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Polygenic traits, population genetics
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PRE-CLASS QUESTIONS. VARIANT of FRIDAY? You breed a self-compatible plant that exhibits three dominant traits:
1. flower color (A –yellow, a - white); 2. leaf shape (B – entire, b - toothed); and 3. seed shape (C- round; c - flat).
The recessive characters for these three are: a) flower color = white (aa); leaf shape = toothed (bb); and seed shape = flat (cc). You conduct a test cross with a homozygous recessive (aabbcc) to genotype your unknown plant.
The resulting F1 generation has the following results: approximately equal quantities of:
1) Yellow flowered, entire leafed, round seeded (A_B_C_; (2) Yellow flowered, toothed leafed, round seeded (A_bbC_ (3) White flowered, toothed leafed, round seeded (aabbC_)(4) White flowered, entire leaved, round seeded (aaB_C_).
• The unknown parental genotype is:• 1: AABBCC 2:AABbCC 3: AABBCc• 4: AaBBCC 5:AaBbCC 6: AaBBCc• 7: AABbCc 8:AABbCc 9: AaBbCc
EPISTASIS
So, did mendelian genetics solve the So, did mendelian genetics solve the inheritanceinheritance problem? problem?
1.1. Biometricians Biometricians vsvs. Mendelians. Mendelians
• Biometricians held that Darwin was right, and that all evolution Biometricians held that Darwin was right, and that all evolution proceeded by the gradual accumulation of small, incremental changes.proceeded by the gradual accumulation of small, incremental changes.
• Thought that mendelian traits, with their large phenotypic effects, were Thought that mendelian traits, with their large phenotypic effects, were special cases.special cases.
2.2. Mendelians believed that natural selection on continuous variation was Mendelians believed that natural selection on continuous variation was ineffective, and could never produce truly new types.ineffective, and could never produce truly new types.
• Mutations of major effect were key in the evolutionary processMutations of major effect were key in the evolutionary process, , especially the evolution of new types/species.especially the evolution of new types/species.
3.3. Quantitative genetics/polygenic traits to the rescue!Quantitative genetics/polygenic traits to the rescue!
• However, maybe this looks more like Darwin’s ideas about blending?
• Let’s take a look back.
Polygenic traitsPolygenic traits
1865 1868 1889 1892 1897 1900 1902 1905 1909 1910 1915
Mendel proposes existence of HEREDITARY PARTICLES
Darwin proposes that gemmules transmit traits
Hugo de Vries develops concept of pangenes (special particles for every hereditary character)
Weissmann proposes that hereditary particles reside on chromosomes
deVries, Correns, & von Seysennegg REDISCOVER MENDEL’S WORK
E. B. Wilson describes behavior of chromosomes during embryo development
Wilhelm Waldeyer identifies CHROMOSOMES
Wilson & students confirm MEIOSIS IS REDUCTION DIVISION FOR GAMETES
Wilson & Sutton identify sex chromosomes
Johannsen distinguishes between GENOTYPE VS PHENOTYPE
Morgan argues against role of chromosomes in inheritance
Morgan eats crow & admits that CHROMOSOMES BEAR PARTICULATE UNITS OF INHERITANCE!
Development of a theory of inheritance Development of a theory of inheritance
• Mendel’s characters were discrete and qualitative (red or white, smooth or wrinkled).
• For most traits, phenotypes vary continuously over a range—quantitative variation, or continuous.
Did mendelian genetics solve the inheritance problem?Did mendelian genetics solve the inheritance problem?
Nilsson-Ehle (1909) to the rescue: effects of multiple Nilsson-Ehle (1909) to the rescue: effects of multiple loci on wheat chaff colorloci on wheat chaff color
• 3 loci (A, B, & C), each with 2 alleles (A,A’; B,B’; C,C’) each 3 loci (A, B, & C), each with 2 alleles (A,A’; B,B’; C,C’) each contributes equally to the color of the chaff in wheat.contributes equally to the color of the chaff in wheat.
Polygenic Traits
How multiple loci lead to continuous/quantitative How multiple loci lead to continuous/quantitative variationvariation
• When a large enough number of genes influence a character, it When a large enough number of genes influence a character, it will have a continuous, normal frequency distribution. will have a continuous, normal frequency distribution.
