HardyWeinberg Tutorial
This tutorial is designed to try and help you understand the basics of HardyWeinberg Equilibrium. Please
work through the entire tutorial.
Question 1
HardyWeinberg equilibrium makes several assumptions.
Which two of the those listed below are not assumptions which must be met for a population to reach
HardyWeinberg Equilibrium?
(a) Sexaul Reproduction
(b) Nonoverlapping Generations
(c) Random Mating
(d) Natural Selection occurs
(e) Population size is small
(f) All of the above are assumptions of HardyWeinberg Equilibrium.
Question 1
You said that (a) Sexual Reproduction was not an assumption of HardyWeinberg Equilibrium.
No, Populations must undergo sexual reproduction for HardyWeinberg Equilibrium to be achieved. Sexual reproduction allows the random reshuffling
of alleles in the next generation which is needed to prevent alleles from being lost due to chance.
Return to question 1
Question 1
You said that (b) Nonoverlapping Generations was not an assumption of HardyWeinberg Equilibrium.
No, nonoverlapping generations is one of the HardyWeinberg Equilibrium asumptions. If the individuals who gave
birth to the current offspring
are still around to mate with their kids inbreeding is more likely to occur which will prevent HardyWeinberg
Equilibrium from being reached.
Return to question 1
Question 1
You said that (c) Random Mating was not an assumption of HardyWeinberg Equilibrium.
No, Populations must undergo random mating for HardyWeinberg Equilibrium to be achieved. If individuals don't
mate randomly they will be actively
choosing mates and thus alleles won't be recombining at random in the offspring. This should lead to directional
changes in
allele frequencies while HardyWeinberg Equilibrium requires random changes.
Return to question 1
Question 1
You said that (d) Natural Selection occurring is not an assumption of HardyWeinberg Equilibrium.
Correct!!, for HardyWeinberg equilibrium to be reached natural selection can not be occuring.
If populations are undergoing natural selection at the locus under consideration allele frequencies will be changing in
a specific direction and changing continuously, HardyWeinberg Equilibrium predicts that allele frequencies will stay constant.
Return to question 1
Go to question 2
Question 1
You said that (a)Population size is small was not an assumption of HardyWeinberg Equilibrium.
YES! HardyWeinberg Equilibrium assumes that population size is very large.
Return to question 1
Go to question 2
Question 1
You said that (f) All of the above are assumptions of HardyWeinberg Equilibrium.
No, look again, one of the choices is not an assumption of HardyWeinberg Equilibrium.
Return to question 1
Question 2
Which of the equations below is HardyWeinberg Equilibrium based
on?
(a) p + pq + q = 1
(b) p + 2pq + q = 1
(c) p + p^{2}q^{2} + q = 1
(d) p^{2} + 2pq +q^{2} = 1
(e) p^{2} + 2p^{2}q^{2} +q^{2} = 1
Question 2
You said that (a) p + pq + q = 1is the equation associated with HardyWeinberg Equilibrium.
No, this is the general format of the equation but it is a bit more complicated than this one.
You can see how the equation works by looking at the table below.

p

q

p

p^{2}

pq

q

qp = pq

q^{2}

Adding the cells of the table together then provides you with, p^{2} + pq + pq + q^{2} = p^{2} +
2pq + q^{2}.
Return to question 2
Question 2
You said that (b) p + 2pq + q = 1is the equation associated with HardyWeinberg Equilibrium.
No, this is the general format of the equation but it is a bit more complicated than this one.
You can see how the equation works by looking at the table below.

p

q

p

p^{2}

pq

q

qp = pq

q^{2}

Adding the cells of the table together then provides you with, p^{2} + pq + pq + q^{2} = p^{2} +
2pq + q^{2}.
Return to question 2
Question 2
You said that (c) p + p^{2}q^{2} + q = 1is the equation associated with HardyWeinberg Equilibrium.
No, this is the general format of the equation but it is a bit more complicated than this one.
You can see how the equation works by looking at the table below.

p

q

p

p^{2}

pq

q

qp = pq

q^{2}

Adding the cells of the table together then provides you with, p^{2} + pq + pq + q^{2} = p^{2} +
2pq + q^{2}.
Return to question 2
Question 2
You said that (d) p^{2} + 2pq +q^{2} = 1is the equation associated with HardyWeinberg Equilibrium.
Yes, this is the equation that is associated with HardyWeinberg Equilibrium. In this equation p^{2} represents the frequency of individuals
who are homozygous for one allle, 2pq is the frequency of heterozygous individuals, and q^{2} is the frequency of individuals
who are homozygous for the other allele.
You can see how the equation works by looking at the table below.

p

q

p

p^{2}

pq

q

qp = pq

q^{2}

Adding the cells of the table together then provides you with, p^{2} + pq + pq + q^{2} = p^{2} +
2pq + q^{2}.
Return to question 2
Return to question 3
Question 2
You said that (e) p^{2} + 2p^{2}q^{2} +q^{2} = 1is the equation associated with HardyWeinberg Equilibrium.
No, this is the general format of the equation but it is a bit less complicated than this one.
You can see how the equation works by looking at the table below.

