Proving that the teacher is wrong

  • Thread starter jaumzaum
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Hello guys. I want serious help here.

I'm in first year of medical school and I jut had my second test in Genetics. One question of the test gave us a heredogram and asked us what was the most probable type of genetic disease the family had. There were 2 types of disease that would fit the heredogram. Recessive autosomal and Dominant X-linked.

I put both and explained why both them fit. But a lot of my colleagues put only one of them. He said after that he wanted X-linked, because the heredogram had more trends of X-linked diseases. I just don't bite that.

This is not the way we calculate probability. For example a autosomal disease can happen in 44 out of the 46 chromosomes while the X-linked only in the X chromossome (1.5 in average). Also, what if recessive diseases are more common than dominant. What if most of the people with X-linked diseases are infertile or if most of them dies. All of these reduces the chance of a given heredogram being Dominant X-linked. We can't say something one based on the trends.

The heredogram is the below.

03193CD2-FB05-417F-9E7C-19211740ADAE.jpeg


Do you agree with me? I really need some opinions of other people so we can contest the teacher. Otherwise half of the classroom will get a zero.
 
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epenguin

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Not immediately obvious but I think the indicative characteristic pattern the teacher meant you to pick up is in the fact that 3/6 of female descendants have the disease but only 1/7 of the male descendants.

Your considerations on fertility and survival apply equally to both modes of inheritance. (such an effect does not look to be very strong here. This happens in the cases of late onset diseases.)

For your information I do not think that heredogram is a word of English, though it is clear what it means. When I searched for it on Wikipedia all the articles that came were in Spanish. But continue to use it maybe it will catch on!
 
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Thank you @epenguin!
Now I understood what he wanted. He wanted to know the model that "fits more" the heredogram (I really don't know the word for this kkkk). But this is not the same thing as asking for the model that is more likely. For example, consider we just found out a new disease type that has only 7 cases in the whole world. All cases are female, all affected females have only daughters, that happens to be blond, deaf, have septum deviation and dies before 25 yo. You as a geneticist have just been presented with a random heredogram, the heredogram has a affected female, an non affected male, and 4 daughters, all of them blind, deaf, with septum deviatio and dead before 25 yo. This new model is the one that fits more with the]at heredogram right? But is it the more likely? Is it the one that you will start searching for?

Do you see my point? I really don't know all that probabilities I said above about fertility and all other stuff. But probably the frequency of autosomal are greater than X linked (based on chromosome numbers). My point is that for asking which one is more likely, it is not that simple to answer.

Do you think he will be right if he consider only people who wrote Dominant X linked? Or do you think it's more fare to consider both diseases?
 

epenguin

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Ignore my first sentence, I must have made a mistake in my reasoning.

Both types of inheritance fit the heredogram equally well as far as I can see.

Your reasoning that the number of autosomal chromosomes is much greater than that of Y and therefore you would expect more dominant autosomal diseases than X-linked dominant diseases seems OK.

I'm sorry, I really do not know what your teacher has in mind, are you sure you have understood? Your colleagues were wrong to put only one type, maybe that is what he meant.
 

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TeethWhitener

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heredogram (I really don't know the word for this kkkk).
Genogram. https://en.wikipedia.org/wiki/Genogram

I thought this would be fairly obvious when I first looked at it, but having thought about it some more, it's actually a really interesting statistics problem. (I'm assuming the filled symbols are the affected individuals and the empty symbols are the unaffected individuals.)

One difficulty is that you don't know (for instance) whether the non-descendants are homozygous or heterozygous. So if you start by assuming the trait is recessive autosomal, you know that the genotype of the male in the first generation is rr, but you don't know whether the female is RR or Rr. Ditto for the males married to the affected females in the second generation. If the first generation female is Rr, the second generation male with no affected children is RR, and the second gen male with affected children is Rr, then the observed pattern of inheritance is well within the realm of possibility.

