Hardy-Weinberg Chi Square Example
Note: Remember that frequencies range from 0 to 1!!
Question 1a: In a certain population of newts, being poisonous (P) is dominant over not being
poisonous (p). You count 200 newts, and 8 are not poisonous. What are the allele frequencies of
the parent population?
1.
p2 + 2pq + q2 = 1
2.
PP Pp pp
3.
Poisonous not poisonous
4.
192
8
2
So you know that q = 8/200 = .04. To find q, simply take the square root of q2: q = √.04 = 0.2
p = 1-q = 1 – 0.2 = 0.8
Question 1b: Fifty newts are washed downstream after a big storm and colonize a new pond.
What do you expect the frequency and number of each genotype to be?
You would expect the allele frequencies to remain the same, so q = 0.2 and p = 0.8. To find the
genotype frequencies, fill in the Hardy-Weinberg equation.
p2 = (0.8)2 = 0.64
2pq = 2*0.2*0.8 = .32
q2 = (0.2)2 = .04
To find the number of each genotype, multiply the total population by the genotype frequency.
PP (p2) = 50*.64 = 32
Pp (2pq) = 50*0.32 = 16
pp (q2) = 50*0.04 = 2
Question 1c: You count the new population of newts and find 21 homozygous poisonous newts,
23 heterozygous poisonous newts, and 6 homozygous non-poisonous newts. Is this what you
expect? (Test using Chi square). If it is not, what are the new allele frequencies?
1. For this, you set up the same Chi square table as for the Punnett square crosses, with one
difference: instead of an expected ratio, you will have an expected frequency.
O
Efreq
E
O-E
(O-E)2
(O-E)2/E
2
Poisonous PP (p )
21
Poisonous Pp (2pq)
23
2
Not poisonous pp (q ) 6
Total
50
2. You already calculated the expected frequency and the expected number of newts for each
genotype, so write those numbers in the table.
O
Efreq
E
O-E
(O-E)2
(O-E)2/E
PP (p2)
21
p2=0.82=.64
50*.64=32
Pp (2pq)
23
2pq=2*0.2*0.8 =.32 50*0.32=16
pp (q2)
6
q2 = 0.22 = .04
50*0.04 = 2
Total
50
3. Now complete the rest of the table (or you may simply use the Chi square formula directly).
O
Efreq
E
O-E
(O-E)2
(O-E)2/E
2
PP (p )
21
0.64
32
-11
121
121/32=3.78
Pp (2pq) 23
0.32
16
7
49
49/16=3.06
2
pp (q )
6
0.04
2
4
16
16/2=8
Total
50
3.78+3.06+8=14.8
4. You have 2 alleles (two categories), so your degrees of freedom is one less than that: 2-1=1.
The table value for 1 d.f. is 3.841. Your calculated value (14.8) is larger than the table value
(3.841), therefore the differences between your expected and observed values are NOT due to
chance. You must reject your hypothesis that the observed and expected numbers are statistically
the same.
5. To calculate the new allele frequencies, use the Hardy-Weinberg formula.
1.
p2 + 2pq + q2 = 1
2.
PP Pp pp
3.
Poisonous not poisonous
4.
21 23
6
Because you know both how many homozygous dominant newts there are, and how many
homozygous recessive newts there are, you can calculate both p and q from your observed
values.
p2 = 21/50 = .42
so p = √.42 = .65
2
q = 6/50 = .12
so q = √.12 = .35
Double check and make sure that p + q = 1. (.65 + .35 = 1)
Note: in some problems you may know only the phenotypes instead of the genotypes. In that
case, combine the p2 and 2pq rows of your table for the appropriate phenotype.
For example, in the previous problem, if you knew only that there were 44 poisonous newts and
6 newts that were not poisonous, your table would look like this:
O
Efreq
E
O-E
(O-E)2
(O-E)2/E
50*0.96
poisonous
44
p2 + 2pq
2
(PP + Pp)
=0.8 +2*0.8*0.2 = 48
(p2 + 2pq)
=0.96
Not poisonous 6
q2 = 0.22 = 0.04 50*0.04
(pp)
=2
2
(q )
Total
50
Degrees of
Freedom
1
2
3
4
Chi square
value
3.841
5.991
7.815
9.488
Hardy-Weinberg practice problems.
1. Walking through the forest, you find a large population of toadstools. From your extensive
knowledge of the kingdom fungi, you know that the allele for being spotted (S) is dominant over
the allele for being plain (s). In this population of 1007, you find 14 toadstools that are not
spotted. What are the allele frequencies?
In a different forest, you find a somewhat smaller population of 549. Through genetic testing,
you determine that there are 308 homozygous spotted, 206 heterozygous, and 34 homozygous
plain toadstools. Is this what you expected? If not, what are the allele frequencies of this
population?
2. You that know the allele for fuzzy peaches is recessive and that its frequency is 0.7. In an
orchard of 500 trees, you pick one peach from each tree. How many fuzzy peaches do you
expect to find? (hint: fuzzy peaches’ genotype is ff).
You count 250 fuzzy peaches and 250 bald peaches. Is this the same as you expected? Test
using Chi square analysis.
3. In a population of Venus fly traps in North Carolina, you observe 150 that can eat large flies,
and 400 that can eat only small flies. If the ability to eat large flies (s) is recessive, what is the
allele frequency?
Back in the MI, you visit a specialty store that sells Venus fly traps. Of the 56 plants in the store,
how many do you expect to eat large flies?
You count and see that 25 can eat large flies. Is this what you expected? (Test with the Chi
square.) If not, what are the new allele frequencies?
4. NASA finally sends a manned flight to Mars. The astronauts discover that Martian rocks are
alive and can reproduce. In a sample of 290 rocks, they count 130 red rocks and 160 brown
rocks (red is dominant over brown). What are the allele frequencies for red and brown?
The astronauts bring a group of 20 rocks back with them for further study. Of these rocks, 10 are
red and 10 are brown. Given the parental population’s allele frequencies, is this what you would
expect (test with Chi square)? If not, what are the new allele frequencies?
5. On your Spring Break in Mexico, you take a break to count the starfish in a tidal pool. You
notice that there is a rare recessive trait that causes the starfish to have 6 legs instead of 5. Of the
27 starfish you can find, only 1 has 6 legs. What are the allele frequencies for this population?
Two years later, you return to the same beach and count the starfish again. This time you find 4
starfish out of 31 that have 6 legs. Are the genotype frequencies the same as two years earlier?
Test using Chi square. If it is not, calculate the new allele frequencies.
(Given that the current mutation rate is 0 and that there is no natural selection acting on these
starfish, what is this random change in allele frequency called?)
Seven years later, you come back to the beach. This time you can find no starfish with 6 legs.
Given the most recent allele frequencies you know, is this what you would expect?
(What is the complete loss of an allele called?)