CN101210266A - Measuring method for relativity of interaction and genetic character between genome genetic markers - Google Patents

Abstract

The invention relates to a novel genetics detection method-purely detecting susceptibility epistasis (PDSE), belonging to the fields of biotechnology and computer. The inventive method and equipment can determine the interactions among genes or genetic markers, and the correlation of the interaction and specific trait, such as certain complex genetic disease.

Description

The measuring method of interaction between genome genetic markers and inherited character dependency

Technical field

The present invention relates to biotechnology and computer realm, more specifically the present invention relates to a kind of brand-new genetics detection method (PDSE, purely detecting susceptibility epistasis) and relevant equipment.The inventive method and equipment can be determined interaction and this interaction and the specific trait between gene or between genetic marker, for example dependency of certain complex inheritance disease.

Background technology

At the various trait phenotype, study its hereditary mechanism, be one of main means of genetics research.Since the complicacy of biosystem, many common proterties, such as complex inheritance disease, and often not by genetic locus decision one by one, the mutual relationship between genetic locus can have influence on the susceptibility of disease.Many common diseases all are the complex inheritance diseases, and for example mental disorder, diabetes, hypertension or the like constitute tremendous influence to human beings'health, also genetic research, analysis have been proposed great challenge.

At present, research for genetic marker site and inherited character dependency, mainly that usefulness is linkage analysis (Lee YA, Wahn U, Kehrt R, Tarani L, Businco L, Gustafsson D, AnderssonF, Oranje AP, Wolkertstorfer A, v Berg A, Hoffmann U, Kuster W, WienkerT, Ruschendorf F, Reis A.A major susceptibility locus for atopicdermatitis maps to chromosome 3q21.Nat Genet.2000Dec; 26 (4): 470-3.) and association analysis.Yet the shortcoming of these methods is to be difficult to judge between two or more genes or whether to have interaction between the genetic marker, whether more is difficult to judge between this interaction and the specific trait (as certain genetic diseases) dependency.

Generally speaking, this area press for exploitation a kind of can be effectively, purely at the susceptible interaction/epistatic effect detection method between gene/genetic marker.

Summary of the invention

Purpose of the present invention just provide a kind of can be effectively, purely at susceptible interaction/epistatic effect measuring method and relevant device between gene/genetic marker.

Particularly, the object of the present invention is to provide a kind of detection method that can detect the interaction relationship (comprising various interactions such as epistatic effect) between the genetic locus.This detection method comprises: sample collecting, the gene type at detected genetic marker site, genetic marker interphase interaction analysis, genetic marker interphase interaction are in the steps such as correlation analysis of proterties.Utilize detection method provided by the present invention, can detect between the gene relevant or the interaction between genetic marker, and then develop the medicine of corresponding inherited character or genetic diseases prediction, detection reagent and treatment disease based on these results with various inherited character, genetic diseases.

In a first aspect of the present invention, provide a kind of definite genome genetic markers whether to have the method for dependency, wherein, described genetic marker is positioned at two sites, and first site has A and two kinds of forms of a, and second site has B and two kinds of forms of b, and the method comprising the steps of:

(a) measure in first colony each individuality in the form and the form on second site in first site,

Be designated as N1 with having the genotypic sample number of AA genotype and BB simultaneously;

Be designated as N2 with having the genotypic sample number of Aa genotype and BB simultaneously;

Be designated as N3 with having the genotypic sample number of aa genotype and BB simultaneously;

Be designated as N4 with having the genotypic sample number of AA genotype and Bb simultaneously;

Be designated as N5 with having the genotypic sample number of Aa genotype and Bb simultaneously;

Be designated as N6 with having the genotypic sample number of aa genotype and Bb simultaneously;

Be designated as N7 with having the genotypic sample number of AA genotype and bb simultaneously;

Be designated as N8 with having the genotypic sample number of Aa genotype and bb simultaneously;

Be designated as N9 with having the genotypic sample number of aa genotype and bb simultaneously;

And, measure in second colony each individuality in the form and the form on second site in first site,

Be designated as Q1 with having the genotypic sample number of AA genotype and BB simultaneously;

Be designated as Q2 with having the genotypic sample number of Aa genotype and BB simultaneously;

Be designated as Q3 with having the genotypic sample number of aa genotype and BB simultaneously;

Be designated as Q4 with having the genotypic sample number of AA genotype and Bb simultaneously;

Be designated as Q5 with having the genotypic sample number of Aa genotype and Bb simultaneously;

Be designated as Q6 with having the genotypic sample number of aa genotype and Bb simultaneously;

Be designated as Q7 with having the genotypic sample number of AA genotype and bb simultaneously;

Be designated as Q8 with having the genotypic sample number of Aa genotype and bb simultaneously;

Be designated as Q9 with having the genotypic sample number of aa genotype and bb simultaneously;

(b) calculate the relation of interphase interaction in twos tolerance parameter value between each genotype in first colony according to N1, N2, N3, N4, N5, N6, N7, N8 and N9, and calculate the relation of interphase interaction in twos tolerance parameter value between each genotype in second colony according to Q1, Q2, Q3, Q4, Q5, Q6, Q7, Q8 and Q9;

(c) according in twos interphase interaction relation tolerance parameter value and this genotype the in twos interphase interaction relation tolerance parameter value in second colony of a genotype in first colony, calculate this genotypic dependency statistical parameter value;

(d) according to normal distribution probability, calculate each genotypic significance tolerance p value by described dependency statistical parameter value, and compare with the significance threshold alpha,

Wherein significance tolerance p value represents then that less than the significance threshold alpha there is dependency in two genome genetic markers in this genotype; And significance tolerance p value represents then that less than the significance threshold alpha there is not dependency in two genome genetic markers in this genotype.

In another preference, described first colony is the case group, and individuality wherein all suffers from identical disease; And second colony is a control group, and individuality does not wherein suffer from the described disease of first colony.

In another preference, described significance threshold alpha is 0.05.

In another preference, in step (b), according to following formula calculate between the AA genotype and BB genotype in first colony, the relation of interphase interaction in twos tolerance parameter value r value between AA genotype and the Bb genotype, between AA genotype and the bb genotype, between Aa genotype and the BB genotype, between AA genotype and the Bb genotype, between Aa genotype and the bb genotype, between aa genotype and the BB genotype, between aa genotype and the Bb genotype, between aa genotype and the bb genotype:

And, according to the relation of the interphase interaction in twos tolerance parameter value r value between the AA genotype and BB genotype that calculate according to following formula in second colony, between AA genotype and the Bb genotype, between AA genotype and the bb genotype, between Aa genotype and the BB genotype, between AA genotype and the Bb genotype, between Aa genotype and the bb genotype, between aa genotype and the BB genotype, between aa genotype and the Bb genotype, between aa genotype and the bb genotype:

And, in step (c), calculate each genotypic dependency statistical parameter value λ according to following formula _Genotpye:

In the formula,

λ _GenotpyeRepresent a certain genotypic dependency statistical parameter value,

n _{First colony}The individual number of representing first colony;

n _{Second colony}The individual number of representing second colony;

r _{First colony}Represent that described genotype is at the first intragroup relation of the interphase interaction in twos tolerance parameter value;

r _{Second colony}Represent that described genotype is at the second intragroup relation of the interphase interaction in twos tolerance parameter value.

