Midterm Foundations (Quiz 1, 2,3)

On a Monday morning, students at a local college begin arriving at the Health Services Clinic with severe cough that included some bloody phlegm. Many of the students reported attending a party over the weekend at which they smoked a batch of marijuana. The clinician was concerned that a specific batch of marijuana may be responsible for their respiratory illness. The clinician immediately began asking other students attending clinic without the bloody phlegm whether they were at the same party, whether they had recently smoked marijuana, and if so, whether it was a new batch of marijuana to them. They assembled the data into the following table: Smoked new batch Bloody phlegm + Other reason for visit Yes 12 2 No 4 9 What epidemiological design does this most closely resemble? What epidemiological design does this most closely resemble? A. Case control B. Cohort (retrospective) C. Cohort (prospective) D. RCT E. Case series

A. had exposed and nonexposed populations. causation was unknown. Trying to establish correlation to causation. Had the outcome (bloody phlegm) and wanted to establish if the bad pot brownies caused it. They picked their controls. Answer: a. Case control. Why? The study was conducted using two samples: those with the outcome (Cases) and those without (controls). Because it is now a controlled study, it can't be a case series. Because there is no randomization, it can't be an RCT. Because subjects were sampled on the basis of outcome (bloody phlegm +/-), it can't be a cohort study because cohorts are assembled on the basis of exposure.

The concentration of chemotherapeutic agents in some cancer cells is reduced because of the expression in the cell membrane of efflux pumps. Which of the following pairings MOST ACCURATELY describes the type of transporter involved and the source of energy driving efflux? A. SLC family member : K+ gradient across cell membrane B. SLC family member : Na+ gradient across cell membrane C. SLC family member : ATP hydrolysis D. ABC family member : Na+ gradient across cell membrane E. ABC family member : ATP hydrolysis

ABC cassettes correlate with MDR1 / p-glycoproteins which pump drugs out. They are primary transportes/need atp/active transport. SLCs are facilitative/secondary transporters (symport). (uses conc gradient generated by primary transport). Correct Answer: E Feedback: The transporters responsible for removal of chemotherapeutic agents from the brain are members of the ABC family of transporters; they utilize energy from the conversion of ATP to ADP to drive the drug out of the cells against a concentration gradient. Only answers D and E indicate that the relevant transporter system is in the ABC family. Of these answers, only Answer E correctly identifies the source of energy for this transport. Answer E is correct. Answers A, B and C are incorrect since all they propose that the relevant transporter is a member of the SLC family of transporters. SLC transporters use an ionic gradient to provide the energy to transport their substrates against a concentration gradient for the substrate, but that is not relevant to the transport of chemotherapeutic agents out of cells.

MacConkey agar is a commonly used nutritive medium that incorporates bile salts, neutral red dye, and a carbohydrate. These components render the medium A. differential based on glucose fermentation. B. enriched for growth of fastidious organisms. C. selective for Gram-negative organisms and differential based on lactose fermentation. D. selective for Gram-positive organisms. E. uninhibitory to common Gram-positive and Gram-negative species.

AGAR GOES FROM TAN TO MAGENTA/BRIGHT PINK IF LACTOSE FERMENTING C correctly describes the properties of MacConkey agar. It contains bile salts that inhibit the growth of bacteria that do not survive well in the gut such as the Gram-positive organisms of the skin and respiratory tract. The bulk of facultative organisms in the gut are Gram-negative bacilli of the Enterobacteriaceae family, so MacConkey selects for them. Additionally, MacConkey agar contains lactose as one of the carbon sources. Organisms that can ferment lactose will produce acids as a by-product. Acidification of the colonies and surrounding medium causes lactose fermenters to appear bright pink or magenta because the pH indicator neutral red is pink at acid pH. Gram negative organisms that do not ferment lactose will grow but do not change the color of the medium so they are clear or tan on MacConkey agar. Thus the agar differentiates Gram-negatives that ferment or do not ferment Lactose.

A renal transplant patient is started on a daily oral dose of 1 g of cyclosporine. Immediately after the first dose, the serum concentration is determined to be 100 ng/ml. The next day, just prior to the second dose, the serum concentration is determined to be 75 ng/ml. What will the concentration of cyclosporine be at steady state? A. 150 ng/ml B. 200 ng/ml C. 300 ng/ml D. 400 ng/ml E. 500 ng/ml

Accumulation factor 1/ the amount eliminated 100-75= 25 so 1/.25 1/25 equals = .04 x 100 = 4. So its 4x the conc. Multiply starting dose by factor 100 x 4 = 400 answer is D The correct answer is D. To answer this question you need to use the accumulation factor, which is 1/fraction eliminated in one dosing interval. In this case the fraction eliminated in one dosing interval is 0.25. So the accumulation factor is 1/0.25 and the answer is therefore 4 x the concentration after the first dose, or 4 x 100 = 400 ng/ml. Incorrect Feedback: The correct answer is D. To answer this question you need to use the accumulation factor, which is 1/fraction eliminated in one dosing interval. In this case the fraction eliminated in one dosing interval is 0.25. So the accumulation factor is 1/0.25 and the answer is therefore 4 x the concentration after the first dose, or 4 x 100 = 400 ng/ml.

A hormone is known to activate phospholipase C with the subsequent release of calcium from internal stores. The release of calcium most likely occurs as a result of an increase in the concentration of which of the following intracellular messengers? Calcium cAMP cGMP Diacylglycerol Inositol 1,4,5-trisphosphate

Activation of phospholipase C leads to the cleavage of phosphatidylinositol 4,5-biphosphate (PIP2) to produce in diacylglycerol (DAG) and inositol 1,4,5-trisphosphate (IP3). IP3 can diffuse through the cytoplasm binding to receptors in the endoplasmic reticulum to cause the release of calcium that is stored there. google Phospholipase C (PLC) enzymes convert phosphatidylinositol-4,5-bisphosphate into the second messengers diacylglycerol and inositol-1,4,5-triphosphate. The production of these molecules promotes the release of intracellular calcium and activation of protein kinase C, which results in profound cellular changes. (too far down in the pathway) The neutral lipids diacylglycerols (DAGs) are involved in a plethora of metabolic pathways. They function as components of cellular membranes, as building blocks for glycero(phospho)lipids, and as lipid second messengers. IP3's main functions are to mobilize Ca2+ from storage organelles and to regulate cell proliferation and other cellular reactions that require free calcium. In smooth muscle cells, for example, an increase in concentration of cytoplasmic Ca2+ results in the contraction of the muscle cell.

Which best describes the purpose of reflection? A. To criticize oneself for mistakes made in the past B. To develop deeper understanding of oneself and situations for future encounters C. To improve performance without self-evaluation D. To ignore past experiences and approach situations with a blank slate

Answer: B) To develop a deeper understanding of oneself and situations for future encounters Explanation: "Reflection is a 'metacognitive process that occurs before, during and after situations with the purpose of developing greater understanding of both self and the situation so that future encounters with the situation are informed from the previous encounters.'" [1] Reflection is critical for leadership development and adult learning. Reference: Sandars J. The use of reflection in medical education: AMEE Guide No. 44. Med Teach. 2009 Aug;31(8):685-95. doi: 10.1080/01421590903050374. PMID: 19811204.

It is your turn to interview the next patient in your medical interviewing course. You quickly recall some of the tips you were given on effective information gathering skills. Select the best answer that reflects how you want to approach your questions for the patient. A. Offer your patient choices to select from with your questions B. Ask one question at a time C. Sound professional and use medical terms/jargon D. Swiftly move through the interview to stay on task

B The correct answer is B. Remember, the goal is to effieciently, yet completely, acquire relevant and reliable data from your patient. Some approaches to your questions include: Language must be readily understood - avoid jargon, enunciate- Questions must be singular and deliberate, not rapid-fire series. Takeyour time. Ask one question at a time.- Do not put words in the patient's mouth- Questions should aim to promote FREELY FLOWING content from thepatient, followed by PROGRESSIVE FOCUS Start with open ended questions and then get more specific but still open-ended. Avoid leading questions or providing patients with choices to select from.

What are cohort studies assembled on?

Basis of risk or exposure. NOT ON OUTCOME!!!

Receptors for therapeutic drugs typically display which of the following properties? A. Bind ligands and often generate second messengers B. Mobilize release of intracellular calcium by phosphorylation of G proteins C. Metabolize ligands to inactive compounds​​​​​​​ D. Are never located in the nucleus of the cell​​​​​​​ E. Can be saturated by antagonists but not by agonists

Bind ligands and often generate second messengers Correct Answer: A Feedback: Answer A correctly identifies the drug-binding property of receptors (without binding there is no drug action), and notes one possible consequence of a drug-receptor interaction (it does not state that ALL receptors generate 2nd messengers). Answer B is incorrect since not all drug receptors result in the mobilization of intracellular calcium when activated. C is incorrect since receptors do not metabolize drugs. D is incorrect since some receptors may locate to the nucleus of cells to induced their pharmacological effects; an example would be nuclear receptors such as steroid hormone receptors. E is incorrect, because the occupancy of all receptors can be saturated by either agonists or antagonists if given in sufficient quantity.

The ability to have difficult conversations while continuing to communicate, negotiate, plan strategically, and build a team all reside in which competency domain? A. Self-management B. Team management C. Influence and Communication D. Systems-based practice E. Executing towards a vision

C

Drug Y is eliminated by a Phase II reaction in the liver and subsequent kidney excretion. Normally, about 50% of Drug Y in the plasma is eliminated in 4 hours. In a patient who has been taking excessive amounts of Drug Y, you observe that the rate of elimination is holding constant despite increases in the measured Drug Y plasma concentration. A. Elimination mechanisms should still be able to compensate for the excessive amounts of Drug Y. Therefore, the patient likely has kidney malfunction, resulting in decreased elimination of Drug Y. B. Excessive amounts of Drug Y have likely damaged the liver, preventing it from performing Phase II reactions that are necessary for elimination. C. The rate-limiting elimination mechanism is likely saturated; thus, the rate of elimination is following zero-order kinetics. D. This is normal. The rate of elimination for Drug Y is expected to remain constant over the range of therapeutically effective concentrations.

C - although phase II is not rate limiting, I guess it gets too saturated before hand? Couldnt make it to phase II Answer C is best because Drug Y normally exhibits first-order elimination kinetics, which can be deduced by the statement that "50% of Drug Y in the plasma is eliminated in 4 hours". This is just another way of saying it has a 4-hour half life. For first-order kinetics, the rate of elimination increases with plasma concentration. For zero-order kinetics, the rate of elimination is constant, regardless of plasma concentration. The excessive amounts of Drug Y have likely resulted in saturation of elimination pathways, resulting in apparent zero-order elimination. Statement D is false. Statements A and B do not have sufficient supporting evidence.

