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Homework Help: Probability of a colour blind grandson

  1. Jul 30, 2015 #1

    Suraj M

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    1. The problem statement, all variables and given/known data
    Here is the question as it is.
    A colour blind man marries a woman with normal sight who has no history of colour blindness in her family. What is the probability of their grandson being colour blind?
    (1) 0.25
    (2) 0.5
    (3) 1
    (4) Nil
    2. Relevant equations
    --

    3. The attempt at a solution
    Im sorry, but i hope the images will do,
    IMG_1732.JPG
    Ignore the zero i've written, that branch is anyway not needed..

    IMG_1733.JPG

    Is this correct?
    Just wanted to check, they haven't mentioned anything else, so have to consider everything, right?
    But my teacher says that we dont need to consider all these cases for questions like these.
    please help me out
    thank you
     
    Last edited: Jul 30, 2015
  2. jcsd
  3. Jul 30, 2015 #2
    It's like none of there sons gonna be colour blind, half of daughters will be normal and half carrier. From the carrier daughters,whatever husband they get only 1/2 sons will be colour blind. That's the solution in genetics. You calculate the probability using this data in mathematical equation. Maybe your answer is not correct as half of sons of the daughters are colour blind, from carrier daughters only. And carrier daughters are half of the total daughters.
     
  4. Jul 30, 2015 #3
    Suppose they have four children, two daughters and two sons(only one daughter is carrier). Each of the children gets four children same way, two sons and two daughters. So, total grandsons are 4*2=8.

    Again only a single son of the carrier daughter is probable to be a colour-blind.
     
  5. Jul 30, 2015 #4
    I am sorry, I mixed up. Both the daughters will be carrier. If they have 4 children each, two daughters and two sons then 1 of the sons will be colour blind, no matter what husband they get. So, two daughters will have two colour blind sons. Finally there are 2 out of 8 grandsons who are colour blind. The probability is 2/8 or 0.25
     
  6. Jul 30, 2015 #5

    Suraj M

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    2 out of 4 right?, there are just 4 grandsons?
     
  7. Jul 30, 2015 #6
    8 grandsons

    In post 2, I said how. Its not 8 grandsons, its 8 probable types of grandsons
     
  8. Jul 30, 2015 #7
    That's what you do in genetics. There are two types of genes coming from father, two types from mother, total makes 4 types. Since 2 of these types are female, so male types are just 2.
     
  9. Jul 30, 2015 #8

    Suraj M

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    I don't understand, whatever be the genotype of the male that the carrier marries, the probability of her son to be colourblind is 1/2 ???
    so shouldn't 0.5
     
  10. Jul 30, 2015 #9

    RUber

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    You are assuming that whomever the son or daughter marries is also neither colorblind nor a carrier, right?
    So this is the minimum probability. If you add in the chance of the son or daughter marrying someone with the gene, the odds go up a little bit.
     
  11. Jul 30, 2015 #10

    RUber

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    There should be no chance for the son to have the gene, since it is on the X chromosome, right?
     
  12. Jul 30, 2015 #11
    Yes, probability for her colour blind son is 0.5, but the question is not her probability to have colour blind sons, its her parents probability to have colour blind grandsons
     
  13. Jul 30, 2015 #12
    No, I am assuming about the daughters as it will be all the same then. No assumption to who the sons are married to.

    Here, the sons can be married to three types of women- normal, colourblind and carrier. I just assumed it to be normal.

    If yoy consider all the three cases then for each son

    1) having normal wife colour blind son is 0 out of 2, Then probability 2/8

    2) For carrier, it is 1 out of 2. For bolth sons its 1 out of 2.

    Probability 3/8

    3) colour-blind woman then it is 2 out of 2. For both sons its 4 out of 4

    Probability 5/8

    Then final probability should be calculated by mathematics
     
  14. Jul 30, 2015 #13
    The no.2 is a mistake. For both sons it will be 2 out of 4, so total probabilities is 4 out of 8.
     
  15. Jul 30, 2015 #14

    RUber

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    That's not how probability works. I am sure that the probability of a non-carrier woman is not the same as a colorblind woman. Without additional information about the population of possible wives, you should probably go for the minimal case where the son's wife would be a non-carrier.

    Example: Assume 90% of potential wife population is non-carrier, 9% is carrier, 1% is colorblind
    The probability of a son's son being colorblind is (.09)*(.5) + (.01)*(1) = .055.

    Similarly assume potential husband population is similarly distributed with 90% non-colorblind and 10% colorblind.
    Then the daughter's son's odds of being colorblind are (.9)*(.5) + (.1)*(.5) = .5.

    Then the odds of a grandson being colorblind would be p(son)*p(colorblind son of son) + p(daughter)*p(colorblind son of daughter) = (.5)*(.055) + (.5)*(.5) = .2775.

    Since no additional information about the population statistics are provided in the problem, you should only assume non-colorblind, non-carriers, which would give an answer just slightly smaller than .2775.
     
  16. Jul 30, 2015 #15

    Suraj M

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    Oh sorry i forgot to give you the options, i edited the post, do see it,
    so 0.25 is your answer?
    I think we should assume a mendelian population.
     
    Last edited: Jul 30, 2015
  17. Jul 30, 2015 #16
    I could get it that there was some problems in my later calculation but I couldn't understand where.

    But the part with the husband is not correct. Whatever husband the daughter gets the probable colour blind son for each daughter is always 0.5 as fathers don't give the gene to the sons.
     
  18. Jul 30, 2015 #17
    0.25 is the answer just when you consider that the sons have normal wives.

    I think its preferable answer if you go with your teacher you need not consider all the things (like what wives they marry to). But if you consider with all these then you need extra data and also calculate the probability of sons son to be colour blind in the process RUber did
     
  19. Jul 30, 2015 #18

    Suraj M

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    Agreed on that, but can you ignore the male? what if he marries a colourblind woman or a carrier?
     
  20. Jul 30, 2015 #19

    Suraj M

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    Well this question appeared in a national exam (AIPMT) on saturday, the answers aren't out yet, but the answers ive come across by experts is 0.5 and Nil.
    I have seen both these answers,
     
  21. Jul 30, 2015 #20
    Then its not possible to say how they do it, or what conditions they are really asking for. Maybe they skip the part of the sons and calculated just the grandsons of the daughters as sons aren't responsible to colour blind grandsons. Then it will be 1/2. If you get the solutions then let me know.
     
  22. Jul 30, 2015 #21

    Suraj M

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    Wouldn't what Ive done in the attempt make sense? My teacher thinks the answer is 0.5!
     
  23. Jul 30, 2015 #22
    With the normal sons, the blue marked genotypes, are you considering them as wives? Then why their probability 1/4, 1/2, and 1/4?

    If you consider the population it will be according to RUbers calculation, otherwise taking all equal it maybe 1/3 each. Maybe you took the probability of wives same as children.
     
  24. Jul 30, 2015 #23

    Suraj M

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    In a mendelian population, 50% of women are carriers 25% homozygous( each). thats why i took 1/4 1/2 1/4.
     
    Last edited: Jul 30, 2015
  25. Jul 30, 2015 #24
    Your math approach is correct. If Mendelian population can be considered then it maybe a solution.

    Though I am not sure if Mendelian population can be considered. Of course every 1 out of 4 women are not colour blind. The population is much less.

    If I were to answer my bet would have been 0.25. But Nil and 0.5 may also be answer.
     
  26. Jul 30, 2015 #25

    RUber

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    Suraj, I would recommend against the Mendelian population assumption, unless it is simply an academic exercise or you have been instructed to do so.

    In the event that you do make that assumption, you would have:
    P(grandson is colorblind) = (.5) * ( (.25)(0) + (.5)(.5) + (.25)(1)) +(.5)(.5) = .5. So I would concur with your math and fireflies.

    The true probability, based on realistic population prevalence of the condition, would be much close to .25.
     
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