Jurg Ott / 2 November 2013

ott@rockefeller.edu

LIPED program for linkage analysis

Table of Contents

1. INTRODUCTION 3

2. WORKING WITH SEVERAL LOCI 4

3. PENETRANCES 5

4. NON-AUTOSOMAL LINKAGE 5

5. INPUT FILE 6

5.1. Problem description 6

5.2. Format for locus symbols, allele symbols and phenotype symbols 7

5.3. Format to read input item 5.12 (symbols for a pair of alleles and values of penetrances) 7

5.4. Format for pedigree data 7

5.5. Symbols, to be read with Format as on input item 5.4 7

5.6. Number of alleles per locus 8

5.7. Number of phenotypes per locus 8

5.8. Locus type, indicated by the variable KONT 8

5.9. Output options as indicated by the variable IAU(i) 8

5.9.a (optional) Recombination fractions when IAU(1)=9 in input item 5.9 10

5.10. Locus description 11

5.11. Gene frequencies (dummy values required with a value of 1 in col.7 of input item 5.1) 11

5.12. Mode of inheritance 11

5.13. Pedigree information 12

5.14. Pedigree data 12

5.14a (optional) Identification of "doubled" individuals 12

5.14b (optional) Haplotype frequencies 12

5.15. (optional) Recombination fractions 13

5.16. Direction of further analysis 13

6. DIMENSIONS 14

7. COMPLEX PEDIGREES 15

8. MUTATION 17

9. QUANTITATIVE PHENOTYPES 17

10. AGE-DEPENDENT PENETRANCE 18

10.1  Age classes with different penetrances 18

10.2  Age-of-onset distributions 19

UNAFFECTED INDIVIDUALS 19

AFFECTED INDIVIDUALS 19

UNKNOWN DISEASE STATUS 19

10.3 Lognormal distribution of age of onset 19

10.4  Straight-line curves for age of onset (locus type 3) 22

11. CALCULATION OF GENETIC RISKS 24

12. LIKELIHOOD AT A SINGLE LOCUS 25

13. HELPFUL HINTS 25

14. EXAMPLES 26

Eample 1 26

Example 2 27

Example 3 27

Example 4 27

Example 5 27

15. LITERATURE 27

1. INTRODUCTION

The LIPED program (for LIkelihoods in PEDigrees) estimates the recombination fraction by calculating pedigree likelihoods for various assumed values of the recombination fraction. The algorithm is based on Elston and Stewart (1971) with some extensions. Its first application (to the large Alaska pedigree, Schrott et al. 1972) resulted in mild evidence for linkage of familial hypercholesterolemia to the C3 polymorphism (Ott et al 1974), which was later confirmed by various authors. This disease locus (LDLR, previously FH and FHC) is now located on chromosome 19p13.3. The program contained one error (in the likelihood calculation for quantitative traits), which was pointed out to me by Dr. Robert Elston. Details on the theory underlying LIPED and its likelihood calculations may be found in Thompson (2011). LIPED is still used today (see poster presented at HGM 2012 in Sydney) although rarely and in updated form.

This manual describes the PC version (June 1995) of the LIPED computer program for genetic linkage analysis. Only two loci can be handled at a time, for example, a disease locus and a marker locus. Originally written in Fortran IV (Ott 1974), LIPED requires input in fixed format (numbers must be provided in a fixed number of spaces or columns). The code is essentially as originally written, with some additions such as proper treatment of age of onset data. This describes an updated version (June 1995) of LIPED suitable for use on PCs. Only minor modifications were made in the latest revision. The program has been compiled with Microsoft Fortran PowerStation 4.0. for Windows.

KNOWN BUGS: For a pedigree consisting of a single individual, LIPED does not calculate a likelihood. This "problem" may be avoided by including two parents with unknown phenotypes. In practice, this bug is irrelevant for linkage analyses.

Files included:

To initiate the program, type LIPED. It will assume that input is furnished in the file, LIPED.DAT.
When LIPED is used for research, the appropriate literature reference is Ott (1974) or Ott (1976) or Ott (1999).

2. WORKING WITH SEVERAL LOCI

Two kinds of loci are distinguished in the LIPED program: main locus (internal number zero) and marker loci (numbered from 1 to NMARK). Lod scores can be computed for any combination main locus vs. marker locus. With input item 16 (see section 5, Input File, below), any one of the marker loci can be declared to represent a new main locus so that lod scores may also be computed among marker loci. If more than one marker locus is present, the program creates two temporary disk files that will be deleted on program termination. Note the following restriction: with a single marker locus, any number of pedigrees may be analyzed in a single run. However, with more than one marker locus, only a single pedigree may be analyzed in a run. One way to overcome this restriction is as follows. If several independent families are presented to the program as one single pedigree, LIPED will recognize this and carry out the proper calculations, the resulting lod score being the sum over the individual families; however, individual lods for the families will not be recognizable by the user. Analyzing several independent families as a single large pedigree requires a substantial amount of memory. If an error occurs (MNP or MLIST too small), you may have to analyze the families in the usual manner as separate pedigrees.

Generally, to analyze several pedigrees with phenotypes at more than two loci, one might proceed as follows. First, one decides on the two loci to be analyzed. If one of these is the main locus, then the comparison main locus vs. marker is identified on one line of input item 9. Otherwise, all numbers in input item 9 are set equal to 0, and a comparison among markers is defined on a new line of input item 9 that appears after a line containing 5000 which immediately follows the pedigree data. Thus far, one has decided on the 2 loci to be compared. Now, one must tell LIPED where on the line (in which columns) to read the phenotypes of these loci, which is done with FORTRAN Format expressions (see beginning of section 5) furnished in input item 4.

If you interrupt LIPED while it is still running and if more than one marker locus has been defined, scratch files named "ZZ..." or "for..." will remain on the disk. These files would be deleted when the program terminates normally. You may simply delete them.

3. PENETRANCES

Penetrance is defined as the probability of occurrence of a particular phenotype given the presence of a certain genotype. Accordingly, with respect to a disease, penetrance is the probability of being affected given a certain genotype.

