## 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 8 months ago 8 months ago

1. aaronq

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

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

so what is p

4. aaronq

p = frequency of the dominant allele in the population

5. KarolinaJosephine

yes I know but I don't know what it is

6. aaronq

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

but I don't know how. that's the reason I posted it.

8. KarolinaJosephine

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

they gave you q^2 right? find q, since p+q=1 -> p=1-q

10. KarolinaJosephine

I have $p^{2}+2pq+q^{2}=1$ and p+q=1

11. aaronq

look at the last thing i wrote

12. KarolinaJosephine

wouldn't that give a negative number?

13. aaronq

how? taking the squared root of the percentage of homozygous recessive individuals wouldn't give you a negative number.

14. KarolinaJosephine

idk I'm all confused

15. aaronq

what are you confused about? $q ^{2}= a \to \sqrt{q ^{2}}=\sqrt{a}$ $q = \sqrt{a}$

16. KarolinaJosephine

and how did you come up with that

17. aaronq

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

I'm about to say F it. I don't understand any of this.

19. aaronq

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

watched it. still confused.

21. asmagul

you have to study genetics and hardy Weinberg theorem

22. KarolinaJosephine

well I know that much.

23. aaronq

what don't you understand?

24. KarolinaJosephine

all of it

25. aaronq