# The Hardy Weinberg Equilibrium

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Alright so basically I'm doing grade 12 U Biology in order to upgrade before heading to college, although im entirely stuck on how to this. I remember doing punnet squares in grade 11 but this just seems a little complex, and the text books really don't help that much.

And here's the questions I am stuck on. I'm totally lost. Like I tried it out and to find 2pq I didnt end up with 1.

26. A population has two alleles, B and b. The allele frequency of B is 0.73. What is the heterozygotes frequency, if the population is in the H-W equilibrium? (10 marks)

Marking guide

• Calculation indicate that heterozygotes frequency, if the population is equal to 2pq in the Hardy=Weinberg equation (2 Marks)
• Substitutes correct number into equation "p+q=1" ( 2 marks)
• Obtains correct result using equation "p+q=1" (2 marks)
• Obtains correct final result (2 marks)
• States the final result in a meaningful sentence (2 marks)

27.

a) Using the data in the data in the diagram provided in figure 14.4 of a theoretical bacterial or insect species, graph the phenotypes of the resistant and non-resistant forms over the three generations a), c), and e).

b ) Calculate the p and q frequencies for each of the three generation's starting populations, assuming the resistant allele is recessive.

Figure 14.4

Marking guide

• Accurately counts and then calculates the percentage of resistant and non-resistant forms in the three generations show in a), c), and e) of figure 14.4 (5 marks)
• Correctly represents data from above calculations in graph form (5 marks)
• Above graph has appropriate labels (1 mark)
• Accuratly calculates the p and q frequencies of all three generations (3 x 3 =9 marks)

Heres the example from the text book

Edited by Ninjakat

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And here's the questions I am stuck on. I'm totally lost. Like I tried it out and to find 2pq I didnt end up with 1.

26. A population has two alleles, B and b. The allele frequency of B is 0.73. What is the heterozygotes frequency, if the population is in the H-W equilibrium? (10 marks)

2pq does not have to equal 1. The Hardy-Wienberg Equation states $p^{2}+2pq+q^{2}=1$. So in this problem B=p=.73. The equation p+q=1 allows ups to find q=1-p=1-.73. Then to find the frequency of heterozygotes we just do $2pq=2(.73)(1-.73)$

As for 27 could you post the required figure, 14.4?

Edited by DJBruce

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yep sorry i forgot to add it!

Alright for that one the answer for 26 i dont quit understand at all.. Like imagine teaching this too someone whos never done math or bio before and explain in lamens terms and in small steps.

thanks

Edited by Ninjakat

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2pq does not have to equal 1. The Hardy-Wienberg Equation states $p^{2}+2pq+q^{2}=1$. So in this problem B=p=.73. The equation p+q=1 allows ups to find q=1-p=1-.73. Then to find the frequency of heterozygotes we just do $2pq=2(.73)(1-.73)$

As for 27 could you post the required figure, 14.4?

So if 2pq= 2 (0.73) (1-0.73)

=2 (0.73) (0.27)

= 2 x 0.1971

= 0.3942

so is the answer 0.39 ?

And what exactly do they want me to show for my work?

Now i think i just need help with 27 :\

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Marking guide

[/b]

• Calculation indicate that heterozygotes frequency, if the population is equal to 2pq in the Hardy=Weinberg equation (2 Marks)
• Substitutes correct number into equation "p+q=1" ( 2 marks)
• Obtains correct result using equation "p+q=1" (2 marks)
• Obtains correct final result (2 marks)
• States the final result in a meaningful sentence (2 marks)

So from your grading scale you'll wanna define what p is in this case. You'll then wanna show how you solved for q in the equation, p+q=1. You should then define what 2pq is equal to, followed by substituting in the actual values for p and q, and evaluating it. Once you evaluate it you'll wanna explain your answer in context in a sentence.

As for 27 the first part asks you to sketch a graph of the phenotypes of the population over time. So you could make two line graphs where you plot the number of organisms with one phenotype as one line and the number of individuals for the other phenotype as the other line.

As for the other part you need the frequency of the homozygous recessive will be equal to $q^{2}$. So for each population calculate the frequency of homozygous recessive then take the square root of that to find q. Once you know q you can solve the equation p+q=1 for p. You then just have to do that for each population.

Edited by DJBruce

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As for 27 the first part asks you to sketch a graph of the phenotypes of the population over time. So you could make two line graphs where you plot the number of organisms with one phenotype as one line and the number of individuals for the other phenotype as the other line.

How do i get the penotypes and population? like what numbers am I using?

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The phenotypes of the organisms are either:

- Resistant to toxin

- Non-resistant to toxin

To find the phenotypes of each individual look at what type of dot represents them. To calculate the phenotype frequency for the recessive trait count all dots showing the recessive trait and divide that by the total number of dots. For example in the initial population, square a, there are 56 total dots, individuals, 3 of the dots are colored meaning that they show resistant to the toxin so 3. The other 53 dots, individuals, show the non-resistant to toxin phenotype. To calculate the homozygous recessive frequency we would take the number showing the recessive trait, 3 and divide it by the total population of 56. So:

$\frac{3}{56}=q^{2}$

$\sqrt{\frac{3}{56}}=q$

So now you know q and then should be able to find p. I hope this helps.

Edited by DJBruce

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I dont get it.. i get 18.6

Doesnt P+q = 1..

I dont understand what the hell im doing.

If I square 18.6 I get 4.31 which is still wrong

Edited by Ninjakat

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I dont get it.. i get 18.6

For what do you get 18.6?

Doesnt P+q = 1..

I don't understand what the hell im doing.

Yep

$p+q=1$

That means these are also true:

$1-p=q$

$1-q=p$

Don't get down on yourself you can do it.

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HI, Im working on the same course and Im having a hard time.