ADVANCED PROBLEMS IN SPLICING VARIANTS AND TRANSCRIPTOME ANALYSIS.

ADVANCED PROBLEMS IN SPLICING VARIANTS AND

TRANSCRIPTOME ANALYSIS

1 ALTERNATIVE SPLICING AND ISOFORM QUANTIFICATION

Problem 1. Consider a gene with three exons and two known splice variants:

In an RNA-seq experiment, you observe the following read counts:

a) Calculate the Percent Spliced In (PSI) value for Exon 2.

b) Estimate the relative abundance of each isoform.

c) Calculate the expected number of reads for each exon and junction if the total number of

reads is doubled, assuming the same relative abundances.

Solution 1. a) The Percent Spliced In (PSI) value for Exon 2 is calculated as:

PSI = Inclusion reads

Inclusion reads +Skipping reads ×100%

Inclusion reads = (Junction 1-2 + Junction 2-3) / 2 = (250 + 250) / 2 = 250 Skipping reads =

Junction 1-3 = 450

PSI = 250

Exon 1: 2000 reads Exon 2: 600 reads Exon 3: 1600 reads Junction 1-2: 500 reads Junction 2-

3: 500 reads Junction 1-3: 900 reads

Therefore: a) The PSI value for Exon 2 is 35.71%. b) The relative abundance of Isoform 1 is

35.71% and Isoform 2 is 64.29%. c) The expected read counts when doubled are as listed

above.

2 DIFFERENTIAL SPLICING ANALYSIS

Problem 2. In a study comparing splicing patterns between two conditions (A and B), you

observe the following PSI values for a particular exon across three replicates:

Condition A: 0.65, 0.70, 0.68 Condition B: 0.45, 0.50, 0.48

Assume the PSI values follow a normal distribution.

Solution 2. a) For Condition A: Mean (μA) = (0.65 + 0.70 + 0.68) / 3 = 0.6767 Standard

deviation (σA) = sqrt([(0.65 - 0.6767)² + (0.70 - 0.6767)² + (0.68 - 0.6767)²] / 2) = 0.0252

For Condition B: Mean (μB) = (0.45 + 0.50 + 0.48) / 3 = 0.4767 Standard deviation (σB) =

sqrt([(0.45 - 0.4767)² + (0.50 - 0.4767)² + (0.48 - 0.4767)²] / 2) = 0.0252

b) Performing a two-sample t-test:

t = (μA - μB) / sqrt((σA² + σB²) / 3) = (0.6767 - 0.4767) / sqrt((0.0252² + 0.0252²) / 3) =

10.0

Degrees of freedom = 2 + 2 = 4

The critical t-value for α = 0.05 and df = 4 is 2.776. Since |10.0| > 2.776, we reject the null

hypothesis.

To quantify these effects precisely, we would need to perform a power analysis and recalculate

Therefore: a) Condition A: μA = 0.6767, σA = 0.0252; Condition B: μB = 0.4767, σB = 0.0252

b) There is a significant difference in PSI values between conditions (p < 0.05) c) The effect

size (Cohen’s d) is 7.92, indicating a very large effect d) Increasing sample size would decrease

p-value and increase statistical power

3 RNA-SEQ DATA NORMALIZATION AND DIFFERENTIAL EXPRESSION

Problem 3. In an RNA-seq experiment comparing gene expression between two conditions

(Wild-type and Mutant), you observe the following read counts for three genes across two

replicates:

d) Calculate the False Discovery Rate (FDR) using the Benjamini-Hochberg procedure with α =

0.05.

Solution 3. a) To calculate DESeq2 size factors:

1. Calculate geometric mean for each gene: Gene A: (100 * 120 * 500 * 550)^(1/4) ≈ 234.38

Gene B: (2000 * 2200 * 1800 * 2000)^(1/4) ≈ 1994.99 Gene C: (50 * 60 * 40 * 45)^(1/4) ≈

48.33

2. Calculate ratios and take median for each sample: WT Rep1: median([100/234.38,

2000/1994.99, 50/48.33]) ≈ 1.0025 WT Rep2: median([120/234.38, 2200/1994.99, 60/48.33])

≈ 1.1025 Mut Rep1: median([500/234.38, 1800/1994.99, 40/48.33]) ≈ 0.9023 Mut Rep2:

median([550/234.38, 2000/1994.99, 45/48.33]) ≈ 1.0025

c) Log2 fold changes and statistical tests:

