CN107557458A - A kind of method and device of effective detection genotype - Google Patents
A kind of method and device of effective detection genotype Download PDFInfo
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Abstract
The invention discloses a kind of method and device of effective detection genotype, this method includes：If the first, 1, pregnant woman's genotype is AA, fetus has 2 kinds of situations, has the value of 2 kinds of different mutator signal/wild type gene signals；If the 2, pregnant woman's genotype is Aa, it is such in the case of fetus have 3 kinds of different values；If the 3, pregnant woman's genotype is aa, it is such in the case of fetus have 2 kinds of different values；2nd, sample to be tested is obtained, sample to be tested is subjected to digital pcr reaction, obtain the value of mutator signal/wild type gene signal, compared with the value of mutator signal/wild type gene signal in step 1, error is within 10%, then belong to gene type corresponding to the value, the quick gene for judging pregnant woman and fetus.The method and device of effective detection genotype of the present invention, it can simply and rapidly judge fetus and the gene type of pregnant woman.
Description
Technical field
The present invention relates to technical field of gene detection, and in particular to a kind of method and device of effective detection genotype.
Background technology
Gene is the basic genetic unit for controlling character.Current most widely used genetic test is neonate's heredity disease
Detection, the diagnosis of genetic disease and the auxiliary diagnosis of some common diseases of disease.There is more than 1000 kinds of genetic disease to lead at present
Cross technique of gene detection and make diagnosis, so as to reduce the birth rate of defect youngster.
Digital pcr is a kind of nucleic acid molecules absolute quantitation technology, independent of standard curve and sample for reference, is directly detected
The copy number of target sequence.Therefore, this detection mode has the sensitivity outstanding than traditional qPCR, specificity and accurate
Property.Therefore, the application field especially suitable for that can not be differentiated very well by Ct values：It is relative to copy number variation, abrupt climatic change, gene
Expression study (the uneven expression of such as allele), the checking of two generation sequencing results, miRNA expression analysis, slender cellular gene expression
Analysis etc..
Authorization Notice No. is that CN104182655B (Application No. 201410439489.2) Chinese invention patent discloses
A kind of method for judging fetus genotype, including step in detail below：A. maternal blood DNA sample is extracted, carries out extron
Sequencing, obtains initial data；B. Quality Control is carried out to initial data, compared with reference sequences, detected SNP, annotate and count；C. base
In pregnant woman and four kinds of combinations of the genotype of fetus, the SNP of four kinds of combinations is calculated with greatest hope value-based algorithm
The mixed Gauss model of the minimum gene frequency in site；D. three critical values of mixed Gauss model are calculated；E. lead to
The magnitude relationship crossed between the minimum gene frequency of more each SNP site and three critical values judges fetus genotype.Should
Although technical scheme can judge fetus genotype, complex.
The content of the invention
The invention provides a kind of method and device of effective detection genotype, can simply and rapidly judge parental generation and
The gene type of filial generation.
A kind of method of effective detection genotype, comprises the following steps：
Sample to be tested is obtained, sample to be tested is subjected to digital pcr reaction, obtains mutator signal/wild type base
Because of the value of signal；
The ratio for detecting mutator signal/wild type gene numerical value is 0, (CV ＜ 10%), i.e. test sample gene
Type is determined as AA；
The ratio for detecting mutator signal/wild type gene numerical value is 1, and (CV ＜ 10%), i.e. test sample judge
For Aa；
Ratio >=3 of mutator signal/wild type gene numerical value are detected, (CV ＜ 10%), i.e. test sample gene
Type is determined as aa.
Described sample to be tested is single individual pure sample sheet, such as blood DNA, oral secretion DNA, amniotic fluid or navel
The direct sample such as blood DNA.
Test sample need to be only control of the target sample without normal sample, it is assumed that cls gene to be checked is by allele control
System, its wild type are denoted as A, and saltant type is denoted as a.
