Urinalysis & Medical Microscopy
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Questions and Answers
- 1.
If ammonia is not produced by the distal convoluted tubule, the urine pH will be:
- A.
Acidic
- B.
Basic
Correct Answer
B. BasicExplanation
If ammonia is not produced by the distal convoluted tubule, the urine pH will be basic. Ammonia is a weak base and helps in maintaining the pH balance of urine. It combines with hydrogen ions in the urine, resulting in the formation of ammonium ions (NH4+). These ammonium ions help to neutralize the acidity of urine, making it more basic. Therefore, if ammonia is not produced by the distal convoluted tubule, the urine pH will not be balanced by the presence of ammonium ions, leading to a more basic pH.Rate this question:
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- 2.
The primary chemical constituents of normal urine are:
- A.
Protein, sodium, and water
- B.
Urea, water, and protein
- C.
Urea , chloride, and water
- D.
Urea, bilirubin, and glucose
Correct Answer
C. Urea , chloride, and waterExplanation
Urea is a waste product formed in the liver from the breakdown of proteins. It is the main component of urine and helps to eliminate nitrogenous waste from the body. Chloride is an electrolyte that helps maintain fluid balance and is excreted in urine. Water is the solvent that carries these waste products and helps to maintain hydration. Therefore, the primary chemical constituents of normal urine are urea, chloride, and water.Rate this question:
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- 3.
A person exhibiting oliguria would have a daily urine volume of:
- A.
200 - 400 mL
- B.
600 - 1000 mL
- C.
1000 - 1500 mL
- D.
> 1500 mL
Correct Answer
A. 200 - 400 mLExplanation
Oliguria is a condition characterized by a decreased urine output. A person exhibiting oliguria would have a daily urine volume of 200 - 400 mL. This means that the person is producing a significantly lower amount of urine compared to the normal range of urine output, which is typically around 800 - 2000 mL per day. Oliguria can be a sign of various underlying medical conditions, such as dehydration, kidney dysfunction, or urinary tract obstruction. Monitoring urine output is important in assessing kidney function and overall hydration status.Rate this question:
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- 4.
An unpreserved specimen collected at 8 a.m. and remaining at room temperature until the afternoon shift arrives can be expected to have:1. Decreased glucose and ketones 2. Increased bacteria and nitrite 3. Decreased pH and turbidity 4. Increased cellular elements
- A.
1, 2, and 3
- B.
1, 2, and 4
- C.
1 and 2 only
- D.
4 only
Correct Answer
C. 1 and 2 onlyExplanation
An unpreserved specimen collected at 8 a.m. and remaining at room temperature until the afternoon shift arrives can be expected to have decreased glucose and ketones because these substances can degrade over time at room temperature. Additionally, the specimen can be expected to have increased bacteria and nitrite due to the growth of microorganisms in the absence of preservation methods.Rate this question:
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- 5.
What three changes will affect the results of the microscopic examination of urine if it is not tested within 2 hours?
- A.
Decreased bacteria, decreased red blood cells, decreased casts
- B.
Increased bacteria, increased red blood cells, increased casts
- C.
Increased bacteria, decreased red blood cells, decreased casts
- D.
Decreased bateria, increased red blood cells, increased casts
Correct Answer
C. Increased bacteria, decreased red blood cells, decreased castsExplanation
If urine is not tested within 2 hours, bacteria in the urine can multiply, leading to an increased presence of bacteria. Red blood cells and casts, on the other hand, may decrease in number over time due to degradation or settling. Therefore, the correct answer is that there will be an increased presence of bacteria, along with decreased red blood cells and casts.Rate this question:
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- 6.
Persons taking diuretics can be expected to produce:
- A.
Oliguria
- B.
Polyuria
- C.
Proteinuria
- D.
Pyruria
Correct Answer
B. PolyuriaExplanation
People taking diuretics can be expected to produce polyuria. Diuretics are medications that increase urine production by promoting the excretion of excess fluid and electrolytes from the body. This leads to an increased volume of urine being produced, resulting in polyuria. Oliguria refers to decreased urine production, which is the opposite of what diuretics do. Proteinuria refers to the presence of protein in the urine, and pyuria refers to the presence of pus in the urine, neither of which are directly related to diuretic use.Rate this question:
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- 7.
The concentration of a normal urine specimen can be estimated by which of the following?
- A.
Colour
- B.
Clarity
- C.
Foam
- D.
