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Anonymous commented at 2013-08-06 19:56:58 » #1375047
I doubt most people on a hentai board are really too concerned about the hows, whys and so on, but here we are. =P
Anyway, you are correct that there's no guarantee. It's perfectly possible for an inbred child to be healthier than some other random child off the street, but I'm not so sure you're correct about the likelihood of having no defects is above 50%. It's just that most defects are really subtle, and one may not even be aware of them. Even if the child has, say, slight impediments in its understanding of language and speech, it may not even be picked up at all, as the variation from the norm is so small that it's still within a certain range of what's considered normal development. They're still defects, but they may often not be in the way of leading a perfectly normal life.
5 Points
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I doubt most people on a hentai board are really too concerned about the hows, whys and so on, but here we are. =P
Anyway, you are correct that there's no guarantee. It's perfectly possible for an inbred child to be healthier than some other random child off the street, but I'm not so sure you're correct about the likelihood of having no defects is above 50%. It's just that most defects are really subtle, and one may not even be aware of them. Even if the child has, say, slight impediments in its understanding of language and speech, it may not even be picked up at all, as the variation from the norm is so small that it's still within a certain range of what's considered normal development. They're still defects, but they may often not be in the way of leading a perfectly normal life.
5 Points
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Anonymous commented at 2013-08-07 18:09:48 » #1375688
Some small mistakes here, but all over, I'm impressed you took the time to sit down and try to go through the math. Let's see here then:
As you said P(homozygous child) = 0.25 so long as both parents have received the same recessive gene from grandparents. Then, you get your terms somewhat mixed up, and make some mistakes.
First of all, let's just rid ourselves of gene B, as it's really not necessary, and only serves to confuse things. For the sake of this problem, A and B are basically the same, as they both express a healthy phenotype. (phenotype = expressed trait, genotype = allele combination) Now, for simplicity's sake, we assume the gene shows 100% penetrance, and there's no codominance. Otherwise we can just forget all about trying to actually throw math into things, as biology is fucking complex. We now have two possible phenotypes overall, and only one in the parents, as we assume they have a healthy phenotype. This does not, however, mean that they are not carriers of the recessive gene. Let's look at the grandparents for a bit now.
You assumed in your solving of the problem that both grandparents were carriers of the recessive gene, but this is rather unlikely, so I'll assume that only one carries the given recessive gene we're looking for. Now if we put up one of your punnet squares...
x A c
A AA Ac
A AA Ac
We see that there's a 50% probability per parent of inheriting the recessive gene from the grandparents, and no probability of being ill. If the grandparents had both been carriers however, it would look like this:
x A c
A AA Ac
c Ac cc
Seeing as we assume the parents are healthy, there is actually 2/3 probability per parent of being carriers of the gene, not 1/3. You forgot that two combinations gave rise to the same carrier genotype, and thus, the rest of your reasoning is somewhat incorrect. (Also, I can't quite follow the rest of your argument. I think you should've ended up with 1/9 * 1/4 = ~2.8% following your reasoning, but it would help if you could explain your reasoning in an answer post later) I will now keep working under the assumption that only one grandparent was a carrier, however, as this will usually be the case, and can still easily give rise to a homozygous child.
P(parent is carrier) = 50%
P(both parents are carriers) = 50% * 50% = 25%
P(child homozygous with given defect from grandparent) = P(both parents are carriers) * P(homozygous child given carrier parents) = 25% * 25% = 6.25%
So, there's actually only a 6.25% probability of the child having any given defect, but this still adds up to a rather high number, considering we have some 20000 genes or so. As I've just shown, we do not have to work under the assumption that both grandparents are carriers, as even one carrier grandparent is enough for the child to have a measurable probability of suffering from said defect. Because of this, we actually -can- multiply by a rather large number, as we don't have to look for genes that both grandparents possess, but rather all bad genes that one or both of them possess. Finally, just to put the final nail in the coffin, I took the liberty of checking wikipedia for some quick statistics, and came up with this:
"Children of parent-child or sibling-sibling unions are at increased risk compared to cousin-cousin unions. Studies suggest that 20-36% of these children will die or have major disability due to the inbreeding. A study of 29 offspring resulting from brother-sister or father-daughter incest found that 20 had congenital abnormalities, including four directly attributable to autosomal recessive alleles." Source: en.wikipedia.org/wiki/Incest#Inbreeding
20/29 = ~69% chance of having congenital abnormalities, and 20-36% chance of major disability or death.