In another preference, in step (d), calculate each genotypic significance tolerance p value according to following formula:

p _genotpye＝2×(1-Ф(|λ _geneotpye|))，

In the formula,

Ф is the standardized normal distribution probability function;

λ _GenotpyeRepresent a certain genotypic dependency statistical parameter value;

p _GenotpyeRepresent a certain genotypic significance tolerance p value.

In another preference, in step (d), calculate corresponding to each genotypic significance tolerance p value by computer simulation method.

In another preference, in step (b), calculate the relation of the interphase interaction in twos between each genotype tolerance parameter value in first colony with following formula:

In the formula,

OR _{First colony: genotpye}Represent the in twos interphase interaction relation tolerance parameter value of a certain genotype in first colony;

D1 represents to have in first colony this genotypic individual number;

D4 represents that first colony does not have genotypic individual number fully;

D2 and D3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in first colony;

And, calculate the relation of the interphase interaction in twos between each genotype tolerance parameter value in second colony with following formula:

In the formula, OR _{Second colony: genotpye}Represent the in twos interphase interaction relation tolerance parameter value of a certain genotype in first colony,

W1 represents to have in second colony this genotypic individual number;

W4 represents that second colony does not have genotypic individual number fully;

W2 and W3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in second colony;

And, in step (c), calculate described genotypic dependency statistical parameter value according to following formula;

$\mathrm{\σ}=\sqrt{\frac{1}{D1}+\frac{1}{D2}+\frac{1}{D3}+\frac{1}{D4}+\frac{1}{W1}+\frac{1}{W2}+\frac{1}{W3}+\frac{1}{W4}}$

In the formula,

Z _GenotypeRepresent a certain genotypic dependency statistical parameter value;

And, in step (d), calculate each genotypic significance tolerance p value according to following formula:

p _genotpye＝2×(1-Ф(|Z _geneotpye|))，

In the formula,

Ф is the standardized normal distribution probability function;

Z _GenotpyeRepresent a certain genotypic dependency statistical parameter value;

p _GenotpyeRepresent a certain genotypic significance tolerance p value.

In another preference, in step (b), in first colony, by following 2 * 2 tabulations,

D1 D2

D3 D4

In the formula

D1 represents to have in first colony this genotypic individual number;

D4 represents that first colony does not have genotypic individual number fully;

D2 and D3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in first colony;

With Pearson chi-square value χ ² _{First colony: genotype}Value is calculated the mutual relationship of this genotype in first colony;

In second colony, by following 2 * 2 tabulations,

W1 W2

W3 W4

In the formula

W1 represents to have in second colony this genotypic individual number;

W4 represents that second colony does not have genotypic individual number fully;

W2 and W3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in second colony;

With Pearson chi-square value χ ² _{First colony: genotype}Value is calculated the mutual relationship of this genotype in second colony;

And, for first colony and second colony that merge, by following 2 * 2 tabulations,

T1 T2

T3 T4

In the formula

T1 represents to have this genotypic individual number in first colony and second colony;

T4 represents that first colony and second colony do not have genotypic individual number fully;

T2 and T3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in first colony and second colony;

With Pearson chi-square value χ ² _{Total group: genotype}Value is calculated the mutual relationship of this genotype in first colony and second colony;

And, in step (c), calculate a certain genotypic dependency statistical parameter value χ according to following formula ² _Genotype

χ ² _Genotype=χ ² _{First colony: genotype}+ χ ² _{Second colony: genotype}-χ ² _{Total group: genotype}

In the formula, χ ² _GenotypeRepresent a certain genotypic dependency statistical parameter value.

In another preference, in step (b), in first colony, by following 2 * 2 tabulations,

D1 D2

D3 D4

In the formula

D1 represents to have in first colony this genotypic individual number;

D4 represents that first colony does not have genotypic individual number fully;

D2 and D3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in first colony;

With log-likelihood ratio chi-square value χ ² _{First colony: genotype}Value is calculated the mutual relationship of this genotype in first colony;

In second colony, by following 2 * 2 tabulations,

W1 W2

W3 W4

In the formula

W1 represents to have in second colony this genotypic individual number;

W4 represents that second colony does not have genotypic individual number fully;

W2 and W3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in second colony;

With log-likelihood ratio chi-square value χ ² _{First colony: genotype}Value is calculated the mutual relationship of this genotype in second colony;

And, for first colony and second colony that merge, by following 2 * 2 tabulations,

T1 T2

T3 T4

In the formula

T1 represents to have this genotypic individual number in first colony and second colony;

T4 represents that first colony and second colony do not have genotypic individual number fully;

T2 and T3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in first colony and second colony;

With log-likelihood ratio chi-square value χ ² _{Total group: genotype}Value is calculated the mutual relationship of this genotype in first colony and second colony;

And, in step (c), calculate a certain genotypic dependency statistical parameter value χ according to following formula ² _Genotype

χ ² _Genotype=χ ² _{First colony: genotype}+ χ ² _{Second colony: genotype}-χ ² _{Total group: genotype}

In the formula, χ ² _GenotypeRepresent a certain genotypic dependency statistical parameter value.

In another preference, in step (d), calculate corresponding to each genotypic significance tolerance p value by computer simulation method.

In a second aspect of the present invention, provide a kind of and be used for determining whether genome genetic markers exists the equipment of dependency, wherein, described genetic marker is positioned at two sites, and first site has A and two kinds of forms of a, and second site has B and two kinds of forms of b, and described equipment comprises:

(a) in first colony that an input unit, described input unit will be measured each individuality in the form in first site and the form on second site as input,

Wherein, be designated as N1 with having the genotypic sample number of AA genotype and BB simultaneously;

Be designated as N2 with having the genotypic sample number of Aa genotype and BB simultaneously;

Be designated as N3 with having the genotypic sample number of aa genotype and BB simultaneously;

Be designated as N4 with having the genotypic sample number of AA genotype and Bb simultaneously;

Be designated as N5 with having the genotypic sample number of Aa genotype and Bb simultaneously;

Be designated as N6 with having the genotypic sample number of aa genotype and Bb simultaneously;

Be designated as N7 with having the genotypic sample number of AA genotype and bb simultaneously;

Be designated as N8 with having the genotypic sample number of Aa genotype and bb simultaneously;

Be designated as N9 with having the genotypic sample number of aa genotype and bb simultaneously;

And, in second colony that described input unit will be measured each individuality in the form in first site and the form on second site as input,

Be designated as Q1 with having the genotypic sample number of AA genotype and BB simultaneously;

Be designated as Q2 with having the genotypic sample number of Aa genotype and BB simultaneously;

Be designated as Q3 with having the genotypic sample number of aa genotype and BB simultaneously;

Be designated as Q4 with having the genotypic sample number of AA genotype and Bb simultaneously;

Be designated as Q5 with having the genotypic sample number of Aa genotype and Bb simultaneously;

