Science Saturday 13!
SCIENCE SATURDAY IS FINALLY BACK!
Today I'm going to continue with the gene linkage theme from last time and talk about lod scores. Prepare for some possible (probable) boredom-- but if you ever take a genetics course (or, should it strike your fancy, go into genetics as a career), it will be heckly useful to you.
Lod stands for Log of Odds. It expresses the likelihood that a specific combination of parental and recombinant offspring occur when two genes sit apart at a specific distance. It's really useful to use with humans because you can't always just perform testcrosses efficiently (or morally) to map out genetic distance-- I mean, you might happen to have hot and productive sex with someone who is completely recessive for the genes of interest, but that's not always the case. Also, you wouldn't get a big sample size to work with like you would with organisms like drosophila so you're usually working backwards through a large pedigree that contains many affected individuals. I think that it'll make more sense if I explain it through the formula for it.
Log(WxYz/Ab)
W = .5-recombination frequency (i.e., parental frequency)
x = # parental individuals
Y = Recombination frequency
z = # recombinant individuals
A = recombination frequency given that the genes are unlinked
b = all individuals
Variables W and Y assume linkage at certain distance. So, let's say you want to work this out for linkage at 20 centimorgans (cM). Genetic distance = crossover frequency = (1/2) recombination frequency, so 20 cM = .2 crossover = .1 recombination, so Y = .1; W = .4.
Variables x and z depend on the number of individuals in the generation. So if a woman has five children, and you know that three are recombinant and two are parental, then z = 3 and x = 2.
Variables A and b are always the same. If two genes are unlinked, then each possible combination of genes has a .25 chance of happening. So A = .25. If we have the same woman and her five children as before, then b = 5.
Now we can figure it out! YAY!
Log((.42 x .13)/.255) = -.78558.
So now you have a probability that these genes are unlinked at 20 cM.
INTERPRETING YOUR LOD SCORE!
A lod score of -2 suggests that there is probably not linkage and so you need to adjust your distance; a lod score of 3 suggests that there is most definitely linkage. Remember how logarithms work; this is base-ten, so a 3 means that you've got a 1/1000 chance that the genes are unlinked; a -2 means you've got a 100/1 chance that the genes are unlinked.
WHAT YOU SHOULD DO WHEN YOU GET A LOD SCORE SHOWING A PROBABLE LACK OF LINKAGE:
Try again with a different distances and choose the one that's closest to three.
20 cM: -.78558
30 cM: -.37329
40 cM: -.13237
50 cM: 0
60 cM: .04372
70 cM: -.00531
80 cM: -.1835
90 cM: -.6321
100 cM: 1.505
If you get 100 cM, the genes are far enough apart that they may as well be on different chromosomes, which is SERIOUS unlinkage (which is now a word). So these two traits are, as far as certainty can be had in genetics, certainly unlinked.
WHAT IF YOU'RE NOT SURE ABOUT LINKAGE?
There's a way to factor in different possibilities of recombination- like, if you have a set of 5 children, and you know for sure that two are recombinant and two are parental, but there's this one that you're not sure about, and you're testing at 20 cM again, you have .5((.43 x .12)/.255) + .5((.42 x .13)/.255), and then you take the log of that. If there are three possibilities for the distribution of parental/recombinant offspring, then .5 becomes .333 and you add up three possibilities before taking the log. If you don't know about any of the five, then you'll have six possibilities (all parental; 4 P, 1 R; 3 P, 2 R; 2 P, 3 R; 1 P, 4 R; all recombinant) so you'll have to add up .1666((.45 x .10)/.255) + .1666((.44 x .11)/.255) + .1666((.43 x .12)/.255) + .1666((.42 x .13)/.255) + .1666((.41 x .14)/.255) + .1666((.40 x .15)/.255) AND THEN take the log of that.
Fortunately, there is software that can do this for you.
Hooray!
