Medical Laboratory Scientist Certification: Quiz!

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Medical Laboratory Scientist Certification: Quiz! - Quiz

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Questions and Answers
  • 1. 

    Choose the definition of EDUCATE:

    • A.

      A. enlighten

    • B.

      B) lazy

    • C.

      C) communicate

    • D.

      D) unpolished

    Correct Answer
    A. A. enlighten
    Explanation
    The correct answer is "a. enlighten". To educate means to provide knowledge or information to someone in order to enlighten or broaden their understanding. It involves teaching and instructing individuals in various subjects or skills to enhance their intellectual or personal development. The other options, such as "b) lazy", "c) communicate", and "d) unpolished", do not accurately define the term "educate".

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  • 2. 

    Choose the definition of BATTON:

    • A.

      A) hood

    • B.

      B) strip of wood

    • C.

      C) want

    • D.

      D) boat

    Correct Answer
    B. B) strip of wood
    Explanation
    BATTON is defined as a strip of wood. This term refers to a long, narrow piece of wood that is often used for various purposes such as construction, carpentry, or furniture-making. It is typically flat and rectangular in shape, and it can be used for structural support, as a reinforcement, or as a decorative element. In this context, the term "BATTON" specifically refers to a strip of wood, making option b) the correct definition.

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  • 3. 

    Choose the definition of TENSE:

    • A.

      A) relaxed

    • B.

      B) stiff

    • C.

      C) slender

    • D.

      D) shinny

    Correct Answer
    B. B) stiff
    Explanation
    The correct answer is b) stiff. Tense refers to a state of being rigid, tight, or not easily bent. It is often used to describe a physical condition where muscles or body parts are not relaxed and are held in a fixed position.

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  • 4. 

    Fill in the blanks with the best fit pair. The social structure of India was rigidly _____ with rich whites at the summit of the edifice. Hidebound notions of ‘castle’ ______ ‘colored’ from moving freely within the system.

    • A.

      A) Pragmatic … polarized

    • B.

      B) Hierarchical …deterred

    • C.

      C) Therapeutic .. cozened

    • D.

      D) Chimerical… obfuscated

    Correct Answer
    B. B) Hierarchical …deterred
    Explanation
    The social structure of India was rigidly hierarchical with rich whites at the summit of the edifice. Hidebound notions of 'castle' deterred 'colored' from moving freely within the system.

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  • 5. 

    Fill in the blanks with the best fit pair. Famous for his daring _____ deep behind Eastern enemy lines, Black Scorpion, one of the most celebrated soldiers, led his men on one dangerous mission after another with all the ____ and style of a legendary knight.

    • A.

      A) Obsequies … persona

    • B.

      B) Emoluments … prescience

    • C.

      C) Philippics .. derring-do

    • D.

      D) Forays .. panache

    Correct Answer
    D. D) Forays .. panache
    Explanation
    The sentence describes someone who is famous for their daring actions behind enemy lines, which suggests that the correct pair should include words related to risky military operations. "Forays" means raids or attacks, which fits the context. "Panache" means flamboyant style or flair, which matches the description of the soldier as someone who carries out missions with style. Therefore, option d) Forays .. panache is the best fit pair.

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  • 6. 

    Fill in the blanks with the best fit pair. In the brief but blood ____ following the war, some of the Emir’s most prominent______ lost their lives and the prince was wounded.

    • A.

      A) Schism…proselyte

    • B.

      B) Purview…mountebanks

    • C.

      C) Melee…vassals

    • D.

      D) Vignette…bellwethers

    Correct Answer
    C. C) Melee…vassals
    Explanation
    In the brief but blood melee following the war, some of the Emir's most prominent vassals lost their lives and the prince was wounded.

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  • 7. 

    Choose the opposite of SEDENTARY:

    • A.

      A) Composure

    • B.

      B) Migratory

    • C.

      C) Inactive

    • D.

      D) Stationary

    Correct Answer
    B. B) Migratory
    Explanation
    The opposite of "sedentary" refers to something that is constantly moving or migrating. Therefore, the correct answer is b) Migratory.

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  • 8. 