A kind of method of effective detection genotype, comprises the following steps：
First, there are two individual hybrid dnas of genetic affinity, be parental generation and filial generation, total dissociative DNA total amount is 1, its
In, filial generation dissociative DNA accounting is B%, and the total amount of filial generation dissociative DNA is 1*B%, and the total amount of parental generation dissociative DNA is 1* (1-
B%)；
If the 1, parental gene type is AA, filial generation has 2 kinds of situations：
A：Progeny genotypes are AA, then now in total dissociative DNA mutator amount be 0, detect mutator signal/
The value of wild type gene signal also should be 0；
B：Progeny genotypes are Aa, then filial generation mutator amount is 1*B%/2, and filial generation wild type gene amount is also 1*
B%/2, value=(1*B%/2)/((1* of mutator signal/wild type gene signal is now detected in total dissociative DNA
B%/2)+1* (1-B%))；
The 2nd, if parental gene type is Aa, then the wild type and mutated genes amount of parental generation can be denoted as 1* (1-B%)/2,
Filial generation has 3 kinds of situations in the case of such：
A：Progeny genotypes are AA, i.e., filial generation wild type gene amount is 1*B%, and mutated genes amount is 0, now total free
Value=(1* (1-B%)/2)/((1* (1-B%)/2)+(1* of mutator signal/wild type gene signal is detected in DNA
B%))；
B：Progeny genotypes are Aa, i.e. filial generation mutated genes and wild type gene amount are denoted as 1*B%/2, now total trip
From value=((1*B%/2)+(1* (1-B%)/2))/((1* that mutator signal/wild type gene signal is detected in DNA
B%/2)+(1* (1-B%)/2))=1；
C：Progeny genotypes are aa, then filial generation mutated genes amount is 1*B%, and wild type gene amount is 0, now total trip
From value=((1*B%)+(1* (1-B%)/2))/(1* (1- that mutator signal/wild type gene signal is detected in DNA
B%)/2)；
The 3rd, if parental gene type is aa, then parental wild-type gene dosage is 0, and mutated genes amount is denoted as 1* (1-B%),
Filial generation has 2 kinds of possibility in the case of such：
A：Progeny genotypes are Aa, i.e., filial generation mutator and wild gene amount are respectively 1*B%/2, now total dissociative DNA
In detect value=((1*B%/2)+(1* (1-B%)))/(1*B%/2) of mutator signal/wild type gene signal；
B：Progeny genotypes are aa, i.e., filial generation mutated genes amount is 1*B%, and wild type gene amount is 0, now total free
Without wild type gene in DNA, therefore, the value for detecting mutator signal/wild type gene signal is infinity, without meter
Calculate；
2nd, sample to be tested is obtained, sample to be tested is subjected to digital pcr reaction, obtains mutator signal/wild type
The value of gene signal, compared with the value of mutator signal/wild type gene signal in step 1, error within 10%,
Then belong to gene type corresponding to the value, the quick gene for judging parental generation and filial generation.
In step 1, B% is 8%~30%.
In step 2, described sample to be tested is maternal blood cfDNA etc., however it is not limited to such sample.
A kind of method of effective detection genotype, comprises the following steps：
First, the total dissociative DNA total amount of maternal blood is 1, wherein, fetus dissociative DNA accounting is B%, fetus dissociative DNA
Total amount is 1*B%, and the total amount of pregnant woman's dissociative DNA is 1* (1-B%)；
If the 1, pregnant woman's genotype is AA, fetus has 2 kinds of situations：
A：Fetus genotype is AA, then now in total dissociative DNA mutator amount be 0, detect mutator signal/
The value of wild type gene signal also should be 0；
B：Fetus genotype is Aa, then fetus mutator amount is 1*B%/2, and fetus wild type gene amount is also 1*
B%/2, value=(1*B%/2)/((1* of mutator signal/wild type gene signal is now detected in total dissociative DNA
B%/2)+1* (1-B%))；
The 2nd, if pregnant woman's genotype is Aa, then the wild type and mutated genes amount of pregnant woman can be denoted as 1* (1-B%)/2,
Fetus has 3 kinds of situations in the case of such：
A：Fetus genotype is AA, i.e., fetus wild type gene amount is 1*B%, and mutated genes amount is 0, now total free
Value=(1* (1-B%)/2)/((1* (1-B%)/2)+(1* of mutator signal/wild type gene signal is detected in DNA
B%))；
B：Fetus genotype is Aa, i.e. fetus mutated genes and wild type gene amount are denoted as 1*B%/2, now total trip
From value=((1*B%/2)+(1* (1-B%)/2))/((1* that mutator signal/wild type gene signal is detected in DNA
B%/2)+(1* (1-B%)/2))=1；
C：Fetus genotype is aa, then fetus mutated genes amount is 1*B%, and wild type gene amount is 0, now total trip
From value=((1*B%)+(1* (1-B%)/2))/(1* (1- that mutator signal/wild type gene signal is detected in DNA
B%)/2)；
The 3rd, if pregnant woman's genotype is aa, then pregnant woman's wild type gene amount is 0, and mutated genes amount is denoted as 1* (1-B%),
Fetus has 2 kinds of possibility in the case of such：
A：Fetus genotype is Aa, i.e., fetus mutator and wild gene amount are respectively 1*B%/2, now total dissociative DNA
In detect value=((1*B%/2)+(1* (1-B%)))/(1*B%/2) of mutator signal/wild type gene signal；
B：Fetus genotype is aa, i.e., fetus mutated genes amount is 1*B%, and wild type gene amount is 0, now total free
Without wild type gene in DNA, therefore, the value for detecting mutator signal/wild type gene signal is infinity, without meter
Calculate；
2nd, sample to be tested is obtained, sample to be tested is subjected to digital pcr reaction, obtains mutator signal/wild type
The value of gene signal, compared with the value of mutator signal/wild type gene signal in step 1, error within 10%,
Then belong to gene type corresponding to the value, the quick gene for judging pregnant woman and fetus.