Be designated as Q6 with having the genotypic sample number of aa genotype and Bb simultaneously;

Be designated as Q7 with having the genotypic sample number of AA genotype and bb simultaneously;

Be designated as Q8 with having the genotypic sample number of Aa genotype and bb simultaneously;

Be designated as Q9 with having the genotypic sample number of aa genotype and bb simultaneously;

(b) one is used for calculating the relation of interphase interaction in twos tolerance parameter value between each genotype of first colony according to N1, N2, N3, N4, N5, N6, N7, N8 and N9, and calculates the device of the relation of the interphase interaction in twos tolerance parameter value between each genotype in second colony according to Q1, Q2, Q3, Q4, Q5, Q6, Q7, Q8 and Q9;

(c) one is used for calculating the device of this genotypic dependency statistical parameter value according to the relation of the interphase interaction in twos tolerance parameter value and the relation of the interphase interaction in twos tolerance parameter value of this genotype in second colony of a genotype in first colony; With

(d) one according to normal distribution probability, calculates each genotypic significance tolerance p value by described dependency statistical parameter value, and the device that compares with the significance threshold alpha,

Wherein significance tolerance p value represents then that less than the significance threshold alpha there is dependency in two genome genetic markers in this genotype; And significance tolerance p value represents then that less than the significance threshold alpha there is not dependency in two genome genetic markers in this genotype.

Embodiment

The inventor has developed a kind of brand-new genetics detection method (PDSE, purely detecting susceptibility epistasis) and relevant device first through extensive and deep research.The inventive method comprises that mainly sample collecting, the gene type at detected genetic marker site, genetic marker interphase interaction analysis, genetic marker interphase interaction are in the steps such as correlation analysis of proterties, can determine interaction and this interaction and specific trait between gene or between genetic marker, for example the dependency of certain complex inheritance disease.

Usually, the present invention includes following concrete steps:

1. sample collecting: collect and have the sample population case of certain inherited character to be detected, and the dna sample (for example, from blood sample, extract DNA, from the oral mucosa sample, extract DNA or the like) that does not have the sample population control of this inherited character;

2. at the gene type in detected genetic marker site:, all will carry out gene type in the DNA sample of in each part step 1, gathering for genetic marker site to be tested.The experimental technique of somatotype can use methods of genotyping any practicality, existing, for example: order-checking, Taqman or the like;

3. genetic marker interphase interaction analysis in twos: the gene type result who supposes to have obtained case and each sample of control sample group in A and two two equipotential genotype of B genetic marker site.For the A site, allelotype represents that with A, a genotype is represented with AA, Aa, aa; For the B site, allelotype represents that with B, b genotype is represented with BB, Bb, bb.

With frequency of genotypes AA and BB is example, when interaction situation in case crowd of research AA and BB, need in case, count have simultaneously AA genotype and the genotypic sample number M1 of BB, simultaneously have non-AA genotype (Aa or aa) and the genotypic sample number M2 of BB, have a sample number M3 of AA genotype and non-BB genotype (Bb or bb) and have non-AA genotype (Aa or aa) simultaneously and the sample number M4 of non-BB genotype (Bb or bb) simultaneously.At this time, use $r_{\mathrm{Case}:\mathrm{AABB}}=\frac{M1\×M4-M2\×M3}{\sqrt{M1\×M2\×M3\×M4}}$ Calculate AA and the BB mutual relationship in Case.By that analogy, can obtain the mutual relationship r between other genotype among the Case _Case:AaBB, r _Case:aaBB, r _Case:AABb, r _Case:AaBb, r _Case:aaBb, r _Case:AAbb, r _Case:Aabb, r _Case:aabbIn Control, repeat same calculating, also can obtain the mutual relationship tolerance between each genotype among the Control.

The genetic marker interphase interaction is in the correlation detection of proterties: be the example explanation with frequency of genotypes AA and genotype BB still, when obtaining r by step 3 _Case:AABBAnd r _Control:AABBAfter, can calculate $z_1=\frac{1}{2}\ln \left(\frac{1+r_{\mathrm{Case}:\mathrm{AABB}}}{1-r_{\mathrm{Case}:\mathrm{AABB}}}\right),$ $z_2=\frac{1}{2}\ln \left(\frac{1+r_{\mathrm{Control}:\mathrm{AABB}}}{1-r_{\mathrm{Control}:\mathrm{AABB}}}\right),$ Use then ${\mathrm{\λ}}_{\mathrm{AABB}}=\frac{z_1-z_2}{\sqrt{\frac{1}{n_{\mathrm{Casee}}-3}+\frac{1}{n_{\mathrm{Contrill}}-3}}}$ As the statistical parameter of correlation detection, wherein n _CaseAnd n _ControlRepresent in the step 2 in two genetic marker sites all Case of somatotype success and the dna sample number of Control respectively.λ _AABBThe conformance with standard normal distribution is by calculating p _AABB=2 * (1-Ф (| λ _AABB|)), wherein Ф is the standardized normal distribution probability function; Perhaps calculate p with the method for computer simulation permutation _AABBFor corresponding significance threshold alpha (for example α=0.05), compare p _AABBCan judge with the size of α whether the interaction between frequency of genotypes AA and the genotype BB is relevant significantly with corresponding inherited character.Similar detection can be generalized between other genotype (for example Aa and BB, aa and BB, or the like).

The inventive method can realize by manual method, also can pass through computer realization.In addition, the inventive method method also can realize by the specific equipment of the present invention.

In the present invention, described equipment can make up by software, hardware, firmware or its and realize.

The inventive method can be used in the analyzing gene group interaction relationship between any two genetic marker sites and the dependency between this interaction and any inherited character.If detected inherited character is a genetic diseases, that detected result just helps prevention, the early diagnosis and therapy of disease probably.

The major advantage of the inventive method is:

(a) based on the experimental technique that detects the DNA gene type:, can not changed the therefore detected result interference that also can not receive environmental factors because the DNA genotype is inherent by the residing environment of object;

(b) analytical procedure is simple and easy to use.

Below in conjunction with specific embodiment, further set forth the present invention.Should be understood that these embodiment only to be used to the present invention is described and be not used in and limit the scope of the invention.The experimental technique of unreceipted actual conditions in the following example, usually according to normal condition, people such as Sambrook for example, molecular cloning: laboratory manual (New York:Cold Spring Harbor Laboratory Press, 1989) condition described in, or the condition of advising according to manufacturer.

Embodiment 1

The inherited character that is detected in the present embodiment is the complex inheritance disease, schizophrenia.

Step 1: carry out the collection and the genome extracting of sample:

From Shanghai, Jilin gathered 713 schizophreniacs' (case group) blood sample and 689 normal peoples' (Control) blood sample.All blood samples extract DNA via conventional phenol chloroform method from people's blood, concentration correction is to 10ng/ul.

Step 2: the genotype that detects gene locus

Site to be detected in the present embodiment is rs4145621 (its allelotrope site is to being C/T) and two single base mutation sites of rs1415263 (its allelotrope site is to being C/T).The experimental technique that gene type adopted among the embodiment is the Taqman classifying method, and service platform is 7900 type quantitative real time PCR Instruments of ABI company.The primer probe is from the Assay On Demand service of ABI company, and reaction reagent is all bought the company from ABI.

