Mathematics Model of Inheritance Three Different Traits in Genetics with Matrix Approach

ABSTRACT


INTRODUCTION
The decline of hereditary traits has a molecular basis, that is, the exact replication of Deoxyribonucleic Acid (DNA) and produces copies of genes that can be passed on from the parent to the offspring (Campbell, 2002).A single gene governs many traits in plants, animals, and microbes.Genes in diploid individuals are allele pairs and each of their offspring.The inheritance of a trait that can be known from parents to offspring genetically is called heredity.This inheritance law follows an orderly pattern and repeats itself from generation to generation (Crowder, 2007).The development of genetic science is characterized by discovering genes and chromosomes, which subsequently develop towards molecular genetics.The discovery of recombinant DNA opened up the development of genetic engineering.In gene therapy efforts, genetic engineering technology provides hope in various fields such as the pharmaceutical industry, health, society and humanity, agriculture, and law (Nusantari, 2018).
Crossbreeding is the mating between genetically different individuals or populations to produce a combination of traits from the parent or the recombination of genes in the offspring.Crosses can occur between individuals of different species and individuals within a single species, commonly known as crosses between strains or accessions (Alianto & Huda, 2008).With the crossing, we can get new seeds of good quality because of the combination of traits and can also give rise to new types due to the crossing of the two different traits or

METHOD
The research method used in this n research is a literature method, a study or activity to collect information relevant to the topic or problem that is the object of the study.Information in this whitish study is obtained from books and journals.

Research Procedure
The procedure for this study is as follows: 1) Make a cross between large seeds with a smooth texture and round shape (AABBCC) and small seeds with a wrinkled texture and elongated shape (aabbcc) so that the first offspring will be obtained first Filial (F 1 ) is large seeds with a smooth texture and round shape (AaBbCc).Further, make an intermediate cross AaBbCc and AaBbCc so will be obtained a second Filial (F 2 ) 2) Conducting a reverse cross between possible offspring the entire set of possible combinations of these alleles AA,Aa,BB,Bb,CC,Cc (AoBoCo) and parental parents (AABBCC and aabbcc).

3)
Formed a table of genotype opportunities from the results of reverse crossing between offspring AoBoCo and parental parents (AABBCC and aabbcc) 4) Forms a linear equation from the table of chances based on the genotype of the selected parent.

5)
We are forming equations in matrix notation in such a way that a matrix is obtained with the elements present on the matrix according to the table of opportunities of each genotype.

6)
Determines the matrix's eigenvalues and the eigenvectors corresponding to the eigen values.

7)
Forming a matrix  of eigenic vectors thenn determines the inverse of the matrix  .

8)
Furthermore, the matrix t can be diagnosed i so that it cannot form an explicit equation and look for the limit value of the explicit equation.

9)
Perform numerical simulations for the parental crossover of broods (AABBCC and aabbcc) with typed offspring AoBoCo .10) Conclude from the results of such studies.

RESULTS AND DISCUSSION
The Probability of Crossover Genotypes 3 Different Traits Table 1.Punnet-Square Crosses of 3 Different Traits . 1 Tabel Punnet-Square persilangan 3 sifat beda Based on the eight crosses above shows the crossing and the possibilities of the resulting offspring.Furthermore, it will present the chances of a possible genotype in offspring for the entire possible combination of the parent's genotype based on the crossover results.
From the matrix above, the inverse of the matrix can be determined with the help of Software Maple 13, and then the matrix is obtained as follows.

Mathematical Models with Matrix Approach
From the matrix  above, the inverse of the matrix  can be determined with the help of Maple 13 Software, and then the matrix is obtained as follows : Based on the equation   =    0 , then it can be obtained By using Maple 13 Software, the Equation System formed : ) Equations ( 1) is an explicit equation for the probability of all eight genotypes in the n-th generation in terms of the probability of the initial genotype.Because ( 1 2 )  it tends to approach 0 towards n the infinite, the limit of the equation above is as follows. ) So the chances of hereditary genotypes  going to infinity are obtained, namely: From the approach of the limit above to  go towards infinity obtained the valuea n = 1 and b n = 0 c n = 0 , , d n = 0, e 0 = 0, f 0 = 0, g 0 = 0, h 0 = 0, so that at that limit all descendants in the population will have a genotype AABBCC

Numerical Analysis Numerical Analysis For Crossing and 𝐀𝐀𝐁𝐁𝐂𝐂𝐀𝐨𝐁𝐨𝐂𝐨
In AABBCC and AoBoCo crosses, numerical analysis can be performed by taking  = 1, 2, … ,30.

