Population genetics ppt slide for test problems and summarized stud

1. Population Genetics

2. • GENETIC POPULATION – a group of sexually or
potentially interbreeding individuals
A. Important attributes of a population
1. Gene/Allelic Frequencies – proportion of
different alleles of a gene in a population
F(A)=2 (# of homozygotes)+heterozygotes
2 x total # of individuals
2. Genotypic Frequencies – the proportion of
different genotypes in a population
3. Gene Pool – the sum total of genes in the
reproductive gametes of a population

3. B. The Hardy-Weinberg Equilibrium
“For a stable population, the gene frequencies
remain constant from generation to generation.”
In Equation: 2 alleles: p+q=1 where: F(A)=p; F(a)=q
Hardy-Wienberg Equation in binomial Expansion:
p2 + 2pq + q2 =1
Where: p2 = f (AA)
2pq= f (Aa)
q2 = f (aa)

4. C. Qualifications of a Stable Population
1. Population must be infinitely large and is
composed of sexually producing diploid
individuals and that there are always a couple
acting as parents.
2. Mating is PANMICTIC (random)
3. Rate of mutation and rate of migration is non-
existing or at least constant.
4. Gene involved has no selective value/ the
individual concerned will have equal chance of
survival and reproduction. (No Natural
Selection)
5. No Genetic Drift

5. D. Gene Frequencies of Co-dominant traits
(easier to determine because genotypic ratio =
phenotypic ratio)
Sample Problem 1: MN blood series
Individuals with the following blood type are as in
the surveyed data. Calculate the gene
frequencies and genotypic frequencies of MN
blood series in the population.
M=245
MN =210
N = 45
ANSWERS
F(MM)
=245/500=0.49
f(MN)=
0.42;
f(NN)=
0.09
f(M)=
0.7
F(N)=
0.3

6. E. Gene frequencies of Dominant traits
Difficulty: genotypic frequency is not equal to
phenotypic frequency
Sample Problem 2: PTC taster (bitter sweet) and
controlled by dominant (T). In 1,000
individuals there were 700 tasters, 300
nontasters. Calculate the genotypic
frequencies and the number of individuals
belonging to each group.
p+q
(p+q)
2
f(t)=q=sqt
q
2
=sqt300/1000
=
0.5477
f(T)=
p=1-q
=
1-0.5477=0.4523
F(TT)=p
2
=0.2046
f(Tt)=2pq=0.4954
F(tt)=0.300
#
individuals
belongig
to
each
genotupe:
f(TT)=205;
f(Tt)=495;
f(tt)=300

7. Sample problem 3
• Among the Hopi Indians of Arizona, albinism is
remarkably common. In 1996, 26 cases of
albinism was observed in a total population of
6,000 Hopis. (a) Calculate the frequency of the
trait (b)Determine the frequency of the albino
gene. (c) How many heterozygotes are there
(carriers of the gene)?
a)
q
2
=26/6000=0.0043
b)
q=sqt0.0043=0.0658
c)
2pq=0.122
x
6000=732

8. Sample Problem 4
A sample of 1,522 persons living in London
disclosed of 464 of M, 733 type MN and 325
type N. Calculate the gene frequency of M
and N.
F(M)=0.545633
or
0.55
F(N)=0.454336
or
0.45

9. Sample Problem 5
• A sample if 100 persons disclosed 84 PTC
tasters.
a) calculate gene frequencies of T and t.
b) How many heterozygotes are there?
c) How many homozygote dominants are
there?
a)
f(t)=0.4
F(T)=0.6
b)
#
of
Tt=
48
c)
#
of
TT
=
36

10. E. Case of DI and MULTIHYBRID
DIHYBRID: Aa gametes A and a : freq p &q
Bb “ B and b : “  r & s
independent genes : (p + q ) (r +s )
gametes produced : (pr + ps + qr + qs)2 Hardy-
Weinberg Law : (pr + ps + qr + qs)2 =1
MULTIHYBRID : Product Rule of Probability
f(AAbbCc) = p2s22zy
F. MULTIPLE ALLELES
Frequencies and corresponding genotypes:
(p + q+ r)2 =1
G. SEX-LINKED GENES
FEMALES: 2 X chromosomes  behave as autosomal p2+2pq+q2=1
MALES –hemizygous p+q=1 where: p=A-males; q=a-males
H. SEX-INFLUENCED : determine the sex in which the trait is expressed
in dominant form e.g: baldness: males: p2 2pq Females p2

11. If an X-linked recessive disorder is in Hardy-
Weinberg equilibrium and the incidence in
males equals 1 in 100, what will be the
expected incidence of affected homozygous
females?
Sample Problem 6
ANSWER
1/10,000