‹#›
•P O P U L A T I O N
•G E N E T I C S
![3D1862506](3D"slide0001_image002.gif")
CHW population
(limiting conditions)=
(limiting conditions)=
-large =3D more than 1000
individuals
-panmictic
-no mutations
-no selection
-no migration
-monogenic traits - at least two different alleles
-non-overlapping of generations
![](3D"slide0093_image003.gif")
1
2
2=
=
3;
)=
=
3;
(=
=
3;
)=
=
3;
(=
=
3;
2=
=
3;
)=
=
3;
(=
=
3;
=3D=
3;
+=
3;
+=
3;
aa=
3;
Aa=
3;
AA=
3;
q
pq
p
Castle=
-Hardy-Weinberg
law
=
Applied on panm=
ictic
population under the assumption of limiting conditions
B=
asic
relation for a system with two alleles i=
n a
given gene
![](3D"slide0094_image004.gif")
1
)=
3;
(=
3;
)=
(=
3;
=3D
+
a=
A=
q=
3;
p=
3;
Castle=
-Hardy-Weinberg
law
![](3D"slide0094_image005.gif")
)=
(=
a=
q=
3;
1=
3;
)=
(=
=3D
-
A=
p=
3;
approximation
1
![](3D"slide0094_image006.gif")
)=
=
3;
(=
=
3;
A=
=
3;
p
2=
=
3;
)=
=
3;
(=
=
3;
=3D=
3;
Aa=
3;
pq
q
2
, if
approaches
.
4
![](3D"slide0029_image007.gif")
Task 1=
/p.139 =
211; frequencies of
alleles in MN blood group system
Phenotype
Number of persons
M
406
MN
744
N
332
5
Task 1=
/p.139 =
211; frequencies of
alleles in MN blood group system
•Solution:
•direct =
c=
alculation
of the f=
req=
uency o=
f one of a=
l=
leles =
a=
ccording
to formula [=
3] on
p. 138
•
•
•
•Calculation of the frequency of the second=
a=
ll=
ele as
addition to 1.
![](3D"slide0095_image008.gif")
in the sample
individuals
all
number of
2<=
/div>
)<=
/div>
(<=
/div>
tes
of hetero=
zygo
number
)<=
/div>
(<=
/div>
of homozygotes
number
2<=
/div>
´
+=
3;
´
=3D=
3;
Aa=
=
3;
AA=
=
3;
p<=
/span>=
3;
6
Task 1/p.139 – =
frequencies of alleles in MN system
![](3D"slide0096_image009.gif")
0
744
812
alleles M
![](3D"slide0096_image010.gif")
1 556
1 408
![](3D"slide0096_image011.gif")
1 482
Total
![](3D"slide0096_image012.gif")
664
744
0
alleles N
![](3D"slide0096_image013.gif")
332
N
744
406
persons
Number of
MN
M
phenotype
![](3D"slide0096_image014.gif")
![](3D"slide0096_image015.gif")
![](3D"slide0096_image016.gif")
![](3D"slide0096_image017.gif")
![](3D"slide0096_image019.gif")
![](3D"slide0096_image021.gif")
7
Task 2a/p. 139 – Estimation of gene frequencies in the dominant system
3=
0 % of
persons unable to recognize the bitter taste of PTC in a given population
(non-tasters)
(Tt) =
=3D
0.495
![](3D"slide0099_image022.gif")
![](3D"slide0099_image023.gif")
10
![](3D"slide0099_image024.gif")
genotypes
![](3D"slide0099_image025.gif")
phenotype
Estimation of gene frequencies in two-allele polymorphism
(T, t=
)
in the population sample of students
in the population sample of students
p=
henotyping
of PTC tasting in students in the classroom: o=
ne
drop of saturated solu=
tion
of phenylthiocarbamide (PTC) on the tip of the
tongue
=3D
•
•
![](3D"slide0100_image026.gif")
![](3D"slide0100_image027.gif")
![](3D"slide0100_image028.gif")
11
![](3D"slide0031_image029.gif")
Task 4/p. 140 – estimates of
frequencies of
deleterious (recessive) alleles
Disease
Abbrev.
population
frequency
phenylketonuria
PKU
1/8100
cystic fibrosis (mucoviscidosis)
CF
1/2500
12
![](3D"slide0097_image030.gif")
Task 4/p. 140 – estimates of
frequencies of
deleterious (recessive) alleles
S=
olution: the estimate c=
alculated according to formula [4] on p. 139 (top),
13
![](3D"slide0098_image031.gif")
Task 4/p. 140 – estimates of frequencies =
of deleterious (recessive) alleles
estimate
49/50 =3D 1
89/90 =3D 1
p =3D 1 - q
1/50
1/90
q
1/2500
1/8100
Freque=
ncy in popula=
tion
CF
PKU
diseas=
e
2pq =3D 2q=
2 x 1 x 1/50
=3D 1/25
=3D 1/25
2 x 1 x 1/90
=3D 1/45
=3D 1/45
˙
˙
˙
14
Small population
-less than 1000 individuals
Gene drift
Simple model of gene drift I
![3D"MPj0177=](3D"slide0122_image032.jpg")
•modelling of random use of alleles (gametes)
•random mechanism – dice throw
<=
/div>
•small population N =3D 3, and original p =3D q =3D ½
Drift – changes in gene
frequencies in small populations elicited by rand=
om
processes
•allele A picked out … side 1, 2 or 3 cast
•allele a picked out … side 4, 5 or 6 cast<=
span
lang=3DEN-US style=3D'font-family:"Times New Roman CE";mso-ascii-font-fami=
ly:"Times New Roman CE";
mso-hansi-font-family:"Times New Roman CE";font-size:75%;color:yellow;
mso-color-index:3;mso-special-format:lastCR;display:none'>
•6 throws each time – results of series mark in the table
•According to the result (ratio, new frequencies) pA=
/ =
qa change the rules for next series of throws (generation)
•continue until one allele fixed
12
=
=
Simple model of gene drift II
= – Task 14/p. 148
= – Task 14/p. 148
![](3D"slide0123_image034.gif")
13
=
Simple model of gene drift III
– Task 14/p. 148
– Task 14/p. 148
An
example of results after the firts series of throws and the change of t=
he
rules for next series
![](3D"slide0124_image036.gif")
14
=
<=
span
style=3D'font-size:91%'>Simple model of gene drift IV
– Task 14/p. 148
– Task 14/p. 148
An
example of results after the firts series of throws and the change of t=
he
rules for next series
![](3D"slide0125_image038.gif")
15
Simple
model of gene drift IV
– Task 14/p. 148
– Task 14/p. 148
•<=
/span>w=
ork in
pairs, one is throwing, the other is recording
•<=
/span>r=
emind
you the change of rules, where according to the =
r=
esult
the gene frequencies are changing for next s=
eries
of throws and consequently which sides of the d=
ice
will represent the allele A or a
•<=
/span>c=
ontinue
in experiment until the point of fixation of one a=
llele
•n=
ow,
swap your roles, this way one pair will finish and e=
valuate
at least two experiments, each experiment and r=
esult
visualize in a graph
•f=
or
whole classroom calculate average time of fixation as
weighted mean of results of all partial experiments, t=
he
ratio of fixed A and a alleles should be approx. 1 : 1
16
Simple model of gene drift V – Task 14/p. 148
C=
onclusion
of experiments
![](3D"slide0127_image040.gif")
![](3D"slide0127_image041.gif")
17
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=
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