1. Basic Genetics (1) Mendelian genetics
How does a gene transmit from a parent to its progeny (individual)?
(2) Population genetics
How is a gene segregating in a population (a group of individuals)?
(3) Quantitative genetics
How is gene segregation related with the phenotype of a character?
(4) Molecular genetics
What is the molecular basis of gene segregation and transmission?
(5) Developmental genetics
(6) Epigenetics (genetic imprinting)

2. Genomic Imprinting
The callipygous animals 1 and 3 compared to normal animals 2 and 4 (Cockett et al. Science 273: , 1996)

3. Mendelian Genetics  Probability Population Genetics  Statistics
Quantitative genetics  Molecular Genetics
Statistical Genetics  Mathematics with biology (our view)
Cutting-edge research into the interface among genetics, evolution and development
(Evo-Deve)
Wu, R. L. Functional mapping of complex traits. Nature Reviews Genetics (submitted).

4. Mendel’s Laws Mendel’s first law
There is a gene with two alleles on a chromosome location (locus)
These alleles segregate during the formation of the reproductive cells, thus passing into different gametes
Diploid
Gene A
A| a |
  Centromere
A | a |
Probability ½ ½
Gamete Gamete
A pair of chromosomes

5. Mendel’s second law
There are two or more pairs of genes on different chromosomes
They segregate independently (partially correct)
Diploid
A|a|, B|b|
A|, B| A|, b| a|, B| a|, b|
Probability ¼ ¼ ¼ ¼
Four two-gene gametes

6. What about three genes?

7. Linkage (exception to Mendel’s second law)
There are two or more pairs of genes located on the same chromosome
They can be linked or associated (the degree of association is described by the recombination fraction)
High linkage Low linkage
A A B

8. How the linkage occurs? – consider two genes A and B1 2 3 4 A a A A a a A a A a A a A a B b B B b b B B b b B B b b
Stage 1: A pair of chromosomes, one from the father and the other from the mother
Stage 2: Each chromosome is divided into two sister chromatids
Stage 3: Non-sister chromatids crossover
Stage 4: Meiosis generates four gametes AB, aB, Ab and ab –
Nonrecombinants (AB and ab) and
Recombinants (aB and Ab)

9. How to measure the linkage? – based on a design
Parents AABB × aabb
Gamete AB ab
F AaBb × aabb
Gamete AB Ab aB ab ab
Backcross AaBb Aabb aaBb aabb
Observations n1 n2 n3 n4
Gamete type Non-recom/ Recom/ Recom/ Non-recom/
Parental Non-parental Non-parental Parental
Define the proportion of the recombinant gametes over the total gametes as
the recombination fraction (r) between two genes A and B
r = (n2+n3)/(n1+n2+n3+n4)

10. Several concepts Genotype and Phenotype
Locus (loci), chromosomal location of a gene
Allele (A, a), a copy of gene
Dominant allele, one allele whose expression inhibits the expression of its alternative allele
Recessive allele (relative to dominant allele)
Dominant gene (AA and Aa are not distinguishable, denoted by A_)
Codominant gene (AA, Aa and aa are mutually distinguishable)
Genotype (AA, Aa or aa)
Homozygote (AA or aa)
Heterozygote (Aa)
Phenotype: trait value

11. Chromosome and Meiosis Chromosome: Rod-shaped structure made of DNA
Diploid (2n): An organism or cell having two sets of chromosomes or twice the haploid number
Haploid (n): An organism or cell having only one complete set of chromosomes
Gamete: Reproductive cells involved in fertilization. The ovum is the female gamete; the spermatozoon is the male gamete.
Meiosis: A process for cell division from diploid to haploid (2n  n) (two biological advantages: maintaining chromosome number unchanged and crossing over between different genes)
Crossover: The interchange of sections between pairing homologous chromosomes during meiosis
Recombination, recombinant, recombination fraction (rate, frequency): The natural formation in offspring of genetic combinations not present in parents, by the processes of crossing over or independent assortment.

