Print Topic - Archive

BTD Forums  /  Live Right 4 Your Type  /  Is secreter status get passed on genetically -type
Posted by: 1592 (Guest), Tuesday, December 4, 2007, 2:48am
Wondering if anyone knows the genetic workings of Secreters.  Do I as an O type father pass on my Secreter status to my O type daughter.  Are there dominants in the Secreter gene etc.  Or is it somehow randam with only a test to determine.

Thanks in advance for any info on this.

Jake :) :) 8)
Posted by: Linda, Tuesday, December 4, 2007, 2:56am; Reply: 1
Secretor gene is dominant and the non-secretor gene is recessive.  You could still carry the non-secretor gene and pass it along to your children.  Your wife could  be a non-secretor or a secretor also carrying a non-secretor gene.  So, children would have to be tested either way.
Posted by: 1592 (Guest), Tuesday, December 4, 2007, 3:40am; Reply: 2
Thanks Linda  :) :) :B :-/
Posted by: Kristin, Tuesday, December 4, 2007, 5:20pm; Reply: 3
The only situation where children would not need to be tested to determine secretor status is if both parents test as non-secretors. Nonnies can only pass on the nonnie gene.
Posted by: 1592 (Guest), Wednesday, December 5, 2007, 4:16am; Reply: 4
I figure the odds of each child of mine being a non-secreter at 2.24% since I have tested as a secreter.

I am either SS or Ss.  I cannot be ss (20% of the population)

I figure SS at 64% and Ss at 16% based roughly on the 80/20 split.

I figure then my chance to pass on an s gene at 8% and multiply it times the chances my partner is an ss (20%) equals 1.6%, added to 8% times 16% times one-half equals .064%

or 2.24%

Does this make sense or am I missing something?

Jake :) :B 8)
Posted by: Don, Wednesday, December 5, 2007, 4:29am; Reply: 5
I think you missed something.

80 % of the population are secretors. I have not seen the breakdown of that 80%, as you show it, of SS vs. Ss. Have you? I haven't really looked for it, but for all I know 90% of the secretors may be Ss.

That would throw you calculations off quite a bit.
Posted by: honeybee, Wednesday, December 5, 2007, 4:32am; Reply: 6
so nonnies can only pass on nonnie gene? That is interesting! Thanks
Posted by: Lloyd, Wednesday, December 5, 2007, 4:35am; Reply: 7
Quoted from 1592


Does this make sense or am I missing something?

Jake :) :B 8)


You are missing something. You know you are SS or Ss, but the assumption of the ratio between the two is innacurate. The whole workup would take a Bayesian analysis in order to get a reasonable number and I'm not up to it tonight. I will post if/when I do, unless someone beats me to it.  :-/

The 80/20 is based on random donors. Your case is not random.  ;)
Posted by: 1592 (Guest), Wednesday, December 5, 2007, 4:47am; Reply: 8
well, I'll look forward to some analysis that makes more sense.  But as of this moment I think the 80/20 split is the most reasonable to assume for the 80% who are secreters.  It just doesn't make sense to me for the SS versus the Ss of the 80% to stray much.  I've passed the problem on to my oldest child who majored in math in college and I'll see what he thinks in the next day or two.

Best to all,

Jake
Posted by: Lloyd, Wednesday, December 5, 2007, 5:49am; Reply: 9
Okay, I did a quick and dirty version that is close enough to work with. First a 3 x 3 grid was set up to express the various combinations of SS, Ss and ss. Then to get the results to be stable with a 20% population of nonsecretors I assigned .20 to ss, .30 to SS and .50 to Ss as an intuitve starting place. The grid returned numbers close enough to the starting values that it will be suitable for our purposes, the actual numbers should be very close if we are using 20% as the nonsecretor population. Don't forget, that is a rough number to start with and the actual numbers will be slightly different.

Based on that, someone who tests as a sectretor will be SS 3/8 of the time and Ss 5/8 of the time. Their partner will be ss 1/5 of the time and Ss 1/2 the time. Which would leave:

ss = (( 5/8 )/2 ) *( 1/5+( 1/2 )/2 ) = .140625 or about 14% of the time. (when one parent is known to be a secretor and the other parent is random.)

Double checking this by running the grid as a 2 x 3 (no ss from one donor) gives a comparable number. Not quite QED but close enough.  ;D
Posted by: Lola, Wednesday, December 5, 2007, 6:19am; Reply: 10
Quoted Text
unless someone beats me to it

guess who did? ;D
Posted by: Lloyd, Wednesday, December 5, 2007, 5:10pm; Reply: 11
A more rigorous mathematical approach (and simpler) is that for ss to remain at .2, each donor must contribute an s (sqrt .2) on average. Which leads to the conclusion that Ss = ((sqrt .2)-.2))*2 = .494427, leaving SS at .305573, numbers sufficiently close to my original guess.

We have to remember that there are other considerations, such as ss is not equal to exactly .2 in the population, higher or lower frequency of miscarriage in some combinations, cultural and other factors that affect weighting in sub-groups. Which makes the whole thing a huge Bayesian problem. The quick and dirty mathematical approximation is good enough for our purposes.

Lola, these numbers have been figured out countless times by countless people since Mendelian genetics were introduced, for a variety of genetic problems, before you or me were born. Combinatorial probablity is pretty adavanced these days.  :B
Posted by: 1592 (Guest), Thursday, December 6, 2007, 4:39am; Reply: 12
Thanks Lloyd,
My son came up with 13.8%, pretty much the same way you did.  So approximately 6 to 1 for each of them.  I think I'll just assume they're secreters.

Best,

Jake :)  :) ;D 8)
Posted by: Kristin, Thursday, December 6, 2007, 3:20pm; Reply: 13
Quoted from honeybee
so nonnies can only pass on nonnie gene? That is interesting! Thanks


Yes... nonnies can only pass on the nonnie gene, not the secretor gene. Secretors can pass on either one... if they are Ss.
Print page generated: Saturday, October 25, 2014, 5:46am