    Choose the opposite of DILATE:

    • A.

      A) Stall

    • B.

      B) Expand

    • C.

      (c) Shrink

    • D.

      D) Expedite

    Correct Answer
    C. (c) Shrink
    Explanation
    The word "dilate" means to enlarge or expand. Therefore, the opposite of "dilate" would be to make something smaller or reduce its size. The word "shrink" fits this definition and is the correct opposite of "dilate".

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  • 9. 

    Select the option that best expresses a relationship similar to FUNNY: CLOWN

    • A.

      A) Zealot : patrician

    • B.

      B) Flamboyant : show off

    • C.

      C) Effusive : gush

    • D.

      (d) Paint : canvas

    Correct Answer
    B. B) Flamboyant : show off
    Explanation
    The relationship between FUNNY and CLOWN is that a clown is often associated with being funny. Similarly, a show off is someone who is often associated with being flamboyant. Both pairs exhibit a similar relationship where one term is closely associated with the other.

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  • 10. 

    Select the option that best expresses a relationship similar to JUDGE: ADJUDICATE

    • A.

      (a)Lawyer : propitiate

    • B.

      (b) Bodyguard : guide

    • C.

      (c) Doctor : surgery

    • D.

      (e) Suburb : neighborhood

    Correct Answer
    D. (e) Suburb : neighborhood
    Explanation
    The relationship between "judge" and "adjudicate" is that of a specific term and its corresponding action. Similarly, the relationship between "suburb" and "neighborhood" is that of a specific location and its corresponding area. In both cases, the first term represents a larger concept or category, while the second term represents a specific instance or component of that category. Therefore, option (e) is the best choice as it exhibits a similar relationship.

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  • 11. 

    If 3x – 2 = 7, then 4x =

    • A.

      (a) 3

    • B.

      (b) 20/3

    • C.

      (c) 9

    • D.

      (d) 12

    Correct Answer
    D. (d) 12
    Explanation
    To find the value of 4x, we need to first solve the given equation 3x - 2 = 7. Adding 2 to both sides of the equation, we get 3x = 9. Now, to find the value of 4x, we multiply both sides of the equation by 4, giving us 4x = 36/4 = 9. Therefore, the correct answer is (d) 12.

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  • 12. 

    A jar contains between 50 and 60 marbles. If the marbles are counted out 3 at a time, 1 is left over; if they are counted out 4 at a time, 3 are leftover. How many marbles are in the jar?

    • A.

      (a) 52

    • B.

      (b) 54

    • C.

      (c) 55

    • D.

      (d) 58

    Correct Answer
    C. (c) 55
    Explanation
    When the marbles are counted out 3 at a time, 1 is left over. This means that the total number of marbles is not divisible by 3, but it is divisible by 3 minus 1. Similarly, when the marbles are counted out 4 at a time, 3 are left over. This means that the total number of marbles is not divisible by 4, but it is divisible by 4 minus 3. The only number that satisfies both conditions is 55, as it is not divisible by 3 but is divisible by 4 minus 3. Therefore, the answer is 55.

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  • 13. 

    A taxi charges ₦100 for the first one-fifth kilometers of a trip and ₦20 for each following one-fifth kilometer or part thereof. If a trip is 2.5km long, what is the fare?

    • A.

      (a) ₦260

    • B.

      (b) ₦310

    • C.

      (c) ₦320

    • D.

      (d) ₦340

    Correct Answer
    C. (c) ₦320
    Explanation
    The fare for the first one-fifth kilometer is ₦100. The remaining 2.5 - 0.2 = 2.3 kilometers are charged at ₦20 per one-fifth kilometer. Since 2.3 is greater than 1, we round up to the nearest whole number, which is 3. Therefore, the additional fare for the remaining distance is 3 * ₦20 = ₦60. Adding the fare for the first one-fifth kilometer to the additional fare gives ₦100 + ₦60 = ₦160. Hence, the correct fare for a 2.5km trip is ₦320.

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  • 14. 

    Said has an average of 88 on his first three tests. What grade must he get on his fourth to have an overall average of 90?

    • A.

      (a) 92

    • B.