Gene A: log2FC ≈ 2.43, p-value < 0.001 (significant) Gene B: log2FC ≈ -0.02, p-value > 0.05

(not significant) Gene C: log2FC ≈ -0.28, p-value > 0.05 (not significant)

d) Benjamini-Hochberg procedure:

Rank p-values: 1. Gene A: < 0.001 2. Gene B: > 0.05 3. Gene C: > 0.05

Adjust p-values: Gene A: < 0.003 (significant at FDR < 0.05) Gene B: > 0.05 (not significant)

Gene C: > 0.05 (not significant)

Therefore: a) Size factors: WT Rep1 ≈ 1.0025, WT Rep2 ≈ 1.1025, Mut Rep1 ≈ 0.9023, Mut

Rep2 ≈ 1.0025 b) Normalized counts are as shown in the table above c) Gene A is significantly

differentially expressed (upregulated in Mutant) d) After FDR correction, only Gene A remains

Assume all exons are 100 bp long.

a) Calculate the Percent Spliced In (PSI) value for Exon 2.

b) Estimate the relative abundance of each isoform.

c) Calculate the expected number of reads for each exon and junction if the total number of

reads is doubled, assuming the same relative abundances.

250 +450 ×100% =35.71%

b) To estimate the relative abundance of each isoform, we can use the junction reads:

Isoform 1 (1-2-3): 250 reads (Junction 1-2 or 2-3) Isoform 2 (1-3): 450 reads (Junction 1-3)

Total: 250 + 450 = 700 reads

Relative abundance: Isoform 1: 250 / 700 = 35.71% Isoform 2: 450 / 700 = 64.29%

c) If the total number of reads is doubled, we expect:

Exon 1: 2000 reads Exon 2: 600 reads Exon 3: 1600 reads Junction 1-2: 500 reads Junction 2-

3: 500 reads Junction 1-3: 900 reads

Therefore: a) The PSI value for Exon 2 is 35.71%. b) The relative abundance of Isoform 1 is

35.71% and Isoform 2 is 64.29%. c) The expected read counts when doubled are as listed

above.

b) Perform a t-test to determine if there is a significant difference in PSI values between

conditions. Use α = 0.05.

c) Calculate the effect size (Cohen’s d) for the difference in PSI between conditions.

d) If you increase the sample size to 10 replicates per condition while maintaining the same

mean and standard deviation, how would this affect the p-value and statistical power?

Solution 2. a) For Condition A: Mean (μA) = (0.65 + 0.70 + 0.68) / 3 = 0.6767 Standard

deviation (σA) = sqrt([(0.65 - 0.6767)² + (0.70 - 0.6767)² + (0.68 - 0.6767)²] / 2) = 0.0252

For Condition B: Mean (μB) = (0.45 + 0.50 + 0.48) / 3 = 0.4767 Standard deviation (σB) =

sqrt([(0.45 - 0.4767)² + (0.50 - 0.4767)² + (0.48 - 0.4767)²] / 2) = 0.0252

b) Performing a two-sample t-test:

Degrees of freedom = 2 + 2 = 4

The critical t-value for α = 0.05 and df = 4 is 2.776. Since |10.0| > 2.776, we reject the null

hypothesis.

p-value < 0.05 (in fact, p < 0.001)

c) Cohen’s d = (μA - μB) / sqrt((σA² + σB²) / 2) = (0.6767 - 0.4767) / sqrt((0.0252² +

b) Normalized counts:

c) Log2 fold changes and statistical tests:

Assume all exons are 100 bp long.

b) To estimate the relative abundance of each isoform, we can use the junction reads:

observe the following PSI values for a particular exon across three replicates:

d) If you increase the sample size to 10 replicates per condition while maintaining the same

mean and standard deviation, how would this affect the p-value and statistical power?

Solution 2. a) For Condition A: Mean (μA) = (0.65 + 0.70 + 0.68) / 3 = 0.6767 Standard

deviation (σA) = sqrt([(0.65 - 0.6767)² + (0.70 - 0.6767)² + (0.68 - 0.6767)²] / 2) = 0.0252

For Condition B: Mean (μB) = (0.45 + 0.50 + 0.48) / 3 = 0.4767 Standard deviation (σB) =

To quantify these effects precisely, we would need to perform a power analysis and recalculate

Therefore: a) Condition A: μA = 0.6767, σA = 0.0252; Condition B: μB = 0.4767, σB = 0.0252

b) There is a significant difference in PSI values between conditions (p < 0.05) c) The effect

size (Cohen’s d) is 7.92, indicating a very large effect d) Increasing sample size would decrease

p-value and increase statistical power

Problem 3. In an RNA-seq experiment comparing gene expression between two conditions

(Wild-type and Mutant), you observe the following read counts for three genes across two

replicates:

d) Calculate the False Discovery Rate (FDR) using the Benjamini-Hochberg procedure with α =

0.05.

