Question

# You have exposed some algae to a high-salinity environment and used high-throughput sequencing to measure gene...

You have exposed some algae to a high-salinity environment and used high-throughput sequencing to measure gene expression levels. You have a list of 50 genes that are known to be involved in helping cells respond to high salinity and want to know whether more of those genes show a change in their expression (i.e., are either up-regulated or down-regulated). To do so, you can compare the observed number of up-regulated genes from that list to the number that might be expected.

 Number observed Number Expected Chi-square statistic Increase in expression 15 16.67 No change in expression 15 16.67 Decrease in expression 20 16.67

1-

onsider a case where you expect the 50 genes to be equally likely to increase, decrease, or show no change in their expression levels. Calculate the chi-square statistics for each set of genes and enter it below.

 Number observed Number Expected Chi-square statistic Increase in expression 15 16.67 Answer No change in expression 15 16.67 Answer Decrease in expression 20 16.67 Answer

2-What is the overall chi-squared statistic?

3-How many degrees of freedom are there?

4-Calculate the probability of the chi-squared statistic to 2 decimal places & would you reject the null hypothesis that these observed counts are no different from expected? Explain why in one or two sentences.

H0: observed counts are no different from expected

H1: observed counts are significantly different from expected

 Number observed Number Expected Chi-square statistic=(o-e)2/e Increase in expression 15 16.67 0.16730054 No change in expression 15 16.67 0.16730054 Decrease in expression 20 16.67 0.66520096

The overall statistic can be obtained by adding the individual statistic

We have total 3 observations, i.e. n=3

Hence the degrees of freedom =n-1 = 3-1 = 2

Answer(4): The p-value for above test is 0.61

The p-value is greater than α=0.05 which indicates that we do not have enough evidence against null hypothesis to reject it, so we fail to reject the null hypothesis and we can conclude that these observed counts are no different from expected.

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