# Thread: Hardy-Weinberg equations...How do you do them?

1. Below is an example of a question that may show up on a test:

A small lizard living on the sandy plains of India is preyed upon by mongooses (small, carnivorous mammal). Its defense includes freezing to avoid attracting the predators eye. In addition, it is cryptically colored. The color alleles are B and b. Genotype BB is a dark brown with light brown stripes, Bb is medium brown with light brown stripes, bb is uniformly very light grey. A preliminary count shows that BB occurs at a frequency in the population of 36%. Calculate the gene frequencies of B and b. Next, use the Hardy Weinberg equation to calculate the expected frequencies of the other two genotypes.  2.

3. I won't answer homeworks for you , but I will explain H-W equilibrium.

Let's start with the definition. For any population where the following is NOT the case:

non-random mating, mutations, selection, limited population size, random genetic drift and gene flow

Then the population will be H-W Equilibrium. What this means, is that you have a certain proportion of alleles in the population, and if random mating occurs and these alleles don't mutate, then on average (average requires a large population - otherwise allelic extinctions can occur by chance) the proportion of alleles will remain the same. Makes sense right? I'm not introducing new alleles, all mating does is jumble them up. It's like I have two boxes of red and blue balls and I continually switch balls back and forth between them: the total proportion of red and blue doesn't change.

So, for a two allelic system (B or b), you can have BB, Bb, bB and bb right? Also Bb and bB are the same (ignoring imprinting). So if I know the proportion of people with BB, the probability (= proportion of people in a population) of having this is P(B) * P(B). I needed to pick two blue balls out of the box with probability P(B) for each.

But how do I know what P(b) is? I'm not going to tell you, but think about it: alleles in this population are only allowed to be P(B) or P(b) - so knowing P(B), how do I work out P(b)?

From here you should be able to work out Bb/bB (remember there are TWO options here!) and bb.

Let me know how things go.   4. i believe the equation is, p squared + 2pq + q squared  5. But how do I know what P(b) is? I'm not going to tell you, but think about it: alleles in this population are only allowed to be P(B) or P(b) - so knowing P(B), how do I work out P(b)?
P(b) must be 64% then? And then I plug it in to the equation (p squared + 2pq + q squared)?

From here you should be able to work out Bb/bB (remember there are TWO options here!) and bb.
What are they?  6. I'm afraid that's not the right answer.

Think about it, a genotype (ie. BB) is made up of TWO alleles. We want to calculate the frequency of the alleles, not the genotype.

We are told that the genotypic frequency of BB is 36%. We need to work out from this the proportion of alleles in the population that are B as opposed to b. So I pick *two* alleles out of a box and I get BB 36% of the time. So if I pick one allele, what is the probability it is B?

Hint: P(BB) = P(B)^2

The equation given by madison is good and correct, but it's much better to understand how to use these things and what they conceptually mean than to try and blindly apply a formula.

Let me know if this helps   7. All I know is:

BB (p squared)

Bb (2pq)

bb= q squared

Think about it, a genotype (ie. BB) is made up of TWO alleles. We want to calculate the frequency of the alleles, not the genotype.
Understood.

So I pick *two* alleles out of a box and I get BB 36% of the time. So if I pick one allele, what is the probability it is B?

Hint: P(BB) = P(B)^2
I don't get it...I'm not mathematically inclined..  8. I'm not even taking this level of Biology yet. I just want to learn how to work these equatons now in order to get a head start. I just need to see one example completed (preferrably the one above because it has you calculate both the gene frequencies of B and b as well as expected frequencies) and explained fully in lay-person language to unterstand.  9. The equation are simply derived from squaring a binomial, i.e., . (Note that since the sum of p and q is one, than squaring the binomial doesn't change the numerical value of the expression since one to any power is one). The equations allow us to relate genotypic and phenotypic frequencies, and to use them you must first find q, or more generally q^2, which you can take the square root of to find q.

In this case, homozygous dominant BB has a frequency of 0.36, therefore B as a frequency of 0.6. Therefore, allele b has a frequency of 0.4, since . Therefore, the frequencies are the following:

B: 0.6
b: 0.4
BB: 0.36
bb: 0.16
2Bb: 0.48  10. I get the first part and thanks :-D but how did you get bb: 0.16
2Bb: 0.48?

