Estudo dirigido 1 evolução

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Hardy-­‐Weinberg	
  Estudo	
  dirigido	
  1	
  	
  1)	
  Calcule	
  as	
  frequência	
  esperada	
  do	
  genótipo	
  abaixo	
  e	
  a	
  probabilidade	
  de	
  se	
  encontrar	
  um	
  indivíduo	
  com	
  esse	
  mesmo	
  genótipo,	
  levando	
  em	
  consideração	
  a	
  seguinte	
  tabela	
  com	
  as	
  frequências	
  de	
  cada	
  alelo.	
  Na	
  sua	
  opinião	
  a	
  identificação	
  feita	
  por	
  esse	
  conjunto	
  de	
  marcadores	
  é	
  confiável?	
  
	
  	
  	
  2)	
  Teste	
  a	
  hipótese	
  nula	
  de	
  Hardy-­‐Weinberg	
  para	
  uma	
  amostra	
  populacional	
  de	
  459	
  indivíduos	
  com	
  os	
  seguintes	
  genótipos:	
  MM	
  =	
  144,	
  MN	
  =	
  201,	
  NN	
  114.	
  3)	
  Um	
  sítio	
  de	
  clivagem	
  da	
  enzima	
  de	
  restrição	
  BanI	
  está	
  localizado	
  dentro	
  de	
  um	
  íntron	
  longo	
  do	
  gene	
  que	
  codifica	
  a	
  álcool	
  desidrogenase	
  em	
  D.	
  melanogaster.	
  Este	
  sítio	
  foi	
  localizado	
  em	
  29	
  dos	
  60	
  cromossomos	
  isolados	
  de	
  uma	
  população	
  amostrada	
  em	
  Raleigh,	
  Carolina	
  do	
  Norte	
  (Keitman	
  e	
  Aguadé,	
  1986).	
  Considere	
  que	
  B	
  e	
  b	
  representam	
  a	
  presença	
  e	
  ausência,	
  respectivamente,	
  do	
  sítio	
  de	
  BanI	
  no	
  cromossomo.	
  Assumindo	
  o	
  equilíbrio	
  de	
  Hardy-­‐Weinberg,	
  calcule	
  as	
  freqüências	
  esperadas	
  dos	
  genótipos	
  BB,	
  Bb	
  e	
  bb.	
  4)	
  No	
  sistema	
  sanguíneo	
  Ss,	
  relacionado	
  ao	
  sistema	
  MN,	
  três	
  fenótipos	
  correspondentes	
  aos	
  genótipos	
  SS,	
  Ss	
  e	
  ss	
  podem	
  ser	
  definidos.	
  Numa	
  amostra	
  de	
  1000	
  britânicos,	
  o	
  número	
  observado	
  de	
  cada	
  genótipo	
  para	
  o	
  sistema	
  Ss	
  foi	
  99	
  SS,	
  418	
  Ss	
  e	
  483	
  ss.	
  Estime	
  a	
  freqüência	
  alélica	
  de	
  S	
  (p)	
  e	
  s	
  (q)	
  e	
  verifique	
  se	
  as	
  frequênias	
  genotípicas	
  estão	
  em	
  equilíbrio	
  Hardy-­‐Weinberg.	
  5)	
  Numa	
  amostra	
  de	
  1617	
  bascos	
  da	
  Espanha,	
  o	
  número	
  de	
  indivíduos	
  com	
  sangue	
  do	
  tipo	
  A,	
  B,	
  O	
  e	
  AB	
  foi	
  724,	
  110,	
  763	
  e	
  20,	
  respectivamente.	
  As	
  melhores	
  estimativas	
  para	
  as	
  freqüências	
  alélicas	
  são	
  p1	
  =	
  0.2661	
  (para	
  IA),	
  p2	
  =	
  0.0411	
  (para	
  IB)	
  e	
  p3	
  =	
  0.6928	
  (para	
  IO).	
  Calcule	
  os	
  números	
  esperados	
  para	
  os	
  quatro	
  genótipos	
  e	
  conduza	
  um	
  teste	
  para	
  verificar	
  se	
  freqüências	
  observadas	
  seguem	
  o	
  princípio	
  de	
  Hardy-­‐Weinberg.	
  
