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Liever eerst lezen? Hieronder staat elke vraag uit deze oefentoets. Antwoorden en uitleg blijven ingeklapt tot je ze opent.

1. 1. A globular protein folds in water. Which amino acid side chain is most likely to end up buried in the hydrophobic core?

   • A. Aspartate
   • B. Lysine
   • C. Valine
   • D. Serine
   Toon antwoord en uitleg

   Antwoord: C. Valine

   Folding in water pushes nonpolar side chains inward and leaves charged and polar ones on the surface where they can interact with water. Valine has a branched hydrocarbon side chain with nothing to hydrogen bond, so the core is where it costs the least energy. Aspartate and lysine carry charges at physiological pH and serine has a hydroxyl group, so all three prefer the surface.

2. 2. A competitive inhibitor is added to an enzyme assay. What happens to the apparent Km and the Vmax?

   • A. Km increases and Vmax is unchanged
   • B. Km decreases and Vmax is unchanged
   • C. Km is unchanged and Vmax decreases
   • D. Km increases and Vmax decreases
   Toon antwoord en uitleg

   Antwoord: A. Km increases and Vmax is unchanged

   A competitive inhibitor binds the active site, so it fights substrate for the same spot. That means you need more substrate to reach half maximal velocity, which shows up as a higher apparent Km. Enough substrate still outcompetes the inhibitor completely, so Vmax is untouched. Pure noncompetitive inhibition is the mirror image: Vmax falls and Km stays put.

3. 3. Complete the sentence.

   In oxidative phosphorylation, the electron transport chain uses the energy released by electron transfer to pump protons into the _____. ATP synthase then makes ATP as those protons flow back down their gradient into the _____.

   Opties voor invulveld 1: mitochondrial matrix, intermembrane space, cytosol

   Opties voor invulveld 2: mitochondrial matrix, intermembrane space, cytosol

   Toon antwoord en uitleg

   Antwoord: intermembrane space, mitochondrial matrix

   The electron transport chain is a series of redox reactions, and the energy it releases is stored as a proton gradient across the inner mitochondrial membrane rather than as ATP directly. ATP synthase is the only part that turns that gradient back into chemical energy. This is why an uncoupler that lets protons leak across the membrane keeps electron transport and oxygen consumption running while ATP output collapses.

4. 4. A buffer is made from a weak acid and its conjugate base. Its buffering capacity is greatest when:

   • A. The solution pH is two units below the pKa
   • B. The solution pH is two units above the pKa
   • C. The solution pH equals the pKa
   • D. The weak acid is present in large excess over the conjugate base
   Toon antwoord en uitleg

   Antwoord: C. The solution pH equals the pKa

   The Henderson-Hasselbalch equation says pH equals pKa plus the log of the base to acid ratio, so pH equals pKa exactly when that ratio is one. With equal amounts of both forms the buffer has the most room to absorb either added acid or added base before one form runs out. Practically, a buffer is useful roughly one pH unit on either side of its pKa, which is why bicarbonate and phosphate cover different ranges in the body.

5. 5. Which statements about eukaryotic gene expression are correct?

   Selecteer alles wat van toepassing is.

   • A. Introns are spliced out of the primary transcript before the mature mRNA leaves the nucleus
   • B. The genetic code is degenerate, so several codons can specify the same amino acid
   • C. RNA polymerase requires a primer to begin synthesis
   • D. A missense mutation changes a codon so a different amino acid is inserted
   • E. Ribosomes read the mRNA from the 3 prime end toward the 5 prime end
   Toon antwoord en uitleg

   Antwoord: A. Introns are spliced out of the primary transcript before the mature mRNA leaves the nucleus, B. The genetic code is degenerate, so several codons can specify the same amino acid, D. A missense mutation changes a codon so a different amino acid is inserted

   Splicing, capping and polyadenylation all happen in the nucleus, and only the processed mRNA is exported. Degeneracy is why the third base of a codon often tolerates a change with no effect on the protein, which is the definition of a silent mutation, while a missense mutation swaps the amino acid and a nonsense mutation creates a stop codon. RNA polymerase can start a chain from scratch, unlike DNA polymerase, and ribosomes read mRNA 5 prime to 3 prime.

