regions transcription. Short, discontinuous DNA fragments synthesized on the lagging strand during Okazaki DNA replication. They are later joined together by DNA ligase to form a Fragments continuous strand. Specific DNA sequences recognized and cut by restriction enzymes (restriction Restriction endonucleases). These enzymes cleave DNA at specific nucleotide sequences, Sites often producing staggered or blunt ends, depending on the enzyme. Short Tandem Short sequences of DNA, typically 2-6 base pairs in length, that are repeated in tandem multiple times. These repeats are highly polymorphic and are often used Repeats in DNA profiling and forensic analysis. 12. Ectoderm: - Attract, think of the things people find attractive. Hair, skin, eyes, brain, mammary glands Epithelial and communication systems (nervous system) Nervous tissue NOT connective tissue, like blood Mesoderm: - Means you get around, main things you need for movement. Cardiovascular system Muscle Bone Ligament Cartilage Blood Kidneys Heart Hemopoietic stem cells Endoderm: - Endternal organs, internal organs Inner lining of the gut Pancreas Portions of the liver GI Respiratory Urinary Hemopoietic stem cells are the precursor for blood cells, hence, that is why it is derived from the mesoderm. The Lac operon, found in bacteria, regulate the expression of genes involved in the metabolism of lactose. Operon: unit of genomic DNA containing a cluster of genes that re under control of a single regulatory signal(promoter). The genes are co-transcribed into a single mRNA strand and translated together or undergo trans-slicing to have single mRNA. Genes in an operon are expressed either all together or not at all. The lac operon is composed of 3 main components: a. Promoter a. A DNA sequence where RNA polymerase binds to initiate transcription of the lac genes. b. Operator a. A DNA sequence that is adjacent to the promoter that acts as a biding site for the lac repressor protein. c. Structural genes a. lacZ i. codes for an enzyme called beta-galactosidase. ii. This enzyme breaks lactose into glucose and galactose. b. lacY i. codes for lactose permease ii. lactose permease is a protein that helps bring lactose into the cell. c. lacA i. codes for thiogalactoside transacetylase. The lac operon genes are not always transcribed, rather, they are only transcribed when lactose is present and glucose is not. This is a way for the cell to save energy by only producing enzymes when they are needed. When breaking down lactose, we only need LacZ and LacY. LacZ cuts the lactose into pieces and LacY brings the lactose into the cell. When glucose is readily available to the cell, the repressor protein is constitutively expressed, it is transcribed at baseline. Then the regulatory protein binds to the lac operator, and that interferes with and represses the binding of RNA polymerase, which wants to bind to the lac promoter. This prevents and represses the transcription of these genes for lactose metabolism. When glucose is not readily available to the cell, an alternate source of energy is available, in the form of lactose. First, lactose will enter the cell at a slow rate. The metabolite of lactose, called allolactose, binds to the repressor, and alters the confirmation of the repressor protein, its shape, and it causes it to loosen up and fall off the operator. With the repressor now gone, RNA polymerase is now free to sort of roll down and transcribe the 3 genes, leading to higher levels of the encoded protein. Then you have lactose permease which allows more lactose to enter the cell, then you have lacZ which can break down lactose into glucose and galactose to further the cell’s needs. When glucose and lactose are both present the cell prefers glucose. The transport of glucose into the cell blocks the transport of lactose which is a process called inducer exclusion. This leads to decreased lactose which allows the repressor protein to bind to the operator again which prevents the transcription of the rest of the lacyes s operon genes Interaction between inducer and repressor that mediate gene expression Cell expends energy as necessary Y is the repressor protein that represses the binding of RNA polymerase. The cell wants to repress the binding of RNA polymerase because there is o lactose to break down, so there is no need to make enzymes to do it. Therefore, the cell uses a repressor protein to stop RNA polymerase from making those enzymes. X indicated in the promoter region, is the RNA polymerase. This is located here because it isa special sequence of DNA that tells RNA polymerase where to begin making the RNA copy. Without the promoter, RNA polymerase would not know where to start. The lac operon is considered inducible because its normally off, but can be turned on (or induced) when lactose is present. 15. Vessel resistance= 𝑉𝑙𝑒𝑛𝑔𝑡ℎ 𝑉𝑟𝑎𝑑𝑖𝑢𝑠 ^4 Diameter and length of artery halved: Vessel resistance= = (𝑉𝑙𝑒𝑛𝑔𝑡ℎ /2) 𝑉𝑟𝑎𝑑𝑖𝑢𝑠 )^4 16 ( (𝑉𝑙𝑒𝑛𝑔𝑡ℎ/2) ( 𝑉𝑟𝑎𝑑𝑖𝑢𝑠 )^4 16 =8(L/r^4) (16) (𝑉𝑙𝑒𝑛𝑔𝑡ℎ /2) 𝑉𝑟𝑎𝑑𝑖𝑢𝑠 )^4 2 ( 16. A is incorrect because this implies that the substrates are amino acids, which does not have to be the case. If they were amino acids, it is not the function of the enzyme to alter the primary structure of the enzymes. Altering the primary structure, or AA sequence, of the substrate would not be how it facilitate the reaction, rather, it would be the reaction itself. B is incorrect because enzymes DO NOT alter the equilibrium of a given reaction. This is standard and needs to be memorized for the MCAT. C is correct because larger enzymes may contain an active site that contains acidic or basic amino acids that give their microenvironment a different pH than that surrounding the enzyme. This can be used to facilitate a specific chemical reaction. D is incorrect because enzymes LOWER the activation energy. An action potential generated in the central nervous system is first propagated DOWN the axon of the efferent (motor) neuron to the synaptic terminal. Then, acetylcholine is released into the synapse and binds to nicotinic receptors on the postsynaptic membrane of the neuromuscular junction. This triggers further depolarization and propagation of the action potential along the muscle fiber membrane into T-tubules, which initiates the release of Ca2+ from the sarcoplasmic reticulum. The released calcium binds to troponin, which induces a conformational change in tropomyosin that exposes the myosin binding sites. Sarcoplasmic reticulum: specialized smooth ER found in muscle cells, particularly in skeletal and cardiac muscle fibers. The sarcoplasmic reticulum is a network of membranous tubules and vesicles that are located within muscle cells and it surrounds myofibrils, which are the contractile units of muscle fibers. Specificity: measures the ability to of a test to correctly identify individuals who do not have the disease or condition being tested for. It gauges the tests capacity to avoid false positives. A highly specific test has very few false positives Sensitivity: the ability of a diagnostic test to correctly identify individuals who have a particular disease or condition. It measures the test ability to detect true positives. A highly sensitive test will RARELY produce false-negative results, meaning it effectively captures most individuals who truly have a condition. Mnemonic: seNsitivity means very few false negatives sPecifity means very few false positives. A is correct because since the question states that it is highly sensitive, this means there should be virtually no false negatives. In other words, nearly every actual instance of CPT-I activity will be detected by radiolabeled carnitine incorporation. B is incorrect because high sensitivity indicates that an assay can effectively detect instances of CPT-I activity when it occurs. However, it does not mean that every radiolabeled carnitine incorporation represents CPT-I activity. The question indicates that the test is highly sensitive, meaning there should be virtually no false negatives. In other words, nearly every actual instance of CPT-I activity will be detected by radiolabeled carnitine incorporation. Section: Psychological and Social 1. The key word here is spoken words. The Broca’s area is the region of the bran responsible for the production of speech. Occipital Lobe Fore brain Visual processing area of the brain Associated with visuospatial processing, processing written language, distance and depth perception, color determination, object and face recognition, and memory formation. Damage can cause: Specific areas of the lobe can lead to blind spots or visual field defects; one side of the lobe may result in a loss of vision in the opposite visual field of each eye. Can cause difficulties in any of the things mentioned above. Broca’s Area Forebrain, frontal lobe Plays a role in production of speech and language comprehension. Generation of speech and grammatical processing. Wernicke’s area Cerebrum Forebrain, temporal lobe Damage to Broca’s area causes Broca’s aphasia, which is characterized by difficulty in producing speech, limited vocabulary and halting speech patterns while maintaining relatively intact comprehension. Involved in comprehension of language, including spoken and written language. Differs from Broca’s area, which is responsible for the production of speech. The cerebrum is the largest and most prominent part of the human brain, comprising about 85% of its total weight. It is located in the upper part of the cranial cavity and is divided into two hemispheres: the left hemisphere and the right hemisphere. These hemispheres are connected by a bundle of nerve fibers called the corpus callosum, which allows for communication between them. 2. When solving this, although I did not know Parkinson’s was specifically diagnosed this way, I rationalized what would be able to be diagnosed this way. Disorder Description Correctness A mood disorder characterized by Incorrect. Dysthymic disorder is not Dysthymic depression that is less severe than major associated with dopamine-producing disorder depressive disorder. It would most neurons. likely be clinically diagnosed. Incorrect. Autism is not associated with A developmental disorder characterized insufficient dopamine or degeneration of by impaired social and communication Autism dopaminergic neurons. An 18fluorodopa skills. Diagnosis is based on PET scan would be of little diagnostic symptoms rather than analytical tests. value. Characterized by neritic plaques and neurofibrillary tangles throughout the Incorrect. Alzheimer's disease is not Alzheimer's entire cortex. Not predominantly predominantly associated with the loss of disease associated with the loss of function of function of dopamine-producing neurons. dopamine-producing neurons. Caused by deterioration and death of Correct. Parkinson's disease is associated dopamine-secreting cells in the with the loss of dopamine-producing Parkinson's substantia nigra of the midbrain. PET neurons. A PET scan using a tracer that disease scan with a tracer binding to binds to dopaminergic neurons is likely dopaminergic neurons is appropriate. measuring the activity of these neurons. 