Computes the (joint) genotype probability distribution of one or several pedigree members, possibly conditional on partial marker data.

1 2 | ```
oneMarkerDistribution(x, ids, partialmarker, theta=NULL, grid.subset=NULL,
loop_breakers=NULL, eliminate=0, verbose=TRUE)
``` |

`x` |
A |

`ids` |
A numeric with ID labels of one or more pedigree members. |

`partialmarker` |
Either a single integer indicating the number of one of |

`theta` |
The recombination fraction between marker and disease locus. Only relevant if at least one individual is affected by disease. In that case an error is raised if |

`grid.subset` |
A numeric matrix describing a subset of all marker genotype combinations for the |

`loop_breakers` |
A numeric containing IDs of individuals to be used as loop breakers. Relevant only if the pedigree has loops. See |

`eliminate` |
A non-negative integer, indicating the number of iterations in the internal genotype-compatibility algorithm. Positive values can save time if |

`verbose` |
A logical. |

A named array (of dimension `length(ids)`

) giving the joint marker genotype distribution for the `ids`

individuals, conditional on 1) the marker allele frequencies given in `partialmarker`

,
2) non-missing alleles in `partialmarker`

, and 3) the disease model of `x`

(if the pedigree is affected).

Magnus Dehli Vigeland

`twoMarkerDistribution`

, `allGenotypes`

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 | ```
x = nuclearPed(2)
x_aff = swapAff(x, c(1,3))
x_aff = setModel(x_aff, model=1) # dominant model
snp = marker(x, 1, c(1,1), 2, c(1,0), alleles=1:2, afreq=c(0.1, 0.9))
res1 = oneMarkerDistribution(x, ids=3:4, partialmarker=snp)
res2 = oneMarkerDistribution(x_aff, ids=3:4, partialmarker=snp, theta=0.5)
# should give same result, since theta=0.5 implies that marker is independent of disease.
stopifnot(all.equal(res1, res2))
#### Different example for the same pedigree. A marker with 4 alleles:
m2 = marker(x, 3:4, c('C','D'), alleles=LETTERS[1:4])
oneMarkerDistribution(x, ids=1, partialmarker=m2)
# Same as above, but computing only the cases where individual 1 is heterozygous.
# (The numbers 5:10 refer to the 6 last rows of allGenotypes(4),
# which contain the heterozygous genotypes.)
oneMarkerDistribution(x, ids=1, partialmarker=m2, grid.subset=matrix(5:10, ncol=1))
#### Expanding on the previous example:
# Joint genotype probabilities of the parents, but including only the combinations
# where the father is heterozygous and the mother is homozygous:
grid = expand.grid(5:10, 1:4)
oneMarkerDistribution(x, ids=1:2, partialmarker=m2, grid.subset=grid)
#### Something else:
# The genotype distribution of an individual whose half cousin is homozygous
# for a rare allele.
y = halfCousinPed(degree=1)
snp = marker(y, 9, c('a','a'), alleles=c('a', 'b'), afreq=c(0.01, 0.99))
oneMarkerDistribution(y, ids=8, partialmarker=snp)
#### X-linked example:
z = linkdat(Xped, model=4) # X-linked recessive model
z2 = swapAff(z, 1:z$nInd, 1) # disease free version of the same pedigree
snpX = marker(z, c(5,15), c('A','A'), alleles=c('A', 'B'), chrom=23)
r1 = oneMarkerDistribution(z, ids=13, partialmarker=snpX, theta=0.5) # results: A - 0.8; B - 0.2
r2 = oneMarkerDistribution(z2, ids=13, partialmarker=snpX) # should be same as above
r3 = oneMarkerDistribution(z, ids=13, partialmarker=snpX, theta=0) # results: A - 0.67; B - 0.33
stopifnot(all.equal(r1,r2), round(r1[1], 2)==0.8, round(r3[1], 2) == 0.67)
``` |

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