ASCP Question
Acetest® can be used to test for ketones in: - Urine - Urine and serum - Urine, serum, and whole blood - Urine, serum, whole blood, and CSF
Urine, serum, and whole blood
Which of the following statement is INCORRECT regarding the bone marrow biopsy specimen? - Bone marrow cellularity can be accurately determined - Cellular morphologic detail can be determined. - Bone marrow architecture can be examined - Prussian blue staining could be used to evaluate iron stores, and diagnosis of anemia.
Cellular morphologic detail can be determined. Cellular morphologic details CANNOT be determined via bone marrow biopsies. Bone marrow cellularity can be determined accurately by comparing the amount of hematopoietic tissue with the amount of adipose tissue. Bone marrow core biopsy could be examined to reveal bone marrow architecture. Estimation of quantity of iron and its distribution is part of bone marrow examination. The Prussian blue stain provides the most direct means of assessing body iron stores.
What is the increase in the risk percentage for developing antibodies against red cell antigens (RBC alloimmunization) for patients who are characterized as chronically transfused patients? 1% - 4% 2% - 8% 5% - 10% 20% - 50%
2%-8%
Which group of conditions increases the risk of HbS polymerization? - Acid pH, dehydration, decreased level of 2,3-DPG - Alkaline pH, dehydration, increased level of 2,3-DPG - Acid pH, dehydration, increased level of 2,3-DPG - Alkaline pH, dehydration, decreased level of 2,3-DPG
Acid pH, dehydration, increased level of 2,3-DPG The risk of HbS polymerization is enhanced by a low (acid) pH, a state of dehydration, and increased levels of 2,3-DPG. Increased temperature (above 37°C) also adds to the risk
That portion of an enzyme which when separated from its cofactor is called a(n): - Holoenzyme - Isoenzyme - Coenzyme - Apoenzyme
Apoenzyme A holoenzyme is an enzyme system composed of an apoenzyme and a coenzyme. The inactive enzime portion of the holoenzyme is called apoenzyme. The apoenzyme requires the tightly binding of the coenzyme (the non-protein molecule, often a vitamin) to become an active enzyme. An isoenzyme is a related enzyme with a different chemical structure.
Glycine-HCL/EDTA treatment of red cells can destroy which of the following antigens, allowing for confirmation of a suspected antibody and detecting additional antibodies? - Kidd - Bg and Kell - Ss - Duffy
Bg and Kell
Illustrated in this photograph is a positive slide coagulase test (Image B) compared to a negative control (Image A). The clumping of the bacterial cells, as seen in Image B, results from the bacterial production of: - Coagulase-reacting factor - Free coagulase - Bound coagulase - Fibrinogen
Bound coagulase The slide coagulase test is based on the detection of bound coagulase, also known as clumping factor, that is produced on the bacterial cell wall. This factor reacts directly with the fibrinogen in the plasma, resulting in the production of fibrin strands that in turn cause the cells to rapidly agglutinate. Coagulase reacting factor is a plasma substance with which the free coagulase produced by the bacterial cells reacts to form the clot seen in the tube coagulase test.
Suppose a patient with stiffness in her fingers has a positive antinuclear antibody (ANA) test with a centromere pattern at a 1:1280 titer. What is the most likely diagnosis?
CREST syndrome Centromere antibodies are present in over half of patients with the CREST syndrome, an acronym named after the major features of this autoimmune disorder: calcinocis, Raynaud's phenomenon, esophageal dysmotility, sclerodactyly, and telangiesctasia. While the ANA result alone is not diagnostic, this patient is having clinical symptoms of stiff fingers that fits nicely with the centromere ANA pattern. In Systemic lupus erythematosus (SLE), the ANA pattern is usually homogeneous or speckled; in rheumatoid arthitis and Sjogren's syndrome, it is usually specked.
