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Calcutate the expected values, X2 value and the p value from the following data set:

Phenotype          Observed
abc               200
+++               200
ab+                 50
++c                 50
a++                 125
+bc                 125
a+c                 25
+b+                 25

Submitted: 489 days and 2 hours ago.

Category: Math

Value: $9

Status: AWAITING CUSTOMER ACTION

Optional Information:
Subject: Principles of Geneti

Already Tried:
Nothing much. Just the myanswers section from yahoo.

Posted by Sk1llz 489 days and 2 hours ago.

Info Request

Do you know the probabilities/distribtuion for/of abc, +++, ab+, ++c, a++, +bc, a+c, +b+?

489 days and 2 hours ago.

Reply

Reply to Sk1llz's Post: no

Posted by Sk1llz 489 days and 2 hours ago.

Info Request

Well, do you know the expectation of them?

489 days and 1 hours ago.

Reply

No. I don't know the expectation of them

489 days and 1 hours ago.

Reply

For the previous request, I think I know the probabilities/distribution of them. Here they are:

.95 .90 .80 .70 .50 .30 .20 .10 and it breaks away from .05 .01 .001

489 days and 1 hours ago.

Reply

I just need the expected values for each degree of freedom and then I can do the rest. Finding the answers for the X2 value and the p value is just and option. But if you could, I would be really thankful to you.

Posted by Sk1llz 489 days and 1 hours ago.

Info Request

Those can't be the probabilities. The sum of the probabilities must be equal to 1.

Sincerely,
Sk1llz

489 days and 1 hours ago.

Reply

The probability is 1/2 and 1/2

Posted by Sk1llz 489 days and 1 hours ago.

Info Request

No it can't be, we need 8 different probabilties whose sum equals 1. Unless there is something I am missing. We have done a total of 200 + 200 + 50 + 50 + 125 + 125 + 50 + 50 = 800 observations out of how many?

We have observed 200 abc's, but out of how many? If you can answer that question then the probabilities ".95 .90 .80 .70 .50 .30 .20 .10" makes sense.

Sincerely,
Sk1llz

489 days and 1 hours ago.

Reply

It doesn't say anything about out of how many.

489 days and 1 hours ago.

Reply

The probabilities could be 1/3, 1/3, and 1/3 since we are dealing with 3 genes.

Posted by Sk1llz 489 days and 1 hours ago.

Info Request

Does it sound familiar if I say then maybe we have to assume they all have the same frequency. In other words p = 1/8 = 0.125 for all phenotypes?

489 days and 1 hours ago.

Reply

The original question was, "You testcross a fly thta is heterozygous for three genes, A, B, and C. The wild type allele is designated with a "+" and the mutant allele is designated with a lower case letter.

489 days and 1 hours ago.

Reply

Yes. It sounds familiar.

Posted by Sk1llz 489 days and 1 hours ago.

Info Request

I tried finding the chi-square test statistic when assuming they all have the same frequency, but it yields a test-statistic that is equal to 375 which has a p-value of 5*10^(-77) which says this assumption is insane. So we're back to square one and searching for the frequency.
I'm not into genetics, I only know pure math and statistics. So it might be that the sentence "You testcross a fly thta is heterozygous for three genes, A, B, and C. The wild type allele is designated with a "+" and the mutant allele is designated with a lower case letter" says something about the frequency.

489 days ago.

Reply

I guess it does.

489 days ago.

Reply

I guess you can't solve this problem.

Posted by Sk1llz 488 days and 12 hours ago.

Info Request

Not unless I'm given more information. We need to find a way to find the expectation, but since I'm a mathematician without any biological education I have no understanding about genetics.

Sincerely,
Sk1llz

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Math

Have your own Math question?

Question

Calcutate the expected values, X2 value and the p value from the following data set: Phenotype Observed abc 200 +++ 200 ab+ 50 ++c 50 a++ 125 +bc 125 a+c 25 +b+ 25

Optional Information

Optional Information: Subject: Principles of GenetiAlready Tried: Nothing much. Just the myanswers section from yahoo.

Info Request

Do you know the probabilities/distribtuion for/of abc, +++, ab+, ++c, a++, +bc, a+c, +b+?

Reply

Reply to Sk1llz's Post: no

Info Request

Well, do you know the expectation of them?

Reply

No. I don't know the expectation of them

Reply

For the previous request, I think I know the probabilities/distribution of them. Here they are: .95 .90 .80 .70 .50 .30 .20 .10 and it breaks away from .05 .01 .001

Reply

I just need the expected values for each degree of freedom and then I can do the rest. Finding the answers for the X2 value and the p value is just and option. But if you could, I would be really thankful to you.

Info Request

Those can't be the probabilities. The sum of the probabilities must be equal to 1.Sincerely,Sk1llz

Reply

The probability is 1/2 and 1/2

Info Request

Reply

It doesn't say anything about out of how many.

Reply

The probabilities could be 1/3, 1/3, and 1/3 since we are dealing with 3 genes.

Info Request

Does it sound familiar if I say then maybe we have to assume they all have the same frequency. In other words p = 1/8 = 0.125 for all phenotypes?

Reply

The original question was, "You testcross a fly thta is heterozygous for three genes, A, B, and C. The wild type allele is designated with a "+" and the mutant allele is designated with a lower case letter.

Reply

Yes. It sounds familiar.

Info Request

Reply

I guess it does.

Reply

I guess you can't solve this problem.

Info Request

Not unless I'm given more information. We need to find a way to find the expectation, but since I'm a mathematician without any biological education I have no understanding about genetics.Sincerely,Sk1llz

Related Math Questions

Calcutate the expected values, X2 value and the p value from the following data set:

Phenotype Observed
abc 200
+++ 200
ab+ 50
++c 50
a++ 125
+bc 125
a+c 25
+b+ 25

Optional Information:
Subject: Principles of Geneti

Already Tried:
Nothing much. Just the myanswers section from yahoo.

For the previous request, I think I know the probabilities/distribution of them. Here they are:

.95 .90 .80 .70 .50 .30 .20 .10 and it breaks away from .05 .01 .001

Those can't be the probabilities. The sum of the probabilities must be equal to 1.

Sincerely,
Sk1llz

Not unless I'm given more information. We need to find a way to find the expectation, but since I'm a mathematician without any biological education I have no understanding about genetics.

Sincerely,
Sk1llz