How many genes? Mapping mouse traits, cont.

1
How many genes? Mapping mouse traits, cont.

• Lecture 3, Statistics 246
• January 27, 2004

2
Inferring linkage and mapping markers

• We now turn to deciding when two marker loci are
  linked, and if so, estimating the map distance
  between them. Then we go on and create a full
  (marker) map of each chromosome, relative to
  which we can map trait genes. With these
  preliminaries completed, we can map trait loci.

3
The LOD score

• Suppose that we have two marker loci, and we
  dont know whether or not they are linked. A
  natural way to address this question is to carry
  out a formal test of the null hypothesis H r1/2
  against the alternative K rlt 1/2, using the
  marker data from our cross. The test
  statistic almost always used in this context is
  log10 of the ratio of the likelihood at the
  maximum likelihood estimate to that at the
  null, r1/2, i.e.

4
Calculating the LOD score

• Recall that the (log) likelihood here is based
  on the multinomial distribution for the
  allocation of n132 intercross mice into their
  nine 2-locus genotypic categories. As we saw
  earlier, it can be written
• and so we take the difference between this
  function evaluated at and at r1/2, which is
• where qi is 1/16, 1/8 or 1/4, depending on i.

5
Null probabilities of 2-locus genotypes
L1 L2 A H B
A 1/16 1/8 1/16
H 1/8 1/4 1/8
B 1/16 1/8 1/16
This is just putting r 1/2 in an earlier
table.
Exercise Suggest some different test statistics
to discriminate between the null H and the
alternative K. How do they perform in comparison
to the LOD?
6
Using the LOD score

• Normal statistical practice would have us
  setting a type 1 error in a given context (cross,
  sample size), and determining the cut-off for the
  LOD which would achieve approximately the desired
  error under the null hypothesis.
• This approach is rarely adopted in genetics,
  where tradition dictates the use of more
  stringent thresholds, which take into a account
  the multiple testing common on linkage mapping.
  It was originally motivated by a Bayesian
  argument, and in fact, Bayesian approaches to
  linkage analysis are increasingly popular. Let
  us use of Bayes formula in the form
• log10 posterior odds log10 prior odds
  LOD,
• where the odds are for linkage. With 20
  chromosomes, which we might assume approx the
  same size, and not too long, the prior
  probability of two random loci being on the same
  chromosome and hence linked, is about 1/20. In
  order to overcome these prior odds against
  linkage, and achieve reasonable posterior odds,
  say 1001, we would want a LOD of at least 3.

7
Linkage groups

• And so it has come to pass that a LOD must be
  gt3 to get peoples attention. Well be a little
  more precise later.
• The next step is to define what are called
  linkage groups. These partition the markers into
  classes, every pair of markers being either
  closely linked (i.e. r ? 0), or being connected
  by a chain of markers, each consecutive pair of
  which is closely linked. In practice, we might
  define closely linked to be something like
• a) lt c1, and b) LOD( ) gt c2, where
  e.g. c1 0.2, c2 3.

8
Forming linkage groups, cont.

• When one tries to form linkage groups, it is
  not unusual to have to vary c1 and c2 a little,
  until all markers fall into a group of more than
  just one marker. When this is done, it is hoped
  that the linkage groups correspond to
  chromosomes. If the chromosome number of the
  species is known, and that coincides with the
  number of linkage groups, this is a reasonable
  presumption. But much can happen to dash this
  hope one may have two linkage groups
  corresponding to different arms of the same
  chromosome, and not know that one can have a
  marker at the end of one chromosome linked to a
  marker at the end of another chromosome, though
  this should be rare if there is plenty of data
  and so on.

9
Ordering linkage groups

• Next we want to order the markers in a
  linkage group( ideally, on a chromosome). How do
  we do that? An initial ordering can be done by
  starting one of the markers, M1 say, on the most
  distant pair, here distance being recombination
  fraction, or map distance. Call M2 the closest
  marker to M1 and continue in this way.
• Now we want to confirm our ordering. One way
  is to calculate a (maximized) log likelihood for
  every ordering, and select the one with the
  largest log likelihood. But if we have (say) 11
  markers on a chromosome, this is 11! 4?107
  orders. What people often do is take moving
  k-tuples of markers, and optimize the order of
  each, e.g. with k 3 or 4. Whichever strategy
  one adopts, multi (i.e. gt2) locus methods are
  needed.

