AAMC Bio Question Review

If chromosomal duplication before tetrad formation occurred twice during spermatogenesis, while the other steps of meiosis proceeded normally, which of the following would result from a single spermatocyte? A. One tetraploid sperm B. Four diploid sperm C. Four haploid sperm D. Eight haploid sperm

**** Read Q Carefully. It stated that the chromosome where duplicated. So you have to double the amount of cells that are produced. **** Answer Key: B The question asks the examinee to predict the result if chromosomal duplication occurred twice before tetrad formation and all other steps in spermatogenesis occurred normally. If all other steps occur normally, then four sperm would be produced. Because replication occurred twice instead of once prior to tetrad formation, each sperm would have twice the normal amount of DNA. As a result, four diploid sperm would be produced instead of four haploid sperm. A, C, and D are incorrect because they describe results that contain an incorrect number of sperm or ploidy number for the sperm produced. Thus, B is the best answer.

Rates of endocytosis vary from cell type to cell type. What cell would be predicted to have the highest rate of endocytosis? A. A macrophage B. An erythrocyte C. An osteoblast D. A neuron

Answer Key: A Endocytosis is the process by which cells engulf extracellular material. Macrophages specialize in this activity. They clear cell and tissue debris and foreign objects from the body, so option A is the correct answer. Erythrocytes specialize in transport of oxygen and do not engage in exocytosis or endocytosis, so option B is incorrect. Osteoblasts form bone tissue. They actively secrete bone matrix, so would not engage in much endocytosis, therefore, answer choice C is incorrect. Neurons secrete neurotransmitter by exocytosis, but do not engage in endocytosis, so answer choice D is also incorrect.

A wound that penetrates the rib cage and lets air into the right pleural cavity stops air flow into the right lung because the: A. lung cannot be expanded. B. rib cage cannot be expanded. C. diaphragm cannot be lowered. D. air dries and stiffens the lung.

Answer Key: A Non-forced inspiration occurs when the diaphragm and muscles between the ribs (intercostal muscles) contract. The ribs attached at the sternum and spine lift in what is called a "bucket-handle" motion and the diaphragm flattens. These motions increase the volume of the thoracic cavity. Between the lung and the thoracic wall is a space, the pleural cavity. As the volume of the thoracic cavity increases, atmospheric pressure presses against the inside of the flexible, elastic lung tissue forcing it to conform to the thoracic wall. Air thus enters the lung as it expands with the thorax. If the thoracic wall is pierced (causing what is known as a sucking chest wound), air enters the pleural cavity. The air pressure inside and outside the lung balance causing the lung to elastically collapse and preventing air intake. The rib cage can still expand (so choice B is incorrect), the diaphragm can still contract (so choice C is incorrect), but the lung cannot be expanded in response, so the correct response is A. Choice D is incorrect because the effect is immediate and it would take some time for the lung to dry and stiffen, if indeed that would happen at all.

In the human "knee-jerk" reflex, the knee is struck and the lower leg jerks forward. Which of the following represents the complete pathway that the nerve impulse travels in effecting this response? A. Sensory neuron, motor neuron B. Sensory neuron, brain, motor neuron C. Sensory neuron, associative neuron, brain, associative neuron, motor neuron D. Sensory neuron, associative neuron, motor neuron, associative neuron, motor neuron

Answer Key: A The knee-jerk reflex is a simple monosynaptic stretch reflex. A tap to the tendon that connects the quadriceps to the patella activates a sensory neuron that directly synapses with a motor neuron in the spinal cord, causing the quadriceps to contract. Thus, A is the best answer.

The normal path of sperm movement from the male testis to the point of fertilization in the female is: A. epididymis, vas deferens, urethra, vagina, cervix, uterus, fallopian tube. B. epididymis, vas deferens, ureter, cervix, uterus, fallopian tube. C. epididymis, vas deferens, urethra, vagina, uterus, ovary. D. interstitial cells, epididymis, vas deferens, vagina, uterus, ovary.

Answer Key: A The question asks the examinee to identify the correct order of structures through which sperm cells travel from the male testis to the point of fertilization in the female. As sperm cells leave the testis they travel through the epididymis to the vas deferens and into the urethra. The sperm then enter the female's vagina, travel through the cervix and uterus, and enter the fallopian tube, where fertilization most commonly takes place. Of the options listed, only A correctly identifies both the structures and the order. Thus, A is the best answer.

Which of the following procedures would be LEAST likely to prevent bacterial synthesis of the superantigen protein? A. Adding tRNA nucleotides that can bind to mRNA and bacterial ribosomes B. Adding a repressor protein that binds to the operator site of the bacterial superantigen gene C. Adding a specific complementary nucleic acid sequence that can bind to mRNA transcribed from the superantigen gene D. Adding a stop codon within the bacterial superantigen gene

Answer Key: A The stem asks students to identify the procedure that would be LEAST likely to prevent synthesis of the superantigen. Foils B (binding of the operator), C (binding of the mRNA), and D (insertion of a stop codon) would all be effective ways to prevent synthesis of the superantigen. Adding tRNA nucleotides would be unlikely to prevent synthesis of the antigen. Therefore, the key is A.

An organism is likely to be a bacterium rather than a virus if it: A. reproduces via fission. B. has a rigid outer covering. C. lacks a nuclear membrane. D. contains both RNA and protein.

Answer Key: A Viruses reproduce by replicating within a host cell and thus cannot be said to reproduce by fission, as bacteria do. Many bacteria do have rigid cell walls, but many viruses have rigid capsids, so the presence of a rigid outer covering does not distinguish the two. Neither bacteria nor viruses have a nuclear membrane and both bacteria and viruses can contain RNA and protein. Thus, the best distinction between a bacterium and a virus is reproduction by fission, answer choice A.

Suppose that an extract from a muscle cell contains only the following: all the enzymes of the glycolytic pathway, including the enzyme that converts pyruvate to lactate; phosphate and other salts; NAD+ and ADP. When the extract is incubated anaerobically and glucose is introduced, neither pyruvate nor lactate is produced. What must be added in order for pyruvate to be made? A. O2 B. ATP C. NADH D. Acetyl-coenzyme A

Answer Key: B ATP must be added. The overall glycolytic pathway yields two ATP molecules, but in order to gain these two ATP molecules by the end of the process, some ATP must be consumed at the beginning. The first reaction, the phosphorylation of glucose, for instance, requires ATP. If this is lacking in the reaction mixture, glycolysis will not be initiated. Once initiated with a small amount of ATP, however, more ATP is produced and the reaction will become self-sufficient. Therefore, the correct answer is option B. Oxygen and acetyl-coenzyme A are needed only for oxidative metabolism, so options A and D are incorrect. NAD is supplied and the process will produce NADH, but NADH is not needed for the reaction sequence to proceed, so option C is incorrect.

Which of the following structures is derived from the same germ cell layer as the heart? A. Eye B. Bone C. Spinal cord D. Liver

Answer Key: B In gastrulation, the three cell regions, or germ layers, are formed in the embryo. From the outer layer (ectoderm) come cells of the nervous system and epidermis, from the inner layer (endoderm) come cells of the lining of the digestive tube and associated organs, and from the middle layer (mesoderm) come the blood cells, connective tissues (bones, muscles, and tendons), and several organs (kidney, heart, gonads). Thus, the eye and spinal cord are formed from ectoderm, the liver is formed from endoderm, and the heart and bone are formed from mesoderm. Answer choice B is the best answer.

Actin filaments within cells can be identified experimentally by the use of a labeled molecule that binds specifically to actin and not to other cell substances. Which of the following would be best to use as the labeled molecule? A. ATP B. Myosin C. Albumin D. Myoglobin

Answer Key: B Myosin binds to actin during contraction and will remain bound until phosphorylated by ATP. Actin filaments can be "decorated" with portions of the myosin molecule. Although ATP provides energy for muscle contraction, it provides energy for many other cell activities as well and would therefore be nonspecific. In addition, ATP binds to myosin, not to actin, in the contraction cycle. Albumin is a protein found in blood plasma. Myoglobin is the oxygen-binding pigment in muscle. Neither of these molecules is likely to have any specific affinity for actin. Thus, answer choice B is the best answer.

Which of the following hormones is LEAST directly regulated by the anterior pituitary? A. Cortisone B. Epinephrine C. Progesterone D. Thyroxin

Answer Key: B The correct answer is option B, epinephrine. Epinephrine is secreted by the adrenal medulla and its secretion is controlled by the autonomic nervous system. Secretion of the other three hormones is regulated by secretion of anterior pituitary hormones. Cortisone secretion is regulated by adrenocorticotropic hormone (ACTH), progesterone by luteinizing hormone (LH), and thyroxin by thyroid stimulating hormone (TSH).

Synthesis of antibody proteins in eukaryotic cells is associated with what organelle? A. Nucleus B. Mitochondrion C. Endoplasmic reticulum D. Golgi apparatus

Answer Key: C As secreted proteins, antibodies are translated by ribosomes attached to the rough endoplasmic reticulum. Thus, C is the best answer.

The mineral component of human bone is a salt that consists primarily of all of the following EXCEPT: A. calcium. B. phosphate. C. potassium. D. hydroxyl groups.

Answer Key: C The inorganic component of bone consists of submicroscopic deposits of calcium phosphate similar to hydroxyapatite (Ca10[PO4]6[OH]2), so it would be expected to contain calcium, phosphate and hydroxyl groups. Therefore, the correct answer states that the component not found in the salt is potassium, answer choice C.

When muscles in the skin contract and cause the hair of an animal to "stand on end," the skin could be functioning as a regulator of: A. pH. B. salt excretion. C. body temperature. D. skeletal muscle tone.

Answer Key: C The question asks the examinee to identify a likely regulatory function of an animal's skin when its muscles contract, causing the animal's hair to stand on end. A major function of hair is to help mammals regulate body temperature by reducing the amount of heat lost to the environment. Hair provides insulation. When the hair stands on end, trapping air within the layer of hair, then its insulating effect becomes even more effective. Thus, as C indicates, a primary function of this action is to help regulate body temperature. The contraction of skin muscles that causes the hair to stand on end would have little effect on skin pH (A), salt excretion (B), or skeletal muscle tone (D). Thus, C is the best answer.

A researcher has a short polynucleotide strand with the following base sequence: AUCCCUGG. This strand must be: A. template DNA. B. DNA. C. RNA. D. a peptide.

Answer Key: C The question shows a short polynucleotide containing A, U, C, and G. Given that it is a polynucleotide containing uracil (U), one can conclude that it is RNA. Therefore, the key is C.

Translation of antibody proteins in eukaryotic cells is associated with what organelle? A. Nucleus B. Mitochondrion C. Endoplasmic reticulum D. Golgi apparatus

Answer Key: C The stem asks which of the organelles listed is associated with the translation of antibody proteins. Of the organelles listed, only the endoplasmic reticulum is associated with translation of proteins. Synthesis is accomplished by the ribosomes, which are associated with the endoplasmic reticulum. The nucleus contains the genetic information, the mitochondria are involved in cellular respiration, and the Golgi apparatus is involved in the processing and/or export of proteins. Therefore, the key is C.

During prokaryotic protein synthesis, translation begins as soon as the newly synthesized mRNA strand begins to extend from the DNA strand. This situation differs from that in eukaryotes, because eukaryotes: A. carry out translation without using ribosomes. B. transcribe mRNA molecules without using DNA. C. destroy most mRNA as soon as it is synthesized. D. localize the processes of transcription and translation in the nucleus and cytoplasm, respectively.

Answer Key: D Eukaryotes have membrane-bound organelles including the nucleus, which contains the DNA. Transcription of DNA into RNA occurs in the nucleus. The RNA is then transported to the cytoplasm where ribosomes translate it into proteins. Thus, D is the best answer.

After the gall bladder is removed from a patient, the patient will most likely have reduced ability to digest: A. protein. B. starch. C. sugar. D. fat.

Answer Key: D The gall bladder stores bile produced by the liver and secretes it into the small intestine as needed. Bile acts as an emulsifier, facilitating fat digestion. When the gall bladder is removed, a patient will have reduced ability to digest fats.

Albumin is the major blood osmoregulatory protein. The most likely effect of a sharp rise in the level of serum albumin is: A. a drop in blood pressure. B. an increase in immune response. C. an efflux of albumin into the interstitial fluids. D. an influx of interstitial fluid into the bloodstream.

Answer Key: D The question asks the examinee to identify the most likely effect of a sharp rise in the level of serum albumen, a major blood osmoregulatory protein. A sharp rise in osmotically active albumin in the serum would increase the flow of interstitial fluid into the bloodstream and result in an increase in blood pressure, not a decrease. Therefore, A is incorrect, and D is correct. B is incorrect because the immune response does not depend on the level of serum albumen. C is incorrect because albumen would not normally pass through capillary walls. Thus, D is the best answer.

HIV (human immunodeficiency virus) is a retrovirus, an RNA virus that can insert itself into the human genome. This virus can reproduce in host cells because it contains: A. enzymes that destroy T cells. B. viral DNA that is compatible with human DNA. C. core proteins rather than DNA. D. reverse transcriptase.

Answer Key: D The question states that HIV is a retrovirus. For the virus to reproduce itself in host cells, its RNA must be converted to DNA, which requires the enzyme reverse transcriptase. Therefore, the key is D.

