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Louis Kessler's Behold Blog

Cast and Crutches - Sat, 31 Dec 2016

On Monday Dec 19, it felt like I might have sprained my ankle. It was my first squash game of the day and I planted my foot and I felt a tiny pop in my ankle. It wasn’t too bad and I tried to continue but less than a minute later I felt another pop in the ankle. Now it felt like a badly sprained ankle.

I immediately told the guys that’s it for me today, got some ice and iced it for 40 minutes while my buddies finished up their matches. It didn’t feel too bad after the icing and fortunately I was carpooling with Rob and it was his turn driving and he took me home.

That night it didn’t swell up much, but it hurt like crazy. The next morning, my wife took me into our nearby Sports Medicine clinic. They took X-rays and could see on the X-ray (I’ll try to get it and post it here), that I had an avulsion fracture of the ankle. That means the place the peroneal tendons (which go from the outside of foot back to the calf) are attached to the ankle broke off taking a tiny piece of the bone with it. The X-ray can’t see the tendons, but it did show the bit of bone.  

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The peroneal tendons (which I never knew existed two weeks ago) are like two rubber bands glued to the bottom of the fibula. Both the rubber bands and the “glue” are stronger than the bone and with pressure, something had to give. Now the two tendons were held in place at the top and bottom but not in the middle and were floating. The doctor put me in a solid cast which could bear weight, and made an appointment with the foot surgeon to see if I would need an operation.

So a week of hobbling around in that cast and crutches, with my wife and daughters driving and otherwise helping me. On Wednesday Dec 28, I met with the foot surgeon. She told me this is unlikely to heal well on its own. I had peroneal tendon subluxation which means the tendons were out of place.

Yesterday, Dec 30, I went in for day surgery. I had my wisdom teeth removed 40 years ago, but this is my first surgery in a hospital. I got changed and they set me up on an IV in the waiting room. I would be getting a spinal, which would freeze my lower half and I’d be awake. They asked me if I wanted to have a sedative applied so that I wouldn’t remember the surgery. i turned it down because I wanted to experience it.

When surgery time came, they rolled me into the operating room. There were two doctors and about 4 others there, including the anaesthesiologist who would then administer the spinal. Barely felt that. The IV previously was worse. The operation was to take about 45 minutes. I couldn’t see anything because they had a big canopy over me. I couldn’t make out much of what the doctors were saying because they were talking very quietly. I only had the anaesthesiologist who came around to check on me every few minutes. The spinal was great. I felt nothing! When they came to cut a small groove in my fibula so that they could place the tendons, I got to enjoy a few minutes of listening to what sounded like a circular saw, but again I felt absolutely nothing and couldn’t tell anything was being done to me. My anaesthesiologist asked me if the saw sound bothered me, but I said no, it was just like being at the dentist.

The operation took about 1 hour and 15 minutes, which was 30 minutes longer than expected because when they got in, they found the tendons were quite frayed and needed repair, which they did. It actually was a good thing I did the surgery because the state of the tendons would have caused me problems in the not too distant future if I had chosen not to have the surgery.

imageAfter surgery, I was taken to the recovery room for an hour and I was watched closely. My surgeon came in and said everything went well and told me about the frayed tendons that they repaired. They also put me in a new non-weight bearing cast which I would have to be in for 6 weeks and then a walking boot for 4 weeks. I had appointments set up for 2 weeks and 6 weeks from now with the surgeon to check progress, but she said I can come in 5 weeks instead of 6, and we’d see if I could switch to the walking boot to make RootsTech easier for me. Either way, crutches or boot, I wasn’t going to miss RootsTech. Hopefully it will be the boot. Crutches would make quite a scene as I go up onstage to present Double Match Triangulator in the Innovator Showdown semi-finals.

After the recovery room, my wife joined me back to the staging area to ensure that the spinal wears off which takes a few hours. After I could get up on my own steam, I was released and my wife drove me home.

