Bio UWorld Only

Kinase and Phosphatase

Post-translational modifications occur after protein production. Protein phosphorylation is a reversible post-translational modification that can alter protein conformation, interaction, and function. Kinases are enzymes from the transferaseclass that catalyze the addition of phosphate groups (phosphorylation) from ATP or GTP to target proteins. In contrast, phosphatases are hydrolase enzymes that carry out the catalytic removal of phosphate groups (dephosphorylation) from a specific substrate by using water to remove a molecule of inorganic phosphate.

The initial filtration step in the glomerulus of the mammalian kidney occurs primarily by: A) Passive flow due to a pressure difference. B) Passive flow resulting from a countercurrent exchange system. C) Active transport of water, followed by movement of electrolytes along a resulting concentration gradient. D) Active transport of electrolytes, followed by passive flow of water along the resulting osmolarity gradient.

A - There are three pressures that work together to regulate filtration in the glomerulus: glomerular capillary pressure, capsular hydrostatic pressure, and blood colloid osmotic pressure. The glomerular capillary pressure will force filtrate from a capillary into Bowman's capsule; the other two forces promote movement of the filtrate in the opposite direction. B:It is not a countercurrent exchange system; it is the difference in pressures that regulates the formation of the filtrate. C: Water is not actively transported from the glomerulus into Bowman's capsule. D: Electrolytes are not actively transported from the glomerulus into Bowman's capsule.

What are microfilaments composed of?

Actin

Scientists have recently discovered a drug that inhibits PTH. Which of the following is most likely to be an effect of the drug? A) Decrease in osteoclasts and decrease in bone hydroxyapatite levels B) Decrease in osteoclasts and increase in bone hydroxyapatite levels C) Decrease in osteoblasts and decrease in bone hydroxyapatite levels D) Decrease in osteoblasts and increase in bone hydroxyapatite levels

B - PTH stimulates osteoclasts, which consume bone. Therefore, inhibiting PTH would decrease osteoclasts (choices C and D are incorrect). Hydroxyapatite is the form in which Ca2+ is stored in bones. Without osteoclasts, Ca2+ would remain in the bone, increasing hydroxyapatite levels (choice A is incorrect).

In general, telomeres are NOT important to bacterial cells because most bacterial chromosomes: A) Do not replicate. B) Are circular. C) Replicate quickly and efficiently. D) Are composed of single-stranded DNA.

B - The term telomere refers to a distal end of a chromosome. Telomeres contain non-coding repeats. As opposed to eukaryotic cells, bacterial DNA is constituted by one single circular molecule, thus it does not contain telomeres. A: Bacterial chromosomes replicate and their DNA copies are distributed to their daughter cells like in eukaryotic cells. C: Lack of telomeres is not due to the speed of replication. D: Bacterial DNA is double stranded like the eukaryotic DNA.

Which of the following hormones is(are) directly required for spermatogenesis? I. Luteinizing hormone (LH) II. Follicle-stimulating hormone (FSH) III. Inhibin IV. Testosterone A) IV only B) I and IV only C) II and IV only D) I, II, and III only

C - Sertoli cells support and nourish the spermatocytes and promote the process of spermatogenesis. Spermatogenesis would not occur without Sertoli cells. The two hormones that directly stimulate Sertoli cells are FSH and testosterone.

The stereochemical designators α and β distinguish between: A) enantiomers at an epimeric carbon atom. B) enantiomers at an anomeric carbon atom. C) epimers at an anomeric carbon atom. D) epimers at a non-anomeric carbon atom.

C - The α versus βdesignation distinguishes between molecules with multiple chiral centers, but differ only in the configuration of the site known as the anomeric carbon atom.

Which of the following correctly pairs a cell type with its function? A) Osteoclasts are responsible for the creation of new and growing bone. B) Osteoclasts are mature osteoblast that will eventually become surrounded by the secreted matrix. C) Osteoblasts are responsible for secretion of the bony matrix. D) Osteoblasts are responsible for bone resorption.

C - osteoblast serve to promote the creation of new bone, osteoclasts are responsible for bone reabsorption recycling. Osteocytes are mature osteoblast cells that eventually become surrounded by the matrix that they secrete.

Which blood vessel has the lowest pressure? A) Arterioles B) Capillaries C) Venules D) Veins

D

Under physiological conditions, increased activity of succinyl-CoA synthetase will most likely result in: I. Increased levels of succinyl-CoA. II. Increased levels of succinate. III. Increased levels of GTP. A) I only B) II only C) I and III only D) II and III only

D - Increased activity of succinyl-CoA synthetase will result in greater levels of the reaction products, succinate and GTP. Succinyl-CoA is the substrate of the reaction and its levels will likely decrease with increased succinyl-CoA function.

After the depletion of hepatic glycogen in newborns, which compounds can be used as precursors to sustain the blood glucose level? I. Acetyl-CoA II. Lactate III. Oxaloacetate IV. α-Ketoglutarate A) I and II only B) II and III only C) I, II,and IV only D) II, III, and IV only

D - Only lactate, oxaloacetate, and α-ketoglutarate are used as starting materials in gluconeogenesis.

The cell type in the male reproductive system that is most analogous to the female ovum is the: A) Spermatogonium. B) Primary spermatocyte. C) Spermatid. D) Spermatozoon

D - The mature ovum is the female gamete that has completed meiosis and contains the haploid number of maternally derived chromosomes. This makes it most analogous to spermatozoa, the mature male gametes that contain the haploid number of paternally derived chromosomes.

Defects of lower motor neurons

Defects of lower motor neurons are associated with four primary abnormalities, or signs. These can be thought of as the result of a lack of receptivity by the lower motor neuron. Defects in the lower motor neuron weaken the resulting signal to the muscle it meets. Hyporeflexia: Diminished receptor sensitivity in muscles resulting in decreased stretch reflexes Hypotonia: Decreased muscle tension and tightness, which inhibits muscles from being able to stretch fully Fasciculations: Spontaneous, involuntary muscle twitching Muscular atrophy: degradation of skeletal muscle

Slow twitch muscles

Slow twitch muscles are slow to fatigue, therefore featured in postural muscles that remain contracted for extended periods of time. Therefore, they are likely to contain large amounts of myoglobin, a red protein and stores oxygen within muscle.

Transporter proteins

The passage notes that P-gp is an ABC transporter protein with ATPase activity. These transporters use ATP to actively transport molecules such as drugs out of the cell.

Kidney Function

-Control of extracellular fluid volume & regulation of blood pressure via the renin-angiotensin system -Regulation of osmolarity (ie, excretion or retention of water) -Regulation of ion concentration (Na+, K+, Ca2+) by balancing dietary ion intake with urinary excretion -Regulation of pH via bicarbonate buffer system (excretion/retention of HCO3− and H+) -Excretion of waste (creatinine, ammonia, urea, foreign substances) -Production of renin (enzyme involved in blood pressure regulation) & hormones (erythropoietin for erythrocyte production & calcitriol for increased Ca2+ levels)

Epilepsy may result in motor seizures due to massive synchronous firing of neurons in a small area of the cerebral cortex (the epileptic focus). Excitation spreads from the focus, involving an increasingly larger area of the cortex. A drug for the treatment of epilepsy would be most effective if it caused which of the following changes in the epileptic focus? A) An increase in the neuron-firing threshold B) An increase in extracellular Na+ concentration C) A decrease in axon-membrane permeability to negative ions D) A decrease in the length of the depolarization stage

A - An increase in the threshold would make it more challenging for a neuron to fire an action potential. This will reduce the frequency of downstream neuron depolarizations, resulting in relief of the symptoms. B: An increase in the concentration of Na+ outside the cell will not help reduce the symptoms, because there will still be a flux of ions inside the cell when the firing starts, causing continuous depolarization. C: Anion influx hyperpolarizes a neuron, reducing likelihood of action potential firing. Therefore, a decrease in axonal-membrane permeability to negative ions will not provide a relief as it will increase action potential firing. D: A decrease in length of the depolarization stage will allow for a higher frequency of action potential firing. This would worsen the situation instead of improving it.

In the pentose phosphate pathway, which enzyme catalyzes the production of 6-phosphogluconolactone? A) Glucose 6-phosphate dehydrogenase B) 6-Phosphogluconolactonase C) Phosphogluconate dehydrogenase D) Glycerol 3-phosphate dehydrogenase

A - Glucose 6-phosphate dehydrogenase catalyzes the conversion of glucose 6-phosphate to 6-phosphogluconolactone in the pentose phosphate pathway.

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

A - The heart and blood vessels both differentiate from the mesoderm.

Which of the following organs produces bile? A) Liver B) Stomach C) Pancreas D) Gallbladder

A - The liver is an accessory organ to the digestive tract that produces bile (choice A is correct). The gallbladder is a small pouchlike structure that is responsible for storing and secreting bile but does not significantly contribute to its production (choice D is incorrect). The stomach and pancreas release a different set of enzymes that contribute to chemical digestion (choices B and C are incorrect).

Tetanic contraction

A condition promoted by high frequency stimulation of an associated motor unit.

ATP hydrolysis

ATP is converted to ADP & phosphate energized myosin heads (removal of phosphate) need energy

Autophosphorylation

ATP must be included so that autophosphorylation would be possible.

The finches observed by Darwin on the Galapagos Islands are an example of adaptive radiation. In order to set up conditions that would produce adaptive radiation, it would be necessary to place members of: A) One species in one rapidly changing environment. B) One species in several different environments. C) Several very similar species in the same environment. D) Several unlike species in one environment to compete for the same resources.

B - Adaptive radiation involves the divergence of one species into multiple species over time, which can occur when subgroups of the original species are separated or isolated in different environments so that these subgroups evolve independently of one another. B is correct because it describes this situation. A is not the best answer because a single rapidly changing environment does not encourage adaptive radiation. Rather, as the environment changes, the species would be more likely to collectively evolve into a single new species adapted to the new environment.

A resident of a famine area who appears undernourished and extremely emaciated has eaten only starches for the past 3 months. A urine analysis shows that a large amount of nitrogen is being excreted. This is most likely evidence of: A) An abnormally high rate of glycogen breakdown in the liver. B) Breakdown of the body's own structural proteins to provide energy. C) Utilization of the last remaining fat reserves to provide energy. D) Incomplete reabsorption of nitrogenous products due to kidney failure.

B - In starvation, the body uses up its stores of carbohydrate and lipids, and then begins to break down body proteins for metabolic energy. A byproduct of the metabolism of the amino acids from protein is nitrogen.

Which statement best explains why the microfilament lengths do NOT change when the sarcomere shortens in a muscle contraction? A) The - ends of the microfilaments are capped by Z lines, and the actin subunit concentration is kept above 1 µM in muscle cells. B) The - ends of the microfilaments are capped by Z lines, and the + ends are capped by another protein. C) The actin subunit concentration is kept above 4 µM in muscle cells. D) The - ends polymerize and the + ends depolymerize at the same rate.

B - Within a sarcomere, the microfilament length remains stable. Because one end of the microfilament is anchored in the Z line, actin monomers are prevented from being added to or subtracted from that end. This rules out the possibility of treadmilling. Therefore, to retain a stable length, both ends of the microfilament must be capped.

An ulcer that penetrated the wall of the intestine would allow the contents of the gastrointestinal tract to enter: A) The perineum. B) The peritoneal cavity. C) The pleural cavity. D) The lumen of the intestine.

B- If an ulcer penetrated the walls of the intestine, this would allow the contents of the gastrointestinal tract to enter the peritoneal cavity. Membranes surround this cavity, which would prevent further transport of the gastrointestinal contents through the rest of the body. An ulcer in the small intestine would not allow the contents to enter the lumen because this is the normal place in which the contents are found.

Which of the following structures develops into the peripheral nervous system? A) Notochord B) Neural folds C) Neural crest cells D) Neural tube

C

The initial source of energy replacement in the liver of newborn infants is formed by glycosidic bonds between glucose molecules through: A) α(1→4) linkage linearly and β(1→6) linkage at branch points. B) β(1→6) linkage linearly and α(1→4) linkage at branch points. C) α(1→4) linkage linearly and α(1→6) linkage at branch points. D) α(1→6) linkage linearly and α(1→4) linkage at branch points.

C - The glucose polymer in liver (glycogen) is formed by glycosidic bonds between glucose molecules through α(1→4) linkage linearly and α(1→6) linkage at branch point.