A markedly elevated 5-hydroxyindole acetic acid (5-HIAA) determination is presumptive evidence for: - Lead poisoning - Pheochromocytoma - Porphyria - Carcinoid tumor
Carcinoid tumor 5-Hydroxyindoleacetic acid, or 5-HIAA, is the main metabolite of serotonin and testing is most frequently performed for the diagnosis of carcinoid tumors. 5-HIAA can be increased in urine as much as ten fold for patients with serotonin-producing carcinoid tumors. Lead poisoning does not produce increased levels of 5-HIAA. Pheochromocytomas are catecholamine-producing tumors that derive from chromaffin cells within the adrenal glands and do not increase 5-HIAA levels. Porphyrias are a group of diseases caused by deficiencies in heme biosynthesis.
Parasites that belong to the category Sporozoa are also known as: - Flukes - Digenea - Hemoflagellates - Coccidia
Coccidia Coccidial parasites do not possess specific organelles for locomotion and thus belong to the Sporozoa. Coccidia are a group of protozoal parasites where asexual replication occurs outside of the human host and sexual reproduction occurs inside the human host. Digenea is a class of parasites that includes the flukes. Hemoflagellates belong to the Mastigophora and are flagellates found in blood and tissue.
Which of the following is true of genotype screening in pharmacogenomics? - Genotype screening gives a better overall picture of drug metabolism than measuring metabolism with probe drugs. - Genotyping does not take into account drug interactions which can affect metabolism. - Genotyping typically involves measuring only one mutation site or polymorphism. - Genotyping has no known impact on drug metabolism.
Genotyping does not take into account drug interactions which can affect metabolism. Genotyping, while more robust and definitive, cannot factor in environmental or health variables that could affect drug metabolism. Probe drug analysis does factor in these variables, but it is more complex and tedious. Genotyping typically involves measuring many polymorphisms. For example, a laboratory that offers CYP2D6 profiling may measure twelve of the most common and significant mutation sites on this enzyme.
The LD (lactate dehydrogenase) molecule is made of a combination of polypeptides that can be separated into isoenzymes that help determine the location of an elevated LD. LD 3 is found primarily in the lung, lymphocytes, spleen, and pancreas but what polypeptides make up LD 3? HMMM MMMM HHMM HHHM
HHMM The correct answer is HHMM for LD 3. HMMM is LD 4 and found in the liver; MMMM is LD 5 and found in skeletal muscle; and HHHM is LD 2 and found in the heart and red blood cells.
This suspicious form shown in the trichrome stained image to the right measured 13 µm and was recovered from a stool sample There was no peripheral chromatin.
Iodamoeba butschlii trophozoite The size of this Iodamoeba butschilii trophozoite helps to distinguish it from that of the other amoeba; these trophozoites usually measure between 10-20 µm. Other important identifying characteristics include its single, blunt pseudopod, large eccentric karyosome, large nucleus, and the lack of peripheral chromatin. The trophozoite form of Entamoeba coli is larger, usually measuring 20-25 µm, has several blunt and broad pseudopods, a single nucleus with a large eccentric karyosome, and coarse irregular peripheral chromatin. Entamoeba histolytica has an average size of 15-20 µm and a finger-like pseudopod with a single nucleus, central karyosome and finely granular cytoplasm. Ingested red blood cells may be present in the cytoplasm. Endomilax nana shares some similar nuclear characteristics with Iodamoeba butschili which include a blunt pseudopod, a large nucleus and karyosome, and the lack of peripheral chromatin. An important differentiating characteristic is the significantly smaller size of Endolimax nana measuring from 8-10 µm.
