AAMC Qbank1 Bio GM

Sarah noted that her skin blood vessels were usually constricted to conserve body heat in the cold environment of the mountains. However, her skin blood vessels would occasionally dilate for short periods of time. What would be the most probable physiological purpose for this periodic vasodilation? A Maintain normal skin tone B Maintain sufficient oxygenation of cells C Reduce excessive blood pressure D Maintain normal muscle tone

According to the item, Sarah noticed that her skin blood vessels were usually constricted to conserve body heat in the cold environment of the Colorado mountains where she went skiing. Occasionally, however, her vessels would dilate for short periods of time to enable a sufficient supply of blood (and oxygen) to her cells. Due to the physical exertion of skiing, her cells had an increased need for oxygen. C is not answer b/c see pic

During the initial skin diving session, when her heart and breathing rates were increased, Sarah noticed that she produced more urine than usual. This was most probably a result of: A increased blood pressure caused by her excitement or anxiety. B reduced blood pressure caused by her excitement or anxiety. C absorption of water from the ocean. D inability to cool the skin through evaporative water loss.

According to the passage, Sarah was in excellent physical condition prior to her trip to the Caribbean Sea to go skin diving. After her first diving experience, she noticed an elevated pulse rate and ventilation rate. According to the item, she also noticed that she produced more urine than usual. The increased urine production can be explained by an increased blood pressure caused by adrenaline, released in response to excitement or anxiety—the fight or flight response.

After Sarah's accident, her attending physician detected the protein myoglobin in her urine. What type of injury is consistent with this observation? Broken bone Damaged muscle Damaged kidney A I only B III only C I and III only

After Sarah's accident, the physician detected myoglobin in Sarah's urine. Myoglobin is the substance that holds oxygen in the muscles and organs. The physician's observation is consistent with an injury to muscle or organs, but not bone. II and III are both correct according to AAMC, and they claim that myoglobin is found in muscles AND organs. 519 (131/129/131/128) Myoglobin is found in skeletal muscles and the heart. When myoglobin enters the nephron, it interacts with proteins and precipitates, causing an obstruction usually in the distal convoluted tubule. This damages renal cells, which leads to a release of ROS that in turn oxidize ferrous oxide into ferric oxide, causing a hydroxyl free radical. These two processes combined can lead to acute renal failure if left untreated. So yeah, if you are seeing myoglobinuria, out of those choices, II and III would be consistent.

The concentration of the protein cyclin rises and falls during the cell cycle as shown in Figure 1. Figure 1Changes in the concentration of cyclin during phases of the cell cycle What mechanism could account for this oscillation of cyclin protein concentration? _ A Replication of the cyclin gene during S phase of interphase B Segregation of chromosomes carrying the cyclin genes during mitosis C Translation of cyclin mRNA in interphase and proteolysis of cyclin protein in mitosis D Translation of cyclin mRNA in mitosis and proteolysis of cyclin protein in interphase

A Replication of the cyclin gene during S phase of interphase Jack Westin Advanced Solution: While this could explain the increase in cyclin levels during interphase, we are not given any clues into why there is oscillation of cyclin concentration. Ideally, we get an answer choice that explains the fluctuation in cyclin concentration. B Segregation of chromosomes carrying the cyclin genes during mitosis Jack Westin Advanced Solution: This is similar to answer choice A because we don't get any clear reasoning behind the rise and fall of cyclin concentration. We want an answer choice that explains the increase in cyclin concentration during interphase, and the decrease during mitosis. C Translation of cyclin mRNA in interphase and proteolysis of cyclin protein in mitosis Jack Westin Advanced Solution: This answer choice is consistent with what we're looking for: Translation of cyclin mRNA in interphase corresponds to the increased cyclin concentration; proteolysis (the breakdown of protein) of cyclin protein in mitosis means decreased cyclin concentration. This is a superior answer choice to answer choices A and B. D Translation of cyclin mRNA in mitosis and proteolysis of cyclin protein in interphase Jack Westin Advanced Solution: This answer choice does explain why there are fluctuations in cyclin protein concentration, but the reasoning here is backwards. We have translation of cyclin mRNA in interphase. How do we know that? The increased cyclin concentration. We have proteolysis of cyclin in mitosis, and how do we know that? The breakdown of cyclin and the decreased cyclin concentration. We're left with our best answer, answer choice C: Translation of cyclin mRNA in interphase and proteolysis of cyclin protein in mitosis.

Inflation of the lungs in mammals is accomplished by:

.Diaphram contract .intercostal muscles (rib muscles) also help to expand chest cavity • expansion causes negative pressure Diaphragm contracts and moves downward. External intercostal muscles( contract) lift and expand ribcage.

room temperature

25 degrees Celsius Room temperature can vary depending on cultural and regional norms, as well as personal comfort preferences. However, it is generally considered to be around 20-25 degrees Celsius (68-77 degrees Fahrenheit).

Human body temperature

37 degrees Celsius 98.6 degrees Fahrenheit

Symptoms of burning, itching, and pain occur when DDT is absorbed through the skin because: A motor neurons are depolarized. B motor neurons are hyperpolarized. C sensory neurons are depolarized. D sensory neurons are hyperpolarized.

Ans: c Symptoms of burning, itching, and pain occur when DDT is absorbed through the skin because: Jack Westin Advanced Solution: We can go back to the passage, then use our general knowledge to finally decide the correct answer:Above I've added paragraph 3 from the passage. We're told burning and itching occur when dissolved DDT is applied to skin. We're told sufficient levels of exposure can lead to pain. And ultimately, we're given a hypothesis that tries to explain why these symptoms occur. That sentence is the key to answering our question. We asked the cause of the unpleasant symptoms, and this sentence says, "According to one hypothesis, these symptoms occur because DDT becomes incorporated into nerve cells, allowing Na+ to diffuse freely through axonal membranes." We know from our content review that after resting potential is established, cells can be depolarized. What are we seeing here? These sodium ions can diffuse freely, and that's what happens during depolarization. DDT is affecting neural impulses, or action potentials. And we're seeing these neurons depolarized. A motor neurons are depolarized. Jack Westin Advanced Solution: We said the diffusion of sodium cations happens during depolarization. What are motor neurons? Motor neurons transmit information from the brain to effector cells in the body. Is that what's happening here? We're likely dealing with a different type of neuron considering the sensations listed in the question stem. Sensory neurons convey information from tissues and organs into the central nervous system. So essentially converting external stimuli into electrical impulses. Let's see if we can get a better answer choice. B motor neurons are hyperpolarized. Jack Westin Advanced Solution: We're not happy with the first part of this answer choice because it mentions motor neurons and we want an answer choice that talks about sensory neurons. What happens during hyperpolarization? So after depolarization, voltage-gated sodium ion channels are closed, and there's movement of potassium. This is repolarization. Following repolarization, there is an overshoot in the potential of the cell because of this movement of potassium. We have an even greater negative charge than in the re

At the end of his initial hospital stay, a few E. coli cells remained in the patient's colon, even though he was taking antibiotics. These cells were most likely present because: A the antibiotics caused drug-resistance mutations to occur in the bacterial DNA. B the bacteria in the patient developed an immune reaction to the antibiotics. C the patient's colon cells became increasingly resistant to the antibiotics during his hospitalization. D chance mutations in a few E. coli before the treatment made these cells and their descendants antibiotic-resistant.

At the end of his initial hospital stay, a few E. coli cells remained in the patient's colon, even though he was taking antibiotics. These cells were most likely present because: Jack Westin Advanced Solution: For this question, we'll have to explain how resistance to antibiotics in bacteria works. Quick overview of resistance: how are bacteria able to acquire antibiotic resistance? There're a few reasons. The short generation time, random mutations, and genetic recombination allow bacteria to have a higher likelihood of acquiring antibiotic resistance. Note: the patient has only a few E. coli cells that remain in the colon. Let's use this information and see if any of the answer choices are consistent with this thinking. A the antibiotics caused drug-resistance mutations to occur in the bacterial DNA. Jack Westin Advanced Solution: This answer choice is possible, but it's unlikely. We want to know the most likely reason why these few E. coli cells are present. Antibiotics are not a cause of mutations. If this answer choice instead mentioned undergoing conjugation with antibiotic-resistant cells, this answer would be more likely to be correct. We're not confident in the antibiotics causing mutations. B the bacteria in the patient developed an immune reaction to the antibiotics. Jack Westin Advanced Solution: This answer choice is factually incorrect. Bacteria don't work the same way as animals where they can develop an immune reaction to pathogens. This answer choice is also unlikely. We can still keep answer choice A as our best answer, because it's more plausible, even if it's not likely. C the patient's colon cells became increasingly resistant to the antibiotics during his hospitalization. Jack Westin Advanced Solution: This answer choice is out of scope. We're focused on the E. coli cells, not the patient's colon cells in this question. Even if this were factually true, it doesn't answer the specific question being asked. We can eliminate answer choice C. D chance mutations in a few E. coli before the treatment made these cells and their descendants antibiotic-resistant. Jack Westin Advanced Solution: This answer choice is possible, and it's likely. Mutations can occur randomly. And when there are antibi

endocrine regulation of calcium in blood

Endocrine control o Remodeling of bone is subject to endocrine control ▪ high levels of PTH signal osteoclasts to begin breaking down bone so that calcium can be released in the blood ▪ Bone's function in Mineral Homeostasis o Osteoblasts use calcium from the blood to form new osteon ( calcitonin) o osteoclasts break down bone and release calcium into bloodstream ( enhanced by parathyroid hormone and Vitamin D ( counterintuitive, I know ))

If blood pressure doubled and the resistance to blood flow increased by 50%, the amount of blood pumped by the heart would have: A increased by 1/3. B increased by 1/2. C decreased by 1/3. D decreased by 1/2.

