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Can you help me out with a genetics problem?Moderator: BioTeam
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Can you help me out with a genetics problem?Look at this:
The annual plant Haplopappus Gracilis has two pairs of chromosomes 1 and 2. In this species the probability that two traits a and b selected at random will be on the same chromosome of H. Gracilis? The answer in the book : The probability is the product of the probabilty that both a and b are on chromosome 1 and the probability that they both are on chromosome 2. So, 1/2 * 1/2 = 1/4 (25%). What I didn't really understand is how they reached this conclusion ...I would LOVE to see what the reasoning is. I've tried combinations and probabilities, but I've been thinking about this for so long that I seem to be awfully entangled right now. PLEASE PLEASE help me out!!!!
The question asks what is the probability of having a and b on the same chromosome. This shows that you need to multiply the probability of have a to the probability of having b. I'm assuming a and b are recessive which would mean you would have a 1/2 chance of recessive a and a 1/2 chance of recessive b. If the question would have stated a or b then you would have added the probabilities. Hope that's not too confusing.[/u]
LOL...
You see Crissie only a human has 23 pairs of chromosomes. Your plant has only 2 pairs of chromosomes. So you calculate the probability by a mathematical equation probability=number of favorable cases/number of possible cases favorable casesonly one: a and b are on the same chromosome. possible cases4: 1)a and b are on the given chromosome 2)a and b are both on the other chromosome 3)a is on chromosome 1, b is on chromosome 2 4)a is on chromosome 2, b is on chromosome 1 so probability is 1/4. given as a percentual value it is 1/4 *100 =25% A and b are genes, not alleles, so it does not matter wether the alleles that make them up are recessive or dominant. Bafta in continuare "As a biologist, I firmly believe that when you're dead, you're dead. Except for what you live behind in history. That's the only afterlife"  J. Craig Venter
Hey, thanks!
I thought about this...Let me explain everything ...My book from Missouri used the first Happloppopus example to explain a formula ... They say the probability to have 2 traits on the same chromosome is n pairs of chromosomes (1/2) I wanted to know how they ended up with this formula. The next question was ...what is the probability to have 2 genes on the same chromosome if there are 23 pairs  that is , in the case of a human being...They used the formula that I mentioned right above and I just wanted to know what I could do to come up with the formula all by myself...
Re: Can you help me out with a genetics problem?
here is my already famous 'probabilytree' again: _________________traits ________________/_____\ _________Chr 1(1/2)__Chr2(1/2) (trait a on chromosome 1 or 2 in first _________ /___________________\_________species) _________/____________________\ ___Chr 1(1/2)_Chr 2(1/2)_Chr 1(1/2)_Chr 2(1/2)(trait a on chromosome _________________________________________1 or 2 in next species) so the probability that the traits are on chromosome 1 = 1/2 * 1/2 = 1/4 now assume we have n chromosomes, than the probability to find gene a on a particular chromosome = 1/n and also on the other one. > formula is: 1/(n^2) because 1/n * 1/n = 1/(n*n) = 1/(n^2) thus the chance that a trait is found on chromosome 1 of there is a total of 2 chromosomes is 1/(2^2) = 1/4 if there are 23 chromosomes: P = 1/(23^2) = 1/529
how can that be > because we have 23 chromosomes, 0,5^23 can never be the answer.
> the chance that a gene is on a chromosome of interest is 1/23 > the chance that the gene is on the same chromosome in another species with the same amount of chromosomes, is also 1/23 > the total chance is therefore 1/23 * 1/23 and not 1/2 * 1/2 * ....... 1/2 * 1/2. that's impossible. The fact that when we have 2 chromosomes the probability to encounter a gene on chromosome 1 = 1/2 doesn't mean that the probabilty is the same when we have 23 chromosomes. Then the probability that gene a is on chromosome 1 is 1 of the 23 = 1/23.
If you have 23 chromosomes the probability of 2 genes being on the same one is not 1/2, it is much lower(1/529) calculated by sdekivit
Think of it like this probability=number of favorable cases/number of possible cases favorable cases23: a and b are on the same chromosome. They can be on any of the 23 chromosomes. possible cases: 23*23=529 So it is 23/529=1/23 Now if you were to choose the chromosome from the beginning the probability is that calculated by sdekivit=1/529, since you have only one favorable case.(if they ask, what is the probability of the genes being both on chromosome number 3)? "As a biologist, I firmly believe that when you're dead, you're dead. Except for what you live behind in history. That's the only afterlife"  J. Craig Venter
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