It took until 10 p.m. or so until the full spinal wore off. Then the pain started and it was bad. I took two Tylenol 3’s which the doctor gave me but that hardly seemed to do anything. I barely slept last night. In the morning, my wife who is a pharmacist, recommended two Ibuprofens. Within an hour, that did the trick and most of today has been pain free. I must have had swelling from the operating I couldn’t see under the cast and those helped relieve this.

So, what a way to start a new year. At least I will be on my road to recovery now and I’ll do what I need to do to make it to RootsTech. This is likely the end of my 30 years of playing squash. My last major injury from squash was a completely torn achilles 5 years ago on the same leg, which was fully healed and unrelated to this new injury. I’ve had other aches and pains and nagging injuries from squash, but these two, with the out-of-action time that they cause to the rest of my life, are just not worth it anymore to me. This one just came too close to ruining RootsTech for me. So lots of walking, biking and swimming should do well to keep up my fitness levels. I will miss my squash guys and their camaraderie over the years though.

A side benefit of the next six weeks: I’ll have lots of time at my computer to work on Behold and do genealogy stuff.


Update:  Monday, Jan 16, 2017

imageLast Friday, two weeks after the operation. They took the cast off and removed the stitches. They put a rectangular bandaid over it and a sock over that, and no longer did I have to wear a cast, but I’m in a walking boot – actually the same one that I got for my achilles tear 5 years ago, so for up to 8 more weeks, I’m a stormtrooper again. Four weeks non-weight bearing with the boot with crutches, and 4 weeks after that still in the walking boot weight bearing without crutches. I’ll wear the walking boot during the day and at night.

Up to today, I was taking a shower sitting on the edge of the tub with my right foot out of the tub with the cast and then the boot in a bag so it wouldn’t get wet. On Friday the doctor told me that starting today, I could take off the boot a few times a day to shower and to do exercises to bend my foot up and back (but not side to side). So today I took off the bandage and washed my right foot and leg for the first time in 2 1/2 weeks. It felt so good to do that.

The incision was a slight curve about 3 inches long. You can see where the stitches were removed but it’s all healing nicely. It’s a bit tender but doesn’t hurt and there’s very little swelling. My right calf muscle is going to pot though. I’m sure I’ll bring that back up to snuff not too long after the boot comes off and biking season starts.image

My next appointment is in 2 1/2 weeks which will be 5 weeks from the operation. That was supposed to be at 6 weeks when I can start weight-bearing, but my doctor moved it up because I was going to be out of town at RootsTech the next week. She’ll evaluate then and tell me if I’ll be taking crutches to Salt Lake CIty.

Funny that my right leg feels fine. It’s the upper quad in my left leg that I partially tore 3 months earlier that’s giving me more problems, especially at night. But all the extra work my left leg is doing now to support all of me is helping it.

Probability of No X Segments Matching - Sun, 25 Dec 2016

Okay. Let’s do what we did last post for autosomal this time for the X chromosome.

I’ll assume you already know the unique pattern of how the X chromosome get’s passed down, where males get their one X from their mother and females get one of their Xs from their mother and the other from their father. The mother’s is from both of her parents and since the X chromosome (according to FamilyTreeDNA) is 196 cM, that means it recombines with an average of about 1.96 crossovers, which I will round to be 2.. The father’s is passed intact only to his daughter without recombining.

So a son only gets one X chromosome from his mother which will have on average 2 crossovers. A daughter gets one from her mother with 2 crossovers and one from her father with zero crossovers.

This is interesting. That means is a 50% chance of 2 crossovers if it is a son, and that leaves a 25% chance of 2 crossovers and a 25% of zero crossovers if it is a daughter. That works out to 75% chance of 2 and 25% chance of zero giving an expected value of 1.5 crossovers per generation.