	
  
22 CHAPTER 2
··
Current forensic DNA profiles use 10–13 loci to
estimate expected genotype frequencies. Problem 2.1
gives a 10-locus genotype for the same individual 
in Table 2.2, allowing you to calculate the odds 
ratio for a realistic example. In Chapter 4 we will
reconsider the expected frequency of a DNA profile
with the added complication of allele-frequency 
differentiation among human racial groups.
Testing for Hardy–Weinberg
A common use of Hardy–Weinberg expectations is
to test for deviations from its null model. Populations
with genotype frequencies that do not fit Hardy–
Weinberg expectations are evidence that one or
more of the evolutionary processes embodied in the
assumptions of Hardy–Weinberg are acting to det-
ermine genotype frequencies. Our null hypothesis 
is that genotype frequencies meet Hardy–Weinberg
expectations within some degree of estimation error.
Genotype frequencies that are not close to Hardy–
Weinberg expectations allow us to reject this null
hypothesis. The processes in the list of assumptions
then become possible alternative hypotheses to
explain observed genotype frequencies. In this sec-
tion we will work through a hypothesis test for
Hardy–Weinberg equilibrium.
The first example uses observed genotypes for the
MN blood group, a single locus in humans that has
Calculate the expected genotype frequency
and odds ratio for the 10-locus DNA profile
below. Allele frequencies are given in 
Table 2.3.
D3S1358 17, 18
vWA 17, 17
FGA 24, 25
Amelogenin X, Y
D8S1179 13, 14
D21S11 29, 30
D18S51 18, 18
D5S818 12, 13
D13S317 9, 12
D7S820 11, 12
What does the amelogenin locus tell us and
how did you assign an expected frequency
to the observed genotype? Is it likely that
two unrelated individuals would share this
10-locus genotype by chance? For this
genotype, would a match between a crime
scene sample and a suspect be convincing
evidence that the person was present at the
crime scene?
Problem box 2.1
The expected genotype 
frequency for a DNA profile
The loci used for human DNA profiling are a
general class of DNA sequence marker known
as simple tandem repeat (STR), simple
sequence repeat (SSR), or microsatellite loci.
These loci feature tandemly repeated DNA
sequences of one to six base pairs (bp) and
often exhibit many alleles per locus and high
levels of heterozygosity. Allelic states are
simply the number of repeats present at 
the locus, which can be determined by
electrophoresis of PCR amplified DNA
fragments. STR loci used in human DNA
profiling generally exhibit Hardy–Weinberg
expected genotype frequencies, there is
evidence that the genotypes are selectively
“neutral” (i.e. not affected by natural
selection), and the loci meet the other
assumptions of Hardy–Weinberg. STR loci 
are employed widely in population genetic
studies and in genetic mapping (see reviews
by Goldstein & Pollock 1997; McDonald &
Potts 1997).
This is an example of the DNA sequence
found at a microsatellite locus. This sequence
is the 24.1 allele from the FGA locus (Genbank
accession no. AY749636; see Fig. 2.8). The
integral repeat is the 4 bp sequence CTTT 
and most alleles have sequences that differ 
by some number of full CTTT repeats.
However, there are exceptions where alleles
have sequences with partial repeats or 
stutters in the repeat pattern, for example 
the TTTCT and CTC sequences imbedded 
in the perfect CTTT repeats. In this case, 
the 24.1 allele is 1 bp longer than the 
24 allele sequence. (continued)
Box 2.1 DNA profiling
9781405132770_4_002.qxd 1/19/09 2:22 PM Page 22
22 CHAPTER 2
··
Current forensic DNA profiles use 10–13 loci to
estimate expected genotype frequencies. Problem 2.1
gives a 10-locus genotype for the same individual 
in Table 2.2, allowing you to calculate the odds 
ratio for a realistic example. In Chapter 4 we will
reconsider the expected frequency of a DNA profile
with the added complication of allele-frequency 
differentiation among human racial groups.
Testing for Hardy–Weinberg
A common use of Hardy–Weinberg expectations is
to test for deviations from its null model. Populations
with genotype frequencies that do not fit Hardy–
Weinberg expectations are evidence that one or
more of the evolutionary processes embodied in the
assumptions of Hardy–Weinberg are acting to det-
ermine genotype frequencies. Our null hypothesis 
is that genotype frequencies meet Hardy–Weinberg
expectations within some degree of estimation error.
Genotype frequencies that are not close to Hardy–
Weinberg expectations allow us to reject this null
hypothesis. The processes in the list of assumptions
then become possible alternative hypotheses to
explain observed genotype frequencies. In this sec-
tion we will work through a hypothesis test for
Hardy–Weinberg equilibrium.
The first example uses observed genotypes for the
MN blood group, a single locus in humans that has
Calculate the expected genotypefrequency
and odds ratio for the 10-locus DNA profile
below. Allele frequencies are given in 
Table 2.3.
D3S1358 17, 18
vWA 17, 17
FGA 24, 25
Amelogenin X, Y
D8S1179 13, 14
D21S11 29, 30
D18S51 18, 18
D5S818 12, 13
D13S317 9, 12
D7S820 11, 12
What does the amelogenin locus tell us and
how did you assign an expected frequency
to the observed genotype? Is it likely that
two unrelated individuals would share this
10-locus genotype by chance? For this
genotype, would a match between a crime
scene sample and a suspect be convincing
evidence that the person was present at the
crime scene?
Problem box 2.1
The expected genotype 
frequency for a DNA profile
The loci used for human DNA profiling are a
general class of DNA sequence marker known
as simple tandem repeat (STR), simple
sequence repeat (SSR), or microsatellite loci.
These loci feature tandemly repeated DNA
sequences of one to six base pairs (bp) and
often exhibit many alleles per locus and high
levels of heterozygosity. Allelic states are
simply the number of repeats present at 
the locus, which can be determined by
electrophoresis of PCR amplified DNA
fragments. STR loci used in human DNA
profiling generally exhibit Hardy–Weinberg
expected genotype frequencies, there is
evidence that the genotypes are selectively
“neutral” (i.e. not affected by natural
selection), and the loci meet the other
assumptions of Hardy–Weinberg. STR loci 
are employed widely in population genetic
studies and in genetic mapping (see reviews
by Goldstein & Pollock 1997; McDonald &
Potts 1997).
This is an example of the DNA sequence
found at a microsatellite locus. This sequence
is the 24.1 allele from the FGA locus (Genbank
accession no. AY749636; see Fig. 2.8). The
integral repeat is the 4 bp sequence CTTT 
and most alleles have sequences that differ 
by some number of full CTTT repeats.
However, there are exceptions where alleles
have sequences with partial repeats or 
stutters in the repeat pattern, for example 
the TTTCT and CTC sequences imbedded 
in the perfect CTTT repeats. In this case, 
the 24.1 allele is 1 bp longer than the 
24 allele sequence. (continued)
Box 2.1 DNA profiling
9781405132770_4_002.qxd 1/19/09 2:22 PM Page 22
Genotype frequencies 25
larger if the expected number is 8 (it is 50%). Adding
all of these relative squared differences gives the 
total relative squared deviation observed over all
genotypes.
(2.8)
We need to compare our statistic to values from 
the χ2 distribution. But first we need to know 
how much information, or the degrees of freedom 
(commonly abbreviated as df ), was used to estimate
the χ2 statistic. In general, degrees of freedom are
based on the number of categories of data: df = no. of
classes compared – no. of parameters estimated – 1
 