Change in P= CO x TPR 2P/1.5TRR = CO 2/1.5 = 4/3 = 1/3 + 3/3 = 1/3 + 1 Math1: I apologize for the confusion earlier. Let's clarify the math regarding the increase by 50% and its equivalence to a factor of 1.5. When we say "increase by 50%," we mean adding half of the original value to itself. Mathematically, an increase by 50% can be calculated as follows: Increase = Original Value * 0.5 And the new value after the 50% increase would be: New Value = Original Value + Increase = Original Value + Original Value * 0.5 = Original Value * 1.5 So, an increase by 50% is indeed equivalent to multiplying the original value by 1.5. This is where the factor of 1.5 comes from in the context of a 50% increase. In the case of resistance increasing by 50%, the resistance becomes 1.5 times the original resistance, which corresponds to the factor of 1.5. Similarly, the doubling of blood pressure means the pressure becomes 2 times the original pressure. I apologize for the confusion in my previous explanations. Thank you for your patience. Given this clarification, the correct answer is: A. increased by 1/3. Math 2: I apologize for any confusion caused by my previous responses. Let's clarify the relationship between a 1.33 increase and a 1/3 increase. When we say "increased by 1/3," we mean that the quantity has grown to one-third more than its original value. Mathematically, an increase by 1/3 can be calculated as follows: Increase = Original Value * (1/3) And the new value after the 1/3 increase would be: New Value = Original Value + Increase = Original Value + Original Value * (1/3) = Original Value * (1 + 1/3) = Original Value * 4/3 So, an increase by 1/3 is equivalent to multiplying the original value by 4/3, which is approximately 1.33. In the context of the problem you presented, if the amount of blood pumped by the heart increased by a factor of 1.33, it means the blood flow increased by one-third. The options provided in the problem might not use the exact fraction "1/3" but instead approximate it as "1/2" or "1/3" to simplify the choices. In summary, the increase by a factor of 1.33 is indeed very close to an increase by 1/3, and this is likely why the provided answer choices include options related to both f

Control of heart rate, muscle coordination, and appetite is maintained by the: A hypothalamus, cerebrum, and brain stem, respectively. B brain stem, hypothalamus, and cerebrum, respectively. C cerebellum, hypothalamus, and brain stem, respectively. D brain stem, cerebellum, and hypothalamus, respectively.

Control of heart rate, muscle coordination, and appetite is maintained by the brain stem, cerebellum, and hypothalamus, respectively.

If DDT accumulates in the liver, all of the following bodily functions may be significantly impaired EXCEPT: A absorption of fats in the small intestine. B production of bile. C detoxification of poisons. D regulation of blood pressure

If DDT accumulates in the liver, all of the following bodily functions may be significantly impaired EXCEPT: Jack Westin Advanced Solution: Be careful with the verbiage here. We're going to find an answer choice that doesn't reflect a function of the liver. It says "all of the following will be impaired EXCEPT." The liver plays a vital role in the digestion of fats, and detoxifying the blood. It produces bile that's required for the breakdown of fatty components of the food in the duodenum. The liver is mostly composed of hepatocytes which are involved in the synthesis of cholesterol, bile salts, and phospholipids as well. When there is excess glucose in the blood, it's stored in the liver as glycogen. A absorption of fats in the small intestine. Jack Westin Advanced Solution: We said the liver produces bile while helps in the digestion of fats, and absorption in the duodenum. This goes against what we're looking for in this question. Make sure you remember the "EXCEPT" in the question stem. B production of bile. Jack Westin Advanced Solution: We actually just mentioned this function in our previous answer choice, and in our prediction. The liver produces bile to break down fats. Again, this directly contradicts our breakdown of liver functions because this is a function of the liver. We're looking for a function that's not related to the liver. C detoxification of poisons. Jack Westin Advanced Solution: This is another answer we mentioned in our prediction. We said the liver detoxifies the blood-it removes ammonia from blood. It also metabolizes alcohol, drugs, and toxins. Another answer that doesn't fit our criteria and similar to answer choices A and B. D regulation of blood pressure. Jack Westin Advanced Solution: This is the only answer choice that wasn't explicitly mentioned in my breakdown of liver functions. Regulation of blood pressure isn't a significant function of the liver, so accumulation of DDT in the liver would not significantly affect blood pressure regulation. That means we're left with our correct answer, answer choice D: regulation of blood pressure.

" The two divisions of the autonomic nervous system (ANS) are the parasympathetic (PS) and the sympathetic (S) divisions. Structurally, the S ganglia appear near the spinal cord in the thoracic and lumbar regions and connect with each other to form the sympathetic trunk." Based on information in the passage, would the S or the PS division of the ANS be expected to produce a more rapid systemic (whole-body) response to a stimulus? A The S division, because its ganglia are interconnected B The S division, because it secretes norepinephrine C The PS division, because its ganglia are not interconnected D The PS division, because it secretes acetylcholine

Jack Westin Advanced Solution: Based on information in the passage, would the S or the PS division of the ANS be expected to produce a more rapid systemic (whole-body) response to a stimulus? We're looking for a specific response, and the response has to be quick, and throughout the body. We're going to focus on the structure of the two divisions the author presented early in the passage. I want to pull up part of the passage here to make sure the breakdown of the question is clear: I want to bring our attention to something near the middle of this paragraph. It says: the S ganglia appear near the spinal cord in the thoracic and lumbar regions and connect with each other to form the sympathetic trunk. Then soon after we have the PS ganglia lie in or near the organs they connect to but do not connect with each other. What does this tell us? When we're trying to produce a rapid systemic response, we want as many pathways as possible. The sympathetic ganglia connect with each other, meaning exactly what we're looking for. More pathways by which we can get a response to a stimulus. That's the biggest difference the author presents to us in the passage. aamc: Solution: The correct answer is A. All of the answers presented are correct statements, but only one contains information consistent with a mechanism that would increase the speed of the systemic response to a stimulus. Multiple pathways to ganglia lead to more rapid response of the sympathetic system. Multiple pathways are also more consistent with a systemic, rather than localized, response. The type of neurotransmitter involved is irrelevant and the fact that the parasympathetic ganglia are not interconnected is a reason to reject the choices which refer to that system. Thus, answer choice A is the best answer.

When viewing an X ray of the bones of a leg, a doctor can tell if the patient is a growing child, because the X ray shows: _ A cartilaginous areas in the long bones. B bone cells that are actively dividing. C the presence of haversian cells. D shorter-than-average bones.

Jack Westin Advanced Solution: In other words, how can a doctor distinguish between a growing child's leg bones, and an adult's leg bones?In growing, child bones we see epiphyseal plates at the ends of long bones. These plates are just the area of growth in a long bone. During development, bones get longer at epiphyseal plates by conversion of excess cartilage to bone, through ossification. Ossification, is just the process of bone formation. That means we expect cartilage near the ends in a growing child. We'd have that cartilage be fully converted to bone in adults, where we wouldn't see cartilage in the X-ray. A cartilaginous areas in the long bones. Jack Westin Advanced Solution: This answer choice matches our breakdown. We mentioned that during development, bones get longer at epiphyseal plates. That happens by conversion of excess cartilage to bone, through ossification. That means in a child, we'd still have that cartilage present. We like this answer choice for now. B bone cells that are actively dividing. Jack Westin Advanced Solution: This answer choice presents an option that's present in a growing child. But is this specifically what we're looking for? Think back to the question stem. We want to know: how can a doctor distinguish between a growing child's leg bones, and an adult's leg bones? Dividing bone cells are doing to be found in both children, and adults. We can eliminate answer choice B. C the presence of haversian cells. Jack Westin Advanced Solution: I believe that should be canals*. These canals are essentially tubes that contain blood vessels and nerves. But this is similar to answer choice B. We're not going to find these in only a growing child's leg bones, or only an adult's leg bones. Haversian canals are found in both. Answer choice A still remains superior. D shorter-than-average bones. Jack Westin Advanced Solution: This answer choice is too extreme. While it might be true that children generally have shorter bones, that's not a surefire way to distinguish between a growing child's bone and an adult's bone. It's not always true a child will have shorter than average bones. Just like it's not always true an adult will have average, or longer than average bones. We're left with ou

What is the net volume of fresh air that enters the alveoli each minute, assuming that the breathing rate is 10 breaths/min, the tidal volume is 800 mL/breath, and the nonalveolar respiratory system volume (dead space) is 150 mL? A 65 mL B 95 mL C 6500 mL D 7850 mL