3. First, I was looking at which part of the brain was marked, to see if I knew what section was responsible for what. Here, I struggled with remembering what is responsible for what. Therefore: a. First image: i. cerebellum b. Second image: i. Pre-frontal cortex in the frontal lobe c. Third image: i. Primary somatosensory cortex located on the anterior portion of the parietal lobe. d. Fourth Image: i. Occipital lobe In the question when it states that subject is instructed to concentrate on an image and at the same time their finger will be pricked, you are supposed to understand that the focus of the image is going to become irrelevant and rather the somatosensory cortex located on the anterior portion of the parietal lobe is going to light up because that is responsible for the sensation of the prick from the pin. Primary somatosensory cortex: - Often referred to as S1 - Region of the brain involved in processing tactile sensations such as touch, pressure, vibration, pain, and temperature. - Receives information from the body’s skin, muscles, joints, and other sensory receptors, allowing for the perception and interpretation of various somatosensory stimuli. 4. PET Scan: used to visualize and measure various physiological processes in the body including metabolism, blood flow, and neurotransmitter activity. They involve the use of a radioactive tracer, known as radiopharmaceutical, which is injected into the body. The tracer emits positrons, which are (+) charged particles. As the tracer decays, it emits gamma rays, which are detected by the PET scan. - Method of subtraction in context of PET scans: You obtain 2 separate PET scans then subtract the data from one scan from the data of the other scan. This is used to compare brain activity under different conditions or in response to different stimuli. Steps for method of subtraction: a. Baseline scan a. Obtain a Pet scan that is while the individual is at rest or engaged in a neutral task b. Captures the brain under NORMAL conditions b. Task-specific scan a. Second scan conducted while the individual is performing a specific task such as reading, speaking, listening to music, or solving problems. b. Captures the brain activity associated with the performance of the task. c. Subtraction a. The data obtained from the baseline scan is subtracted from the data obtained from the task specific scan by voxel, which is small 3D pixels representing brain activity. b. This process reveals the areas of the brain that exhibit increased or decreased activity during the performance of the task compared to the baseline condition. This method allows you to isolate and visualize changes in brain activity associated with specific tasks, behaviors, or experimental conditions. Option a Option b Incorrect because butterfly has 9 items (letters) and the asterisk does not really have a cardinality (rather than 1) so it is plausible that the darked region of scan 2-scan1 is responsible for assessing cardinality of an object. Correct because scan 1 the subject is told to fixate on an asterisk, and in scan 2 the subject is told to concentrate on the word butterfly. When conducting the method of subtraction (scan 2-scan 1) you will have brain activity present in scan 2 but absent in scan 1. With that being said, we would conclude that anything involved in the difference in an asterisk and the word butterfly such as size, number of things and meaning, is something we would be looking for. An INVALID conclusion would be performed identically by the brain for both asterisk and the word butterfly. Option c Option d This option says the function of determining what elements are to be fixated on in the visual field is performed identically for asterisk as it for the word butterfly. Incorrect because BUTTERFLY is an English word, asterisk is not, so it is plausible that the darkened region of scan 2-scan1 is responsible for determining whether something is an English word or not. Incorrect butterfly has 2 symbols that are identical (two T’s) and different symbols (every letter other than T) while the asterisk is one entity. Therefore, it is plausible that the darkened region of scan2-scan1 is responsible for establishing wther 3 symbols are identical or different. 5. Option A is a trick answer and is something to be aware of when solving questions like this. A high self-esteem is defined as the overall evaluation of his or herself self-worth. When reading his quote, “I believed I could have been one of those kids that was lost in the streets without the discipline instilled in me by my parents and other relatives.” An argument here is that he could have a high-selfesteem since he did not become what he stated in the quote BUT then you can rebuttal that and say that anyone can make a positive statement about themselves and not have a high self-esteem. You need to look for the more concrete, testable, things being covered here, NOT the mediocre answer. Option B is incorrect because we can define a learned helplessness as an extreme phenomenon where people, after being repeatedly exposed to situations they cannot control, will not even attempt to exert control in a similar situation that they can control. In the quote, nothing in this excerpt indicates that he has a learned helplessness rather, he stated that the discipline worked for him and that he is going to reevaluate how he disciplines his son moving foreword. He is in control of disciplining his son and there is no learned helplessness identified in the excerpt. Option C is incorrect because we can define self-efficacy as a persons belief in how effective he/she is in performing a task. With the option saying low self-efficacy, that assumes that one does not believe in thesmlf for performing the task, which is not idenfitied here ta all. Option D, is correct because the lochys of control is the way humans perceive the cause of the outcome. The internal locus of control is when one believes they can control the outcome with their actions and efforts, and those ith an external locus oc control, are more passive, attributing outcomes to ouside sources rather than their own efforts. Peter attributes his past success to his parenting, which is an external source. Attributing outcomes to external factors such as other people, luck, ec is the hallmark of peope with external loci of control. Individuals with an external locus of control tend to attirubrte success or fialures to factors beyond their control, leading to a sense of powerlessness or dependence on external circumstances. 6. The fundamental error of attribution is when judging other ›actions, humans tend to underestimate the importance of the situation and overestimate the importance of the character of the actor, which we can also refer to this as actor-observer bias. You are looking at who one is, rather than what one did. a. If someone cuts you off in traffic, you might assume they are rude or aggressive person (internal attribution) rather than considering that they might be in a hurry due to an emergency (situational attribution) When analyzing the options, A is correct because the editorial quote addresses Peterson as a person, rather than the situation. In the fundamental error of attribution, you are looking at who one is rather than what they did. When answering this question you are looking for an answer choice that describes Peterson’s character, rather than his action. 7. 8. Cultural Relativism: States that an individual’s beliefs and behaviors should be understood and judged within the context of their own culture, rather than by the standards of another culture. This implies that different cultures have different moral codes, social norms, values, and that they should be respected and understood within their specific cultural contexts. Ethnocentrism: The tendency to judge other cultures based on the standards and values of one’s own culture. Cultural relativism encourages individuals to suspend judgment and avoid imposing their own cultural norms onto others. Scapegoating: individuals or groups are unfairly blamed, singled out, or made the target of hostility and aggression for problems, faults, or issues that may be beyond their control or may have multiple causes. It involves attributing blame or responsibility to a specific individual or group, often without sufficient evidence or justification. The reason why option A and B are incorrect is because there was no cultural relativism applied here. Peterson was judged on his cultural upbringing and parenting techniques and scapegoating was applied to the professional sanctions that were against him because he was targeted for responsibility that were beyond his control, hence it was cultural. 