# POPULATION GENETIC ANALYSIS OF DNA SEQUENCE VARIATIONS

Guided by:

## Abstract

## Abbreviations

DNA | Deoxyribonucleic acid |

mRNA | messenger RNA |

IGSR | International Genome Sample Resource |

d.f. | degrees of freedom |

l.o.s. | level of significance |

HWE | Hardy-Weinberg Equilibrium |

## INTRODUCTION

The 1000 Genomes Project ran between 2008 and 2015, creating the largest public catalogue of human genetic variations and genotype data. It aims to provide a deep characterisation of human genome sequence variation as a foundation for investigating the relationship between genotype and phenotype. The project was funded by the Wellcome Trust to maintain and expand the resources. The IGSR was set up to ensure accessibility and usage of the dataset in future and also to expand the dataset with new populations that were previously not a part of 1000 Genomes. The final analysis of the 1000 Genomes, as in October 2015, incorporates 26 populations from Asia (South Asian & East Asian), Africa, Europe and America. In this study, six populations have been considered for analysis viz. three South Asian (SAS) populations: Gujarati Indian from Houston, Texas (GIH), Indian Telugu from the UK (ITU), Bengali from Bangladesh (BEB) and three European (EUR) populations: Finnish in Finland (FIN), Iberian population in Spain (IBS) and Toscani in Italia (TSI).

This study is a part of Population Genetics which deals with genetic differences within and between populations. Statistical methodologies like data cleaning, testing of hypotheses, chi-square applications and graphical representation have been used to get an analytical insight of the data. The dataset for this project has been downloaded from the 1000 Genomes website pertaining to DNA variations on chromosome 1(a long chromosome) and 21(a short chromosome) for the six populations under study.

The rationale behind this study is to analyse the genotype data of the selected populations using statistical methodologies, study the various characteristics of the data, test relevant hypotheses that could answer questions about the variations in the genetic level and infer the structure of genomes across the populations for the two chromosomes.

## METHODOLOGY

The project makes a comparative study of three South Asian and three European populations. The steps mentioned below are followed for these two sets of populations:

1. DATA COLLECTION

Data have been downloaded from the 1000 Genomes website pertaining to DNA variations on Chromosome 1 & 21 for the six populations. These data contain the genotype counts of 3000 genetic loci for each population.

Numbers of individuals for each population differ and are tabulated below:

2. IDENTIFICATION OF DATA FOR ANALYSIS

This step prepares the data for further analysis. This study is based on the biallelic loci, i.e., if two alleles occur at a particular locus, it is biallelic for that population. So, as the data contains genetic loci with multiallelic sites and multi nucleotide variants which are removed and the remaining loci are carried forward for analysis.

3. ANALYSIS OF MONOMORPHIC LOCI

In a population, if only one allele occurs at a site or locus, it is monomorphic or monoallelic in that population. The alleles with frequency less than 0.01 or greater than 0.99 are monomorphic in nature. Otherwise, it is polymorphic in nature. A monomorphic locus occurs in the same form across the population thus, does not account for any variation. So, these loci are excluded from the study as they provide no information about variation. However, an analysis can be made on the distribution of the monomorphic loci in one population with respect to the other populations.

4. IDENTIFICATION OF POLYMORPHIC LOCI

The loci with variable proportions of both the alleles are polymorphic in nature. These loci provide information about genetic variation and are of great importance. The loci which are polymorphic across all the three populations are selected for the next step of the study.

5. TESTING FOR HARDY-WEINBERG EQUILLIBRIUM USING CHI- SQUARE TEST FOR GOODNESS OF FIT

Hardy-Weinberg Principle states that the allele frequencies in a population are stable and constant from generation to generation. One of its assumptions is that mating is random for which the choice of male and female gametes are independent trials. Thus, it is based on the outcome of repeated and independent trials. One of the most important implications of Hardy Weinberg Principle is that we can calculate the expected genotype frequencies for the future generations. However, in practice, there are various evolutionary influences like natural selection, mutation, gene flow etc. that defy the Hardy-Weinberg Principle. So, it is necessary to have an idea about the deviation from the equilibrium and understand the validity of the model. We use chi-square test of goodness of fit for this purpose.

