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  1. #1
    unocutie21 is offline Newbie
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    Urgent Bio/Genetics help

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    Hi exam in less than 2 weeks. I need help with hardyWeinberg equations. When they ask for gene frequency and stuff like that with Sickle cell, and Cystic Fibrosis, PKU. I hear there are always questions like that on Step 1. So if anybody had Q & A that i could do. I would appreciate it. I have also heard that u can memorize the answers cause it is always the same like CF is 1 out of 2000. So anything they give will be related to that. So any help anyone can give would be great
    thanks in advance.

  2. #2
    jameslynton's Avatar
    jameslynton is offline Super Moderator 48 points
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    The Hardy-Weinberg are just quadatric equations to solve - like think algebra I. Here are some links:

    http://jon.visick.faculty.noctrl.edu...utlines/19.pdf

    http://darwin.eeb.uconn.edu/eeb348/l...erg/node3.html

    I serious doubt these are on the tests now. You would be better served reviewing bio ethics and what do you do/say next questions over these. Your best bet is to not look at a review book for genertics but look at a good cell biology - biochemistry text book like - Lippencott where they discuss inhertance. Would also review the NM rule for calculating probility would be a swell thing to know also. For example if each parent has a 1/30 chance of CF then 1/30 * 1/30 = 1/900 for the child.
    or one parent has a 1/4 and the other 1/64 then 1/4 * 1/64 = 1/256
    Much more handy dandy - for figuring chances in genetics.
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  3. #3
    maixungphong's Avatar
    maixungphong is offline Junior Member
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    hemophilia

    Quote Originally Posted by jameslynton
    The Hardy-Weinberg are just quadatric equations to solve - like think algebra I. Here are some links:

    http://jon.visick.faculty.noctrl.edu...utlines/19.pdf

    http://darwin.eeb.uconn.edu/eeb348/l...erg/node3.html

    I serious doubt these are on the tests now. You would be better served reviewing bio ethics and what do you do/say next questions over these. Your best bet is to not look at a review book for genertics but look at a good cell biology - biochemistry text book like - Lippencott where they discuss inhertance. Would also review the NM rule for calculating probility would be a swell thing to know also. For example if each parent has a 1/30 chance of CF then 1/30 * 1/30 = 1/900 for the child.
    or one parent has a 1/4 and the other 1/64 then 1/4 * 1/64 = 1/256
    Much more handy dandy - for figuring chances in genetics.

    Here is a question on the USMLE.

    A woman is a known carrier for a rare autosomal recessive disease (incidence 1/10,000). What is the probability of having an affected child if she marries her step sister?
    USMLE ma`y lai. dday tao ddap. ma`y che^'t

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    maixungphong's Avatar
    maixungphong is offline Junior Member
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    Sorry, I missed the question.

    Here is a question on the USMLE.

    A woman is a known carrier for a rare autosomal recessive disease (incidence 1/10,000). What is the probability of having an affected child if she marries her first cousin?
    USMLE ma`y lai. dday tao ddap. ma`y che^'t

  5. #5
    IMG SURVIVOR's Avatar
    IMG SURVIVOR is offline Moderator 536 points
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    and the answer is??????????
    Moderator: USMLE AND Residency Forums.

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    why even bother with the obvious. Just know where you are need it and where you can help the most.

  6. #6
    Unregistered-h Guest

    Where the couisin comes from?

    It depends on the cousin...
    1/4 if the cousin is a carrier to 1/40 000 if the cousin is the cousin because of the other line of the family...
    Let's say, she is a carrier because she inhereted it from her mother. If we asssume that her cousin is son of her mother sister or brother, he has more probabilitites of being a carrier. How probable?: it depends on other information that we do not have.
    If he was a carrier, then the probability of an affected child would be 1/4

    If she is a carrier because she inhereted it from her mother, but the cousin is a son of his father sister or brother, then his cousin has the same probability of having the gen as general population 1/100
    (if the incidence of the disease is 1/10 000 I assume that the gen is carried by 1/100 , since 1/100 AND 1/100 is 1/10 000)
    Following probabilities theory:
    1/4 times 1/100 = 1/400


    PS: I would like that somebody posted the real answer. I am just guessing based on Probability theory and common sense but I am not sure that this is thereal answer (common sense is the less common of the senses, as you know).
    PS:2 I do know for sure that if she marries her step sister the probability of both having an affected child is 0

    h

  7. #7
    jameslynton's Avatar
    jameslynton is offline Super Moderator 48 points
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    Hold on where did you get 1/4 from?

    two cases:
    Case 1 cousin is carrier at same rate as female
    1/10,000 that would give a 1/100,000 chance of expression

    Case two cousin not a carrier
    chance 0 because gene is recessive and you need to inhert two recessive genes for it to be expressed.

  8. #8
    wcb22 is offline Elite Member
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    Quote Originally Posted by jameslynton
    The Hardy-Weinberg are just quadatric equations to solve - like think algebra I. Here are some links:

    http://jon.visick.faculty.noctrl.edu...utlines/19.pdf

    http://darwin.eeb.uconn.edu/eeb348/l...erg/node3.html

    I serious doubt these are on the tests now. You would be better served reviewing bio ethics and what do you do/say next questions over these. Your best bet is to not look at a review book for genertics but look at a good cell biology - biochemistry text book like - Lippencott where they discuss inhertance. Would also review the NM rule for calculating probility would be a swell thing to know also. For example if each parent has a 1/30 chance of CF then 1/30 * 1/30 = 1/900 for the child.
    or one parent has a 1/4 and the other 1/64 then 1/4 * 1/64 = 1/256
    Much more handy dandy - for figuring chances in genetics.
    you are correct that the carrier rate is about 1/30 in caucasians, however, the hardy weinburg equation would put the homozygous rate 1/3600. p2+2pq+q2=1 = disease prevalence, and p+q=1

    another example, sickle cell anemia is a lot better, because the numbers are nice.

    carrier rate in african americans is 1/10 more or less. 2pq = heterozygote prevalence, so (1/20)^2 = 1/400, or the homozygous prevalence.

    someone mentioned the 1/10,000 disease, you can quickly deduce that the heterozygote prevalence = 1/50.

    2pq=1, 1/(2*50) = 1/100, then (1/100)^2 = 1/10,000, the homozygous prevalence.

    i keep 3 numbers in my head at all times, and it makes things easy.

    sickle cell anemia, 1/10 carrier in african americans, 1/400 homozygous
    cystic fibrosis, actually around 1/25 caucasian carrier, 1/2500 homozygous
    something like PKU, 1/50 carrier, 1/10,000 homozygous.

    it gives you a nice reference range to see if you are right or not.
    Last edited by wcb22; 08-16-2006 at 08:01 PM.
    M.D., PGY-3 Internal Medicine

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