Jack Westin Advanced Solution: The author gives us specific values that we'll use to solve for net volume per minute.The MCAT makes math questions a very specific way. The actual addition, subtraction, multiplication, and division isn't hard. This isn't testing how well you can do mental math. The test-maker wants to know if you can take the values given to you, and apply them properly. You're going to find that you'll likely be able to come up with any of the 4 answer choices listed if you manipulate the numbers in the question stem. That's why it's important to know exactly what the question's asking us and what each number represents, so we can solve the question properly.We're solving for net volume per minute here. First thing we're given is tidal volume. 800 mL per breath, means 800 mL is the volume of air delivered to the lungs with each breath taken, or each respiratory cycle. But we want to know the volume of fresh air that enters the alveoli per minute. We're told we have anatomical dead space of 150 mL. That's the volume of air that is not entering the alveoli and exchanging with blood. Since no gas exchange happens, it's dead space, and has to be excluded from the net air entering the alveoli per minute. We can subtract the nonalveolar respiratory system volume, that 150 mL, from the tidal volume, 800 mL. That 650 mL represents the net volume of fresh air that enters the alveoli per breath. But that's not what we're looking for. We want to know net volume of fresh air that enters the alveoli each minute. We have 10 breaths per minute, multiplied by our net volume of 650 mL per breath. Note the units: breaths will cancel out. We're left with 6500 mL per minute. Our answer choice needs to be a volume per minute, so we're happy with our final units.We did no approximating, and no rounding here. We solved for an exact value, which corresponds to answer choice C. That means answer choices A, B, and D are all incorrect, but why would anyone incorrectly pick A, B, or D? This goes back to what I said prior to the breakdown of the question. The test-maker wants to know if you can take the values given to you, and apply them properly. Answer choice A involves dividing our net volume of 650 mL per breath by

A student postulated that the sodium pump directly causes action potentials along neurons. Is this hypothesis reasonable? A No; action potentials result in an increased permeability of the plasma membrane to sodium. B No; the myelin sheaths of neurons prevent movement of ions across the plasma membranes of the neurons. C Yes; sodium is transported out of neurons during action potentials. D Yes; action potentials are accompanied by the hydrolysis of ATP.

Solution: The correct answer is A. The question requires the examinee to determine whether the sodium pump directly causes action potentials. The massive influx of sodium through voltage gated ion channels causes the depolarization of the neuron that occurs during the action potential (A). While some Na+ movement does occur down the length of the axon during the action potential (it occurs in the nodes of Ranvier, which are unmyelinated), this movement does not depend on the sodium pump nor does it cause action potentials (B). The sodium pump functions to move Na+ out of, not into, the cell (C). While the sodium pump does use energy released by the hydrolysis of ATP to move sodium and potassium ions across cell membranes, this energy is not used to cause the influx of sodium responsible for causing action potentials (D). Thus, A is the best answer.

Which of the following changes would NOT interfere with the repeated transmission of an impulse at the vertebrate neuromuscular junction? _ A Addition of a cholinesterase blocker B Addition of a toxin that blocks the release of acetylcholine C An increase in acetylcholine receptor sites on the motor end plate D Addition of a substance that binds to acetylcholine receptor sites

Jack Westin Advanced Solution: To answer this question, we have to consider which changes will not interrupt a nerve impulse at the neuromuscular junction, or in other words, between a motor neuron and a muscle fiber. Key word here is not. We expect three of these answer choices will interfere with the repeated transmission of the impulse and one answer choice will not.Neuromuscular junction is a chemical synapse between a motor neuron and a muscle fiber. Hence the name, neuromuscular. We're focused on repeated transmission of an impulse at this junction, and muscle action.Let's break down what happens. A motor neuron attaches to a myocyte at a motor end plate, forming a neuromuscular junction. The action potential of the neuron releases acetylcholine into the synaptic cleft. Because we're dealing with muscle action, we also know we're dealing with the neurotransmitter acetylcholine; that's the major neurotransmitter in the parasympathetic division of the autonomic nervous system. The acetylcholine binds membrane-bound receptors on the motor end plate, which activates ion channels. A Addition of a cholinesterase blocker Jack Westin Advanced Solution: On the surface, this doesn't match exactly what we discussed in our breakdown. A cholinesterase blocker means we don't have the normal breakdown of acetylcholine in the synaptic cleft. We have excess acetylcholine, and we'd have continuous transmission. That also means receptors are no longer available to react to impulses, but rather we have continuous binding. B Addition of a toxin that blocks the release of acetylcholine Jack Westin Advanced Solution: This answer choice directly contradicts our breakdown. If we block the release of acetylcholine, we would interfere with the transmission of an impulse at the junction. No presynaptic release of acetylcholine means we don't have anything binding to the postsynaptic receptors. Answer choice A is actually a better answer choice than this. Answer choice A proposed continuous stimulation because of the presence of acetylcholine. Answer choice B is proposing no acetylcholine release whatsoever. Without acetylcholine we don't get transmission of the impulse. That means answer choice A is superior. C An increase in acet

DNA polymerase catalyzes the replication of chromosomal DNA in bacteria as shown below. A double-stranded DNA molecule contains bases with a ratio of (A + T)/(G + C) = 3:1. This molecule is replicated with DNA polymerase in the presence of the four deoxynucleoside triphosphates with a molar ratio of (A + T)/(G + C) = 1:1. What is the expected ratio of (A + T)/(G + C) in the double-stranded daughter DNA molecule? _ A 1:3 B 1:1 C 2:1 D 3:1

Jack Westin Advanced Solution: To answer this question, we're going to focus on DNA replication being semi-conservative. According to the semiconservative replication model, the two original DNA strands will separate during replication. These two strands will serve as a template for a new DNA strand. That means each newly synthesized double helix is a combination of one old and one new DNA strand. We still have complementary pairs. The ratio of adenine and thymine to the ratio of guanine and cytosine are going to remain the exact same. The deoxynucleoside triphosphates themselves serve to build the new strands of DNA. The ratio of the dNTPs isn't going to change the ratio in the daughter DNA molecule. That means the 3:1 ratio is going to remain exactly the same. We can pick the only answer choice that is consistent with that, which is answer choice D.

brain stem consists of

Midbrain, pons, medulla (also called medulla oblongata

Most bacterial cells and human cells are alike in: A the ability to produce ATP via ATP synthase. B the chemical composition of their ribosomes. C their enclosure within cell walls. D the shape of the self-replicating structures that carry their DNA.

Most bacterial cells and human cells are alike in: Jack Westin Advanced Solution: Here we have another content question in disguise. As always, these can be tricky because they are located in a passage, which might mislead you into thinking you will need to use passage information to solve. A the ability to produce ATP via ATP synthase. Jack Westin Advanced Solution: This is correct. Metabolically, eukaryotes and prokaryotes carry out most of the same metabolic pathways, including glycolysis, anaerobic glycolysis, the citric acid cycle, and the electron transport chain. Where eukaryotes and prokaryotes differ is in the location of these pathways. In eukaryotes, glycolysis is located inside of the cytosol while the rest of the listed pathways are located somewhere in the mitochondria. In prokaryotes, all of these pathways occur in the cytosol. In terms of the electron transport chain, which is coupled to ATP synthase, the ETC & ATP Synthase of eukaryotes is located in the inner mitochondrial membrane, whereas the ETC & ATP Synthase of prokaryotes is located in the cell's plasma membrane. B the chemical composition of their ribosomes. Jack Westin Advanced Solution: Human cells and prokaryotes contain very chemically different ribosomes. C their enclosure within cell walls. Jack Westin Advanced Solution: Human cells do not have cell walls, unlike many bacteria. D the shape of the self-replicating structures that carry their DNA. Jack Westin Advanced Solution: The shape of eukaryotic chromosomes is linear, whereas the shape of prokaryotic chromosomes is circular.

Most proteins in present-day mitochondria are made by cytoplasmic ribosomes from mRNA transcribed from nuclear genes. Can this fact be reconciled with the hypothesis described in the passage? A Yes; the transfer of genes from symbionts to the eukaryotic nucleus could have occurred during the last billion years of evolution. B Yes; this difference from bacteria is unimportant, because the many similarities between bacteria and mitochondria provide sufficient evidence in favor of the hypothesis. C No; the fact that mitochondrial proteins are made in the cytoplasm is convincing evidence that mitochondria do not have a bacterial origin. D No; because bacteria can make all their own proteins and mitochondria cannot, this disproves the hypothesis.