The result has obtained 713 schizophreniacs, and (the case group, (control group is Control) in the detected result in above-mentioned two sites for blood sample case) and 689 normal peoples.

Step 3: below the somatotype result of two sites in case group and control group all be listed in.For brevity, rs4145621 is regarded as the A site, wherein detected result is that C then counts A, and detected result is that T then counts a; Rs1415263 is regarded as the B site, and wherein detected result is that C then counts B, and detected result is that T then counts b;

In the case group:

	AA	Aa	aa
				BB	0	2	48
Bb	4	148	73
				Bb	168	167	22

In the control group:

	AA	Aa	aa
				BB	0	0	39
Bb	0	130	95
				bb	169	152	40

According to the method for calculation described in the summary of the invention, be example promptly with frequency of genotypes AA and BB, when interaction situation in case crowd of research AA and BB, in case, count have simultaneously AA genotype and the genotypic sample number M1 of BB, simultaneously have non-AA genotype (Aa or aa) and the genotypic sample number M2 of BB, have a sample number M3 of AA genotype and non-BB genotype (Bb or bb) and have non-AA genotype (Aa or aa) simultaneously and the sample number M4 of non-BB genotype (Bb or bb) simultaneously.

Then, use $r_{\mathrm{Case}:\mathrm{AABB}}=\frac{M1\×M4-M2\×M3}{\sqrt{M1\×M2\×M3\×M4}}$ Calculate AA and the BB mutual relationship in Case.

By that analogy, calculate the mutual relationship r between other genotype among the Case respectively _Case:AaBB, r _Case:aaBB, r _Case:AABb, r _Case:AaBb, r _Case:aaBb, r _Case:AAbb, r _Case:Aabb, r _Case:aabb

In Control, repeat same calculating, obtain the mutual relationship tolerance between each genotype among the Control.

In case group and the control group genotype between mutual relationship tolerance (r) result as follows:

In the case group:

	AA	Aa	aa
				BB	-0.18	-0.27	0.51
Bb	-0.42	0.23	0.17
				bb	0.51	-0.08	-0.45

In the control group:

	AA	Aa	aa
				BB	-0.16	-0.23	0.42
Bb	-0.46	0.19	0.24
				bb	0.52	-0.07	-0.44

Step 4: convert the r value to the z value, and calculating parameter λ and significance tolerance p value.

In this step, at first will convert the z value to from the r value that step 3 calculates.

Be the example explanation still, when obtaining r by step 3 with frequency of genotypes AA and genotype BB _Case:AABBAnd r _Control:AABBAfter, be calculated as follows out $z_1=\frac{1}{2}\ln \left(\frac{1+r_{\mathrm{Case}:\mathrm{AABB}}}{1-r_{\mathrm{Case}:\mathrm{AABB}}}\right),$ $z_2=\frac{1}{2}\ln \left(\frac{1+r_{\mathrm{Control}:\mathrm{AABB}}}{1-r_{\mathrm{Control}:\mathrm{AABB}}}\right).$

The result is as follows:

In the case group:

	AA	Aa	aa
				BB	-0.18	-0.28	0.57
Bb	-0.45	0.24	0.18
				bb	0.56	-0.08	-0.48

In the control group:

	AA	Aa	aa
				BB	-0.16	-0.24	0.44
Bb	-0.49	0.19	0.25
				bb	0.58	-0.07	-0.47

Then, use ${\mathrm{\λ}}_{\mathrm{AABB}}=\frac{z_1-z_2}{\sqrt{\frac{1}{n_{\mathrm{Case}}-3}+\frac{1}{n_{\mathrm{Controll}}-3}}}$ As the statistical parameter of correlation detection, wherein n _CaseAnd n _ControlRepresent in the step 2 in two genetic marker sites all dna sample numbers of the case group (case) of somatotype success and control group (Control) respectively.

The dependency statistical parameter λ that changes out is as follows:

	AA	Aa	aa
				BB	-0.40	-0.69	2.23
Bb	0.69	0.77	-1.22
				bb	-0.30	-0.11	-0.24

λ _AABBThe conformance with standard normal distribution is by calculating p _AABB=2 * (1-Ф (| λ _AABB|)), wherein Ф is the standardized normal distribution probability function.For example, λ _AABBValue be-0.40, therefore according to standardized normal distribution find Ф (| λ _AABB|) be 0.66, be 0.69 thereby calculate the p value.For other λ values, calculate with same procedure.

In addition, also available computers mimic method (permutation) is calculated p _AABBIn brief, this analogue method goes out λ by the computer program stochastic simulation exactly _AABBProbability distribution function, obtain current λ then _AABBPairing p value.

Calculate significance tolerance p value by λ:

	AA	Aa	aa
				BB	0.69	0.49	0.03
Bb	0.49	0.44	0.22
				bb	0.76	0.91	0.81

The significance threshold alpha is set in 0.05.

It should be noted that, genotype is that the p value between aa and the BB is 0.03, be set in 0.05 less than the significance threshold alpha, this shows that the mutual relationship between aa and the BB genotype has significant difference in case group and control group, and the interaction and the disease of these two genes have dependency.

Embodiment 2

The process of embodiment 2 is identical with embodiment 1 basically, and difference is method of calculation.

In the present embodiment, carry out genetic marker interphase interaction analysis in twos:

Suppose to have obtained the gene type result of case and each sample of control sample group in A and two two equipotential genotype of B genetic marker site.For the A site, allelotype represents that with A, a genotype is represented with AA, Aa, aa; For the B site, allelotype represents that with B, b genotype is represented with BB, Bb, bb.

With frequency of genotypes AA and BB is example, when interaction situation in case crowd of research AA and BB, at first need in case, count have simultaneously AA genotype and the genotypic sample number D1 of BB, simultaneously have non-AA genotype (Aa or aa) and the genotypic sample number D2 of BB, have a sample number D3 of AA genotype and non-BB genotype (Bb or bb) and have non-AA genotype (Aa or aa) simultaneously and the sample number D4 of non-BB genotype (Bb or bb) simultaneously.Then, use ${\mathrm{OR}}_{\mathrm{Case}:\mathrm{AABB}}=\frac{D1\×D4}{D2\×D3}$ Calculate AA and the BB mutual relationship in Case.

Similarly, need in control, count have simultaneously AA genotype and the genotypic sample number N1 of BB, simultaneously have non-AA genotype (Aa or aa) and the genotypic sample number N2 of BB, have a sample number N3 of AA genotype and non-BB genotype (Bb or bb) and have non-AA genotype (Aa or aa) simultaneously and the sample number N4 of non-BB genotype (Bb or bb) simultaneously.Then, use ${\mathrm{OR}}_{\mathrm{Control}:\mathrm{AABB}}=\frac{N1\×N4}{N2\×N3}$ Calculate AA and the BB mutual relationship in Control.