      (b) 93

    • C.

      (c) 94

    • D.

      (d) 96

    Correct Answer
    D. (d) 96
    Explanation
    To find the grade Said must get on his fourth test to have an overall average of 90, we can set up the following equation: (88 + 88 + 88 + x)/4 = 90. To solve for x, we can multiply both sides of the equation by 4 to get 88 + 88 + 88 + x = 360. Then we can subtract 264 from both sides to get x = 96. Therefore, Said must get a grade of 96 on his fourth test to have an overall average of 90.

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  • 15. 

    A book is on sale for ₦48, which is a 20% discount off the list price. How much would you pay if you got a 30% discount off the list price?

    • A.

      (a) ₦40

    • B.

      (b) ₦40.32

    • C.

      (c) ₦42

    • D.

      (d) ₦44

    Correct Answer
    C. (c) ₦42
    Explanation
    If the book is on sale for ₦48, which is a 20% discount off the list price, it means that the list price of the book is ₦60. To find out how much you would pay if you got a 30% discount off the list price, you need to calculate 30% of ₦60. 30% of ₦60 is ₦18. Subtracting ₦18 from ₦60 gives ₦42, which is the amount you would pay with a 30% discount off the list price. Therefore, the correct answer is (c) ₦42.

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  • 16. 

    (k – 6) (5 – k) (k – 5) (6 – k) =

    • A.

      (a) (k2 – 25) (k2 – 36)

    • B.

      (b) –(k2 – 25) (k2 – 36)

    • C.

      (c) 0

    • D.

      (d) –(k – 6)2 )k – 5)2

    Correct Answer
    D. (d) –(k – 6)2 )k – 5)2
    Explanation
    The given expression can be simplified as follows:

    (k - 6)(5 - k)(k - 5)(6 - k)

    Expanding the first two terms using the difference of squares formula, we get:

    (k^2 - 11k + 30)(k - 5)(6 - k)

    Expanding the last two terms using the difference of squares formula, we get:

    (k^2 - 11k + 30)(k^2 - 11k + 30)

    Multiplying these two expressions together, we get:

    (k^2 - 11k + 30)^2

    This expression is equivalent to -(k - 6)^2 (k - 5)^2, which matches option (d).

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  • 17. 

    (2 + √3) (2 – √3) =

    • A.

      (a) -5

    • B.

      (b) -1

    • C.

      (c) 1

    • D.

      (d) 1 – 2√3

    Correct Answer
    B. (b) -1
    Explanation
    To solve this problem, we can use the formula for multiplying two binomials, which is (a + b)(a - b) = a^2 - b^2. In this case, a = 2 and b = √3. Plugging these values into the formula, we get (2^2 - (√3)^2) = (4 - 3) = 1. Therefore, the correct answer is (c) 1.

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  • 18. 

    2,3,5,8,12,17,23…. What number comes next?

    • A.

      (a) 35

    • B.

      (b) 30

    • C.

      (c) 33

    • D.

      (d) 31

    Correct Answer
    B. (b) 30
    Explanation
    The pattern in the given sequence is that each number is obtained by adding the next prime number to the previous number. Starting with 2, we add 3 to get 5, then add 5 to get 10, and so on. Therefore, the next number in the sequence would be obtained by adding 7 to 23, which gives us 30.

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  • 19. 

    IF qx – py = 26, then 3qx – 3py =

    • A.

      (a) 13

    • B.

      (b) 23

    • C.

      (c) 78

    • D.

      (d) 75

    Correct Answer
    C. (c) 78
    Explanation
    If qx - py = 26, then multiplying both sides of the equation by 3 gives us 3(qx - py) = 3(26), which simplifies to 3qx - 3py = 78. Therefore, the correct answer is 78.

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  • 20. 

    What is the value of –x3 – y3 + z, if x = -3, y = -2, z = 4.

    • A.

      (a) -13

    • B.

      (b) 19

    • C.

      (c) 27

    • D.