In step 1, B% is 8%~30%.In step 1, B% be 8%~12%, further preferably, B% be 9%~
11%, most preferably, B% 10%.In step 2, sample to be tested be pregnant woman's second trimester peripheral blood extraction dissociative DNA, institute
The second trimester stated is pregnant 13 weeks to pregnant 27 weeks.
In step 2, further preferably, error is within 5%.
Common Pregnant Women second trimester fetus dissociative DNA content average energy reaches 10%, then if total dissociative DNA amount is calculated as 1,
Then pregnant woman's dissociative DNA and fetus dissociative DNA account for 0.9 and 0.1 respectively.
Most preferably, a kind of method of effective detection genotype, comprises the following steps：
First, the total dissociative DNA total amount of maternal blood is 1, wherein, the total amount of fetus dissociative DNA is 0.1, pregnant woman's dissociative DNA
Total amount be 0.9；
If the 1, pregnant woman's genotype is AA, fetus has 2 kinds of situations：
A：Fetus is normal genotype AA, then its wild type can be denoted as 0.1, saltant type 0, now total dissociative DNA
The value of middle mutator signal/wild type gene signal is 0/ (0.9+0.1)=0；
B：Fetus carries mutator Aa to be recessive, then its wild type can be denoted as 0.05, saltant type 0.05, now
The value of mutator signal/wild type gene signal is 0.05/ (0.9+0.05)=0.053 in total dissociative DNA；
The 2nd, if pregnant woman's genotype is Aa, then the wild type and mutated genes amount of pregnant woman can be denoted as 0.45, such feelings
Fetus has 3 kinds of situations under condition：
A：Fetus genotype is AA, i.e., fetus wild type gene amount is 0.1, and mutated genes amount is 0, now total free
Value=0.45/ (0.45+0.1)=0.82 of mutator signal/wild type gene signal is detected in DNA；
B：Fetus genotype is Aa, i.e. fetus mutated genes and wild type gene amount are denoted as 0.05, now total free
Value=(0.45+0.05)/(0.45+0.05)=1 of mutator signal/wild type gene signal is detected in DNA；
C：Fetus genotype is aa, then fetus mutated genes amount is 0.1, and wild type gene amount is 0, now total free
Value=(0.45+0.1)/0.45=1.22 of mutator signal/wild type gene signal is detected in DNA；
The 3rd, if pregnant woman's genotype is aa, then pregnant woman's wild type gene amount is 0, and mutated genes amount is denoted as 0.9, such feelings
Fetus has 2 kinds of possibility under condition：
A：Fetus genotype is Aa, i.e., fetus mutator and wild gene amount are respectively 0.05, are now examined in total dissociative DNA
Measure value=(0.9+0.05)/0.05=19 of mutator signal/wild type gene signal；
B：Fetus genotype is aa, i.e., fetus mutated genes amount is 0.1, and wild type gene amount is 0, now total free
Without wild type gene in DNA, therefore, the value for detecting mutator signal/wild type gene signal is infinity, without meter
Calculate；
2nd, the dissociative DNA of pregnant woman's second trimester peripheral blood extraction is obtained, carries out digital pcr reaction, obtains mutator letter
Number/value of wild type gene signal, compared with the value of mutator signal/wild type gene signal in step 1, error exists
Within 10%, then belong to gene type corresponding to the value, the quick gene for judging pregnant woman and fetus.
