BIO227 Homework 1
Chapters 1 and 2
Solutions
1. Assume a locus with 3 alleles, denoted 1, 2 and 3. Given parents with genotypes 12 and 23,
list the possible offspring genotypes and give the probability of each genotype in the list,
conditional on parental genotypes. What is the probability of the genotypes not on the list?
Pr(12)=Pr(13)=Pr(22)=Pr(23)=1/4
Pr(11)=Pr(33)=0
2. Problem 2a and 2b from text (page 28).
Given pure strains AABB and aabb, all F1 offspring are AaBb. These offspring can further
generate offspring with the following ratio: AABB (1), AABb (2), AAbb (1), AaBB (2), AaBb
(4), Aabb(2), aaBB (1), aaBb (2), aabb (1).
AABB (1), AABb (2), AaBB (2), and AaBb (4) all correspond to the same phenotype AB (9).
Similarly, AAbb and Aabb correspond to Ab (3), aaBB and aaBb correspond to aB (3), and aabb
coorespond to ab (1). Hence, the ratio 9:3:3:1.
3. Problem 6 from text (page 29).
Pr(affected)=Pr(two mutations)=0.000001. As the probability of having two mutations is so
small, we assume that neither of the parents of the affected person has two mutations. Therefore,
both parents must be heterozygous. The probability that they each transfer the disease allele to
another offspring is 1/2*1/2=1/4.
4. Problem 7 from text (page 29).
Let 𝑋~𝑁(𝜇, 1) be the trait when there are two mutations, and 𝑌~𝑁(0,1) be the trait when there
is 0 or 1 mutation. We want Pr (𝑋 > 𝑌), or equivalently, Pr (𝑋 − 𝑌 > 0). Since X and Y are
independent, 𝑋 − 𝑌~𝑁(𝜇, 2) and Pr 𝑋 − 𝑌 > 0 = Φ 𝜇/ 2 . Therefore, the two probabilities
are 0.64 and 0.92 respectively.
5. Problem 11 from text (page 30).
See Slide 31 of Lecture 2 for a graphical illustration.
6. We know that P(Aa offspring | Aa × Aa parents) =1/2. What is the probability that these same
parents produce 2 Aa offspring, given our understanding of the biologic basis of inheritance?
Explain your answer.
Given the genotype of the parents, the genotypes for the two offspring are independent of each
other. Therefore, the probability that both offspring are Aa is 1/2*1/2=1/4.
7. Consider the following penetrance model for an additive mode of inheritance on the log scale:
log P(disease | X) = a+bX
where X gives the number of disease (D) alleles in an individual’s genotype (i.e. 0, 1 or 2) and
P(disease | X) means the probability that the individual is diseased, conditional on their
genotype. This model is said to be additive on the log scale because disease depends only on the
number of disease alleles.
a) What is the interpretation of a and b in the model above for log P?
a is the log probability of disease for individuals with genotype 0. Equivalently, exp(a) is
the probability of disease for individuals with genotype 0.
b is the increase in log probability of disease comparing individuals with genotype x+1 to
individuals with genotype x. Equivalently, the probability of disease for individuals with
genotype x+1 is exp(b) times that for individuals with genotype x.
b) The multiplicative relative risk model is defined by
RR2 = RR12
where
RR1 = P(disease|Dd)/P(disease|dd)
and
RR2 = P(disease|DD)/P(disease|dd).
Show that the additive genetic mode of inheritance on the log scale is equivalent to the
multiplicative relative risk model.
Under the additive genetic mode of inheritance,
log P(disease|DD)=a+2b
log P(disease|Dd)=a+b
log P(disease|dd)=a
It follows that
RR1 = exp(a+b)/exp(a)=exp(b)
RR2 = exp(a+2b)/exp(a)=exp(2b)
Therefore, RR12=RR2.
8. Suppose we observe an offspring whose parents are Dd × Dd. Also assume any disease
model such that the P(disease) increases with the number of D alleles. Would you expect
P(offspring is Dd | Dd × Dd parents and offspring is diseased) still equals ½? Just explain your
reasoning. No need for any formal calculations.
Without knowing the disease status of the offspring, the probability that the offspring is Dd is ½.
However, knowing that the offspring is diseased may change this probability, because the
probability of disease increases with the number of D alleles. For example, if Pr(disease|DD) is
close to 1 while Pr(disease|Dd) and Pr(disease|dd) are close to 0, knowing the offspring is
diseased makes it much more likely that the offspring is DD.
More specifically, using Bayes rule we have that
Pr (diseased|Dd)Pr(Dd)
Pr (diseased)
This equation shows that, unless Pr(diseased|Dd)=Pr(diseased), that is, diseased and being
heterozygote are independent events, then Pr(Dd|disease) will not equal Pr(Dd).
Pr (Dd|diseased)=