The duration of action of a drug metabolized by CYP3A4 is likely to be increased by which of the following? A. chronic administration of phenobarbital before and during therapy withthe drug. B. chronic therapy with cimetidine before and during therapy with the drug. C. displacement from tissue binding sites by another drug. D. increased cardiac output. E. chronic administration of rifampin

CYP3A4 - what is metabolized by many drugs. inc bioavailability, drug not broken down. indicates inhibitor. cimetidine is an inhibitor. As well as grapefruit. Who are my inhibitors? macrolide, triazole antifungals, ritonavir, fluoxetine, cimetidine Who are inducers? its the seizure meds, stomach meds, certain antibacterials, w/ other drugs statins, w/other drugs warfarin, w/ other drugs digoxin , they up it benapyrene, carbamazepine(epilepsy,mania, TN), phenobarbital epilepsy phenytoin (epilepsy), rifampin, , omeprazole (1A2) barbituates, phenytoin, primidone, rifampin (antimycobacterium) (2C9) carbamazepine, phenobarbital, phenytoin, rifampin (2c19) ethanol (2e1) barbituates, phenytoin, carbamazepine, rifampin, St john warts (3A4)

A renal transplant patient is started on a daily oral dose of 1 g of cyclosporine. Immediately after the first dose, the serum concentration is determined to be 100 ng/ml. The next day, just prior to the second dose, the serum concentration is determined to be 75 ng/ml.If the elimination half-life of cyclosporine is 69 hr, and the volume of distribution in this patient is 300 L, what is the clearance? A. 0.3 L/hr B. 0.69 L/hr C. 3 L/hr D. 6.9 L/hr E. 30 L/hr

Cl= Ke x Vd Ke = 0.693/69hr .001 x 300 L = 3 L The correct answer is C. The relationship between clearance (CL) and volume of distribution (Vd) is (CL) = keVd, where ke is the rate elimination constant. Since ke = 0.693/t1/2, if the elimination half-life is 69 hr, then ke = 0.693/69 or 0.01. Thus if Vd is 300 L, then CL = 0.01 x 300 = 3 L/h. This is the hardest question you are likely to see on an NBME exam.

A hospitalized patient with pneumonia is given a standard dose regimen of codeine to suppress cough. After several doses his respiration is severely depressed and he becomes comatose. Treatment with an opiate antagonist, naloxone, produces recovery of respiration followed by a rapid return of consciousness. Which of the following statements is the best explanation of these responses? A. Ultra-rapid metabolism by CYP2D6​​​​​​​ B. Impaired metabolism by CYP2D6 C. Ultrarapid metabolism by CYP2C19 D. Impaired metabolism by CYP2C19 E. Fast acetylation by N-acetyltransferase-2.

Correct Answer: A CODEINE IS CONVERTED TO MORPHINE BY CYP2D6 \ when broken down more active. Enzyme was induced to cause more breakdown and too much effective morphine was active. Also be mindful of the antagonist. This was important and was a hint that something was activated that had to be inhibited by this antagonist The D6 - makin mo drugs when metabolized /codeine to morphine C19 - metabolized drugs, like a chinese cargo plane, bombing the CPY , less drugs present Codeine is converted to morphine by CYP2D6. With very rapid metabolism by this enzyme, the rate of conversion is accelerated and the levels of morphine may rise to toxic levels, inhibiting respiration, anoxia, and potentially to coma. The actions of morphine are rapidly antagonized by its antagonist, naloxone. Answer A is therefore a good explanation of the sequence of events experienced by the patient, and is the correct answer. Answer B is incorrect since slow CYP2D6 metabolism results in very low levels of morphine and reduced effectiveness of codeine as a cough suppressant and/or analgesic. Answers C, D and E are all incorrect since they refer to rates of metabolism by enzymes that are not involved in the metabolism of codeine. ALSO side note CYP2C19 is the principal enzyme involved in the hepatic metabolism of drugs such as antimalarial (proguanil), oral anticoagulants (R-warfarin), chemotherapeutic agents (cyclophosphamide), anti-epileptics (S-mephenytoin, diazepam, phenobarbitone), antiplatelets (clopidogrel), proton pump inhibitors (omeprazole, ... anti fungals these are inhibitors.... the C is for cimetine nd being an inhibitor.

An enthusiastic MS-I student while in a lab discovers that some cells are showing giant cell formation during culture. She extracts the cells and performs an assay and finds that virus infecting the cells is in fact an + Single stranded RNA virus. However, when she performs a PCR (Polymerase chain reaction) she finds the cells actually have viral DNA. She concludes that the virus uses A. RNA-dependent RNA polymerase B. RNA-dependent DNA polymerase C. DNA-dependent RNA polymerase D. DNA-dependent DNA polymerase

Correct: The presence of viral DNA derived from a +ssRNA virus indicates that the RNA was reverse transcribed to DNA. Reverse transcription is carried out by the reverse transcriptase enzyme which is also called RNA-dependent DNA polymerase.

You are interviewing a patient who reports that they have had a cough for the last two weeks. After you explored his presenting complaint and finished the majority of your history of present illness (HPI), what else might you obtain to better determine whether there are any additional relevant symptoms they might be experiencing? A. Social history B. Past surgical history C. Review medication list D. Review of Systems

D can go from head to toe and see if theres additional symptoms the pt may have not remember/or thought related The correct answer is D. The ROS is a longer series of questions relating to common but significant symptoms that are often overlooked by patients because they are so common. These are asked at the end of the HPI for such symptoms that are relevant to the HPI, but also at the end of the medical history for other symptoms in an effort to be comprehensive about any other missed symptoms. This assures that all other major areas of a patient's health are covered.

Gastric acid keeps the stomach pH at around 1.4, which facilitates protein digestion and kills many microorganisms that are ingested with food. Gastric acid is secreted by the parietal cells, which have an intracellular pH of around 7.4. What is the ratio of hydrogen ion concentration between the stomach and the intracellular environment of the parietal cells? A. 7.4 / 1.4 B. 1.4 / 7.4 C. 10^6 D. 10^-6

Definitions. pH = -log [H+]; [H+]= 10-pH If the stomach pH = 1.4, [H+] = 10-1.4 If the intracellular pH = 7.4, [H+] = 10-7.4 The ratio of [H+] between stomach and intracellular environment is 10-1.4/10-7.4 = 10(-1.4 + 7.4) = 10^6 Therefore, the correct answer is C.

Simple diffusion?

Diffusion that doesn't involve a direct input of energy or assistance by carrier proteins. SO STRAIGHT ACROSS THE MEMBRANE oxygen is an example

Which one of the following agents have a positive effect on O2 binding to hemoglobin? A. 2,3-BPG B. CO2 C. H+ D. CO E. O2

E is correct. What does oxygen binding by hemoglobin shows positive cooperativity mean? Hemoglobin displays positive cooperativity since the binding of the first ligand increases the affinity for the next, and so on. Such sigmoidal curves are characteristic of cooperative transitions between two distinct states that involve the making (or disruption) of numerous weak (non-covalent) interactions. hemoglobin has a 2 alpha (like myoglobin) and two beta chains. when binding with oxygen, this helps a conformation change occur by helping to expose Fe ion/cofactor more. Helps other subunits bind better. Type of homotrophic allostericism A is incorrect, because 2,3-BPG is a negative allosteric regulator

Your patient weighs 200 lbs (91 kg). Estimate the total body water volume in this patient. A. 5L B. 54L C. 91L D. 120L E. 200L

Facts: . Liters is equivalent to kg, use kg. 60% of the body weight is water. ex.) 70kg is 150 pounds To get from pounds total to kg (if the kg was listed) divide by 2.2. (so 1kg is equal to 2.2 pounds). So 91kg x 0.6 = 54L. That is how you find the bodies total water weight (which is 60% of the total weight) Quiz answer (legit one) t: Total body water accounts for approximately 60% of the total body weight. The patient weight 91 kg and therefore there is approximately 91 * 0.6 = 54 kg of water. Under normal conditions, 1 L of water weighs 1 kg. Therefore total body water is estimated to be about 54 L. SIDE NOTE: dont forget the 60/40/20 rule, 60 percent of total weight is the total water weight, and 40% of that is intracellular water and the other 20% is extracellular water

The rate-limiting step is determines the rate of the reaction. So I knew K3 was at the end. V0 is defined as the rate of the overall reaction. We know that in any multi-step reactions, the rate of the overall reaction is primarily dictated by the rate of conversion of the slowest step (also known as the rate-limiting step), which, in this reaction scheme, is represented by ES -> E + P. The rate constant for the ES -> E + P reaction step is given as k3 , therefore, the rate of the conversion of ES to E + P should be given by k3 [ES], the product of the rate constant (k3) and the concertation of the reactants (in this case, [ES]). Therefore, "E" is the right answer K1/ k2 represents dissociation constant (Kd) of the interconversion between E + S and ES k1 [E][S] represents the rate of formation of the ES complex from E + S k2 [ES] represents the rate of dissociation of ES complex to E + S k1 [ES] does not represent a valid reaction

For the following enzyme-catalyzed reaction, if the conversion of ES to E + P is the rate limiting step, how would you define V0? A. k1 / k2 B. k1 [E] [S] C. k2 [ES] D. k3 [ES] E. k1 [ES]

Plasma osmolarity = easy way? Na * 2 = plasma osmolarity (extracellular osmolarity) Real way 2*Na + glucose/18 + bun/2.8 = plasma osmolarity Plugging in the numbers from the test result, (2x143) + 86/18 + 18/2.8 = 297 mosm/L. Remember that a safe estimate of the plasma osmolarity determined by just multiplying the PNa by 2. In this case, that would have given a result of 286 mosm/L which is pretty close. This becomes less accurate when blood glucose rises significantly and contributes more to the total plasma osmolarity. remember plasma osmolarity is THE SAME as interstitial fluid/extracellular fluid, both are in the tissues. You misread this as intracellular which would be v different .

Give the best estimate of the osmolarity of the interstitial fluid given the following blood chemistry results: 86 mosm/L 143 mosm/L 247 mosm/L 297 mosm/L

Question 460 / 1 pts A mutation in the SLC2A1 gene results in GLUT1 deficiency syndrome, a condition giving rise to a variety of neurological symptoms and usually inherited as an autosomal dominant trait. The major function of GLUT1 is glucose mobilization into brain neurons from surrounding glia cells to replace glucose as it is utilized. Which one of the following mechanisms is involved in GLUT1-mediated glucose permeation across the neuronal membranes? A. Passive diffusion B. Facilitated diffusion C. Secondary active transport D. Primary active transport E. Channel-mediated diffusion

I knew it was facilitated diffusion, but I didnt know how channel mediated diffusion was different. Inherent in facilitated diffusion was the carrier protein I GUESSS. So channels are mainly for a bunch of ions to pass thru. Facilitated diffusion, especially for Glucose uses carrier proteins bc its specific to transporting glucose and it is moved from an area of high to low conc OF glucose (so with the gradient) (REMEMBER - here are two main types—sodium-glucose linked transporters (SGLTs) and facilitated diffusion glucose transporters (GLUT)—which can be divided into many more subclasses.) Test Answer Incorrect: Glucose is a relatively large polar molecule that is repelled by the nonpolar interior of membranes. Therefore it requires other mechanisms to cross lipid bilayers. The uptake of glucose by neuronal cells is down a concentration gradient, and so does not require an active transport mechanism, ruling out (c) and (d). Both channels and carrier proteins (facilitator proteins) move solutes down the concentration gradient. However, channels are mostly involved in moving ions, and in a rapid fashion. Carrier proteins move polar molecules down the concentration gradient through a process called facilitated diffusion. Human GLUT1 is a carrier protein involved in the facilitated diffusion of glucose.

