Pleaseee help:)

Consider two genes on the same chromosome 3 Morgans distance apart. Given a crossover occurs, what is the probability it has occured at 1 Morgan?

Thanks

I just read it....I can't see what I did wrong? If you can help me in anyway I would appreciate it.

Thanks

math25 said:

Consider two genes on the same chromosome 3 Morgans distance apart. Given a crossover occurs, what is the probability it has occured at 1 Morgan?

Thanks

Click to expand...

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.

Well. I am not sure if I can use poisson distribution to do this problem?

I know that 3 Morgan=300CMorgans
So, average rate is Lambda=3

I guess, the problem that I have is the time interval....I am not sure how to calculate probability that it occurred at 1 Morgan.

Thanks

The Domain of the Poisson Distribution is \(\displaystyle [0,\infty)\). Is this applicable?

No. because the distance is 0 to 3 ???

Sorry, I am totally lost...

Here is my 2nd attempt to do this problem...

f(x)=1/3 for 0<x<d

P(one occurrence)=3e^-3

F(X)=1-P(X>X)
=1-((e^-x *x^0)/o!) ((e^-(d-x)*(d-x)^1)/1!)
=1-(e^-x * e^-(d-x) * (d-x))
1-(d-x)e^-d

Again, I am stuck....

f(x0=(dFc(x)/dx)=e^-3 / 3e^-3 = 1/3 for 0<x<3

I'm afraid I don't know how Morgans work.

It the likelihood Fo crossing over inversely proprtional to the Morgan?

p(1) = 1/Total
p(2) = 1/[2(Total)]
p(3) = 1/[3(Total)]

Total = 11/6

p(1) = 6/11
p(2) = 6/22
p(3) = 6/33

Checking 6/11 + 6/22 + 6/33 = (36 + 18 + 12)/66 = 66/66 = 1

Okay, so there are your probabiltiies for that scenario.

Is it just as likely?

p(1) = 1/3
P(2) = 1/3
P(3) = 1/3

I believe that each crossover has equal probability....and crossover occurs at rate 1 per Morgan, so in this case lambda=3

anyone?

Okay, we're almost there.

These Morgans appear to be discrete measures. Can there be 2.5 Morgans? If not, then your continuous model would be inappropriate.

It seems to me that you have 3 Morgans and it is a discrete distribution.
Crossing over can occur at 0, 1, 2, or 3. If each is equally probably, then p(1) = 1/4 = p(2) = p(3) = p(0)

I know a genetics guy. I'll ask. Sorry, it's not always the mathematics that's the trouble spot.

The Morgans, or cM (centi morgans) are strange animals. Is it a distance on a chromosome or is it a probability of crossing over?

Anyway, whether it is a DNA strand or a probability that stretches to 100%, I am feeling better about the Poisson. It's close enough.

With lambda = 3 we have p(1) = e^(-3)(3^1)/1! = 3*e^(-3)

I'm going with that until I see my genetics guy.

I guess, its both, the distance between two chromosomes, for example 3 Morgans =300 CMorgans. Also, the average rate of crossover is lambda=3

It is possible to have 2.5 Morgans....which makes me think its continuous?

Thank you for your help and please if you can ask him asap...I need this for tomorrow morning

1) Morgans are old school. No one does this, anymore, because you can just sequence the gene and KNOW how far apart things are.
2) A Morgan isn't really a distance, it's a % of appearance in the off-spring. It is used to estimate the distance.
3) No one believes that Morgans are particularly useful for distances over 30, since that far apart it's hard to tell from which direction the crossing over has ocuurred.

Having said that, as long as \(\displaystyle \lambda\) is small, maybe < 5 (actually, I would be okay a little larger than 5, but I'm getting pretty nervous by 10), the Poisson approximation seems acceptable. For \(\displaystyle \lambda = 5\), we have \(\displaystyle p(30) = 2.366*10^{-14}\).

So, with \(\displaystyle \lambda = 3\), we have:

p(0) = 0.04979
p(1) = 0.14936
p(2) = 0.22404
p(3) = 0.22404
p(4) = 0.16803
p(5) = 0.10082
p(6) = 0.05041

Yes, you can have fractional Morgans, say 2.5, but these cannot lead to fractional distances. Distances are strictly integer.

So, with \(\displaystyle \lambda = 2.5\), we have:

p(0) = 0.08208
p(1) = 0.20521
p(2) = 0.25652
p(3) = 0.21376
p(4) = 0.13360
p(5) = 0.06680
p(6) = 0.02783

I think that's it.

What you have is probability of the number of crossover that will occur for example P (one crossover)=0.149

In my problem, it given that crossover will occur, but I need to find probability that it will occur at 1 and that's where I have hard time calculating that it will happen at distance 1.

Thanks for all your help

I remain confident that is what is shown. I cannot tell the difference between the two statements. Make sure you show us the solution. We all need more familiarity with Morgans. Strange little hybrids, these Morgans.