Transcript 13 Genetics - One Cue Systems

Introduction
Science of genetics can be divided into 4 subdisciplines:
• Transmission genetics: how are genes passed from one
generation to the next? Study object: individual
• Molecular genetics: how relates DNA to phenotype? Study
object: cell
• Population genetics: how behave traits in populations?
Study object: population
• Quantitative genetics: what are rules for polygenic
inheritance? Study object: population
In this course we will study
• Complex transmission genetics and we will use statistical
methods for the analysis of genetic experiments
• Population and quantitative genetics
• Extra-nuclear genetics
Laboratoriumtechniek 1
Laws of Probability
1. Rule of multiplication
Rule of multiplication = The
probability that
independent events will
occur simultaneously is the
product of their individual
probabilities
2. Rule of addition
Rule of addition = The
probability of an event that
can occur in two or more
independent ways is the
sum of the separate
probabilities of the different
ways
Question: In a Mendelian cross
between pea plants that are
heterozygous for flower color
(Pp), what is the probability
that the offspring will be
homozygous recessive?
Question: In a Mendelian cross
between pea plants that are
heterozygous for flower color (Pp),
what is the probability of the
offspring being a heterozygote?
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Laws of Probability
3. Conditional probability
The probability of an outcome that depends
on a specific condition related to that outcome
Question: In the F2 of a monohybrid cross with
plants with purple and white flowers: what is the
probability that a plant with purple flowers is
heterozygous?
Pa probability of being heterozygous = 1/2
Pb probability of the condition (purple) = 3/4
Pc = Pa/Pb = 2/3
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Laws of Probability
4. The binomial theorem
• The probability of two alternative outcomes during a number
of trials.
Question: In a family of four, what is the probability of having two
boys and two girls?
N
Binomial
1
2
3
4
5
(x
(x
(x
(x
(x
+
+
+
+
+
y)1
y)2
y)3
y)4
y)5
Expanded binomial
= 1x + 1y
= 1x2 + 2xy + 1y2
= 1x3 + 3x2y + 3xy2 + 1y3
= 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4
= 1x5y0 + 5x4y1 + 10x3y2 + 10x2y3 + 5x1y4 + 1x0y5
N=4, x=0.5, y=0.5
P(4 boys)= 1x4 = 1/16
P(2 boys and 2 girls)= 6x2y2 = 6/16
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Laws of Probability
• The general formula for the binomial expansion is:
Probability = N!
·(px)(qN-x)
x!(N-x)!
N=total number of events
p is the probability of event P
q=probability of the alternative event
x is the number of times event P occurs
! means factorial.
The total number of events (N) is 4 children. The number of times (x) that
event P (male child) occurs is 2. The probabilities p and q are both 0.5.
Thus:
4! ·(0.52) (0.52) = 24/4×1/16 =6/16=3/8
(2!)(2!)
Of all families with 4 children 3 out of 8 are predicted to have two boys
and two girls.
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Testing hypotheses using the 2-statistic
Mendel raised 705 purple-flowered
and 224 white-flowered plants in
the F2 generation.
Hypothesis: Is this a ratio of 3:1?
1. Calculate expected values based
O
E
|O–E|
on the result total (929):
705 696.75 8.25
3 : 1 = 696.75 : 232.25
224 232.25 8.25
2. Calculate the differences between 929 929
observed (O) and expected (E)
2
3. Standardise the differences:
|O – E |2
2
 =  
E
4. Calculate degrees of freedom (DF)
= number of differences – 1
5. Lookup in 2 table at DF
6. If p<0.05 then reject hypothesis
2
(8.25-0.5)
(8.25-0.5)2
=  + 
696.75
232.25
The correction for continuity of
0.5 is only applied when DF=1
2 = 0.086 + 0.259 = 0.345
At DF=1 0.345 lies between 0.016
and 0.455 with 0.900>p>0.500,
so accept hypothesis
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Testing hypotheses using the 2-statistic
Mendel raised the F2 generation of a dihybrid cross and found
yellow/round 315, green/round 108, yellow/wrinkled 101 and
green/wrinkled 32.
Hypothesis: Is this a ratio of 9:3:3:1?
•
class
round/yellow
round/green
wrinkled/yellow
wrinkled/green
Total
observed
315
108
101
32
556
expected
312.75
104.25
104.25
34.75
556.00
deviation
+2.25
+3.75
-3.25
-2.75
+0.00
• Standardise the differences:
|O – E |2
2 =   = 0.470
E
• Calculate degrees of freedom (DF)
= number of differences – 1 = 3
• Lookup in 2 table at DF: P>0.9
• If P<0.05 then reject hypothesis.