Most proteins in present-day mitochondria are made by cytoplasmic ribosomes from mRNA transcribed from nuclear genes. Can this fact be reconciled with the hypothesis described in the passage? Jack Westin Advanced Solution: To answer this question, I want to address how proteins are made by mitochondria, and in the cytoplasm after being transcribed from nuclear genes. Scientists hypothesized mitochondria evolved from bacteria that entered primitive eukaryotic anaerobes. Even though most proteins in present-day mitochondria are made in the cytoplasm from mRNA transcribed from nuclear genes, the two are not mutually exclusive. Their symbiotic relationship has been evolving for millions (and probably billions) of years, so it makes sense that protein synthesis can change to a more efficient method. What's a possible, more efficient method? Having proteins transcribed from nuclear genes instead of from mitochondria directly. A Yes; the transfer of genes from symbionts to the eukaryotic nucleus could have occurred during the last billion years of evolution. Jack Westin Advanced Solution: This is a viable explanation, which is all we need to answer this question. Evolution over billions of years can lead to a lateral transfer of genes, just like is proposed in the question stem. Like I mentioned when I was breaking down the question, it's possible the primitive eukaryotic host cell nucleus acquired some mitochondrial genes over these years. That includes genes for making mitochondrial proteins. B Yes; this difference from bacteria is unimportant, because the many similarities between bacteria and mitochondria provide sufficient evidence in favor of the hypothesis. Jack Westin Advanced Solution: This answer choice is unreasonable. If there is evidence against a theory, the theory has been disproven. This is in spite of the similarities. This answer choice is essentially changing the question, and changing the criteria by which the hypothesis is being tested. We can eliminate answer choice B. C No; the fact that mitochondrial proteins are made in the cytoplasm is convincing evidence that mitochondria do not have a bacterial origin. Jack Westin Advanced Solution: This answer choice is implying mitochondria don't have a

Plant cell vs animal cell

Plant cells and animal cells are both types of eukaryotic cells, meaning they have a distinct nucleus and membrane-bound organelles. While they share many common organelles, there are some key differences between the two cell types in terms of organelles. Here's a comparison of some of the major organelles found in plant and animal cells: **Organelles Found in Both Plant and Animal Cells:** 1. **Nucleus:** The control center of the cell that contains genetic material (DNA) and controls cellular activities. 2. **Mitochondria:** Organelles responsible for producing energy through cellular respiration. 3. **Endoplasmic Reticulum (ER):** A network of membranes involved in protein synthesis and lipid metabolism. There are rough ER (with ribosomes) and smooth ER (without ribosomes). 4. **Golgi Apparatus:** Involved in processing, packaging, and transporting proteins and lipids. 5. **Lysosomes:** Contain enzymes that break down waste materials and cellular debris. 6. **Cytoplasm:** The gel-like substance within the cell that contains organelles and provides a medium for cellular processes. 7. **Cytoskeleton:** A network of protein filaments and tubules that provide structural support and enable cell movement. **Organelles Unique to Plant Cells:** 1. **Cell Wall:** A rigid structure outside the cell membrane that provides support and protection to plant cells. Animal cells lack cell walls. 2. **Chloroplasts:** Organelles containing chlorophyll that are responsible for photosynthesis, the process by which plants convert sunlight into energy. 3. **Large Central Vacuole:** A large membrane-bound sac filled with water, enzymes, and other substances. It plays a role in maintaining turgor pressure, storing nutrients, and waste disposal. 4. **Plasmodesmata:** Channels that pass through the cell walls of plant cells, allowing for communication and transport between adjacent cells. **Organelles Common in Animal Cells but Not Always Present in Plant Cells:** 1. **Centrioles:** Organelles involved in cell division and the formation of the spindle fibers during mitosis. They are generally absent in plant cells, except in certain specialized cells. While plant and animal cells share many organelles and functions, the presence of

Dewlaps that reflect UV light would evolve by natural selection only if: A individuals with UV-reflective dewlaps produced more offspring than did individuals without them. B individuals with UV-reflective dewlaps were better able to communicate than individuals without them. C individuals with UV-reflective dewlaps were less subject to predation than individuals without them. D individuals with UV-reflective dewlaps mated more frequently than did individuals without them.

Solution: The correct answer is A. Although many different types of adaptations may help an individual organism survive, they will not be passed on to the next generation unless the organism produces offspring, passing on the genes that cause the advantageous phenotype. To evolve by natural selection and become a general characteristic of the species, the genes that cause dewlaps to reflect UV light must become a significant portion of the gene pool, which will most likely occur if individuals with UV-reflective dewlaps produce more offspring than do individuals without them. Thus, A is the best answer. Jack Westin Advanced Solution: We can answer this question using our general knowledge of evolution, and natural selection. Even though this can be tangentially related to the passage, this could almost be a standalone question.Natural selection is the reproduction of individuals with favorable traits. These individuals survive environmental change because of those traits. Natural selection is an inevitable outcome of three principles:most characteristics are inherited;more offspring are produced than can survive;offspring with more favorable characteristics will survive and have more offspring than those individuals with less desirable traits.We want to explain how dewlaps that reflect UV light would fit within this definition, and cause more offspring to be produced and survive.

The extreme hyperglycemia of these animals suggests that major changes in the normal glucose regulatory mechanisms occur during freezing. Which of the following observations would support this hypothesis? A Suppression of insulin secretion during freezing episodes B Suppression of glucagon secretion during freezing episodes C Slowing of glycogen catabolism in the liver during freezing episodes D Increased sensitivity of all pancreatic endocrine responses during freezing episodes

Solution: The correct answer is A. Hyperglycemia normally elicits insulin secretion. Support for the observation that major changes in the normal glucose regulatory mechanisms occur would, therefore, be supported by the observation that insulin secretion was suppressed during hibernation. Glucagon secretion is suppressed by hyperglycemia. Glycogen break-down contributes to glucose levels; a slowing of glycogen catabolism would still contribute to glucose levels but at a slower rate. Changing the sensitivity of all pancreatic endocrine responses would not be a major change in mechanism, but would instead be a change in threshold of activity. Thus, answer choice A is the best response.

It was hypothesized that the decrease in blood flow to the skin resulted from a change in the activity of the sympathetic nerves to the skin. Which of the following observations would support this hypothesis? A A change in the norepinephrine content of blood draining from the skin B In vitro contraction of the smooth muscle in skin blood vessels in response to acetylcholine C A lack of epinephrine receptors in skin blood vessels D In vivo dilation of the skin blood vessels

Solution: The correct answer is A. One hypothesis suggests that the decrease in blood flow to the skin results from a change in the activity in the sympathetic nerves to the skin. This hypothesis would be supported if researchers observed a change in the norepinephrine content of blood draining from the skin. reddit: Yes. Both of them do have sympathetic effects. the NE is released primarily by neurons whereas Epi is mostly from adrenal glands. They both bind to adrenergic receptors (adrenaline, adrenergic, see what they did there?) and have similar effects. They aren't exactly the same, but don't worry about that too much, you'll probably not have a question asking anything where you need to differentiate the two

From which germ layer(s) do the tissues of the heart and blood vessels differentiate? Ectoderm Mesoderm Endoderm _ A II only B III only C I and II only D I and III only

Solution: The correct answer is A. The heart and blood vessels both differentiate from the mesoderm. Thus, A is the best answer.

n which organelle of a eukaryotic cell is the pyrimidine uracil, as part of uridine triphosphate (UTP), incorporated into nucleic acid? _ A The nucleus B The Golgi bodies C The ribosomes D The endoplasmic reticulum

Solution: The correct answer is A. The nitrogenous base, uracil, combined with the sugar ribose and phosphate makes up the nucleotide uridine. It is found in RNA, but not in DNA. The corresponding DNA nucleotide is thymine. Uridine is incorporated into RNA in the nucleus where transcription of DNA into RNA takes place. RNA is manufactured in the nucleus from a DNA template. Therefore, the correct answer is answer choice A. RNA is necessary for protein synthesis by free ribosomes and those attached to endoplasmic reticulum, but it is not synthesized there, so answer choices C and D are incorrect. Proteins are packaged for export in the Golgi apparatus, but it too takes no part in RNA synthesis, so answer choice B is incorrect.

In human females, mitotic divisions of oogonia that lead to formation of presumptive egg cells (primary oocytes) occur between: _ A fertilization and birth only. B fertilization and puberty only. C birth and puberty only. D puberty and menopause only.

Solution: The correct answer is A. The question asks the examinee to identify the stages in the human female life cycle between which all the mitotic divisions that lead to primary oocytes occur. These stages occur in the following order: fertilization, birth, puberty, menopause. All of the mitotic divisions that form primary oocytes occur prior to birth. Thus, A is the best answer.

Assuming that the vertebrates were all of comparable size, which of the following vertebrates would be expected to have the strongest and heaviest bones? _ A A land-dwelling mammal B A water-dwelling mammal C A flying bird D An amphibian

Solution: The correct answer is A. The question asks the examinee to identify which vertebrate is likely to have the strongest and heaviest bones. Of the animals listed, a land-dwelling animal would be expected to have the heaviest bones (A), since enhanced bone density would be required to withstand the load bearing activity that results from the impact of gravity on land-dwelling animals. Less dense bones would be present in water-dwelling creatures (B) and amphibians (D), since the impact of gravity would be ameliorated by life in an aquatic environment. Likewise less dense bones would be a necessity for flight (C). Thus, A is the best answer.

In regard to their relative size, the described objects are: A smaller than all known eukaryotic cells. B approximately the size of a typical coccus bacterium. C larger than a human red blood corpuscle. D larger than all known bacteriophages.

Solution: The correct answer is A. Virus structure cannot be visualized with a light microscope. The limit of resolution of a light microscope is about 200 nm. The virions described in the passage are below that limit, 100nm. Bacteria and eukaryotic cells such as the human red blood cell are well above that size so they can be seen with a light microscope, so choices B and C are incorrect. Bacteriophages are viruses in bacteria so they must be much smaller than bacteria, so choice D is not correct. The correct answer is A, they are smaller than all known eukaryotic cells.