Suppose, in a population of* N* individuals, we study two alleles A and a, then the possible genotype combinations are AA, Aa and aa.

Let they have certain genotypes frequencies, f(AA), f(Aa) and f(aa) which are calculated after observation. Let the probability of occurrence of A and a in a genetic locus be *p *and *q* respectively, such that *p + q= *1.

According to Hardy-Weinberg Principle, the probabilities *p *and *q* are constant from generation to generation. So, for homozygotes, the probability of two A-bearing gametes coming together is

P(AA) = *p *x* p=p ^{2} *

and for a-bearing gametes the probability is

P(aa) =*q *x *q=q ^{2}.*

For heterozygotes, the probability is

P(Aa) = *(p *x *q) + (q* x* p) =2pq.*

Therefore, the expected genotype frequencies are given as:

f_{E }(AA) = *p ^{2}N*

f_{E }(Aa) = *2pqN*

f_{E }(aa) = *q ^{2}N*

*Chi-square test of goodness of fit *enables us to find if the deviation of the experiment from theory is just by chance or due to inadequacy of the theory to fit the observed data.

Let, for a particular genetic locus, O_{i } be the observed genotype frequency and E_{i} be the expected genotype frequency.

We are to test the null hypothesis,

H_{o}: The genetic locus under study is in Hardy-Weinberg Equilibrium,

against the alternative hypothesis,

H_{1}: The genetic locus under study is not in Hardy-Weinberg Equilibrium.

We can tabulate the data as:

GENOTYPE | O | E |

AA | f(AA) | f |

Aa | f(Aa) | f |

Aa | f(aa) | f |

Total | N | N |

The test-statistic χ^{2} = $\displaystyle\sum_{i=1}^3\frac{(O_i-E_i)^2}{E_i}$ ; $\displaystyle\sum_{i=1}^3O_i=\overset3{\underset{i=1}{\displaystyle\sum\;E_i}}=N$

where χ^{2 }follows chi-square distribution. The d.f. associated with χ^{2} is given by,

n= number of independent observations – number of parameters estimated from the data - 1

Here, the number of independent observations is 3, as there are three genotype frequencies. Number of estimated parameters is 1, as we have estimated *p* from the observed data.

Therefore, d.f. is n=3-1-1=1

The p value associated with χ^{2 } is the probability that chance alone produce a deviation between the observed and expected values. So, a larger p value implies that chance alone could account for the deviation and it strengthens our confidence in the validity of the model. However, a smaller p value clearly indicates that there is certainly some cause other than chance that causes the deviation. Hence, validity of the model is undermined.

For testing the Hardy Weinberg Principle, we consider p=0.0001. So, if p<0.0001, we reject the null hypothesis and conclude that the particular locus is not in Hardy Weinberg Equilibrium. Again, if p>0.0001, we may accept the null hypothesis and infer that the locus is in Hardy Weinberg Equilibrium.

In this project, testing of Hardy Weinberg Equilibrium has been done for all the six populations. The loci which follow the principle and are common to all the three populations (BEB, GIH & ITU in SAS and FIN, IBS & TSI in EUR) are carried forward for further analysis.

6. TEST OF INDEPENDENCE OF GENOTYPE FREQUENCIES USING CONTINGENCY TABLES

Hypothesis testing based on contingency tables define the relationship between the row and column attributes. Suppose we have an *r* x *s *contingency table, i.e., a table with *r *rows and *s* columns of two attributes. We need to test whether the row attributes are differently distributed over the column attributes.

