Diagnostic Exams - Biology Flashcards
β-oxidation is a means of creating acetyl-CoA from fatty acids. This acetyl-CoA can then enter the Krebs cycle. Each turn of the β-oxidation cycle produces one acetyl-CoA molecule and a fatty acid two carbons shorter than it was at the beginning of the cycle. Stearic acid is an 18-carbon saturated fatty acid. How many turns of the β-oxidation cycle would it take to completely break down stearic acid into acetyl-CoA groups?
A. 4
B. 5
C. 8
D. 9
C. Each turn of the β-oxidation cycle produces one acetyl-CoA and a fatty acid two carbons shorter than before. Stearic acid, with 18 carbons, would ultimately produce 9 acetyl-CoA. If stearic acid enters the β-oxidation cycle, then after one turn of the cycle we would have one acetyl-CoA and a 16C fatty acid. After two turns of the cycle we would have two acetyl-CoA and a 14C fatty acid. This would continue in this manner until after the 8th turn of the cycle we would produce our 8th acetyl-CoA molecule, and all that would be left over would be another two-carbon acetyl-CoA molecule (the 9th acetyl-CoA). This last 2-carbon molecule does not need to go through the β-oxidation cycle again, so only 8 turns of cycle are necessary.
Concepts tested
Biochemistry: Protein/Fat Metabolism
How many NADH molecules are produced from a single glucose molecule during cellular respiration?
A. 6
B. 8
C. 10
D. 12
C. A single glucose molecule generates a total of 10 NADH during cellular respiration. 2 NADH are made in glycolysis, 2 NADH at the pyruvate dehydrogenase complex, and 6 NADH in the Krebs cycle (choice C is correct and choices A, B, and D are wrong).
Concepts tested
Biochemistry: Glycolysis/Cellular Respiration
In the absence of oxygen, fermentation occurs in order to allow glycolysis to continue. Pyruvate, an end-product of glycolysis, is converted to either lactic acid or ethanol. The conversion of pyruvate to lactic acid is a(n):
A. oxidation.
B. reduction.
C. decarboxylation.
D. isomerization.
B. In order to keep running glycolysis in the absence of oxygen, NAD+ must be regenerated from NADH. In lactic acid fermentation, the oxidation of NADH occurs via the reduction of pyruvate to lactic acid (choice B is correct and choice A is wrong). In alcoholic fermentation (the conversion of pyruvate to ethanol) pyruvate must first be decarboxylated, since ethanol is a two-carbon structure, whereas pyruvate is a three-carbon structure. However, lactic acid is also a three-carbon structure, so no decarboxylation is needed in lactic acid fermentation (choice C is wrong). Isomerization is simply a rearrangement of atoms, and this is not the case here (choice D is wrong).
Concepts tested
Biochemistry: Carbohydrate Metabolism
High levels of ATP would:
A. stimulate phosphofructokinase and inhibit pyruvate kinase, thus stimulating glycolysis.
B. stimulate phosphofructokinase and stimulate pyruvate kinase, thus stimulating glycolysis.
C. inhibit phosphofructokinase and stimulate pyruvate kinase, thus inhibiting glycolysis.
D. inhibit phosphofructokinase and inhibit pyruvate kinase, thus inhibiting glycolysis.
D. High levels of ATP indicate that the cell does not need to run glycolysis, and instead can use glycolytic intermediate in gluconeogenesis (choices A and B are wrong). Phosphofructokinase produces fructose-1,6-bisphosphate, driving the cell toward glycolysis; high ATP levels would inhibit this enzyme to inhibit glycolysis. Pyruvate kinase catalyzes the final step in glycolysis, the conversion of phosphoenolpyruvate to pyruvate. This enzyme must be inhibited by ATP to inhibit glycolysis (choice C is wrong and choice D is correct).
Concepts tested
Biochemistry: Glycolysis/Cellular Respiration
Which of the following is an example of reciprocal regulation of glycogen metabolism?
A. High levels of citrate stimulate phosphofructokinase and inhibit fructose-1,6-bisphosphatase.
B. High levels of AMP inhibit glycolysis and stimulate gluconeogenesis.
C. Glucagon inhibits glycogen phosphorylase and stimulates glycogen synthase.
D. Insulin stimulates glycogen synthase and inhibits glycogen phosphorylase.
D. Reciprocal regulation occurs when a single molecule stimulates a pathway in one direction while inhibiting the pathway in the opposite direction. Insulin is released when blood sugar is high; it stimulates glycogen synthase (to store glucose as glycogen) and inhibits glycogen phosphorylase (the first enzyme in glycogen breakdown, choice D is correct). Glucagon is antagonistic to insulin; it is released when blood sugar is low to stimulate glycogen phosphorylase and glycogen breakdown while inhibiting glycogen synthase (choice C is wrong). While citrate and AMP do reciprocally regulate the indicated enzymes and pathways, how they regulate those pathways is not described correctly. Citrate inhibits phosphofructokinase to slow down glycolysis and stimulates fructose-1,6-bisphosphatase to increase gluconeogenesis (choice A is wrong); AMP acts to stimulate glycolysis (because ATP is low) and inhibit gluconeogenesis (choice B is wrong). And in either case, this is not regulation of glycogen metabolism.