According to Figure 1, at approximately what plasma concentration of glucose is the Tm (320 mg/min) reached? A 6.5 mg/mL B 10.0 mg/mL C 11.5 mg/mL D 12.5 mg/mL

Solution: The correct answer is B. Above 10 mg/mL, glucose begins to be found in the urine. The Tm for glucose is therefore 10 mg/mL, answer choice B. This can be read from the graph by looking at the concentration in the plasma where the concentration in the urine is zero. In other words, where the clearance line for glucose crosses the axis.

" Just read" Equal concentrations of 8 mg/mL of Substance A and glucose are found in a volunteer's plasma. Based on Figure 1, which substance will the kidney clear from the plasma more rapidly? A Substance A, because the slope of the clearance line for Substance A is higher than that for glucose B Substance A, because Substance A reaches its Tm at a lower plasma concentration than does glucose C Glucose, because glucose reaches its Tm at a higher plasma concentration than does Substance A

Solution: The correct answer is B. The Tm is a characteristic of the individual substances in the tubule system and a measure of how efficiently each substance can be reabsorbed. A high Tm indicates a high capacity for reabsorption of substances in the kidney tubules. In figure 1, the Tm for each substance can be read as the concentration in plasma when the concentration in the urine is zero. In this case it looks like the Tm for Substance A is a little over 6 mg/mL and that of glucose is 10 mg/mL. So substance A has a lower Tm. This means the tubules will not reabsorb it very efficiently. Much of it will be spilling into the urine, thus being eliminated from the body. The question asks which will clear from the blood more rapidly at a concentration of 8 mg/mL and why. The answer is that all glucose will be reabsorbed at that concentration, none will appear in the urine and none will be cleared from the plasma. Glucose has such a high Tm, that all of the glucose will be reabsorbed into the bloodstream, perhaps to reenter the kidney tubule again. A rate of 8mg/mL is above the Tm of substance A, so there will be some substance A in the urine at this plasma concentration. The answer to the question depends on the value of Tm, not the slope of the clearance line, so answer choices A and D can be eliminated. Substance A will clear more rapidly than glucose, therefore, answer choice B is correct.

People who are born without sweat glands are likely to die of heat stroke in the tropics. This indicates that, under tropical conditions, the human body may: A gain, rather than lose, heat by evaporation. B gain, rather than lose, heat by radiation. C need to use different mechanisms than in temperate zones to maintain body temperature. D be better able to regulate body temperature than under temperate conditions

Solution: The correct answer is B. The question asks the examinee to explain why individuals without sweat glands are likely to die of heat stroke in the tropics. If people lack sweat glands, they are unable to make sweat nor to capitalize upon the evaporative cooling of sweat (A). These individuals are forced to rely nearly solely on vasodilation (radiation) for responding to elevated external temperatures. That individuals without sweat glands are likely to suffer heat stroke in the tropics, indicates that radiation alone is ineffective for cooling under these conditions. In fact, the human body may gain heat by absorbing radiation from the sun leading to an elevation in body temperature (B). These individuals are not effectively able to manage their body temperature under tropical conditions, given that they are likely to die of heat stroke (D). Given that the same mechanisms are used for cooling the human body in temperate and tropical zones, it is not likely these individuals are using different mechanisms to maintain body temperature in temperate zones (C). Thus, B is the best answer.

All of the following occur during normal inspiration of air in mammals EXCEPT: _ A elevation of the rib cage. B relaxation of the diaphragm. C reduction of pressure in the pleural cavity. D contraction of the external intercostal rib muscles.

Solution: The correct answer is B. The question asks the examinee to identify which one of four actions is not associated with normal inspiration in mammals. B is the best answer because the diaphragm does not relax during inspiration. The diaphragm contracts and pulls downward, causing air to enter the lungs. A, C, and D all describe actions that do occur during inspiration. Elevation of the rib cage increases the volume inside the chest cavity (A), reduction of pressure inside the pleural cavity causes air to move into the lungs (C), and contraction of the external intercostal rib muscles helps the chest expand (D). Thus, B is the best answer.

Radioactively labeled uracil is added to a culture of actively dividing mammalian cells. In which of the following cell structures will the uracil be incorporated? _ A Chromosomes B Ribosomes C Lysosomes D Nuclear membrane

Solution: The correct answer is B. The question asks the examinee to predict the cell structure in which radioactively labeled uracil will become incorporated if added to a culture of actively dividing mammalian cells. Uracil is a component of RNA. Therefore, one would expect to find the radioactively labeled uracil in cell structures that contain RNA. B is the best answer because ribosomes contain rRNA and proteins. A is incorrect because chromosomes, by definition, consist primarily of proteins and DNA. Although DNA is composed of nucleic acids, DNA contains thymine instead of uracil. C and D are incorrect because RNA is not an integral component of either lysosomes (C) or the nuclear membrane (D). Thus, B is the best answer.

If the dose of Streptococcus Strain A required to cause infection is 1 x 105 bacteria and that of Streptococcus Strain B is 5 x 104 bacteria, which of the following statements describes the relative potencies of these strains? A Strain A is five times as potent as Strain B. B Strain A is one-fifth as potent as Strain B. C Strain A is twice as potent as Strain B. D Strain A is half as potent as Strain B.

Solution: The correct answer is D. Since it takes fewer strain B bacteria to cause an infection than strain A (50,000 versus 100,000, respectively), then strain B is more potent. How much more potent is it? Since 100,000 is twice 50, strain B is twice as potent. Therefore, the correct answer is option D, which states that the strain A is half as potent as strain B.

The outer layers of human skin are composed of dead cells impregnated with keratin and oil, which make the epidermis relatively impermeable to water, yet humans sweat freely in hot temperatures. This occurs because: _ A the salt in sweat allows it to diffuse through the skin. B sweat glands have special channels through the skin. C an osmotic gradient in sweat moves it through the skin. D sweating occurs in only those areas of the body where the skin is water permeable.

Solution: The correct answer is B. The sweat glands secrete onto the surface of the skin through channels continuous with the most superficial layer of the skin, the epidermis. These channels prevent water loss by isolating the water-permeable, sweat-secreting cells from dry surface air. The openings of the sweat glands on to the surface of the epidermis are pores. The correct answer is B. All the other answers require some movement of water through the epidermis itself, which is relatively impermeable.

The two primary factors that normally determine the level of blood pressure are: A the blood concentration of L-NMMA and norepinephrine. B the cardiac output and the resistance to blood flow. C the blood volume and the amount of L-arginine in the diet. D the heart rate (heartbeats/minute) and the cardiac stroke volume.

Solution: The correct answer is B. Two factors that normally determine the blood pressure are the cardiac output and the resistance to blood flow. Cardiac output (stroke volume x heart rate) determines the amount of blood pumped into the system by the heart per unit time. The resistance to blood flow is primarily determined by the caliber of the small arteries, arterioles, and precapillary sphincters. Thus blood pressure equals total peripheral resistance times cardiac output, a relationship analogous to Ohm's law for electrical circuits. L-NMMA is not normally present, as implied in the passage by the statement that this substance was developed for use as an inhibitor for NO synthase. Thus, answer choice B is the best answer. The two primary factors that normally determine the level of blood pressure are: Jack Westin Advanced Solution: This could easily be a standalone or discrete question even. It's only slightly related to the topics in the passage, so it's included here.Blood pressure is the pressure of the blood against the walls of blood vessels. What determines this value? We have heart rate and stroke volume that determine cardiac output. We also have the tension in the blood vessels that can either reduce or increase blood flow. Vasoconstriction can reduce blood flow and increase blood pressure, for example. A the blood concentration of L-NMMA and norepinephrine. Jack Westin Advanced Solution: This is a very narrow answer, but these two factors do affect one of the main factors we mentioned. Both can affect blood flow. The presence of both will typically lead to vasoconstriction, and ultimately an increase in blood pressure. B the cardiac output and the resistance to blood flow. Jack Westin Advanced Solution: This sounds like a better, more neutral, and more broad answer. What do I mean by that? Answer choice A only provided two ways by which blood pressure can be affected. Answer choice B encompasses both by including "the resistance to blood flow." In addition to the resistance, we also have cardiac output, which considers heart rate and stroke volume. Even though answer choice A is listing two factors that can affect blood pressure, this is a much more complete answer. C the blood volume and the

Based on Figure 1, at what free actin subunit concentration (or range of concentrations) will the microfilament treadmill? A 0.25 µM B 1.0 µM C 1.5 µM D Any concentration between 1.0 µM and 4.0 µM

Solution: The correct answer is C. Figure 1 shows that at a free actin concentration of 1.5µM, the rate at which actin subunits are added to the + end of the microfilament is equal to the rate at which actin filaments are removed from the - end. This fits the definition of treadmilling. Thus, C is the best answer.

" Figure 1 Changes in the + and - ends of a microfilament as a function of actin concentration" At what concentration of free actin will the + end of the microfilament grow faster than the - end? A Exactly at 1 µM B Only between 1 µM and 4 µM C At any concentration greater than 1 µM D At any concentration

Solution: The correct answer is C. Figure 1 shows that at free actin concentrations greater than 1µM, actin is added to the + end of a microfilament. At concentrations lower than 1µM, the + end loses actin subunits. The - end of the microfilament does not begin to add actin until the free actin concentration is greater than 5.5µM. The rate at which actin is added to the + end is greater than the rate it is added to the - end, implying that at any concentration greater than 1µM, the + end of the microfilament grows faster than the - end. Thus, C is the best answer.

