Hardy Weinberg Theorem

[cheiney]

Here's a problem that I've been having difficulties with solving from the AP Biology class. The Hardy Weinberg Theorem is used to determine gene frequency of specific alleles. It states: p^2+2pq+q^2=1. p=dominant allele q=recessive allele, pq=heterozygote. Heres an example just so you understand how it works:

In a population of 200 mice, 182 are homozygous dominant and 18 are homozygous recessive. To find the frequency, you would put the number of recessive or dominant traits over the total population, and then square root it. For recessive, it would be the square root of 18/200, which happens to be 0.3, or 30%. 30% of the total alleles are recessive. That means 70% of the alleles are dominant.

This is the question, I'm having problems with, which shows codominance:

There are 500 individuals. 195 of these individuals are phenotypically blood type A, 70 blood type B, 25 blood type AB, and 210 blood type O. (No Rh factor taken into account). Find the gene frequency of each allele.

Note: A can be Ia Ia or Ia i. B can be Ib Ib or Ib i. AB must be IaIb, and O must be ii.

 

[Brian Le]

Oh goodness, I was falling asleep today when my teacher showed us this... But let's take a look see. But be cautioned that my suggestion MAY be wrong.

If you have codominance, couldn't you just put it into the variable 2pq? Or you can just first calculate the frequency of what is easily calculatable, such as blood type O, blood type A, and blood type B. Once those three gene frequencys have been calculated, then the remainding percentage should be the codominant frequency. Or you can put blood type AB as 2pq, because blood type AB is technically heterozygous.

Again, I do not know if that is right, but from what I remember during lecture today, that is one way I would approach it.

 

[cheiney]

It doesn't work out, because AB isn't really heterozygous. AB is homozygous dominant, because both the A allele and the B allele are dominant. The only thing I can really calculate out is the O frequency, which should be the square root of 210/500, which is .648, or 64.8%

 

[Brian Le]

Very true, I forgot that. I failed the genetics unit >.>. When is this homework due? I can try asking my teacher tomorrow.

 

[cheiney]

It's not for homework, it's simply an extra credit question. Try bringing it in tomorrow and see if he/she can do it.

 

[qqwref]

There are 500 individuals. 195 of these individuals are phenotypically blood type A, 70 blood type B, 25 blood type AB, and 210 blood type O. (No Rh factor taken into account). Find the gene frequency of each allele.

Note: A can be Ia Ia or Ia i. B can be Ib Ib or Ib i. AB must be IaIb, and O must be ii.

Here's how I'd do it. (There are many different ways to deal with these numbers; a proper analysis would require some advanced statistics, but this would probably be good enough for an extra credit assignment.)

Consider the individuals that only have Ia or i. They are all type A or type O, right? There are a total of 195 + 210 = 405 of those individuals, and in that population we have p^2+2pq+q^2=1 where p is the probability of an Ia (as opposed to an i). So then q^2 = 210/405 and q is 72% (p is 28%).
Similarly for the individuals that only have Ib or i, they are all type B or type O, so we get a total of 300 individuals and q^2 = 210/300 and q is 83.666% (p is 16.333%).

We want the actual probabilities of Ia, Ib, and i, though... so these aren't the real probabilities just yet. What they tell us are the ratios of the probabilities. Using the notation P(Ia) for the probability (gene frequency) of the single allele Ia, what we know from the above paragraph is that P(Ia)/P(i) = 0.28/0.72, and that P(Ib)/P(i) = 0.16333/0.83666. Since P(Ia)+P(Ib)+P(i) = 1, the sum P(Ia)/P(i) + P(Ib)/P(i) + P(i)/P(i) is 1/P(i) (and P(i)/P(i) is 1). So we get P(i) = 0.631 and then P(Ia) = 0.245 and P(Ib) = 0.123.

These probabilities would give an estimate of 185 type A, 85 type B, 30 type AB, and 200 type O. Since this resembles the numbers we were given, the actual probabilities should be pretty close to these.

 

[Brian Le]

cheiney: I brought it to my teacher and he figured something out, but he didn't finish it because, according to him, "the math is too complex for my standards". He's gonna bring it to a Calculus teacher to see what he can about the problem.

But from what I can remember, you can use the following equation in solving the problem: p^2 + 2pr + q^2 + 2qr + r = 1.

p = dominant, q = recessive, and r = blood type O

The first part, p^2 + 2pr represents the allele for blood type A. p^2 = allele IaIa, 2pr = Iai (remember, little i represents blood type O).

q^2 = allele IbIb. 2qr = allele Ibi.

Finally, r = ii. Or was it IaIb, I do not remember.

Sorry if this wasn't very helpful, but I'll be sure to get the answer as soon as possible. In the meantime, try asking your teacher (unless they don't want to help you).

 

[cheiney]

There are 500 individuals. 195 of these individuals are phenotypically blood type A, 70 blood type B, 25 blood type AB, and 210 blood type O. (No Rh factor taken into account). Find the gene frequency of each allele.

Note: A can be Ia Ia or Ia i. B can be Ib Ib or Ib i. AB must be IaIb, and O must be ii.

Here's how I'd do it. (There are many different ways to deal with these numbers; a proper analysis would require some advanced statistics, but this would probably be good enough for an extra credit assignment.)

Consider the individuals that only have Ia or i. They are all type A or type O, right? There are a total of 195 + 210 = 405 of those individuals, and in that population we have p^2+2pq+q^2=1 where p is the probability of an Ia (as opposed to an i). So then q^2 = 210/405 and q is 72% (p is 28%).
Similarly for the individuals that only have Ib or i, they are all type B or type O, so we get a total of 300 individuals and q^2 = 210/300 and q is 83.666% (p is 16.333%).

We want the actual probabilities of Ia, Ib, and i, though... so these aren't the real probabilities just yet. What they tell us are the ratios of the probabilities. Using the notation P(Ia) for the probability (gene frequency) of the single allele Ia, what we know from the above paragraph is that P(Ia)/P(i) = 0.28/0.72, and that P(Ib)/P(i) = 0.16333/0.83666. Since P(Ia)+P(Ib)+P(i) = 1, the sum P(Ia)/P(i) + P(Ib)/P(i) + P(i)/P(i) is 1/P(i) (and P(i)/P(i) is 1). So we get P(i) = 0.631 and then P(Ia) = 0.245 and P(Ib) = 0.123.

These probabilities would give an estimate of 185 type A, 85 type B, 30 type AB, and 200 type O. Since this resembles the numbers we were given, the actual probabilities should be pretty close to these.

The logic behind this looks correct, and it seems like the rounding of the numbers in each step is what would cause the different numbers. I'll try using this to see what my teacher says and I'll give you feedback. Thanks for the help, I appreciate it.