p

q

p

p^{2}

pq

q

qp = pq

q^{2}

Adding the cells of the table together then provides you with, p^{2} + pq + pq + q^{2} = p^{2} +
2pq + q^{2}.
Return to question 2
Question 3
When a population is in HardyWeinberg Equilibrium,
(a) allele frequencies decline at a steady rate
(b) allele frequencies stay the same
(c) allele frequencies increase at a steady rate
(d) genotype frequencies decline at a steady rate
(e) genotype frequencies stay the same
(f) genotype frequencies increase at a steady rate
Question 3
You said that (a) allele frequencies decline at a steady rate when in HardyWeinberg Equilibrium.
No, in HardyWeinberg Equilibrium allele frequencies should not be changing. Equilibrium implies a steady state,
one that is unchanging, so if allele
frequencies are decreasing the system can't be at equilibrium.
Return to question 3
Question 3
You said that (b) allele frequencies stay the same when in HardyWeinberg Equilibrium.
Yes! In HardyWeinberg Equilibrium allele frequencies should not be changing. Equilibrium implies a steady state,
one that is unchanging, so if allele
frequencies are changing the system can't be at equilibrium.
Return to question 3
Go to question 4
Question 3
You said that (c) allele frequencies increase at a steady rate when in HardyWeinberg Equilibrium.
No, in HardyWeinberg Equilibrium allele frequencies should not be changing. Equilibrium implies a steady state,
one that is unchanging, so if allele
frequencies are increasing the system can't be at equilibrium.
Return to question 3
Question 3
You said that (d) genotype frequencies decline at a steady rate when in HardyWeinberg Equilibrium.
No, HardyWeinberg Equilibrium deals with alleles not genotypes. Try again.
Return to question 3
Question 3
You said that (d) genotype frequencies stay the same when in HardyWeinberg Equilibrium.
It is likely that if genotype frequencies are constant that the population is in HardyWeinberg Equilbrium,
but HardyWeinberg Equilibrium deals with alleles not genotypes, so there a better answer. However, you
are semicorrect because typically, when allele frequencies are
constant genotype frequencies also stay the same. In reality, however, genotype frequencies can change while allele frequenices
stay the same. Can you figure out how this could occur? Try again.
Return to question 3
Question 3
You said that (d) genotype frequencies increase at a steady rate when in HardyWeinberg Equilibrium.
No, HardyWeinberg Equilibrium deals with alleles not genotypes. Try again.
Return to question 3
Question 4
Assuming that the rest of the Hardy Weinberg assumptions are
met, how many generations of random mating are required to bring
a population into HardyWeinberg Equilibrium?
(a) 0  random mating never brings a population into HardyWeinberg Equilibrium
(b) 1
(c) 10
(d) 100
Question 4
You said that it takes(a) 0  generations of random mating to bring a population in to HardyWeinberg Equilibrium because random mating never brings
a population into HardyWeinberg Equilibrium.
No, random mating will eventually bring a population into HardyWeinberg Equilibrium if the other assumptions are met. Try again.
Return to question 4
Question 4
You said that (b) 1generation of random mating will bring a population into HardyWeinberg Equilibrium.
YES! One generation of random mating is sufficient to bring a population into HardyWeinberg Equilibrium.
Return to question 4
Go to question 5
Question 4
You said that (c) 10generations of random mating will bring a population into HardyWeinberg Equilibrium.
No, random mating brings populations into HardyWeinberg Equilibrium much quicker than 10 generations.
Return to question 4
Question 4
You said that (d) 100generations of random mating will bring a population into HardyWeinberg Equilibrium.
No, random mating brings populations into HardyWeinberg Equilibrium much quicker than 100 generations.
Return to question 4
Question 5
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Suppose that the deformed frogs that have made the press recently are the result of a parasite
infection and that in a nearby Indiana pond, we know that there is a parasite resistance allele in
the population. The allele has a frequency of 20% and is a dominant allele. What percentage of the
population is predicted to be resistant to the parasite should it be introduced into the pond?
a. 20%
b. 4%
c. 36%
d. 64%
e. 80%
f. 32%
Question 5
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You said that the expected frequency of resistant frogs is (a) 20%.
No, because it would mean that the allele frequency that we are given is the
same as the genotype frequency. Frogs are diploid of course and any genotype is determined by the
union of two alleles. According to the HardyWeinberg law, this union is the product of the allele
frequencies.
we are given that p = 0.20, therefore q = 0.80 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.2)^{2}

2(0.2*0.8)

(0.8)^{2}

0.04

0.32

0.64

Because both RR and Rr individuals are phenotypically resistant to the parasite (i.e., R is
dominant to r), the percentage of resistant individuals is .04 + .32 = .36 or 36% and
alternative c is correct.
Return to question 5.
Question 5
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You said that the expected frequency of resistant frogs is (b) 4%.
Not quite. You are part way there though becuase this is the frequency of the individuals homozygous for the "R" allele.
However this answer is only true if either the allele were codominant,
not dominant (that is, an Rr individual has an intermediate level of resistance not complete
resistance) or if the allele is actually recessive rather than dominant. We were given that it was
dominant.
we are given that p = 0.20, therefore q = 0.80 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.2)^{2}

2(0.2*0.8)

(0.8)^{2}

0.04

0.32

0.64

Because both RR and Rr individuals are phenotypically resistant to the parasite (i.e., R is
dominant to r), the percentage of resistant individuals is .04 + .32 = .36 or 36% and
alternative c is correct.
Return to question 5.
Question 5
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You said that the expected frequency of resistant frogs is (c) 36%.
YES! Because the allele is dominant you know that homozygous and heterozygous individuals for the "R" allele are
resistant.
we are given that p = 0.20, therefore q = 0.80 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.2)^{2}

