Basic Genetics

(1) Mendelian geneticsHow does a gene transmit from a parent to its progeny (individual)?

(2) Population geneticsHow is a gene segregating in a population (a group of individuals)?

(3) Quantitative geneticsHow is gene segregation related with the phenotype of a character?

(4) Molecular geneticsWhat is the molecular basis of gene segregation and transmission?

(5) Developmental genetics(6) Epigenetics (genetic imprinting)

Genomic Imprinting

The callipygous animals 1 and 3 compared to normal animals 2 and 4 (Cockett et al. Science 273: 236-238, 1996)

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).

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 GameteA pair of chromosomes

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

What about three genes?

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

B

How the linkage occurs? – consider two genes A and B

A a

B b

A

B

A

B

a a

b b

A

B

a

b

a A

B b

A

B

a

b

a A

B b

1 2 3 4

Stage 1: A pair of chromosomes, one from the father and the other from the motherStage 2: Each chromosome is divided into two sister chromatidsStage 3: Non-sister chromatids crossoverStage 4: Meiosis generates four gametes AB, aB, Ab and ab – Nonrecombinants (AB and ab) and Recombinants (aB and Ab)

How to measure the linkage? – based on a design

Parents AABB × aabbGamete AB ab

F1 AaBb × aabbGamete AB Ab aB ab ab

Backcross AaBb Aabb aaBb aabbObservations 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)

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

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.

 

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)

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

Calculations of allele frequencies and genotype frequencies

Genotypes Counts Estimates genotype frequenciesAA 224 PAA = 224/294 = 0.762Aa 64 PAa = 64/294 = 0.218aa 6 Paa = 6/294 = 0.020

Total 294 PAA + PAa + Paa = 1

Allele frequenciespA = (2214+64)/(2294)=0.871, pa = (26+64)/(2294)=0.129, pA + pa = 0.871 + 0.129 = 1

Expected genotype frequenciesAA pA

2 = 0.8712 = 0.769Aa 2pApa = 2 0.871 0.129 = 0.224Aa pa

2 = 0.1292 = 0.017

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

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…

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 ratio1. AA AA AA2. AA Aa ½AA + ½Aa3. AA aa Aa4. Aa AA ½AA + ½Aa5. Aa Aa ¼AA + ½Aa + ¼aa 6. Aa aa ½Aa + ½aa7. aa AA Aa8. aa Aa ½Aa + ½aa9. aa aa aa

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

PAA(t+1) = 1[PAA(t)]2 + ½ 2[PAA(t)PAa(t)] + ¼[PAa(t)]2 = [PAA(t) + ½PAa(t)]2

Similarly, we havePaa(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 havePAA(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)

(1) Genotype (and allele) frequencies are constant from generation to generation,

(2) Genotype frequencies = the product of the allele frequencies, i.e., PAA = pA

2, PAa = 2pApa, Paa = pa2

For a population at Hardy-Weinberg disequilibrium (HWD), we have• PAA = pA

2 + D• PAa = 2pApa – 2D• Paa = pa

2 + D

The magnitude of D determines the degree of HWD.• D = 0 means that there is no HWD.• D has a range of max(-pA

2 , -pa2) D pApa

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:

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.

Example 1AA Aa aa Total

Obs 224 64 6 294

Exp n(pA2) = 222.9 n(2pApa) = 66.2 n(pa

2) = 4.9 294

Test statisticsx2 = (obs – exp)2 /exp = (224-222.9)2/222.9 + (64-66.2)2/66.2 +

(6-4.9)2/4.9 = 0.32is 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

Example 2AA Aa aa

Total Obs 234 36 6 276

Exp n(pA2) n(2pApa) n(pa

2) = 230.1 = 43.8 = 2.1 276

Test statisticsx2 = (obs – exp)2/exp = (234-230.1)2/230.1 + (36-

43.8)2/43.8 + (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.

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 = pb

2

PAA+PAa+Paa = 1 PBB+PBb+Pbb=1

pA + pa = 1 pB + pb = 1

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.

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

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

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 ispAB(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)

WhenceD(t+1) = (1-r)t D(1)

Estimate and test for LDAssuming random mating in the population, we have joint probabilities of the two genes

BB (PBB) Bb (PBb) bb (Pbb)_______________________________________________________________________________________AA (PAA) pAB

2 2PABPAb PAb2

n22 n21 n20

Aa (PAa) 2PABPaB 2(PABPab+PAbPaB) 2PAbPab

n12 n11 n10

aa (Paa) PaB2 2PAbPab Pab

2

n02 n01 n00

________________________________________________________________________________________

Multinomial pdfH1: D 0 log f(pij|n)= log n!/(n22!…n00!) + n22 log pAB

2 + n21log (2pABpAb) + n20 log pAb2

+ …Estimate pAB, pAb, paB (pab = 1-pAB-pAb-paB) pA, pB, D

H0: D = 0log f(pi,pj|n)= log n!/(n22!…n00!) + n22log(pApB)2 + n21log(2pA

2pBpb)+n20log(pApb)2

+ …

Estimate pA and pB.