Passage::: Calcitonin, which decreases bone resorption; however, its effect is small (more like fine-tuning). Under what condition would the level of calcitonin tend to increase? A. When there is a dietary deficiency of calcium B. When there is a dietary deficiency of vitamin D C. When the level of calcium in the plasma is high D. When the level of parathyroid hormone is too low

NOTE::: Bone resorption is resorption of bone tissue, that is, the process by which osteoclasts break down the tissue in bones and release the minerals Solution: The correct answer is C. Calcitonin reduces bone resorption. Bone resorption occurs when the level of calcium in the blood plasma is low, but resorption is not needed when the level of calcium is high. Therefore, resorption would be reduced by calcitonin under conditions in which the level of calcium in the plasma is high. Thus, C is the best answer.

Passage::: Scientists have hypothesized that mitochondria evolved from aerobic heterotrophic bacteria that entered and established symbiotic relationships with primitive eukaryotic anaerobes. Many structural and functional similarities between mitochondria and present-day bacteria support this hypothesis. They are approximately the same size, reproduce by similar means, and contain non-histone-bound DNA. They contain the tRNAs, ribosomes, etc., necessary for transcription and translation, and they show some similarities in base sequences of rRNAs. In addition, the inner membranes of mitochondria have enzymes and transport systems similar to those on the plasma membranes of bacteria. One similar system is the electron transport system (ETS). Electron transport in both mitochondria and bacteria is accomplished using three large protein complexes, each composed of multiple polypeptides (Figure 1). According to the hypothesis described in the passage, the bacteria that entered primitive eukaryotic cells were able to carry out which of the following functions that the primitive eukaryotic cells could NOT? A. Glycolysis B. Krebs cycle and electron transport C. Cell division D. Transcription and translation

Primitive::: relating to, denoting, or preserving the character of an early stage in the evolutionary or historical development of something. Solution: The correct answer is B. The primitive eukaryotic cells are described in the passage as anaerobes, so it is likely that they could breakdown sugars by glycolysis. That means answer choice A is incorrect. The bacteria are described as aerobic, so the passage suggests that the ability to use oxygen was acquired with the bacteria. Electron transport in mitochondria and bacteria is described and the electron transport system is needed for eukaryotes to use oxygen. The acquisition of aerobic bacteria as partners would have provided them with the ability to carry out aerobic metabolism through use of the Krebs cycle and electron transport. Answer choice B is therefore the correct choice. The anaerobic, eukaryote-precursor cells must have engaged in cell division and the processes of transcription and translation, so answer choices C and D are incorrect.

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.

Solution: The correct answer is A. 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.

The initial increase in heart and breathing rates during the skin diving trip was probably a result of: A. activation of the sympathetic autonomic nervous system by the new experience. B. activation of the parasympathetic autonomic nervous system by the new experience. C. hypoxia caused by the inability of her blood hemoglobin concentration to supply sufficient oxygen for the strenuous exercise of swimming at sea level. D. elevated core body temperature caused by swimming in warm tropical waters.

Solution: The correct answer is A. 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. The most likely explanation for her body's response was the activation of her sympathetic autonomic nervous system—the "fight or flight" response caused by adrenaline.

Which of the following statements explains most plausibly why host antibodies are ineffective against H. pylori? A. Antibody proteins may be denatured in the harsh environment of the stomach. B. Antibodies are not generally effective against bacteria. C. H. pylori infection may suppress the activity of the immune system. D. Antibodies are not secreted from host tissues into extracellular spaces.

Solution: The correct answer is A. According to the passage, stomach ulcers and some forms of gastric cancer may be linked to H. pylori infections in the stomach. Without treatment by antibiotics, such infections can be persistent. The most plausible explanation for why host antibodies are ineffective against H. pylori is that antibody proteins may be denatured (destroyed) in the harsh (acidic) environment of the stomach. If the antibody proteins are denatured, they will not function properly.

URL for image::: https://s3.amazonaws.com/wmx-api-production/courses/1659/images/BS-11-1-PT4-173.gif The above diagram represents the neural pathway that causes an individual to retract a stubbed toe. If one were to modify this diagram to represent the pathway involved in feeling pain in the stubbed toe, where could additional neurons be placed? A. At II and III B. At II and IV C. At III and IV D. At I and IV

Solution: The correct answer is A. Additional neurons would be placed at synapses II and III, which are located in the central nervous system. It is from these sites that ascending tracts to the brain, where pain is perceived, originate. The efferent nerve terminal, represented by IV, is located in the muscle and is a motor, not a sensory, site. Impulses arrive at IV from the central nervous system. Although the afferent leg of pain information (diagrammed as I) is part of the pain-perception pathway, additional pathways would be added in the spinal cord. Without such ascending pathways, a reflex could occur, but the sensation of the pain would be absent. Occasionally these ascending tracts are cut as a means of alleviating pain. Thus, answer choice A is the best answer. ******* It is A. Reflex arcs are simply sensory--->interneuron>Motor neuron. To cognitively process the sensation of pain of the stubbed toe, you could add more neuron's between 2 and 3, which could serve as interneurons to other neural pathways.

Would an increase in the level of plasma aldosterone be expected to follow ingestion of excessive quantities of NaCl? A. No; aldosterone causes Na+ reabsorption by kidney tubules. B. No; aldosterone causes Na+ secretion by kidney tubules. C. Yes; aldosterone causes Na+ reabsorption by kidney tubules. D. Yes; aldosterone causes Na+ secretion by kidney tubules.

Solution: The correct answer is A. Aldosterone, which is produced by the adrenal cortex, causes Na+ reabsorption by kidney tubules. Such a mechanism decreases Na+ levels in the urine. The steroid aldosterone does not cause Na+ secretion into the urine. Because ingestion of excessive NaCl would trigger Na+secretion into the urine, plasma-aldosterone levels would not increase. Rather, the body would rely on those homeostatic mechanisms that excreted the excess Na+. Thus an increase in plasma aldosterone would not be expected to follow ingestion of large quantities of NaCl. The best answer choice is A. *************** BETTER EXPLANATION:::: You're correct that aldosterone will increase the reuptake of Na+ and reduce Na+ excretion, causing blood pressure to increase. However, when you intake excessive NaCl, you will have high blood pressure already (increased blood osmolarity). Thus, aldosterone concentrations would decrease to oppose the increase in blood pressure that is caused by excessive NaCl. ALSO_________ Yeah aldosterone on top of blood that already has Na in it will just make the body's blood pressure raise even more which is not good lol

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, Ais the best answer. ****** Natural Selection::: the process whereby organisms better adapted to their environment tend to survive and produce more offspring.

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.

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. ***** Hyperglycemia refers to high levels of sugar, or glucose, in the blood. It occurs when the body does not produce or use enough insulin, which is a hormone that absorbs glucose into cells for use as energy. High blood sugar is a leading indicator of diabetes.

An extra S phase occurs during amitotic division in a small macronucleus to minimize fluctuations in DNA content. This is most likely triggered by the presence of: A. low concentrations of DNA in the macronucleus. B. centromeres in the macronucleus. C. high concentrations of DNA in the micronucleus. D. mitotic enzymes in the micronucleus.

Solution: The correct answer is A. In answering this question it helps to know what mitosis is all about. Having worked hard to memorize prophase, metaphase, anaphase, telophase and interphase students still may not be too clear on the function mitosis serves for the dividing cell. In metaphase the replicated chromosomes line up on the metaphase plate and divide, one copy moving to each daughter cell. The elaborate processes involved in mitosis ensure that both daughter cell gets exactly the same chromosomes. The passage suggests that there might be some alternative to mitosis: amitotic division. When building a new macronucleus Tetrahymena does not bother with mitosis which would equally divide the chromosomes. In stead it makes an amitotic, approximate division into two uneven nuclei. An extra S (synthesis) phase results in an additional doubling of the DNA. The extra S phase ensures each nucleus contains at least one copy of each vital gene. This would minimize fluctuations only if DNA levels were low. The passage states that the extra S phase occurs only in small macronuclei. For the extra S phase to be advantageous to the organism it must, in some way, be responsive to low DNA levels, so A is the correct answer. Choice C is incorrect because it suggests the strange hypothesis that high concentrations of DNA in the micronucleus would trigger the S phase in the macronucleus. Choices B and D are incorrect because they suggest that the presence of components involved in mitosis (the centriole, B, and mitotic enzymes, D) might trigger amitotic division in the macronucleus.

What would be the result of complete removal of the parathyroid glands? A. Severe neural and muscular problems due to deficiency of calcium in the plasma B. An increase in calcitonin production to compensate for calcium deficiency in the plasma C. A drastic change in the ratio of mineral to matrix tissue in bones D. Calcification of some organs due to accumulation of calcium in the plasma

Solution: The correct answer is A. Removal of the parathyroid gland would lead to hypocalcemia, a condition of low blood calcium, resulting from the lack of parathyroid hormone. This would cause increased neuromuscular excitability because of the change in membrane potential, which under normal physiological conditions, is partially kept in balance with extracellular calcium. Typically, the person would eventually die from severe respiratory muscle spasms. Thus, A is the best answer.

Passage: Plasma clearance is affected by the tubular transport maximum (Tm) of a substance. The Tm is the maximum rate of transport (mg/min) at which a substance can be reabsorbed by the kidney. That is, if the filtration rate of a substance exceeds its Tm, the substance will begin to appear in the urine. The Tm for glucose averages 320 mg/min in an adult human. Under normal conditions, the tubular load of glucose (the amount/min that filters into the kidney tubules) is approximately 125 mg/min. The amount of glucose in the urine under these conditions is approximately: A. 0 mg/min. B. 125 mg/min. C. 195 mg/min. D. 515 mg/min.

Solution: The correct answer is A. The amount of glucose in the fluid in the capsular space as it enters the tubule system would be the same as the amount of glucose in the plasma, but by the time the fluid reaches the distal tubule much of the glucose will have been reabsorbed. If the amount being filtered per minute is less than Tm, all of it will have been reabsorbed. If it is more than Tm, then the amount will be linearly related to the amount in the plasma. In the case proposed in this question, the amount being filtered per minute (125mg/min) is less than the Tm for glucose given in the passage (320 mg/min), so all of the glucose should be reabsorbed. The correct answer, therefore, is 0 mg/min, answer choice A.

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.

Solution: The correct answer is A. 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. ****** So increased blood pressure would result in increased glomerular filtration rate so the body can lower its blood pressure hence why increased urine output. ------ The glomerular filtration rate is directly proportional to the pressure gradient in the glomerulus, so changes in pressure will change GFR. GFR is also an indicator of urine production, increased GFR will increase urineproduction, and vice versa. ALSO*******a decrease in blood pressure, the nephron increases sodium andwater reabsorption, to increase ECFV and thereby increase blood pressure.

The persistence of a pulse for several hours after the onset of ice formation (Figure 1) is crucial to the cryoprotective role of glucose because: A. circulating blood distributes the glucose throughout the body tissues. B. circulating blood equilibrates the temperature throughout the body. C. a beating heart warms body tissues and slows ice formation. D. a beating heart requires a constant supply of glucose as an energy source.

Solution: The correct answer is A. The cryoprotective role of glucose depends upon the persistence of a pulse for several hours after the onset of ice formation for distribution of the glucose throughout the body tissues by the circulatory system. Although blood flow conducts heat between internal parts of the body and the skin efficiently, temperature equilibration is not the reason that the pulse contributes to the cryoprotective capability of glucose. Nor is the heat generated by a beating heart the critical factor in the cryoprotective role of glucose. That a beating heart does require a constant energy source is true, but this does not explain the relationship between pulse persistence and the cryoprotective role of glucose. Thus, answer choice A is the best answer. *****A cryoprotectant is a substance used to protect biological tissue from freezing damage

A virgin female Drosophila mates and produces 34 daughters and 38 sons. Eighteen of these sons sire only daughters, while the remainder sire approximately equal numbers of daughters and sons. What are the genotypes of the original female and the male with whom she mated? A. XiXs and XsY B. XiXs and XiY C. XiXi and XsY D. XsXs and XiY

Solution: The correct answer is A. The male parent must have been XsY or there would only have been daughters in the offspring of the first generation, so options B and D are incorrect. If the female had been XiXi then all its sons would have had only daughters, but twenty of the sons have both sons and daughters, so option C cannot be correct. The correct answer is option A, and the pedigree of that mating is shown below in Figure 2. https://s3.amazonaws.com/wmx-api-production/courses/1659/images/BS-11-1-PT6-200-S.gif Figure 2:: Pedigree of the cross that would yield the offspring described in the question.

In 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.

