Answer Happy

PROBLEM 6. In the blood type ABO system there is one gene with 3 alleles that give four different phenotypes O, A, B, AB of 6 different genotypes. Let 90, 9A, 9B be probabilities of the genotypes. Then the relation between the phenotype blood type probabilities and the genotypes are given by Table 4. Table 3: The relation between phenotypes and genotypes for the bloodtypes phenotype genotype р 9 0 00 ро 9. A AA or AO PA gả + 29ago B BB or BO PB 98 +29890 AB AB PAB 29AGB Σ 1 1 Table 4: The AOB blood types frequencies in the Norwegian population. The numbers are in percent Po PA PB PAB 40.0 48.0 8.0 4.0 a) Formulate the log likehood equations for the gene frequencies given data X = (Xo, Хл, Хв, Хлв). b) Use R and find the (90, 9A, 9B) that fit the Norwegian data. You can do this by searching over a fine grid on [0, 1] x [0, 1] for obtaining the minimum of the object

function Q defined by = def S = {0, A, B, AB}, Q(90,91) (p) – P1(90,9x)), JES 9B = 1- 90 - 9A, (90,9x) = argmin Q. - - As an example, for J = A we have PA – PA(90,9A) = 0.48 – 9Ả – 29A9o. Note that and Pb(90,9N) = (1 - 90 - 9A)2 + 2(1 – 90 - 91)9o. c) Simulate n = 1000 data points each of them consisting of two genes by sampling c with replacement from g - distribution. Then you convert each pair of genes to a phenotype blood type with help of Table 4. In this way you get a sample of n = 1000 blood types. d) Find the MLE of g from the data you have simulated. You may do the same method as under b) but this time with the true log likelihood as object function and do maximation, i. e. (90,9A) = argmax log f(X|p) and GB 1-go-9A e) Apply the method defined by b) with the maximum likelihood estimator pas input. Call the new estimator ğ= (go, ğa,ĪB). Why is this estimator inferior to the one obtained by d). Comment your numerical findings. f) Is ğ consistent? The answer needs no calculations. def