Cell Injury, Adaptations, and Death Flashcards

1
Q
  1. A 50-year-old male dies during a car accident. At autopsy, he is noted to have a moderately stenotic
    aortic valve. The heart weighs 500 grams. The cause
    of death from the motor vehicle accident is entirely
    blunt force injuries of the head and neck, including an
    atlanto-occipital dislocation. No internal injuries below
    the neck are identified. There is no history of heart failure.
    A section of the myocardium from the left ventricle
    would reveal which of the following processes?
    A. Hypertrophy
    B. Hyperplasia
    C. Atrophy
    D. Metaplasia
    E. Reversible cell injury
    F. Irreversible cell injury
A
  1. Correct: Hypertrophy (A)
    Because of the stenotic aortic valve, increased pressure
    is placed on the left ventricular myocardium, which
    responds by increasing the size of the cells, which
    is hypertrophy (A). The myocardium has adapted
    to the increased pressure, and since then there has
    been no heart failure and no reversible or irreversible
    cell injury (E, F). Cardiac myocytes are essentially
    not capable of division and are increasing in size,
    not decreasing, so neither hyperplasia nor atrophy is
    occurring (B, C), and there is no switch in type of cell
    (e.g., cardiac muscle to another form of mesenchymal
    tissue), so metaplasia is not occurring (D).
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2
Q
  1. A 25-year-old male, who is an offensive lineman
    for the local college football team, dies during a car accident.
    At autopsy, the injuries, being multiple rib fractures
    and lacerations of the lungs and heart, with 1500
    mL of blood found in the left pleural cavity, are confined
    to the chest. At autopsy, a biopsy of his left biceps femoris
    muscle would reveal which of the following?
    A. Physiologic hyperplasia
    B. Pathologic hyperplasia
    C. Physiologic hypertrophy
    D. Pathologic hypertrophy
    E. Physiologic atrophy
    F. Pathologic atrophy
A
  1. Correct: Physiologic hypertrophy (C)
    As the decedent is athletic, there is a stimulus causing
    enlargement of the biceps femoris muscle, and
    as athletic activity is a normal activity, the resultant
    enlargement of the skeletal muscle cells would
    be considered physiologic and not pathologic. The
    skeletal muscle cells cannot divide, so they would
    respond to increased workload by increasing in size
    (hypertrophy), and not by increasing in number
    (hyperplasia) (C). (A-B, D-E) are incorrect based on
    the previously discussed information.
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3
Q
  1. A 27-year-old male is in a car accident and sustains
    a fracture of the 2nd thoracic vertebra, with
    resultant damage to the spinal cord at the same level,
    leading to paraplegia, requiring him to be wheelchair
    bound. Five years following the accident, he develops
    a neoplasm of the right lower extremity, requiring a
    wide local excision, which includes a small superficial
    segment of underlying skeletal muscle. When viewing
    the microscopic slides, which of the following
    would the pathologist identify in the skeletal muscle?
    A. Physiologic hyperplasia
    B. Pathologic hyperplasia
    C. Physiologic hypertrophy
    D. Pathologic hypertrophy
    E. Physiologic atrophy
    F. Pathologic atrophy
A
  1. Correct: Pathologic atrophy (F)
    Damage to the spinal cord can lead to deinnervation
    of the lower extremities. Deinnervation is one mechanism
    causing atrophy, and, as the deinnervation
    was due to a traumatic injury, the resultant atrophy
    would be considered pathologic (F) and not physiologic
    (E). As there is neither an increase in the size
    of the skeletal muscle cells, nor an increase in their
    number, the other answers are incorrect (A-D).
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4
Q
  1. A pathologist is examining the biopsy of an individual’s
    left ventricular myocardium and notices that
    the cardiac myocytes are enlarged. Of the following
    conditions, which would best explain this finding?