Concepts tested
Biochemistry: Carbohydrate Metabolism
β-Oxidation is a means of creating acetyl-CoA from fatty acids. This acetyl-CoA can then enter the Krebs cycle. Each turn of the β-oxidation cycle produces one acetyl-CoA molecule and a fatty acid two carbons shorter than it was at the beginning of the cycle. Additionally, one NADH and one FADH2 are generated per turn. Lauric acid is an 12-carbon saturated fatty acid. Including those produced in the Krebs cycle, how many total NADH and FADH2 molecules would be generated from the complete β-oxidation of lauric acid and subsequent entry of the acetyl-CoA into the Krebs cycle?
A. 24 NADH and 12 FADH2
B. 23 NADH and 11 FADH2
C. 12 NADH and 12 FADH2
D. 11 NADH and 11 FADH2
B. Each turn of the β-oxidation cycle produces one acetyl-CoA and a fatty acid two carbons shorter than before. Lauric acid, with 12 carbons, would ultimately produce 6 acetyl-CoA. If lauric acid enters the β-oxidation cycle, then after one turn of the cycle we would have one acetyl-CoA and a 10C fatty acid. After two turns of the cycle we would have two acetyl-CoA and an 8C fatty acid. This would continue in this manner until after the 5th turn of the cycle we would produce our 5th acetyl-CoA molecule, and all that would be left over would be another two-carbon acetyl-CoA molecule (the 6th acetyl-CoA). This last 2-carbon molecule does not need to go through the β-oxidation cycle again, so only 5 turns of cycle are necessary. Thus, 5 NADH and 5 FADH2 would be generated during β-oxidation. When the 6 acetyl-CoA molecules go through the Krebs cycle, and additional 18 NADH would be generate (3 per turn of the Krebs cycle) and and additional 6 FADH2 would be generated (1 per turn), for a total of 23 NADH and 11 FADH2.
Concepts tested
Biochemistry: Protein/Fat Metabolism
Some amino acids can be converted to pyruvate via several biochemical pathways. Pyruvate can then enter the cellular respiration pathways, either by decarboxylation to acetyl-CoA or by carboxylation to oxaloacetate. For a single pyruvate molecule, first converted to acetyl-CoA, then traveling through the Krebs cycle, how many NADH molecules are produced?
A. 3
B. 4
C. 6
D. 8
B. The decarboxylation of pyruvate to acetyl-CoA nets 1 NADH, and as that acetyl-CoA travels through the Krebs cycle, an additional 3 NADH are generated, resulting in a total of 4 NADH per pyruvate (choice B is correct and choices A, C, and D are wrong).
Concepts tested
Biochemistry: Glycolysis/Cellular Respiration
Eukaryotes running aerobic respiration net 30 ATP per glucose, while prokaryotes net 32 ATP. Why?
A. Prokaryotic glycolysis does not require the input of 2 ATP for the phosphorylation of glucose and fructose-6-P.
B. Prokaryotes generate more pyruvate from glucose than do eukaryotes.
C. In eukaryotes, the electrons from glycolytic NADH must be shuttled from the cytosol into the mitochondrion, and bypass the first proton pump.
D. Eukaryotes have a more efficient ATP synthase.
C. In eukaryotes, glycolysis occurs in the cytosol, while the PDC, Krebs cycle, and electron transport occur in the mitochondria. The NADH from glycolysis is oxidized in the cytosol (so that NAD+ continues to remain available for glycolysis), and the electrons are shuttled into the electron transport chain. However, the electrons bypass the first proton pump (NADH dehydrogenase) and are delivered to coenzyme Q, the second molecule in the electron transport chain. This results in the movement of fewer protons out of the mitochondrial matrix, and thus less ATP made when the protons reenter the matrix through the ATP synthase. Prokaryotes run glycolysis, PDC, and the Krebs cycle in the cytosol, with all NADH immediately available to the first pump in their transport chain (which is located in their cell membrane; choice C is correct). Prokaryotic glycolysis is the same as eukaryotic, requiring 2 ATP to phosphorylate glucose and fructose-6-P (choice A is wrong), and generating 2 pyruvate per glucose (choice B is wrong). If eukaryotic ATP synthase were more efficient, they would make more ATP than prokaryotes, not less (choice D is wrong).
Concepts tested
Biochemistry: Glycolysis/Cellular Respiration
Which of the following does NOT occur during starvation?
β-oxidation in the mitochondrial matrix provides acetyl-CoA to feed into the Krebs cycle.
Ketone bodies are converted into acetyl-CoA and the acetyl-CoA is converted into glucose.
Fatty acid synthesis in the cytoplasm produces NADH to drive electron transport and oxidative phosphorylation.
A. I only
B. II only
C. II and III only
D. I, II, and III
C. Item I DOES occur during starvation: β-oxidation of fatty acids provides acetyl-CoA that can turn the Krebs cycle (choices A and D can be eliminated). Both of the remaining answers include Item II, so Item II must NOT occur during starvation: while ketone bodies (used for fuel by the nervous system during times of starvation) can be converted into acetyl-CoA once they reach their target organs/cells, the acetyl-CoA is not converted into glucose. First, we lack the enzymes necessary to do that, and second, it isn’t necessary as the acetyl-CoA can enter the Krebs cycle directly. Item III does NOT occur during starvation: fatty acids are broken down, not synthesized (choice B can be eliminated and choice C is correct; both Items II and III do not occur during starvation).