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the 2-statistic
Laboratoriumtechniek 8
Population genetics
Questions studied in Population
genetics:
• how much genetic variation is
found in natural populations and
what processes control the amount
of variation?
• What processes are responsible for
producing genetic divergence
between populations?
• How do population characteristics
like breeding system, fecundity,
and age structure influence the
gene pool?
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Genetic structure of populations
To study the genetic structure of a Mendelian population one
must describe the gene pool by calculating the genotypic and
allelic frequencies
• Genotypic frequencies
tiger moth:
f(BB)= 452/497=0.909 (top 2)
f(Bb)= 43/497= 0.087 (middle 4)
f(bb)= 2/497= 0.004 (bottom 1)
• Allelic frequencies:
tiger moth:
p=f(B)=(2×452+43)/(2×497)=0.953
q=f(b)=(2×2+43)/(2×497)=0.047
or
p=f(B)=f(BB)+½f(Bb)=0.953
q=f(b)=f(bb)+½f(Bb)=0.047
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Genetic structure of populations
Allelic frequencies with multiple alleles A1, A2 and A3
p=f(A1)=
q=f(A2)=
r=f(A3)=
(2×count of A1A1)+(A1 A2)+(A1 A3)
(2×total number of individuals)
(2×count of A2A2)+(A2 A1)+(A2 A3)
(2×total number of individuals)
(2×count of A3A3)+(A3 A1)+(A3 A2)
(2×total number of individuals)
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Genetic structure of populations
Allelic frequencies at an X-linked locus:
p=f(XA)=
q=f(Xa)=
(2×XAXA females)+(XA Xa females)+(XA Y males)
(2×number of females)+(number of males)
(2×XaXa females)+(XA Xa females)+(Xa Y males)
(2×number of females)+(number of males)
If the numbers of males and females are equal:
p=f(XA)=
2
1
1
A
A
A
a
— [f(X X ) + — f(X X )] + — f(XAY)
3
2
3
q=f(Xa)=
2
1
1
a
a
a
A
— [f(X X ) + — f(X X )] + — f(XaY)
3
2
3
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The Hardy-Weinberg Law
The Hardy-Weinberg Law says that:
• In an infinitely large, randomly mating population, free from
mutation, migration, and natural selection,
• the frequencies of the alleles do noy change over time,
• genotypic frequencies remain in the proportions p2 (frequency
of AA), 2pq (frequency of Aa) and q2 (frequency of aa),
p2 + 2pq + q2 = 1
Assumptions:
If population size is limited: chance deviations cause genetic
drift
Random mating not necessary for all traits: humans mate
preferentially for height, IQ, skin colour, etc. but random for
blood type.
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The Hardy-Weinberg Law
Female gametes
When is a population in genetic
equilibrium?
1. The allele frequencies will not
change from generation to generation
2. The genotypic frequencies will be in
the proportions p2, 2pq and q2 after
one generation of random mating
Male gametes
A(p)
a(q)
A(p)
AA(p2)
Aa(pq)
a(q)
Aa(pq)
aa(q2)
Why?
Each generation of zygotes produces
A and a in proportions p and q.
Female gametes
If p=0.6 and q=0.4 then
AA=0.36 (p2), Aa=2×0.24 (2pq), and
aa=0.16 (q2)
Total=1.00
Male gametes
A(0.6)
a(0.4)
A(0.6)
AA(0.36)
Aa(0.24)
a(0.4)
Aa(0.24)
aa(0.16)
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The Hardy-Weinberg Law
Example:
In the US the frequency of
tyrosinase-negative albinism is 1 in
40,000, or 0.000025.
The trait is recessive, so the
genotype is aa.
So q2 equals 0.000025, thus
q=0.005 and p=0.995.
The heterozygote frequency is
therefore
2pq=2×0.995×0.005=0.00995
(almost 1%!)
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The Hardy-Weinberg Law
Extension for more than 2 alleles.
• 2 alleles: p2 + 2pq + q2 =
(p + q)2
• 3 alleles: (p + q + r)2 =
p2 + q2 + r2 + 2pq + 2pr + 2qr
• 4 alleles: (p + q + r + s)2
• etc.