Suppose, we need to test whether the genotype frequency distribution of G_{1}, G_{2} and G_{3} is the same across a set of given populations P_{1}, P_{2} and P_{3}. Let, I_{1}, I_{2 }and I_{3} be the number of individuals from the populations P_{1}, P_{2} and P_{3 }respectively. Here, genotype frequency and number of individuals from a population are two independent attributes.

i.e., the null hypothesis to be tested at each genetic locus is:

H_{o}: There is no significant difference in the distribution of genotype frequencies across the three populations.

Against the alternative hypothesis,

H_{1}:_{ } There is significant difference in the distribution of genotype frequencies across the three populations.

So, we get a 3x3 contingency table as follows:

Genotype
Individuals | G | G | G | Total |

I | (I | (I | (I | (I |

I | (I | (I | (I | (I |

I | (I | (I | (I | (I |

Total | (G | (G | (G | M |

where, (I_{i}) = number of individuals in the i^{th} population

(G_{j}) = number of individuals possessing the j^{th} genotype

(I_{i }G_{j}) = number of individuals with j^{th} genotype from the i^{th} population

M = total frequency

These are the observed frequencies.

Under the null hypothesis, the attributes are independent, so the expected frequencies are:

P[I_{i}] = Probability that an individual is from i^{th} population = $\frac{(I_i)}M$

P[G_{j}] = Probability that an individual possesses j^{th} genotype= $\frac{(G_j)}M$

P[I_{i}G_{j}] = Probability that an individual has j^{th} genotype and belongs to the i^{th} population= $\frac{(I_iG_j)}M$

Since, I_{i }and G_{j} are independent,

P[I_{i}G_{j}] = P[I_{i}] . P[G_{j}] = $\frac{(I_i)}M$ . $\frac{(G_j)}M$

(I_{i}G_{j})_{E} = Expected number of individuals with j^{th} genotype from the i^{th} population

$=$M . $\frac{(I_i)}M$. $\frac{(G_j)}M$ = $\frac{(I_i).(G_j)}M$

where χ^{2 }follows chi-square distribution. The d.f. associated with χ^{2} is given by,

n= (r-1)(s-1)

for a 3 x 3 contingency table, d.f. is = (3-1)(3-1)= 4

We test the hypothesis at 5% l.o.s. But, for the calculation of the p-value we use Bonferroni correction.

WHY BONFERRONI CORRECTION?

Bonferroni correction is an adjustment made to the p-value. The p-value is divided by the number of observations. In genetic studies, several dependent or independent statistical tests are done on a large dataset. Suppose, we have a dataset of one million SNPs for a significance test and let the p- value is 0.05. Then we would have 50,000 SNPs which are significant for the test, which could be for no reason. The chances of obtaining false positive results increases.

Due to this correction, the observations with larger deviations are detected and are significant for the given test. However, smaller effects escape notice along with those with no effect that are not rejected.

Thus, if for a certain genetic locus the p value is less than the critical p value, the null hypothesis is rejected and we can conclude that there is significant difference in the distribution of genotype frequencies across the three populations. If the p value is more than the critical p value, the null hypothesis may be accepted and it can be inferred that there is no significant difference in the distribution of genotype frequencies across the three populations.

7. HISTOGRAM OF ALLELE FREQUENCIES

In order to study the allele frequency distribution across different populations, the alternate allele is considered and the allele frequency distribution for both the chromosomes is plotted as a histogram. The allele frequencies are along the X-axis and their frequency of occurrence is along Y-axis. This helps to compare the structure of the genotype composition of the different populations.

8. z- TEST FOR DIFFERENCE OF PROPORTIONS

We apply the z-test for difference of proportions to test the following null hypothesis,

Hypothesis 1

H_{o}: There is no significant difference in the distribution of polymorphic loci in the two sets of population across chromosome 1 and 21.

Against the alternative hypothesis,

H_{1}: There is significant difference in the distribution of polymorphic loci in the two sets of population across chromosome 1 and 21.

Also we apply t-test for difference of means to test the null hypothesis,

Hypothesis 2

H_{o}: There is no significant difference in the distribution of polymorphic loci in the two sets of population for a particular chromosome.