Concepts tested
Biochemistry: Protein/Fat Metabolism
Blood pressure is affected by many factors. Which of the following would increase blood pressure?
An increase in peripheral blood vessel diameter
An increase in heart rate due to physical exertion
An increase in blood volume due to aldosterone
A. I only
B. II only
C. II and III only
D. I, II, and III
C. Item I is false: an increase in vessel diameter would reduce peripheral resistance. Since peripheral resistance and blood pressure are directly proportional, this would lead to a decrease in blood pressure (choices A and D can be eliminated). Since both remaining answer choices include Item II, it must be true: an increase in heart rate would lead to an increase in cardiac output. Cardiac output and blood pressure are directly proportional, so this would increase blood pressure. Item III is true: consider the heart and blood vessels as a “container” for blood in the body. If the volume of blood in the “container” increases but the container size remains the same, the pressure inside the container must increase (choice B can be eliminated and choice C is correct).
Concepts tested
Cardiovascular System: Blood Pressure Renal System: Renal Regulation of Blood Pressure
Which of the following statements is true about arteries but not about veins?
A. They have valves to maintain flow in a single direction.
B. They carry only oxygen-rich blood.
C. They are a low-pressure system of vessels.
D. They have a muscle layer to regulate blood flow.
D. The walls of arteries contain a layer of muscle that can adjust the diameter of the artery to regulate blood flow to different regions of the body. Veins lack this muscle layer (choice D is true about arteries but not about veins and is the correct answer choice). Veins have valves, but arteries do not (choice A is true about veins, not arteries and can be eliminated). Most arteries carry oxygen-rich blood, but the pulmonary arteries carry oxygen-poor blood; likewise, most veins carry oxygen-poor blood, but the pulmonary veins carry oxygen rich blood (choice B is true of neither arteries nor veins and can be eliminated). Arteries are a high-pressure system and veins are a low pressure system (choice C is true about veins and can be eliminated).
Concepts tested
Cardiovascular System: Blood Vessels
Cardiac autorhythmic cells (and to some extent cardiac muscle cells) have the ability to trigger their own action potential independent of any neural input or other stimulation. Which of the following is responsible for this autorhythmic property?
A. K+ leak channels
B. Na+ leak channels
C. Na+/K+ ATPase
D. Slow voltage-gated Na+ channels
B. Cardiac autorhythmic cells contain Na+ leak channels that allow Na+ to enter the cell according to its gradient. As Na+ enters, the cell potential rises (depolarizes) until it reaches the threshold for slow voltage-gated Na+ channels; this triggers the action potential itself (choice D is wrong). K+ leak channels allow K+ to leave the cell, making it more negative (hyperpolarizing it) and less likely to fire an action potential (choice A is wrong). The Na+/K+ ATPase establishes the Na+ and K+ gradients necessary for heart function, but do not themselves trigger action potentials (choice C is wrong).
Concepts tested
Cardiovascular System: Cardiac Action Potential/Conduction System
Heart murmurs are extra, abnormal sounds (beyond the normal closure of the valves) produced during the cardiac cycle. They can be caused by stenotic (stiffened) valves, or by valves that do not close properly and allow regurgitation. Murmurs are classified as diastolic or systolic depending on when the additional sound is produced. A heart murmur caused by a failure of the AV valves to close properly would most likely be classified as a:
A. diastolic murmur, because this would allow flow from the atria to the ventricles during diastole.
B. systolic murmur, because this would allow regurgitation of blood from the ventricles to the atria during systole.
C. diastolic murmur, because this would allow regurgitation of blood from the arteries to the ventricles during diastole.
D. systolic murmur, because this would allow additional blood to flow from the atria to the ventricles during systole.
B. The AV valves close at the beginning of systole to prevent regurgitation of blood into the atria while the ventricles are contracting. If the AV valves failed to close properly, blood from the high-pressure ventricles would flow back into the low-pressure atria during systole and would produce an abnormal murmur (choice B is correct). Flow from the atria to the ventricles during systole would be prevented by the pressure gradient (choice D is wrong). Blood normally flows from the atria to the ventricles during diastole, through the open AV valves; this would not produce a murmur (choice A is wrong), and these valves do not separate the ventricles and the arteries so this would not affect blood flow between those regions (choice C is wrong).
Concepts tested
Cardiovascular System: Heart Anatomy
A woman with blood type A+ has four children. The blood types of her children are O+, AB–, A+, and AB+. Which of the following statements is true?
A. Her husband cannot have type AB blood.
B. Her husband must have type B– blood.
C. Her husband is homozygous for the IB allele.
D. Her husband must be heterozygous for the Rh allele.
A. In order to produce children with blood type O, both parents must donate a recessive i allele, thus the husband cannot have type AB blood (choice A is true), and he cannot be homozygous for IB (choice C is wrong). In order to produce children that are Rh–, both parents must donate the recessive r allele. Since she is A+, then she must be heterozygous for the Rh factor (Rr), but the husband could be Rh– (homozygous recessive rr) or Rh+ and heterozygous (Rr, choices B and D are wrong).
Concepts tested
Genetics: Mendelian Genetics/Probability Cardiovascular System: Blood
Which of the following are involved in carrying blood to the kidneys?
