Question

nyia Ayla) Exercise 4.5 Following antibody identification, your patient possesses anti-K, anti-Jk", and anti-E. How many RBC units do you need to screen to find two antigen-negative RBC units? The antigens occur in the population with the following frequencies: E+ = 30%; K+ = 9%; Jk(a+) - 77%. Exercise 4.6 You performed a population study on your class and found that 84% are D-positive. Using the Hardy-Weinberg calculation, determine the percentage of your classmates who are heterozygous for D antigen. Exercise 4.7 A pregnant woman has a positive antibody screen. Antibody identification revealed anti-K as the antibody specificity. The father's red cells were tested for the K antigen. The father's phenotype is K+(K-positive). A molecular red cell phenotype was ordered on the father to determine the probability of HDFN due to anti-K. What information would be available through the molecular testing of the father?

Answer 1

Note: In case of Co-dominant inheritance, each allele will have different frequencies to be present in population. However frequencies are ratios from population. so (frequency of presence of allele E+)+ (frequency of absence of allele E+) =1

Patient possess antibodies for K, JK and E i requires RBC without antigens for these. in population. I Antigen Given frequencies. Frequencies of ge absence of antigen. Ě + 30% = 0. 3 1 0.3 0.7 E- k + 9% - 0.09 - 0.09 0.91 K- Jk Cat) 77% = 0.77 1 0.77= 0.23 JkCa) Note: co-dominant i individuals inheritance. carry more than I type allele To find antigen negative unit, all antigens should be absent. :. (frequency of) & ( frequency of) (frequency of) FL Clic - 0.7 x 0.91 X 0,23 = unit 0.146 without is frequency to find RBC antigen J, Ska and E.

Frequency of finding 1st antigen negative RBC unit will be 0.146.

In 100 units screened 14.6 will be antigen negative RBC.

??? United to be screened for finding 2 antigen negative units=(2 x 100)/14.6 = 13.69

minimum 14 units of RBC should be screen to find required units