2(0.2*0.8)

(0.8)^{2}

0.04

0.32

0.64

Because both RR and Rr individuals are phenotypically resistant to the parasite (i.e., R is
dominant to r), the percentage of resistant individuals is .04 + .32 = .36 or 36% and
alternative c is correct.
Go to question 6.
Question 5
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You said that the expected frequency of resistant frogs is (d) 64%.
No, it looks like you got confused and estimated the frequency of susceptible frogs because 64% corresponds
to the frequency of the susceptible genotypes, rr not the resistant ones.
we are given that p = 0.20, therefore q = 0.80 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.2)^{2}

2(0.2*0.8)

(0.8)^{2}

0.04

0.32

0.64

Because both RR and Rr individuals are phenotypically resistant to the parasite (i.e., R is
dominant to r), the percentage of resistant individuals is .04 + .32 = .36 or 36% and
alternative c is correct.
Return to question 5.
Question 5
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You said that the expected frequency of resistant frogs is (e) 80%.
No, this is an allele frequency, not a genotype frequency and it is for the wrong allele, the "r" allele.
we are given that p = 0.20, therefore q = 0.80 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.2)^{2}

2(0.2*0.8)

(0.8)^{2}

0.04

0.32

0.64

Because both RR and Rr individuals are phenotypically resistant to the parasite (i.e., R is
dominant to r), the percentage of resistant individuals is .04 + .32 = .36 or 36% and
alternative c is correct.
Return to question 5.
Question 5
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You said that the expected frequency of resistant frogs is (e) 32%.
No, you are part way there, because this is the frequency of the heterozygotes who are resistant, but don't forget that
homozygotes for the "R" allele are also resistant.
we are given that p = 0.20, therefore q = 0.80 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.2)^{2}

2(0.2*0.8)

(0.8)^{2}

0.04

0.32

0.64

Because both RR and Rr individuals are phenotypically resistant to the parasite (i.e., R is
dominant to r), the percentage of resistant individuals is .04 + .32 = .36 or 36% and
alternative c is correct.
Return to question 5.
Question 6
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What if the allele had a frequency of 20%; however, we were wrong, the resistant allele is a recessive allele not a dominant
allele. What percentage of the population is predicted to be resistant to the parasite should it be introduced into the pond?
a. 20%
b. 4%
c. 36%
d. 96%
e. 80%
f. 32%
Question 6
Click on the frogs to learn more about them
You said that the expected frequency of resistant frogs is (a) 20%.
No, because it would mean that the allele frequency that we are given is the
same as the genotype frequency. Frogs are diploid of course and any genotype is determined by the
union of two alleles. According to the HardyWeinberg law, this union is the product of the allele
frequencies.
we are given that p = 0.80, therefore q = 0.20 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.8)^{2}

2(0.2*0.8)

(0.2)^{2}

0.64

0.32

0.04

Return to Question 6.
Question 6
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You said that the expected frequency of resistant frogs is (b) 4%.
YES! Since the allele is recessive only individuals who are homozygous for the "r" allele will be resistant.
we are given that p = 0.80, therefore q = 0.20 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.8)^{2}

2(0.2*0.8)

(0.2)^{2}

0.64

0.32

0.04

Go to Question 7.
Question 6
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You said that the expected frequency of resistant frogs is (c) 36%.
No, this would be the answer if the allele were dominant, but because it is recessive only individuals who are homozygous for the
"r" allele (rr) are resistant.
We are given that p = 0.80, therefore q = 0.20 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.8)^{2}

2(0.2*0.8)

(0.2)^{2}

0.64

0.32

0.04

Go to question 6.
Question 6
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You said that the expected frequency of resistant frogs is (d) 96%.
No, it looks like you got confused and estimated the frequency of susceptible frogs because 96% corresponds
to the frequency of the susceptible genotypes, RR and Rr not the resistant rr ones.
We are given that p = 0.80, therefore q = 0.20 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.8)^{2}

2(0.2*0.8)

(0.2)^{2}

0.64

0.32

0.04

Return to Question 6.
Question 6
Click on the frogs to learn more about them
You said that the expected frequency of resistant frogs is (e) 80%.
No, this is an allele frequency, not a genotype frequency and it is for the wrong allele, the "R" allele.
We are given that p = 0.80, therefore q = 0.20 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.8)^{2}

2(0.2*0.8)

(0.2)^{2}

0.64

0.32

0.04

Return to Question 6.
Question 6
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You said that the expected frequency of resistant frogs is (e) 32%.
No, this is the frequency of heterozygotes in the population. Remember that we are dealing with a recessive allele here, so
heterozygotes are susceptable.
We are given that p = 0.80, therefore q = 0.20 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.8)^{2}

2(0.2*0.8)

(0.2)^{2}

0.64

0.32

0.04

Return to Question 6.
Question 7
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If the resistant allele had a frequency of 20% and was recessive allele, what percentage of the population
would be susceptible to the parasite but still carry an allele for resistance?
a. 20%
b. 4%
c. 36%
d. 96%
e. 80%
f. 32%
Question 7
Click on the frogs to learn more about them
You said that the expected frequency of resistant frogs is (a) 20%.
No, this is just the frequency of the resistant allele. Frogs are diploid of course and any genotype is determined by the
union of two alleles. According to the HardyWeinberg law, this union is the product of the allele frequencies.
we are given that p = 0.80, therefore q = 0.20 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.8)^{2}