A. Abdominal aorta and renal vein
B. Inferior vena cava and renal vein
C. Abdominal aorta and renal artery
D. Inferior vena cava and renal artery
C. The abdominal aorta carries blood away from the heart and into the trunk of the body. The renal artery branches off the aorta to supply the kidneys with blood (choice C is correct). The renal vein and inferior vena cava are both involved in returning blood back to the circulatory system after it has been filtered (choices A, B, and D are incorrect).
Concepts tested
Renal System: Kidney Functions
The exocrine functions of the pancreas include chemical digestion of each of the following EXCEPT:
A. polypeptides.
B. carbohydrates.
C. nucleotides.
D. triglycerides.
C. Proteins (polypeptides) are chemically digested by pancreatic proteases such as trypsin and chymotrypsin (choice A is digested by pancreatic enzymes and can be eliminated). Some carbohydrate chemical digestion starts in the mouth, but much of it is done in the duodenum by enzymes such as pancreatic amylase (choice B can be eliminated). Triglycerides (lipids) are not digested until they reach the small intestine; this is mediated by pancreatic lipases after bile from the liver/gallbladder has emulsified the fat molecules (choice D can be eliminated). However, pancreatic nucleases digest DNA and RNA, not nucleotides (choice C is not digested by pancreatic enzymes and is the correct answer choice).
Concepts tested
Digestive System: Accessory Organs
Which of the following is NOT an accurate statement concerning the hydrochloric acid produced in the stomach?
A. HCl is produced by the parietal cells.
B. HCl converts the zymogen pepsinogen into the active enzyme pepsin.
C. HCl engages in acid hydrolysis of proteins, aiding in their elementary digestion.
D. HCl is part of the innate immune mechanisms present in the duodenum.
D. Hydrochloric acid is made by parietal cells in the stomach (choice A is an accurate statement and can be eliminated), causes zymogen conversion (choice B is an accurate statement and can be eliminated), and facilitates non-specific acid hydrolysis of proteins (choice C is an accurate statement and can be eliminated). However, while it plays a role in innate immune function of the stomach, HCl is neutralized by aqueous bicarbonate upon entry into the duodenum (choice D is not an accurate statement and is the correct answer choice).
Concepts tested
Digestive System: Alimentary Canal
All of the following are functions of the liver EXCEPT:
A. storage of glycogen.
B. synthesis of blood proteins.
C. bile production.
D. secretion of digestive enzymes.
D. The liver has a number of functions in the body, including synthesis of bile (choice C can be eliminated), glycogen storage and metabolism (choice A can be eliminated), synthesis of blood proteins (such as albumin, fibrinogen, angiotensinogen, lipoproteins, etc., choice B can be eliminated), amino acid metabolism, production of urea, vitamin storage, detoxification, etc. However, the liver does not secrete digestive enzymes (choice D is not a function of the liver and is the correct answer choice).
Concepts tested
Digestive System: Accessory Organs
Which of the following is a true statement?
A. The release of secretin, triggered by high duodenal pH, causes the release of bicarbonate and enzymes from the pancreas.
B. The release of secretin, triggered by low duodenal pH, causes the release of bicarbonate and enzymes from the pancreas.
C. The release of secretin, triggered by low gastric pH, causes the release of bicarbonate and enzymes from the pancreas.
D. The release of secretin, triggered by high gastric pH, causes the release of bicarbonate and enzymes from the pancreas.
B. Secretin is released from duodenal cells when acidic chyme enters the duodenum from the stomach and duodenal pH drops (choice A is wrong). This causes the release of bicarbonate (and enzymes) from the pancreas to help neutralize that acid, bringing duodenal pH into a more neutral range. This provides a more optimal environment for the pancreatic enzyme to function in, as they do not work well at low (acidic) pH (choice B is true). Since the hormone is released by duodenal cells, and since the pH change must occur in the duodenum, it should not be affected by gastric pH (choices C and D are wrong).
Concepts tested
Digestive System: Accessory Organs
Which of the following is true about the renin-angiotensin axis?
A. The juxtaglomerular cells in the afferent arteriole are baroreceptors that sense low pressure, ACE converts angiotensinogen to angiotensin I, and renin converts angiotensin I to angiotensin II.
B. The juxtaglomerular cells in the afferent arteriole are baroreceptors that sense high pressure, ACE converts angiotensinogen to angiotensin I, and renin converts angiotensin I to angiotensin II.
C. The juxtaglomerular cells in the afferent arteriole are baroreceptors that sense low pressure, renin converts angiotensinogen to angiotensin I, and ACE converts angiotensin I to angiotensin II.
D. The juxtaglomerular cells in the afferent arteriole are baroreceptors that sense high pressure, renin converts angiotensinogen to angiotensin I, and ACE converts angiotensin I to angiotensin II.
C. Juxtaglomerular cells are baroreceptors and secrete renin in response to low blood pressure (choices B and D are wrong). Renin converts the zymogen angiotensinogen into angiotensin I. Angiotensin I is further converted to angiotensin II by ACE (angiotensin converting enzyme, choice A is wrong and choice C is correct). Angiotensin II is a powerful vasoconstrictor that quickly increases blood pressure.
Concepts tested
Renal System: Renal Regulation of Blood Pressure
Which of the following describes the route via which urine leaves the kidney and is voided from the body?