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

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

Glycogen phosphorylase

Glycogen phosphorylase is the enzyme that catalyzes the rate-limiting step in glycogen breakdown (glycogenolysis)

In order from smallest to largest levels of organization: sarcomere --> myofibril --> muscle fiber --> fasciculus

Sarcomere describe the range of the film is in the smallest functional unit of muscle. Essentially, myofibrils are collections of parallel sarcomeres forming a filament. A muscle fibers a multinucleate muscle cell containing numerous myofibrils, while a fasciculus is a unit of muscle that contains multiple fibers wrapped and connective tissue.

Transcription Factors

Transcription factors regulate the expression of other genes by binding to DNA, not RNA.

HIV

Viral DNA integrated into a host cell genome by integrase would originate from a retrovirus. mRNA transcribed from retroviral DNA is either used to synthesize viral proteins, or used as the RNA genome for progeny viruses. Thus, the sequence of the nucleotide in the original viral genome will be the same as that of the transcribed mRNA. It is a Scientific Reasoning and Problem Solving question because you are asked to recall that HIV is a retrovirus with an RNA genome and use your understanding of reverse transcriptase to use a given transcribed mRNA sequence to determine the original viral genome sequence.

Activin is a small signaling proteins with a number of biological roles. One of its functions involves increase of FSH binding to the FSH receptor in the developing follicle. In males, an active and receptor agonist is most likely to cause: A) Increase spermatogenesis. B) Decreased spermatogenesis. C) Desensitization of Leydig cells. D) Desensitization of Sertoli cells.

A - Activin essentially magnifies the effects of FSH on its target cells. In males, FSH acts on Sertoli cells to stimulate spermatogenesis. An activin receptor agonist as a molecule that stimulates the activin receptor, exactly like endogenous activity. C,D: According to this question, activin has a sensitizing effect. Other words, the normal effects of FSH or increase of the presence of Active.

The translation of tau mRNA would most likely involve: I. An enzyme with the ability to hydrolyze peptide bonds. II. tRNA molecules carrying amino acids corresponding to codons on the mRNA molecule. III. A ribosomal complex capable of reading the target mRNA in a 3′ to 5′ direction. A) I only B) II only C) I and II only D) I and III only

B - Translation, the energy-requiring process by which ribosomes (with E, P, and A sites) make proteins from mRNA, occurs in eukaryotes in the following stages: Initiation: The small 40S ribosomal subunit complexes with initiation factors and binds the mRNA 5′ cap to scan the mRNA for the start codon (5'-AUG-3'), which codes for the amino acid methionine. An initiator transfer RNA (tRNA) molecule "charged" with methionine contains a complementary anticodon (3'-UAC-5') that base pairs with the AUG codon on mRNA (Number II). Subsequently, the 60S subunit is recruited and binds the initiator tRNA at the P site, marking the formation of the translation complex. Elongation: The ribosome continues to elongate the polypeptide chain by reading each mRNA codon in a 5′ to 3′ direction (Number III). During this step, a new charged tRNA with a complementary anticodon enters the A site. The enzyme peptidyl transferase then transfers the growing polypeptide chain from the tRNA at the P site to the new tRNA at the A site by catalyzing the peptide bond between adjacent amino acids on these tRNAs (Number I). The ribosome moves one codon forward, allowing the uncharged tRNA at the P site to translocate to the E site, where it is ejected. Another charged tRNA enters the A site, and this process is repeated until the ribosome encounters a stop codon. Termination: The polypeptide is released from ribosomes when a stop codon (UAA, UAG, or UGA) is detected in the mRNA at the A site, indicating the end of translation. Release factors induce peptidyl transferase to cleave the ester (not peptide) bond between the polypeptide and the final tRNA, causing disassociation of the translation complex.

Which organ synthesizes a compound that facilitates the mechanical digestion of lipids? A) Pancreas B) Salivary glands C) Gallbladder D) Liver

D

Renin angiotensin system

The renin angiotensin system regulates blood pressure by stimulating the release of aldosterone, which increases sodium reabsorption. Water naturally follows the ion movement, thus increasing blood pressure and decreasing urine output. If this system were disabled, blood pressure could still be regulated by anti-diuretic hormone, which operates through a mechanism independent of the renin angiotensin system. Instead of directly involving sodium reabsorption, 88 functions by increasing the permeability of the collecting duct to water.

Cardiac muscle versus skeletal muscle

One of the hallmarks of cardiac muscle tissue is an action potential with a plateau and SD polarization. Specifically repolarization is delayed due to the action of slow to open voltage Kates calcium channels. This maintains regular beat of the heart. Skeleton muscle, on the other hand, like such a plateau, and the opening of potassium channels is able to rapidly re-polarize the cell.

More muscles

Smooth muscle cells contain both thick myosin and then actually filaments. However, the lack of Sarku American organization results in the characteristic lack of striation in a smooth appearance. Unlike skeleton cardiac muscle, smooth muscle also contains intermediate filaments are anchored to dance party is distributed along the cytoplasm of the cell.

Physiological functions of skeletal muscle

Temperature regulation and movement of fluid within the body Shivering results in the generation of heat through metabolism, moreover, blood is passed through the valves and veins in part due to contraction of skeletal muscle. So, most of the driving force for fluid flow in the veins is generated from the pumping of the heart. Skeletal muscle also contributes to the flow fluid through the lymphatic system. Smooth muscle within the walls of blood vessels is responsible for the regulation of diameter and therefore for the resistance to fluid flow.

Doctors are studying a patient who shows signs of incomplete digestion. Tests show that many of the carbohydrates, proteins, and lipids in the patient's chyme are not digested after passing through the duodenum. Which of the following could be a cause of the problem? A) Incomplete mechanical digestion B) Pancreatic dysfunction in insulin production C) Overdigestion by pepsin D) Failure of enteropeptidase secretion

D - Complete digestion of carbohydrates, proteins, and lipids in chyme is reliant on pancreatic enzymes (choice D is correct). The pancreas is responsible for insulin production, but insulin is not involved in extracellular nutrient digestion (choice B is incorrect) Pancreatic enzyme secretion and activation in the duodenum are dependent on trypsin activity. Trypsin is activated by enteropeptidase. Thus, a failure of enteropeptidase secretion would prevent pancreatic enzymes from functioning, creating the described symptoms.

When concentrated urine is being produced, in which of the following regions of the kidney will the glomerular filtrate reach its highest concentration? A) Proximal convoluted tubule B) Distal convoluted tubule C) Cortical portion of the collecting duct D) Medullary portion of the collecting duct

D - The collecting duct is the final structure in which water reabsorption occurs, which concentrates filtrate. The medullary portion of the collecting duct is the last portion of the tubules where reabsorption can occur. In the portion of the tubule that follows, there will be no more reabsorption. Thus, the medullary portion of the collecting duct contains the most concentrated glomerular filtrate that will correspond to the urine.

Which of the following does NOT originate from the endoderm? A) The liver B) The epithelial lining of the digestive tract C) The epithelial lining of the respiratory tract D) The epithelial lining of the mouth

D - The endoderm gives rise to the epithelial lining of the digestive and respiratory tract, as well as its associated viscera (choices A, B, and C are incorrect). The endoderm does not give rise to the epithelial lining of the mouth or anus, which are derived from the ectoderm (choice D is correct).

The process of culturing bacteria often involves inoculation of cells on a noncellular, agar-based medium. Such a methodology would NOT result in growth of animal viruses because animal viruses: A) Are obligate parasites. B) Lack DNA. C) Assimilate carbon. D) Require essential vitamin supplements for growth.

A - Animal viruses can only infect animal cells, not bacteria. Thus, animal viruses cannot grow in agar plates with bacteria. B: Some animal viruses contain DNA. The inability to culture animal viruses on agar plates is not due to the structure of their genome, but rather to their selective capability of infection. C: Carbon assimilation is not the reason why viruses cannot be grown in bacteria seeded agar plates, but rather because animal viruses cannot infect bacteria. D: It is the animal cells infected by the animal viruses that will require essential vitamin supplements for growth, and not the viruses per se. Animal viruses will infect only animal cells, thus, they cannot be grown in bacteria seeded agar plates.

Based on the passage, which of the following is LEAST likely to be a symptom of diabetes mellitus? A) Loss of appetite B) Sweet-tasting urine C) Unexplained weight loss D) Feelings of fatigue

A - Based on the passage, diabetes affected individuals will use proteins and lipids as a source of glucose, thus increasing, rather than decreasing, their appetite. B: High levels of glucose are a symptom of diabetes. This is due to the fact that kidneys can reabsorb glucose up to 320 mg/min (tubular maximum). The levels of glucose in the filtrate that are in excess will pass into urine, leading to a sweet taste. C: Unexplained weight loss might occur as the body uses proteins and lipids as a source of energy. D: Feeling of fatigue might occur as the body uses amino acids and lipids as a source of energy.

An increase of blood calcitonin would most likely stimulate production of which of the following cells? A) Osteoblasts B) Osteoclasts C) Osteocytes D) Osteons

A - Calcitonin stimulates osteoblasts to build up bones in response to high blood Ca2+ levels. Osteoclasts consume bone and are stimulated by PTH (choice B is incorrect). Osteocytes are mature bone cells found in lacunae (choice C is incorrect). Osteons are bone matrix units found in long bones (choice D is incorrect).

Which of the following processes facilitates the high variability and diversity in antibody structure? A) Recombination of DNA B) DNA methylation C) Alternative splicing D) Post-translational modification

A - DNA recombination is the process through which variable regions on antibodies are produced. DNA methylation is involved in epigenetics and gene silencing (choice B is incorrect). Alternative splicing refers to mRNA processing (choice C is incorrect). Post-translational modifications are on proteins after they are created and not how genetic variation is made (choice D is incorrect).

A biologist use them eyes and head of a single thick filament, which is protruding in the form of a cross bridges near, but not attached to, a thin filament. The cross bridge is currently positioned at 90 degree angle to the thin filament. This moment in time occurs immediately after: A) ATP is hydrolyzed a form ADP and inorganic phosphate. B) ATP is released from myosin. C) ATP binds myosin. D) None of the above.

A - The questions stem mentions to actin and myosin are not attached, but that they are position perpendicularly, implying that they are preparing for future binding. Hydrolysis of ATP into ADP and inorganic phosphate restores the myosin head to a 90° angle, cocking it for a new attachment of the ensuing power stroke. B: This occurs immediately after myosin viaduct and before the power stroke. C: ATP binding of myosin does detach it from action, but it's bent until ATP is hydrolyzed.

Phosphatase

A phosphatase is an enzyme that removes a phosphate group from its substrat

Muscles

A single unit muscle, which is composed of cells that are connected via gap junctions, this allows the action potential propagate between cells, leading to simultaneous contraction. On the other hand, a multi unit muscle, in which each sells innervated individually and contracts on his own, as usual for precise construction. Single units with muscle is found in the lining of blood vessels, the G.I. tract, the uterus.

Layers of the skin

As the most exterior layer of skin the epidermidis is showing at the top. The dermis is the tissue layer located immediately below the epiderm is. It is composed mainly of connective tissue, such as collagen, and contains the bases of sebaceous glands and many sweat glands. The hypodermis is located immediately below the dermis, but consist of the fatty tissue underneath.

What is the threshold value for a neuron to initiate an action potential? A) +35 mV B) -50 mV C) -70 mV D) 0 mV

B - -50 mV is the threshold value at which an action potential is generated (choice B is correct). Repolarization is initiated when the cell membrane reaches +35 mV (choice A is incorrect). The resting membrane potential of the cell is equal to -70 mV (choice C is incorrect). There is no special event that occurs when the membrane reaches 0 mV (choice D is incorrect).

An acetylcholine inhibitor in the heart would NOT affect what heart measurement? A) Heart rate B) Hemoglobin saturation C) Cardiac output D) Vagal tone

B - Acetylcholine is the neurotransmitter released by the vagus nerve to suppress the constant pulse by the SA node. If the acetylcholine was inhibited, the vagal tone would decrease (choice D is incorrect), the heart rate would increase (choice A is incorrect), and the cardiac output would also increase (choice C is incorrect), but hemoglobin saturation would remain relatively unchanged.

The influx of calcium ions into the neuron most directly causes which of the following to occur? A) Hyperpolarization of the neuron B) Release of neurotransmitters into the synaptic cleft C) Sequestration of neurotransmitters into storage vesicles D) The repolarization of the neuron

B - Action potentials trigger the influx of calcium ions which cause the release of neurotransmitters into the synaptic cleft (choice B is correct). The neurotransmitters would be stored in vesicles prior to this event (choice C is incorrect). Hyperpolarization and repolarization are due to the efflux of potassium ions (choices A and D are incorrect).