Against the null hypothesis,

H_{1}: There is significant difference in the distribution of polymorphic loci in the two sets of population for a particular chromosome.

The test statistic is given by,

z= $\displaystyle \begin{array}{l}\frac{\vert\overline{p_1}-\overline{p_2}\;\vert}{\sqrt{\displaystyle\frac{\overline p(1-\overline p)}n}}\\\end{array}$

where $\overline{p_1}$& $\overline{p_2}$ are sample proportions. n is the sample size, $\overline p$is the pooled sample proportion, z $\sim$N(0,1).

The calculations, statistical tests of significance and graph plots for the selected loci have been done using R programming and MS Excel.

## RESULTS AND DISCUSSION

SAS POPULATION

After removing the multiallelic and multi-nucleotide variants, we have 2867 out of 3000 loci in chromosome 1 and 2897 out of 3000 loci in chromosome 21 which are biallelic and the analysis is carried forward with these loci.

1) Analysis of monomorphic and polymorphic loci

CHROMOSOME 1 (2867 loci under study)

BEB population

Monomorphic loci: 2608 (90.1%)

GIH | ITU | Count |

Monomorphic | Monomorphic | 2557 (98%) |

Monomorphic | Polymorphic | 12 (0.48%) |

Polymorphic | Monomorphic | 10 (0.38%) |

Polymorphic | Polymorphic | 29 (1.12%) |

Polymorphic loci: 259 (9.9%)

GIH population

Monomorphic loci: 2616 (91.24%)

BEB | ITU | Count |

Monomorphic | Monomorphic | 2557 (97.74%) |

Monomorphic | Polymorphic | 10 (0.38%) |

Polymorphic | Monomorphic | 37 (1.14%) |

Polymorphic | Polymorphic | 12 (0.48%) |

Polymorphic loci: 251 (8.76%)

ITU population

Monomorphic loci: 2643 (92.2%)

BEB | GIH | Count |

Monomorphic | Monomorphic | 2557 (88.26%) |

Monomorphic | Polymorphic | 29 (1%) |

Polymorphic | Monomorphic | 37 (1.27%) |

Polymorphic | Polymorphic | 20 (0.69%) |

Polymorphic loci: 224 (7.8%)

CHROMOSOME 21 (2897 loci under study)

BEB population

Monomorphic loci: 2571 (88.74%)

GIH | ITU | Count |

Monomorphic | Monomorphic | 2543 (98.91%) |

Monomorphic | Polymorphic | 15 (0.58%) |

Polymorphic | Monomorphic | 8 (0.31%) |

Polymorphic | Polymorphic | 5 (0.19%) |

Polymorphic loci: 326 (11.26%)

GIH population

Monomorphic loci: 2579 (89%)

BEB | ITU | Count |

Monomorphic | Monomorphic | 2543 (98.60%) |

Monomorphic | Polymorphic | 15 (0.58%) |

Polymorphic | Monomorphic | 15 (0.58%) |

Polymorphic | Polymorphic | 6 (0.23%) |

Polymorphic loci: 318 (11%)

ITU population

Monomorphic loci: 2571 (88.74%)

BEB | GIH | Count |

Monomorphic | Monomorphic | 2543(98.91%) |

Monomorphic | Polymorphic | 8 (0.31%) |

Polymorphic | Monomorphic | 15 (0.58%) |

Polymorphic | Polymorphic | 5 (0.19%) |

Polymorphic loci: 326 (11.26%)

2) Number of Polymorphic loci common to BEB , GIH & ITU Populations

Chromosome 1 | 190 (6.62%) |

Chromosome 21 | 300 (10.35%) |

3) Number of loci that are in Hardy Weinberg Equilibrium for all the three populations

Chromosome 1 | 150 (78.94%) |

Chromosome 21 | 140 (46.67%) |

4) Contingency test is applied for the loci obtained under step (3)

p value is determined using Bonferroni Correction.

p=0.05/150=0.0003 (for Chromosome 1)

p=0.05/140=0.0003 (for Chromosome 21)

After the contingency test it is found that there are 4 loci in Chromosome 1 and 21 locus in Chromosome 21 for which p < 0.0003.

