BioBio Chem Final

During peak cardiac activity, heart muscle utilizes the anaerobic pathway for only 10% of its energy as opposed to up to 85% for maximally contracting skeletal muscle. With respect to skeletal muscle, cardiac muscle would: consume less glucose per ATP produced.

Aerobic respiration includes glycolysis, the PDC, the Krebs cycle, and electron transport, to produce a total of 32 ATP per glucose (assuming the malate-aspartate shuttle is used to transfer the electrons from glycolytic NADH into the mitochondria). By contrast, anaerobic respiration involves only glycolysis and fermentation of pyruvate to lactate or alcohol, to produce 2 ATP per glucose. Anaerobic respiration is therefore much less efficient and must consume much more glucose to produce the same amount of ATP. Since cardiac muscle uses mostly aerobic respiration, it will be more efficient than skeletal muscle and uses less glucose to produce an equivalent amount of ATP (choice C is correct). Conversion of pyruvate to lactate occurs during fermentation, which cardiac muscle carries out less of, not more (choice A is wrong). The amount of contractile force generated per ATP is independent of the source of ATP used (choice B is wrong). Cardiac muscle uses more aerobic respiration, so it will use more oxygen per ATP produced, not less (choice D is wrong). ========================= Since cardiac muscle = aerobic respiration, it will be more efficient than skeletal muscle and uses less glucose to produce an equivalent amount of ATP (choice C is correct) skeletal muscle = fermentation = less efficient = use more glucose

The loss of net production of ATP during glycolysis that occurred in the presence of arsenate and Pi was due to: decreased formation of 1,3-bisphosphogylcerate.

Arsenate competes with Pi as a substrate for glyceraldehyde-3-phosphate dehydrogenase. In the presence of arsenate, 1-arseno-3-phosphoglycerate is produced instead of 1,3-bisphosphoglycerate. =================== glyceraldehyde-3-phosphate dehydrogenase = formation of 1,3-bisphosphogylcerate.

When normal human cells are grown in culture, they will divide a limited number of times - typically 50 rounds of mitosis. After this number is reached, the cells become apoptotic. This cell death is a result of: chromosomal telomeres shortening after each round of division.

As cells undergo mitosis, the telomeres, the portion of the DNA on the ends of the chromosomes, gets progressively shorter. Eventually the DNA loses its telomeres entirely and is unable to reproduce. The cell then dies, undergoing apoptosis. The position of telomeres on the chromosome is shown below.

A mutation in the thyroid hormone receptor which blocks hormone binding would cause which of the following? Elevated secretion of TRH.

As shown in Figure 1 and described in the passage, thyroid hormone causes feedback inhibition of TRH and TSH secretion. Thyroid hormone receptor is responsible for hormone signaling, including feedback inhibition by thyroid hormone. If thyroid hormone receptor is defective, this will prevent feedback inhibition, causing TRH and TSH levels to become elevated (choice B is correct). ================================ block = elevation -------------------- > Thyroid-stimulating hormone releasing hormone (TRH) (hypothalamus ) > thyroid-stimulating hormone (TSH) (anterior pituitary) > TSH stimulates release thyroid hormone (T3 and T4). (non-polar)

Information in the passage indicates that signaling via CB1 in mice decreases the sensory response initiated by the activation of what receptor type? Nociceptors

B is correct. According to the passage, binding of endocannabinoids to CB1 produces "pain-diminishing analgesic" effects. Activation of sensor neurons known as nociceptors by noxious stimuli mediates the perception of pain.

According to the passage, which nerves would be most likely to have a large number of gap junctions? A. Afferent motor neurons B. Defensive reflex neurons C. Parasympathetic neurons D. Sensory neurons

B is correct. According to the passage, electrical synapses are faster than chemical ones and are continuous in their activity. We can infer that their speed is the primary reason why they would be preferred by the body over the traditional chemical synapse. The best answer is thus related to a process that requires speed and/or continuous activity. ================================

If a lung is punctured, what effect will this likely have on breathing? The punctured lung will be unable to inflate, affecting inhalation.

B is correct. During typical breathing, the diaphragm contracts, flattening it and increasing the volume of the intrapleural space. As a result, the lungs expand and the pressure inside them decreases. The resulting pressure differential between the lungs and the atmosphere causes air to rush in, a process termed inspiration. If the lungs are punctured, air will flow freely between the lung and the intrapleural space, and the lung will not expand (since no pressure differential can be maintained). Expansion of the thoracic cavity should still lead to some air flow into the lung, however, although it will also mix freely with the thoracic cavity's air, making it more difficult to absorb oxygen within the damaged lung while also making it harder to maintain a pressure differential with the other lung. Thus, B is definitely true.

The reactions catalyzed by xanthine oxidase in Figure 2 are reversible. If Keq for the conversion of hypoxanthine, water, and oxygen to xanthine and hydrogen peroxide is approximately 1, which change in an equilibrium mixture of the five compounds will NOT necessarily increase the ratio of xanthine to hypoxanthine?

B is correct. For the reaction described, Keq is approximately 1. As we do not know whether this is an endothermic or exothermic reaction, changing the temperature may drive the reaction forward, or it may drive it backwards. In that case, a change in temperature might not necessarily lead to a change in concentration of reactants and products. So increasing temperature may increase this ratio, but it may decrease it as well! ============================= change temperature does not change the concentrations, but kinetics

Which components of cells are physically connected by a gap junction? The cytoskeleton of one to the cytoskeleton of the other

B is correct. The answer is given directly in paragraph 2, which mentions "...two hemichannels (connexons), one embedded in each cell's plasma membrane and anchored to the cytoskeleton of the cells." This means that the cytoskeleton of one cell is connected to the cytoskeleton of another. A schematic of a gap junction is shown below.

Which of the following is NOT a part of the respiratory system? The lacteals

B is correct. This is a straightforward anatomy question with no information from the passage needed. The lacteals (shown below) are structures in the intestines associated with absorbing fat into the lymphatic system. =========================== A: The epiglottis is part of the respiratory system. It is a flap of cartilage that covers the trachea to ensure that food properly remains in the digestive tract. C: The larynx, or voice box, is also part of the respiratory system. It is located immediately above the trachea. D: The right bronchus (and left bronchus) are both part of the respiratory system.

If pharmacologists wished to make morphine available to brain tissue, which of the following changes could be made to its molecular structure to allow for the best chance to use the drug as a direct brain treatment? Replace the alcoholic protons with acetyl groups

B is correct. This question asks us to consider the structure of morphine given in the passage. What needs to change for morphine to enter the brain while maintaining its function? 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. So, replacing them with acetyl groups would allow the drug to become more lipophilic. ========================== acetyl groups = convert hydrophilic to become hydrophobic

Based on the information presented in the passage, which of the following is most likely true about infections of the CNS? Bacterial infections of the brain are more serious because antibiotics or other treatments have a hard time passing the blood-brain barrier.

B is correct. This question asks us to determine why bacterial infections of the brain are more serious than normal infections. The passage states that large molecules have a hard time getting past the blood-brain barrier. Thus, choice B is correct.

The pKa values for cysteine are shown below. At what pH will a solution of 2.2 M cysteine be isoelectric?

B is correct. We need to determine the pH at which Cys will be isoelectric, which means having no net electric charge. There are two ways to solve this question. First, we can use the given pKa values to determine the charge that Cys will carry in each solution. Alternatively, we can recall that an amino acid will be in its zwitterion (i.e. isoelectric) form when the pH of the environment is equal to the pKa of the amino acid. This means we need to take the average of the two most relevant pKa's on cysteine. While Cys is not commonly listed as an acidic amino acid, its side chain includes a thiol (-SH) group, which can be deprotonated/made negative. A way to figure this out is to note that the structure of cysteine at pH = 7 shows that the side group is protonated. So it must be that even though the pKa is 8.33, the thiol is able to act as an acid (reminder, a pKa > 7 does NOT mean the group is basic). For amino acids with acidic side chains, the isoelectric point is the average of the two most acidic (lowest) pKa's.

Which of the following would be true about cis-oleic acid, a monounsaturated fatty acid with the formula CH3(CH2)7(CH)2(CH2)7COOH? It will generate approximately 119 ATP after 8 rounds of β-oxidation, followed by the Krebs cycle and the electron transport chain. Recognize the ability to treat this question as a 2x2 elimination. cis-Oleic acid has 18 carbons and thus will undergo 8 rounds of β-oxidation (eliminate choices A and C). This will generate 9 molecules of acetyl-CoA, 8 molecules of NADH and 7 molecules of FADH2 since it is a monounsaturated fatty acid. Each of the 9 acetyl-CoAs will go through the Krebs cycle and this will generate 27 NADH (which will give 67.5 ATP), 9 FADH2 (which will give 13.5 ATP) and 9 GTPs (9 ATP equivalents). This means the acetyl-CoAs alone generate 90 ATP equivalents. Since the NADH and FADH2 made in β-oxidation will generate even more ATP, the total will exceed 90 (eliminate choice D and choice B is correct).

Beta oxidation > # of carbons > #/2 = # acetyl-CoA > # acetyl-CoA - 1 = # of rounds & NADH > # acetyl-CoA - 2 = # of FADH2 ========================= Kreb cycle 1 acetyl-CoA = 3 NADH, 1 FADH2, 1GTP =================== ETC 1 NADH = 2.5 ATP 1 FADH2 = 1.5 ATP 1 GTP = 1 ATP

According to the data in shown in Figure 1, the Km for the demethylation of the histone by KDM4C is most nearly: 120 μM

C is correct. For an enzyme-catalyzed reaction, the Michaelis constant (Km) is the substrate concentration required to reach one-half of the maximal velocity, or Vmax. First go to Figure 1. he question mentions KDM4C, so we are concerned with the graph on the right. The Vmax looks to be just over 1, maybe about 1.1-1.2 μM/min (the test makers will not expect pinpoint accuracy), so half of that value gives us 0.5-0.6 μM/min. Examining the graph, we can see that this velocity corresponds to a point somewhat greater than 100 μM but significantly less than 200 μM. Choice C is the only answer within this range. ==================================== 1)looking for the max Velocity/2 at Y 2) find the concentration at the X corresponding Y

Meiosis I results in: C. 2 haploid cells with 23 chromosomes, each chromosome consisting of 2 sister chromatids.

C is correct. Meiosis I results in 2 haploid cells, each with 23 chromosomes consisting of 2 sister chromatids per chromosome. In the male, the sister chromatids are split into 4 gametes during meiosis II. For females, meiosis I results in a secondary oocyte (a gamete) and a polar body. Penetration of the secondary oocyte by a sperm brings on anaphase II. Telophase II produces a zygote and a second polar body. Remember for the MCAT: mitosis results in diploid daughter cells, while meiosis results in haploid cells to produce gametes.

Which of the following enzymes should the researchers add to the cell samples if they want to reverse the general catalytic effects of protein kinase A? Protein phosphatase 1

C is correct. Protein kinase A has an extremely wide variety of specific effects, but this question asks about its general catalytic function. As a kinase, PKA functions to add a phosphate group to its substrate. The opposite of this activity is the removal of phosphate from a substrate, a function which is performed by phosphatase enzymes. We do not need to know the exact reactions catalyzed by each enzyme to answer this question; nomenclature alone shows us that only choice C represents a phosphatase. ============================== kinase = add phosphate phosphatase = remove phosphate

Those species that are capable of both sexual and asexual reproduction will typically prefer sexual reproduction because it: creates more variation in the next generation.