How does the absorption of vitamins and minerals vary? A) All types of nutrients are absorbed into the bloodstream B) All types of nutrients are absorbed into the lymphatic system C) Vitamin K is absorbed through lacteals while vitamin C is absorbed through microvilli D) Vitamin B is absorbed through lacteals while vitamin E is absorbed through microvilli

C

What is the purpose of the acrosomal reactions? A) To protect sperm cells from an acidic environment B) To generate energy for sperm cells C) To allow sperm cells to penetrate the zona pellucida D) To prevent polyspermy

C

Which of the following is not a primary function of the liver? A) Filtering toxins such as ammonia B) Glycogenolysis and gluconeogenesis C) Removing urea from the bloodstream D) Albumin production

C

Which part of the nephron is most directly affected by the endocrine dysfunction? A) Bowman's capsule because the adrenal cortex secretes antidiuretic hormone to promote liquid filtration B) The loop of Henle because the adrenal cortex secretes antidiuretic hormone to promote liquid filtration C) The distal convoluted tube because the adrenal cortex secretes aldosterone to promote water absorption D) The collecting tube because the adrenal cortex secretes antidiuretic hormone to promote water absorption

C

Which of the following cell types are most likely to respond to IL-6 (a cytokine)? A) Helper T cells and plasma B cells B) Helper T cells and memory B cells C) Cytotoxic T cells and plasma B cells D) Cytotoxic T cells and memory B cells

C - IL-6 is a type of cytokine. Cytokines signal plasma cells to secrete more antibodies, as well as killer T cells to kill infected cells (choice C is correct).

Which of the following processes assist in preventing polyspermy? A) Fertilization B) Acrosomal reactions C) Cortical reactions D) Mitosis

C - The cortical reactions help prevent polyspermy. Fertilization is the joining of a sperm cell and egg cell (choice A is incorrect). The acrosomal reactions allow the sperm cell to penetrate the zona pellucida (choice B is incorrect). Mitosis is a cellular process through which cells divide to form identical daughter cells (choice D is incorrect).

A student remembers a bird's chirp sound for a few seconds before forgetting. This is an example of retrieving information from: A) Short-term memory B) Working memory C) Procedural memory D) Echoic memory

D - Echoic memory allows one to remember something that was just recently heard (choice D is correct). Procedural memory is a type of long-term memory that involves remembering how to do things, such as tying a shoe or riding a bike (choice C is incorrect). Short-term memory holds information for a very short amount of time—often only a few seconds (choice A is incorrect). Working memory is a form of short-term memory (choice B is incorrect).

The inflation of the lungs in normal inspiration involves: I. Contraction of the diaphragm II. Reduction of intrapleural pressure III. Elevation of the rib cage A) I and II only B) I and III only C) II and III only D) I, II, and III

D - The lungs are not directly attached to the thoracic walls; they are separated by a thin, fluid-filled space known as the pleural cavity. When the thoracic cavity (chest cavity) expands during inspiration, the pressure in the intrapleural space decreases and pulls on the lungs, causing them to expand (Number II). This mechanism is known as negative pressure breathing. Normal (quiet) inspiration is primarily initiated by the diaphragm. The diaphragm is a sheet of muscle below the lungs, and it is normally curved upwards. When the diaphragm contracts, it pulls down on the lungs and causes them to expand (Number I). In addition, the external intercostal muscles (a group of muscles at the rib cage) can contract to elevate the rib cage and further expand the lungs (Number III).

An enzyme is more effectively inhibited by uncompetitive inhibitors when: I. The substrate concentration is decreased. II. The substrate concentration is increased. III. The inhibitor concentration is increased. A) I only B) III only C) I and III only D) II and III only

D - Uncompetitive inhibitors bind their target enzymes only when the substrate is first bound to the enzyme. Since at higher substrate concentrations, the substrate-enzyme complex are more abundant, the uncompetitive inhibitor will work most effectively when the substrate concentration is the highest. Additionally, an increase in the inhibitor concentration results in increased enzyme binding and inhibition.

Antibodies

The constant region of antibodies does not bind antigens. Therefore, in this scenario, the antibody used in Experiment 3 does not bind TNF directly. Instead, the constant region determines antibody class and function by dictating which body cells and proteins the antibody can interact with in order to facilitate antigen destruction. The amino acid sequence of the constant region is largely conserved across all antibodies of the same class.

Colon Function

The food eaten during a meal is first digested in the stomach and passed on as a semifluid mass (chyme) into the small intestine. Chyme is a mixture of water, hydrochloric acid, digestive enzymes, and nutrients from the ingested food (eg, proteins, carbohydrates, fats). The small intestine proceeds to absorb the nutrients and most of the water from chyme and passes the remaining undigested material into the large intestine. The large intestine is composed of three subdivisions: the cecum, colon, and rectum. The colon functions to absorb electrolytes (eg, sodium, chloride) and additional water from the mass of undigested material. As water is absorbed by the colon, the undigested material concentrates into feces (waste matter), which is stored in the rectum for subsequent excretion. Based on the passage, Caco-2 cells are derived from the colonic segment of the human gut and, accordingly, this segment absorbs water.

Joints

The joint between the humerus and ulna is filled with synovial fluid and is this considered a synovial joint. Most joints that move or allow for rotation can be characterized as this particular type. Fibrous joints contain fibers connective tissue and do not move. One example is the joint between the plates of the school. Cartilaginous joints, which contain cartilage, allow very little movement. The connection between vertebral discs as one of these joints.

SDS-PAGE

The proteins shown in Figure 1 were analyzed via sodium dodecyl sulfate polyacrylamide gel electrophoresis (SDS-PAGE), which is used to separate proteins by size (molecular weight). During this procedure, samples are first heated with SDS, a denaturing detergent that coats proteins with a negative charge proportional to their length (size). SDS binds hydrophobic regions in the protein core, unfolding and extending the polypeptide into a linear arrangement of amino acids (ie, primary structure). This detergent also solubilizes the proteins and releases them from all associations with other molecules. Proteins are generally smaller than DNA molecules and are therefore loaded on a highly crosslinked polyacrylamide gelinstead of an agarose gel (has larger pores). An electric current is applied and negatively coated proteins travel through the gel matrix and toward the anode, denoted as the positive electrode in electrophoresis. (Note that this convention is opposite that used in electrochemical cells, in which the anode is negative.) Smaller proteins travel faster than larger ones, creating lanes of size-separated protein bands. These bands are then stained for visualization using Coomassie blue or the more sensitive silver stain, both of which bind tightly to proteins. Band intensity is an indicator of protein abundance.

Western Blot

Western blot is a laboratory technique that first separates proteins by size (gel electrophoresis) and then transfers them to a blotting membrane. A primary antibody specific for the protein of interest is used as a probe. The membrane is washed to clear unbound antibodies and then treated with a secondary antibody that specifically binds to the primary antibody. Depending on the type of secondary antibody used, detection of the proteins of interest is achieved either by fluorescence or by a reporter enzyme. In general, proteins with the shortest amino acid sequences have lower molecular weights and migrate quickly through the gel. Band intensity (thickness) demonstrates the relative levels of expression of the target proteins.

Loop of Henley

While the descending limb of the loop of Henley is permeable to water, it is generally impermeable to an A+ and other ions. As filtrate progresses through the descending loop of Henley, it's osmolarity increases as water flows out of the descending limb and into the interstitial. The ascending limb of the loop of Henle is fairly impermeable to water informational to salt. The collecting duct is generally impermeable to water, except in the presence ADH.

Which of the following correctly pairs a cellular process with the location in which that process occurs in a prokaryotic cell? A) Transcription, cytoplasm B) ATP synthesis, mitochondria C) Degradation of macromolecules, lysosomes D) Modification of carbohydrates on transmembrane proteins, Golgi complex

A - Prokaryotic cells do not contain nuclei, membrane-bound organelles in which nuclear genes are transcribed in eukaryotic cells. Consequently, transcription occurs in the cytoplasm of bacterial cells.

ellagra also results from a deficiency of nicotinamide, which is synthesized from tryptophan. Nicotinamide nucleotides are neither oxidized nor reduced during which step of cellular respiration? A) Glycolysis B) Chemiosmosis C) Citric acid cycle D) Electron transport chain

B - Chemiosmosis is the only step of cellular respiration where NAD+ is neither reduced to form NADH, nor is NADH oxidized to form NAD+.

Which of the following statements correctly describes the distinction between the exocrine and endocrine portions of the testis? A) The exocrine portion secretes only peptides; the endocrine portion secretes only steroids. B) The exocrine portion releases its products into ducts; the endocrine portion releases its products into the blood. C) The exocrine portion secretes only cellular elements; the endocrine portion secretes only chemical substances. D) The exocrine portion is the target tissue for the products of the endocrine portion

B - Exocrine glands secrete their products through ducts; endocrine glands release their products into the bloodstream.

Which type of interaction does NOT contribute to the stabilization of the tertiary structure of a protein? A) Disulfide bond B) Phosphodiester bond C) Hydrogen bond D) Salt bridge

B - Phosphodiester bonds link adjacent nucleotides in DNA. They do not contribute to the stabilization of protein structure.

During the production of insulin, the translated polypeptide is cleaved into the mature form and secreted from the cell. The cleavage most likely takes place in which of the following locations? A) Nucleus B) Ribosomes C) Endomembrane system D) Cytoplasm

C - The endomembrane system is the portion of the cells that is in charge of modifying proteins that will be secreted. Thus, it is most likely that insulin cleavage will occur in the endomembrane system. D: Insulin cleavage is most likely to occur in the endomembrane system, rather than the cytoplasm, because insulin is a secreted protein.

Which event is directly mediated by a ligand-gated ion channel? A) Release of Ca2+ from the sarcoplasmic reticulum of a muscle fiber to initiate muscle contraction B) Influx of Na+ across the axon membrane of a somatic neuron during action potential propagation C) Influx of Na+ across the motor end plate resulting in the depolarization of the muscle fiber membrane D) Re-entry of Ca2+ back into the sarcoplasmic reticulum of a muscle fiber to end muscle contraction

C - the influx of Na+ across the motor end plate occurs when Na+ ion channels bind the ligand acetylcholine.

Which of the following molecular interactions plays the biggest role in protein secondary structure? A) Van der Waals B) Hydrophobic interactions C) Salt bridges D) Hydrogen bonds

D- Hydrogen bonds Hydrogen bonding plays the biggest role in secondary structure, which includes alpha helices and beta sheets. In secondary structure, hydrogen bonding allows bonds to form between backbone carbonyl oxygens and nitrogens, which lead to the observed secondary structures. Choices A-C all play a role in tertiary and quaternary structure.

A male African clawed frog (Xenopus laevis) and a female Western clawed frog (Xenopus tropicalis) mate and produce multiple hybrid embryos that die early in development. Given this, death of the hybrid embryo is most likely caused by: A) Improper segregation of sister chromatids during embryonic mitosis B) Sperm and egg incompatibility C) Reduced production of gametes in the parental species D) Improper pairing of homologous chromosomes during meiosis in the embryo

A - Hybrid offspring are produced when members of two different species interbreed. The mating of hybrid offspring with members of either parental species may cause genetic leakage, or the transfer of genetic information between species. However, in some cases, two species may be unable to produce hybrid embryos or hybrids may be inviable (unable to survive) or infertile. The inability of two species to produce viable or fertile hybrid offspring is called reproductive isolation, which may be prezygotic (before zygote formation) or postzygotic (after zygote formation). In this scenario, mating of a male African clawed frog (Xenopus laevis) and a female Western clawed frog (Xenopus tropicalis) produces hybrid embryos that die early in development, making this an example of postzygotic reproductive isolation. Mitosis is essential for embryonic growth and production of the organism's somatic cells. Failure of sister chromatids to segregate properly (eg, due to chromosomal nondisjunction) during embryonic mitosis would generate cells containing missing or extra genetic information. Accordingly, the ability of these cells to divide and differentiate into functional somatic cells in the embryo would be impaired. This would inhibit zygote development and likely cause early embryonic death. (Choice B) Sperm and egg incompatibility in the two species would prevent fertilization and the formation of an embryo. Because mating of these frog species produced embryos, the sperm and egg must be compatible. (Choice C) Gametes (eggs, sperm) form via meiosis during gametogenesis. Reduced sperm and egg production in parental species would decrease the probability of fertilization and zygote formation following a mating event. However, the frogs are described as having produced multiple hybrid embryos, and reduced gamete formation in the parental species would not affect the development of any embryos formed after mating. (Choice D) During meiosis I, homologous chromosomes pair up and are segregated into cells that divide to form gametes in meiosis II. This process ensures the production of viable gametes containing correct chromosomal numbers. Errors in this process would impair gametogenesis in the hybrid offspring and potentially lead to infertility. However, impaired gametogenesis in the hybrid offspring would not cause embryonic death.