And that seems to make sense, since if you got up the female line via mother-mother-mother-mother…, you’ll get 2 crossovers each generation.If you go up the most possible male line which is father-mother-father-mother…, you’ll get zero,2,zero,2,… crossovers which average 1 crossover each generation. So 1.5 seems like it could very well be the average over all lines.

For autosomal, we started with the 23 chromosomes pairs and increased them by 34 segments each generation since both pairs total about 3400 cM. Here for the X chromosome, we’ll start with 1 and increase by 1.5 segments per generation. It’s okay if we use fractional segments here because we’re dealing with averages.

For autosomal, we doubled the number of ancestors each generation. The X chromosome grows not by doubling, but via a Fibonacci sequence. As a lover of mathematics, I must say it’s nice to get good old Fibonacci into DNA. A Fibonacci sequence starts with 1 and 1 and then the next number is always the sum of the previous two, so it’s 1, 1, 2, 3, 5, 8, 13, 21,… A male starts with one X chromosome parent, whereas a female starts with two, so they are offset with one another and an overall average can be taken.

Now lets put the generational levels together:

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There you see the segments growing 1.5 per generation, the male and female Fibonacci sequences and their average that represents the expected number of ancestors.

The “P(NoMat)” column is the probability of no segments matching a specific ancestor given that there are N ancestors and S segments and is calculated as:

(1 – 1 / N) ** S

Finally, we can work out the expected number of ancestors that match on the X chromosome by multiplying the number of Ancestors by the probability of matching (which is 1 – the probability of not matching). For higher generations, this number is the same as the number of segments, because it is very unlikely that such a distant ancestor will contribute more than one segment each.

N * P(NoMat)

What this table says is that after 13 generations of X chromosomes, you will have on average 20.5 segments. 95.93% of the 493.5 possible X ancestors will not contribute meaning the 20.5 segments come from 20.1 ancestors, so there is still a chance one or two of them may contribute more than one segment.

Comparing the probabilities of not matching with autosomal is interesting:

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With autosomal, it takes 9 generations before there’s less than a 50% chance that an ancestor won’t pass you a segment. For the X chromosome, it only takes 6 generations for less than a 50% chance. And there’s even a small chance that you won’t inherit an X-segment after 1 generation. This could happen if the X chromosome from the mother’s side has no crossovers and comes just one of her parents. See the section: The X Doesn’t Recombine as Expected.

Back to statistics: The Poisson distribution can approximate the number of crossovers per generation. Assuming we are talking about the mother’s X chromosome which has an average of 2 crossovers, a Poisson distribution wiith mean of 2 can give a reasonable estimate of the expected chance of each number of crossovers in one generation on the X chromosome:

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One thing left to do. Like we did for autosomal in my last blog post, we also want to determine the average segment length of a match. So we get this:

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Comparing average segment length of an autosomal match with that of an X chromosome match (above) gives:

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This shows that autosomal matching segments at any generation are on average a bit longer than X chromosome matching segments.

So now I have everything I need to program this into Behold. Behold will be working with the actual ancestors and know whether it’s a male or female and will take this into account. This will enable to Behold will give more accurate information than what I’ve shown above which are just averages. Also, Behold will correctly add the probabilities and compute the expected lengths when there’s pedigree collapse and one ancestor is an ancestor on multiple sides. This should be really useful information that I don’t believe is available anywhere else.

My calculations and assumptions above and in my previous post are as far as I can tell, correct for the averages. I would love to get these two posts peer-reviewed by some genetic genealogists and/or genetic researchers. With encouragement, I could turn these posts into a submission for a publication like the Journal of Genetic Genealogy. I’d be happy to have any problems pointed out and will make any clarifications or corrections that are necessary.

Probability of No Autosomal Segments Matching - Mon, 19 Dec 2016

Back to Behold, but still DNA.

I am adding some DNA features to Behold that I know I need and are not in any genealogy programs currently out there.

