Bio/Biochem

Which of the following is LEAST likely to be observed in a patient experiencing hyperventilation? A. Hypoxia B. Net exhalation of CO2 C. Increased blood pH D. Increased hemoglobin O2 affinity

A. This question asks us to determine the effects of hyperventilation. During hyperventilation, there is a loss of CO2 and an increase in O2 in the blood. Hypoxia is another term for oxygen deprivation, which is the opposite of what would occur here.

Which biomolecule is most likely to act as a hydride donor during an enzyme catalyzed reaction? NAD+ NADH H3O+ H2O

Answer B is correct. A hydride is the anion of hydrogen (H:-). This species participates in many biological reactions but is too unstable to exist under most physiological conditions. Specialized coenzymes such as NADH and NADPH serve as hydride donors, while NAD+ and NADP serve as hydride acceptors. This makes Answer B correct. Answer A is incorrect because NAD+ is likely to act as a hydride acceptor, not a hydride donor. Answers C and D are both incorrect because water and hydronium are likely to donate or accept protons, but not hydride ions.

What type of linkage cannt be digested by the human body

B (1-4)

A pharmacologist wishes to convert morphine into a form available to brain tissue. Which of the following changes could be made to morphine's molecular structure to improve its ability to cross the BBB while maintaining its therapeutic effects? A. Attach a glucose molecule to the nitrogen atom B. Replace the proton of each alcohol with an acetyl group C. Oxidize all double bonds in the molecule D. Reduce all double bonds in the molecule

B. This question asks us to consider the structure of morphine given in the passage. We are told in the passage that lipid-soluble molecules are able to pass through, while hydrophilic molecules are not. Morphine is hydrophilic in part because of its hydroxyl groups. Although the C=O bond in the acetylated molecule is polar, it would not donate hydrogen bonds. As a result, the acetylated compound would be more lipophilic than morphine. Diacetylated morphine is called heroin and, while the effects and pharmacology of heroin are somewhat distinct from those of morphine, heroin does indeed cross the BBB more readily.

Defects in which of the following proteins would most likely lead to loss of structural integrity in skin epithelial cells? A. Microtubules B. Titin C. Intermediate filaments D. Myosin

C. Intermediate filaments such as keratin are directly responsible for structural integrity in skin.

In the first trial of the PCR procedure, only one primer was added to the mixture. What was the most likely outcome during this trial? A. Both of the DNA strands were linearly amplified. B. Only one strand of the DNA was exponentially amplified. C. Only one strand of the DNA was linearly amplified. D. PCR amplification was unable to proceed.

C. When performing PCR, if only one primer is added to the mixture, then that primer will bind to one of the strands and initiate replication of that strand to produce the complementary strand. For one double-stranded template DNA molecule, after one cycle, we are left with two of the complementary strand while still having only one of the strands where the primer binds. In the next cycle, the primer again binds to the strand to which it is complementary. Therefore, only the complementary strand will be replicated again; this time we have three of the complementary strand. As you can see here, one of the strands is being replicated linearly (1 → 2 → 3 →...n), while the other strand is not replicated at all. After 30 cycles, for example, assuming that we began with one copy of each strand, we would have 30 copies of one strand and only one copy of the other. This differs from PCR that includes two primers (which is typical), where we would expect exponential amplification (230 copies of each strand).

The ADH active site features a Zn2+ cation coordinated to cys46, cys174, his67, and to the oxygen atom of the ethanol substrate. Within the ADH active site, the pKa values of cys46, cys174, and his67 are: decreased due to charge destabilization of their protonated forms by Zn2+. decreased due to charge stabilization of their protonated forms by Zn2+. increased due to charge stabilization of their protonated forms by Zn2+. increased due to charge destabilization of their protonated forms by Zn2+.

Choice A is correct. The coordination of a cation such as Zn2+ to a neutral Lewis base increases the positive charge and destabilizes the Lewis base. This significantly increases the acidity of any protons attached to the cationic atom of the Lewis base and decreases their associated pKa values. Answer A is the correct choice because Zn2+ destabilizes the protonated form of cysteine or histidine, making the proton more acidic (lower pKa). Answer B is incorrect because it is the deprotonated form, not the protonated form, that would be stabilized by coordination to Zn2+. Answer C is incorrect because the pKa is decreased, not increased, and also because the protonated form is destabilized, not stabilized. Answer D is incorrect because the pKa is decreased, not increased.

Large stores of glycogen in skeletal muscle are critical during prolonged periods of exercise, because: myosin hydrolyzes ATP during muscle contraction. actin requires ATP for polymerization. muscle glycogen prevents hypoglycemia. exposure of the myosin binding site requires ATP.

Choice A is correct; ATP is required as part of the sliding filament model of muscle contraction, which is obtained by hydrolyzing stored glycogen to generate glucose and then metabolizing that glucose. Specifically, hydrolysis of ATP is necessary to cock the myosin head, and the binding of ATP to the myosin head is necessary to release the myosin head from the actin filament in preparation for the following contraction cycle. Answer B is a true statement, but it is not the correct answer since polymerization of actin is not the reason large amounts of ATP are required during extended exercise. Answer C is not correct since muscle glycogen is not responsible for the regulation of blood sugar. Answer D confuses ATP with calcium and troponin/tropomysin. Under normal conditions, the myosin binding site located on the actin filament is covered by tropomyosin. When calcium binds troponin, troponin undergoes structural changes that move tropomyosin, exposing the binding site.

Many viruses that invade human hosts have developed the ability to inhibit normal cellular transcription by expressing inhibitor molecules that act in the: nucleoplasm. mitochondria. cytoplasm. endoplasmic reticulum.

Choice A is the correct answer. Transcription occurs in the nucleoplasm of the nucleus. RNA polymerase uses the DNA template of the chromosomes to produce a pre-mRNA. The pre-mRNA is then processed to mRNA and moves to the cytoplasm for translation. Translation, not transcription, occurs in the cytoplasm, which eliminates Answer C. Translation of mRNA that encodes proteins for export is often associated with the rough endoplasmic reticulum, but this is translation, not transcription, which eliminates Answer D. Mitochondrial transcription and translation are labeled as "mitochondrial" to distinguish them from cellular process, eliminating Answer B.

Over-activated microglia may exacerbate neurodegeneration through: cell-mediated adaptive immunity. humoral adaptive immunity. non-specific innate immunity. non-specific adaptive immunity.

Choice C is correct. There are two types of immunity in animals, innate immunity and adaptive immunity. There are also two kinds of adaptive immunity: cell-mediated and humoral immunity. In the final paragraph, the passage indicates researchers concluded that microglial cells enhanced the inflammatory response. Because inflammation is non-specific and part of an innate immune response, choice C is correct. Answers A and B can be eliminated because they refer to adaptive immunity. Choice D refers to non-specific adaptive immunity, which does not exist; innate immunity is described as non-specific, while adaptive immunity is always pathogen-specific.

To perform the immunocytochemistry measurement, researchers dissolved an experimental antibody in aqueous solution, then added the solution to the cell culture where it bound the target p53 protein. p53 most likely associated with the antibody as a result of: covalent bonds. hydrophobic interactions. the overlap of π orbitals. hydrogen bonds.

Choice D is the correct answer. In general, protein-protein binding interactions, such as those interactions associated with receptor binding, antigen-antibody binding, or binding of protein subunits in quaternary structure, are the result of non-covalent interactions (e.g., hydrophobic interactions, hydrophilic interactions, electrostatic attractions and hydrogen bonds). This eliminates Answer A because it suggests the use of covalent bonding, as well as Answer C because p overlap is an aspect of covalent bonding. Answer B is unlikely because the antibody and its target are in aqueous environments.

Researchers hypothesized that p53 levels were high in de-pigmented tissues because: p53 plays a role in apoptosis. p53 is an inhibitor of angiogenesis. p53 plays a role in inhibiting the cell cycle in DNA-damaged cells. cancer incidence was lower than expected in de-pigmented patches.

Choice D is the correct answer; Given passage information, the researchers based their hypothesis on the information that skin cancer levels are lower in vitiligo de-pigmented patches relative to cancer levels in de-pigmented epithelium caused by other melanin-deficiency disorders. They suspected that a tumor suppressor gene, such as p53, might be elevated and protecting these tissue from mutagenic sun rays since they do not have the normal protection provided by the pigment melanin. The passage also states that the roles of p53 in apoptosis (Answer A), angiogenesis inhibition (Answer B), and cell cycle checkpoints (Answer C) are important in cancer prevention. However, the unexpectedly low incidence of cancer is what more directly led researchers to conduct this experiment. This fact is rather clearly stated in paragraph 2.

Phosphodiesterase

Cleaves phosphodiester bonds

he Hardy-Weinberg equation is shown. Assuming q represents the recessive allele in a hypothetical population, which variable can be determined using a population survey? p² + 2pq + q² = 1 p p2 2pq q2

D

Eantiomers

D and L mirror images, must be nonsuperimposable and have a chiral carbon

If the DNA of a representative species from each of the major kingdoms was examined, the sequences coding for which of the following would be expected to be most similar? A. Photosynthesis B. Cholesterol synthesis C. Protein degradation D. Transcription

D. DNA sequences that are common among different species, phyla, or even kingdoms are called conserved sequences. Conserved sequences tend to remain that way due to the fact that they code for a vital function that is common among disparate species. While the enzymes for DNA synthesis, transcription, and translation do vary between species, they are by far the most conserved of the provided answers with regard to structure and function.

Southern blot

DNA

What does calcitonin do?