So. I found this really awesome site that takes you through the process: http://www.ncbi.nlm.nih.gov/books/bv.fcgi?rid=hmg.box.1391
Today I'm going to continue with the gene linkage theme from last time and talk about lod scores. Prepare for some possible (probable) boredom-- but if you ever take a genetics course (or, should it strike your fancy, go into genetics as a career), it will be heckly useful to you.
Lod stands for Log of Odds. It expresses the likelihood that a specific combination of parental and recombinant offspring occur when two genes sit apart at a specific distance. It's really useful to use with humans because you can't always just perform testcrosses efficiently (or morally) to map out genetic distance-- I mean, you might happen to have hot and productive sex with someone who is completely recessive for the genes of interest, but that's not always the case. Also, you wouldn't get a big sample size to work with like you would with organisms like drosophila so you're usually working backwards through a large pedigree that contains many affected individuals. I think that it'll make more sense if I explain it through the formula for it.
Log(WxYz/Ab)
W = .5-recombination frequency (i.e., parental frequency)
x = # parental individuals
Y = Recombination frequency
z = # recombinant individuals
A = recombination frequency given that the genes are unlinked
b = all individuals
Variables W and Y assume linkage at certain distance. So, let's say you want to work this out for linkage at 20 centimorgans (cM). Genetic distance = crossover frequency = (1/2) recombination frequency, so 20 cM = .2 crossover = .1 recombination, so Y = .1; W = .4.
Variables x and z depend on the number of individuals in the generation. So if a woman has five children, and you know that three are recombinant and two are parental, then z = 3 and x = 2.
Variables A and b are always the same. If two genes are unlinked, then each possible combination of genes has a .25 chance of happening. So A = .25. If we have the same woman and her five children as before, then b = 5.
Now we can figure it out! YAY!
Log((.42 x .13)/.255) = -.78558.
So now you have a probability that these genes are unlinked at 20 cM.
INTERPRETING YOUR LOD SCORE!
A lod score of -2 suggests that there is probably not linkage and so you need to adjust your distance; a lod score of 3 suggests that there is most definitely linkage. Remember how logarithms work; this is base-ten, so a 3 means that you've got a 1/1000 chance that the genes are unlinked; a -2 means you've got a 100/1 chance that the genes are unlinked.
WHAT YOU SHOULD DO WHEN YOU GET A LOD SCORE SHOWING A PROBABLE LACK OF LINKAGE:
Try again with a different distances and choose the one that's closest to three.
20 cM: -.78558
30 cM: -.37329
40 cM: -.13237
50 cM: 0
60 cM: .04372
70 cM: -.00531
80 cM: -.1835
90 cM: -.6321
100 cM: 1.505
If you get 100 cM, the genes are far enough apart that they may as well be on different chromosomes, which is SERIOUS unlinkage (which is now a word). So these two traits are, as far as certainty can be had in genetics, certainly unlinked.
WHAT IF YOU'RE NOT SURE ABOUT LINKAGE?
There's a way to factor in different possibilities of recombination- like, if you have a set of 5 children, and you know for sure that two are recombinant and two are parental, but there's this one that you're not sure about, and you're testing at 20 cM again, you have .5((.43 x .12)/.255) + .5((.42 x .13)/.255), and then you take the log of that. If there are three possibilities for the distribution of parental/recombinant offspring, then .5 becomes .333 and you add up three possibilities before taking the log. If you don't know about any of the five, then you'll have six possibilities (all parental; 4 P, 1 R; 3 P, 2 R; 2 P, 3 R; 1 P, 4 R; all recombinant) so you'll have to add up .1666((.45 x .10)/.255) + .1666((.44 x .11)/.255) + .1666((.43 x .12)/.255) + .1666((.42 x .13)/.255) + .1666((.41 x .14)/.255) + .1666((.40 x .15)/.255) AND THEN take the log of that.
Fortunately, there is software that can do this for you.
Hooray!
So. I found this really awesome site that takes you through the process: http://www.ncbi.nlm.nih.gov/books/bv.fcgi?rid=hmg.box.1391