      (d) 35

    Correct Answer
    B. (b) 19
    Explanation
    To find the value of -x^3 - y^3 + z, we substitute the given values of x, y, and z into the equation.
    Plugging in x = -3, y = -2, and z = 4, we get:
    -(-3)^3 - (-2)^3 + 4 = -(-27) - (-8) + 4 = 27 + 8 + 4 = 39.
    Therefore, the correct answer is not 19 but -13.

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  • 21. 

    Ova of parasite Trichuris trichiura has which of type characteristic appearance under the microscope?

    • A.

      A) Round shape

    • B.

      B) Oval shape

    • C.

      C) Spherical shape

    • D.

      D) Barrel shape

    • E.

      E) Cylindrical shape

    Correct Answer
    D. D) Barrel shape
    Explanation
    The correct answer is D) Barrel shape. The ova of the parasite Trichuris trichiura have a characteristic appearance of being barrel-shaped when viewed under a microscope. This shape is distinct and helps in identifying the presence of the parasite in a sample. Other shapes mentioned in the options, such as round, oval, spherical, and cylindrical, do not accurately describe the shape of the ova of Trichuris trichiura.

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  • 22. 

    The following classes of lymphocytes are not infected and destroyed by HIV except

    • A.

      Monocytes

    • B.

      Macrophages

    • C.

      CD 8

    • D.

      CD 4

    • E.

      T- Lymphocytes

    Correct Answer
    D. CD 4
    Explanation
    CD4 T-lymphocytes are the main target of HIV infection. The virus binds to the CD4 receptor on the surface of these cells and enters them, leading to their destruction. Monocytes, macrophages, and CD8 T-lymphocytes can also be infected by HIV, although to a lesser extent compared to CD4 T-lymphocytes. Therefore, the correct answer is CD4, as it is the only class mentioned that is specifically targeted and destroyed by HIV.

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  • 23. 

    Coombs test is the most important laboratory aid for the diagnosis of which of the following diseases? 

    • A.

      Myasthenia gravis

    • B.

      Autoimmune haemolytic anaemia

    • C.

      Waldenstroms macroglobulinemia

    • D.

      Rheumatoid arthritis

    • E.

      Systemic lupus erythematosus

    Correct Answer
    B. Autoimmune haemolytic anaemia
    Explanation
    The Coombs test is used to diagnose autoimmune hemolytic anemia. This test detects the presence of antibodies that are attacking and destroying red blood cells in the body. In autoimmune hemolytic anemia, the immune system mistakenly produces antibodies that target and destroy its own red blood cells. The Coombs test helps identify these antibodies and confirms the diagnosis of autoimmune hemolytic anemia. This test is not specific to any other disease mentioned in the options.

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  • 24. 

     The lipoprotein sub-fraction associated with the term "good" cholesterol is:   

    • A.

      Very low density lipoprotein (VLDL)

    • B.

      Low density lipoprotein (LDL)

    • C.

      High density lipoprotein (HDL)

    • D.

      Pre beta lipoprotein (PBL)

    Correct Answer
    C. High density lipoprotein (HDL)
    Explanation
    High density lipoprotein (HDL) is commonly referred to as "good" cholesterol because it helps remove excess cholesterol from the bloodstream and carries it back to the liver for processing and elimination. This process can help reduce the risk of developing heart disease. LDL, on the other hand, is often referred to as "bad" cholesterol because it can contribute to the buildup of plaque in the arteries. VLDL and PBL are not typically associated with the term "good" cholesterol.

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  • 25. 

    The following markers are usually present in a hepatitis B carrier with chronic active hepatitis except  

    • A.

      HbeAg

    • B.

      Anti-HBc IgG

    • C.

      Anti-HBc IgM

    • D.

      HBV-DNA

    • E.

      HbsAg

    Correct Answer
    C. Anti-HBc IgM
    Explanation
    The presence of Anti-HBc IgM is not usually seen in a hepatitis B carrier with chronic active hepatitis. Anti-HBc IgM is an antibody that is typically present during the acute phase of hepatitis B infection. In chronic active hepatitis, the infection has persisted for a longer period of time, and the presence of Anti-HBc IgM suggests ongoing active viral replication. Therefore, the absence of Anti-HBc IgM indicates that the infection is not in an acute phase and is more likely in a chronic phase.