    A. Pulmonary hypertension
    B. Systemic hypertension
    C. Lambl’s excresences on the aortic valve
    D. A probe patent fossa ovalis
    E. Acute pericardial hemorrhage
A
  1. Correct: Systemic hypertension (B)
    The pathologist is viewing hypertrophy of the cardiac
    myocytes. While identification of enlarged
    cells themselves is a difficulty, enlarged rectangular
    nuclei (i.e., boxcar nuclei) serve as a marker of cardiac
    myocyte hypertrophy. Increased blood pressure
    in the systemic vessels would put strain on the left
    ventricle and lead to hypertrophy (B). Pulmonary
    hypertension would cause similar changes in the
    right ventricular myocardium (A). Lambl’s excrescences
    are incidental small nodules on the valve
    leaflets that are of no physiologic consequence (C).
    A probe patent fossa ovalis is usually of no physiologic
    significance (D). However, if a left to right shunt
    developed, this condition could potentially lead to
    volume overload in the right atrium and ventricle,
    with resultant hypertrophy and dilation, but the left
    ventricle would not be affected. An acute pericardial
    hemorrhage would not explain hypertrophy, as
    cardiac myocyte hypertrophy does not develop over
    such a short period of time, but instead requires a
    longer exposure to the stimulus (E).
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5
Q
  1. A 2-year-old female aspirates a small coin, unbeknownst
    to her caregivers. Over the next few days,
    she develops a cough and becomes less responsive
    and sleeps longer periods. Her parents also notice
    that she feels warm to the touch, but do not take her
    temperature. Six days following the coin being swallowed,
    her parents find her unresponsive in her bedroom.
    They call 9-1-1 and she is pronounced dead at
    the hospital. Autopsy reveals a widespread bronchopneumonia
    in the right lung and a coin lodged in the
    right mainstem bronchus, nearly completely blocking
    it. A microscopic section of the bronchus from
    where the coin was lodged reveals stratified squamous
    epithelium, which appears essentially normal,
    similar to that seen in the inner lining of the esophagus.
    Which of the following processes has occurred
    in the bronchus?
    A. Hypertrophy
    B. Hyperplasia
    C. Metaplasia
    D. Atrophy
    E. Irreversible cellular injury
A
  1. Correct: Metaplasia (C)
    The bronchus is usually lined by respiratory epithelium
    (pseudostratified columnar epithelium);
    however, when the nature of the stimulus to the
    epithelium changes, which, in this case is trauma
    caused by the pressure from the coin, the epithelium
    can change to a different form to better handle the
    abnormal stimulus. The transition from one epithelium
    type to another is termed metaplasia (C). In
    the lung, squamous metaplasia most commonly is
    a result of cigarette smoking, and, in this situation,
    metaplasia can lead to dysplasia and finally to carcinoma,
    accounting for the presence of squamous cell
    carcinomas in the lung. If the coin is removed, the
    epithelium could transition back to respiratory epithelium;
    therefore, the change is reversible, and not
    irreversible cellular injury. (A-B, D-E) are incorrect
    based on the previously discussed information.
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6
Q
  1. After developing hematuria and flank pain, a
    55-year-old male is diagnosed with a renal cell carcinoma
    in his right kidney. To treat the tumor, the
    patient’s right kidney is resected. Five years later, he
    dies in a car accident. At autopsy, there is an absence
    of the right kidney, and the left kidney weighs 300
    grams (normal weight is 150-200 grams for an adult
    male). Of the following processes, which is most
    likely occurring in the left kidney?
    A. Physiologic hypertrophy
    B. Pathologic hypertrophy
    C. Physiologic hyperplasia, hormonal type
    D. Physiologic hyperplasia, compensatory type
    E. Pathologic hyperplasia
A
  1. Correct: Physiologic hyperplasia,
    compensatory type (D)
    Because of the absence of the right kidney, the left
    kidney has a higher workload, and in response, the
    left kidney increases the number of cells in its structure
    to handle this increased workload and thus,
    the organ weight has increased. The cells in the kidney
    are capable of division, and thus hyperplasia
    can occur (unlike in the cardiac or skeletal muscle).
    The hyperplasia is physiologic in nature because
    it is stimulated by an increased workload, and it is
    to compensate for the loss of the other kidney (D).
    (A-C, E) are incorrect based on previously discussed
    information.