Which one of the following adipocyte products is an important messenger in metabolism, signaling the hypothalamus that there are changes in fat stores? - Leptin - Resistin - IL-6 - Angiotensinogen
Leptin Leptin signals the hypothalamus that there are changes in fat stores. Leptin is synthesized and released from adipose cells in response to adipose tissue changes. It reduces intracellular lipid levels in many types of body cells and thus improves insulin sensitivity. It is an appetite suppressant and inhibitor of fatty liver formation. Resistin increases insulin resistance and enhances adhesion molecules present on endothelial cells. Resistin is another inflammatory cytokine that is increased in obesity. It is synthesized and secreted by macrophages and adipocytes. IL-6 responds to tissue injury. IL-6 increases insulin resistance by inhibiting insulin receptor signal transduction in liver cells. It also increases other inflammatory cytokines, interleukin-1 (IL-1) and TNF-a, and stimulates the liver to produce C-reactive protein (CRP). Adipose tissue and liver cells produce angiotensinogen, a precursor of angiotensin II. Besides increasing blood pressure, angiotensin II may stimulate adipose cell formation and thus increase adipose mass.
Patients may sometimes develop antibodies to which antigen without having a red cell stimulation?
Lua Anti-Lutheran a, or Anti-Lua, may sometimes be seen as a naturally occurring antibody. An individual must undergo a red cell stimulation to produce an antibody to Fya. Pregnant females will sometimes produce a transient naturally-occurring Anti-Lea, but an antibody to Leb is neither naturally-occurring nor common. Individuals may sometimes demonstrate a naturally-occurring Anti-M. Anti-N, however, is neither naturally-occurring nor common.
In which of the following conditions would you expect to find Howell-Jolly bodies? - Iron deficiency anemia - Splenomegaly - Diabetes - Megaloblastic anemia
Megaloblastic anemia Howell-Jolly bodies are associated with megaloblastic anemia, as well as in other conditions where normal mitosis may be interrupted. Howell-Jolly bodies may be observed on a Wright-stained peripheral blood smear from a patient who has a hypo-functioning spleen or has had his/her spleen removed. Howell-Jolly bodies are not associated with iron deficiency anemia or diabetes, although hypochromic microcytic cells may be seen. While splenomegaly is associated with a number of anemias, enlargement of spleen does not result in increased Howell-Jolly bodies. The spleen, in fact, would remove them. In diabetes, no red cell anomalies are seen, except for falsely elevated MCV in some patients due to hyperglycemia causing transient swelling of red cells while being counted in an instrument.
All of the following are methods employed for measuring the specific gravity of urine, EXCEPT? Refractometry Hydrometry Osmolality Urine reagent strips
Osmolality Osmolality is dependent on the number of particles of solute in a unit of solution and is often measured by freezing point depression. Specific gravity is usually directly related to osmolality, but is more easily measured. It provides a rough measure of the urine concentrating power of the kidney. The other methods are used to measure specific gravity in urine samples.
An electrode has a silver/silver chloride anode and a platinum wire cathode. It is suspended in KCl solution and separated from the blood to be analyzed by a selectively permeable membrane. Such an electrode is used to measure which of the following? pH PCO2 PO2 HCO3
PO2 Such an electrode would be used in an ABL blood gas analyzer and it would be used to measure PO2. The electrode would consist of a silver anode, platinum cathode, & Ag/AgCl reference band in an electrolytic solution of KCl. An electrode used to measure pH would consists of a small blub located at the tip of the electrode made of layers of hydrated and non-hydrated glass. The electrode is inside and sits in a chloride ion buffer solution. An electrode used to measure PCO2 would consist of a plastic jacket filled with sodium bicarbonate buffer and a gas-permeable membrane.f HCO3 is not measured directly using an electrode, it is calculated.
What procedure utilizes leukapheresis to collect the buffy coat from whole blood?
Photopheresis Photopheresis utilizes leukapheresis to collect the buffy coat layer from whole blood. These cells are treated with 8-methoxypsoralen, exposed to ultraviolet A light and then reinfused into the patient. Photopheresis has been shown to be efficacious and has been approved by the Food and Drug Administration for the treatment of cutaneous T-cell lymphoma. Plasmapheresis is the removal and retention of the plasma, with return of all cellular components to the patient. Therapeutic apheresis (TA) involves the removal of a specific blood component, with return of the remaining blood constituents to the patient. However, with TA the component being removed is considered pathological or contributing to the patient's underlying disease state. Erythrocytapheresis, or red cell exchange, removes a large number of red blood cells from the patient and returns the patient's plasma and platelets along with compatible allogeneic donor red blood cells.