A. Ureter, involuntary urinary sphincter, bladder, voluntary urinary sphincter
B. Ureter, bladder, involuntary urinary sphincter, voluntary urinary sphincter
C. Ureter, bladder, voluntary urinary sphincter, involuntary urinary sphincter
D. Bladder, ureter, involuntary urinary sphincter, voluntary urinary sphincter
B. Urine leaves the kidneys via the ureters and travels to the bladder where it is stored prior to excretion (choices A and D are wrong). When voiding, urine passes through the internal urinary sphincter, which is composed of smooth muscle and is therefore under autonomic (involuntary) control and then the external urinary sphincter, which is composed of skeletal muscle under voluntary control (choice C is wrong and choice B is correct).
Concepts tested
Renal System: Kidney Functions
An elderly patient presents with chronic renal failure and metabolic acidosis. What impact would this have on oxygen delivery to tissues?
A. Increased oxygen delivery due to hemoglobin’s increased oxygen affinity
B. Increased oxygen delivery due to hemoglobin’s decreased oxygen affinity
C. Decreased oxygen delivery due to hemoglobin’s increased oxygen affinity
D. Decreased oxygen delivery due to hemoglobin’s decreased oxygen affinity
B. Metabolic acidosis (decreased blood pH) will decrease hemoglobin’s affinity for oxygen (choices A and C can be eliminated) and subsequently increase oxygen delivery to tissues (choice D can be eliminated and choice B is correct). Other causes of decreased oxygen affinity include elevated body temperature and increased production of BPG.
Concepts tested
Renal System: Renal Regulation of pH
The nephrons of the kidney are responsible for filtering blood and modifying the filtrate to produce urine. Which of the following provides the best description of the initial composition of filtrate in a healthy individual?
A. Water, glucose, ions
B. Water, ions, cells
C. Water, cells, proteins
D. Water, proteins, glucose
A. The glomerular basement membrane is only permeable to water and small, hydrophilic molecules, such as glucose, amino acids, and ions (choice A is correct). Whole cells or whole proteins should not be able to cross through this filter (choice B, C, and D are wrong).
Concepts tested
Renal System: Nephron Structure/Function
Which of the following are the functions of the large intestine?
A. Digest and absorb nutrients and water, store feces
B. Absorb nutrients and store feces
C. Digest macromolecules and absorb nutrients
D. Reabsorb water and store feces
D. The large intestine does not participate in any digestion (choices A and C are wrong), and only very minimal nutrient absorption (some vitamins, choice B is wrong). The main function of the large intestine is to reabsorb the large amounts of water that enter from the small intestine. This helps prevent dehydration by compacting and solidifying waste products into feces, which is then stored in the large intestine until elimination (choice D is correct).
Concepts tested
Digestive System: Alimentary Canal
The mouth is NOT responsible for which of the following functions?
A. Mastication
B. Bolus creation
C. Saccharide absorption
D. Carbohydrate digestion
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
What is selectively reabsorbed by the proximal convoluted tubule?
A. Water and ions only
B. Glucose and amino acids only
C. Water, ions, glucose, amino acids
D. Water, ions, glucose, proteins
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
Low blood pressure and low blood volume trigger which of the following?
A. An increase in ADH followed by a subsequent decrease in aldosterone
B. An increase in ADH followed by a subsequent increase in aldosterone
C. An increase in aldosterone followed by a subsequent decrease in ADH
D. An increase in aldosterone followed by a subsequent increase in ADH
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
A mutation leading to a loss of function in which of the following hormones would have the greatest impact on enzyme activation in the stomach?
A. Gastrin
B. Cholecystokinin
C. Erythropoeitin
D. Secretin
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
What role does the macula densa (in the distal tubule) play in regulating blood pressure?
A. High filtrate osmolarity triggers the macula densa to stimulate the JG cells and dilate the afferent arterioles.
B. Low filtrate osmolarity triggers the macula densa to stimulate the JG cells and dilate the afferent arterioles.
C. High filtrate osmolarity triggers the macula densa to stimulate the JG cells and constrict the afferent arterioles.
D. Low filtrate osmolarity triggers the macula densa to stimulate the JG cells and constrict the afferent arterioles.
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
Which of the following types of renal vasculature are NOT responsible for returning components to the circulatory system as part of filtrate adjustment?
A. Efferent arterioles
B. Afferent arterioles
C. Peritubular capillaries
D. Vasa recta
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
Which of the following statements about bile is/are true?
Bile emulsifies lipids for easier digestion.
The amphipathic nature of bile allows it to digest lipids.
Bile is produced by the liver and gallbladder.
A. I only
B. I and III only
C. II and III only
D. I, II, and III
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
The renal medulla is made up primarily of:
A. glomeruli.
B. nephrons.
C. proximal tubules.
D. collecting ducts.
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
Which of the following does NOT describe the esophagus?
A. The opening of the esophagus is protected by the epiglottis.
B. The upper esophagus is composed of skeletal muscle.
C. Peristalsis begins in the esophagus.
D. The esophagus is separated from the stomach by the cardiac sphincter
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
All of the following are true of the bacteria living in the large intestine EXCEPT:
A. the bacteria compete with consumed pathogens that have evaded other innate immune system defenses.
B. the bacteria are mostly aerobic with some facultative anaerobes.
C. the bacteria produce vitamin K, which is necessary for proper blood clotting.
D. the bacteria are considered to be mutualistic symbionts because the human host derives benefit while the bacteria obtain a steady source of food in the form of undigested material passing through the gut
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
Which route describes how filtrate moves through the nephron?