The chemical gramicidin inserts into membranes and creates an artificial pathway for proton movement. Based on Figure 1, if mitochondria are treated with gramicidin, the rate of ATP synthesis will most likely: A increase, because of increased proton movement back into the mitochondria. B decrease, because of a decreased rate of hydrogen-atom donation by NADH. C decrease, because the proton gradient will rapidly reach equilibrium. D not be altered, because sufficient protons will remain between the membranes to generate ATP.

Solution: The correct answer is C. Hydrogen ions (H+) are protons. The provision of a channel for proton flow across the membrane would allow hydrogen ions to flow across the membrane until equilibrium had been achieved between the concentrations on each side of the membrane. Because ATP synthesis is driven by a flow of hydrogen ions down a concentration gradient, ATP production will decrease and eventually stop as equilibrium is established (not increase as suggested in answer choice A or remain unchanged as suggested in answer choice D). The decrease has nothing to do with the rate of hydrogen ion donation by NADH, answer choice B. Answer choice C is the correct answer.

" Escherichia coli, a bacterial species that inhabits the colon," The patient's ruptured appendix required treatment with antibiotics because he had a bacterial infection caused by: A M. tuberculosis. B E. coli entering the colon from the appendix. C E. coli entering the abdominal cavity from the appendix. D E. coli entering the appendix from the colon.

Solution: The correct answer is C. The passage indicates that the ruptured appendix preceded infection with M. tuberculosis, so the infective agent would have had to be the E. coli bacteria, a species common to the colon. This item further relies on anatomical knowledge of the gastrointestinal tract for an answer. The passage states that E. coli are found within the colon, where they function in digestion and vitamin production. The appendix is continuous with the colon so that bacteria can move between these two structures; a ruptured appendix would allow E. coli into the abdominal cavity, which is not normal. Thus, answer choice C is the best answer. The patient's ruptured appendix required treatment with antibiotics because he had a bacterial infection caused by: Jack Westin Advanced Solution: We're going to think through the effects of a ruptured appendix, and how that may cause infection. The test-maker gives us a clue in the question stem and tells us the patient had a bacterial infection. The issue with an appendix rupture is that bacteria the fine when it's in the appendix or moving between the appendix and colon. In fact, we read in the passage: that same bacteria benefits humans by breaking down food and producing vitamins. Ultimately, the bacterial infection was problematic because of the bacteria infecting other organs. New research does suggest the appendix might be home to what we consider "good" bacteria, but we're not focused on that for this question. We're focused only on what's in the passage, what's asked in the question stem, and the topics listed on AAMC's content outline. A M. tuberculosis. Jack Westin Advanced Solution: This answer choice is unreasonable. The patient didn't contract M. tuberculosis until after leaving the hospital from his initial stay. The ruptured appendix happened before tuberculosis, so initial reaction is we're not liking this answer choice. B E. coli entering the colon from the appendix. Jack Westin Advanced Solution: This answer choice is more plausible, and in fact likely happened. When the appendix ruptures, E. coli are going to enter other organs, like we said. The problem here is, E. coli inhabit the colon. The author mentions in the first paragraph that E.

Which of the following processes is LEAST directly influenced by adrenergic drugs? A Peristalsis B Secretion of digestive enzymes C Enzymatic breakdown of food molecules D Nutrient delivery to muscles and organs

Solution: The correct answer is C. The question asks the examinee to identify the bodily process among the options listed that is LEAST directly influenced by adrenergic drugs. According to the passage, adrenergic drugs mimic activation of the sympathetic nervous system; therefore, the best answer will be the process that is LEAST directly controlled by the sympathetic nervous system. The sympathetic nervous system directly inhibits peristalsis (A) and secretion of digestive enzymes (B). It also increases the blood glucose concentration and causes dilation of the blood vessels that supply the deep muscles and internal organs, which aids nutrient delivery (D) to these tissues. The sympathetic nervous system does not directly affect the activity of digestive enzymes (C) after they have been secreted. Thus, C is the best answer.

To be an effective therapy, an antisense gene that is incorporated into a genome that contains the target gene must be: A on the same chromosome as the target gene but not necessarily be physically adjacent. B on the same chromosome as the target gene and must be physically adjacent. C regulated in a similar manner as the target gene. D coded on the same strand of DNA as the target gene. just read answer

Solution: The correct answer is C. The question focuses on antisense genes that are incorporated into the genome of their target gene and asks the examinee to identify which antisense gene characteristic would be required for the gene to be therapeutically effective. To provide effective therapy, this antisense gene would need to be regulated in a manner similar to the manner in which the target gene is regulated so that the antisense RNA is produced at the same time that the sense mRNA is produced. This would ensure that the antisense RNA is available to bind with the sense mRNA, thereby preventing its subsequent translation. As a result, C is correct. A, B, and D are incorrect because the key feature determining whether an antisense drug will work is the timing of the expression of the antisense gene. This is controlled by specific regulatory elements, not necessarily the location of the antisense gene relative to the target gene. Thus, C is the best answer. Reddit: If you want the antisense RNA to have a therapeutic effect, it would need to be expressed whenever the target mRNA that it's meant to block is to be expressed. Thus the target gene and the antisense would have to be regulated in a similar manner. I.e same promoter Just because a gene is physically near another doesn't mean if one is expressed the other will be. Gene expression is based on a handful of regulatory processes.

When the environmental temperature is 33° C, vasodilation of cutaneous blood vessels helps to regulate the body temperature of a human by: A slowing blood flow through the skin. B maintaining an even distribution of heat throughout the body. C radiating excess body heat into the environment. D preventing needed body heat from being lost to the environment.

Solution: The correct answer is C. This question asks the examinee to identify the impact of vasodilation on the maintenance of body temperature. When vasodilation occurs, the walls of blood vessels relax, allowing more blood to enter the area. The presence of increased blood within dilated vasculature in cutaneous tissue allows heat to escape from the surface of the body into the environment (C). Vasodilation increases, not decreases, bloodflow to the skin (A). This process does not produce an even distribution of heat, but rather facilitates the disproportionate movement of blood (and heat) to the surface of the body so that heat can be released (B). Vasodilation typically functions to facilitate, and not to prevent, heat loss (D). Thus, C is the best answer.

Increased vasoconstriction has an important role in which of the following situations? A Causing the decrease in blood pressure associated with fainting B Increasing blood flow to muscle during exercise C Increasing blood flow to skin during blushing D Maintaining blood pressure during a hemorrhage

Solution: The correct answer is D. Although the passage relates to an experiment that studies mechanisms of vasoconstriction, the answer to this question relies on an examinee's background knowledge of the cardiovascular system. As blood is lost from the circulation, reduction in vessel size helps maintain the necessary pressure to keep the blood circulating to all body tissues. Vasoconstriction, the narrowing of a vessel, restricts blood flow to an organ and can increase blood pressure, whereas vasodilation has the opposite effect. Increased vasoconstriction is important in maintaining blood pressure during a hemorrhage. Vasodilation increases blood flow to both the muscle during exercise and the skin during blushing. Thus, answer choice D is the best answer.

The enzyme pepsin, which catalyzes the hydrolysis of proteins in the stomach, has a pH optimum of 1.5. Under conditions of excess stomach acidity (pH of 1.0 or less), pepsin catalysis occurs very slowly. The most likely reason for this is that below a pH of 1.0: _ A pepsin is feedback-inhibited. B pepsin synthesis is reduced. C the peptide bonds in pepsin are more stable. D the three-dimensional structure of pepsin is changed.

Solution: The correct answer is D. Proteins are long one-dimensional strings of amino acids. But, for a protein to function properly, it must have a very specific three-dimensional structure. This three-dimensional structure of a protein is stabilized by covalent bonds and noncovalent interactions between different regions of the linear peptide. This three-dimensional structure can be disrupted by heating or by changing the pH. The disorganization of proteins by such agents is called denaturation. An example of denaturation is the hardening of an egg during cooking. Enzymes are proteins that act as organic catalysts, speeding chemical reactions but not being consumed in them. Their function is highly dependent on a precise three dimensional structure, especially at the site of catalysis, known as the active site. The lowering of pH described in the question is likely to have caused the enzyme pepsin to lose its three-dimensional shape and thus its catalytic activity as described in answer choice D, the correct answer. A pepsin is feedback-inhibited. Jack Westin Advanced Solution: While this may or may not be true, we always have to consider every answer within the context of the question being asked. Feedback inhibition is not the reason pepsin catalysis is occurring more slowly at the very low pH below 1.0. B pepsin synthesis is reduced. Jack Westin Advanced Solution: Pepsinogen and gastric lipase are secreted by chief cells. Acid converts inactive pepsinogen to pepsin which would still be taking place. Be careful with the verbiage here. We're told catalysis occurs very slowly, not that there is a lack of pepsin in the stomach. C the peptide bonds in pepsin are more stable. Jack Westin Advanced Solution: The stability of peptide bonds isn't going to cause catalysis to occur more slowly. In fact, more stable bonds would mean the 3D shape/folding is less altered than we would expect. This answer choice is implying the opposite of what we're looking for in our correct answer. Answer choices A and B are still superior to C because they don't directly contradict what we're expecting in the low pH situation. D the three-dimensional structure of pepsin is changed. Jack Westin Advanced Solution: This answer choice

In eukaryotes, oxidative phosphorylation occurs in the mitochondrion. The analogous structure used by bacteria to carry out oxidative phosphorylation is the: _ A cell wall. B ribosome. C nuclear membrane. D plasma membrane.