Fluorometers are designed so that the path of the excitation light is at a right angle to the path of the emitted light. What is the purpose of this design? - Prevent loss of emitted light - Prevent loss of the excitation light - Focus emitted and excitation light upon the detector - Prevent incident light from striking the detector
Prevent incident light from striking the detector The angle is set at 90 degrees in fluorometers in order to only detect the emitted light from the sample substance after it has been excited with a light beam. This helps to avoid measuring the ultraviolet light emitted from the light beam itself, or any other incident light. Excitation is the light that is absorbed by the sample (higher energy, shorter wavelength) while fluorescent light is that light that is emitted as energy leaves the sample (lower energy, longer wavelength). By having the photodetector at a right angle, only the emitted/fluoresced light will be detected.
Which adsorption technique removes cold (IgM) antibodies, particularly anti-I specificities? - Cold autoadsorption - Warm autoadsorption - Differential (allogeneic) - Rabbit erythrocyte stroma (RESt)
Rabbit erythrocyte stroma (RESt) RESt removes cold (IgM) antibodies, particularly anti-I specificities. However, it may adsorb anti-B and other IgM antibodies. Cold autoadsorption uses patient red cells to remove cold autoantibodies to determine whether alloantibodies are present. Warm autoadsorption uses patient red cells are used to remove warm autoantibodies to determine whether alloantibodies are present. Allogeneic adsorption uses known phenotyped red cells to separate specificities: 1) warm autoantibodies from alloantibodies 2) alloantibodies with several specificities.
The measurement of 17-ketosteroids (17-KSs) in urine is performed to assess PRIMARILY which organ(s)? - Testes - Adrenal cortex - Testes and adrenal cortex - Anterior pituitary
Testes and adrenal cortex 17-ketosteroids are metabolites of testosterone and adrenal steroid hormones. The majority of testosterone is produced in the testes and the adrenal cortex. About 1/3 of total urine 17-KSs in men are metabolites of testosterone with the rest derived from steroids produced by the adrenal glands. In women, just about all of the 17-KSs are derived directly from the steroids produced by the adrenal glands.
A prothrombin time (PT) specimen was collected at an outpatient clinic and will not be picked up by the testing laboratory's courier until several hours later. How should the specimen be stored until it is picked up by the courier?
The tube should remain unopened and be kept at room temperature (20-25°C) Blood collected for a prothrombin time PT test that willl be completed within 24 hours of collection should remain uncentrifuged in an unopened tube. Room temperature storage (20° - 25° C) is recommended. Temperatures between 2° - 4° C may result in cold activation of factor VII that would alter the PT result. The tube should not be opened to remove the plasma prior to testing if the test will be completed within 24 hours of collection. Plasma should be removed and frozen at -20° C or lower if the PT test will not be completed within 24 hours. Whole blood however should never be frozen because the red cells will lyse and the PT test result would be drastically altered. The specimen does not require protection from light.
All of the following affect the relative centrifugal force of a centrifuge, EXCEPT?
Time The amount of time a sample is spun does not affect the relative centrifugal force. The Relative Centrifugal Force (RCF) is equal to (1.12 x 10-5 )(radius of centrifuge)(revolutions per minute).
In reference to semen analysis microscopic examination, differentiation and enumeration of "round cells" which are often present in semen count are refered as: Sperm vitality Seminal fluid fructose White blood cells (WBCs) Bacteria
White blood cells Round cells refer to WBCs or immature sperm cells in a semen sample. These cells can be differentiated on a stained smear, using 1000X magnification, or more precisely by performing a peroxidase assay. An increased number of WBCs present indicates inflammatory conditions associated with infection and poor sperm quality. Sperm vitality and seminal fluid fructose are the types of additional testing done when abnormalities suspected in sperm morphology and functionality. The presence of Bacteria in seminal analysis may be indicative of an infection.