Which one of the following is a feature of transcription but not a feature of replication? A. It results in formation of phosphodiester linkages B. It is primer-independent C. It proceeds in the 5'-->3' direction D. It depends on a DNA template E. It uses precursors containing guanine

ITS b BUT THE TEST IS INCORRECT AND SAYS A IS THE ANSWER

Consider a lipid-soluble drug that is administered by intravenous injection. This drug leaves the blood stream to enter various body compartments, is metabolized in the liver, and the drug and its metabolite are excreted by the kidneys. If the blood concentration of the drug is plotted against time, which of the following would best describe the graph? A. a straight line declining eventually to a blood concentration of zero B. a horizontal line C. a slow initial decline, with a faster rate of decline at later time points D. a fast initial decline, followed by a slower rate of decline at later time points E. an initial rising phase reaching a peak, followed by a liner decline at later time points

IV - it is a huge slide bc it bypasses first pass metabolism. It is absorbed instantaneously so you start balls to the wal high and decline down at a curve. GI absorption on the other hand will have that peak going up to show how much is absorbed then its a gradual downhill from there. Correct Answer: D Feedback: The drug is administered intravenously so we do not need to consider continued absorption that occurs after oral administration; the question ask about the usual mode of elimination. Elimination of most drugs is a first-order process, in which the rate of elimination is proportional to the concentration in blood. Initially the blood concentration is high, but this declines as the drug is removed from the blood stream; thus the change of blood concentration with time will be an initial rapid decline with progressive slowing of the rate of decline, as described in Answer D, the correct answer. Answer A would only be correct is the drug completely saturated the elimination system at all concentration, an unlikely event. All drugs are eliminated by some mechanism, so Answer B, which implies zero elimination, is very unlikely. Answer C is also very unlikely; elimination does not increase in speed as the blood concentration falls. Answer E describes the change in blood concentration with time after oral administration, but that is not relevant to this question. Thus, answers A, B, C, and E are all incorrect.

The major components of bacterial cells can be separated from one another by a variety of cell fractionation techniques. When this is done, the bulk of the Lipid A present is found in which fraction? A. Capsule B. Cytoplasm C. Cytoplasmic membrane D. Outer membrane E. Periplasmic space

Lipid A is embedded in the inner leaflet of the bacterial outer membranes (gram positives have no outer membrane and no lipid A).

You want to estimate the change in vascular resistance in a skeletal muscle during exercise. Before exercise begins you measure the following parameters: Mean arterial pressure = 102 mmHg Mean venous pressure = 2 mmHg Blood flow to muscle = 500 mL/min Once exercise begins in earnest, the same parameters are :Mean arterial pressure = 107 mmHg Mean venous pressure = 7 mmHg Blood flow to muscle = 5 L/min Based on these results, how much did vascular resistance change. A. No change B. 16-fold increase C. 16-fold decrease D. 10-fold increase E. 10-fold decrease

My answer observe that the pressures stayed the same (subt map and mvp and they equal 100 on both accounts). Flow rate tho? it inc from 500 ml/min to 5000ml/min. that is a 10 fold change. Plug it into the resistance formula P= Q*R --> P/Q = R (inverse relationship means R is decreasing 10 fold.) Pressure decreases 10 fold. Test answer You want to estimate the change in vascular resistance in a skeletal muscle during exercise. Before exercise begins you measure the following parameters:Mean arterial pressure = 102 mmHgMean venous pressure = 2 mmHgBlood flow to muscle = 500 mL/minOnce exercise begins in earnest, the same parameters are:Mean arterial pressure = 107 mmHgMean venous pressure = 7 mmHgBlood flow to muscle = 5 L/minBased on these results, how much did vascular resistance change. A. No change B. 16-fold increase C. 16-fold decrease You Answered D. 10-fold increase Correct Answer E. 10-fold decrease

Using a really small pressure measuring device you measure the following parameters inside a capillary: Pc = 30 mmHg πif = 8 mmHg Pif = 1 mmHg πc = 4 mmHg Using Starling's Law, determine whether fluid movement occurs into or out of the capillary. A. IN B. OUT

My answer: remember forces out - forces in. Forces Out Pc = forces outwards against the vessel/outside interstitial fluid pi if = force of the (small) proteins inside the vessel forcin fluid to leave the vessel Forces In Pif = force of the fluid pushing inwards against the vessel pi c = force of the fluid entering the vessel Oncotic pressure: What is the difference between oncotic and osmotic pressure? In summary while osmotic pressure refers to the general force that drives water across a semipermeable membrane, oncotic pressure specifically refers to the osmotic pressure exerted by proteins in the blood plasma. Both concepts are important for understanding the movement of fluids in the body. (30+8)- (1+4) = 33 goin outwards. Why? Osmotic pressure is the force that drives the movement of water molecules from a region of low solute concentration to a high solute concentration, while oncotic pressure is the force exerted by proteins in the blood that draws water into the blood vessels Their answer The correct answer is B. Close examination of the values and comparison to the normal values presented in lecture will show that πc is lower than normal. This is something that might be observed if the concentration of plasma proteins is reduced.

What is mindfulness? A. The process of reflecting on past experiences B. The ability to be present, in the moment, without judgment C. Criticizing oneself for mistakes made in the past D. Approaching the current situation through the lens of past experiences

Not about reflecting, its about being in the NOW in a good beneficial way Answer: B) The ability to be present, in the moment, without judgment Explanation: In order to reflect, one must first learn how to be present, in the moment, without judgment. This is mindfulness. Mindfulness is key to therapeutic presence which is a way of being with patients that promotes healing and connection. There are many benefits of mindfulness practice including improved attention and performance, lower levels of anxiety, and an increased sense of well-being. Your ability to be truly present will be essential in promoting safe, inclusive and equitable healthcare for your future patients. When we practice mindfulness and engage in rigorous reflection, we can start to see and understand people as they are, not as we are. This is the heart of reflective practice. Source: Canvas in the RP 101 Introduction: Objectives and Background

In a study of the association between prediagnostic IgE levels and risk of glioma, (J Natl Cancer Inst (2012) doi: 10.1093/jnci/djs315), 77 of 594 case subjects (patients with glioma) had elevated IgE in a prediagnostic serum sample. 198 of 1,177 control subjects (patients without glioma) had elevated IgE. The odds ratio for the association between glioma and elevated IgE is: A. 0.16 B. 0.34 C. 0.74 D. 0.77 E. 0.81

Odds ratios exposed / total group - exposed Do it for both groups then divide both the odds (which is the ratio part) and get .74 C is correct: the odds a case was exposed is 77 / (594 - 77) = .149. The odds a control subject was exposed is 198 / (1,177 - 198) = 0.202. The odds ratio is .149 / .202 = .74. a. incorrect: this is the risk of exposure among all subjects b. incorrect: this is the proportion of cases among all subjects d. incorrect: this is the proportion of cases exposed, divided by the proportion of controls exposed. The odds ratio is the odds a case was exposed divided by the odds a control was exposed. e. incorrect: this is the proportion of cases among the exposed divided by the proportion of cases among the unexposed. In a case-control study, the proportion of cases is set by the experimenter and is not a true measure of incidence or prevalence, so this measure is not generally used for case-control studies.

Desensitization can be homologous or heterologous in nature; homologous desensitization refers to the loss of response solely to agonists that act at a particular GPCR subtype, whereas heterologous desensitization refers to a more generalized effect involving the simultaneous loss of agonist responsiveness at multiple GPCR subtypes even in the absence of agonist occupation of the other receptors. Desensitization - a rapid loss of receptor function, usually with a time course of msec to sec I misunderstood the question, it was just W being exposed to cells for a long ass time. So of course the receptors were desensitized and the receptors lost function (esp if it was over the msec to sec time). Also, it would be heterologous bc these chemicals are not from the signaling pathway itself, its an outside molecule/drug. Test Answer Correct answer: D Feedback: Panel II shows that after 60 mins of exposure the response to W is much reduced relative to the dose-response curve in the absence of prior drug exposure as depicted in Panel I. At the same time the response to Drug X is unchanged from Panel I. The selective reduction of response to drug W while drug X response is unchanged by the

Panel II of the figure shows concentration-response curves for other cells that were exposed to a 100 µM concentration of drug W for 60 mins, briefly washed, then tested with increasing concentrations of drugs W or X. Which term best describes the loss of maximum stimulatory activity of drug W following prior exposure to the drug? A. competitive antagonism B. irreversible antagonism C. negative allosteric modulation D. homologous desensitization E. heterologous desensitization

Rule 1# shift right is decreasing the receptor activity/effect/its antagonising Shift left, its enhancing the receptor activity/effect/ its an agonist Rule 2# If you look at the example graphs on competitive antagonists vs noncompetitive antagonists, competitive just shift it right (bc remember you can overcome it by adding more substrate and the Vmax/ Bmax effect increases - so the result is the same its just a curve moved to the right) Noncompetitive tho? The curve begins to lie down to take a nap, moving to the right. The noncompetitive curve starts to flatten out. This is because the Kd is not affected, but the Bmax/Vmax is decreasing. This means it will not reach its full effect. In this picture, it shows both curves but says really only the left one is effective at decreasing the NT effect on the receptor. So we know 1 its an antagonist, and 2 its a noncompetitive antagonist. ALSO the question legit says that it really cant inhibit the binding at the main binding site well anyway. TEST EXPLANATION Correct Answer: C Feedback: The left panel shows that drug B reduces the maximum biologic response produced by drug A while increasing its EC50 only marginally. The right panel sh

Question 390 / 1 pts You are attending a presentation in which the speaker is describing a new drug (drug B in the figure) designed to treat a common neurological condition. The left figure shows that drug B can reduce the effect of the neurotransmitter A, which binds to a receptor X critically associated with the condition. The right figure shows that drug B has very limited effect on the binding of transmitter A to the receptor. You conclude that drug B is: A. a reversible competitive antagonist at the site on receptor X to which A binds B. an irreversible antagonist of A at the site on receptor X to which A binds C. an allosteric non-competitive antagonist of transmitter A D. an allosteric potentiator of the actions of transmitter A E. a partial agonist acting at the same site as transmitter A

