Diagnostic Exams - Biology Flashcards

Question

The mouth is NOT responsible for which of the following functions? A. Mastication B. Bolus creation C. Saccharide absorption D. Carbohydrate digestion

Answer 1

C. The mouth breaks up food via chewing (i.e. mastication; choice A is a function of the mouth and can be eliminated). It then forms the fragments into a smooth ball, or bolus, and lubricates that ball with saliva (choice B is a mouth function and can be eliminated). Salivary amylase (ptyalin) is responsible for carbohydrate digestion into disaccharides (choice D is a function and can be eliminated), however no nutrient absorption occurs in the mouth (choice C is not a function of the mouth and is the correct answer choice). Concepts tested Digestive System: Alimentary Canal

Answer 2

C. Selective reabsorption begins immediately after filtration and starts in the proximal convoluted tubule. Much of the water and ions in the filtrate are returned to the circulatory system (choice B is wrong). Glucose and amino acids are also reabsorbed (choice A is wrong and choice C is correct). A healthy kidney does not allow proteins to enter the filtrate, as they are too big to pass through the glomerular basement membrane (choice D is wrong). Concepts tested Renal System: Nephron Structure/Function

Answer 3

D. First, aldosterone levels increase (choices A and B are wrong). This allows sodium reabsorption from the distal nephron. The resulting increase in plasma osmolarity triggers antidiuretic hormone (ADH, or vasopressin) release (choice C is wrong and choice D is correct). This induces expression of water channels on the cells of the distal tubule and collecting duct, allowing water reabsorption. Concepts tested Renal System: Renal Regulation of Blood Pressure

Answer 4

A. Gastrin is secreted from G cells in the stomach, and stimulates acid and pepsinogen secretion, as well as increases stomach motility. Hydrochloric acid (HCl) cleaves the zymogen pepsinogen to active pepsin; a mutation leading to loss of gastrin function would impair HCl secretion and thus pepsin activation (choice A is correct). CCK (cholecystokinin) triggers release of bile and also works to regulate the flow of chyme through the pyloric sphincter and into the small intestine, but does not impact gastric enzyme activation (choice B is wrong). Erythropoeitin stimulates red blood cell production in the bone marrow and is made by the kidneys, thus it is not relevant to this question (choice C is wrong). Secretin triggers the release of pancreatic exocrine secretions (including digestive enzymes and aqueous bicarbonate) into the small intestine, and is not involved in gastric enzyme activation (choice D is wrong). Concepts tested Digestive System: Alimentary Canal

Answer 5

B. The macula densa on the distal convoluted tubule monitors filtrate osmolarity. If the filtrate osmolarity decreases (choices A and C are wrong), this indicates a drop in filtration rate. The macula densa then stimulates the juxtaglomerular cells (in order to activate the renin-angiotensin axis to increase blood pressure), while also causing the afferent arterioles to dilate (choice D is wrong and choice B is correct). The combination of increased systemic blood pressure and increased blood flow to the glomerulus increases filtration rate. Concepts tested Renal System: Renal Regulation of Blood Pressure

Answer 6

B. The afferent arterioles carry blood toward the glomerulus, but do not function in returning substances to the circulatory system after filtration (choice B is the correct answer). The efferent arterioles carry blood away from the glomerulus and ultimately branch into the peritubular capillaries. These vessels interlace with the nephron and pick up reabsorbed material (choices A and C return components to the circulatory system and can be eliminated). The vasa recta is part of the peritubular capillaries; it runs in parallel with the renal tubule and is also involved in returning material to the circulatory system (choice D can be eliminated). Concepts tested Renal System: Nephron Structure/Function

Answer 7

A. Item I is true: bile is amphipathic (has both polar and non-polar regions); this allows it to interact with both lipids and the hydrophilic intestinal contents. The lipids are emulsified, allowing pancreatic lipases easier access to them for digestion. Bile is sometimes referred to as intestinal soap, because it emulsifies fat in the intestines much like soap emulsifies fat on your hands when you wash them (choice C can be eliminated). Item II is false: bile is not an enzyme and does not digest lipids. It only makes it easier for the pancreatic enzymes to digest (choice D can be eliminated). Item III is false: bile is made only by the liver. It is stored and concentrated in the gallbladder (choice B can be eliminated and choice A is correct). Concepts tested Digestive System: Accessory Organs

Answer 8

D. The renal medulla is divided into regions called pyramids, which are composed of collecting ducts (choice D is correct). Glomeruli are the beginning of the nephron, where filtration occurs, and proximal tubules are found directly after glomeruli; both are located in the renal cortex (choices A and C are wrong). While portions of the nephron are found in the renal medulla, "nephron" is a broad term and not the best answer choice here (choice D is better than choice B). Concepts tested Renal System: Nephron Structure/Function

Answer 9

A. The trachea (not the esophagus) is protected by the epiglottis, as it prevents food or liquids from entering the lungs when swallowing (choice A does not describe the esophagus and is the correct answer choice). The upper portion of the esophagus is made of skeletal muscle so that swallowing can be initiated voluntarily (choice B describes the esophagus and can be eliminated), but this merges quickly into smooth muscle, which provides the peristalsis to move the bolus down to the stomach (choice C describes the esophagus and can be eliminated). The cardiac sphincter separates the esophagus and the stomach (choice D describes the esophagus and can be eliminated). Concepts tested Digestive System: Alimentary Canal

Answer 10

B. The gastrointestinal tract has a limited supply of oxygen within it, especially once the large intestine is reached. Bacteria residing there must be facultative or obligate anaerobes (choice B does not describe the bacteria in the gut and is the correct answer choice). The bacteria compete with pathogens, providing an additional immune mechanism (choice A is true and can be is eliminated) and also produce vitamin K, which is involved in blood clotting (choice B is true and can be eliminated). The fact that both humans and bacteria benefit from this arrangement makes this a mutualistic relationship (choice D is true and can be eliminated). Concepts tested Digestive System: Alimentary Canal

Answer 11

D. The filtrate is first formed in Bowman's capsule, as components of blood pass from the glomerular capillaries through the glomerular basement membrane (choices A and C are wrong). The filtrate then enters the renal tubule, the first portion of which is the proximal convoluted tubule (PCT). From the PCT, the filtrate moves through the loop of Henle, and finally the distal convoluted tubule (DCT, choice B is wrong and choice D is correct). Concepts tested Renal System: Nephron Structure/Function

Answer 12

B. Water, ions, urea, and glucose are all small enough and hydrophilic enough to be filtered into the nephron at the glomerulus. However, glucose is completely reabsorbed into the blood by a healthy individual, so urine does not contain glucose (choice A is wrong). Proteins are too large to be filtered and so never enter the nephron (choices C and D are wrong). Concepts tested Renal System: Nephron Structure/Function

Answer 13

C. The majority of peptidases in the duodenum come from the pancreas and are secreted as zymogens. Because they work in the duodenum they must be activated there, so stomach acid can't really play a role (choices A and D are wrong). Enterokinase, a duodenal enzyme, cleaves and activates the pancreatic zymogen trypsinogen; the active enzyme trypsin then cleaves other pancreatic zymogens to activate them (choice C is correct). The release of pancreatic enzymes is triggered primarily by secretin (a duodenal hormone); cholecystokinin (another duodenal hormone) can also help (choice B is wrong). Concepts tested Digestive System: Digestive Enzymes

Answer 14

A. The loop of Henle establishes a concentration gradient in the medulla, so that as the collecting duct passes through it, water can be reabsorbed (thus concentrating the urine; choice A is correct). ADH makes the collecting duct permeable to water, facilitating this reabsorption (choice C is wrong), but if the gradient did not exist, there would be no driving force for the movement of water out of the collecting duct. The selective reabsorption of ions under the control of hormones takes place at the distal convoluted tubule (DCT). For example, this is where aldosterone and parathyroid hormone have their effect (choice B is wrong). Only sodium is selectively reabsorbed by the loop of Henle. Potassium in general is secreted into the urine (choice D is wrong). Concepts tested Renal System: Nephron Structure/Function

Answer 15

B. Item I is true: non-competitive inhibitors do not bind at the active site, they bind at allosteric sites and prevent the enzyme from catalyzing the reaction (choice C can be eliminated). Item II is true: by "turning off" the enzymes to which it is bound, the inhibitor lowers the effective enzyme concentration, and lowering the enzyme concentration lowers Vmax (choice A can be eliminated). Item III is false: non-competitive inhibition cannot be overcome by adding more substrate. This is only true of competitive inhibition, where the inhibitor and substrate both bind to the active site (choice D can be eliminated and choice B is correct). Concepts tested Biochemistry: Enzymes and Enzyme Inhibition

Answer 16

B. Vmax depends only on the enzyme and the concentration of the enzyme. Item I would not affect Vmax: decreasing the substrate concentration would decrease V (the rate of product formation), but would not affect Vmax, which is a theoretical maximum rate of product formation, assuming you have enough substrate to get there (choice C can be eliminated). Item II would not affect Vmax: competitive inhibitors reduce V, but given enough substrate the reaction will ultimately reach the same Vmax as an uninhibited reaction (choice A can be eliminated). Item III would definitely affect Vmax (and can therefore be eliminated): more enzyme means that more product can be formed, and Vmax would increase (choice D is wrong and choice B is correct). Concepts tested Biochemistry: Enzymes and Enzyme Inhibition

Answer 17

B. Sucrose is a disaccharide formed by joining a glucose molecule to a fructose molecule (choice B is correct and choice C is wrong). Maltose is already a disaccharide (choice A is wrong), formed by joining two glucose molecules together (choice D is wrong). Concepts tested Biochemistry: Biomolecules: Carbohydrates

Answer 18

D. Statement I is true and therefore an incorrect choice: protein kinases and enzyme phosphorylases attach phosphate groups to enzymes and are often used to regulate enzyme activity (choices A and C can be eliminated). Both remaining answer choices include Statement II, so it must be a false statement: while enzymes do make reactions go faster, they do this by reducing the energy of activation, not increasing it. Statement III is also false and a correct choice: enzymes do not change the equilibrium of a reaction, they only help the reaction reach equilibrium more quickly (choice B can be eliminated and choice D is correct). Concepts tested Biochemistry: Enzymes and Enzyme Inhibition

Answer 19

C. Competitive inhibitors can resemble either the substrate of a reaction or the transition state of a reaction (choice D is a true statement and can be eliminated), and as such, bind at the active site of an enzyme (choice B is a true statement and can be eliminated). If the substrate concentration is significantly increased, then it becomes more likely to bind substrate at the active site than to bind inhibitor, and the inhibited reaction can reach the same Vmax as the uninhibited reaction (choice C is a false statement and the correct answer choice). Km is the substrate required to reach 1/2 Vmax. Because you need more substrate to run the inhibited reaction at the same rate (V) as the uninhibited reaction, Km increases (choice A is a true statement and can be eliminated). Note that this Km, the Km measured in the presence of an inhibitor, is called the "apparent Km." Concepts tested Biochemistry: Enzymes and Enzyme Inhibition

Answer 20

A. Km is the substrate concentration required to push a reaction to half of its maximum rate of product formation (choice C is wrong), and is a measure of affinity between an enzyme and its substrate. However a low Km indicates a high enzyme-substrate affinity, not the other way around. In other words, a low Km means that only a small amount of substrate is required to get to 1/2 Vmax, thus there must be a high affinity between the enzyme and the substrate. If Km is high, a lot of substrate is needed to reach 1/2 Vmax, indicating a low enzyme-substrate affinity (choice B is wrong). Km is unaffected by enzyme concentration; just because the amount of enzyme changes, does not mean its affinity for its substrate changes (choice D is wrong). Concepts tested Biochemistry: Enzymes and Enzyme Inhibition

Answer 21

A. Enzymes (or in this case, transport proteins) that display cooperative binding have sigmoidal curves (choice D is wrong). In order to display cooperative binding, a protein must be made up of more than one subunit. Since myoglobin is made of only a single subunit, it cannot display cooperative binding and would have a simple saturation curve (choice A is correct and choices B and C are wrong). Concepts tested Biochemistry: Enzymes and Enzyme Inhibition

Answer 22

C. Item I is true: starch is the polysaccharide used by plants to store glucose (choice B can be eliminated). Item II is true: glycogen is the polysaccharide that animals use to store glucose (choice A can be eliminated). Item III is false: cellulose is a polymer of glucose, but is used primarily for plant structure (choice D can be eliminated and choice C is correct). Concepts tested Biochemistry: Biomolecules: Carbohydrates