12. Molecular markers
Genetic markers are DNA sequence polymorphisms that show Mendelian inheritance
Marker types
- Restriction fragment length polymorphism
(RFLP)
- Amplified fragment length polymorphism
(AFLP)
- Simple sequence repeat (SSR)
- Single nucleotide polymorphism (SNP)

13. Population Genetics
Different copies of a gene are called alleles; for example A and a at gene A;
These alleles form three genotypes, AA, Aa and aa;
The allele (or gene) frequency of an allele is defined as the proportion of this allele among a group of individuals;
Accordingly, the genotype frequency is the proportion of a genotype among a group of individuals

14. Calculations of allele frequencies and genotype frequencies
Genotypes Counts Estimates genotype frequencies
AA PAA = 224/294 = 0.762
Aa PAa = 64/294 = 0.218
aa Paa = 6/294 = 0.020
Total PAA + PAa + Paa = 1
Allele frequencies
pA = (2214+64)/(2294)=0.871, pa = (26+64)/(2294)=0.129,
pA + pa = = 1
Expected genotype frequencies
AA pA2 = = 0.769
Aa 2pApa = 2   = 0.224
Aa pa2 = = 0.017

15. Genotypes Counts Estimates of genotype freq.
AA nAA PAA = nAA/n
Aa nAa PAa = nAa/n
aa naa Paa = naa/n
Total n PAA + PAa + Paa = 1
Allele frequencies
pA = (2nAA + nAa)/2n
pa = (2naa + nAa)/2n
Standard error of the estimate of the allele frequency
Var(pA) = pA(1 - pA)/2n

16. The Hardy-Weinberg Law
In the Hardy-Weinberg equilibrium (HWE), the relative frequencies of the genotypes will remain unchanged from generation to generation;
As long as a population is randomly mating, the population can reach HWE from the second generation;
The deviation from HWE, called Hardy-Weinberg disequilibrium (HWD), results from many factors, such as selection, mutation, admixture and population structure…

17. Mendelian inheritance at the individual level
(1) Make a cross between two individual parents
(2) Consider one gene (A) with two alleles A and a  AA, Aa, aa
Thus, we have a total of nine possible cross combinations:
Cross Mendelian segregation ratio
1. AA  AA  AA
2. AA  Aa  ½AA + ½Aa
3. AA  aa  Aa
4. Aa  AA  ½AA + ½Aa
5. Aa  Aa  ¼AA + ½Aa + ¼aa
6. Aa  aa  ½Aa + ½aa
7. aa  AA  Aa
8. aa  Aa  ½Aa + ½aa
9. aa  aa  aa

18. Mendelian inheritance at the population level
A population, a group of individuals, may contain all these nine combinations, weighted by the mating frequencies.
Genotype frequencies: AA, PAA; Aa, PAa; aa, Paa
Cross Mating freq. (t) Mendelian segreg. ratio (t+1)
AA Aa aa
1. AA  AA PAA(t)PAA(t)  1 0 0
2. AA  Aa PAA(t)PAa(t)  ½ ½ 0
3. AA  aa PAA(t)Paa(t)  0 1 0
4. Aa  AA PAa(t)PAA(t)  ½ ½ 0
5. Aa  Aa PAa(t)PAa(t)  ¼ ½ ¼
6. Aa  aa PAa(t)Paa(t)  0 ½ ½
7. aa  AA Paa(t)PAA(t)  0 1 0
8. aa  Aa Paa(t)PAa(t)  0 ½ ½
9. aa  aa Paa(t)Paa(t)  0 0 1

19. PAA(t+1) = 1[PAA(t)]2 + ½ 2[PAA(t)PAa(t)] + ¼[PAa(t)]2
Similarly, we have
Paa(t+1) = [Paa(t) + ½PAa(t)]2
PAa(t+1) = 2[PAA(t) + ½PAa(t)][Paa(t) + ½PAa(t)]
Therefore, we have
[PAa(t+1)]2 = 4PAA(t+1)Paa(t+1)
Furthermore, if random mating continues, we have
PAA(t+2) = [PAA(t+1) + ½PAa(t+1)]2 = PAA(t+1)
PAa(t+2) = 2[PAA(t+1) + ½PAa(t+1)][Paa(t+1) + ½PAa(t+1)] = PAa(t+1)
Paa(t+2) = [Paa(t+1) + ½PAa(t+1)]2 = Paa(t+1)

