Question

A 57 year-old female was brought to the emergency room following what appeared to be a cardiac arrest on the street. A bystander gave CPR until EMS personnel arrived. The following are her laboratory results: LD (100-220 U/L) Total CK (24-170 U /L CK-MB mass (0-5 ng/m) Day 1- Admission 10:00 pm 215 165 U/L 4 ng/ml Day 2 Day 2 5:30 am 9:00pm 5:45am 218 400 385 5200 U/L 170 U/L 191 ng/mL 300 ng/mL 12 ng/mL Day 5 1. Calculate the CK Relative Index for the second set of lab results. (1 pt) 2. Based on her lab data, did the patient have a myocardial infarction? Provide the lab data that supports your answer. (2 pts) 3. Draw the LDIisoenzyme electrophoresis densitometer pattern expected with an acute myocardialinfarction. 3. Draw the LD peaks.) (2 pts) e, as well as the individual isoenzyme 4. Draw the CK isoenzyme electrophoresis densitometer pattern expected with an acute myocardial infarction. (Make sure to label the anode and cathode, as well as the individual isoenzyme peaks.) (2 pts)

Answer 1

1. CK Relative index for the second set of lab results is

The formula to find out CK relative index = CK -MB/Total CK×100.

Second set CK-MB = 300ng/ml

Total CK = 5200U/L

Then , CK relative index = 300 / 5200 ×100 = 5.76

2. Yes , the patient have a myocardial infarction based on the CK-MB relative index value . The normal CK relative index is between 2.5 to 3 . When both CK -MB is high( 300 ng/ml )and also the ratio of CK - MB to total CK is more than 2.5 to 3 (i .e 5.76) . It means the heart is damaged .