In step 2, further preferably, error is within 5%.
A kind of device of effective detection genotype, including：
The module of storage and calculation step one (is used for the module for realizing step 1)；
Digital pcr device；
And calculation step two and with numeric ratio in step 1 compared with module (be used to realize the calculating mould of step 2
Block).
Compared with prior art, the invention has the advantages that：
The method and device of effective detection genotype of the present invention, it is applicable to based on a variety of single bases that dissociative DNA is template
Genotype detection and analysis because of mutation.The calculation is simple, is exaggerated CV values, reduces existing platform to CV (variation lines
Number variable coefficient abbreviation) it is worth requiring, feasibility is strong.The method and dress of effective detection genotype of the present invention
Put, can simply and rapidly judge fetus and the gene type of pregnant woman.
Embodiment
Embodiment 1
Suitable for target individual to need the individual that is detected, for example, blood DNA, oral secretion DNA and amniotic fluid or
The direct sample such as bleeding of the umbilicus DNA.Now there was only 3 kinds of situations：
1 detects that mutator signal/wild type gene signal is 0, i.e. target gene type is determined as AA；
2 detect that mutator signal/wild type gene signal is 1, i.e. target gene type is determined as Aa；
3 detect that mutator signal/wild type gene signal is determined as aa for infinity, i.e. target gene type.
The genotype of SLC26A4 919-2 genes is detected, samples sources are amniotic fluid/bleeding of the umbilicus DNA.Pass through the spy of digital pcr
Specific amplification, an amniotic fluid DNA detect that mutator signal/wild type gene signal=33.608/0=33.608 is (unlimited
Greatly), its genotype results be determined as aa, another bleeding of the umbilicus DNA detect mutator signal/wild type gene signal=
33.568/32.495=1.03 (it is approximately equal to 1, meets CV<8%), its genotype is determined as the result of determination of two samples of Aa. all
It is consistent with clinical sequencing result.
Embodiment 2
This method is applicable to based on genotype detection of the dissociative DNA for a variety of single gene mutations of template and analysis.
Common Pregnant Women second trimester if cffDNA (fetus dissociative DNA) content average energy reaches 10%, then if total free
Amount of DNA is calculated as 1, then parent cfDNA (dissociative DNA) and fetus cfDNA account for 0.9 and 0.1 respectively.Ratio in model 2 is then specific
Change as follows：
1. if pregnant woman does not carry mutator, then its wild type can be denoted as 0.9, saltant type 0.In such situation
Under, fetus can only be following 2 kinds of situations：
A：Fetus is normal genotype, then its wild type can be denoted as 0.1, saltant type 0, now be dashed forward in total cfDNA
The ratio of modification and wild type is 0/ (0.9+0.1)=0；
B:Fetus carries mutator to be recessive, then its wild type can be denoted as 0.05, saltant type 0.05, always now
The ratio of saltant type and wild type is 0.05/ (0.9+0.05)=0.053 in cfDNA；
2. if pregnant woman is the normal pregnancies of carrying mutator, then its wild type can be denoted as 0.45, saltant type 0.45.
In such cases, fetus is likely to occur following 3 kinds of situations：
A. fetus is normal genotype, then its wild type can be denoted as 0.1, saltant type 0, now be dashed forward in total cfDNA
The ratio of modification and wild type is 0.45/ (0.45+0.1)=0.82；
B：Fetus carries mutator to be recessive, then its wild type can be denoted as 0.05, saltant type 0.05, always now
The ratio of saltant type and wild type is (0.45+0.05)/(0.45+0.05)=1 in cfDNA；
C:Fetus is pure and mild recessive genotype, then its wild type is denoted as 0, saltant type 0.1, is now dashed forward in total cfDNA
The ratio of modification and wild type is (0.45+0.1)/0.45=1.22；
3. if pregnant woman is pure and mild recessive genotype pregnant woman, then its wild type can be denoted as 0, saltant type 0.9.In such feelings
Under condition, fetus is likely to occur following 2 kinds of situations：
A:Fetus carries mutator to be recessive, then its wild type can be denoted as 0.05, saltant type 0.05, always now
The ratio of saltant type and wild type is (0.9+0.05)/0.05=19 in cfDNA；
B:Fetus is pure and mild recessive genotype, then its wild type is denoted as 0, saltant type 0.1, now nothing in total cfDNA
Wild type DNA, mutant DNA content is high, without ratio calculated.