6. 6. In the proximal tubule of the nephron, glucose crosses the apical membrane into the cell against its own concentration gradient by moving together with sodium. This is best described as:

   • A. Simple diffusion
   • B. Facilitated diffusion
   • C. Primary active transport
   • D. Secondary active transport
   Toon antwoord en uitleg

   Antwoord: D. Secondary active transport

   Secondary active transport means the carrier itself burns no ATP. It rides an ion gradient that some other pump paid for. Here the sodium potassium ATPase on the basolateral side spends ATP to keep intracellular sodium low, and the sodium glucose cotransporter uses that inward sodium gradient to drag glucose uphill. Block the ATPase and the sodium gradient collapses, and glucose reabsorption stops even though the cotransporter is untouched.

7. 7. Complete the sentence.

   In an infrared spectrum, a broad absorption centered near 3300 reciprocal centimeters points to an _____ bond, while a strong sharp band near 1700 reciprocal centimeters points to a _____.

   Opties voor invulveld 1: O to H, C to H, C to C double

   Opties voor invulveld 2: carbon carbon double bond, carbonyl group, nitrile group

   Toon antwoord en uitleg

   Antwoord: O to H, carbonyl group

   Infrared spectroscopy reports bond vibrations, and the two most useful signals for a first pass are the broad hydroxyl stretch and the sharp carbonyl stretch. The hydroxyl band is broad because hydrogen bonding gives the bond a range of strengths across the sample. Seeing both a broad hydroxyl band and a carbonyl band in one spectrum is the classic hint that you are looking at a carboxylic acid.

8. 8. Which carboxylic acid derivative is most reactive toward nucleophilic acyl substitution?

   • A. An acyl chloride
   • B. An ester
   • C. An amide
   • D. A carboxylate ion
   Toon antwoord en uitleg

   Antwoord: A. An acyl chloride

   Reactivity in this family tracks two things: how good the leaving group is and how much the attached atom donates electron density into the carbonyl. Chloride is a weak base and a good leaving group, and it donates very little, so the acyl chloride sits at the top. Nitrogen donates strongly, which makes amides the most stable of the neutral derivatives, and a carboxylate is already deprotonated and negatively charged, so a nucleophile has little reason to attack it.

9. 9. A reaction has a negative delta G under the given conditions. Which statements are correct?

   Selecteer alles wat van toepassing is.

   • A. The reaction must proceed quickly
   • B. The reaction is spontaneous as written under those conditions
   • C. The reaction must be exothermic
   • D. An enzyme could speed it up without changing the value of delta G
   • E. Under standard conditions, a negative standard delta G means the equilibrium constant is greater than one
   Toon antwoord en uitleg

   Antwoord: B. The reaction is spontaneous as written under those conditions, D. An enzyme could speed it up without changing the value of delta G, E. Under standard conditions, a negative standard delta G means the equilibrium constant is greater than one

   Delta G is about thermodynamics, not speed. A spontaneous reaction can take years if the activation energy is high, and a catalyst lowers that barrier without touching the free energy difference between reactants and products. Spontaneity also does not require heat release, since a positive entropy change at high enough temperature can make delta G negative for an endothermic reaction. Standard delta G is related to the equilibrium constant by delta G naught equals minus RT ln K, so a negative value puts K above one.

10. 10. Circulating thyroid hormone rises because the thyroid gland itself is overproducing. In an intact feedback loop, what happens to TSH released by the anterior pituitary?

    • A. It rises, because hormone levels and TSH move together
    • B. It falls, because the high hormone level suppresses the pituitary and the hypothalamus
    • C. It stays the same, because TSH is regulated only by the hypothalamus
    • D. It rises first and then stays permanently elevated
    Toon antwoord en uitleg

    Antwoord: B. It falls, because the high hormone level suppresses the pituitary and the hypothalamus

    The hypothalamus releases TRH, the anterior pituitary releases TSH, and the thyroid releases its hormones, which then feed back to shut down the two levels above. When the gland at the end of the chain is the source of the excess, the feedback signal is strong and TSH goes down. That inverse pattern is the point of a negative feedback loop, and the same logic runs the hypothalamus, pituitary and adrenal axis with cortisol.

