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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
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
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.
p = frequency of the dominant allele in the population
yes I know but I don't know what it is
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.
but I don't know how. that's the reason I posted it.
I have a paper with some equations on it so I have that but I don't know what to do with it
they gave you q^2 right? find q, since p+q=1 -> p=1-q
I have \[p^{2}+2pq+q^{2}=1\] and p+q=1
look at the last thing i wrote
wouldn't that give a negative number?
how? taking the squared root of the percentage of homozygous recessive individuals wouldn't give you a negative number.
idk I'm all confused
what are you confused about? \[q ^{2}= a \to \sqrt{q ^{2}}=\sqrt{a}\] \[q = \sqrt{a}\]
and how did you come up with that
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
I'm about to say F it. I don't understand any of this.
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
watched it. still confused.
you have to study genetics and hardy Weinberg theorem
well I know that much.
what don't you understand?
re-read the question and narrow down what you don't understand, i can't help you otherwise.
I never said you alone had to help me.