Penetrances are needed to describe the relation between genotypes and phenotypes. In the following three simple and common cases, only full penetrance (values 0 or 1 only) is assumed to occur. Assume a locus with 2 alleles, T and t. When this is a disease locus, let T be the dominant disease allele and consider the phenotypes AFF for affected and NA for unaffected.

     ---------------------------------------------
                         Phenotypes
              ------------------------------------
     Geno-    Dominant    Recessive     Codominant
     type      disease      disease        case
               AFF  NA      AFF  NA     TT  Tt  tt
     ---------------------------------------------
      T T       1    0       1    0      1   0   0
      T t       1    0       0    1      0   1   0
      t t       0    1       0    1      0   0   1
     ---------------------------------------------

4. NON-AUTOSOMAL LINKAGE

To code for X-linked inheritance in LIPED, tables such as the one above are used to represent the relation between genotypes and phenotypes. They apply directly for females. For males, for example, the genotype T/T is interpreted as T/y (hemizygote), and all lines corresponding to heterozygote genotypes are disregarded. Therefore, it is not necessary to distinguish male and female phenotypes. For example, TT can serve as a phenotype for either sex.

To analyze loci on the Y-chromosome, it is easiest to code Y-linkage as a special case of autosomal linkage, but precautions must be taken. For details, see Ott (1986); note that the methods described in that reference apply only to full penetrance.

5. INPUT FILE

For most input quantities, their location (column numbers) on an input line is fixed and must strictly be adhered to. For some input quantities, the user is flexible and can determine with so-called Format expressions where on a line that quantity will be found by the program. For each of the Format expressions, below, a recommendation is given that will accommodate most situations occurring in practice. It provides four spaces (columns) for each input quantity. A short explanation of FORTRAN Formats is as follows.

An input quantity is read either with an A-Format (alphanumeric) or an F-Format (floating point quantities) where it is determined by the program which of the two forms much be used for each input quantity (described in this section). For example, (A4) means that an input quantity such as a phenotype symbol should occupy 4 spaces (columns). If several input quantities must be read by the program, each requires its own Format, for example, (A4, A4, A4) or, equivalently, (3A4). To skip reading over a number of spaces, the X-Format is used. For example, the Format expression (2A4, 8X, A4) means that the program will read input quantity 1 in columns 1-4, quantity 2 in columns 5-8, and quantity 3 in columns 17-20. For alphanumeric quantities (A format), the position within the space provided is critical; preferably the same number of spaces is always used for the same quantity and it is right-justified within the allotted space.

The input file must consist of the following "lines", here being numbered as input item 5.1, item 5.2, etc.

5.1. Problem description

 Col.1-2
NMARK, number of marker loci in addition to the main locus. LIPED stops when NMARK<1 is encountered.

 Col. 4
= 0 (a value of 1 prints internal information not generally interpretable)

 Col. 5
= 0 usual setting
= 1 to prevent underflows as much as possible. This should be used only

when an underflow has occurred since underflows
are unlikely with double precision calculations.


 Col. 6
= 0 for autosomal loci
= 1 for loci on the X-chromosome

 Col. 7
= 0 if gene frequencies (rather than haplotype frequencies) will be read
= 1 if haplotype frequencies are to be read (input item 5.14b). Note that in

this case, dummy gene frequencies must still be provided (input item 5.11).


 Col. 8-20 Mutation rate at main locus (see MUTATION below).

 Col.21-80 Text

5.2. Format for locus symbols, allele symbols and phenotype symbols

 Col. 1-80
Format to read input item 5.10 (symbols for loci, alleles and phenotypes). Only A-format is allowed; the maximum length is A4 for locus and allele symbols, and A8 for phenotype symbols. EXAMPLE: (20A4). The format for the phenotype symbols must correspond in length to the one used to read the actual phenotypes of the pedigree data (input item 5.14).

5.3. Format to read input item 5.12 (symbols for a pair of alleles and values of penetrances)

 Col. 1-80  Two A-formats and then F-formats. EXAMPLE: (2A4,21F4.0)

5.4. Format for pedigree data

 Col. 1-80
Format to read input item 5.5, 5.14 and 5.14a. A-format only. EXAMPLE: (25A4). The first four items must not contain more than 4 characters each whereas phenotypes can be up to 8 characters long, ie, may be read with an A8 Format, for example.

5.5. Symbols, to be read with Format as on input item 5.4

At place of  Provide symbol for
item no.
----------------------------------------------------------------
 
3           no parent (e.g.: blank)
 
4           male sex (e.g.: m). The first different sex symbol encountered
            
in input item 14 will be considered the symbol for female sex.
 
5           unknown phenotype at main locus (e.g.: blank)
 
6, etc.     unknown phenotype at marker locus 1, etc.
----------------------------------------------------------------

5.6. Number of alleles per locus

 Col. 1-2  Number of alleles at main locus (locus 0)
 Col. 3-4  Number of alleles at marker locus 1, etc.

5.7. Number of phenotypes per locus

 Col. 1-2  Number of phenotypes at main locus
 Col  3-4  Number of phenotypes at marker locus 1, etc.

For loci of type KONT = 1, 2, or 3, the number of phenotypes is predetermined and will be set by the program. Here you may simply use 1 for number of phenotypes when KONT > 0.

5.8. Locus type, indicated by the variable KONT

 Col. 1-2 KONT for main locus
 Col. 3-4 KONT for marker locus 1, etc., where he following values for KONT apply:

  1. for a locus with discrete phenotypes but penetrances possibly other than 0 and 1. This is more general but runs slower than KONT = 0.

  2. for a locus with discrete phenotypes and penetrances of 0 and 1 only.

  3. for a locus with quantitative phenotypes following a conditional normal distribution (see section on quantitative phenotypes).

  4. for a locus with age-dependent penetrances following a lognormal distribution (see section on age-dependant penetrance).

  5. for a locus with straight-line age-dependant penetrances.

5.9. Output options as indicated by the variable IAU(i)

 Col. 2 IAU(1) for main locus versus marker locus 1
 Col. 4 IAU(2) for main locus versus marker locus 2, etc., where the values of IAU(i) have the following effect (below, rm and rf = male and female recombination fractions, respectively):

  1. to do checks only and compute likelihood at rm = rf = 0.5

  2. when no computation is desired for main locus versus marker locus i

  3. if the values of the recombination fractions are to be read by input item 5.15 below. In this case, one set (input line) of values rm and rf after each pedigree, for each likelihood to be computed. Note that after input item 5.16, additional lines of input of this type may be given to allow for comparisons among marker loci.