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7
Q
  1. A 90-year-old male with severe dementia has
    been bedridden for 5 years. Physical examination
    reveals no decubitus ulcers on his back, buttocks,
    or lower extremities; however, his upper and lower
    extremities are reduced in circumference, the lower
    extremities more so than the upper extremities,
    with apparent significant loss of muscle mass. Which
    of the following combinations of molecular mediators
    is most responsible for the decreased size of the
    skeletal muscles?
    A. Cyclooxygenase-Thromboxane A2
    B. 12-Lipooxygenase-Lipoxin A4
    C. Plasmin-C3a
    D. p53-Bax
    E. Ubiquitin-proteosome
A
  1. Correct: Ubiquitin-proteasome (E)
    The patient has disuse atrophy of the upper and
    lower extremities. The two primary processes occurring
    in atrophy are decreased protein synthesis and
    increased protein degradation. Protein degradation
    is accomplished by binding of ubiquitin to the substances
    to be degraded followed by its subsequent
    destruction by proteasomes (E). Cyclooxygenase and
    thromboxane A2 and 12-lipooxygenase-lipoxin A4
    function in inflammation (A, B), and plasmin-C3a
    in the complement cascade (C), both of which could
    be active in the muscle to some degree but are not
    the main source of protein degradation, and p53-Bax
    functions in apoptosis (D), which might be occurring
    to some small degree but is not the main cause of the
    protein degradation.
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8
Q
  1. A 56-year-old male with a history of smoking
    and hypertension develops an occlusive thrombus in
    his left anterior descending coronary artery following
    rupture of an atherosclerotic plaque. He survives
    the event, but therapeutic lysis of the thrombus is
    not accomplished until 1 hour after its formation. He
    is told that he sustained a subendocardial myocardial
    infarct. If a biopsy of the subendocardial myocytes
    was performed 24 hours after the occlusion of the
    vessel, of the following, which intracellular change
    would be expected to be identified?
    A. Swelling of the mitochondria
    B. Dilation of the rough endoplasmic reticulum
    C. Nuclear clumping of chromatin
    D. Lipid vacuoles in the cytoplasm
    E. Karyorrhexis of nuclei
A
  1. Correct: Karyorrhexis of nuclei (E)
    After 1 hour of ischemia, some of the affected myocytes
    would have irreversible damage, in this case,
    represented by the subendocardial myocardial
    infarct. Of the choices, only karyorrhexis of the
    nuclei is characteristic of irreversible injury, i.e.,
    necrosis (E), whereas the other choices are seen
    with reversible injury (A-D). Other microscopic features
    of irreversible ischemic injury (i.e., necrosis)
    include increased eosinophilia of the cytoplasm, and
    other nuclei changes including karyolysis (i.e., fading
    of the chromatin) and pyknosis (i.e., shrinkage
    of the nucleus). Karyorrhexis is fragmentation of the
    nucleus. As the thrombus was lysed relatively early,
    myocytes closer to the epicardium may have had
    signs of reversible injury, but not irreversible injury.
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9
Q
  1. The husband of a 62-year-old female found her
    unresponsive on the couch in their apartment when
    he returned home from work. Despite the efforts of
    emergency responders, she was pronounced dead at
    the hospital. An autopsy reveals a well-demarcated
    wedge-shaped yellow lesion in the cortex of the left
    kidney with preservation of normal gross architecture.
    Of the following, what is the most likely etiology
    for this autopsy finding?
    A. Mycobacterium tuberculosis infection
    B. Klebsiella bacterial infection
    C. Renal cell carcinoma
    D. Metastatic colonic adenocarcinoma
    E. Thrombus of distal branch of left renal artery
A
  1. Correct: Thrombus of distal branch of left
    renal artery (E)
    The gross description is that of an infarct, which has
    coagulative necrosis. In coagulative necrosis, the normal
    organ architecture is preserved in the beginning
    phases of the development of the infarct. A thrombus
    of a branch of the renal artery can produce an infarct
    of the cortex of the kidney (E). Infarcts are typically
    wedge-shaped, essentially exhibiting the downstream
    effects of a blockage of the arterial system,
    with the amount of organ affected increasing moving
    distal from a single point, the tip of the wedge.
