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KarolinaJosephine Group Title

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

  • one year ago
  • one year ago

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  1. aaronq Group Title
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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

    • one year ago
  2. aaronq Group Title
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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.

    • one year ago
  3. KarolinaJosephine Group Title
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    so what is p

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

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

    • one year ago
  6. aaronq Group Title
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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.

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

    • one year ago
  8. KarolinaJosephine Group Title
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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

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

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

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

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

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

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

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

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

    • one year ago
  17. aaronq Group Title
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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

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

    • one year ago
  19. aaronq Group Title
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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

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

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

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

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

    • one year ago
  24. KarolinaJosephine Group Title
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    all of it

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

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

    • one year ago
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