  4. to compute lods at rm = rf = 0, .001, .05, .10, .20, .30, .40

  5. to compute lods at rm = rf = 0, .001, .05, .10, .15, ...

  6. to compute lods at values of rm and rf shown below.

  7. to compute lods at values of rm and rf shown below

  8. to compute lods at values of rm and rf shown below

  9. to compute lods at values of rm and rf shown below

  10. to compute lods at values of rm and rf shown below

  11. to read recombination fraction values from input item 5.9a once for all pedigrees and to sum the lod scores over pedigrees. Allowed only with NMARK = 1 on input item 5.1, i.e., when no more than one marker locus is specified.

  12. to compute lods at rm = 0, .001, ... (as with option 2) whereas rf = 2rm(1 – rm) [Ott, 1991, equation (8.7), p. 175], that is, the female map distance (Haldane) is assumed to be twice the male map distance.

  13. to compute lods at rm = 0, ... (as with option 3) whereas rf = 2rm(1 – rm)

Below is a graphic representation of the values of rm and rf
at which lods will be computed depending on the value of IAU(i):

IAU(i) = 4 (22 points)
rm
.5  |x   x   x   x   x   x   x   x
.4  |                        x   x
.3  |                    x       x
.2  |                x           x
.1  |            x               x
.05 |        x                   x
.001|    x                       x
 0  |x                           x
    +-----------------------------
     0 .001 .05 .1  .2  .3  .4  .5 rf

IAU(i) = 5 (34 points)
rm
.5  |x   x   x   x   x   x   x   x   x   x   x   x
.45 |                                        x   x
.4  |                                    x       x
.35 |                                x           x
.3  |                            x               x
.25 |                        x                   x
.2  |                    x                       x
.15 |                x                           x
.1  |            x                               x
.05 |        x                                   x
.001|    x                                       x
 0  |x                                           x
    +---------------------------------------------
     0 .001 .05 .1  .15 .2  .25 .3  .35 .4  .45 .5 rf

IAU(i) = 6 (9 points)        IAU(i) = 7 (16 points)
rm                         rm
.5 |   x     x     x      .5 |  x     x     x     x
.3 |   x     x     x      .35|  x     x     x     x
.1 |   x     x     x      .2 |  x     x     x     x
   +------------------    .05|  x     x     x     x
     0.1   0.3   0.5 rf      +------------------------
                              0.05  0.20  0.35  0.50 rf

IAU(i) = 8 (64 points)
rm
.5  |x   x   x   x  x  x  x  x
.4  |x   x   x   x  x  x  x  x
.3  |x   x   x   x  x  x  x  x
.2  |x   x   x   x  x  x  x  x
.1  |x   x   x   x  x  x  x  x
.05 |x   x   x   x  x  x  x  x
.001|x   x   x   x  x  x  x  x
 0  |x   x   x   x  x  x  x  x
    +-------------------------
     0 .001 .05 .1 .2 .3 .4 .5 rf

Option 8 is useful for approximate factorization of joint male and female lods into sex-specific lods.

5.9.a (optional) Recombination fractions when IAU(1)=9 in input item 5.9

Each pedigree will be analyzed at the rm,rf-values provided here.
 Col. 1- 5 Value for male recombination fraction

 Col. 6-10 Value for female recombination fraction. These two values are read with Format 2F5.4. For example, an input line for rm = 0.1 and rf = 0.45 may look like this: b1000b4500, or bbb.1bb.45, where b stands for blank (space). For each likelihood to be computed, one such line of values must be provided. To terminate the set of rm,rf-values, enter 60000 as the last line. Maximum number of lines including the terminating line is equal to MT.

The following input items, no. 5.10 through no. 5.12, must be repeated for each locus in the order main locus, marker locus 1, marker locus 2, etc.

5.10. Locus description

To be read with Format as provided on input item 5.2. The following items are expected:
 - name of locus                (at most 4 characters)
 - symbol for allele 1          (at most 4 characters)
 - symbol for allele 2, etc.    (at most 4 characters)
 - symbol for phenotype 1       (at most 8 characters)
 - symbol for phenotype 2, etc. (at most 8 characters)

5.11. Gene frequencies (dummy values required with a value of 1 in col.7 of input item 5.1)

 Col. 1- 8 Population frequency of allele 1

 Col. 9-16 Population frequency of allele 2, etc. These values are read with format 10F8.4, that is, every ten numbers must be on a single line. Each number occupies at most 8 spaces with an implied decimal point between the first four and last four spaces. For example, bbbb9500 is equivalent to bbbbb.95 and represents 0.95.

5.12. Mode of inheritance

To be read with the Format as provided on input item 5.3. As many lines of input item 5.12 are expected as there are genotypes at the given locus. In the case of X-linkage, this refers to the female genotypes. For males, with X-linkage, a genotype A/A is interpreted as A/y while heterozygote genotypes such as A/a are disregarded. On each line (for each genotype), the following items are expected:
 - symbol for first allele
 - symbol for second allele; these two define a genotype
 - probability of observing phenotype 1 under the given genotype
 - probability of observing phenotype 2 under the given genotype, etc.

The above applies to loci with KONT = 0 or KONT = -1 on input item 5.8. For quantitative phenotypes (KONT = 1), four items are expected for each genotype, two alleles (defining the genotype) plus a mean and a standard deviation. For age-dependent penetrances (KONT = 2 or KONT = 3), each line must contain two allele symbols plus six parameters, ie. three parameters for females and three parameters for males (the sex-specific three parameters are defined in chapter 10). Note that in this microcomputer implementation of LIPED, all phenotypes must be read with A-Formats even though they may be quantitative measurements or age values.

The following input items, no. 5.13 through no. 5.16, are to be repeated for each pedigree except that input item 5.14b is needed only once, after the first pedigree:

5.13. Pedigree information

 Col. 1- 4 Number of individuals in pedigree. Count a "doubled" individual as 2 persons (see section on complex pedigrees).