20. Concluding remarks
A population with [PAa(t+1)]2 = 4PAA(t+1)Paa(t+1) is said to be in Hardy-Weinberg equilibrium (HWE). The HWE population has the following properties:
(1) Genotype (and allele) frequencies are constant from generation to generation,
(2) Genotype frequencies = the product of the allele frequencies, i.e., PAA = pA2, PAa = 2pApa, Paa = pa2
For a population at Hardy-Weinberg disequilibrium (HWD), we have
PAA = pA2 + D
PAa = 2pApa – 2D
Paa = pa2 + D
The magnitude of D determines the degree of HWD.
D = 0 means that there is no HWD.
D has a range of max(-pA2 , -pa2)  D  pApa

21. Chi-square test for HWE
Whether or not the population deviates from HWE at a particular locus can be tested using a chi-square test.
If the population deviates from HWE (i.e., Hardy-Weinberg disequilibrium, HWD), this implies that the population is not randomly mating. Many evolutionary forces, such as mutation, genetic drift and population structure, may operate.

22. Example 1
AA Aa aa Total
Obs
Exp n(pA2) = n(2pApa) = n(pa2) =
Test statistics
x2 =  (obs – exp)2 /exp = ( )2/ ( )2/ (6-4.9)2/4.9 = 0.32
is less than
x2df=1 ( = 0.05) = 3.841
Therefore, the population does not deviate from HWE at this locus.
Why the degree of freedom = 1? Degree of freedom = the number of parameters contained in the alternative hypothesis – the number of parameters contained in the null hypothesis. In this case, df = 2 (pA or pa and D) – 1 (pA or pa) = 1

23. Example 2
AA Aa aa Total
Obs
Exp n(pA2) n(2pApa) n(pa2)
= = =
Test statistics
x2 =  (obs – exp)2/exp = ( )2/ ( )2/ (6-2.1)2/2.1 = 8.8
is greater than x2df=1 ( = 0.05) = 3.841
Therefore, the population deviates from HWE at this locus.

24. Linkage disequilibrium
Consider two loci, A and B, with alleles A, a and B, b, respectively, in a population
Assume that the population is at HWE
If the population is at Hardy-Weinberg equilibrium, we have
Gene A Gene B
AA: PAA = pA2 BB: PBB = pB2
Aa: PAa = 2pApa Bb: PBb = 2pBpb
Aa: Paa = pa2 bb: Pbb = pb2
PAA+PAa+Paa = PBB+PBb+Pbb=1
pA + pa = pB + pb = 1

25. But the population is at Linkage Disequilibrium (for a pair of loci)
But the population is at Linkage Disequilibrium (for a pair of loci). Then we have
Two-gene haplotype AB: pAB = pApB + DAB
Two-gene haplotype Ab: pAb = pApb + DAb
Two-gene haplotype aB: paB = papB + DaB
Two-gene haplotype ab: pab = papb + Dab
pAB+pAb+paB+pab = 1
Dij is the coefficient of linkage disequilibrium (LD) between the two genes in the population. The magnitude of D reflects the degree of LD. The larger D, the stronger LD.

26. pA = pAB+pAb
= pApB + DAB + pApb + DAb
= pA+DAB+DAb  DAB = -DAb
pB = pAB+paB
= pB+DAB+DaB  DAB = -DaB
pb = pAb+pab
= pb+DaB+Dab  Dab = -DaB
Finally, we have DAB = -DAb = -DaB = Dab = D.
Re-write four two-gene haplotypes
AB: pAB = pApB + D
Ab: pAb = pApb – D
aB: paB = papB – D
ab: pab = papb + D
D = pABpab - pAbpaB
D = 0  the population is at the linkage equilibrium