11. 11. A vessel behaves as a rigid tube with steady laminar flow. If its radius drops to half while the pressure difference across it stays the same, the flow rate becomes about:

    • A. One half of the original
    • B. One quarter of the original
    • C. One eighth of the original
    • D. One sixteenth of the original
    Toon antwoord en uitleg

    Antwoord: D. One sixteenth of the original

    Poiseuille's law says flow rate is proportional to the fourth power of the radius for a given pressure difference, viscosity and length. Halving the radius therefore divides flow by two to the fourth, which is sixteen. That fourth power dependence is why radius changes dominate resistance in the circulation, and why small changes in vessel diameter matter far more than changes in length or viscosity.

12. 12. A student has a headache, takes a painkiller, and the headache goes away. She now reaches for the painkiller sooner whenever a headache starts. In operant terms this is:

    • A. Positive reinforcement
    • B. Negative reinforcement
    • C. Positive punishment
    • D. Classical conditioning
    Toon antwoord en uitleg

    Antwoord: B. Negative reinforcement

    In operant conditioning, reinforcement means the behavior becomes more frequent and punishment means it becomes less frequent. Positive and negative describe whether something was added or taken away, not whether it felt good. Here an unpleasant stimulus was removed and the behavior increased, which is negative reinforcement by definition. Classical conditioning would involve pairing two stimuli to produce an involuntary response, not strengthening a voluntary action by its consequence.

13. 13. The smallest change in a stimulus that a person can detect half of the time is called the:

    • A. Absolute threshold
    • B. Signal detection criterion
    • C. Difference threshold
    • D. Point of sensory adaptation
    Toon antwoord en uitleg

    Antwoord: C. Difference threshold

    The difference threshold, also called the just noticeable difference, is about telling two stimuli apart, while the absolute threshold is the smallest stimulus that can be detected at all half the time. Weber's law adds that the difference threshold is a constant proportion of the original stimulus rather than a fixed amount, so a heavier weight needs a bigger change before you notice it. The signal detection criterion is a separate idea about how willing someone is to say yes when they are unsure.

14. 14. Researchers record the diets of 2,000 healthy adults, then follow the same people for 20 years and record who develops heart disease. Which statements about this study are correct?

    Selecteer alles wat van toepassing is.

    • A. It is a prospective cohort study
    • B. Participants were randomly assigned to a diet
    • C. Participants dropping out over the 20 years can bias the result
    • D. It can measure how many new cases of heart disease appear in each diet group
    • E. It shows that diet causes heart disease
    Toon antwoord en uitleg

    Antwoord: A. It is a prospective cohort study, C. Participants dropping out over the 20 years can bias the result, D. It can measure how many new cases of heart disease appear in each diet group

    The exposure was measured before the outcome appeared and the group was followed forward, which makes this a prospective cohort study. That design can measure incidence, meaning new cases over time, which case control studies cannot do directly. What it cannot do is establish causation, because nobody assigned the exposure and people who eat differently may differ in many other ways. Attrition matters because the people who leave a long study are often not a random sample of those who stay.

15. 15. Complete the sentence.

    Blaming a stranger's rudeness on their personality while ignoring what is going on around them is the _____. Explaining your own good grade by your ability but your own bad grade by an unfair test is the _____.

    Opties voor invulveld 1: fundamental attribution error, self-serving bias, just world hypothesis

    Opties voor invulveld 2: actor observer bias, self-serving bias, halo effect

    Toon antwoord en uitleg

    Antwoord: fundamental attribution error, self-serving bias

    Attribution questions turn on two axes: whether the cause is placed inside the person or in the situation, and whether the person being judged is you or somebody else. The fundamental attribution error is over-weighting personality when explaining other people. The self-serving bias protects your own self-image by taking internal credit for success and external blame for failure. Both are on the MCAT because they shape how patients and clinicians read each other.

Wat de drie natuurwetenschappelijke onderdelen echt vragen

De MCAT heeft vier onderdelen. Drie ervan zijn natuurwetenschappelijk: Chemical and Physical Foundations of Biological Systems, Biological and Biochemical Foundations of Living Systems, en Psychological, Social, and Biological Foundations of Behavior. Het vierde is Critical Analysis and Reasoning Skills, dat puur uit lezen bestaat en zijn eigen pagina heeft. Elk natuurwetenschappelijk onderdeel telt op dit moment 59 vragen in 95 minuten, en ruwweg 15 van die 59 zijn discrete vragen die op zichzelf staan zonder tekst. Onderdeelscores lopen van 118 tot 132, en de vier onderdelen tellen op tot een totaal tussen 472 en 528, met 500 midden op de schaal. De AAMC kan het format veranderen, dus controleer de actuele getallen op aamc.org voordat je er een planning omheen bouwt.