Solution: The correct answer is D. The inner membrane of a mitochondrion is analogous to the plasma membrane of a prokaryote. The enzymes for oxidative phosphorylation are embedded in the inner membrane. The endosymbiotic theory suggests that mitochondria are descendents of prokaryotes that were engulfed by endocytosis into a vesicle lined with a membrane derived from the cell membrane of the eukaryote host. This is the outer membrane. The inner membrane is the plasma membrane of the endosymbiotic prokaryote, so answer choice D is correct. Answer choice A is incorrect because the cell wall of bacteria does not resemble a plasma membrane. Ribosomes have nothing to do with oxidative phosphorylation, so answer choice B is incorrect. Bacteria lack a nuclear membrane, thus answer choice C is also incorrect.

An effective and efficient method for the delivery of an antisense gene could be: A orally as an emulsified product. B microinjection into individual body cells. C intravenously as a nonantigenic, blood-stable product. D infection of an embryo by a virus modified to carry the gene.

Solution: The correct answer is D. The question asks the examinee to identify an effective and efficient method for the delivery of an antisense gene. For an antisense gene to work, it must be incorporated into the cell in which it will perform its job so that its product is available to hybridize with the sense mRNA that needs to be blocked. Of the options listed, the best way to deliver the antisense gene into all the cells of the individual would be to infect an embryo with a virus that carries the antisense gene. The appropriate virus could become incorporated into the genome of the embryonic cells, thus causing all cells derived from these embryonic cells to contain the antisense gene. Therefore, D is the best answer. A is incorrect because the gene would most likely be destroyed in the acidic environment of the stomach. B is not the best option because microinjection of a gene into all cells would be extremely impractical. C is incorrect because even though the gene could be injected as a blood-stable product, it is not likely that the drug would enter its intended target cells. Thus, D is the best answer.

Delayed ovulation, as a cause of tubal pregnancy, would most likely be associated with delayed secretion of which of the following hormones? A Progesterone B Estrogen C HCG D Luteinizing hormone

Solution: The correct answer is D. The question requires the examinee to identify that luteinizing hormone (D) is the hormone responsible for triggering ovulation. While the sex hormones progesterone (A) and estrogen (B) are either secreted in response to the luteinizing hormone surge or actually trigger the luteinizing hormone surge, respectively, they are not directly involved in triggering ovulation. HCG (C) is the pregnancy hormone, but it doesn't have a role in the typical ovulatory cycle. Thus, D is the best answer.

Just read If acetylcholine is removed from the circulation faster than norepinephrine is, which of the following autonomic processes would be most rapidly inactivated? A Dilation of the pupils B Dilation of blood vessels in the skeletal muscles C Rise in blood pressure D Stimulation of digestive secretions

Solution: The correct answer is D. This question requires some prior knowledge of the function of the autonomic nervous system. The sympathetic division of the autonomic nervous system arouses the animal for "fight or flight." Blood is directed away from the digestive system and toward the skin and muscles; the digestive system becomes less active; the heart beats faster and with greater force; the bronchi of the lungs relax to let in more air; and the pupils of the eyes dilate. Rapid removal of acetylcholine from the circulation would inactivate parasympathetic stimulation of digestive secretion consonant with rapid arousal of the animal in an emergency. Thus, answer choice D is the best answer.

Which of the following tissues have cells that are in direct contact with the external environment or elements of the external environment? The lining of the reproductive tract The lining of the respiratory tract The lining of the gastrointestinal tract _ A I and II only B I and III only C II and III only D I, II, and III

Solution: The correct answer is D.. Jack Westin Advanced Solution: Which of the following tissues have cells that are in direct contact with the external environment or elements of the external environment?The lining of the reproductive tractThe lining of the respiratory tractThe lining of the gastrointestinal tractTo answer this question, we'll have to consider all three options listed above. One thing I want to point out for this question: all three of the choices are listed in exactly three answers. If we find that any of the three options does not have cells in direct contact with the external environment or elements of the external environment, we'll eliminate those three options and have our correct answer. Alternatively, if all three tissues have these cells, then D becomes our correct answer.Right away you might notice a common denominator between all three options. Mucous membranes line the reproductive, respiratory, and digestice tracts. Why is that? It acts as a barrier and stops pathogens from entering the body. Think of nostrils and mouth, genital area, anus, and lips. Those are all places where we have some contact with the external environment or elements of the external environment. What does that mean for us here? All three of the answer choices match the criteria set forth in the question stem. All three have cells that are in direct contact with the external environment or elements of the external environment. That means we can eliminate answer choices A-C which only list two of the three options. We're sticking with the best answer here: Answer choice D.

In Experiment 2, the increased blood pressure resulting from the higher-than-normal concentration of ADH most likely affected the urinary output of Substance A by increasing the: A glomerular filtration rate. B Tm of solutes. C water reabsorption from the tubules. D concentrating ability of the loop of Henle.

The best answer is that increased blood pressure will affect the glomerular filtration rate, answer choice A. Tm is a characteristic that depends on the characteristics of the cells lining the renal tubules and independent of blood pressure, so answer choice B is not correct. Water resorption and concentrating ability are the same, so answer choices C and D are essentially the same. Increasing blood pressure should increase flow of fluid through the kidney system and decrease, rather than increase, water reabsorption, so these answer choices are incorrect.

" Figure 1 Effect of acetylcholine on tension in rabbit aorta with and without endotheliumNE = norepinephrine added; ACH = acetylcholine; at washout, NE and ACH are removed " The concentration range within which muscle tension is most sensitive to acetylcholine (Figure 1) is: A less than 10-8 M. B near 10-7 M. C greater than 10-6 M. D much wider in the ring without endothelium than in the ring with endothelium.

The concentration range within which muscle tension is most sensitive to acetylcholine (Figure 1) is: Jack Westin Advanced Solution: We need to reference Figure 1 once again, and this is actually something we've touched on while going through our previous questions.One more time we have Figure 1 from the passage. We want to point out the range where muscle tension is most sensitive to ACH. We've discussed this at length already. The ring without endothelium isn't sensitive to ACH in the concentrations given in our figure. The ring with endothelium is most sensitive to ACH at a concentration of 10^-7 Molar. Notice the steep slope after the 10^-7 Molar ACH is added. We want our answer to be closest to that number. A less than 10-8 M. Jack Westin Advanced Solution: Tension actually increased after the addition of this ACH. B near 10-7 M. Jack Westin Advanced Solution: This matches our initial prediction, but let's keep going through the other answers to be thorough. C greater than 10-6 M. Jack Westin Advanced Solution: Tension was already decreased significantly when this ACH was added. We're sticking with answer choice B. D much wider in the ring without endothelium than in the ring with endothelium. Jack Westin Advanced Solution: The ring without endothelium was not sensitive to ACH at all. Correct answer here is still answer choice B.

In eukaryotic cells, the process of incorporating uridine nucleotides into nucleic acid polymers occurs in which of the following structures of the cell? _ A Nucleus B Lysosome C Ribosome D Golgi body

The question asks the examinee to identify the cellular structure in eukaryotic cells in which the process of incorporating uridine nucleotides into nucleic acid polymers occurs. Uridine nucleotides are incorporated into RNA, and the question is therefore asking where transcription occurs. In a eukaryotic cell, transcription occurs in the nucleus (A). Thus, A is the best answer.

If oligonucleotides such as mRNA were not degraded rapidly by intracellular agents, which of the following processes would be most affected? A The production of tRNA in the nucleus B The coordination of cell differentiation during development C The diffusion of respiratory gases across the cell membrane D The replication of DNA in the nucleus

The question asks the examinee to identify the process most likely to be affected if oligonucleotides, such as mRNA, were not degraded rapidly by intracellular agents. The destruction of mRNA prevents continuous protein production, allowing the cell to change its protein expression over time. B is the best answer because the coordination of cell differentiation during development is extremely sensitive to the timing of mRNA turnover. A and D are not the best answers because the exact timing of mRNA turnover is less critical to the successful completion of tRNA production (A) and DNA replication (D). C is incorrect because it is unlikely that an accumulation of mRNA would affect the diffusion of respiratory gases across the cell membrane. Thus, B is the best answer.

The most effective method for producing an increase in the total amount of water lost through the skin during a certain period would be: _ A inhibiting kidney function. B decreasing salt consumption. C increasing water consumption. D raising the environmental temperature.

Water is lost through the skin primarily as a means to keep the body at normal temperatures. Therefore, raising the environmental temperature would cause a person to perspire, releasing water to the environment where it can evaporate and cool down the body. Thus, D is the best answer.

Which of the following changes in flow rate or in solute concentrations would NOT occur if the blood inflow rate were increased, increasing the pressure in the dialysis chamber? A The blood volume reaching the outflow tube per unit time would increase. B The osmotic concentration of proteins in the dialysate fluid would increase. C The osmotic concentration of proteins in the blood outflow would increase or remain unchanged. D The filtration rate across the dialysis membrane would increase.