27. How does D transmit from one generation (1) to the next (2)?
D(2) = (1-r)1 D(1) …
D(t+1) = (1-r)t D(1)
t, D(t+1)  r

28. Proof to D(t+1) = (1-r)1 D(t)
The four gametes randomly unite to form a zygote. The proportion 1-r of the gametes produced by this zygote are parental (or nonrecombinant) gametes and fraction r are nonparental (or recombinant) gametes. A particular gamete, say AB, has a proportion (1-r) in generation t+1 produced without recombination. The frequency with which this gamete is produced in this way is (1-r)pAB(t).
Also this gamete is generated as a recombinant from the genotypes formed by the gametes containing allele A and the gametes containing allele B. The frequencies of the gametes containing alleles A or B are pA(t) and pB(t), respectively. So the frequency with which AB arises in this way is rpA(t)pB(t).
Therefore the frequency of AB in the generation t+1 is
pAB(t+1) = (1-r)pAB(t) + rpA(t)pB(t)
By subtracting is pA(t)pB(t) from both sides of the above equation, we have
D(t+1) = (1-r)1 D(t)
Whence
D(t+1) = (1-r)t D(1)

29. Estimate and test for LD
Assuming random mating in the population, we have joint probabilities of the two genes
BB (PBB) Bb (PBb) bb (Pbb)
_______________________________________________________________________________________
AA (PAA) pAB2 2PABPAb PAb2
n22 n21 n20
Aa (PAa) 2PABPaB 2(PABPab+PAbPaB) 2PAbPab
n12 n11 n10
aa (Paa) PaB2 2PAbPab Pab2
n02 n01 n00
________________________________________________________________________________________
Multinomial pdf
H1: D  0
log f(pij|n)
= log n!/(n22!…n00!)
+ n22 log pAB2 + n21log (2pABpAb) + n20 log pAb2
+ …
Estimate pAB, pAb, paB (pab = 1-pAB-pAb-paB)  pA, pB, D
H0: D = 0
log f(pi,pj|n)
+ n22log(pApB)2 + n21log(2pA2pBpb)+n20log(pApb)2
Estimate pA and pB.

30. Chi-square Test of Linkage Disequilibrium (D)
Test statistic
 x2 = 2nD2/(pApapBpb)
  is compared with the critical threshold value obtained from the chi-square table x2df=1 (0.05). n is the number of individuals in the population.
If x2 < x2df=1 (0.05), this means that D is not significantly different from zero and that the population under study is in linkage equilibrium.
 If x2 > x2df=1 (0.05), this means that D is significantly different from zero and that the population under study is in linkage disequilibrium.

31. Four gametes are observed as follows: Gamete AB Ab aB ab Total
Example
(1) Two genes A with allele A and a, B with alleles B and b, whose population frequencies are denoted by pA, pa (=1- pA) and pB, pb (=1- pb), respectively
(2) These two genes are associated with each other, having the coefficient of linkage disequilibrium D
Four gametes are observed as follows:
Gamete AB Ab aB ab Total
Obs n=2000
Gamete frequency pAB pAb paB pab =474/2000 =611/2000 =142/2000 =773/2000
=0.237 =0.305 =0.071 =

32. Estimates of allele frequencies pA = pAB + pAb = 0.237 + 0.305 = 0.542
 pB = pAB + paB = = 0.308
pb = pAb + pab = = 0.692
The estimate of D
D = pABpab – pAbpaB =  –  =
Test statistics
x2 = 2nD2/ (pApapBpb) =21000 /(0.5420.4580.3080.692) = is greater than x2df=1 (0.05) =
Therefore, the population is in linkage disequilibrium at these two genes under consideration.

33. A second approach for calculating x2:
Gamete AB Ab aB ab Total
Obs n=2000
Exp 2n(pApB) 2n(pApb) 2n(papB) 2n(papb)
= = = =
 x2 =  (obs – exp)2 /exp
= ( )2/ ( )2/ ( )2/ ( )2/633.2 =
= 2nD2/ (pApapBpb)

34. Measures of linkage disequilibrium
D, which has a limitation that its value depends on
the allele frequencies
D = 0.02 is considered to be
large for two genes each with diverse allele frequencies, e.g., pA = pB = 0.9 vs. pa = pb = 0.1
small for two genes each with similar allele frequencies, e.g., pA = pB = 0.5 vs. pa = pb = 0.5

35. To make a comparison between gene pairs with
different allele frequencies, we need a new normalized measure.
The range of LD is
 max(-pApB, -papb)  D  min(pApb, papB)
The normalized LD (Lewontin 1964) is defined as
 D' = D/ Dmax,
 where Dmax is the maximum that D can have, which is
 Dmax = max(-pApB, -papb) if D < 0,
or min(pApb, papB) if D > 0.
For the above example, we have D' = /min(pApb, papB) = /min(0.375, 0.141) = 0.496

