Bio340 – What is the expected long-term equilibrium frequency

13 Jun Bio340 – What is the expected long-term equilibrium frequency

Question
Bio340 –

Problem Set 3 – Part 1

Obermiller – Summer 2016

1) A rare human disease – Snyder Robinson Syndrome – is due to a genetic defect in
enzyme called SMS. In a certain, isolated population, a heterozygote child carrying one
mutant (defective) allele and one wild-type allele of SMS gene has 5% less chance of
surviving into an reproductive adult compared to a homozygous child with two wild-type
alleles. A homozygote child with two mutant alleles suffers 28% decline in survival relative
to a wild-type homozygote. (Once becoming adults, individuals with three genotypes all
leave the same number of offspring, on average.) A deleterious mutation that destroys the
normal function of SMS occurs during meiosis with probability 10 -4. (10 points total; 5 pts.
ea).
a. What is the expected long-term equilibrium frequency of the defective SMS allele in
this population?

b. Given the answer to the above question, what is the expected number of children who
are homozygous for defective alleles if there are 10 million children in this population?

2) A certain plant population has maintained the Hardy-Weinberg equilibrium. The current
frequencies of alleles A and a are 0.7 and 0.3. If these plants suddenly stop outcrossing and
start reproducing by self-fertilization, what will be the frequencies of each genotype in the
next generation? (9 points; 3 ea. genotype freq.)

1

Bio340 –

Problem Set 3 – Part 1

Obermiller – Summer 2016

3) The following table shows the result of a QTL mapping experiment. The phenotypes of
two inbred lines (P1 and P2) are different. A male from P1 line and a female from P2 line were
crossed to produce F1 offspring. Then, an F1 male was back-crossed a P2 female to produce
the BC offspring. There are three microsatellite markers (M1, M2 and M3) used in this
experiment. The marker genotypes for P1 and P2 lines and F1 and BC offspring are shown
along with their phenotypic distribution. (12 pts).
Generation

Marker genotype*

No. of progenies

Phenotype

(M1 – M2 – M3)

(mean std. deviation)

P1

4/4 – 8/8 – 12/12

100 5

P2

10/10 – 15/15 – 9/9

140 5

F1

4/10 – 8/15 – 9/12

120 5

BC

10/10 – 15/15 – 9/9 (a)

87

138 10

4/10 – 15/15 – 9/9 (b)

9

123 10

10/10 – 8/15 – 9/9 (c)

11

138 11

10/10 – 15/15 – 9/12 (d)

92

140 11

4/10 – 8/15 – 9/9 (e)

94

122 10

4/10 – 15/15 – 9/12 (f)

12

121 9

10/10 – 8/15 – 9/12 (g)

8

136 10

4/10 – 8/15 – 9/12 (h)

85

122 10

* repeat number of microsatellites – for example, 6/10 means that the genotype of the marker locus is (AT)6/(AT)10,
if this locus is a microsatellite with AT repeats.

a

Which genotypes in the BC generation (labeled a to h in the table) were produced by
meiotic recombination between M1 and M2?

b Are markers M1 and M2 on the same chromosome? Why?

c

Which marker is closely linked to a locus where a mutation that produced a
phenotypic difference between P1 and P2 lines is located? Briefly explain.