More than judge it is based on cffDNA contents averagely to account for the 10% of total dissociative DNA content to calculate, as long as now PCR
Stability is fine, and CV is not more than 8%, then can easily carry out the detection of genotype.By the present embodiment as can be seen that this method will
0.45,0.5 and 0.55 (0.05 difference) when the ratio that genotype judges by detecting saltant type originally has been amplified to present
After the amplification of 0.82,1 and 1.22 (0.18 and 0.22 difference) difference, existing platform can be detected.
If the ratio is more than 10%, it is above-mentioned it is various in the case of gap is bigger between the ratio that is calculated, CV is worth will
Ask lower, detection is more accurate simple.
Embodiment 3
This method is applicable to based on genotype detection of the dissociative DNA for a variety of single gene mutations of template and analysis.
Common Pregnant Women second trimester if cffDNA (fetus dissociative DNA) content average energy reaches 20%, then if total free
Amount of DNA is calculated as 1, then parent cfDNA (dissociative DNA) and fetus cfDNA account for 0.8 and 0.2 respectively.Ratio in model 2 is then specific
Change as follows：
1. if pregnant woman does not carry mutator, then its wild type can be denoted as 0.8, saltant type 0.In such situation
Under, fetus can only be following 2 kinds of situations：
A：Fetus is normal genotype, then its wild type can be denoted as 0.2, saltant type 0, now be dashed forward in total cfDNA
The ratio of modification and wild type is 0/ (0.8+0.2)=0；
B:Fetus carries mutator to be recessive, then its wild type can be denoted as 0.1, saltant type 0.1, always now
The ratio of saltant type and wild type is 0.1/ (0.8+0.1)=0.111 in cfDNA；
2. if pregnant woman is the normal pregnancies of carrying mutator, then its wild type can be denoted as 0.4, saltant type 0.4.
In the case of such, fetus is likely to occur following 3 kinds of situations：
A. fetus is normal genotype, then its wild type can be denoted as 0.2, saltant type 0, now be dashed forward in total cfDNA
The ratio of modification and wild type is 0.4/ (0.4+0.2)=0.667；
B：Fetus carries mutator to be recessive, then its wild type can be denoted as 0.1, saltant type 0.1, always now
The ratio of saltant type and wild type is (0.4+0.1)/(0.4+0.1)=1 in cfDNA；
C:Fetus is pure and mild recessive genotype, then its wild type is denoted as 0, saltant type 0.2, is now dashed forward in total cfDNA
The ratio of modification and wild type is (0.4+0.2)/0.4=1.5；
3. if pregnant woman is pure and mild recessive genotype pregnant woman, then its wild type can be denoted as 0, saltant type 0.9.In such feelings
Under condition, fetus is likely to occur following 2 kinds of situations：
A:Fetus carries mutator to be recessive, then its wild type can be denoted as 0.1, saltant type 0.1, always now
The ratio of saltant type and wild type is (0.8+0.1)/0.1=9 in cfDNA；
B:Fetus is pure and mild recessive genotype, then its wild type is denoted as 0, saltant type 0.2, now nothing in total cfDNA
Wild type DNA, mutant DNA content is high, without ratio calculated.
Embodiment 4
This method is applicable to based on genotype detection of the dissociative DNA for a variety of single gene mutations of template and analysis.
Common Pregnant Women second trimester if cffDNA (fetus dissociative DNA) content average energy reaches 30%, then if total free
Amount of DNA is calculated as 1, then parent cfDNA (dissociative DNA) and fetus cfDNA account for 0.7 and 0.3 respectively.Ratio in model 2 is then specific
Change as follows：
1. if pregnant woman does not carry mutator, then its wild type can be denoted as 0.7, saltant type 0.In such situation
Under, fetus can only be following 2 kinds of situations：
A：Fetus is normal genotype, then its wild type can be denoted as 0.3, saltant type 0, now be dashed forward in total cfDNA
The ratio of modification and wild type is 0/ (0.7+0.3)=0；
B:Fetus carries mutator to be recessive, then its wild type can be denoted as 0.15, saltant type 0.15, always now
The ratio of saltant type and wild type is 0.15/ (0.7+0.15)=0.176 in cfDNA；
2. if pregnant woman is the normal pregnancies of carrying mutator, then its wild type can be denoted as 0.35, saltant type 0.35.