Example: allozym-polymorphism in
Mytilus edulis
Allele
frequency

LAP98
LAP96
LAP94
p=0.52
q=0.31
r=0.17

LAP=leucine aminopeptidase
Assignment: calculate the genotypic
frequencies
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The Hardy-Weinberg Law
Testing for Hardy-Weinberg equilibrium using
Example tiger moth:
observed genotypic frequencies:
f(BB)= 452/497=0.909
f(Bb)= 43/497= 0.087
f(bb)= 2/497= 0.004
Allelic frequencies:
p=f(B)=0.953
q=f(b)=0.047
f(BB)= p2 × N =
f(Bb)= 2pq × N =
2
expected
observed
|O–E|
451.4
452
0.6
44.5
43
1.5
1.1
2
0.9
f(bb)= q2 × N =
The number of degrees of freedom is 1, because p and q are not independent,
so the correction for continuity is applied
2
(0.6-0.5)2
(1.5-0.5)2
(0.9-0.5)2
=  +  +  = 0.168
451.4
44.5
1.1
The change P for this result is >0.5, so there is no significant deviation from
the Hardy-Weinberg equilibrium.
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The Hardy-Weinberg Law
The Hardy-Weinberg Law can be used to estimate allelic
frequencies
Example:
In Arizona (US) live the Hopi Indians.
In this tribe 26 cases of albinism were
observed in a population of 6000.
So q2 = 26/6000 = 0.0043 and q= 0.065.
Therefore p=0.935.
The frequency of heterozygotes is
2pq = 2×0.935×0.065 = 0.122, thus one
out of eight Hopi’s carries an allele for
albinism.
Photograph taken in 1900
Is this calculation allowed?
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Genetic variation
The genetic structure of populations varies in space and time.
In mussel we observe cline:
change of allelic frequencies along
a geographical transect.
In humans 15% of the variation
is found between different populations, 85% is shared across
populations.
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Genetic variation
How is genetic variation measured?
• The genetic basis of most traits is too
complex to assign genotypes to
individuals. Only a few traits behave in a
Mendelian fashion.
• Cytological variation (chromosome
morphology and banding pattern) was
observed in salivary glands of fruit flies.
• Use of starch gel electrophoresis to study
protein polymorphism made it possible to
determine genotypes of many individuals
at many loci.
• By measuring variation at the DNA- or
RNA-level.
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Genetic variation
Allozyme
electrophoresis
•
•
•
•
proportion of
polymorphic loci =
frequency of loci
with more than one
allele in a population
heterozygosity HI=
proportion of an
individal’s loci that
are heterozygous
allozymes=alleles
for different
enzymes
isozymes=enzymes
of like structure and
function coded by
different loci
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Genetic variation
Electrophoretic pattern
•
Monomer
11 homozygous
12 heterozygous
22 homozygous
b) dimer
11 homozygous
12 heterozygous
22 homozygous
c) tetramer
11 homozygous
22 heterozygous
22 homozygous
Neutral mutation model states that observed variation in enzymes
is neutral to natural selection.
Study assignments:
“Measuring Genetic Variability in Natural Population by Allozyme
Electrophoresis”
See Blackboard External Links
“Flash animation Measuring Genetic Variation”
See Blackboard Course Documents Week 2
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Genetic variation
Measuring genetic
variation at the DNA
level by Restriction
Fragment Length
Polymorphism (RFLP)
•
•
•
•
Amplify DNA-sequence
of interest of a number
of individuals with PCR
Use restriction enzymes
to cut amplified
fragments into smaller
fragments
Separate restriction
fragments on agarose
gel
RFLP’s are inherited like
alleles for other traits
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Genetic variation
• Individual nucleotide heterozygosity HI in eukaryotes varies
from 0.001 to 0.02
• In the human genome nucleotide heterozygosity HI is
estimated 0.0008, which means that an individual is
heterozygous at about 1 in every 1250 nucleotides
• The highest diversity occurs at sites that do not change the
amino acid sequence, which are called synonymous changes
• Non-synonymous changes may affect a protein’s function and
are subjected to natural selection and have been eleminated
from the population
• In many regions of the DNA there are short, identical
segments, called Short Tandem Repeats (STR) or
microsatellites, the number of which varies from person to
person. They are used in forensics, parent testing, etc.
Exercises: 22.1-22.14 with a *
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Forces that change gene frequencies
• The simplest way to calculate the population heterozygosity HT
for a single locus is as:
HT = 1 – pi2
i
where pi is the frequency of the i-th of k alleles. [Note that p1,
p2, p3 etc. may correspond to what you would normally think of
as p, q, r, s etc.].
