You conduct a Case/Control study comparing the frequency of a single SNP (with alleles C and T) among individuals who have developed high blood pressure as compared with control individuals of a similar age who show no sign of this condition. You obtain the following results:
Case | Controls | |
C | 1025 | 725 |
T | 902 | 922 |
a. | Does a statistically significant association exist between this SNP and the risk of developing high blood pressure, if this is the only SNP that you test? |
b. | Would this association still be significant if this was one of a million SNPs that you tested in your study? Explain. |
c. | Which allele (C or T) is associated with a higher risk of developing high blood pressure? |
d. | Calculate the allelic odds ratio for the risk allele. What does the allelic odds ratio mean about a person’s chance of developing high blood pressure? |
a.
To determine:
The existence of any statistical association between the SNP and the risks of developing hypertension, in case only the one SNP under consideration is being tested.
Introduction:
Conclusions of genetic studies are drawn based on statistical evaluations. Chi-square test is one of the most commonly used statistical tools that are used for the analysis of the obtained data.
Explanation of Solution
SNPs act as tools that can be used for the identification of various genetic and non-genetic disease at gene levels.
In the given case, the significance of the provided data is tested using the chi-square tests. For this, the null and the alternate hypothesis are:
The null hypothesis: there is no independence between the SNP and the risk of development of hypertension.
The alternate hypothesis: the SNP and the risk of development of hypertension are independent of each other.
Assuming that the level of significance for the case is
By using the above formula for the given data, the obtained results are as follows:
Cases | Controls | Total | ||
C | Observed | 1025 | 725 | 1750 |
Expected | 943.55 | 806.45 | 1750.00 | |
81.45 | -81.45 | 0.00 | ||
(O – E)2/E | 7.03 | 8.23 | 15.26 | |
T | Observed | 902 | 922 | 1824 |
Expected | 983.45 | 806.45 | 1824.00 | |
-81.45 | 81.45 | 0.00 | ||
(O – E)2/E | 6.75 | 7.89 | 14.64 | |
Total | Observed | 1927 | 1647 | 3574 |
Expected | 1927.00 | 1647.00 | 3574 | |
0.00 | 0.00 | 0.00 | ||
(O – E)2/E | 13.78 | 16.12 | 29.90 |
In the given case:
It can be seen that the p-value is lesser than the significance level of 0.05. This shows that the null hypothesis is rejected, and there is no dependence on the risk of development of hypertension on the SNP under consideration.
Hence, there exists a statistically significant association between the SNP and the risk of development of hypertension.
b.
To determine:
The significance of the association in case the SNP under consideration is amongst the million SNPs that were tested.
Introduction:
SNPs (also called single nucleotide polymorphism) are the most common type of genetic variations that can be \found in different individuals. SNPs represent the locations in the genome of an organism where there is a change of a single nucleotide between two individuals of the same species.
Explanation of Solution
In cases of the SNP under consideration being amongst the millions of SNPs that are being tested, even then the association between the SNP and the risk of development of hypertension will be significant. It is so because the one significant SNP causing hypertension will be a part of approximately 3 billion other base pairs.
c.
To determine:
The allele (C or T) that is associated with a high risk of developing hypertension.
Introduction:
Different diseases are associated with certain base-pairs of a particular gene. Activation of inactivation of a certain gene can cause the disease in different individuals. In such cases, genetic therapies are adopted to cure or treat the patient.
Explanation of Solution
To know the risk caused by each of the alleles, it can be done by calculating the level of the proportion of each allele. Level of the proportion of C and T are as follows:
Allele | C |
Cases | 1025 |
Controls | 725 |
Total | 1750 |
Level of proportion | 0.59 |
This gives a level of the proportion of allele C as 0.59
Allele | T |
Cases | 902 |
Controls | 922 |
Total | 1824 |
Level of proportion | 0.49 |
This gives a level of the proportion of allele T as 0.49
Level of the proportion of allele C (0.59) is greater than the level of the proportion of allele T (0.49). Hence, high risk of hypertension is associated with the allele C
d.
To determine:
The allelic odds ratio for these alleles and the information that allelic odd ratio gives about the chance of a person developing hypertension.
Introduction:
The odd allelic ratio is the indication of the association of a particular allele of a gene to the chances of development of the disease caused by that particular gene. Comparative study of two individuals carrying the different alleles for the same gene is done to find the allelic odd ratios.
Explanation of Solution
Allelic odd ratios can be calculated using the formula:
Substituing up the values form the given data in the above formula:
This gives allelic odd ratio for C to be 0.70
The total odd ratio can be calculated by the formula:
Based on the above calculation, the results and the information about a person developing high blood pressure is represented as follows:
Odd ratio | 1.44 |
95% CI | 1.266 to 1.649 |
Percentage statistics | 5.46 |
Significance level | P < 0.0001 |
Thus from the above table, it can be concluded that the person has a very less chance of developing the problem of hypertension.
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