I rest my case.
Tl;dr:
The guy above me used some faulty reasoning, and there is a substantial probability of 1st generation offspring from sibling or parent/child unions having congenital abnormalities, major disability and high mortality.
10 Points
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Some small mistakes here, but all over, I'm impressed you took the time to sit down and try to go through the math. Let's see here then:
As you said P(homozygous child) = 0.25 so long as both parents have received the same recessive gene from grandparents. Then, you get your terms somewhat mixed up, and make some mistakes.
First of all, let's just rid ourselves of gene B, as it's really not necessary, and only serves to confuse things. For the sake of this problem, A and B are basically the same, as they both express a healthy phenotype. (phenotype = expressed trait, genotype = allele combination) Now, for simplicity's sake, we assume the gene shows 100% penetrance, and there's no codominance. Otherwise we can just forget all about trying to actually throw math into things, as biology is fucking complex. We now have two possible phenotypes overall, and only one in the parents, as we assume they have a healthy phenotype. This does not, however, mean that they are not carriers of the recessive gene. Let's look at the grandparents for a bit now.
You assumed in your solving of the problem that both grandparents were carriers of the recessive gene, but this is rather unlikely, so I'll assume that only one carries the given recessive gene we're looking for. Now if we put up one of your punnet squares...
x A c
A AA Ac
A AA Ac
We see that there's a 50% probability per parent of inheriting the recessive gene from the grandparents, and no probability of being ill. If the grandparents had both been carriers however, it would look like this:
x A c
A AA Ac
c Ac cc
Seeing as we assume the parents are healthy, there is actually 2/3 probability per parent of being carriers of the gene, not 1/3. You forgot that two combinations gave rise to the same carrier genotype, and thus, the rest of your reasoning is somewhat incorrect. (Also, I can't quite follow the rest of your argument. I think you should've ended up with 1/9 * 1/4 = ~2.8% following your reasoning, but it would help if you could explain your reasoning in an answer post later) I will now keep working under the assumption that only one grandparent was a carrier, however, as this will usually be the case, and can still easily give rise to a homozygous child.
P(parent is carrier) = 50%
P(both parents are carriers) = 50% * 50% = 25%
P(child homozygous with given defect from grandparent) = P(both parents are carriers) * P(homozygous child given carrier parents) = 25% * 25% = 6.25%
So, there's actually only a 6.25% probability of the child having any given defect, but this still adds up to a rather high number, considering we have some 20000 genes or so. As I've just shown, we do not have to work under the assumption that both grandparents are carriers, as even one carrier grandparent is enough for the child to have a measurable probability of suffering from said defect. Because of this, we actually -can- multiply by a rather large number, as we don't have to look for genes that both grandparents possess, but rather all bad genes that one or both of them possess. Finally, just to put the final nail in the coffin, I took the liberty of checking wikipedia for some quick statistics, and came up with this:
"Children of parent-child or sibling-sibling unions are at increased risk compared to cousin-cousin unions. Studies suggest that 20-36% of these children will die or have major disability due to the inbreeding. A study of 29 offspring resulting from brother-sister or father-daughter incest found that 20 had congenital abnormalities, including four directly attributable to autosomal recessive alleles." Source: en.wikipedia.org/wiki/Incest#Inbreeding
20/29 = ~69% chance of having congenital abnormalities, and 20-36% chance of major disability or death.
I rest my case.
Tl;dr:
The guy above me used some faulty reasoning, and there is a substantial probability of 1st generation offspring from sibling or parent/child unions having congenital abnormalities, major disability and high mortality.
10 Points
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Anonymous commented at 2013-08-07 20:10:11 » #1375743
The problem with your reliance on that study is a very, VERY small sample size. There's also no information about whether the sample was geographically or racially varied. The 69% and 20-36% probabilities are probably not very statistically reliable as the results could have been skewed by the very small sample size.
2 Points
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The problem with your reliance on that study is a very, VERY small sample size. There's also no information about whether the sample was geographically or racially varied. The 69% and 20-36% probabilities are probably not very statistically reliable as the results could have been skewed by the very small sample size.