Odour
Correct Answer
A. ColourExplanation
The concentration of a normal urine specimen can be estimated by its color. The color of urine can vary depending on the concentration of waste products, such as urea and creatinine, as well as the presence of other substances like blood or bilirubin. A darker color generally indicates a higher concentration, while a lighter color suggests a more diluted urine sample. Therefore, by assessing the color of urine, one can get an estimate of its concentration.Rate this question:
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- 8.
The normal yellow colour of unrine is produced by:
- A.
Bilirubin
- B.
Haemoglobin
- C.
Uribilinogen
- D.
Urochrom
Correct Answer
D. UrochromExplanation
Urochrom is the pigment responsible for the normal yellow color of urine. It is produced as a result of the breakdown of red blood cells and the metabolism of hemoglobin. Bilirubin is a yellow pigment produced from the breakdown of heme in the liver and is responsible for the yellow color of jaundice. Haemoglobin is a protein found in red blood cells that carries oxygen, but it does not directly contribute to the color of urine. Urobilinogen is a colorless compound that is converted into urochromes, which give urine its yellow color. Therefore, the correct answer is Urochrom.Rate this question:
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- 9.
A strong odour of ammonia in a urin specimen could indicate:
- A.
Ketones
- B.
Normal
- C.
Phenylketonuria
- D.
Urinary tract infection
Correct Answer
D. Urinary tract infectionExplanation
A strong odor of ammonia in a urine specimen could indicate a urinary tract infection. Ammonia is a byproduct of the breakdown of urea in the urine, and a strong odor suggests that there may be an overgrowth of bacteria in the urinary tract. This can be a sign of infection, as bacteria produce ammonia as they break down urea. Other symptoms of a urinary tract infection may include frequent urination, pain or burning during urination, and cloudy or bloody urine. It is important to seek medical attention if these symptoms are present.Rate this question:
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- 10.
Testing of miceoalbuminuria is valuable for monitoring patients with:
- A.
Hypertension
- B.
Diabetes mellitus
- C.
Cardiovascular disease risk
- D.
All of the above
Correct Answer
D. All of the aboveExplanation
Testing for microalbuminuria is valuable for monitoring patients with hypertension, diabetes mellitus, and cardiovascular disease risk. Microalbuminuria refers to the presence of small amounts of albumin in the urine, which can be an early indicator of kidney damage. Hypertension, diabetes mellitus, and cardiovascular disease are all conditions that can lead to kidney damage, so monitoring microalbuminuria can help identify and manage these complications in patients with these conditions. Therefore, testing for microalbuminuria is valuable for monitoring patients with all of the above conditions.Rate this question:
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- 11.
The three intermediate products of fat metabolism incluse all of the following except:
- A.
Acetoacetic acid
- B.
Ketoacetic acid
- C.
Beta-hydroxybutyric acid
- D.
Acetone
Correct Answer
B. Ketoacetic acidExplanation
The three intermediate products of fat metabolism include acetoacetic acid, beta-hydroxybutyric acid, and acetone. Ketoacetic acid is not an intermediate product of fat metabolism.Rate this question:
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- 12.
The principle of the reagent strip tests for glucose is the:
- A.
Peroxidase activity of glucose
- B.
Glucose oxidase reaction
- C.
Double sequential enzyme reaction
- D.
Dye-binding of glucose and chromogen
Correct Answer
C. Double sequential enzyme reactionExplanation
The principle of the reagent strip tests for glucose is the double sequential enzyme reaction. This means that two enzymes are involved in the reaction. The first enzyme, glucose dehydrogenase, catalyzes the oxidation of glucose to produce gluconic acid and hydrogen peroxide. The second enzyme, peroxidase, catalyzes the reaction between hydrogen peroxide and a chromogen, resulting in a color change that indicates the presence of glucose. This double sequential enzyme reaction allows for a more accurate and reliable measurement of glucose levels in a sample.Rate this question:
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- 13.
Ketonuria may be caused by all of the following except:
- A.
Bacterial infection
- B.
Diabetic acidosis
- C.
Starvation
- D.
Vomiting
Correct Answer
A. Bacterial infectionExplanation
Ketonuria is the presence of ketones in the urine, which is usually a sign of increased fat metabolism. Bacterial infection is not a direct cause of ketonuria. Diabetic acidosis, starvation, and vomiting can all lead to an increase in ketone production and subsequent ketonuria. However, bacterial infection does not directly affect ketone production and therefore does not cause ketonuria.Rate this question:
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- 14.
The principle of the reagent strip test for blood is based on the:
- A.