Increases bone formation

Do viruses undergo cell division?

No as they're non-living

Fischer to Haworth projection.

R OH on fischer is below in haworth L ON on fischer is above in haworth

Northern blot

RNA

Which is heavier: thiamine or uracil

Thymine

Hydrophobic=Lipophilic

-does not like water -does like other lipids Usually benzene rings

nucleotides in base pair

2 nucleotides per base pair

Passive immunity could result from: I. receiving a vaccine. II. contracting a viral or bacterial infection. III. receiving an injection of antibodies.

3 only Passive immunity is recieving antibodies (milk) Active immunity is making your own antibodies

Cell differentiation is mediated primarily by: A. gene expression levels. B. differing genetic material. C. relative age of the cell. D. DNA ethylation.

A. Cell differentiation occurs primarily through different gene expression levels.

Researchers noted that when a lysine residue on a histone is acetylated, its side chain becomes neutral in charge. Combined with the information in the passage, which of the following conclusions will the researchers most likely reach? A. Histone deacetylation generally decreases gene expression. B. Histone acetylation tends to promote a more condensed chromatin structure. C. Histone deacetylation reduces the extent to which the gene in question will be replicated. D. Histone acetylation favors the formation of heterochromatin rather than euchromatin.

A. Lysine is + charged and interacts with - DNA to tightly bind to histones and reduce gene expression. C is incorrect because replication occurs no matter what, only transcription is affected by acetylation

Which hormone is responsible for aiding in the secretion of IgA in breast milk? A. Prolactin B. Calcitonin C. Epinephrine D. LH

A. Prolactin is the hormone responsible for stimulating the mammary glands to produce milk.

What is directly responsible for allowing the alveoli to remain inflated during exhalation? A. Surfactant that decreases surface tension B. Surfactant that increases surface tension C. Negative intrapleural pressure D. Positive pressure in the pleural cavity

A. Surfactant in the lungs is responsible for lowering the surface tension and preventing alveolar collapse during expiration.

Which of the following is most likely true of the 5'-UTR region of the FAM83H gene? A. It is transcribed, but is typically not translated or is only partially translated. B. It is always both transcribed and translated. C. It is neither transcribed nor translated. D. It is cleaved from the pre-mRNA transcript as part of a post-transcriptional modification.

A. The 5'-untranslated region (5′-UTR) is the region of mRNA that is directly upstream from the initiation codon. This region is important for the regulation of translation of a transcript; from this information alone, we know that the 5'-UTR must be transcribed. However, as it is the "untranslated" region, we can conclude that this region is not translated or only partially translated into a protein.

Start codon

AUG

Calcium is a necessary structural or functional component in which of the following processes? I. Bone tissue formation II. Blood clotting III. Muscle contraction IV. Nerve signal transmission

All 4

Eukaryotic plasma membranes usually contain: I. carbohydrates. II. steroids. III. phospholipids. IV. proteins.

All 4

Allosteric effector

Allosteric effectors are molecules that bind to an enzyme at a site other than the enzyme's active site and regulate its function.

After leaving the stomach, in which order does food travel through the regions of the small intestine? Duodenum → jejunum → ileum Duodenum → ileum → jejunum Ileum → duodenum → jejunum Ileum → jejunum → duodenum

Answer A is correct. After leaving the stomach, chyme first enters the duodenum, which rules out Answers C and D. The chyme then travels from the duodenum to the jejunum, which rules out answer B. Answer A is the correct order. Proceeding from the stomach to the large intestine, the order of the segments of the small intestine is duodenum, jejunum, then ileum.

Muscles that rely exclusively on glycolysis fatigue more quickly than muscles that utilize aerobic respiration. Does this finding support a role for lactic acid accumulation in activity-related MMF? Yes, because glycolysis produces lactic acid. No, because glycolysis produces lactic acid. Yes, because glycolysis produces ATP. No, because glycolysis produces ATP.

Answer A is correct. Muscles that rely exclusively on glycolysis are known as fast, glycolytic fibers. Per glucose metabolized, or per ATP produced, they will produce much more lactic acid than slow, oxidative fibers that preferentially rely on aerobic respiration. Thus, the information in the stem provides evidence in support of a correlation between increased lactic acid (and/or the associated decrease in pH) and more rapid muscle fatigue. Answer B should be dismissed because it suggests that the relationship just described does NOT support a role for lactic acid. Answer C correctly states that glycolysis produces ATP, but the fact that glycolysis produces ATP does not support the role of lactic acid in muscle fatigue. This is the logical equivalent to saying that you have evidence that Bob committed the crime because you saw Fred at the crime scene. Answer D is incorrect because both glycolysis and aerobic respiration produce ATP. The fact that glycolysis produces ATP does not disprove that glycolytic fibers experience lactic acid buildup.

For an enzyme catalyzed reaction, initial reaction velocity is linearly related to which variable? Solution pH Enzyme concentration Temperature Substrate concentration

Answer B is correct. The relationship between product formation and initial enzyme concentration is proportional, while the relationship between product formation and initial substrate concentration is not. In the Michaelis-Menten Equation, the numerator includes a product of [S] and the denominator includes a sum of [S], which means the relationship between velocity and [S] cannot be linear. This eliminates Answer D. Enzyme activity has a maximum pH below and above which activity drops, eliminating Answer A. Enzyme activity has a maximum temperature below and above which activity drops, eliminating Answer C. Furthermore, enzymes quickly become denatured at elevated temperatures. Answer D is incorrect because, as the substrate concentration increases, the initial reaction velocity levels off. Students should be familiar with the common graph of V0 vs [S], which is hyperbolic, not linear. It approaches an asymptote at Vmax.

Which adult tissue is embryologically derived from the ectoderm? Gastrointestinal epithelium Sensory epithelium Peritoneal membrane Nuclear membrane

Answer B is correct. The sensory epithelium of the eyes, ears, and nose, along with all of the major structures of the brain, spinal cord, and peripheral nervous system, are derived from the ectoderm. For future reference, a few counterintuitive ectoderm structures students may not recognize include the hair and nails, the subcutaneous glands, and the enamel of the teeth. Answer A is false because gastrointestinal epithelium is derived from the endoderm. The endoderm also gives rise to a few epithelial linings students may not recognize, such as respiratory epithelium, the linings of the urinary bladder and urethra, and the epithelial cells that surround the tympanic cavity and auditory tube. The liver and pancreas are also derived from endoderm, though many students mistakenly remember these organs as mesoderm; usually because most simplified models for remembering germ layers suggest that the "organs" come from the mesoderm. Answer C is false because the peritoneum and most of the major organs are derived from mesoderm. Answer D is false because the nuclear membrane is a cell-level structure, not a tissue.

If all three of the molecules shown are mixed together in a single solution, which species will participate in hydrogen bonding? I. CH4 II. CH2O III. CH3OH I only III only II and III only I, II, and III

Answer C is the correct answer. Hydrogen bonds are formed between electronegative atoms such as oxygen, and a hydrogen atom that is covalently bound to an electronegative atom (F, O, or N). Methanol (III) would be the only molecule capable of forming hydrogen bonds with itself, but because they are all mixed together, the carbonyl oxygen in formaldehyde can act as a hydrogen bond acceptor (but not a hydrogen bond donor). This makes options II and III both correct and Answer C the best choice. Methane would not be miscible with formaldehyde and methanol and is not capable of hydrogen bonding under any circumstance because it lacks the necessary electronegative atom.

Glycoproteins lose oligosaccharides over time. A decrease in the number of carbohydrates is used by the cell to target older glycoproteins for degradation. If most glycoproteins are not replaced following turnover, should this process have influenced the design of this study? No, because additional proteins will be translated and glycosylated to replace the degraded glycoproteins. No, because the loss of a few carbohydrates will have no effect on the binding of glycoproteins to the column. Yes, the researchers should have only sampled glycoproteins from one-month-old insects from each species. Yes, the researchers should have only sampled glycoproteins from very young insects from each species.

Answer D is correct. Because glycoproteins lose sugar units as they age, and because the experiment is looking specifically at proteins with sugars attached, it is essential to study the insects at an early life stage. This will ensure that as many glycoproteins as possible are isolated, since as the insects age more and more glycoproteins will undergo degradation. Answer A is false because the stem specifically gives the assumption that most glycoproteins are not replaced after they are degraded. Answer B is false because there is not sufficient information in the passage to conclude that most glycoproteins have enough sugar units to survive age-related losses and still have at least one mannose residue to bind the column. Further, even if a protein still has sugars attached to it, the stem states that a decrease in the number of sugars can cause a glycoprotein to be targeted for degradation. In that case, some glycoproteins may well have been recycled by the insect's cells before the experiment began. Answer C is false because one-month old is an arbitrary age that may be early or late in the lifespan of an insect. Some insects live for only a few hours or days.

Under normal physiological conditions, DSBs in DNA are repaired by which mechanism? Nucleotide-excision repair Base-excision repair Mismatch repair Homologous recombination

Answer D is correct; all the other repair mechanisms listed are for single-stranded DNA. Answer D is the only repair mechanism for double-stranded DNA. Answers A and C are both mechanisms wherein a small region of a single strand of the DNA is removed and replaced. Answer B is a mechanism where a single base is removed and replaced.