Then, can calculate ${\mathrm{\σ}}_{\mathrm{AABB}}=\sqrt{\frac{1}{D1}+\frac{1}{D2}+\frac{1}{D3}+\frac{1}{D4}+\frac{1}{N1}+\frac{1}{N2}+\frac{1}{N3}+\frac{1}{N4}},$

And $Z_{\mathrm{AABB}}=\frac{\ln \left({\mathrm{OR}}_{\mathrm{Case}:\mathrm{AABB}}\right)-\ln \left({\mathrm{OR}}_{\mathrm{Control}:\mathrm{AABB}}\right)}{{\mathrm{\σ}}_{\mathrm{AABB}}},$

Z _AABBThe conformance with standard normal distribution is by calculating p _AABB=2 * (1-Ф (| Z _AABB|)), wherein Ф is the standardized normal distribution probability function; Perhaps calculate p with the method for computer simulation permutation _AABBFor corresponding significance threshold alpha (for example α=0.05), compare p _AABBCan judge with the size of α whether the interaction between frequency of genotypes AA and the genotype BB is relevant significantly with corresponding inherited character.

Similar detection can be generalized between other genotype, calculates p with the method for symbolic substitution _AaBB, p _AaBB, p _AABb, p _AaBb, p _AaBb, p _AAbb, p _Aabb, p _Aabb, and judge its significant correlation.

The result:

For the case group (case) that constitutes by 713 schizophreniacs among the embodiment 1 and 689 control groups (Control) that the normal people constitutes, detected result according to rs4145621 (its allelotrope site is to being C/T) and two single base mutation sites of rs1415263 (its allelotrope site is to being C/T), and calculate according to aforesaid method, the result who obtains shows, mutual relationship between aa and the BB genotype has significant difference in case group and control group, the interaction and the disease of these two genes have dependency.

	AA	Aa	aa
				BB	0.72	0.59	0.02
Bb	0.49	0.23	0.35
				bb	0.96	0.26	0.61

Embodiment 3

The process of embodiment 3 is identical with embodiment 1 basically, and difference is method of calculation.

In the present embodiment, followingly carry out genetic marker interphase interaction analysis in twos:

Suppose to have obtained the gene type result of case and each sample of control sample group in A and two two equipotential genotype of B genetic marker site.For the A site, allelotype represents that with A, a genotype is represented with AA, Aa, aa; For the B site, allelotype represents that with B, b genotype is represented with BB, Bb, bb.

With frequency of genotypes AA and BB is example, when interaction situation in case crowd of research AA and BB, need in case, count have simultaneously AA genotype and the genotypic sample number D1 of BB, simultaneously have non-AA genotype (Aa or aa) and the genotypic sample number D2 of BB, have a sample number D3 of AA genotype and non-BB genotype (Bb or bb) and have non-AA genotype (Aa or aa) simultaneously and the sample number D4 of non-BB genotype (Bb or bb) simultaneously.For D1,2 * 2 tabulations that D2, D3, D4 constitute (D1 1 row 1 of being expert at, D2 1 row 2 of being expert at, D3 2 row 1 of being expert at, D4 are expert at 2 row 2) are with conventional Pearson chi-square value χ ² _Case:AABBValue is calculated AA and the mutual relationship of BB in Case.

Similarly, in control, count have simultaneously AA genotype and the genotypic sample number N1 of BB, simultaneously have non-AA genotype (Aa or aa) and the genotypic sample number N2 of BB, have a sample number N3 of AA genotype and non-BB genotype (Bb or bb) and have non-AA genotype (Aa or aa) simultaneously and the sample number N4 of non-BB genotype (Bb or bb) simultaneously.For N1,2 * 2 tabulations that N2, N3, N4 constitute (N1 1 row 1 of being expert at, N2 1 row 2 of being expert at, N3 2 row 1 of being expert at, N4 are expert at 2 row 2) are with conventional Pearson chi-square value χ ² _Control:AABBValue is calculated AA and the mutual relationship of BB in Control.

Then, for D1+N1,2 * 2 tabulations that D2+N2, D3+N3, D4+N4 constitute (D1+N1 1 row 1 of being expert at, D2+N2 1 row 2 of being expert at, D3+N3 2 row 1 of being expert at, D4+N4 are expert at 2 row 2) calculate Pearson χ ² _{Combined:AABB}

At last, calculate χ according to following formula ² _AABB:

χ ² _AABB＝χ ² _Case:AABB+χ ² _Control:AABB-χ ² _{Combined:AABB}

χ ² _AABBMeet degree of freedom and be 1 card side and distribute, by consulting that card side shows or obtaining corresponding p with the method for computer simulation permutation _AABB

For corresponding significance threshold alpha (for example α=0.05), compare p _AABBCan judge with the size of α whether the interaction between frequency of genotypes AA and the genotype BB is relevant significantly with corresponding inherited character.Similar detection can be generalized between other genotype, calculates p with the method for symbolic substitution _AaBB, p _AaBB, p _AABb, p _AaBb, p _AaBb, p _AAbb, p _Aabb, p _Aabb, and judge its significant correlation.

The result:

For the case group (case) that constitutes by 713 schizophreniacs among the embodiment 1 and 689 control groups (Control) that the normal people constitutes, detected result according to rs4145621 (its allelotrope site is to being C/T) and two single base mutation sites of rs1415263 (its allelotrope site is to being C/T), and calculate according to aforesaid method, the result who obtains shows, mutual relationship between aa and the BB genotype has significant difference in case group and control group, the interaction and the disease of these two genes have dependency.

	AA	Aa	aa
				BB	0.63	0.79	0.048
Bb	0.69	0.63	0.65
				bb	0.67	0.86	0.71

Embodiment 4

The process of embodiment 4 is identical with embodiment 1 basically, and difference is method of calculation.

In the present embodiment, followingly carry out genetic marker interphase interaction analysis in twos:

Suppose to have obtained the gene type result of case and each sample of control sample group in A and two two equipotential genotype of B genetic marker site.For the A site, allelotype represents that with A, a genotype is represented with AA, Aa, aa; For the B site, allelotype represents that with B, b genotype is represented with BB, Bb, bb.

With frequency of genotypes AA and BB is example, when interaction situation in case crowd of research AA and BB, in case, count have simultaneously AA genotype and the genotypic sample number D1 of BB, simultaneously have non-AA genotype (Aa or aa) and the genotypic sample number D2 of BB, have a sample number D3 of AA genotype and non-BB genotype (Bb or bb) and have non-AA genotype (Aa or aa) simultaneously and the sample number D4 of non-BB genotype (Bb or bb) simultaneously.For D1,2 * 2 tabulations that D2, D3, D4 constitute (D1 1 row 1 of being expert at, D2 1 row 2 of being expert at, D3 2 row 1 of being expert at, D4 are expert at 2 row 2) are with log-likelihood ratio chi-square value χ ² _Case:AABBValue is calculated AA and the mutual relationship of BB in Case.