If the first generation female is RR, then the chance of having 3 out of 4 rr children is a Poisson problem, with the answer given by ##\frac{1}{6e}## (about 1 in 16), still not completely out of the realm of possibility. Edit: Haha, wow, it really is early. Ignore that.

It's too early, and my statistics talents are too meager, to tackle this problem completely, but maybe someone else who's into statistics can make more progress (@StatGuy2000 ?).
 

TeethWhitener

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If we assume recessive autosomal, the first generation must be Male: rr, Female Rr, giving a 50% chance that the offspring will be affected. 3 of the 4 offspring are affected, so a binomial distribution gives the likelihood of this outcome at 25%. We also know (still assuming recessive autosomal) that the second gen male with affected kids must have an Rr genotype. The likelihood of having 2 out of 3 affected kids is 37.5%.

The real kick in the teeth is the second gen male with no affected kids. If he's RR, then the chance of this happening is 100%. If he's Rr, the chance falls to about 1 in 16. Whether he's RR or Rr depends on the prevalence of the recessive gene in the population, which is totally unknown. This is similar to your point:
But probably the frequency of autosomal are greater than X linked (based on chromosome numbers).
I'm not sure if this is true or not, but if we assume the disease is truly rare--to the point where we have never seen it before and don't know its inheritance pattern--we can tentatively posit that the second gen male with no affected kids is RR with high probability. (Of course, the fact that two people in the family happened to marry dominant heterozygous carriers randomly seems like a fairly low probability event in itself.)

That's probably as far as I can get without making a(n even bigger) fool of myself.
 

epenguin

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Yes, recessive autosomal is also consistent with the heredogram.It was late at night and I did not consider any other hypothesis than those of the OP.

He is also trying to take into account frequencies in the general population.

probability of the male in question (The partner of the third daughter) carrying r is this far as we know (i.e. if no consanguinity) is the same as that in the general population, i.e. low and so this is a relatively low probability explanation. But TWGot in before me just before I finished the sentence.
 
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epenguin said:
For your information I do not think that heredogram is a word of English, though it is clear what it means. When I searched for it on Wikipedia all the articles that came were in Spanish. But continue to use it maybe it will catch on!
In English we call it a genogram.
 
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Whether he's RR or Rr depends on the prevalence of the recessive gene in the population
Actually it's easier than that.
If the disease is recessive autosomal, all affected are rr.
For the children to be rr, both parents have to give one alele r to them
In that case, the father (1st gen) is rr and the mother is Rr. That way, the son (2nd gen) have to be Rr.
The right father (in the second gen) have to be Rr (same reasoning). But the left father can be both Rr as RR.
 

Ygggdrasil

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The more common term in genetics for the diagram is a pedigree (for example, see https://www.ncbi.nlm.nih.gov/books/NBK21257/).

One could approach the question in problem in a Bayesian manner: calculate the likelihood of the observed phenotypes given that the disease is either sex-linked dominant or autosomal recessive and use these values to infer the likelihood that the disease is sex-linked dominant or autosomal recessive given the observations. However, this runs into two issues: 1) as @TeethWhitener notes, calculating the likelihood for the autosomal recessive case requires knowing the prevalence of carriers in the general population (an unknown), and 2) as @jaumzaum noted, the calculation requires some prior knowledge of the relative number of autosomal recessive conditions versus sex linked dominant conditions (maybe this could be found by querrying some database).

Without doing calculations, I'd say that while the pedigree is consistent with both hypotheses (an autosomal recessive disorder or a sex-linked dominant disorder), if the disease is a rare disease then the sex-linked dominant hypothesis is more likely. The sex-linked dominant hypothesis requires no additional carriers to have entered the family whereas the autosomal recessive hypothesis requires one non-related carrier to have entered the family. However, if carriers for the disease are not rare (e.g. because, for example, all members of the family are part of a small community with a limited gene pool), more calculations would be needed to determine which hypothesis is more likely.
 

epenguin

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But autosomal dominant also does not require additional carriers to have entered the family, and the student reasonably expects this type of inherited disease to be more common than X-linked dominant (though I have not been able to find confirmation yet that this is factually so).
 