10
Likelihoods for 3-locus data

• Suppose that we have 3 markers M1 , M2 and
  M3 in that order. How do we calculate the log
  likelihood of the associated 3-locus marker data
  from our intercross?
• Recalling the discussion preceding the
  Punnett square of the last lecture, the parental
  haplotypes here are a1a2a3 and b1b2b3 while
  are would no fewer than 6 forms of recombinant
  haplotypes
  
• the four single recombinants a1a2b3 , a1 b2
  b3 , b1b2a3 and b1a2a3 ,
  and the two double recombinants a1b2 a3 and
  b1a2b3 .
• Proceeding as before, we calculate the
  probability of each of these in terms of the
  recombination fractions r1 and r2 across
  intervals M1-M2, and M2-M3, respectively. For
  simplicity, we assume the Poisson model, with
  independence of recombination across disjoint
  intervals. For example, a1a2a3 would have
  probability (1- r1)(1- r2)/4, a1a2b3 would have
  probability (1- r1)r2/4, while a1b2 a3 would
  have probability r1r2 .
• We would do this for every one of the 8
  paternal and 8 maternal haplotypes, and then
  collect them up to assign probabilities for each
  of the 33 3-locus genotypes (AAA, AAH, , BBB),
  and maximize the multinomial likelihood in the
  parameters r1 and r2 . This is just as in the
  2-locus case.

11
Multilocus linkage loci gt3

• It should have become clear by now that the
  strategy just outlined is not going to work too
  easily when there are (say) 11 loci in a linkage
  group.
• In that case, haplotypes are strings of the
  form a1a2b3 a10b11 , where there are just 2
  parental and 210-2 distinct recombinant
  haplotypes. The number of parental haplotype
  combinations is the square of this number, and
  they must be mapped into 311 11-locus genotypes,
  and a multinomial MLE carried out to estimate 10
  recombination fractions. What can be done?
• In 1987 the first large scale human genetic
  map was published, and at the same time a new
  algorithm was announced for both human pedigrees
  and experimental crosses, such as our intercross.
  This algorithm made use of hidden Markov models,
  and for the first time allowed full likelihood
  calculations in our current context without the
  exponential blow-up just described.

12
Multilocus mapping no details

• Im not going to cover this topic in detail this
  year, as I discussed it a few years ago, and
  those interested can read it there
• www.stat.berkeley.edu/users/terry/Classes/s260.199
  8/index.html
• We will meet hidden Markov models again pretty
  soon, as they are have become a common feature of
  statistical genetics and computational biology
  since the early 1980s.
• Now suppose that we have ordered our marker
  loci as just described, either by maximizing the
  likelihood within linkage groups over all orders,
  or by doing so in moving windows of size 3-5. How
  do we look at the result?

13
Checking the map, after removal of bad markers
Top triangle is a transform of the recombination
fraction, namely -4(1log2r ). Bottom triangle
contains the LOD scores at the maximum likelihood
estimate of recombination fraction. Notice the
bad bits in the top LH and bottom RH corners.
est.rf, plot.rf (from an R package)
14
Checking existing genetic maps

• As indicated earlier, the markers in our cross
  came from MIT, and they were already mapped.
  Most researchers would simply use the
  pre-existing map, as this would usually (but not
  always) be based on many more recombinations than
  could be expected in a single cross. Why might we
  not just do the same?
• Well, existing maps are rarely completely
  error-free, and one should always look at ones
  own data.
• An added benefit of looking at ones own data
  in relation to an existing map is that this
  should bring to light markers with a large
  numbers of genotyping errors, assuming the map is
  correct.

15
Interplay between error detection and maps

• Genotyping errors in mouse crosses can usually
  only be detected with the appearance of unusual
  numbers of close recombination events
• This depends entirely on the quality of the map
• The availability of the mouse genome sequence
  allows us to check genetic maps against the
  physical maps we locate the (unique) PCR primers
  for our microsatellite markers. This has brought
  a new era in quality of maps (includes human
  genetic maps!).
• The next slide depicts the genetic map we used.

16
Locations of our markers
After a commercial, we move on to mapping coat
color genes.

17
R
18
R/qtl
Authors Karl Broman, Hao Wu, Gary Churchill,
Saunak Sen, Brian Yandell
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Benefits of using R/qtl

• Lots of graphics
• Good error detection with accompanying graphics
• Single and two qtl mapping (and interaction
  terms)
• Choice of several input formats
• Includes Mapmaker format
• Many alternatives for mapping methods
• Many different models for phenotypes, e.g.
  standard normal, nonparametric model, binary
  traits

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Why map coat color genes in our C57/BL6 x NOD F2
intercross?