 Col. 5- 8 Number of (pairs of) doubled individuals; = 0 for simple pedigrees

 Col. 9-68 optional remarks

5.14. Pedigree data

To be read with Format as given on input item 5.4. For each individual, the following items must be given:
                                                max.length
 - symbol identifying the individual (ID)         4
 - ID for one of the parents *)                   4
 - ID for the other of the parents *)             4
 - symbol for individual's sex                    4
 - phenotype at main locus                        8
 - phenotype at marker locus 1, etc.              8

 *) Note that each individual must either have two parents in the pedigree, or both parents' ID may be replaced by the symbol for no parent. If you have information on only one parent, you must provide an ID for the other parent who will then have unknown phenotypes.

5.14a (optional) Identification of "doubled" individuals

Applies only to complex pedigrees (number greater than zero in col. 5-8 of input item 5.13). For simple pedigrees, no input item 5.14a is expected. To be read with Format as given on input item 5.4. For each pair of doubled individuals, the following two items are required:
 - ID of first member of pair of doubled individuals
 - ID of second member of pair of doubled individuals

5.14b (optional) Haplotype frequencies

This information is needed only once, after the first pedigree, and only with a value of 1 in col. 7 of input item 5.1. The haplotype frequencies are read with format 10F8.4 (as are the gene frequencies). These values must be given in the following order. Consider a main locus and a marker locus where n is the number of alleles at the marker locus. Then, the order of that haplotype corresponding to the i-th allele at the main locus and the j-th allele at the marker locus is given by n(i – 1) + j. As an example with 2 alleles at the main locus and 3 alleles at the marker locus, the haplotypes are numbered as follows:

       j=1  j=2  j=3
  ------------------
  i=1   1    2    3
  i=2   4    5    6

 Note: there is no check that the haplotype frequencies sum to 1.

5.15. (optional) Recombination fractions

This information is needed only when IAU(i) = 1 on input item 5.9. Then, as many likelihood calculations will be carried out as there are lines of input item 5.15. Each line is read with format 2F5.4 (cf. input item 5.9a):

 Col. 1- 5 value for male recombination fraction

 Col. 6-10 value for female recombination fraction.
As the terminating line, enter 60000. For IAU(i) = 1 and i > 1, multiple sets of recombination fractions (i of them), separated by 60000, must be entered.

5.16. Direction of further analysis

 Col. 1-4 Value to determine what action to take next.

 = 5000 if new lines of input item 5.9 are to be read (allowed only if no more than one pedigree is present in this problem) thus allowing for linkage analyses between marker loci. The new lines are expected immediately after input item 5.16. On each line, there must be as many values as there are marker loci. The program will scan these values, IAU(i),i = 1,2,.., and the first marker locus with associated value IAU different from zero will be considered the new main locus. From then on, on that line, the values of IAU have the same meaning as on input item 5.9.

Multiple lines of input item 5.9 may follow a single 5000 value on input item 5.16. For example, consider a total of 5 loci, that is, one main locus (locus no. 0) and 4 marker loci (numbered 1 through 4):
 
_
 
_2_2_2_2   ← original input item 5.9
 
_
 
5000
 
_1_2_2_2   |
 
_0_1_2_2   | extra lines of input item 5.9
 
_0_0_1_3   |
 
9000

Each of these extra lines of input item 5.9 has one field for each of the original marker loci. For instance, the following extra line, _0_1_2_2 would mean: "now take marker locus 2 as the new main locus, and pair it with marker locus 3 (using option 2), then with marker locus 4 (using option 2)", which could be extended to all marker loci. Note that whenever option 1 is specified, a corresponding set of lines of input item 5.15 is expected immediately after the line containing option 1. To terminate the set of extra lines of input item 5.9, enter a line with 8000 or 9000 (same meaning as below) in col. 1-4.

 = 7000 if a new pedigree is to be read. Then, new lines of input item 5.13 etc. are expected. Note that this is allowed only when no more than one marker locus is present.

 = 8000 if a new problem is to be analyzed. Then, new lines of input item 5.1 etc. are expected.

 = 9000 to terminate this run.

6. DIMENSIONS

In the CONSTANT.INC (CONSTANT.FOR in Win version) file, constants are given which are used for dimensioning arrays. Example values are as follows.

MLIST =  50 headsibs (nuclear families)
MMARK =  30 marker loci in addition to the main locus
MNAL  =   5 alleles at any locus
MNDI  =   5 pairs of doubled individuals
MNFE  =  21 phenotypes at any locus
MNP   =  25 genotype vectors stored in memory
MNPT  = 200 individuals in a pedigree
MT    =  20 pairs of theta values after item 5.9, including the terminating 60000 line.

To change these, simply adjust the values of the constants in the parameter statements and recompile the program.

The following information is for programmers only and is not needed for general program use. With the abbreviations,
 KK  = MNAL*MNAL
 KK1 = MNAL*(MNAL+1)/2,
 KK2 = KK*(KK+1)/2,
the array dimensions are given as follows (the arrays not listed below have fixed dimensions):

FENO1(MNPT,KK1)      IAD(KK,KK)     LIST(MLIST)     PHI(KK2)
FENO2(MNPT,KK1)      IAU(MMARK)     NAL(MMARK)      PHIS(KK)
GEN(MNAL)            ID(MNPT)       NC(MNPT)        PHPROB(MNFE)
GENO(MNP,KK2)        IGENO(MNP)     NF(MMARK)       THV1(MT)
GF1(MNAL)            ISEX(MNPT)     NM(MNPT)        THV2(MT)
GF2(MNAL)            KONT(MMARK)    NS(MNPT)        THVS(MT)
GVX1(MNFE,KK1)       LDI(2,MNDI)    PHE1(MNFE)      UNK(MMARK)
GVX2(MNFE,KK1)       LGC(MNDI)      PHE2(MNFE)
HOLD(MNDI,KK2)       LGENO(MNPT)    PHEPED(MMARK)

Note: MNFE must have a value of at least 8.