A. Loop of Henle, PCT, glomerulus, DCT
B. Glomerulus, PCT, DCT, loop of Henle
C. Loop of Henle, glomerulus, PCT, DCT
D. Glomerulus, PCT, loop of Henle, DCT
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
Urine formation begins with filtration of the blood, after which the filtrate is modified via reabsorption and secretion as it travels along the nephron. Water, ions, and small hydrophilic molecules are all filtered into the nephron. Which of the following provides the best description of the composition of urine in a healthy individual?
A. Water, ions, urea, glucose
B. Water, ions, urea
C. Water, proteins, ions, urea
D. Water, urea, proteins, glucose
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
How are peptidases activated in the duodenum?
A. Trypsinogen from the pancreas is converted to trypsin by stomach acid; trypsin then activates the other pancreatic enzymes.
B. Enterokinase converts trypsinogen from the pancreas into trypsin; trypsin then triggers the release of other pancreatic enzymes.
C. Enterokinase converts trypsinogen from the pancreas into trypsin; trypsin then activates the other pancreatic enzymes.
D. Trypsinogen from the pancreas is converted to trypsin by stomach acid; trypsin then activates the brush border enzyme enterokinase.
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
The function of the loop of Henle is to:
A. create a concentration gradient in the medulla to facilitate the reabsorption of water at the collecting duct.
B. reabsorb ions such as sodium and calcium under the control of various hormones.
C. reabsorb water under the control of ADH.
D. selectively reabsorb potassium and sodium to allow the reabsorption of water and urine concentration.
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
A non-competitive inhibitor:
binds to an allosteric site. reduces the Vmax of a reaction. can be overcome by adding more substrate. A. I only B. I and II only C. II and III only D. I, II, and III
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
Vmax is the maximum rate of product formation for a given enzyme. Which of the following would NOT affect Vmax?
Decreasing the substrate concentration Adding a competitive inhibitor Increasing the enzyme concentration A. I only B. I and II only Correct Answer (Blank) C. II and III only D. I, II, and III
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
Which of the following monosaccharides are linked together to form sucrose?
A. Glucose and maltose
B. Glucose and fructose
C. Fructose and galactose
D. Glucose and glucose
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
Enzymes are proteins that catalyze specific chemical reactions within a cell. Which of the following statements about enzymes is/are NOT true?
A common means of regulating enzyme activity is through phosphorylation of the enzyme.
Enzymes increase the energy of activation for a reaction, thereby making it go faster.
Enzymes shift the equilibrium of a reaction towards products.
A. I only
B. II only
C. I and II only
D. II and III only
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
Which of the following statements is NOT true about competitive inhibition?
A. The Km of the uninhibited reaction is lower than the Km of the inhibited reaction.
B. Competitive inhibitors bind at the active site of an enzyme.
C. Even if you greatly increase substrate concentration, the Vmax of the inhibited reaction will never reach the same Vmax of the uninhibited reaction.
D. Competitive inhibitors can resemble the transition state of a reaction.
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
Which of the following statements about Km is true?
A. Km is the substrate concentration required to reach 1/2 Vmax, and is a measure of affinity between an enzyme and its substrate.
B. Km is a measure of affinity between an enzyme and its substrate; a high Km indicates a strong affinity.
C. Km is the substrate concentration required to reach Vmax, and a low Km indicates a high enzyme-substrate affinity.
D. Km is affected by enzyme concentration; if the enzyme concentration is reduced, Km decreases.
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
Hemoglobin is an oxygen-carrying protein found in red blood cells. It is made up of four protein subunits that display cooperative binding. Myoglobin is also an oxygen-carrying protein, however it is found in muscle cells and it is made of only a single protein subunit. How would the saturation curves for hemoglobin and myoglobin compare?
A. Hemoglobin would have a sigmoidal curve while myoglobin would have a simple curve.
B. Both hemoglobin and myoglobin would have sigmoidal curves, but the curve of myoglobin would be right-shifted compared to the curve of hemoglobin.
C. Hemoglobin would have a simple curve while myoglobin would have a sigmoidal curve.
D. Both hemoglobin and myoglobin would have simple curves, but myoglobin would display only 1/4 the saturation level of hemoglobin.
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
Polysaccharides can be used for many different functions. Which of the following is/are polysaccharides that are used primarily for glucose storage?
Starch Glycogen Cellulose A. I only B. II only C. I and II only D. I, II, and III
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
Which of the following is made up of pluripotent cells?
A. The mesoderm
B. The zygote
C. The morula
D. The inner cell mass
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
Labor and delivery depends on each of the following EXCEPT:
A. uterine smooth muscle excitability.
B. oxytocin release from the posterior pituitary.
C. uterine contractions and cervical pressure.
D. prolactin release from the anterior pituitary.
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
When is meiosis II of oogenesis considered complete?
A. After the first polar body is secreted and disintegrates
B. After the primary oocyte completes cytokinesis
C. After the secondary oocyte is fertilized
D. When four ootids are generated from one oogonium
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
Which of the following is a correct timeline for first trimester fetal development?
