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KarolinaJosephine

  • one year ago

can anyone help with this? it seems like such a simple question but my brain just isn't grasping it. in the US approximately one child in every 10,000 is born with phenylketonuria a syndrome that affects individuals homozygous for the recessive allele. given this incidence what percent of the population are heterozygous carriers of the recessive PKU allele? give answer to the nearest hundredth of a percent

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  1. aaronq
    • one year ago
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    The only equations you need for these types of problems: p + q =1 p = frequency of the dominant allele in the population q = frequency of the recessive allele in the population and p^2 + 2pq + q^2 =1 p^2 = percentage of homozygous dominant individuals q^2 = percentage of homozygous recessive individuals 2pq = percentage of heterozygous individuals

  2. aaronq
    • one year ago
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    you know it's recessive and they give you a number of individuals, you can convert it to a percentage of homozygous recessive individuals (q^2). You're trying to find out what 2pq is.

  3. KarolinaJosephine
    • one year ago
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    so what is p

  4. aaronq
    • one year ago
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    p = frequency of the dominant allele in the population

  5. KarolinaJosephine
    • one year ago
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    yes I know but I don't know what it is

  6. aaronq
    • one year ago
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    that's the challenge in the question, if i told you the answer i wouldn't be of much help would i? use the formulas i wrote.

  7. KarolinaJosephine
    • one year ago
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    but I don't know how. that's the reason I posted it.

  8. KarolinaJosephine
    • one year ago
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    I have a paper with some equations on it so I have that but I don't know what to do with it

  9. aaronq
    • one year ago
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    they gave you q^2 right? find q, since p+q=1 -> p=1-q

  10. KarolinaJosephine
    • one year ago
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    I have \[p^{2}+2pq+q^{2}=1\] and p+q=1

  11. aaronq
    • one year ago
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    look at the last thing i wrote

  12. KarolinaJosephine
    • one year ago
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    wouldn't that give a negative number?

  13. aaronq
    • one year ago
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    how? taking the squared root of the percentage of homozygous recessive individuals wouldn't give you a negative number.

  14. KarolinaJosephine
    • one year ago
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    idk I'm all confused

  15. aaronq
    • one year ago
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    what are you confused about? \[q ^{2}= a \to \sqrt{q ^{2}}=\sqrt{a}\] \[q = \sqrt{a}\]

  16. KarolinaJosephine
    • one year ago
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    and how did you come up with that

  17. aaronq
    • one year ago
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    it's math... one child in every 10,000 is born with phenylketonuria a syndrome that affects individuals homozygous.. q^2 = percentage of homozygous recessive individuals so solve for q, just take the squared root of both sides

  18. KarolinaJosephine
    • one year ago
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    I'm about to say F it. I don't understand any of this.

  19. aaronq
    • one year ago
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    so, you have q^2=1/10000 to solve for q, you take the squared root of both sides: \[q ^{2}=\frac{ 1 }{ 10000 } \to q=\sqrt{\frac{ 1 }{ 10000 }} \to q=0.01\] so p=1-q = 0.99 2ab is the percentage of heterozygous individuals (i.e. carriers) 2(0.99)(0.01)= 0.0198 1.98 % Watch this video, i think he explains it well. http://www.youtube.com/watch?v=xPkOAnK20kw

  20. KarolinaJosephine
    • one year ago
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    watched it. still confused.

  21. asmagul
    • one year ago
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    you have to study genetics and hardy Weinberg theorem

  22. KarolinaJosephine
    • one year ago
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    well I know that much.

  23. aaronq
    • one year ago
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    what don't you understand?

  24. KarolinaJosephine
    • one year ago
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    all of it

  25. aaronq
    • one year ago
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    re-read the question and narrow down what you don't understand, i can't help you otherwise.

  26. KarolinaJosephine
    • one year ago
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    I never said you alone had to help me.

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