Cancer as a mini-evolutionary process.

Transcription

1 Department of Mathematics, National Tsing-Hua University January 6, 2010 Cancer as a mini-evolutionary process. Yoh Iwasa Department of Biology, Kyushu University

2 Evolution: slow change of organisms mutation: mistake of reproduction natural selection: mutant with a higher survival and a faster reproductive rate will replace the old type.

3 Colon cancer arises in a crypt A crypt consists of cells. Apoptosis on top of crypt 36 hours A small number of stem cells replenishes the whole crypt. 7 The colon contains 10 crypts.

4 Tumorigenesis includes multiple steps of mutations tti of stem cells: loss of tumor suppressors oncogenes angiogenesis (induction of blood vessels) metastasis etc. Carcinogenesis is an Evolutionary Process.

5 (1) Chromosomal Instability (2) Tissue structure (3) Chronic myeloid leukemia

6 Tumor Suppressors p53 Rb APC Tumor Suppressors prevents cell division and causes apoptosis,..... if something is wrong.

7 Loss of Both Copies of a Tumor Suppressor Gene is the First Step toward Cancer. 2u u + p 0 + / + + / / TSG TSG TSG Cancer initiation Escape apoptosis

8 Chromosomal Instability (CIN) Normal cells CIN cells

9 Can Chromosomal Instability Enhance the Risk of Cancer?

10 Effect of Chromosomal Instability y( (CIN) 2u u + p + / + + / 0 / TSG TSG TSG without t CIN 2n c u 2n c u Cancer initiation TSG +/+ CIN 2u TSG +/ CIN fast p TSG CIN Cancer initiation with CIN Loss Of Heterozygosity

11 Fixation of mutant fraction Nuρ() r >> O(N) cell generations can be approximated by a Markovian transition

12 Moran Process 1. Select a cell for reproduction 2. Cell division 4. Add the new cell 3. Remove a cell

13 Fixation probability ρ( ) ρ r N: large N: small r fitness 1/N r fitness Deleterious mutations can be fixed, if N is small.

14 Fixation of an intermediate mutant cell generations

15 Tunneling: The Second Mutation Spreads without the Fixation of the First One cell generations

16 explicit formula for tunneling rate ( 1 r)+ 1 r R tunnel Nu 1 1+ r ( ) r ru Nu r ρ () a ρ() r + 1 Nu 1 u 2 ρ ( a ) N + ( )ru 2 ρ a () () ρ r deleterious mutation if 1 r >> 2 u 2 ρ() a if 1 r << 2 u 2 ρ ( a ) neutral mutation +

17 Intermediate mutant is deleterious Fixation probability of Type 2 mutant X tunneling 1-step process 0.0 fixation of type 1 2-step process N Small compartment: 2-step evolution Large compartment: Tunneling (1-step evolution)

18 Fixation probability of Type 2 mutant Intermediate mutant is neutral X tunneling 1-step process N fixation of type 1 2-step process Small compartment: 2-step evolution Large compartment: Tunneling + 2-step evolution

19 2u 2Nu ( u + p0) ρ( a) X 0 X1 X 2 N ( u + p0) ρ( a) TSG + /+ TSG + / TSG / Nu c ρ(r) 2 r Nu u p ( ) 1 r ar c ρ Nu c ρ(r) rp Nu c ρ( ar) 1 r Y0 Y1 2 2u fast TSG + /+ TSG + / TSG / CIN CIN CIN 2Nu pρ( a) Y 2 r Nuu p ( ) 1 r ar c ρ r<1

20 Can Chromosomal Instability Enhances the Risk of Cancer?

21 TSG + / 2u Chromosomal Instability u + p 0 / + + / Cancer initiation / TSG TSG without CIN 2n c u 2n c u TSG +/+ CIN 2u TSG +/ CIN fast p TSG CIN Cancer initiation with CIN Loss Of Heterozygosity CIN occurs before the loss of TSG.

22 Does the Tissue Structure t Change the Cancer Risk?

23 Colon cancer arises in a crypt A crypt consists of cells. Apoptosis on top of crypt 36 hours A small number of stem cells replenishes the whole crypt. 7 The colon contains 10 crypts.

24 Small Compartments Large Compartments Mutants with higher fitness (gate-keeper) Mutants with lower fitness (care-taker) Small compartments reduce the risk of mutants with higher fitness, but enhance the risk of CIN.