A single-cell diploid organism acquires a deleterious point mutation before entering mitosis. If one of the daughter cells reproduces asexually, and the other daughter cell goes through meiosis to produce gametes that fuse with gametes from other individuals, which cell lineage will be most evolutionarily successful? Neither cell lineage will have an advantage because both cell lines carry the deleterious mutation. The asexually-reproducing cell lineage, because only one copy of the mutation is present per cell. The sexually-reproducing lineage, because crossing over events will eventually erase the mutation. The sexually-reproducing lineage, because only one-half of the offspring will have the mutation.

Answer D is the best choice. The point mutation will affect only one set of chromosomes in a diploid organism. Asexual reproduction will transmit the mutation to all progeny, for all subsequent generations. By contrast, sexual reproduction will result in only half the progeny in the first generation inheriting the mutation. These progeny will thrive and produce subsequent generations of mutation-free offspring. Answer A is incorrect because although both lines will carry the mutation, it will be found among fewer individuals in the sexually-reproducing cell lineage. Answer B is wrong because the sexually-reproducing lineage will have greater evolutionary fitness. Answer C is incorrect because crossing-over does not eliminate point mutations, it only exchanges their location with similar DNA from non-sister chromatids (homologues).

Consuming a meal high in fat is likely to cause an increase in the: I. discharge of bile by the gall bladder. II. activity of pancreatic lipase in the small intestine. III. absorption of triglycerides into villi capillaries. I and II only I and III only II and III only I, II, and III

Answer choice A is correct. Statements I and II are true, but Statement III is false. Statement I is correct because the discharge of bile from the gall bladder is stimulated by the presence of fat in the duodenum. Statement II is true because pancreatic lipase is the primary enzyme responsible for the hydrolysis of fats in the small intestine. Therefore, an increase in the amount of fat in the intestine following the consumption of a high-fat meal will indeed increase the enzymatic activity of pancreatic lipase. This eliminates Answer B, because it does not include Statement II. Statement III is false because fats are not absorbed directly into the capillaries of the villi. Instead, they are absorbed into small lymphatic vessels called lacteals. Lacteals combine to form larger lymphatic vessels which later empty into the thoracic duct. Answers B, C and D are incorrect because they include Statement III.

Sweat pores in the skin release water and various other substances onto the surface of the skin. What physical property of water allows sweating to reduce body temperature? Lower density in the solid phase High heat capacity Low heat capacity High surface tension

Answer choice B is correct. Water has a number of properties that make it uniquely suited to serve as "the solvent of life." Most of these are due to the strong intermolecular forces among water molecules and its polarity. The one that is most relevant to cooling the body is water's high heat capacity—when water molecules acquire enough energy to evaporate from the surface of the skin, they have absorbed a relatively large amount of heat and therefore, take away a large amount of heat. This is known as evaporative cooling. Answers A and D describe properties of water, but are not relevant to evaporative cooling. Answer C is incorrect as water has a high, not a low, heat capacity.

Ftt is an obligate intracellular bacterium. An antibody response can still be effective against Ftt because antibodies: enhance the rate of phagocytosis of Ftt. bind to Ftt plasma B-cells, stimulating antibody production. bind to Ftt and prevent it from entering the cell. cross the Ftt cell membrane and prevent translation of Ftt proteins.

Answer choice C is correct. The primary function of antibodies is to activate complement, enhance phagocytosis, and neutralize infectious agents by preventing binding to the receptor on the target cell. Although answer choice A is correct in that antibodies can enhance the phagocytosis of bacteria into the cell, this would not be effective given that Ftt are already obligate intracellular bacteria. Ftt is a prokaryotic cell, therefore it does not have its own immune system, or plasma B-cells that produce antibodies—making choice B incorrect. The best answer is that the antibodies will neutralize the infection by preventing F. tularensis from interacting with its cellular receptor and thereby preventing its entry into the host cell. Answer choice D is incorrect because antibodies are not widely known to cross the cell membrane and have no known function in preventing translation. This is even less likely for a bacterium with a cell wall.

What is the most probable impact of a point mutation that substitutes the terminal guanine of a 5' exon with an alternate nucleotide? The mRNA will be prematurely terminated. The protein will be nonfunctional due to an amino acid change. The intron will not be spliced from the mRNA transcript. There will be no effect, because this is a silent mutation.

Answer choice C is the correct answer. A guanine at the 5' position is required for RNA splicing. If a different nucleotide were present, the process would not function and the intron would not be removed. This makes choice C the best answer. This will not create a stop codon, making choice A incorrect. Answer B may be tempting because failure of the splicing mechanism may well result in a non-functional protein. However, given both answers, B and C, answer C is a more complete and accurate description that emphasizes the primary importance of guanine in the splicing mechanism. This is particularly important given that the importance of a guanine substitution was specifically referenced in the stem. Examinees should expect AAMC questions to feature distractors that may be true, or could be true, but are not as complete, or as correct, as the best answer. Answer D must be incorrect because, as already described, failure of exon splicing will have multiple deleterious effects.

Most scientists consider a single confirmed case of polio to be an epidemic. This assumption is supported by passage information because: polio is a serious disease that results in paralysis or death. the fecal-oral transmission route causes WPV to spread rapidly. WPV is generally epidemic in countries with poor sanitation. polio is asymptomatic in 90% of WPV infections.

Answer choice D is correct. According to the passage, polio is asymptomatic in 90% of carriers, therefore if there is one confirmed case, that is strong evidence that there is more widespread WPV infection in the population. Answer choice A is a true statement, but it is not a logical reason to think that one case is indicative of an epidemic. This is an example of a common AAMC trend in which a true statement is nevertheless incorrect because it does not logically address the actual question posed by the stem. Answer B is incorrect because fecal-to-oral transmission is far less likely to occur than are other methods of viral transmission, such as aerosol transmission. Because it is less likely that feces will contaminate via the oral route, it is not logical to state that this would cause WPV to "spread rapidly." Polio is not necessarily epidemic in countries with poor sanitation. Even if this were the case, there is no information in the passage to suggest this conclusion, so Answer C should be dismissed.

Which expression gives the probability that a mother and father will have four children, two girls and two boys, in any birth order? (½)4 4 (½) 2 (½)4 6 (½)4

Answer choice D is the correct answer. Each birth is an independent event. For each "birth event" the probability of having a boy is ½ and the probability of having a girl is also ½. This type of problem would be more simple had the question asked "What is the probability of having four children, all of them boys?" The probability of having four boys in a row is (1/2)4 or 1/16. However, the problem changes when the question states that the births can appear in any order. In this case, one must first predict all of the possible birth orders by which these parents could have two boys and two girls: GGBB, GBGB, GBBG, BBGG, BGBG, BGGB It may be worthwhile for the examinee to try for a moment to think of a seventh or eighth possible birth order—it will become apparent that these six permutations represent all of the possible orders for having two girls and two boys. Next, one must calculate the possibility of each birth order. Because the probabilities of having a boy or a girl are both ½, the probability of the first order is ½ x ½ x ½ x ½ = (½)4 = 1/16. Notice that the math will be identical for the other five birth orders. The final key to this question is to recognize that six birth orders are six different events, ANY of which could satisfy the probability requirement of two boys + two girls, in any order. When multiple events are possible, and any of those events satisfy the question, the probabilities of each event must be ADDED. To clarify, think of a simple example. If you draw a ball from a bowl every morning and the bowl contains 33% red balls, 33% blue balls, and 33% green balls, the chance of drawing a green ball is clearly 1/3. But what if the question is "What is the probability of drawing either a green OR a blue ball?" It is now apparent that 66% of the balls in the bowl satisfy this requirement, so your odds are 2/3. Notice that this is the same as adding the probability of drawing a green ball to the probability of drawing a blue ball: 1/3 + 1/3 = 2/3. In the same way, we must add the probabilities of these two parents having two girls and two boys in any of these six possible birth orders. We established previously that the probability of one birth order is 1/16. Therefore, the chance of any of them happening is: 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6(1/16). This is equivalent to Answer D: 6(1/2)4. This proof eliminates Answers A, B and C.

Which of the following is NOT a part of the respiratory system? A. The epiglottis B. The lacteals C. The larynx D. The right bronchus

B. Lacteals are part of the intestine

Following puberty, the testes begin producing large amounts of testosterone. After production, the testosterone: A. is stored in secretory granules until needed to be secreted in response to declining plasma testosterone concentrations. B. diffuses into the circulatory system and is transported around the body while bound to a plasma protein. C. is confined to the nucleus of the cells where it is produced, where it upregulates genes associated with secondary sex characteristics in males. D. is transported to target tissues, where it binds to cell surface receptors and triggers a second messenger cascade.

B. Remember that testosterone (shown below) is a steroid hormone. As such, it can freely diffuse through cell membranes. Also, as a hormone, testosterone is transported throughout the body in the circulatory system, not confined to the nuclei of cells in the testes. Testosterone must bind to a transport protein since it is not hydrophilic and thus cannot simply dissolve in the blood plasma directly.

Based on information in the passage, what type of inhibition best describes the action of NADPH on G6PD? I. Competitive II. Allosteric III. Irreversible A. I only B. II only C. II and III only D. I, II, and III

B. The passage states that high levels of NADPH inhibit G6PD, and Equation 1 shows that the substrate of G6PD is G6P. Because the structures of G6P and NADPH are very different, it is unlikely that NADPH competes with G6P at the active site; thus, this is not competitive inhibition (I). The passage also never suggests that the inhibition of G6PD is irreversible; rather, it is likely to be dynamic based on the amount of NADPH available (III). Therefore, NADPH most probably binds to a site that is not the active site, which is characteristic of allosteric inhibition (II). A generic model of allosteric inhibition is included below.