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.

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 474 611 142 773 2n=2000Gamete frequency pAB pAb paB pab

=474/2000 =611/2000 =142/2000 =773/2000

=0.237 =0.305 =0.071 =0.386 1

Estimates of allele frequencies

pA = pAB + pAb = 0.237 + 0.305 = 0.542

pa = paB + pab = 0.071 + 0.386 = 0.458

 pB = pAB + paB = 0.237 + 0.071 = 0.308

pb = pAb + pab = 0.305 + 0.386 = 0.692The estimate of DD = pABpab – pAbpaB = 0.237 0.386 – 0.305 0.071 = 0.0699

Test statistics

x2 = 2nD2/ (pApapBpb) =210000.06992/(0.5420.4580.3080.692) = 184.78 is greater than x2

df=1 (0.05) = 3.841.Therefore, the population is in linkage disequilibrium at these two genes under consideration.

A second approach for calculating x2: Gamete AB Ab aB ab TotalObs 474 611 142 773 2n=2000

Exp 2n(pApB) 2n(pApb) 2n(papB) 2n(papb)

=334.2 =750.8 =281.8 =633.2 2000

  x2 = (obs – exp)2 /exp = (474-334.2)2/334.2 + (611-750.8)2/750.8 + (142-281.8)2/281.8 + (773-633.2)2/633.2

= 184.78

= 2nD2/ (pApapBpb)

Measures of linkage disequilibrium

 

(1) 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

(2) 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' = 0.0699/min(pApb, papB) = 0.0699/min(0.375, 0.141) = 0.496

(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 = 0.0699/(pApbpapB) = 0.3040. 

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.

 

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: 199-204.

 

Dawson, E., G. R. Abecasis, S. Bumpstead, Y. Chen et al. 2002 A first-generation linkage disequilibrium map of human chromosome 22. Nature 418: 544-548.

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)

(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).

Quantitative geneticsMany 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

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.

Number of Genes Number of Genotypes

1 3

2 9

5 243

10 59,049

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 valueAABB 1 12 AABb 2 11 AAbb 1 10 AaBB 2 10 AaBb 4 9 Aabb 2 8 aaBB 1 8 aaBb 2 7 aabb 1 6

A grapical format is used to present the above results:

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

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.

• 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: 102-109.

Genetic Parameters: Means and (Co)variancesOne-gene model

Genotype aa Aa AAGenotypic value G0 G1 G2

Net genotypic value -a 0 d a

origin=(G0+G1)/2a = additive genotypic valued = dominant genotypic value

Environmental deviation E0 E1 E2

Phenotype orPhenotypic value Y0=G0+E0 Y1=G1+E1 Y2=G2+E2

Genotype frequency P0 P1 P2

at HWE =q2 =2pq =p2Deviation from population mean -a - d - a -

=-2p[a+(q-p)d] = (q-p)[a+(q-p)d] = 2q[a+(q-p)d]

-2p2d +2pqd -2q2dLetting =a+(q-p)d =-2p-2p2d =(q-p)+2pqd =2q-2q2d

Breeding value -2p (q-p) 2qDominant deviation -2p2d 2pqd -2q2d

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

= 2pq2 + (2pqd)2

= 2a (or VA) + 2

d (or VD) Additive genetic variance, Dominant genetic

variance,depending on both on a and d depending only on d

Phenotypic variance 2P = q2Y0

2 + 2pqY12 + p2Y2

2 – (q2Y0 + 2pqY1 + p2Y2)2

DefineH2 = 2

g /2P as the broad-sense heritability

h2 = 2a / 2

P 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.

What is = a+(q-p)d?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 AaFrequency q pValue change d-(-a) a-d

= q[d-(-a)]+p(a-d) = a+(q-p)d

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 1 - - a

AA × Aa 4p3q ½(a+d) ½ ½ - ½(a+d)

AA × aa 2p2q2 0 - 1 - d

Aa × Aa 4p2q2 d ¼ ½ ¼ ½d

Aa × aa 4pq3 ½(-a+d) - ½ ½ ½(-a+d)

aa × aa q4 -a - - 1 -a

________________________________________________

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

= pq2

= ½2a

 The regression of offspring on midparent values isb = Cov(OP¯)/2(P¯)

= ½2a / ½2

P

= 2a /2

P

= h2

where 2(P¯)=½2P is the variance of midparent value.

IMPORTANT

The regression of offspring on midparent values can be used to measure the heritability!

This is a fundamental contribution by R. A. Fisher.

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 + 2

a/4

Monozygotic twins Cov(MT) = 2a + 2

d

Nephew and uncle Cov(NU) = 2a/4

First cousins Cov(FC) = 2a /8

Double first cousins Cov(DFC) = 2a/4 + 2

d/16

Offspring and midparent Cov(O) = 2a/2

____________________________________________________