Passage::: An XsY male is standard: he sires equal numbers of sons and daughters. An XiY male expresses the sex ratio trait: he sires only daughters. Total reproductive output is not affected; XsY males and XiY males sire equal numbers of offspring. If none of the Xi-bearing genotypes (XiY, XiXi, or XiXs) is selected against, then the frequency of Xi is expected to increase to 100%, unless other genes act to suppress expression of e and f. Occasionally, XiY males sire viable but sterile sons of normal appearance. Genetic analyses show that all these sons are XO, inheriting their X chromosome from their mother and lacking a Y chromosome. If all genotypes are equally fit and if there are no genetic modifiers of the sex ratio trait, what will be the ultimate fate of a population in which 50% of the X chromosomes are currently Xi and 50% are Xs? A. Extinction B. Stable population size, with a predominance of females C. Stable population size, with all individuals producing a 50:50 sex ratio D. Stable population size, with some individuals producing an excess of females and some producing an excess of males

Solution: The correct answer is A. The passage states that if none of the Xi genotypes are selected against, then the Xi chromosome will increase to 100%. If all the males are XiY, then only females will be produced and the population will become extinct. Answer A is correct. The population will not be stable, but will increasingly become female, so options B and C are incorrect. Option D is also incorrect, not only because it assumes a stable population size, but also because it supposes that some individuals will produce an excess of males. -------- A stationary population is a special example of a stable population with a zero growth rate, neither growing nor shrinking in size, and is equivalent to a life tablepopulation. ... By definition, stable populations have age-specific fertility and mortality rates that remain constant over time. ********* "If none of the Xi-bearing genotypes (XiY, XiXi, or XiXs) is selected against, then the frequency of Xi is expected to increase to 100%," This was tricky, but that line pretty much sells it for me. If Xi is going to persist you are going to run out of viable males since XiY only produces viable females.

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.

Solution: The correct answer is A. The question asks the examinee to identify a characteristic common to most bacterial and human cells. Of the options listed, only A, the ability to produce ATP via ATP synthase is common to both bacterial and human cells. Both types of cells possess a membrane-embedded electron transport chain capable of generating a H+ gradient, which drives synthesis of ATP via ATP synthase. This ATP synthesis takes place on the plasma membrane of bacteria and on the inner mitochondrial membrane in human cells. The chemical composition of prokaryotic and eukaryotic ribosomes (B), although similar, is distinct enough that several types of antibiotics are able to preferentially target prokaryotic ribosomes over eukaryotic ribosomes. Of human and bacterial cells, only bacterial cells have cell walls (C), and most bacterial chromosomes are circular whereas human chromosomes are linear (D). Thus, A is the best answer. ****** ATP synthase is an important enzyme that provides energy for the cell to use through the synthesis of adenosine triphosphate (ATP).

The one aspect of ectopic pregnancy common to all the causes described in the passage is that the zygote fails to: A. implant in the uterus. B. leave the ovary. C. reach the fallopian tube. D. begin its development.

Solution: The correct answer is A. The question asks the examinee to identify the aspect common to all ectopic pregnancies, regardless of cause. In all pregnancies, normal or ectopic, the egg leaves the ovary (B), reaches the fallopian tube (C), is fertilized (C), and begins its development. In ectopic pregnancies, however, the embryo is not effectively transported to the uterus and it implants in the fallopian tubes, rather than in the uterus (A). Thus, A is the best answer.

Muscles with striated fibers are the primary muscle type in: A. the heart. B. the uterus. C. arteries and veins. D. the small intestine.

Solution: The correct answer is A. The question asks the examinee to identify the body organ or structure from the options listed in which the primary muscle type contains striated muscle fibers. Skeletal and cardiac muscles contain striated muscle fibers. Smooth muscles do not. Of the options listed, only A, the heart, is made of cardiac muscle and therefore has striated muscle fibers. The primary muscle type of the uterus (B), arteries and veins (C), and the small intestine (D) is smooth muscle. Thus, A is the best answer.

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

Solution: The correct answer is A. 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.

Passage::: Set Point Hypothesis The brain regulates body weight just as a thermostat maintains a constant room temperature. The brain adjusts metabolism and behavior to maintain a predetermined body weight. Genes also influence the set point, which can increase with age—but only to the extent dictated by inheritance. Diet and exercise cannot reset the set point over the long term. Settling Point Hypothesis:: Body weight is determined by the interaction of two factors—metabolism and genes—with the environment. Depending on genotype, various metabolic feedback loops may allow weight to be stabilized at a new level. Thus, in an environment where high-calorie food is plentiful, individuals with a genetic predisposition to obesity will tend to become more overweight than those without such a predisposition. What do both the set point hypothesis and the settling point hypothesis seek to explain? A. How multiple, interacting factors determine body weight B. How individual factors acting alone influence body weight C. How metabolism and the environment influence body weight D. How the environment and behavior influence body weight

Solution: The correct answer is A. The question asks the examinee to identify the characteristic that both the set point and the settling point hypotheses try to explain. A is correct because both hypotheses describe multiple factors that regulate body weight. B is incorrect because both hypotheses list multiple interacting factors that influence body weight, not individual factors acting independently. C is incorrect because only the settling point hypothesis seeks to explain the effects of metabolism and environment together. D is not the best answer because both hypotheses discuss the importance of factors in addition to environment and behavior on body weight. Thus, 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.

Solution: The correct answer is A. The question asks the examinee to identify the characteristic that differentiates growing, developing long bones from adult bones. Long bones grow via endochondral ossification, which requires cartilaginous growth plates at the ends of long bones, that thicken as cartilage and later become ossified (A). Dividing bone cells (B) and haversion canals (C) can be present in fully ossified adult bones. Some adults who are short in stature may have fully ossified long bones that are shorter than those of a developing child (D). Thus, A is the best answer.

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.

Passage: Excessive sweating may upset homeostasis by impairing water and salt regulation. During dehydration, the kidneys may reduce their urinary output from the normal level of 1.0-1.5 L H2O/day to as little as 0.5 L H2O/day, and renal salt excretion may decline to near zero. Excessive sweating also reduces the volume of blood available for delivering O2 to the internal body tissues. failure during severe dehydration is most likely due to: A. inadequate blood volume for effective filtration. B. inability to produce sufficient urine. C. buildup of salts in the distal tubules. D. increased body temperature.

Solution: The correct answer is A. This question asks the examinee to identify the cause of kidney failure during times of extreme dehydration. As the passage states, severe dehydration greatly reduces the volume of filtrate moving through the nephrons of the kidney. If fluid volume is too drastically reduced, the kidney will be unable to effectively do its job of filtering and maintaining homeostasis within bodily fluids. Thus, A is the best answer. ******** The reason why the answer is not C is because the passage specifically mentions that during bouts of dehydration, the kidneys release 'near zero' salts. With these questions. remember that the passage is key. It holds all the information required for questions like this other than knowing what is the pushing force of filtration in the kidneys. Blood pressure. At first sight of the question after a nice thorough reading of the passage, a light bulb should have went off and said, "Hmm, choice A seems very reasonable and can very well be the answer. B seems plausible but not instantly wrong. C is instantly wrong from the passage. D is minimally plausible and can be eliminated. Between A and B, A seems the most likely choice due to passage information and minor nuances in the logic AAMC uses in their questions

When erythrocytes are placed in distilled water, the volume of each erythrocyte increases because the: A. gradient of ions causes water to enter the cells. B. contractile filaments of the cytosol open pores in the plasma membrane. C. sodium pump transports sodium out of the erythrocytes more rapidly than normal. D. erythrocyte's DNA produces degradative enzymes.

Solution: The correct answer is A. This question asks the examinee to identify why the volume of an erythrocyte increases when it is placed in distilled water. The volume of the erythrocyte does not increase because pores open in the plasma membrane (B). As the passage states, it is the pressure that builds up within the interior of the cell, as a result of the entry of water, that causes pores to form in the membrane of the cell. The volume of the erythrocyte does not increase because the Na+pump extrudes sodium in response to the cell's placement in distilled water (C). If this were to occur, the extrusion of sodium would reduce the osmotic pressure that exists within the cell, which would reduce the volume of water that enters the cell. The volume of the erythrocyte does not increase because of the production of degradative enzymes (D). If degradative enzymes were produced, the integrity of the cell membrane would be compromised and cells would fragment, rather than swell. When erythrocytes, which contain ions and proteins, are placed in distilled water, the osmotic pressure exerted by the ions and proteins within the erythrocytes causes water to enter the cells. Thus, A is the best answer.

Do the results presented in Figure 2 support the hypothesis that glucose has a cryoprotective role in frogs? A. Yes; survival and protection against hemolysis are promoted by exogenous glucose. B. Yes; death caused by freezing is directly proportional with the extent of hemolysis. C. No; injected glucose lowered blood hemoglobin levels, suggesting that the survival rates are not related to the treatment. D. No; injected saline promoted hemolysis, suggesting death was most likely related to circulatory collapse.

Solution: The correct answer is A. Under normal conditions hemoglobin is carried in red blood cells and is not found in the plasma. Ice formation can disrupt cell structure and cause the red blood cells to rupture or lyse, releasing hemoglobin into the plasma. One can infer from the information given that plasma hemoglobin is an operational measure of the extent of hemolysis (rupture of red blood cell). The information in the figure shows a positive correlation between survival and glucose concentration and a negative correlation between plasma hemoglobin and glucose concentration. Both of these correlations support a protective effect of high blood-glucose concentration in protecting the animal from the effects of cold. There is also a negative correlation between survival and plasma hemoglobin, but this correlation does not address the issue of a causal role for glucose concentration in protecting the animal. Thus, answer choice A is the best answer.

Passage: After conjugation, the four sexually produced progeny divide by binary fission. Each daughter cell receives an exact copy of the micronuclear DNA and an uneven (but approximately equal) amount of DNA from the amitotic macronucleus. To minimize fluctuations in DNA content, small macronuclei undergo an additional S phase before division, and large macronuclei eliminate an S phase. The macronuclei of the asexual progeny in Tetrahymena and the cytoplasm of the ova-producing cells of female vertebrates share a common feature in that both: A. undergo uneven division. B. contain uneven amounts of nuclear material. C. regulate their contents by adding or skipping an S phase. D. are apportioned at mitosis.

Solution: The correct answer is A. When a vertebrate oocyte divides in meiosis the cytoplasm is distributed to the two daughter cells. Most of it goes to the daughter cell destined to be the ovum. The other daughter cells, the polar bodies, are cast off with little cytoplasm. The amount of nuclear material in each, however, is the same, so choice B is incorrect. The distribution of cytoplasm is uneven, as is the distribution of macronuclear DNA during the amitotic cell division in Tetrahymena. The answer is A, that the common feature is that both cytoplasmic division in ova and division of nuclear material in Tetrahymena are both uneven. Answer C is incorrect, Tetrahymena is unusual in having an amitotic division that requires an extra S phase; vertebrate ova have no such phase. Answer D is incorrect because there is no mitosis in the division of the macronucleus. ***The secondary oocyte divides unevenly in order to conserve the cytoplasm in the one functional ovum.

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.

Solution: The correct answer is A. The endosymbiotic theory described in the passage is widely accepted today, so there must be some way of using the theory to reconcile the fact that mitochondrial proteins are made in the cytoplasm. We know, for example, that lateral transfer of genes from one genome to another is widespread among living organisms. It would have been an easy matter for mitochondrial genes to be acquired by the host cell nucleus and to disappear from the mitochondrial genome. It is likely that this exchange would have resulted in a net gain in efficiency for the composite organism. Answer choice A is therefore the most plausible answer.

According to Figure 1, at approximately what plasma concentration of glucose is the Tm (320 mg/min) reached? Figure 1: https://s3.amazonaws.com/wmx-api-production/courses/1659/images/BS-11-1-PT3-BS-P009.gif 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.

One type of metabolic feedback loop that influences weight control involves the regulation of glucose levels in the blood. Which organ in the digestive system participates in this regulation by breaking down glycogen? A. Stomach B. Liver C. Pancreas D. Small intestine

Solution: The correct answer is B. The question asks the examinee to identify which organ in the digestive system participates in regulation of glucose levels by breaking down glycogen. In order to regulate blood glucose levels, glycogen is made, stored, and broken down in the liver (B), not the stomach (A), the pancreas (C), or the small intestine (D). Thus, B is the best answer.

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

Solution: The correct answer is B. 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.

Passage: There is a relationship between H. pyloriinfection and cancer. Infected individuals have a two-fold increased risk of gastric cancer, although >75% of patients with active infections do not develop cancer. Genetic studies of H. pylori have identified genes that are expressed in different strains of this bacterium. One gene, vacA, encodes a toxin. Expression of another gene, cagA, leads to inflammation and may be related to the genesis of gastric cancer. Although many individuals develop antibodies against H. pylori antigens, these antibodies rarely eradicate the infection; evidently, this pathogen has developed effective ways to elude host defenses. One difference between different strains of H. pylori is that they: A. attack different hosts. B. express different genes. C. exhibit different degrees of resistance to antibiotics. D. exist in either developed or developing countries.

Solution: The correct answer is B. According to the passage, there is more than one strain of H. pylori. One strain expresses the gene vacA that encodes a toxin. The other strain encodes a gene cagA that leads to inflammation and might be related to the genesis of gastric cancer. Thus, a significant difference between the strains is that the strains express different genes.

Why do calcium supplements often include vitamin D? A. Vitamin D is needed to prevent rickets. B. The activated form of vitamin D stimulates the absorption of calcium into the blood. C. The activated form of vitamin D enhances the action of calcitonin. D. The activated form of vitamin D enhances the uptake of calcium by bone tissue.

Solution: The correct answer is B. Activated vitamin D acts on the small intestine to stimulate the absorption of calcium into the bloodstream. The inclusion of vitamin D in calcium supplements would ensure that vitamin D is present in the body to help promote this absorption. Thus, B is the best answer.

The chemical valinomycin inserts into membranes and causes the movement of K+ into the mitochondria. Based on Figure 1, if mitochondria are treated with valinomycin, the rate of ATP synthesis in the mitochondria will most likely: A. decrease, because the K+ will compete with protons at the active site on ATP synthetase. B. decrease, because movement of K+ into the mitochondrial compartments will disrupt proton movement into the intermembrane space. C. increase, because the net positive charge in the mitochondria will cause increased movement of protons into the intermembrane space. D. increase, because the additional positive charge will further activate ATP synthetase.

Solution: The correct answer is B. Any disruption of mitochondria is likely to decrease ATP production since they are a major cellular source of that molecule. Answer choices C and D can not be right because they propose an increase in ATP production. There is no information in the question to suggest that valinomycin will cause K+ to compete with H+ for an active site on ATP synthetase. Furthermore, one would suspect that ATP or precursor molecules such as phosphate and ADP would occupy the active site on the ATP synthase molecule. Answer choice A, therefore, does not seem plausible. The question does provide the information that valinomycin increases the flow of K+ across the membrane. An influx of another positively charged ion into the compartment would disrupt the electrochemical gradient responsible for the necessary flow of protons. Answer B, therefore, seems more plausible than any other choice. ***** Having positive ions in the intermembrane space will make it harder for protons to be pumped into the intermembrane space. Fewer protons in the intermembrane space means fewer protons available to move through ATP synthetase, so C and D are definitely not correct. There is nothing to suggest that K+ could occupy the active site of ATP synthetase. K+ is HUGE compared to a proton. B is correct.

Passage: The five species are closely related and live in Puerto Rico. Three species (A, B, and C) live in open unshaded fields, and the other two species (D and E) live in the understory of a closed canopy forest. Figure 1::: https://s3.amazonaws.com/wmx-api-production/courses/1659/images/BS-11-1-GID%200822.jpg Which of the following conclusions about dewlap reflectance is supported by information in the passage? A. Lizard habitat is determined by dewlap reflectance for each species. B. High UV dewlap reflectance is most important in brightly lit habitats. C. High dewlap reflectance is most important in dimly lit habitats. D. Dewlap reflectance is highest at the blue end of the visible spectrum.

Solution: The correct answer is B. Figure 1 shows that the three species of lizards that live in unshaded fields possess dewlaps that are significantly more capable of reflecting UV light than do the two species of lizards that live in the shaded understory. This supports the conclusion that high dewlap reflectance is most important in brightly lit habitats. Thus, B is the best answer.

The posttranslational modification of some of the eukaryotic cell's most abundant proteins is thought to affect the ability of those proteins to condense DNA into 30‑nm fibers. Given this, these proteins are most likely: A. tubulins. B. histones. C. transcription activators. D. DNA polymerase subunits.

Solution: The correct answer is B. Histones are among the most abundant proteins in eukaryotic cells. These small, basic proteins come together with DNA to form nucleosomes, the bead-like, primary structural elements of chromatin. Interactions among histone proteins in separate nucleosomes allow those nucleosomes to form the highly compacted 30-nm chromatin fibers. Posttranslational modifications that alter the charge on histone proteins affect their ability to condense DNA. Thus, B is the best answer.

The sequence of events in the human menstrual cycle involves close interaction among which organs? A. Hypothalamus-thyroid-ovary B. Hypothalamus-pituitary-ovary C. Pituitary-thyroid-ovary D. Pituitary-adrenal glands-ovary

Solution: The correct answer is B. Neither the thyroid gland nor the adrenal glands are particularly involved in the menstrual cycle. The hypothalamus exerts control over the pituitary hormones involved in menstruation by secreting hormone-releasing factors into the pituitary portal circulation. The gonadotropic hormones FSH and LH produced by the pituitary and the ovarian hormones estrogen and progesterone all have a role in regulating the human menstrual cycle. Therefore, close interaction among the hypothalamus, the pituitary, and the ovary is necessary for the human menstrual cycle. Thus, answer choice B is the best answer.

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? Figure 1: https://s3.amazonaws.com/wmx-api-production/courses/1659/images/BS-11-1-PT3-BS-P009.gif 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 D. Glucose, because the slope of the clearance line for glucose is lower than that for 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.

H. pylori infection may cause increased proliferation of mucosal cells in the stomach. This may lead to gastric cancer if: A. genetic mutations occur in proliferating germ cells. B. genetic mutations occur in proliferating somatic cells. C. the immune system fails to recognize bacterial antigens. D. crowded mucosal cells are likely to remain in interphase.

Solution: The correct answer is B. The passage discusses the connection between H. pylori infection and increased risk of gastric cancer. If H. pylori infection causes increased proliferation of mucosal cells in the stomach, this could lead to gastric cancer if genetic mutations occur in proliferating somatic cells that line the stomach. ----Proliferating means to increase rapidly in numbers; multiply

Passage:::: .........Negative effects of nonspecific stimulation by superantigens occur because the activation of so many T cells causes the release of massive levels of cytokines. This increased cytokine release is probably responsible for many of the acute problems seen in TSS, and also in some autoimmune diseases such as arthritis, multiple sclerosis, and rheumatic fever. Staphylococcus and Streptococcus bacteria cause problems in acute infections such as toxic shock syndrome primarily by: A. multiplying to produce large numbers of bacteria. B. stimulating exaggerated immune responses. C. causing autoimmune reactions. D. inhibiting metabolic enzymes with toxins.

Solution: The correct answer is B. The passage does not suggest that the problems associated with Staphylococcus and Streptococcus infections are due to large numbers of bacteria in the system, so A can be rejected. There is much discussion in the passage, however, about the effects of these organisms on the immune response, that they produce superantigens and activate 1 in 20 T cells rather than 1 in 100,000. Therefore, option B is the correct answer. The passage indicates that autoimmune diseases might also be caused by increased cytokine release, but autoimmune disorders by definition are caused by immune reactions to antigens produced by the body itself, not by antigens from other organisms. Option C, has to be rejected for this reason. There is no suggestion that Staphylococcus and Streptococcus exert any effect through inhibition of metabolic enzymes, so option D is incorrect.

If the e and f genes are expressed, the Xi chromosome will be prevented from reaching 100% frequency if selection pressures cause which of the following to be true? A. XsXs flies have the lowest fitness of any genotype. B. XsXs flies have the highest fitness of any genotype. C. XiY flies and XsY flies have equal fitness. D. XiXs flies and XsXs flies have equal fitness.

Solution: The correct answer is B. The passage says that, if none of the Xi bearing genotypes is selected against, the frequency of Xi will increase to 100%. Since not being selected against is equivalent to having equal fitness, options C and D are incorrect. If fitness is not equal among all the genotypes and the XsXsgenotype has the highest fitness, this would increase the proportion of Xs in relation to Xi in the population counteracting the drive to increase Xi at the expense of Xs, so option B is the correct answer.

Based on experiments similar to these, researchers have proposed the following overall equation for the sodium pump. 3 Na+(inside) + 2 K+(outside) + ATP4− + H2O → 3 Na+(outside) + 2 K+(inside) + ADP3− + Pi 2− + H+ Based on Reaction A, if all the energy produced from glycolysis were used to remove Na+ from a cell, how many molecules of Na+ would be removed per molecule of glucose? A. 3 B. 6 C. 9 D. 12

Solution: The correct answer is B. The question asks the examinee to determine the number of molecules of Na+ that could be removed from a cell using the energy produced from glycolysis. Since glycolysis produces two ATP molecules per glucose molecule, and 3 Na+molecules are extruded for each ATP molecule utilized by the sodium pump, a total of 6 molecules of Na+ are extruded per molecule of glucose. Thus, B is the best answer.

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.

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

Solution: The correct answer is B. The question asks the examinee to identify the experiment that would best test the hypothesis that urease is necessary for colonization of the stomach by H. pylori. The most effective way to demonstrate necessity is to test for the occurrence of a specific outcome in the absence of the agent whose necessity is being tested. In this case, the best test would be to examine whether H. pylori can colonize the stomach in the absence of urease. Of the options listed, only B uses this approach by exposing uninfected animals to a strain of H. pylori that lacks urease. Thus, B is the best answer.

All of the following are functions of mammalian skin EXCEPT: A. sensation. B. respiration. C. protection from disease. D. protection against internal injury.

Solution: The correct answer is B. The question asks the examinee to identify the function from the options listed that is not a function of mammalian skin. Mammalian skin is involved with touch sensation (A) and with protection from disease (C) and internal injury (D). However, in mammals, skin is not responsible for exchanging oxygen and carbon dioxide with the environment (B). Thus, B is the best answer.

If some but not all of the offspring from repeated matings of the same pair of fruit flies show the recessive traits of vestigial wings (vv) and ebony color (ee), which of the following could have been the genotypes of the individuals mated? A. One VVEE, one vvee B. Both VvEe C. One vvEE, one VVee D. Both vvee

Solution: The correct answer is B. The question asks the examinee to identify the genotypes of fruit flies that, when mated, produce some offspring that display the recessive phenotypes of vestigial wings (vv) and ebony color (ee) and some offspring that display the dominant phenotypes for those traits. To display both recessive phenotypes, the offspring must have inherited a recessive allele for each gene from both parents. This implies that each parent must have at least one copy of a recessive allele for both genes. However, some of the offspring did not display the recessive phenotypes for these traits. This implies that at least one parent has a dominant allele for the vestigial gene and at least one parent has a dominant allele for the ebony gene. Of the options listed, only B, both parental genotypes VvEe, satisfies these requirements. Thus, B is the best answer.

From 4% to 10% of all maternal deaths in the United States each year result from ectopic pregnancy. The most likely cause of death in these cases is: A. severe hormonal imbalance. B. loss of blood when the fallopian tube ruptures. C. infection in the region of the pregnancy. D. inadequate nutrition due to fetal use of maternal nutrients

Solution: The correct answer is B. The question asks the examinee to identify the immediate, life-threatening complication that can occur with an ectopic pregnancy: namely, the risk of hemorrhage, due to rupture of the fallopian tubes in response to implantation and growth of the embryo in this region of the reproductive tract (B). While other complications can result from, or are associated with, an ectopic pregnancy, none pose the immediate, potentially lethal threat that the risk of hemorrhage poses. Thus, B 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

Solution: The correct answer is B. 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. ******** While the loss of some proteins could affect the production of tRNA, it is not AS BAD when compared loosing the ability for cells to differentiate during development (not developing certain organs, a heart, or a brain, etc.).

The sodium pump would be most active in cells of which of the following structures? A. Veins B. Loop of Henle C. Lungs D. Bone marrow

Solution: The correct answer is B. The question asks the examinee to identify the structure in which the sodium pump is most active. Of the structures listed, Na+ reabsorption is of utmost importance to the function of the loop of Henle (B), where Na+ is reabsorbed from the filtrate moving through the nephron. This is ultimately the mechanism whereby the kidneys concentrate urine. While cells of the other structures may express Na+ pumps to help maintain ion gradients across their cell membranes, those cells are not nearly as dependent on the Na+ pump for their primary function as is the loop of Henle. Thus, B is the best answer. Also, Bone marrow is the spongy tissue inside some of your bones, such as your hip and thigh bones. It contains stem cells. The stem cells can develop into the red blood cells that carry oxygen through your body, the white blood cells that fight infections, and the platelets that help with blood clotting.

DDT would most likely initiate cancer or cause a mutation if which of the following structures is damaged? A. Nuclear envelope B. Chromosome C. Ribosome D. Histone

Solution: The correct answer is B. The question asks the examinee to identify the structure that when damaged would most likely lead to cancer or result in a mutation. Mutations are heritable changes in the sequence of the nucleic acid component of chromosomes (B), and mutations that lead to unregulated cellular growth can lead to cancer. The other options listed do not mention chromosomes. Thus, B is the best answer. **** Chromosome mutation can cause cancer!! Each gene has a specific function in the body. Some genes control cell division. Whenmutations occur in these genes, a cell may begin to divide without control. ... Over time, a number of mutations may occur in a single cell, allowing it to divide and grow in a way that becomes a cancer.

The concentration range within which muscle tension is most sensitive to acetylcholine (Figure 1) is: Figure 1: https://s3.amazonaws.com/wmx-api-production/courses/1659/images/BS-11-1-PT4-BS-P007.gif 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.

Solution: The correct answer is B. The relaxation of the intact blood vessel is dramatic at about 10-7 M ACH. The steepness of the curve means slight changes in ACH concentration are producing large changes in muscle tension. Below 10-8 muscle tension is increasing despite ACH. At concentrations above 10-6 muscle tension is fully relaxed and there is no response to ACH. In the blood vessel with endothelium removed muscle tension continues to increase in all ranges of ACH studied suggesting there is no relaxation in response to ACH. Thus, answer choice B is the best answer.

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. ******* B. Sweat glands have special channels through skin Explanation - The upper layer of skin is made up of epidermis and dead cells which is impermeable to water. However, we are able to sweat during hot weather because the sweat glands have specific opening on the surface of the skin called as pores. The sweat is discharged through these pores. Sweat also has dissolved salts in it but it does not allows it to diffuse through the skin rather it helps in maintaining the osmoregulation of the body.

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. *** Cardiac output: The amount of blood the heart pumps through the circulatory system in a minute. The amount of blood put out by the left ventricle of the heart in one contraction is called the stroke volume. ****The stroke volume and the heart rate determine the cardiac output.