Answer 23

D. The cells of the inner cell mass differentiate into the three primary germ layers and have the ability to become any cell in the body; they are considered to be pluripotent cells (choice D is correct). The mesoderm is one of the three primary germ layers and can become many, but not all cell types in the body; it is considered to be multipotent (choice A is wrong). The zygote and morula are totipotent; they can become any of the cells in the body and can also form the trophoblast (choices B and C are wrong). Concepts tested Development: Embryology/Fetal Development

Answer 24

D. Choices A, B, and C all describe physiological components of the birthing process and can be eliminated. Prolactin is required for lactation, which occurs postpartum (labor does not depend on choice D so it is the correct answer choice). Concepts tested Endocrine System: Hormones and Hormone Functions Reproductive Systems: Female Reproductive System

Answer 25

C. The secondary oocyte starts meiosis II before ovulation but freezes in metaphase II. Meiosis II is completed (and the second polar body is secreted) only if fertilization occurs (choice C is correct). Note that only one ootid is generated from a oogonium, because of polar body secretion (choice D is wrong). The disintegration of the first polar body is not part of meiosis II (choice A is wrong), and the division of the primary oocyte is part of meiosis I (choice B is wrong). Concepts tested Reproductive Systems: Oogenesis

Answer 26

C. Gametes fuse in fertilization to form the zygote, which undergoes cleavage to form the morula. Next the blastocyst forms, which includes the trophoblast (choice B can be eliminated). Placental development starts once the blastocyst has implanted in the uterine wall (or endometrium; choice D is wrong). The developing human is called an embryo for the first two months, and is called a fetus after this (choice A is wrong). Choice C presents the correct order of development. Concepts tested Development: Embryology/Fetal Development

Answer 27

A. Oogonia mature into primary oocytes, which undergo meiosis I to generate a secondary oocyte and a polar body. If fertilized, the secondary oocyte undergoes meiosis II to generate an ootid (which matures into an ovum) and a second polar body (choice A is correct). Two rounds of cytokinesis are still required, one for each telophase (I and II; choice B is wrong). Each cytokinesis event generates one polar body, not two (choice C is wrong). Oocytes undergo meiosis, not mitosis. In female humans, oogonia perform mitosis in the fetal stage, to generate a lifetime supply of primary oocytes. In contrast, spermatogonia can perform mitosis through the life of a human male; this is one of the reasons females go through menopause and males do not (choice D is wrong). Concepts tested Reproductive Systems: Spermatogenesis Reproductive Systems: Oogenesis

Answer 28

D. Fertilization occurs in a fallopian tube (choice A is wrong). The sperm's bindin protein binds an oocyte receptor to allow plasma membrane fusion and injection of the sperm nucleus (choice B is wrong). Fast and slow blocks function to decrease the likelihood of multiple sperm fusing with an oocyte (choice C is wrong). The spermatozoa must penetrate both the corona radiata (made of granulosa cells) and the zona pellucida to access the oocyte (choice D is correct). Concepts tested Development: Embryology/Fetal Development

Answer 29

B. Progesterone helps with endometrium development, in preparation for pregnancy (choice B is correct). It provides negative feedback to the brain, to decrease GnRH, FSH, and LH. Falling LH levels ultimately cause the corpus luteum to disintegrate, progesterone levels to drop, and menstruation to occur (choice C is wrong). Estrogen's positive feedback loop occurs before ovulation and triggers the LH surge that causes ovulation (choice D is wrong). The chorion develops from the trophoblast of a blastocyst (developing embryo), and secretes hCG to maintain the pregnancy (choice A is wrong). Concepts tested Reproductive Systems: The Menstrual Cycle

Answer 30

A. The ectoderm differentiates to form all nervous system tissue, and most epithelial components are derived from the endoderm (choice B is a correct match and can be eliminated). Cleavage is the development of the morula from the zygote (choice C is a correct match and can be eliminated). Gastrulation is germ layer development (choice D is a correct match and can be eliminated). However, blastulation is the development of the blastocyst, which is composed of the trophoblast and an inner cell mass (choice A is an incorrect match and the correct answer choice). Concepts tested Development: Embryology/Fetal Development

Answer 31

A. Fertilization of both oocytes from a double ovulation is in fact what leads to fraternal twins, however the two ovulations must occur within 24 hours of each other, not a week apart (choice A is not true and is the correct answer choice). The LH surge that causes ovulation is the result of the positive feedback effect that estrogen has at high levels (choice B is true and can be eliminated), and the LH surge causes both ovulation and the formation of the corpus luteum after ovulation (choice C is true and can be eliminated). The drop in estrogen and progesterone as the corpus luteum degenerates (around day 27-28) is what triggers menstruation (choice D is true and can be eliminated). Concepts tested Reproductive Systems: The Menstrual Cycle

Answer 32

A. A block in prophase II of spermatogenesis would mean that the secondary spermatocytes could not complete meiosis II and would accumulate (choice B is wrong). This would lead to fewer spermatids (choice A is correct). Since spermatogenesis would be decreased, there would be less gamete migration from the seminiferous tubule basement membrane to its lumen (choice D is wrong). Finally, primary spermatocytes are the cells going through meiosis I, not meiosis II. Since the block is in meiosis II (i.e. after the primary spermatocytes have become secondary spermatocytes), the number of primary spermatocytes should be unaffected (choice C is wrong). Concepts tested Reproductive Systems: Spermatogenesis

Answer 33

A. Many maternal blood components, such as oxygen, nutrients, some antibodies, some drugs and toxins, and some viruses, can pass from the maternal blood supply into the developing fetus. In contrast, few molecules can pass from fetal blood into maternal circulation (choice A is correct). While choice B is an accurate statement, it does not answer the question (choice B can be eliminated). Fertilization does occur in the fallopian tubes, but this does not explain the differential exchange between fetus and mother (choice C can be eliminated). The chorion secretes hCG and other hormones in the first trimester, and ultimately becomes the placenta, but again, this does not explain the fact that more things can pass from mother to baby than from baby to mother (choice D can be eliminated). Concepts tested Reproductive Systems: Female Reproductive System

Answer 34

C. The Wolffian ducts regress in females (choice B and D are wrong), and the Mullerian ducts develop into female internal genitalia, such as the fallopian tubes, uterus, and cervix (choice C is correct). Note that mammary glands develop mostly from the ectoderm, because they are similar to glands found in skin (choice A is wrong). Concepts tested Reproductive Systems: Female Reproductive System

Answer 35

C. Many components of semen are generated by the accessory glands (bulbourethral glands, seminal vesicles, and prostate; choice C is correct). Sperm are made in the seminiferous tubules of the testes, mature in the epididymis and are stored in the vas deferens (or ductus deferens). During ejaculation, they travel with the rest of semen through the ejaculatory duct and then the urethra (choice A is wrong). The testes contain the seminiferous tubules and the epididymis, but not the bulbourethral glands (choice B is wrong). While the urethra does function in urination, it plays no role in solid waste (feces) excretion (choice D is wrong). Concepts tested Reproductive Systems: Male Reproductive System

Answer 36

D. Sperm are ejaculated into the female vagina (choice C is wrong), swim through the cervix into the uterus, and travel through a fallopian tube to an oocyte (choice D is correct). Sperm never enter the ovary (choices A and B are wrong). Note that choice A describes the path of a fetus during the birthing process.

Answer 37

B. Multipotent cells can become some, but not all cell types in the body. The cells of the three primary germ layers are considered multipotent; for example, some cells in the mesoderm could become muscle, but not skin or bone (choice B is correct). Totipotent cells, like the zygote or morula, can become any cell type in the body (choice A is wrong), including the trophoblast (choice C is wrong). Specialized (determined) cells can only become one cell type in the body (choice D is wrong). Concepts tested Development: Stem Cells

Answer 38

B. Choices A, C and D are accurate statements and can be eliminated. Smooth muscle contractions of the uterus and abdominal skeletal muscle contractions help push the baby out (choice B is false and the correct answer choice). Concepts tested Reproductive Systems: Female Reproductive Syste

Answer 39

A. Spermatids are made in the seminiferous tubules of the testes, mature in the epididymis and are stored as spermatozoa in the vas deferens until ejaculation. Fructose in the semen provides nutrients, and alkaline secretions help neutralize the low pH in the male urethra (from urine passing through) and the female vagina (which functions as part of the innate immune system, similar to the low pH in the stomach). Overall, the best answer choice is A. Neither component provides lubrication (choices B and D are wrong) and temperature control is a function of the scrotum (choice C is wrong). Concepts tested Reproductive Systems: Male Reproductive System

Answer 40

D. Repeated vomiting can raise blood pH (metabolic alkalosis) and increase hemoglobin’s affinity for oxygen (choice B is wrong). This is seen as a left-shift in the oxygen saturation curve for hemoglobin (choice D is correct and choice C is wrong). Note that answer choices B and C describe decreased oxygen affinity which may be due to situations such as acidosis, increased temperature, or increased CO2. Cooperativity does not change significantly with the shift described by the Bohr effect; the curve would still be sigmoidal (choice A is wrong). Concepts tested Cardiovascular System: Blood Digestive System: Alimentary Canal

Answer 41

C. This question is essentially asking what function the scalene muscles play in respiration and, as you are not required to know this function, is best answered by process of elimination. The elasticity of the chest wall, which is partially responsible for preventing collapse of the lungs, is a function primarily of bone and other connective tissue, and not any particular muscle group (choice A can be eliminated). Passive expiration is a passive process and no muscle contraction is required (choice D can be eliminated), and there is no reason to believe the function of the diaphragm will be directly affected by the impaired function of the scalene muscles (choice B can be eliminated). This leaves choice C: the scalene muscles, in addition to the sternocleidomastoid and others, are involved in inspiration (choice C is correct). Concepts tested Respiratory System: Ventilation and pH Regulation

Answer 42

D. The alveolar ducts are a portion of the respiratory zone where gas exchange occurs (choice D is correct). They branch further out into individual alveoli where the vast majority (approximately 90%) of gas exchange occurs (choices A and B are wrong). The respiratory zone is composed of the respiratory bronchioles, alveolar ducts, and alveoli (choice C is wrong). Concepts tested Respiratory System: Conduction and Respiratory Zones

Answer 43

A. The outermost layer of the skin is the epidermis, made up primarily of cells called keratinocytes (choice A is correct). The dermis is beneath the epidermis and contains collagen and elastic fibers which help support the skin (choice B is wrong). The hypodermis, also known as the subcutaneous tissue, is primarily fat (choices C and D are wrong). Concepts tested Skin: Layers

Answer 44

C. Without surfactant, the smaller airways (alveolar ducts/alveoli) collapse and are held together by the surface tension of water. This results in reduced oxygen and carbon dioxide exchange, which leads to respiratory acidosis (choices A and B can be eliminated and choice C is correct). Larger airways, such as bronchi, are structurally supported by connective tissue and are much less likely to collapse (choice D is wrong). Concepts tested Respiratory System: Conduction and Respiratory Zones Respiratory System: Ventilation and pH Regulation

Answer 45

C. After aspirating the toy car, a portion of the lung (or the entire lung) being supplied by that bronchus will no longer participate in gas exchange. This results in a smaller quantity of oxygen being absorbed (decreasing blood oxygen levels, choices A and B can be eliminated) and an increased quantity of carbon dioxide accumulating (choice D can be eliminated and choice C is correct). Concepts tested Respiratory System: Ventilation and pH Regulation

Answer 46

D. The normal buffer of blood, bicarbonate, has been removed and carbonic anhydrase, which facilitates the reaction of carbon dioxide and water to form carbonic acid, has been inhibited. This results in blood pH being independent of carbon dioxide concentration (choices A and B are not correct) and no significant shift in the hemoglobin saturation curve would take place (choice D is correct). If the pH of the synthetic blood was changed, we would still observe the expected shifts in the hemoglobin/oxygen saturation curve as no changes in hemoglobin were made in this setup (choice C is not correct). Concepts tested Cardiovascular System: Blood

Answer 47

A. Hyperventilation results in reduced blood carbon dioxide levels, which leads to an increase in blood pH and an increase in hemoglobin’s affinity for oxygen (choice A is correct). The other answer choices all result in a decrease in oxygen affinity (choices B, C and D are wrong). Concepts tested Respiratory System: Ventilation and pH Regulation

Answer 48

C. Anatomical dead space is the portion of the respiratory tree where gas exchange does not take place, also known as the conduction zone. The conduction zone includes the nose, pharynx, larynx, trachea, bronchi, bronchioles, and terminal bronchioles (choice C is correct). The respiratory zone includes the respiratory bronchioles, alveolar ducts, and alveoli (choices A, B and D are wrong). Concepts tested Respiratory System: Conduction and Respiratory Zones

Answer 49

D. Items I and II are true: hyperventilation causes the loss of excess carbon dioxide, which shifts the equilibrium of the respiratory equation (CO2 + H2OEquilibrium arrow.H2CO3Equilibrium arrow.H+ + HCO3–) to the left. As H+ and HCO3– combine to form H2CO3, blood pH increases (choices A and C can be eliminated). Item III is true: as blood pH increases, the oxygen saturation curve of hemoglobin shifts to the left. A left-shifted curve means that hemoglobin's affinity for oxygen is increased, and it will be more saturated (choice B can be eliminated and choice D is correct). Concepts tested Respiratory System: Ventilation and pH Regulation

Answer 50

D. Pulmonary surfactant decreases water’s surface tension (decreasing the cohesive forces of water) to help prevent collapse of the small airways in the lung (choice D is correct, and choices B and C are wrong). Pulmonary surfactant does not have a direct impact on alveolar elasticity although the decrease in cohesive force effectively reduces alveolar collapsing forces (i.e., a decrease in alveolar elasticity; choice A is wrong). Concepts tested Respiratory System: Conduction and Respiratory Zones

Answer 51

B. Bacteria are very hardy cells. While the original culture was in stationary phase, all it takes is more food to get the cells growing again (choice C is wrong). Initially, the new culture will undergo a lag phase as the bacteria replicate their genomes and get ready to divide (choice A is wrong), but then the cells will start undergoing binary fission and the culture density will increase exponentially. This is termed the log phase (choice B is correct). If the culture were to reach stationary phase again after undergoing logarithmic growth, it is more likely to be turbid (not clear) due to the high concentration of bacteria. Furthermore, depending on media and other conditions, we cannot be certain of the culture reaching the stationary phase again overnight (choice D is wrong). Concepts tested Microbiology: Bacteria

Answer 52

D. Gram-negative bacteria have a thin cell wall (choices A and C are wrong) and an outer membrane which leads to a light pink staining. The bacterial cell wall is made of peptidoglycan; note that the plant cell wall is made of cellulose and the fungal cell wall is made of chitin (choice B is wrong). Bacilli means rod-shaped, cocci means round, and spirochetes or spirilla mean spiral-shaped (choice D is correct). Concepts tested Microbiology: Bacteria

Answer 53

B. In the lysogenic cycle, the viral genome inserts into the host genome. The host genome continues to be transcribed and translated, but the viral genome is silent due to viral-encoded repressor proteins (choice A is wrong). Both the host and viral genomes are copied during DNA replication (choice B is correct). The viral genome is activated and excised when the host cell is stressed, and will enter the lytic or productive cycle. If the virus waits until the host cell is dead, it will certainly not be able to reproduce (choice C is wrong). Finally, excision of the viral genome is relatively imprecise. Some of the host genome can also be excised and packaged into viral particles. This new bit of DNA can be transferred to the next host on subsequent infection (transduction, choice D is wrong). Concepts tested Microbiology: Viruses and Subviral Particles

Answer 54

B. While there are many similarities between the lytic and productive viral cycles, there are many differences also. In both cycles, the virus takes over certain parts of the host cell, to allow production of viral proteins and to replicate the viral genome (choice B is correct). At the end of a lytic cycle the host cell lyses and dies to allow release of viral particles, however in the productive cycle, viruses bud out of the host cell; this allows the host to survive, and generates viruses with an outer envelope (choice A is only true of the lytic cycle and can be eliminated). The viral genome integrates into the host genome in the lysogenic cycle (choice C is not true of either the lytic or productive cycles and can be eliminated). Hydrolase is a lytic cycle early gene, and lysozyme is a lytic cycle late gene (choice D is only true of the lytic cycle and can be eliminated). Hydrolase degrades the host genome to generate dNTP building blocks. Lysozyme degrades the bacterial cell wall, allowing bacterial lysis. Concepts tested Microbiology: Viruses and Subviral Particles

Answer 55

A. A plaque is a clear area on a plate otherwise covered in bacterial cells. It can be caused by addition of a toxin, antibiotic, or virus; each of these can kill bacteria and would generate a clear area. Since whatever was in the plaque did not outright kill the actively growing culture, it is not likely a toxin or antibiotic (choice C is wrong). Most likely, the plaque was a region of dead bacteria due to viral infection, and contains active virus. Infection of new cultures with active virus can sometimes result in new strains, if the virus transfers some DNA from its previous host; this process is known as transduction (choice A is the most likely scenario). Since a plaque is typically an area of dead bacteria, choice D is wrong; dead bacteria cannot undergo conjugation. Random genomic mutations are possible but less likely than transduction (choice B is wrong). Concepts tested Microbiology: Bacteria

Answer 56

D. All viruses are very specific regarding their host cell. This specificity comes from the requirement of a virus to bind a host cell receptor to allow viral entry (choice A is wrong). Attachment can also be called adsorption, and penetration can also be called eclipse (choice B is wrong). Prokaryotic viruses lack an envelope and therefore cannot undergo membrane fusion with their host. Instead, they rely on injection of viral contents (choice C is wrong). Animal viruses can enter their host cell via membrane-fusion (where the envelope of the virus fuses with the host plasma membrane) or endocytosis (choice D is correct). In both cases, receptor binding is still required, and the viral genome is uncoated after entry into the host. A few animal viruses insert their genome without being taken up entirely, but this is rare. Concepts tested Microbiology: Viruses and Subviral Particles

Answer 57

C. The Hfr strain has the F factor in the genome, male (F+) bacteria contain the F factor as a plasmid, while the female bacteria (F–) has no F factor. Both male (F+) and Hfr strains can mate with female strains (choice D is wrong). The Hfr cell keeps a copy of its genome (choice B is wrong) and passes a copy to the female cell, through the conjugation bridge (or sex pilus). The female cell can receive a portion of the Hfr chromosome, or a copy of the whole thing (if the cells stay connected for a longer time). The F factor is the final gene transferred during conjugation; if the F factor doesn’t make it over to the female cell, she will stay female (but may acquire new genetic traits). If the F factor is transferred, it can end up in a plasmid (the cell would then be male, F+) or the genome (the cell would then be Hfr). While choice A is possible, choice C is more likely. Concepts tested Microbiology: Bacteria

Answer 58

B. (+) RNA viral genomes can serve as templates for transcription, but (–) RNA genomes are complementary to a transcript template (choice C is true and can be eliminated). As a consequence, both must encode an RNA-dependent RNA polymerase to replicate the genome (choice A is true of (+) and (–) RNA viruses and can be eliminated), but a (–) RNA virus must also carry a copy of this protein, to generate functional mRNA from the genome (choice B is false and the correct answer choice). Since a (+) RNA genome can serve as a template for protein production (i.e. mRNA), if a (+) RNA viral genome was injected into a host cell, it would be immediately infective (choice D is true and can be eliminated). Concepts tested Microbiology: Viruses and Subviral Particles

Answer 59

D. All these statements are true and accurate except choice D. Cyanobacteria perform photosynthesis, but since bacteria contain no membrane-bound organelles (including chloroplasts), the photosynthetic machinery is located in the cell membrane and the cytoplasm (similar to the cell respiration machinery). Note that it is thought that chloroplasts in plants and algae evolved from an endosymbiotic relationship with cyanobacteria. Concepts tested Microbiology: Bacteria

Answer 60

C. Chemotrophs get their energy from chemicals, not light (choice D is wrong). Heterotrophs rely on organic molecules made by other organisms; they cannot make them from CO2 (choice B is wrong). A leucine auxotroph cannot synthesize the amino acid leucine; to survive, leucine must be provided for the organism in its medium (choice A is wrong and choice C is correct). Concepts tested Microbiology: Bacteria

Answer 61

D. Auditory hair cells respond to vibrations of the basilar membrane and bend when vibrations occur. Thus they are classified as mechanoreceptors (choice D is correct). Thermoreceptors respond to temperature (choice A is wrong), nociceptors respond to pain (choice B is wrong), and chemoreceptors respond to chemicals (choices A, B, and C are wrong). Concepts tested Nervous System: Sensory Systems

Answer 62

D. The experiment describes two neurons synapsing with a third neuron from which membrane potential is recorded. When one neuron is excited, a depolarization is recorded, and when the second neuron is excited at the same time, a larger depolarization is recorded. This additive effect (two neurons synapsing on a single neuron and generating a larger response) is known as spatial summation (choice D is correct). Temporal summation involves the additive response due to repeated stimulation of a single neuron (choice C is wrong). Long-term depression is a decrease in neural excitability (choice A is wrong) and neural adaptation is a change in response over time (usually a decrease in firing) with a constant stimulus (choice B is wrong). Concepts tested Nervous System: Neurons and Action Potentials

Answer 63

B. By increasing the permeability of the neuron membrane to potassium, the researcher shifted the resting membrane potential in a more negative direction. This pushes the neuron further from threshold and it would require more stimulation in order to generate action potentials, resulting in a probable decrease in firing rate (choice B is correct and choice A is wrong). The Na+/K+ ATPase is an active transporter, using ATP to move Na+ ions out of the cell and K+ ions into the cell; its action would be unaffected by an increase in membrane potassium permeability (choice C is wrong). The rapid depolarization seen in an action potential is due to the opening of voltage-gated sodium channels. Assuming the cell could get to threshold, the opening of these channels should not be affected by an increase in potassium permeability (choice D is wrong). Concepts tested Nervous System: Neurons and Action Potentials

Answer 64

D. The corpus callosum connects the left and right hemispheres of the brain and in select patients, severing this connection can decrease the severity of recurrent seizures (choice D is correct). Severing the brain stem, which is critical for numerous basic vital functions (e.g. breathing), would kill the patient (choice A is wrong). The cerebral cortex, which is the outer layer of the brain, is critical for higher level thought and self-awareness (choice B is wrong), and the thalamus is involved in relaying sensory and motor information to and from the brain (choice C is wrong). Concepts tested Nervous System: Central Nervous System

Answer 65

B. Loud sounds produce large amplitude vibrations (choices C and D can be eliminated). High pitched (high frequency) sounds vibrate the stiffer region of the basilar membrane near the oval window, while low pitched (low frequency) sounds vibrate the more flexible apex of the basilar membrane (choice B is correct and choice A is wrong). Concepts tested Nervous System: Sensory Systems

Answer 66

B. Increased time required for inactivation would result in a lengthened absolute refractory period which could decrease the maximal rate, or frequency, of firing (choice B is correct and choices A and D are wrong). Increasing the time required for inactivation would not alter the maximal sodium conductance rate (choice C is wrong). Note that answer choices A and D are identical and both cannot be correct. Concepts tested Nervous System: Neurons and Action Potentials

Answer 67

D. Continuous myelination of the axon would decrease the ability of an action potential to reach the axon terminus (choice D is correct). Under normal conditions, myelin surrounds the plasma membrane and prevents ion flow, resulting in an increase in membrane resistance (choices A and C are wrong). Breaks in the myelin sheath along the axon, known as nodes of Ranvier, are where voltage-gated sodium channels exist at high concentrations. These nodes effectively boost the signal as the action potential travels along the axon. Without these nodes, the ability of that initial action potential to reach the axon terminus would be greatly impaired. The threshold of voltage-gated sodium channels is unlikely to change with myelination (choice B is wrong). Concepts tested Nervous System: Neurons and Action Potentials

Answer 68

B. Rod cells are unique in that they are depolarized at rest by the constant influx of Na+. Stimulation by light causes the Na+ channel to close (choices C and D are wrong) and the rod cell to hyperpolarize (choice B is correct and choice A is wrong). Concepts tested Nervous System: Sensory Systems