Conn's syndrome, also known as primary hyperaldosteronism, is most likely to cause which symptom? A. High renin concentration B. Low blood potassium C. Low blood sodium D. Hypotension

B. This question is asking you to recall the effects of aldosterone and how it achieves those effects. Aldosterone increases H2O and Na+ reabsorption from the kidney while exchanging K+ ions for Na+ ions. The triggers for and results of aldosterone secretion are shown below.

In the U.S. population, the frequency of the allele for colorblindness, Xc, is 8%. Which of the following is the frequency of colorblind women and colorblind men in the population, respectively? A. 8%, 0.64% B. 0.64%, 4% C. 0.64%, 8% D. 64%, 8%

C. For a woman to be colorblind, she must get two copies of the Xc allele. Since the frequency of this allele is 0.08, the odds of being homozygous for the colorblind allele are 0.0064, or 0.64%. Males only need a single copy of the Xc allele to be colorblind (since they only contain one allele copy, due to their small Y chromosome), making the odds of a male being colorblind 8%.

Which of the following would be LEAST useful in cellular movement? A. Flagella B. Actin polymerization C. Microtubule depolymerization D. Cilia

C. In order for cells to travel to the site of injury, they need to migrate. Microtubule depolymerization is responsible for separating chromosomes during anaphase of mitosis or meiosis I or II. It does not contribute to overall cell migration.

The concentration of intracellular signaling molecules fluctuates rapidly in dividing cells during the cell cycle. Which of the following experimental techniques would be best to elucidate the mechanism of regulation for these proteins? A. RT-PCR and Southern blot B. Southern blot and northern blot C. Western blot and RT-PCR D. Western blot and Southern blot

C. Rapidly dividing cells undergo mitosis under the influence of specific signaling molecules. These molecules are expressed when their genes are transcribed, then are translated into proteins. In order to gain the best understanding of how a signaling protein's levels are regulated, both the protein and mRNA levels would need to be studied. Western blotting gives us information about the amount of protein expressed in a cell, while RT-PCR gives us information about the amount of RNA expressed.

Which of the following is NOT a function of the blood-brain barrier? A. Protection of the brain from the action of systemic peptide hormones B. Maintenance of a stable chemical equilibrium for the brain C. Protection of the brain from carbon dioxide poisoning D. Allowing more glucose to enter the brain than any other tissue

C. The passage states that the blood-brain barrier protects the brain from harmful agents that are large or polar. Carbon dioxide (shown below) is both small and nonpolar, making it highly lipid-soluble. Thus, it will freely pass through the BBB.

In severe diabetic hyperglycemia (high blood sugar), insulin cannot effectively induce the uptake of glucose by cells. Assuming otherwise healthy nephrons, chronic hyperglycemia directly leads to the presence of which of these molecules in the urine? I. Proteins II. Glucose III. Ketone bodies A. I only B. I and II only C. II and III only D. I, II, and III

C. This question requires outside knowledge about glucose metabolism. If cells cannot take up glucose, it will remain in the blood and eventually be excreted in the urine when it builds up to the point that it cannot be reabsorbed by the nephron (II). In a state of extended hyperglycemia, the body relies on fat metabolism to generate energy, which produces ketone bodies that are also excreted in the urine (III).

In humans, the quadriceps muscle is a skeletal leg muscle under voluntary control. As such, it contracts in response to signals from the motor cortex in the cerebrum. If such a muscle contracts involuntarily, that indicates: A. a pathological hyperpolarization of the nerve innervating the muscle. B. an appropriate modulation of Ca2+ ions in the muscle tissue. C. the operation of a reflex arc. D.significant sodium efflux from the neuron.

C. Voluntary muscles may contract involuntarily due to a reflex arc. The classic example is the patellar tendon reflex, in which sudden stretching of the patellar tendon leads to an involuntary contraction of the quadriceps. Such contraction occurs before the signal has even reached the brain. A typical reflex arc is shown below. This arc contains a sensory neuron, which carries sensory information from peripheral receptors toward the spinal cord, and a motor neuron, which carries a signal from the spinal cord to an effector muscle. This particular reflex arc also contains an interneuron, which is a neuron within the spinal cord that synapses on both the sensory and motor neurons, connecting them.

NK cells are innate immune cells and Cytotoxic T Lymphocytes (CTLs) are adaptive immune cells. In what other way do NK cells and CTLs most likely differ? CTLs require pathogen processing and presentation, whereas NK cells do not. NK cells are pathogen-specific, whereas CTLs are not. NK cells are leukocytes, whereas CTLs are lymphocytes. NK cells cannot target tumor cells, whereas CTLs can.

Choice A is correct. Adaptive immune cells, including CTLs, require activation and pathogen surface presentation. This makes answer choice A correct. Answer B should be rejected because the reverse is actually true. Adaptive cells (CTLs) are pathogen-specific, while innate cells (NK cells) target infected or cancerous cells generally. Answer C should be rejected because both NK cells and CTLs are lymphocytes that target intracellular infections or tumor-bearing cells. The term leukocyte is a general term for all white blood cells. A lymphocyte is a type of leukocyte. Other examples of leukocytes include: neutrophils, basophils, eosinophils, monocytes, and macrophages. Answer D is contradicted by the information in the first paragraph, which states that NK cells target cancerous and virus-infected cells. This question illustrates the value of restating the question stem. The question becomes much easier for most students if one simplifies the question stem to what it is really saying: "How do innate and adaptive immune cells differ?" Many AAMC question stems can be simplified in a similar manner.

Biologists introduced 35S into bacteriophage proteins and inoculated cultured bacteria with the labeled phages. From what bacterial centrifugation fraction will the radiolabel be detected? Viral coat proteins from the extracellular fraction Bacterial proteins from the bacterial cytosol fraction Viral coat proteins from the intracellular fraction Bacterial proteins from both the intracellular and extracellular fractions

Choice A is correct. Identifying the correct answer requires two pieces of information. First, 35S will be incorporated into the viral protein, primarily the protein coat. It will not be incorporated into viral nucleic acid, which contains no sulfur. Second, when bacteriophage viruses infect cells, the protein coat remains outside the cell and the viral nucleic acids are injected into the cell. Thus, the radioactivity would be recovered from the protein coat, outside the cell, and not inside the cell. This information allows the other three choices to be eliminated: it will not be inside the bacterial cells, as suggested in Answer B; and it will not be found in the bacterial cell at all, which eliminates Choices C and D.

During embryological development, Agrin released from a somatic motor neuron causes acetylcholine receptors throughout the sarcolemma to migrate to the motor end plate. With which cellular structure is Agrin most likely to interact during NMJ formation? Myocyte plasma membrane Nerve cell plasma membrane Myocyte cytoplasm Nerve cell cytoplasm

Choice A is the correct answer. The acetylcholine receptors are located in the sarcolemma, the cell membrane of the muscle cell. It is therefore reasonable to assume that inducing the receptors to cluster at the motor end plate requires that Agrin interact with those receptors directly, or with the membrane in which they are located. Choices B and D can be eliminated because the Agrin protein is released from the axon terminal, and because the sarcolemma is not found in axons. Choice C can be eliminated because Agrin is a protein unlikely to travel to the muscle cell cytoplasm, given that neuromuscular junctions are found on the cell surface.

Phosphorylation is a common mechanism for activating or inactivating an enzyme. Phosphorylation most likely alters enzymatic function because the charge on the phosphate: creates a repulsive interaction that stabilizes the enzyme. creates a repulsive interaction that causes a conformational change in the enzyme. is attracted to negatively charged amino acid side chains in the active site. increases the diffusion rate of the enzyme into hydrophobic regions of the cell.

Choice B is correct; the addition of a phosphate can induce a conformational change that could either activate or inactivate the enzyme or binding interaction. The examinee should also recognize that phosphate bears a negative charge. Answer A is not logical because a repulsive interaction would not stabilize a protein structure. Repulsive interactions, especially in a folded protein, would more likely be destabilizing. Answer C is not correct since a negatively-charged phosphate would be repelled, not attracted to, negatively-charged amino acids. Answer D is not correct since phosphate is charged and therefore hydrophilic, not hydrophobic.

Which physiological mechanism is LEAST likely to result in a significant decrease in blood acidity? Exhalation of carbon dioxide Release of H3PO4 into the blood stream Conversion of bicarbonate to carbon dioxide in the blood Excretion of protons into the urine by the kidneys

Choice B is the best answer. The major physiological mechanisms used to control blood pH include exhalation of carbon dioxide, conversion of carbon dioxide into bicarbonate in the blood, and the excretion of protons through the kidneys. In the precise way they are stated here, all three mechanisms will decrease blood acidity (i.e., increase blood pH). Answer A will decrease blood acidity according to Le Chatelier's Principle as it is applied to the following equation: CO2 + H2O ⇄ H2CO3 ⇄ H+ + HCO3- Exhaling carbon dioxide removes it from this equilibrium, driving the reaction to the left and consuming H+ ions. Fewer H+ ions would decrease blood acidity (increase pH), so Answer A is incorrect. Answer C is incorrect because this is the reverse of the reaction shown above, which would consume H+ ions and therefore decrease blood acidity. Answer D will decrease blood acidity by directly removing H+ ions from the blood. Answer B is the correct answer to this "LEAST" question type because the release of phosphoric acid into the blood would directly increase blood acidity, rather than decreasing it. However, this could be tempting to exclude if the student recalls that the phosphate buffer system plays a minor role in maintaining blood pH. However, this buffer is represented by the equation given below, and adding phosphoric acid to the blood would still increase blood acidity. Phosphoric acid will not only dissociate into dihydrogen phosphate and a hydrogen ion, but the resulting dihydrogen phosphate will result in the formation of additional protons according to Le Chatelier's Principle: H2PO4- ⇄ HPO42- + H+

Which experimental result from the study of transgenic rats pre-disposed to AD would most strongly support the theory that a reduction in the number of dendritic spines causes AD? The number of dendritic spines on cortical neurons is inversely correlated with the severity of AD symptoms. Increasing the dosage of a drug that promotes dendritic spine formation is associated with a significant decrease in AD symptoms. Symptom-free rats with a reduced number of dendritic spines on their cortical neurons eventually develop AD. A drug that inhibits dendritic spine formation on cortical neurons is associated with a significant increase in AD symptoms.

