1. ClinGen Gene Curation: Segregation Analysisok
Biocurator Training Module
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2. Segregation Analysis Training Module
The use of segregation studies in which family members are genotyped to determine if a variant co-segregates with a disease can be a powerful piece of evidence to support or refute a gene-disease relationship.
Expert reviewers may choose to specify the most appropriate way to approach segregation scoring within their disease domain.
If a LOD score has been provided by the authors, this should be documented and used to assign points in the segregation matrix.
If a LOD score has been provided by the authors, do NOT use the formula(s) in the SOP to estimate a new LOD score.
If the curator/expert do not agree with the published LOD score, do not assign any points and record your reasoning.
To complete this module, you will need:
The Gene Clinical Validity Curation Process Standard Operating Procedure, Version 5 ok
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3. + + + + - + - + Click here when done I-1 I-2 II-1 II-4 II-6 II-9 II-2
Estimating a LOD score for an autosomal dominant condition with adult onset
On the pedigree below, click on the individuals that you would use to count a segregation, starting from individual I-1. Filled-in circles/squares represent affected individuals. + indicates genotype positive individuals. Tip: before clicking on a circle or square, make sure you see the hand icon while hovering over it.
I-1
I-2 +
II-1
II-4
II-6
II-9 + +
II-2
II-5
(+)
II-7
II-8 +
II-10
II-3
III-4
III-7 - + - +
III-1
III-2
III-3
III-6
III-5
III-8 ok
Click here when done

4. + + + + - + - + I-1 I-2 II-1 II-4 II-6 II-9 II-2 II-5 (+) II-7 II-8
Estimating a LOD score for an autosomal dominant condition with adult onset
On the pedigree below, click on the individuals that you would use to count a segregation, starting from individual I-1. Filled-in circles/squares represent affected individuals. + indicates genotype positive individuals. Tip: before clicking on a circle or square, make sure you see the hand icon while hovering over it.
I-1
I-2 +
II-1
II-4
II-6
II-9 + +
II-2
II-5
(+)
II-7
II-8 +
II-10
II-3
III-4
III-7 - + - +
III-1
III-2
III-3
III-6
III-5
III-8 ok
Although the disease state and genotype of individual II-6 is unknown, he is counted as an obligate carrier since he is between two affected, genotype positive individuals (I-1 and III-5). There are 4 segregations from individual I-1 (II-1, II-4, II-6, and II-9) and another from II-6 to III-5. Individual III-8 is genotype positive but is not affected, so you would not count that as a segregation. Individual III-1 is affected but was not genotyped, so you would not count that as a segregation either. The estimated LOD score for 5 segregations is 1.5.
Z(LOD score) = 𝑙𝑜𝑔 (0.5) 5 = 1.5
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5. + + + + + + + + - + - + + - - - + + Click here when done I-1 I-2 I-1
Summing LOD scores across families for an autosomal dominant condition
Before assigning points for segregation, you should sum the LOD scores from all families meeting size requirements. On the pedigrees below, click on the individuals that you would use to count a segregation, starting from the proband. Filled-in circles/squares represent affected individuals. + indicates genotype positive individuals. +
I-1
I-2
I-1
I-2
I-1
I-2
II-1 + + + + + +
II-1
II-2
II-3
II-2
II-3
II-5
II-4
II-6 +
II-1
II-2
III-1
III-2
III-4 - +
III-3 -
III-1 +
III-2 +
III-3 - - -
III-4
III-5
III-6 ok
IV-1 + +
IV-2
Click here when done