Answer 69

C. The stem of the question informs us that curare has an impact on postsynaptic receptors. This makes it unlikely to have any effect presynaptically, such as neurotransmitter release (choice A can be eliminated) or calcium influx into the presynaptic terminal (choice B can be eliminated). Voltage-gated calcium channels are present on the presynaptic neuron and are also not likely to be affected by a compound which blocks postsynaptic receptors (choice D can be eliminated). The only answer choice which clearly has a postsynaptic action is choice C; often neurotransmitter receptors on postsynaptic neurons are linked to ion channels. The opening of these ion channels impacts ion flow into the postsynaptic cell. If curare blocked the receptors from binding neurotransmitter, the ion channels could not be opened, leading to a decrease in cation flow into the postsynaptic neuron and ultimately, prevention of action potential transmission (choice C is correct). Curare is a competitive inhibitor of acetylcholine which is a neurotransmitter that acts, among other places, at the neuromuscular junction. Note that you did not have to know anything about curare specifically to answer this question. Concepts tested Nervous System: Synaptic Transmission

Answer 70

C. The parasympathetic division is also known as the "rest and digest" division, and leads to effects in the body such as increased digestive activity (choice A Is likely to occur and can be eliminated), reduced air flow (choice B is likely to occur and can be eliminated), and reduced heart rate (choice D is likely to occur and can be eliminated). However distance vision is a feature of the sympathetic nervous system; the parasympathetic division enhances near-object vision (choice C is least likely to occur and is the correct answer choice). Concepts tested Nervous System: PNS/Autonomic Nervous System

Answer 71

A. The medulla contains the main respiratory rhythmicity centers that control ventilation rate, and also contains the central chemoreceptors that would monitor CSF pH (choice A is correct). The hypothalamus maintains body homeostasis, including body temperature, ion balance, hunger/thirst, etc., primarily by controlling the pituitary gland (choice B is wrong). The cerebellum helps to smooth and coordinate body movement (choice C is wrong), and the corpus callosum is the bridge between the two hemispheres of the brain (choice D is wrong). Concepts tested Nervous System: Central Nervous System

Answer 72

B. Temporal summation is the repeated activation of a given synapse resulting in a response of greater magnitude. In order to accomplish this, the rate of excitation must outpace the rate of recovery or else the postsynaptic neuron will have returned to its resting state before the second stimulation arrives (choice B is correct and choice A is wrong). As temporal summation results from the repeated stimulation of a single postsynaptic neuron, proximity is not an issue (choice C and D are wrong). Concepts tested Nervous System: Neurons and Action Potentials

Answer 73

D. The region of the brain that processes olfactory sensation (smell) is the temporal lobe. The parietal lobe is for general sensations of pain, pressure, and touch (choice A is wrong), the occipital lobe processes visual sensation (choice B is wrong), and the frontal lobe controls voluntary movement and problem solving (choice C is wrong).

Answer 74

C. Rod cells are unique in that they are depolarized at rest by the constant influx of Na+. Stimulation by light causes the Na+ channel to close and the rod cell to hyperpolarize (choices A and B are wrong). Rod cells synapse with bipolar cells, which synapse with ganglion cells (choice C is correct and choice D is wrong). Concepts tested Nervous System: Sensory Systems

Answer 75

D. When the resident is asked to assist on his first surgery, it triggers a fight-or-flight response, i.e., activates the sympathetic division of the autonomic nervous system. Increased heart rate, bronchial dilation, and increased blood flow to skeletal muscle are all typical responses of sympathetic activity (choices A, B and C are sympathetic responses and can be eliminated). Pupillary constriction is typical of a parasympathetic response (choice D is not a sympathetic response and is the correct answer choice). Concepts tested Nervous System: PNS/Autonomic Nervous System

Answer 76

C. The cerebellum plays an important role in motor control and coordination (choice C is correct), which are impaired in this patient. The cerebrum is involved in many functions, typically associated with a particular lobe. For example, the frontal lobe controls voluntary movement and problem solving, the parietal lobe controls general sensation, the temporal lobe controls hearing and memory, and the occipital lobe controls vision. The frontal lobe is involved in movement, but it generates the signal for movement to occur, while the cerebellum smooths and coordinates the movement. Since the movement is occurring, although in a less coordinated manner, the cerebellum is more likely to be the problem than the cerebrum (choice A is wrong). The hypothalamus is involved in many processes to maintain body homeostasis, including body temperature, hunger/thirst, and sleep (choice B is wrong), and the corpus callosum is the bridge between the two hemispheres of the brain (choice D is wrong). Concepts tested Nervous System: Central Nervous System

Answer 77

A. Pitch is determined by location, and loudness is determined by amplitude (choices C and D can be eliminated). High pitched (high frequency) sounds vibrate the stiffer region of the basilar membrane near the oval window, while low pitched (low frequency) sounds vibrate the more flexible apex of the basilar membrane (choice A is correct and choice B is wrong). Concepts tested Nervous System: Sensory Systems

Answer 78

A. Resting membrane potential is dictated by the relative ion concentrations inside and outside of the cell as well as by their relative permeabilities. At rest, the resting membrane potential of a typical neuron is –70 mV which arises primarily from the potassium leak channels present in the membrane (the equilibrium potential of potassium is –90 mV). There is relatively little sodium permeability across the membrane under resting conditions and an increase would significantly depolarize the membrane (bringing it closer to sodium’s equilibrium potential of +40 mV, choice A is correct). An increase in potassium permeability may hyperpolarize the membrane, but this would have a less significant effect than an increase in sodium permeability (choice B is wrong). Calcium cannot cross the plasma membrane; simply altering its concentration without changing membrane permeability to calcium would not affect resting membrane potential (choices C and D are wrong). Concepts tested Nervous System: Neurons and Action Potentials

Answer 79

D. RNA polymerase II is responsible for the production of mRNA in the nucleus of eukaryotic cells. In neurons, the cell body (or soma) houses the nucleus and the majority of metabolic activity (choice D is correct). The axon terminal is the end of an axon which frequently forms a synapse with another neuron or effector organ (choice A is wrong). The axon hillock is the region where the axon and soma meet and is important in action potential initiation (choice B is wrong). Dendrites are offshoots from the soma which receive input from other neurons (choice C is wrong).

Answer 80

D. The relative refractory period results from the hyperpolarization of the membrane due to the slow deactivation of voltage-gated potassium channels. To extend the relative refractory period, we would need to decrease the rate of deactivation of these channels (i.e., make them deactivate even more slowly). This would result in an extended period of hyperpolarization where a greater stimulus would be required in order to fire an action potential (choice D is correct and choice B is wrong). Changing the rate of voltage-gated sodium channel inactivation would change the absolute refractory period, not the relative refractory period (choices A and C are wrong). Concepts tested Nervous System: Neurons and Action Potentials

Answer 81

B. The sympathetic division of the ANS primarily releases norepinephrine (NE) at the organ level; the only significant exception is the sweat glands, where it releases acetylcholine (ACh; choice B is correct). ACh is the primary neurotransmitter of the parasympathetic division (choice A is wrong). Epinephrine is a hormone; its release into the blood is triggered by the sympathetic nervous system and it can stimulate similar receptors as norepinephrine (choice C is wrong). Serotonin is a neurotransmitter used in the brain; it is not secreted by autonomic neurons (choice D is wrong). Concepts tested Nervous System: PNS/Autonomic Nervous System

Answer 82

B. If the eye-color trait is being inherited differently in males and females, it must be a sex-linked trait. Since both females and males are affected, it must be an X-linked trait. The white-eyed female in the original cross must be XrXr and the male is XRY. All female F1 flies are XRXr (red-eyed) and males are XrY (white-eyed). If these two mate (XRXr × XrY), the F1 offspring will be 25% XRXr (red-eyed females that carry the white-eye allele; choice D would be observed and can be eliminated), 25% XrXr (pure-breeding white-eyed females; choice A would be observed and can be eliminated), 25% XRY (red-eyed males that cannot produce white-eyed daughters) and 25% XrY (white-eyed males that can produce white-eyed daughters; choice C would be observed and can be eliminated). Note that half the F2 generation has red eyes and half has white eyes (choice B would NOT be observed and is the correct answer choice). Concepts tested Genetics: Mendelian Genetics/Probability

Answer 83

D. All humans receive their mitochondrial genome from their mother’s egg, so mitochondrial traits are passed from mothers to all offspring (choices A and B can be eliminated). If the trait does not pass from fathers to sons, it is likely X-linked. X-linked recessive traits can be masked by a mother’s dominant X chromosome in female offspring. If the trait is showing up in all of their daughters, it is most likely an X-linked dominant trait (choice C can be eliminated and choice D is correct). In this case, when a father passes this X chromosome to a daughter, the allele for this trait is expressed phenotypically. Concepts tested Genetics: Mendelian Genetics/Probability

Answer 84

A. Since the ability to taste phenylthiocarbamide is dominant we will assign it an allele frequency of p. From the equation for genotype frequency (p2 + 2pq + q2 = 1), individuals who have this trait will make up p2 + 2pq of the population, which the question states is 70%, or 0.70. The recessive allele leads to tasteless phenylthiocarbamide (choices B and D can be eliminated) and occurs at a frequency of q, so individuals who cannot taste phenylthiocarbamide occur at a frequency of q2. Since the question stem tells you p2 + 2pq = 0.70, q2 = 1 – 0.70 = 0.30, and q would be the square root of this number, or approximately 0.55 (choice A is correct and choice C is wrong). Concepts tested Genetics: Hardy-Weinberg

Answer 85

B. Recombination frequency gives a measure of map distances between genes on the same chromosome (choice B is correct). While sequencing a chromosome can determine how many base pairs apart the genes are, it will not give map unit information directly (choice A is wrong). Determining the length of a chromosome or the ploidy of the cell will give no additional information on map distance (choices C and D are not relevant and can be eliminated). Concepts tested Genetics: Linked Genes

Answer 86

B. Nondisjunction is a meiotic anaphase separation defect. It leads to too many or too few chromosomes (also known as aneuploidy) in gametes and resultant zygotes. Trisomy is when a cell contains three copies of a given chromosome, instead of two, and is an example outcome of nondisjunction (choice B is correct). There is no reason to believe that aneuploidy would result in the loss of a portion of a chromosome (choice A is incorrect), an enzymatic deficiency (choice C is incorrect) or thyroid hormone deficiency (choice D is incorrect). Concepts tested Cell Biology: Mitosis/Meiosis

Answer 87

C. If we assign D = normal and d = hemophiliac, the cross in the question stem is XDXd × XdY. If the offspring is female, she will receive the Xd chromosome from her father, and the probability of this is 0.5. To have a normal phenotype, she must receive the XD chromosome from her mother, and the probability of this is 0.5. Overall, the probability of having a normal daughter is (0.5)(0.5) = 0.25, (choice C is correct). Concepts tested Genetics: Mendelian Genetics/Probability

Answer 88

B. Let’s assign the two parental fish AABB × aabb. The F1 fish would be AB/ab (in linkage notation) and is mated with a ab/ab fish in a testcross. If the two genes are linked and 42 cM (centimorgans, a measure of distance on the chromosome) apart, 42% of offspring would result from crossing over in the heterozygous parent (generating aB or Ab gametes, choices C and D are wrong) and the remaining 58% would have parental combinations of alleles (AB or ab). In other words, we would expect 29% AB/ab and 29% ab/ab (the parental combinations totaling 58%; choice B is correct and choice A is wrong), and 21% Ab/ab and 21% aB/ab (recombinant combinations totaling 42%). Concepts tested Genetics: Linked Genes

Answer 89

D. The cross in the question stem is AaBbCc × AaBBcc. For the first locus, three genotypes will be possible (25% AA, 50% Aa and 25% aa) but only two phenotypes: 75% dominant A and 25% recessive a (choice A is false and can be eliminated). The worms will be 50% BB and 50% Bb at the second locus (choice B is false and can be eliminated) but will all express the dominant B phenotype. For the third locus, the worms will be 50% Cc and 50% cc, and the phenotypic ratios will be the same (choice C is false and can be eliminated). Therefore, the answer must be D; there are two possible phenotypes at the first locus, one possible phenotype at the second locus, and two possible phenotypes at the third locus. (2)(1)(2) = 4 possible phenotypes. Concepts tested Genetics: Mendelian Genetics/Probability