2(0.2*0.8)

(0.2)^{2}

0.64

0.32

0.04

Return to Question 7.
Question 7
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You said that the expected frequency of resistant frogs is (b) 4%.
No, remember this is the frequency of the resistant frogs.
we are given that p = 0.80, therefore q = 0.20 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.8)^{2}

2(0.2*0.8)

(0.2)^{2}

0.64

0.32

0.04

Return to Question 7.
Question 7
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You said that the expected frequency of resistant frogs is (c) 36%.
No, this would be the answer if the allele were dominant, but because it is recessive only individuals who are homozygous for the
"r" allele (rr) are resistant.
we are given that p = 0.80, therefore q = 0.20 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.8)^{2}

2(0.2*0.8)

(0.2)^{2}

0.64

0.32

0.04

Return to Question 7.
Question 7
Click on the frogs to learn more about them
You said that the expected frequency of resistant frogs is (d) 96%.
No, it looks like you got confused and estimated the frequency of all susceptible frogs. 96% corresponds
to the frequency of the susceptible genotypes, RR and Rr not just the ones who also carry an allele for resistance.
we are given that p = 0.80, therefore q = 0.20 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.8)^{2}

2(0.2*0.8)

(0.2)^{2}

0.64

0.32

0.04

Return to Question 7.
Question 7
Click on the frogs to learn more about them
You said that the expected frequency of resistant frogs is (e) 80%.
No, this is the allele frequency for the susceptable allele, not a genotype frequency.
we are given that p = 0.80, therefore q = 0.20 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.8)^{2}

2(0.2*0.8)

(0.2)^{2}

0.64

0.32

0.04

Return to Question 7.
Question 7
Click on the frogs to learn more about them
You said that the expected frequency of resistant frogs is (e) 32%.
YES! Only the heterozygous individuals are susceptable but still carry an allele for resistance.
we are given that p = 0.80, therefore q = 0.20 and we can assign frequencies to the 3 possible
genotypes at this locus:
RR

Rr

rr

p^{2}

2pq

q^{2}

(0.8)^{2}

2(0.2*0.8)