36. (3) Linkage disequilibrium measured as the correlation
between the A and B alleles
R = D/(pApapBpb), r: [-1, 1]
Note: x2= 2nR2 follows the chi-square distribution
with df = 1 under the null hypothesis of D = 0.
For the above example, we have
R = /(pApbpapB) =

37. Application of LD analysis
D(t+1) = (1-r)tD(t),
This means that when the population undergoes random mating, the LD decays exponentially in a proportion related to the recombination fraction.
(1)   Population structure and evolution
  Estimating D, D' and R  the mating history of
population
The larger the D’ and R estimates, the more likely the population in nonrandom mating, the more likely the population to have a small size, the more likely the population to be affected by evolutionary forces.

38. Human origin studies based on LD analysis
Reich, D. E., M. Cargill, S. Bolk, J. Ireland, P. C. Sabeti, D. J. Richter, T. Lavery, R. Kouyoumjian, S. F. Farhadian, R. Ward and E. S. Lander, 2001 Linkage disequilibrium in the human genome. Nature 411:
Dawson, E., G. R. Abecasis, S. Bumpstead, Y. Chen et al A first-generation linkage disequilibrium map of human chromosome 22. Nature 418:

39. LD curve for Swedish and Yoruban samples.
To minimize ascertainment bias, data are only
shown for marker comparisons involving the
core SNP. Alleles are paired such that D' > 0
in the Utah population. D' > 0 in the other
populations indicates the same direction of
allelic association and D' < 0 indicates the
opposite association. a, In Sweden, average
D' is nearly identical to the average |D'|
values up to 40-kb distances, and the overall
curve has a similar shape to that of the Utah
population (thin line in a and b). b, LD
extends less far in the Yoruban sample, with
most of the long-range LD coming from a
single region, HCF2. Even at 5 kb, the average
values of |D'| and D' diverge substantially. To
make the comparisons between populations
appropriate, the Utah LD curves are calculated
solely on the basis of SNPs that had been
successfully genotyped and met the minimum
frequency criterion in both populations
(Swedish and Yoruban) (Reich,te al. 2001)

40. (2) Fine mapping of disease genes
The detection of LD may imply that the recombination fraction between two genes is small and therefore closer (given the assumption that t is large).

41. Quantitative genetics
Many traits that are important in agriculture, biology and biomedicine are continuous in their phenotypes. For example,
Crop Yield
Stemwood Volume
Plant Disease Resistances
Body Weight in Animals
Fat Content of Meat
Time to First Flower IQ
Blood Pressure

42. The following image demonstrates the variation for flower diameter, number of flower parts and the color of the flower Gaillaridia pilchella (McClean 1997). Each trait is controlled by a number of genes each interacting with each other and an array of environmental factors.

43. Number of Genes Number of Genotypes
,049

44. Consider two genes, A with two alleles A and a, and B with two alleles B and b. - Each of the alleles will be assigned metric values - We give the A allele 4 units and the a allele 2 units - At the other locus, the B allele will be given 2 units and the b allele 1 unit
Genotype Ratio Metric value
AABB
AABb
AAbb
AaBB
AaBb
Aabb
aaBB
aaBb
aabb

45. A grapical format is used to present the above results:

46. Normal distribution of a quantitative trait may be due to
Many genes
Environmental effects
The traditional view: polygenes each with small effect and being sensitive to environments
The new view: A few major gene and many polygenes (oligogenic control), interacting with environments

47. Traditional quantitative genetics research: Variance component partitioning
The phenotypic variance of a quantitative trait can be partitioned into genetic and environmental variance components.
To understand the inheritance of the trait, we need to estimate the relative contribution of these two components.
We define the proportion of the genetic variance to the total phenotypic variance as the heritability (H2).
- If H2 = 1.0, then the trait is 100% controlled by genetics
- If H2 = 0, then the trait is purely affected by environmental factors.