These loci have significant difference in the genotype frequency distribution across the three populations.

EUR POPULATION

After removing the multiallelic and multi-nucleotide variants, we have 2867 out of 3000 loci in chromosome 1 and 2893 out of 3000 loci in chromosome 21 which are biallelic and the analysis is carried forward with these loci.

1) Analysis of monomorphic and polymorphic loci

CHROMOSOME 1 (2867 loci under study)

FIN population

Monomorphic loci: 2620 (91.38%)

IBS | TSI | Count |

Monomorphic | Monomorphic | 2577 (98.35%) |

Monomorphic | Polymorphic | 11 (0.42%) |

Polymorphic | Monomorphic | 21 (0.81%) |

Polymorphic | Polymorphic | 11 (0.42%) |

Polymorphic loci: 247(8.62%)

IBS population

Monomorphic loci: 2638 (92.01%)

FIN | TSI | Count |

Monomorphic | Monomorphic | 2577 (97.69%) |

Monomorphic | Polymorphic | 11 (0.42%) |

Polymorphic | Monomorphic | 38 (1.44%) |

Polymorphic | Polymorphic | 12 (0.45%) |

Polymorphic loci: 229 (7.99%)

TSI population

Monomorphic loci: 2645 (92.26%)

FIN | IBS | Count |

Monomorphic | Monomorphic | 2577 (97.42%) |

Monomorphic | Polymorphic | 20(0.76%) |

Polymorphic | Monomorphic | 40(1.52%) |

Polymorphic | Polymorphic | 8(0.30%) |

Polymorphic loci: 222 (7.74%)

CHROMOSOME 21 (2893 loci under study)

FIN population

Monomorphic loci: 2581 (89.21%)

IBS | TSI | Count |

Monomorphic | Monomorphic | 2541(98.45%) |

Monomorphic | Polymorphic | 12(0.46%) |

Polymorphic | Monomorphic | 10(0.39%) |

Polymorphic | Polymorphic | 18(0.7%) |

Polymorphic loci: 312 (10.79%)

IBS population

Monomorphic loci: 2570 (88.83%)

FIN | TSI | Count |

Monomorphic | Monomorphic | 2541 (98.87%) |

Monomorphic | Polymorphic | 12 (0.47%) |

Polymorphic | Monomorphic | 15(0.58%) |

Polymorphic | Polymorphic | 3(0.08%) |

Polymorphic loci: 323 (11.17%)

TSI population

Monomorphic loci: 2574 (88.97%)

FIN | IBS | Count |

Monomorphic | Monomorphic | 2541(98.72%) |

Monomorphic | Polymorphic | 10 (0.38%) |

Polymorphic | Monomorphic | 15 (0.58%) |

Polymorphic | Polymorphic | 8 (0.32%) |

Polymorphic loci: 319 (11.03%)

1) Number of Polymorphic loci common to FIN , IBS & TSI Populations

Chromosome 1 | 188 (6.55%) |

Chromosome 21 | 286 (9.88%) |

2) Number of loci that are in Hardy Weinberg Equilibrium for all the three populations

Chromosome 1 | 134 (71.27%) |

Chromosome 21 | 128 (44.75%) |

3) Contingency test is applied for the loci obtained under step (2)

p value is determined using Bonferroni Correction.

p =0.05/134=0.0004 (for Chromosome 1)

p=0.05/128=0.0004 (for Chromosome 21)

After the contingency test it is found that there is 1 locus in each Chromosome 1 and 21 for which p < 0.0004.

These loci have significant difference in the genotype frequency distribution across the three populations.

4) 1) z-TEST FOR DIFFERENCE OF PROPORTIONS

For hypothesis 1:

We consider the polymorphic loci in SAS population chromosome 1(p_{1}) & 21(p_{2}).

