QuantitativePopulation Genetics

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Quantitative/Population Genetics
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Problems
1 (1993 Q 10) In a colony of crested porcupines,
the average dorsal spine length in both sexes was
25 cm. The 10 of males with the longest spines
(average 30 cm) were mated to random females.
The average spine length of the progeny was 27
cm. What is the heritability of spine
length? 2 (1993 Q 8) Calculate the risk that a
couple, whose fathers are first cousins, have of
producing a child homozygous for a recessive
disease, relative to the risk for the general
population. Assume that the frequency of the
recessive allele is q 0.01.   3 (1992 Q 10)
The following data refer to body size in Parus
major (the Great Tit). Values are for the male
and female parents, and the mean of two offspring
(one of each sex). Mother Father Offspring 2
1.0 19.0 20.2 19.0 23.0 20.6 18.0 20.0
19.4 17.5 20.5 19.6 20.0 20.0 19.8 2
0.5 21.5 20.4 (all values in
grams) Question Estimate the heritability of
body size in this species.    
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Possible solutions

• Realized heritability is defined as (response) /
  (selection strength), abbreviated R/S.
• In the example S will be
• (the mean of the selected parents) (the
  population mean) (3025)/2 - 25 which is
  2.5 cm
• R will be 9 spine length of progeny mean
  spine length) 27-252.0
• The heritability is then 2.0/2.5 0.8

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Q2 From first principles F of progeny is
1/1281/128 1/64 (remember count no of steps in
C.A. loops including parents ) freq q0.01
therefore q2 10-4 is the risk to the
population at large The progeny has 1/64 of its
genome in homozygous condition and the remaining
63/64 has the same homozygosity as the population
at large. So the risk for this individual will
be 0.01x 1/64. plus 63/64 x 10-4. 2.547 x
10-4 The relative risk is (risk to
individual) / (risk to population) 2.55 From
equation relative risk is qF q2(1-F) /
q2. which is estimated above.
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Calculation of Inbreeding Coefficients in
pedigrees

• Locate all common ancestors (CA) of parents
• Trace all independent pathways leading from one
  parent to the other through a CA
• For each path, the probability of identity is
  given by 0.5 at each step (generation), and by
  0.5 x (1 FA) in the CA where FA is the
  inbreeding coefficient of the CA

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Heritability estimates Draw the graph (Use
mid-parent)
Regression slope of line gives h2 value b
(20.5-19.5)/(21-19) 1/2 h20.5
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More Problems ?!
4 (1992 Q 11) According to the Greek tragedians,
Oedipus unwittingly married his own mother,
Jocasta, and they had four children Eteocles,
Polynices, Antigone and Ismene. What value of F,
the inbreeding coefficient, would you estimate
for each of these children, and what proportion
of genes do they share with each other?   5 (1993
Q 9) One mutant allele at the dull locus in
Drosophila is caused by the insertion of a
mariner transposable element into the middle of
the dull gene. The element excises spontaneously
from the gene at a frequency of 1 per
generation, making the gene functional again.
Starting with a population homozygous for the
mariner insertion, how many generations will
elapse before homozygotes for functional dull
genes are found at a frequency of 5 ? (Assume
that the population is large, mating is random,
and the dull mutation does not affect the fitness
of the flies.)   6 (1996 Q10) About 70 percent of
all white North Americans can taste the chemical
phenylthiocarbamide, and the remainder cannot.
The ability to taste is determined by the
dominant allele T, and the inability to taste is
determined by the recessive allele t. If the
population is assumed to be in Hardy-Weinberg
equilibrium, what are the genotypic and allelic
frequencies in this population?  
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Q4 Draw the pedigree and count the steps 2 steps
from any of the progeny giving inbreeding Fx
(0.5)2 0.25 For second part, the Fx of any
progeny is 0.5 x the proportion of shared genes
of parents. Can calculate Fx of offspring between
2 of the progeny using Paths method (3
paths). Find Fx1/4 1/16 5/16 so proportion
of shared genes 10/16 0.625
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Q5. From Hartls Primer in Popn
genetics Element transposes out at rate 1 per
generation p (0.99)t freq of dull- alleles q1
- (0.99)t freq of dull alleles When are dull
homozygotes at 5 ie when is 1 - (0.99)t 2 gt
5 answer when t25.1 generations Q6 Tasters
are 30 of population Allele is recessive q2
0.3 ? q0.55, p0.45
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7 (1996 Q12) Cattle in Africa may be divided into
those that are susceptible to sleeping sickness
and those that possess innate genetic resistance
to the disease. Two candidate genes, TNF and FU,
each with two alleles, were typed in both
susceptible and resistant populations. Resulting
allele frequency counts are given
below ___________________________________________
__________________________ TNF locus FU
locus ____________________ ____________________
Population Allele 1 Allele 2 Allele 1 Allele
2 ____________ ________ ________ ________ _______
_   Resistant 203 12 130
120   Susceptible 216 34 101
114 ______________________________________________
_______________________   Is there any
statistical evidence that an allele at either
locus may be associated with disease
resistance? Solution Use Chi squared
S(observed expected)2/expected For TNF have 215
resistant and 250 susceptible cattle total465
419 Allele 1 and 46 allele 2 (again total
465) Expect proportion 215/465 cattle to be
resistant and 250/465 susceptible For allele 2
expect 250/465 x 46 24.7 susceptibles (observe
34) diff 9.3 subtotal Chi squared 3.5 All 4
categories Chi squared 8.4 significant _at_
1d.f. ( 2 x 2 tables) For FU Chi squared 0.54
not significant