• If we want the gene diversity over several loci we need double
summation and subscripting as follows:
HT = 1 – pij2
i j
•
2 alleles: if p=q=0,5 then HT = 1–(0,25+0,25) = 0,5
• 10 alleles: if p1=p2=p3 … p10=0,1 then HT = 1–(10×0,01) = 0,9
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Forces that change gene frequencies
For many populations the conditions for Hardy-Weinberg
equilibrium do not hold, because:
• Mutation occurs
• They are small
• Natural selection favours/dismisses alleles
• Mating may be non-random
• Migration takes place
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Forces that change gene frequencies
Mutations are heritable random changes in the DNA of a locus
• New alleles occur only as a result of mutation
• Mutations provide raw genetic material on which natural
selection acts
• Kinds of mutations:
- base substitutions:
Original Nucleotide
Base
transitions A«»G and C«»T are
substitutions
A
T
C
more common than transversions
A
4.4
6.5
Mutant
A«»T, G«»T, C«»G and A«»C
Nucleotide
T
4.7
21.0
- Deletions and insertions
C
5.0
8.2
- Chromosomal mutations
G
9.4
3.3
4.2
(inversions and translocations)
see Flash animation Translocations on Blackboard
G
20.7
7.2
5.3
-
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Forces that change gene frequencies
Estimates of mutation rates
• Using sequence analysis
– 1/1.000.000.000 mutations per base pair per cell division
– Varies 1000-fold within genome: high at mutational
hotspots
• Using single base mutations with phenotypic effect
– 1/100.000 per gamete (humans, mice, Drosophila)
• By keeping DNA non-expressed for many generations
– Mutation rate = 1 per zygote, mostly mildly deleterious
and recessive
• Using amino acid chang
– 4.2 amino acid altering mutations per ind. per generation
– 1.6 of these are deleterious
• Mutation rates in males > females
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Forces that change gene frequencies
• A mutation can be neutral, detrimental, or beneficial, which
depends on the specific environment e.g. DDT-resistance in
insects
• Forward mutation rate u is higher than reverse mutation
rate v
• Forward mutation from A to a lowers p and the reverse
mutation form a to A increases p. At equilibrium:
v
u
p =  and q = 
u+v
u+v
Laboratoriumtechniek 29
Forces that change gene frequencies
• How fast do mutations change allele frequencies?
Example:
p=0.9 and q=0.1
u=5×10–5 and v=2×10–5
After one generation:
p=vq–up=(2×10–5×0.1)–(5×10–5×0.9)=–0.000043
At equilibrium p=0.286
This means that a change from 0.50 to 0.49 takes 680 generations
A change from 0.30 to 0.29 takes 17,200 generations!
Laboratoriumtechniek 30
Forces that change gene frequencies
Random genetic drift
Laboratoriumtechniek 31
Forces that change gene frequencies
Random genetic drift
Laboratoriumtechniek 32
Forces that change gene frequencies
Random genetic drift is caused by sampling error when
gametes are drawn randomly from a small population.
Example:
A small population with 12 individuals p=0.5 and q=0.5 has genotypes
3AA, 6Aa and 3 aa. Suppose an accident kills 25% of the population,
which all happen to be aa.
This chance is .......
In the next generation p becomes ......
and q becomes .......
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Forces that change gene frequencies
Bottlenecks and founder effects
• When populations remain small over
many generations drift plays an
important role in allelic frequencies.
Scattered populations in isolated
habitats each undergo drift.
• Founder effect occurs when a
population is established with a small
number of breeding individuals.
Example:
Tristan da Cunha was settled in 1817 by
William Glass and wife. A few more settlers
joined them. In 1855 population size was
100 with 26% of all genes from the Glasses.
In 1961 after a volcanic eruption the whole
population of 300 was evacuated to
England. 14% of their genes were still
Glass’s.
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Forces that change gene frequencies
• Bottleneck effect occurs after a drastic reduction in population
size.
Example:
On Tristan da Cunha in 1857 to population dropped from 103 to 33. In
1885 15 men drownded, many left the island and the population
decreased further from 106 to 59. Many alleles were lost as a result.
• Main effect of genetic drift is fixation: only one allele is left.
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Forces that change gene frequencies
• Genetic drift is random, individual populations do not change in
the same direction. This divergence increases over generations.