χ2
2 2 221 6
181 61
43 2
518 80
21 6
360
=
−
+ +
−( . )
.
( . )
.
( . )
..
.
58
7 46=
for the χ2 test itself. In this case df = 3 − 1 −1 = 1 for
three genotypes and one estimated allele frequency
(with two alleles: the other allele frequency is fixed
once the first has been estimated).
Figure 2.9 shows a χ2 distribution for one degree
of freedom. Small deviations of the observed from the
expected are more probable since they leave more
area of the distribution to the right of the χ2 value. 
As the χ2 value gets larger, the probability that the
difference between the observed and expected is just
due to chance sampling decreases (the area under
the curve to the right gets smaller). Another way 
of saying this is that as the observed and expected 
get increasingly different, it becomes more improb-
able that our null hypothesis of Hardy–Weinberg is 
actually the process that is determining genotype
frequencies. Using Table 2.5 we see that a χ2 value 
of 7.46 with 1 df has a probability between 0.01 
and 0.001. The conclusion is that the observed
genotype frequencies would be observed less than
1% of the time in a population that actually had
Hardy–Weinberg expected genotype frequencies.
Under the null hypothesis we do not expect this much
difference or more from Hardy–Weinberg expecta-
tions to occur often. By convention, we would reject
chance as the explanation for the differences if the 
χ2 value had a probability of 0.05 or less. In other
words, if chance explains the difference in five trials
out of 100 or less then we reject the hypothesis that
the observed and expected patterns are the same.
The critical value above which we reject the null
hypothesis for a χ2 test is 3.84 with 1 df, or in nota-
tion χ20.05, 1 = 3.84. In this case, we can clearly see an
excess of heterozygotes and deficits of homozygotes
and employing the χ2 test allows us to conclude that
Hardy–Weinberg expected genotype frequencies are
not present in the population.
··
0 1 2 3 4 5 6 7 8 9 10
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
 
