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High School Biology : Hardy-Weinberg Equilibrium

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Example Questions

Example Question #81 : Genetics And Evolution

In a population of rabbits, the dominant B allele produces brown fur and the recessive b allele produces white fur. There are 544 brown rabbits and 306 white rabbits in the population.

How many of the brown rabbits would you expect to be homozygous dominant (BB)?

Possible Answers:

272

340

Cannot be determined from the information given

172

136

Correct answer:

136

Explanation:

In the Hardy-Weinberg equation, the value of the term p^2 is the percentage of homozygous dominant (BB) individuals in a population. Thus, to answer this question we will need to solve for p^2 .

$p^{2}+ 2pq + q^{2} = 1$

With the information we are given, we can calculate q^2 , which is the percentage of the population that is homozygous recessive (bb). We know the total number of homozygous recessive rabbits (306 white rabbits) and the total number of rabbits in the population (306 + 544 = 850). We can thus divide the number of white rabbits by the population total to find q^2 :

$q^{2} = \frac{bb(white)}{Total} = \frac{306}{850} = 0.36$

36% of the population is homozygous recessive. We then take the square root of q^2 to find :

$q = \sqrt{q^{2}} = \sqrt{0.36} = 0.60$

This value tells us that 60% of all the alleles in the population are recessive (b).

The Hardy-Weinberg equation is based off the idea that the total number of alleles in a population is the sum of the dominant and recessive alleles, as shown by this equation:

p + q = 1

We can use this equation to solve for using :

p = (1-q) = (1-0.60) = 0.40

This value tells us that 40% of all the alleles in the population are dominant (B).

We can now square to obtain p^2 . Recall that this will be the frequency of homozygous dominant individuals in the population:

$p^{2} = (0.40)^{2} = 0.16$

The result is 16%, meaning 16% of the rabbits in the population are homozygous dominant. We can then multiply 16% by the total population to find the number of rabbits who are homozygous dominant:

$BB = (p^{2} \cdot Total) = (0.16 \cdot 850) = 136$

Thus, 136 homozygous dominant rabbits would be expected in this population.

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Example Question #1 : Understanding Hardy Weinberg Calculations

A population of frogs, which is at equilibrium, carries a dominant allele for black spots (B) and a recessive allele for no spots (b). 76 frogs have no spots, and 399 frogs have black spots.

How many of the frogs are heterozygous (Bb) for the black spot trait?

Possible Answers:

190

285

171

228

Correct answer:

228

Explanation:

In the Hardy-Weinberg equation, the value of the term 2pq is the percentage of heterozygous (Bb) individuals in a population. Thus, to answer this question we will need to solve for 2pq .

$p^{2}+ 2pq + q^{2} = 1$

With the information we are given, we can calculate q^2 , which is the percentage of the population that is homozygous recessive (bb). We know the total number of homozygous recessive rabbits (76 without spots) and the total number of rabbits in the population (76+ 399= 475). We can thus divide the number of frogs without spots by the population total to find q^2 :

$q^{2} = \frac{bb(spotless)}{Total} = \frac{76}{475} = 0.16$

16% of the population is homozygous recessive. We then take the square root of q^2 to find :

$q = \sqrt{q^{2}} = \sqrt{0.16} = 0.40$

This value tells us that 40% of all the alleles in the population are recessive (b).

The Hardy-Weinberg equation is based off the idea that the total number of alleles in a population is the sum of the dominant and recessive alleles, as shown by this equation:

p + q = 1

We can use this equation to solve for using :

p = (1-q) = (1-0.40) = 0.60

This value tells us that 60% of all the alleles in the population are dominant (B).

Now that we have and , we can calculate 2pq . Recall that this will be the frequency of heterozygous individuals in the population:

2pq = 2 (0.60) (0.40) = 0.48

The result is 48%, meaning 48% of the frogs in the population are heterozygous. We can then multiply 48% by the total population to find the number of frogs who are heterozygous::

$Bb = (2pq \cdot Total) = (0.48)( 475) = 228$

Thus, 228 heterozygous frogs would be expected in this population.

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Example Question #1 : Understanding Hardy Weinberg Calculations

For a particular gene, the allele is dominant to the allele . If in a population the allele frequency for is 0.85, what is the frequency of individuals that are heterozygotes? Assume Hardy-Weinberg equilibrium conditions are met.

Possible Answers:

0.15

0.7225

0.1275

0.0225

0.255

Correct answer:

0.255

Explanation:

There are two equations for Hardy-Weinberg equilibrium:

A + a = 1

A^2 + 2Aa + a^2 = 1

Using the first equation, we can substitute in 0.85 for and solve to get a = 0.15 .

We can then use the second equation to find the frequencies of each genotype.

A^2 = frequency of genotype

2Aa = frequency of genotype

(Why do we multiply by 2? Because we must count both and as heterozygotes.)

a^2 = frequency of genotype

Thus we get 2Aa=2(0.85)(0.15) = 0.255 .