• This is the basis of the Neutral theory of molecular evolution:
a new mutation that is neutral with respect to natural selection
will most likely be lost as a result of genetic drift. Occasionally
the mutation will drift to fixation. As mutation is a recurring
event, a gene will accumulate differences over time by chance
alone.
In this way the genes of two related lineages can be compared
and used to estimate the date since they last shared a common
ancestor.
Following this method humans and chimpanzees shared a
common ancestor 6 million years ago.
What is your opinion about this method?
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Forces that change gene frequencies
• Can we predict Genetic drift?
Depends on effective population size = number of individuals
that contribute gametes to the next generation
4×Nf×Nm
pq
Ne =  , variance =  , standard error s=
Nf+Nm
2Ne
√
pq

2Ne
In which Nf=number of females, Nm= number of males
Example:
In a population of 70 females where only 2 males fertilise all the
females, the females each contribute 1/2×1/70= 0.007 and each male
1/ ×1/ = 0.25 of the gametes
2
2
Ne = (4×70×2)/(70+2) = 7.8, which means that drift occurs in about
the same rate as in a population with 4 females and 4 males!
If p=0.8 then sp= √pq/2Ne = 0.1
Laboratoriumtechniek 37
Forces that change gene frequencies
Genghis Khan a Prolific Lover, DNA Data Implies
Hillary Mayell
for National Geographic News
February 14, 2003
Genghis Khan, the fearsome Mongolian warrior of the 13th
century, may have done more than rule the largest
empire in the world; according to a recently published
genetic study, he may have helped populate it too.
An international group of geneticists studying
Y-chromosome data have found that nearly 8 percent of
the men living in the region of the former Mongol
empire carry y-chromosomes that are nearly identical.
That translates to 0.5 percent of the male population in
the world, or roughly 16 million descendants living
today.
Laboratoriumtechniek 38
Forces that change gene frequencies
•
The amount of drift is inversely proportional to population size.
Example:
If the initial allele frequency p0=0.3
and N=10 then it takes 2.4×10
generations on average for the allele
to be lost.
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Forces that change gene frequencies
• Balance between mutation and random genetic drift.
Mutation introduces new variation, genetic drift removes it.
• The Infinite alleles model states that every mutation generates a
novel allele. In a large gene of 10,000 base pairs the chance
that two mutations generate the same allele is very small. (300
amino acid protein ‡ 900 nucleotides ‡ 4900 = 10542 possible
sequences)
• In this situation the forces of mutation and genetic drift balance
each other. The chance of drawing two alleles and having them
be different is
4Ne×u
HT = 
1+4Ne×u
HT is equal to the heterozygosity
in the total population.
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Forces that change gene frequencies
Migration
• Immigration introduces new alleles into populations: gene flow
• Gene flow changes allelic frequencies in the recipient population
•
Example:
px = f(A)x = 0.8 in population x and py = f(A)y = 0.3 in population y
In each generation some individuals migrate from x to y. After migration
population y consists of two groups of individuals: a proportion m of
migrants with px = 0.8 and 1–m residents with py = 0.3.
The frequency of A in y after migration is p’y = mpx + (1–m)py
p= m(px–py)
• Through exchange of alleles
different populations remain
similar, divergence is reduced
Example: Monarch butterfly
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Forces that change gene frequencies
Natural selection
• In contrast to mutation, genetic drift and migration, natural
selection results in adaptation.
Adaptation is the proces by which traits evolve that make
organisms more suited to their environment.
• Natural selection is the dominant force in evolution. The concept
comes from Darwin (who had no clue about transmission
genetics).
• Natural selection can be defined as differential reproduction of
genotypes. It simply means
that individuals with certain
alleles produce more offspring
than others. Therefore those
alleles increase in frequency
in the next generation.
•
Example Industrial melanism,
described around 1900.
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Forces that change gene frequencies
• Estimation of fitness must be done
with great care!
Example:
Starlings have an optimal number of
eggs they could raise to mature
offspring. Laying more eggs, they
produced fewer succesful chicks.
So, assigning higher fitness values to
birds that laid mote eggs would be
incorrect.
• At times natural selection can
elimate genetic variation, and at
other times it maintains variation. It
can change allelic frequencies or
prevent them from changing. It can
produce genetic divergence between
populations or maintain genetic
uniformity.
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Forces that change gene frequencies
• Natural selection is measured in terms of Darwinian fitness W,
which is defined as the relative reproductive ability of a
genotype.