Is it 2Bb= 2(6x4), and bb= 4x4?  11. Originally Posted by gottspieler
I get the first part and thanks :-D but how did you get bb: 0.16
2Bb: 0.48?

Is it 2Bb= 2(6x4), and bb= 4x4?
You are correct. Although, you should be careful in your notation and write your significant digits after the decimal place. For the case of bb, for example, I took , since b is equal to 0.4. For 2Bb I simply did the arithmetic, as you noted, which was equivalent to the expression .

Feel free to write back if you have any other difficulties.  12. 2Bb: 0.48
This doesn't tell me the frequency of Bb, it's the frequency of Bb times itself...so would I need to divide by 2 for the frequency of Bb?  13. Originally Posted by gottspieler
2Bb: 0.48
This doesn't tell me the frequency of Bb, it's the frequency of Bb times itself...so would I need to divide by 2 for the frequency of Bb?
No. The frequency of the heterozygous genotype is denoted , and the formula for finding it is . The first term in the above trinomial is , which is the frequency of the homozygous dominant genotype. The final term, , is the frequency of the homozygous recessive genotype. Therefore, the second term, 2Bb has to be the frequency of the heterozygous genotype, since there are three possible genotypes. The formula for the heterozygous gentype than is . Than you can find the frequency of the heterozygous genotype by multiplying the product of the recessive and dominant allele by two.

Let's try another example:

"The frequency of two alleles in a gene pool is 0.4 (A) and 0.6 (a). What is the percentage of the heterozygous individuals in the population, assuming that the population is in Hardy-Weinberg equilibrium?"

In this case, we have two alleles, A and a, with frequencies 0.4 and 0.6 respectively. In this question, we are asked to find the heterozygous genotype frequency, which can be found by taking twice the product of the dominant and recessive allele, or in thise case, 2Aa. Since , our heterozygous genotype frequency is 0.48, or in other words fourty-eight percent of the population.

As a final note, the method for finding the frequency of the heterozygous individuals in the population might be confusing to some, since the formula for finding the homozygous dominant and homozygous recessive frequencies are the same as their notations, while that for heterozygous frequency is not. By this I mean that the formula for finding the frequency of BB (the dominant genotype) is B^2, which is equal to BB, and the formula for finding bb (the homozygous recessive genotype) is b^2, which is equal to bb, while the formula for finding the frequency of the Bb (the heterozygous genotype) is 2Bb. You must simply look past that.  14. Thanks. I get it now. I won't need this until I take Evolution 4111 which will probably be the semester right before I graduate with a B.S. in Bio (wish me luck..lol) but knowing it now will soften the blow of what I'll be dealing with in the future concerning population genetics...  15. Originally Posted by gottspieler
Thanks. I get it now. I won't need this until I take Evolution 4111 which will probably be the semester right before I graduate with a B.S. in Bio (wish me luck..lol) but knowing it now will soften the blow of what I'll be dealing with in the future concerning population genetics...
Alright. For practice, try this problem:

"The allele for the ability to roll one's tongue is dominant over the allele for the lack of this ability. In a population of 5000 individuals, 45% show the recessive phenotype (bb). How many individuals would you expect to be homozygous dominant (BB) and heterozygous for this trait?"  16. Originally Posted by Ellatha Originally Posted by gottspieler
Thanks. I get it now. I won't need this until I take Evolution 4111 which will probably be the semester right before I graduate with a B.S. in Bio (wish me luck..lol) but knowing it now will soften the blow of what I'll be dealing with in the future concerning population genetics...
Alright. For practice, try this problem:

"The allele for the ability to roll one's tongue is dominant over the allele for the lack of this ability. In a population of 5000 individuals, 45% show the recessive phenotype (bb). How many individuals would you expect to be homozygous dominant (BB) and heterozygous for this trait?"

Would I start by finding the square root of 45 to find the b allele frequency(if so, how can I write it as a percent and figure out B(.67 I guess))? Then by default I'd know B. Then find 2Bb. Then what's left over is BB?  17. I am glad to see that you understand it; congratulations.

Regarding your questions... 0.45 indicates 45%. From this, you know that B + (0.45)^(1/2) = 1, than subtracting (0.45)^(1/2) from both sides would yield B. You could also find BB by squaring B, as a note. You will probably be ahead of your class when you begin learning more extensively about Hardy-Weinberg.  18. I greatly appreciate your help.  Bookmarks
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