9. Formal norm Informal norm More Sanction Explicit, specific, and publically available. Ex) Laws Imprecise, implicit rule or expectation of behavior for members of a society. Ex) handshakes A norm that is regarded by society as highly important (moral) Ex)being faithful to ones spouse Methods for punishing or reinforcing behavior that are in accordance or discordance with societal norms Aspect Definition(action potential) Initiation Threshold Rising Phase Falling Phase Refractory Period Propagation Description Brief, rapid changes in electrical potential across the membrane of neurons or muscle cells. Triggered by depolarization of the neuron's membrane, often in response to a stimulus. Depolarization must reach a certain threshold level to initiate an action potential. Voltage-gated sodium channels open, allowing influx of sodium ions and rapid depolarization of the neuron. Voltage-gated potassium channels open, allowing efflux of potassium ions and repolarization of the neuron. Following an action potential, a brief period where the neuron is temporarily unable to generate another action potential. Helps regulate signal propagation. Action potentials travel along the neuron's axon, often through saltatory conduction in myelinated neurons. It is important to know that action potentials are all-or-nothing: - The firing rate of a neuron can only be increased by increasing the frequency of action potentials. - There is no such thing as large or small action potentials, the way neurons increase their firing activity is due to increasing the frequency of action potentials, not amplitude. Cortical neurons Geniculate neurons Half Length 496 3/10/24 1. The volatile aromatics produced during the cyclization reactions at 120°C have lower heats of combustion than aliphatic (straight chain, saturated hydrocarbons) of comparable molecular weight. The most likely reason for this is: a. the carbon-carbon bonds of an aromatic compound have a lower bond energy than the carbon-carbon bonds of an aliphatic compound and release more energy when combusted. b. aromatic compounds are stabilized by resonance and have a higher precombustion energy than aliphatic compounds. c. aromatic compounds are stabilized by resonance and have a lower pre-combustion energy than aliphatic compounds. i. Aromatic compounds obviously have a cyclic structure and due to their structure they possess delocalized pi electrons, which means they are not confined to a specific set of atoms but instead are spread out over the entire conjugated system of the aromatic ring. ii. Delocalized Pi electrons allows for resonance stabilization, which further means that electrons are not confide to specific bonds but are rather spread out over the whole ring. iii. The increased stability due to resonance makes it less likely for aromatic compounds to undergo combustion compared to aliphatic compounds. iv. The owner pre-combustion energy refers to the energy required to initiate the combustion reaction. v. In aromatic compounds, their resonance stabilization lowers the energy barrier required to break bonds and initiate combustion compared to aliphatic compounds. d. the carbon-carbon bonds of an aromatic compound have a lower bond energy than the carbon-carbon bonds of an aliphatic compound and release less energy when combusted. A combustion reaction in chemistry is when a substance reacts rapidly with oxygen gas (O2) and typically releases heat and light energy. Usually, the reaction often produces CO2 and H20 as the primary products. The fuel for a combustion reaction is typically a hydrocarbon compound such as: - methane CH4 propane C3H8 octane C8H18 Combustion is initiated by providing sufficient energy to break the chemical bonds in the fuel molecules. This energy can come from various sources: heat, spark, flame. Combustion is vital for everyday activities such as heating, cooking, transportation, and energy production. HOWEVER, incomplete combustion can lead to the form of carbon monoxide. Aliphatic: compounds characterized as having straight or branched carbon chains. - Saturated aliphatic compounds have only single bonds between carbon atoms. Unsaturated aliphatic compounds contain double or triple bonds between the carbon atoms. 2. Based on the information presented in the passage, which of the following chemicals would be effective in suppressing smoke and flamepsread in thermally degrading PVC? a. BH3 b. CH3i. This is incorrect because this would be a lewis base. CH3- has a lone pair of electrons, hence the negative charge. c. H2 d. NH3 In the passage it states the following: “Charring of PVC reduces heat transmission, and char is highly thermally stable. Therefore, a chemical additive that catalyzes this cross-linking will reduce the overall smoke and flame production from thermally degrading PVC. One major class of additives found to promote cross-linking reactions is Lewis acids, notably MoO3.” Lewis acids are electron pair ACCEPTORS. That is any substance that can accept a pair of electrons from another entity to form a new bond. - Lewis acids have vacant orbitals that can accept electron pairs. All cations are Lewis acids since they are able to accept electrons. Remember, cations (+ charged ions), the charge indicates the # of electrons that the ion has lost. You can use electron configuration to identify an empty orbital: B in BH3 electron configuration: 1S^2 2S^2 2P^1, there is an empty orbital hence the p only has 1 electron filled, rather than 6. S P D F 2 6 10 14 UV/Visible spectroscopy is a technique that measures the absorbance or transmittance of UV or visible light by a sample. This reflects the principle that molecules can absorb light of specific wavelengths, leading to electronic transitions within the molecule. Highest Occupied Molecular Orbital: the molecular orbital that contains the highest energy electrons in a molecule under normal conditions. Lowest Unoccupied Molecular Orbital: This orbital lies just above the HOMO and is th first orbital available for electrons to enter upon excitation. How Conjugation Affects Energy Levels: Conjugated systems have alternating single and double bonds which allows pi electrons to delocalize over the entire system. The delocalization does the following: 1. Stabilize the molecule by lowering the overall energy of the molecular system 2. Reduces the energy gap between HOMO and LUMO because the Pi electrons are more spread out, the energy difference between the state where they reside (HOMO) and the first excited state they can jump to (LUMO) decreases. Conjugation and Light Absorption 1. Lower energy difference a. As conjugation increases the energy gap between the HOMO and the LUMO narrows. b. The energy of a photon must match this gap for absorption to occur. 2. Higher wavelength absorption a. Since energy and wavelength are inversely related (E=hc/ λ), a smaller energy gap means the molecule absorbs photons of longer wavelengths. Therefore in the context of light, longer wavelengths correspond to lower energy. Increased Conjugation=Increased UV absorbance UV/Visible spectroscopy detects the degree of conjugation. Highly conjugated molecules=lower energy differences between their highest occupied molecular orbital and their lowest unoccupied molecular orbitals, and therefore absorb. Lower energy (higher wavelength) photons. In the passage it states that conjugated polyenes form when PVC begins thermally degrading therefore the more conjugation present the more UV / visible light the compound will absorb therefore in this question we are looking for a graph that shows an increase rather than a decrease. Carboxylic Acid Derivatives R can be: 1. Halogen---→ acyl halide a. F,Cl,Bt,I 2. Alkoxy group--→ester a. OH 3. Amino group--→amide a. NH2 4. Carboxylate-→ anhydride Section: Biology from the passage we must look at the following to help answer: Biopsies of the tumors indicated that for a majority of the patients, the tumor cells had many morphological traits of the exocrine tissue of origin (neoplastic), while in a minority, they resembled the tissue of origin only partially (dysplastic), or had lost much of the original tissue morphology (anaplastic). This information correlates to figure 4, which shows the survival trend according to the tumor differentiation Option A: Neoplastic adenocarcinoma (well differentiated) in the duodenum vs. neoplastic adenocarcinoma (well differentiated) at the head of the pancreas. From figure 3, there is sufficient information to interpret that the duodenum location has a lower survival rate compared to the head of the pancreas, Option B: Neoplastic adenocarcinoma (well differentiated) in the ampulla of vater vs. anaplastic (poorly differentiated) located at the head of the pancreas. Here, we need to incorporate figure 4. We know that anaplastic is considered poorly differentiated and has a significantly lower survival rate in comparison to the neoplastic, which is well differentiated. Then, you can also look at figure 3 and see that the location of ampulla vater has a higher percentage than tumors in the head of the pancreases. Here I did not consider the figure 4, hence, this question was a comprehension/interpretation assessment. Although rare, endocrine neoplasms of the pancreas do exist. A patient with such a condition may: b. be unable to effectively digest fats. i. The pancreas plays a critical role in digestion through its exocrine function, producing pancreatic enzymes like lipase that help in the digestion of fats. While issues with the pancreas can lead to difficulties in digesting fats, this is more directly related to the exocrine function of the pancreas rather than its endocrine function. ii. Endocrine neoplasms specifically affect the hormone-producing part of the pancreas, which is more directly involved with blood glucose regulation than fat digestion. Thus, while pancreatic disorders can lead to problems with fat digestion, this option doesn't directly relate to the endocrine neoplasm's impact. c. experience an increase in basal metabolic rate. i. The basal metabolic rate is a measure of the amount of energy, expressed in calories, that your body needs to maintain basic physiological functions while at rest. ii. These basic functions include: 1. Breathing 2. Circulation 3. Cell production 4. Nutrient processing 5. Protein synthesis 6. Ion transport iii. It is hormones from the thyroid, not the pancreas, that play the key role in regulation of basal metabolic rate. d. have an elevated pH in the duodenal lumen. i. although the pancreas does play a role in raising the pH of the intestinal lumen via bicarbonate secretion, this is exocrine functionality, not endocrine. Endocrine implies the secretion of hormones into the blood. e. have difficulty regulating blood glucose levels. i. Insulin reduces blood glucose. ii. Glucagon