    Preservation of architecture does not occur with liquefactive
    necrosis (as could be seen with a bacterial
    pyelonephritis) or caseous necrosis (as could be seen
    with a tuberculosis infection), and by their nature,
    tumors do not have preservation of architecture
    within their boundaries (A-D).
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10
Q
  1. In his autopsy report, a pathologist describes a
    5 x 3 cm focus of liquefactive necrosis. The source of
    the liquefactive necrosis was a thrombus occluding an
    artery, and the autopsy was performed 2 days following
    the event that produced the liquefactive necrosis.
    In which of the following sections of the autopsy
    report would this lesion most likely be described?
    A. Central nervous system
    B. Cardiovascular system
    C. Respiratory system
    D. Hepatobiliary system
    E. Genitourinary system
A
  1. Correct: Central nervous system (A)
    In most organ systems, a thrombus will lead to ischemic
    injury followed by coagulative necrosis (B-E);
    however, in the brain, a thrombus and the resultant
    ischemic injury produce liquefactive necrosis (A).
    It should be noted that after an extended period of
    time, areas of coagulative necrosis in organs other
    than the brain can become liquefactive necrosis,
    but this change would not be expected within only
    2 days.
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11
Q
  1. A 37-year-old male with a history of chronic
    alcohol abuse and gallstones is brought to the emergency
    room by a friend. The patient has been complaining
    of severe abdominal pain for 3 days following
    a bout of increased alcohol consumption. Laboratory
    testing in the emergency room indicates an elevated
    amylase and lipase. Despite treatment, the patient
    dies. An autopsy of the individual most likely will
    reveal which of the following in the greater omentum?
    A. Coagulative necrosis
    B. Liquefactive necrosis
    C. Gangrenous necrosis
    D. Caseous necrosis
    E. Fat necrosis
    F. Fibrinoid necrosis
A
  1. Correct: Fat necrosis (E)
    Based on the clinical history (i.e., abdominal pain
    with history of gallstones and alcohol abuse) and
    laboratory testing (elevated amylase and lipase), the
    individual most likely has acute pancreatitis, with
    alcohol use and gallstones being two risk factors for
    this condition, and with the symptomatology and
    laboratory testing supporting this diagnosis. The
    release of pancreatic enzymes damages adipose cells
    in the peritoneal cavity (e.g., the omentum), and the
    subsequently released fatty acids join with calcium
    leading to small chalky white patches. This change
    is termed fat necrosis (E). While examination of the
    pancreas itself could reveal coagulative necrosis or,
    depending on the time course of the inflammation,
    liquefactive necrosis (A, B), these conditions would
    not be the main condition identified in the omentum.
    Gangrenous necrosis is found in the extremities,
    and caseous necrosis is associated with tuberculosis
    most commonly (C, D). A variety of conditions are
    associated with fibrinoid necrosis, including some
    forms of vasculitis, but it would not be present in the
    omentum in this clinical scenario (F).
  2. Correct: Sodium-potassium pump (C)
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12
Q
  1. A pathologist is examining a section of kidney
    and notes hydropic change in the proximal convoluted
    tubule epithelial cells. Damage to or decreased
    function of which of the following cellular structures
    is most important in directly causing this finding?
    A. Rough endoplasmic reticulum
    B. Smooth endoplasmic reticulum
    C. Sodium-potassium pump
    D. Phagocyte oxidase
    E. Cytoskeletal proteins
A
  1. Correct: Sodium-potassium pump (C)
    Hydropic change, i.e., cellular swelling, is a sign of
    reversible cellular injury. The sodium-potassium
    pump normally pumps sodium out of the cell and
    potassium into the cell. With ischemia, one cause
    of reversible cellular injury, the amount of ATP produced
    is reduced, and without ATP, the sodiumpotassium
    pump will not function, allowing sodium
    to enter the cell. Water follows sodium into the cell,
    causing the cell to swell, and producing hydropic
    change (C). (A-B, D-E) are incorrect based on the
    previously discussed information.
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13
Q
  1. For his research, a scientist exposes cultured
    cells to certain forms of radiation. Following the exposure,
    he can examine intracellular proteins for evidence
    of cross-linking. Of the following, which has the
    highest propensity for causing such protein changes?