7. COMPLEX PEDIGREES

In a so-called simple pedigree, tracing the inheritance of genes by going backwards through the generations (upwards in the pedigree) always leads to the same pair of founder parents.  Pedigrees for which this is not the case are called complex pedigrees.  In particular, pedigrees with the following features are examples of complex pedigrees:  (1) both members of a pair of parents have themselves parents in the pedigree;  (2) consanguinity loop, i.e., parents are related;  (3) marriage loop, e.g., two brothers are married to two sisters, or an individual has been married twice, the two spouses being related with each other but not with the individual who married twice.  An example of the last kind is pedigree 1, below, where [] refers to a female and () refers to a male:

Pedigree 1:  marriage loop

          [1]--.--(2)
               |
        .--------------.
        |              |
      (3)--.--[4]--.--(5)
           |       |
          (6)     (7)

Without special measures, LIPED analyzes only simple pedigrees. The analysis of complex pedigrees is possible by manipulating the pedigree in a certain way so that it "appears" to LIPED as a simple pedigree. This manipulation consists of replacing a particular individual by two individuals as shown in the example below (pedigree 2), and by identifying the two individuals actually corresponding to the same individual in the original pedigree (input item 5.14a). Note that such a "doubling" of individuals is necessary for breaking up loops, and also whenever more than one of the two parents has parents in the pedigree.

When an individual has been "doubled", the number of individuals in the pedigree must be increased by 1 thus counting a pair of doubled individuals as two persons. Up to MNDI individuals may be "doubled" so that, e.g., multiple consanguineous loops can be accommodated. At the end of a pedigree with doubled individuals, the two individuals corresponding to the one original person must be identified (input item 5.14a).

Pedigree 2: Example of a pedigree, manipulated for processing by LIPED

       Original pedigree                 Manipulated pedigree, acceptable to LIPED

[1.1]--.--(1.2)--.--[1.3]        [1.1]--.------(1.2)----.--[1.3]
       |         |                      |               |
       |         |                      |               |
     (2.1)--.--[2.2]                 (2.1a)  (2.1b)-.-[2.2]
            |                                       |
            |                                       |
          [3.1]                                   [3.1]

Here are some important notes regarding "doubling" of individuals:

  1. Individuals can only be doubled, not tripled. For example, an individual who is an offspring and is also married twice with children from each marriage cannot be manipulated as described.

  2. Computation time generally increases drastically with the number of pairs of doubled individuals. When one has a choice among several candidates to be doubled, it is recommended to take an individual with as much phenotypic information as possible in order to exclude as many genotypes as possible. For example, in pedigree 1, above, any one of individuals 3, 4 or 5 could be chosen for doubling. In the presence of one doubled individual, the QLIK routine for calculating the likelihood is executed for each genotype of that individual, except for those genotypes known to be incompatible with the individual's phenotypes or the phenotypes of his or her offspring. Analogously, for several pairs of doubled individuals, QLIK is called a maximum of m times, where m is calculated as follows. Let n be the number of haplotypes at the two loci jointly, i.e., n is the product of the number of alleles at the two loci under consideration. The number of joint genotypes is then given by g = n(n + 1)/2, so that m = gNDI, where NDI is the number of pairs of doubled individuals. For example, with 2 and 3 alleles at the respective two loci, one has n = 6 haplotypes and g = 21 genotypes. With NDI = 3 pairs of doubled individuals, QLIK may be called up to m = 9261 times. The present version of PC-LIPED counts these calls and displays them on the screen.

  3. Whenever the genotype of an individual can unequivocally be inferred with certainty (including phase), such an individual may be represented as multiple individuals in the pedigree if necessary, and this individual must not be counted as a so-called doubled individual (treat it as separate multiple individuals). The likelihood will then not be correct but the lod score will be unaffected by such a manipulation. For example, if an individual is known to be A/A at locus 1 and B/b at locus 2, the joint genotype is known to be AB/Ab. Note that for doubly heterozygous individuals it will not generally be possible to make use of this feature even though phase may be known, as there is no easy way to identify phases in LIPED on the basis of phenotypes.

8. MUTATION

Mutation is allowed for at the current main locus only and is assumed to occur with a constant rate from any of the alleles no. 2, 3,... towards the first allele, with the mutation rate being specified in col. 8-20 of input item 5.1. Backmutation is assumed to be negligible. Also, in the computation of the likelihood, it is assumed that a mutation occurs only in one or the other of two parents, but not simultaneously in both parents.

WARNING: when processing a disease locus with mutation and subsequently, in the same run, testing marker versus marker, then the mutation rate keeps applying to the current main locus unless a new run is carried out with the mutation rate set equal to zero.

Note that only simple pedigrees can be processed by LIPED unless special steps are taken to code for a complex pedigree (see that section above).

9. QUANTITATIVE PHENOTYPES

At any locus, quantitative rather than qualitative phenotypes can be read. For a locus with quantitative phenotypes, the following special rules must be observed.

Input
item  Explanation
----------------------------------------------------------------
   3  With two F-formats, read one mean and one standard deviation for each genotype.

   7  Set the number of phenotypes equal to 2.  The program will correct wrong numbers.

   8  Set KONT equal to 1.

  10  Two phenotype symbols will be read by the program but they are not used in any way.

  12  After the symbol for the second allele, two items are expected, the mean and the standard

 deviation of the phenotype distribution given the particular genotype specified by the two alleles.


  14  The phenotype values must not occupy more than 8 spaces each.
----------------------------------------------------------------

10. AGE-DEPENDENT PENETRANCE

10.1  Age classes with different penetrances

Age-dependant penetrance refers to the fact that a carrier of a disease gene may not exhibit the disease at birth but only later in life, that is, the penetrance (= probability of showing a certain phenotype given a genotype) depends on the age of an individual.

The easiest way of implementing age-dependant penetrance is by forming age classes and having different penetrances in these classes. For affecteds, irrespective of their age, only one class is required (provided that no phenocopies are allowed for), but unaffecteds must be grouped into age classes. For example, in a given disease, if all gene carriers beyond 10 years of age have expressed the disease, a suitable assumption is that penetrance rises linearly from 0 at age 0 to 100% at age 10, as pictured below:

  Penetrance
        |
     1  |     -------------
        |   /
        |  /
        | /
        |/
     0  +------------------age
        0    10    20

One might then form 6 classes as follows, where AFF stands for the 'affected' phenotype, and NA1, NA2, etc. stands for unaffected in age class 1, 2, etc.; NA5 denotes unaffected individuals older than 10 years who are taken to be known not to carry the disease gene. The disease is assumed dominant, the disease being T. Note that the probability of being unaffected is 1 minus the probability of being affected.