C is correct. Sexual reproduction involves "shuffling the genetic deck" when recombining the genes of two parents. This significantly increases variation in the next generation. This increase in variability helps improve the survival of the whole species by allowing it to adapt more quickly to changing selection pressures. A: Sexual reproduction improves the fitness of the whole species, not any individual member of the species.

Which of the following molecules is/are most likely to have selective proteins in the BBB to facilitate its passage into the brain? A. Antibodies B. Starch C. Amino acids D. Urea

C is correct. The correct answer will be a molecule or substance that is essential to brain function. Amino acids are necessary for the production of proteins, which are essential for the function of any cell. A: In general, antibodies are too large to cross the BBB. B: While select monosaccharides (e.g. glucose, fructose) can cross the BBB, large polysaccharides like starch are unable to cross.

Which of the following is NOT a function of the blood-brain barrier? Protection of the brain from carbon dioxide poisoning

C is correct. 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.

Troponin isoenzymes are used as an alternative biomarker in the diagnosis of heart attacks. In which of the following muscle types does the troponin complex function in contraction? I and III only

C is correct. Troponin is a complex of three proteins (troponin I, troponin C, and troponin T) required for muscle contraction in skeletal muscle and cardiac muscle, but not smooth muscle. I and III are true; II is not. Choice C is then correct.

Uracil is usually found in: I. tRNA. II. ribosomes. III. single-stranded DNA. A. I only B. II only C. I and II only D. I, II, and III

C is correct. Uracil is found in any structure made of RNA. Both tRNA and ribosomes are made of RNA. Single-stranded DNA has thymine rather than uracil. ========================== ribosome contains ribosomes

In a population of Amish people, the frequency of the recessive autosomal allele for polydactyly is 1.2%. What percent of the population are heterozygotes for the polydactyly allele? 2.37%

C is correct. We can use the Hardy-Weinberg equation to solve this question. Remember that the total number of alleles in the population has to add up to 1: A + a = 1 And the total number of genotypes in the population must also add up to 1: AA + 2Aa + aa = 1 We're told that a = 0.012. (Note that the question gave the frequency of the recessive autosomal allele, not the frequency of individuals in the population that are homozygous recessive!) By the first equation above, A = 0.988. The carriers are the heterozygotes with the genotype Aa. Their frequency is: 2Aa = 2 x 0.988 x 0.012 = 0.988 x 0.024 ========================================== requency of the recessive autosomal allele = a frequency of individuals in the population that are homozygous recessive = aa

Based on mass, which of the following molecules will most easily pass through a gap junction? An Ala-Leu dipeptide

C is correct. We could do the math for this question and determine the exact mass of each of the answer choices. However, we want to work smarter, not harder, so let's see if we can determine the smallest (by mass) molecule here. By far, the smallest molecule listed is the two-amino-acid chain of alanine and leucine. The heaviest amino acid, tryptophan, weights 204 Da. Thus, two bound residues must weigh 408 Da or less. This is much smaller than the other answer choices. If you did the math, you would find that each amino acid has 2 carbon, 2 oxygen, 1 nitrogen, and 4 hydrogen atoms. Alanine has a methyl side chain (1 C, 3 H), while leucine has an isobutyl side chain (4 C, 9 H), for an additional 5 carbon and 12 hydrogen atoms and a total of 9 C, 2 O, 2 N, and 20 H. This is by far the smallest molecule presented. ================================= smallest molecule = pass through a gap

Diffusion capacity of carbon monoxide (DLCO) is another parameter used for diagnosis of pulmonary diseases because carbon monoxide is diffusion limited. According to the passage, which of the following would be true in emphesyma compared to a normal patient? The DLCO would be lower

Carbon monoxide is diffusion limited; therefore, the rate of diffusion of CO is only limited by factors affecting diffusion across the membrane (choice D can be eliminated). The factors would include thickness of the diffusion barrier and the effective surface area. In emphesyma, the effective surface area is reduced due to loss of alveolar sacs, thus DLCO would be decreased (choice B is correct; choice A and C can be eliminated). ================= surface area is reduced due to loss of alveolar sacs = Diffusion capacity of carbon monoxide (DLCO) decreased

Which of the following is the LEAST characteristic of the Krebs cycle? Most of the energy is produced through substrate-level phosphorylation.

Choice B is the only false statement (which is the correct answer for this question). Only two of the 20 ATP made from the Krebs cycle are from substrate-level phosphorylation.

A sample of bacteria has a doubling time of 30 minutes. If a student begins with 1 g of the bacteria, at what time will the sample weigh 583.8 g? Between 4.5 and 5 hours

D is correct. If the mass of the bacteria were to double every 30 min, beginning with 1 g, then the sequence of mass doubling can be represented as below in the simplest terms: 1 2 4 8 16 32 64 128 256 512 1024 This shows that the mass would be 583.8 g during the final interval represented above, which occurs after 270 min (4.5 hours) have elapsed but before 300 min (5 hours) have passed. A, B, C: In all of these choices, not enough time has elapsed for the bacterial mass to reach 583.8 g. Give feedback on this question

Folate is a common supplement given to pregnant women to prevent neural tube defects. The most likely reason for this is: folate plays a role in ectodermal induction.

D is correct. Neural tube defects are defects in the central nervous system. The nervous system is derived from the ectoderm. It can be concluded from the given information that folate is important for neurulation, or the induction of the ectoderm to differentiate into the nervous system. ====================================== The ectoderm primarily gives rise to the nervous system and epidermis (skin), as well as related structures like hair, nails, and sweat glands, and the linings of the mouth, anus, and nostrils. The process through which the nervous system is formed from the ectoderm is known as neurulation = neural tube. The mesoderm generates many of the structures present within the body, including the musculature, connective tissue (including blood, bone, and cartilage), the gonads, the kidneys, and the adrenal cortex. The endoderm is basically responsible for the interior linings of the body, including the linings of the gastrointestinal system, the pancreas and part of the liver, the urinary bladder and part of the urethra, and the lungs. ================================= ectoderm = neurulation/ neural tube = nervous system, hair, nails, mouth, anus, and nostrils mesoderm = connective tissue, kidneys, gonads, adrenal cortex. endoderm = gastrointestinal system (pancreas, liver, bladder, lungs)

n glycolysis, how many moles of inorganic phosphate are consumed in the oxidation of 3 moles of glucose to pyruvate, as pictured in Figure 1? 6 For every glucose molecule, 2 molecules of glyceraldehyde-3-phosphate are produced. Every molecule of glyceraldehyde-3-phosphate requires one molecule of Pi to be converted into 1,3-bisphosphoglycerate. For 3 moles of glucose, 6 moles of glyceraldehyde-3-phosphate are produced, requiring 6 moles of Pi.

1 glucose = 2 X glyceraldehyde-3-phosphate 1 glyceraldehyde-3-phosphate = requires one molecule of Pi

Which of the following represents the correct sequence for embryogenesis? Fertilization → blastulation → gastrulation → neural tube formation → somite formation Fertilization is the first step, followed by a series of rapid cell cleavages to form a hollow ball of cells called the blastulam (eliminate choices A and B). Next comes the gastrula, in which cells move into the interior of embryo to form the three germ layers (eliminate choice C). Gastrulation is followed by the formation of the neural tube, which will form the nervous system, followed by the formation of other organs and tissues, such as the somites that will differentiate into bones and muscle. This makes choice D the only possible correct order of events.

> Fertilization > blastulam (a series of rapid cell cleavages to form a hollow ball of cells) > gastrula (cells move into the interior of embryo to form the three germ layers) > neural tube, which will form the nervous system > formation of other organs and tissues, such as the somites that will differentiate into bones and muscle.

The passage states that gabapentin is an effective mediator for neuropathic pain. Which of the following provides the best explanation of the direct effect of GBP on transmission of pain signals? GBP decreases the capacity of the pre-synaptic terminal to allow Ca2+ influx, decreasing glutamate release. By preventing the transport and expression of Ca2+ channels on the pre-synaptic membrane, GBP effectively decreases the capacity of the terminal to allow Ca2+ influx; therefore, preventing a crucial step in the classical model of neurotransmission and decreasing release of glutamate (choice C is correct). GBP does not directly block the influx of Ca2+ into the pre-synaptic terminal (eliminate choice A). GBP does not block Ca2+ influx and does not inhibit the action potential (eliminate choice B). While GBP does effectively decrease the influx of Ca2+ into the pre-synaptic terminal, its most direct action is decreasing expression of Ca2+ channels on the pre-synaptic membrane, thereby decreasing the capacity of the membrane to facilitate Ca2+ influx (eliminate choice D).

> inhibitor binds to the receptor > pre-synaptic terminal (Ca+influx ↓) > glutamate release (↓)

Glucose-6-phosophate dehydrogenase (G6PDH) deficiency is an inheritable metabolic disorder that can result in the destruction of red blood cells, and in severe cases, can lead to kidney and liver failure. Which of the following best describes the reason why G6PDH deficiency can result in premature red blood cell destruction? G6PDH deficiency results in a deficiency of NADPH, which results in increased damage from reactive oxygen species and causes premature cell death. Glucose-6-phosphate dehydrogenase (G6DPH) catalyzes the first step in the pentose phosphate pathway, where NADPH is produced. NADPH plays a major role in preventing damage due to oxidative stress/reactive oxygen species. A deficiency of G6PDH prevents the production of NADPH, thus individuals with this deficiency would experience increased damage from reactive oxygen species (choice C is correct). G6PDH deficiency would not be expected to alter the intracellular osmolarity (choice A is incorrect) or to increase the intracellular pH (choice D is also incorrect). Choice B is incorrect: macrophages are not involved in the death of red blood cells in G6PDH cells.

> pentose phosphate pathway > Glucose-6-phosphate dehydrogenase > NADPH (damage decreases due to oxidative stress/reactive oxygen species)

Test subjects not adhering to instructions to limit salt intake to 6 grams per day likely demonstrated which of the following in testing data?Increased DPB due to the resulting increased osmotic pressure of the blood High salt intake will increase the salinity of the blood which will draw water into circulation by osmosis, elevating the observed blood pressure (answer choice A is correct and choice C is incorrect).

> salt (Na+) (increases) > blood pressure (increases) > the water absorbed in the cell (increases) > Osmosis

In humans, the fusion of the plasma membranes of a sperm and an ovum is followed first by which of the following? Release of the second polar body from the fertilized ovum Human ova do not complete the second meiotic cell division, including the formation of the second polar body, until after fertilization (choice A is correct). This process occurs immediately after fertilization and is a prerequisite for the first cell division of the zygote (choice C is wrong). Implantation happens a few days later, and gastrulation occurs at a stage many cell divisions after the first cell division (choices B and D are wrong).

> sperm and an ovum fusion > Release of the second polar body from the fertilized ovum > first cell division > Gastrulation > Implantation in the uterus

Excess calcitonin in the blood would most likely result in which of the following abnormalities? Abnormally dense bones Calcitonin reduces blood calcium levels through three processes: decreasing calcium reabsorption by the kidney, decreasing calcium absorption by the small intestine, and activating osteoblasts, which store calcium as bone. This last function increases bone density (choice B is correct; eliminate choice A). Osteoclasts degrade bone and are stimulated by parathyroid hormone, not calcitonin (eliminate choice C). Decreasing the collagen content in bones would also result in weaker bones (eliminate choice D), and in any case, the amount of collagen is not dependent on calcium or its regulatory hormones.