Similarly, in control, count have simultaneously AA genotype and the genotypic sample number N1 of BB, simultaneously have non-AA genotype (Aa or aa) and the genotypic sample number N2 of BB, have a sample number N3 of AA genotype and non-BB genotype (Bb or bb) and have non-AA genotype (Aa or aa) simultaneously and the sample number N4 of non-BB genotype (Bb or bb) simultaneously.For N1,2 * 2 tabulations that N2, N3, N4 constitute (N1 1 row 1 of being expert at, N2 1 row 2 of being expert at, N3 2 row 1 of being expert at, N4 are expert at 2 row 2) are with log-likelihood ratio chi-square value χ ² _Control:AABBValue is calculated AA and the mutual relationship of BB in Control.

Then, for D1+N1,2 * 2 tabulations that D2+N2, D3+N3, D4+N4 constitute (D1+N1 1 row 1 of being expert at, D2+N2 1 row 2 of being expert at, D3+N3 2 row 1 of being expert at, D4+N4 are expert at 2 row 2) calculate log-likelihood ratio χ ² _{Combined:AABB}

At last, calculate χ according to following formula ² _AABB:

χ ² _AABB＝χ ² _Case:AABB+χ ² _Control:AABB-χ ² _{Combined:AABB}。

χ ² _AABBMeet degree of freedom and be 1 card side and distribute, by consulting that card side shows or obtaining corresponding p with the method for computer simulation permutation _AABBFor corresponding significance threshold alpha (for example α=0.05), compare p _AABBCan judge with the size of α whether the interaction between frequency of genotypes AA and the genotype BB is relevant significantly with corresponding inherited character.Similar detection can be generalized between other genotype, calculates p with the method for symbolic substitution _AaBB, p _AaBB, p _AABb, p _AaBb, p _AaBb, p _AAbb, p _Aabb, p _Aabb, and judge its significant correlation.

The result:

For the case group (case) that constitutes by 713 schizophreniacs among the embodiment 1 and 689 control groups (Control) that the normal people constitutes, detected result according to rs4145621 (its allelotrope site is to being C/T) and two single base mutation sites of rs1415263 (its allelotrope site is to being C/T), and calculate according to aforesaid method, the result who obtains shows, mutual relationship between aa and the BB genotype has significant difference in case group and control group, the interaction and the disease of these two genes have dependency.

	AA	Aa	aa
				BB	0.62	0.71	0.048
Bb	0.71	0.62	0.66
				bb	0.66	0.88	0.71

Embodiment 5

The inherited character that is detected in the present embodiment is the complex inheritance disease, schizophrenia.

Step 1: carry out the collection and the genome extracting of sample:

From Shanghai, Jilin gathered 377 schizophreniacs' (case group) blood sample and 288 normal peoples' (Control) blood sample.All blood samples extract DNA via conventional phenol chloroform method from people's blood, concentration correction is to 10ng/ul.

Step 2: the genotype that detects gene locus

Site to be detected in the present embodiment is rs3902902 (its allelotrope site is to being C/T) and two single base mutation sites of rs4646642 (its allelotrope site is to being C/T).The experimental technique that gene type adopted among the embodiment is the Taqman classifying method, and service platform is 7900 type quantitative real time PCR Instruments of ABI company.The primer probe is from the Assay On Demand service of ABI company, and reaction reagent is all bought the company from ABI.

The result has obtained 377 schizophreniacs, and (the case group, (control group is Control) in the detected result in above-mentioned two sites for blood sample case) and 288 normal peoples.

Step 3: below the somatotype result of two sites in case group and control group all be listed in.For brevity, rs3902902 is regarded as the A site, wherein detected result is that C then counts A, and detected result is that T then counts a; Rs4646642 is regarded as the B site, and wherein detected result is that C then counts B, and detected result is that T then counts b;

In the case group:

	AA	Aa	aa
				BB	89	64	49
Bb	91	65	36
				bb	116	85	49

In the control group:

	AA	Aa	aa
				BB	76	59	42
Bb	71	45	35
				bb	95	60	47

According to the method for calculation described in the summary of the invention, be example promptly with frequency of genotypes AA and BB, when interaction situation in case crowd of research AA and BB, in case, count have simultaneously AA genotype and the genotypic sample number M1 of BB, simultaneously have non-AA genotype (Aa or aa) and the genotypic sample number M2 of BB, have a sample number M3 of AA genotype and non-BB genotype (Bb or bb) and have non-AA genotype (Aa or aa) simultaneously and the sample number M4 of non-BB genotype (Bb or bb) simultaneously.

Then, use $r_{\mathrm{Case}:\mathrm{AABB}}=\frac{M1\×M4-M2\×M3}{\sqrt{M1\×M2\×M3\×M4}}$ Calculate AA and the BB mutual relationship in Case.

By that analogy, calculate the mutual relationship r between other genotype among the Case respectively _Case:AaBB, r _Case:aaBB, r _Case:AABb, r _Case:AaBb, r _Case:aaBb, r _Case:AAbb, r _Case:Aabb, r _Case:aabb

In Control, repeat same calculating, obtain the mutual relationship tolerance between each genotype among the Control.

In case group and the control group genotype between mutual relationship tolerance (r) result as follows:

In the case group:

	AA	Aa	aa
				BB	-0.11194	-0.1023	0.297554
Bb	0.08214	0.039989	-0.18226
				bb	0.028771	0.056631	-0.12087

In the control group:

	AA	Aa	aa
				BB	-0.16545	0.166513	0.027795
Bb	0.076546	-0.07524	-0.017
				bb	0.089142	-0.09426	-0.01163

Step 4: convert the r value to the z value, and calculating parameter λ and significance tolerance p value.

In this step, at first will convert the z value to from the r value that step 3 calculates.

Be the example explanation still, when obtaining r by step 3 with frequency of genotypes AA and genotype BB _Case:AABBAnd r _Control:AABBAfter, be calculated as follows out $z_1=\frac{1}{2}\ln \left(\frac{1+r_{\mathrm{Case}:\mathrm{AABB}}}{1-r_{\mathrm{Case}:\mathrm{AABB}}}\right),$ $z_2=\frac{1}{2}\ln \left(\frac{1+r_{\mathrm{Control}:\mathrm{AABB}}}{1-r_{\mathrm{Control}:\mathrm{AABB}}}\right).$ The result is as follows:

In the case group:

	AA	Aa	aa
				BB	-0.11241	-0.10266	0.306834
Bb	0.082326	0.040011	-0.18432
				bb	0.028778	0.056691	-0.12146

In the control group:

	AA	Aa	aa
				BB	-0.16699	0.168078	0.027802
Bb	0.076696	-0.07538	-0.017
				bb	0.089379	-0.09454	-0.01163

Then, use ${\mathrm{\λ}}_{\mathrm{AABB}}=\frac{z_1-z_2}{\sqrt{\frac{1}{n_{\mathrm{Case}}-3}+\frac{1}{n_{\mathrm{Controll}}-3}}}$ As the statistical parameter of correlation detection, wherein n _CaseAnd n _ControlRepresent in the step 2 in two genetic marker sites all dna sample numbers of the case group (case) of somatotype success and control group (Control) respectively.