(0.2)^{2}

0.64

0.32

0.04

Go to question 8.
Question 8
Red foxes have three different types of coat color, red, cross,
and black/silver (see the image) which is controlled by two loci
each with two alleles. We will only be concerned with the "A"
locus here.
Suppose you went to Yellowstone National Park and counted how
many foxes you saw of each colorphase. You talk with one of
the Rangernaturalists at the end of the trip, telling her that
you saw 58 redphase, 32 crossphase, and 10 black/silverphase
foxes. The naturalist wants to know more about the these foxes
so she asks you a series of questions.
What was the frequency of the "A" allele?
Question 8
You said that the frequency of the "A" allele is (a) 7%.
No, remember that to find the number of "A" alleles you should count how many "A" alleles there are and then divide
this by the total number of alleles in the system. There are 58 redphase foxes each with 2 "A" alleles, and 32 crossphase
foxes each with 1 "A". So there are a total of 148 "A" allleles {2 x 58 (from redphase foxes) + 1 x 32 (from crossphase foxes)}.
There is a total of 100 foxes and since each fox has 2 alleles there is a total of 200 alleles in the system. So finally you
have the number of "A" alleles = 148/200 = 0.74.
Return to Question 8
Question 8
You said that the frequency of the "A" allele is (b) 10%.
No, remember that to find the number of "A" alleles you should count how many "A" alleles there are and then divide
this by the total number of alleles in the system. There are 58 redphase foxes each with 2 "A" alleles, and 32 crossphase
foxes each with 1 "A". So there are a total of 148 "A" allleles {2 x 58 (from redphase foxes) + 1 x 32 (from crossphase foxes)}.
There is a total of 100 foxes and since each fox has 2 alleles there is a total of 200 alleles in the system. So finally you
have the number of "A" alleles = 148/200 = 0.74.
Return to Question 8
Question 8
You said that the frequency of the "A" allele is (c) 26%.
No, remember that to find the number of "A" alleles you should count how many "A" alleles there are and then divide
this by the total number of alleles in the system. There are 58 redphase foxes each with 2 "A" alleles, and 32 crossphase
foxes each with 1 "A". So there are a total of 148 "A" allleles {2 x 58 (from redphase foxes) + 1 x 32 (from crossphase foxes)}.
There is a total of 100 foxes and since each fox has 2 alleles there is a total of 200 alleles in the system. So finally you
have the number of "A" alleles = 148/200 = 0.74.
Return to Question 8
Question 8
You said that the frequency of the "A" allele is (d) 38%.
No, remember that to find the number of "A" alleles you should count how many "A" alleles there are and then divide
this by the total number of alleles in the system. There are 58 redphase foxes each with 2 "A" alleles, and 32 crossphase
foxes each with 1 "A". So there are a total of 148 "A" allleles {2 x 58 (from redphase foxes) + 1 x 32 (from crossphase foxes)}.
There is a total of 100 foxes and since each fox has 2 alleles there is a total of 200 alleles in the system. So finally you
have the number of "A" alleles = 148/200 = 0.74.
Return to Question 8
Question 8
You said that the frequency of the "A" allele is (e) 55%.
No, remember that to find the number of "A" alleles you should count how many "A" alleles there are and then divide
this by the total number of alleles in the system. There are 58 redphase foxes each with 2 "A" alleles, and 32 crossphase
foxes each with 1 "A". So there are a total of 148 "A" allleles {2 x 58 (from redphase foxes) + 1 x 32 (from crossphase foxes)}.
There is a total of 100 foxes and since each fox has 2 alleles there is a total of 200 alleles in the system. So finally you
have the number of "A" alleles = 148/200 = 0.74.
Return to Question 8
Question 8
You said that the frequency of the "A" allele is (f) 58%.
No, this is the frequency of redphase foxes in the population, but other foxes have a redphase allele also.
Remember that to find the number of "A" alleles you should count how many "A" alleles there are and then divide
this by the total number of alleles in the system. There are 58 redphase foxes each with 2 "A" alleles, and 32 crossphase
foxes each with 1 "A". So there are a total of 148 "A" allleles {2 x 58 (from redphase foxes) + 1 x 32 (from crossphase foxes)}.
There is a total of 100 foxes and since each fox has 2 alleles there is a total of 200 alleles in the system. So finally you
have the number of "A" alleles = 148/200 = 0.74.
Return to Question 8
Question 8
You said that the frequency of the "A" allele is (g) 74%.
YES! You must have remembered that to find the number of "A" alleles you count how many "A" alleles there are and then divide
this by the total number of alleles in the system. {2 x 58 (from redphase foxes) + 1 x 32 (from crossphase foxes)}.
There is a total of 100 foxes and since each fox has 2 alleles there is a total of 200 alleles in the system. So finally you
have the number of "A" alleles = 148/200 = 0.74.
Go to question 9
Question 9
Red foxes have three different types of coat color, red, cross,
and black/silver (see the image) which is controlled by two loci
each with two alleles. We will only be concerned with the "A"
locus here.
Suppose you went to Yellowstone National Park and counted how
many foxes you saw of each colorphase. You talk with one of
the Rangernaturalists at the end of the trip, telling her that
you saw 58 redphase, 32 crossphase, and 10 black/silverphase
foxes. The naturalist wants to know more about the these foxes
so she asks you a series of questions.
What was the frequency of the "a" allele?
Question 9
You said that the frequency of the "a" allele is (a) 7%.
No, it looks like you squared the real frequency when you shouldn't have.
There are two ways to figure out the frequency of the "a" allele. In the last question you said that the frequency of the "A"
allele was 0.74 (or 74%). Since there are only 2 different types of alleles in the population and the total of those two frequencies
must = 1, (p + q = 1), you can use the equation 1.0  p = q.
or
Remember that to find the number of "a" alleles you should count how many "a" alleles there are and then divide
this by the total number of alleles in the system. There are 10 black/silverphase foxes each with 2 "a" alleles, and 32 crossphase
foxes each with 1 "a". So there are a total of 52 "a" allleles {2 x 10 (from redphase foxes) + 1 x 32 (from crossphase foxes)}.
There is a total of 100 foxes and since each fox has 2 alleles there is a total of 200 alleles in the system. So finally you
have the number of "A" alleles = 52/200 = 0.26.
Return to Question 9
Question 9