>Calcitonin increases > osteoblasts increases > calcium storage increases > Calcium reabsorption decreases in kidney and small intestine > Calcium decreases in the blood stream

In the absence of acetylcholine signaling, the overall tone of smooth muscle in artery walls will be reduced, reducing blood pressure (choice B is correct). Transcription is not related to smooth muscle contraction (eliminate choice A), and the drug acts on smooth muscle, not cardiac muscle (eliminate choice C). If the smooth muscle in blood vessels fails to contract, then blood vessels would dilate. This would increase, not decrease, renal blood flow (eliminate choice D).

>In the absence of acetylcholine signaling（X ) >the overall tone of smooth muscle in artery walls (will be reduced >blood pressure (reducing -------------------------------- >Vasodilation (less narrowing of the blood vessels ↓) >muscular wall of the vessels （less contraction ↓）, thus retaining body heat ↓ increasing vascular resistance. ↓ >renal blood flow (increase ↑) -------------------------------- > Vasoconstriction （more narrowing of the blood vessels ↑） > muscular wall of the vessels （more contraction ↑）, > thus retaining body heat （↑） > increasing vascular resistance. （↑） > flow of blood（decreased ↓） ================ narrowing, contraction, body heat, resistance and blood flow

Aldosterone binds to aldosterone receptors to regulate transcription of a specific set of genes. The enzyme that synthesizes mRNA is RNA polymerase II, so it is this enzyme that would be most directly affected by the activated aldosterone receptor (choice B is correct). DNA polymerase is used in replication, not transcription (choice A is incorrect), and while aldosterone may ultimately affect the activity of the Na+/K+ ATPase and renin, these would be indirect effects (choices C and D are incorrect).

>RNA polymerase II (regulate transcription of a specific set of genes) >DNA polymerase (used in replication) > aldosterone may ultimately affect the activity of the Na+/K+ ATPase and renin (indirect effects)

A student finishes an experiment involving several bacteria which are highly pathologic in humans. She wishes to dispose of the agar plates and micropipette tips she used. Which of the following procedures should she carry out? C. Place all materials in an open metal container and autoclave the container.

The most effective technique for sterilizing used laboratory materials is using an autoclave and should be the standard procedure followed here. An autoclave brings the materials to a temperature over 120ºC and a pressure over 2 atm, which is enough to kill almost anything.

Acetylcholine binding to AChR directly leads to which of the following effects? Influx of Na+

The nicotinic AChR in NMJ is a ligand-gated cation channel on the sarcolemma (choice B can be eliminated). Therefore, when acetylcholines binds to it, there is an influx of cations, especially Na+ (choice A is correct), which causes depolarization of the muscle fiber (choices C and D can be eliminated).

What is the FRC for an individual with the following lung volumes: total lung capacity (TLC) = 5700 mL, vital capacity (VC) = 4700 mL, inspiratory reserve volume (IRV) = 3000mL and tidal volume (VT) = 500 mL? A. 3500 mL B. 1000 mL C. 2200 mL Correct Answer (Blank) D. 1200 mL

The passage states that FRC = ERV + RV. TLC = VC + RV rightwards arrow RV= TLC - VC = 5700 mL - 4700 mL = 1000 mL VC = VT + IRV + ERV rightwards arrow ERV = VC - VT - IRV = 4700 mL - 500 mL - 3000 mL = 1200 mL FRC = ERV + RV = 1200 mL + 1000 ml = 2200 mL Thus choice C is correct and choices A, B and D are eliminated.

What type of cell is the origin of an acoustic neuroma? Schwann cell

The passage states that an acoustic neuroma is a cancer of the myelin-producing cells of the vestibulocochlear nerve, a cranial nerve. Cranial nerves are part of the peripheral nervous system, so they are myelinated by Schwann cells (choice D is correct) rather than oligodendrocytes (eliminate choice C). Microglia have an immune role in the CNS (eliminate choice A), while astrocytes carry out various maintenance and regulatory activities in the CNS (eliminate choice B). ============== myelin-producing = Schwann Microglia = immune astrocytes = nutrients

A person who suffers partial destruction of the lateral lemniscus on the left side would be expected to suffer which of the following? Bilateral hearing loss

The passage states that the first decussation of the auditory pathway occurs at the connection between the cochlear nucleus and the superior olivary nucleus. Thus, any lesion that occurs past that point will result in binaural, rather than monaural, hearing loss (choice A is incorrect). By the same token, total sensorineural deafness would require bilateral destruction of the hearing pathway (choice C is incorrect). Auditory processing and association occur in the higher brain, after processing by the primary auditory cortex (choice D is incorrect). Bilateral hearing loss (choice B) is correct; a lesion on one side of the brain will result in partial but not complete hearing loss, since auditory information ascends bilaterally. ============================== partial destruction of the lateral lemniscus on the left side = Bilateral hearing loss (but not completely)

In which cellular organelle are you most likely to find an abnormal accumulation of deposited glycogen in patients affected by Pompe disease? Lysosomes

The passage states that the organelle lacking sufficient functional acid maltase and in which abnormal glycogen accumulates contains an acidic environment. Of the four choices, lysosomes have the most acidic environment making choice A true and the correct answer (eliminate choices B, C and D).

Which statement best explains why the lung volume in Figure 1 never drops below 15 ml/kg? A. Since PV is constant, pressures below that value will cause ruptures in the lung tissue. B. At this volume, maximum contraction of the diaphragm has forced out as much air as possible from the lungs. C. The residual volume is such that the lungs are at their minimum volume under maximum intrapleural pressure. D. Below this volume, spontaneous lung collapse will occur, and the body has evolved safeguards.

The residual volume is such that the lungs are at their minimum volume under maximum intrapleural pressure. D: This is true because if air was forcibly removed from the lungs, below a certain point, the lungs would collapse under intrapleural pressure. However, this will not occur spontaneously from exhaling with too much force. That leaves only C, the sole working statement. Pressure and volume are inversely related, so under maximum pressure, the lungs will reach a minimum volume. This volume of air remaining in the lungs is termed the residual volume.

During oxidative stresses, HbS can cause the affected red blood cells to sickle. The spleen is very vascular and is responsible for removing old or damaged red blood cells. Recurrent sickling can cause blockage of blood flow in the spleen, resulting ultimately in death of splenic cells. Which of the following would result from destruction of the spleen?mpaired ability to fight off encapsulated bacteria

The spleen is one of the major organs of the immune system. It is rich with lymphocytes, both B and T, and serves a major resource in the body's natural defenses against pathogens, especially encapsulated bacteria (choice B is correct).

In miRNA-directed gene silencing, a small RNA binds to an mRNA and directs degradation of the mRNA or prevents translation of the mRNA. Which of the following terms describes the process through which binding occurs?

This is a question of definition; only hybridization describes a process of binding through complementary nucleotides.

Which of the following is most likely NOT a symptom of acylcarnitine translocase deficiency? Hyperglycemia

This question asks us to determine the likely effects of a malfunction of acylcarnitine translocase. This enzyme is essential for the catabolism of fatty acids to occur. This means that without it, there will be a great abundance of fatty acids in the body and less energy available. It will also mean that glucose will be relied upon much more heavily as a source of ATP. This implies that there will be little glucose in the blood, and that hyperglycemia will not be a symptom of acylcarnitine translocase deficiency. =============================== Hyperglycemia = high glucose

All of the following are true of eukaryotic mRNA EXCEPT that: it is polycistronic

Transcription always occurs in the 5' to 3' direction, and eukaryotic mRNA is usually capped before translation (choices A and B are true and therefore eliminated). Eukaryotic mRNA is smaller than the original RNA transcript due to the fact that the introns must be removed and the exons spliced together (choice C is true and eliminated). However, eukaryotic mRNA is monocistronic, meaning that only a single type of protein can be translated from a given mRNA (choice D is incorrect and therefore the correct answer). Polycistronic mRNA is a feature of prokaryotes, which must be more efficient with the limited amount of DNA they have to work with and can translate several different proteins from the same mRNA. ========================================= eukaryotic mRNA = monocistronic (a single type of protein can be translated ) Polycistronic mRNA = Polycistronic (translate several different proteins from the same mRNA)

According to the passage, which of the following changes would be LEAST likely to occur after treatment of tumor cells with Intercalator1? An increase in methylation of CpG sites Since this is a "LEAST" question, the correct answer is the choice that is a false statement. According to the passage, Intercalator1 is an intercalating agent that is a target for an anti-cancer drug. It prevents the transcription and translation of DNMT1 (choice B is true and can be eliminated), which would decrease methylation of CpG sites (choice C is a false statement, and is thus the correct choice).

anti-cancer drug = prevents the transcription and translation = decrease methylation ===================== methylation = gene replication

HIV, the virus that causes AIDS, is a retrovirus. Retroviruses have an RNA genome that, after entry into the host cell, is reverse transcribed to DNA, then incorporated into the host genome. RNA copies of the viral genome are produced using the normal host machinery, then packaged into viral capsid proteins for release. Which one of the following is most likely to contribute to the development of drug-resistant HIV? Frequent random errors in transcription by host-cell enyzmes Drug resistance develops as mutant versions of the virus are produced and begin infecting new host cells. The change must be able to be passed on to viral progeny. If the viral proteins fold differently in the presence of drug than in its absence, this might account for the mechanism of action of the drug, but does not explain how drug resistance develops, nor could this be a heritable change (choice C is wrong). Changes in the tertiary structure of the viral RNA by the drug is another temporary, non-heritable change (choice D is wrong). Choices A and B are both plausible mechanisms for creating new, heritable versions of the virus. However, mutation of the virus after insertion into the host-cell genome would require that an error in DNA replication be made—a rather unlikely event, given the proofreading ability of DNA polymerase. Viral progeny are produced through transcription (creating a new RNA viral genome off the permanent DNA version inserted into the host-cell genome), and the question states that this occurs using the normal host machinery. Since the normal host RNA polymerases have no proofreading function, it is more likely that errors will occur here, leading to mutant virus (choice B is better than choice A).

gene mutation is the strongest

Q 80

gluconeolysis = happens first from storage gluconeogenesis = happens second from non-glucose resources

Q 47

higher binding affinity/ more stable = lower Kd/ km higher Kd = lower binding affinity = less stable

The blood vessel that carries the most highly oxygenated blood in the fetus is the: umbilical vein The most highly oxygenated blood would be the blood returning from the point of oxygenation to the systemic circulation. In the adult this would be the pulmonary vein, but the point of oxygenation in the fetus is the placenta, not the lungs (choice D is correct and choice A is wrong). The ductus arteriosus does carry oxygenated blood but is found after the umbilical vein, so it does not carry the most highly oxygenated blood (choice B is wrong). The umbilical artery carries deoxygenated blood from the fetus to the placenta (choice C is wrong).

highly oxygenated fetus = highly oxygenated blood = umbilical vein

Which of the following describes the most likely mechanism by which gemcitabine inhibits DNA synthesis? After addition to the growing DNA strand, gemcitabine's atypical structure inhibits DNA polymerase's recognition of the growing strand. According to the passage, gemcitabine is a nucleoside analog, with 2 fluorine atoms at the 2' position. Because the 5' end of gemcitabine is unaltered, it can be added onto the growing DNA strand. The presence of the abnormal fluorine atoms, however, will prevent DNA polymerase from adding additional nucleotides (choice B is correct). Note: After gemcitabine is added to the growing DNA strand, one additional nucleotide is added but not more than one. This is not a fact needed to answer the question though. While fluorine can act as a leaving group, if the 3' OH were to attack the 2' carbon it would form a phosphoester bond as opposed to a phosphodiester bond (choice A is incorrect). It is the 3' OH on the growing DNA strand that nucleophilically attacks the 5' triphosphate of the incoming nucleotide. The lack of 2' OH, which is not present in DNA, would not affect replication (choice C is wrong). Cisplatin, not gemcitabine forms cross-bridges between the DNA strands (choice D is wrong).