Answer 90

D. If the male produces offspring with four differing phenotypes in a testcross, he must be heterozygous for both genes (SsCc, where S = stubbled, s = long bristles, C = curly wings, c = straight wings). Crossing that male with a female of the same genotype is SsCc × SsCc. The probability of long bristles (ss) is 0.25. The probability of straight wings (cc) is 0.25. The probability of a long-bristled straight-winged offspring is (1/4)(1/4) = 1/16. Note also that this is a dihybrid cross and offspring are produced in a 9:3:3:1 ratio; 9 dominant/dominant (stubbled/curly), 3 dominant/recessive (stubbled/straight), 3 recessive/dominant (long/curly), and 1 recessive/recessive (long/straight). So only 1 of the 16 has the phenotype in question. Concepts tested Genetics: Mendelian Genetics/Probability

Answer 91

C. If the Adams family parents are pure-breeding (or homozygous), then choice A is true and can be eliminated. All offspring would be IAIBRR and these individuals are called universal acceptors because they can accept all blood types without complications. If the Wong family parents are heterozygous for the ABO blood group (IAi × IBi), and if the Wong husband is heterozygous for the Rh antigen (Rr), then both choices B and D are possible and both can be eliminated (note that the Wong mother must be homozygous recessive for the Rh antigen as she is Rh negative). Some of the children could be blood type O– (iirr) which are universal donors and a blood type different from either parent. In each family, some grandparents could be blood type O (choice C is not true and is the correct answer choice). For example, IBRr (type B+) and iirr (type O–) grandparents could give rise to the Wong family mother. Concepts tested Genetics: Mendelian Genetics/Probability

Answer 92

D. The question states that in this population, polydactyly is caused by a dominant allele. In the Hardy Weinberg equation for allele frequency (p + q = 1), p is the frequency of the dominant allele. If p = 0.02, then q = 0.98. Plugging these numbers into the equation for genotype frequency (p2 + 2pq + q2 = 1), q2 (the frequency of autosomal recessive individuals, and thus UNaffected individuals) is equal to (0.98)2 = 0.96; or in other words, 96% of the population is NOT affected. Therefore, 4% of the population IS affected. Concepts tested Genetics: Hardy-Weinberg

Answer 93

A. The centrosome is duplicated before meiosis, during S-phase (choice B is wrong). Crossing over happens in prophase I, also before metaphase I (choice C is wrong). The spindle (made of microtubules) starts attaching to chromosomes in prometaphase I (in between prophase and metaphase) and is fully attached by metaphase I (choice A is correct; note that the attachment of the spindle fibers and subsequent pulling of the chromosomes is what allows homologous chromosomes to line up across from each other at the metaphase plate). Cytokinesis begins near the end of anaphase and finishes at the end of telophase (choice D is wrong). Concepts tested Cell Biology: Mitosis/Meiosis

Answer 94

C. Since the ATPase moves more ions out of the cell than in, and since in the absence of leak channels these ions have no way to cross the cell membrane, the cytoplasm would ultimately become hypotonic, meaning it has less particles when compared to the extracellular environment (choice C is correct). Hypertonic (or hyperosmotic) would mean that the cytoplasm had more particles than the extracellular environment (choices A and D are wrong), and isotonic would mean they had the same number of particles (choice B is wrong). Concepts tested Cell Biology: Osmosis/Diffusion/Colligative Properties

Answer 95

A. In secondary active transport, energy is used to establish an electrochemical gradient, and that gradient then provides the driving force to move other molecules against their concentration gradient. The use of energy is one step removed from the molecule movement, hence the designation of "secondary." The Na+-glucose cotransporter moves glucose against its gradient as Na+ moves down its gradient. The Na+ gradient is established by the Na+/K+ ATPase, a primary active transporter (uses ATP directly, choice D is wrong). Ligand-gated ion channels are opened by the binding of a ligand to a receptor; movement of the ion then occurs passively down its gradient (choice B is wrong). K+ leak channels are an example of facilitated diffusion; K+ moves across the membrane, down its gradient, through the channel (choice C is wrong). Concepts tested Cell Biology: The Cell Membrane

Answer 96

A. Colligative properties, such as boiling point elevation, freezing point depression, vapor pressure depression, and osmotic pressure only depend on the number of particles in solution. If Solution X has a greater boiling point elevation than Solution Y, it must have more particles than (i.e., be hypertonic to) Solution Y (choice A is correct and choice C is wrong). In osmosis, water moves from a hypotonic solution (e.g., Solution Y) into a hypertonic solution (e.g., Solution X, choice D is wrong). Only knowing boiling point elevation does not tell us anything about the solution’s ability to dissociate, i.e., whether or not it is an electrolyte. Solution X might be 1 M MgCl2 and Solution Y might be 1 M NaCl. Both would be electrolytes, and Solution X would still have more particles (choice B is wrong). Concepts tested Cell Biology: Osmosis/Diffusion/Colligative Properties

Answer 97

D. If recombination is to occur, it happens during prophase I of meiosis, not during the cell cycle, which is loosely composed of mitosis and interphase (choice D is not a cell cycle event and is the correct answer choice). The genome is replicated during S phase (choice A is a cell cycle event and can be eliminated). Transcription and translation (i.e., protein synthesis) happen during G1 and G2 (choices B and C are cell cycle events and can be eliminated).

Answer 98

B. The component constituent to the amino acid chain is the signal sequence on the N-terminus; the trafficking is not done from the C-terminus (choices A and C are wrong). The signal sequence is bound by the signal recognition particle to ensure movement to the ER lumen (choice B is correct). Single-strand binding proteins are involved in DNA replication, not protein trafficking (choice D is wrong). Concepts tested Cell Biology: Secretory Pathway

Answer 99

D. Osmosis is the passive movement of a solvent down its concentration gradient, but this movement will be impacted by the overall mix of solutes in a given solution. If two solutions are hypertonic relative to two separate ions, but overall have equivalent ion concentration (i.e. are isotonic to each other), osmosis will not occur (choice A is wrong). Similarly, if two solutions are hypotonic relative to two separate ions, but overall have equivalent ion concentrations (are isotonic to each other), osmosis will not occur (choice B is wrong). An isotonic balance will not promote osmosis (choice C is wrong), but a hypertonic solution next to a hypotonic solution (regardless of the specific ion) will (choice D is correct). Concepts tested Cell Biology: Osmosis/Diffusion/Colligative Properties

Answer 100

B. Red slow-twitch muscle fibers have many mitochondria and a good blood supply so they are very fatigue-resistant. Item I is true: the posture muscles of the back need to be fatigue-resistant because they are in use consistently (choice C can be eliminated). Item II is true: the thigh muscles of marathon runner participate in endurance activity and need fatigue-resistant fibers (choice A can be eliminated). Item III is false: the calf muscles of pole-vaulters are in use for a very short period of time and likely contain white fast-twitch fibers (Type IIB fibers; choice D can be eliminated and choice B is correct). Concepts tested Muscular System: Skeletal Muscle Fiber Types

Answer 101

D. During the primary bone growth and development of childhood, growth hormone causes the deposition of cartilage at the epiphyseal plate to lengthen the bone (choices A and B are wrong). This cartilage is soon replaced with the long-term bone components of collagen and hydroxyapatite (a calcium-phosphate crystalline solid) (choice C is wrong and choice D is correct). Concepts tested Skeletal System: Connective Tissue/Bone Skeletal System: Bone Structure/Turnover

Answer 102

A. All types of muscle cells contain the proteins actin and myosin (choices B and C are shared and can be eliminated). Calcium plays a role in the regulation of all muscle types (choice D is shared and can be eliminated). Troponin is a regulatory protein in cardiac and skeletal muscle, but not in smooth muscle, which utilizes calmodulin instead (choice A is not shared and is the correct answer choice). Concepts tested Muscular System: Cardiac and Smooth Muscle

Answer 103

D. Calcium regulates the contraction of skeletal muscle fibers (choices A and B are wrong) because of its interaction with troponin. When calcium levels rise in the cell, troponin (bound to tropomyosin) binds calcium and changes shape; this shape change shifts the position of tropomyosin and uncovers the myosin binding sites on actin, allowing contraction to proceed (choice C is wrong and choice D is correct). Concepts tested Muscular System: Sliding Filament Theory/Excitation-Contraction Coupling

Answer 104

B. The A band represents the complete length of the thick filament (myosin) which is neither shortened nor lengthened during contraction (choices A and D are wrong). The Z lines demarcate the individual sarcomeres and are also not altered by contraction (choice C is wrong). The H zone is the region measuring thick filament only with no overlap and this does change as sliding occurs; similarly, the I band is the region measuring the thin filament only (actin) and also changes as sliding occurs (choice B is correct). Concepts tested Muscular System: Skeletal Muscle Structure and Sarcomeres Muscular System: Sliding Filament Theory/Excitation-Contraction Coupling

Answer 105

C. All types of muscle depend on voltage-gated sodium channels for depolarization (choice A is wrong) and voltage-gated potassium channels for repolarization (choice B is wrong). Potassium leak channels are important in establishing the resting membrane potential in all muscle types (choice D is wrong). However, only cardiac muscle cells have voltage-gated calcium channels; these are used to extend depolarization and create the distinctive plateau phase seen in cardiac muscle action potentials (choice C is correct). Concepts tested Muscular System: Cardiac and Smooth Muscle

Answer 106

A. Since the question asks for structures from largest to smallest, the biggest structures are the bundles of muscle tissue called fascicles (choices C and D are wrong). Fascicles are composed of fine muscle fibers called myofibers which are individual muscle cells that contain myofibrils. Finally, the myofibrils are subdivided into many sarcomeres (choice B is wrong and choice A is correct). Concepts tested Muscular System: Skeletal Muscle Structure and Sarcomeres

Answer 107

C. Connective tissues are classified as "loose" or "dense" based on the amount of collagen they contain; those with a lot of collagen are dense tissues, and tend to play more supportive roles. Ligaments connect bones to bones and are made primarily of collagen, thus they are a type of dense connective tissue (choice C is not loose connective tissue and is the correct answer choice). Adipose tissue, or fat, contains virtually no collagen and is a type of loose connective tissue (choice A can be eliminated). Collagen is found in both loose and dense connective tissue; the difference between the two is the amount of collagen (choice B is a component of loose connective tissue and can be eliminated). The extracellular matrix is the mix of proteoglycans and glycosaminoglycans that exists as the tissue between cells, It is responsible for holding on to water (among other functions); it is a component of loose connective tissue and many other tissues (choice D can be eliminated). Concepts tested Skeletal System: Connective Tissue/Bone

Answer 108

A. Mature bone is made rigid by the inclusion of hydroxyapatite, a calcium-phosphate crystalline solid (choices C and D are found in bone and can be eliminated). Collagen forms the framework for mature bone and this framework is then hardened by the crystalline solid (choice B is found in bone and can be eliminated). Cartilage is deposited in growing bone, but is replaced with solid bone matrix as bone matures (choice A is not found in fully mature bone and is the correct answer choice). Concepts tested Skeletal System: Bone Structure/Turnover

Answer 109

D. The epiglottis is found in the respiratory tract, above the opening to the trachea. When swallowing, the epiglottis must fold downward to seal off the trachea thus preventing food and liquid from entering. This requirement for flexibility makes the epiglottis most likely to be made of elastic cartilage. Articular cartilage needs to be strong but somewhat flexible and is made of hyaline cartilage (choice A is wrong), and both the intervertebral disks and the pubic symphysis must be very strong and are made of fibrous cartilage (choices B and C are wrong). Concepts tested Skeletal System: Connective Tissue/Bone

Answer 110

C. Parathyroid hormone and calcitriol (the active form of vitamin D) are involved in stimulating the osteoclasts to phagocytically destroy bone, thus freeing up calcium to be transported by the bloodstream to other parts of the body (note however that calcitriol's effect on bone is minor). They also promote the absorption of calcium from the gut (choice C is correct). Calcitonin inhibits osteoclasts, decreasing bone resorption (choices A and B are wrong), and cortisol is involved in long-term stress responses (choice D is wrong). Concepts tested Endocrine System: Hormones and Hormone Functions Skeletal System: Bone Structure/Turnover

Answer 111

C. Female (or F–) bacteria can undergo conjugation with both male (F+) and Hfr strains (choice A is wrong). Hfr strains can only mate with female bacteria, not other Hfr strains (choice B is wrong). Conjugation increases the number of F+ cells. However, when bacteria divide by binary fission, plasmid distribution in the two daughter cells is relatively unregulated. This can result in a male parental cell giving both male and female daughter cells (choice C is correct). The conjugation bridge (or sex pilus) is formed by the F+ (or Hfr) cell. During conjugation, nucleic acids (either a plasmid or a chromosome) are transferred to the female cell across this bridge (choice D is wrong). Concepts tested Microbiology: Bacteria