The gene loci for the alpha globin chains are adjacent to the locus for which other globin chain? - Beta - Delta - Epsilon - Zeta
Zeta The order of globin gene loci on chromosome 16 is Zeta, Alpha 2, and Alpha 1. Beta globin gene, delta globin gene, and epsilon globin genes are found on chromosome 11.
Which marker is most useful for detection of gestational trophoblastic disease?
hCG (human chorionic gonadotrophin) hCG is used for pregnancy testing, but it is also the most useful marker for the detection of gestational trophoblastic disease.The main clinical use of CEA is a marker for colorectal cancer.AFP is often elevated in patients with hepatocellular carcinoma and germ cell tumors.CA-125 is a serological marker of ovarian cancer.
Which fungi has a growth rate of 2-21 days and the morphological features seen in the tissue form are round spherules 30-60 µm in diameter containing 2-5 µm endospores? - Blastomyces spp. - Coccidioides spp. - Histoplasma capsulatum - Paracoccidioides brasiliensis
Coccidioides spp. Coccidioides spp. have a growth rate of 2-21 days. The recommended morphological features of the tissue form are round spherules 30-60 µm in diameter containing 2-5 µm endospores. Empty spherules are also commonly seen. Blastomyces spp. have a growth rate of 2-30 days. The recommended morphological features of the tissue form are 8-15 µm broad-based budding cells with contoured walls. Cytoplasmic granulation is often obvious. Histoplasma capsulatum has a growth rate of 3-45 days. The recommended morphological features of the tissue form are 2-5 µm small, oval to spherical budding cells. These are often seen inside of mononuclear cells. Paracoccidioides brasiliensis has a growth rate of 21-28 days. The recommended morphological features of the tissue form are 10-24 µm multiple budding cells (bud size = 1-2 µm). Buds are attached to the parent cell by a narrow neck.
The following describes a clot-based test to assess Protein C function, as well as the results of someone's test who has a protein C deficiency.
A partial thromboplastin time (PTT) is performed using C-depleted normal plasma, a venom activator of protein C (such as Protac), and a heparin neutralizer. The clotting time would not be prolonged in someone with a protein C deficiency. The correct answer is the third choice. The addition of C-depleted plasma to the patient's plasma ensures that all other factors in the test are brought to normal levels except for protein C. The addition of the venom activates the protein C. Heparin neutralizer is used in case the person is on therapeutic heparin, which would falsely prolong the PTT. A person with normal protein C levels would have a prolonged PTT because their normal protein C levels would interfere with coagulation. Someone with protein C deficiency would have shorter clotting times. However, APC resistance, Lupus anticoagulant, and thrombin inhibitors such as argatroban can falsely prolong the PTT. The first choice indicates the correct reagents, but a protein C deficient plasma would not substantially prolong the PTT. The second and fourth choices indicate the wrong reagents. Low phospholipid reagents are not appropriate for this test. Also, adding protein C enriched plasma would bring protein C levels to normal which would defeat the purpose.
The top image is characteristic of the clinical presentation of oropharyngeal thrush that may be observed in patients with AIDS. Specimen samples sent to the mycology laboratory for culture often grow yeast colonies on agar plates within 2 - 3 days of incubation at 30° C. Observed in the bottom photograph is the surface of a niger seed agar plate growing two different yeast species. The one to the left represents the clinical isolate and the one to the right is a control strain. It is important to recognize the colonies of the clinical isolate as it may be potentially resistant to commonly used anti-fungal agents. Select the species of yeast involved in this infection.
