A hapten is an incomplete antigen.
All of the following cellular antigens are important to an immunohematologist except:

• A. Histocompatibility antigens
• B. Blood group antigens
• C. Autoantigens
• D. Haptens

Haemophilus influenzae is both X and V factor dependent for growth, and is the most common cause of obstructive laryngitis and epiglottitis. H. parainfluenzae is only V-factor dependent and causes only mild pharyngitis, rarely obstructive in nature. H. aphrophilus can be part of the normal upper respiratory flora in humans, does not cause pharyngitis, and is not X or V-factor dependent for growth. Although H. haemolyticus is both X and V-factor dependent, it is beta hemolytic on blood agar and is generally considered non pathogenic.
Acute obstructive epiglottitis, both in adults and children, is caused by the bacterial species shown in this split screen photograph. The colonies recovered on chocolate agar (upper frame) required both hematin and NAD, (as shown by colony growth only between the X and V strips in the lower frame). The most likely identification is:

• A. Haemophilus influenzae
• B. Haemophilus haemolyticus
• C. Haemophilus aphrophilus
• D. Haemophilus parainfluenzae

The colonies seen growing on the bird seed agar appear smooth and have a distinct reddish-brown pigmentation. The active ingredient in bird seed (Guizotia abyssinica) agar is caffeic acid, which is extracted and placed in an agar containing 1% glucose. Of the cryptococci, and other species of yeasts, Cryptococcus neoformans selectively produces the enzyme phenoloxidase, which oxidizes the caffeic acid in the medium to melanin, producing the red-brown pigmentation.
The other yeast species included in this exercise do not possess phenoloxidase activity and therefore remain non-pigmented when grown on bird seed agar.
The colonies shown in this photograph were grown on Guizotia abyssinica (bird seed) agar at 30°C for 72 hours. The most likely identification is:

• A. Candida parapsilosis
• B. Saccharomyces cerevisiae
• C. Cryptococcus laurentii
• D. Cryptococcus neoformans

One method of calculating a glomerular filtration rate is using creatinine and urine volume to determine creatinine clearance. The equation is as follows:
Creatinine Clearance = (urine creatinine X urine flow rate) / plasma creatinine; where urine flow rate = volume in mL /24 hours x h/60 min) In this case = creatinine clearance = 124 X (2200/24 x hour/60) / 2 = 94.7 or 95 ml/min What is the glomerular filtration rate for a patient with a serum creatinine of 2 mg/dL, if the urine creatinine was 124 mg/dL and the urine volume was 2.2 L/24 hrs?

• A. 13.6 mL/min
• B. 95 mL/min
• C. 136 mL/min
• D. 9.5 mL/min
• E. 1.36 mL/min