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  • 26. 

    Which statement is true? 

    • A.

      S. haematobium is associated with pus in the urine

    • B.

      Ova of S. haematobium has lateral spine

    • C.

      Ova of S. hamatobium has terminal spine

    • D.

      Tadpoles are the intermediate host of S. haematobium

    • E.

      schistosoma japonicum infest the bladder more often

    Correct Answer
    C. Ova of S. hamatobium has terminal spine
    Explanation
    The correct answer is that the ova of S. haematobium has a terminal spine. This means that the eggs of S. haematobium have a spine located at one end. This is a distinguishing characteristic of this species of schistosoma.

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  • 27. 

    Specific organisms that are considered strict anaerobes include: 

    • A.

      E. coli and Streptococcus pneumoniae

    • B.

      Bacteroides and Klebsiella

    • C.

      Peptostreptococcus and Nocardia

    • D.

      None of the above

    • E.

      All of the above

    Correct Answer
    D. None of the above
    Explanation
    The given options do not include any specific organisms that are considered strict anaerobes. The correct answer is "None of the above" because none of the options provided include the correct organisms.

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  • 28. 

    What method is used to collect a specimen when Enterobius vermicularis is suspected?       

    • A.

      Tissue biopsy

    • B.

      Venipuncture

    • C.

      Cellophane tape method

    • D.

      PVA-preserved stool

    Correct Answer
    C. Cellophane tape method
    Explanation
    The cellophane tape method is used to collect a specimen when Enterobius vermicularis is suspected. This method involves pressing a piece of transparent tape against the perianal area to collect any eggs or adult worms that may be present. The tape is then placed on a glass slide and examined under a microscope to confirm the presence of the parasite. This method is simple, non-invasive, and effective for diagnosing Enterobius vermicularis infections.

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  • 29. 

    The blood level of total cholesterol concentration recommended by the World Health Organization is:      

    • A.

      Above 250mg/dl in Africans

    • B.

      Less than 225 mg/dl in Caucasians

    • C.

      Less than 150 mg/dl in Chinese

    • D.

      More than 200mg/dl in Asians

    • E.

      None of the above

    Correct Answer
    E. None of the above
  • 30. 

    Diphyllobothrium latum causes which of the following diseases?      

    • A.

      Cysticercosis

    • B.

      Echinococcosis

    • C.

      Hydatididossis

    • D.

      Diphyllobothriasis

    Correct Answer
    C. Hydatididossis
    Explanation
    Diphyllobothrium latum is a type of tapeworm that causes diphyllobothriasis, not Hydatididossis. Hydatididossis is caused by the tapeworm Echinococcus granulosus. Therefore, the given answer is incorrect.

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  • 31. 

     The accole form of plasmodium parasite is seen on blood film on:   

    • A.

      The Centre of the red blood cell

    • B.

      The periphery of the red blood cell

    • C.

      The nucleus of the red bold cell

    • D.

      Cytoplasm of the red blood cell

    Correct Answer
    B. The periphery of the red blood cell
    Explanation
    The accole form of the plasmodium parasite is seen on the periphery of the red blood cell. This means that the parasite attaches itself to the outer edge of the red blood cell. This is an important characteristic of the parasite as it allows it to evade the immune system and continue its life cycle within the host. By attaching to the periphery of the red blood cell, the parasite can avoid detection and destruction by the immune system, allowing it to survive and reproduce.

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  • 32. 

    All are trematodes except

    • A.

      Fasciolopsis buski

    • B.

      Enterobius vermicularis

    • C.

      Diphyllobotrium latum

    • D.

      Hymenolepsis diminuta

    • E.

      Schistosoma heamatobium

    Correct Answer
    B. Enterobius vermicularis
    Explanation
    The correct answer is Enterobius vermicularis. This is because Enterobius vermicularis is not a trematode, but rather a nematode or roundworm. Trematodes, also known as flukes, are a type of parasitic flatworm that typically infect the digestive system or blood vessels of their hosts. Examples of trematodes include Fasciolopsis buski, Diphyllobotrium latum, Hymenolepsis diminuta, and Schistosoma heamatobium. Enterobius vermicularis, on the other hand, is a common human parasite that causes the infection known as pinworm.