• the locations of these genes are known
• even with a modest number of mice we should be
  able to map these genes easily
• it is a useful check that everything is as it
  should be with our data
• and finally, it is a good exercise for us.
• Exercise. Look up the agouti and albino loci at
  the Mouse Genome Informatics database.

21
Recall our earlier Punnett square
22
Segregation data at a random marker

• Phenotype by genotype at D12Mit51
• (complete data only)
• A B H
• Agouti 19 18 35
• Black 8 3 18
• White 9 7 12

23
Mapping a segregating trait

• We turn now to mapping the two coat color
  genes segregating in our cross, beginning with
  the albino locus, and then the agouti locus. To
  do so, we need a genetic model, that is, we need
  to know or guess the relation between genotypes
  at our trait loci and phenotypes, which is
  embodied in the notion of a penetrance function.
• Looking at the preceding table, the albino
  trait segregates just as though governed by a
  recessive gene, so we postulate a locus with a
  recessive and a dominant allele for it. Although
  this is not precisely the case for the non-agouti
  trait, it is almost, and we do likewise.
• Later we will consider their interaction.

24
Probabilities of albino-marker genotypes (?4)

• Recall that the NOD mouse (A) is homozygous
  for the albino allele, while the C57/BL6 (B) is
  homozygous for the non-albino allele. We can
  collapse an earlier table to get

Colour M A H B
Albino (1-r)2 2r(1-r) r2
Full color 1-(1-r)2 2 - 2r(1-r) 1-r2
Here r is the rec. fr. between a marker and the
albino locus.
25
Segregation data at the marker closest to Tyrc

• Phenotype by genotype at D7Mit126
• _at_ 50 cM (the Tyrc locus is at 44 cM)
• A B H
• Agouti 3 19 47
• Black 0 10 19
• White 21 0 1

26
Mapping the albino locus
Plot of LOD score at each marker along the genome
27
Chromosome 7 genotypes for the albino mice.
A homozygous NOD, B homozygous B6, H
heterozygote. Genotypes are read down.
Pale blue shading is conserved NOD
haplotype. D7Mit128 is near the Tyrc locus,
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Honesty in advertising, and LOD thresholds

• There is more material in preparation here.
• Please revisit this space in a day or so.

29
Approximate probabilities of agouti-marker
genotypes (?4)

• Recall that the C57/BL6 (B) is homozygous for
  non-agouti, while the NOD (A) is homozygous
  agouti. Ignoring the 1/16 of the intercross who
  would exhibit the non-agouti trait (and be black)
  if they werent albino, we get the following
  approximate table, where 1/16 of the mice will be
  misclassified. Here r is the recombination
  fraction between a marker and the agouti locus.

Colour M A H B
Non-black 1-r2 2-2r(1-r) 1- (1-r)2
Black r2 2r(1-r) (1-r)2
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Segregation data at the marker closest to the
agouti locus

• Phenotype by genotype at D2Mit48
• _at_ 87 cM (agouti locus is at 89 cM)
• A B H
• Agouti 24 2 46
• Black 0 28 1
• White 5 6 14

31
Mapping the agouti locus
Plot of LOD score at each marker along the genome
32
Chromosome 2 genotypes for the black progeny.
Mauve shading indicates conserved C57/BL6
haplotype. Marker D2Mit48 is very close to the
agouti locus.
33
Conclusion single locus mapping

• agouti locus (A,a alleles) on Chr 2 at 89.9 cM
• albino locus (C,c alleles) on Chr 7 at 44 cM (now
  known as Tyrc gene)
• In the data set
• at 89 cM on Chr 2 with a LOD score gt 20
• Marker D2M48 (8th marker on Chr 2)
• at 43 cM on Chr 7 with a LOD score gt 20
• Marker D7M126 (4th marker on Chr 7)

The method worked for agouti, even though 1/16th
of the mice were misclassified
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Acknowledgement

• These last 3 lectures would not have been
  possible without the very substantial input of
  Melanie Bahlo and Tom Brodnicki of the Walter
  Eliza Hall Institute of Medical Research,
  Melbourne Australia.
• Tom (together with people from the WEHI mouse
  facility) carried out the cross, and did all the
  phenotyping, while Melanie did all the data
  analysis presented, and contributed a lot to the
  presentation. Overall, responsibility for the
  presentation (especially all the errors!) remains
  mine.

35
General exercise
Go through the last 3 lectures and redo all the
calculations as you can for the case of a
backcross rather than an intercross. You will
find it all simpler, and in every case, closed
form expressions appear, where we needed
iterative methods for the intercross.