Re: Another attempt to explain genetics

[Guest guest]

Dave, you are amazing! I am in awe. Thank you for making that make sense. Cheryl

ID PSC/AIH, UC, fibromyalgia,etc.

Subject: Another attempt to explain genetics

> Here's another attempt to explain genetics, and the difference

> between a simple genetic disorder (1 gene) and a complex genetic

> disorder (3 genes).

>

> So my guess would be that there may be 3 susceptibility genes

> involved in PSC using this recessive gene model. Note that the

> disease would appear to spring out of nowhere ... neither parent has

> the disease.

>

> Best regards,

>

> Dave

> (father of (22); PSC 07/03; UC 08/03)

>

>

>

[Guest guest]

Dave, you are amazing! I am in awe. Thank you for making that make sense. Cheryl

ID PSC/AIH, UC, fibromyalgia,etc.

Subject: Another attempt to explain genetics

> Here's another attempt to explain genetics, and the difference

> between a simple genetic disorder (1 gene) and a complex genetic

> disorder (3 genes).

>

> So my guess would be that there may be 3 susceptibility genes

> involved in PSC using this recessive gene model. Note that the

> disease would appear to spring out of nowhere ... neither parent has

> the disease.

>

> Best regards,

>

> Dave

> (father of (22); PSC 07/03; UC 08/03)

>

>

>

[Guest guest]

Dave, Thanks for that! It really is not all that different from

balanced/unbalanced

translocations. Thanks for that explanation -- it make sense to me now.

There is an auto-immune component in my extended family. I was the 1st person

dx'ed

with any kind of auto-immune disease, but my mom and her sisters all have

various auto-

immune thyroid issues, as do two of their female 1st cousins who are related

paternally.

One of my aunts has systemic lupus and another has PBC. We are an

epidemiologist's

dream!

Colleen

> Only the aa, bb, cc child (1 in 64 = 1.56%) would develop PSC.

> Because the prevalence of the disease in the general population is

> estimated to be 8 to 14 per 100,000 persons = 0.008% to 0.014%

> the " heritability coefficient " would be about 100 if you use the

> upper value of 0.014% (the " heritability coefficient " equals the

> prevalence among siblings divided by the general population

> prevalence). This is similar to the data described by Karlsen for PSC

> (Karlsen TH, Schrumpf E, Boberg KM 2007 Genetic epidemiology of

> primary sclerosing cholangitis. World J. Gastroenterol. 13: 5421-

> 5431).

>

> So my guess would be that there may be 3 susceptibility genes

> involved in PSC using this recessive gene model. Note that the

> disease would appear to spring out of nowhere ... neither parent has

> the disease.

[Guest guest]

,

But if one parent has PSC (ie they have aa, bb, cc) then that parent will be contributing bad genes and the chances are a lot higher that the one well parent will contribute bad genes and that their children will get it. I suspect its more than four genes.

Ian (52) PSC 89

Here's another attempt to explain genetics, and the difference between a simple genetic disorder (1 gene) and a complex genetic disorder (3 genes).Suppose that two parents both carry 1 copy each of a recessive/mutant

gene for a disease (such as cystic fibrosis), but have 1 dominant/normal copy of the gene. Both parents would be disease-free because you need 2 copies of the recessive/mutant gene to develop cystic fibrosis. The parents would be " carriers " . Using CFTR to

denote the normal or dominant gene, and cftr to denote the recessive/mutant gene, the two parents would each be of genotype CFTR/cftr.Their children can randomly inherit one version of the gene from each

parent. So the two parents could have the following types of children: CFTR/CFTR, CFTR/cftr, cftr/CFTR or cftr/cftr. Only the cftr/cftr children (1 in 4, or 25%) would develop cystic fibrosis.In this example, the prevalence of the disease amongst siblings would

be 25%, whereas the prevalence in the general population is 1 in 2,500 to 3,500 Caucasian newborns ( = 0.03 to 0.04%), so the " heritability coefficient " would be about 700 (i.e. 25 divided by 0.035).

Consider now that to develop PSC one needed 3 different mutant genes. Let's use A, B and C to denote the normal/dominant versions of the genes, and a, b and c to denote the mutant/recessive versions of the

genes. Let's suppose that both parents were Aa, Bb, Cc. The children could be any of these 64 possibilities:AA, BB, CCAA, BB, CcAA, BB, cCAA, BB, ccAA, Bb, CCAA, Bb, CcAA, Bb, cC

AA, Bb, ccAA, bB, CCAA, bB, CcAA, bB, cCAA, bB, ccAA, bb, CCAA, bb, CcAA, bb, cCAA, bb, ccAa, BB, CCAa, BB, CcAa, BB, cCAa, BB, ccAa, Bb, CCAa, Bb, CcAa, Bb, cC