A. Fetus – morula – blastocyst – trophoblast
B. Zygote – trophoblast – morula – embryo
C. Fertilization – morula – blastocyst – fetus
D. Zygote – morula – placenta – blastocyst
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
Oogenesis is different from spermatogenesis in that:
A. only one ootid is generated from one oogonium.
B. only one round of cytokinesis occurs.
C. two polar bodies are secreted at the end of each telophase.
D. oocyte mitosis is more efficient than spermatocyte mitosis, leading to a regenerating pool of oogonium.
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
During fertilization:
A. the sperm and oocyte plasma membrane fuse to form a zygote in the uterus.
B. the sperm
’
s bindin protein facilitates nuclear fusion.
C. both fast and slow blocks increase the likelihood of a second sperm fusing with the oocyte membrane.
D. the sperm must penetrate both the corona radiata and the vitelline layer.
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
High progesterone levels right after ovulation:
A. are the result of high steroid hormone production by the chorion.
B. function to fine-tune maturation of the endometrium in preparation for implantation.
C. drop after high FSH levels negatively feedback to the brain.
D. stimulate high estrogen production in a positive feedback loop.
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
Which of the following is an incorrect matching of developmental phase and characteristics?
A. Blastulation – development of the blastocyst, including the trophoblast and the outer cell mass
B. Differentiation – the ectoderm develops into neural tissue, while the endoderm develops into epithelial tissue
C. Cleavage – cell divisions without an increase in cell volume, resulting in the morula
D. Gastrulation – development of the three primary germ layers (ectoderm, endoderm and mesoderm)
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
ll of the following are true statements EXCEPT:
A. fraternal twins occur from the fertilization of both oocytes of a double ovulation; one that occurs about midcycle (day 14) and one that occurs about a week later.
B. ovulation is triggered by an LH surge that is the result of the positive feedback of estrogen on the anterior pituitary gland.
C. LH causes ovulation and the formation of the corpus luteum.
D. menstruation is triggered by a drop in estrogen and progesterone levels.
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
A block in prophase II of spermatogenesis would result in:
A. limited numbers of spermatids.
B. a decreased population of secondary spermatocytes.
C. an accumulation of primary spermatocytes.
D. increased gamete migration from the seminiferous tubule basement membrane to its lumen.
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
Which of the following explains why incompatible fetal blood type does not induce an immune response in a pregnant female, but a HIV-positive woman can pass the disease on to her new baby?
A. The placenta is a complex capillary network that allows many maternal blood components to pass into fetal circulation, but only waste products to pass into maternal circulation.
B. The fetus receives half its genomic information from each parent, but mitochondrial genetic material is inherited solely from the mother.
C. Fertilization occurs in the fallopian tube and the embryo can be exposed to maternal factors during migration to the uterus.
D. The chorion is specific and limiting in terms of which factors can pass into the maternal or fetal circulatory systems.
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
The Mullerian ducts in females:
A. develop into the ovary and fetal mammary tissue.
B. regress and the Wolffian ducts develop into female internal genitalia.
C. develop into internal genitalia such as the fallopian tubes, uterus, and cervix.
D. regress and the Wolffian ducts develop into female external genitalia.
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
The male reproductive system includes accessory glands, gonads and other genital structures (both internal and external). Which of the following is true?
A. Semen must pass through the urethra, then the ductus deferens on its way out of the body.
B. The testes contains the seminiferous tubules and bulbourethral glands, and the scrotum aids in temperature control.
C. Accessory organs such as the seminal vesicles and prostate help with semen production.
D. In addition to helping with ejaculation, the urethra also functions in urine and solid waste excretion.
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
Which of the following describes the path of travel for sperm in the female reproductive tract?
A. Ovary, fallopian tube, uterus, birth canal
B. Vagina, ovary, fallopian tube, uterus
C. Cervix, uterus, vagina, fallopian tube
D. Vagina, cervix, uterus, fallopian tube
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.
Multipotent cells can become
A. all cell types in the body, but not the trophoblast.
B. some, but not all, cell types in the body.
C. all cell types in the body, including the trophoblast.
D. one specific cell type in the body.
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
Which of the following is a FALSE statement about the birthing process?
A. Cervical dilation is the first stage of labor.
B. Uterine and abdominal smooth muscle contractions help push the baby through the birth canal. C. The placenta is delivered after the baby, in the third stage of labor.
D. Uterine contractions minimize bleeding and prolactin promotes milk production after the baby is born.
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
The primary function of fructose and buffers in semen is to help spermatozoa survive the path from the:
A. vas deferens to the fallopian tube, by providing nutrients and pH regulation.
B. seminiferous tubules to the fallopian tube, by providing nutrients and lubrication.
C. epididymis to the fallopian tube, by providing nutrients and temperature control.
D. sustenacular cells to the fallopian tube, by providing lubrication and pH regulation.
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
A patient with bulimia begins to develop lethargy which is believed to be related to metabolic alkalosis. Which of the following best explains this finding?
A. AA decrease in the cooperativity of oxygen binding to hemoglobin
B. An increase in oxygen delivery due to hemoglobin’s decreased oxygen affinity
C. A right-shift in the oxygen saturation curve for hemoglobin
D. A left-shift in the oxygen saturation curve for hemoglobin
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
A patient has suffered nerve damage which has impaired the contraction of his scalene muscles and results in difficulty breathing. Which of the following is most likely to be affected?
A. Chest wall elasticity
B. Diaphragm function
C. Inspiration
D. Passive expiration
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
What best characterizes alveolar ducts, which follow the respiratory bronchioles?