inhibits DNA synthesis = inhibits DNA polymerase's recognition

A rapid mechanism is thought to govern the localization of AQP5 in response to changes in extracellular osmolarity. If this mechanism is independent of both PKA activity and S156 phosphorylation, which of the following will most likely be observed? B. HEK cells exposed to the most hypotonic conditions will display the greatest degree of AQP5 membrane localization, allowing water to flow into the cells.

is correct. According to the passage, aquaporins "transport water in response to osmotic gradients." An osmotic gradient is simply a concentration difference; in this case, it is the difference in solute concentration between the cell interior and its external environment. A very hypotonic, or dilute, environment will create a greater osmotic gradient than an isotonic environment, which would be similar in concentration to the cell. Thus, we can infer that aquaporins, including AQP5, will localize to the plasma membrane to a greater extent in hypotonic than in isotonic conditions. (Note that no choices even included the possibility of hypertonic conditions.) Water tends to flow down its concentration gradient, from areas of low solute (and high water) concentration to areas of high solute (and low water) concentration. Here, then, water should flow from the hypotonic exterior of each cell into the more solute-rich interior, as depicted below. Note that even if you do not know exactly what is happening here, you can eliminate choices that violate basic science, leaving only choice B. =============================== hypotonic = water to flow into the cells = lyse/ burst hypertonic = shrink

Protostomes undergo spiral and determinant morula cleavage, where the two first divisions of the zygote are equal and the others are unequal. This results in some small and some large cells early on in morula formation, and means cell fate is decided early in the process. Deuterostomes develop differently. They perform radial cleavage, which results in totipotent cells that are all the same size. Which of the following is supported by this information? Humans are deuterostomes. Based on information in the question, the protostome morula is more differentiated (or specialized) than the deuterostome morula (choices B and D can be eliminated). This means the deuterostome morula cells are more stem cell like (choice C can be eliminated). Cells in the human morula are all the same and are totipotent (choice A is correct).

more differentiated = more specialized cells

In transposition of the great arteries, a congenital birth defect, the aorta of the newborn is connected to the right ventricle and the pulmonary artery is attached to the left ventricle. This would result after birth in circulation of: deoxygenated blood through the systemic vasculature and oxygenated blood through the pulmonary vasculature. In the normal circulation, the right ventricle pumps blood through the pulmonary artery to the lungs. Blood returns from the lungs to the left atrium and is pumped by the left ventricle through the aorta to the rest of the body. If blood from the left ventricle passes to the pulmonary artery instead of the aorta and is then returned to the left atrium, oxygenated blood will loop around through the pulmonary system without passing to the rest of the body. At the same time, the deoxygenated blood from the systemic circulation will pass to the right atrium, then back out to the systemic circulation through the aorta rather than the pulmonary artery, causing deoxygenated blood to loop through this system (choice B is correct and choice A is wrong). Choices C and D are wrong since they have all blood either oxygenated or deoxygenated.

pulmonary artery = deoxygenated = right ventricle = pulmonary circulation aorta = oxygenated = left ventricle = systemic circulation

The melting temperature of the 15-mer oligonucleotide 5'-GACTGCGGTTTTAAC-3' is 42 °C. Which of the following oligonucleotides would be found primarily in a random coil state at 42 °C? The melting temperature is the temperature at which half of the DNA strands are in the random coil or single-stranded state. As the given oligonucleotide has a melting temperature of 42°C, then the correct answer must be an oligonucleotide with a melting temperature with a lower melting temperature in order to be primarily in the random coil state at 42°C. In order to have a lower melting temperature, the oligonucleotide must contain fewer GC nucleotides than the given oligonucleotide (Choice C is correct; Choices A, B, and D are wrong). This is because GC base pairing involves three hydrogen bonds while AT only involves two. 5'-GACTGTAATGGTAAC-3'

random coil state at melting point = less GC content

Which of the following would be the most effective method to determine if transcription of a gene had been silenced?RT-PCR RT-PCR uses reverse transcriptase to make cDNA from mRNA. The cDNA can then amplified, and then tested for. This is the only method listed that can be used to test for the presence of mRNA and determine if transcription of the gene is occurring or not (choice C is correct). Both DNA Fingerprinting and DNA Sequencing are methods used to analyze DNA sequences; this does not provide information about transcription (choices A and D can be eliminated). While ELISA could be used to test for the presence of a protein product, it would not be the best test to use in this situation (choice B is incorrect). The absence of a protein could be due to a variety of reasons, including problems with translation, protein degradation, etc. Thus, RT-PCR would be a better method to specifically test for gene expression (i.e., transcription).

reverse transcriptase = gene expression

All of the following contribute to speciation EXCEPT: A. geographic isolation of populations. B. genetic diversity. C. natural selection. D. maintenance of Hardy-Weinberg equilibrium. Speciation requires two populations to become reproductively isolated so that they can no longer interbreed. Geographic isolation is the easiest way to allow two populations to diverge and become reproductively isolated (choice A is true and eliminated). To become reproductively isolated, the populations must diverge genetically, evolving differently, and this would require genetic diversity in the population to begin (choice B is true and eliminated). Natural selection could drive the changes in isolated populations that cause them to evolve into two species (choice C is true and eliminated). If Hardy-Weinberg is maintained, then there can be no changes in the allele frequencies in the gene pool of a population and no evolution, which would not allow speciation (choice D is false and the correct answer choice here).

speciation = evolution = no equilibirum

Since fluoroquinolones diffuse into cells with water, they are able to enter all cells, eukaryote and prokaryote alike; they diffuse through special water channel proteins called porins (eliminate choice A). Both eukaryotes and prokaryotes utilize DNA polymerase and helicase (eliminate choices B and D). However, only prokaryotes

use DNA gyrase to supercoil their DNA; eukaryotes wind DNA around histones through the action of other topoisomerases (choice C is correct). ======================= gyrase = supercoil prokaryotes DNA topoisomerases = supercoil eukaryotes DNA

Spermatids are the sperm precursors that have completed meiosis but have not yet fully matured. A mature sperm has fully completed meiosis, and this is true of spermatids as well; it is the ova that are frozen in meiosis II until after fertilization (eliminate choice A). Recombination occurs in both oogenesis and spermatogenesis during meiotic prophase I, which the spermatid already completed (choice B is correct). The spermatid has passed through two reductive divisions in meiosis to end up with only one copy of the genome (eliminate choice C). The cell does have a nucleus. In fact, the sperm has virtually no cytoplasm, but the genome is still packaged into a nucleus (eliminate choice D).

》Spermatids （have completed meiosis but have not yet fully matured） 》sperm （fully completed meiosis）

chloride moves inward from the interstitium (eliminate choice A). The more CO2, the more bicarbonate which is formed and transported into the interstitium in exchange for chloride (choice B is correct). If more bicarbonate is produced and excreted into the interstitium, more chloride will enter the cell and be secreted into the lumen (eliminate choice C). The more CO2 in the cell, the more H+ will be produced and driven into the lumen (eliminate choice D).

》anti-transport 》CO2 （more） 》bicarbonate HCO3-（more into cell/lumen） 》H+ interstitium （more into cell/lumen） 》chloride （more from interstitium to cell/lumen） 全部increases into cell， 除了cl into interstitium （decreases in the lumen）

Stimulation of the iris dilator muscle is a result of activation of: A. sympathetic motor neurons.

Dilation of the pupils is a classic fight-or-flight response. The fight-or-flight response is part of the sympathetic nervous system. ===================================== fight-or-flight = sympathetic motor neurons Parasympathetic activation causes the pupils to contract, not dilate = normal state

Fructose-6-phosphate is shuttled from glycolysis to the pentose phosphate pathway to support both nucleotide and fatty acid biosynthesis.

F Glucose-6-phosphate (not fructose-6-phosphate) is shuttled from glycolysis to the pentose phosphate pathway (choice A is wrong) ======================= (reactant) Glucose-6-phosphate = pentose phosphate pathway

A bacteriologist initiated an E. coli culture from one E. coli cell and hypothesized that some of the progeny in the culture would be genetically different from the original parent cell. Is this hypothesis true?

F No; bacteria are asexual organisms, and in the absence of mutation, all progeny of any one cell are genetically identical to the parent cell. Correct Answer (Blank) Bacteria reproduce asexually, by one cell replicating its genome and then splitting into two cells that are genetically identical to the original cell. The only potential sources of genetic variation in bacteria are mutation and the transfer of genetic information through conjugation, transduction, or transformation, none of which are linked to reproduction. In the absence of mutation, all progeny of a cell will be identical to the original cell (choice C is correct).

High glycolysis rates can support the pentose phosphate pathway, which generates ribose-2-phosphate for nucleotide catabolism.

F The pentose phosphate pathway generates ribose-5-phosphate (not ribose-2-phosphate) and this allows nucleotide anabolism (not catabolism, choice C is wrong). ====================== (product) ribose-5-phosphate = pentose phosphate pathway

A FISH probe should be made of a segment of double-stranded DNA which is complementary to the gene being studied, covalently linked to a fluorescent protein.

F This is similar to how probes are used in Southern and Northern blots. The probe must be single-stranded DNA or RNA if it is going to hybridize or bind to the chromosomes (which are denatured first, also similar to the process in Southern or Northern blotting), and should be covalently linked to the fluorescent molecule (choice C is correct) ====================== probe = must be single-stranded DNA or RNA Southern = DNA Northern = RNA

Which of the following conditions leads to a reduction in water reabsorption by the kidney? Decreased permeability of the collecting duct to water

Factors that reduce water resorption would tend to increase urinary output, decrease blood volume, and decrease blood pressure. High levels of vasopressin (ADH) would increase water resorption and plasma volume (choice A is wrong). The greater the osmolarity of the interstitial fluid in the medulla of the kidney, the greater the water resorption as filtrate passes into the medulla in the descending loop of Henle (choice B is wrong). As filtrate passes through the collecting duct, water can be reabsorbed and the urine concentrated. If the collecting duct is impermeable to water, however, then less water will be reabsorbed, and the filtrate will remain dilute (choice C is correct). ACTH is the hormone that controls secretion of aldosterone (as well as cortisol) by the adrenal cortex. An ACTH-secreting tumor will cause elevated secretion of aldosterone, increased Na+ reabsorption, and a subsequent increase in water reabsorption (choice D is wrong). ============================= ACTH = aldosterone = water resorption medulla = water resorption collecting duct = water resorption

Which of the following best represents O 2 saturation vs. O 2 concentration for fetal hemoglobin (HbF) and maternal (or adult) hemoglobin (HbA)? Fetal hemoglobin must have a higher oxygen affinity than maternal hemoglobin to be able to bind oxygen at lower partial pressures than maternal hemoglobin (choice A is correct and choice B is wrong).

Fetal hemoglobin = higher oxygen affinity = shift to the left

Vinca alkaloids are a class of anti-cancer drugs derived from the periwinkle plant. Once absorbed into a cell, they interfere with the polymerization of microtubules. These drugs can prevent cancer from spreading by disrupting metaphase, thereby halting tumor growth.