Discrete phenotypic classesDiscrete phenotypic classes Continuous phenotypic variationContinuous phenotypic variation
When multiple genes control the expression of a trait When multiple genes control the expression of a trait
• Mendel’s characters were discrete and qualitative.• For more complex characters, phenotypes vary continuously
over a range—quantitative variation, or continuous.• (Quantitative variation is usually due to both genes and
environment (environment can determine expression).) Most quantitative characters are controlled by multiple
genes: they are POLYGENICPOLYGENIC
PleiotropyPleiotropy
Examples: leg length, arm length; PKU ( phenylketonuria),
Pleiotropy
Now where to?
• Quantitative Characters• Moving from individual genotypes to
characteristics of populations– Remember, natural selection acts of the
individual to create changes in the distribution of characters in the population.
– In order to understand selection, we need to be able to characterize the population.
• Hence, POPULATION GENETICS
Key Concepts
• We can look at a population as a big mixed set of alleles and we can calculate allele frequencies if we know genotype frequencies.
• We can also calculate expected genotype frequencies IF we know allele frequencies.
• There are some simple rules that guide us: frequencies sum to 1; the multiplication and addition rule of probabilities still apply.
COLORCOLOR Package 1Package 1
RedRed 21 (0.123)21 (0.123)
GreenGreen 33 (0.193)33 (0.193)
YellowYellow 25 (0.146)25 (0.146)
BlueBlue 30 (0.175)30 (0.175)
OrangeOrange 33 (0.193)33 (0.193)
BrownBrown 29 (0.170)29 (0.170)
TOTAL (n)TOTAL (n) 171171
Population Genetics. Population Genetics. 1. Multiplicative Rule, Additive Rule1. Multiplicative Rule, Additive Rule
What is the probability of drawing:
Two red:
Green and yellow:
Two green:
Orange and brown:
Just to hammer this home:
A red, green and a blue**
**-math problem, not a genetics problemSimple conditional probabilities
0.5 0.5
0.5 0.25 0.25
0.5 0.25 0.25
Red Green Yellow Blue Orange Brown
0.123 0.193 0.146 0.175 0.193 0.17
Red 0.123 0.015129 0.023739 0.017958 0.021525 0.023739 0.02091
Green 0.193 0.023739 0.037249 0.028178 0.033775 0.037249 0.03281
Yellow 0.146 0.017958 0.028178 0.021316 0.02555 0.028178 0.02482
Blue 0.175 0.021525 0.033775 0.02555 0.030625 0.033775 0.02975
Orange 0.193 0.023739 0.037249 0.028178 0.033775 0.037249 0.03281
Brown 0.17 0.02091 0.03281 0.02482 0.02975 0.03281 0.0289
Convince yourself.
A1A1
A1A2
A1A1
A1A1
A1A1
A1A1
A1A1
A1A1 A1A1
A1A1
A1A1
A1A1
A1A1
A2A2
A2A2
A2A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A1
A1A1
A1A1
A2A2 A2A2
N = 50 individualsN = 50 individuals
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2
A1A2 A1A2
A1A2
A1A2
Population GeneticsPopulation GeneticsAllele Frequencies, Genotype Frequencies,Allele Frequencies, Genotype Frequencies,
5
30
15
Finding p and q
p= freq of A1= q = freq A2=
p + q = 1
If we can estimate the frequency of alleles (p, q), then we can estimate the probability of encountering a genotype.
Text Fig 22.7Text Fig 22.7
p p = _________________= _________________
q = _________________q = _________________
Mix into giant bag (gene pool)...Mix into giant bag (gene pool)...
An example from An example from the textbook...the textbook...
Population GeneticsPopulation GeneticsAllele Frequencies from Genotype FrequenciesAllele Frequencies from Genotype Frequencies
Text Fig 22.7Text Fig 22.7
p = 0.55p = 0.55
q = 0.45q = 0.45
Generation IIGeneration II
AA =___________AA =___________
Aa = ___________Aa = ___________
Aa = ___________Aa = ___________
Population GeneticsPopulation GeneticsThe Hardy-Weinberg EquilibriumThe Hardy-Weinberg Equilibrium
0.55(2) = 0.3025; 2*.45*.55 = 0.495; 0.45(2) = 0.2025
Make Sense?