 -------------------------------------------
                         Phenotypes
                ----------------------------
 Genotype       AFF  NA1  NA2  NA3  NA4  NA5
 -------------------------------------------
   T T           1   .88  .63  .38  .13   0
   T t           1   .88  .63  .38  .13   0
   t t           0    1    1    1    1    1
 -------------------------------------------

10.2  Age-of-onset distributions

Rather than forming age classes, the distribution of the age at disease onset may be assumed to follow a certain distribution. In LIPED, two such distributions are implemented, the lognormal and a straight-line distribution. Below, F denotes the distribution (cumulative sum) of age at onset whereas f denotes the corresponding density (histogram).

Whatever the age of onset distribution used, to represent in a single number the various pieces of phenotypic information (age at onset, present age, affection status) at a disease locus, the following conventions must be observed in LIPED. In principle, the phenotype to be provided in the input to LIPED is an individual's present age (or age last seen) or the age at onset, taken with a minus sign for unaffecteds, and taken to be positive for affecteds. Present age and age at onset are distinguished as outlined, below.

UNAFFECTED INDIVIDUALS

The phenotype is an individuals present age (or age last seen), taken with a minus sign (the sign distinguishes affecteds from unaffecteds). Example: unaffected, present age is 56, phenotype given in program is -56. If present age is unknown, a guess must be used, for example, based on ages of sibs or parents.

AFFECTED INDIVIDUALS

If actual age at disease onset is unknown, the phenotype is a person's present age. Example: 56. If present age is unknown, a guess must be used, based on ages of relatives.

If age at disease onset is known, it is entered into LIPED by the following coding scheme: The phenotype to be provided is obtained by adding 500 to the age at onset. Example: age at disease onset is 23 years; phenotype to be provided is 523.

NOTE: actual age at disease onset is relevant only when disease can occur under different genotypes with different penetrances. If this is not so (it usually is not), then present age may be given for all affecteds.

UNKNOWN DISEASE STATUS

The phenotype is given as 0. Alternatively, on input item 5.5, one may define any other code for unknown phenotype, for example, blank.

10.3 Lognormal distribution of age of onset

It is often meaningful to assume that age of onset is lognormally distributed, that is, that LN(age of onset) follows a normal distribution where LN denotes natural logarithm (a simpler assumption for age-of-onset distribution is covered in section 10.4, below). Mean and standard deviation for the lognormal and normal distributions are defined and connected with each other as follows:

            Age (orig. values)        LN(age)
            (lognormal distr.)     (normal distr.)
--------------------------------------------------
mean              μ                    u
std. dev.         σ                    s
--------------------------------------------------

For ease of presentation, define m = exp(u) and w = exp(s2). Then one has:

μ = m √w
σ = m √[w(w – 1)] = μ √(w – 1)

u = 2 LN(μ) – 0.5 LN(μ2 + σ2)
s = √[LN(μ2 + σ2) – 2 LN(μ)] = √[2{LN(μ) – u}],

where LN denotes natural logarithm. Also, with given mean, μ, of the raw data, and standard deviation, s, of the transformed data, one obtains the mean of the transformed data as u = LN(μ) – 0.5s2.

Some example values are given in the following table:

     --------------------------------------------
     Original scale      LN scale (normal distr.)
     -------------       ------------------------
        μ      σ             u          s
     --------------------------------------------
       20      5            2.97       0.25
       20     10            2.88       0.47
       20     15            2.77       0.67

       40      5            3.68       0.12
       40     10            3.66       0.25
       40     15            3.62       0.36
     --------------------------------------------

The LOGNORM program (included) transforms values u and s into the corresponding values of μ and σ, and vice versa.

If age at onset for an (affected) individual is known, the corresponding likelihood is simply f(age at onset), where f is the lognormal density. If age at onset is unknown, then the likelihood is F(age) where 'age' denotes current age, or age last seen, and F is the lognormal distribution function. For unaffecteds, the likelihood is equal to 1 – F(age). If the final penetrance, t, is less than 100% then f and F above are multiplied by t.

Lognormal age-dependent penetrance is modeled in analogy to quantitative phenotypes (see previous section) except that here, 6 parameters must be specified (3 for females and 3 for males). For each genotype (input item 5.3), these are
  - the mean, u, of LN(age of onset)
  - the standard deviation, s, of LN(age of onset)
  - the limiting penetrance, t, when age is very high, for females, followed by the analogous three parameters for males.

Depending on the values of the 6 parameters given (input item 5.3) for each genotype, the following 2 situations can be distinguished. Assume a disease locus with two alleles, a dominant disease allele, D, and a normal allele, d.

1. Age of onset follows a lognormal distribution with parameters u and s, where the final penetrance attained (at high age) is equal to t. For example (parameters taken to be the same for males and females), one may have on input item 5.3:

D D 3.35 0.17 1.0 3.35 0.17 1.0 → u = 3.35, s = 0.17, final penetrance 100%

D d 3.35 0.17 0.6 3.35 0.17 0.6 → susceptible individuals express disease with max. penetrance of 60% when they are very old

d d  3.0 0.1 0.0  3.0 0.1 3.0 → genotype d/d not susceptible to disease; values of u and s are irrelevant (likelihood is zero for affecteds and 1 for unaffecteds)

2. Penetrance does not depend on age but is a fixed value (t for affecteds, 1-t for unaffecteds). To accommodate this situation, set s = 0.0. The value of u is then irrelevant. For example, one may have

d d  0.0 0.0 0.01  0.0 0.0 0.01 → t = 0.01, that is, d/d genotypes express the disease with probability 1%, irrespective of age (likelihood is 0.01 for affecteds and 0.99 for unaffecteds); the value of u is irrelevant. This case should be used with great care since it does not differentiate between age of onset known or unknown.

In summary, for a locus with age-dependent (lognormal) penetrance, the following special rules must be observed.

Input
item Explanation
-----------------------------------------------------------
  3 Provide six F-formats to read on each line (genotype) one mean, one standard deviation and one final penetrance for each sex (note that these Formats will apply to all loci).