25 Moran Process 1. Select a cell for reproduction 2. Cell division 4. Add the new cell 3. Remove a cell

26 1. Select a cell for reproduction 2. Cell division 3. Shift the others Linear Process No Somatic Selection CIN is important 4. The last one falls off the edge

27 (1) Compartmentalization (2) Stem cells/non-stem cells Somatic Selection is Suppressed Risk via High Fitness Mutants is Reduced Risk via Low Fitness Mutants is Enhanced (e.g. CIN)

28 Dynamics of Chronic Myeloid Leukemia (CML)

29 LT HSC Fig. 1 ST HSC MPP CMP CLP MEP GMP NKP CBP CTP ERP MKP Pro - B Pro - T Erythrocyte Megakaryocyte Granulocyte Monocyte Dendrocytes NK B lymphocyte T lymphocyte Platelets

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33 Imanitib ibis a very effective drug. Under treatment cancer cells decrease in two phases

34 After the stop of treatment, the number of cancer cells quickly increases and exceeds the original level. Why?

35 Resurgence Resistant cells

36 normal cells stem cells regulated at a constant level progenitor cells differentiated cells transition is accompanied by expansion in number (finite time cell divisions) terminally differentiated cells observed cell number

37 normal cells leukemic cells stem cells stem cells progenitor cells progenitor cells differentiated cells differentiated cells terminally differentiated cells terminally differentiated cells without treatment

38 normal cells stem cells leukemic cells stem cells progenitor cells progenitor cells differentiated cells differentiated cells terminally differentiated cells terminally differentiated cells with treatment

39 Population Dynamic Model?x 0 = [ λ( x 0 ) d 0 ]x 0?y 0 = [ r y ( 1 u) d 0 ]y 0?x 1 = a x x 0 d 1 x 1?y 1 = a y y 0 d 1 y 1?x 2 = b x x 1 d 2 x 2 y?y 2 = b y y 1 d 2 y 2?x = c x d x y?y = c d 3 x y y 2 3 y 3 normal cells leukemic cells

40 Two rates of cancer cells decrease correspond to turnover rates of PC and DC d 2 = 0.05 d 1 = d 2 = 20 days 1 =125 days d 1

41 After the stop of treatment, the number of cancer cells quickly increases and exceeds the original level. Stem cells are insensitive to the drug

42 normal cells leukemic cells resistant tumor stem cells SC SC progenitor cells PC PC differentiated cells DC DC terminally differentiated cells TC with treatment TC

43 Population Dynamic Model?x = λ x ) d ]x?y = 0 [ ( 0) 0] 0 y [ 0 r y ( 1 u) ) d 0 ]y 0?z 0 = ( r z d 0 )z) 0 + r y y 0 u?x 1 = a x x 0 d 1 x 1?y 1 = a y y 0 d 1 y 1?z 1 = a z z 0 d 1 z 1?x 2 = b x x 1 d 2 x 2?y 2 = b y y 1 d 2 y 2?z 2 = b z z 1 d 2 z 2?x 3 = c x x 2 d 3 x 3?y 3 = c y y 2 d 3 y 3?z 3 = c z z 2 d 3 z 3 resistant cells

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47 Model for the number of resistant cells when sensitive cell number reaches M nu umber of le eukem mic ste em ce lls the detectable number u time M sensitive cells mutation rate resistant cells

48 M 1 probability of = 1 exp[ R Rx p ] resistance x x=1 expected number of new mutations survivorship of one lineage until the detection time x :the number of sensitive cells when the mutant is produced

49 R x is the expected number of new mutations when the number of sensitive cells is x. R x = rux 1 d /r 0 f x (t)dt r: division rate df M 1 d: death rate u: mutation rate f 1 ( 0)=1, f x ( 0)=0 for x =2, 3,...,M 1 df 1 dt = 2df (r + d) f 2 1 df x dt = r(x 1) f x 1 + d(x +1) f x +1 (r + d)xf x df = r(m 2) f M 2 (r + d)(m 1) f M 1 dt the probability that t there are x sensitive cells at time t

50 generating g function: g (ξ,t) = E[ξ Z (t ) Z(0) = 1] g (ξ,0) = ξ g( (ξ,t + Δt) ) = aδtg g( (ξ,t) 2 + bδt 1+ (1 (a + b)δt)g ) )g(ξ,t) g = (a bg )(1 g ) t (a b )t g (ξ,t) = (ξ 1)(b a)e( (ξ b a) (ξ 1)e (a b )t (ξ ba) M = x exp[(r d)t] t = 1 r d log( M x ) b b p x x = 1 1 a a M ( a b) r d ( ) a: division rate b: death rate M: detection size