Chromosome 1 | Chromosome 21 |

259 | 326 |

251 | 318 |

224 | 326 |

From the data we get,

n_{1 }= 2867, n_{2} = 2897

The sample proportion ${\overline p}_1$ = $\frac{259+251+224}{3\;\times\;2867}$ = 0.085339

The sample proportion $\displaystyle {\overline p}_2$ = $\frac{326+318+326}{3\;\times\;2897}$ = 0.11161

The pooled proportion $\overline p$ = $\begin{array}{l}\frac{(3\;\times\;n_1\;\times\;\overline{p_1}\;)\;+\;(3\;\times\;n_2\;\times\overline{p_2}\;)}{3\;\times\;(\;n_1\;+n_2)}\\\end{array}$ = $\frac{734+970}{17292}$ =0.0985

The value of calculated z is,

z= 11.5931

The critical value of z is 3.3459. Since, calculated z is greater than critical value of z, it is significant. Hence, we reject the null hypothesis and conclude that there is significant difference in the distribution of polymorphic loci of SAS population across chromosome 1 and 21.

Again we consider polymorphic loci in EUR population for Chromosome 1(q_{1}) and 21(q_{2})

Chromosome 1 | Chromosome 21 |

247 | 312 |

229 | 323 |

222 | 319 |

From the data we get,

m_{1 }= 2867, m_{2} = 2897

The sample proportion $\displaystyle {\overline q}_1$ = $\frac{247+229+222}{3\;\times\;2867}$ = 0.08115

The sample proportion $\displaystyle {\overline q}_2$ = $\frac{312+323+319}{3\;\times\;2897}$ = 0.1099

The pooled proportion $\displaystyle \overline q$= $\begin{array}{l}\frac{(3\;\times\;m_1\;\times\;\overline{q_1}\;)\;+\;(3\;\times\;m_2\;\times\overline{q_2}\;)}{3\;\times\;(\;m_1\;+m_2)}\\\end{array}$= $\frac{698+955}{17292}$= 0.9559

The value of calculated z is,

z= 26.99

The critical value of z at 5% l.o.s. is 3.3459. Since, calculated z is greater than critical value of z, it is significant. Hence, we reject the null hypothesis and conclude that there is significant difference in the distribution of polymorphic loci of EUR population across chromosome 1 and 21.

In hypothesis 2,

For Chromosome 1, let a_{1} and a_{2} denote the polymorphic loci of SAS and EUR populations.

From the data we get,

n_{1 }= 2867, m_{1} = 2867

The sample proportion $\displaystyle {\overline a}_1$ = $\frac{259+251+224}{3\;\times\;2867}$ = 0.0853

The sample proportion $\displaystyle {\overline a}_2$ = $\frac{247+229+222}{3\;\times\;2867}$ = 0.08115

The pooled proportion $\displaystyle \overline a$= $\begin{array}{l}\frac{(3\;\times\;n_1\;\times\;\overline{a_1}\;)\;+\;(3\;\times\;m_1\;\times\overline{a_2}\;)}{3\;\times\;(\;n_1\;+m_1)}\\\end{array}$= $\frac{734+698}{17202}$= 0.083

The value of calculated z is,

z= 2.0047

The critical value of z at 5% l.o.s. is 3.3459. Since, calculated z is less than critical value of z, it is not significant. Hence, we , may accept the null hypothesis and conclude that there is no significant difference in the distribution of polymorphic loci of chromosome 1 across SAS and EUR population.

For Chromosome 21, let b_{1} and b_{2} denote the polymorphic loci of SAS and EUR populations.