In such cases, fetus is likely to occur following 3 kinds of situations：
A. fetus is normal genotype, then its wild type can be denoted as 0.3, saltant type 0, now be dashed forward in total cfDNA
The ratio of modification and wild type is 0.35/ (0.35+0.3)=0.538；
B：Fetus carries mutator to be recessive, then its wild type can be denoted as 0.15, saltant type 0.15, always now
The ratio of saltant type and wild type is (0.35+0.15)/(0.35+0.15)=1 in cfDNA；
C:Fetus is pure and mild recessive genotype, then its wild type is denoted as 0, saltant type 0.3, is now dashed forward in total cfDNA
The ratio of modification and wild type is (0.35+0.3)/0.35=1.857；
3. if pregnant woman is pure and mild recessive genotype pregnant woman, then its wild type can be denoted as 0, saltant type 0.7.In such feelings
Under condition, fetus is likely to occur following 2 kinds of situations：
A:Fetus carries mutator to be recessive, then its wild type can be denoted as 0.15, saltant type 0.15, always now
The ratio of saltant type and wild type is (0.7+0.15)/0.15=5.667 in cfDNA；
B:Fetus is pure and mild recessive genotype, then its wild type is denoted as 0, saltant type 0.3, now nothing in total cfDNA
Wild type DNA, mutant DNA content is high, without ratio calculated.
This tests the checking for having selected a relatively stable system to carry out theoretical model, and specific data are as shown in table 1：
Table 1
As shown in upper table, according to the ratio of mutator signal/wild type gene signal, the first pregnant woman's gene is AA,
Fetus is also AA, and second of pregnant woman's gene is AA, and fetus is also Aa, and the third pregnant woman's gene is Aa, and fetus is also AA, the 4th kind
Pregnant woman's gene is Aa, and fetus is also Aa, and the 5th kind of pregnant woman's gene is Aa, and fetus is also aa, and the ratio in the case of this 5 kinds can be clear and definite
Distinguish, therefore, this method feasibility can be verified.
Claims (8)
- A kind of 1. method of effective detection genotype, it is characterised in that comprise the following steps：Sample to be tested is obtained, sample to be tested is subjected to digital pcr reaction, obtains mutator signal/wild type gene letter Number value；The ratio for detecting mutator signal/wild type gene numerical value is 0, i.e. test sample genotype is determined as AA；The ratio for detecting mutator signal/wild type gene numerical value is 1, i.e. test sample is determined as Aa；Ratio >=3 of mutator signal/wild type gene numerical value are detected, i.e. test sample genotype is determined as aa.
- 2. the method for effective detection genotype according to claim 1, it is characterised in that described sample to be tested is single The pure sample sheet of one individual.