51 Probability of Resistance P =1 exp MuF 1 d r where F = ba 1 (b a)y a b ( ) ( r d) dy d i i i M : detection size u : mutation rate r, d : division/death rate (sensitive cells) a, b : division/death rate (resistant cells)

52 Simulation results fit the formula. P P 0.3 r = a = 3 d = b = u = 10 5 M=10 4 a b r d = 2 d =b=1 u= M (detection size) r (division rate)

53 Slow growth implies higher risk of resistant cells small r higher risk of resistance longer time until diagnosis more mutations more cell division

54 Mean number of resistant cells (conditional to one or more resist. cells)

55 Distribution of resistant it tcell number z:large Pr [ y 1 < Y y 2 ]=] 1 1 ( Fy 1 ) 1 α Fy 2 ( ) 1 α z: small Pr[ Y = y]= ( 1 ba) 2 αf ( ) y y 1 z 1 α 1 z 1 ( ba)z 0 ( ) y +1 dz

56 Cancer Progression is Somatic Evolution.

57 Collaborators Martin A. Nowak (PED, Harvard Univ) Franziska Michor (Sloan-Kettering Cancer Research Inst.) Steven A Frank (UC Irvine) Robert May y( (University of Oxford) Dominik Wodarz (UC Irvine) Natalia L Komarova (UC Irvine) Bert Vogelstein (Johns Hopkins Univ.) Christoph Lengauer (Johns Hopkins Univ.) Tim. P. Hughes (Inst.Med.Vet.Sci. Adelaide) Susan Branford (Inst.Med.Vet.Sci. Adelaide) Neil ilpshh(ucla) P. Shah Charles L. Sawyers (UCLA) Hiroshi Haeno (Kyushu University) i

58 Probability of drug resistance and number of resistance in CML Poisson distribution of infected cell number (viral infection) i Drug resistance requires two mutations Age distribution for CML incidence

59 Drug resistant virus: HIV hepatitis B virus influenza virus simple herpes virus Pathogenic virus.. (1) produces enzymes decomposing the drug (2) changes the structure of target molecules (3) interrupt the drug delivery

60 When virus infection is diagnosed, what is the probability for resistant strain(s) already to exist in the host.

61 Proliferation of Virus target cell Virus viral particles from a single infected cell can infect multiple cells. multiple l infected an infected cell Viral particles cells Skip

62 R x : mean number of mutants when the number of infected cells is x f x 1 f x f x+1 r r r r d f 1 0 d d ( ) = 1, f x ( 0 ) = 0 for x = 2, 3,..., M 1 d R rux x = 1?g 0 f x (t)dt x-1 cells x cells x+1 cells probability of x infected cells at t df 1 dt = (d + r e λ ) 2 f 2 {d + r(1 λe λ )} 1 f 1 (x k +1) df x 1 x dt = r λ ) (x k +1)! e λ k f k + (d + r e λ ) (x +1) f x+1 k=1 {d + r(1 λe λ )} x f x M 2 (M k df M 1 λ = r ) dt (M k)! e λ k f k k=1 {d + r(1 λe λ )} (M 1) f M 1 probability of extinction of infected cells re λ(1?g ) + d = (r + d)?g

63 q x : probability for the lineage of resistant cell line until the onset of drug use. M ν i = ν i = ν (1 g x q ) x i! e 1 g x 1 e i= 0 ( ) x 1 0 R x 1 g x = g r(λ 1) d log M x g x : probability bilit of extinction of resistant virus 0 i time t q x M = x exp[{r(λ 1) d}t] 1 t = log M r(λ 1) d x dg dt = ae ν (1 g ) + b (a + b)g, g (0) = 0

64 Probability that one or more cells are infected by drug resistant virus at diagnosis (1 point mutation is enough for resistance) q x 1 P=1 exp rux R M 1 x p f (t)dt (1 exp ν(1 g ( p ( ( g )? 1?g 0 x x x=1 M : population p size at diagnosis u : mutation rate r, d : cell division/mortality (wild type) a, b (resistant strain) mean production number: (wild type) λ ν (resistant

65 Probability of drug resistance is high when..., mutation rate u is high infected ed cell number at diagnosis M is large reproductive rate r is small P r=1.5 r=5 r : reproductive rate of sensitve type u(mutation rate) t) P r= r=5 M (cell number)

66 Risk of drug resistance is high, when.. mutation rate is high population p size at diagnosis is large population growth rate of virus is slow Risk of drug resistance is reduced b id tif i ti t d t ti d by identifying patients and starting drug treatment as early as possible.