You said that the frequency of the "a" allele is (b) 10%.
No.
There are two ways to figure out the frequency of the "a" allele. In the last question you said that the frequency of the "A"
allele was 0.74 (or 74%). Since there are only 2 different types of alleles in the population and the total of those two frequencies
must = 1, (p + q = 1), you can use the equation 1.0  p = q.
or
Remember that to find the number of "a" alleles you should count how many "a" alleles there are and then divide
this by the total number of alleles in the system. There are 10 black/silverphase foxes each with 2 "a" alleles, and 32 crossphase
foxes each with 1 "a". So there are a total of 52 "a" allleles {2 x 10 (from redphase foxes) + 1 x 32 (from crossphase foxes)}.
There is a total of 100 foxes and since each fox has 2 alleles there is a total of 200 alleles in the system. So finally you
have the number of "A" alleles = 52/200 = 0.26.
Return to Question 9
Question 9
You said that the frequency of the "a" allele is (c) 26%.
YES! Which one of the 2 different ways described below did you get to this answer?
There are two ways to figure out the frequency of the "a" allele. In the last question you said that the frequency of the "A"
allele was 0.74 (or 74%). Since there are only 2 different types of alleles in the population and the total of those two frequencies
must = 1, (p + q = 1), you can use the equation 1.0  p = q.
or
Remember that to find the number of "a" alleles you should count how many "a" alleles there are and then divide
this by the total number of alleles in the system. There are 10 black/silverphase foxes each with 2 "a" alleles, and 32 crossphase
foxes each with 1 "a". So there are a total of 52 "a" allleles {2 x 10 (from redphase foxes) + 1 x 32 (from crossphase foxes)}.
There is a total of 100 foxes and since each fox has 2 alleles there is a total of 200 alleles in the system. So finally you
have the number of "A" alleles = 52/200 = 0.26.
Return to Question 9
Go to question 10
Question 9
You said that the frequency of the "a" allele is (b) 38%.
No.
There are two ways to figure out the frequency of the "a" allele. In the last question you said that the frequency of the "A"
allele was 0.74 (or 74%). Since there are only 2 different types of alleles in the population and the total of those two frequencies
must = 1, (p + q = 1), you can use the equation 1.0  p = q.
or
Remember that to find the number of "a" alleles you should count how many "a" alleles there are and then divide
this by the total number of alleles in the system. There are 10 black/silverphase foxes each with 2 "a" alleles, and 32 crossphase
foxes each with 1 "a". So there are a total of 52 "a" allleles {2 x 10 (from redphase foxes) + 1 x 32 (from crossphase foxes)}.
There is a total of 100 foxes and since each fox has 2 alleles there is a total of 200 alleles in the system. So finally you
have the number of "A" alleles = 52/200 = 0.26.
Return to Question 9
Question 9
You said that the frequency of the "a" allele is (e) 55%.
No.
There are two ways to figure out the frequency of the "a" allele. In the last question you said that the frequency of the "A"
allele was 0.74 (or 74%). Since there are only 2 different types of alleles in the population and the total of those two frequencies
must = 1, (p + q = 1), you can use the equation 1.0  p = q.
or
Remember that to find the number of "a" alleles you should count how many "a" alleles there are and then divide
this by the total number of alleles in the system. There are 10 black/silverphase foxes each with 2 "a" alleles, and 32 crossphase
foxes each with 1 "a". So there are a total of 52 "a" allleles {2 x 10 (from redphase foxes) + 1 x 32 (from crossphase foxes)}.
There is a total of 100 foxes and since each fox has 2 alleles there is a total of 200 alleles in the system. So finally you
have the number of "A" alleles = 52/200 = 0.26.
Return to Question 9
Question 9
You said that the frequency of the "a" allele is (f) 58%.
No.
There are two ways to figure out the frequency of the "a" allele. In the last question you said that the frequency of the "A"
allele was 0.74 (or 74%). Since there are only 2 different types of alleles in the population and the total of those two frequencies
must = 1, (p + q = 1), you can use the equation 1.0  p = q.
or
Remember that to find the number of "a" alleles you should count how many "a" alleles there are and then divide
this by the total number of alleles in the system. There are 10 black/silverphase foxes each with 2 "a" alleles, and 32 crossphase
foxes each with 1 "a". So there are a total of 52 "a" allleles {2 x 10 (from redphase foxes) + 1 x 32 (from crossphase foxes)}.
There is a total of 100 foxes and since each fox has 2 alleles there is a total of 200 alleles in the system. So finally you
have the number of "A" alleles = 52/200 = 0.26.
Return to Question 9
Question 9
You said that the frequency of the "a" allele is (g) 74%.
No. This is the frequency of the "A" allele.
There are two ways to figure out the frequency of the "a" allele. In the last question you said that the frequency of the "A"
allele was 0.74 (or 74%). Since there are only 2 different types of alleles in the population and the total of those two frequencies
must = 1, (p + q = 1), you can use the equation 1.0  p = q.
or
Remember that to find the number of "a" alleles you should count how many "a" alleles there are and then divide
this by the total number of alleles in the system. There are 10 black/silverphase foxes each with 2 "a" alleles, and 32 crossphase
foxes each with 1 "a". So there are a total of 52 "a" allleles {2 x 10 (from redphase foxes) + 1 x 32 (from crossphase foxes)}.
There is a total of 100 foxes and since each fox has 2 alleles there is a total of 200 alleles in the system. So finally you
have the number of "A" alleles = 52/200 = 0.26.
Return to Question 9
Question 10
Red foxes have three different types of coat color, red, cross,
and black/silver (see the image) which is controlled by two loci
each with two alleles. We will only be concerned with the "A"
locus here.
Suppose you went to Yellowstone National Park and counted how
many foxes you saw of each colorphase. You talk with one of
the Rangernaturalists at the end of the trip, telling her that
you saw 58 redphase, 32 crossphase, and 10 black/silverphase
foxes. The naturalist wants to know more about the these foxes
so she asks you a series of questions.
Assuming that the population mates randomly, what frequency do
you expect the "A" allele to be at in the next generation?
Question 10
You said that the frequency of the "A" allele in the next generation would be (a) 7%.