- A kind of 3. method of effective detection genotype, it is characterised in that comprise the following steps：First, there are two individual hybrid dnas of genetic affinity, be parental generation and filial generation, total dissociative DNA total amount is 1, wherein, son It is B% for dissociative DNA accounting, the total amount of filial generation dissociative DNA is 1*B%, and the total amount of parental generation dissociative DNA is 1* (1-B%)；If the 1, parental gene type is AA, filial generation has 2 kinds of situations：A：Progeny genotypes are AA, then now mutator amount is 0 in total dissociative DNA, detects mutator signal/wild The value of type gene signal also should be 0；B：Progeny genotypes are Aa, then filial generation mutator amount is 1*B%/2, and filial generation wild type gene amount is also 1*B%/2, Value=(1*B%/2)/((1*B%/2)+1* of mutator signal/wild type gene signal is now detected in total dissociative DNA (1-B%))；The 2nd, if parental gene type is Aa, then the wild type and mutated genes amount of parental generation can be denoted as 1* (1-B%)/2, such In the case of filial generation have 3 kinds of situations：A：Progeny genotypes are AA, i.e., filial generation wild type gene amount is 1*B%, and mutated genes amount is 0, now total dissociative DNA In detect value=(1* (1-B%)/2)/((1* (1-B%)/2)+(1* of mutator signal/wild type gene signal B%))；B：Progeny genotypes are Aa, i.e. filial generation mutated genes and wild type gene amount are denoted as 1*B%/2, now total free Value=((1*B%/2)+(1* (1-B%)/2))/((1* of mutator signal/wild type gene signal is detected in DNA B%/2)+(1* (1-B%)/2))=1；C：Progeny genotypes are aa, then filial generation mutated genes amount is 1*B%, and wild type gene amount is 0, now total free Value=((1*B%)+(1* (1-B%)/2))/(1* (1- of mutator signal/wild type gene signal are detected in DNA B%)/2)；The 3rd, if parental gene type is aa, then parental wild-type gene dosage is 0, and mutated genes amount is denoted as 1* (1-B%), such In the case of filial generation have 2 kinds of possibility：A：Progeny genotypes are Aa, i.e., filial generation mutator and wild gene amount are respectively 1*B%/2, are now examined in total dissociative DNA Measure value=((1*B%/2)+(1* (1-B%)))/(1*B%/2) of mutator signal/wild type gene signal；B：Progeny genotypes are aa, i.e., filial generation mutated genes amount is 1*B%, and wild type gene amount is 0, now total dissociative DNA Middle no wild type gene, therefore, the value for detecting mutator signal/wild type gene signal is infinity, without calculating；2nd, sample to be tested is obtained, sample to be tested is subjected to digital pcr reaction, obtains mutator signal/wild type gene The value of signal, compared with the value of mutator signal/wild type gene signal in step 1, error then belongs within 10% In gene type corresponding to the value, the quick gene for judging parental generation and filial generation.
- 4. the method for effective detection genotype according to claim 3, it is characterised in that in step 1, B% be 8%~ 30%.
- A kind of 5. method of effective detection genotype, it is characterised in that comprise the following steps：First, the total dissociative DNA total amount of maternal blood is 1, wherein, fetus dissociative DNA accounting is B%, the total amount of fetus dissociative DNA For 1*B%, the total amount of pregnant woman's dissociative DNA is 1* (1-B%)；If the 1, pregnant woman's genotype is AA, fetus has 2 kinds of situations：A：Fetus genotype is AA, then now mutator amount is 0 in total dissociative DNA, detects mutator signal/wild The value of type gene signal also should be 0；B：Fetus genotype is Aa, then fetus mutator amount is 1*B%/2, and fetus wild type gene amount is also 1*B%/2, Value=(1*B%/2)/((1*B%/2)+1* of mutator signal/wild type gene signal is now detected in total dissociative DNA (1-B%))；The 2nd, if pregnant woman's genotype is Aa, then the wild type and mutated genes amount of pregnant woman can be denoted as 1* (1-B%)/2, such In the case of fetus have 3 kinds of situations：A：Fetus genotype is AA, i.e., fetus wild type gene amount is 1*B%, and mutated genes amount is 0, now total dissociative DNA In detect value=(1* (1-B%)/2)/((1* (1-B%)/2)+(1* of mutator signal/wild type gene signal B%))；B：Fetus genotype is Aa, i.e. fetus mutated genes and wild type gene amount are denoted as 1*B%/2, now total free Value=((1*B%/2)+(1* (1-B%)/2))/((1* of mutator signal/wild type gene signal is detected in DNA B%/2)+(1* (1-B%)/2))=1；C：Fetus genotype is aa, then fetus mutated genes amount is 1*B%, and wild type gene amount is 0, now total free Value=((1*B%)+(1* (1-B%)/2))/(1* (1- of mutator signal/wild type gene signal are detected in DNA B%)/2)；The 3rd, if pregnant woman's genotype is aa, then pregnant woman's wild type gene amount is 0, and mutated genes amount is denoted as 1* (1-B%), such In the case of fetus have 2 kinds of possibility：A：Fetus genotype is Aa, i.e., fetus mutator and wild gene amount are respectively 1*B%/2, are now examined in total dissociative DNA Measure value=((1*B%/2)+(1* (1-B%)))/(1*B%/2) of mutator signal/wild type gene signal；B：Fetus genotype is aa, i.e., fetus mutated genes amount is 1*B%, and wild type gene amount is 0, now total dissociative DNA Middle no wild type gene, therefore, the value for detecting mutator signal/wild type gene signal is infinity, without calculating；2nd, sample to be tested is obtained, sample to be tested is subjected to digital pcr reaction, obtains mutator signal/wild type gene The value of signal, compared with the value of mutator signal/wild type gene signal in step 1, error then belongs within 10% In gene type corresponding to the value, the quick gene for judging pregnant woman and fetus.