 
 P
ro
ba
bi
lit
y 
of
 o
bs
er
vi
ng
 a
s 
m
uc
h 
or
 m
or
e
 d
iff
er
en
ce
 b
y 
ch
an
ce
 
χ2
0.05,1
 = 3.84
χ2
Observed
 = 7.46
χ2 value
Figure 2.9 A χ2 distribution with one degree of freedom. 
The χ2 value for the Hardy–Weinberg test with MN blood
group genotypes as well as the critical value to reject the null
hypothesis are shown (see text for details). The area under 
the curve to the right of the arrow indicates the probability 
of observing that much or more difference between the
observed and expected outcomes.
Table 2.5 χ2 values and associated cumulative probabilities in the right-hand tail of the distribution for 1–5 df.
Probability
df 0.5 0.25 0.1 0.05 0.01 0.001
1 0.4549 1.3233 2.7055 3.8415 6.6349 10.8276
2 1.3863 2.7726 4.6052 5.9915 9.2103 13.8155
3 2.3660 4.1083 6.2514 7.8147 11.3449 16.2662
4 3.3567 5.3853 7.7794 9.4877 13.2767 18.4668
5 4.3515 6.6257 9.2364 11.0705 15.0863 20.5150
9781405132770_4_002.qxd 1/19/09 2:22 PM Page 25
20
C
H
APTER 2
··
Table 2.3 Allele frequencies for nine STR loci commonly used in forensic cases estimated from 196 US Caucasians sampled randomly with respect to
geographic location. The allele names are the numbers of repeats at that locus (see Box 2.1). Allele frequencies (Freq) are as reported in Budowle et al. (2001),
Table 1, from FBI sample population.
D3S1358 vWA D21S11 D18S51 D13S317 FGA D8S1179 D5S818 D7S820
Allele Freq Allele Freq Allele Freq Allele Freq Allele Freq Allele Freq Allele Freq Allele Freq Allele Freq
12 0.0000 13 0.0051 27 0.0459 <11 0.0128 8 0.0995 18 0.0306 <9 0.0179 9 0.0308 6 0.0025
13 0.0025 14 0.1020 28 0.1658 11 0.0128 9 0.0765 19 0.0561 9 0.1020 10 0.0487 7 0.0172
14 0.1404 15 0.1122 29 0.1811 12 0.1276 10 0.0510 20 0.1454 10 0.1020 11 0.4103 8 0.1626
15 0.2463 16 0.2015 30 0.2321 13 0.1224 11 0.3189 20.2 0.0026 11 0.0587 12 0.3538 9 0.1478
16 0.2315 17 0.2628 30.2 0.0383 14 0.1735 12 0.3087 21 0.1735 12 0.1454 13 0.1462 10 0.2906
17 0.2118 18 0.2219 31 0.0714 15 0.1276 13 0.1097 22 0.1888 13 0.3393 14 0.0077 11 0.2020
18 0.1626 19 0.0842 31.2 0.0995 16 0.1071 14 0.0357 22.2 0.0102 14 0.2015 15 0.0026 12 0.1404
19 0.0049 20 0.0102 32 0.0153 17 0.1556 23 0.1582 15 0.1097 13 0.0296
32.2 0.1122 18 0.0918 24 0.1378 16 0.0128 14 0.0074
33.2 0.0306 19 0.0357 25 0.0689 17 0.0026
35.2 0.0026 20 0.0255 26 0.0179
21 0.0051 27 0.0102
22 0.0026
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