  7 Set the number of phenotypes equal to 1 (the program will set the correct number of phenotypes).

  8 Set KONT equal to 2.

 10 Six phenotype symbols will be read by the program, but they are not used in any way.

 12 After the symbol for the second allele, six items are expected: mean, standard deviation and final penetrance for females, and the analogous three parameters for males.

 14 The value for the phenotype (age) must not occupy more than 8 spaces (4 recommended), the actual number of spaces used being determined by the format statement given in input item 5.4. A positive age value refers to an affected individual, a negative age figure identifies an unaffected individual. Phenotypes are coded following the rules given in section 10.2, above.
-------------------------------------------------------------

10.4  Straight-line curves for age of onset (locus type 3)

       F
      
|
    
t |          --------
      
|        /.
      
|       / .
      
|      /  .
      
|     /   .
      
|    /    .
    
0 +------------------ age
          
A1    A2

F is the probability of being affected, that is, the penetrance (or likelihood) is equal to F for an affected individual (age at onset unknown) and equal to 1 – F for an unaffected individual. According to the figure, above, the age-of-onset curve is defined as

    
/ 0                   if a ≤ A1
F = |  t(a - A1)/(A2 - A1) if A1 < a < A2
    
\ t                   if a ≤ A2

where "a" is an individual's present age, or age last seen. If age at onset is known (for an affected individual) then the likelihood (density) is equal to f = t/(A2 – A1) if the age of onset is between A1 and A2, and equal to zero otherwise. If age at onset is considered a random variable, according to the present definition and with t = 1, it follows a uniform distribution with mean (A2 – A1)/2 and standard deviation (A2 – A1)/3.464.

For a locus of type 3 (straight line age of onset), coding is very similar to the conventions used for lognormal age of onset (specific instructions are given below). The phenotypes are the ages of each individual, taken to be positive for affected individuals and taken with a minus sign for unaffected individuals. Zero will be interpreted as unknown, but any other symbol may also be designated to represent unknown phenotype. For affected individuals with known age at onset, enter a number equal to 500 plus age at onset as the phenotype (see section 10.2, above).

As in lognormal age-dependent penetrance, 6 parameters must be specified but here, they have the following meaning. For each genotype (input item 5.3), they are (see graph, above)
 - the age, A1, at which penetrance becomes positive
 - the age, A2, at which penetrance reaches its final values
 - the limiting penetrance, t, when age is very high, for females, followed by the analogous three quantities for males.

Depending on the values of the 6 parameters given (input item 5.3) for each genotype, the following 2 situations can be distinguished. Assume a disease locus with two alleles, a dominant disease allele, D, and a normal allele, d.

1. Age of onset follows a straight-line distribution with parameters A1 and A2, where the final penetrance attained (at high age) is equal to t. For example (parameter values taken to be the same for females and for males), one may have on input item 5.3:

D D  10 60 1.0  10 60 1.0 for individuals with D/D genotype, susceptibility to disease starts at age 10 and penetrance reaches its maximum of 100% at age 60.

D d  10 60 0.6  10 60 0.6 susceptible individuals express disease with max. penetrance of 60% when they are 60 years or older.

d d  10 11 0.0  10 11 0.0 genotype d/d not susceptible to disease; values of A1 and A2 are irrelevant (given genotype d/d, likelihood is zero for affecteds and 1 for unaffecteds).

2. Penetrance does not depend on age but is a fixed value (t for affecteds, 1 – t for unaffecteds). To accommodate this situation, set A2 = 0.0. The value of A1 is then irrelevant. For example (same parameter values for males and females), one may have

d d  0.0 0.0 .01  0.0 0.0 .01 t = 0.01, that is, d/d genotypes express the disease with probability 1%, irrespective of age (likelihood is 0.01 for affecteds and 0.99 for unaffecteds); the value of A1 is irrelevant. This case should be used with great care since it does not differentiate between age of onset known or unknown.

In summary, for a locus with straight-line age-dependent penetrance, the following special rules must be observed.

Input
item   Explanation
-----------------------------------------------------------
 
3  On each line (genotype) provide six F-formats to read one starting age (A1), one finishing age (A2) and one final penetrance (t) for each sex. Note that these Formats will apply to all loci.

 
7  Set the number of phenotypes equal to 1 (the program will set the correct numbers).

 
8  Set KONT equal to 3.

 
10  Six phenotype symbols will be read by the program, but they are not used in any way.

 
12  After the symbol for the second allele, six items are expected: starting age (A1), finishing age (A2) and final penetrance (t) for females, and the analogous three parameters for males.

 
14  The value for the phenotype (age) must not occupy more than 8 spaces (4 recommended), the actual number of spaces used being determined by the format statement given in input item 5.4. A positive age value refers to an affected individual, a negative age figure identifies an unaffected individual. Phenotypes are coded following the rules given in section 10.2, above.
-----------------------------------------------------------

11. CALCULATION OF GENETIC RISKS

To calculate conditional genotype probabilities for a specific individual, given all the family data, one must carry out several likelihood computations and combine their results as follows. For example, consider an individual with phenotype 'unaffected' and penetrances as given in the table below, where D is the disease allele and d is the normal allele at the main locus.

 -------------------------------------
            Penetrance for phenotypes
           ---------------------------
 Genotype  affected  unaffected    XDd
 -------------------------------------
   D D       0.9        0.1         0
   D d       0.6        0.4        0.4
   d d        0          1          0
 -------------------------------------

For this unaffected individual, one wants to compute the risk that he or she has genotype D/d. To obtain this risk, one runs LIPED twice, each time with a different phenotype assigned to this individual, that is, in run 1, the individual has phenotype unaffected, and in run 2, the individual has phenotype XDd. Denote the resulting likelihoods (not lod scores) by L(ua) and L(XDd). Then, the risk to this individual of having genotype D/d is given by L(XDd)/L(ua). Note that other programs, such as the MLINK program of Dr. Mark Lathrop, can compute genetic risks directly.