Sickle cell anemia results from a point mutation replacing a glutamate residue by a valine residue at position 6 in one of the subunits of the oxygen carrier protein hemoglobin. This mutation severely affects oxygen carrying capacity of the hemoglobin molecules. Hemoglobin binds oxygen through a tightly coordinated heme group at its active site. This function of hemoglobin is mediated by a highly specialized protein structure known as A. Rossman Fold B. Leucine Zipper C. Globin Fold D. beta-Barrel E. Zinc Finger

A is incorrect because Rossman Fold is meant of NAD+ binding (the alpha beta alpha barrels/ called the sandwich fold) in lactate dehydrogenase (changes pyruvate to lactate) B is incorrect because Leucine Zipper is designed for protein-protein interaction (2 alpha helices, the AA interact every 7th residue, help with dimerization but do NOT help with DNA binding. (heptad repeat) C is the correct answer, since globin fold has evolved for heme binding (2 alpha helices) helps oxygen bind cooperatively /positively D. Beta-barrel structure not found in hemoglobin, it is designed for carrying hydrophobic cargo in aqueous medium spans thru the cell membrane, hydrophobic core, has antiparrallel beta strands, 2 beta sheets put together like a barrell like structure, binding site for retinol/vitamin A protects from hydrophilic environment E. Zinc finger motif is designed for Zinc-dependent DNA binding. Function is DNA binding. Has alpha and beta sheets. Finger/penis lookin structure have 2 his and 2 cys coordinating/holding Zinc in place. In many transcription factors. (fem testicularization) Extras helix loop helix - 2 alpha helices joined by loop, Ca+ binding , loop region binds the Ca (part of muscle troponin c) helix turn helix - 2 alpha helices joined by small loop turn interacts with the major groove of DNA/DNA binding, one of the helices interacts with dna (not the loop)

C is correct grapefruit increases bioavailability Correct answer: C Feedback: Grapefruit juice contains an agent that inhibits the enzyme CYP3A4, both in the liver and in the gut wall - both sites where atorvastatin is subject to metabolism by CYP3A4. As a result, drinking a significant amount of grapefruit juice will inhibit the metabolism of atorvastatin, modestly increasing the peak blood concentration of the drug and slowing its elimination by metabolism to inactive products. The only correct answer is C, both peak concentration and elimination half-life will be increased. Answer A is incorrect since the elimination half-life will be increased, not reduced; Answer B is incorrect since peak concentration will not be reduced. Answer D is incorrect since both peak concentration and half-life will be increased not reduced. Answer E is incorrect since there will be a change in both peak concentration and half-life. more: grapefruit, cpy3A4 - inhibitor - inactivate heme via alkylation isofrolole - CYP1A - inhibitor ethanol - CPY2E1 - inducer - inc prod of CYP enzymes , dec drug in blood plasma levels charcoal, tobacco, green veggies CYP1A - inducers - drug not as effective

Question 61 / 1 pts A 50-year-old salesman who regularly takes atorvastatin (metabolized by CYP3A4) for control of cholesterol is on a road trip where he enjoys a large glass of grapefruit juice every morning (not an item in his usual; diet). What effect might the grapefruit juice have on his peak blood level of atorvastatin following each dose, and on the elimination half-life of the drug? A. Peak concentration will be increased and elimination half-life will be reduced B. Peak concentration will be reduced and elimination half-life will be increased C. Peak concentration and elimination half-life will be increased D. Peak concentration and elimination half-life will be reduced E. Peak concentration and elimination half-life will be unaffected

Negative sense RNA viruses

RNA genomes that need to be converted into the proper form to be made into proteins require synthesis of an RNA strand complementary to the negative sense RNA strand -used as template for protein synthesis -must carry an RNA replicase in virions to make complementary strand

If in an enzyme-catalyzed reaction (E + S <--> ES --> E + P) with starting substrate concentration of 5 mM, the initial velocity (V 0) reaches 20% of the Vmax, then the calculated K m for the substrate is which of the following? A. 2mM B. 4mM C. 16mM D. 20mM E. 25mM

So we are technically "given" the Vo, Vmax, initial substrate conc. Find the Km. Bc they gave us all this and they're initial concentrations (not final or simply plots or overall rates) we need to use the michaelis menten equation. First to make sense out of the Vo is 20% if the Vmax, do the math to find the fraction. 100/20 = 5. So the fraction is Vo is 1/5 parts (with Vmax being the total of 5 parts) Then juggle the michealis menten equation around. Vo= Vmax*S/ Km + S ---> Vo/Vmax = S/Km + S Plug in 1/5 = 5 / Km +5 Move top down 1/25 = 1/ km+5 Butterfly (if the tops the same I guess?) 25 = km + 5 Subtract 25-5 = 20mM 20 mM = Km Test answer Incorrect: The rate (V0) of an enzyme-catalyzed reaction (E + S <-> ES -> E + P) is given by V0 = (Vmax . [S])/ (Km + [S]). If (V 0) = 20% of the Vmax, then one can write V 0 / Vmax = 1/5. Rearranging the equation above, we get V 0 / Vmax = [S]/ Km + [S] Replacing V 0 / Vmax by 1/5 and [S] by 5 mM, we get 1/5 = 5 mM / Km + 5 mM 1/25 mM = 1/ Km + 5mM 25 mM = Km + 5mM Km = 25 mM - 5mM = 20 mM Therefore, "D" is the right answer Google What are the requirements for the Michaelis-Menten equation? Michaelis Menten Equation Vmax describes the maximum reaction velocity, and [S] is the substrate concentration. The variable Km is the Michaelis constant. It is equal to the substrate concentration when the reaction velocity (V) is equal to half the maximum reaction velocity, or half Vmax.

You are interviewing a patient in your medical interviewing course. After appropriately opening the discussion with a personal greeting that shows genuine interest, displaying welcoming nonverbal behavior, obtaining the patient's initial concerns and negotiating the agenda of the visit, you begin to gather more detailed information, mindful of the importance of also building the relationship in order to optimize the quality and quantity of information. Which of the following is one of the 3 major components of the EEC element, "build a relationship"? A. Listening B. Facilitating expression of feelings C. Eliciting patient concerns D. Effective transitioning E. Use of open-ended questioning

The 3 components of building a relationship on the EEC are: Listening, Empathy, and Nonverbal Behavior. This was from the dotcom stuff remember LEN for building a relationship. Correct Answer is A.- Listening B. While effective facilitation ofexpression of feelings will enhance and build therelationship, this component is more important inunderstanding the patient's perspective. C. Eliciting patient concerns- Eliciting patient concerns is moreimportant in understanding the patient's perspective. D. Effective transitioning- Effective transitioning is animportant component of "gathering information". E. Use of open-ended questioning- Open ended questioning is an important component of "gathering information.

In the michelis menten, you can find Km by Vmax/2 In the lineweaver burke plot, you can find Km by locating where the line intersects on that X axis and by dividing 1/S = Km (S is where the X axis intersection is) So, 1/-4 = 0.25 Correct: In a double reciprocal plot of an enzyme catalyzed reaction, the Km is given by the negative reciprocal of the X-axis intercept (-{1/Km}) of the straight line (representing the rate of the reaction), which in this particular plot is at - 4 mM-1 point mark on the X-axis. The negative reciprocal of the value - 4mM-1 is 0.25mM (option D). None of the other values (0.5 mM, 1 mM, 2.5 mM, 4 mM) represent the negative reciprocal of the X-axis intercept (- 4mM-1).

The calculated Km for the substrate in the above reaction will be A. 0.25 mM B. 0.5 mM C. 1 mM D. 2.5 mM E. 4 mM

The Receiver-Operator Characteristic (ROC) curve helps determine: A. the trade off for test characteristics between sensitivity and specificity B. the best time of day to perform a laboratory test based upon circadium rhythms. C. how the accuracy of a test degrades over time D. the trade off for test characteristics between precision and accuracy.

The correct answer is A. Most tests yield values on a continuum. The ROC curve helps us determine the the trade off for the test characteristics between sensitivity and specificity. Best cutoff is usually near the "shoulder" of the ROC curve and better tests have more area under the ROC curve.

Question 310 / 1 pts An enlisted soldier presents to the the sick call tent during a training exercise out in the field. They are complaining of a productive cough and have a fever. When you listen to their lung fields you note that their breath sounds are absent in the right lower lung field. You are concerned that they might have pneumonia with possible effusion (infected fluid) surrounding the lung. You don't have any access to radiology where you are but you know of some examination techniques that might help you determine if they might have pneumonia or an effusion. Using the ball of your hands positioned at the bases of their lower lung fields, you ask the patient to repeat the words "ninety-nine" or "one-one-one" to compare the vibrations from side to side. What are you trying to detect upon examination that would suggest pneumonia or fluid in the lung? A. Crackles B. Fremitus C. Rhonchi D. Hyperresonance

The correct answer is B- fremitus. Fremitus refers to the palpable vibrations that are transmitted through the bronchopulmonary tree to the chest wall as the patient is speaking and is normally symmetric. Asymmetric decreased fremitus raises the likelihood of unilateral pleural effusion, pneumothorax, or neoplasm , which decreases transmission of low-frequency sounds. Asymmetric increased fremitus occurs in unilateral pneumonia which increases transmission through consolidated tissue. Crackles and rhonchi are sounds your HEAR upon AUSCULTATION but are not what you would feel upon palpation. You can also percuss the lower lung fields by tapping with your plexor finger to look for different sounds. Hyperresonant sounds are loud and high pitched which typically mean there is air trapped in the lungs (possible pneumothorax or hyperinflated lungs due to chronic obstructive lung disease or asthma).

Failure to decolorize during the Gram stain procedure will result in bacteria that appear to be: A. Gram negative B. Gram positive C. Gram refractory D. Gram variable

The correct answer is B. Decolorizing is to remove unbound crystal violet. Failure to remove crystal violet results in purple bacteria. Purple color in the Gram stain indicates Gram+.

Which of the following properties of selective agar allows it to differentiate various types of bacteria? A. Demonstrates hemolysis B. Inhibits growth of some bacteria C. Makes the colonies different colors D. Makes the colonies smell different

The correct answer is B. Selective agar inhibits growth of some bacteria. Example, mannitol salt agar has a high salt content which Staphylococcal species tolerate, but which many bacteria do not. FUN FACT - mackonkey contains bile salts so it replicates a gut like envt- gram postiive do not like this and dont grow in these kind of salt shit

You are seeing a patient who arrived to the ED after being involved with a motor vehicle accident. They lost consciousness for a short while so you want to do a thorough neurological exam. You ask them to shift their gaze from a far object to a near object. What would you expect to see in a normal patient? A. Pupils enlarge B. The lens becomes less convex C. Pupils constrict D. The near object becomes blurry

The correct answer is C (pupils constrict). This is referred to as Near Reaction. This response is mediated by the oculomotor nerve (CN III). Coincident with this pupillary constriction, but not part of it, are (1) convergence of the eyes, a bilateral medial rectus movement; and (2) accommodation, an INCREASED convexity of the lenses caused by contraction of the ciliary muscles. In accommodation, the change in shape of the lenses bring near objects into focus (to maintain a clear image).