A patient found to have an abnormally low number of helper T cells in her blood contracts a bacterial infection. During the infection, which of the following is most likely NOT affected by the patient's low count of helper T cells? A) Expression of major histocompatibility complex proteins that present bacterial antigens on the membrane of infected host cells B) Proliferation of B lymphocytes in response to the bacterial infection C) Levels of circulating antibodies against the invading bacterium D) Activation of apoptosis-inducing cytotoxic T cells

A - In cell-mediated immunity, T lymphocytes (cells) recognize and mount immune responses against foreign antigens displayed by other cells on specialized surface proteins called major histocompatibility complex (MHC) proteins. MHC proteins on any given cell display fragments of any proteins present within that cell. Accordingly, cells containing foreign pathogens, such as certain immune cells or infected cells, will generally display foreign protein fragments (antigens) on their MHC proteins. Helper T cells are T cells that recognize foreign antigens displayed by MHC proteins of other immune cells, such as: B lymphocytes, cells that bind and engulf a foreign antigen. The antigen is broken down into fragments within the B lymphocyte, and these fragments are transported to be displayed on MHC proteins present on the cell membrane. A helper T cell binds the foreign antigen presented by the B lymphocyte and releases signaling molecules (cytokines) that induce division of the B lymphocyte into many identical cells, some of which differentiate to secrete antibodies against the invading pathogen. macrophages and dendritic cells, which engulf foreign antigens (via phagocytosis) and present these antigen fragments on the MHC proteins found on their cell membrane. Upon binding these antigens, helper T cells release cytokines that enhance other immune responses, including the phagocytotic activity of macrophages and the activation and proliferation of cytotoxic T cells, immune cells that release toxins to induce apoptosis in infected cells. In this scenario, a patient with an abnormally low number of helper T cells contracts an infection. Consequently, a low count of helper T cells would affect the activation of cytotoxic T cells and B lymphocytes (which produce antibodies against the bacterium) (Choices B, C, and D). However, because expression of MHC proteins is solely dependent on a cell's transcriptional and translational machinery, cells would still be able to display bacterial antigens on their MHC proteins regardless of helper T cell count.

A patient is found to have reduced blood flow through pulmonary vessels and excess fluid accumulation in the abdomen and legs. Which of the following conditions would best explain this patient's symptoms? A) Reduced blood flow through the heart valve separating the right atrium and right ventricle B) Partial blockage of the heart valve separating the left ventricle and aorta C) Increased volume of blood pumped from the right ventricle D) Backflow of blood from the left ventricle into the left atrium

A - The heart, consisting of four chambers (two atria and two ventricles), continuously contracts to pump blood throughout the circulatory system. The left atrium receives oxygenated blood returning from the lungs and transfers this blood into the left ventricle, a highly muscular chamber that pumps blood throughout the entire body via systemic arteries. Deoxygenated blood returns to the heart via systemic veins and fills the right atrium, which transfers blood into the right ventricle. Right ventricular contraction pumps deoxygenated blood to the lungs via the pulmonary arteries. Specialized valves facilitate unidirectional blood flow through the heart. Atrioventricular (AV) valves allow blood to flow from the atria to the ventricles, and semilunar (SL) valves allow blood to flow from the ventricles into the arteries. In this scenario, a patient exhibits reduced pulmonary blood flow and excess fluid accumulation in the abdomen and legs. These symptoms are best explained by reduced blood flow through the heart valve separating the right atrium and ventricle. This condition would decrease the volume of blood loaded into the right ventricle prior to contraction, reducing blood flow from the right ventricle into the pulmonary arteries. In addition, the reduced volume of blood flowing into the right ventricle and through the pulmonary circuit would cause a backup of blood in the right atrium and systemic vessels. Accordingly, blood volume and hydrostatic pressure in systemic veins would increase, causing excessfluid leakage from systemic vessels into the surrounding interstitial fluid and subsequent fluid accumulation (edema) around tissues. (Choice B) Partial blockage of the valve separating the left ventricle and aorta would likely decrease blood flow through systemic (not pulmonary) vessels. (Choice C) An increased volume of blood pumped from the right ventricle would lead to increased (not decreased) blood flow through pulmonary vessels. (Choice D) Backflow of blood from the left ventricle to the left atrium would cause blood to back up into pulmonary vessels. This would increase the volume of blood and hydrostatic pressure in pulmonary vessels, likely causing excess fluid leakage from vessels of the pulmonary (not systemic) circuit.

One function of the bulbourethral glands is to: A) Secrete a mucus that is thick and alkaline B) Facilitate sperm maturation C) Transfer mature sperm to the urethra D) Produce the largest portion of seminal fluid

A - The male reproductive system is responsible for producing sperm (ie, gametes) and includes the following structures: Testes: reproductive glands that produce sperm during spermatogenesis. Epididymis: a long, tightly coiled tube on the posterior of each testis; immobile sperm produced by the testes become mature and motile within this tube. Mature sperm are stored in the epididymis until release (Choice B). Ductus (vas) deferens: a long, muscular tube that transfers mature sperm to the urethra (Choice C). Seminal glands: accessory glands that produce the largest portion of seminal fluid (Choice D). Prostate: a gland that produces prostatic fluid-containing enzymes necessary to prevent the coagulation of sperm in the vagina. Bulbourethral (Cowper's) glands: glands that secrete thick, alkaline mucus to lubricate the tip of the penis. The alkalinity of the mucus neutralizes acids in the urine to protect the sperm from the acidic environment of the urethra. Penis: an exterior organ that functions in sexual intercourse by which ejaculation of semen (mixture of sperm and seminal fluid) occurs.

Which of the following skin functions will most likely be impaired in tissue with superficial burns? I. Keratinocyte maturation II. Immune surveillance III. Shock absorption A) I and II only B) I and III only C) II and III only D) I, II, and III

A - The outermost layer of the epidermis, the stratum corneum, is composed of 20-30 layers of these dead keratin-filled cells and functions as a physical barrier to protect the organism against pathogens, ultraviolet light, water loss, and injury due to abrasion or puncture. As stated in the passage, a superficial burn damages only the epidermis; as a result, stem cell production would be affected (Number I). Epidermal Langerhans cells are immune (dendritic) cells that recognize and ingest antigens before migrating to nearby lymph nodes to present these antigens to T cells, activating the adaptive immune response. The adaptive immune response is highly targeted to eliminate a specific pathogen, whereas the innate immune response is generalized to attack any foreign substance (Number II). The dermis (middle layer) is composed of connective tissue and contains blood vessels, sensory receptors, hair follicles, and sweat (sudoriferous) and oil (sebaceous) glands. The hypodermis, or the subcutaneous layer beneath the dermis, is composed primarily of adipose tissue that protects the internal organs by acting as a shock absorber and insulator (Number III).

The sarcomere, the smallest functional unit in strited muscle, contains thick filaments that: A) Are composed of the muscle protein myosin and located in the A band of the sarcomere. B) Are composed of the muscle proteins titin and located in the I band of the sarcomere. C) Contain a double helix of actin and myosin and that are located in the H zone. D) Are composed of elastin and are located along the Z line.

A - Thick filament in a sarcomere of striated muscle are composed of the protein myosin. These filaments have protuberances known as myosin heads along both ends which engage in the myosin and ATPase activity that drives cross bridge formation cycle. Thick filaments are located in the A band of the sarcomere and interact with the adjacent thin filaments, which are made up of Actin. B: Titan is located along the junctions of the sarcomere and the I band, but it does not make up thick filaments.

Which of the following sequences accurately describes the pathway of communication between neurons? A) Axon, soma, dendrite, synapse B) Axon, synapse, dendrite, soma C) Dendrite, axon, soma, synapse D) Synapse, soma, axon, dendrite

B - An electric signal in the pre-synaptic neuron travels down a thin nerve fiber called an axon which conducts the signal to the axon terminals. From there, chemical messengers called neurotransmitters are released into the synaptic cleft(synapse), the region between the axon terminals and the dendrites of the next neuron. Neurotransmitters bind to receptors on the dendrites of the post-synaptic neuron, altering the electric potential of the cell. Lastly, the change in electric potential spreads to the cell body (soma). The information transmitted from the pre-synaptic to post-synaptic neuron can have either an excitatory or inhibitory effect, increasing or decreasing the likelihood of the signal being transmitted, respectively. If the summed excitatory and inhibitory signals exceed a certain threshold in the post-synaptic neuron, the electric signal is transmitted down the axon and the cycle continues as this neuron communicates with the next neuron.

Which part of the immune system is most responsible for producing antibodies in response to an infection? A) T cells in humoral immunity B) B cells in humoral immunity C) T cells in cell-mediated immunity D) B cells in cell-mediated immunity

B - Antibodies are produced by plasma cells, which are a type of B cell (choices A and C are incorrect). While plasma cells are cells that secrete antibodies, B cells are part of the humoral immune response (choice D is incorrect).

What process prevents T cell mediated autoimmune disorders? A) Eliminating hyperactive immune cells through positive selection B) Eliminating hyperactive immune cells through negative selection C) Eliminating hypoactive immune cells through positive selection D) Eliminating hypoactive immune cells through negative selection

B - Autoimmune disorders are caused by immune cells that attack self-cells, meaning they are hyperactive (choices C and D are incorrect). The process of eliminating these hyperactive cells is known as negative selection, while positive selection kills immune cells that do not elicit an immune response (choice A is incorrect).

Which of the following is NOT a function of bone? A) Production of red blood cells B) Blood filtration C) Energy storage D) Hematopoieses

B - Blood filtration is the primary function of kidneys (choice B is correct). Hematopoiesis, or red blood cell development, takes place in spongy bone (choices A and D are incorrect). Yellow bone marrow in the medullary cavity of the diaphysis stores energy in the form of adipose tissue (choice C is incorrect).

Chain skeletal muscle contraction, the dimensions of the various parts of the circle mirror change in what manner? I. The A band shortens. II. The I band shortens. III. The H zone lengthens. A) I only B) II only C) II and III D) I, II, and III

B - During contraction, the I band and H band shorten, while the A band is unchanged.

The regulation of respiratory rate is normally most sensitive to: A) PO2 in the blood B) PCO2 in the blood C) PO2 in the alveoli D) PCO2 in the alveoli

B - The regulation of respiratory rate primarily depends on the pH of the blood as measured by central and peripheral chemoreceptors. These receptors directly detect the [H+] in the blood, which is dependent on the partial pressure of CO2in the blood through the bicarbonate buffer system.

One function of catecholamines released by the adrenal glands during times of stress is to: A) Stimulate secretion of melatonin by the pineal gland. B) Increase dilation of bronchioles in the lungs. C) Inhibit glucose reabsorption by the kidneys. D) Promote the release of digestive enzymes

B - In response to a stressor, the adrenal medulla releases catecholamines (norepinephrine and epinephrine) that induce the following physiological changes: Redirection of blood flow throughout the body via vasoconstriction and vasodilation Vasoconstriction (narrowing) of blood vessels and reduced blood supply to organs that carry out nonessential functions (eg, stomach, intestines, kidneys) conserve oxygen and nutrients for organs that are necessary for immediate survival. Vasodilation (widening) of blood vessels increases blood flow to organs essential for immediate survival (eg, heart, skeletal muscle). Increased heart rate and cardiac muscle contractility promote increased blood flow to the brain, lungs, and skeletal muscles, allowing an organism to process and execute a response to the stressful stimulus. Dilation of airways (bronchioles) enables increased respiratory function and oxygen delivery to tissues.

A physician finds that a patient experiences excessive bleeding after sustaining a cut on the arm. The excessive bleeding is LEAST likely to be caused by: A) Reduced liver function B) High numbers of macrophages near the cut C) Reduced proliferation of platelets D) High levels of serum antibodies against clotting factors

B - In this scenario, a patient with a cut experiences excessive bleeding, indicating an impaired ability to form clots and prevent blood loss from damaged vessels. This symptom could have several causes, including reduced liver function, which would likely decrease clotting factor production (Choice A), and reduced platelet proliferation (Choice C). In addition, because antibodies mark cellular materials for destruction by immune cells, heightened levels of serum antibodies against clotting factors would promote destruction of these factors, impairing clot formation (Choice D).