Basically, I want to know the expected (i.e. mean) amount of autosomal, X, Y and mt DNA that each person will share with main person (or people) selected for the family. This is a centimorgan (cM) amount. It is straightforward to figure out, since the expected autosomal amount gets halved every generation, Y and mt only get passed through the male and female lines respectively, and X, although slightly more complicated, is manageable with females getting all their father’s and half their mother’s and males getting half their mother’s.

In addition to that, I want to know the probability of no segments matching. This is important, because if you have a 5th cousin, and you know that there’s, say, a 50% chance that they will not match at all, then you should only expect that half of the 5th cousins that DNA tested will match you somewhere. And fewer than half of them will show up as matches with your DNA testing company because the companies have a minimum match criteria before they claim two people match, and they need to do that to prevent too many false positive random matches.

I took a look to see if I could find the theoretical probabilities that I needed. I found at the ISOGG page on Cousin Statistics two tables:

I found it very interesting that these two tables give the same information but with slightly different numbers. For instance 4th cousins are 9 generations (DNA-wise) apart sharing on average (1/2)^9 = 1/512 of their DNA. And a person with their 7xgreat grandparent also shares 1/512 of their DNA. But the 1st table gives 30.70% for 4th cousins, and the second gives 37.43% for 7xgreat grandparents. I would have thought these two numbers should be the same, and I can’t check the original article these were derived from because I’m not a PubMed author and don’t know any PubMed author’s who can invite me.

None the less, the numbers in these tables are reasonably close to each other. So now I just need a method to calculate them for any degree of generational distance. I love when I get to do something statistical which was part of my education and my work. Not too often have I had to use my statistics education for genealogy, so here’s my chance.

Let’s go to Jim Bartlett’s blog post: Crossovers by Generation. Take some time to read it and learn something like I did. I’m going to reproduce Jim’s Table 3:

05D Figure 3

The important columns are the one’s marked “Segments” and “Number of Ancestors”. Because there are on average 34 crossovers per generation, the number of segments grows linearly, 34 per generation. But the number of ancestors is growing exponentially, doubling every generation. After 9 generations, there are more ancestors than segments. By generation 13, there are 8192 ancestors, but only 465 segments. That means at most only 465 out of those 8192 ancestors will match, and that’s if none of them match on more than one segment. That already tells you that at least (8192 – 465) / 8192 = 94.32% of your 13th generational relatives will not match you.

Now let’s use some statistics. The statistical probability of no segments matching given that there are N ancestors and S segments is: 

(1 – 1 / N) ** S

What that says is that for generation 13, there is a 8191/8192 chance of a person not matching in one segment, and the non-match has to be in all 465 segments. Calculate this out and it comes to 94.48%.

Let’s do that for a bunch of generational levels and compare that to Table 1 and Table 2:

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Hmmm. Not too bad. In fact the Statistical calculation comes very close to the Table 1 numbers. So close, that when I plot the three sets of values, you see a  small difference only with the Table 2 numbers but the other two are right on top of each other.

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Excellent. So now I have validated that these numbers are close enough and that I can therefore use them.

One last thing left to do. The mean amount of autosomal DNA passed down is always halved each generation. On average, that means with a 13 generational difference, the expected DNA shared is 1 / 8192 = 0.01% which would work out to just 1 cM. That’s an awfully small match to be detected.

But that average includes all the ancestors who don’t match at all. We know that 94.48% or 7740 of the 8192 do not match. Better is to show the expected DNA matching when the two people do match. This would then be just 1 / 450 = 0.22% which would work out to an expected average match of 15 cM for the 450 people 13 generations apart that do match.

Let’s try this for the whole range of generations::

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So now I have what I need. Behold is going to show:

  1. The probability of having a DNA match (e.g. 5.52%)
  2. The average match length if they do match (e.g. 15.0 cM)

Let me of course add 100 caveats. These are approximate values. The actual percentages may vary. Matching cM may vary greatly, etc., etc., blah, blah.