6. Z(LOD score) = 𝑙𝑜𝑔 10 1 (0.5) 4 = 1.2 (x2) = 2.4
Summing LOD scores across families for an autosomal dominant condition
Before assigning points for segregation, you should sum the LOD scores from all families meeting size requirements. On the pedigrees below, click on the individuals that you would use to count a segregation, starting from the proband. Filled-in circles/squares represent affected individuals. + indicates genotype positive individuals. +
I-1
I-2
I-1
I-2
I-1
I-2
II-1 + + + + + +
II-1
II-2
II-3
II-2
II-3
II-5
II-4
II-6 +
II-1
II-2
III-1
III-2
III-4 - +
III-3 -
III-1 +
III-2 +
III-3 - - -
III-4
III-5
III-6 ok
IV-1 + +
IV-2
The first two families have 4 segregations each. The third family has 2 segregations and cannot be included (there must be 4 or more segregations to count). The estimated LOD score for each of the first two families is 1.2, for a total of 2.4. According to Figure 6 in version 5 of the SOP, you would assign point depending on the sequencing method(s) used.
Z(LOD score) = 𝑙𝑜𝑔 (0.5) 4 = 1.2 (x2) = 2.4
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7. Z(LOD score) = 𝑙𝑜𝑔 10 1 (0.25) 3−1 (0.75) 5 = 1.82
Estimating a LOD score for an autosomal recessive condition
For each pedigree presented, indicate whether or not you would use the family to calculate an estimated LOD score using the formula in the SOP. Filled-in circles/squares represent affected individuals.
I-1
I-2
I-1
I-2
I-1
I-2
mut/WT
mut/WT
mut/WT
mut/WT
II-1
II-2
II-3
II-4
II-5
II-6
II-7
II-8
II-1
II-2
II-3
II-4
II-5
II-1
II-2
WT/WT
mut/mut
mut/WT
a/b
mut/mut
mut/WT
mut/mut
mut/WT
mut/WT
WT/WT
mut/mut
a/b
a/b
WT/WT
WT/WT
YES NO
YES NO
YES NO
At least 3 affected individuals must be present in the family to count.
There are 3 affected individuals and 5 unaffected individuals, all genotyped. You would not use the parents in the calculation of the estimated LOD score because unaffected individuals can only be counted if at the same risk as the affected individuals to inherit 2 altered alleles. Using the formula in the SOP:
Z(LOD score) = 𝑙𝑜𝑔 (0.25) 3−1 (0.75) 5 = 1.82
There are 3 affected individuals who have been genotyped. However, the parents were not genotyped so you cannot be sure that the variants are in trans without further information. ok
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8. Not enough affected individuals
Summing LOD scores across families for an autosomal recessive condition
Before assigning points for segregation, you should sum the LOD scores from all families meeting size requirements. On the pedigrees below, click on the individuals that you would use to estimate a LOD score. Filled-in circles/squares represent affected individuals.
I-1
I-2
I-1
I-2
I-1
I-2
WT/a
WT/a
WT/a
WT/a
WT/a
WT/a
a/a
a/a
a/a
WT/a
WT/WT
a/a
a/a
WT/a
WT/a
a/a
a/a
a/a
II-1
II-2
II-3
II-4
II-1
II-2
II-3
II-1
II-2
II-3
II-4
II-5
3 affected, 1 unaffected
Z(LOD score) = 𝑙𝑜𝑔 (0.25) 3−1 (0.75) 1
Z(LOD score) = 1.32
Not enough affected individuals
to count this family
3 affected, 2 unaffected
Z(LOD score) = 𝑙𝑜𝑔 (0.25) 3−1 (0.75) 2
Z(LOD score) = 1.45
See Calculations
Total estimated LOD score = 2.75; Assign segregation point depending on the sequencing method(s) used.
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9. Click here when done + - + + + + - + + + + +
Estimating a LOD score for an X-linked condition (X-linked recessive with mild expression in some females)
On the pedigree below, starting with individual I-1, click on the individuals that you would use to estimate a LOD score. Filled-in circles/squares represent affected individuals (dark) and mildly affected individuals (light). + indicates genotype positive individuals. X* indicates the affected allele.
I-1 + -
I-2
XX* XY
Obligate carrier + + XY
XX*
X*Y
(XX*) XY
II-1
II-2
II-3
II-4
II-5 + + - + +
XX*
X*Y XY XX
X*Y
XX* XY
III-1
III-2
III-3
III-4
III-5
III-6
III-7
Do not count if unaffected
Do not count if unaffected + + +
XX*
XX*
X*Y
IV-1
IV-2
IV-3
Do not count because she had a 100% chance of inheriting the affected allele from her father
Click here when done

10. + - + + + + - + + + + + > NEXT
Estimating a LOD score for an X-linked condition (X-linked recessive with mild expression in some females)
On the pedigree below, starting with individual I-1, click on the individuals that you would use to estimate a LOD score. Filled-in circles/squares represent affected individuals (dark) and mildly affected individuals (light). + indicates genotype positive individuals. X* indicates the affected allele. + -
I-1
I-2
There are 8 segregations. Segregations can be counted as long as there is an equal probability of inheriting either X from the mother. Individual IV-1 was not counted because she could have only gotten the affected X from the father (III-5). Using the formula from the SOP for autosomal dominant/X-linked diseases:
Z(LOD score) = 𝑙𝑜𝑔 (0.5) 8 = 2.4
XX* XY
Obligate carrier + + XY
XX*
X*Y
(XX*) XY
II-1
II-2
II-3
II-4
II-5 + + - + +
XX*
X*Y XY XX
X*Y
XX* XY
III-1
III-2
III-3
III-4
III-5
III-6
III-7
Do not count if unaffected
Do not count if unaffected + + +
XX*
XX*
X*Y
IV-1
IV-2
IV-3
Do not count because she had a 100% chance of inheriting the affected allele from her father
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11. Email Jen McGlaughon at jen_mcglaughon@med.unc.edu
Questions or comments?
Jen McGlaughon at