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Example Question #51 : Genetics Principles

If p=0.9 and q=0.1 what will be the final genotypic frequencies of a population be under Hardy-Weinberg equilibrium?

Possible Answers:

$\text{Homozygous dominant}=0.9$

$\text{Heterozygous}=0.1$

$\text{Homozygous recessive}=0$

$\text{Homozygous dominant}=0.07$

$\text{Heterozygous}=0.12$

$\text{Homozygous recessive}=0.88$

$\text{Homozygous dominant}=0.81$

$\text{Heterozygous}=0.18$

$\text{Homozygous recessive}=0.01$

$\text{Homozygous dominant}=0.11$

$\text{Heterozygous}=0.88$

$\text{Homozygous recessive}=0.01$

Correct answer:

$\text{Homozygous dominant}=0.81$

$\text{Heterozygous}=0.18$

$\text{Homozygous recessive}=0.01$

Explanation:

Recall that under Hardy-Weinberg conditions, the allele frequencies remain constant. The formulas for Hardy-Weinberg equilibrium are:

p+q=1

and

p^2+2pq+q^2=1

Here, p^2 is the frequency of the homozygous dominant genotype, 2pq is the frequency of the heterozygous genotype, and q^2 is the frequency of the homozygous recessive genotype. Since we have all the variables we need, and we want to find the genotypic frequencies (not the phenotypic) we plug into the second equation.

0.9^2+2(0.9*0.1)+0.1^2=1

p^2=0.81

2pq=0.18

q^2=0.01

Be sure to check to make sure these frequencies add up to 1.

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Example Question #11 : Hardy Weinberg Equilibrium

If p=0.3 and q=0.7, what will be the final genotypic frequencies of a population be under Hardy-Weinberg equilibrium?

Possible Answers:

$\text{Homozygous dominant}=49\%$

$\text{Heterozygous}= 81\%$

$\text{Homozygous recessive}= 1\%$

$\text{Homozygous dominant}=30\%$

$\text{Heterozygous}= 21\%$

$\text{Homozygous recessive}= 49\%$

$\text{Homozygous dominant}=9\%$

$\text{Heterozygous}= 70\%$

$\text{Homozygous recessive}= 21\%$

$\text{Homozygous dominant}=9\%$

$\text{Heterozygous}= 42\%$

$\text{Homozygous recessive}= 49\%$

Correct answer:

$\text{Homozygous dominant}=9\%$

$\text{Heterozygous}= 42\%$

$\text{Homozygous recessive}= 49\%$

Explanation:

Recall that under Hardy-Weinberg conditions, the allele frequencies remain constant. The formulas for Hardy-Weinberg equilibrium are:

p+q=1

and

p^2 + 2pq + q^2 = 1

0.3^2+2(0.3*0.7)+0.7^2=1

p^2=0.09

2pq=0.42

q^2=0.49

Be sure to check to make sure these frequencies add up to 1.

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Example Question #52 : Genetics Principles

Tail length in a population of aardvarks is determined by one gene, where L=long tails and l=short tails. If the frequency of L in the population is 0.4, determine the expected frequencies of each possible genotype: homozygous dominant (LL), heterozygous (Ll), and homozygous recessive (ll).

Possible Answers:

LL = 0.16

Ll = 0.68

ll = 0.16

LL = 0.4

Ll = 0.0

ll = 0.6

LL= 0.24

Ll= 0.58

ll= 0.18

There is not enough information given to determine an answer.

LL = 0.16

Ll = 0.48

ll = 0.36

Correct answer:

LL = 0.16

Ll = 0.48

ll = 0.36

Explanation:

To begin, we must know that we are working with the Hardy-Weinberg equations:

p+q=1

$p^{^{2}}+2pq + q^{2}=1$

We can designate the dominant allele (L) as and the recessive allele (l) as .

Since the frequency of L was given (0.4) we know the the frequency of l must be:

1 - 0.4 =0.6

Now we have both allele frequencies and can plug them into the equation:

$(0.4)^{^{2}}+2(0.4)(0.6)+(0.6)^{2} = 1$

So for $p^{^{2}}$ , LL (homozygous dominant), is equal to 0.16

for 2pq , Ll (heterozygous), is equal to 0.48

for $q^{2}$ , ll (homozygous recessive), is equal to 0.36

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High School Biology : Hardy-Weinberg Equilibrium

Study concepts, example questions & explanations for High School Biology

All High School Biology Resources

Example Questions

Example Question #81 : Genetics And Evolution

Example Question #1 : Understanding Hardy Weinberg Calculations

Example Question #1 : Understanding Hardy Weinberg Calculations

Example Question #51 : Genetics Principles

Example Question #11 : Hardy Weinberg Equilibrium

Example Question #52 : Genetics Principles

All High School Biology Resources

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