No, if the population mates at random the allele frequencies won't change between generations. That
is specifically what is so special about HardyWeinberg Equilibrium, it allows you to know what the allele frequencies will
be in the next generation, and from that the genotype frequencies.
Return to Question 10.
Question 10
You said that the frequency of the "A" allele in the next generation would be (b) 10%.
No, if the population mates at random the allele frequencies won't change between generations. That
is specifically what is so special about HardyWeinberg Equilibrium, it allows you to know what the allele frequencies will
be in the next generation, and from that the genotype frequencies.
Return to Question 10.
Question 10
You said that the frequency of the "A" allele in the next generation would be (c) 26%.
No, if the population mates at random the allele frequencies won't change between generations. That
is specifically what is so special about HardyWeinberg Equilibrium, it allows you to know what the allele frequencies will
be in the next generation, and from that the genotype frequencies.
Return to Question 10.
Question 10
You said that the frequency of the "A" allele in the next generation would be (d) 38%.
No, if the population mates at random the allele frequencies won't change between generations. That
is specifically what is so special about HardyWeinberg Equilibrium, it allows you to know what the allele frequencies will
be in the next generation, and from that the genotype frequencies.
Return to Question 10.
Question 10
You said that the frequency of the "A" allele in the next generation would be (e) 55%.
No, if the population mates at random the allele frequencies won't change between generations. That
is specifically what is so special about HardyWeinberg Equilibrium, it allows you to know what the allele frequencies will
be in the next generation, and from that the genotype frequencies.
Return to Question 10.
Question 10
You said that the frequency of the "A" allele in the next generation would be (f) 58%.
No, if the population mates at random the allele frequencies won't change between generations. That
is specifically what is so special about HardyWeinberg Equilibrium, it allows you to know what the allele frequencies will
be in the next generation, and from that the genotype frequencies.
Return to Question 10.
Question 10
You said that the frequency of the "A" allele in the next generation would be (g) 74%.
YES! When the population mates at random the allele frequencies don't change between generations. That
is specifically what is so special about HardyWeinberg Equilibrium, it allows you to know what the allele frequencies will
be in the next generation, and from that the genotype frequencies.
Return to Question 10.
Go to question 11.
Question 11
Red foxes have three different types of coat color, red, cross,
and black/silver (see the image) which is controlled by two loci
each with two alleles. We will only be concerned with the "A"
locus here.
Suppose you went to Yellowstone National Park and counted how
many foxes you saw of each colorphase. You talk with one of
the Rangernaturalists at the end of the trip, telling her that
you saw 58 redphase, 32 crossphase, and 10 black/silverphase
foxes. The naturalist wants to know more about the these foxes
so she asks you a series of questions.
Assuming that the population is not in HardyWeinberg and it mates randomly, what is the expected
frequency of the redphase foxes in the next generation?
Question 11
You said that the frequency of redphased foxes after a generation of random mating would be (a) 7%.
No, remember that you can calculate the frequency of the genotypes in the next generation using the formula
p^{2} + 2pq + q^{2} = 1. In this case p represents the frequency of the "A" allele. Since the
frequency of the "A" allele is 0.74, you can determine the expected frequency of the redphase foxes by squaring
0.74. I.e., Frequency of redphase foxes in the next generation = 0.74^{2}.
Return to Question 11.
Question 11
You said that the frequency of redphased foxes after a generation of random mating would be (b) 10%.
No, remember that you can calculate the frequency of the genotypes in the next generation using the formula
p^{2} + 2pq + q^{2} = 1. In this case p represents the frequency of the "A" allele. Since the
frequency of the "A" allele is 0.74, you can determine the expected frequency of the redphase foxes by squaring
0.74. I.e., Frequency of redphase foxes in the next generation = 0.74^{2}.
Return to Question 11.
Question 11
You said that the frequency of redphased foxes after a generation of random mating would be (c) 26%.
No, remember that you can calculate the frequency of the genotypes in the next generation using the formula
p^{2} + 2pq + q^{2} = 1. In this case p represents the frequency of the "A" allele. Since the
frequency of the "A" allele is 0.74, you can determine the expected frequency of the redphase foxes by squaring
0.74. I.e., Frequency of redphase foxes in the next generation = 0.74^{2}.
Return to Question 11.
Question 11
You said that the frequency of redphased foxes after a generation of random mating would be (d) 38%.
No, remember that you can calculate the frequency of the genotypes in the next generation using the formula
p^{2} + 2pq + q^{2} = 1. In this case p represents the frequency of the "A" allele. Since the
frequency of the "A" allele is 0.74, you can determine the expected frequency of the redphase foxes by squaring
0.74. I.e., Frequency of redphase foxes in the next generation = 0.74^{2}.
Return to Question 11.
Question 11
You said that the frequency of redphased foxes after a generation of random mating would be (e) 55%.
YES! You must have used the equation p^{2} + 2pq + q^{2} = 1. In this case p represents the
frequency of the "A" allele. Since the frequency of the "A" allele is 0.74, you can determine the expected
frequency of the redphase foxes by squaring 0.74.
I.e., Frequency of redphase foxes in the next generation = 0.74^{2}.
Go to question 12.
Question 11
You said that the frequency of redphased foxes after a generation of random mating would be (f) 58%.
No, this was the frequency of the redphase foxes in the generation before the random mating. While the allele
frequencies haven't changed, the genotype frequency did.
Remember that you can calculate the frequency of the genotypes in the next generation using the formula
p^{2} + 2pq + q^{2} = 1. In this case p represents the frequency of the "A" allele. Since the
frequency of the "A" allele is 0.74, you can determine the expected frequency of the redphase foxes by squaring
0.74. I.e., Frequency of redphase foxes in the next generation = 0.74^{2}.