SAS | EUR |

326 | 312 |

318 | 323 |

326 | 326 |

From the data we get,

n_{2 }= 2897, m_{2} = 2897

The sample proportion $\displaystyle {\overline b}_1$ = $\frac{326+318+326}{3\;\times\;2897}$ = 0.11161

The sample proportion $\displaystyle {\overline b}_2$ = $\frac{312+323+319}{3\;\times\;2897}$ = 0.1099

The pooled proportion $\displaystyle \overline b$= $\displaystyle \begin{array}{l}\frac{(3\;\times\;n_2\;\times\;\overline{b_1}\;)\;+\;(3\;\times\;m_2\;\times\overline{b_2}\;)}{3\;\times\;(\;n_2\;+m_2)}\\\end{array}$= $\frac{970+955}{17382}$= 0.1107

The value of calculated z is,

z= 0.1107

The critical value of z at 5% l.o.s. is 3.3459. Since, calculated z is less than critical value of z, it is not significant. Hence, we, may accept the null hypothesis and conclude that there is no significant difference in the distribution of polymorphic loci of chromosome 21 across SAS and EUR population.

5) HISTOGRAM

SAS POPULATION (CHROMOSOME 1)

EUR POPULATION (CHROMOSOME 1)

SAS POPULATION (CHROMOSOME 21)

EUR POPULATION (CHROMOSOME 21)

## CONCLUSION AND RECOMMENDATIONS

The analysis of the genotype data for the different populations indicates that for a particular chromosome, the genetic structures within the South Asian populations as well as the European populations are quite similar. Analysis of the allele frequencies for each evaluated locus demonstrated that they did not differ significantly. From the results of contingency tests, we can infer that the genotype frequency distribution is independent of the geographical location of the individuals. Also, it is seen that for a particular chromosome, the distribution of alleles is almost the same across the six populations. This is because of natural selection of genes for a particular chromosome. However, if we consider the genetic structures of Chromosome 1 and 21 for each population set, there are significant differences for the two sets of populations. This can be explained by the phenomenon of natural selection that act differently in the two sets of chromosomes. These data can be further studied to trace the ancestries of these populations. However, further modifications and effective statisticalmeasures are desired for extracting more information.

## REFERENCES

1. Hartl D.L., Clark A.G. (2007): Principles of Population Genetics, Fourth Edition, Sinauer Associates, Inc. Publishers (Sunderland, Massachusetts)

2. Zheng-Bradley X., Flicek P. (2016): Applications of the 1000 Genomes Project resources

3. The 1000 Genome Project Consortium (2010): A map of human genome variation from population scale sequencing

4. Elston R., et.al (2015): Genetic Terminology

5. Das R., Upadhyai P. (2017): Application of geographic population structure (GPS) algorithm for biogeographical analysis of populations with complex ancestries: a case study of South Asians from 1000 genomes project”

http://www.usq.edu.au/library/help/referencing/apa

http://owl.english.purdue.edu/owl/resource/560/01/

## ACKNOWLEDGEMENTS

I am very thankful to the **Indian Academy of Sciences (Bangalore)**,** Indian National Science ****Academy (New Delhi)**, **National Academy of Sciences (Allahabad) **for providing me the **Focus Area Summer Research Fellowship Programme-2018 **and allowing me to work under such experienced professionals. I am highly obliged to my project guide, **Partha P. ****Majumder**, Distinguished Professor at **National Institute of Biomedical Genomics, Kalyani **for giving me the privilege to work under his supervision and guiding me throughout the project. I am also very thankful to Ms. Vijay Laxmi Ray, Ms. Srija Mukhopaddhay and Mrs. Chandrika Bhattacharya for their continuous guidance and patience.

I would take a moment to thank Bhargob Kakoty, my friend who had informed me about this fellowship. I am highly grateful to Dibyajyoti Bora, Assistant Professor at Cotton University for recommending me to apply for this fellowship. I am also thankful to Dr. Kamal Barman and Dr. Bandana Sharma, Associate Professors (Dept. of Statistics) ,who had continuously supported me to carry out this project. I have gained a lot of knowledge and experience from this summer project which will be highly useful for my future studies. I am highly thankful to Almighty God for always blessing me. I am obliged to my parents who are always by my side to help me, my friends who are my constant supporters, my fellow trainees for encouraging me for the completion of this project.

Thank you!

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