De stof zelf komt uit de inhoudscategorieën die de AAMC publiceert. Wat de meeste mensen verrast, is hoeveel biochemie er langskomt, en dat het in twee verschillende onderdelen langskomt. Aminozuren, enzymkinetiek en metabolisme kunnen in het biologieonderdeel worden getoetst en nog eens in het onderdeel scheikunde en natuurkunde. De natuurkunde is smal en toegepast, vooral vloeistoffen en druk, thermodynamica, optica en schakelingen, en ze is bijna altijd ingebed in een levend systeem in plaats van in een blok dat van een helling glijdt.

Het onderdeel psychologie en sociologie is een echt stofonderdeel met een eigen vocabulaire, geen onderdeel gezond verstand, en daar lopen mensen op stuk. Conditionering, geheugensystemen, waarnemingsdrempels, attributietheorie en onderzoeksopzet hebben allemaal precieze definities, en de vragen draaien om die definities. In alle drie de onderdelen zijn de meeste vragen geen simpele reproductie. Ze geven je een situatie, een grafiek of een set resultaten en vragen wat daaruit volgt, dus een term kennen is maar de helft van wat je nodig hebt.

Hoe je deze oefentoets gebruikt

Elke vraag hieronder is met opzet een discrete vraag. Discrete vragen toetsen het zuiverst of je de stof echt kent, want er staat geen tekst naast die je stilletjes context aanreikt. Kun je discrete vragen koud beantwoorden, dan worden teksten een stuk makkelijker. Kun je dat niet, dan verbergt oefenen met teksten het gat in plaats van het te dichten. De twee verwante quizzen hieronder behandelen de gemengde tekstvorm en het leesonderdeel, dus doe ze alle drie en vergelijk waar je punten laat liggen.

Maak hem in een keer zonder aantekeningen, met ongeveer 90 seconden per vraag, en lees daarna de uitleg bij elke vraag, ook bij de vragen die je goed had. Goed gokken en het weten zijn niet hetzelfde, en maar een van de twee overleeft de examendag. Behandel elke fout als een onderwerp en niet als een feit. Had je de buffervraag en de thermodynamicavraag fout, herhaal dan zuur-base en evenwicht samen, want het is dezelfde redenering toegepast op twee verschillende grootheden.

Nog iets over wat deze vragen zijn. Het StudyPDF-team heeft alle 15 geschreven op basis van de inhoudscategorieën die de AAMC publiceert. Het zijn geen AAMC-vragen, ze zijn niet overgenomen uit een vragenbank of voorbereidingsboek, en StudyPDF heeft geen band met de AAMC. Het is examenoefening, geen medisch advies, en niets op deze pagina mag worden gebruikt om een beslissing over een echte patiënt te nemen.

Zo bouwt StudyPDF volledige oefentoetsen uit je eigen materiaal

Vijftien vragen laten zien welke inhoudscategorie zwak is. Ze kunnen geen hele MCAT-voorbereiding dragen, en een algemene vragenbank sluit zelden aan bij de nadruk van de voorbereidingsboeken en collegeaantekeningen waar je echt uit studeert.

StudyPDF werkt in plaats daarvan vanuit je eigen materiaal. Je uploadt een hoofdstuk uit je voorbereidingsboek, je collegeslides biochemie of het routeoverzicht dat je zelf maakte, en Bo, de studie-agent, schrijft uit precies die pagina's volledige oefentoetsen. Elke vraag is geworteld in jouw materiaal en elke uitleg noemt waar het antwoord vandaan komt, dus als iets fout lijkt, ga je meteen naar de bron om het te controleren. Je kunt zo vaak als je wilt nieuwe toetsen maken, ze richten op een enkel gebied zoals enzymkinetiek of onderzoeksopzet, en bijhouden welke begrippen steeds fout blijven gaan.

Je hebt geen bestand nodig om te beginnen. Heb je niets om te uploaden, noem dan een onderwerp, bijvoorbeeld de citroenzuurcyclus of operante conditionering, en Bo schrijft daar een oefenset over. Beginnen is gratis.