if someone is antiK antiJka and antiE antibodies how many RB

if someone is anti-K, anti-Jka and anti-E antibodies. how many RBC units are needed to find 2 antigen-negative units? the antigens occur in the population at these frequencies: E, 44%; K, 9%; Jka, 85%

Solution

Frequency of E antigen in the population is 44%.Therefore, frequency of population negative for E antigen is E(-) = 100 - 44 = 56% or 0.56. Similarly, K(-) = 100 - 9 = 91% or 0.91 and Jka (-) = 100 - 85 = 15% or 0.15 Number of individuals who will have E, K and Jka negative blood will be 0.56 * 0.91 * 0.15 = 0.08 or 8% of individuals (i.e. out of 100 units tested, 8 units will be antigen negative for all the three antigens mentioned here). However, we require only two antigen negative units. Therefore, number of RBC units needed will be : 2/0.08 = 25 RBC units will be needed.