Formation of microtubules is crucial for formation of the metaphase plate. Microtubules polymerize from the centrioles outward. They contact centromeres (to become kinetochore fibers) and "push" the chromosomes towards the center of the cell to form the metaphase plate. Without proper microtubule polymerization, metaphase (and the rest of mitosis) cannot occur. If mitosis cannot occur, then neither can tumor growth. Note that microtubules are also required during prophase in forming the mitotic spindle; however, some parts of prophase can still occur in the absence of microtubule formation (DNA condensation, loss of the nuclear membrane). This makes choice D a better choice than choice B, since virtually all of metaphase depends on proper microtubule polymerization. Pseudopod formation requires the growth of microfilaments (actin fibers), not microtubules (choice A is wrong), and transcription (RNA polymerization) does not require microtubules at all (choice C is wrong).

A physician treating a patient with Von Gierke's disease notes elevated lipid, lactic acid and uric acid levels in the patient's blood. Which of the following additional findings regarding the patient is the physician most likely to also note? Decreased blood glucose concentrations

Glucose-6-phosphatase, the enzyme that the passage states is deficient in Von Gierke's disease patients, hydrolyzes glucose-6-phosphate, resulting in the release of a phosphate group and free glucose for export from the cell. When this enzyme is absent, glucose-6-phosphate, a product of the breakdown of glycogen in the liver, will remain trapped within the liver and will not pass into circulation. As a result, Von Gierke's patients are at an increased risk of hypoglycemia and require regular administration of exogenous glucose to meet metabolic demands. This is consistent with choice D which is the correct answer. ===================================== Glucose-6-phosphatase = hydrolyzes glucose-6-phosphate = glucose increases

Like α(1→4)-glycosidase, which of the following enzymes acts to cleave α(1→4) linkages between glucose molecules? Glycogen phosphorylase

Glycogen phosphorylase catalyzes the rate limiting step of glycogenolysis—the release of glucose-1-phosphate from glycogen via its action on terminal α(1→4) linkages. This makes choice A true and the correct answer. Glucose-6-phosphatase hydrolyzes glucose-6-phosphate, resulting in the release of a phosphate group and free glucose for export from the cell. Choice B is false and an incorrect answer. Phosphoglucomutase is involved in both glycogenolysis and glycogenesis by interconverting glucose 1-phosphate and glucose 6-phosphate, thus choice C is false and the incorrect answer. Pyrophosphatases are members of the acid hydrolase family of enzymes that cleave diphosphate bonds. Choice D is false and an incorrect answer. ============================= Glycogen phosphorylase = glycogenolysis = release of glucose-1-phosphate from glycogen

Overdosing on acetylsalicylic acid can cause a metabolic acidosis as well as hyperventilation. Which of the following is true? Little compensation is needed to correct the acid-base disturbance

Hyperventilation causes a decrease in CO2 concentration. Remembering that CO2 is acid (see Equation 1 and the formation of H+), hyperventilation creates a respiratory alkalosis. The combination of both respiratory alkalosis (which results in a higher blood pH) and metabolic acidosis (which results in a lower blood pH) would result in a near normal blood pH. As a consequence, there will be little compensation required to maintain a normal level pH in the blood (choice B is correct).

uestion 57 Which of the following do NOT have proteins with a nuclear localization signal? I. E. coli II. Homo sapiens III. Fungi IV. Archaea

I and IV only

In another experiment, a mutase was used to transform the glycolytic intermediate 3-phosphoglycerate into 1-phosphoglycerate, which was immediately removed from the pathway and not able to become a substrate in subsequent steps. One would expect to find levels of ATP produced under these conditions to be: identical to levels produced by the arsenate-poisoned system

If 3-phosphoglycerate is converted into 1-phosphoglycerate, then no energy will be extracted as ATP later in the pathway by pyruvate kinase. Thus, the net ATP production per glucose will be 0 ATP. Per glucose, the ATP produced at each step is: -1 Hexokinase -1 Phosphofructokinase +2 Phosphoglycerate kinase 0 Pyruvate kinase (normally +2) This is the same amount of ATP as in the presence of arsenate (choice C is correct and choice B is wrong).

An ectopic TSH-secreting tumor is suspected in a hyperthyroidic (high thyroid-hormone concentration) patient with goiter. Administration of a high dose of thyroid hormone confirms this, due to the fact that: [TSH] in the systemic circulation remains high.

In a normal individual, high levels of thyroid hormone would suppress the release of TRH by the hypothalamus and TSH by the anterior pituitary. Injection of even more thyroid hormone should suppress the release of TRH and TSH even further. If this does not occur, and TRH and/or TSH levels remain high, it is likely that something outside the normal axis is involved. The passage states that ectopic tumors do not respond to normal feedback inhibition pathways; so, if an ectopic TSH-secreting tumor were present, the injection of thyroid hormone would not reduce the level of TSH in the system (choice D is correct and choice A is wrong). Since the tumor only secretes TSH, TRH levels should respond normally to the additional thyroid hormone—that is, they should go down due to negative feedback (choices B and C are wrong). ===================================== normal: injection X = decreases X sick: injection X = no decreases X

Which of the following events takes place during Phase 0? An influx of extracellular Na+

In both cardiac and skeletal muscle, as in neurons, the initial depolarization (Phase 0) is caused by the opening of voltage-gated sodium channels. When the channels open, sodium flows into the cell down the gradient maintained as part of the resting membrane potential (choice D is correct). Calcium and potassium flux do not occur in this period, and voltage-gated channels are opening, not closing (choices A, B, and C are wrong). ====================== influx of extracellular Na+ = depolarization = both cardiac and skeletal muscl (Calcium and potassium flux do not occur in this period)

Which one of the following mutations would be most likely to convert a proto-oncogene into an oncogene? Missense mutation

In order for the mutation to have the described effect, it must modify the protein without completely eliminating it or destroying its effect. A missense mutation converts a codon for one amino acid into a codon for a new amino acid, resulting in a small change within the protein's primary sequence, and an alteration (but usually not a total elimination) of the protein's function (choice B is correct). A silent mutation converts a codon for an amino acid into a new codon for the same amino acid and has no effect on the protein product (eliminate choice A). A nonsense mutation creates a stop codon out of an amino acid codon, resulting in truncation of the protein and (usually) a loss of its function (eliminate choice C). A deletion mutation eliminates one or more base pairs, altering the reading frame and drastically changing the amino acid sequence of the protein (eliminate choice D).

Which of the following best describes the cascade that leads to cellular apoptosis after exposure to intra- or extracellular death signals? Initiator caspases cluster together and activate each other. The activation of initiator caspases leads to the activation of effector caspases, which cleave proteins at aspartic acid sites, triggering apoptosis.

Initiator caspases cluster together and initiate apoptosis (choice C is correct). ============== proto-oncogene = A normal cellular gene that has the potential to become an oncogene. oncogene = a gene that in certain circumstances can transform a cell into a tumor cell. Tumor suppressor gene = ntioncogene, is a gene that protects a cell from one step on the path to cancer

Which of the following findings are likely to be seen on the ABR of a patient with bilateral acoustic neuromas? Decreased wave I amplitude Increased distance between waves I and III Increased distance between waves III and V

Item I is correct: In the legend for Figure 1, the text states that amplitude on ABR is correlated with number of neurons firing, while latency is correlated with speed of transmission. The passage states that acoustic neuromas decrease both number of functional neurons and speed of action potential propagation. Thus, bilateral acoustic neuromas will result in decreased wave I amplitude due to decreased numbers of functional neurons (eliminate choices B and D). Item II is correct: Since speed of action potential propagation within the vestibulocochlear nerve is slowed, measuring the time it takes signals to propagate through the vestibulocochlear nerve to another point in the brain will reveal a longer latency period (choice A is eliminated and choice C is correct). Item III is incorrect: Once a signal is within the brain, there is no reason to believe it will propagate any slower than it would in a normal individual. ==================== signal is within the brain = same speed as normal

Thirteen amino acids, including methionine, valine and proline, are glucogenic in humans. This means their α-keto acid carbon skeleton is converted to pyruvate during amino acid catabolism. After deamination, valine can therefore: Be converted into CO2 and H2O to generate ATP. Generate at least three NADH and two FADH2. Enter gluconeogenesis to generate glucose. Be converted into CO2 and H2O to generate ATP. Generate at least three NADH and two FADH2. Enter gluconeogenesis to generate glucose.

Item I is true: Amino acids are catabolized via deamination into α-keto acids and ammonia. Based on the information in the question stem, the α-keto acid formed from valine will be converted to pyruvate. Pyruvate can keep going through cellular respiration to generate CO2, H2O and ATP. Eliminate choice B. Item II is false: Pyruvate is converted into one acetyl-CoA (during which 1 NADH is made), and the acetyl-CoA would then generate three NADH and only one FADH2 as it cycles through the Krebs cycle. Eliminate choice C. Item III is true: Pyruvate can also enter gluconeogenesis to generate glucose. Eliminate choice A and choice D is correct. ====================== catabolized Amino acids = deamination into α-keto acids and ammonia Pyruvate = gluconeogenesis = generate glucose Pyruvate = glycolysis/ Kreb cycle = generate ATP/CO2/H2O

Human T-cell lymphotropic virus (HTLV) is very similar to HIV. Which of the following would be true regarding HTLV? HTLV exhibits a lysogenic life cycle. HTLV is a retrovirus. HTLV destroys the antibody-producing cells. I and II only

Item I is true: HIV exhibits a lysogenic life cycle since it integrates into the host's genome. If HTLV is similar to HIV, then it must also be lysogenic (choice C can be eliminated). Item II is true: If HTLV is similar to HIV, then it must also be a retrovirus (choice B can be eliminated). Item III is false: HIV and HTLV infect the host's T-cells. The B-cells are the antibody-producing cells of the body. There is no discussion of HIV infecting B cells (choice D can be eliminated and choice A is correct).

Which of the following developmental processes occur among annelids? The blastopore develops into the mouth Spiral cleavage Indeterminate cleavage The blastopore develops into the mouth Spiral cleavage

Item I is true: annelids are protostomes, in which the cells near the blastopore develop into the mouth (eliminate choice C). Item II is true: they undergo spiral cleavage (eliminate choice B). Item III is false: cleavage is determinate (choice A is correct; eliminate choice D). ======================== blastopore = develops mouth

In gastroesophageal reflux disease (GERD), gastric secretions exit the stomach via the lower esophageal sphincter and come in contact with the lower part of the esophagus. How would this affect the esophageal epithelium? Increased proliferation of the stratified squamous epithelium Conversion to columnar epithelium similar to that in the stomach Introduction of bicarbonate-secreting cells

Item I is true: in the short term, gastric acid would corrode the outermost layer of epithelium in the lower esophagus, increasing the production of cells (choice B can be eliminated). Item II is true: over time, the acid will destroy the epithelium and expose the submucosa. Columnar epithelium found in the stomach will gradually encroach on this area and become the dominant form of epithelium, since it is not sensitive to acid (choice A can be eliminated). Item III is false: There are specialized areas known as Brunner's glands in the proximal duodenum that produce bicarbonate to neutralize the acidic chyme as it passes from the stomach to the duodenum. However, there is no reason to assume that these cells would traverse the stomach to the lower esophagus (choice D can be eliminated and choice C is correct).