increases blood glucose iii. Somatostatin inhibits both insulin and glucagon as well as human growth hormone. iv. A tumor arising from insulin, glucagon, it somatostatin-secreting cells would likely lead to dysregulation of blood glucose. Exocrine and Endocrine refer to the ways in which glands secrete substances, and these concepts apply to various glands throughout the body. Here are some differences: Endocrine Exocrine Secrete hormones directly into the Secrete substances onto an epithelial surface bloodstream, without using ducts. by way of a duct. This category includes glands that secrete: - sweat These hormones travel through the blood to target organs or cells, where they elicit - saliva specific responses affecting growth, metabolism, reproduction, and other vital functions. The primary secretions are hormones, which are chemical messengers that regulate various physiological processes. - tears breast milk digestive enzymes These secretions can be released onto the skin surface or into cavities of the body, such as mouth or intestines. Example: Salivary glands secrete saliva into the mouth, aiding in digestion and hygiene. The 3 main endocrine hormones of the pancreas: a. insulin b. glucagon c. somatostatin 3. Which of the following correctly lists the relative positions where trypsinogen, bile, and chyme enter the small intestine following a Whipple procedure, ordered from most proximal to most distal? a. Chyme, trypsinogen, bile. b. Chyme, bile, trypsinogen. c. Trypsinogen, chyme, bile. d. Trypsinogen, bile, chyme. The surgical procedure for treatment of pancreatic cancer is a pancreaticoduodenectomy, or Whipple's procedure. It involves removal of the distal half of the stomach, gallbladder, distal portion of the common bile duct (CBD), as well as the head of the pancreas, duodenum, proximal jejunum, and lymph nodes. Trypsinogen: - An inactive precursor (zymogen) of the enzyme trypsin. It is produced and secreted by the pancreas. - Bile: - Once secreted into the small intestine, trypsinogen is activated to trypsin by an enzyme called enteropeptidase, which is produced by the cells lining the duodenum. Trypsin plays a crucial role in digesting proteins by breaking them down into smaller peptides. an emulsifying agent synthesized by the liver and stored in the gallbladder. Aids in the digestion and absorption of fats in the small intestine. Bile salts emulsify fats, making them easier for digestive enzymes to break down. Chyme: - acidic fluid composed of gastric juices and undigested food that churns in the stomach. It is the material that needs to be further broken down by digestive enzymes and bile in the small intestines. 4. Right atrium: - Received deoxygenated blood from the body through the superior and inferior vena cava Once it collectes the deoxygenate blood it will pump it into the right ventricle through the tricuspid valve when it contracts. Right ventricle: - Pumps the deoxygenated blood received from the right atrium into the lungs via the pulmonary artery for oxygenation. This is facilitated by the contraction of the ventricle and the opening of the pulmonary valve to allow blood flow into the pulmonary circulation. Left atrium: - Receives oxygenated blood from the lungs through the pulmonary veins Primary role is to collect the oxygenated blood and upon contraction, pump it into the left ventricle through the mitral valve. Left ventricle: - Thicker walls than the right ventricle due to the higher force needed ot pump blood throughout the body. Responsible for pumping the oxygenated blood received from the left atrium into the systemic circulation via the aorta, supplying the body with oxygen and nutrients. Superior vena cava: - Located in upper thoracic cavity Return deoxygenated blood from the body’s upper part (above the diaphragm) to the right atrium of the heart. Blood moves into the right ventricle, which then pumps it to the lungs for oxygenation. Inferior Vena Cava: - large vein that returns deoxygenated blood from the lower half of the body to the heart. Tricuspid valve: - Regulates blood flow from the right atrium to the right ventricle. When the right atrium contracts, the tricuspid valve opens to allow deoxygenated blood to flow into the right ventricle. When the right ventricle contracts, the valve closes to prevent blood from flowing back into the right atrium. Pulmonary semilunar valve: - - Regulate blood flow from the right ventricle to the pulmonary artery, and to the lngs. When the right ventricle contracts, the valve opens, allowing deoxygenated blood to be pumped into the lungs for oxygenation. When the right ventricle relaxes, the valve closes to prevent blood from flowing back into the heart. To begin, we can first skim the passage and draw this from the following text: Once 1,25-dihydroxycholecalciferol is synthesized, it exhibits a "hormonal" effect on intestinal epithelial cells by upregulating the expression of calcium-binding proteins, which migrate to enterocyte brush borders and transport calcium into the cytoplasm. The rate of calcium absorption is directly proportional to the quantity of this calcium-binding protein. We know that the 1,25-dihydroxycholecalciferol is going to be located on/in a enterocyte NOT a hepatocyte. Enterocytes are absorptive cells lining the inner surface of the small intestine and colin. They play a crucial role in the digestion and absorption of nutrients. Hepatocyte: cells of the liver. They have many functions: - Play a central role in the metabolism of carbs, proteins, and lipids. These cells detoxify carious metabolites and drugs, transforming them into less harmful compounds that can be easily excreted. Produce bile Store vitamins and minerals and release them as needed. Synthesize cholesterol and convert it into bile. Chemical Structure Types of Hormones Steroid Hormones Peptide/Protein Hormones Derived from Chains of amino acids, cholesterol, lipidwater-soluble. soluble. Mechanism of action Pass through cell membranes to bind to intracellular receptors, influencing gene expression directly by acting on the DNA in the nucleus; Bind to receptors on the surface of target cells, initiating a cascade of intracellular events through second messangers (cAMP, Ca2+) Primary sources Adrenal cortex (cortisol aldosterone), gonads (testosterones, estrongenes and progesterone) and lacenta (estrogens and progesterone. Cortisol, aldosterone, testosterone, Pancreas (insulin, glucagon), anterior pituitary (growth hormone, FSH, LH) , posterior pituitary (ADH), and others Examples Amino Acid Derivative Hormones Derived from specific amino acids, primarily tyrosine and tryptophan. Thyroid hormones act similar to steroid hormones by entering cells and affecting gne expression, while caetcholamines (epinephrine and norepinephrine) act more like peptide hormones by binding to cell surface and receptors. Thyroid gland (T3 and T4), adrenal medulla (epinhephrine and norepinephrine) , and pineal gland (melatonin)l estrogens, progesterones We know that vitamin D has fat-soluble properties, hence it is 1 of 5 fat-soluble vitamins(A,D,E, and K_Theerefore, due to its fat soluble properties we can assume that it will act as a steroid hormone, diffusing into the cell before it binds otl the receptor. Magoosh FL #3 490 3/18/24 Section: Chemical and Physical The small peak at 99.04 in the mass spectrum most likely corresponds to: a. A double charged (+2) version of the molecular ion. i. If there was a double charge, then the molecular ion would appear at 49 on the mass spectrum, not 99, because of the m/z. It would be 99.04/2 because you are dividing the mass by the charge. b. A protonated version of the molecular ion. i. Also, incorrect because the horizontal axis on mass spectra is the mass to charge ratio. A protonated version would have a peak at about 49.5. c. An isotopic version of the molecular ion. i. The molecular ion in mass spectroscopy is the peak that corresponds to the entire unknown molecule with a charge of +1. For MCAT purposes, it is always the largest peak that is furthest to the right. d. The result of a side reaction between the molecular ion and the acetone solvent. Typically, the charge (z) for most organic molecules in mass spectrometry is +1, because the molecules are usually ionized by the loss of a single electron. Mass spectroscopy: A. Break a compound into different fragments and ionize them. We are breaking our compound. B. In mass spectroscopy, when your sample is bombarded with electrons, the atoms get ionized, and they gain a charge. C. The amount an ion is deflected in a mass spectrometer is determined by its mass to charge ratio(z). a. Ions with higher mass-to-charge ratio are deflected less. b. Ions with lower mass-to-charge ratio are deflected more. D. Unlike UV, IR, and NMR: a. Compound is destroyed. b. No electromagnetic wave absorption Which of the following was most likely the contaminant? From 4,000-1,400 is the functional group region. 