IF p = IF p = __________, and q = , and q = __________
f(f(AAAA) = f() = f(AAaa) = f() = f(a,aa,a) =
Population GeneticsPopulation GeneticsSome Some FUNFUN!! Problems Problems
pp22 pqpq
pqpq qq22
IF p = IF p = __________, and q = , and q = __________
f(f(AAAA) = f() = f(AAaa) = f() = f(aaaa) =
IF p = IF p = __________, and q = , and q = __________
f(f(AAAA) = f() = f(AAaa) = f() = f(aaaa) =
IF p = IF p = __________, and q = , and q = __________
f(f(AAAA) = f() = f(AAaa) = f() = f(aaaa) = (HWeq)
IF p = IF p = __________, and q = , and q = __________
f(f(AA11AA11) = f() = f(AA11AA22) = f() = f(AA22AA22) =
Population GeneticsPopulation GeneticsThe Hardy-Weinberg EquilibriumThe Hardy-Weinberg Equilibrium
pp22 pqpq
pqpq qq22
p2 p*q p*r
p*q q2 q*r
p*r q*r r2
p q r p q r p + q + r = _______
p
q
r
p =
q =
r =
p
q
r
3 alleles!!!3 alleles!!!
H-W Equilibrium: a fundamental theorem of population geneticsH-W Equilibrium: a fundamental theorem of population genetics
• The H-W theorem makes a set of The H-W theorem makes a set of nullnull predictions about expected allelic and predictions about expected allelic and genotypic frequencies genotypic frequencies
• If the observed values of genotypic frequencies do not match those expected If the observed values of genotypic frequencies do not match those expected at H-W equilibrium, at H-W equilibrium, THEN SOME EVOLUTIONARY PROCESSES MUST BE THEN SOME EVOLUTIONARY PROCESSES MUST BE ACTING ON THE GENE POOLACTING ON THE GENE POOL
Population GeneticsPopulation GeneticsThe Hardy-Weinberg EquilibriumThe Hardy-Weinberg Equilibrium
Which populations are at Which populations are at Hardy-Weinberg Equilibrium?Hardy-Weinberg Equilibrium?
1. p = 0.75; q = 0.25
2. p = 0.75; q = 0.35
3. AA = 214; Aa = 428; aa = 214
4. AA = 320; Aa = 160; aa = 20
5. f(AA) = 0.35; f(Aa) = 0.70; f(aa) = 0.35
6. f(AA) = 0.55; f(Aa) = 0.30; f(aa) = 0.15
7. f(AA) = 0.64; f(Aa) = 0.32; f(aa) =0.04
?NoYesYesNo
No
Yes
Things that follow from the H-W equilibrium:Things that follow from the H-W equilibrium:
1.1. At H-W equilibrium, genotypic frequencies will be determined At H-W equilibrium, genotypic frequencies will be determined solely by allelic frequenciessolely by allelic frequencies
2.2. Note that when there are 2 alleles, and p = q = 0.5, the Note that when there are 2 alleles, and p = q = 0.5, the expected ratio of homozygotes to heterozygotes is 1:2:1, as it expected ratio of homozygotes to heterozygotes is 1:2:1, as it was for Mendelian inheritancewas for Mendelian inheritance
Population GeneticsPopulation GeneticsThe Hardy-Weinberg EquilibriumThe Hardy-Weinberg Equilibrium
H-W Equilibrium: a fundamental theorem of population geneticsH-W Equilibrium: a fundamental theorem of population genetics
• The H-W theorem makes a set of The H-W theorem makes a set of nullnull predictions about expected allelic and predictions about expected allelic and genotypic frequencies genotypic frequencies
• If the observed values of genotypic frequencies do not match those expected If the observed values of genotypic frequencies do not match those expected at H-W equilibrium, at H-W equilibrium, THEN SOME EVOLUTIONARY PROCESSES MUST BE THEN SOME EVOLUTIONARY PROCESSES MUST BE ACTING ON THE GENE POOLACTING ON THE GENE POOL
Population GeneticsPopulation GeneticsThe Hardy-Weinberg EquilibriumThe Hardy-Weinberg Equilibrium
Coming up this week
Tuesday
• Sexual Selection– Why do we have amazing
examples of gaudy animals?
• Kin Selection– Why do individuals help
their kin? Should they?
• (Exam results)– Exams handed back in
labs– Pop quizzes??
Wednesday and Friday
• Why populations may not be at Hardy Weinberg equilibrium– Natural selection– Mutation– Non-random mating– Drift– etc.
That is it for today.