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.

Homework Answers

Answer #1

Answer(1):

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

Answer(2):

The overall statistic can be obtained by adding the individual statistic

Answer(3):

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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