Ygggdrasil

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But autosomal dominant also does not require additional carriers to have entered the family, and the student reasonably expects this type of inherited disease to be more common than X-linked dominant (though I have not been able to find confirmation yet that this is factually so).
That's true, an autosomal dominant disease would also be consistent with the observations.
 
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maybe this could be found by querrying some database
@Ygggdrasil do you know any database where I could calculate this information?

Now I really wanted to know kkkk
I think it woul be something nice for a geneticist to have in mind before analyzing any type of pedigree
 

Ygggdrasil

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@Ygggdrasil do you know any database where I could calculate this information?

Now I really wanted to know kkkk
I think it woul be something nice for a geneticist to have in mind before analyzing any type of pedigree
The first resource that comes to mind is the Online Mendelian Inheritance in Man (OMIM) database: https://omim.org/about
 
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Hey all, I've got a B.S. in Biology, but genetics is something greatly interesting to me, and I took all the classes I could about genetics as I chased my degree.

Unfortunately, I side with the teacher because of the exact phrasing of the question, and I'll mention that I dislike how the question was asked as well.

If I was debating for the O.P.'s teacher, the sticking points I'd use is that the teacher asked for which form of inheritance is "most likely", not which was possible or under which conditions could either happen. The way the O.P. answered the question, it sounds like he explained why either was POSSIBLE, which the teacher seemed to expect was a GIVEN, which is why the teacher asked WHICH was most likely.

It's easiest to calculate the odds of X-linked dominant: If you can pass the condition on you also show you have it and if you're male with it mated to a female without it, 100% of your daughters and 0% of your sons will have it; if you're female with it mated to a male without it, 50% of all of your children of either gender will inherit it.

Generation 1 shows the only possible outcome, that can only be used to calculate possible or not but it can be treated as a 100% possibility.

Generation 2 shows 7 children born who should have had a 50% chance to inherit it (it doesn't matter that they are born to different affected mothers, as IF this condition is X-linked dominant every cross of an affected mother to an unaffected male produces children that each have a 50% chance to inherit), the odds of a 50% chance happening twice out of 7 trials is a 16.4% chance (https://www.quora.com/If-you-toss-a-coin-7-times-what-is-the-probability-of-getting-two-tails has the short form of why, statistics and specifically probability studies explains why in as much detail as desired).

The 100% possibility of the first generation multiplies with the 16.4% chance of getting this second generation result; the odds of this being X-linked dominant can be called 16.4%.

It is vastly more difficult to calculate the odds of the disease being autosomally recessive because we have no information about population frequency, but we see 3 unaffected partners, two of whom produce affected children. Affected kids from a cross "proves" an unaffected heterozygous parent in the crossing (or a de novo or chromosomal cross over error... but lets pretend those don't exist at this level of genetics class), but the lack of an affected kid does not prove there's not an unaffected heterozygous parent in the third cross, that produced 4 unaffected kids. There's probability involved in both how common heterozygous recessives are in the population and in the actual genetics of the unaffected parent with the unaffected kids (if heterozygous, there's a 6.25% chance all 4 kids in that family would still be unaffected).

Even if we assume the chance that the unaffected family has a homozygous normal parent, and couldn't be affected, there were still 7 kids that had a 50% chance of inheriting the mutation, and 5 got it - the chance of that is the same as our X-linked dominant's chance of 2 out of 7 being affected with a 50% chance, it's the inverse of that outcome, so 16.4% still.

But the odds going into the situation are not 100%, as they were with the dominant gene, because we have to factor in the hard to figure out odds of recessive heterozygous parents being involved, and include the harder to determine odds on the unaffected family actually being children who could have inherited but didn't - reducing the odds well below the 16.4% of the dominant possibility.

Thus I unfortunately side with the teacher, using everything I know about statistics and genetics, even though I'd rather have seen the teacher ask you to describe why either is possible or something similar to that. But the answer the teacher was looking for was to tell him or her which was most likely (hope you were asked to explain why, too?) and the answer I'd expect to see considered correct is the X-linked dominant.
 