𝑛
Table 2 shows that if it is approached with  = 1, 2, … ,30, then for a n , it will close the value of 1 in the 25 th to nth generation, it will be AABBCC genotype.For b n , c n and d n in the 22nd generation to the nth generation, there were no more descendants with the genotype AABBCc, AABbCC, and AaBBCC.For e n , f n and g n in the 11 th generation to the nth generation, there were no more offspring with genotypes AABbCc, AaBBCc and AaBbCC.For h n , in the 7 th generation to the nth generation, there ara no more descendants of the AaBbCc genotypes.Implementations for Table 2 show that in the first generation to a n is as large 0,5787037 or 57% , the second generation is obtained a n is as large 0,7702546 or 77%, and the third generation a n is as large 0,880136 or 88%.The fourth generation is obtained a n as large 0,938793 or 93%.So, the crossing AABBCC and AoBoCo in the fourth generation have already reached the level of trust 93%.This shows that superior offspring can be obtained by making four crosses, namely typed offspring with a AABBCC level of trust 93%.More details of the crossing and AABBCCAoBoCo can beseen in Figure 1.Following

Figure 1. Illustration of the nth genotype Cross of aabbcc and AoBoCo
Figure 1 above shows that a n will be stable or convergent at numbers 1 to infinity.This figure shows that the offspring of the AABBCC genotype will always exist in each iteration.Thus the genotype of the cross between AABBCC and AoBoCo in the nth generation will produce offspring with the AABBCC genotype (a n ).Meanwhile, b n , c n , d n , e n , f n , g n , h n will be stable from 0 to infinity.This figure shows offspring with genotypes AABBCc, AABbCC, AaBBCC, AABbCc, AaBBCc, AaBbCC, and AaBbCc will not appear in every iteration.This indicates that the AABBCC crosses with AoBoCo in the nth generation there are no more descendants of AABBCc, AABbCC, AaBBCC, AABbCc, AaBBCc, AaBbCC, AaBbCc.So it can be concluded that the inheritance of traits from crosses between AABBCC and AoBoCo in the nth generation will produce offspring with the AABBCC genotype (normal homozygous).

Numerical Analysis For Crossing 𝐚𝐚𝐛𝐛𝐛𝐜𝐜 𝐚𝐧𝐝 𝐀𝐨𝐁𝐨𝐂𝐨
In aabbcc and AoBoCo crosses, numerical analysis can be performed by taking  = 1, 2, … ,30.So the illustration can be seen in the table below.

PYTHAGORAS: Jurnal
Table 3 shows that if approached with  = 1, 2, … ,30,, then for a n , it will close the value 1 in the 25 th to nth generation, it will have the genotype AaBbCc.For b n , c n and e n in the 22 nd generation to the nth generation, there are no more descendants with genotypes AaBbcc, Aabbcc, and aaBbCc.For d n , f n and g n in the 11 th generation to the nth generation, there are no more descendants with genotypes Aabbcc, aaBbcc, and aaBbCc.For h n , in the 7 th generation to the nth generation, there are no more descendants of the aabbcc genotype.
The implementation for Table 3 can be seen that in the first generation for a n of 0.5787037 or 57%, in the second generation a n of 0.7702546 or 77% is obtained, in the third generation a_n is obtained of 0.880136 or 88%, in the fourth generation, it is obtained a n of 0.938793 or 93%.So it shows that the cross between AABBCC and AoBoCo in the fourth generation has reached a 93% confidence level.This indicates that by doing four crosses, superior offspring can be obtained, namely offspring with the AABBCC genotype with a confidence level of 93%.For more details, the cross between AABBCC and AoBoCo can be seen in Figure 2 2 above shows that a n will be stable or convergent at numbers 1 to infinity.This figure shows that the offspring of the AaBbCc genotype will always exist in every iteration.Thus, the genotype of a cross between aabbcc and AoBoCo in the nth generation will produce offspring with the genotype of AaBbCc.Meanwhile, b n , c n , d n , e n , f n , g n , h n , and h n will be stable from 0 to infinity.This figure shows the offspring with genotype AaBbcc, AabbCc, Aabbcc, aaBbCc, aaBbcc, aabbCc, and aabbcc will not appear in every iteration.This indicates that in aabbcc crosses with AoBoCo in the nth generation, there are no more descendants of AaBbcc, AabbCc, Aabbcc, aaBbCc, aaBbcc, aabbCc, aabbcc.Produce offspring with the AaBbCc genotype (normal heterozygous).

CONCLUSSION
The following conclusions can be obtained based on the discussion and analysis in the previous chapter.1. Inheritance of traits with three distinct characteristics meets the Linear Equation.In general, the formulation of the distribution of genotypes is as follows.

Figure 2 .
Figure 2. Illustration of the nth genotype Cross of aabbcc and AoBoCo Figure2above shows that a n will be stable or convergent at numbers 1 to infinity.This figure shows that the offspring of the AaBbCc genotype will always exist in every iteration.Thus, the genotype of a cross between aabbcc and AoBoCo in the nth generation will produce offspring with the genotype of AaBbCc.Meanwhile, b n , c n , d n , e n , f n , g n , h n , and h n will be stable from 0 to infinity.This figure shows the offspring with genotype AaBbcc, AabbCc, Aabbcc, aaBbCc, aaBbcc, aabbCc, and aabbcc will not appear in every iteration.This indicates that in aabbcc crosses with AoBoCo in the nth generation, there are no more descendants of AaBbcc, AabbCc, Aabbcc, aaBbCc, aaBbcc, aabbCc, aabbcc.Produce offspring with the AaBbCc genotype (normal heterozygous).
The finished behavior of the result of the crossover is in the form of an explicit equation formulated as follows.  =    0 =  −1    0 where  = 1, 2, 3, … with, explicit equations for the fractions of the eight genotypes in the nth generation of backcrosses between AABBCC and AoBoCo are as follows.

Table 3 . Genotype Illustration Results of Crossing aabbcc and AoBoCo
Where table 4.3 can be made a matrix model, that is.