Bilateral occlusion of the middle cerebral artery, which provides blood to the lateral temporal and parietal lobes, would be expected to cause an abnormality in which of the following components of a patient's ABR? Decreased wave III amplitude Decreased wave V amplitude Decreased wave VII amplitude

Item III is correct: The question describes a stroke which would obliterate cortical processing of auditory information in the auditory cortex and cause central deafness. Wave VII corresponds to the primary auditory cortex; since a stroke results in neuron death, it will show decreased amplitude (choices A and B can be eliminated). Items I and II are not correct: The rest of the pathway is not affected by the stroke, so amplitude of these areas should be normal (choice C is eliminated and choice D is correct). =================== neuron death = decreased amplitude

Increased levels of which of the following foods would likely be recommended for individuals on a Standard diet, but NOT for individuals on a Paleo diet? Bananas and squashes Whole grain breads Cheese and yogurt Whole grain breads

Item III is false. Cheese, being high in fat, would unlikely be part of a Standard diet (choice D can be eliminated and choice A is the correct answer choice).

Muscle fibers are composed of small contractile units called sarcomeres. During contraction, which of the following occurs within a sarcomere? Myosin filaments shorten. Actin filaments shorten. Overlap between actin and myosin filaments increases.

Items I and II are false: During contraction, neither myosin nor actin filaments get shorter. The overlap between them increases to make the sarcomere shorter (eliminate choices A, C and D). Item III is true: The overlap between the fibers increases as part of contraction. ======================= neither myosin nor actin filaments get shorter = overlap between them increases to make the sarcomere shorter

If necessary to design a new experiment, which of the following best explains why researchers could use measurements of intracellular lactate levels (ILL) in cancer cells to assess efficacy of cancer drugs? Low ILL would indicate that glycolysis is significantly inhibited.

Lactate is a product of fermentation, and the first paragraph states that cancerous cells rely heavily on fermentation for energy production. Inhibiting glycolysis or fermentation would reduce ILL and indicate that the metabolism of the cancer cell is being effectively inhibited by the drug. lactate is part of the glycolysis

After recovering from a severe illness caused by O157:H7, a person who could previously eat dairy products without difficulty has become lactose intolerant. Which of the following is the most likely cause of this? The disruption of the membrane during infection has also impacted the brush border enzymes.

Lactose is a found in dairy products. It is a disaccharide consisting of glucose and galactose. Lactose is hydrolyzed by lactase, which is a brush border enzyme. The passage states that the infection by O157:H7 damages the epithelium; by doing so the brush border and the brush border enzymes are also damaged (choice A is correct). ====================== Lactose = sugar destroy = border enzyme

Decreased number of alveoli in the lungs leads to respiratory distress because: reduced surface area in the lungs reduces the rate at which O2 and CO2 can diffuse through the lung epithelium.

Many biological processes depend on a very high surface area-to-volume ratio. The large number of tiny alveolar sacs in the lungs, the presence of microvilli in the small intestine, and the folding of the inner mitochondrial membrane are all examples in which an increased surface area allows for biological processes to take place more quickly. When the individual alveoli break down (as, for example, in emphysema) the lungs lose the necessary surface area to allow for effective diffusion of respiratory gases. The alveoli and their larger environment (the human lungs) are shown below.

The following characteristics are shared between skeletal and smooth muscle with the exception of: Potential exhibits a plateau

Neither skeletal nor smooth muscle contraction is initiated by an action potential with a plateau; this feature is only characteristic of cardiac action potentials (answer choice B does not indicate a similarity and is the correct answer). The upstroke in the action potential for both skeletal and smooth muscle is triggered by an inward Na+ current (answer choice A states a similarity and is not the correct answer). Once received by either skeletal or smooth muscle, the action potential does, indeed, cause voltage-gated Ca2+ channels to open; in skeletal muscle, these channels are located on the sarcoplasmic reticulum while, in smooth muscle, they are located in the cell membrane (answer choice C is wrong). Indeed, Ca2+ is the molecular substrate for triggering contraction in both skeletal and smooth muscle; binding with troponin triggers skeletal muscle contraction and calmodulin binding activates myosin light-chain kinase in smooth muscle (answer choice D is wrong). ====================== Potential exhibits a plateau = cardiac action binding with troponin = skeletal muscle contraction binding with calmodulin = smooth muscle contraction ====================== Ca2+ receptor = sarcoplasmic reticulum = skeletal Ca2+ receptor = cell membrane = smooth muscle

Which of the following describes how the brain uses parallel processing to process a visual stimulus? Visual areas V1-V5 are utilized to analyze different aspects of an image simultaneously. Parallel processing describes the simultaneously processing of different aspects of a stimulus (choice B is correct),

Parallel processing = simultaneity

Most biological unsaturated fatty acids are cis and contain non-conjugated double bonds. Because of this, additional steps are required in β-oxidation. These most likely include: Combining two double bonds via a reductase enzyme (which uses NADPH as a reducing agent), then changing the resultant cis double bond to trans via an isomerase enzyme.

Polyunsaturated fatty acids require both an isomerase and a reductase enzyme to complete β -oxidation; this also requires the reducing agent NADPH (choice D is correct). ===================== >beta-oxidation >unsaturated fatty acids = 1st reduced = then isomerase to trans (from cis) = become conjugated

PDC deficiency is a rare genetic disease. Which of the following conditions would be expected in patients with the disease?Lactic acidosis In PDC deficiency, the pyruvate cannot be converted to acetyl-CoA and therefore, the cells can only undergo glycolysis and anaerobic respiration. The end product of anaerobic respiration in humans is lactic acid. Thus, patients with PDC deficiency have lactic acidosis (choice B is correct). Glucose is still being utilized for energy so there is no reason to assume its circulating levels would increase (eliminate choice C) and PDC does not play a role in processing lipids (eliminate choice D). A lack of PDC activity would not increase the body's pH (eliminate choice A).

Pyruvate dehydrogenase complex deficiency = glycolysis and anaerobic respiration = lactic acid (increases) ===================== Pyruvate dehydrogenase complex = pyruvate converted to acetyl-CoA

An error is made during production of an RNA transcript which results in a change in a single amino acid in the final protein. This error is most likely a result of: the poor editing ability of RNA polymerase.

RNA polymerases have no editing function since RNA is a transient molecule and its information is not passed on to offspring (choice B is correct). The question states that the error is made during production of the RNA, so the change in amino acid sequence reflects the new nucleotide sequence and is not due to incorrect translation of the mRNA (choice A is wrong). Frameshift mutations are very severe and would result in all of the amino acids from the error on being changed (choice C is wrong), and DNA is not translated by RNA polymerase, it is transcribed (choice D is wrong). RNA polymerases = transcription = but not genes

Aβ peptides can negatively regulate various steps of acetylcholine (ACh) synthesis, release, and signaling. Which of the following would be the best target for pharmacological intervention? An allosteric inhibitor of acetylcholinesterase

Since the Aβ peptides negatively regulate ACh synthesis, release, and signaling, any kind of pharmacological intervention should be targeted at increasing ACh or its activity. An inhibitor of acetylcholinesterase (AChE) would prolong the life of ACh in the synaptic cleft, thus increasing its effect on the postsynaptic cell (choice C is correct).

In an experiment, various concentations of streptomycin were added to a culture of E. coli. At low concentrations of streptomycin, increased amount of misreading of the mRNA was observed. At high streptomycin concentrations, the initiation of protein translation was inhibited. Which of the following statements is consistent with these results? Streptomycin inhibits the small subunit (30S) of the ribosome.

Since the errors are observed in translation and not in transcription, it is likely the ribosome that is being affected and not RNA polymerase (choice C is wrong). The role of small subunit (30S) of ribosome is to initiate translation by recognizing the first AUG start site and recruiting the tRNAfmet and the large subunit. It is also important for proofreading and maintaining fidelity during the translation process. Therefore, inhibition of small subunit would result in decreased initiation of protein translation and mRNA misreading (choice A is wrong and choice B is correct). =============== translation = small subunit (30S)

Changes in prion proteins take place at the alpha-helix and beta-sheet levels of protein structure; this is the secondary structure of a protein.

T ================== alpha-helix and beta-sheet levels = secondary

fructose-2,6-bisphosphate, It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating phosphofructokinase and inhibiting fructose-1,6-bisphosphatase

T ======================= > gluconeogenesis > reactant ( fructose-1,6-bisphosphATE ) > product ( fructose-6-phosphate ) > enzyme ( fructose-1,6-bisphosphatase) ======================= > glycolysis > reactant (fructose-6-phosphate) > product (fructose-1,6-bisphosphATE) > enzyme (PFK = phosphofructokinase-1) > secondary enzyme (fructose-2,6-bisphosphate)

Q 10

- means deficiency + means adding

Pyruvate kinase catalyzes the last step of glycolysis (eliminate choice C), where phosphoenolpyruvate is converted to pyruvate. Because it is a kinase, pyruvate kinase transfers an inorganic phosphate from phosphoenolpyruvate to ADP to form a molecule of ATP

>kinase (transfers an inorganic phosphate from somewhere to ADP form am ATP) > pyruvate kinase (phosphate from phosphoenolpyruvate) > phosphoenolpyruvate (converted to pyruvate) > Pyruvate (catalyzes the last step of glycolysis)

The hormones secreted by the adrenal glands include epinephrine from the medulla, and cortisol, aldosterone, and low levels of sex steroids from the adrenal cortex. As stated in the passage, aldosterone increases potassium secretion and also increases sodium reabsorption (choices A and B are true and thus are eliminated). The increased Na+ reabsorption leads to increased water reabsorption, thereby increasing the blood volume (choice C is true and thus is eliminated). The loss of aldosterone would cause water and sodium loss, decreased blood volume and decreased blood pressure. In response to decreased blood pressure, renin secretion would increase, not decrease (choice D is incorrect and thus the correct answer choice).

>medulla (epinephrine) > adrenal cortex (cortisol, aldosterone, and low levels of sex steroids) >aldosterone ( increases potassium secretion and also increases sodium reabsorption) > increased Na+ reabsorption (leads to increased water reabsorption) >thereby increasing the blood volume ================================= > In response to decreased blood pressure (renin secretion would increase) 【只有这个是反比例】

The conformational change of a regulatory protein after the binding of a repressor most likely represents an alteration of the protein's: tertiary structure.

A conformational change in a protein is a change in the larger scale folding of a protein, with changes in the relative positions in space of amino acids located far from each other in the linear polypeptide chain. The tertiary structure of a protein involves large-scale structure within a polypeptide chain that is stabilized by interactions between amino acids that can be distant from each other in the linear sequence; thus, this is the most likely level of structure altered during a conformational change (choice D is correct). ========================== tertiary structure = conformational change

The researchers participating in the experiments described above were assigned to conduct new experimentation, also on Leigh syndrome-positive mouse specimens. As a precursor step, it was decided to create a new population of Leigh syndrome-positive mice. What would be the option most likely to be successful in creating this population? Induction of intentional mutations in the mitochondria of pre-fertilization ova

A is correct. The information in the second paragraph of the prompt establishes that Leigh syndrome is passed on from mother to child through cytoplasmic components. Mitochondria are the most likely culprits, as they have their own DNA and they been implicated in a number of genetic diseases that are passed on from mother to child. This narrows the choices to A, B and C. Here, (A) is correct because in a fertilized egg, the mitochondria of ova predominate due to their much larger numbers compared to the mitochondria of sperm, as well as other reasons such as sperm mitochondria's association with ubiquitin protein which causes them to be targeted for destruction post-fertilization.