Given that scar tissue forms when collagen-producing cells replace normal tissue after an injury, all of the following processes would be impaired in the area of scarred skin that forms over a partial-thickness burn EXCEPT: A) Resistance to ultraviolet radiation B) Insulation C) Hair growth D) Salt excretion

B - Normal tissue that has been damaged is replaced by scar tissue, which is made primarily of collagen-producing cells (fibroblasts). When scar tissue replaces the dermis and epidermis, the following normal skin functions are affected: Protection from ultraviolet (UV) radiation: Melanin, a dark pigment produced by epidermal melanocytes, absorbs and prevents UV light from damaging the DNA of other cells in the body. Melanin is transferred from the melanocytes to the keratinocytes, cells that produce the structural protein keratin. The lack of melanocytes (and consequently the lack of melanin) in the scar tissue would reduce the affected area's resistance to the harmful effects of UV radiation (Choice A). Hair growth: Hair strands are composed of dead keratinized cells and are produced in the hair follicles, structures that contain both epidermal and dermal components. Therefore, scarring that replaces the epidermis and dermis would impair hair growth (Choice C). Fingernails, toenails, and calluses are also protective keratinized structures derived from the skin. Sweating: Sweat glands arise from the dermis and secrete sweat (a mixture of water, salts, trace nitrogenous waste, and antimicrobial proteins) through ducts in the epidermis. Because scar tissue does not contain sweat glands, the scarred area would be incapable of salt excretion (Choice D). (Choice B) The subcutaneous layer of skin, also called the hypodermis, is composed of adipose tissue that buffers rapid fluctuations in body temperature by acting as a thermal insulator. Based on the passage, the subcutaneous layer is notdamaged in partial-thickness burns, so the insulation function of skin would not be affected.

SOD mRNA in P. leiognathi encodes a signal sequence that directs the transport of the SOD protein for secretion. This signal sequence will direct SOD proteins to which structure in P. leiognathi? A) Smooth endoplasmic reticulum B) Plasma membrane C) Mitochondrial outer membrane D) Golgi body

B - P. leiognathi is a unicellular prokaryotic organism, and therefore lacks membrane-bound organelles such as the Golgi body and the smooth endoplasmic reticulum (SER) present in eukaryotes. As a result, in P. leiognathi, this signal sequence guides the protein destined for secretion (ie, SOD) directly to protein channels found on the plasma membrane. The channel contains a recognition sequence that interacts with the signal sequence on the polypeptide to induce translocation of the peptide across the plasma membrane and into the extracellular environment.

A male patient at a fertility clinic is noted to have unusually low sperm count. What types of cells should be suspected to be dysfunctional? A) Leydig cells B) Sertoli cells C) Granulosa cells D) Theca cells

B - Sertoli cells are support cells that assist in the synthesis of sperm, and should be suspected to be dysfunctional if the patient is unable to produce motile sperm (choice B is correct). Leydig cells produce testosterone (choice A is incorrect). Granulosa cells form the zona pellucida (choice C is incorrect). Theca cells produce androstenedione, an estrogen precursor (choice D is incorrect).

A recent study of a model organism showed that HMG-5, another member of the HMG family, mainly localized to the end of chromosomes. The site of HMG-5 localization would correspond to regions in humans that: A) Contain multiple copies of different DNA sequences B) Contain heterochromatin similar to that of centromeres C) Are subsequently lengthened with each round of cell division D) Are actively replicated by the enzyme DNA polymerase

B - Telomeres are regions at chromosomal ends that are repeatedly truncated with each round of cell division. Centromeresjoin two sister chromatids and are required for proper chromosome division during mitosis. Despite their different chromosomal locations, both telomeres and centromeres are composed of heterochromatin, a tightly condensed complex of DNA wrapped around histones. Because of its structurally restrictive form, heterochromatin is transcriptionally inactive, meaning that proteins responsible for regulating gene expression cannot access the tightly packed DNA. As a result, hetechromatic regions are often gene-poor and contain repetitive DNA. (Choice A) Telomeres contain repeats of only TTAGGG, a single DNA sequence of six nucleotides that is added by the enzyme telomerase. Multiple repeats of differing DNA sequences are not present in telomeres. (Choice C) Telomeres are shortened with each round of cell division. However, only embryonic stem cells (not somatic cells) express telomerase and therefore have very long telomeres; this allows them to proliferate indefinitely in a controlled manner. (Choice D) DNA polymerase, which is responsible for carrying out DNA synthesis, cannot replicate chromosomal ends.

A humanistic psychologist would most likely support which of the following statements? A) Individuals are all part of a collective conscious through which knowledge is maintained. B) Individuals are unique in their own way. C) A child is most likely to misbehave due to a lack of proper neurotransmitters. D) Dreams contain information that individuals repress.

B - The humanistic perspective emphasizes individuality and uniqueness (choice B is correct). Choice A reflects Jung's theory, Choice C reflects the biological perspective, and Choice D reflects Freud's psychoanalytic perspective.

Healthy subjects show a shallow increase in oxygen-carrying capacity at 0-20 mm Hg, but a sharp increase at 20-50 mm Hg. Which statements explain this effect? I. Oxygen binding induces a change from the R state to the T state II. The affinity of hemoglobin for oxygen increases with O2 pressure III. Hemoglobin exhibits positive binding cooperativity IV. Hemoglobin exhibits negative binding cooperativity A) I and IV B) II and III C) I, II, and III D) I, II, and IV

B - The sigmoidal (S-shaped) curve of the ODC plot indicates positive cooperativity of binding (Number III). Hemoglobin consists of four subunits, each of which can exist in either the T state ("tense," low oxygen affinity) or the R state ("relaxed," high oxygen affinity). Oxygen binding to one subunit stabilizes its R state. The bound subunit then "positively cooperates" with other subunits, inducing them to change from the T to the R conformation and making additional oxygen binding easier. Therefore, overall oxygen affinity increases with O2 pressure (Number II). (Number I) Oxygen binding induces a change from the T state to the R state, not from R to T. (Number IV) Negative cooperativity would result if oxygen binding to one subunit stabilized the T state of other subunits. Many cell receptors, for example, exhibit negative cooperativity for their ligands. In graphical form, this would be reflected as a relatively steep initial binding curve followed by an abrupt change to a shallower slope.

Assume that some individuals who inherit these SNPs express increased cardiac vascularization and increased arterial oxygen saturation. Which of the following can be concluded? A) Individuals in this population have inherited structural abnormalities B) The population is no longer in Hardy-Weinberg equilibrium C) The trait must be sex-linked recessive D) The trait must be codominant

B - To meet the Hardy-Weinberg conditions, a population cannot experience changes in allele frequency. Thus, the population cannot undergo natural selection (choice B is correct). A structural abnormality is a change in a localized region of one or more chromosomes; there is not enough evidence that this has occurred (choice A is incorrect). While there is some evidence that this trait is recessive, there is not enough evidence that this trait follows non-Mendelian genetics (choices C and D are incorrect).

Which of the following is most likely to be found within the cell membranes of intestinal epithelial cells? A) Peptidoglycan B) Cholesterol C) Cytoskeletal filaments D) Cellulose

B - he cell membrane is described by the fluid mosaic model as a nonrigid (fluid) phospholipid bilayer through which diverse proteins are scattered (the mosaic) and are able to migrate laterally. In addition to phospholipids, animal cell plasma membranes also contain cholesterol, glycolipids (lipids with attached sugar groups), glycoproteins (proteins with attached sugar groups), and other proteins. Cholesterol, which is found in eukaryotic but not prokaryotic cell membranes, functions to regulate the fluidity of the bilayer. Due to their unique structure, cholesterol molecules decrease fluidity at higher temperatures and increase fluidity at lower temperatures. As eukaryotic cells, intestinal epithelial cell membranes contain cholesterol. (Choice A) All cells, eukaryotic and prokaryotic, have a plasma membrane, but some cell types are surrounded by an additional enveloping layer known as the cell wall. Unlike the plasma membrane, the cell wall is a rigid structure that defines the shape of the cell it surrounds and provides added protection. Animal cells, including intestinal epithelial cells, do not have cell walls. However, nearly all prokaryotic cells have a cell wall. The major structural component of the bacterial cell wall is peptidoglycan, a polymer of sugars and short polypeptides. (Choice C) Cytoskeletal filaments are found in the cytoplasm of cells and can interact directly with various components of the cell membrane (eg, by anchoring membrane proteins in place). However, they are not themselves part of the plasma membrane. (Choice D) Cellulose, a polysaccharide, is the primary component of the cell wall in plant cells. Cellulose is not found in the membranes of animal cells.

Along with many other effects, surgical lesion of the anterior pituitary in an adult male would likely cause which of the following? A) Increased serum testosterone. B) Increased GnRH production. C) Increased spermatogenesis. D) Increased osteogenesis.

B - like most hormone axes in the body, the hypothalamus pituitary axis is subject to negative feedback. Pituitary lesion will reduce or eliminate the secretion of FSH and LH. GnRH secreting neurons in the hypothalamus will respond to the drop in serum hormone levels by increasing GnRH secretion. A: Testosterone production is stimulated by LH action on Leydig cells in the testes. Without Elledge production, synthesis of testosterone is likely to drop dramatically. C: Spermatogenesis is promoted by the action of FSH on the testes Sertoli cells. When FSH production is limited, spermatogenesis is likely to decrease.

The epidermidis, or most exterior layer of the skin, can be best described as: A) A stratified columnar endothelium. B) A stratified squamous epithelium. C) A simple squamous epithelium. D) A stratified cuboidal epithelium.

B - stratified, or multi layer, epithelium is best suited for protective purposes, especially in tissues like the skin where the most external layers are constantly being lost. In addition, epidermal cells are squamous, or a flattened or thin. The epidermis is not composed of columnar, or tall, rectangular cells. That would more closely reflect the shape of the simple epidermal cells lining the small intestine. Additionally the term endothelium generally describes a specialized epithelium that lines blood vessels, not that of the skin. The skin, with his vital protective role, could not be composed of a simple, single layered, epithelium. Such a lining would be to permeable, both the substances on the exterior and interior of the body. Epidermal cells are squamous not cuboidal, or box like in

The reduced virulence of lac+ S. typhimurium has been correlated with an inability to generate the torque that turns the flagellum. Which structure in the flagellum is most likely being affected by lac genes? A) Hook B) Filament C) Basal body D) Cell membrane

C - The virulence of S. typhimurium decreases due to impaired generation of flagellar rotation. Flagella in prokaryotes consist of three basic parts: a basal body, a hook, and a long helical tube called a filament. The component that generates torque is the basal body, a transmembrane unit that works as a molecular motor. Cells that are unable to generate torque likely have a defect in flagellar basal bodies. (Choice A) The hook is a flexible structure that connects the basal body and the filament. It transmits torque generated by the basal body but does not generate any rotation itself. (Choice B) Filaments, which are mostly composed of flagellin, are the recipients of the rotation generated by the basal body. They respond to rotation and propel bacteria but do not generate rotational motion. (Choice D) The cell membrane helps hold the flagellum in place but does not generate flagellar motion.

A self-directed antibody known as rheumatoid factor (RF) has been found to be involved in RA-mediated inflammation. RF is an antibody that targets a patient's own IgG antibodies, which results in the formation of immune complexes that induce an inflammatory response. High concentrations of self-antibodies such as RF are NOT typically found in the human body because: A) RF functions as a complement protein B) B cells and T cells against self-antigens are destroyed C) Antigen-presenting cells cannot present self-antigens D) T-cell receptor recombination cannot produce self-identifying receptors

B - which immature T cells and B cells possessing receptors that bind to self-antigens are destroyed. Elimination of these cells occurs either by programmed cell death (apoptosis) or by becoming unresponsive to antigens (anergic). According to the question, rheumatoid factor (RF) is an antibody that mediates an autoimmune response by targeting a patient's own IgG antibodies. RF is not usually made in high concentrations because B cells and T cells that mediate self-recognition and antibody production are typically destroyed during negative selection. (Choice A) Complement proteins are blood proteins that increase the effectiveness of antibodies by helping to recruit and activate phagocytes. The question describes RF as an antibody that targets self-IgG antibodies; it is not a complement protein. (Choices C and D) Rearrangement of DNA leads to the expression of surface antibodies/receptors that can identify self-antigens (B cells and T cells). Identification of self-antigens by B cells enables subsequent presentation of the self-antigens. However, these self-recognizing cells are normally destroyed during immune cell maturation.

