(Solved) Calculating probabilities in pedigrees of X-linked conditions

freeblue

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10 years ago

Calculating probabilities in pedigrees of X-linked conditions

I understand this chapter for the most part, but I really can't solve this one question. Sorry, I couldn't get the question all in one screenshot so I had to use two! Thanks in advance for the help!

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padre

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Pjjjw

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9 years ago

Thanks

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ldiaz2012

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9 years ago

Thanks for the great explanations! really helpful Smiling Face with Open Mouth

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padre

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9 years ago

Given the feedback, I'm marking this as solved, unless someone else can come up with something.

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vynovegil

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9 years ago

A good method for calculating probabilities in pedigrees is to consider the requirements for a certain outcome.

For question 1, the following are required for IV-3 to have condition A (aa):
II-5 has the genotype Aa (probability = 2/3).
If II-5 has the genotype Aa (accounted for by the above probability), then he passes an a allele to III-4 (probability = 1/2).
If III-4 has the genotype Aa (accounted for by the above probability), then she passes an a allele to IV-3 (probability = 1/2).
II-7 has the genotype Aa (probability = 2/3).
If II-7 has the genotype Aa (accounted for by the above probability), then he passes an a allele to III-5 (probability = 1/2).
If III-5 has the genotype Aa (accounted for by the above probability), then he passes an a allele to IV-3 (probability = 1/2).

All of these requirements are needed in sequence, so you apply the product rule (2/3 x 1/2 x 1/2 x 2/3 x 1/2 x 1/2 = 1/36).

Similarly, for question 2, the following are required for IV-3 to have condition B:
II-4 passes an X b chromosome to III-4 (probability = 1/2).
If III-4 has the genotype X B X b (accounted for by the above probability), then she passes an X b chromosome to IV-3 (probability = 1/2).
III-5 passes a Y chromosome to IV-3 (probability = 1/2).

All of these requirements are needed in sequence, so you apply the product rule here, too (1/2 x 1/2 x 1/2 = 1/8).

Once the individual probabilities are known, the sum and/or product rules can be used for various combinations (both conditions, either condition, etc.)

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padre

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gw234

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9 years ago

Thank You

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grumbelcakes

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9 years ago

Thank you

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PhillipePet

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8 years ago

No, we dont. Or I dont. Ive read your all your posts since we started posing the question, and I can find no answer.

You may think youve answered it, but in that case, you need to link to it, so that we can figure out why your answer doesnt look like one to us.

Do you understand that you cannot begin to calculate a probability until a you know what event or class of events you are calculating the probability of and b under what conditions the probability is to be calculated for?

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padre

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8 years ago

My post has the accepted answer...

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Loloboi

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#10

8 years ago

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shaneisbased

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#11

5 years ago

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BennyDooney

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#12

5 years ago

Thank you!!

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Noharmony

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#13

5 years ago

Thank you!

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