Answer 112

A. A retrovirus is an example of a (+) RNA virus (choices B and D can be eliminated) that undergoes the lysogenic cycle. To do this, the RNA genome must be converted to DNA by an RNA-dependent DNA polymerase. This enzyme is commonly called reverse transcriptase (choice A is correct and choice C is wrong). Note that the virus isn't absolutely required to carry a copy of this enzyme (although many do), since the (+) RNA genome can be used as a template for translation. Concepts tested Microbiology: Viruses and Subviral Particles

Answer 113

C. During the lysogenic viral cycle, the viral genome integrates into the host genome but remains mostly silent (choice A is not the best answer). Viral hydrolase breaks down the host genome and this does not occur in the lysogenic cycle (choice B is wrong). With every round of host cell division, the viral genome is replicated (choice C is correct). Bacterial lysis is a characteristic of the lytic cycle, not the lysogenic cycle (choice D is wrong). Concepts tested Microbiology: Viruses and Subviral Particles

Answer 114

A. Transformation is the process by which bacterial cells acquire genetic information from the external environment; this is what occurred in this experiment. Transfection is a similar process, but for eukaryotic cells (choice C is wrong). Transduction requires the presence of a lysogenic phage, and there is no indication in the question stem that one was used here (choice D is not supported). Selective media is used for the growth of selected microorganisms. It can contain an antibiotic which will select for any cell that expresses a resistance gene to this antibiotic (choice A is correct). Many strains of bacteria like growing at approximately body temperature and there is no information to support that this temperature will allow for plasmid selection (choice B is wrong). Concepts tested Microbiology: Bacteria

Answer 115

C. Bacterial cells divide via binary fission and not mitosis (choices A and D are incorrect). If they are uracil auxotrophs, they must be supplied with uracil as they cannot synthesize their own (choice C is correct and choice B is wrong). Concepts tested Microbiology: Bacteria

Answer 116

D. Phototrophs get their energy from the sun and mesophiles live at moderate temperatures such as room or body temperature (choice D is correct). Heterotrophs rely on organic compounds generated by other organisms and psychrophiles live at cold temperatures (choice A is incorrect). Thermophiles live in hot springs (choice B is incorrect). Chemoautotrophs are common in deep sea vents and these organisms build organic molecules from CO2 and use chemicals (commonly hydrogen sulfide) for energy. Since there would be little light in a deep sea vent, this is unlikely to be the home of a phototroph (choice C is incorrect). Concepts tested Microbiology: Bacteria

Answer 117

D. In both the lytic and the productive viral reproduction cycles, the virus takes over parts of the host cell (choice C is wrong). In the lytic cycle, the host genome is commonly degraded to generate free dNTP building blocks. However, the host genome typically stays intact during the productive cycle (choice A is wrong). Viral genome integration into the host genome occurs in the lysogenic cycle, not the lytic or the productive cycles (choice B is wrong). The virus particles are released when the host cell bursts in the lytic cycle, but escape by budding out of the host cell in the productive cycle (choice D is correct). Concepts tested Microbiology: Viruses and Subviral Particles

Answer 118

B. All viruses are very particular about the specifics of their host cell. This specificity comes from the requirement of a virus to bind a host cell receptor, to allow viral entry (choice A is wrong). Receptor binding allows a prokaryotic virus to inject its genome into the host cell (choice D applies to bacterial viruses and can be eliminated). In contrast, animal viruses usually enter the host cell via membrane-fusion (where the envelope of the virus fuses with the host plasma membrane) or endocytosis (choice B is correct). In both cases, receptor binding is still required, and the viral genome is uncoated after entry into the host. Lysozyme is an example of a late gene for prokaryotic viruses, and facilitates viral exit at the end of a lytic cycle. It is not used for viral entry, and in any case, does not apply to animal viruses (choice C is wrong). Concepts tested Microbiology: Viruses and Subviral Particles

Answer 119

A. Gram-positive bacteria have a thick peptidoglycan cell wall (choices B and D are wrong). Facultative anaerobes can survive without oxygen, but will use it if it is present (choice A is correct and choice C is wrong). Obligate anaerobes die in the presence of oxygen, and tolerant anaerobes will survive in oxygen but will not use it. Concepts tested Microbiology: Bacteria

Answer 120

C. Choice C is a true statement and the correct answer choice. The bacterial genome is a circular, double-stranded DNA molecule (choice A is wrong). While many peptide chains can be made from a polycistronic mRNA, this is done by a 70S ribosome in bacteria. The eukaryotic ribosome is 80S (choice B is wrong). The peptidoglycan cell wall prevents lysis due to osmotic pressure, but doesn’t generate osmotic pressure gradients (choice D is wrong). Many eukaryotic cells have aquaporin water channels in the plasma membrane to facilitate osmosis. Concepts tested Microbiology: Bacteria

Answer 121

C. Choice A describes a typical eukaryotic genome and choice B describes a typical prokaryotic genome (both choices are wrong). Choice C is a true statement, and the correct answer choice. Since viral genomes are so variable, there is no reason to believe choice D is true. Concepts tested Microbiology: Viruses and Subviral Particles

Answer 122

B. Cells undergoing meiosis are haploid after the first cytokinesis event (choices A and C are wrong). In prophase II, the sister chromatids have not separated yet (choice B is correct and choice D is wrong). Concepts tested Cell Biology: Mitosis/Meiosis

Answer 123

A. Diffusion is the passive movement of a solute down its concentration gradient (choices B and D are wrong). In order for the movement to be passive, it must be thermodynamically favorable and will increase the entropy or measurement of disorder in the system (choice C is wrong and choice A is correct). Concepts tested Cell Biology: Osmosis/Diffusion/Colligative Properties

Answer 124

C. Cytokinesis refers to physical process of cell division once the replicated genome has been appropriately divided (choices A, B, and D are wrong). This division starts during anaphase and is completed during telophase at which time the nuclear membranes will re-form (choice C is correct). Concepts tested Cell Biology: Mitosis/Meiosis

Answer 125

D. The portion of the Golgi complex closest to the endoplasmic reticulum is its cis stack so this is the first step in trafficking (choice D is correct). From there, proteins travel to the medial stack and finally the trans stack, the Golgi component farthest from the endoplasmic reticulum. At this point, the protein can be released via exocytosis or anchored to the cell membrane, depending on its function (choices A, B, and C are wrong). Concepts tested Cell Biology: Cellular Organelles and Structures

Answer 126

A. Cellular specialization typically leads to a decrease in cell division, i.e., highly differentiated cells do not divide. There is no point to replicating DNA (the S phase of the cell cycle) if division is not going to occur (choice A is correct). Though not dividing, specialized cells can still grow (choice B is wrong), genomes may be packaged or unpackaged as needed in order to access required genes (choice C is wrong), and transcription (and translation) of needed cellular products will also continue (choice D is wrong). Choices B, C, and D all occur during the G1 phase of the cell cycle. Concepts tested Cell Biology: Mitosis/Meiosis

Answer 127

D. Both mitochondria and prokaryotes have a single, distinct chromosome (choice D is correct). Mitochondria are a type of membrane-bound organelle, but do not contain such structures (choice A is wrong). Maternal inheritance refers to the inheritance of the mitochondria and its genome exclusively from the mother (as are all organelles); such a pattern does not exist in prokaryotes (choice B is wrong). Both mitochondria and prokaryotes have ribosomes (choice C is wrong). Concepts tested Cell Biology: Cellular Organelles and Structures

Answer 128

B. Item I is false: apoptosis is programmed cell death, triggered when cell damage is critical. Apoptosis is used to destroy potential cancer cells before they become a problem (choices A and C can be eliminated). Both remaining answer choices include Item II so it must be true: p53 is a tumor suppressor gene, halting cell growth and division when the cell is damaged, or triggering apoptosis if the damage is too severe. Preventing the expression of p53 would prevent these effects and could lead to cancer. Item III is false: protooncogenes are genes that when damaged or mutated, become oncogenes, triggering inappropriate cell division and cancer. If Ras protein expression is prevented, then cancer is less likely (choice D can be eliminated and choice B is correct). Concepts tested Cell Biology: Cancer

Answer 129

D. The alpha and beta tubulin proteins are used to compose microtubules (choice D is correct). The rolled sheet is then anchored in the microtubule organizing center; the free end can be lengthened or shortened as needed. Microfilaments are composed of actin (choice B is wrong), intermediate filaments are a mix of heterogeneous polypeptides (choice C is wrong), and intermediate tubules are not an actual cytoskeletal component (choice A is wrong). Concepts tested Cell Biology: Cellular Organelles and Structures

Answer 130

B. The use of an integral membrane protein makes this facilitated diffusion (choice B is correct) since simple diffusion does not require a transport protein (choice A is wrong). Isotonic and hypotonic do not describe types of diffusion, but rather refer to comparative solute concentrations between two locations (choices C and D are wrong). Concepts tested Cell Biology: The Cell Membrane

Answer 131

B. Item I is true: adenylate cyclase can be directly activated by stimulatory (Gs) G-protein linked receptors (choice C can be eliminated). Item II is true: phospholipase C is another enzyme that can be directly activated by G-proteins (choice A can be eliminated). Item III is false: some tyrosine kinases are cytoplasmic and can be activated as part of the rise in cAMP due to activation of adenylate cyclase by a G-protein. However, this is an indirect activation by a G-protein, and the question specifically asks which are directly activated (choice D can be eliminated and choice B is correct). Concepts tested Cell Biology: G Proteins/Signal Transduction

Answer 132

D. Exocytosis is the process by which the cell uses vesicles to move material out of the cell (choice D is correct). Phagocytosis and pinocytosis are large-scale and small-scale uptake, respectively (choices A and C are wrong), while clathrin-coated pits are involved in receptor-mediated endocytosis (choice B is wrong). Concepts tested Cell Biology: Secretory Pathway

Answer 133

A. The nucleolus assembles ribosome subunits from folded rRNA and various protein components (choice A is correct). Protein synthesis happens on the ribosomes themselves (choice B is wrong). ATP production is the job of the mitochondria (choice C is wrong) while peroxisomes break down lipids not needed by the cell (choice D is wrong). Concepts tested Cell Biology: Cellular Organelles and Structures

Answer 134

A. Cell membranes are composed of a lipid bilayer where the outer facing head groups are polar and the inner facing, intercalating tails are nonpolar. This creates the hydrophobic barrier that is the hallmark of membranes in aqueous environments. Cell membranes are interspersed with proteins, but the number of proteins varies based on the cell type so it is not always equal to the lipid content (choice A is false and is the correct answer choice). Increased lipid saturation makes the membrane more rigid, thus decreasing the ability of proteins within it to move (choice B is true and can be eliminated). To invert, lipids or proteins would have to drag their hydrophilic regions (the parts interacting with the polar edges of the membrane) through the hydrophobic center of the lipid bilayer, and this is not thermodynamically favorable (choice C is true and can be eliminated). Lipid rafts are a common component of membranes and a property of the mosaic model (choice D is true and can be eliminated). Concepts tested Cell Biology: The Cell Membrane

Answer 135

C. Active transport is used to move molecules against their concentration gradients (choices A and B are wrong), hence the need for energy in order to make such movement possible. In primary active transport, that movement is directly linked to the hydrolysis of ATP (choice C is correct and choice D is wrong). Concepts tested Cell Biology: Osmosis/Diffusion/Colligative Properties

Answer 136

C. Although both solutions have the same molarity, they do not have the same number of particles. Because NaCl ionizes in solution into two particles, a sodium ion and a chloride ion, a 1 M NaCl solution actually contains 2 moles of particles. Osmotic pressure is directly related to the number of particles in a solution, so if Solution B has twice the number of particles than Solution A, they do not have the same osmotic pressure (choice A is wrong). Because Solution B has more particles, it has less room for water, and Solution A has the higher “water concentration.” Thus, if they were placed on opposite sides of a semipermeable membrane, water would flow by osmosis from Solution A (fewer particles and more water) to Solution B (more particles and less water). The volume of Solution B would increase (choice B is wrong). Boiling point elevation and freezing point depression are also related to the particle concentration and the van’t Hoff factor. Solution B, with twice the number of particles as Solution A, would have twice the boiling point elevation (choice C is correct) and twice the freezing point depression (choice D is wrong). Concepts tested Cell Biology: Osmosis/Diffusion/Colligative Properties