A skin graft candidate with full-thickness burns on approximately 50% of the total BSA is at risk for all of the following EXCEPT: A) Extensive fluid loss B) Bacterial infection C) Increased sensitivity to touch D) Vitamin D deficiency

C - According to the passage, the damage associated with a full-thickness burn extends through the epidermis and dermis and into the subcutaneous layer (hypodermis). Therefore, this type of burn involves injury to the epidermal and dermal sensory receptors, resulting in impaired touch sensation. (Choice A) Intercellular tight junctions and lipids between epidermal cells form a waterproof barrier that limits the escape of water from the body. Sebum, an oily substance secreted onto the skin surface by sebaceous (oil) glands, also plays a minimal role in waterproofing the skin. An individual with full-thickness burns would be less able to retain fluids and electrolytes because the physical barrier that normally seals water inside the body (the skin's outer layer) has been damaged or lost. (Choice B) The skin protects the body from pathogens by acting as a physical barrier that blocks their entry and by housing immune cells that attack invading microbes. Full-thickness burns destroy these immune defenses and therefore increase the individual's risk for infection. (Choice D) The body obtains vitamin D through dietary absorption in the intestine and through the ultraviolet light-mediated conversion of epidermal cholesterol into a vitamin D precursor. Because this individual has suffered damage to approximately 50% of the total BSA, the skin is no longer able to contribute as effectively to the overall level of vitamin D in the body.

Which of the following would show up on an ECG? A) Leaky Valve B) Stroke Volume C) Fibrillation D) Blood Pressure

C - An ECG does not tell everything about the heart, but it does tell information about the electrical impulses in the heart. The only choice that has to do with contraction is fibrillation.

Which part of an antibody binds to antigens? A) The constant region, consisting of only the heavy chain B) The constant region, consisting of both the heavy and light chain C) The variable region, consisting of both the heavy and light chain D) The variable region, consisting of just the heavy chain

C - Antibodies have both a heavy and a light chain, as well as a variable and a constant region. Antibodies bind to antigens in the variable region, which allows them to bind various antigens (choices A and B are incorrect). The paratope of the antibody is composed of both the heavy and light chains (choice D is incorrect).

When Caco-2 cells are cultured in hypoxic conditions, HIF-1 (a transcription factor) is most likely located in the: A) Tight junctions B) Cytoplasm C) Nucleus D) Lysosomes

C - Based on the passage, HIF-1 is a transcription factor that controls the expression of genes that establish and maintain epithelial barrier integrity under normal oxygen conditions. However, when cells are cultured in hypoxic conditions, the integrity of the epithelial barrier is compromised. This will signal HIF-1 components to localize to the nucleus and form the heterodimer that binds the target DNA sequence, promoting transcription of tight junction genes in an attempt to restore barrier function. (Choice A) HIF-1 regulates the expression of tight junction genes but does not interact directly with tight junctions. (Choice B) The cytoplasm is the thick fluid that fills a cell and includes all materials (eg, organelles, enzymes) within the cell but outside the nucleus. Although transcription factors are translated in the cytoplasm, they function in the nucleus. Therefore, in hypoxic conditions HIF-1 would be in the nucleus, not the cytoplasm. (Choice D) Lysosomes are membrane-bound organelles that degrade many macromolecules (eg, proteins, nucleic acids, carbohydrates, lipids). These cells are being cultured in hypoxic conditions, meaning that the epithelial barrier has sustained damage. Because HIF-1 activity is required to maintain barrier integrity, HIF-1 will not be hydrolyzed by lysosomes.

Which of the following statements is NOT true about hyaline cartilage? A) It lacks innervation. B) It gives rise to bone through endochondral ossification C) It serves as an attachment that holds bones to muscle. D) It is avascular and receives nutrients from surrounding fluids

C - Cartilage is a firm but flexible connective tissue that lacks blood vessels and nerves. Chondrocytes (cartilage cells) secrete chondrin, which is the specialized extracellular matrix that makes up cartilage. The most common type of cartilage is hyaline cartilage, which plays a role in bone development and lines the ends of articulating bones.

Following digestion in the small intestine, certain lipids are absorbed by intestinal epithelial cells and packaged into large droplets. These lipid droplets are released from intestinal epithelial cells and transported into the lymphatic system to ultimately drain into a large vein near the heart. Which series shows the order in which these lipid droplets are transported from the intestine to the bloodstream? A) Lymph capillaries → interstitial fluid → lymph vessels → lymph duct → vein B) Interstitial fluid → lymph vessels → lymph duct → lymph capillaries → vein C) Interstitial fluid → lymph capillaries → lymph vessels → lymph duct → vein D) Lymph capillaries → interstitial fluid → lymph duct → lymph vessels → vein

C - Certain lipids digested in the small intestine are absorbed by intestinal epithelial cells and packaged into large droplets. However, their large size prevents them from crossing directly into the capillaries surrounding intestinal cells. Instead, these lipid droplets are transported through the lymphatic system and into the bloodstream as follows: 1. Lipid droplets are released from the epithelial cells into the interstitial fluid. 2. Lymph capillaries collect lipid droplets from the interstitial fluid. 3. Lymph containing the lipid droplets then flows from the capillaries into increasingly larger lymph vessels. 4. Lymph is then transported into a large tubule structure called a lymph duct, which drains into a large vein near the heart. As a result, the lipid droplets within the lymph enter the bloodstream and circulate throughout the body (Choices A, B, and D).

Upregulation of which enzyme could improve oxygen release into the tissues of cirrhotic patients? A) Phosphoglycerate kinase B) Glyceraldehyde-3-phosphate dehydrogenase C) Bisphosphoglycerate mutase D) Phosphoglycerate mutase

C - Hemoglobin's affinity for oxygen is decreased in the presence of 2,3-bisphosphoglycerate (2,3-BPG) within the red blood cell. 2,3-BPG binds hemoglobin at an allosteric site and stabilizes the deoxygenated conformation. This reduces oxygen binding and increases O2 release into tissues (right shift on oxygen dissociation curve). Therefore upregulation of an enzyme that produces 2,3-BPG could improve oxygen release. Mutases facilitate reactions that change the location of a functional group within a molecule. The enzyme bisphosphoglycerate mutase converts 1,3-BPG to 2,3-BPG by transferring a phosphate from carbon 1 to carbon 2 within a bisphosphoglycerate molecule. Upregulation of this enzyme would therefore increase the concentration of 2,3-BPG and could facilitate oxygen release into the tissues of cirrhotic patients. (Choice A) Kinases add and, in some cases, remove phosphate groups. Phosphoglycerate kinase interconverts 1,3-bisphosphoglycerate and 3-phosphoglycerate and is not involved in production of 2,3-BPG. (Choice B) Glyceraldehyde-3-phosphate dehydrogenase converts glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate (1,3-BPG). It is not involved in oxygen release by hemoglobin. (Choice D) Although phosphoglycerate mutase does have mutase activity, it interconverts 3-phosphoglycerate and 2-phosphoglycerate. It acts on phosphoglycerates, not bisphosphoglycerates.

Following intron splicing in the nucleus, a mature mRNA molecule consists of 500 nucleotides and is translated into a 110 amino acid protein. Given this, which of the following statements regarding this mRNA molecule would be true? A) 110 nucleotides of the mRNA code for amino acids B) The mRNA was transcribed from a 500 base pair gene C) 170 nucleotides of the mRNA do not code for amino acids D) The protein translated from the mRNA has a molecular weight of approximately 5.5 kDa

C - In this scenario, a mature mRNA molecule consisting of 500 nucleotides was translated into a 110 amino acid protein. Because each amino acid is encoded by a three-nucleotide codon, the number of nucleotides translated into this protein can be calculated as follows: 110 AA x 3 nucleotides per codon=330 nucleotides Therefore, 330 nucleotides in this mRNA molecule code for amino acids (Choice A). The number of nucleotides that do not code for amino acids can then be calculated as follows: 500 total nucleotides−330 coding nucleotides=170 noncoding nucleotides. Therefore, 170 nucleotides in this mRNA molecule, including the stop codon and UTRs, do not code for amino acids.

If mutant HMGN2 proteins were unable to bind their substrate and then erroneously mobilize to nucleoli, these proteins would be found in the sites of: A) Lipid synthesis B) ATP synthesis C) Ribosome production D) Ribosome attachment

C - In this scenario, mutant HMGN2 proteins erroneously mobilize to nucleoli. In eukaryotic cells, the nucleolus is a dense, round body found within the nucleus that contains ribosomal DNA. Here, RNA polymerase I functions exclusively to transcribe the pre-ribosomal RNA (rRNA) gene into a single template that is subsequently processed into mature rRNA. As such, the function of RNA polymerase I is restricted to the nucleolus. In addition, the nucleolus is involved in the maturation and assembly of ribosomal subunits. Ribosomal proteins are synthesized in the cytoplasm from mRNA and transported into the nucleolus, where they combine with the newly transcribed rRNA to form precursors to the 40S and 60S subunits. These subunits are then shuttled out of the nucleus via nuclear pores and fully mature in the cytoplasm. (Choice A) The smooth endoplasmic reticulum (ER) has varying metabolic functions depending on the cell type; examples include lipid synthesis (testes/ovaries), drug/poison detoxification (liver), and calcium ion storage (muscle). However, ribosome synthesis is not a function of the ER. (Choice B) Mitochondria mainly function as the "powerhouses" of the cell by producing ATP, the cell's energy currency, via cellular respiration. Mitochondria also regulate cellular metabolism. (Choice D) The rough ER has long, folded membranes coated with attached ribosomes that translate proteins destined for secretion into the rough ER lumen.

Several techniques have been developed to treat blood disorders caused by genetic mutations. Which of the following methods would be most effective in the treatment of a patient homozygous for a deleterious mutation in a gene coding for a subunit of hemoglobin? A) Replacement of the mutant gene with a functional copy in red blood cells B) Treatment with a drug that increases red blood cell production C) Introduction and activation of a functional hemoglobin gene in bone marrow stem cells D) Single blood transfusion from a patient homozygous for the wild-type hemoglobin gene

C - One method of treating genetic mutations is gene therapy, a technique in which a functional gene is introduced into a patient's cells to replace a mutant gene. The new gene can then be transcribed and translated into functional proteins. A common gene therapy method uses retroviral vectors to insert functional gene copies into the stem cells of patients exhibiting specific genetic mutations. In this scenario, a patient is homozygous for a deleterious mutation in a gene coding for a hemoglobin subunit. Expression of this mutant gene would likely produce a portion of the hemoglobin protein with impaired function. Because RBCs are produced from stem cells in the bone marrow, the most effective treatment for this patient would be gene therapy to introduce and activate a functional hemoglobin gene in bone marrow stem cells. This would likely lead to the production of RBCs that contain hemoglobin with normal oxygen-binding abilities. (Choice A) The mutant gene could not be replaced by a functional copy in mature RBCs because mature RBCs lacknuclei and do not contain DNA. (Choice B) Treatment with a drug that increases RBC production would increase the differentiation of bone marrow stem cells into mature RBCs. However, because this treatment does not introduce a functional copy of the hemoglobin gene, these cells would still express the mutation as well as functionally impaired hemoglobin. (Choice D) A blood transfusion from a patient homozygous for the wild-type hemoglobin gene may introduce RBCs expressing functional hemoglobin into the patient's bloodstream. However, all aging RBCs are eventually destroyed by phagocytic cells (ie, macrophages) in the spleen. Because RBCs with wild-type hemoglobin would eventually be destroyed, the patient would require multiple blood transfusions to maintain a population of healthy RBCs.