In which of the following are the acrosomal enzymes synthesized? Rough endoplasmic reticulum

Acrosomal enzymes are located inside a membrane-bound vesicle and are going to be secreted. It is likely that they are processed through the secretory pathway, and to do so they must be synthesized in the rough ER (choice A is correct),

Q 117

All layers vein, arteries, capillaries = have a layer

Diabetes insipidus is a condition in which an individual fails to produce antidiuretic hormone. If a person suffering from diabetes insipidus were to consume large amounts of a sugar-containing beverage, which of the following would most likely be observed after one hour? Production of dilute urine

An individual who cannot make ADH will be unable to reabsorb water from the collecting duct of the nephron, resulting in urine that is very dilute (choice B is correct, and choice A is wrong). Since diabetes insipidus is strictly a disorder involving ADH production, no effect would be seen on plasma glucose levels (very low after one hour due to the effects of insulin) or urine glucose levels, which should always be zero in a normal individual. Do not confuse this disorder with diabetes mellitus, a disorder involving insulin in which blood and urine glucose levels are abnormally high (choices C and D are wrong). ============= ADH = antidiuretic hormone = increases water absorption

In a male individual with Down's Syndrome (trisomy 21), how many chromosomes would be visible at metaphase I of spermatogenesis? 47

An individual with trisomy 21 has an extra copy of chromosome 21 (three total copies). During metaphase I, the developing gametes are still diploid (separation of homologues has not yet occurred), so this individual would have the normal 46 chromosomes plus the extra copy of chromosome 21, for a total of 47 chromosomes (choice D is correct; eliminate choices A, B and C).

Which of the following correctly lists a pair of analogous structures and a pair of homologous structures, respectively? The wing of a bee and the wing of a bird; the arm of a human and the flipper of a walrus

Analogous structures are those structures that evolved independently to carry out the same function. Thus, the wing of a bee and the wing of a bird are analogous structures. Homologous structures are those that have a similar evolutionary history, arising from the same source, even if they now have different functions. The forelimbs of mammals (human arm, walrus flipper, bat wing) would all be homologous despite their different functions. Thus, this choice correctly indicates a pair of analogous structures and then a pair of homologous structures. ==================================== Analogous structures = same function Homologous structures = similar evolutionary history, different functions

The enzyme lysozyme degrades peptidoglycan. If lysozyme is mixed in blood agar prior to addition of Strep. viridans, which of the following would be observed? No colonies

Bacterial cell walls are made of peptidoglycan and are susceptible to degradation by lysozyme. Gram-positive strains (such as Strep. viridans) are particularly vulnerable since their cell wall is unprotected by membrane like the cell walls of Gram-negative strains. Thus the cell wall of the Strep. viridans will be damaged by the lysozyme, so that the bacteria will not survive and no colonies will form (choice D is correct). Note that the hemolysis observed around some colonies is caused by the ability of bacteria to lyse red blood cells and is not related to lysozyme activity (choices A, B, and C are wrong).

If a drug is applied to a rod cell which blocks sodium channels in the absence of light, this will result in: the hyperpolarization of the retinal rod plasma membrane.

Blocking the sodium channels in the absence of light would have the same effect on the cell as a burst of light. The membrane would hyperpolarize due to the fact that sodium can no longer enter the cell to depolarize it (choice A is correct; eliminate choice B). Sodium concentration would decrease, not increase, because the channel is blocked (eliminate choice C), and, finally, the conversion of all trans-retinal to 11-cis-retinal would occur only in the presence of light (eliminate choice D).

According to the data in Figure 1, what is the probability that a male Robertsonian translocation carrier who mates with a normal female will produce a viable offspring? 1/2

C is correct. According to Figure 1, there are 6 possible outcomes for the gamete production in a 14/21 translocation carrier.

Which of the following biomolecules is an example of a nucleoside? Adenosine

D is correct. A nucleoside is composed of a nitrogenous base and a five-carbon sugar (ribose or deoxyribose). Adenosine, shown below, is such a molecule. For adenosine, the nitrogenous base is adenine, while the five-carbon sugar is ribose (note the presence of both a 2' and a 3' hydroxyl group).

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

D is correct. 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.

What type of control does siRNA exert on G6P expression? A. Transcriptional control B. Promotion C. Repression D. Post-transcriptional control

D is correct. Due to its structure, siRNA is only able to bind to other RNA strands, not to DNA or protein. Therefore, it must interfere with gene expression after transcription has already occurred, but before translation. Specifically, it prevents the translation of mRNA corresponding to the target protein. ============================ binds to mRNA = Post-transcriptional control

Which of the following experimental findings would best confirm that a patient's type II diabetes was caused by a genetic defect and not a poor diet?Isolation of one of the patient's insulin receptors reveals a lysine-histidine substitution in the insulin-recognition site

Detection of an amino acid substitution in the insulin receptor insulin recognition site would be a strong indicator that a genetic malformation resulted in the production of an ineffective receptor (choice C is correct). ============== substitution = genetic malformation

Isolate genomic DNA from a cancer cell line and sequence the gene for pyruvate carboxylase, to confirm which amino acid is altered in the mutant form.

F The question stem states that a frameshift mutation in pyruvate carboxylase has already been found. There would therefore be little point in sequencing the gene again. In addition, "confirm[ing] which amino acid is altered in the mutant form" is a better match to confirming a point mutation rather than a frameshift mutation (choice A is wrong). ====================== pyruvate carboxylase = pyruvate oxaloacetate = gluconeogenesis = glucose increases oxaloacetate = either glucose ↓ or ↑

If the researchers had used a luminescent probe to visualize the location of FAS itself in Experiment 3, where would it predominantly be found? Cytosol

Fatty acid synthesis takes place in the cytosol and requires the presence of FAS (choice A is correct). No part of the fat metabolism takes place in the nucleus of the cell (eliminate choice B). The mitochondrial matrix is the location for beta-oxidation (fatty acid break down; eliminate choice C). Elongation and desaturation of fatty acids takes place inside the smooth ER using separate enzymes. FAS is not used during extensions of fatty acids beyond carbon-16 nor is it involved in the introduction of any double bonds (eliminate choice D). ============= Fatty acid synthesis = takes place in the cytosol

Based on Figure 1, what is the net reaction for the transformation of glucose to pyruvate?

Glucose + 2 Pi + 2 ADP + 2 NAD+Equilibrium2 pyruvate + 2 ATP + 2 NADH + 2 H+ + 2 H2O

Q 19

HIV, host mRNA = viral original gene (because it uses reverse transcriptase)

2,4-dinitrophenol (2,4-DNP) is a highly toxic substance which was sold to the public as a weight loss drug in the 1930s. It acts by permeabilizing the inner mitochondrial membrane (IMM) to ions. Which of the following is true of 2,4-DNP's effects on oxidative phosphorylation? It causes a decrease in the electromotive potential built up by the electron transport chain.

If the inner mitochondrial membrane became permeable to ions, then hydrogen ions would not need to go through the ATP synthase in order to re-enter the matrix. This would dissipate the proton gradient established by the electron transport chain, and decrease its potential to generate ATP (choice D is correct). FADH2 and NADH would still be consumed by the transport chain proteins, since the electron transport chain is not shut down; it's just that the proton gradient would be more easily dissipated (choice C is wrong). The dissipation of the gradient would result in heat production; note that this is similar to what happens in brown fat, used by hibernating animals to stay warm (choice B is wrong). Electrons do not flow through the ATP synthase (choice A is wrong). =================== permeable to ions = dissipate the proton gradient

Isoforms are 》 different forms of the same protein 》and can be due to gene duplication or alternative splicing. 》 However, the ribosome has no function in splicing (eliminate choice D).

Isoforms > different forms of the same protein > gene duplication or alternative splicing > ribosome has no function in splicing

Which of the following laboratory techniques could be used to detect prions? PCR ELISA Western blotting

Item I is false: PCR is used to amplify segments of DNA. Prions are proteins, and thus, PCR would not be an appropriate laboratory technique (choices A and C can be eliminated). Both remaining choices include Item II so it must be true: ELISA uses antibodies to test for proteins, and could be (and is) used to detect prion proteins. Item III is also true: Western blotting uses antibody probes to detect proteins (after separating them on a gel using gel electrophoresis and transferring them to a filter). Thus, western blotting can be (and is) used to detect prion proteins (choice B can be eliminated and choice D is correct).

L-DOPA is a common medication administered to treat the symptoms of Parkinson's disease. What is the likely mechanism of L-DOPA? L-DOPA increases dopamine concentration.

L-DOPA is the precursor to the neurotransmitters dopamine, norepinephrine (noradrenaline), and epinephrine (adrenaline), collectively known as the catecholamines. L-DOPA is used to increase dopamine concentrations in the treatment of Parkinson's disease. ================================ low dopamine = Parkinson's disease

Q 91

LH = testosterone FSH and testosterone = spermatogenesis

Which of the following would most likely become a major concern of mortality in MG patients? Respiratory distress

MG is a disease affecting skeletal muscle; therefore, all skeletal muscles in the body are susceptible (eliminate choice A). A very important skeletal muscle in humans is the diaphragm; in severe MG patients, death can occur due to respiratory distress or arrest. Paralysis of facial muscles and weakness of upper limbs can occur; however, they are generally not the primary cause of mortality (eliminate choices B and C; choice D is correct).

Q 6

Phosphorylation = remove a H from OH, and replace a phosphate into it

All of the following are components of plasma EXCEPT: erythrocytes

Plasma is the cell-free portion of blood. Proteins and salts are present in plasma, but cells—such as erythrocytes—are not, by definition. ==================== Plasma = portion of blood (erythrocytes)

The graph below shows the membrane potential of a rod-cell plasma membrane: Which one of the following is represented at time t = 1 sec? Hyperpolarization The resting potential of rod cells is much higher than that of normal cells due to the constant influx of sodium. Hyperpolarization is defined as the membrane potential moving away from rest potential in the negative direction, and this is what is occurring at time t = 1 sec on the graph (choice A is correct). Depolarization is movement of the membrane potential away from rest in the positive direction (eliminate choice B), and repolarization is a return to rest potential (eliminate choice C). There is no such as thing as nonpolarization (eliminate choice D).

Repolarization typically results from the movement of positively charged K+ ions out of the cell. The repolarization phase of an action potential initially results in hyperpolarization, attainment of a membrane potential, termed the afterhyperpolarization, that is more negative than the resting potential.

Retinal isomerase is used to alter the conformation of retinal. Retinal isomerase is most likely manufactured by: dehydration synthesis

Retinal isomerase, an enzyme, is a protein, and proteins are synthesized by the condensation of amino acids. Condensation involves dehydration.

Q 70

SDS-PAGE = weight not charge

One effect of acetylation in this context is that the charge of the otherwise-positive lysine residue is effectively neutralized. The aptamer designed by the scientists binds selectively to histone proteins whose surrounding DNA: is bound less tightly than the DNA of histone proteins without acetylation of the lysine residues.

Since DNA is negatively charged, it would bind more tightly to histone proteins when the lysine residues are positively charged (in other words, when the histone is NOT acetylated). When acetylated, the lysine residues are no longer charged so the normal DNA around the histone protein would be bound less tightly, and the aptamer can get access to the histone and bind to it ============ acetylated = no longer charged = less tightly

A series of cell lines were generated, each missing expression of one or more IDH isoform. When grown in the presence of 14C-D-glucose: large amounts of 14C-lactate is detectable in the growth media of IDH3-null lines but not IDH1 or IDH2-null lines.

Since IDH3 is important in the TCA cycle, cells that lack IDH3 will resort to anaerobic metabolism to survive. This will generate large amounts of lactate (choice C is correct).

Glutamate can be used to produce Krebs cycle intermediates such as oxaloacetate and citrate and these molecules can be used to generate lipids, amino acids and nucleotides.