• The genotype that produces the most offspring has fitness W=1.
• Selection coefficient s is the the relative intensity of selection
against a genotype = 1–W
Example:
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Forces that change gene frequencies
Possible outcomes of natural selection
• W11= W12= W22. No selection, allelic frequencies do not change.
• W11= W12<1 and W22= 1. Selection operates against the
dominant allele.
• W11= W12= 1 and W22<1. Selection operates against the
recessive allele.
• W11< W12< 1 and W22=1. Selection operates without effects of
dominance, the heterozygote has intermediate fitness.
• W11 and W22<1 and W12= 1. Selection is favouring the
heterozygote.
• W12< W11 and W22= 1. Selection operates is favouring both
homozygotes.
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Forces that change gene frequencies
Laboratoriumtechniek 46
Forces that change gene frequencies
Effectiveness of selection against a
recessive lethal genotype at different
initial allelic frequencies
Fitnesses of the genotypes AA, Aa and
aa are 1, 0.5, and 0.5 for the dominant
case, 1, 0.75 and 0.5 for the additive
case, and 1, 1 and 0.5 for the recessive
case
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Forces that change gene frequencies
• Heterozygote superiority = heterosis = overdominance
• In this case both alleles are maintained in the population at an
equilibrium, depending on the relative fitnesses of the
homozygotes.
If the selection coefficient of AA is s and the selection coefficient
of aa is t, then at equilibrium:
peq=f(A)=t/(t+s) and qeq=f(a)=s/(t+s)
•
Example:
Sickle-cell anemia is common in
areas with malaria.
Hb-A/Hb-A has normal blood
Hb-A/Hb-S has sickle-cell trait
Hb-S/Hb-S has sickle-cell disease
Heterozygotes are relatively
resistent to the malaria parasite.
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Forces that change gene frequencies
Balance between mutation and natural selection
• Natural selection reduces the frequency of deleterious alleles:
q=–spq2
• Mutation produces new alleles that can be harmful: q=up
• At equilibrium both forces are in balance: spq2 =up
• So at equilibrium: qeq=(u/s)
• If the recessive homozygote is lethal: qeq=u
•
Example:
mutation rate u=10–6 and s=0.1
at equilibrium the frequency of the allele will be qeq=u/s=0.0032
• Most deleterious alleles remain within the population at low
frequency because of equilibrium between mutation and natural
selection
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Forces that change gene frequencies
Assortative mating
• Many populations do not mate randomly for some traits
• Positive assortative mating occurs when similar phenotypes mate
preferentially
• Negative assortative mating occurs when dissimilar phenotypes
mate preferentially
• Assortative mating does not alter the allelic frequencies, but it
may influence the genotypic frequencies
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Forces that change gene frequencies
Inbreeding
• Inbreeding involves preferential mating between relatives
• Inbreeding is measured as F, the coefficient of inbreeding.
The greater F, the greater the reduction in heterozygosity.
expected heterozygosity – observed heterozygosity
F = 
expected heterozygosity
• In random mating F=0
• In self-fertilisation
F=0.5, the heterozygote frequency
decreases by 50%
in each generation
• Inbreeding leads to
reduced fitness:
inbreeding depression
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Forces that change gene frequencies
migration
genetic drift
natural selection
Non-random mating
Changes in allelic frequency in a
population
Genetic divergence between
populations
Change in genetic variation within
populations
mutation
Summary of effects of evolutionary forces on genetic structure
+
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o
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Exercises: 22.15-22.34 with a *
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The role of genetics in conservation biology
The role of genetics in conservation biology
• Reduction of habitat size decreases population size: the gene
pool gets smaller which may affect long term survival
• Population viability analysis estimates the population size that is
necessary for survival
• Breeding programmes or action against habitat loss?
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Speciation
Speciation is division of one species into two new species
• A species is a group of interbreeding organisms (or that
exchange hereditary material)
• Population subdivision may be weak or extreme to the point that
two populations never interbreed
• If isolation takes place for a long period, fixation of different
alleles occurs, leading to
• Postzygotic isolation: hybrids are infertile or have low fitness
• Individuals that engage interspecific mating suffer disadvantage
• Prezygotic isolation:
- temporal isolation by changing the mating period
- ecological isolation by exploiting different niche
• Genetic means to prevent the formation of hybrids:
- behavioral incompatibility: both species avoid eachother
- mechanical isolation: their genetalia do not fit
- gametic isolation: fusion of the gametes fails
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Speciation
Speciation can be allopatric or sympatric
• Allopatric speciation occurs when populations become
geographically isolated, are exposed to divergent selection and
evolve independantly.
