Tutorial #4
by Ma’ayan Fishelson
Changes made by Anna Tzemach
Hardy-Weinberg equilibrium
A||a
Other Possibilities: (a||A), (A||A),(a||a)
What is the probability that a person chosen at random from
the population would have each one of these genotypes ?
Hardy-Weinberg equilibrium: If the allele frequencies are
P(a)=p and P(A)=q, then, under some assumptions:
P(Aa)= 2pq, P(AA)= p2, P(aa)= q2.
Corresponds to the random mating of
two gametes, an egg & a sperm.
Hardy-Weinberg equilibrium
This model relies on the following assumptions:
a.
b.
c.
d.
e.
f.
g.
Infinite population size.
Discrete generations.
Random mating.
No selection.
No migration.
No mutation.
Equal initial genotype frequencies in the two sexes.
Hardy-Weinberg equilibrium
Let the initial genotype frequencies be:
P(A/A) = u, P(A/a) = v, P(a/a) = w.
The next generation satisfies:
Mating Type
Nature of offspring
Frequency
A/A x A/A
A/A
u2
A/A x A/a
½A/A + ½A/a
2uv
A/A x a/a
A/a
2uw
A/a x A/a
¼A/A + ½A/a + ¼a/a
v2
A/a x a/a
½A/a + ½a/a
2vw
a/a x a/a
a/a
w2
P(A/A) = u2 + uv + ¼v2 = (u + ½v)2
P(A/a) = uv + 2uw + ½v2 + vw = 2(u + ½v)(½v + w)
P(a/a) = ¼v2 + vw + w2 = (½v + w)2
Hardy-Weinberg equilibrium
If we define the frequencies of the alleles as:
•
p = P(A) = u + v/2
•
q = P(a) = v/2 + w
then, the genotype frequencies are:
•
P(A/A) = p2
•
P(A/a) = 2pq
•
P(a/a) = q2
Second generation respects the same distribution:
P(AA)= (p2 + ½2 pq)2 = [p(p+q)]2 = p2
P(Aa)= 2(p2 + ½2pq) (½2pq +q2) =2p(p+q)q(p+q)= 2pq
P(aa) = (½2pq + q2)2 = [q(p + q)]2 = q2
Linkage equilibrium
A a
B b
A a
B b
a a
b b
A a
b b
There is no
association
between alleles
at any 2 loci.
Linkage equilibrium: P(Aa,Bb) = P(Aa)P(Bb)
Hardy-Weinberg equilibrium: P(Aa) = P(A)P(a), P(Bb)=P(B)P(b)
Convergence to equilibrium is at the rate of (1 – θ)n.
θ = (θf + θm) / 2, where θf and θm are the female
and male recombination fractions, respectively.
THE HARDY-WEINBERG LAW
• p+q=1
• p2 + 2pq + q2 = 1
• p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
• p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
Question #1
•
You have sampled a population in which you know
that the percentage of the homozygos recessive
genotype (aa) is 36%. Calculate the following:
A.
B.
C.
D.
E.
The frequency of the “aa” genotype.
The frequency of the “a” allele.
The frequency of the “A” allele.
The frequencies of the genotype “AA” and “Aa”.
The frequencies of the two possible phenotypes
if “A” is completely dominant over “a”.
Question #1- Solution
•
•
•
•
•
The frequency of the "aa" genotype. Answer: 36%, as given in the problem itself.
The frequency of the "a" allele. Answer: The frequency of aa is 36%, which means
that q2 = 0.36, by definition. If q2 = 0.36, then q = 0.6, again by definition. Since q
equals the frequency of the a allele, then the frequency is 60%.
The frequency of the "A" allele. Answer: Since q = 0.6, and p + q = 1, then p = 0.4;
the frequency of A is by definition equal to p, so the answer is 40%.
The frequencies of the genotypes "AA" and "Aa." Answer: The frequency of AA is
equal to p2, and the frequency of Aa is equal to 2pq. So, using the information above,
the frequency of AA is 16% (i.e. p2 is 0.4 x 0.4 = 0.16) and Aa is 48% (2pq = 2 x 0.4
x 0.6 = 0.48).
The frequencies of the two possible phenotypes if "A" is completely dominant over
"a." Answers: Because "A" is totally dominate over "a", the dominant phenotype will
show if either the homozygous "AA" or heterozygous "Aa" genotypes occur. The
recessive phenotype is controlled by the homozygous aa genotype. Therefore, the
frequency of the dominant phenotype equals the sum of the frequencies of AA and
Aa, and the recessive phenotype is simply the frequency of aa. Therefore, the
dominant frequency is 64% and, in the first part of this question above, you have
already shown that the recessive frequency is 36%.
Question #2
• A temp. dependant enzyme that affects the survival of
organisms has 2 alleles (A,B). In a population of 1000
individuals, these alleles have the following
frequencies:
• A = 0.45
• B = 0.55
The population is exposed to a high temp. which
proves to be lethal for all individuals homozygous
for the A allele.
• Given that the population is maintained at this higher
temp., what will be the genotypic and allelic frequencies
of the next generation ?
Question #2 - Solution
•
•
•
•
•
•
• A = 0.45
• B = 0.55
AA = 0.45*0.45 = 0.2025
AB = 2*0.45*0.55 = 0.495
BB = 0.55*0.55 = 0.3025
Expose
AB = 0.495/(0.495 +0.3025) = 0.62
BB = 0.3025/(0.495+0.3025) = 0.38
Question #2 – Solution
Second generation
Mating
Type
Nature
of
offsprin
g
Frequency
A/BxA/B
¼ A/A + ½ 0.62*0.62 =
A/B+ ¼ BB 0.3844
A/BxB/B
½ A/B + ½
B/B
2*0.62*0.38
= 0.4712
B/BxB/B
B/B
0.38*0.38 =
0.1444
• A/A = ¼ *0.3844 =
0.0961
• A/B = ½ *0.3844 +
½*0.4712 = 0.6634
• B/B = 0.1444 + ¼
0.3844 =0.2405
Hardy-Weinberg Equilibrium
for X-linked Loci
• Suppose we have two alleles, A1 and A2 with
equilibrium frequencies p1 and p2.
• The female genotypes have the following
frequencies:
A1/A1
p12
A1/A2
2p1p2
A2/A2
p22
• In males the genotypes A1 and A2 have the
frequencies p1 and p2.
Question #3
• The red cell antigen Xg(a) is an X-linked
dominant allele with a frequency of
approximately p=0.65 in Caucasians.
• How many males and how many females
carry the antigen ?
Question #3 - Solution
• dominant allele with a frequency of
approximately p=0.65
• In males the genotypes A1 and A2
have the frequencies p1 and p2 =>
0.65=> 65%
• In females => AA + Aa => 0.65^2 +
2*0.65*0.35 = 0.4225 + 0.455 =
0.8775
Question #4
• Only about 70% of all white North Americans can
taste the chemical phenylthiocabamide.
• The ability to taste is determined by the
dominant allele T.
• The inability to taste is determined by the
recessive allele t.
• Assume that the population is in Hardy-Weinberg
equilibrium.
• What are the genotypic and allelic frequencies
in this population ?
Question #5 – Linkage
Equilibrium
• The degree of Linkage Disequilibrium D can
be measured using the following equation:
D = PABPab – PaBPAb
• What value of D do we expect to see when
the population is in Linkage Equilibrium ?