STAT/BIOSTAT 550, 2014: Homework 5: Due May 16.


1. (based on Crow and Kimura, Ch 4, #15)
An individual B has phenylketonuria, a rare recessive condition, for which the allele frequency q is 0.01. What is the probability that B's relative C has phenylketonuria if
(a) C is a nephew of B
(b) C is a double first cousin of B
(c) C is a quadruple half first cousin of B
(Note the kinship coefficient between B and C is the same for these three relationships)

2. Consider a male with inbreeding coefficient f1 who is unrelated to a female individual who has inbreeding coefficient f2. The individuals have two offspring B and C.
(a) Show that the probabilities ki (i=0,1,2) that B and C have i genes IBD are
k2 = (1 + f1)(1 + f2) /4
k0 = (1 - f1)(1 - f2) /4
k1 = 1 - k0 - k2
and that the coefficient of kinship between B and C is
psi(B,C) = (2 + f1 + f2) /8
(b) Suppose the first child has a recessive trait, the allele for which has frequency q. If nothing is known about the trait in the parents, what is the probability the second child will have this trait also?
(c) Suppose the first child has a recessive trait, the allele for which has frequency q. If nothing is known about the trait in the parents, what is the probability the second child will be an (unaffected) carrier of the recessive allele?
(d) Suppose now it is known that neither parent has the trait, but again the first child is affected. What is the probability the second child will have this trait also? What is the probability the second child will be an unaffected carrier of the recessive allele?

3. Suppose a pair of full sibs F and G share neither of their genes IBD (an event which has probability 1/4).
Given this, a third unilateral non-inbred relative H may share a gene IBD
with each of F and G, with F but not with G, with G but not F, or with neither.
That is, the four ibd states can be written
1 2 3 4 1 3, 1 2 3 4 1 5, 1 2 3 4 3 5, and 1 2 3 4 5 6.
Given that F and G share no genes ibd, if H is an aunt of F and G, the conditional probabilities of these four states are 1/4 each,
while if H is a half-sister of F and G the probabilities are 0,1/2,1/2,0 respectively.
You do not have to show this!.

Suppose at a marker locus the genotypes of F, G and H are ab, cd, and ac, respectively
where the a,b,c,d alleles have population frequencies qa qb qc and qd.
Find the probability of these data if
(a) H is an aunt of sibs F and G.
(b} H is a half-sister of sibs F and G.

4. (a) Regardless of genetic interference, show that if the probability of more than one crossover in a small sement of chromosome in an offspring gamete is negligible, then the recombination probability between loci at the two ends of the segment is approximately the genetic distance.
(b) Regardless of genetic interference, show that recombination probabilities are subadditive. That is, if r1 and r2 are the recombination probabilities in adjacent intervals, then the recombination r between the end points of the combined interval is no greater than r1+r2. When does r = r1+r2?
(c) (Lange, Chapter 7: number 1) In this part of the question, assume absence of genetic interference; i.e. Haldane's map function.
Suppose loci are ordered, 1,2,3,...,L along a chromosome, and rj is the recombination frequency between locus j and locus j+1. Show that the recombination frequency r between locus 1 and locus L satisfies the formula (1- 2 r) is the product of the (1- 2 rj).
(Hint: convert to genetic distance.)
(d) The Kosambi map function is r(d) = (e4d -1)/2(e4d+1). Suppose the recombination frequency between the ends of an interval is r. Show that the genetic length of the interval is a larger number of Haldane centiMorgans than it is Kosambi centiMorgans. Using the fact that the Kosambi map function expresses positive interference, reconcile the fact that Kosambi centiMorgans are ``larger'' with the fact that, in both cases, a Morgan (Haldane or Kosambi) is the genetic distance within which 1 crossover is expected.