Candida dubliniensis Candida dubliniensis is the correct response. Distinctive is the appearance of rough colonies with extending "feet" from the outer margins as seen in the streak to the left on the surface of the niger seed agar plate, in contrast, to the colonies with smooth borders, as observed in the control streak of Candida albicans on the right. Clinical cases of oropharyngeal thrush may be observed in patients with AIDS, and niger seed ("bird seed") agar plates should be set up to identify C. dubliniensis, as this organism can be misidentified as Candida albicans. Carbohydrate assimilation profiles, inoculation to CHROMagar, growth requirements, and the use of species-specific DNA probes may be performed to make this distinction. Candida dubliniensis will not grow at 45° C while Candida albicans will grow at 45° C. Candida glabrata is incorrect because this yeast is not one of the fungal species producing oropharyngeal thrush. Colonies on non-selective agar are somewhat slower growing, maturing in 3 - 5 days. Niger seed agar would not be used for the differential identification of C. glabrata. Colonies that might grow would not have the feet-like extensions from the outer margins of the colonies.Candida parapsilosis is incorrect because this yeast is not one of the fungal species producing oropharyngeal thrush. Niger seed agar would not be used for the differential identification of C. parapsilosis; colonies that might grow would not have the feet-like extensions from the outer margins.Candida albicans is incorrect. Candida albicans can produce oropharyngeal thrush; however, the colonies on niger seed agar would have smooth margins without peripheral feet-like extensions from the outer borders, illustrated as the "control" yeast in this exercise. In addition, this yeast species will grow at 45° C while Candida dubliniensis will not grow at 45° C.
For which of these conditions or procedures there may be an increased number of megakaryocytes in the bone marrow, but a decreased number of circulating platelets? - Folic acid deficiency - Aplastic anemia - Radiation Therapy - Wiskott-Aldrich syndrome
Folic acid deficiency Pancytopenia is often seen with megaloblastic anemias that are caused by folic acid or vitamin B12 deficiency. Thrombopoiesis (as well as erythropoiesis and granulopoiesis) is ineffective. The bone marrow will contain normal, or even increased megakaryocytes, but the number of platelets entering the peripheral circulation is decreased. In aplastic anemia, megakaryocytes are decreased in number in the bone marrow, leading to a decreased number of circulating platelets. Radiation therapy causes bone marrow hypoplasia. Platelets as well as all other cell lines are depressed. The effect is transient; once the therapy has ended, the marrow will regenerate. In Wiskott-Aldrich syndrome, platelets are very small and thrombocytopenia is present.
The image to the right was recovered from a sputum sample. The form in the image measured 112 µm by 55 µm. Which of the following is the correct identification?
Paragonimus westermani Paragonimus westermaniegg is the correct answer because out of all the parasites listed, this species is the only fluke egg known to be found in human sputum. Diagnosis is made by the characteristic size (80-120 µm x 45 - 60 µm), shape (oval), color, (yellow-brown), and the presence of an operculum. Schistosoma mansoni egg is incorrect because this parasite is found in the venules of the large intestines. The eggs measure 115-180 µm x 40-75 µm, have a large lateral spine, no operculum, and are embryonated. Pneumocystis jiroveci cyst is incorrect because this organism is an atypical fungus that is found in immunocompromised patients. Diphyllobothrium latum (freshwater fish tapeworm) egg is incorrect because this parasite is diagnosed by its characteristic size (58 - 75 µm x 40 - 50 µm), shape (oval), color, (yellow-brown), presence of an operculum, and a small knob at the abopercular end. This is the only cestode to have an aquatic life cycle. Fish is the reservoir host and humans are the definitive host. Eggs or proglottids are found in the intestines.
The qualitative differences between A1 and A2 phenotypes includes all of the following EXCEPT: - The formation of anti-A1 in A subgroups. - The amount of transferase enzymes. - The length of the precursor oligosaccharide chains. - The lack of agglutination of patient red cells with anti-A reagent.