With X-linked recessive deleterious traits, for a female founder individual (no parents in pedigree), the prior probability, q, of being a carrier of the disease gene is a multiple of the mutation rate, u. For example, in Duchenne muscular dystrophy (DMD), q = 4u (Murphy and Chase, "Principles of Genetic Counseling"). In the likelihood calculation of pedigree data, on the other hand, the prior probability of a founder's genotype is determined solely by the gene frequency, p. For example, the prior probability that a founder is heterozygous is given by 2p(1 – p). Therefore, to implement the prior probability, q, that a woman is heterozygous for an X-linked recessive deleterious gene, in the likelihood calculation, one must choose the gene frequency of the deleterious gene, p, such that q = 2p(1 – p) or, approximately, p = q/2 (in DMD, thus, p = 2u).

12. LIKELIHOOD AT A SINGLE LOCUS

In some applications, the likelihood at a single (disease) locus only is needed. For example, one may want to estimate from family data gene frequencies or age-of-onset parameters at a single locus. In LIPED, single-locus calculations are accom-modated easiest by defining a dummy second locus with a single allele of frequency 1.

13. HELPFUL HINTS

In a pedigree to be processed by LIPED, any individual must have either both parents in the pedigree, or be a founder individual, that is, have both parents unknown (not in pedigree). Note that siblings cannot be recognized as such unless their parents are also in the pedigree. If such parents are not actually known, they will must still be present in the pedigree, possibly with all phenotypes coded as unknown.

When there is at least one known recombination in a pedigree but the value of the recombination fraction is set equal to zero, then the likelihood will be equal to zero, and the log likelihood equal to -∞. On output, -∞ is represented as -99.99.

When the likelihood is equal to zero, either because recombinants are present while the recombination fraction (θ, theta) is set to zero or because of a genetic inconsistency (incompatible genotypes of some individuals), LIPED will report this with the message, "L(rm, rf = 0 at rm, rf =...", and will print the male and female recombination fractions at which the likelihood is zero, and sequential number and ID code of the individual at which this was first detected. An incompatibility then exists among the indicated individual and his or her spouse(s) and descendants. Note that additional incompatibilities not yet detected may exist in the given pedigree. Using θ = 0, this helps to find recombinations in a pedigree. In families with loops (eg, inbreeding), this scheme does not work, and LIPED will report a zero likelihood only at θ = 0.5 but cannot pinpoint where this was first detected.

In the locus descriptions, when there are several alleles, the number of possible phenotypes may become quite large. For the analysis, however, it is not necessary to list all phenotypes that might possibly occur. One only needs to identify those phenotypes that are actually present in at least one family member.

When you request output on a disk file, the input used will be appended to the output file. If you do not want this, it is easiest to proceed as follows. Pretend running an additional problem, that is, your last input line is 8000 rather than 9000, and add an additional input line containing 0 (= number of marker loci) in column 2. LIPED will then stop without appending the input file to the output file. This method is used in example 5 (file EX5.DAT), below.

14. EXAMPLES

Eample 1

Th EX1.DAT file contains input corresponding to a family pedigree with the structure as shown in section 7, "Complex pedigrees" (pedigree 2), above. Two dominant loci are used with two alleles each, where A > a and B > b. The gene frequencies are p = P(A) = 0.4 and q = P(B) = 0.3. Calculation of the pedigree likelihood by first principles yields

L(θ) = (1 – p)5 p(1 – q)3 q2 (1 – θ)[1 + q(1 – θ)2 + qθ2]/8,

where θ denotes the recombination fraction. With this, one obtains, for example, log [L(0.5)] = –4.16106927 and log[L(0.2)] = –3.93702064, which agrees with the output given by LIPED. The lod score at θ = 0.2 is thus 0.224.

Example 2

The EX2.DAT file shows an example with 3 pedigrees and the use of output option 9, ie, summation of lod scores over pedigrees. The second pedigree in this set of 3 pedigrees requires much more computer time per lod score than either pedigree 1 or 3.

Example 3

Compare linkage relationships among 6 gene markers, here labelled main locus and marker loci 1 through 5. The first comparison made takes more computer time per lod score than the other comparisons. This example shows the use of various output options in combination with locus comparisons.

Example 4

The EX4.DAT file shows a published pedigree with Norrie's disease (X-linked recessive) and 2 marker loci. Analysis is disease versus each marker and marker versus marker.

Example 5

Mode of inheritance of disease locus in example 1 is changed such that penetrance rises linearly from age 0 to 10. As no individual is in that age range, the output is the same as in example 1.

15. LITERATURE

Ban Y, Davies TF, Greenberg DA, Concepcion ES, Tomer Y (2002) The influence of human leucocyte antigen (HLA) genes on autoimmune thyroid disease (AITD): results of studies in HLA-DR3 positive AITD families. Clin Endocrinol (Oxf) 57:81-88 [a recent example for the use of the LIPED program]

Cheung KH, Nadkarni P, Silverstein S, Kidd JR, Pakstis AJ, Miller P, Kidd KK (1996) PhenoDB: an integrated client/server database for linkage and population genetics.
Comput Biomed Res 29:327-337 [an example of the use of the LIPED program]

Elston RC, Stewart J (1971) A general model for the analysis of pedigree data.
Hum Hered 21:523-542

Ott J (1974) Estimation of the recombination fraction in human pedigrees: efficient computation of the likelihood for human linkage studies.
Am J Hum Genet 26:588-597

Ott J, Schrott HG, Goldstein JL, Hazzard WR, Allen FH Jr, Falk CT, Motulsky AG (1974) Linkage studies in a large kindred with familial hypercholesterolemia.
Am J Hum Genet 26, 598-603

Ott J (1976) A computer program for linkage analysis of general human pedigrees.
Am J Hum Genet 28:528-529

Ott J (1986) Y-linkage and pseudoautosomal linkage.
Am J Hum Genet 38:891-897

Ott J (1999)
Analysis of Human Genetic Linkage, 3rd edition. Johns Hopkins University Press, Baltimore

Schrott HG, Goldstein JL, Hazzard WR, McGoodwin MM, Motulsky AG (1972) Familial hypercholesterolemia in a large kindred. Evidence for monogenic mechanism.
Annals of Internal Medicine 76:711-720 [description of the Alaska pedigree]

Terwilliger JD, Ott J (1994)
Handbook of Human Genetic Linkage. Johns Hopkins University Press, Baltimore

Thompson E (2011) The structure of genetic linkage data: From LIPED to 1 M SNPs.
Hum Hered 71:86-96