    A. Superoxide
    B. Hydrogen peroxide
    C. Hydroxyl radical
    D. Nitric oxide
    E. Oxidized glutathione
A
  1. Correct: Hydroxyl radical (C)
    Superoxide, hydrogen peroxide, hydroxyl radical,
    and nitric oxide are all reactive oxygen species (ROS),
    which can cause cross-linking of proteins, lipid peroxidation,
    and DNA damage. Oxidized glutathione is
    produced in the breakdown of hydrogen peroxide.
    The most reactive of the four ROS listed is hydroxyl
    radical (C). (A-B, D-E) are incorrect based on the previously
    discussed information.
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14
Q
  1. A scientist is studying the effects of an experimental
    medication, IST-151, on cultured hepatocytes.
    The medication, unfortunately, results in the
    death of the hepatocytes and appears to do so predominantly
    through activation of an intracellular
    endonuclease, which damages the nucleus of the cell,
    leading to death. Of the following mechanisms of cellular
    injury, which one is the most direct in mediating
    the previously described effects (i.e., activation of
    the endonuclease)?
    A. Mitochondrial damage resulting in decreased
    production of ATP
    B. Mitochondrial damage leading to increased
    production of reactive oxygen species
    C. Calcium influx into the cell
    D. Breakdown of lysosomal membranes
    E. DNA damage leading to activation of proapoptotic
    proteins
A
  1. Correct: Calcium influx into the cell (C)
    The four main mechanisms by which cellular injury
    occurs are 1) mitochondrial damage, 2) entry of calcium
    into the cell, 3) damage to the plasma membrane
    (causing dissolution of the cell), or 4) DNA
    damage and/or protein misfolding, which can lead
    to apoptosis. Calcium influx into the cell can activate
    a variety of enzymes including phospholipases,
    proteases, endonucleases, and ATPases (C). (A-B,
    D-E) are incorrect based on the previously discussed
    information.
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15
Q
  1. A 46-year-old female receives radiation therapy
    of the neck for a neoplasm of the thyroid gland.
    After therapy, it is noted that her thyroid gland has
    markedly decreased in size, leading to hypothyroidism,
    and she must be placed on a thyroid replacement
    therapeutic drug regimen. Of the following,
    which statement is most characteristic regarding the
    process by which the thyroid gland decreased in size?
    A. The plasma membranes of the thyroid follicular
    cells lysed
    B. Prominent calcium influx led to marked
    activation of phospholipase
    C. Ischemic injury induced by the radiation led to
    coagulative necrosis
    D. Few, if any, inflammatory cells would have been
    seen histologically
    E. Damage to cytoskeletal elements led to
    dissolution of the cells
A
  1. Correct: Few, if any, inflammatory cells
    would have been seen histologically (D).
    Radiation causes damage to DNA, which can trigger
    the process of apoptosis. During apoptosis, the
    cells break down into fragments, each with an intact
    membrane, and little, if, any inflammation is triggered
    (D). (A-C, E) are incorrect. See previously discussed
    information.
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16
Q
  1. A scientist is studying apoptosis. By the
    introduction of a naturally occurring chemical, she
    wishes to shorten the life span of cultured cells that
    are derived from a human liver and that have been
    exposed to radiation. Which of the following effects,
    if caused by the introduced chemical, would produce
    her desired outcome?
    A. Increased concentration of bcl-2
    B. Increased concentration of bcl-xL
    C. Increased concentration of BH3 proteins
    D. Decreased concentration of Bax
    E. Decreased concentration of Bad
A
  1. Correct: Increased concentration of BH3
    proteins (C)
    BH3 proteins normally detect damaged DNA or
    misfolded proteins and promote the activation of
    Bax and Bad, which are pro-apoptotic mediators,
    leading to the death of the cell. If the chemical that
    the researcher was using increased the concentration
    of BH3 proteins, apoptosis would be promoted
    and the life expectancy of cells previously exposed
    to radiation would be shorter (C). Bcl-2 and Bcl-xL
    inhibit apoptosis, and therefore, increasing their
    concentration would lengthen the life span of the
    cell (A-B). Similarly, Bax and Bad promote apoptosis,
    and decreasing their concentration would
    inhibit apoptosis and lengthen the life span of the
    cells (D-E).