Passage::: Vitamin C, which is required for the synthesis of bone matrix and is therefore needed for bone formation. Parathyroid hormone, which acts on bone tissue to encourage the formation and activity of osteoclasts (which break down bone cells) and to impair new bone formation. A low level of calcium in the plasma will trigger an increase of: I. osteoclast activity. II. parathyroid hormone. III. vitamin C. A. I only B. I and II only C. I and III only D. II and III only

Solution: The correct answer is B. When the level of calcium in the blood plasma is low, the body responds by mobilizing stores of calcium from the bones via the activity of parathyroid hormone. Parathyroid hormone will increase the number of osteoclasts, which break down bone cells. Therefore, one would expect an increase in both parathyroid hormone and osteoclast activity in order to increase the level of calcium in the blood plasma (options I and II). However, vitamin C (option III) promotes bone formation, a process that would further lower the calcium level in the plasma. Thus, B is the best answer.

In mammals, which of the following events occurs during mitosis but does NOT occur during meiosis I? A. Synapsis B. The splitting of centromeres C. The pairing of homologous chromosomes D. The breaking down of the nuclear membrane

Solution: The correct answer is B. One of the key differences between mitosis and meiosis occurs during their respective anaphases. During anaphase of mitosis, sister chromatids are pulled apart at the centromeres, each becoming an independent chromosome in the two diploid daughter cells. During anaphase I of meiosis I, homologous pairs of chromosomes are separated into the two daughter cells. However, each chromosome still consists of two sister chromatids joined to each other at the centromere. It is not until anaphase II of meiosis II that the centromere is split and the sister chromatids separate. Thus, B is the best answer. ******** Important::: Synapsis (also called syndesis) is the pairing of two homologous chromosomes that occurs during meiosis. It allows matching-up of homologous pairs prior to their segregation, and possible chromosomal crossover between them. Synapsis takes place during prophase I of meiosis.

In a mating of two Tetrahymena strains that are homozygous in their macronuclei and heterozygous in their micronuclei for a recessive gene, what percentage of the F1 generation will express the recessive phenotype? A. 0% B. 25% C. 50% D. 100%

Solution: The correct answer is B. The macronuclei do not participate in mating so only the genotypes of the micronuclei need to be considered. If we call the recessive gene rand its dominant allele R, then a cross between two heterozygote strains Rr will produce the genotypes RR:Rr:rr in a ration of 1:2:1. The recessive gene will be masked so that its expression will not be observable if a dominant gene is present in the genotype. Only in the double recessive (rr) case will the presence of the recessive gene be observed. The answer then is that on the average 25% of the offspring will demonstrate the recessive phenotype (choice B).

The prolonged increase in heart and breathing rates during the snow skiing trip was probably a result of: A. activation of the sympathetic autonomic nervous system by the new experience. B. activation of the parasympathetic autonomic nervous system by the new experience. C. hypoxia caused by insufficient blood hemoglobin concentration to supply oxygen for exercise at the low oxygen pressure found at high altitudes. D. depressed core body temperature (hypothermia) caused by exposure to cold temperatures at high altitudes.

Solution: The correct answer is C. According to the passage, Sarah went skiing in the mountains of Colorado. At first, she noticed an elevated pulse rate and ventilation rate. As the week progressed, these rates dropped, but were still higher than usual. This prolonged increase in heart rate and breathing rate was most likely the cause of hypoxia (insufficient oxygen to the body cells) caused by insufficient blood hemoglobin to supply oxygen for exercise at the low oxygen pressure found at high altitudes.

According to the passage, the cagA gene product will cause: A. the disruption of host cell enzymatic activity. B. the disruption of host cell protein synthesis. C. the movement of leukocytes into mucosal tissue. D. the vasoconstriction of arterioles in the mucosal layer.

Solution: The correct answer is C. According to the passage, expression of the cagA gene leads to inflammation of the mucosal lining of the stomach. Thus, the cagA gene product triggers the movement of leukocytes into the mucosal tissue—because leucocytes gravitate toward an inflammation.

In a laboratory population of Drosophila, all the males are XsY. Among the females, 15% are XiXi, 50% are XiXs, and 35% are XsXs. Assuming random mating, what proportion of male flies in the next generation will be XiY? A. 12% B. 30% C. 40% D. 65%

Solution: The correct answer is C. All the males in the next generation will acquire a Y chromosome from their male parent, so the contribution of the male can be ignored in solving this problem. All the XiXi females will have XiY sons, so 15% of the XiY flies in the next generation will come from this type of female. Half of the sons the XiXs females produce will be XiY. Since XiXs females make up 50% of the population, 25% of the males in the next generation will come from this type of female. None of the sons of XsXs females will be XiY. The total number of male flies that are XiY is 40%; therefore, answer choice C is correct.

To support the symbiotic hypothesis presented in the passage, mitochondria should be similar to bacteria in which of the following ways? A. They should use 80S ribosomes. B. They should be incapable of binary fission. C. They should have circular DNA. D. They should be capable of anaerobic respiration.

Solution: The correct answer is C. Answer choice C is the correct answer because it states that the fact that mitochondria and bacteria both have circular DNA is a reason to think they are related. Such a characteristic is likely to have been highly conserved over time since it involves the basic material of life. Changes in the topology of the DNA molecule would involve substantial changes in the way the molecule replicated and the way it was transcribed. The fact that both have circular DNA supports the symbiotic hypothesis presented in the passage. Alternative choices A and B presented are not actually true characteristics of bacteria. Bacteria do have 70S ribosomes, not the 80S ribosomes of eukaryotes as stated in answer choice A and they do reproduce by binary fission, which choice B denies. The loss of ability to carry out anaerobic respiration (answer D) might be a product of the long history of association with the host cell. It is not unusual for symbiotic organisms to lose abilities that are compensated for by host functions.

Which of the following recombinant processes depends on the F factor plasmid? A. Transformation B. Transduction C. Conjugation D. Translocation

Solution: The correct answer is C. Bacteria can exchange genes by three processes: conjugation, transformation and transduction. The process of conjugation involves production of a special conjugation pilus (sex pilus) by one bacterium and transfer through it of DNA to another bacterium. It requires special genes for the pilus and these are usually present on a plasmid, a separate extragenomic strand of DNA not incorporated into the bacterium's own DNA. This plasmid is referred to as the fertility or F factor. Conjugation is a feature of Gram-negative bacteria. It confers the advantages of sexual reproduction on the bacterium. The plasmid benefits by being able to move from one host bacterium to another through the conjugation pilus. Conjugation is the process described in the stem, so choice C is correct. Choice A and B are the other two processes that do not require a plasmid. In transformation bacteria take up DNA from their surroundings, the media in which they are immersed. Transduction is the process whereby genes are transferred by a virus. Choice D is also incorrect; translocation, is a general term used in biology to describe movement from one place to another (genes on a chromosome, proteins in a cell, sap in a tree).

From the data in Figure 1, one can conclude that the sensitivity of aortic smooth muscle to acetylcholine is: Figure 1: https://s3.amazonaws.com/wmx-api-production/courses/1659/images/BS-11-1-PT4-BS-P007.gif A. decreased by the presence of norepinephrine. B. increased by the presence of norepinephrine. C. increased at least 10 times by the presence of the endothelium. D. greatest at 10-8 M, with or without endothelium.

Solution: The correct answer is C. Figure 1 allows the conclusion that the sensitivity of aortic smooth muscle to ACH is increased at least 10 times in the presence of endothelium. Relaxation occurs in the ring with endothelium at 10-7M ACH but in the other ring does not occur even with 10-6 M ACH; these concentrations differ by a factor of 10. Both rings show the same response to norepinephrine. There is no response to ACH at 10-8 M with or without endothelium. Thus, answer choice C is the best answer. *** Read answer choices and graph very carefully.

If a person's gallbladder is removed, the person should restrict the consumption of: A. proteins. B. polysaccharides. C. triglycerides. D. lactose.

Solution: The correct answer is C. The gall bladder is an organ that stores bile produced by the liver. The major dissolved components of bile are breakdown products of hemoglobin such as bilirubin and bile salts. The bile salts are amphipathic; that is, they have a hydophobic portion that is soluble only in fats and oils and a hydrophilic portion that is soluble in water. Bile salts allow dietary fats (and oils), which do not dissolve in the watery digestive juices, to form an emulsion of tiny droplets dispersed in the digestive juices. One end of a bile salt molecule associates with the fats and the other with the aqueous solution in the gut. The dispersion of fats in the aqueous digestive juices and aids digestion by making them available to digestive enzymes. Since triglycerides are hydrophobic fats, bile would aid in their digestion, so the answer is C, triglycerides. Removal of the gall bladder would have no effect on digestion of proteins (choice A), polysaccharides (choice B) or lactose (choice D).

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. ***** If the proton gradient is at equilibrium with the free energy of electron transport, then the electrons cannot be transported through the ETC.

The lipases catalyze the hydrolysis of fats and other carboxylic acid esters. The lipases illustrate the fact that: A. some enzymes are molecules other than proteins. B. most enzymes interact with only one specific substrate molecule. C. some enzymes interact with several different substrate molecules that have similar chemical linkages. D. some enzymes interact with many biologically active substrate molecules of dissimilar structures and linkages.

Solution: The correct answer is C. Lipases catalyze the hydrolysis of fats and other carboxylic esters—similar to fats, but not fats. Lipase's ability to catalyze the hydrolysis of fats and similar molecules reveals that some enzymes interact with several different substrate molecules that have similar chemical linkages.

A lower-than-normal blood pressure will cause which of the following effects on the rate of plasma clearance of Substance A? A. An increase, because the concentration of Substance A in the urine will increase B. An increase, because the ADH levels will be very low C. A decrease, because the decreased rate of urine output will allow more reabsorption by the kidney D. A decrease, because ADH levels will be very high

Solution: The correct answer is C. Low blood pressure could have an effect on ADH levels (as suggested in answer choices B and D), which could, in turn, affect the amount of substance A in the urine, but there is not enough information in the passage to decide if such an effect exists and, if it does, whether its effect would overcome the certain affect of lowering the blood pressure. The best answer is answer choice C: that low blood pressure decreases the glomerular filtration rate, allowing more time for reabsorption and decreasing the amount of substance A in the urine. Blood pressure is the source of the energy that forces fluid into the capsular space. If the heart stopped and the blood in the glomerular capillaries had no hydrostatic pressure, fluid in the space around the glomerulus would flow back into the capillary bloodstream. This would occur because the protein-rich blood would be hypertonic with respect to the protein-poor fluid in the capsular space so that the fluid would flow down the osmotic gradient into the blood.

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.

Solution: The correct answer is C. One characteristic that distinguishes eukaryotic cells from prokaryotic cells is that eukaryotic cells contain membrane-bound organelles such as mitochondria. Thus, C is the best answer.

In Figure 1, which of the following observations led to the conclusion that intact endothelium is necessary for the relaxation of smooth muscle when acetylcholine (ACH) is applied? Figure 1: https://s3.amazonaws.com/wmx-api-production/courses/1659/images/BS-11-1-PT4-BS-P007.gif A. The tension initially changes when norepinephrine (NE) is added. B. The tension decrease occurs more quickly in the ring without endothelium. C. The tension decreases upon addition of 10-7 M ACH, only in the ring with endothelium. D. The tension decreases during washout in the ring without endothelium.

Solution: The correct answer is C. That the tension decreases upon addition of 10-7 M ACH only in the ring with endothelium shows that intact endothelium is necessary for ACH to exert its effect. Even at the higher concentration of 10-6M ACH the aorta without endothelium did not show any decrease in tension. Both rings show the same response to norepinephrine, but this is a contraction, not a relaxation. There is a slow decrease in tension during washout in the ring without endothelium due to removal of the norepinepherine by bathing the preparation in norepinepherine-free, but that decrease is unrelated to the investigation of the effect of acetylcholine. Thus, answer choice C is the best answer. **** The endothelium serves as a permeable barrier for the blood vessel and is involved in the regulation of blood flow. Within basic research, endothelial cells are pivotal to applications related to wound healing, angiogenesis, inflammatory processes, blood brain barriers, diabetes and other cardiovascular diseases.

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

Solution: The correct answer is C. The amount of air entering the lungs in a single breath, or tidal volume, is given as 800 mL/breath. Of that 800 mL only 650 mL reaches the alveoli per breath (800 mL of air inhaled minus 150 mL of nonalveolar respiratory volume). Therefore the net volume of air that reaches the alveoli each minute is equal to 650 mL/breath multiplied by 10 breaths/min, or 6500 mL. Thus, C is the best answer. ***** Let me give this a try, so you know what tidal volume means. How much air is going into the lungs. In this case, its 800 ml. The question is how much air enters the alveoli. This is not the same thing as tidal volume. Remember alveoli is at the bottom of the lungs. The deadspace is the air thats occupying the trachea or your windpipe. The air here does not reach the alveoli. You have to subtract this 150 from 800 because this air will not reach the alveoli. When you exhale, it will be the first air to be pushed out. So 650 ml/breath reaches the alveoli * 10 breaths = 6500 ml

Passage::: Plasma clearance refers to the capacity of the kidney to remove a substance from the plasma. It is determined by comparing the concentrations of the substance in the plasma and the urine, and then calculating the rate at which the substance appears in the urine. Plasma clearance is affected by the tubular transport maximum (Tm) of a substance. The Tm is the maximum rate of transport (mg/min) at which a substance can be reabsorbed by the kidney. That is, if the filtration rate of a substance exceeds its Tm, the substance will begin to appear in the urine. The Tm for glucose averages 320 mg/min in an adult human. 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.