A 70 kg man who has been exhibiting symptoms of a stroke for 30 minutes is brought to the emergency room. You determine that administration of the thrombolytic agent alteplase is the appropriate course of treatment. The volume of distribution of alteplase is 0.1 L/kg, and its elimination half-life is 5 minutes. The product instructions state that an intravenous bolus dose of 70 μg/kg should be administered, followed by an intravenous infusion of 700 μg/kg over the next 60 minutes. What is the serum concentration immediately after the bolus dose? A. 70 μg/L B. 460 μg/L C. 700 μg/L D. 4.9 mg/L E. 7.0 mg/L

The correct answer is C, 700 μg/L. Remember that volume of distribution (the apparent volume into which the drug will be distributed) is equal to the amount of drug in the body divided by the drug's concentration in plasma at time zero, or Vd = amount of drug in the body/drug concentration in plasma. Therefore, the serum concentration = drug amount/Vd or 70μg/kg divided by 0.1 L/kg or 700 μg/L. Vd = drug/ drug conc plasma (in units its just mg/L) so it was just the 70 ug/kg IV bolus immediate divided by kg/0.1L = 700 if you have the right units and a formula that matches, you got everything DONT NEED THE WHOLE RoA equation if Vd and drug

Vo is the overall rate of the reaction which is the left side ledger. You can find Vmax from this by looking at the plateau of the graph. Vmax is only that flat top line across the highest plateau part of the graph saying you made it! your saturated and working your hardest! In the michealis menten graph, the Vmax is that top mark. In the lineweaver burke plot, you look at where the line intersects the Y axis. It is 1/vmax however. So to find Vmax, you must find that point of intersection on the Y axis and put it under 1. So in this problem its V max = 1/0.4 = 2.5 Their answer In a double reciprocal plot of an enzyme catalyzed reaction, the value of the Y-axis intercept of the straight line (representing the rate of the reaction) represents the 1/Vmax. In this particular plot the Y-axis intercept is at 0.4 µM/min (D), which equals to 1/Vmax. Solving 1/Vmax = 0.4 µM/min, we get Vmax = 1/0.4 µM/min = 2.5 µM/min. Therefore, "D" is the right answer.

The figure below represents the double reciprocal plot of an enzyme-catalyzed reaction carried out at different substrate concentrations. What is the calculated Vmax of the reaction? A. 0.1 µM/min B. 0.4 µM/min C. 1 µM/min D. 2.5 µM/min E. 5.0 µM/min

go up in a line from the top dots that are linear. Get 10 mg/L for plasma conc Vd = drug/plasma conc 100/10 = 10L Answer A

The graph depicts the plasma concentration of Drug M after a single 100 mg dose given orally. Drug M is 100% bioavailable. What is the approximate Volume of Distribution (Vd) of Drug M? A. 10 L B. 50 L C. 100 L D. 800 L E. 1000 L

Correct Answer = C. A short sequence in U1 is complementary to the 5' splice site. This mutations makes this 5' splice insufficiently complementary to U1 for it to bind, hence blocking one of the initial steps of splicing. U1 cant recognize it. GOOGLE In the canonical pathway, U1 snRNP recognizes the pre-mRNA at the 5′ splice site (5′SS); then, U2 snRNP stably binds the branch site (BS) region to form a prespliceosome. U4/5/6 tri-snRNP then joins, after rearrangements, U1 and U4 snRNPs are released, and the remaining U2/5/6 core forms the catalytic spliceosome.

The sequence of the sense strand of one 5' splice site in the SQY gene is shown below. In a mutant form of SQY, three bases in the intron are mutated, as indicated in the figure inhibiting splicing of this intron. Which of the following is the best explanation for the inhibition of splicing in this case? A. an intronic splicing silencer has been mutated. B. an exonic splicing enhancer has been mutated. C. the 5' splice site is no longer complementary to U1 snRNA. D. the 5' splice site is no longer complementary to U2 snRNA. E. U2AF no longer binds to the 5' splice site.

when looking at the template strand, if there is not a noncoding strand, make it underneath. The while its 5--/ /-->3 prime and the one underneath is 3-- / /--> 5, pick the tip left 5' corner sequence and the bottom right 5" corner sequence So it would be 5--/ /-->3 3-- / /--> 5, align both picks to the right direction the question is showing 5-->3 for both and you got the answer Test Answer E PCR amplification of a DNA can only be done using those terminal sequences whose 3' ends facetowards the region to be amplified and not using those whose 3'-ends face away from the regionto be amplified. For examples, in the figure below wherein both strands of the template DNA areshown, only the red sequences (p and s) and not the black sequences (q and r) meet this criterion. p q5'-ATGCATCAGCAACACTCCGT--------//--------TAACCTTCCTTTTGACTTAT-3'3'-TACGTAGTCGTTGTGAGGCA--------//--------ATTGGAAGGAAAACTGAATA-5' r sThe primer set-E contains p and s in the proper orientation. Therefore 'e' is the correct answer.Primer set-A includes p and reverse-oriented q. Primer set-B has sequences p and q. Primer set C has q and reverse-oriented p. Primer set-D has s and reverse-oriented p. Therefore, A thro

The sequence of the termini of one of the strands of a template DNA fragment and the sequences of five primer sets are given above. Which one of the five primer sets will be useful in amplification of the template DNA by PCR? A. Primer set A B. Primer set B C. Primer set C D. Primer set D E. Primer set E

partial agonists are antagonsits to full agonists. We can see in graph C, it can only go to half the graph while A's graph can go to the full Bmax (vmax). Then when they come together, C mellows A out to 50%. This is how we know its a partial agonist. TEST ANSWER Correct Answer: E Feedback: Drug C has a lower maximum response than A, and reduces the effects of A when given together. Since it produces an agonist effect when given alone, it is not a competitive antagonist (answer A is incorrect), not a non-competitive antagonist (answer B is incorrect). C cannot produce as large a maximum response as A, yet it acts at the same receptor; it is therefore not a full agonist (answer C is incorrect). An inverse agonist produces a reduction in baseline response, but drug C gives as modest agonist effect; it is therefore not an inverse agonist (answer D is incorrect). The only correct answer is E, partial agonist; drug C is a weak agonist that inhibits the actions of a full agonist - i.e. a partial; agonist.

The speaker also describes a second new compound (drug C) to treat the same condition. The figure above shows the pharmacological properties of drug C. Drug C also binds to receptor X. You conclude that drug C is: A. a competitive antagonist at receptor X B. an allosteric non-competitive antagonist at receptor X C. a full agonist at receptor X D. an inverse agonist at receptor X E. a partial agonist at receptor X

What dose of Drug Z should be injected intravenously every 5 hours to obtain an average steady-state plasma drug concentration of 10 mg/L? The label on Drug Z indicates a Clearance (CL) of 20 L/h and a Volume of Distribution (Vd) of 100 L. A. 1 mg B. 100 mg C. 200 mg D. 500 mg E. 1000 mg

Vd = drug dose/ plasma conc mg/mg/L 10x100 = 1000mg Dosing Rate = CL * Css; at steady-state, the dosing rate is equal to the clearance rate ALSO Yes, you can use the volume of distribution (Vd) equation to determine the dose of Drug Z that should be injected intravenously every 5 hours to obtain an average steady-state plasma drug concentration of 10 mg/L. The formula you can use is: 10 mg/L⋅20 L/h⋅5 h1=1000 mgDose=110mg/L⋅20L/h⋅5h​=1000mg

What is the difference between oncotic and osmotic pressure?

What is the difference between oncotic and osmotic pressure? In summary while osmotic pressure refers to the general force that drives water across a semipermeable membrane, oncotic pressure specifically refers to the osmotic pressure exerted by proteins in the blood plasma. Both concepts are important for understanding the movement of fluids in the body.

Which of the following agents, when used in combination with other anti-HIV drugs permits dose-reductions? A. cimetidine B. efavirenz C. ketoconazole D. procainamide E. quinidine F. ritonavir G. succinylcholine H. verapamil

Who are my inhibitors? macrolide (ABX), triazole antifungals (FUNGAL DRUGS), ritonavir(HIV PROTEASE INHIBITOR), fluoxetine (SSRI), cimetidine (H2 RECEPTOR ANTAGONIST) - who fits the bill best? ritonavir - HIV/AIDS is VIRulent Ritonavir inhibits hepatic CYP P450 drug metabolism; it is used in low doses in combination drug regimens for the treatment of HIV, permitting dose reductions of other protease inhibitors such as indinavir. The answer is F. Some of the other listed drugs may inhibit some CYP P450 enzymes, but are not used in combination therapies for HIV. Others do not inhibit P450 enzymes.

A patient is prescribed an anticoagulant after experiencing a stroke. He is given the recommended dose of warfarin and twice-weekly measures of prothrombin time (also known as INR) are commenced. After a week of treatment he has a highly elevated INR and reports severe bruising in his thigh in response to a slight knock. The most likely explanation for the excessive bruising is that A. he has a polymorphism for the enzyme vitamin K epoxide reductase complex 1 (VKORC1), lowering sensitivity to warfarin inhibition of the clotting process. B. he carries a fast metabolizer allele for the enzyme CYP2C9. C. he carries a slow metabolizer allele for the enzyme CYP2C9. D. he carries a fast metabolizer allele for the enzyme N-acetyl transferase-2 (NAT-2). E. he is a heavy smoker.

a lot of drug still present, the drug is not being broken down well. He a slow metabolizer C9 - metabolizes drugs, The symptoms described, markedly increased clotting time with evidence of internal bleeding (bruising) after a minor knock, suggest that this patient is hypersensitive to warfarin. Genetic variants in the enzymes metabolizing warfarin (mainly CYP2C9) and the enzyme on which it acts to modify the clotting process (VKORC1) are relatively common. A. Incorrect, since the VKPRC1 polymorphisms he expresses lowers the sensitivity to warfarin, it does not increase sensitivity. B. Incorrect, since a CYP2C9 polymorphism that induced fast metabolism would lower his sensitivity to warfarin, not increase sensitivity. C. Correct; a slow CYP2C9 metabolizer is likely to be hypersensitive to warfarin. D. Incorrect. Warfarin sensitivity is unrelated to the activity of NAT-2. E. Incorrect. Smoking tends to induce CYIA1 enzymes, with little overall effect on warfarin metabolism; induction by smoking can not explain increased warfarin levels.