Aa, Bb, ccAa, bB, CCAa, bB, CcAa, bB, cCAa, bB, ccAa, bb, CCAa, bb, CcAa, bb, cCAa, bb, ccaA, BB, CCaA, BB, CcaA, BB, cCaA, BB, ccaA, Bb, CCaA, Bb, CcaA, Bb, cC

aA, Bb, ccaA, bB, CCaA, bB, CcaA, bB, cCaA, bB, ccaA, bb, CCaA, bb, CcaA, bb, cCaA, bb, ccaa, BB, CCaa, BB, Ccaa, BB, cCaa, BB, ccaa, Bb, CCaa, Bb, Ccaa, Bb, cC

aa, Bb, ccaa, bB, CCaa, bB, Ccaa, bB, cCaa, bB, ccaa, bb, CCaa, bb, Ccaa, bb, cCaa, bb, ccOnly the aa, bb, cc child (1 in 64 = 1.56%) would develop PSC. Because the prevalence of the disease in the general population is

estimated to be 8 to 14 per 100,000 persons = 0.008% to 0.014% the " heritability coefficient " would be about 100 if you use the upper value of 0.014% (the " heritability coefficient " equals the

prevalence among siblings divided by the general population prevalence). This is similar to the data described by Karlsen for PSC (Karlsen TH, Schrumpf E, Boberg KM 2007 Genetic epidemiology of primary sclerosing cholangitis. World J. Gastroenterol. 13: 5421-

5431).So my guess would be that there may be 3 susceptibility genes involved in PSC using this recessive gene model. Note that the disease would appear to spring out of nowhere ... neither parent has the disease.

Best regards,Dave (father of (22); PSC 07/03; UC 08/03) -- Ian Cribb former P.Eng. (resigned in good standing)

[Guest guest]

,

But if one parent has PSC (ie they have aa, bb, cc) then that parent will be contributing bad genes and the chances are a lot higher that the one well parent will contribute bad genes and that their children will get it. I suspect its more than four genes.

Ian (52) PSC 89

Here's another attempt to explain genetics, and the difference between a simple genetic disorder (1 gene) and a complex genetic disorder (3 genes).Suppose that two parents both carry 1 copy each of a recessive/mutant

gene for a disease (such as cystic fibrosis), but have 1 dominant/normal copy of the gene. Both parents would be disease-free because you need 2 copies of the recessive/mutant gene to develop cystic fibrosis. The parents would be " carriers " . Using CFTR to

denote the normal or dominant gene, and cftr to denote the recessive/mutant gene, the two parents would each be of genotype CFTR/cftr.Their children can randomly inherit one version of the gene from each

parent. So the two parents could have the following types of children: CFTR/CFTR, CFTR/cftr, cftr/CFTR or cftr/cftr. Only the cftr/cftr children (1 in 4, or 25%) would develop cystic fibrosis.In this example, the prevalence of the disease amongst siblings would

be 25%, whereas the prevalence in the general population is 1 in 2,500 to 3,500 Caucasian newborns ( = 0.03 to 0.04%), so the " heritability coefficient " would be about 700 (i.e. 25 divided by 0.035).

Consider now that to develop PSC one needed 3 different mutant genes. Let's use A, B and C to denote the normal/dominant versions of the genes, and a, b and c to denote the mutant/recessive versions of the

genes. Let's suppose that both parents were Aa, Bb, Cc. The children could be any of these 64 possibilities:AA, BB, CCAA, BB, CcAA, BB, cCAA, BB, ccAA, Bb, CCAA, Bb, CcAA, Bb, cC

AA, Bb, ccAA, bB, CCAA, bB, CcAA, bB, cCAA, bB, ccAA, bb, CCAA, bb, CcAA, bb, cCAA, bb, ccAa, BB, CCAa, BB, CcAa, BB, cCAa, BB, ccAa, Bb, CCAa, Bb, CcAa, Bb, cC

Aa, Bb, ccAa, bB, CCAa, bB, CcAa, bB, cCAa, bB, ccAa, bb, CCAa, bb, CcAa, bb, cCAa, bb, ccaA, BB, CCaA, BB, CcaA, BB, cCaA, BB, ccaA, Bb, CCaA, Bb, CcaA, Bb, cC

aA, Bb, ccaA, bB, CCaA, bB, CcaA, bB, cCaA, bB, ccaA, bb, CCaA, bb, CcaA, bb, cCaA, bb, ccaa, BB, CCaa, BB, Ccaa, BB, cCaa, BB, ccaa, Bb, CCaa, Bb, Ccaa, Bb, cC

aa, Bb, ccaa, bB, CCaa, bB, Ccaa, bB, cCaa, bB, ccaa, bb, CCaa, bb, Ccaa, bb, cCaa, bb, ccOnly the aa, bb, cc child (1 in 64 = 1.56%) would develop PSC. Because the prevalence of the disease in the general population is

estimated to be 8 to 14 per 100,000 persons = 0.008% to 0.014% the " heritability coefficient " would be about 100 if you use the upper value of 0.014% (the " heritability coefficient " equals the

prevalence among siblings divided by the general population prevalence). This is similar to the data described by Karlsen for PSC (Karlsen TH, Schrumpf E, Boberg KM 2007 Genetic epidemiology of primary sclerosing cholangitis. World J. Gastroenterol. 13: 5421-