Return to Question 11.
Question 11
You said that the frequency of redphased foxes after a generation of random mating would be (g) 74%.
No, This is what the allele frequency would be, so you are half way there. Remember that you can calculate the frequency of the genotypes in the next generation using the formula
p^{2} + 2pq + q^{2} = 1. In this case p represents the frequency of the "A" allele. Since the
frequency of the "A" allele is 0.74, you can determine the expected frequency of the redphase foxes by squaring
0.74. I.e., Frequency of redphase foxes in the next generation = 0.74^{2}.
Return to Question 11.
Question 12
Red foxes have three different types of coat color, red, cross,
and black/silver (see the image) which is controlled by two loci
each with two alleles. We will only be concerned with the "A"
locus here.
Suppose you went to Yellowstone National Park and counted how
many foxes you saw of each colorphase. You talk with one of
the Rangernaturalists at the end of the trip, telling her that
you saw 58 redphase, 32 crossphase, and 10 black/silverphase
foxes. The naturalist wants to know more about the these foxes
so she asks you a series of questions.
Assuming that the population mates randomly, what is the expected
frequency of the crossphase foxes in the next generation?
Question 12
You said that the frequency of crossphased foxes after a generation of random mating would be (a) 7%.
No, remember that you can calculate the frequency of the genotypes in the next generation using the formula
p^{2} + 2pq + q^{2} = 1. In this case p represents the frequency of the "A" allele and q represents the "a" allele.
The expected frequency of the heterozygous foxes (Crossphase foxes Aa) is found by multiplying 2*p*q, so 2 * 0.74 * 0.26 = the
expected frequency of the crossphase foxes in the next generation.
Return to Question 12.
Question 12
You said that the frequency of crossphased foxes after a generation of random mating would be (b) 19%.
No, it looks like you just multiplied the frequency of the "A" allele and the "a" allele together, 0.74 * 0.26 = 0.19.
Remember that you can calculate the frequency of the genotypes in the next generation using the formula
p^{2} + 2pq + q^{2} = 1. In this case p represents the frequency of the "A" allele and q represents the "a" allele.
The expected frequency of the heterozygous foxes (Crossphase foxes Aa) is found by multiplying 2*p*q, so 2 * 0.74 * 0.26 = the
expected frequency of the crossphase foxes in the next generation.
Return to Question 12.
Question 12
You said that the frequency of crossphased foxes after a generation of random mating would be (c) 26%.
No, remember that you can calculate the frequency of the genotypes in the next generation using the formula
p^{2} + 2pq + q^{2} = 1. In this case p represents the frequency of the "A" allele and q represents the "a" allele.
The expected frequency of the heterozygous foxes (Crossphase foxes Aa) is found by multiplying 2*p*q, so 2 * 0.74 * 0.26 = the
expected frequency of the crossphase foxes in the next generation.
Return to Question 12.
Question 12
You said that the frequency of crossphased foxes after a generation of random mating would be (d) 38%.
YES! You must have remembered that you calculate the frequency of the genotypes in the next generation using the formula
p^{2} + 2pq + q^{2} = 1, 2 * 0.74 * 0.26 = 0.38.
Go to question 13.
Question 12
You said that the frequency of crossphased foxes after a generation of random mating would be (e) 55%.
No, remember that you can calculate the frequency of the genotypes in the next generation using the formula
p^{2} + 2pq + q^{2} = 1. In this case p represents the frequency of the "A" allele and q represents the "a" allele.
The expected frequency of the heterozygous foxes (Crossphase foxes Aa) is found by multiplying 2*p*q, so 2 * 0.74 * 0.26 = the
expected frequency of the crossphase foxes in the next generation.
Return to Question 12.
Question 12
You said that the frequency of crossphased foxes after a generation of random mating would be (f) 58%.
No, remember that you can calculate the frequency of the genotypes in the next generation using the formula
p^{2} + 2pq + q^{2} = 1. In this case p represents the frequency of the "A" allele and q represents the "a" allele.
The expected frequency of the heterozygous foxes (Crossphase foxes Aa) is found by multiplying 2*p*q, so 2 * 0.74 * 0.26 = the
expected frequency of the crossphase foxes in the next generation.
Return to Question 12.
Question 12
You said that the frequency of crossphased foxes after a generation of random mating would be (g) 74%.
No, remember that you can calculate the frequency of the genotypes in the next generation using the formula
p^{2} + 2pq + q^{2} = 1. In this case p represents the frequency of the "A" allele and q represents the "a" allele.
The expected frequency of the heterozygous foxes (Crossphase foxes Aa) is found by multiplying 2*p*q, so 2 * 0.74 * 0.26 = the
expected frequency of the crossphase foxes in the next generation.
Return to Question 12.
Question 13
Red foxes have three different types of coat color, red, cross,
and black/silver (see the image) which is controlled by two loci
each with two alleles. We will only be concerned with the "A"
locus here.
Of course you wanted to see if your predictions were correct, so you return to Yellowstone the next year and count the
foxes again. This time you find 137 redphase, 56 crossphase, and 7 black/silverphase foxes.
Based on this information do you think that the red fox population in Yellowstone is in HardyWeinberg equilibrium.
Justify your answer using the expected frequency of redphase foxes in the next generation (you calculated this in a
previous question)?
Yes
No
Question 13
You said that you thought this fox population was in HardyWeinberg Equilbrium.
No, The frequency of the "A" allele is 82.5% {(2*137+56)/400} and the frequency of the redphase foxes is 68.5% (137/200).
Both of these are much greater than the expected frequency of the "A" allele (74%) and redphase genotypes (55%). This
suggests that the population is not in HardyWeinberg Equilibrium, because the allele frequencies are changing.
Return to question 13.
Question 13
You said that you didn't think that this fox population was in HardyWeinberg Equilbrium.
Correct, The frequency of the "A" allele is 82.5% {(2*137+56)/400} and the frequency of the redphase foxes is 68.5% (137/200).
Both of these are much greater than the expected frequency of the "A" allele (74%) and redphase genotypes (55%). This
suggests that the population is not in HardyWeinberg Equilibrium, because the allele frequencies are changing.
CONGRATULATIONS YOU FINISHED THE TUTORIAL!!