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Solutions for exercises using Familias
Thore Egeland [1] There are also some videos available for some other exercises.
Exercise S1 Answers: a) We can write [pic] Consider first the numerator. The only possible genotype for the
child, given the mother and the AF, is A/B, and therefore the
probability is 1. For the denominator, the father must have passed on
the A allele, and therefore the probability is[pic]. Hence [pic]. The
standard interpretation is
"The data is 20 times more likely assuming AF to be the father
compared to the alternative that some unknown man is the father".
b) The buttons labeled 1, 2, 3 and 4 below will be used as explained under.
[pic] How to do it:
1. General DNA data: Click Add to enter an allele system. In the new
window, enter allele system name S1 and the two alleles A and B, both
with frequencies 0,05. Enter the C allele with frequency 0,9. Press
OK.
2. Persons: Enter the persons: mother, AF and child and their gender.
3. Case-Related DNA data: Double-click each person to enter his or her
DNA data. In the new window, enter the appropriate allele system (use
the pull-down menu) and the observed alleles for this person, then
press Add and OK.
4. Pedigrees: Click Add to enter the pedigree corresponding to
hypothesis H1. Enter H1 as Pedigree name. Enter the mother as the
parent of the child in the pull-down menu and click Add. Enter the AF
as the parent of the child in the pull-down menu and click Add. Click
OK to finish the definition of the pedigree corresponding to
hypothesis H1. Click Add in the pedigree window once more to add the
pedigree corresponding to hypothesis H2. Enter H2 as Pedigree name.
Press LR. (Normally one would answer Yes when asked to save.) This
gives relative values. It remains to specify the denominator, H2 in
this case, by marking the line corresponding to H2. Press the button
Relative or absolute values. The pedigree window should now appear as
[pic]
c) Repeat step 1 above to define the marker S2 and step 3 to define the
genotypes for all individuals. Steps 2 and 4 need not be repeated. The
LR is then obtained as before and the answer is: LR=200.
d) [pic]
h) LR=(1+3*0.02)/(2*0.02+(1-0.02)*0.05)=11.91
i) To be discussed in class. We note now, however, that Hardy-Weinberg
equilibrium is required for the LR derivation for each marker. This
assumption is not needed when we use theta-correction. Furthermore,
linkage disequilibrium is needed. We have also assumed that there are
no mutations or silent alleles. Exercise S2
a) The numerator, [pic]and therefore LR=0 as confirmed by Familias.
b) The General DNA data is used to specify the mutation model and should
be completed as below [pic] Answer LR=4.07E-03=0.00407
LR=0.01*(0.212+0.292)/(2*0.212*0.292)=0.00407 Exercise S3
The pedigree corresponding to H1, Figure 3, is specified as follows [pic] Most answers are given in the rightmost column of the below table from
the report generated by Familias. "D7S820, gives a very large LR, namely
11189" since 11.1 is so rare. If this marker is omitted the new LR is
530437141/11189=47407.
Likelihood ratio versus H2: 530437140,978139 -------------------------------------------------------
| System: | Likelihood: | LR versus H2: |
-------------------------------------------------------
| D3S1358 | 0,002107356328 | 24,3295665373587 |
-------------------------------------------------------
| D21S11 | 0,00106135425 | 4,42583732057416 |
-------------------------------------------------------
| D18S51 | 0,0008102440125 | 26,7722473604827 |
-------------------------------------------------------
| D7S820 | 5,037122916E-07 | 11189,4499975311 |
-------------------------------------------------------
| D16S539 | 0,000811482 | 0,899280575539568 |
-------------------------------------------------------
| CSF1PO | 0,0130390037415 | 2,25229008488432 |
-------------------------------------------------------
| F13B | 0,001423488512 | 5,88485536018151 |
-------------------------------------------------------
| LPL | 0,0007365230775 | 1,37960642346607 |
-------------------------------------------------------
Exercise S4
a) LR(grand father/unrelated) = 0.98 for D3S1358.
b) LR(grand father/unrelated) = 0.0085 for all markers.
c) Whether autosmal and haplotype markers can be combined and in case how
is a big question and there appears to be no consenus Exercise S5
a) W=20/(1+20)=0.952
b) W=200/(1+200)=0.995
c) Same answers as above.
d) This is a again a big question. Most recommendations are in favor of LR.
The problem with W is to decide on the prior. W may be easier to
interpret. Exercise S6 a) LR=10. Comment: The formula provided follows from [pic]
The same result is obtained in this particular case if inbreeding is
ignored.
b) LR (H1/H2)= 10, LR(H1/H3)=1.905. We are asked to verify that
[pic]
c) [pic] Exercise S7 Most answers are given in problem.
Regarding question f. We can add the following: If a model is unstable,
allele frequencies change with generations. This adds some intuition as to
why introducing an extra person, say a grandparent, typically changes
results slightly for an unstable model. Regarding question g. Consider the first model listed in Familias
Probability decreasing with range (stable) which gave LR=6.4E-03=0.0064.
Remove alleles so that only 14,15, 16,17 and a 'Rest allele' remains. Then
LR=7.8E-03=0.0078, a change. Again this effect does not have to with
Familias; there are two different models, one with 8 alleles and one with
alleles (which constrains mutations within those five alleles) and
different models typically give different results. Exercise S8 a) [pic]
b) The formula in Exercise S2 follows by setting [pic] and the numerical
result LR=0.0029 follows.
c)
[pic]
Exercise S9 a) LR(H1: father/H2:not father)=0 from Familias.
b) LR=0 for PENTA_E as can be seen from report from Familias.
c) LR= 4515433
d) See Familias file ExS9Brother.txt which gives LR=1.4.
e) To be discussed. Exercise S10 a) LR=0.791/0.209=3.8 Details on Familias implementation, including
definition of pedigrees, can be seen from ExS10.txt.
b) To be discussed.
Exercise S11
The LR is 1.36 How to do it:
Enter the allele system setting the silent allele frequency to 0.05. Enter
the persons and their DNA data as usual. Construct the pedigrees manually.
Use LR and Relative and Absolute values as previously A theoretical argument, not requested, is given below: Let pA = pB = 0.1
and pS = 0.05. According to http://dna-view.com/patform.htm, [pic] which coincides with Familias. Exercise S12 See pages 36-37 of the manual. Exercise S13 LR=0.0068 Exercise S14 Explained in problem. -----------------------
[1] Thore Egeland, Sept 28, 2012
Norwegian University of Life Sciences
P.O Box 5003
N-1432 Aas Norway
Thore.Egeland@umb.no
http://arken.umb.no/~theg/alcala