The lack of agglutination of patient red cells with anti-A reagent. Qualitative differences for A1 and A2 phenotypes includes the following: differences in the precursor oligosaccharide chains (in length and complexity of branching), small differences in transferase enzymes (decreased in A2 subgroup), and the formation of anti-A1 in the serum of A2 phenotype individuals. Both A1 and A2 patient red cells react with anti-A reagent. Dolichos biflorus or anti-A1 lectin reagent is used to differentiate between A1 and A2 phenotypes. This lectin reagent agglutinates with A1 patient red cells but does NOT agglutinate with A2 patient red cells.
Which of the following best describes a hemoglobinopathy? - Any problem involving hemoglobin destruction. - Any problem associated with hemoglobin production. - A deletion of the loci of one or more hemoglobin chains. - A substitution of an amino acid in the hemoglobin chain.
A substitution of an amino acid in the hemoglobin chain. A hemoglobinopathy is a substitution of an amino acid in the hemoglobin chain. When an amino acid substitution occurs, a new protein is formed. Hemoglobin destruction or problems with hemoglobin production are not exclusive to hemoglobinopathies. Other anemias that are not considered hemoglobinopathies may also involve increased hemoglobin destruction or problems with hemoglobin production. However, other anemias demonstrating hemoglobin production problems do not form an abnormal hemoglobin protein. Deletion of a hemoglobin chain loci results in decreased production of that chain (thalassemia), and not production of a different hemoglobin protein.
What is the cause of iron overload in hereditary hemochromatosis? - Absorption of excessive amounts of iron in the small intestine - Ingestion of excessive amounts of iron from diet or supplements - Inability of the body to excrete normal amounts of dietary iron - Failure of developing red blood cells to incorporate iron into protoporphyrin IX
Absorption of excessive amounts of iron in the small intestine Hereditary hemochromatosis is a genetic disorder typically involving a deficiency of hepcidin due to a mutation in the hepcidin gene or genes whose products regulate the expression of hepcidin. These mutations cause increased iron absorption in the small intestine, leading to iron overload. Ingestion of excessive amounts of dietary iron results in secondary hemochromatosis. Iron is recycled when cells die; the body does not have a mechanism for iron excretion. Failure of developing red blood cells to incorporate iron into protoporphyrin IX results in a sideroblastic anemia.
Which enzyme is responsible for the rate-limiting step of steroid hormone synthesis? 21-hydroxylase CYP450 F-zone enzymes 17-hydroxylase
CYP450 CYP450, cytochrome P 450, is the enzyme responsible for the rate-limiting step converting cholesterol to pregnenolone in steroidogenesis (which is also the first step). 17- and 21-hydroxylases are enzymes that hydroxylate their specific carbons in steroidogenesis but not the rate-limiting step, while F-zone enzymes convert pregnenolone into further steroid products.
A culture was taken of a conjunctival exudate. The bacterial species, represented by the tiny colonies seen on this chocolate agar plate were recovered after 48 hours incubation at 35°C. The isolate was cytochrome oxidase positive and highly susceptible to penicillin. What is the most likely identification of this organism? - Moraxella lacunata - Moraxella catarrhalis - Neisseria subflava - Haemophilus aphrophilus
Haemophilus aphrophilus Moraxella lacunata has been known since the turn of the century as an agent of acute conjunctivitis. The colonies on chocolate agar are typically pinpoint, as illustrated here, and are highly susceptible to low concentrations of penicillin. As the species name indicates, M. lacunata may pit the agar; however, this property is best seen in a serum medium such as Loeffler's and is usually not seen on chocolate agar. Moraxella catarrhalis produces colonies considerably larger than the pinpoint colonies seen here after 48 hours incubation, and most strains are penicillin resistant from the production of beta lactamases. The colonies of Neisseria subflava are also larger than those seen here, have a yellow pigmentation, and many strains are penicillin resistant. Haemophilus aphrophilus produces tiny colonies similar to those seen here; however, this species is cytochrome oxidase negative.