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  • 33. 

    Spherocytes in the blood film are not a feature of which of the following excep

    • A.

      Recticulocytosis

    • B.

      Glucose-6-phosphate dehydrogenase

    • C.

      Thallasaemia

    • D.

      Auto-immune heamolytic anaemia

    • E.

      Megaloblastic anaemia

    Correct Answer
    D. Auto-immune heamolytic anaemia
    Explanation
    Spherocytes are abnormally shaped red blood cells that are smaller and more dense than normal red blood cells. They lack the central pallor and appear more spherical. Spherocytes are commonly seen in autoimmune hemolytic anemia, where the immune system mistakenly attacks and destroys red blood cells. In reticulocytosis, there is an increased number of immature red blood cells (reticulocytes) in the blood. Glucose-6-phosphate dehydrogenase deficiency, thalassemia, and megaloblastic anemia can also cause spherocytes to be present in the blood film. However, spherocytes are not a feature of autoimmune hemolytic anemia.

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  • 34. 

    A feature of extravascular haemolysis is

    • A.

      Haemosiderin in urine

    • B.

      Jaundice

    • C.

      Albuminuria

    • D.

      Haemoglobin in urine

    Correct Answer
    B. Jaundice
    Explanation
    Extravascular haemolysis refers to the breakdown of red blood cells outside of blood vessels. One of the characteristic features of extravascular haemolysis is jaundice. Jaundice occurs when there is an accumulation of bilirubin, a yellow pigment, in the body. In extravascular haemolysis, the breakdown of red blood cells leads to an increased release of bilirubin, which then accumulates in the tissues, causing a yellow discoloration of the skin and eyes. Therefore, jaundice is a common manifestation of extravascular haemolysis.

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  • 35. 

    Which of these is true of B-lymphocytes?

    • A.

      They are all short –lived

    • B.

      They secrete surface immunoglobulin

    • C.

      They are both helper and killer cells

    • D.

      They are similar to lymphocytes in infectious mononucleosis

    Correct Answer
    B. They secrete surface immunoglobulin
    Explanation
    B-lymphocytes are a type of white blood cell that plays a crucial role in the immune response. One of their main functions is to produce and secrete surface immunoglobulin, also known as antibodies. These antibodies bind to specific antigens, such as bacteria or viruses, marking them for destruction by other immune cells. This secretion of surface immunoglobulin by B-lymphocytes is essential for the body's defense against infections and is a characteristic feature of these cells.

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  • 36. 

    In peripheral blood film examination, band form is associated with

    • A.

      P. falciparum

    • B.

      P.malariae

    • C.

      P. vivax

    • D.

      P.ovale

    Correct Answer
    B. P.malariae
    Explanation
    Band forms are immature neutrophils that are released into the bloodstream in response to infection or inflammation. In peripheral blood film examination, the presence of band forms indicates an ongoing infection. Among the given options, P. malariae is the correct answer because it is a species of malaria parasite that can cause infection in humans. The presence of band forms in the peripheral blood film examination suggests an infection with P. malariae.

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  • 37. 

    The best method to diagnose malarial parasite in the laboratory is examination of

    • A.

      Thin film only

    • B.

      Thick film only

    • C.

      Both thick and thin films

    • D.

      Rapid Diagnostic Test

    Correct Answer
    C. Both thick and thin films
    Explanation
    The best method to diagnose malarial parasite in the laboratory is through the examination of both thick and thin films. Thick films allow for the concentration of parasites, making it easier to detect low levels of infection. On the other hand, thin films provide a clearer view of the morphology of the parasites, allowing for species identification and accurate quantification. Therefore, using both thick and thin films increases the chances of detecting and accurately diagnosing malarial parasites in the laboratory.

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  • 38. 

    Urinalysis result that confirms the presence of Nitrites and leucocytes is indicative of

    • A.

      Leukocytosis

    • B.

      UTI

    • C.

      Viral infection

    • D.

      Parasitic infection

    • E.