Solution: The correct answer is C. The glomerulus is a tuft of capillaries that bulges into the capsular space (also referred to in some texts by its eponym, Bowman's space), which is a potential space lined with simple squamous epithelium. Fluid is expressed from the blood across the capillary endothelium. It enters first the capsular space, then the kidney tubule. The part of the tubule closest to the glomerulus is the proximal tubule, the next part is the U-shaped loop of Henle and the last part of the tubule is the distal tubule. Suppose one gradually increased the rate at which fluid was expressed from the bloodstream (the glomerular filtrate rate) and measured the concentration of some substance reabsorbed by the kidney as it left the distal tubule as urine. The concentration might not rise at first, because the cells lining the tubule might be completely reabsorbing the substance and putting it back into the blood stream. Eventually, however, the rate of flow would reach a point at which it exceeds the rate at which the tubule cells could reabsorb the substance. The fluid would be flowing too rapidly through the tubule for the cells to reabsorb all the substance. The rate of flow through the tubule at which the substance begins to be observed in the urine is Tm. At that point the rate of flow of fluid through the tubule begins to exceed the capacity of the kidney tubule cells to reabsorb the substance. This description of Tm is given only in answer choice C, the correct answer.

What mechanism could account for this oscillation of cyclin protein concentration? Figure 1. Concentration of the protein cyclin rises and falls during the cell cycle. https://s3.amazonaws.com/wmx-api-production/courses/1659/images/BS-11-1-BS168.jpg 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

Solution: The correct answer is C. The graph shown in the question indicates that the concentration of cyclin rises and falls in a regular manner throughout the cell cycle, reaching a peak just at the beginning of mitosis, gradually declining during mitosis, reaching a minimum at the end of mitosis, and gradually increasing during interphase. The mechanism that can best account for this oscillation in the concentration of cyclin is translation of cyclin mRNA (creating the protein from mRNA template) followed by proteolysis (destruction) of cyclin protein during mitosis.

Some of the DNA sequences that are eliminated during macronuclear differentiation (Figure 1, Step 6) may be sequences involved in: A. transcription. B. translation. C. meiosis. D. ribosome production.

Solution: The correct answer is C. The macronucleus does not go through the process of meiosis. Genes involved in meiosis are therefore superfluous in this genome. The correct answer is choice C, that DNA sequences involved in meiosis may be eliminated. On the other hand the primary use of the macronucleus is to provide the proteins for the cell's day-to-day functioning. The functions of transcription (choice A), translation (choice B) and ribosome production (choice D) must be coded for by macronuclear genes because they are necessary for it to direct protein synthesis. ***** Basically, mitosis is important for cell function and survival. Meiosis is not important so not all the cells that are produced will live. As a result, the macronuclear differentiation that is produced by meiosis that are deleted by be sequences involved in meiosis.

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. cells may induce neighboring cells to differentiate. cell differentiation is an "all or none" phenomenon.

Solution: 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.

Passage:::: Because this unique type of binding activates approximately 20% of the T lymphocytes, as opposed to 1 in 100,000 T cells activated by conventional antigenic stimulation, superantigens are considered nonspecific stimulators. According to the passage, superantigens increase the number of activated T cells over activation levels observed with conventional antigens by a factor of: A. 20. B. 5,000. C. 20,000. D. 100,000.

Solution: The correct answer is C. The passage states that 20 percent of T lymphocytes are activated by superantigens, while one in 100,000 is activated by conventional antigens. Taking 20 percent of 100,000 yields 20,000, so the correct response is answer choice C.

Passage: The relaxing substance was found to be nitric oxide (NO), which is derived from the amino acid L-arginine. To study the physiological role of NO, competitive inhibitors of the enzyme responsible for NO synthesis (NO synthase) were developed; L-NMMA is one such inhibitor. Thus, a serendipitous observation led to the discovery of a new function of vascular endothelium, the regulation of blood flow. Assume that NO is continuously synthesized. The addition of a saturating concentration of L-NMMA to a relaxing aortic ring with intact endothelium would probably: A. increase its sensitivity to acetylcholine. B. cause the ring to dilate (reduce its tension). C. cause the ring to contract (increase its tension). D. prevent norepinephrine from increasing ring tension.

Solution: The correct answer is C. The passage states that L-NMMA is an inhibitor of NO synthetase, which is responsible for the synthesis of nitric oxide (NO). NO serves as the relaxing compound. If NO synthesis in intact endothelium was inhibited by addition of a saturating concentration of L-NMMA, one would expect that the tension of the aortic ring would increase. Therefore, the key is C. ***** Read the question carefully. So NO (nitric oxide) acts as a muscle relaxer. It says in the question the NO was already present in the muscle tissue. Then, L- NMMA was added to the muscle tissue. L- NMMA is an inhibitor of NO. So, if you inhibit the agent that was relaxing the muscle, what happens to the muscle?? It becomes constricted. Thus, the ring will contract (increasing tension). Simple :-) ,, think simple, read, re-read, and think carefully.

Presumably, hyperglycemia promotes cellular dehydration because: A. glucose, as an energy source, accelerates the osmotic work performed by plasma membranes. B. glucose, as an energy source, accelerates plasma membrane ion exchange pumps. C. glucose molecules raise the osmotic pressure of the extracellular space. D. glucose molecules are exchanged for water molecules across the plasma membrane.

Solution: The correct answer is C. The passage suggests that glucose release from glycogen stores raises the osmotic pressure of body fluids and dehydrates cells protecting them from freezing. The glucose acts as an antifreeze, lowering the freezing point of the frog's body fluids. Freezing is gradual in that water moves osmotically out of cells into the concentrated unfrozen extracellular fluid. Ice formation can thereby be restricted to extracellular water; ice crystal formation within cells is lethal. An animal that can permit some freezing of its body fluids and sustain life is said to be freeze-tolerant. Although glucose is an energy source and can be transported across the plasma membrane, the protection of glucose in this harsh environment is due to its effects on the diffusion of water (i.e., its osmotic effects). Water diffusion does not require ion-exchange pumps. Glucose is not exchanged for water molecules across the membrane. Thus, answer choice C is the best response.

Passage: When DDT is dissolved in oil and applied to human skin, it initiates local sensations of prickling, burning, and itching. At sufficient levels of exposure, generalized symptoms of pain, tiredness, irritability, anxiety, and weakness ensue. According to one hypothesis, these symptoms occur because DDT becomes incorporated into nerve cells, allowing Na+ to diffuse freely through axonal membranes. 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.

Solution: The correct answer is C. The question asks the examinee to determine how DDT absorption through the skin causes symptoms of burning, itching, and pain. These three sensations are detected by sensory neurons (C and D), not motor neurons (A and B). In addition, the action potentials that are involved in the transmission of these sensations result from a depolarization (C), not a hyperpolarization (D), of the sensory neuron membrane. Thus, C is the best answer.

It is now generally accepted that H. pylori can cause ulcers. Proof of this most likely depended on the demonstration that: A. people with stomach ulcers have antibodies to H. pylori. B. healthy individuals have antibodies to H. pylori. C. ulcers could be produced in healthy organisms by infecting them with H. pylori. D. the organism can be passed from mother to fetus during pregnancy.

Solution: The correct answer is C. The question asks the examinee to identify a key criterion that an organism must meet before it can be declared the infectious agent responsible for causing a particular disease or condition. One of these criteria is the demonstration that the putative infectious agent is capable of causing the disease or condition in an organism that was healthy prior to exposure to the agent. C correctly identifies this criterion by stating that proof that H. pylori can cause stomach ulcers depended on demonstrating that ulcers can be produced in healthy organisms by infecting them with H. pylori. Although A, B, and D may or may not be true, they do not demonstrate causality. Thus, C is the best answer.

Passage: Normally produced messenger RNA (mRNA) molecules are known as the sense RNA. Antisense nucleic acids, which are complementary to a portion of the sense mRNA, can be synthesized. The antisense molecules will bind specifically to the sense mRNA and prevent the production of the natural gene product. .... ..... In this manner antisense and sense RNA molecules could be produced simultaneously—in effect, preventing the production of the gene product permanently. When used as described in the passage, antisense drugs prevent: A. DNA replication. B. RNA transcription. C. RNA translation. D. cell replication.

Solution: The correct answer is C. The question asks the examinee to identify the effect of an antisense drug. When antisense mRNA molecules are used as described in the passage, the antisense molecules bind specifically to the sense mRNA, preventing the process of translation. As a result, C is the best answer. A, B, and D are not the best answers because the passage describes the effect of antisense RNA binding to mRNA, and mRNA is directly involved in translation. However, because RNA can hybridize with either RNA or DNA, a role for antisense RNA in the regulation of DNA replication, transcription, and even cell replication may be possible, but these processes are not the focus of the passage. Thus, C is the best answer.

Which of the following nucleotide sequences describes an antisense molecule that can hybridize with the mRNA sequence 5′-CGAUAC-3′? A. 5′-GCTATG-3′ B. 5′-GCUAUG-3′ C. 3′-GCUAUG-5′ D. 3′-GCAUAG-5′

Solution: The correct answer is C. The question asks the examinee to identify the sequence of an antisense molecule that could hybridize with the mRNA sequence 5′-CGAUAC-3′. When the RNA molecules hybridize, the antisense molecule would line up in an antiparallel fashion with the sense molecule, meaning its 3′ end would line up with the 5′ end of its complement. The nitrogenous bases would form the following pairs: A with U, G with C. The only correct sequence is C, which is 3′-GCUAUG-5′. A, B, and D do not have the correct sequence with the correct polarity. Thus, C is the best answer. **************** However, RNA can form duplexes just as DNA does. All that is needed is a second strand of RNA whose sequence of bases is complementary to the first strand; e.g.,5´ C A U G 3´ mRNA3´ G U A C 5´ Antisense RNA The second strand is called the antisense strand because its sequence of nucleotides is the complement of message sense. When mRNA forms a duplex with a complementary antisense RNA sequence, translation is blocked. This may occur becausethe ribosome cannot gain access to the nucleotides in the mRNA orduplex RNA is quickly degraded by ribonucleases in the cell (see RNAi below). Review: https://www.biology-pages.info/A/AntisenseRNA.html Also***** Antisense is a term that's used to describe one of the two strands of DNA, OR actually in some cases also RNA.

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.

Solution: The correct answer is D. The question asks the examinee to identify the bodily function that would NOT be significantly impaired by DDT accumulation in the liver. DDT would be expected to affect all liver functions, and, therefore, all options associated with a major liver function are incorrect. The detoxification of poisons and the production of bile, which facilitates fat absorption in the small intestine by breaking large fat droplets into smaller ones, are major liver functions. Therefore, A, B, and C are not correct. The production of factors involved in blood pressure regulation (D) is not a major function of the liver. Thus, D is the best answer.

Set Point Hypothesis The brain regulates body weight just as a thermostat maintains a constant room temperature. The brain adjusts metabolism and behavior to maintain a predetermined body weight. Genes also influence the set point, which can increase with age—but only to the extent dictated by inheritance. Diet and exercise cannot reset the set point over the long term. Settling Point Hypothesis Body weight is determined by the interaction of two factors—metabolism and genes—with the environment. Depending on genotype, various metabolic feedback loops may allow weight to be stabilized at a new level. Thus, in an environment where high-calorie food is plentiful, individuals with a genetic predisposition to obesity will tend to become more overweight than those without such a predisposition. Which hypothesis implies that a person can deliberately alter his or her own body maintenance weight? A. The set point hypothesis because a thermostat can be reset B. The set point hypothesis because the set point can change with age C. The settling point hypothesis because, with the correct genotype, one's metabolism may allow weight to stabilize at a new level D. The settling point hypothesis because diet and exercise cannot reset the set point

Solution: The correct answer is C. The question asks the examinee to identify which hypothesis implies that a person can deliberately alter his or her own body maintenance weight. A and B are incorrect because the set point hypothesis suggests that the set point cannot be reset through behavior modification. C correctly states that the settling point hypothesis allows the set point to be reset through behavior modification. D is incorrect because it states that diet and exercise cannot reset the set point, which contradicts the premise of the settling point hypothesis. Thus, C 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

Solution: The correct answer is C. The question asks what process would not interrupt the repeated transmission of a nerve impulse to a muscle cell. The impulse is transmitted by release of the neurotransmitter acetylcholine from the presynaptic membrane and its reception by membrane-bound proteins on the postsynaptic membrane. Answers B and D are incorrect, because they involve substances that would interfere with the presynaptic release or postsynaptic reception of neurotransmitter. Addition of a cholinesterase blocker would produce a buildup of acetylcholine in the synapse and prevent receptors from responding to impulses, so option A is incorrect. The only change that would not result in blockage of the impulse transmission is option C, which is the correct answer. An increase in acetylcholine receptors on the postsynaptic membrane would enhance transmission, not interfere with it. ******* Acetylcholine can be found in all motor neurons, where it stimulates muscles to contract. From the movements of the stomach and heart to the blink of an eyelash, all of the body's movements involve the actions of this important neurotransmitter. ---- Cholinesterase inhibitors, also known as anti-cholinesterase, are chemicals that prevent the breakdown of the neurotransmitter acetylcholine or butyrylcholine.