1,400 and below is the finger print region, which is unique to a compound. IR spectroscopy: - - Used to identify functional groups in molecules by measuring the vibrations of chemical bonds. Energy from light causes a bond to stretch Transmittance is how much light went through your sample If you have 100% transmittance that means all that light when through your sample and nothing was absorbed For frequency less than 100% that means that some of the light was absorbed by your molecule Compounds are subjected to IR rays and the wavelengths of absorbance are observed Changes in dipole moment are measured Key absorption bands to look for in an IR spectrum include: O-H stretch around 3200-3600 cm^-1 for alcohols and phenols, which is usually broad. N-H stretch around 3300 cm^-1 for amines and amides, which can be either sharp or broad. C=O stretch (carbonyl group) typically around 1650-1750 cm^-1, which is very distinctive. For the MCAT remember 1700cm-1 is the peak that indicates the presence of a carbonyl group. C=C stretch for alkenes around 1640-1680 cm^-1. C-N stretch for aliphatic amines around 1200-1350 cm^-1. Aromatic C=C stretch around 1450-1600 cm^-1, usually shown as multiple peaks. What is the energy of the infrared radiation associated with the absorption peak at 3000 cm-1 in figure 1? (Note: Planck's constant is 6.6 × 10 -34) a. 7 × 10-19 Joules. b. 6 × 10-20 Joules. c. 1 × 10-19 Joules. d. 2 × 10-21 Joules. The MCAT is testing you on the formula for the energy of a photon: E=hc/λ =(6.6x10^-34)(3.8x10^8m/s)/3.3x10-6m =6x10^-20 h=plancks constant c=speed of light in a vacuum (3.8x10^8 m/s) λ=wavelength of the light Wave numbers are RECIPROCAL wavelengths, so since the IR spectrum absorbs 3000cm-1 (wavenumbers) then the wavelength of the light will be 1/3000=3.3x10^-4cm =3.3x10^-6m - - NMR spectroscopy Used to determine the structure of organic compounds. NMR gives us information about the types of atoms in a molecule, their numbers, positions, and even how they are connected. Molecules are exposed to magnetic fields resonate and align spins of certain nuclei with or against the field. When these nuclei are exposed to a radio frequency, they absorb energy and flip their spin. The energy difference is what NMR measures. Works best with odd number of nucleons Typically used with odd numbers of nucleons, so NMR typically uses the hydrogen nucleus because there is only 1 nuclei Shielding is when electrons around a nucleus protect it from the magnetic field, electron withdrawing groups deshielded a nucleus A chemical shift( is a measure of how shielded a nucleus is , the further downfield(to the left) the less shielded HNMR Spectroscopy: - Peak number corresponds to the # of chemically different hydrogen nuclei Peak area represents the relative abundance of nuclei Peak position reveals the chemical environment (shielding) of the nuclei Coupling (splitting) is chemically different nuclei on adjacent carbons which will cause each other’s peaks to split If the procedure in the passage was repeated with ethyl bromide, the 1 H-NMR spectrum would consist of: one 3H triplet and one 2H quartet. one 2H triplet and one 3H quartet. one 3H singlet and one 2H doublet. one 2H singlet and one 3H doublet. This is testing us on whether we know how to draw ethyl bromide and predict its coupling pattern. Remember, coupling is splitting, which is chemically different nuclei on adjacent carbons which will cause each other’s peak to split. There are 2 sets of equivalent hydrogens: 1. The 2 hydrogens on C1 2. The 3 hydrogens on C2 When looking at C1, we see that the 2 hydrogens on C1 have 3 adjacent hydrogens, so by using the n+1 rule, their coupling pattern would be a quartet (3 adjacent hydrogens+1) When looking at C2, the three hydrogens have 2 adjacent hydrogens, so their coupling pattern will be a triplet (2adjacent hydrogens +1) The n+1 rule is a formula for predicting the # of peaks based on the neighboring hydrogens n is the number of neighboring protons. P=vapor pressure Hvap=molar heat of vaporization T=absolute temperature R=universal gas constant (0.00821latm/molk) C=arbitrary constant We can see that the equation is linear, hence: Y= dependent variable = ln(p) M=slope of the line=Hvap X=independent variable=1/T B=y-intercept=C Since the slope is negative, shown below, that is how we can know that the graph should be linearly decreasing. What I did: when I was solving this question, I did not directly refer to the colligative properties but rather I referred to the passage, specifically in the second paragraph, which states in order to determine the effect of temperature on the vapor pressure of a pure liquid, a chemist places 4.6 milliliters of n-hexane in a sample tube. The sample tube is immersed in an ethylene glycol(antifreeze) bath capable of maintaining the sample at any temperature within the ranges of -20 to 80°. After I read that I looked back at the question and when it says that the normal melting point was -95.3°C and the normal boiling point was 68.7° since they were looking for what would happen with a 4.9 mL sample I figured that the answer choice had to reflect a boiling point and a melting point that was greater than the normal points since it was.03 greater than the sample indicated in the passage. Non-volatile solute: a substance that does not easily vaporize. When dissolved in a solvent, it does not contribute to the vapor pressure of the solution. The 4.9mL sample of n-hexane with 0.05g of a non-volatile solute dissolved in it can be solved via the colligative properties. According to the boiling point elevation property, when you add a non-volatile solute to a solvent, it raises the boiling point of the solvent. Therefore, the boiling point went from 68.7 to 69.5. The freezing point depression describes how the freezing point of a solvent decreases with a non-volatile solute is added. It is the process of going from liquid to solid and how this transition temperature is lowered by the presence of solute particles. Hence, if it takes a lower degree for something to freeze we can insinuate that the melting point is also decreased. Colligative properties are physical properties of solutions that are dependent upon concentration of particle in the solution, not chemical particle identity. There are 4 properties: 1. Vapor pressure depression (raoults law) a. When we create a solution, the vapor pressure of the solution is less than the vapor pressure of the pure solvent. b. PA=XAPA° i. PA=Vapor pressure of a solution ii. XA=mole fraction of the solvent iii. PA°=vapor pressure of the pure solvent c. As we increase the solute concentration the mole fraction of the solvent decreases and the vapor pressure of the solution decreases 2. Boiling point elevation a. Opposite of freezing point depression b. When you add a non-volatile solute to a solvent, it raises the boiling point of the solvent. This is because the solute particles disrupt the solvent particles, which makes it harder for them to escape into a gas (which is what happens when something boils). c. 3. Freezing point depression a. When a solutions freezing point is lower than that of the pure solvent. This happens because the solute particles interfere with solvent molecules ability to form a solid structure. b. 4. Osmotic pressure a. Attractive force that particles dissolved in a solution exert for their solvent their self. b. The more solute particles we have dissolved in a solution, the greater the osmotic pressure. c. In a solution, osmotic pressure is the pressure that needs to be applied to STOP the flow of solvent across a semi-permeable membrane. d. Solution concentration measurements: 1. Molarity(M) a. Mole’s solute/liters solution 2. Molality (m) a. Mole’s solute/kg solvent Solute: what gets dissolved (like sugar in tea) Solvent: what does the dissolving (like the water in the tea) Von’t Hoff Factor(i):a measure of the # of particles a substance splits into when it dissolves. Example: If you dissolve table salt (NaCl) in water, it breaks down into two pieces: Na+ and Cl= therefore the Von’t Hoff factor is 2. To convert a temperature from Celsius (°C) to Fahrenheit (°F), you can use the following formula: °F=(°C×95)+32°F=(°C×59)+32 Here’s a step-by-step guide on how to do the conversion: 1. Multiply the Celsius temperature by 9/5 (or 1.8). 2. Add 32 to the result from step 1. Vapor pressure: - - - Vapor pressure is determined by the temperature and the identity of the liquid, not the surface area between the liquid and the gas. Since the temperature is staying the same, we can ignore the fact that the surface area increased because it will not affect it. If we know vapor pressure at one temperature and want to calculate it at another temperature, we would use the clausis-claperyon equation, and when looking at the equation we can even see how there is not a variable for surface area. Review on vapor pressure: - Definition: Pressure that is exerted by a vapor in equilibrium with its liquid (or solid) phase at a given temperature in a closed system. It represents how much molecules of the substance escape from the liquid to the gas phase. Vapor pressure is determined by the temperature and the identity of the liquid, not the surface area between the liquid and the gas. Vapor pressure INCREASES with temperature because more molecules have enough kinetic energy to escape into the vapor phase. - Substances with weaker intermolecular forces (London dispersion, dipole- dipole, and hydrogen bonding) have HIGHER vapor pressures at a given temperature, which is because as molecules can escape more easily into the gas phase. Clausius-Clapeyron equation describes HOW the vapor pressure of a substance changes with temperature. The equation shows the relationship between vapor pressure and temperature. Boiling point is defined as the temperature at which the vapor pressure is equal to the atmospheric pressure. Vapor pressure and atmospheric pressure DO NOT have a reciprocal relationship with boiling point. Mandelate racemase is a member of the enolase superclass. What is (are) the likely final product(s) of the reaction shown below: This question is testing our understanding of what racemic means. A racemic mixture contains equal amounts of ( R) and (S) enantiomers. Therefore, in this question, we can expect to figure out that mandelate racemase catalyzes the racemization of S or R mandelate into a mixture of both. When you see a reaction involving the enolase enzyme: 1. A dehydration reaction will occur. a. Enolase promotes the removal of a H20 molecule from the substrate. b. The removal of water by enolase results in the formation of an enol intermediate. 