Which of the following statements describes the meaning of "therapeutic index"? A. A catalog of medicines available in a drug formulary B. The ratio of the mean lethal dose of a drug in 50% of a specific population to the mean toxic dose. C. The ratio of the mean toxic dose of a drug in 50% of a specific population to the mean therapeutic dose. D. The logarithm to the base 2 of the mean therapeutic dose of a drug in a specific population.

a relative measure of drug safety. It is a ratio of median toxic dose or mediam lethal dise divided by median effective (therapeutic) concentration. The higher the TI, the higher the safety of the drug. Formula: TD50/ED50 OR LD50/ED50. TD 50 is the toxic dose - enough of the drug to produce a toxic effect. LD50 dose of the drug that induces death causing death in 50% of the animals ED 50 dose of the drug giving a response equal to 50% of the max response OR a given effect/response in 50% of the population

A 14 year old boy playing outside has an encounter with a member of the Apis mellifera species (a honey bee). During the encounter, he is stung. Within minutes, the area becomes erythematous, pruritic and edematous. Based on your knowledge of Starling's Law of Capillary Transport, what is the most likely cause of edema in this boy? A. Decreased capillary hydrostatic pressure (Pc) B. Increased interstitial fluid hydrostatic pressure (Pif) C. Increased capillary colloid osmotic pressure (Pc) D. Decreased interstitial fluid colloid osmotic pressure (Pif) E. Increased capillary permeability constant (K)

allergic reaction releases histamines. Inflammation at the site can change the capillary permeability constant What do we mean by capillary permeability? noun. The property or capability of capillary walls to allow the selective flow of substances and cells into and out of the vessel Tests answers Starling's Law of Capillary Fluid Movement Net fluid movement = K * { (Pc + Pif) - (Pif + Pc) } In this case, the only one of the possible answers that could cause an increase in the amount of fluid that exits the capillaries and accumulates in the interstitium is increased capillary permeability constant (K). This can be a result of histamine released locally in response to a bee sting.

Grapefruit juice contains factors that modulate the activity of cytochrome P450 drug metabolizing enzymes. Which of the following effects of grapefruit juice is most significant clinically? A. inhibition of the metabolism of drugs by CYP3A4 B. inhibition of the metabolism of drugs by CYP1A1 C. induction of CYP2E1, resulting in increased toxicity of some frequentlyused drugs. D. induction of CYP1A1 activity, resulting in the enhanced conversion ofpro-carcinogens to active carcinogens. E. impaired conversion of codeine to morphine.

anotha one, CYP3A4 inhibitor, grapefruit and cimetidine. Ingestion of grapefruit juice in sufficient quantity inhibits predominantly the CYP3A4/5 isoform of P450.A is correct since this is the action of the active principle(s) of grapefruit juice.B. Incorrect, since there is less inhibition of this isoform, and many fewer drugs are metabolized by this isoform.C. Incorrect, since grapefruit juice does not induced P450 enzymes.D. Incorrect, since grapefruit juice does not induced P450 enzymes.E. Incorrect, since grapefruit is not a strong inhibitor of CYP2D6 the isoform responsible for the conversion of codeine tomorphine.

Difference between blood, plasma, and serum?

blood - all of it plasma- all of it except formed elements (RBC, WBC, platelets) serum - all of it except formed elements (RBC, WBC, platelets) AND clotting factors

A 54-year-old woman takes an oral dose of aspirin, a weak acid to relieve muscle pain. A few minutes later she experiences significant gastric irritation and pain, and takes several spoons of sodium bicarbonate as an antacid. Apart from the episode of copious eructation (belching) that this induces, what effect will the sodium bicarbonate have on the absorption of aspirin from the stomach? A. The rate of aspirin absorption will not be changed. B. Blood levels of aspirin will increase more rapidly. C. Blood levels of aspirin will increase more slowly. D. Bicarbonate will complex with aspirin in the stomach to neutralize the charge thus accelerating its absorption. E. The production of carbon dioxide gas will result in precipitation of aspirin in the gastric fluid, delaying absorption.

c IS CORRECT the henderhassle balch ratios are what pulls the weak acids out of the stomach and into the more basic blood plasma. If the bicarb neutralizes it even more, there will be much less/drive to move it into the blood. Also it changes it to be ionized which hurts it from crossing thru the membrane. Ingestion of sodium bicarbonate will increase gastric pH from its normally acidic level; increasing pH towards neutral levels will increase the degree of ionization of a weak acid such as aspirin, thus reducing the percent in a form able to cross lipid membranes readily. The effect will be to delay the absorption of aspirin. C is correct. A, B, D and E are incorrect.

chemical name code name adopted name proprietary name

chemical name: precise but usually too long code name : manufacturer's initial code; used in preclinical studies until potential therapeutic use is established adopted name: US adopted name; selected by AMA, APhA and the US Pharmaxopeia (approved name) THIS IS THE DRUG NAME I WILL USE Proprietary name: describes the medicine; may include more than one drug, composition may change

Which of the following gene transfer mechanisms requires homologous recombination? he following gene transfer mechanisms requires homologous recombination? A. Conjugation B. Generalized transduction C. Transformation (of a complete replicon/plasmid) D. Transposition E. Specialized transduction

conjugation: A temporary union of two organisms for the purpose of DNA transfer. (cell to cell transfer of DNA) CORRECT ANSWER Generalized transduction: DNA is transferred from a donor cell to a recipient via a bacteriophage. Bacterial DNA packaged on accident in the viral phage. Bc random fragments (not like specialized where they were picked and specifically incorporated) , the donor DNA is not a replicon, its stable inheritance of the donor DNA requires recombination, failure of donor DNA to recombine with the recipient =abortive transduction where donor DNA is degraded. Transformation: modification of a cell or bacterium by the uptake and incorporation of exogenous DNA (a complete plasmid/replicon means it can replicate in recipient/ that the DNA can replicate itself independently from a single origin, dont need no help replicating - competence, environment, and proximity matter Transposition: "jumping genes" Do not require homology (the state of having the same or similar relation, relative position, or structure.) between the donor site and the recipient site. Need transposases. Independent of the Rec A protein. (What is the role of RecA in DNA repair? RecA is the major enzyme involved in homologous genetic recombination and recombinational repair. The protein coats single-stranded DNA (ssDNA) and causes it to invade double-stranded DNA (dsDNA), thereby catalyzing strand exchange between DNA molecules (37).) Specialized transduction: DNA is transferred from a donor cell to a recipient via a bacteriophage. Due to a specific mutation in the phage, it cuts out bacterial DNA, has it become a permanent part of the phage DNA, and continuously expresses it. Test answer Homologous recombination is the reciprocal exchange between donor and recipient DNA that are homologous in nucleotide sequence. Generalized transduction relies on sequence homology between the chromosomal DNA and the genetic material to be newly incorporated. Conjugation and plasmid transformation utilize replicons, which do not require any homologous genetic material as they do not enter the host microbe's chromosome. Transposition does not require sequence homology for gene transfer to occur.

3 means for genetic recombination/horizontal gene transfer in bacteria

conjugation: A temporary union of two organisms for the purpose of DNA transfer. Transformation:(genetics) modification of a cell or bacterium by the uptake and incorporation of exogenous DNA Transduction: DNA is transferred from a donor cell to a recipient via a bacteriophage

What do case series not have?

controls. They have no controls. Just a big ol complied group of ppl with similar symptoms/conditions.

Succinylcholine has a very short duration of action because its elimination/metabolism occurs primarily by which of the following mechanisms? A. ​​​​​​High first-pass metabolism in the liver B. Very rapid filtration of the unmetabolized drug by the kidney with no reabsorption C. Metabolism by CYP450 enzymes in the intestinal wall D. Metabolism by plasma cholinesterases E. Metabolism by nerve terminal choline oxidases

correct answer is D succinylcholine is a skeletal muscle relaxant, these drugs dont enter neurons thats not a thing i guess. It is acted on by my homies in the nicotinic receptors /end plate motor receptors. Which makes sense if it is a skeletal muscle relaxant. Correct Answer: D Feedback: Succinylcholine is an ester that is rapidly metabolized by plasma cholinesterase. The only correct answer is D. Answer A is incorrect since the drug is largely destroyed in plasma before it reaches the liver. Answer B is incorrect since the drug is largely destroyed in plasma before it reaches the kidney. Answer C is incorrect since the drug is not given by the oral route; metabolism in the gut wall is irrelevant. Answer E is incorrect since the drug, carrying a positive charge at physiological pH, does not enter into neurons; its action is on the extracellular acetylcholine binding site of nicotinic cholinergic receptors. Oxidation of choline in nerve terminals is not relevant to the elimination of succinylcholine.

Intrinsic host factors that inhibit establishment of infection

cough and gut movement Answer: B and C. Antibiotics, pasteurization and wearing a face mask may decrease establishment of infection but are not host factors. Other host factors include skin as a barrier, acidic pH of stomach, saliva (anti-microbe factors present), shedding of infected cells.

Affinity is a a measure of which of the following?

drug binding

After a training exercise on an extremely hot summer day, a young recruit complains of dizziness and appears confused. After speaking to the patient you learn that he had consumed 6 L of water during the 1 hour training exercise. You order a blood chemistry analysis. The results show: Na = 130 mmol/L (normal 135-145 mmol/L) K = 8.4 mmol/L (normal 3-5 mmol/L) Glucose = 50 mg/dL (normal 70-100 mg/dL) BUN = 12 mg/dL (normal 8-26 mg/dL) Estimate the patient's plasma osmolarity. A. 130 mosm/L B. 200 mosm/L C. 267 mosm/L D. 297 mosm/L E. 322 mosm/L

easy way? Na * 2 = plasma osmolarity (extracellular osmolarity) Real way 2*Na + glucose/18 + bun/2.8 = plasma osmolarity Correct answer is: (2*130) + (50/18) + (12/2.8) = 267 mosm/L. The plasma osmolarity is low in this patient and the dizziness and confusion may be the result of this electrolyte abnormality.

Expression of a eukaryotic gene is abolished by a deletion of 10 bases that are 60 bases upstream of its normal transcription start site. Which of the following is the most likely mechanism leading to loss of gene expression? A. loss of a 5' splice site. B. deletion of the CAAT box. C. deletion of the sigma factor binding site. D. alteration of the translational reading frame. E. deletion of translation start site.

expressing euk gene w/ gene deletion? affects dna to mrna so transcription. What is 20 upstream TATA box what is 60-80 up? CAAT box. What does rna polymerase II need to recognize/ be stabilized enough to bind to the dna? A promoter sequence (remember RNA pol dont need no primer) So, if we cant even make the transcript, we cant get a HnRNA. Also a sigma factor binding site is in bacteria's promoter site. TEST ANSWER Answers a, d, and e are wrong because the -60 base isn't transcribed, and answer c is wrong because eukaryotes don't have sigma factor.