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.

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. ****** The passage said Genes that produce an antisense sequence can also be synthesized and added to the genome of organisms. In this manner antisense and sense RNA molecules could be produced simultaneously—in effect, preventing the production of the gene product permanently. Problem::: You need to read the passage carefully to answer the questions.

In the macronucleus, the genes for rRNA are located extrachromosomally. This suggests that the rRNA genes are: A. nonlinear. B. nonfunctional. C. self-replicating. D. rearranged.

Solution: The correct answer is C. The stem of this item states that the rRNA genes are extrachromosomal and asks what this suggests. Genes are strands of DNA, so his must mean that they are strands of DNA that are not part of the cell's chromosomes. These strands still must self-replicate just as the chromosomes do if they are to be passed from one generation to the next, so choice C is the correct answer. All nucleic acids are a linear arrangement of the component nucleotides, so choice A is incorrect. The rRNA genes must be functional or the organism could not produce protein, so choice B is incorrect. Finally, the fact that they are extrachromosomal give no clue to their arrangement, so choice D can also be rejected.

In addition to the skin and circulatory systems, which of the following organ systems is most likely to be affected by TSS? A. The musculoskeletal system B. The digestive system C. The lymphatic system D. The respiratory system

Solution: The correct answer is C. The symptoms mentioned in the passage suggest involvement of the circulatory system (hypotension) and skin (rash, desquamation), so the question is asking what other bodily system might be involved in toxic shock syndrome. The other system that is most prominently discussed is the lymphatic system, so choice C must be correct. The other systems are not discussed in the passage and there is no other reason to think they might be involved, so the musculoskeletal system (answer A), digestive system (answer B), and respiratory system (answer D) can be eliminated from consideration.

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. ***** Think about it like this, when you're exercising, your body temperature increases and in order to maintain homeostasis, your body will try to reduce the body temp by sweating. vasodilation helps facilitate this by opening up the pores on your skin and cool downs your body.

The liver is different from many other organs in that it can at least partially regenerate following illness or damage. This regeneration is accomplished primarily through: A. fission. B. meiosis. C. mitosis. D. cell growth.

Solution: The correct answer is C. Unlike other organisms, the liver can partially regenerate after illness or damage. This regeneration is accomplished by mitosis. Mitosis is the process whereby human body cells (not gametes) reproduce.

Graph Observation: 0 until 12 heart rate is present. From 12 until 26 hours heart rate is not present. During what time period of the freeze-thaw episode shown in Figure 1 were the frog's tissues relying entirely on anaerobic respiration? A. Before 0 hours B. Between 0 and 6 hours C. Between 0 and 24 hours D. Between 12 and 26 hours

Solution: The correct answer is D. According to Figure 1, the heart rate was 0 between 12 and 26 hours after the onset of freezing. During that period the heart was not circulating oxygenated blood to the tissues so the cells would have had to rely entirely on anaerobic respiratory mechanisms for energy. Recall that anaerobic metabolic pathways can produce energy without using oxygen by glycolysis. Before 0 hours, the freeze-thaw episode had not begun. From 0 to 12 hours the heart rate indicates that oxyhemoglobin was being circulated in the blood for oxidative metabolism by the cells, so that anaerobic respiration would have been unnecessary. Thus, answer choice D is the best answer.

Most people infected with H. pylori do not develop gastric cancer because they: A. do not incorporate bacterial genes in their chromosomes. B. have robust immune systems that defeat early cancers. C. eradicate the infection before any tumors develop. D. tolerate the infection without developing tumors.

Solution: The correct answer is D. According to the passage, infection by H. pyloriincreases one's risk of gastric cancer, but less than 25% of people affected ultimately develop such cancer. Most people affected with H pylorido not develop cancer because they tolerate the infection without developing tumors. The DNA is not damaged and uncontrolled cell division does not occur. ****** Tolerance to infections is defined as the ability of a host to limit the impact of parasites, pathogens or herbivores on host health, performance, and ultimately on fitness.Tolerance is not equivalent to resistance.

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

Solution: The correct answer is D. 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.

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.

If these lizards use UV light in communication, a mutation that eliminated UV photoreceptors would probably cause the LEAST disadvantage to: Figure 1. Reflectance spectra (percent reflectance compared to that from a magnesium carbonate white standard) from the dewlaps of five Anolis species https://s3.amazonaws.com/wmx-api-production/courses/1659/images/BS-11-1-GID%200822.jpg A. species A. B. species B. C. species D. D. species E.

Solution: The correct answer is D. Assuming that the lizards use the UV-reflectivity of the dewlap primarily as a means of intraspecies communication, species E would most likely be least affected by a mutation that eliminated UV photoreceptors. Its dewlaps are the least UV-reflective of the five lizard species, which indicates that species E is least likely to rely heavily upon this form of communication in the first place. Thus, D is the best answer.

Passage: The five species are closely related and live in Puerto Rico. Three species (A, B, and C) live in open unshaded fields, and the other two species (D and E) live in the understory of a closed canopy forest. Male lizards have a dewlap, a large fold of skin under the throat that they can fan out like a flag. Flashing the dewlap plays an important role in lizard communication such as territorial displays, warning signals, and courtship. If Anolis lizards have X-Y chromosomal sex determination, the locus of a gene for the UV reflectance pigment: A. must be on the X chromosome. B. must be on the Y chromosome. C. must be on an autosome. D. could be on a sex chromosome or on an autosome.

Solution: The correct answer is D. Based on the information presented, the gene encoding UV-reflectance pigment could be on a sex chromosome or an autosome. The fact that the pigment is expressed in the dewlap, a structure found only in males, is not sufficient to eliminate any chromosome as the location of this gene. Thus, D is the best answer. ************ EXPLANATION:::: Consider something like beard colors in humans. Some men have a red coloration in their beard. Most women do not express the phenotype "bearded", however that is not enough to say that beard color is on the Y-Chromosome, it could be anywhere, it is just only able to result in a phenotype where there is in fact a beard present. I understand your frustration because this question seems almost deliberately misleading, however if you look at the details I think its a fair, albeit challenging question. The key is in the fact that females lack the structure altogether, and therefore we would be unable to tell their genotype on the basis of phenotype.

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.

Solution: The correct answer is D. Control of heart rate, muscle coordination, and appetite is maintained by the brain stem, cerebellum, and hypothalamus, respectively.

Inbreeding can reduce the fitness of a population in the short term because it causes an increase in the: A. genetic diversity of the population. B. levels of aggression in the population. C. rate of spontaneous mutations. D. incidence of expression of deleterious recessive traits.

Solution: The correct answer is D. Deleterious genes are generally rare because they tend to be eliminated through natural selection. Only when an organism is homozygous (has two copies of a gene, one from each parent) does a recessive gene reveal its presence. Because recessive genes can be masked by dominant genes, they are less exposed to natural selection. So most organisms carry many deleterious recessive genes. The chances of having offspring that are homozygous for a given recessive gene are rare when a mate is chosen randomly from the population. The chances of getting a pair of deleterious recessive genes increase enormously when the mate is a relative, because relatives are likely to have a similar genotype. The answer is therefore choice D. Inbreeding decreases rather than increases genetic diversity (choice A), its effect on aggression is hard to determine (Choice B) and it has no effect on spontaneous mutations (choice C), so these are incorrect choices.

When an initially heterozygous macronucleus undergoes repeated binary fission, the result will be: A. the loss of macronuclear chromosomes. B. an increased rate of crossing over in the macronucleus. C. the production of a macronucleus with a genetic origin distinct from the micronucleus. D. a variable allele distribution in the macronucleus.

Solution: The correct answer is D. In the passage it states that the macronucleus is 45-ploid. If haploid cells have one copy of each chromosome and diploid cells have two copies of each chromosome, then one can infer that 45-ploid cells have 45 copies of each chromosome. Polyploidy, having more than two copies of each chromosome, is unusual in animals, but important in plants and microorganisms like Tetrahymena. The amitotic division of the macronucleus will result in uneven distribution of chromosomes, hence an unpredictable genome. Answer D is therefore correct; the allele distribution will be uneven. Given the large numbers of identical chromosomes in the macronucleus, however, repeated divisions are unlikely to result in the loss of a chromosome, so choice A is incorrect. There is no crossing over in the production of a macronucleus, because the phenomenon of crossing over is a very complex and specialized event that occurs only when meiosis will be taking place, so B is incorrect. C is incorrect because all macronuclei trace their origin ultimately to their micronucleus. ***** 24. C is wrong because it is asking about the macronucleus, not the micronucleus! D is the correct answer; the last paragraph explicitly mentions that "...an uneven (but approximately equal) amount of DNA from the amiotic macronucleus)"

Inflation of the lungs in mammals is accomplished by: A. diffusion of gases. B. active transport of gases. C. positive pressure pumping action. D. negative pressure pumping action.

Solution: The correct answer is D. Inflation of lungs in mammals is accomplished by negative pressure pumping action. Because the lung stays in contact with the thoracic wall as it enlarges due to contraction of the diaphragm and the external intercostal muscles, a pressure that is lower than atmospheric pressure (negative pressure) is generated within the alveolar sacs. Although diffusion of gases is important in the movement of oxygen into the blood from the alveoli and the movement of carbon dioxide out of the blood into the alveoli, it does not explain the inflation of the lung alveoli at inspiration. Active transport describes a situation in which the net movement of a substance occurs against the energy gradient. Thus, answer choice D is the best answer.

Two neighboring lizard populations would be considered separate species if: A. one population inhabited the forest and the other lived in a field. B. one population had a UV-reflective dewlap and the other did not. C. they did not communicate with each other. D. they did not interbreed and produce fertile offspring.

Solution: The correct answer is D. One of the key factors that determines a species is the ability to successfully breed and produce fertile offspring. Two organisms that do not meet this criteria are considered separate species. Thus, 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.

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.

Which of the following statements best explains why Xi has the potential to increase to 100% frequency in gene pools that contain it? A. XiXs flies have the highest fitness of any genotype. B. XiXi flies tend to migrate and introduce the Xichromosome into new populations. C. XiXi flies pass X chromosomes to all their offspring, but XsXs flies pass their X chromosomes to only half their offspring. D. XiY flies pass their X chromosome to all their offspring, but XsY flies pass their X chromosome to only half their offspring.

Solution: The correct answer is D. Some pairs of alleles confer a higher fitness when an organism is heterozygous, but there is no information in the passage to suggest that heterozygous XiXs Drosophilaare more fit than homozygous XiXi or XsXs Drosophila, so Option A cannot be correct. Option B is not correct because there is nothing in the passage to suggest differences in migration between genotypes; the only difference mentioned is in the transmission of chromosomes. It is the XiY male, not the XiXi female, that passes on all its X chromosomes to its daughters, so option C is incorrect. The correct answer is option D, which states that XiY males have no sons, only daughters all of whom have the Xi gene.

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

Solution: The correct answer is D. The double stranded daughter DNA molecule would be an exact duplicate of the parent molecule. It would have the same (A + T)/(G + C) ratio. The correct answer is therefore 3:1, choice D.

Passage::: Two common diseases affect bone metabolism: 1.Rickets affects children, causing inadequate mineralization of new bone matrix. The ratio of mineral to organic matter is lower than normal. This abnormality can result in distortion of the bones, especially long bones, leading to bowed legs. It is caused by insufficient vitamin D activity. 2.Osteoporosis usually affects older people, most often postmenopausal women. There is more bone resorption than bone formation, resulting in a reduction of bone mass but a normal ratio of mineral to organic matter. The causes are not definitely known, although weight-bearing activity promotes bone mass increase, and lack of activity causes bone loss. Which of the following conditions could produce rickets? I. Metabolic deficiency of parathyroid hormone II. Impairment of conversion of vitamin D to its active form III. Inability of the active form of vitamin D to act on its target tissue A. I only B. I and II only C. I and III only D. II and III only

Solution: The correct answer is D. The passage states that rickets is caused by insufficient vitamin D activity. Insufficient vitamin D activity would reduce the ability of the body to absorb ingested calcium from the small intestine. To maintain calcium levels in the blood plasma, parathyroid hormone would promote the breakdown of bone tissue, causing the bones to become weak. If there were a metabolic deficiency of parathyroid hormone, the body would be unable to break down bone tissue (option I), causing a higher than normal ratio of mineral to organic matter in the bones instead of a lower than normal ratio. However, if the body were unable to convert vitamin D to its active form, or if vitamin D were unable to act on its target tissue, overall vitamin D activity would be impaired (options II and III), which can lead to rickets. Thus, D is the best answer.

An intravenous infusion causes a sharp rise in the serum level of albumin (the major osmoregulatory protein in the blood). This will most likely cause an: A. increase in the immune response. B. increase in tissue albumin levels. C. outflow of blood fluid to the tissues. D. influx of tissue fluid to the bloodstream.