After secundary contact postzygotic isolation prevents gene flow,
reinforced by (following) prezygotic isolation.
• In sympatric speciation gene flow is limited by some form of
isolation, but the populations are not totally separated.
The most likely cause is selection to use different resources.
Example:
The fruitfly Rhagoletis polmonella used to lay its eggs in hawthorn. In
1864 Rhagoletis was found in apples. Today we can discriminate between
an “apple race” and a “hawthorn race”. Behaviour stimulates divergence:
- females prefer to lay eggs in the fruit in which they grew up
- males prefer to mate on the fruit from which they emerged
Breeding times have diverged, enzym patterns are different.
• Fast sympatric speciation is possible by polyploidisation
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Speciation
Genetic basis for speciation
• Younger species rely mainly on postzygotic barriers
• Often male (XY) hybrids are sterile, not the females (XX). Why?
- Possibly because many genes are involved in the fertility of
males.
- By pleiotropic effects of genes that were selected during
adaptation to different environments
• Sexual selection
Example: 800 Drosophila species in Hawaii
are the result of allopatric speciation caused
by founder effect.
Remarkable morphology of males is byproduct of genes controlling courtship, females
are often similar. Females choose only their
own type.
In Lake Victoria in Africa >500 cichlid species evolved in 13.000 years.
Males of related species only differ in colour. Female choice of male
colour causes reproductive isolation.
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Basic quantitative genetics
Quantitative traits are affected
by alleles at many loci.
• Many polygenic phenotypic
traits vary continuously
among individuals (e.g. body
size, running speed,
competitive ability, ...). Traits
often “bell-shaped” frequency
distribution.
• Population genetics cannot be
used; segregation too
complex, genes are unknown,
phenotype not predictable
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Basic quantitative genetics
• Is phenotypic variation result
of
genetic make-up or
environment?
• In general: P = E + G
this simplification does not hold
when there is interaction
between G and E (G×E>0).
• If G×E=0 then variance
var(P)=var(E)+var(G).
In controlled experiments
var(E) or (G) can be made 0,
which allows to estimate the
source of the variance.
• Question: how do you make
var(G) zero? And var(E)?
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Basic quantitative genetics
• Broad sense
heritability:
H2=var(G)/var(P)
• Does not distinguish
between genetic or
maternal factors or
interactions that influence
the result.
• Is measured by twin
studies or crosses
between highly inbred
lines that yield virtually
the same heterozygotes.
• Is of limited value for
estimating the response
to selection (e.g. if there
is heterosis)
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Basic quantitative genetics
• Var(G) = var(A)+var(D)+var(I)
in which:
var(D) = variance due to dominance at one locus
var(I) = variance due to epistasis between loci
var(A) = additive genetic variance, responds to selection
• Narrow sense heritability:
h2=var(A)/var(P)
• h2 is estimated from the response
to selection itself, in so-called
truncation selection:
only individuals with an interesting
trait above a certain (truncation)
value are selected to breed the next
generation.
S is the selection differential,
R is the selection response, and
h2 = R/S
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Basic quantitative genetics
• Can also be measured
following the procedure
to the right.
h2 is equal to the slope of
regression function
(= 0.75 in the example)
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Basic quantitative genetics
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Basic quantitative genetics
Question:
A botanist measured seed weight and estimated its narrowsense heritability in three widely separated Phlox populations.
Mean seed
weight
Narrow sense
heritability
Population A
15 mg
0.60
Population B
12 mg
0.65
Population C
17 mg
0.58
Because the heritabilities were high and approximately equal, he
concluded that the differences in seed weight between the
populations were largely due to genetic differences. Do you
agree? Explain.
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Basic quantitative genetics
Warnings regarding heritability estimates
• Represents contribution of genetic variation in trait: Does not
tell how much of trait is genetic or environmental. What does
it mean:
“intelligence in humans is determined 70% by genes and
30% by environmental conditions”
To have a trait you need both. You can only tell to what
extent differences between individuals are genetic or
environmental.
• Depends on population and environment (inbred lab
population: lower heritability; Wild type under constant
environment: high heritability): h² estimate only valid for
populations and conditions tested.
• High heritability for a trait does not imply that populations
differences in this trait are genetically determined.
• An individual does not have heritability; a population does.