The dependency statistical parameter λ that changes out is as follows:

	AA	Aa	aa
				BB	0.928131	-4.60421	4.745321
Bb	0.095737	1.962359	-2.84549
				bb	-1.0306	2.571954	-1.86784

Calculate significance tolerance p value by λ:

	AA	Aa	aa
				BB	0.353332	4.20×10 -06	2.12×10 -06
Bb	0.923709	0.049737	0.004447
				bb	0.302712	0.010132	0.061797

The significance threshold alpha is set in 0.05, and genotype is that the p value between BB and the Aa is 4.20 * 10 ^-06, be set in 0.05 less than the significance threshold alpha; Genotype is that the p value between BB and the aa is 2.12 * 10 ^-06, be set in 0.05 less than the significance threshold alpha; Genotype is that the p value between Bb and the Aa is 0.049737, is set in 0.05 less than the significance threshold alpha; Genotype is that the p value between Bb and the aa is 0.004447, is set in 0.05 less than the significance threshold alpha; Genotype is that the p value between bb and the aa is 0.010132, is set in 0.05 less than the significance threshold alpha, and the interaction and the disease of these two genes have dependency.

All quote in this application as a reference at all documents that the present invention mentions, just quoted as a reference separately as each piece document.Should be understood that in addition those skilled in the art can make various changes or modifications the present invention after having read above-mentioned teachings of the present invention, these equivalent form of values fall within the application's appended claims institute restricted portion equally.

Claims (10)

1. whether a definite genome genetic markers exists the method for dependency, and wherein, described genetic marker is positioned at two sites, and first site has A and two kinds of forms of a, and second site has B and two kinds of forms of b, it is characterized in that, comprises step:

(a) measure in first colony each individuality in the form and the form on second site in first site,

Be designated as N1 with having the genotypic sample number of AA genotype and BB simultaneously;

Be designated as N2 with having the genotypic sample number of Aa genotype and BB simultaneously;

Be designated as N3 with having the genotypic sample number of aa genotype and BB simultaneously;

Be designated as N4 with having the genotypic sample number of AA genotype and Bb simultaneously;

Be designated as N5 with having the genotypic sample number of Aa genotype and Bb simultaneously;

Be designated as N6 with having the genotypic sample number of aa genotype and Bb simultaneously;

Be designated as N7 with having the genotypic sample number of AA genotype and bb simultaneously;

Be designated as N8 with having the genotypic sample number of Aa genotype and bb simultaneously;

Be designated as N9 with having the genotypic sample number of aa genotype and bb simultaneously;

And, measure in second colony each individuality in the form and the form on second site in first site,

Be designated as Q1 with having the genotypic sample number of AA genotype and BB simultaneously;

Be designated as Q2 with having the genotypic sample number of Aa genotype and BB simultaneously;

Be designated as Q3 with having the genotypic sample number of aa genotype and BB simultaneously;

Be designated as Q4 with having the genotypic sample number of AA genotype and Bb simultaneously;

Be designated as Q5 with having the genotypic sample number of Aa genotype and Bb simultaneously;

Be designated as Q6 with having the genotypic sample number of aa genotype and Bb simultaneously;

Be designated as Q7 with having the genotypic sample number of AA genotype and bb simultaneously;

Be designated as Q8 with having the genotypic sample number of Aa genotype and bb simultaneously;

Be designated as Q9 with having the genotypic sample number of aa genotype and bb simultaneously;

(b) calculate the relation of interphase interaction in twos tolerance parameter value between each genotype in first colony according to N1, N2, N3, N4, N5, N6, N7, N8 and N9, and calculate the relation of interphase interaction in twos tolerance parameter value between each genotype in second colony according to Q1, Q2, Q3, Q4, Q5, Q6, Q7, Q8 and Q9;

(c) according in twos interphase interaction relation tolerance parameter value and this genotype the in twos interphase interaction relation tolerance parameter value in second colony of a genotype in first colony, calculate this genotypic dependency statistical parameter value;

(d) according to normal distribution probability, calculate each genotypic significance tolerance p value by described dependency statistical parameter value, and compare with the significance threshold alpha,

Wherein significance tolerance p value represents then that less than the significance threshold alpha there is dependency in two genome genetic markers in this genotype; And significance tolerance p value represents then that less than the significance threshold alpha there is not dependency in two genome genetic markers in this genotype.

2. as the described method of claim l, it is characterized in that described first colony is the case group, individuality wherein all suffers from identical disease; And second colony is a control group, and individuality does not wherein suffer from the described disease of first colony.

3. as the described method of claim l, it is characterized in that described significance threshold alpha is 0.05.

4. as the described method of claim l, it is characterized in that, in step (b), according to following formula calculate between the AA genotype and BB genotype in first colony, the relation of interphase interaction in twos tolerance parameter value r value between AA genotype and the Bb genotype, between AA genotype and the bb genotype, between Aa genotype and the BB genotype, between AA genotype and the Bb genotype, between Aa genotype and the bb genotype, between aa genotype and the BB genotype, between aa genotype and the Bb genotype, between aa genotype and the bb genotype:

And, according to the relation of the interphase interaction in twos tolerance parameter value r value between the AA genotype and BB genotype that calculate according to following formula in second colony, between AA genotype and the Bb genotype, between AA genotype and the bb genotype, between Aa genotype and the BB genotype, between AA genotype and the Bb genotype, between Aa genotype and the bb genotype, between aa genotype and the BB genotype, between aa genotype and the Bb genotype, between aa genotype and the bb genotype:

And, in step (c), calculate each genotypic dependency statistical parameter value nine λ according to following formula _Genotpye:

In the formula,

λ _GenotpyeRepresent a certain genotypic dependency statistical parameter value,

n _{First colony}The individual number of representing first colony;

n _{Second colony}The individual number of representing second colony;

r _{First colony}Represent that described genotype is at the first intragroup relation of the interphase interaction in twos tolerance parameter value;

r _{Second colony}Represent that described genotype is at the second intragroup relation of the interphase interaction in twos tolerance parameter value.

5. method as claimed in claim 4 is characterized in that, in step (d), calculates each genotypic significance tolerance p value according to following formula:

p _genotpye＝2×(1-Φ(|λ _geneotpye|))，

In the formula,

Φ is the standardized normal distribution probability function;

λ _GenotpyeRepresent a certain genotypic dependency statistical parameter value;

p _GenotpyeRepresent a certain genotypic significance tolerance p value.

6. the method for claim 1 is characterized in that, in step (b), calculates the relation of the interphase interaction in twos between each genotype tolerance parameter value in first colony with following formula:

In the formula,

OR _{First colony: genotpye}Represent the in twos interphase interaction relation tolerance parameter value of a certain genotype in first colony;

D1 represents to have in first colony this genotypic individual number;

D4 represents that first colony does not have genotypic individual number fully;

D2 and D3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in first colony;

And, calculate the relation of the interphase interaction in twos between each genotype tolerance parameter value in second colony with following formula:

In the formula, OR _{Second colony: genotpye}Represent the in twos interphase interaction relation tolerance parameter value of a certain genotype in first colony,

W1 represents to have in second colony this genotypic individual number;

W4 represents that second colony does not have genotypic individual number fully;

W2 and W3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in second colony;

And, in step (c), calculate described genotypic dependency statistical parameter value according to following formula;

$\mathrm{\σ}=\sqrt{\frac{1}{D1}+\frac{1}{D2}+\frac{1}{D3}+\frac{1}{D4}+\frac{1}{W1}+\frac{1}{W2}+\frac{1}{W3}+\frac{1}{W4}}$

In the formula,

Z _GenotypeRepresent a certain genotypic dependency statistical parameter value;

And, in step (d), calculate each genotypic significance tolerance p value according to following formula:

p _genotpye＝2×(1-Φ(|Z _geneotpye|))，

In the formula,

Φ is the standardized normal distribution probability function;

Z _GenotpyeRepresent a certain genotypic dependency statistical parameter value;

p _GenotpyeRepresent a certain genotypic significance tolerance p value.