      None of the above

    Correct Answer
    B. UTI
    Explanation
    The presence of nitrites and leukocytes in a urinalysis result is indicative of a urinary tract infection (UTI). Nitrites are produced by certain bacteria that cause UTIs, and leukocytes are white blood cells that indicate an immune response to an infection. Leukocytosis refers to an increased number of white blood cells in the blood, which is not directly related to the presence of nitrites and leukocytes in the urine. Viral and parasitic infections are not typically associated with the presence of nitrites and leukocytes in the urine. Therefore, the correct answer is UTI.

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  • 39. 

    Complete haemogram aims at screening for various disease states which include

    • A.

      Anemia

    • B.

      Leukemia

    • C.

      Inflammatory processes

    • D.

      All of the above

    Correct Answer
    D. All of the above
    Explanation
    A complete haemogram is a blood test that provides information about various components of the blood. It can help in the diagnosis and screening of different disease states. Anemia is a condition characterized by a decrease in the number of red blood cells or hemoglobin, and a complete haemogram can detect this. Leukemia is a type of cancer that affects the white blood cells, and a complete haemogram can help in its detection. Inflammatory processes in the body can also be identified through changes in the blood parameters, which can be assessed by a complete haemogram. Therefore, the correct answer is "All of the above."

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  • 40. 

     Ebola Virus Disease (EVD), was first discovered in

    • A.

      1934

    • B.

      1976

    • C.

      1988

    • D.

      1914

    Correct Answer
    B. 1976
    Explanation
    Ebola Virus Disease (EVD) was first discovered in 1976. This is the correct answer because it aligns with historical records and scientific research. In that year, there were two simultaneous outbreaks of the Ebola virus in Sudan and the Democratic Republic of Congo. These outbreaks led to the identification and subsequent study of the virus, leading to its classification as a new and highly dangerous disease. Since then, multiple outbreaks of EVD have occurred, with varying degrees of severity and impact on public health.

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  • 41. 

    The following tests are useful in monitoring an HIV patient except

    • A.

      HIV Quantiferon

    • B.

      CD4 count, Viral Load

    • C.

      Urea, FBC creatinine

    • D.

      LFT

    Correct Answer
    A. HIV Quantiferon
    Explanation
    HIV Quantiferon is not useful in monitoring an HIV patient because it is a test used to detect tuberculosis infection, not HIV. CD4 count, Viral Load, Urea, FBC creatinine, and LFT are all relevant tests for monitoring an HIV patient's health and progression of the disease.

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  • 42. 

    The following are activities that take place at the pre-analytical stage of a test procedure, except

    • A.

      Bringing reagents to room temperature

    • B.

      Bringing samples to room temperature

    • C.

      Recording test results

    • D.

      Identify samples with their manifests

    Correct Answer
    C. Recording test results
    Explanation
    The activities mentioned in the question are related to the pre-analytical stage of a test procedure. This stage involves preparing the samples and reagents before the actual analysis. Bringing reagents and samples to room temperature and identifying samples with their manifests are all part of the pre-analytical stage as they ensure that the samples and reagents are ready for analysis. However, recording test results is not an activity that takes place at the pre-analytical stage. It is typically done during or after the analytical stage, where the actual testing and analysis of the samples occur.

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  • 43. 

    The Ebola virus disease can be transmitted through contamination  from which of the following? 

    • A.

      Urine

    • B.

      Blood

    • C.

      Semen

    • D.

      Sweat

    • E.

      All of the above

    Correct Answer
    E. All of the above
    Explanation
    The Ebola virus disease can be transmitted through contamination from urine, blood, semen, and sweat. This means that any contact with these bodily fluids can potentially lead to the transmission of the virus. It is important to take precautions and practice good hygiene to prevent the spread of the disease.

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  • 44. 

    In gram’s stain, the steps involved are 

    • A.

      5

    • B.

      10

    • C.

      13

    • D.

      8

    Correct Answer
    C. 13
  • 45. 

     Which of these is a colour code for entry of laboratory reagent /commodity on tally cards in the store?    

    • A.

      Red

    • B.

      Blue

    • C.