2. Remember that the role of enolase as a promoter of dehydration and enol formation is important in various biochemical pathways and is often a key intermediate in a variety of metabolic and synthetic reactions because they have high reactivity. Which of the following graphs most accurately depicts the initial rate Vo of an enzyme-catalyzed reaction as a function of initial substrate concentration? The Michaelis-Menten equation describes the rate of an enzymatic reaction as a function of substrate concentration. The equation is: 𝑉= 𝑉𝑚𝑎𝑥 [𝑠] 𝐾𝑚+[𝑠] V=initial reaction rate Vmax= max reaction rate when the enzyme is saturated with substrate [s]=substrate concentration Km=Michaelis constant, which is equal to the substrate concentration at which the reaction rate is half of Vmax as S increases, Vo will also increase, but it will level off, asymptotically approaching vmax. Hence, when choosing, you want a hyperbolic shape, and one that is increasing. Nothing decreasing and nothing linear. Entropy: measure of disorder or randomness in a system - Low entropy means there is more order and less randomness High entropy means there is less order and more randomness Water boiling in a pot Water boiling turns into a gas, which is random An amorphous solid glass window does not have a regular, repeating pattern, so its random. Amorphous means without shape. So it is used to describe solids where particles are not arranged in a regular, repeating pattern. Dry ice Is a solid, where molecules are tightly packed and do not move as much. A regular repeating pattern of molecules, which we a call a crystallin structure, are orderly. Co and co2 Gas mixture, which is quite random. Solids are more orderly than liquids or gas. In a spring, the maximal potential energy is ALWAYS equal to the maximal kinetic energy, so these two quantities equal and solve for K, which is the principle of the conservation of energy. That tells us how energy changes from potential to kinetic as the spring moves. - At the MAX stretch: all energy is potential (PE=1/2kx^2) At MAX speed, all energy is kinetic (KE=1/2mv^2) Potential energy: energy that is stored and ready to be used. Kinetic energy: energy that is in motion When we ignore air resistance and friction, we are not accounting for forces that oppose motions, which are forces that slow things down. Therefore, if we are measuring the speed of an object falling, it might be slower than we expect because of these unaccounted forces. The reason that she would calculate a value of G that is less than the actual value of G because when you account for air resistance and friction, it is going to slow things down. Option A is incorrect because when we talk about motion under gravity, the distance an object falls is NOT a straight line (linear), rather, it is a parabola because the object speeds up as it falls. Steroids are organic compounds with four ring cores. I knew this answer just off of context clues and knowing common steroids. Placenta is an organ that develops in the uterus during pregnancy. It provides oxygen and nutrients to the baby and it will remove waste products from the baby's blood. The umbilical arteries are two blood vessels in the umbilical cord. They carry deoxygenated nutrient depleted blood from the baby back to the placenta. The umbilical vein is the other blood vessel in the umbilical cord it carries oxygenated nutrient rich blood from the placenta to the baby Which of the following graphs most accurately represents the half-life of caffeine in the body as a function of time for the duration of one menstrual cycle? (Note: Assume that caffeine intake in constant over the entire cycle). The second paragraph says that an increase in the half-life of caffeine occurs as the result of elevated estrogen levels. Therefore, caffeine half-life over the course of one menstrual cycle should mirror estrogen levels. Definition Type Produced by Roles Estrogen Steroid hormone Primarily the ovaries, and in smaller amounts by the adrenal glands and fat tissues Regulates menstrual cycle Promotes the development and the maintenance of female secondary sexual characteristics (breat development and wider pelvis) Prepares the endometrium for potential implantation of a fertilized egg. Progesterone Steroid hormone Corpus luteum in the ovary AFTER ovulation and in significant amounts by the placenta during pregnancy Prepares the endometrium for the implantation of a fertilized egg Supports early pregnancy by maintain a suitable environment the developing fetus If pregnancy occurs, progesterone levels remain high to inhibit further ovulation Luteinizing hormone Peptide hormone Anterior pituitary gland Triggers ovulation, the release of an egg from the ovary LH surge leads to the transformation of the ruptured follicle into the corpus luteum, which then produces progesterone. Follicle-stimulating hormone Gonadotropin, which means it acts directly on the gonads (ovaries in females and testes in males) Synthesized and At the beginning of secreted by the the menstrua l cycle, anterior pituitary gland FSH facilitates the maturation of a small # of ovarian follicles, one of these will become the dominant follicle that releases an egg during ovulation. As a follicle matures, they produce estrogen and FSH indirectly stimulates the production of estrogen by the growing follicles. The corpus luteum forms from the ovarian follicle after ovulation. Once the mature egg is released from the follicle during ovulation, the ruptured follicle undergoes luteinization, transforming into the corpus luteum. Function Hormone Production: The primary function of the corpus luteum is to produce progesterone, along with smaller amounts of estrogen. Progesterone is essential for preparing the endometrium (the lining of the uterus) for potential implantation of a fertilized egg and for maintaining early pregnancy. Support of Pregnancy: If fertilization occurs and the egg implants in the uterine lining, the corpus luteum receives a signal from the developing placenta (human chorionic gonadotropin, hCG) to continue its hormone production, supporting the pregnancy until the placenta can take over progesterone production, usually around the 10th week of gestation. Life Cycle If fertilization does not occur, the corpus luteum degenerates into a scar tissue called the corpus albicans within about two weeks after ovulation. This degeneration leads to a decrease in progesterone and estrogen levels, which triggers menstruation, marking the beginning of a new menstrual cycle. “The known biological effects of caffeine include central nervous system stimulation, increased secretion of catecholamines, increased heart rate and relaxation of smooth muscle.” Here you have to identify whether you can identify which hormone is catecholamine. Catecholamines are small amines with catechol groups. For the MCAT know the 3 catecholamines: - Epinephrine Norepinephrine Dopamine I solved this based off of thinking how caffeine would affect each of the following, not based on the groups. MCD-diet produces NASH in rats by overloading the hepatocytes with free fatty acids and forcing the liver to enlarge. CPT-1, the mitochondrial gateway for fatty acid entry into the matrix, is the main controller of the hepatic mitochondrial B-oxidation flux. Section: Psychological and Social Matriarchy: social group where women (particularly elder women) possess most of the power and prestige. Exogamy: marrying outside of a particular group Matrilineal descent: kinship pattern where someone or something is believed to belong to the same descent group as the mother. Monogamy: marriage pattern where two people are married to each other and only to each other. Answer analysis: We know that it is not matriarchy because that is when in a social group a women possess most of the power and prestige. In the text it states that the villages chief is male, so it is not likely that the Mwene-Ditu is a matriarchy. Dramaturgy: - - A theory by Erving Goffman A sociological perspective that views social interactions as performances, where people present themselves in ways that will be acceptable to others. It involves 2 parts: o Front stage (public interactions) ▪ We follow certain social rules and roles when we are in public, and we try to manage how others see us o Back stage (private behaviors) ▪ When we can be ourselves without worrying about how others see us. This usually happens when we are alone or with very close people. This theory helps us understand what we might act differently in different situations. A dramaturgical perspective is a way of looking at social interactions, similarly, to seeing life like a play. In this perspective people are seen as actors who perform different roles in different situations. Longitudinal Implies that the subjects of the study are observed or measured over a period of time Case Implies that one person or situation is being given significant focus and attention Exploratory Do the study and then formulate a hypothesis. In contrast to a confirmatory study, which confirmatory studies formulate a Experimental Most rigorous research design Involves random assignment of participants into experimental and control groups, and hypothesis then do the study. the manipulation of one variable that the researcher wishes to investigate while controlling the others. Standard deviation is a measure of the amount of variation or dispersion in a set of values. The total area under the curve sums to 1, which corresponds to the total probability of all possible outcomes. Since the standard deviation is 15 and in the experiment, they want to know the likelihood of being gifted, that is going to correspond to approximately 2% of the area under the IQ distribution curve (i.e. the 98th percentile). Piagets 4 stages of cognitive development: Stage Sensimotor Preoperational Approximate age Birth-2 2-6 Concrete operational 7-11 Formal operational 12-adult Hallmark Object permeance Language development, egocentrism, pretend play Logical thinking, conservation Abstract reasoning Altius Half Length 3/24/24 Only asymmetric carbon here Section: chem/phys What stereochemical designation can be assigned to the asymmetric carbon center in L-dopa? a. b. c. d. Z S E R E and Z configurations: used to describe the geometry around double bounds, particular in alkenes. E configuration mans that the highest propriety substituents on each carbon of a double bond are on OPPOSITE sides. Z configuration means that the highest priority substituents on each carbon of the double bond are on the SAME side. R and S configuration describe the arrangements of atoms around a chiral center, which is defined as a carbon atom bonded to 4 different groups. R configuration is when substituents are clockwise, and the chiral center has the lowest priority group pointing AWAY from you. The S configuration is when the substituents are arranged in a counterclockwise direction around the chiral center when the lowest priority group is pointed away from the observer. Enantiomers: when molecules are mirror