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KarolinaJosephine

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

  • 9 months ago
  • 9 months ago

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  1. aaronq
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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

    • 9 months ago
  2. aaronq
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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.

    • 9 months ago
  3. KarolinaJosephine
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    so what is p

    • 9 months ago
  4. aaronq
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    p = frequency of the dominant allele in the population

    • 9 months ago
  5. KarolinaJosephine
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    yes I know but I don't know what it is

    • 9 months ago
  6. aaronq
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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.

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

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

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

    • 9 months ago
  11. aaronq
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    look at the last thing i wrote

    • 9 months ago
  12. KarolinaJosephine
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    wouldn't that give a negative number?

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

    • 9 months ago
  14. KarolinaJosephine
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    idk I'm all confused

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

    • 9 months ago
  16. KarolinaJosephine
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    and how did you come up with that

    • 9 months ago
  17. aaronq
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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

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

    • 9 months ago
  19. aaronq
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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

    • 9 months ago
  20. KarolinaJosephine
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    watched it. still confused.

    • 9 months ago
  21. asmagul
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    you have to study genetics and hardy Weinberg theorem

    • 9 months ago
  22. KarolinaJosephine
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    well I know that much.

    • 9 months ago
  23. aaronq
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    what don't you understand?

    • 9 months ago
  24. KarolinaJosephine
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    all of it

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

    • 9 months ago
  26. KarolinaJosephine
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    I never said you alone had to help me.

    • 9 months ago
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