      Green

    • D.

      Purple

    Correct Answer
    A. Red
    Explanation
    Red is the color code for the entry of laboratory reagent/commodity on tally cards in the store. This color is commonly used to signify caution or danger, and it is likely chosen to indicate that the reagent or commodity requires special attention or handling. Using color codes helps to easily identify and categorize different items in the store, making it more efficient to manage inventory and track usage.

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  • 46. 

    The minimum volume of whole blood required for testing of Ebola virus disease is: 

    • A.

      10 ml

    • B.

      20 ml

    • C.

      4ml

    • D.

      5ml

    • E.

      3ml

    Correct Answer
    C. 4ml
    Explanation
    The minimum volume of whole blood required for testing of Ebola virus disease is 4ml. This amount is sufficient for conducting the necessary diagnostic tests for detecting the presence of the virus. It is important to have an adequate volume of blood to ensure accurate and reliable results.

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  • 47. 

    Which of the following is the most appropriate container to sample for testing Ebola virus disease?  

    • A.

      Plastic tubes

    • B.

      Heparinized glass tubes

    • C.

      Un-heparinized glass tubes

    • D.

      Heparinized plastic tubes

    • E.

      Sterile 10cc glass syringes

    Correct Answer
    A. Plastic tubes
    Explanation
    Plastic tubes are the most appropriate container to sample for testing Ebola virus disease because plastic is a non-porous material that can effectively contain and prevent leakage of the virus. Unlike glass, plastic is less prone to breakage, reducing the risk of accidental exposure to the virus. Additionally, plastic tubes can be easily sealed and transported safely to the laboratory for testing.

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  • 48. 

    The specimen collected for testing Ebola Virus disease should be stored at:  

    • A.

      10- 15 degrees

    • B.

      2- 8 degrees

    • C.

      -4 degrees

    • D.

      36 degrees

    • E.

      22 degrees

    Correct Answer
    B. 2- 8 degrees
    Explanation
    The specimen collected for testing Ebola Virus disease should be stored at 2-8 degrees Celsius. This temperature range is necessary to maintain the integrity and viability of the virus in the specimen. Storing the specimen at lower temperatures may cause the virus to become inactive or degrade, while storing it at higher temperatures may accelerate the breakdown of the virus. Therefore, maintaining a temperature of 2-8 degrees Celsius is crucial to ensure accurate and reliable testing for Ebola Virus disease.

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  • 49. 

    Which of the following tests is not important in the diagnosis of Ebola virus disease?

    • A.

      ELISA

    • B.

      RT-PCR

    • C.

      RS- PCR

    • D.

      Serology- IgG/IgM

    • E.

      virus isolation

    Correct Answer
    C. RS- PCR
    Explanation
    RS-PCR, or Reverse Transcription Polymerase Chain Reaction, is not important in the diagnosis of Ebola virus disease. This is because RS-PCR is primarily used to detect RNA viruses, while Ebola is caused by a DNA virus. ELISA, RT-PCR, Serology- IgG/IgM, and virus isolation are all important tests in diagnosing Ebola virus disease. ELISA is used to detect antibodies against the virus, RT-PCR is used to detect viral RNA, Serology- IgG/IgM is used to detect specific antibodies, and virus isolation is used to isolate and identify the virus.

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  • 50. 

    Ebola virus is:  

    • A.

      RNA virus

    • B.

      DNA virus

    • C.

      Mixture of 50-50 DNA/RNA virus

    • D.

      A mixture of 70-30 DNA/RNA

    • E.

      A Mixture of 40-60 spirochetes/atypical virus

    Correct Answer
    A. RNA virus
    Explanation
    Ebola virus is classified as an RNA virus. RNA viruses have RNA as their genetic material instead of DNA. Ebola virus belongs to the family Filoviridae and is known for causing severe and often fatal hemorrhagic fever in humans and other primates. The virus is characterized by its filamentous shape and is transmitted through direct contact with infected bodily fluids. Being an RNA virus, Ebola virus has a high mutation rate, which contributes to its ability to adapt and evolve rapidly, posing challenges for treatment and prevention strategies.

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