images of each other but cannot be superimposed on one another. Always involve molecules that are chiral, meaning they have a certain form of asymmetry that prevents them from being superimposed on their mirror images. Which enzymatic transformation from Figure 1 involves strict control of the stereochemistry of a newly formed bond? a. b. c. d. Adrenalin from noradrenalin Noradrenalin from dopamine L-dopa from L-tyrosine L-dopaquinone from L-dopa Choice B is the only choice that involves the creation of a stereocenter. The only change here is the addition of CH3, which does not affect any stereochemistry. No new stereocenter is formed and the asymmetric carbon atom of the amino acid group is carried through unchanged. During the biosynthesis of adrenalin, why is molecular oxygen utilized as a reactant in steps 1 and 3 only? a. The O-O bond in molecular oxygen provides the energy required to drive steps 1 and 3. b. Molecular oxygen is present in the atmosphere; therefore, it is included in steps 1 and 3. c. In both steps 1 and 3, the product contains an extra oxygen atom derived from molecular oxygen. d. Molecular oxygen generates nucleophilic OHions, which attack precursor to form Lepinephrine. Answer A is not correct because when oxygen is bound to itself is pretty stable so using that energy to drive the steps would most likely not be the case. Answer B is not correct because although it is present in the atmosphere, it does not explain why it is utilized as a reactant in steps 1 and 3 only. Answer C makes sense because it adds to L-tyrosine and also onto dopamine so that there is an extra alcohol group. Answer D is not true because this is not going on, you do not get OH- from molecular ions. Melanin is an important skin pigment that protects the skin from harmful radiation. The melanin chromophore absorbs light over a wide range of wavelengths because it contains: a. Carbonyl and amine groups 1. although this is true that they do contain carbonyl and amine groups, it does not refer to the extended pi system, which is the sole reason why melanin is particularly effective at absorbing light across a wide range of wavelengths. b. An extended pi system c. An extended pi system coupled to electron donating amine groups and electron withdrawing carbonyl groups. d. An extended pi system coupled to electron withdrawing amine groups and electron donating carbonyl groups. Conjugated pi systems on its own will absorb light due to the possibility of electron transitions throughout the conjugated system. Therefor we can say that increasing/decreasing amount of conjugation can alter the wavelengths. Amine groups are electron donating because nitrogen atom has a lone pair of electrons that can be donated into the system, which then stabilizes the structure. This enhances the delocalization of pi electrons across the molecule, which is crucial for light absorption. Carbonyl is electron withdrawing. They are characterized as EWG due to the electronegativity of oxygen. The oxygen pulls electron density towards itself, which, then helps stabilize the extended pi-system but does Not act as electron donor in the same way amine groups do. The pi system refers to a series of alternating single and double bonds that allow for delocalization of electrons across the structure. This delocalization means that electrons can move more freely over the length of the molecules, which in turn allows the molecules to absorb light at various wavelengths more efficiently. Which structure represents D-beta-hydroxybutyrate? The D-configuration and L-configuration are used to describe the configurations of carbohydrates and amino acids. These are incorrect because if you look at COO- in reaction one, the carbon that it is attached to is not attached to the OH directly, hence, this structure is incorrect. The attachment demonstrated here is correct. The difference between C and D is the configurations one is L and one is D. To determine the difference: you need to see what side the OH is on. For D it would have to be on the right, and for L it would have to be on the left. The carbon that is most oxidized needs to be on the top. You would have to flip both up 180 degrees to the top (CH2COO-). When you rotate to have the most oxidized carbon at the top, you will find that in C the OH group will end up on the right and in D the OH group will end up on the left. Hence, that is why C is correct. BHB is converted into acetyl-coA in 3 steps, beginning with the reversal of reaction 1. Which compound is NOT a product of one of these steps? a. Butanoic acid a. Not likely to be one of the products hence it is not mentioned at all in this passage. b. NADH c. Acetoacetate d. Acetoacteyl-CoA acetoacetate “The redox reaction that forms BHB from acetoacetate is catalyzed by beta-hydroxybutyrate dehydrogenase and requires the coenzyme NAD+ (reaction 1).” Based on table 1, E°’ for reaction 1 is: a. b. c. d. -0.666V +0.666V -0.026V +0.026V E° for a redox reaction is the sum of E° for the oxidation half reaction and for the reduction half reaction. -0.346 -0.320-→+0.32 (since we are doing the reverse of that reaction because really it should be NADH-→NAD+ + H+ +2e-0.346+0.320=-0.026 ** I had the correct idea I just forgot to flip the sign Stearic acid is found in animal fat and has the chemical formula C17 H35 CO2 H. How many Boxidation steps must stearic acid undergo until it is completely broken down into two-carbon fragments? a. b. c. d. 10 9 8 7 B-oxidation is a process where fatty acids, like steric acid, are broken down in the body. Each step of B-oxidation It is tempting to think that it would be 9 cuts since there is 18 carbons but when you start doing the 2 carbon cuts, when you get to the last carbons you really only need 2 cuts to get the last 4. With the NADH, you are taking the hydride that attacks the oxygen, giving it a negative charge, then you need to add some H+ to get the alcohol in the products. You are taking hydrogen ions out of the solution, out of the cell, so that is increasing the pH because you are lowering the amount of hydrogen ions. Also B and D are incorrect choices because they are the same thing, therefore, it cannot be either of those options since they are the same. Cannot have 2 right answers. C is incorrect because it indicates an increase in the concentration of a reactant (acetoacetate) as opposed to an increase in the concentration of a product. Destruction of motor neurons in the spinal cord that innervate skeletal muscles in the appendages is likely to cause which symptoms in persons with WPV? a. b. c. d. Spastic paralysis with retention of reflexes Flaccid paralysis with retention of sensation Flaccid paralysis and partial loss of sensation Spastic paralysis and loss of all sensation In the passage it states that wild poliovirus (WPV) causes flaccid paralysis when it invades motor neurons of the central nervous system. Spastic paralysis when damage the brain and NOT the spinal cord. Flaccid paralysis when damage the spinal cord. The reason spastic paralysis occurs when there is traumatic injury to the brain is because you can injure something that controls motor function. if you injure something that controls motor function, reflexes occur at spinal cord and not brain. Even though you cannot move your legs, they can still have reflexes. For flaccid, when you hit a reflex, it will stop at the spinal cord and therefore you will not be able to have any reflex response. Sensation has nothing to do with motor neurons, motor neurons only affect movement. When answering this question, I knew that spastic paralysis was not an option because it explicitly stated that flaccid paralysis was caused by wild polio virus. So from a logical standpoint A&D are removed then I became stuck between being C because I did not know anything about sensation but again logically it would not make sense to say partial loss of sensation because nothing in the text stated anything about sensation. Also, on a biological note that is now learned sensation has nothing to do with motor neurons. A healthy person who previously received a polio vaccine and is verified to have high blood serum GMT levels for P1, contracts polio a short time later. Which explanation best accounts for this outcome? a. b. c. d. P2 and P3 serotype proteins have different coding sequences than the P1 serotype. Individuals with high blood serum GMT have high amounts of virus in their blood. this person already had high levels of the P1 virus before receiving the vaccine. Patients with high blood serum GMT are the group most likely to contract polio. GMT= geometric mean titer, a measure of blood serum antibody concentration Option A is demonstrated in table 1, as there is different serotypes of the virus and we can define serotypes as distinct variations within a species of bacteria or virus, classified based on the differences in their surface proteins. Hence, when we see locations of amino acids, this shows the slightly different capsid protein on its surface. Therefore, we can then say that if an individual has a high blood serum for serotype P1, but contracts polo a short time later, we can say that P2 AND P3 serotype proteins have different coding sequences than the P1 serotype. Option B would be incorrect because when someone has a high blood serum they typically have antibodies against the virus therefore the virus would be killed off by now. Option C would be incorrect because the question states that the individual was healthy prior to receiving the vaccine. Option D would be incorrect because someone with high GMT are clearly going to be the ones least likely to contract polio since they have high antibody levels. Among the genes that code for the amino acid sequences in table one comma what is the likelihood of two genes being separated by a crossover event during prophase I of meiosis? a. The A1 and A2a genes are most likely to be separated. b. The A2a and A2b genes are most likely to be separated. c. Both the A2a and A3a genes, and the A1 and A2b genes, are equally likely to be separated. d. None of the genes in table 1 are likely to be separated. Viruses are not made up of cells, viruses do not obtain food to survive, viruses cannot reproduce, hence, we can say that viruses are NOT alive. Obligate intracellular (obligation to live in the cell) parasites (infect) Can exist outside the cell=virion Viruses can be made of: DNA (ds and ss) RNA (ds and ss) Nucleic acids live in the capsid, which is made up of individual proteins called capsomeres, which repeat in a crazy pattern to give us the protein code. Protein coat can take on 2 different shapes: helical and the nucleic acids live in the inside. More commonly, icosahedron, rounded cube with 20 faces. The viral envelope is a lipid bilayer around the protein coat, which is similar to what you would see in cell membrane. These membranes are made from the cells that the virus infects. May contain proteins and glycolipids. Carbohydrates attach to the membrane that participate in cell signaling and that is how a lot of these viruses can identify and attach to the cells they are going to infect. Since viruses are NOT alive they do not replicate from cell division. They require living host for replication. 