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Basic quantitative genetics
Evolutionary implications of quantitative genetics
• When much additive variation: response to selection rapid and
strong
• Artificial selection proves that when selection directional, major
changes possible
• Yet, usually only small changes under natural conditions. Why?
– Continuous, directional selection rare due to fluctuations in
direction of selection
– Pattern of temporal fluctuation may seem to represent
average stability
– Directional selection may cause negative side-effects, which
slow down or halt evolution (trade-offs): After relaxation of
selection: return to original state.
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Basic quantitative genetics
Integration of population and quantitative genetics
• Integration likely because of progress in molecular biology
– Discovery of many alleles with quantitative effects at single
loci
– Link between DNA sequences and phenotypic traits
– Progress in computational techniques = possible to analyse
multi-gene traits
• QTL analysis (Quantitative trait loci analysis)
1. Develop molecular markers (variable DNA sequence).
Associate these markers with extreme values for a quantitative,
phenotypic trait
2. Localise and characterise gene in vicinity of marker
• Conclusions: quantitative traits often controlled by few majoreffect genes and many small-effect genes (modifiers).
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Genetic problems
Q1
In a population of 2000 gaboon vipers, a genetic difference with respect to
venom exists at a single locus. The alleles are incompletely dominant.
The population shows 100 individuals homozygous for the t allele
(genotype tt, non-poisonous), 800 heterozygotes (genotype Tt, mildlypoisonous) and 1100 homozygous for the T-allele (genotype TT, lethally
poisonous).
a. What is the frequency of the t-allele in the population?
b. Are the genotypes in Hardy-Weinberg equilibrium?
Q2
Approximately one normal allele in 30,000 mutates to the X-linked recessive
allele for hemophilia in each human generation. Assume for the purpose
of this problem that one h allele in 300,000 mutates to the normal
alternative in each generation. What allelic frequencies would prevail at
equilibrium under mutation pressure alone?
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Non-Mendelian inheritance
Some traits do not behave in a Mendelian fashion:
• Maternal inheritance
A type of uniparental inheritance in which all progeny have the genotype
and phenotype of the female parent
• Maternal effect
The effect of the maternal parent's genotype on the phenotype of her
offspring
• Genetic imprinting
The phenomenon in which there is differential expression of a gene
depending on whether it was maternally or paternally inherited. Paternal
imprinting means that an allele inherited from the father is not
expressed in offspring. Maternal imprinting means that an allele
inherited from the mother is not expressed in offspring
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Non-Mendelian inheritance
• Maternal inheritance
Example: four-o-clocks
• Maternal effect
Example: Lymnea peregra
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Non-Mendelian inheritance
• Genetic imprinting
DNA is methylated, during
formation of the gametes,
causing the allele not to
be expressed.
As the embryo grows, the genes
act differently depending on
whether they came from the
mother or the father.
Paternal genes speed up growth
of embryos, maternal genes limit
growth.
Raises many questions. Why?
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Non-Mendelian inheritance
• Genetic imprinting
Boys with Prader-Willi (left) and
Angelman (right) syndromes
Both of these conditions are caused
by deletions or other mutations in the
same region of chromosome 15.
However, part of this region is
imprinted (or inactivated) on the
maternally inherited chromosome
(the PWS region), and part is
imprinted on the paternal
chromosome (the AS gene, which is
called UBE3A).
If, for example, there is a deletion in
this region on the paternal
chromosome 15, neither chromosome
15 of the child will express the genes
in the PWS region (the copy inherited
from the mother is inactivated). As a
result, the child will have PWS.
However, if the deletion is on the
maternal chromosome 15, then
neither chromosome 15 of the child
will express the genes in the AS
region, and hence, this child will have
AS.
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Non-Mendelian inheritance
•Genetic imprinting
Model of methylation-dependent
repression from the H19 ICR
At the paternal H19 allele, DNA
methylation (filled circles) at the
Imprinting Control Region (ICR) is
recognised by MeCP2. MeCP2 is
capable of recruiting Sin3a and HDACs
via its Transciptional Repression
Domain (blue ellipse). HDACs act
locally to deacetylate the tails of
histones proximal to H19, resulting in
chromatin compaction and silencing of
the H19 gene.
At the maternal H19 allele the absence
of DNA methylation at the ICR
prevents recruitment of the
MeCP2/Sin3a/HDAC complex. The
histones in the region therefore remain
acetylated and the chromatin is
accessible to factors necessary for
transcription of the H19 gene.
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