Answer 137

A. The most energy-intensive process in translation involves the attachment of amino acids to their corresponding tRNAs, also known as tRNA loading (or tRNA charging, or amino acid activation). This requires two high-energy phosphate bonds per aa-tRNA pair (choice A is correct). Initiation requires only 1 GTP (choice B is wrong). Elongation includes A-site binding and translocation (for amino acid number two and onwards); one GTP is hydrolyzed to bring an aa-tRNA into the A site, and one GTP is required for translocation (choice C is wrong). Termination involves the binding of a release factor to a stop codon and does not directly require any energy (choice D is wrong). In this example, it would require 100 ATP to charge a sufficient number of tRNAs to generate the peptide, 1 GTP for initiation, 49 GTP for A-site binding, and 49 GTP for translocation, but note that the answer to this question does not depend upon the number of amino acids in the peptide. Concepts tested Molecular Biology: Translation

Answer 138

A. Any alteration that would decrease the expression of the genes necessary for lactose transport and breakdown would negatively impact the cell’s ability to utilize lactose as an energy source. If the lactose binding site on the repressor could no longer bind lactose, the repressor would be able to bind to the operator region on DNA, but the presence of lactose would no longer cause dissociation of the repressor, and transcription of the necessary genes would not occur (choice A is correct). If a mutation occurred in the operator which prevented repressor binding, transcription of the lac operon genes would always occur (choice B is wrong). Similarly, a mutation in the DNA binding site of the repressor that prevented it from binding to DNA would lead to constitutive gene expression (choice C is wrong). Increased expression of the repressor itself would not necessarily decrease transcription of the lac operon genes, assuming that the regulation of the repressor by lactose is not affected (choice D is wrong). Concepts tested Molecular Biology: Mutations and DNA Repair Molecular Biology: Regulation of Gene Expression

Answer 139

C. In order to ensure that replication is completed in a reasonable length of time, eukaryotes possess multiple origins of replication on each chromosome (choice C is correct). Eukaryotes utilize polyadenylation of RNA transcripts and prokaryotic genes are frequently polycistronic. However, neither of these statements (while true), answer the question because they are about transcription instead of DNA replication (choices A and B can be eliminated). Eukaryotic cells have multiple linear chromosomes while prokaryotic cells have a single circular chromosome (choice D is wrong). Concepts tested Molecular Biology: DNA Replication

Answer 140

D. The template strand is bound directly by the RNA polymerase and functions as a template for RNA transcript generation (choice D is correct). The template strand’s sequence is complementary to the transcript, not identical (choice A is wrong). The coding strand is the complementary strand of DNA (the strand opposite the template strand) and it is nearly identical in sequence to the RNA transcript (T in the coding strand and U in the RNA transcript, choice B is wrong). Transcription initiation is dictated largely by the promoter, which is found on the DNA, not codons which would be found on RNA and are necessary for translation (choice C is wrong). Concepts tested Molecular Biology: Transcription

Answer 141

D. Splicing occurs in the nucleus following transcription (choice B is wrong) and results in the removal of introns, shortening the mRNA transcript (choice D is correct). Since it is the RNA transcript that is altered, no changes in the genetic code, or DNA, occur (choice A is wrong). While splicing is a common post-transcriptional process, it does not occur in every gene; some genes lack introns (choice C is wrong). Note that the word “every” is included in this answer choice. When the word every (or never) is used, be cautious, as exceptions to the rule are a possibility. Concepts tested Molecular Biology: Transcription

Answer 142

C. Numerous differences between DNA and RNA are present in the eukaryotic cell, including the presence of a 2’-OH group in RNA, uracil being substituted for thymine in RNA (choices A and B are wrong), and the short half-life of mRNA relative to a cell’s DNA (choice C is correct). While transcripts of RNA are polyadenylated at the 3’ end, DNA is not (choice D is wrong). Concepts tested Biochemistry: Biomolecules: Nucleic Acid

Answer 143

B. The strain is unable to complete homologous recombination. This process uses many proteins and enzymes, including helicases, nucleases, single-stranded DNA binding proteins, DNA polymerase, and ligases, thus this strain could have a mutation in any one of these components (choice B is correct). Homologous recombination is completed using crossing over and formation of a joint molecule and requires that the cell be actively dividing. Since the strain cannot do this, choices A and C can be eliminated. This strain must be relying on nonhomologous end joining, which is not a very specific method of DNA repair (choice D can be eliminated), but is better than nothing. It does not require active cell growth, and involves identifying and stabilizing the ends of a DSB, then ligating two DSBs together. Concepts tested Molecular Biology: Mutations and DNA Repair

Answer 144

B. The proper sequence for the initiation of translation in eukaryotes begins with the loaded tRNAmet (the initiator tRNA) binding to the small ribosomal subunit. This tRNA is the only one that can bind tightly to the small subunit in the absence of mRNA (choice B is correct). The next step is for the mRNA to bind to the small subunit; this occurs with the help of the 5' guanine cap (choice D is wrong). The small subunit then scans along the mRNA until the initiator codon (AUG) is found; at that point the met-tRNAmet associates with the mRNA in the P site (choice A is wrong). The final step in initiation is the binding of the large ribosomal subunit to this complex (choice C is wrong). Concepts tested Molecular Biology: Translation

Answer 145

B. Given that there is no change in protein function, the mutation must be minimally disruptive. Silent mutations result when a change in the DNA sequence produces a codon that specifies the same amino acid as the unmutated sequence (this is possible because of the degeneracy of the genetic code; amino acids are specified by more than one codon). Since there is no change in the amino acid sequence of the protein, there is no change in protein function (choice B is the best answer). The insertion of a DNA base in the sequence would alter the reading frame of the coding region, alter many amino acids, and would very likely impact function (choice A is wrong). A missense mutation (the substitution of one amino acid for another) may or may not allow for protein function, and if it does allow for function the function may or may not be identical to the unmutated protein. Without knowing more about the specific mutation, we cannot be certain if it is a missense or not, making silent mutation a better answer (choice B is better than choice C). A nonsense mutation results in a premature stop codon; this would likely affect the protein’s function (choice D is wrong). Concepts tested Molecular Biology: Mutations and DNA Repair

Answer 146

B. Primase, an RNA polymerase, binds to single-stranded DNA in its generation of an RNA primer during replication (choice B is correct). Helicase binds to double-stranded DNA and facilitates DNA unwinding (choice A is wrong). DNA polymerase III binds to an RNA/DNA duplex (i.e., parental DNA strand with an RNA primer attached) before beginning elongation (choice C is wrong) and DNA ligase binds to double-stranded DNA and anneals the phosophodiester backbone of DNA (choice D is wrong). Concepts tested Molecular Biology: DNA Replication

Answer 147

D. Two transposons in opposite orientations could cause chromosomal inversion, where a section of a chromosome is flipped to the reverse orientation (choice A is true and can be eliminated). Two transposons in the same orientation cause chromosomal deletions and translocations (choice B is true and can be eliminated). If a single transposon inserts into a protein-coding region of the DNA, the protein coding region will be disrupted and proteins levels will likely decrease (choice C is true and can be eliminated). However, if a single transposon inserts into the promoter or regulatory region of a protein coding gene, protein levels can either increase or decrease (choice D is not true of transposons and is the correct answer choice). Concepts tested Molecular Biology: Mutations and DNA Repair

Answer 148

C. Translation is one of the most energy-intensive processes in the cell. While initiation requires only 1 GTP, tRNA loading requires 2 high energy phosphate bond equivalents per amino acid/tRNA pair, and elongation for each subsequent amino acid requires 2 GTP. Termination requires 1 GTP. This can be summarized in the equation 4n, where n is the number of amino acids in a peptide chain, and the equation represents the number of high energy bonds required to make the peptide. In this question, you are given the total quantity of energy consumed and need to determine the length of the peptide. If 4n = 404, then n = 101 (choice C is correct). Concepts tested Molecular Biology: Translation

Answer 149

A. This question is asking for a difference between RNA and DNA and only one of the answer choices provided is a valid difference between the two nucleic acids. RNA contains a 2’ hydroxyl group, possesses uracil in place of thymine, and generally only exists in a single-stranded form in the cell (choice A is correct). Thymine should not be present in RNA and prokaryotes do not possess a nucleus (choices B and C are wrong). Finally, both RNA and DNA possess a single phosphate per nucleotide in their phosphodiester backbones (choice D is wrong). Concepts tested Biochemistry: Biomolecules: Nucleic Acids

Answer 150

B. Item I is false: promoters are simply DNA sequences upstream of the transcribed region; they cannot be substituted at will when more of a particular protein is needed (choices A and C can be eliminated). Since both remaining answer choices include Item II, it must be true: preventing a repressor from binding to a gene regulatory region would increase transcription of that gene, ultimately leading to an increase in protein synthesis for that gene. Item III is false: enhancers increase transcription. If an enhancer is inhibited, transcription rates would be reduced, and there would be a decrease in protein synthesis (choice D can be eliminated and choice B is correct). Concepts tested Molecular Biology: Transcription

Answer 151

B. This mutant strain is unable to complete nonhomologous end joining and is thus relying on homologous recombination for DSB. This is a specific repair process (choice A is wrong) that involves formation of a joint molecule (choice B is correct) and uses DNA polymerase (choice D is wrong). A strain that repairs DNA via homologous end joining will be able to repair DSBs reasonably well, and will not accumulate many chromosomal aberrations over time (choice C is wrong). Concepts tested Molecular Biology: Mutations and DNA Repair

Answer 152

D. The 5’ cap contains the ribosome binding site in eukaryotes and is critical for ribosome association (choice D is correct). Of lesser importance are intron splicing and addition of a 3' poly-A tail (choices B and C are wrong). Note that substitution of uracil for thymine is a characteristic of RNA, not a post-transcriptional modification (choice A is wrong). Concepts tested Molecular Biology: Regulation of Gene Expression

Answer 153

B. If the mutation prevents the repressor from binding to the operator, this would result in constitutive expression of the genes in the lac operon, including the fluorescent β-galactosidase (choices C and D are wrong). This would occur whether or not lactose was present (choice B is correct and choice A is wrong). Concepts tested Molecular Biology: Regulation of Gene Expression

Answer 154

A. Excessive supercoiling may result from unwinding DNA without the aid of topoisomerase which cleaves the DNA backbone to prevent such strain (choice A is correct). Helicase functions to unwind DNA; it produces supercoils in the DNA via unwinding, but these would not be excessive. The excessive nature of these supercoils is a failure in topoisomerase function (choice B is wrong), DNA ligase links the phosphodiester backbone between DNA fragments (choice C is wrong), and single-stranded binding proteins help to prevent re-annealing of unwound DNA (choice D is wrong). Concepts tested Molecular Biology: DNA Replication

Answer 155

D. The mutation described here involves the substitution (choice A is wrong) of a guanine for a cytosine to generate the sequence UGA, which is a stop codon. Generation of an early stop codon is known as a nonsense mutation (choice D is correct). A silent mutation involves the change of a nucleotide in a codon without a change in the amino acid specified by that codon (choice B is wrong). If the amino acid does change, it is known as a missense mutation (choice C is wrong).