67 Current model: dl When leukemic stem cell number reaches M, the patient t is diagnosed d as CML. of leu ukemi ic stem cell ls Num mber Symptoms of CML appear u time M drug sensitive stem cells mutation rate drug resistant stem cells

68 Alternative interpretation: A tissue grows exponentially in development. Mutant cells can be produced in the process, and some descendent may exist at diagnosis. End of development M Numbe er of cells time u normal cells mutation rate mutant cells (later causing disease)

69 Probability of drug resistance and number of resistance in CML Poisson distribution of infected cell number (viral infection) i Drug resistance requires two mutations Age distribution for CML incidence

70 Clinical problems caused by cells - Drug resistance with two specific mutations. CML cells acquires resistance when it becomes resistant to two different drugs (Imatinib and Dasatinib) - Cancer (L387M/F317I etc.) Retinobrastoma is caused when both alleles of tumor suppressor gene RB1 in a cell are inactivated during development.

71 Model r a 1 a 2 Type-0 cell u 1 u 2 Type-1 cell Type-2 cell d b 1 b 2 Type-0: Wild type cells Type-1: Cells with one mutation (intermediate state) Type-2: Cells with two mutations (cancer cells, or drug resistant cells)

72 Parameters number of cell division cell death mutants rate rate Type-0 0 r d u 1 Type-1 1 a b 1 1 u 2 Type-2 2 a2 b 2 Cells proliferate until the total cell number reaches M.

73 Model cell number Type-0 mutation Type-1 mutation Type-2 time

74 P x : Probability that fhe first Type-1 cell that survives is producedwhen there are x Type-0 cells. x 1 ( ) ( ) σr σr x 1 σr P x = e x 1 e x ( = e 1 )σr x 1 e x x =1 When the total cell number reaches M Expected number of mutants produced while there are x Type-0 cells: r ce ell numbe R x u 1 1 dr x 1 Probability the linearge from a newly formed Type-1can survive: σ 1 b 1 a 1 time t

75 cell num mber x Probability that type 1 cells produce type 2 cell(s) bf before the total ttlcell number reaches the final value when cell number reaches the final number Q =1 e u 2y ( 1 b 1 a 1 ) x y The number of type 1 cells when total cell number reaches the final value τ x ( )τ x a 1 b 1 y = e :time from the production of a Type-1 cell to the final time ( xe r d )τ x ( + e a 1 b 1 )τ x = M 1 τ x Expected number of mutations until Type-1 cell number reaches y: u 2 y time t 1 b 1 a 1

76 Probability that one or more Type-2 cell exist M 1 at the final time ( ( ) ) P = e β ( x 1 ) ( 1 ee β )1 ee u 2y 2 y 1 b 1 a 1 x=1 where β = σr x = 1 b 1 a 1 ( )τ x a 1 b 1 y = e ( ) ( ) ( ) u 1 ( 1 dr) and τ is a solution of x ( xe r d )τ x ( + e a 1 b 1)τ x = M r, a 1, a 2 : cell division rate db d, 1, b 2 : mortality u1, u2: mutation rates

77 Parameter dependence Cells with two or more mutations exist more likely High mutation rates u1 and u2 Fast cell division rate a 1 of intermediate mutants Large final population size M Large mortality d

78 Probability distribution of Type-2 cell number red line:predictionby the formula Pr[ z 1 < Z < z 2 ]= Pr[ κ z1 <κ Z <κ ] z2 = L z2 L z1 black dots: computer simulations

79 Probability of drug resistance and number of resistance in CML Poisson distribution of infected cell number (viral infection) i Drug resistance requires two mutations Age distribution for CML incidence

80 Step number n-11 n Cancer incidence occurs when n stochastic events occurs to a patient Probability of incidence on or before age t ) P( t) t n

81 CMLincidence data

82 Step number of CML incidence date is close to three If a single mutation is sufficient for CML, can we have step number three?

83 Moran process 1. choose a single cell randomly. 2. cell division 4. add one cell 3. remove one cell

84 nu umber of le eukem mic ste em ce ells Diagnosis as CML (or detection) occurs at a rate proportional lto the number of CML stem cells... u mutation rate Diagnosis i as CML

85 Probability for a patient to be judged as Pt leukemia on or before t t ()= 1 1+ ecx 1 N ( 1 11 r ) 0 qn c e b t x ( ) bdx b = Nu 1 1 r ( ) τ c = r 1 ( ) τ

86 S b f Step number from incidence age distribution can be quite large.

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