17
Q
  1. A 61-year-old female receives radiation therapy
    of the neck for a neoplasm of the larynx. After therapy,
    it is noted that her thyroid gland has markedly
    decreased in size, leading to hypothyroidism. During
    the decrease in size of the thyroid gland, biopsies of the
    parenchyma would have revealed cells with increased
    eosinophilia, and fragmented nuclei, but essentially no
    surrounding inflammatory reaction. Of the following
    cellular components, which one most directly contributed
    to this lack of an inflammatory reaction?
    A. Phospholipase
    B. Phagosome oxidase
    C. CD95
    D. Phosphatidylserine
    E. Phosphatidylinositol
A
  1. Correct: Phosphatidylserine (D)
    With exposure to radiation and subsequent DNA
    damage, the process of apoptosis can remove the
    irreversibly injured cells. During apoptosis, which
    is an organized breakdown of the cell, little if any
    inflammatory reaction is elicited as the cells are kept
    intact during the process, preventing cellular components
    from leaving the cell and triggering inflammation.
    Phosphatidylserine is expressed on the
    outside of the cellular fragments and is recognized
    by phagocytic cells, which promotes uptake of the
    cellular fragments more quickly, thereby preventing
    a normal inflammatory response (D). Although CD95
    is a cellular mediator in apoptosis, being what the Fas
    ligand binds to, the cellular component most directly
    involved in the removal of the apoptotic cell without
    eliciting an inflammatory reaction is phosphatidylserine
    (C). (A-B, E) are incorrect based on the previously
    discussed information.
18
Q
  1. A 40-year-old chronic alcoholic with no other
    medical history died after an alcohol-related seizure,
    in which he fell down the stairs at his house and fractured
    his neck. An autopsy is performed that reveals
    a diffusely golden-yellow discolored liver. Microscopic
    examination of the liver reveals almost every
    hepatocyte to be filled with one large vacuole or a
    few smaller vacuoles. Which of the following mechanisms
    most likely caused this intracellular inclusion?
    A. Abnormal metabolism
    B. A defect in protein folding
    C. The lack of an enzyme
    D. Ingestion of indigestible material
A
  1. Correct: Abnormal metabolism (A)
    The intracellular accumulation described in the
    question is fat (triglycerides). Accumulation of fat in
    hepatocytes is caused by abnormal metabolism (A).
    Such fat accumulations can be found most commonly
    in individuals with alcoholism, diabetes mellitus, or
    obesity. Hyaline change, resulting from accumulation
    of proteins associated with alcoholism, does not
    appear as described (B), and is instead more ropy
    and eosinophilic. The lack of an enzyme could result
    in fatty accumulation within a cell; however, these
    defects are frequently hereditary and often produce
    complications shortly after birth (C). Also, compared
    to fat accumulation associated with alcohol use, they
    are rare. Fat is not indigestible (D). Given time, the
    fatty accumulations in the hepatocytes would clear.
19
Q
  1. A deceased poorly controlled diabetic is found
    in his messy apartment during a welfare check initiated
    by concerned friends. At the time of autopsy,
    his kidneys are noted to be pale. Vitreous electrolyte
    analysis indicates a vitreous glucose of 576 mg/dL
    and acetone is detected in the blood. A diagnosis of
    diabetic ketoacidosis is made. Microscopic examination
    of the kidneys reveals small clear vacuoles in the
    renal tubular epithelial cells. Of the following, what
    is the most likely contents of these vacuoles?
    A. Lipofuscin
    B. Triglycerides
    C. Hemosiderin
    D. Glycogen
    E. Carbon
A
  1. Correct: Glycogen (D)
    The description given does not fit for lipofuscin, which
    is a granular yellow-brown pigment, hemosiderin,
    which is a more granular or chunky yellow-brown
    pigment, or anthracotic pigment (carbon), which is
    finely stippled and black, or green-black (A, C, E). Diabetics
    are described as developing accumulations of
    glycogen, and not triglycerides, in the kidney, as well
    as other organs, including the heart (B, D).