Autonomous retrotransposons are segments of DNA that encode specialized enzymes. These enzymes allow autonomous retrotransposons to move between different regions of a host genome via RNA intermediates using mechanisms similar to those of retroviruses. Some autonomous retrotransposons are segments of retroviral DNA that have been partially inactivated by mutations. Given this, which of the following mutations would most likely convert active retroviral DNA into an autonomous retrotransposon? A) A mutation that prevents the expression of functional reverse transcriptase B) A mutation that prevents integration of retroviral DNA into the host genome C) A mutation that prevents the expression of envelope proteins D) A mutation that prevents RNA polymerase from transcribing retroviral DNA

C - Retroviruses are a class of enveloped viruses that have RNA genomes. Upon entering a host cell, the envelope and capsid disassemble, releasing viral RNA and proteins into the cytoplasm of the host cell. Viral reverse transcriptase converts viral RNA into double-stranded (ds) DNA that is then transported into the nucleus. A specialized enzyme called integrase facilitates the integration of the viral DNA into the host genome. Once integrated, viral DNA is replicated along with the host genome, and transcription and translation of viral DNA produce RNA and proteins that are assembled into new viruses. The question states that autonomous retrotransposons move via RNA intermediates using mechanisms similar to those of retroviruses. In general, transposons are segments of DNA that move between different areas of the genome and can be classified as either DNA transposons or retrotransposons. The retrotransposon is initially transcribed into mRNA, which is then transported into the cytoplasm and translated to produce reverse transcriptase and integrase. Reverse transcriptase produces dsDNA from the mRNA, and integrase facilitates the integration of the dsDNA into a new area of the genome. According to the question, some autonomous retrotransposons are segments of retroviral DNA that have been partially deactivated by mutations. Because fully functional retroviral DNA must encode the protein components necessary to assemble new viruses, a mutation preventing the expression of envelope proteins would stop the production of infectious retroviruses. However, wild-type (normal) reverse transcriptase and integrase would still allow the DNA segment to move throughout the host genome, converting the active retroviral DNA into an autonomous retrotransposon (Choices A and B). (Choice D) Because retrotransposons move via RNA intermediates, the sequences must be initially transcribed by RNA polymerase. Accordingly, DNA segments that are not transcribed into mRNA are not classified as retrotransposons.

Which of the following is true regarding the organization of neuronal tracts in the spinal cord? A) Afferent neuronal fibers carry motor commands from the brain to the body through tracts in the spinal gray matter B) Afferent neuronal fibers carry sensory information from the body to the brain through tracts in the spinal gray matter C) Efferent neuronal fibers carry motor commands from the brain to the body through tracts in the spinal white matter D) Efferent neuronal fibers carry sensory information from the body to the brain through tracts in the spinal white matter

C - The central nervous system (CNS), including the brain and spinal cord, is composed of white matter and gray matter. Located in the center of the spinal cord, gray matter is composed of unmyelinated neuronal cell bodies and dendrites. In the periphery of the spinal cord, white matter is composed of myelinated and unmyelinated axons that allow for long-distance communication between neurons. White matter consists of afferent and efferent axons: Afferent (ascending) axonal tracts carry sensory information from the body to the brain in the dorsal and lateral columns. Efferent (descending) axonal tracts carry motor commands from the brain to the body in the ventral and lateral columns. (Choices A and B) Afferent neuronal fibers carry sensory information from the body to the brain through tracts in the spinal white matter. (Choice D) Efferent neuronal fibers carry motor commands from the brain to the body through tracts in the spinal white matter.

Cells of the innate immune system and cells of the adaptive immune system differ in that only the latter: A) Mount an immediate response against pathogens B) Are produced by bone marrow stem cells C) Bind only one type of antigen D) Express antigen-presenting proteins on their cell surface

C - The immune system, divided into innate and adaptive immunity, comprises several types of immune cells, all of which are produced from bone marrow stem cells (Choice B). Adaptive immune cells must be activated prior to mounting an immune response, a process that may take several days (Choice A). These adaptive immune cells include: T lymphocytes, which recognize and bind specific foreign antigens displayed by specialized surface proteins called major histocompatibility complex (MHC) proteins on other cells. T lymphocytes include both helper T cells and cytotoxic T cells. B lymphocytes, which are activated by binding and phagocytizing foreign antigens. Within B lymphocytes, these antigens are broken down and MHC proteins bind the antigen fragments, transporting them to the cell surface. Generally, helper T cells bind foreign antigens presented by B lymphocyte MHC proteins and release signaling molecules that stimulate B lymphocytes to divide and differentiate into antibody-secreting plasma cells and memory B cells, which can respond more rapidly in the event of a future infection. (Choice D) Antigen-presenting MHC proteins are expressed on the surface of all nucleated cells (ie, except red blood cells). Accordingly, cells of both the innate and adaptive immune system express MHC proteins.

A staple line leak is an unanticipated opening of surgical staples at an incision. In RYGB, (A Gastric Bypass) a staple line leak at the junction of the biliopancreatic limb and the alimentary limb would cause gastrointestinal tract contents to enter: A) The perineum. B) The peritoneum. C) The peritoneal cavity. D) The pleural cavity

C - The peritoneum is composed of two tissue membranes that line the abdomen: the parietal layer, which lines the abdominal wall, and the visceral layer, which covers the abdominal organs. The peritoneal cavity is a potential space (ie, an area between two adjacent structures that may press together) between the parietal and visceral layers of the peritoneum. The peritoneal cavity is found within the abdomen and contains organs such as the liver, stomach, and intestines. Based on the given scenario, a staple line leak at the junction of the biliopancreatic limb and the alimentary limb would allow gastrointestinal (GI) tract contents to enter the peritoneal cavity.

Intraperitoneal injection of insulin would cause the peptide hormone to enter the bloodstream and exert its effect on target cells by: A) Acting as a second messenger B) Diffusing through the plasma membrane C) Binding an extracellular receptor D) Disassociating from a carrier protein

C - The solubility of a hormone in water affects how it is transported in the blood and how it exerts its physiological effect on target cells. Hormones are primarily divided into three different classes and their chemical structure determines their water solubility: Peptide hormones (eg, insulin, glucagon) are composed of amino acids linked by peptide bonds and are made in the rough endoplasmic reticulum (RER). Peptide hormones range from small to large depending on their number of (charged) amino acids. The overall charge makes peptide hormones water-soluble (hydrophilic), which means they can dissolve in the bloodstream but cannot cross the hydrophobic lipid bilayer of the plasma membrane. Accordingly, peptide hormones act as first messengers and must bind extracellular receptors to induce a signaling cascade via activation of intracellular second messengers (Choice A). Steroid hormones are lipid hormones derived from cholesterol and generated in the smooth ER. These hormones are fat-soluble (hydrophobic) and must therefore be bound to a carrier protein in the bloodstream. However, their hydrophobicity allows them to diffuse through the plasma membrane (lipid bilayer) and bind their receptor in the cytoplasm or nucleus. Steroid hormones are labeled first messengers as they perform the initial signaling that influences the nuclear transcription and cytoplasmic translation of physiologically required proteins. Tyrosine derivatives are derived from the amino acid tyrosine. These hormones function similarly to either steroid hormones (eg, thyroid hormones T3 and T4) or peptide hormones (eg, catecholamines epinephrine and norepinephrine).

A patient with a deficiency in the QRS amplitude would likely have a problem with what? A) SA Node B) AV Node C) Vagal Tone D) Purkinje Fibers

D - As mentioned in the passage, the QRS amplitude measures the contraction of the ventricles. The ventricle contraction is spread evenly by the Purkinje fibers, so that is the most likely problematic destination. The SA node would affect the atrial contraction (choice A is incorrect), the AV node simply sends the signal from the SA node to the bundle of His (choice B is incorrect), and the vagal tone impacts the heart rate as a whole (choice C is incorrect).

Based on Figure 1, HMGN2 (shows a nucleosomal binding region in figure) is most likely to interact with a histone complex that is rich in: A) Aspartate and glutamate B) Aspartate and arginine C) Lysine and glutamate D) Lysine and arginine

D - Based on the passage, the protein HMGN2 contains a nucleosomal binding domain, meaning that it most likely interacts with a nucleosome. During the initial steps of DNA packaging in the cell nucleus, the DNA double helix wraps twice around a histone octamer (eight-protein complex) to form a structural subunit known as the nucleosome. H1, H2A, H2B, H3, and H4 are the five major histone types, but the DNA-wrapped octamer core is composed of eight subunits (ie, two molecules each of H2A, H2B, H3, and H4). These histone proteins have a net positive charge at physiological pH because they are rich in arginine and lysine (positively charged, basic amino acids). This net positive charge facilitates histone binding to negatively charged DNA.

A neurotransmitter released from one neuron causes an influx of negatively charged ions into the cytosol of an adjacent neuron. In the adjacent neuron, the influx of these ions would most likely: A) Depolarize the cell membrane and promote initiation of an action potential B) Hyperpolarize the cell membrane and promote initiation of an action potential C) Depolarize the cell membrane and inhibit initiation of an action potential D) Hyperpolarize the cell membrane and inhibit initiation of an action potential

D - Depending on the neurotransmitter's effect on the post-synaptic neuron, synapses can be classified as excitatory or inhibitory. At an excitatory synapse, the pre-synaptic neuron releases neurotransmitters that cause an influx of positively charged ions (Na+) into the post-synaptic neuron. These ions cause the resting membrane potential of the post-synaptic neuron to become more positive, or to depolarize, which promotes AP initiation. In contrast, inhibitory synapses release neurotransmitters that affect the post-synaptic neuron by causing either an influx of negative ions (Cl−) or an efflux of positive ions (K+). The exit of positive ions or entry of negative ions into the post-synaptic neuron causes the cell's membrane potential to become more negative, or to hyperpolarize, which inhibits AP initiation. In this scenario, a neurotransmitter released from one neuron causes an influx of negatively charged ions into an adjacent (post-synaptic) neuron. This influx would cause the adjacent neuron's membrane potential to become more negative, hyperpolarizing the post-synaptic membrane potential and inhibiting AP initiation (Choices A, B, and C).

Recently fed ob/ob mice show deficient binding of glucagon to its G protein-coupled receptor, which most likely leads to: A) Decreased cAMP to ATP conversion B) Upregulated adenylate cyclase activity C) Increased exchange of GDP for GTP D) Reduced protein kinase A activity

D - Glucagon acts on target cells by binding its G protein-coupled receptor on the cell membrane and inducing the adenylyl cyclase/cAMP second messenger cascade. The G protein "attached" to the transmembrane cell surface receptor is composed of three subunits: alpha, beta, and gamma. When inactive, the alpha subunit (Ga) is bound to GDP, which is then replaced with GTP on ligand-receptor binding. The GTP-bound Ga dissociates from the beta and gamma subunits and proceeds to activate adenylate cyclase, an enzyme that catalyzes the conversion of ATP to cAMP, a second messenger. cAMP then activates protein kinase A, which subsequently phosphorylates the proteins necessary to produce ligand-specific physiological effects (eg, glucagon: induction of glycogenolysis and gluconeogenesis). In the given scenario, the downstream effects of deficient glucagon binding in recently fed ob/ob mice would decreasethe following: exchange of GDP for GTP; adenylate cyclase activity; conversion of ATP to cAMP; and protein kinase A activity. (Choices A and B) Deficient binding of glucagon to its G protein-coupled receptor would decrease adenylate cyclase activity, which would result in reduced conversion of ATP to cAMP (not cAMP to ATP). (Choice C) GDP deactivates the Ga subunit, but GTP activates the Ga subunit. Decreased glucagon binding to its G protein-coupled receptor would lead to decreased activation of the Ga subunit, so a decreased exchange of GDP (deactivating) for GTP (activating) would be observed.

A disease that impedes the functioning of MHC class II would have the least effect on which type of cell? A) CD4 cell B) Macrophage C) Dendritic cell D) CD8+ cell

D - HC II is a protein only present on specialized antigen-presenting cells, which includes most white blood cells as well as helper T cells (choices A, B, and C are incorrect). This leaves CD8+ T cells, which are cytotoxic T cells that would be unaffected by a disease impairing MHC class II.

A patient with an upper motor neuron defect may exhibit which of the following symptoms? A) Hypotonia B) Fasciculations C) Muscular atrophy D) Hypertonia

D - Hypertonia is one of the four primary signs of upper motor neuron defects (choice D is correct). Hypotonia, fasciculations, and muscular atrophy are signs of lower motor neuron defects (choices A, B, and C are incorrect).