Choice B is the correct answer. All of the answers provide some level of support for the conclusion that dendrite morphology is an integral part of AD, but only two of them suggest (but do not prove) causality. Answers A and C are strictly correlations and provide no evidence of causality. Answers B and D provide the strongest evidence because they are not strictly correlative. Answer B is a stronger choice than Answer D because it implies that not only does the increase in dendritic spines result in decreased AD symptoms, but it also suggests a dosage effect: more drug (and therefore more dendritic spines) results in even fewer AD symptoms. A dosage effect is stronger evidence of causation than a simple association between treatment and outcome.

CCCP, a chemical that depolarizes the inner mitochondrial membrane by allowing free passage of hydrogen ions, can be used to study mitochondrial diseases. Treatment of healthy cells with CCCP would likely result in: increased ATP production. decreased ATP production. decreased heat output. decreased O2 consumption.

Choice B is the correct answer. CCCP depolarizes the mitochondrial membrane, disrupting the proton gradient established through the electron transport chain. This would decrease ATP output because without a proton gradient, ATP synthase will not function properly. Answer choice A is not correct because ATP production cannot increase when the proton gradient is disrupted. Answer C is incorrect because heat output increases when the proton gradient is disrupted. Answer choice D is not correct because oxygen will be consumed at an equal or greater rate, as electron transport is still occurring through the other ETC complexes. This question illustrates the faulty logic of one spurious "strategy" promulgated by many MCAT prep companies. In this scenario, three of the answers begin with the same word and only one does not. Some have suggested that this is a good reason to choose option A, the only one of the four that is "different." While that could "work" on some questions, it will cause you to miss most questions, including this one. These are really just "guessing patterns," not strategies for mastering the MCAT. Avoid gimmicks of this type and work on improving your critical thinking—something that will actually earn you more points on the MCAT.

Given a genome content of 50% AT, and assuming random nucleotide variability, the probability of the AUUUA sequence is one in every: 25 bases. 45 bases. 2.0 x 105 bases. 4.0 x 105 bases.

Choice B is the correct answer. If a genome is 50 % A/T, then it is also 50 % G/C. At each position there is therefore a 1 in 4 (25%) chance of having the necessary base out of the four possible bases. Since the sequence is five nucleotides long, one must multiply the individual probabilities: (¼)(¼)(¼)(¼)(¼) = 1/1024. Thus, the probability of finding AUUUA is one in every 1024, or 45 bases. Answer A may be selected if the student thought there was a ½ chance of a given base at each position, but this is incorrect because there are four possible bases. Answer choices C and D are permutations of Answers A and B given in scientific notation and both are numbers far too large.

During periods of intense skeletal muscle activity, pyruvate is converted to lactic acid and: NAD+ is oxidized to NADH NADH is oxidized to NAD+ NAD+ is reduced to NADH acetaldehyde is oxidized to ethanol.

Choice B is the correct answer. The question addresses the reactions of fermentation. In humans, when there is insufficient oxygen for the mitochondria to metabolize all the pyruvate produced, the cell can use glycolysis to produce some ATP. However, this reaction requires continued regeneration of NAD+ to accept electrons; regeneration of NAD+ requires other mitochondrial reactions to take place (NADH must donate its electrons to the electron transport chain to produce NAD+). An alternate pathway converts the pyruvate in the cytoplasm to lactic acid, a reduction that is accompanied by the oxidation of NADH to NAD+. This is choice B. Choices A and D can be rejected because the reactions described are reductions, not an oxidations. The reaction in Choice C is backward because NAD+ must be regenerated, not NADH.

The histones released by ischemic neurons were primarily found in which cellular location prior to the ischemic trauma? Dendrites Soma Synaptic end plate Axon

Choice B is the correct answer. This question requires recognizing the function of a histone and combining that knowledge with an understanding of the structure of a neuron. Histones are the proteins around which DNA is wound in its condensed conformation. In an intact cell (including neurons), histones will be primarily located in the nucleus, associated with the DNA they stabilize. In a neuron, the nucleus is found in the soma, or cell body, of the neuron, making B the only plausible answer. Because DNA is not present in the axon, synaptic end plate (distal part of the axon at the synapse), or the dendrites (projections off of the soma), histones will not be found in these locations to any significant degree, eliminating Answers A, C and D.

Prions are infectious agents composed entirely of protein. In human hosts, prions cannot: be inherited or coded for by a gene. convert normal proteins into pathogenic proteins. cause pathology outside the central nervous system. be acquired through food consumption.

Choice C is correct. Although prion infections are usually associated with the consumption of contaminated meat (Answer D), they can also be inherited and encoded for in a person's DNA (Answer A). Prions convert normal proteins of the brain into pathogenic ones (Answer B). This conversion is exponential and takes years to decades to become pathological. The correct answer is option C because prion pathology is only observed in the central nervous system of the infected person or animal.

The synthesis of glucagon in pancreatic α-cells is most closely associated with which cellular organelle? Nucleus Lysosome Endoplasmic reticulum Mitochondria

Choice C is correct. Because glucagon is a secretory protein, it will be translated at the rough endoplasmic reticulum, with the growing protein chain oriented into the ER lumen as translation proceeds. Initially, the mature mRNA will be bound by a cytosolic ribosome, but a signaling sequence will pause translation and cause translocation to the rough ER. This makes Answer C correct. The recognition that glucagon is a secretory protein is necessary for this question, but does not require prior memorization. Students should recognize glucagon as a hormone that regulates blood sugar and know that it is synthesized for the entire body in the pancreas; therefore it must be secreted by the α-cells that produce it. Answer A may be tempting because the gene coding for glucagon would be found in the nucleus, but synthesis refers to translation of the mRNA transcript to form the actual protein, so choice A is incorrect. Answer B is incorrect, and is also implausible because lysosomes contain proteolytic enzymes. Answer D is incorrect because, although mitochondria do synthesize a few proteins for their own use, they do not synthesize secretory proteins for use outside the cell. The overwhelming majority of all mitochondrial proteins are coded for by the nuclear genome.

Once the insect extracts have passed through the GNA affinity column, the researchers should include which component in the elution liquid to most effectively elute the bound proteins from the column? Negatively charged ions Positively charged ions Oligomannose polymers Non-glycosylated insect proteins

Choice C is correct. Because the column described in the passage is an affinity chromatography column, the proteins are bound to the column based on a specific binding interaction between the resin and some component of the glycoproteins. Because we are told that glycoproteins with oligomannose were selectively bound, the GNA resin must bind oligomannose. To remove the proteins, the researchers will need to put in a high concentration of something that also has an affinity for the resin. In this case, since the column is separating out oligomannose glycosylated proteins, using pure oligomannose will be very effective. Answers A and B are incorrect because these are what you would use to elute from an ion exchange chromatography column. In ion exchange, the column is either positively or negatively charged, and molecules bind if they are of the opposite charge. To elute the molecules, you add a high concentration of an ion of the opposite charge to out-compete the bound molecules and elute them off. The conceptual principle in both cases is the same. Answer D is incorrect because, since the column is separating glycosylated proteins, passing non-glycosylated proteins through the column will not have any effect.

Which macromolecule is capable of acting in vivo as a structural component, a catalyst, and a mediator of cell communication? Nucleic acids Phospholipids Proteins Lipids

Choice C is correct. Proteins are frequently discussed in biochemistry as enzymes, but they also have many non-enzymatic functions, including acting as structural components, components in plasma membranes, cell signaling molecules, antibodies, and so forth. Answer A is not correct because nucleic acids usually contain the information to make proteins, but cannot participate in all of the functions listed. Choice B is not correct because phospholipids are structural parts of the plasma membrane but they do not act as enzymes. Choice D is not correct because lipids do not act as enzymes.

Quorum sensing in S. aureus most likely does NOT include communication between which two entities? Modulatory proteins secreted by one bacterium and transcription factors in another bacterium. One S. aureus bacterium and other S. aureus bacteria growing within the same colony. Quorum protein AI-2 secreted by one bacterium and chemoreceptor neurons in another bacterium. Paracrine signaling molecules secreted by one bacterium and the cell membrane of another bacterium.

Choice C is correct. This question stem includes a qualifier, so we are looking for the answer that is NOT a plausible way in which quorum sensing could occur between S. aureus bacteria. Answer C is impossible; being prokaryotes, bacteria lack organized cellular organelles, much less developed nervous systems with specialized chemoreceptors. This makes C the correct answer because it is clearly NOT how quorum sensing functions. Answers A and D both use slightly different terms to describe how quorum sensing most likely occurs. The passage states that a protein (AI-2) is a signaling molecule produced by bacteria for this purpose. We are also told that the process is paracrine-like and involves stimulus-response sets between bacteria. Therefore, Answer D, secretion of a signaling molecule, followed by binding of that ligand to a receptor on a neighboring bacterium, is highly plausible. Answer A uses very similar logic, but introduces transcription factors. This is plausible as well because the passage indicates that bacteria use this process to "coordinate gene expression." Finally, Answer B simply states that S. aureus bacteria in the same colony communicate with one another, which is central to the definition of quorum sensing.

