STAT/BIOSTAT 550 B (DL): Homework 2.

1. (See P.12)
In the MN blood group system there are two alleles, M and N, which are codominant. In a sample of 1000 people, 400 are MM, 440 are MN, and 160 are NN.
(a) Estimate the population frequency of the M-allele
(b) Assuming Hardy-Weinberg equilibrium, estimate the variance of your estimator
(c) Do you believe this population is in Hardy-Weinberg equilibrium?

2. (See P.17)
At one of the loci of the rhesus blood group system, there are 2 alleles C and c.
a) In the simplest test, with only anti-C reagent, C is dominant to c. In a test of 1000 people, 160 are C-negative (genotype cc). Assuming Hardy-Weinberg Equilibrium, estimate the frequency of the c allele, and the variance of your estimator.
b) An anti-c reagent now becomes available, so that C and c are codominant. It is found that 400 individuals are CC and 440 are Cc. How does this change your estimate of the frequency of the c allele? And of the variance of the estimator?

3. (Example, due to Weir.)
Suppose a population consists of a mixture of two randomly-mating subpopulations of equal size. Consider a locus with three codominant alleles. In one subpopulation, the three alleles have frequencies p=0.6, q=0.3,and r=0.1, while in the other subpopulation they are p=0.4, q=0.1, r=0.5. Which genotypes have a smaller frequency than they would have in a single random-mating population with the same overall allele frequencies?

4. (See P. 18)
In some diallelic blood type system suppose the A and B alleles are codominant. A set of N mother-child pairs are sampled. Suppose there are
n₀₀ AA mothers with an AA child
n₀₁ AA mothers with an AB child
n₁₀ AB mothers with an AA child
n₁₁ AB mothers with an AB child
n₁₂ AB mothers with an BB child
n₂₁ BB mothers with an AB child
n₂₂ BB mothers with an BB child
Suppose the A allele frequency is q, and the B allele frequency is 1-q. Assume Hardy-Weinberg equilibrium. From Table 2.1 (P.17), the probabilities of the above 7 type pairs are then q³, q²(1-q), q²(1-q), q(1-q), q(1-q)², q(1-q)², and (1-q)³.

Then we showed in class (or equation (2.8), P.18), that we can find the the maximum likelihood estimator of q, using all the data: call this MLE q₀.
(a) Suppose we only use the mothers to estimate the allele frequency; show the MLE is
(2(n₀₀+n₀₁) +n₁₀+n₁₁+ n₁₂)/2N.
Show this estimator is unbiased. What would be its disadvantage, by comparison with q₀?
(b) Suppose we use all our individuals to estimate q, disregarding the relationships between mothers and children. Show the estimator of q is
(4 n₀₀+3(n₀₁ + n₁₀)+2 n₁₁ + n₁₂ +n₂₁)/4N
Show this estimator is unbiased. What would be its disadvantage, by comparison with q₀.
(Only a brief comment concerning comparisons is needed in (a) and (b) -- no quantitative assessment.)