You saw an interesting case of a 23 year old male patient who presented with bilateral lower extremity stress fractures following routine exercise. This struck you as very odd. After extensive evaluation to determine the underlying etiology of the cause of his fractures, you determine the patient had low testosterone levels with associated osteoporosis. He otherwise was without any other symptoms of hypogonadism (low testosterone). Your faculty felt that this would be a great case report to write up. What are some characteristics of a case that make it a good candidate for a case report? A. It is interesting and you have never seen this before B. It is a known syndrome that should be documented C. It is a known disease or condition with an unusual presentation or is presenting in an unusual population D. It is a common diagnosis with a common presentation

factors for good case reports? - it is a known condition presenting in an unusual way or unusual pop test answer Correct answer is C: Case reports typically describe an unusual or novel occurrence and as such, remain one of the cornerstones of medical progress and can provide many new ideas in medicine. The most common reasons for publishing a case are; (1) an unexpected association between diseases or symptoms; (2) an unexpected event in the course observing or treating patient; (3) findings that shed new light on the possible pathogenesis of a disease or an adverse effect; (4) unique or rare features of a disease; and (5) unique therapeutic approaches.

Positive Sense RNA Viruses

genome may be directly translated to functional proteins by the ribosomes of the host cell

PROBLEM: Scientists have long suspected that there is an association between allergies and gliomas, a common and often fatal type of brain tumor. METHODOLOGY: In Norway, blood samples from annual check-ups and voluntary donations are saved and labeled with personal identification numbers. Researchers were able to obtain access to blood samples taken from 594 current glioma patients decades before they were diagnosed. They were also able to match them with samples from people who never developed glioma. In each sample, they measured the levels of two types of IgE, a protein that plays an essential role in allergic reactions. Then, they used statistical analysis to see whether elevated levels of IgE were associated with a reduced risk of developing a glioma. Which of the following study designs best describes this study? A. Case series B. Cross-sectional C. Case-control D. Cohort E. Randomized controlled trial

had the outcome first, then got to pick your controls? Also looking for risk factors? - case control! Test Explanation c. correct: study participants were identified on the basis of disease status (glioma vs. no glioma) and followed back in time to determine exposure status (elevated IgE) by analyzing previously collected blood samples a. incorrect: a case series includes only patients with disease, and this study included samples from people who never developed glioma (controls). b. incorrect: a cross-sectional study measures both exposure (elevated IgE) and disease (glioma) at the same point in time. Here, researchers analyzed blood samples that were obtained before diagnosis of glioma, to identify elevated IgE prior to glioma diagnosis. d. incorrect: in a cohort study, study participants are identified on the basis of exposure, and observed to see which participants develop disease. Here, participants were identified on the basis of disease (glioma) and followed back in time to determine exposure status (elevated IgE) e. incorrect: in a randomized controlled trial, the exposure is determined by the investigator. Here, the exposure (IgE level) was observed based on previously collected blood samples.

Many drugs are subject to metabolism in the liver by a family of enzymes that utilize a similar mechanism in which iron plays a critical role. Availability of which of the following is also critical for drug metabolism by enzymes of this family? A. sodium ions B. carbon dioxide C. carbon monoxide D. glutathione E. oxygen

in phase 1, the mechanism uses FE3 to bind to NADPH and the cypp450 complex. AFter reduction, oxygen is used to be reduced to activate the complec and later re-oxidize/reset the complex it be E Correct Answer: E Feedback: The major family of drug metabolizing enzymes in the liver is the cytochrome P450 family (CYP450) of mixed function oxidases. The many enzymes of this type all depend on a critical iron molecule in the heme core of the enzyme. The question clearly describes this class of enzyme. The source of oxygen for their oxidizing activity is molecular oxygen. Answer E is therefore correct. Answer A is incorrect since sodium ions are not usually critical for drug-metabolizing enzymes. Answer B is incorrect since CYP450 enzymes do not utilize carbon dioxide. Answer C is incorrect since CYPO450nzyme do not require carbon monoxide. (Note that CYP40 enzymes can complex with carbon monoxide, which causes the spectral properties of the enzymes in solution change to show increased absorption of UV light at 450 nm wavelength, a feature that gives the family its name, but this fact is not relevant to their role in drug metabolism, not important for National Board type questions, and not important for you to remember). Answer D is incorrect since metabolism by CYP450 enzymes does not require glutathione.

The microbiome consists of which of the following in/on the body? A. All organisms B. Bacteria plus metabolites/nucleic acid C. Gram positive organisms, viruses, and fungi D. Metabolites/nucleic acid and all organisms

microbiota is NOT just bacteria, its fungi, archea, protists, algae, ect along with the metabolites. So its all the organisms+ all the metabolites and NAs. Answer: D. The microbiome consists of the microbiota plus the metabolites generated by the microbiota.

Your decide that you need to start an elderly male patient on a drug that is largely eliminated by renal clearance. This drug has a low therapeutic index. Which recent laboratory value for this patient will be most valuable to you in determining the initial dose of the new drug for this patient? A. plasma sodium concentration B. fasting blood glucose C. urine osmolarity D. serum creatine concentration E. serum chloride concentration

need to look at kidney function to see if itll clear at all - eliquis In the aging patient the clearance of drugs by the kidney tends to decline. It is therefore important to estimate the extent of impairment of kidney function in this patient before prescribing a drug with narrow therapeutic index that is largely cleared by the kidneys. The usual measure of kidney filtration is the creatinine clearance, which can be estimated from serum creatinine levels. Answers A,B,C and E list factors that are not directly informative of the glomerular filtration rate, and are therefore incorrect. Answer D, the correct answer, lists a lab value that can be used to calculate an age-adjusted creatinine clearance estimate (using the Cockcroft-Gault equation) that can be used to adjust the dose of the drug for this patient.

Ambulatory blood pressure monitoring is a test that has been studied for the detection of hypertension. The test has a sensitivity of 80% and a specificity of 80% for the diagnosis of hypertension. You decide to screen 100 consecutive individuals presenting to your clinic with this test. Assume that 20 of these 100 individuals actually have a diagnosis of hypertension. Given this information, what is the POSITIVE PREDICTIVE VALUE of this test? A. 80% B. 50% C. 64% D. 32% E. 20%

omg giiiirrrllll 20 ppl in the + dx row 80 ppl in the - dx row (must subt 20 from 100 to get this) sensitivity is 80%. Mult .8 x 20 for the TP = 16 subt to get the rest. Specificity is 80%. Mult .8 by TN .8 x 80 = 64 subt to get rest Do you side to side top row math and get the answer 50%, B The correct answer is B, 50%. In this case, you have 20 patients with disease. Applying sensitivity and specificity of 80% each, you can construct a 2 x 2 table to calculate the PPV. Disease Pos Disease Neg Total Test Pos 16 16 32 Test Neg 4 64 68 Total 20 80 PPV= TP/ False Pos and True Pos PPV=16/16+16= 16/32= 50%

1 / 1 pts What does intersectionality refer to? A. Being present in the moment without judgment B. The development of greater understanding of oneself and the situation C. The recognition and appreciation of social identities D. The interaction and compounding of multiple social identities to create advantages and disadvantages

overlapping. It is positive and negative. Correct answer is D. Explanation: Intersectionality is a term to describe how social identities overlap, interact and compound to create advantages or disadvantages. Understanding social identity and intersectionality is critically important for healthcare leaders in order to address current healthcare disparities and promote health equity. To learn more about intersectionality watch CDC Video on IntersectionalityLinks to an external site.

Several studies have found that about 90% of beta-adrenoceptors in the heart are spare receptors. Which of the following statements about spare receptors is most correct? A. Spare receptors, in the absence of drug, are sequestered in the cytoplasm. B. Spare receptors may be detected by finding that the drug-receptor interaction last longer than the intracellular effect. C. Spare receptors influence the maximal efficacy of the drug-receptor system. D. Spare receptors activate the effector machinery of the cell without the need for a drug. E. Spare receptors may be detected by the finding that the EC50 is smaller than the KD for the agonist.

partial agonists produce a less than the max response, UNLESS there are spare receptors are present and if a large enough number of them are able to activate. This is reflected by the graph of the EC50 being a smaller number than the Kd (remember spare receptors affect the sensitivity not the efficacy, requires less concentration to make more potent) Test answer A. Incorrect. Although some types of receptors may be sequestered in the cytoplasm under certain conditions, there is no difference between spare receptors and activated receptors with respect to location in the cytoplasm. B. Incorrect. Drugs can potentially bind to the non-activated receptors in the same way as they bind to the receptors that are activated; there would not be any difference in the dissociation rate. C. Incorrect. Spare receptors do not influence the maximum efficacy of a specific receptor-effector pathway; they affect the sensitivity of the system to a drug because the presence of spare receptors increases the probability of a drug-receptor interaction occurring, thus shifting the dose-response curve to the left. D. Incorrect. Spare receptors do not increase the probability of activation of the effector pathway in the absence of drug. E. Correct. The presence of spare receptors may be inferred when the concentration of drug producing 50% of maximal effect (the ED50) is lower than the concentration of drug required to occupy 50% of the receptors (the KD).

Which of the following endogenous chemicals does not participate in the formation of drug conjugates by phase II drug metabolizing enzymes? A. glutathione B. 3'phosphoadenosine-5'-phosphosulfate C. S-adenosylmethionine D. acetyl-coenzyme A Correct! E. NADPH

phase two is all about adding on groups. All 4 above were big complexes which would make them more polar/send them to another location. NADPH is an energy carrier and is considered a cofactor and is critical to phase 1. Acetyl coenzyme A is a cofactor and function group so they dont count. Correct Answer: E Feedback: Phase II drug metabolizing enzymes add a more polar group to the parent drug, usually reducing its pharmacological activity and facilitating its excretion. Conjugation reactions in the liver may utilize all of the listed agents except NADPH, which is a critical component of the Phase I CYP450 enzyme cycle, but not involved in Phase II conjugations. The only correct answer is E. Answers A, B, C, and D are all incorrect.

Which of the following types of drugs act by a non-receptor, non-enzymatic physico-chemical mechanism? A. b-receptor blocking agent B. cholinesterase inhibitor C. hepatic enzyme inducer D. local anesthetic E. heavy metal chelating agent

physiochemical: osmotic action- diuretic, laxatives Chemical reactivity: antiacids/acidifying agents, chelating agents (for heavy metals ect.), complexing agents cholestyramine, disinfectants (oxidizing agents) Enzyme Substrates: false transmitter precursors alpha methyl dopa, nucleotide incorporation into NA (nucleotide mimics) A works with receptors - its an enzyme substrate B acts like a NA to inhibit - enzyme substrate CA Hepatic enyzme inducer - works like NA to mimic - enzyme substrate, D local anesthesia - again works with receptors as mimic - enzyme substrate mech E heavy metal chelating agent - not an enzyme mimic, not thru osmotic action, its fits the question. Only chemically reactive. ITS E .

he following factor has the greatest impact on the probability that a positive test is a true positive: A. Sensitivity B. Specificity C. prevalence D. reliability

prevalence (who has the dx) goes hand and hand with predictive value tests ( my test is positive/negative am I really sick/not sick?) e correct answer is C. When the disease is prevalent, most positives are true positives. When a disease is rare, fewer positives are true positives.