The final electron acceptor during alcohol fermentation in S. cerivisiae is acetaldehyde. Is this process analogous to anaerobic respiration in humans? Yes, because acetaldehyde is the final electron acceptor. Yes, because oxygen is the final electron acceptor. No, because pyruvate is the final electron acceptor. No, because oxygen is the final electron acceptor.

Choice C is the correct answer. In humans, when insufficient oxygen is available (as in muscles during extended exercise), or when the necessary cellular equipment is lacking (as in red blood cells), glucose undergoes anaerobic fermentation with pyruvate acting as the final electron acceptor. Many students confuse this with lactate, which is formed in the final step, but it is the reactant molecule, pyruvate, that is reduced and is therefore labeled as the final electron acceptor. Similarly, oxygen is identified as the final electron acceptor in aerobic respiration, even though water is the molecule formed in the final step. Answers A and B are false because yeast fermentation and human anaerobic fermentation utilize different final electron acceptors. Answer D is false because oxygen is the final electron acceptor in human AEROBIC respiration and the question asks about anaerobic respiration.

An experiment focused on anaerobic respiration in muscles under hypoxic conditions should focus on which cellular compartment? Mitochondrial matrix Mitochondrial intermembrane space Cytoplasm Golgi body

Choice C is the correct answer. In mammalian cells, anaerobic respiration is the alternative pathway taken when oxygen is not available for oxidative phosphorylation, such as during the oxygen debt created during extended periods of exercise. Glycolysis occurs in the cytoplasm and the products from that pathway will be the final electron acceptors to produce lactic acid. Because this occurs in the cytoplasm rather than the mitochondria, Answer C is correct. When oxygen is present, the products of glycolysis are trafficked to the mitochondria for oxidative phosphorylation, eliminating answers A and B. The Golgi body (Answer D) is primarily involved in the modification and trafficking of proteins and lipid production and hence, it is not related to anaerobic metabolism, eliminating answer D.

The germ cells of a newly discovered diploid organism divide without the DNA replication step that normally occurs prior to meiosis. For this organism, which type of chromosome is found at the metaphase plate during Meiosis I? Tetrads, consisting of two pairs of sister chromatids Dyads, consisting of a single pair of sister chromatids Monads, consisting of a single chromatid Bivalents, consisting of four pairs of sister chromatids

Choice C is the correct answer. Prior to Meiosis I, under normal circumstances, all chromosomes (46 in humans, consisting of 23 homologous pairs) are replicated. This results in the formation of dyads, duplicated chromosomes that each consist of two strands of DNA; one strand individually is referred to as a chromatid; and the two chromatids are joined by a single centromere. When two homologous dyads pair up during meiosis I, the complex as a whole is called a tetrad (4 strands of DNA). However, in the organism described in the stem, DNA replication does not occur prior to meiosis. Therefore, for a diploid organism with 2n = 46, there will still be 46 chromosomes at the start of meiosis, but each chromosome will contain only one strand of DNA. There will be no sister chromatids present, during either Meiosis I or Meiosis II. This greatly simplifies the question, eliminating answers A, B and D because they all refer to sister chromatids. Bivalent is another word for tetrad, and both terms are exactly equivalent in meaning. Tetrads/bivalents consist of two pairs of sister chromatids (4 DNA strands), NOT four pairs (8 DNA strands) as is suggested by Answer D. A monad is a chromosome consisting of a single DNA strand.

What are the primary functions of the colon and rectum in the human digestive tract? Absorption of macromolecular nutrients and water Mechanical mixing of waste products, storage and elimination Absorption of water and vitamins, storage and elimination Storage and elimination only

Choice C is the correct answer. The colon is often considered to include the rectum and anus as well as the large intestine. The primary function of the large intestine is to absorb (reclaim) water from the waste and the primary function of the rectum and anus is to store and expel waste. None of the other answers include both of these functions.

Which functional group is LEAST likely to act as a nucleophile in a biological reaction? Thiol Alcohol Carbonyl Amine

Choice C is the correct answer. This is a LEAST question type, so we are searching for the functional group that is NOT a nucleophile, or the one that is least likely to act as one in a biological setting. Carbonyls are frequent participants in biological reactions, but primarily as electrophiles, wherein the electrophilic carbonyl carbon is attacked by a nucleophile. Answers A and B, thiols and alcohols, act similarly and are both nucleophiles. Answer D, amines, act as nucleophiles in all cases other than protonation to form a quaternary amine. Students may be challenged by the fact that alcohols and amines can act as electrophiles in some cases, but this is a LEAST question and it is illogical to select either group as being unlikely to react as a nucleophile when alcohols and amines are the two MOST common nucleophiles found functioning in enzyme active sites.

In what way does smooth muscle contraction differ from skeletal muscle contraction? Only smooth muscle contraction involves the formation of myosin cross bridges. Only smooth muscle contraction utilizes troponin and transverse tubules. For skeletal muscle contraction only, troponin is necessary to expose the actin binding site. For skeletal muscle contraction only, calcium is necessary to initiate contraction.

Choice C is the correct answer. Troponin is required for skeletal muscle contraction, but is not involved in smooth muscle contraction. Answer A is not correct because skeletal muscle cells do utilize cross bridges. Choice B is not correct because it is skeletal muscle contraction which utilizes troponin and T-tubules, not smooth muscle contraction. Choice D is not correct because calcium is required for contraction in both smooth and skeletal muscle.

When antidiuretic hormone is acting on the collecting tubules of the kidney in response to acute dehydration, the osmolarity of the ascending limb of the Loop of Henle will: remain constant. increase. decrease. oscillate.

Choice C is the correct answer. When a person becomes dehydrated, the brain secretes antidiuretic hormone (ADH), which increases water reabsorption at the collecting ducts. Reabsorption occurs through pores (aquaporin proteins) that are opened up in the collecting ducts, allowing water to pass into the medulla. As a result, there will be more water in the medulla. A more dilute medulla shifts the equilibrium between the filtrate in the ascending limb and the medulla, driving more ions out of the ascending limb (Note: recall that the ascending limb is impermeable to water, but highly permeable to ions). This increased removal of ions decreases the osmolarity of the filtrate in the ascending limb, so choice C is the correct answer. Answer A is not the correct answer because the osmolarity would change. Answer B is incorrect because osmolarity is expected to decrease, not increase. Answer D is incorrect because the change in osmolarity would be in one direction; it would not oscillate.

Suppose airborne transmission of H7N9 is verified from chickens to humans, but not to ferrets, even though ferrets have tracheal architecture similar to humans. If infected humans experience symptoms of H7N9, which explanation is most plausible? Human tracheal architecture is rapidly evolving to meet the needs of the virus. Human tracheal architecture is more closely related to birds than to mammals. A mutation in the viral HA protein allows it to bind a receptor in the human trachea. A mutation in the ferret SA protein prevents the HA protein from binding its receptor.

Choice C is the most plausible explanation for infection of humans with a novel influenza virus strain. Influenza virus strains are highly adaptable due to their rapid mutation rate and lack of effective DNA repair mechanisms. Influenza pandemics in the past have been caused by strains that obtained an evolutionary advantage via mutations that allow binding to a different host receptor; this is the scenario described by Answer choice C. There is also a hint in the passage, where it is mentioned that it is a viral HA mutation that allows the virus to switch from avian to human hosts. Choice A is incorrect because the human trachea is not changing and would not evolve to help a virus. Be careful to never assign intent to the organisms or genes in an evolutionary scenario. The MCAT often tempts examinees with wrong answers that makes it sound as if a virus, gene, or organism "intentionally" mutated, or that it evolved as a reaction to its surroundings, or that it evolved "in order to obtain" a desirable advantage. All mutations are random. Evolution occurs by sheer happenstance when an accidental mutation happens to convey an adaptive advantage. Choice B is not plausible because humans are mammals and therefore share more traits with other mammals than with birds. Choice D can be eliminated because although mutating ferrets might obtain an evolutionary advantage, it is highly unlikely that all ferrets will mutate in such a way that they are no longer infected by influenza virus. Further, an important concept for MCAT-2015 is the principle that viruses undergo rapid mutation because they lack many of the DNA repair mechanisms found in higher-order organisms. Therefore, the likelihood of viral mutation >> likelihood of mammalian mutation.

The primary mRNA transcript for dystrophin contains 300,000 fewer nucleotides than the DMD gene from which it is transcribed. The primary mRNA transcript is shorter because: introns have been deleted. exons have been deleted. the poly-A tail was not transcribed. regulatory sequences were not transcribed.

Choice D is correct. Genes contains large regulatory sequences, such as the promoter region, that are not transcribed. However, these regulatory sequences are still considered part of the gene they help to regulate. As a result, the primary transcript will always be shorter than the gene itself. This matches best with answer choice D. Answer choices A and B can be eliminated because both introns and exons are part of the primary transcript. Introns will be removed during post-transcriptional processing, prior to translation. Choice C is incorrect because the poly-A tail is added onto the primary transcript during post-transcriptional processing. In other words, the poly-A tail would actually add nucleotides to the primary mRNA transcript.