Which of the following changes in flow rate or in solute concentrations would NOT occur if the blood inflow rate were increased, increasing the pressure in the dialysis chamber? Jack Westin Advanced Solution: Which of the following changes in flow rate or in solute concentrations would NOT occur if the blood inflow rate were increased, increasing the pressure in the dialysis chamber? Verbiage here is key, and we want to be careful. If we have increased blood inflow rate, that increases the pressure in the dialysis chamber. That means there is now more blood and increased pressure. We want to know which of the 4 changes listed as answer choices would NOT occur. We're going to focus on the transport of blood through the hemodialysis units. That means we'll focus on flow rate, solute concentrations, and blood pressure in general.Increased inflow rate means increased pressure in the dialysis chamber. First and foremost, we know increased inflow rate would like also increase the volume of blood traveling through the unit and exiting back to the patient. Next, we expect that increased flow also means the dialysis or filtration rate would also increase. More volume coming in and higher pressure means more filtration taking place. In terms of solute concentrations, we'd still expect diffusion to take place across the semipermeable membrane. As we get into specific situations and attack our 4 answer choices, we have to remember we want an option that will NOT occur. Never let the fact that you misread a question be the reason you get the question wrong. A The blood volume reaching the outflow tube per unit time would increase. Jack Westin Advanced Solution: This first answer choice contradicts what we're looking for. Why? Because we want an option that will not occur. That means answer choice A is a true statement, but an incorrect answer choice. We have greater blood volume inflow, which also means increased blood outflow if we have a properly functions hemodialysis unit. B is the ans The osmotic concentration of proteins in the dialysate fluid would increase. Jack Westin Advanced Solution: This answer choice is assuming proteins can cross the semipermeable membrane, which is untrue. The author says "Protein molecule

Which of the following experiments would best test the hypothesis that urease is necessary for the colonization of the stomach by H. pylori? A Exposing ulcer patients to antibodies to urease B Exposing uninfected animals to a strain of H. pylori that lacks urease C Exposing ulcer patients to radioactive urea D Measuring urease activity in biopsies of ulcers

Which of the following experiments would best test the hypothesis that urease is necessary for the colonization of the stomach by H. pylori? Jack Westin Advanced Solution: Once again, we'll be focusing on causation and not correlation. We ultimately want to find a method to prove urease is necessary. One way we can do this is by trying to colonize the stomach when there is no urease present. If this is possible, then that invalidates the hypothesis. When we're conducting a test like this, we want to start from an uninfected state. A Exposing ulcer patients to antibodies to urease Jack Westin Advanced Solution: In this case, the patients already have ulcers, so we are not effectively testing the colonization. We want to test in an organism that in uninfected to make sure we're testing causation, not simply correlation. B Exposing uninfected animals to a strain of H. pylori that lacks urease Jack Westin Advanced Solution: This answer choice would be able to effectively show whether urease is necessary for the colonization of the stomach by H. pylori. If the H. pylori are able to colonize, then urease is not necessary. We're essentially able to disprove the hypothesis. If the animals remain uninfected, then that goes with the hypothesis presented. This is a superior answer choice to option A. C Exposing ulcer patients to radioactive urea Jack Westin Advanced Solution: We want to test in an organism that in uninfected. Doing this would simply confirm there is H. pylori. Not test whether urease is necessary for colonization. This could be correlation, not causation. Answer choice B remains the best choice. D Measuring urease activity in biopsies of ulcers Jack Westin Advanced Solution: Again, we want to test in an organism that in uninfected. This could once again be correlation, not causation. Just because there is urease activity, does not mean urease is necessary for the colonization of the stomach by H. pylori. Our best answer choice is answer choice B. The best test would be Exposing uninfected animals to a strain of H. pylori that lacks urease.

Which of the following statements about pulmonary function best describes all of the results graphed in Figure 1? A It was less than one-half normal for at least a week. B It declined four days after the onset of the illness. C It showed a rapid decline, followed by slower recovery. D It showed a slow decline, followed by rapid recovery.

Which of the following statements about pulmonary function best describes all of the results graphed in Figure 1? Jack Westin Advanced Solution: Which of the following statements about pulmonary function best describes all of the results graphed in Figure 1? This question is going to come from our passage, so I can pull up Figure 1 from the passage: We have Figure 1 from the passage here. This is a graph that's allowing us to compare the global score of pulmonary function for three different patients. We see a sharp decrease in pulmonary function initially. All of the patients hit a global score of as low as 50. After hitting that low, each patient here seemingly recovers slowly. It takes more time to get back to that normal global score, but each patient eventually gets back. What are our variables? X-axis is days of illness. And as the days go by, we measure the global score of pulmonary function of each patient. Biggest trend is the sharp decrease in global score in the first few days of illness, then a slow climb back up. All of the 3 patients recover and get back to a normal level. A It was less than one-half normal for at least a week. Jack Westin Advanced Solution: One-half normal is a global score between 200 and 225. Each patient is below that threshold for fewer than 7 days total. This answer choice contradicts Figure 1 above. Not a great option to start. B It declined four days after the onset of the illness. Jack Westin Advanced Solution: This answer choice is only true for one patient. Patients 1 and 2 saw a decline earlier than day 4. This answer choice is including all 3 patients, but is only true for patient 3. Neither A nor B are great options so far. Let's keep looking for something better. C It showed a rapid decline, followed by slower recovery. Jack Westin Advanced Solution: This answer choice matches what I said in my breakdown of the question. The big trend we saw was a sharp decrease in global score in the first few days of illness, but there was a much slower recovery that took more days than the initial decline. We can now eliminate answer choices A and B. I said both contradicted Figure 1 from the passage, and now we have a superior answer in answer choice C. D It showed a slow decl

just read: Why did investigators conclude that the new pathogen is a hantavirus? A The experiments showed that it is related to known hantaviruses. B The experiments showed that it infects lung endothelium. C The patient data showed that the type of disease it caused resembles that caused by known hantaviruses. D No pathogenic bacteria were found in the lung samples from patients.

Why did investigators conclude that the new pathogen is a hantavirus? Jack Westin Advanced Solution: Why did investigators conclude that the new pathogen is a hantavirus? This answer is going to come from Experiments 1 and 2. We're going to focus on the results of the two experiments and how they support the conclusion that the new pathogen is a hantavirus. We have the first two experiments here, and we're going to quickly break down the results. Experiment 1: we have a positive immunologic response with a hantavirus. That shows we have similar antigens-we're getting the same immune response.Experiment 2: we have hybridization between the synthesized gene sequences from two known hantaviruses and nucleic acids from the patients' lung tissues. That gives us even more evidence that the pathogen is closely related to the two known hantaviruses in Experiment 2. Using these results, investigators came to their conclusion that the new pathogen is a hantavirus. A The experiments showed that it is related to known hantaviruses. Jack Westin Advanced Solution: This answer choice is almost verbatim what I just mentioned in the breakdown. The results of Experiment 1 and 2 showed the pathogen is related to known hantaviruses so we were able to conclude the new pathogen is a hantavirus as well. B The experiments showed that it infects lung endothelium. Jack Westin Advanced Solution: This was the result of Experiment 3. This is technically a true statement, but this doesn't exactly answer the question being asked here. There are many pathogens that can affect lung endothelium. Just knowing the new pathogen affects lung endothelium isn't enough to conclude the pathogen is a hantavirus. Answer choice A is still superior. C The patient data showed that the type of disease it caused resembles that caused by known hantaviruses. Jack Westin Advanced Solution: This answer choice contradicts the passage. This new pathogen affected the lungs, while the known hantavirus we were introduced to in the passage affected the kidneys, not the lungs. We can eliminate this answer choice and answer choice A still remains our best answer. D No pathogenic bacteria were found in the lung samples from patients. Jack Westin Advanced Solution: This

diurnal rhythm

a pattern of activity or behavior that follows a day-night cycle

According to the passage, the Tm represents the rate of plasma filtration that just exceeds the: A rate of concentration of the substance in the glomerular filtrate. B rate of concentration of the substance in the urine. C capacity of the kidney tubules to reabsorb the substance. D capacity of the bladder to store and excrete the substance. How do you know you are truly ready? Th

c

Which of the following characteristics clearly marks fungi as eukaryotes? _ A They have cell walls. B They contain ribosomes. C They contain mitochondria. D They exhibit sexual reproduction.

fungi is asexual rep and Mito is a membrane bound organelle so ans is C

In almost all vertebrates, when the optic cup fails to develop in the embryo, the lens also fails to form. This constitutes evidence that: _ A the process of neurulation follows gastrulation. B the eye develops early in vertebrate morphogenesis. C cells may induce neighboring cells to differentiate. D cell differentiation is an "all or none" phenomenon.

olution: The correct answer is C. The optic cup develops from a bulge on the side of the developing brain, which influences the overlying ectoderm to produce the lens. It is therefore an example of cells inducing neighboring cells to differentiate, so option C is the correct answer. The other response choices are irrelevant to the question, so they are not good answers. The absence of the optic cup and lens has no influence on timing of neurulation relative to gastrulation (choice A). The presence or absence of the optic cup and lens has no effect on the timing of eye development (choice B). Cell differentiation is not an "all or none" phenomenon (choice D). In most cases, cells progressively differentiate to achieve their final specification.