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Title: Population Genetics
Description: Population Genetics Mendellian Ratios Examples

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Population genetics emerged in the 1920’s and is the extension
of mendellian genetics to the population and is important in
agriculture, medical and human genetics etc
If you don’t have genetic variation, you cannot breed things and select
for a particular trait
...
What are the contributions of these forces to the
genetic difference in populations
...


Phenotypes: M

MN

N

Genotypes:

MN

NN

MM
180

240

80

[data from

a bunch of gels]

Proportions: 180/500 = 0
...
48

80/500 = 0
...
e
...
6
q = frequency (N) = [2 x 80 + 240] / 2 x 500
= 0
...
36 + 1/2 (0
...
6
the same answer*



q = freq (N) = 0
...
48) = 0
...

P- proportion of polymorphism ; H- level of heterozygosity

*For 3 segregating alleles for any co-dominant technique being used

Genotypes

FF

FM

FS

Obs #

100

150

50

MM
125

MS
0

SS
75

Total
500

p = freq (F) = [2 x 100 + 150 + 50]/ 2x 500
= 0
...
4
r = freq (S) = [2 X 75 +50 + 0] / 2 X 500
=0
...
A
haplotype could be more than 1 gene
...

1
...


2
...
HARDY-WEINBERG LAW / EQUILIBRIUM- discovered simple
mathematics for what happened during sexual reproduction
...

2 alleles: A , a
p = freq (A) , q = freq (a)
pA

qa

pA

p2

AA

pqAa

qa

pqAa

p + q = 1

q2aa

Random Mating
A, a

D= freq [AA] , H = freq (Aa) , R = FREQ (aa)

Construct what you would get with random mating and the
frequency with which they would occur
...

Imagine you have 3 alleles
F
Freq:

p

FF

FM
MS

p2

2pq

M

S

q

r

FS
SS
2pr

MM
q2

2qr

r2


Autotetraploid

A, a

AAAA

(p+q)^4

AAAa AAaa Aaaa

p^4

p^3q^1

aaaa
q^4

For rare alleles, recessive:
q = frequency (d) = 0
...
999

Expected proportions:
DDp^2 = 0
...
001998 dd=q^2 = 0
...
69
q^2 = 39
...
62

p = freq (F) = 2 x 22 +30/ 2 x 100 = 0
...
63

100 x

Calculate X^2 = 127

DF = 3-1 = 2

X^2 = 3
...
Genotype frequencies observed
do to measure the expected frequencies
...
Hetero’s could be dying
...

Flower colour
RR, Rr = Red

rr = white

# of expected phenotype
Proportion
Assume HW

Red
75
0
...
25
q^2

Total
100

q^2 = 0
...
5
p = 1 - q = 0
...
25 0
...
25

DF= 2-1-1 = 0 cannot perform Chi squared test on 0 df
...

Can do a test cross to figure out the genotype of the red flower
...

• Inbreeding affects all genes [Refer to graph on slides] —> Faster decrease
in heterozygosity with self compared to with double 1st cousin
mating
...
Also present in some plants
...










Title: Population Genetics
Description: Population Genetics Mendellian Ratios Examples