48. Fisher (1918) proposed a theory for partitioning genetic variance into additive, dominant and epistatic components;
Cockerham (1954) explained these genetic variance components in terms of experimental variances (from ANOVA), which makes it possible to estimate additive and dominant components (but not the epistatic component);
I proposed a clonal design to estimate additive, dominant and part-of-epistatic variance components
Wu, R., 1996 Detecting epistatic genetic variance with a clonally replicated design: Models for low- vs. high-order nonallelic interaction. Theoretical and Applied Genetics 93:

49. Genetic Parameters: Means and (Co)variances
One-gene model
Genotype aa Aa AA
Genotypic value G0 G1 G2
Net genotypic value -a d a
origin=(G0+G1)/2
a = additive genotypic value
d = dominant genotypic value
Environmental deviation E0 E1 E2
Phenotype or
Phenotypic value Y0=G0+E0 Y1=G1+E1 Y2=G2+E2
Genotype frequency P0 P1 P2
at HWE =q2 =2pq =p2
Deviation from population mean  -a -  d -  a - 
=-2p[a+(q-p)d] = (q-p)[a+(q-p)d] = 2q[a+(q-p)d]
-2p2d +2pqd -2q2d
Letting =a+(q-p)d =-2p-2p2d =(q-p)+2pqd =2q-2q2d
Breeding value -2p (q-p) 2q
Dominant deviation -2p2d 2pqd -2q2d

50. Population mean  = q2(-a) + 2pqd + p2a = (p-q)a+2pqd
Genetic variance 2g = q2(-2p-2p2d)2 + 2pq[(q-p)+2pqd]2 + p2(2q-2q2d)2
= 2pq (2pqd)2
= 2a (or VA) 2d (or VD)
Additive genetic variance, Dominant genetic variance,
depending on both on a and d depending only on d
Phenotypic variance 2P = q2Y02 + 2pqY12 + p2Y22 – (q2Y0 + 2pqY1 + p2Y2)2
Define
H2 = 2g /2P as the broad-sense heritability
h2 = 2a / 2P as the narrow-sense heritability
These two heritabilities are important in understanding the relative contribution of genetic and environmental factors to the overall phenotypic variance.

51. What is  = a+(q-p)d? From Aa to aa From AA to Aa Frequency q p
It is the average effect due to the substitution of gene from one allele (A say) to the other (a).
Event A a contains two possibilities
From Aa to aa From AA to Aa
Frequency q p
Value change d-(-a) a-d
 = q[d-(-a)]+p(a-d)
= a+(q-p)d

52. Midparent-offspring correlation
____________________________________________________________________
Progeny
Genotype Freq. of Midparent AA Aa aa Mean value
of parents matings value a d -a of progeny
AA × AA p4 a a
AA × Aa 4p3q ½(a+d) ½ ½ - ½(a+d)
AA × aa 2p2q d
Aa × Aa 4p2q2 d ¼ ½ ¼ ½d
Aa × aa 4pq3 ½(-a+d) - ½ ½ ½(-a+d)
aa × aa q4 -a a
________________________________________________

53. Covariance between midparent and offspring:
Cov(OP¯)
= E(OP¯) – E(O)E(P¯)
= p4a a + 4p3q ½(a+d) ½(a+d) + … + q4 (-a)(-a) – [(p-q)a+2pqd]2
= pq2
= ½2a
The regression of offspring on midparent values is
b = Cov(OP¯)/2(P¯)
= ½2a / ½2P
= 2a /2P
= h2
where 2(P¯)=½2P is the variance of midparent value.

54. This is a fundamental contribution by R. A. Fisher.
IMPORTANT
The regression of offspring on midparent values can be used to measure the heritability!
This is a fundamental contribution by R. A. Fisher.

55. You can derive other relationships
Degree of relationship Covariance
____________________________________________________
Offspring and one parent Cov(OP) = 2a/2
Half siblings Cov(FS) = 2a/4
Full siblings Cov(FS) = 2a/2 + 2a/4
Monozygotic twins Cov(MT) = 2a + 2d
Nephew and uncle Cov(NU) = 2a/4
First cousins Cov(FC) = 2a /8
Double first cousins Cov(DFC) = 2a/4 + 2d/16
Offspring and midparent Cov(O) = 2a/2