A. They are the terminal component of the respiratory zone.
B. They are the area with greatest gas exchange in the lung.
C. They are the first component of the respiratory zone.
D. They are thin-walled branches of the respiratory zone where gas exchange can occur.
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
emphigus vulgaris results from autoantibodies directed against a protein found between keratinocytes (skin cells). Which skin layer would you expect to be involved in the blisters seen in these patients?
A. Epidermis
B. Dermis
C. Hypodermis
D. Subcutaneous tissue
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
In premature deliveries before 30 weeks gestation, pulmonary surfactant has not yet been produced. What difficulties would be expected in these infants?
A. Decreased CO2 exchange resulting in respiratory alkalosis
B. Decreased O2 exchange resulting in respiratory alkalosis
C. Collapse of smaller airways resulting in respiratory acidosis
D. Collapse of bronchi resulting in respiratory acidosis
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
A toddler aspirates a toy car and is rushed to the emergency room to have it removed. What are the most likely blood gas findings?
A. Increased O2, increased CO2
B. Increased O2, decreased CO2
C. Decreased O2, increased CO2
D. Decreased O2, decreased CO2
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
An exercise physiologist performs an in vitro assay to analyze the Bohr effect in a system where the bicarbonate in the blood has been replaced by a synthetic buffer. Following the addition of acetazolamide (a carbonic anhydrase inhibitor), he increased the partial pressure of carbon dioxide in this model system two-fold. Which of the following changes would be expected immediately?
A. A right-shift in the oxygen saturation curve for hemoglobin
B. A left-shift in the oxygen saturation curve for hemoglobin
C. Hemoglobin is insensitive to pH change
D. No significant shift in the oxygen saturation curve
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
A drug is discovered that markedly increases hemoglobin’s affinity for oxygen. What physiological situation is this drug mimicking?
A. Hyperventilation
B. Rigorous exercise
C. Running a fever
D. Increased production of BPG at high elevation
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
Anatomical dead space is classified as that area of the respiratory pathway where no gas exchange occurs with the blood. Which of the following is part of the anatomical dead space?
A. Alveolar duct
B. Alveolus
C. Terminal bronchioles
D. Respiratory bronchioles
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
A student becomes anxious during his final exam and begins hyperventilating. Which of the following is/are true?
Blood pH increases Blood CO2 levels decrease O2 saturation of hemoglobin increases A. I only B. I and II only C. II and III only D. I, II, and III
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
How does pulmonary surfactant function?
A. By increasing alveolar elasticity
B. By increasing the cohesive force of water
C. By increasing the surface tension of water
D. By decreasing the surface tension of water
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
A scientist has had a stationary-phase culture of E. coli at 4°C for several weeks. If she takes a small amount of this culture and puts it into 100 mL of new media (shaking at 37°C overnight), which of the following will be observed?
A. The culture will immediately start growing exponentially and will be in the stationary phase by morning.
B. After the lag phase, the culture will start growing rapidly.
C. No bacterial growth will be observed because most of the original culture was dead.
D. The culture will be mostly transparent in the morning, as the cells will be in stationary phase.
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
A researcher isolates Gram-negative bacilli bacteria. This organism has a:
A. thick peptidoglycan cell wall and is spiral-shaped.
B. thin chitin cell wall and is round-shaped.
C. thick cellulose cell wall and is rod-shaped.
D. thin peptidoglycan cell wall and is rod-shaped.
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
The viral genome integrates into the host genome during the lysogenic cycle. After this:
A. the host genome is not expressed, due to virus-encoded repressor proteins.
B. the viral genome is silent, but replicated along with the host genome.
C. the virus genome excises and activates once the host cell is dead or dying.
D. excision of the viral genome is very precise and occurs only when the host cell is under stress.
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
The lytic and productive viral cycles are similar in that they both:
A. destroy the host cell to allow viral particle release.
B. involve a virus using host cellular machinery to replicate the viral genome and capsid.
C. require the viral genome to be integrated into the host genome.
D. involve expression of hydrolase to generate dNTP building blocks, and lysozyme to allow viral release.
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
A virologist dips his pipette tip into a plaque on a bacterial plate, then into an actively growing culture of E. coli cells. A new strain of E. coli results. Which of the following best explains what occurred?
A. The new E. coli strain is the result of transduction with a lysogenic phage.
B. The new E. coli strain must have acquired random genomic mutations such as deamination.
C. Antibiotics or toxins in the plaque exerted selective pressure on the bacterial culture.
D. The plaque contained bacterial cells that underwent conjugation with the second culture.
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
Which of the following is true regarding viral entry into a host cell?
A. Viral attachment is relatively random, and this helps viruses evolve quickly.
B. Attachment can also be called eclipse, and penetration can also be called adsorption.
C. Prokaryotic viruses enter their host via receptor-mediated membrane fusion.
D. Animal viruses can enter their host via endocytosis, but this requires a specific receptor on the host surface.
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
If a Hfr cell mates with an F– bacterium, which of the following is true?
A. Both cells will become Hfr strains.
B. The Hfr bacterium becomes female (F–), while the female strain becomes Hfr.
C. The female bacterium could end up female (F–), male (F+), or Hfr.
D. The mating is impossible since only F+ bacteria (male) can perform conjugation with F– (female) cells.
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