2 phases for viral replication: lytic cycle - Lyse means to burst. Destructive and deadly phase you are lysing the cell. Lysogenic - Dormant phase, instead of making viral copies, viral genome enters cells DNA and finds a spot in DNA where it cuts the DNA and sticks itself in. Ex) herpes You are lysing the genes, not the cell. Not all viruses have a lysogenic phase. Silence phase, which then can go into lytic phase to cause an eruption. Viral replication: a. Attachment a. Viral particle moves through the system looking for a receptor. b. When the virus recognizes A receptor it will attach and trigger this cell to take in the virus through receptor mediated endocytosis. c. Particle will push in on cell membrane, taking membrane with it, creating a new membrane around itself. b. Viral Entry a. The virus is now inside the cell with its own lipid bilayer, the same one that was around the cell. c. Replication a. In order to expose viral genome, viral membrane has to break, and the viral genetics is injected into the nucleus. b. When the virus takes over the cell machinery it tricks it into making copies. c. DNA→DNA Copies AND DNA→RNA→proteins d. RNA→RNA copies and RNA-→mRNA for proteins e. Retrovirus RNA(copied to form DNA. Completely overrides central dogma)→DNA→RNA→proteins (enzyme responsible from DNA to RNA is called reverse transcriptase) d. Assembly a. Once cell has made several copies of self, they will start to self-assemble. Here, we are packaging up proteins or DNA surrounded by capsid. e. Release a. Cell burst and releases virus particles, now they are looking for new hosts. b. Some particles may escape organism and get released into environment where they will find new environment and host. Incubation period-no symptoms, while virus replicates Most NK cells are cytolytic, but CD56bright CD16-/dim and CD56 dim CD16- cells are immunomodulatory. Which indicator would be least useful in determining if an NK cell population is cytolytic or immunomodulatory? a. b. c. d. Granzyme A Granzyme B Perforin CD16 “Immunomodulatory NK cells, do not express perforin and granzymes, or express them at significantly lower levels compared to the cytolytic NK cells” “perforin is required for the delivery of granzymes into the cell. In the absence of perforin, NK cells are no longer cytolytic.” If you look at the different figures with figure 1 containing expression of granzyme B, you can see that there is a significant difference in the immunomodulatory and cytolytic, which is good because you want to determine differences. In figure 2, expression of perforin is shown, which shows a drastic difference, which is something you want, so that eliminates option B and C. Figure 3, granzyme A expression, shows differences but not as drastic. Therefore, that would be indication of less differentiation. Which conclusion is most strongly supported by the data presented in figures 1, 2, and 3? a. CD56bright CD16-/dim cells produce granzyme A, so they are cytolytic. a. CD56bright CD16-/dim cells we know are NOT cytolytic, you can identify this in figure 2, as they have nearly no perforin expression, which is needed to be considered cytolytic. b. but CD56dim CD16+ cells produce granzymes A and B, but not perforin, so they cannot be cytolytic. a. No a lot of perforin was expressed in figure 2. c. CD56-CD16+ cells produce more of both granzyme proteins in CFS/ME patients than in healthy controls. a. There is no statistical significance indicated in figure 1. Although there is a difference, there is no label identifying it as statistically significant. Healthy control is producing more than the patient. d. Perforin expression is upregulated in the cytolytic cells compared to the immunomodulatory cells. a. We know that cytolytic cells must have perforin for them to be considered cytolytic. NK cells are innate immune cells and cytotoxic T lymphocytes (CTLs) are adaptive immune cells. In what other ways do NK cells and CTLs most likely differ? a. CTLs require pathogen processing and presentation, whereas NK cells do not. b. NK cells are pathogen-specific, whereas CTLs are not. a. NK cells are NOT pathogen-specific as they are apart of the innate immune system. Also, CTLs are highly specific to the antigens they recognize. c. NK cells are leukocytes, whereas CTLs are lymphocytes. a. Both NK and CTLs are a type of WBC. b. Lymphocytes are a subset of leukocytes that include T cells, B cells, and NK cells. Therefore, CTL and NK are both lymphocytes. d. NK cells cannot target tumor cells, whereas CTLs can. a. They both play a role in targeting and killing tumor cells. NK cells can recognize and kill certain tumor cells without prior exposure while CTLs can target tumor cells presenting specific tumor antigens in the context of MHC molecules. Aspect Innate Immunity Adaptive Immunity Components Includes physical barriers (skin, mucous membranes), phagocytes, NK cells, and complement system. Specificity Non-specific; responds to pathogens in a generic Highly specific; targets specific pathogens way. with precision. Memory Does not have memory; response is similar with each exposure. Response Time Immediate; acts within minutes to hours. Consists of lymphocytes (B cells and T cells) and antibodies. Has memory; response is faster and stronger upon subsequent exposures. Slower; takes days to develop but is more targeted. Function First line of defense; limits pathogen spread and works to eliminate invaders. Eliminates identified pathogens and remembers them for faster response in future. Role in Infection Early response to infection; acts before adaptive immunity. Activated after innate immunity; adapts and targets specific pathogens. Recognition Recognizes pathogens through generic patterns on pathogens (PAMPs). Recognizes specific antigens using receptors on B and T cells. Pathogen Removal Phagocytosis, inflammation, and complement activation. Antibody production, cytotoxic activities by T cells, and immune regulation. Macrophages, dendritic cells, neutrophils, NK Cells Involved cells. Activation Does not require previous exposure to pathogens. B cells (produce antibodies), T cells (helper T cells and cytotoxic T lymphocytes). Requires antigen presentation and is enhanced by previous exposure. IF CS/ME is assumed to be a simple autosomal recessive disease, what is the probability that heterozygous CSF/ME parents will have a child whose NK cells exhibit reduced expression of Granzyme B? a. b. c. d. ¾ ½ ¼ 0 Autosomal recessive: condition in which 2 copies of an abnormal gene must be present for the disease or trait to develop. For the trait or disorder to be expressed, an individual must inherit 2 copies of the mutant gene, one from each parent. If an individual receives one normal gene and one gene for the disease, the person would be a carrier for the disease but usually will not show symptoms. Which graph is consistent with the kinetics of adenylate cyclase in the presence and absence of AA? “AA is a competitive inhibitor of adenylate cyclase (ACT)” Feature Competitive Inhibition Non-Competitive Inhibition Binding Site Inhibitor binds to the active site. Inhibitor binds to an allosteric site. Effect on Substrate Binding Directly competes with the substrate. Does not compete with substrate for binding. Effect on Km Increases Km (apparent decrease in affinity). Km remains the same. Vmax remains the same (can be reached at higher substrate Effect on Vmax concentrations). Decreases Vmax (efficiency of the enzyme is reduced). Inhibitor Structure Often similar to the substrate. Not structurally similar to the substrate. Can be overcome by increasing substrate concentration. Cannot be overcome by increasing substrate concentration. Overcome by Substrate **Competitive inhibitors impair reaction progress by binding to an enzyme, decreasing the number of enzyme molecules available to bind the substrate, in non-competitive inhibitors, the substrate is not prevented from binding to the enzyme. Quaternary coordination means that it coordinates with 4 ligands and there are 4 bound atoms or coordination members. This is indicative of a tetrahedral arrangement, and a tehtrahedron has bond angles of 109.5. Enzymes do not change the position of an equilbriu, therefore here, the overall yield of acetoaldehyde would not be altered. The increase of enzmye catalyst Increases the rate of fth ereaction and that is all. An agonist binds to a molecule to an agonizes an effect. Therefore, it would make sense that anxiolytics would target GABA receptors, increasing the effect of GABA. Neurotransmitter Function Role in CNS Role in PNS Activates muscles; Autonomic nervous system functions. Associated Disorders Acetylcholine Muscle activation, arousal, attention, memory, and learning. Involved in attention and arousal. Dopamine Reward, motivation, mood, and movement regulation. Reward system, movement control. - Parkinson's disease, Schizophrenia. Glutamate Synaptic plasticity, learning, and memory. Cognitive functions like learning and memory. - Alzheimer's disease, Multiple sclerosis. GABA Decreases neuronal excitability, induce calming effect. Balances excitatory and inhibitory signals. - Epilepsy, Anxiety disorders. Alzheimer's disease, Myasthenia Gravis.