2 Points
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Anonymous commented at 2013-08-08 04:51:35 » #1376022
First of all, 29 test subjects really isn't that small a number for a test like this. However, seeing as you were still in doubt, I crunched the numbers through a one proportion z-test, just to check whether the results were statistically relevant. Here's what I found:
Z = ((P1 - P0)/root of: (P0*(1-P0)))*root of: n
Where Z will denote the number of standard deviations from the norm, P1 denotes the observed probability in our sample, P0 is the approximate probability in the population as a whole, and n is the sample size.
P1 = 20/29
P0 = 0.03
n = 29
Z = ((20/29-0.03)/root of: (0.03*(1-0.03)))*root of: 29
Z = 20.82425770878876 or ~20.8
The probability of observing results like these, even if the number of test subjects is as low as 29, is as low as observing a result 20.8 standard deviations from the mean. let's put it this way: 95% of all observations will lie within +-1.96 standard deviations, and it is usually where we draw the line during statistical testing. 20.8 is so far outside of this range, that we can draw the near definite conclusion that these observations are statistically relevant. From this test, I can't say anything about whether or not the real probability of congenital defects is above or below 50%, but if you're really interested, I could check this for you as well. I'll pop by from time to time just to clear up possible misconceptions.
- Resident Nerd
8 Points
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First of all, 29 test subjects really isn't that small a number for a test like this. However, seeing as you were still in doubt, I crunched the numbers through a one proportion z-test, just to check whether the results were statistically relevant. Here's what I found:
Z = ((P1 - P0)/root of: (P0*(1-P0)))*root of: n
Where Z will denote the number of standard deviations from the norm, P1 denotes the observed probability in our sample, P0 is the approximate probability in the population as a whole, and n is the sample size.
P1 = 20/29
P0 = 0.03
n = 29
Z = ((20/29-0.03)/root of: (0.03*(1-0.03)))*root of: 29
Z = 20.82425770878876 or ~20.8
The probability of observing results like these, even if the number of test subjects is as low as 29, is as low as observing a result 20.8 standard deviations from the mean. let's put it this way: 95% of all observations will lie within +-1.96 standard deviations, and it is usually where we draw the line during statistical testing. 20.8 is so far outside of this range, that we can draw the near definite conclusion that these observations are statistically relevant. From this test, I can't say anything about whether or not the real probability of congenital defects is above or below 50%, but if you're really interested, I could check this for you as well. I'll pop by from time to time just to clear up possible misconceptions.
- Resident Nerd
8 Points
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Anonymous commented at 2013-08-08 05:00:54 » #1376026
Also, while there are some differences between populations as regards concentration of bad genes, the difference wouldn't be big enough to completely change the results when they're as clear as these. Skew them a little, maybe, but not radically alter them.
-Resident Nerd
6 Points
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Also, while there are some differences between populations as regards concentration of bad genes, the difference wouldn't be big enough to completely change the results when they're as clear as these. Skew them a little, maybe, but not radically alter them.
-Resident Nerd
6 Points
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Anonymous commented at 2013-08-10 18:34:44 » #1377626
The heck? Does this kinda stuff happen often? When I browse for hentai pictures, I don't really expect comments beyond "This shit is hot!" "No, the artist is a faggot." "I don't care! I just came with the force of a thousand suns.", but this comment section suddenly took a rather unexpected turn, and now it's full of math, math, statistics and what do you know? More math. That said, though, that Nerd dude seemed to stomp those other poor chumps. Anyone know whether he's just talking out of his arse, like most internet experts do after a quick wikipedia session, or if he actually knows his shit?
4 Points
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The heck? Does this kinda stuff happen often? When I browse for hentai pictures, I don't really expect comments beyond "This shit is hot!" "No, the artist is a faggot." "I don't care! I just came with the force of a thousand suns.", but this comment section suddenly took a rather unexpected turn, and now it's full of math, math, statistics and what do you know? More math. That said, though, that Nerd dude seemed to stomp those other poor chumps. Anyone know whether he's just talking out of his arse, like most internet experts do after a quick wikipedia session, or if he actually knows his shit?
4 Points
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