Answer 156

C. Following the binding of a loaded tRNA to a codon, the growing peptide chain is transferred from the tRNA occupying the P site to the tRNA occupying the A site via a peptidyl transfer reaction (i.e., a peptide bond is formed between the last amino acid in the chain and the new amino acid on the tRNA in the A site, choice C is correct). Translocation of the ribosome then occurs along the mRNA (choice A is wrong), allowing the tRNA that had occupied the P site to dissociate (choice B is wrong) and opening the A site for a new tRNA. The release factor will bind when a stop codon appears in the A site; this will allow the release of the nascent peptide and dissociation of the ribosome from the mRNA (choice D is wrong). Concepts tested Molecular Biology: Translation

Answer 157

D. The researcher observes several dozen fluorescent puncta which would indicate several dozen replication forks. Prokaryotic cells only possess a single origin of replication and would only display two replication forks and therefore two puncta (choices A and B are wrong). Eukaryotes possess multiple chromosomes with multiple origins of replication on each chromosome; many more replication forks would be observed during replication (choice D is correct and choice C is wrong). Concepts tested Molecular Biology: DNA Replication Microbiology: Prokaryotes vs. Eukaryotes

Answer 158

D. Cortisol is released from the adrenal cortex (choices A and C are wrong), and because it is a steroid hormone, enters cells, binds to an intracellular receptor (choice B is wrong), and modifies transcription. Note that mobilizing fats for energy and reducing inflammation are among cortisol's effects, but it is not released from the adrenal medulla. Thus, choice D is correct. Concepts tested Endocrine System: Hormones and Hormone Functions

Answer 159

A. Item I is true: the release of ADH when blood pressure is low allows more water to be retained and urine to be concentrated (i.e., less water is eliminated in the urine). This water is returned to the bloodstream, increasing blood volume and subsequently, blood pressure (choice C can be eliminated). Item II is false: ADH is actually made in the hypothalamus, then transported to the posterior pituitary, where it is stored until release (choice B can be eliminated). Item III is false: the release of ADH is triggered by an action potential, not a releasing hormone (RH). Only the anterior pituitary hormones are controlled by releasing hormones (choice D can be eliminated and choice A is correct). Concepts tested Endocrine System: Hormones and Hormone Functions

Answer 160

B. Both peptide and steroid hormones bind to protein receptors to impact a given cell (choice B is correct). Peptide hormones function on a much shorter time scale than steroid hormones (choice A is wrong) and only peptide hormones are synthesized at the RER (choice C is wrong). Steroid hormones, because they are hydrophobic, travel through the blood bound to protein (generally albumin) while peptides generally do not require a carrier (choice D is wrong). Concepts tested Endocrine System: Endocrine System Basics

Answer 161

D. Choices A, B, and C are true statements about testosterone, and can be eliminated. Choice D is false (and thus the correct answer choice); testosterone is made by interstitial cells, but does not feedback significantly to these cells. Concepts tested Endocrine System: Hormones and Hormone Functions

Answer 162

A. With menopause, the ovaries begin to degenerate and fail to produce appropriate levels of estrogen. This decrease in estrogen impacts the hormone feedback loop to GnRH and FSH/LH; the lack of negative feedback by estrogen triggers an increase in GnRH (and subsequently FSH and LH) in an attempt to generate more estrogen (choice A is correct and choices B, C, and D are wrong). In fact, a common diagnostic test for menopause is measuring FSH levels. Concepts tested Endocrine System: Hormones and Hormone Functions

Answer 163

A. Aldosterone increases sodium reabsorption from the filtrate, thus increasing plasma osmolarity (choices B and D are wrong). Antidiuretic hormone (ADH, or vasopressin) induces expression of water channels (aquaporins) on the cells that make up distal tubule and collecting duct. This makes these typically water-impermeable tubes now permeable to water. At the distal tubule this allows water to follow sodium out of the filtrate, and at the collecting duct water moves out of the filtrate due to the gradient previously established by the loop of Henle (choice C is wrong and choice A is correct). Concepts tested Endocrine System: Hormones and Hormone Functions

Answer 164

A. Oxytocin is released from the posterior pituitary (choices B and D are wrong), and its functions include inducing uterine contractions during labor (choice A is correct) and milk ejection during lactation. The actual production of milk is triggered by prolactin (choice C is wrong). Concepts tested Endocrine System: Hormones and Hormone Functions

Answer 165

B. GnRH is released from the hypothalamus and induces FSH and LH secretion from the anterior pituitary (choices A and C are wrong). FSH stimulates the sustentacular (Sertoli) cells in the testes to produce many secretions that affect spermatogenesis (primarily). LH stimulates interstitial (Leydig) cells to produce testosterone (choice B is correct and choice D is wrong). Concepts tested Endocrine System: Hormones and Hormone Functions

Answer 166

C. The body builder is attempting to use human growth hormone to increase muscle mass by taking the hormone orally. Growth hormone, which is produced in the pituitary gland, is a peptide hormone that binds to extracellular receptors (choices B and D are wrong) and would be subject to degradation in the stomach and small intestine. This degradation could explain the observed lack of effect as the hormone is not absorbed in a functional form (choice C is correct). Growth hormone has numerous effects including increasing height in children/adolescents, increasing calcium retention, and increasing muscle mass (choice A is not correct). Concepts tested Endocrine System: Endocrine System Basics

Answer 167

A. Glucagon is made by the α cells within the islets of Langerhans and stimulates the breakdown of glycogen stores in the liver to increase circulating glucose levels (choice A is correct). Insulin is made by the β cells of the pancreas, but it decreases the circulating levels of glucose (choice B is wrong). Epinephrine and cortisol both increase glucose, but they are released by the adrenal medulla and adrenal cortex, respectively (choices C and D are wrong). Concepts tested Endocrine System: Hormones and Hormone Functions

Answer 168

A. The pituitary gland, a commonly discussed endocrine gland, is responsible for producing numerous hormones including FSH, LH, ACTH, TSH, prolactin, GH, and MSH (choice A is correct). ADH and oxytocin are produced in the hypothalamus (choices B, C, and D are not correct). Note: to answer this question, you must know that ACTH is produced in the anterior pituitary and that oxytocin and ADH are not. Concepts tested Endocrine System: Hormones and Hormone Functions

Answer 169

B. This patient is suffering from acute thyroid hormone toxicity and is experiencing symptoms of hyperthyroidism including accelerated heart rate, excessive sweating, and heat intolerance. The elevated levels of T4 (choices C and D are wrong) will result in feedback to the hypothalamus and anterior pituitary, causing a decrease in TSH (choice B is correct and choice A is wrong). Note that these symptoms could also be explained if the patient were ingesting TSH, however the stem of the question informs us that the substance being ingested contained iodide, indicating he was consuming thyroid hormone directly. Concepts tested Endocrine System: Hormones and Hormone Functions

Answer 170

C. If 10 in 1000 births have polydactyly, then the condition is found at a frequency of 1% in this population. In the Hardy-Weinberg equation for genotype frequency (p2 + 2pq + q2 = 1), q2 is the frequency of the autosomal recessive genotype. Since the question states that polydactyly (at least in this study) is an autosomal recessive disorder, q2 = .01, so q (the frequency of the recessive allele) equals 0.1. Concepts tested Genetics: Hardy-Weinberg

Answer 171

B. Only choice B is a true statement. Fitness is defined as how well an organism passes its alleles to future generations. If few offspring survive to reproductive age, this does not correspond to increased fitness (choice A is false). Changes in allele frequency—not gene expression patterns—are the basis for evolution in both species and populations (choice C is false). Divergent selection removes members near the average and favors traits at the extremes. When a population is driven closer to the average trait, stabilizing selection has occurred (choice D is false). Concepts tested Evolution/Speciation: Natural Selection/Speciation

Answer 172

D. Nondisjunction during meiosis I leads to two cells with too many chromosomes (n + 1) and two cells with too few (n – 1). Thus, nondisjunction in meiosis I would only produce two populations of cells (choice C is not correct). Nondisjunction during meiosis II leads to two cells with the correct number of chromosomes (n), one cell with an extra chromosome (n + 1) and one cell that is missing a chromosome (n – 1). This could account for the three populations of cells isolated by the researcher (choice D is correct, note that aneuploidy is just an abnormal number of chromosomes). There is no information to support the fact that meiosis is not being completed (choice A is wrong), or that there is limited mitosis of spermatogonia (this would lead to fewer gametes, but not a change in chromosome number; choice B is wrong). Concepts tested Cell Biology: Mitosis/Meiosis

Answer 173

D. If all the F1 flies have red-eyes, you know the red eye trait is dominant. Let's assign R = red and r = white. The parental (P) cross is female XRXR × male XrY. The F1 flies would be XRXr (red-eyed females that carry the white allele but don't express it) and XRY (red-eyed males). The next cross in the question is between a XRXr female and the XrY male from the parental generation (remember that a backcross is when an individual is crossed to a previous generation). The resultant F2 flies would be 25% XRXr (red-eyed females), 25% XrXr (white-eyed females), 25% XRY (red-eyed males) and 25% XrY (white-eyed males). You can see that regardless of sex, half the offspring will have white eyes and half will have red eyes (choice D is correct). Concepts tested Genetics: Mendelian Genetics/Probability

Answer 174

B. If dry ear wax skips generations, it must be a recessive trait. Classically dominant traits cannot be hidden and then show up in future generations (choices C and D can be eliminated). Furthermore, equal distribution among sexes likely rules out a sex-linked trait; dry ear wax must be an autosomal recessive trait (choice A can be eliminated and choice B is correct). Note that traits that pass from fathers to sons (such as the hairy ear phenotype in this question) are coded by the Y chromosome. Concepts tested Genetics: Mendelian Genetics/Probability

Answer 175

C. Let’s assign alleles as A = agouti, a = chinchilla, B = black, and b = pink. The original cross is AABB × aabb. All F1s are AaBb and a male from this group is then backcrossed (must be to the female parent, so AaBb × aabb). The probability of getting chinchilla fur (aa) is 0.5. The probability of getting black eyes (Bb) is 0.5. The probability of having a female mouse is 0.5. Therefore, overall the answer is (0.5)(0.5)(0.5) = 0.125 (choice C is correct). Concepts tested Genetics: Mendelian Genetics/Probability

Answer 176

B. Linkage occurs when two genes are on the same chromosome (choice D is wrong) and less than 50 map units apart. This leads to decreased recombination, and linked alleles being inherited together (choice B is correct). For example, if genes J and N are linked, and a woman is genotype (Jn/jN), she will make more Jn and jN gametes (parental) than jn and JN gametes (recombinants). Linked genes do not have to function in similar pathways (choice A is wrong). A given gene (and any others it may be linked to) is always found in the same location on chromosomes and chromatids (choice C is wrong). Concepts tested Genetics: Linked Genes

Answer 177

D. Let’s assign B = black and b = brown. For the second gene, E allows pigment expression and e does not (this leads to the golden labs). The cross in the question stem is BbEe (a black lab) × bbee (a golden lab), so choices A and C can be eliminated. The genotypic ratio of the puppies would be 25% BbEe (black lab), 25% Bbee (golden lab), 25% bbEe (brown lab) and 25% bbee (golden lab). Overall, 25% of the puppies would have black fur, 25% would have brown fur, and 50% would have golden fur (due to pigment expression being turned off). Concepts tested Genetics: Mendelian Genetics/Probability

Answer 178

A. Let’s assign the dihybrid mouse a genotype of AaBb. The testcross would be AaBb × aabb and the expected ratios of offspring would be 25% AB/ab, 25% Ab/ab, 25% aB/ab and 25% ab/ab (in linkage notation). The question stem says AB/ab is present at only 8% and ab/ab is present at only 10%. These two combinations must represent the recombinant offspring given their low percentages, Ab/ab and aB/ab must be the parental combination of alleles, and the genotype of the dihybrid parent must have been Ab/aB. In other words, the two genes are linked and the dominant allele of one gene is linked to the recessive allele of the other. To determine how far apart the genes are, we can calculate Rf, where Rf = (# recombinants / total) x 100% = [(8% + 10%) / 100%] x 100% = 18% recombination frequency = 18 cM = 18 map units. Overall, the best answer is choice A. Concepts tested Genetics: Linked Genes

Answer 179

C. If we assign D = normal and d = colorblind, the cross in the question stem is XDXd × XDY. The probability of receiving the colorblind allele from the mother is 0.5. The probability of receiving the Y chromosome from the father is 0.5. Therefore, the probability they will have a colorblind son is (0.5)(0.5) = 0.25 (choice C is correct). Concepts tested Genetics: Mendelian Genetics/Probability

Answer 180

D. The question stem says ALDH2*2 is dominant so we will assign it A; the normal allele is recessive and we will assign it a. In the equation for allele frequency (p + q = 1), p = A = 0.3, so q = a = 1 – 0.3 = 0.7). Since this trait is dominant, both homozygous dominants (AA) and heterozygotes (Aa) will express the phenotype. From the equation for genotype frequency (p2 + 2pq + q2 = 1), the proportion of AA in the population is p2 and the proportion of heterozygotes is 2pq. Therefore the answer is (0.3)2 + 2(0.3)(0.7) = 0.09 + 0.42 = 0.51, or 51% (choice D is correct). Alternatively 1 – q2 can be used in place of the above. Concepts tested Genetics: Hardy-Weinberg

Diagnostic Exams - Biology Flashcards

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