The topoIb catalytic cycle cleaves a covalent bond between phosphorous and which other atom? Nitrogen Carbon Sulfur Oxygen

Choice D is correct. Recall that DNA has a double helix structure. The outer edges are formed by alternating deoxyribose sugar molecules and phosphate groups, which make up the sugar-phosphate (or phosphodiester) backbone. A phosphate group is a phosphorus atom bound to four oxygen atom. The phosphodiester bond is the linkage between the 3' carbon atom of one sugar molecule and the 5' carbon atom of another. The covalent phosphotyrosyl intermediate between any enzyme and DNA requires that the phosphodiester backbone of the DNA must be the target of cleavage. While all four options are atoms found in DNA, oxygen is the only choice found in a phosphate group. D is therefore the correct answer.

Where X represents the conjugate base, which of the acids listed is expected to have the smallest H-X bond dissociation energy? Acid A, pKa = 4.10 Acid B, pKa = 4.75 Acid C, pKb = 9.63 Acid D, pKb = 10.90

Choice D is correct. Solving this question requires understanding several concepts and definitions. First, the stronger the bond is between an acidic hydrogen and its conjugate base, the weaker that acid will be. Strong acids are expected to have relatively weaker H-X bonds (smaller bond dissociation energies) than weaker acids (larger bond dissociation energies). Second, a strong acid is defined by the extent to which it dissociates in water. A measure of its dissociation is its Ka, a type of equilibrium constant. pKa is defined as: pKa = -log Ka. The relationship between Ka and Kb is defined by the dissociation of water: Kw = 1 x 10-14 = KaKb. It therefore follows that pKw = 14 = pKa + pKb. This relationship is important so that the pKb values given in the answer choices can be converted to pKa values for direct comparison. Therefore, the pKa values in the list are: A) pKa = 4.10 (given); B) pKa = 4.75 (given); C) pKa = 14 - 9.63 = 4.37; D) pKa = 14 - 10.90 = 3.10. Since choice D has the smallest pKavalue, it is the strongest acid and will be expected to have the weakest H-X bond. Much of this information is presented as background info for the student. If an examinee easily recognizes that C and D can be converted to pKa values, this question is as simple as picking the answer with the lowest pKa.

S. aureus has a generation time of 20 min. If biofilm density is proportional to colony size, which value gives the optical density (OD560) of the biofilm formed by the SBY2 colony one hour after the measurements taken for Figure 2? Figure 2 OD560 is 2 0.3 0.8 6.0 16.0

Choice D is correct. This answer requires understanding two concepts. First, if the generation time is 20 min, the number of bacteria will double three times in one hour. Second, if the optical density is proportional to the colony size (which may or may not be true in practice), then the OD560 value will increase linearly with the colony size. At the time point measured in Figure 2, SBY2 was associated with an optical density of 2.0. Twenty minutes later (one doubling time for the bacteria), this will have doubled to 4.0. Forty minutes from the original time point it will have doubled again to 8.0, and one hour from the original time point it will have doubled a third time, to 16.0, or Answer D. Answer C is the result of multiplying 2.0 by three instead of doubling it three times. Answers A and B use the same logic as choices C and D, but are misapplied to the SBY3 colony instead of the SBY2 colony.

Which innate and adaptive immune cells, respectively, were most helpful in protecting the Chinese patients from influenza mortality? Neutrophils and B cells Dendritic cells and T helper cells Eosinophils and mast cells Natural killer cells and cytotoxic T cells

Choice D is correct. This is a knowledge-based application question about the function of innate and adaptive immune cells. Answer A proposes neutrophils and B cells as primarily responsible for preventing mortality in the Chinese patients. Neutrophils are phagocytic cells whose function is not closely involved with combating viral infections. B cells will eventually produce antibodies against a virus, but are not the most effective cells for combatting viral infections. Answer B is false because although dendritic cells are important in the activation of T cells, it is cytotoxic T Cells, not helper T cells, that are involved in combatting viral infections. Eosinophils and mast cells, Answer choice C, are both innate immune cells, and therefore this choice violates the requirement in the stem that the answer give an innate and adaptive immune cell, respectively. Finally, Natural killer cells and cytotoxic T cells, Answer choice D, are innate and adaptive cells, respectively, and both play a major role in killing virus-infected cells. This makes D the best answer.

Elevation of IL-6 has been linked to Autism Spectrum Disorders (ASDs) and it is therefore believed that inflammation during early childhood may make a child more prone to ASDs. Which gene product, if found to be upregulated in ASD patients, would most weaken this theory? Another pro-inflammatory cytokine that acts synergistically to prevent inflammation. Another pro-inflammatory cytokine that acts antagonistically to prevent inflammation. An anti-inflammatory cytokine that synergizes the effects of pro-inflammatory cytokines. An anti-inflammatory cytokine that antagonizes the effects of pro-inflammatory cytokines.

Choice D is correct. This question requires using and understanding several pieces of information. One is the definitions of antagonism and synergism. The others are passage data, from which one can conclude that factors that increase inflammation should enhance ID. If this does not happen, the theory would be weakened. Answer A can be eliminated as two pro-inflammatory agents can act synergistically, but would enhance, not prevent, inflammation. Answer B should be eliminated as two pro-inflammatory agents cannot act antagonistically. Answer C should be rejected because, while these two in combination could reduce inflammation, an anti-inflammatory and pro-inflammatory agent cannot act synergistically. Answer D is correct: a pro-inflammatory and anti-inflammatory agent could act antagonistically and would counteract inflammation. If ID were enhanced by the two, it would weaken the theory.

Following ingestion of a meal high in simple carbohydrates, the primary metabolic process at work is: glycolysis gluconeogenesis glycogenolysis glycogenesis

Choice D is the correct answer. Glycogenesis is the process of building glycogen to store glucose for the future. Answer A is one of the ways to consume glucose from the blood stream. Glycolysis only consumes a small amount of glucose prior to the pathway being shut down because the cell has sufficient energy. Because glycolysis consumes such a small amount of glucose prior to shutting down, it is not a major player in the balancing of blood sugar. Glycogen can grow to any size and therefore Answer D plays a much larger role in balancing blood sugar. Answers B and C are not correct since both of these processes will result in increased blood sugar since glucose is released from the cell.

The RNA needed for the RT-PCR assay could have been extracted from which of the following cellular locations? I. Nucleus II. Golgi apparatus III. Endoplasmic reticulum IV. Cytoplasm I only I and IV only I, II, and IV only I, III, and IV only

Choice D is the correct answer. Most RNA is made in the nucleus and is used for protein synthesis at the rough endoplasmic reticulum, or on free-floating ribosomes in the cytoplasm. This is answer choice D, as it includes Statements I (nucleus), III (endoplasmic reticulum) and IV (cytoplasm). The Golgi apparatus (Statement II) receives proteins and other products synthesized in the endoplasmic reticulum, but the Golgi is not involved in protein synthesis and should not contain RNA under normal conditions, eliminating answer choice C. Answers A and B are incorrect because, although they include correct statements, they do not include all of the correct statements.

In its native form, β-A is a soluble protein with extensive α-helical domains. Misfolding of β-A causes an α-helix-to-β-sheet transition that decreases solubility. β-A aggregation is likely the result of which interaction between β-sheet motifs on misfolded β-A proteins ? Charge-charge attractions Charge-charge repulsions Hydrophilic interactions Hydrophobic interactions

Choice D is the correct answer. Soluble proteins have their hydrophobic regions sequestered on the inside of the folded protein, away from the hydrophilic cytosol. When the pleated β-sheet forms, hydrophobic groups are exposed to the polar cytosol and are therefore attracted to the sheets of other unfolded proteins. A favorable energy change is therefore associated with aggregation. None of the other choices correctly attribute the aggregation pattern to the hydrophobic regions in the unfolded protein. Answers A, B, and C all list the interaction as hydrophilic or charged. To the degree that misfolding exposes more polar or charged groups to the cytosol, this would increase β-A solubility. Therefore, Answers A, B and C should be eliminated.

Although the average pKa of the Glu side chain is approximately 4.2, the enolase enzyme contains a glutamate in its active site that is protonated at physiological pH. This Glu residue is protonated because: it is buried in the center of the protein, inhibiting proton dissociation. enolase functions inside of lysosomes, where the pH is naturally much lower. the local environment lacks a basic molecule to act as a proton acceptor. the local environment increases the pKa of the Glu side chain.

Choice D is the correct answer. The local environment causes the pKa of the Glu side chain to become more basic, causing the proton on the Glu side chain to remain bound. Answer A is not correct since the active site of most enzymes is on or near the surface of the protein, making it accessible to the surrounding solvent, not buried inside the core of the protein (which is usually almost exclusively hydrophobic). Answer B is incorrect since enolase is one of the enzymes of glycolysis, which occurs in the cytoplasm, not in lysosomes. Further, a lower pH doesn't change the pKa of the side chain; rather, the pKa value tells us the approximate pH at which the proton will be 50% dissociated. Answer C is not correct since the active site is exposed to the aqueous cellular fluid and water is a common proton acceptor.

Based on the amount of KLF6 mRNA expressed after treatment with actinomycin D (Figure 2), cell lines derived from liver carcinomas exhibit: inhibited transcription of KLF6 mRNA. enhanced transcription of KLF6 mRNA. inhibited degradation of KLF6 mRNA. enhanced degradation of KLF6 mRNA.