5431).So my guess would be that there may be 3 susceptibility genes involved in PSC using this recessive gene model. Note that the disease would appear to spring out of nowhere ... neither parent has the disease.

Best regards,Dave (father of (22); PSC 07/03; UC 08/03) -- Ian Cribb former P.Eng. (resigned in good standing)

[Guest guest]

,

But if one parent has PSC (ie they have aa, bb, cc) then that parent will be contributing bad genes and the chances are a lot higher that the one well parent will contribute bad genes and that their children will get it. I suspect its more than four genes.

Ian (52) PSC 89

Here's another attempt to explain genetics, and the difference between a simple genetic disorder (1 gene) and a complex genetic disorder (3 genes).Suppose that two parents both carry 1 copy each of a recessive/mutant

gene for a disease (such as cystic fibrosis), but have 1 dominant/normal copy of the gene. Both parents would be disease-free because you need 2 copies of the recessive/mutant gene to develop cystic fibrosis. The parents would be " carriers " . Using CFTR to

denote the normal or dominant gene, and cftr to denote the recessive/mutant gene, the two parents would each be of genotype CFTR/cftr.Their children can randomly inherit one version of the gene from each

parent. So the two parents could have the following types of children: CFTR/CFTR, CFTR/cftr, cftr/CFTR or cftr/cftr. Only the cftr/cftr children (1 in 4, or 25%) would develop cystic fibrosis.In this example, the prevalence of the disease amongst siblings would

be 25%, whereas the prevalence in the general population is 1 in 2,500 to 3,500 Caucasian newborns ( = 0.03 to 0.04%), so the " heritability coefficient " would be about 700 (i.e. 25 divided by 0.035).

Consider now that to develop PSC one needed 3 different mutant genes. Let's use A, B and C to denote the normal/dominant versions of the genes, and a, b and c to denote the mutant/recessive versions of the

genes. Let's suppose that both parents were Aa, Bb, Cc. The children could be any of these 64 possibilities:AA, BB, CCAA, BB, CcAA, BB, cCAA, BB, ccAA, Bb, CCAA, Bb, CcAA, Bb, cC

AA, Bb, ccAA, bB, CCAA, bB, CcAA, bB, cCAA, bB, ccAA, bb, CCAA, bb, CcAA, bb, cCAA, bb, ccAa, BB, CCAa, BB, CcAa, BB, cCAa, BB, ccAa, Bb, CCAa, Bb, CcAa, Bb, cC

Aa, Bb, ccAa, bB, CCAa, bB, CcAa, bB, cCAa, bB, ccAa, bb, CCAa, bb, CcAa, bb, cCAa, bb, ccaA, BB, CCaA, BB, CcaA, BB, cCaA, BB, ccaA, Bb, CCaA, Bb, CcaA, Bb, cC

aA, Bb, ccaA, bB, CCaA, bB, CcaA, bB, cCaA, bB, ccaA, bb, CCaA, bb, CcaA, bb, cCaA, bb, ccaa, BB, CCaa, BB, Ccaa, BB, cCaa, BB, ccaa, Bb, CCaa, Bb, Ccaa, Bb, cC

aa, Bb, ccaa, bB, CCaa, bB, Ccaa, bB, cCaa, bB, ccaa, bb, CCaa, bb, Ccaa, bb, cCaa, bb, ccOnly the aa, bb, cc child (1 in 64 = 1.56%) would develop PSC. Because the prevalence of the disease in the general population is

estimated to be 8 to 14 per 100,000 persons = 0.008% to 0.014% the " heritability coefficient " would be about 100 if you use the upper value of 0.014% (the " heritability coefficient " equals the

prevalence among siblings divided by the general population prevalence). This is similar to the data described by Karlsen for PSC (Karlsen TH, Schrumpf E, Boberg KM 2007 Genetic epidemiology of primary sclerosing cholangitis. World J. Gastroenterol. 13: 5421-

5431).So my guess would be that there may be 3 susceptibility genes involved in PSC using this recessive gene model. Note that the disease would appear to spring out of nowhere ... neither parent has the disease.

Best regards,Dave (father of (22); PSC 07/03; UC 08/03) -- Ian Cribb former P.Eng. (resigned in good standing)