- 6. the method for effective detection genotype according to claim 5, it is characterised in that in step 1, in step 1, B% is 8%~30%；In step 2, sample to be tested is the dissociative DNA of pregnant woman's second trimester peripheral blood extraction, and described second trimester is pregnant 13 weeks To pregnant 27 weeks.
- 7. the method for effective detection genotype according to claim 5, it is characterised in that in step 2, and in step 1 Mutator signal/wild type gene signal value compare, error is within 5%.
- A kind of 8. device of effective detection genotype, it is characterised in that including：Storage and the module of calculation step one, described step one include：First, the total dissociative DNA total amount of maternal blood is 1, its In, fetus dissociative DNA accounting is B%, and the total amount of fetus dissociative DNA is 1*B%, and the total amount of pregnant woman's dissociative DNA is 1* (1- B%)；If the 1, pregnant woman's genotype is AA, fetus has 2 kinds of situations：A：Fetus genotype is AA, then now mutator amount is 0 in total dissociative DNA, detects mutator signal/wild The value of type gene signal also should be 0；B：Fetus genotype is Aa, then fetus mutator amount is 1*B%/2, and fetus wild type gene amount is also 1*B%/2, Value=(1*B%/2)/((1*B%/2)+1* of mutator signal/wild type gene signal is now detected in total dissociative DNA (1-B%))；The 2nd, if pregnant woman's genotype is Aa, then the wild type and mutated genes amount of pregnant woman can be denoted as 1* (1-B%)/2, such In the case of fetus have 3 kinds of situations：A：Fetus genotype is AA, i.e., fetus wild type gene amount is 1*B%, and mutated genes amount is 0, now total dissociative DNA In detect value=(1* (1-B%)/2)/((1* (1-B%)/2)+(1* of mutator signal/wild type gene signal B%))；B：Fetus genotype is Aa, i.e. fetus mutated genes and wild type gene amount are denoted as 1*B%/2, now total free Value=((1*B%/2)+(1* (1-B%)/2))/((1* of mutator signal/wild type gene signal is detected in DNA B%/2)+(1* (1-B%)/2))=1；C：Fetus genotype is aa, then fetus mutated genes amount is 1*B%, and wild type gene amount is 0, now total free Value=((1*B%)+(1* (1-B%)/2))/(1* (1- of mutator signal/wild type gene signal are detected in DNA B%)/2)；The 3rd, if pregnant woman's genotype is aa, then pregnant woman's wild type gene amount is 0, and mutated genes amount is denoted as 1* (1-B%), such In the case of fetus have 2 kinds of possibility：A：Fetus genotype is Aa, i.e., fetus mutator and wild gene amount are respectively 1*B%/2, are now examined in total dissociative DNA Measure value=((1*B%/2)+(1* (1-B%)))/(1*B%/2) of mutator signal/wild type gene signal；B：Fetus genotype is aa, i.e., fetus mutated genes amount is 1*B%, and wild type gene amount is 0, now total dissociative DNA Middle no wild type gene, therefore, the value for detecting mutator signal/wild type gene signal is infinity, without calculating；Digital pcr device；And calculation step two and with numeric ratio in step 1 compared with module, described step two includes：2nd, sample to be tested is obtained, sample to be tested is subjected to digital pcr reaction, obtains mutator signal/wild type gene The value of signal, compared with the value of mutator signal/wild type gene signal in step 1, error then belongs within 10% In gene type corresponding to the value, the quick gene for judging pregnant woman and fetus.
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| CN108710782A (en) * | 2018-05-16 | 2018-10-26 | 为朔医学数据科技（北京）有限公司 | Genotype conversion method, device and electronic equipment |
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| CN108531572A (en) * | 2018-03-08 | 2018-09-14 | 北京爱普益医学检验中心有限公司 | It is a kind of it is antenatal detection progeny genotypes method and application |
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