Choice D is the correct answer. The passage states that transcription occurs normally in cancerous cells. However, Actinomycin D blocks transcription. Because transcription is blocked, answers A and B can be eliminated. After the addition of actinomycin, the amount of each type of mRNA remaining at a given time point is determined by the mRNA half-life, or the rate at which mRNA is degraded. The half-life of KLF6 mRNA is much shorter in both liver carcinoma cell lines. Put another way, the amount of KLF6 mRNA declines more rapidly in these cell lines. This rules out Answer C. Answer D is the only choice remaining, but also makes logical sense since the cancerous cell lines have the smallest amount of mRNA at the final time point—indicating mRNA degrades fastest in these cells.

Protein-based vaccines stimulate an immune response because they: bind glycans on the plasma membrane of foreign cells. recruit inflammatory cytokines and increase blood flow. stimulate T cells to produce antigen-specific antibodies. contain antigens similar to the antigens found on pathogens.

Choice D is the correct answer. Vaccines that are protein-based work because the antigens they contain resemble the antigens of the invading pathogen. For example, viral vaccines usually contain viral capsid proteins, and bacterial vaccines contain surface proteins. This allows the host immune system to initiate a primary immune response and prepare for a secondary immune response without an actual pathogen infection. Answer A is not correct because protein-based vaccines do not recognize glycans. This is a main theme from the first paragraph of the passage. Schistosomes are unique because it is the glycans on their membranes that interact with the host immune system, whereas in most cases immune responses are initiated in response to foreign proteins. Furthermore, vaccines are not intended to bind to foreign cells directly; they are intended only to stimulate the host immune system. Answer B is incorrect because it describes common components of a non-specific innate immune response, whereas vaccines utilize the antigen-specific nature of adaptive immunity. Answer C is false because antibodies are produced by B cells, not T cells.

During the repair of a DSB in human DNA, which nucleotide polymer is used as the homologous template? ssDNA in the nucleus DNA reverse transcribed from mRNA The template strand The sister chromatid

Choice D is the correct answer. When homologous recombination is used to repair a double-strand break, the template used is the sister chromatid. During this process, the recombination machinery seeks out DNA that compliments the overhangs on each side of the break, and the sister chromatid will have that exact sequence. The sister chromatid is not likely to be damaged in the same region that the broken DNA on the other chromatid, so incorporating DNA from the sister chromatid is an efficient method. Answer A is incorrect because there is no reserve of ssDNA in the nucleus. Answer B is incorrect because humans do not have a gene for a reverse transcriptase enzyme. Further, processed mRNA will no longer contain all of the bases that were present in the original DNA, so a reverse-transcribed DNA strand would be incomplete. Answer C is incorrect because the template strand is one of the two strands in double-stranded DNA, so in a double-strand break the template strand will have been removed or damaged.

coenzyme vs. prosthetic group

Coenzymes are organic molecules that are required by an enzyme to function, but are not usually covalently bound. They are differentiated from prosthetic groups because they are loosely bound to the enzyme, whereas prosthetic groups are tightly bound (usually, but not universally, via a covalent bond). While coenzymes are always organic non-protein molecules, prosthetic groups can be either organic or inorganic.

What can methylation and deamination not do?

Convert a purine to a pyrimidine or vice versa

A patient is administered a drug which mimics the effects of the antagonist hormone of calcitonin. Which of the following is a likely direct effect of this drug? A. Increased buildup of hydroxyapatite crystals in long bones B. Decreased plasma concentrations of Ca2+ C. Increased metabolic rate D. Increased osteoclast activity

D. Calcitonin decreases plasma concentrations of Ca2+ by exerting a number of effects. One key example is that calcitonin stimulates bone formation by increasing osteoblast activity and decreasing osteoclast activity. The antagonist hormone to calcitonin is parathyroid hormone (PTH), which acts to increase plasma Ca2+ concentrations by (among other things) decreasing bone formation by osteoblasts and increasing bone degradation by osteoclasts. Thus, a drug which mimics parathyroid hormone would increase osteoclast activity.

A student attempts to isolate a solid organic drug compound with a known melting point range of 148-151 °F. An impure sample of this drug is likely to have which melting point range? 148-150 °F 150-155 °F 145-147 °F 144-150 °F

D. Impurities depress the melting point because they cause the solid phase to be more disorganized due to extra particles which makes the bonds easier to break.

During rat embryogenesis, researchers noticed the development of a fluid-filled cavity in cells that had previously undergone morulation. That cavity was most likely a: A. trophoblast. B. gastrula. C. inner cell mass. D. blastocoel.

D. The question is asking us to identify the developmental stage after the morula (a mass of 16 undifferentiated cells). A good way to remember the development of the zygote is the mnemonic "More blasting gas, I'm nervous," which outlines the process of morula to blastula to gastrula to neurulation. The blastocoel is a fluid-filled central region present in the blastocyst during mammalian embryogenesis. The blastocyst consists of an inner cell mass (ICM), along with an outer cell layer called the trophoblast, which surrounds both the ICM and blastocoel. These structures are depicted below.

Two known inhibitors were studied to analyze their effects on the reaction rate catalyzed by the enzyme lysozyme. Equimolar amounts of inhibitor A (hydronium ion) and inhibitor B (hydroxide ion) were mixed in a beaker before being added into the reaction mixture containing the substrates. Then, lysozyme was also added into the reaction mixture. Assuming Figure 1 represents the reaction conditions, where would the resulting enzyme kinetics curve for this experiment fall? A.Above line 1 B.Below line 1 but above line 3 C.Below line 3 D.The same as line 1

D.Be sure to read the question stem carefully! Inhibitor A is a strong acid (H3O+) and inhibitor B is a strong base (OH-). Since these inhibitors are mixed together before being added to the reaction mixture, they would simply neutralize each other and become water before having any effect at all. Thus, the enzyme kinetics would not be affected, and the curve would fall along line 1.

Transcription

DNA to RNA in the nucleus

G2 cell phase

Duplication occurs

Western blot

In a Western blot, researchers are usually trying to measure the expression of a protein, or the amount of a protein product. Students should recall that the proteins in the original tissue sample must first be separated by electrophoresis. If, during this step, an unequal amount of sample is loaded into each lane, the resulting Western blot could show a smaller signal in a given lane, NOT because there was less of the protein being studied, but simply because less sample was loaded.

Arachidonic acid, released during AEA hydrolysis, is NOT a precursor for the synthesis of what class of molecules? A.Prostaglandins B.Catecholamines C.Thromboxanes D.Phospholipids

In paragraph 1 of the passage, it's written that AEA is degraded to ethanolamine and arachidonic acid. Ethanolamine is included as a polar head group in the phospholipid phosphatidylethanolamine, while arachidonic acid is important as a precursor for the biosynthesis of the eicosanoid signaling molecules: prostaglandins, thromboxanes, and leukotrienes. Neither molecule contributes to the synthesis of catecholamines, which are a class of molecules derived from tyrosine that include dopamine and norepinephrine.

What does malonyl CoA do?

Malonyl-CoA, an indicator of ongoing fatty acid synthesis, inhibits β-oxidation by preventing the movement of long-chain acyl groups into the mitochondrial matrix, thereby preventing a futile cycle of fatty acid synthesis followed by immediate β-oxidative catabolism of those newly synthesized fatty acids.

Pentose Phosphate Pathway

Parallel to glycolysis, increase G6PD when glucose is present

Proteases

Proteases are enzymes that break protein bonds via hydrolysis.

What causes Sn2 to be favored?

The SN2 mechanism is favored by polar aprotic solvents, such as acetone or DMSO.

Beta oxidation

The basic logic of beta-oxidation is to chop up extended fatty acid chains into two-carbon units of acetyl-CoA. Step one of beta-oxidation involves forming a C=C double bond between the alpha and beta carbons of the carbonyl group at the head of the acyl-CoA molecule. This is coupled to the formation of FADH2. Then, in step two, an -OH group is added to the beta carbon. The C-OH bond on the beta carbon is oxidized to C=O in step three, and NADH is formed. Then, in step four, the molecule is broken up, yielding an acetyl-CoA group and a shorter acyl-CoA group. This process is easiest to visualize with a saturated fatty acid with an even number of carbons, but special enzymes exist to handle unsaturated fatty acid and odd numbers of carbons.

Which adaptive immunity cell type is most effective at removing a cancerous cell from the body without the assistance of other immune cells? Cytotoxic T Lymphocyte Regulatory T Lymphocyte Helper T Lymphocyte Natural Killer Cell

The correct answer is A since the Cytotoxic T Lymphocytes are very specific for the cells that they kill and they are designed to kill virus-infected or cancerous cells. Regulatory (Answer B) and Helper (Answer C) T Lymphocytes are very specific, but they only aid in the removal of cancer cells by preventing wrongful activation (regulatory) or by activating B cells (helper). Further, the clue given in the stem, "without the assistance of other immune cells," negates the possibility of a helper cell such as the helper T Lymphocyte. Natural killer cells (Answer D) also kill cancerous cells, but they are innate immune cells and the stem specifies adaptive immune cells.

viable autosomal trisomies

Trisomy 21, 18, 13

Enol tautomer

both OH and C=C group present

What does PTH do?

breaks down bones by stimulating osteoclasts

Chemical vs Electrical Synapse

electrical- gap junctions and Some neurons and glial cells chemical- release of neurotransmitter

sense vs antisense strand

mRNA identical to sense strand, antisense becomes mRNA

Oxidized vs reduced

oxidized loses hydrogens, reduced gains hydrogens

Western blot

protein

First order kinetics

rate depends on concentration of substrate; primarily linear

Where are plasmids localized

the nucleus

supraasal layers

uppermost layers