7. the method for claim 1 is characterized in that, in step (b), and in first colony, by following 2 * 2 tabulations,

D1 D2

D3 D4

In the formula

D1 represents to have in first colony this genotypic individual number;

D4 represents that first colony does not have genotypic individual number fully;

D2 and D3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in first colony;

With Pearson chi-square value χ ² _{First colony: genotype}Value is calculated the mutual relationship of this genotype in first colony;

In second colony, by following 2 * 2 tabulations,

W1 W2

W3 W4

In the formula

W1 represents to have in second colony this genotypic individual number;

W4 represents that second colony does not have genotypic individual number fully;

W2 and W3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in second colony;

With Pearson chi-square value χ ² _{First colony: genotype}Value is calculated the mutual relationship of this genotype in second colony;

And, for first colony and second colony that merge, by following 2 * 2 tabulations,

T1 T2

T3 T4

In the formula

T1 represents to have this genotypic individual number in first colony and second colony;

T4 represents that first colony and second colony do not have genotypic individual number fully;

T2 and T3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in first colony and second colony;

With Pearson chi-square value χ ² _{Total group: genotype}Value is calculated the mutual relationship of this genotype in first colony and second colony;

And, in step (c), calculate a certain genotypic dependency statistical parameter value χ according to following formula ² _Genotype

χ ² _Genotype=χ ² _{First colony: genotype}+ χ ² _{Second colony: genotype}-χ ² _{Total group: genotype}

In the formula, χ ² _GenotypeRepresent a certain genotypic dependency statistical parameter value.

8. the method for claim 1 is characterized in that, in step (b), and in first colony, by following 2 * 2 tabulations,

D1 D2

D3 D4

In the formula

D1 represents to have in first colony this genotypic individual number;

D4 represents that first colony does not have genotypic individual number fully;

D2 and D3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in first colony;

With log-likelihood ratio chi-square value χ ² _{First colony: genotype}Value is calculated the mutual relationship of this genotype in first colony;

In second colony, by following 2 * 2 tabulations,

W1 W2

W3 W4

In the formula

W1 represents to have in second colony this genotypic individual number;

W4 represents that second colony does not have genotypic individual number fully;

W2 and W3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in second colony;

With log-likelihood ratio chi-square value χ ² _{First colony: genotype}Value is calculated the mutual relationship of this genotype in second colony;

And, for first colony and second colony that merge, by following 2 * 2 tabulations,

T1 T2

T3 T4

In the formula,

T1 represents to have this genotypic individual number in first colony and second colony;

T4 represents that first colony and second colony do not have genotypic individual number fully;

T2 and T3 represent that respectively part has this genotypic other two kinds of genotypic individual numbers in first colony and second colony;

With log-likelihood ratio chi-square value χ ² _{Total group: genotype}Value is calculated the mutual relationship of this genotype in first colony and second colony;

And, in step (c), calculate a certain genotypic dependency statistical parameter value χ according to following formula ² _Genotype

χ ² _Genotype=χ ² _{First colony: genotype}+ χ ² _{Second colony: genotype}-χ ² _{Total group: genotype}

In the formula, χ ² _GenotypeRepresent a certain genotypic dependency statistical parameter value.

9. as claim 1 or 7 or 8 described methods, it is characterized in that, in step (d), calculate corresponding to each genotypic significance tolerance p value by computer simulation method.

10. one kind is used for determining whether genome genetic markers exists the equipment of dependency, and wherein, described genetic marker is positioned at two sites, and first site has A and two kinds of forms of a, and second site has B and two kinds of forms of b, it is characterized in that, comprising:

(a) in first colony that an input unit, described input unit will be measured each individuality in the form in first site and the form on second site as input,

Wherein, be designated as N1 with having the genotypic sample number of AA genotype and BB simultaneously;

Be designated as N2 with having the genotypic sample number of Aa genotype and BB simultaneously;

Be designated as N3 with having the genotypic sample number of aa genotype and BB simultaneously;

Be designated as N4 with having the genotypic sample number of AA genotype and Bb simultaneously;

Be designated as N5 with having the genotypic sample number of Aa genotype and Bb simultaneously;

Be designated as N6 with having the genotypic sample number of aa genotype and Bb simultaneously;

Be designated as N7 with having the genotypic sample number of AA genotype and bb simultaneously;

Be designated as N8 with having the genotypic sample number of Aa genotype and bb simultaneously;

Be designated as N9 with having the genotypic sample number of aa genotype and bb simultaneously;

And, in second colony that described input unit will be measured each individuality in the form in first site and the form on second site as input,

Be designated as Q1 with having the genotypic sample number of AA genotype and BB simultaneously;

Be designated as Q2 with having the genotypic sample number of Aa genotype and BB simultaneously;

Be designated as Q3 with having the genotypic sample number of aa genotype and BB simultaneously;

Be designated as Q4 with having the genotypic sample number of AA genotype and Bb simultaneously;

Be designated as Q5 with having the genotypic sample number of Aa genotype and Bb simultaneously;

Be designated as Q6 with having the genotypic sample number of aa genotype and Bb simultaneously;

Be designated as Q7 with having the genotypic sample number of AA genotype and bb simultaneously;

Be designated as Q8 with having the genotypic sample number of Aa genotype and bb simultaneously;

Be designated as Q9 with having the genotypic sample number of aa genotype and bb simultaneously;

(b) one is used for calculating the relation of interphase interaction in twos tolerance parameter value between each genotype of first colony according to N1, N2, N3, N4, N5, N6, N7, N8 and N9, and calculates the device of the relation of the interphase interaction in twos tolerance parameter value between each genotype in second colony according to Q1, Q2, Q3, Q4, Q5, Q6, Q7, Q8 and Q9;

(c) one is used for calculating the device of this genotypic dependency statistical parameter value according to the relation of the interphase interaction in twos tolerance parameter value and the relation of the interphase interaction in twos tolerance parameter value of this genotype in second colony of a genotype in first colony; With

(d) one according to normal distribution probability, calculates each genotypic significance tolerance p value by described dependency statistical parameter value, and the device that compares with the significance threshold alpha,

Wherein significance tolerance p value represents then that less than the significance threshold alpha there is dependency in two genome genetic markers in this genotype; And significance tolerance p value represents then that less than the significance threshold alpha there is not dependency in two genome genetic markers in this genotype.