Solution: The correct answer is D. The question asks about the effect of an increase in the level of albumin, one of the major plasma proteins. Because albumin has nothing to do with the immune response, answer choice A is incorrect. The plasma proteins can not cross the walls of blood vessels, but water molecules can. The wall of the artery acts as a semipermeable membrane setting up the conditions needed for osmosis to occur. An increase in plasma albumin will upset the osmotic balance because the blood will become hypertonic with respect to the tissue. Water will have to flow into the bloodstream to reestablish equilibrium. One of the causes of edema, increased fluid in body tissues, is a decrease in the plasma protein level. This occurs, for instance, in starvation when the body is forced to use its albumin as an energy source. An increase in the plasma protein level would have the opposite effect: fluid would enter the bloodstream (answer D).

Which statement below most accurately describes the roles of the proteins actin and myosin during muscular contraction? A. Both actin and myosin shorten, causing the muscle tissue to which they are attached to contract. B. Both actin and myosin catalyze the reactions that result in muscle contraction. C. Actin molecules are disassembled by myosin, leading to a shortening of muscle sarcomeres. D. Bridges between actin and myosin form, break, and re-form, leading to a shortening of muscle sarcomeres.

Solution: The correct answer is D. The question asks the examinee to characterize the roles of actin and myosin during muscle contraction. The sliding filament model describes the interaction of actin and myosin during muscle contraction. According to this model, neuronal impulses cause the release of calcium from the sarcoplasmic reticulum within muscle cells. The calcium then binds to troponin, a molecule that along with tropomyosin, blocks the binding sites for myosin on actin molecules. Calcium binding to troponin causes a shift in the troponin/tropomyosin complex, revealing the binding site for myosin. Myosin then binds to actin, causing a conformational change in myosin that "cocks" the head of the myosin molecule and slides the actin filament relative to myosin. ATP binds to myosin, causing it to detach from actin and "recharge" (rebend again). If another binding site is available on actin, myosin will bind again, and slide the actin filament even further. Thus, D is the best answer.

Aldosterone stimulates Na+ reabsorption by the kidneys. What changes in blood volume and pressure would be expected as a result of aldosterone deficiency? A. Increased volume and increased pressure B. Increased volume and decreased pressure C. Decreased volume and increased pressure D. Decreased volume and decreased pressure

Solution: The correct answer is D. The question asks the examinee to determine the result of an aldosterone deficiency on blood volume and blood pressure, given that aldosterone stimulates Na+reabsorption in the kidney. An aldosterone deficiency would result in lower Na+ reabsorption into the bloodstream. Because H2O passively follows Na+ during reabsorption in the kidney, less Na+ reabsorption would result in less H2O reabsorption into the bloodstream. This would result in decreased blood volume. Blood volume would also be affected by lower blood Na+ levels because there would be less of this ion to osmotically hold water in the extracellular fluid. Decreased blood volume would result in decreased blood pressure as well (D). Thus, D is the best answer. ****** Basically,,, Blood volume affects blood pressure. When there's a greater volume of fluid, more fluid presses against the walls of arteries resulting in a higher pressure. Decrease in volume, decease in blood pressure. Positive correlation between blood pressure and blood volume.

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.

Passage:::: Ectopic pregnancy is defined as the development of a fertilized ovum outside the uterine cavity. Most frequently, ectopic development occurs in the fallopian tube (oviduct). A drug that increases the risk of a tubal pregnancy is most likely to inhibit which one of the following actions? A. Contraction of the uterus B. Secretion of follicle-stimulating hormone C. Onset of menstruation D. Transport of the ovum from ovary to uterus

Solution: The correct answer is D. The question asks the examinee to identify the action if inhibited is most likely to lead to an ectopic pregnancy. The main reason for the occurrence of ectopic pregnancies is the failure of the fertilized egg to be transported from the oviducts to the uterus. When the transport system fails, the embryo implants in the oviduct, rather than in the uterus. Accordingly, a drug that inhibits the transport of the ovum from the ovary to the uterus (D) is most likely to increase the risk of tubal pregnancy. Thus, D is the best answer.

Passage::: Stomach ulcers have long been thought to result from the overproduction of HCl by the gastric mucosal cells. When drugs that neutralize or reduce the production of stomach acid are used to treat patients with ulcers, 95% of the patients have a recurrence of the ulcer within two years. Scientists suspected that another factor was the true cause of ulcers. A spiral-shaped bacterium, Helicobacter pylori (H. pylori), was isolated from biopsies of ulcers; the organisms were found in greater than 95% of people with ulcers. When ulcer patients are treated with a combination of antibiotics and drugs that reduce acid production, there seems to be no recurrence of the ulcers. Based on the passage, which of the following methods would best prevent the recurrence of stomach ulcers? A. Use of drugs that prevent the production of acid B. Use of drugs that neutralize stomach acid C. Use of drugs that inhibit bacterial protein synthesis D. Use of drugs that inhibit acid production and bacterial protein synthesis

Solution: The correct answer is D. The question asks the examinee to identify the best method for preventing the recurrence of stomach ulcers on the basis of the passage information. The passage indicates that both HCl and H. pylori are likely to contribute to the development of stomach ulcers. Therefore, the most effective method for preventing the recurrence of stomach ulcers will address both of these factors. Of the options given, only D addresses both factors: inhibitors of acid production would lower the level of HCl in the stomach and inhibitors of bacterial protein synthesis would kill H. pylori. Thus, D is the best answer.

What type or class of chemical messenger traveling in the blood would most probably link the brain with the digestive tract and fat cells in the control of body weight? A. Neurotransmitters B. Digestive enzymes C. Protein receptors D. Hormones

Solution: The correct answer is D. The question asks the examinee to identify the class of substances that would travel through the bloodstream, linking the brain with the digestive tract. D is the best answer because most hormones travel from the endocrine gland in which they were produced, through the blood, to their target tissue. Neurotransmitters by definition are released at neural synapses, although there are chemicals that act both at synapses and like hormones in the bloodstream. Therefore, A is not the best answer. Digestive enzymes are found in the digestive tract, but they do not travel through the bloodstream. Therefore, B is incorrect. Protein receptors interact with chemical messengers but are not the signal themselves. Therefore, C is not the best answer. Thus, D is the best answer. ***** Hormones are chemical messengers that transfer information and instructions from one set of cells to another. 5. Once a hormone is secreted, it travels from the endocrine gland that produced it through the bloodstream to the cells designed to receive its message. ---- NOTE::: Digestive enzymes can't travel in the blood of they will digest all the organs. ALSO--- 1) Protein hormones (or polypeptide hormones) are made of chains of amino acids. An example is ADH (antidiuretic hormone) which decreases blood pressure. 2) Steroid hormones are derived from lipids. Reproductive hormones like testosterone and estrogen are steroid hormones. ------ Protein hormones are faster than steroid hormones because peptide hormones do not enter the target cell. Instead, they bind to a receptor on the membrane surface. ... Because of this, they typically cause much faster effects than those that bind to internal receptors, which influence creation of new proteins.

Which of the following explanations best accounts for the results in Figure 1? Figure 1::: https://s3.amazonaws.com/wmx-api-production/courses/1659/images/BS-11-1-B13286P1.gif A. Strain A grows faster than strainB. B. Strain A grows faster than strainB and is also resistant to streptomycin. C. Strain B grows faster than strainA and is also resistant to streptomycin. B. Strain B grows slower than strainA and is also resistant to streptomycin.

Solution: The correct answer is D. The question asks the examinee to identify the explanation that best accounts for the results shown in Figure 1. In Figure 1, the slope of the line indicating the exponential growth phase of Strain A is significantly steeper than the slope of the line indicating the exponential growth phase of Strain B. This implies that Strain B grows more slowly than Strain A. The figure also shows that Strain B grows equally well in the presence and absence of streptomycin, indicating resistance to streptomycin, whereas the growth of Strain A is stunted in the presence of streptomycin, indicating streptomycin sensitivity. Of the options given, only D indicates that Strain B grows more slowly than Strain A and that Strain B is resistant to streptomycin. Thus, D is the best answer.

Passage::: Vertebrates that have evolved in deserts are better adapted than humans for maintaining homeostasis in hot, dry environments. When severely dehydrated, humans can produce urine that is 4 times as concentrated as plasma. However, camels can more than double, and kangaroo rats can more than triple, the urine-concentrating capacity of humans. Reptiles, which lack sweat glands, maintain homeostasis by means of lower metabolic rates and scaly, relatively impermeable integuments. When the environmental temperature is 45° C, which of the following organisms will have the highest body temperature? A. Human B. Kangaroo rat C. Camel D. Lizard

Solution: The correct answer is D. The question asks the examinee to identify the organism with the highest body temperature in response to elevated external temperatures. This organism will be the organism that is least able to utilize the cooling mechanisms of vasodilation and sweating. A lizard (D) has an impermeable integument, thereby eliminating vasodilation and sweating as options for cooling. Thus, D is the best answer.

Of the following tissues, which is NOT derived from embryonic mesoderm? A. Circulatory B. Bone C. Dermal D. Nerve

Solution: The correct answer is D. The question asks the examinee to identify the tissue that is NOT of mesodermal origin. Of the tissues listed, the only one that does not arise from the mesoderm during embryonic development is nervous tissue (D). Nervous tissue arises developmentally from ectoderm, not mesoderm. Thus, D is the best answer. ************ ---- The Derms and Hypodermis layers of the skin are made from mesoderm. As they connect the Hyperdermis (outer layer of the skin) to the connective tissue to the entire part of the body.

Passage: DDT is an insecticide that crosses biological membranes readily. Once inside the cell, DDT disrupts the function of many enzymes. For example, DDT uncouples oxidative metabolism from the generation of ATP; cellular respiration continues, but no ATP is produced. The mechanism of this action is unknown. When vertebrates ingest DDT, it accumulates in the adipose tissue and fatty deposits of organs such as the kidney, thyroid, testis, and ovary...... Accumulation of DDT in the testes may cause reduced fertility in males because the uncoupling of oxidative metabolism from ATP production may reduce: A. glucose concentration of semen. B. testosterone concentration of semen. C. blood circulation in the testes. D. sperm motility.

Solution: The correct answer is D. The question asks the examinee to predict why reduced male fertility would be caused by the DDT-induced uncoupling of ATP production from oxidative metabolism. This uncoupling would reduce ATP synthesis in the mitochondria and consequently would affect processes requiring large amounts of ATP. Using the information in the passage and in the question, the examinee cannot predict that reduced ATP levels would directly result in reduced semen glucose concentration (A), reduced semen testosterone concentration (B), or reduced blood circulation in the testes (C). However, sperm motility requires large amounts of ATP as evidenced by the high concentration of mitochondria in the sperm midpiece. Therefore, of the choices given, reduced sperm motility (D) is most likely to result from reduced ATP levels. Thus, D is the best answer.

Phenylketonuria is a genetic disorder caused by a mutation in the gene for the enzyme phenylalanine hydroxylase, which eliminates its enzymatic activity. Could an antisense drug help individuals with this disorder? A. Yes, if it binds to the mRNA of the phenylalanine hydroxylase gene and prevents its translation B. Yes, if it is incorporated into the chromosome and prevents the expression of the phenylalanine hydroxylase gene C. No, because mRNA does not persist in the cytoplasm of the cell D. No, because blockage of phenylalanine hydroxylase gene expression will not remedy the original disorder

Solution: The correct answer is D. The question indicates that phenylketonuria (PKU) is a genetic disorder caused by a mutation in the gene for an enzyme, which results in the elimination of enzymatic activity. The question asks the examinee to consider whether an antisense drug could help individuals who have PKU. An antisense drug works to prevent the expression of undesirable genes but does nothing to remedy the problem of a gene that produces an ineffective product. The only way to cure PKU would be to add a gene or gene product that could lead to the production of effective enzymes to replace the ones that do not function correctly. D is the best answer because a blockage of gene expression would only prevent the production of defective enzymes but would not remedy the problem caused by the absence of effective enzymes. A and B are incorrect because preventing expression of an inactive protein is not a cure. C is incorrect because the stability of the mRNA is irrelevant to the situation. 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. ***** 1. Follicular Phase Follicle stimulating hormone (FSH) is secreted from the anterior pituitary and stimulates growth of ovarian follicles The dominant follicle produces estrogen, which inhibits FSH secretion (negative feedback) to prevent other follicles growing Estrogen acts on the uterus to stimulate the thickening of the endometrial layer 2. Ovulation Midway through the cycle (~ day 12), estrogen stimulates the anterior pituitary to secrete hormones (positive feedback) This positive feedback results in a large surge of luteinizing hormone (LH) and a lesser surge of FSH LH causes the dominant follicle to rupture and release an egg (secondary oocyte) - this is called ovulation

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. B. raising the environmental temperature.

Solution: The correct answer is D. 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.

According to the passage, which of the following parts in the frozen body of a freeze-tolerant frog would contain ice? I. Cytoplasm II. Blood plasma III. Lymph A. II only B. III only C. I and II only D. II and III only

Solution: The correct answer is D. You would expect to find ice in the extracellular fluid - i.e., blood plasma (II) and lymph (III) in the frozen body of a freeze-tolerant frog. However, ice in the cytoplasm (III) would be lethal because ice crystal formation within the cells disrupts structural organization. Thus, answer choice D is the best answer. ***** Lymphatic vessels connect to the subclavian veins, which are part of the bloodcirculatory system and connect to the heart. Their key function is to transport excessive tissue fluid from interstitial spaces throughout the body back to the blood stream.