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282
7
Hey all, I've got a B.S. in Biology, but genetics is something greatly interesting to me, and I took all the classes I could about genetics as I chased my degree.

Unfortunately, I side with the teacher because of the exact phrasing of the question, and I'll mention that I dislike how the question was asked as well.

If I was debating for the O.P.'s teacher, the sticking points I'd use is that the teacher asked for which form of inheritance is "most likely", not which was possible or under which conditions could either happen. The way the O.P. answered the question, it sounds like he explained why either was POSSIBLE, which the teacher seemed to expect was a GIVEN, which is why the teacher asked WHICH was most likely.

It's easiest to calculate the odds of X-linked dominant: If you can pass the condition on you also show you have it and if you're male with it mated to a female without it, 100% of your daughters and 0% of your sons will have it; if you're female with it mated to a male without it, 50% of all of your children of either gender will inherit it.

Generation 1 shows the only possible outcome, that can only be used to calculate possible or not but it can be treated as a 100% possibility.

Generation 2 shows 7 children born who should have had a 50% chance to inherit it (it doesn't matter that they are born to different affected mothers, as IF this condition is X-linked dominant every cross of an affected mother to an unaffected male produces children that each have a 50% chance to inherit), the odds of a 50% chance happening twice out of 7 trials is a 16.4% chance (https://www.quora.com/If-you-toss-a-coin-7-times-what-is-the-probability-of-getting-two-tails has the short form of why, statistics and specifically probability studies explains why in as much detail as desired).

The 100% possibility of the first generation multiplies with the 16.4% chance of getting this second generation result; the odds of this being X-linked dominant can be called 16.4%.

It is vastly more difficult to calculate the odds of the disease being autosomally recessive because we have no information about population frequency, but we see 3 unaffected partners, two of whom produce affected children. Affected kids from a cross "proves" an unaffected heterozygous parent in the crossing (or a de novo or chromosomal cross over error... but lets pretend those don't exist at this level of genetics class), but the lack of an affected kid does not prove there's not an unaffected heterozygous parent in the third cross, that produced 4 unaffected kids. There's probability involved in both how common heterozygous recessives are in the population and in the actual genetics of the unaffected parent with the unaffected kids (if heterozygous, there's a 6.25% chance all 4 kids in that family would still be unaffected).

Even if we assume the chance that the unaffected family has a homozygous normal parent, and couldn't be affected, there were still 7 kids that had a 50% chance of inheriting the mutation, and 5 got it - the chance of that is the same as our X-linked dominant's chance of 2 out of 7 being affected with a 50% chance, it's the inverse of that outcome, so 16.4% still.

But the odds going into the situation are not 100%, as they were with the dominant gene, because we have to factor in the hard to figure out odds of recessive heterozygous parents being involved, and include the harder to determine odds on the unaffected family actually being children who could have inherited but didn't - reducing the odds well below the 16.4% of the dominant possibility.

Thus I unfortunately side with the teacher, using everything I know about statistics and genetics, even though I'd rather have seen the teacher ask you to describe why either is possible or something similar to that. But the answer the teacher was looking for was to tell him or her which was most likely (hope you were asked to explain why, too?) and the answer I'd expect to see considered correct is the X-linked dominant.
Thanks Ekscanta.

I really didn't think the teacher wanted us to do all that calculation in the question.
The second question was to Explain why as yousaid.


The problem with your calculation is that it calculates the likelihood of a given pedigree being referring to a specific disease type, considering all disease types are equally frequent in the population, or ignoring these frequencies. Lets stick to your calculations and let's suppose the odds of being X-linked are 16,4% and the odds of being autosomal recessive 6,25%. Now if I tell you autossomal recessive diseases in general are 20 times more common in the population compared to X-linked, the pedigree now is almost 8 times more likely of bring autosomal recessive rather than X-linked.
 

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