T Paragraph 2 describes how glutamine is converted into the Krebs cycle intermediate α-ketoglutarate, which can then be used to generate other Krebs cycle intermediates to power biomolecule synthesis

If the measurements taken in Step 4 of Experiments 1 and 2 had shown no change in bacterial lactose concentration, then it is likely that within the medium:hydrogen ion concentration had remained unchanged in Experiments 1 and 2.

T These experiments show that a gradient of either potassium or protons can drive the transport of lactose into the cell against its gradient. If there is no lactose transport, it is safe to assume that the concentration in the media of the driving ion will remain the same (choice A is correct). ================ no transport = concentration driving ion will remain the same

Some oncogenic changes shut down the mitochondria because these organelles are involved in apoptosis, which would result in cell death

T if a tumor cell is relying on glycolysis and fermentation instead of mitochondrial cell respiration, it will have less use for the mitochondria in general. Since the mitochondria can initiate apoptosis, less reliance on this organelle could lead to apoptosis resistance, conferring a survival advantage to the cancerous cell

Metabolic differences allow cancer cells to adapt to hypoxic (oxygen-deficient) conditions inside solid tumors.

T tumors can grow quite quickly and often the inside of tumors don't have sufficient blood supply. This can lead to hypoxic conditions inside the tumor.

Which of the following is the most likely sequence of events in the infection of E. coli by phage T1? Production of viral DNA polymerase, replication of viral genome, assembly of infectious virus, translation of viral lysozyme

The correct events are in choice D: the expression of the viral DNA polymerase must come first to then replicate the genome. Infectious virus can then be assembled, and released by cell lysis induced by viral lysozyme.

Which of the following is the most likely origin in the afferent sensory neuron of the α2δ-1 Ca2+ channels implicated in the inhibitory mechanism of GBP on signal transmission? Dorsal root ganglion.

The dorsal root ganglion is the location of the neuron cell body where protein production occurs; thus, this is the most likely origin for channel proteins that would subsequently become embedded in the neuron cell membrane (choice C is correct). ====================== dorsal root ganglion = origin for channel proteins

What is the most likely transmission mechanism for the passage of Leigh syndrome from parent to child? Mitochondrial DNA

The inheritance pattern described in the prompt is one in which females transmit Leigh syndrome to all of their offspring. This means Leigh syndrome is a maternally-transmitted disease. Because it is also a cytoplasmically inherited disorder, it will be transmitted through cytoplasmic components, of which mitochondria are examples.

Atazanavir is an example of a protease inhibitor. It causes all the following events in cells that take it up EXCEPT: direct inhibition of the enzyme that incorporates the viral DNA into the host's DNA

The integration of viral DNA into the host's DNA is mediated by the enzyme integrase, not protease (choice C is wrong and therefore is the correct answer). The protease enzyme is used to cut larger proteins into smaller functional proteins, such as gp160 into gp41 and gp120 (choice D is a correct statement and is eliminated). Without these functional pro-teins, viral replication comes to a halt (choice A is a correct statement and is eliminated). If viral replication stops, then fewer helper T-cells will be infected (choice B is a correct statement and is eliminated). ========================== enzyme integrase Vs protease

To produce the maximum number of ATPs, post-glycolytic pathways in aerobes must receive from glycolysis: NADH and pyruvate.

The post-glycolytic pathways are the Krebs cycle and electron transport, which require oxygen, NADH, and pyruvate. Pyruvate is converted into acetyl-CoA to enter the Krebs cycle (choice D is correct).

During the blastula stage, neighboring cells are in close contact with each other. The formation of the blastula is contemporaneous with the formation of tight junctions. The function of tight junctions is to: seal off the blastocoel and unite connected cells into a tissue.

Tight junctions hold cells together and can form a seal so that fluid does not leak between cells (choice C is correct). ======================== Tight junctions = seal off

Q 81

adaptive radiation = one species into multiple species by different environment

It was observed under the microscope that a small section of bone consisted of concentric rings of fibrous material surrounding an open circular region. The bone section was identified as an osteon. The circular region is expected to contain:

blood vessels.

Q 73

catalytic efficiency = Kcat/ km smaller km, better efficiency

In addition to sister chromatid cohesion, cohesion complexes play an important role in: synapsis The passage also states that cohesion complexes play an important role in bringing together homologous chromosomes during meiosis I so that crossing over can occur; this is synapsis (choice A is correct)

cohesion = crossing over = synapsis

Q 33

competitive km increases, Vmax same noncompetitive km same, Vmax decreases uncompetitive km and Vmax both decrease

Q 113

contraction of diaphragm = more negative pressure and inspiration

Q 110

dilation decreases body heat = cooler

Fetal circulation differs from adult circulation in many ways. One of the main differences is that in the former: a pathway exists for blood to circumvent the lungs The purpose of the ductus arteriosus is to help shunt blood past the inactive fetal lungs (choice B is correct).

ductus arteriosus = blood to circumvent the lungs.

During the exponential phase of bacterial growth, bacteria are reproducing by: binary fission During the exponential phase of bacterial growth, the population is increasing in size. The only choice which allows an increase in population size is choice D, binary fission.

exponential phase = population size increased = binary fission

Q 84

glycogen phorylase = break down

Q 32

hill coefficent greater than 1 = cooperative

If a person drank a large quantity of hypersaline ocean water, the person could die because absorption of salt into the blood will cause it to become:

hypertonic compared with the cytosol of the body's cells, causing osmosis of water out of the cells.

Intercalator1 causes mutations in DNA by: inserting itself between base pairs. The passage states that Intercalator1 is an intercalating agent; this means it inserts itself between base pairs in DNA (choice A is correct).

intercalating agent = insertion

the right circulation that is typically low in oxygen will mix with the left circulation that is typically higher in oxygen, so the overall aortic arterial oxygen saturation will be

less than normal (choice B is correct) and the pulmonary arterial oxygen concentration will be higher than it is normally (choice D is wrong).

Q 29

liver does not store bile, = produces bile gall bladder stores it

When conducting a hearing test, sounds of varying pitches and loudness are created in order to determine if any aspect of hearing has been compromised. Which of the following best describes how differences in pitch are determined by the brain? Differences in pitch are determined by which region of the basilar membrane vibrates, stimulating different auditory hair cells. High pitched sounds vibrate the basilar membrane near the oval window and low pitched sounds vibrate the basilar membrane farthest from the oval window. This is a 2x2 question, where one piece of information can eliminate two answer choices. Differences in pitch are determined by the region of the basilar membrane that vibrates, and loudness is determined by amplitude of the vibration (choices A and C are wrong). High pitched sounds vibrate the region of the basilar membrane near the oval window, and low pitched sounds vibrate the basilar membrane farthest from the oval window (choice B is correct and choice D is wrong).

loudness = amplitude ======================== basilar membrane = pitch high pitched = near the oval window low pitched = farthest from the oval window

Gastrin stimulates acid production, and its release is inhibited by stomach acid. This is a

negative-feedback loop designed to maintain acid within a certain pH range (choice B is correct). ================= Gastrin = acid

The oocytes used in the research study in passage were: diploid cells arrested in prophase I. The oocytes used in the research study were taken from newborns. The cells would be in the period of arrest between the onset of oogenesis in the fetal stage and the onset of the menstrual cycle, when one cell per month will move forward with meiosis I. Since these cells have not completed meiosis I yet, they must still be diploids (choices A and D are eliminated). Primary oocytes are arrested in prophase I (choice B is correct). Choice C is incorrect: metaphase I does not occur until a primary oocyte reenters the meiotic cycle after dictyate.

newborn = completed meiosis I yet = prophase I = Primary oocytes = diploid cells

Q 118

pancreas = proteolytic enzyme

Q 86

phosphgluconate = PPP glucose 6 phosphatase dehydrogenase = make phosphgluconate

All of the following apply to mitotic prophase EXCEPT:

recombination between homologous chromosomes

The events that transpire during the fusion of the acrosomal vesicle with the sperm plasma membrane are most similar to those that occur in the: release of neurotransmitters from synaptic terminals.

the acrosomal vesicle is a membrane-bound vesicle that fuses with another membrane, the plasma membrane, to release its contents. A similar process occurs to release neurotransmitter at synapses, fusing two membranes to release the contents of a vesicle into the exterior of the cell (choice B is correct).

The vitelline layer in sea urchins is analogous to which of the following layers in the human ovum? Zona pellucida

the vitelline layer is analogous to the zona pellucida in human ova, since both are protective acellular layers located just outside the plasma membrane ======================== vitelline layer = Zona pellucida = protective acellular layers

In prokaryotes, genes can exist as operons that are transcribed into a polycistronic mRNA, containing multiple genes in a single transcript. In eukaryotes, transcripts exist only as monocistronic mRNA containing a single gene. What fundamental genetic difference is responsible for this distinction? In eukaryotes, each gene has their own transcription initiation site. Hybridization

this question is best answered through process of elimination. Choices A, B, and C are incorrect, for the reasons listed below. Choice D is scientifically true, and it properly answers the question of why eukaryotic mRNA is monocistronic (one gene per transcript), unlike prokaryotic mRNA.

Q 56

transmembrane = nonpolar

The tumor suppressor gene p53 contains a number of lysine residues that can be acetylated to stabilize the protein and facilitate its activation. All of the following could be the result of acetylation of p53 EXCEPT: progression in the cell cycle from G1 to S phase. The question stem states that acetylation of p53 enhances its activity. p53 is a tumor suppressor gene that prevents movement through the cell cycle in the event of DNA damage (thus giving time for the DNA to be repaired), or that triggers apoptosis in the even that DNA damage is too severe to be repaired. Choices A, C, and D are all consistent with the idea of halting the cell cycle, repairing DNA, or committing cell suicide. However, choice B is an enhancement of the cell cycle, not a halting (choice B is not a result of acetylation of p53 and is the correct answer choice).

tumor suppressor gene = halting the cell cycle

diponectin can bind one of three receptors (ADIPOR1, ADIPOR2 or CDH13) at a target cell and this causes activation of AMP-activated protein kinase (AMPK). Which of the following is most likely true? High adiponectin levels activate AMPK, which stimulates hepatic fatty acid oxidation and ketogenesis. The passage says that adiponectin up-regulates catabolic processes (such as fatty acid oxidation), which means it also likely downregulates anabolic processes (such as lipogenesis, cholesterol synthesis and triglyceride synthesis). This means answer choices A and D are incorrect (eliminate both).

upregulates = catabolic processes = such as fatty acid oxidation (increases) downregulates = anabolic processes (such as lipogenesis, cholesterol synthesis and triglyceride synthesis). (decreases)

Inspiration is the drawing of air into the lungs. The diaphragm contracts and flattens during inspiration (eliminate choices A and C), expanding the chest cavity; the lungs (which are stuck to the inside wall of the chest cavity) expand as well. The expansion of the lungs decreases the pressure in the alveoli, causing air to move into the lungs from the exterior (eliminate choice B and choice D is correct).

》Inspiration （drawing of air into the lungs） 》diaphragm （more contracts and flattens ↑) 》chest cavity/lung（less contracts）（↓） 》alveoli pressure （decreases ↓）

If acetylcholine is blocked, calcium levels will be low, protein kinase activity low, and H+ secretion low (choice C is correct). Calcium will be low (eliminate choice A) and H+ secretion will decrease, not increase (eliminate choices B and D).

》acetylcholine （X） 》calcium levels （↓） 》protein kinase activity（↓） 》H+ secretion（↓）；pH （↑）