0
$\begingroup$

Before anything, I figured this would be better on math than in Biology.stackexchange.

So, I am currently enrolled in Biology in school, in which they are teaching us Hardy-Weinberg equations. My answer is incorrect, and I want to know why.

Context:

  • Hardy Weinberg is in the form $p+q=1 \implies p^2+2pq+q^2=1$

  • $p$ is the frequency of the dominant allele. $q$ is the frequency of the recessive allele. There are in total, $2$ types of alleles.

  • $p^2$ is the amount of the next population (out of $1$) which is homozygous dominant, $2pq$ is the amount of the next population which is heterozygous, and $q^2$ is the amount which is recessive

The question goes as follows:

After graduation, you and $19$ of your closest friends (let's say $10$ males and $10$ females) charter a plane to go on a round-the-world tour. Unfortunately, you all crash land on a deserted island. No one finds you and you start a new population totally isolated from the rest of the world. Two of your friends carry (i.e. are heterozygous for) the recessive cystic fibrosis allele (denoted as little $c$). Assuming that the frequency of this allele does not change as the population grows, what will be the incidence of cystic fibrosis on your island?

The answer states:

There are $40$ total alleles in the $20$ people of which $2$ alleles are for cystic fibrosis. So, $\frac 2{20}=0.05$ of the alleles are for cystic fibrosis. That represents $q$. Thus, $cc$ or $q^2=(0.05)^2=0.0025$ or $0.25$% of the population will be born with cystic fibrosis.

This makes complete sense to me. However, it seems as I am doing something wrong in the following systems of equations to solve for $p$ and $q$.

$$p+q=1$$

$$2pq = 0.1$$

Where $0.1$ represents the $2$ out of $20$ people who are heterozygous.

I solved for $q$, which is the percentage of the recessive allele, and got $q=\dfrac{5\pm 2\sqrt{5}}{10}\approx 0.052786404$, the other solution does not apply.

$q^2 \approx 0.00279$, or around $0.279$% of the next population will be born with cystic fibrosis.

Clearly, solutions are not matching. What have I done wrong? And no, it is not a rounding error.

$\endgroup$
  • $\begingroup$ The mistake is in the statement $2pq=0.1$ $\endgroup$ – Jan Mar 15 '18 at 21:02
1
$\begingroup$

The statement that "$p^2$ is the amount of the next population (out of 1) which is homozygous dominant, $2pq$ is the amount of the next population which is heterozygous, and $q^2$ is the amount which is recessive" is not true initially. You yourself say it is true of "the next generation." For example we have $q=.05$ but the number of recessives is not $20\cdot .05^2,$ but $0.$ Similarly for $p^2$ and $2pq.$

Even after the population has reached steady state, imagine that another plane crashes, and all the passengers happen to be heterozygous. Obviously, the equation is not somehow immediately true with the new values of $p$ and $q,$ before any mating has taken place with the new arrivals.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy