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Old 02-13-2011
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Stats The probability that these genes transfer together?

Genes on the bacterial chromosome have the following linkages in conjugal transfer: x and y, 25% of the time; y and z, 50% of the time. If the gene order is x-y-z, approximately what percentage of the time will x and z be transferred together?

A. 1% of the time
B. 5% of the time
C. 13% of the time
D. 20% of the time
E. 40% of the time
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Old 02-13-2011
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Default C.

My guess is C. 13% of the time

Chance that a Y will come with a Z (.5) * chance that an X will come with that Y (.25)

.25 * .50 = .125 ≈ 13%
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Old 02-13-2011
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Default Not sure

Im not sure how to answer this question
but i was thinking along the lines of

(0.25+0.50) - (0.25*0.50) which equals 62.5

this gives probability of not mutually exclusive events.
Not sure, im waiting for the answer and an explanation!
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Old 02-13-2011
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Default

Quote:
Originally Posted by Hope2Pass View Post
Im not sure how to answer this question
but i was thinking along the lines of

(0.25+0.50) - (0.25*0.50) which equals 62.5

this gives probability of not mutually exclusive events.
Not sure, im waiting for the answer and an explanation!

Doesn't that give the probability of one of two events happening or both (genes transferred ∈ {x,y} ∪ {y,z} ∪ {x,y,z})?

The events for which the probabilities are given aren't mutually exclusive; however, they are also not separate, since x, y, and z are on the same chromosome. {x,y,z} is the only set that could include both x and z.

I was thinking of the probabilities as approximating synteny; that is to say, since y and z together is twice as probable as x and y, then y and z must be about half the distance apart:

------x------y---z----------

So the probability that the whole string would be transferred would approximate the probability of the x-z synteny, which would be the inclusive probability of both the x-y and y-z syntenies. Which, I think, is the product of the two.
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Quote:
Originally Posted by Mondoshawan View Post
My guess is C. 13% of the time

Chance that a Y will come with a Z (.5) * chance that an X will come with that Y (.25)

.25 * .50 = .125 ≈ 13%

Very good.

Correct Ans is C) 13%
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