20
Q
  1. A research scientist wants to conduct an
    experiment using lipofuscin as the substrate to see
    how it forms and what adverse effects its presence in
    the cells might create. He collaborates with a medical
    examiner’s office to obtain tissue for his research.
    All other factors being equal, which of the following
    might be expected to contain the greatest amount of
    lipofuscin in each of the parenchymal cells?
    A. Liver from a 15-year-old
    B. Liver from a 40-year-old
    C. Liver from a 75-year old
    D. Pancreas from a 15-year-old
    E. Pancreas from a 40-year-old
    F. Pancreas from a 75-year-old
A
  1. Correct: Liver from a 75-year-old (C)
    Lipofuscin is a wear-and-tear pigment that accumulates
    with age. The older the individual, the more
    likely their cells will contain the pigment. The heart,
    liver, and brain are the organs most commonly found
    to contain the pigment (C). (A-B, D-F) are incorrect.
    See previous discussion.
21
Q
  1. A forensic pathologist is examining a section of
    skin. Just underneath the dermis is a large collection
    of macrophages, each containing a stippled, or somewhat
    chunky-appearing, yellow-brown pigment. The
    pathologist orders a Prussian blue stain, which causes
    the pigment to appear blue. Of the following, which is
    the most likely etiology for the pigment?
    A. Abnormal protein accumulation in an alcoholic
    B. Previous trauma that resulted in hemorrhage
    C. Wear-and-tear pigment in an older individual
    D. Cigarette smoking
    E. Normal melanin accumulation in a darkly
    pigmented individual
A
  1. Correct: Previous trauma that resulted in
    hemorrhage (B)
    The pigment being described is hemosiderin. Hemosiderin
    contains iron and will stain blue with a Prussian
    stain. Hemosiderin results from the breakdown
    of red blood cells (B). None of the other pigments
    listed are described in this manner, nor do they stain
    in this manner (A, C-E).
22
Q
  1. A 36-year-old male crashes while skiing and
    sustains a large laceration of his thigh, which ultimately
    heals by secondary intention, leaving a large
    scar. Six years later, he notices a firm, but ill-defined,
    mass at the site of the scar. Other than the laceration
    of the thigh, his past medical history is essentially
    negative. He consults a surgeon who removes the
    mass. The mass has a gritty texture to it, and under
    the microscope appears to be fibrosis with scattered
    clusters of rounded and globular basophilic material.
    No cartilage or bone is identified. Which of the following
    processes is occurring at this site?
    A. Metaplastic calcification
    B. Dysplastic calcification
    C. Dystrophic calcification
    D. Metastatic calcification
    E. Disuse calcification
A
  1. Correct: Dystrophic calcification (C)
    The description and scenario fit for dystrophic calcification.
    Deposition of calcium at a site of injury in a
    person without an elevated calcium concentration in
    their blood (i.e., normocalcemic) is dystrophic calcification
    (C). If the individual has hypercalcemia, the
    process is termed metastatic calcification and can
    occur in otherwise normal tissue (D). At this site,
    if the process progressed, bone could form, which
    would be a form of metaplasia. Dysplastic calcification
    and disuse calcification are not terms that are
    usually used (A-B, E).
23
Q
  1. A researcher is studying a form of lung cancer
    in which she identifies a gene for a telomerase in the
    DNA of the tumor cells. Which of the following abilities
    would this gene provide the cancer cells with?
    A. Invade through basement membrane
    B. Invade through wall of blood vessel
    C. Implant in organs to form a metastasis
    D. Increased rate of mitotic activity
    E. Ability to divide for an indefinite length of time
A
  1. Correct: Ability to divide for an indefinite
    length of time (E)
    A telomerase functions to add nucleotides to the end
    of a chromosome, maintaining the length of a telomere.
    When telomeres shorten to a certain point, the
    end of the DNA is determined by the cell to be damaged,
    and cell cycles will stop. With maintenance of
    the telomere length, the cell is able to divide for an
    indefinite length of time (i.e., is immortal) (E). (A-D)
    are incorrect. See previous explanation.