Dehydration caused by frequent loose stools is often observed in patients with certain diseases of the gastrointestinal (GI) tract. Given this, which of the following treatments would be most effective in treating this symptom? A) Use of drugs that increase smooth muscle contractions throughout the GI tract B) Use of drugs that decrease glucose reabsorption from the small intestine C) Use of drugs that increase activity of the parasympathetic nervous system D) Use of drugs that decrease the luminal secretion of chloride ions from intestinal epithelial cells

D - In general, nutrients (ie, solutes) absorbed by intestinal epithelial cells are transported into surrounding interstitial fluid before diffusing into capillaries, increasing the osmotic pressure (solute concentration) in the interstitial fluid and, subsequently, in the blood plasma. Because water generally flows from low to high osmotic pressure via osmosis, increased osmotic pressure within interstitial fluid and blood plasma increases intestinal water absorption. In contrast, some epithelial cells secrete ions into the intestinal lumen, increasing osmotic pressure within the intestine to draw water in and lubricate luminal surfaces. In this scenario, dehydration is caused by frequent loose stools in affected individuals. Loose (watery) stools may be due to decreased water absorption or increased water secretion in the intestinal tract. Chloride (Cl−) secretion from intestinal epithelial cells would increase the osmotic pressure of intestinal contents, drawing more water into the intestine and increasing the water content of stool. Therefore, drugs inhibiting luminal secretion of Cl− would likely decrease water secretion into the intestine and make the stools less watery. (Choices A and C) Parasympathetic signals generally enhanceperistalsis and motility of intestinal contents through the GI tract. Drugs that enhance parasympathetic activity and, consequently, peristalsis would increase the flow of chyme through the GI tract, decreasing the time available for nutrient and water absorption by intestinal cells and increasing (notdecreasing) the excretion of watery stools. (Choice B) A drug that decreases glucose absorption would decrease osmotic pressure of the interstitial fluid and capillary blood plasma surrounding the intestine. Decreased osmotic pressure caused by reduced glucose absorption would likely decrease intestinal water absorption and increase (not decrease) the excretion of water in stool.

Developmental protocols for Caco-2 monolayers involve the use of cells with high proliferation potential that can differentiate in a synchronized manner to form homogenous monolayers. This proliferation of Caco-2 cells is primarily achieved through: A) Meiosis B) Fission C) Completion of the G2 phase of the cell cycle D) Completion of the M phase of the cell cycle

D - In multicellular organisms (eg, eukaryotes), nongametic (somatic) cells can divide and multiply via mitosis whereas gametic (reproductive) cells divide and multiply via meiosis. Biological homeostatic mechanisms function to maintain a balanced ratio of proliferating to dying cells in human tissues and organs. In contrast, cell division in single-celled organisms (eg, prokaryotes) occurs via binary fission. Based on the passage, the researchers used Caco-2 monolayers to carry out their experiments, and these Caco-2 cells share the same morphology (shape) and function as cells lining the small intestine. In addition, the question states that these cells are highly proliferative. Caco-2 cells are nongametic (somatic) cells; therefore, each parental Caco-2 cell would only be able to proliferate (divide and multiply in number) in the manner that a somatic cell can, by completing the M phase (mitosis) of the cell cycle to create two genetically identical daughter cells. (Choice A) In eukaryotes, reproductive cells (gametes) are produced via meiosis, during which a parent cell (2n) divides to produce four genetically distinct daughter cells containing half the original number of chromosomes (n). (Choice B) The cells of multicellular organisms (eg, eukaryotes) cannot divide by binary fission. Binary fission is the process by which single-celled organisms, such as bacteria, reproduce asexually. During this process, the parental cell doubles in size and then divides into two identical daughter cells. (Choice C) Cell growth and repair of DNA replication errors occur during the G2 phase of the cell cycle; division does not occur at this stage.

During a nutrition study, a participant consumes five times the physician-recommended amount of food but is unable to sufficiently absorb nutrients. Given this information, this participant would be LEAST likely to exhibit which of the following conditions? A) Loss in diversity of bacteria that produce fatty acids in the large intestine B) Decreased surface area in the small intestine C) Heightened production of serum antibodies against intestinal proteins D) Increased activity of nutrient transporter proteins in the small intestine

D - Nutrient absorption in the gastrointestinal tract is influenced by several factors, including: The presence and diversity of fatty acid-producing bacteria in the large intestine. These bacteria metabolize undigested carbohydrates from the small intestine into short-chain fatty acids. These fatty acids are absorbed and used as additional energy sources. A decrease in these bacterial species would lead to decreased absorption of additional fatty acids as nutrients (Choice A). The small intestine's surface area. Large folds in the intestine's epithelial lining slow the movement of chyme through the intestinal tract. In addition to these large folds, villi (fingerlike projections) extending from the lining and microvilli (cytoplasmic extensions) of individual absorptive cells maximize the time and surface area available for nutrient absorption. Decreased surface area, potentially caused by structural damage (eg, villi destruction), would impair nutrient absorption (Choice B). The functions of intestinal proteins. Structural proteins, digestive enzymes, and nutrient transporters work synergistically to break down and absorb nutrients within the gastrointestinal tract. Because antibodies mark cellular materials for destruction by cells of the immune system, production of antibodies against intestinal proteins would cause the destruction of these proteins, damaging the intestinal tract and decreasing nutrient absorption (Choice C). (Choice D) Intestinal nutrient transporters facilitate nutrient absorption from the intestinal tract into the body. Therefore, increased nutrient transporter activity would increase (not decrease) nutrient absorption.

Investigators used a ribosome profiling technique to measure the level of SOD protein synthesis in P. leiognathi (eukaryote) and ponyfish (prokaryote) cells during ROS accumulation. P. leiognathi cells were found to contain greater numbers of ribosome/SOD mRNA complexes than ponyfish cells. What is the best explanation for this result? A) P. leiognathi SOD mRNA is capable of being bound by more than one ribosome. B) P. leiognathi SOD mRNA has a greater affinity for ribosomes than ponyfish mRNA. C) mRNA transcribed from the ponyfish SOD gene undergoes immediate degradation. D) mRNA transcribed from the bacterial SOD gene is being translated simultaneously.

D - Prokaryotic cells have no nucleus or nuclear envelope, and due to the lack of this physical barrier between DNA and the ribosomal machinery, transcription and translation occur simultaneously in the cytoplasm. Ribosome profiling was used to measure the level of SOD translation by detecting the number of SOD mRNA/ribosome complexes found within ponyfish (a eukaryote) and P. leiognathi (a prokaryote). The question states that P. leiognathi cells have a greater number of ribosome/SOD mRNA complexes compared to ponyfish cells. This rapid protein synthesis is likely due to the fact that all SOD mRNA transcripts in P. leiognathi quickly interact with ribosomes before being fully transcribed (ie, transcription and translation are coupled).

In anaerobic environments, P. leiognathi can produce energy by using an inorganic final electron acceptor other than oxygen in the electron transport chain. Under anaerobic conditions, which of the following is NOT active in ponyfish cells but active in P. leiognathi? A) Glycolysis B) Gluconeogenesis C) Fermentation D) ATP synthase

D - The Krebs cycle and the electron transport chain are only active in the presence of a final electron acceptor, such as oxygen (aerobic respiration) or inorganic ions (anaerobic respiration). The enzyme ATP synthase harnesses the potential energy of the ETC proton gradient to generate ATP from ADP and inorganic phosphate. It is stated in the question that P. leiognathican utilize the ETC in anaerobic environments because it can use compounds other than O2 as the final electron acceptor. Although ATP synthase does not directly need O2 to generate ATP, the formation of the proton gradient by which it functions is oxygen-dependent in ponyfish cells but not in P. leiognathi. Consequently, under anaerobic conditions, ATP synthase is not active in ponyfish cells but is active in P. leiognathi.

As the two anatomical divisions of the adrenal gland, the adrenal cortex and adrenal medulla share a common feature in that both: A) Synthesize steroid hormones. B) Release signaling molecules into lymph vessels. C) Regulate the synthesis of red blood cells. D) Secrete hormones that alter blood pressure.

D - The adrenal cortex secretes the hormones cortisol and aldosterone, and the adrenal medulla secretes the hormones epinephrine and norepinephrine. Aldosterone, secreted from the adrenal cortex, stimulates Na+ reabsorption in the kidneys, leading to increased water retention due to osmosis. Increased water retention increases blood volume, which in turn leads to an increase in blood pressure. Norepinephrine and epinephrine, secreted from the adrenal medulla, function to mobilize the body under extreme stress and promote rapid information processing in part by maximizing blood flow to organs essential for survival. Norepinephrine and epinephrine achieve this effect by promoting the following changes in blood vessel diameter that lead to altered blood pressure: Vasoconstriction (narrowing) of blood vessels supplying the intestines, kidney, and other abdominal organs decreases blood flow to these organs and conserves oxygen/nutrients for other tissues. Vasodilation (widening) of blood vessels leading to the heart and skeletal muscles increases oxygen/nutrient delivery to these organs. In general, the hormones secreted from both the adrenal medulla and cortex function to alter blood pressure.

Which of the following is true regarding the function of neuroglia in the central and peripheral nervous systems? A) Astrocytes are the epithelial cells of the central nervous system and secrete CSF B) Each Schwann cell myelinates multiple axons in the peripheral nervous system C) Each oligodendrocyte myelinates a single axon in the peripheral nervous system D) Microglia are the macrophages of the central nervous system and consume waste

D - The nervous system is composed of neurons and glial cells. While neurons conduct electric impulses, glial cells serve a variety of vital support functions. For example, microglia are the primary immune cells of the central nervous system (CNS) and act as macrophages by phagocytizing pathogens, damaged cells, and other waste materials. Other roles of glial cells found in the CNS include: Oligodendrocytes form myelin sheaths around axons to reduce ion leakage, decrease capacitance, and increase the speed of action potential propagation along the axon. Each oligodendrocyte myelinates segments of multiple adjacent axons in the CNS (Choice C). Astrocytes make extensive contact with blood vessels and regulate blood flow in coordination with synaptic activity and chemical changes. Astrocytes are important for maintaining the chemical homeostasis of the interstitial space, including regulation of fluid and ion balance, pH, and neurotransmitter concentrations. They are also thought to play important roles in neuron development and structural maintenance, as well as coordination between neurons and other glial cells. Ependymal cells are epithelial cells that line the compartments of the CNS and secrete CSF (Choice A). Functions of glial cells in the PNS include: Schwann cells wrap the axons of some neurons with myelin to increase conduction speed. Unlike oligodendrocytes in the CNS, each Schwann cell forms a myelin sheath for a single neuron in the PNS (Choice B). Satellite cells provide structural support and supply nutrients to neuron cell bodies in sensory, sympathetic, and parasympathetic ganglia (groups of cell bodies). They are thought to play roles similar to those of astrocytes in the CNS.

Molecules are regulate blood calcium include: I. Dihydroxycholecaliferol II. Calcitonin III. Parathyroid hormone A) I B) I and II C) II and III D) I, II, and III

D - dihydroxycholecalciferol is the active form of vitamin D, which works in the kidney to increase calcium reabsorption. Calcitonin decreases serum calcium levels, this action is oppose my parathyroid hormone.

Defects of upper motor neurons

Defects of upper motor neurons are associated with four primary abnormalities, or signs. These signs can be thought of as a result of "overstimulation" of the lower motor neuron: a defect in the upper motor neuron causes an abnormally large amount of stimulatory signals to be sent to the lower motor neuron. Hyperreflexia: Hypersensitive receptors in muscles result in greater stretch reflexes Hypertonia: Increased muscle tension and tightness inhibits muscles from being able to stretch fully Clonus: repeating muscular contractions and relaxations Extensor plantar response: Stimulation of the bottom of the foot causes the toes to extend up, when normally they should exhibit a downward response

Menstrual cycle phases

Progesterone is the dominant hormones are the luteal phase and is represented by the dark as line on the graph. The sudden drop in progesterone at the end of the luteal phase triggers uterine contractions, resulting in menses. In fact, progesterone also drops sharply just before birth, relieving the inhibition of uterine contractions are permitted the maintenance of pregnancy. Oxytocin actually stimulates uterine contractions, making it a poor choice for preventing labor. Prolactin levels do not increase until after birth. Estrogen is also elevated during the luteal phase but not nearly as much as progesterone. Progesterone is more instrumental in preventing contractions.

Prokaryotic Gene Expression

The lacI gene codes for the lac repressor and prevents transcription of the lac genes. When sufficient glucose is available, the repressor helps conserve ATP by inhibiting expression of unnecessary genes. When glucose is depleted and lactose is available, lac genes are again expressed to make use of available energy sources. Gene expression requires energy in the form of ATP, so the ability to inhibit expression of unnecessary genes is beneficial. When glucose is present in abundance, metabolism of lactose is not necessary. Therefore, downregulation of the lac genes would be advantageous for lac+ bacteria in glucose-rich media. The lac repressor, encoded by lacI, binds the operon and inhibits transcription of lac genes. At high lactose concentrations, lactose binds the lac repressor and inhibits its interaction with the operon, allowing genes to be transcribed. When glucose is abundant, on the other hand, the lac repressor becomes active. Therefore lacI provides a significant advantage for lac+ bacteria in glucose-rich media because it prevents unnecessary consumption of ATP.