Which of the following cells is the most common nucleated cell in normal adult bone marrow? - Myeloblast - Promyelocyte - Myelocyte - Metamyelocyte
Metamyelocyte Of the given choices, metamyelocytes are the most abundant form seen in a normal bone marrow. Reference interval is 3-20% Myeloblast reference interval is 0-3%. Promyelocyte reference interval is 1-5%. Myelocyte reference interval is 6-17%.
In which age group is Type I Hereditary Hemochromatosis (HH - the most common form) most likely to be initially detected based on clinical symptoms? - Middle aged adults - Young adults - School age children - Infants
Middle aged adults Most patients with HH become symptomatic during middle age. It is caused by an autosomal recessive gene defect which results in excessive iron absorption in the GI tract. Although the gene is present at infancy, the effects are slowly progressive. Thus, the effects of iron accumulation in organs generally do not manifest at younger ages such as young adulthood, school age children, or infants. Increased needs such as growth, menstruation, etc. help defer onset of symptoms.
Select the most appropriate identification of a non-lactose fermenting Enterobacteriaceae based on the following test results: Indole positive TSI K/A H2S negative Citrate positive PDA positive Lysine decarboxylase negative Ornithine decarboxylase negative Urea negative - Proteus mirabilis - Morganella morganii - Serratia marcescens - Providencia stuartii
Providencia stuartii The identification is Providencia stuartii. The key reactions that differentiate it as this organism are the positive reactions for indole, PDA, citrate, and the negative reaction for urea and the decarboxylase reactions. Proteus mirabilis is a non-lactose fermenting Enterobacteriaceae but it is indole negative, H2S positive, and urea positive. Morganella morganii is a non-lactose fermenting Enterobacteriaceae but it is citrate negative and urea positive. Serratia marcescens is a non-lactose fermenting Enterobacteriaceae but it is indole and PDA negative and lysine decarboxylase positive.
Which antinuclear antibody (ANA) pattern is seen in the image on the right, which represents the result of an ANA test viewed using fluorescent microscopy? Note: (a) points to the nuclei of interphase cells, the primary consideration for discerning the ANA pattern and (b) indicates a metaphase mitotic cell.
There is no discernable pattern In order for the ANA to be positive there must be a clearly discernible pattern in the nuclei of the interphase cells. Observing the chromosomal area and cytoplasm of the metaphase cells may assist in identification of the ANA pattern. The cells on this slide demonstrate weak staining in the nuclei of the interphase (nondividing) cells but no clear pattern. In addition, there is absence of fluorescence in the chromatin region (center of nucleus) in the metaphase (dividing) cells. In contrast, a homogeneous pattern would show uniform staining of the nuclei in the interphase cells and presence of staining in the chromatin of the dividing cells. The centromere pattern is characterized by many discrete speckles in both the nuclei of interphase cells and the chromatin of dividing cells. In the nucleolar pattern, there is prominent staining of the nucleoli in the nuclei of the interphase cells and variable staining in the chromatin of the dividing cells.
All of the following represents organisms with the appropriate specimen type for diagnosis EXCEPT: - Entamoeba hartmanni - stool - Trichomonas tenax- vaginal secretions - Pentatrichomonas hominis - stool - Entamoeba gingivalis- mouth
Trichomonas tenax- vaginal secretions Trichomonas tenax - vaginal secretions is the correct answer because Trichomonas tenax is seen only in preparations from the mouth. It is not found in vaginal secretions. Entamoeba hartmanni - is an intestinal amoeba acquired through contaminated food and water. Therefore, a stool specimen is the most appropriate specimen for ova & parasite testing. Pentatrichomonas hominis - is an intestinal flagellate acquired through contaminated food and water. Therefore, a stool specimen is the most appropriate specimen for ova & parasite testing. Entamoeba gingivalis - is found in the oral cavity.