Eukaryotic organisms

protists, fungi, plants, animals (have a nucleus)

Question 340 / 1 pts Flexible sigmoidoscopy is a test that has been studied for the detection of colorectal cancer. Assume the test has a sensitivity of 80% and a specificity of 80%. You decide to screen 100 consecutive individuals presenting to your clinic with this test. Assume that 20 of these 100 individuals actually have a diagnosis of colorectal cancer. Given this information, what is the NEGATIVE PREDICTIVE VALUE of this test? A. 94% B. 80% C. 64% D. 50% E. 32%

same as b4 make the table 20 80 mult sens w/ TP mult spec w/ TN subt to get the FP and FN now do the side to side BOTTOM row thang (TN/TN+FN) get 94% The correct answer is A, 94%. In this case, you have 20 patients with disease. Applying sensitivity and specificity of 80% each with a prevalence of 20% (20 have disease), you can construct a 2 x 2 table to calculate the NPV. Disease Pos Disease Neg Total Test Pos 16 16 32 Test Neg 4 64 68 Total 20 80 NPV= TN/ False Neg +True Neg NPV=64/4+64= 64/68= 94%

When a new rapid test for dengue was administered to 292 Vietnamese patients with febrile illness, 151 patients tested positive using the rapid test, and all of these patients were subsequently found to have dengue using the "gold standard" test. 141 patients tested negative using the rapid test, and 94 of these patients were subsequently found to have dengue using the "gold standard" test (BMC Infectious Diseases 2010, 10:142). What is the sensitivity of the rapid test? A. 33.3% B. 38.4% C. 61.6% D. 100%

set up my lil table TP/ TP+FN C is correct: From the question, TP = 151, FN = 94. Sensitivity is TP / (TP + FN) = 151 / (151+94) = 61.6%. a. incorrect: this is the negative predictive value. From the question, TP = 151, (TP + FP) = 151, FN = 94, (FN + TN) = 141, so FP = 0 and TN = 47. Sensitivity is TP / (TP + FN) b. incorrect: this is the false negative rate, FN / (TP + FN) or 1-sensitivity. . d. incorrect: this is the specificity (also the positive predictive value). From the question, TP = 151, (TP + FP) = 151, FN = 94, (FN + TN) = 141, so FP = 0 and TN = 47. Sensitivity is TP / (TP + FN)

A series of experiments is performed to determine the mechanism by which a novel anti-cancer agent traverses cell membranes and accumulates within target cells. The net flux depends on the concentration of the drug only and increases as the concentration of the drug increases regardless of the concentration tested. Structurally similar compounds accumulate in the cells at a similar rate. Inhibiting ATP synthesis in the cell does not affect the rate at which the drug accumulates. Which of the following is the most likely mechanism by which this agent enters cells? Simple Diffusion Facilitated Diffusion Ion-gated Coupling Primary Active Transport Secondary Active Transport

simple diffusion this is bc more gets across w/ inc of the conc ONLY. Also structurally similar get across at a similar rate (not selective). Also no ATP available dont change the outcome. So little man needs no help crossing the membrane (energy or transporter alike) TEST ANSWER Correct: Net flux depends on concentration—indicative of a diffusion process.Transport process is not saturable (i.e. flux increases regardless of concentration)—consistent with a non-protein mediated transport processTransport process is not selective (i.e. structurally similar compounds transported similarly)Transport does not require ATP—not an active transport mechanism Therefore the most correct answer is simple diffusion.

A 54-year-old woman with a seizure disorder that is difficult to control is prescribed phenobarbital. For several weeks, this patient has been taking ibuprofen, a drug that is metabolized by greater than 70% by the enzyme CYP2C9, to reduce peripheral joint pain. A week after starting the phenobarbital therapy, the patient complains of a return of the pain in her joints. A likely reason for the return of the joint pain is that chronic phenobarbital treatment A. inhibits the metabolism of ibuprofen by CYP2C9 B. inhibits the metabolism of ibuprofen by CYP3A4 C. activates constitutive androstane receptors (CAR), increasing the expression of CYP2C9 protein D. inhibits retinoic acid receptors (RAR), decreasing the expression of CYP2C9

so if shes feeling it, that means the drug is not being broken down. If we were to inhibit metabolism,, this would only mean theres more drug to go around. Bc the drug is being used up quicker and shes feeling pain AND bc phenobarbital is a known inducer, C makes the most sense. If we inc the expression of CYP2C9, we get more enzymes to break down ibuprofen Barbiturate drugs and phenobarbital in particular, are known to be potent CYP450 enzyme inducers with particularly marked induction of CYP2C9 and CYP3A4. The question informs you that ibuprofen is largely metabolized by CYP2C9. This suggests that phenobarbital might well accelerate the metabolism of ibuprofen. The question asks why the analgesic action of ibuprofen might is reduced a few days after the patient starts taking phenobarbital. Answer C is the correct answer. Answer A is incorrect since if phenobarbital has inhibited the metabolism of ibuprofen, its analgesic action would increase, not b reduced. Answer B is incorrect since the major metabolism of ibuprofen is by CYP2C9, not CYP3A4. Answer D is incorrect since phenobarbital does not directly inhibit retinoic acid receptors and it increases, not decreases the expression of CYP2C9. Answer E is incorrect since phenobarbital does not alter the absorption of ibuprofen from the g.i. tract.

Leukotrienes cause bronchoconstriction (mediated at leukotriene receptors) in a patient with asthma. When albuterol (acting at beta-adrenoceptors) is given to the patient during an asthmatic attack, the bronchodilation usually results. Which of the following terms best describes the action of albuterol in this situation? A. Chemical antagonist B. Non-competitive antagonist C. Partial agonist D. Pharmacologic antagonist E. Physiological antagonist

specifically taking about reversing bronchoconstriction to bronchodilation. Looking for a response to reverse/decrease a bodily effect/response to something. The albuterol can only effect the beta adrenoceptors. It cannot affect the leukotriene receptors. SO THE CHEMICAL DOES NOT TOUCH THE LEUKO RECEPTORS. This would be a physiological antagonist then in the perspective of the leuko receptors who are the ones that cause this constriction. If it was beta adrenoceptors causing the bronchoconstriction, then this would be a pharmacological/chemical antagonsit. BUT IT AINT IN THIS ONE.; \ E IS CORRECT TEST ANSWER Because leukotrienes act via leukotriene receptors and albuterol acts by antagonizing actions at beta-adrenoceptors, albuterol cannot be a pharmacological antagonist of this action of leuckotrienes. The only correct answer is E; B, C and D relate to various forms of pharmacologic antagonism. A is incorrect since there is no direct chemical interaction between leukotrienes and albuterol.

According to Kahneman, which of the following tasks would usually require the operation of "system 2" processes? A. Orienting to the location of a sudden noise B. Driving down an empty road C. Finding a strong move in a chess position for a skilled chess master D. Performing long division in the head

system 1 is fast decision making or habituated behavior (so requiring little thought) system 2 is slow long laborious decision making A is def fast so its 1. B is little thought. C is little though bc chess needs to be quick for professionals D is obv system 2 bc it is long and painstaking. Requires thought/literal mental math Also Kahneman's stuff usually involves decision making.

Which protein domain binds directly to PI 3,4,5-tris-phosphate (PIP3)? A. SH2 domain B. Pleckstrin homology domain (PH) C. kinase domain D. phosphorylated tyrosine

the PH is what allows PIP3 to bind to the lipid membrane TEST ANSWER Correct answer is B a) is incorrect because SH2 domains bind phosphorylated tyrosine residues. c) is incorrect because PIP3 does not get phosphorylated, thus a kinase domain would not have a reason to bind. d) is incorrect because SH2 domains bind to phosphorylated tyrosine, and a phosphorylated tyrosine residue is not what binds to PIP3. THIS IS A KIND OF TYROSINE KINASE PATHWAY w/ insulin binding BUTTTT, its later on where the pathway diverges which is important. SH2 is with the map kinase ras pathway and this is with the PDK/Akt kinase cascade pathway. It is when we get PIP3 (after pip2 and ISR) when PH binds with PIP3 to stick it to the membrane.

Estimate the volume of fluid in the interstitial spaces in a patient who weighs 132 lbs (60 kg). 36 L 24 L 12 L 9 L

there are waaay more terms than I though 60/40/20 - refers to total fluid volume= intracellular fluid+ extracellular fluid, SO total FV/intracellular FV/ extracellular FV (remember not to times it by 60 (.6) then by 40 (.4). Just multiply the kg weight straight-up by the percent your looking for. Now the 75/25 rule? that is looking for: Extracelluar water = interstitial volume + plasma volume AKA 75/25 is (interstitial/ plasma volume) aka aka (in the tissues/intravascular) So bc it was asking in the tissues/interstitial, should have first found my extracellular water content 60kgx 0.2 = 12 L then we can find interstitial by multiplying 12L x 0.75 = 3 Oops! Where did you go wrong? Total body water = 0.6 x 60 kg = 36 L Intracellular fluid volume = 0.4 x 60 kg = 24 L Extracellular fluid volume = 0.2 x 60 kg = 12 L Plasma (intravascular) volume = 0.25 x 12 L = 3 L Interstitial fluid volume = 0.75 x 12 L = 9 L

Question 340 / 1 pts The human androgen dihydroxytestosterone directly interacts with androgen receptors in the cytoplasm and relocalize them to the nucleus, where binding to hormone receptor elements on the chromosomal DNA allows transcription of several genes that are responsible for maintenance of the secondary sexual characters. If the cellular concentration of the hormone reaches 4 times the value of its dissociation constant, what fraction of the receptors do you expect to remain hormone-bound? A. 0.4 B. 0.6 C. 0.8 D. 0.9 E. 1.0

you see changing concentrations, you might as well think MICHELIS MENTENNNN EQUATIONNNN So set it up Bo = Bmax * H / Kd + H. So if its 4x the conc of Kd, then H will be 4 and Kd will be 1. Set it up up Bo/Bmax = 4/1+4 = 4/5 = 0.8 x 100 = 80% THERE EXPLANATION B= (Bmax)(H)/ Kd + [H] Therefore we can write B / Bmax = [H] / (Kd + [H]) = 4/ 1 +4 (where b / Bmax represents fraction of the receptors occupied) B / Bmax = 0.8 or 80% sites will be occupied Diff between Km and Kd? Km can produce catalysis and have up to 3 rate contants. Can produce product. Kd can only bind, so there is only 2 rate constants. No catalysis.