Binding of heme and a chromogenic dye
- B.
Peroxidase activity of heme
- C.
Reaction of peroxide and chromogen
- D.
Diazo activity of heme
Correct Answer
B. Peroxidase activity of hemeExplanation
The correct answer is peroxidase activity of heme. Reagent strip tests for blood rely on the peroxidase activity of heme, which is an enzyme found in red blood cells. When blood is present in a sample, the heme catalyzes a reaction between hydrogen peroxide and a chromogen, resulting in a color change. This color change is then used to indicate the presence of blood in the sample.Rate this question:
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- 15.
All of the following can be detected by the leukocyte esterase reaction except:
- A.
Neutrophils
- B.
Eosinophils
- C.
Lymphocytes
- D.
Basophils
Correct Answer
C. LymphocytesExplanation
The leukocyte esterase reaction is a test used to detect the presence of white blood cells (leukocytes) in urine. Leukocytes are a general term for different types of white blood cells, including neutrophils, eosinophils, lymphocytes, and basophils. The reaction detects the enzyme esterase, which is released by neutrophils, eosinophils, and basophils. However, lymphocytes do not release esterase, so they cannot be detected by the leukocyte esterase reaction.Rate this question:
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- 16.
Differentition among RBCs, yeast and oil droplets may be accomplished by all of the following except:
- A.
Observation of budding in yeast cells
- B.
Increased refractility of oil droplets
- C.
Lysis of yeast cells by acetic acid
- D.
Lysis of RBCs by acetic acid
Correct Answer
C. Lysis of yeast cells by acetic acidExplanation
The given answer is correct because lysis of yeast cells by acetic acid can actually help in differentiating among RBCs, yeast, and oil droplets. Acetic acid is known to cause the lysis of yeast cells, leading to the release of their contents. On the other hand, RBCs and oil droplets are not affected by acetic acid in the same way. Therefore, the lysis of yeast cells by acetic acid can be used as a distinguishing characteristic to differentiate them from RBCs and oil droplets.Rate this question:
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- 17.
The finding of yeast cells in the urine is commonly associated with:
- A.
Cystitis
- B.
Diabetes mellitus
- C.
Pyelonephritis
- D.
Liver disorders
Correct Answer
B. Diabetes mellitusExplanation
Yeast cells in the urine are commonly associated with diabetes mellitus. This is because high levels of glucose in the urine can provide a favorable environment for yeast to grow, leading to a yeast infection. In individuals with diabetes, the elevated blood sugar levels can result in increased glucose in the urine, increasing the risk of yeast overgrowth. Therefore, the finding of yeast cells in the urine can be a sign of uncontrolled diabetes mellitus.Rate this question:
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- 18.
All of the following contribute to the formation of urinary crystals except:
- A.
Protein concentration
- B.
PH
- C.
Solute concentration
- D.
Temperature
Correct Answer
A. Protein concentrationExplanation
Protein concentration does not contribute to the formation of urinary crystals. Urinary crystals are formed when certain substances in the urine, such as calcium or uric acid, become too concentrated and solidify. Factors such as pH, solute concentration, and temperature can affect the solubility of these substances and increase the likelihood of crystal formation. However, protein concentration does not directly influence the formation of urinary crystals.Rate this question:
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- 19.
Urine specific gravity is equally influenced by the presence of glucose and sodium
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
Urine specific gravity is not equally influenced by the presence of glucose and sodium. Glucose presence in urine can significantly affect the specific gravity, while sodium levels have a less significant impact. Therefore, the statement is false.Rate this question:
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- 20.
Disposable containers with a capacity of 50 mL are recommended for the collection of specimens for routine urinalysis
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Disposable containers with a capacity of 50 mL are recommended for routine urinalysis because they provide an adequate volume for the collection of urine samples. This capacity allows for multiple tests to be performed on the sample, including analysis for glucose, protein, and other substances. Additionally, the use of disposable containers helps to prevent contamination and ensures accurate results.Rate this question:
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- 21.
The compensation field is the test area located on the lower end of the test strip
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
The compensation field is indeed the test area located on the lower end of the test strip. This area is designed to measure the level of compensation needed for accurate test results. It helps to adjust for any variations or interferences that may occur during the testing process. Therefore, the statement "True" is the correct answer.Rate this question:
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- 22.
Measuring the compensation field corrects the remission of the test areas and prevents false-negative results.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
Measuring the compensation field does not correct the remission of the test areas or prevent false-negative results. This statement is incorrect.Rate this question:
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