Solution 3. a) To calculate DESeq2 size factors:

1. Calculate geometric mean for each gene: Gene A: (100 * 120 * 500 * 550)^(1/4) ≈ 234.38

Gene B: (2000 * 2200 * 1800 * 2000)^(1/4) ≈ 1994.99 Gene C: (50 * 60 * 40 * 45)^(1/4) ≈

48.33

2. Calculate ratios and take median for each sample: WT Rep1: median([100/234.38,

2000/1994.99, 50/48.33]) ≈ 1.0025 WT Rep2: median([120/234.38, 2200/1994.99, 60/48.33])

≈ 1.1025 Mut Rep1: median([500/234.38, 1800/1994.99, 40/48.33]) ≈ 0.9023 Mut Rep2:

median([550/234.38, 2000/1994.99, 45/48.33]) ≈ 1.0025

c) Log2 fold changes and statistical tests:

Gene A: log2FC ≈ 2.43, p-value < 0.001 (significant) Gene B: log2FC ≈ -0.02, p-value > 0.05

(not significant) Gene C: log2FC ≈ -0.28, p-value > 0.05 (not significant)

d) Benjamini-Hochberg procedure:

Rank p-values: 1. Gene A: < 0.001 2. Gene B: > 0.05 3. Gene C: > 0.05

Adjust p-values: Gene A: < 0.003 (significant at FDR < 0.05) Gene B: > 0.05 (not significant)

Gene C: > 0.05 (not significant)

Therefore: a) Size factors: WT Rep1 ≈ 1.0025, WT Rep2 ≈ 1.1025, Mut Rep1 ≈ 0.9023, Mut

Rep2 ≈ 1.0025 b) Normalized counts are as shown in the table above c) Gene A is significantly

differentially expressed (upregulated in Mutant) d) After FDR correction, only Gene A remains

Assume all exons are 100 bp long.

a) Calculate the Percent Spliced In (PSI) value for Exon 2.

b) Estimate the relative abundance of each isoform.

c) Calculate the expected number of reads for each exon and junction if the total number of

reads is doubled, assuming the same relative abundances.

250 +450 ×100% =35.71%

b) To estimate the relative abundance of each isoform, we can use the junction reads:

Isoform 1 (1-2-3): 250 reads (Junction 1-2 or 2-3) Isoform 2 (1-3): 450 reads (Junction 1-3)

Total: 250 + 450 = 700 reads

Relative abundance: Isoform 1: 250 / 700 = 35.71% Isoform 2: 450 / 700 = 64.29%

c) If the total number of reads is doubled, we expect:

Exon 1: 2000 reads Exon 2: 600 reads Exon 3: 1600 reads Junction 1-2: 500 reads Junction 2-

3: 500 reads Junction 1-3: 900 reads

Therefore: a) The PSI value for Exon 2 is 35.71%. b) The relative abundance of Isoform 1 is

35.71% and Isoform 2 is 64.29%. c) The expected read counts when doubled are as listed

above.

b) Perform a t-test to determine if there is a significant difference in PSI values between

conditions. Use α = 0.05.

c) Calculate the effect size (Cohen’s d) for the difference in PSI between conditions.

d) If you increase the sample size to 10 replicates per condition while maintaining the same

mean and standard deviation, how would this affect the p-value and statistical power?

Solution 2. a) For Condition A: Mean (μA) = (0.65 + 0.70 + 0.68) / 3 = 0.6767 Standard

deviation (σA) = sqrt([(0.65 - 0.6767)² + (0.70 - 0.6767)² + (0.68 - 0.6767)²] / 2) = 0.0252

For Condition B: Mean (μB) = (0.45 + 0.50 + 0.48) / 3 = 0.4767 Standard deviation (σB) =

sqrt([(0.45 - 0.4767)² + (0.50 - 0.4767)² + (0.48 - 0.4767)²] / 2) = 0.0252

b) Performing a two-sample t-test:

Degrees of freedom = 2 + 2 = 4

The critical t-value for α = 0.05 and df = 4 is 2.776. Since |10.0| > 2.776, we reject the null

hypothesis.

p-value < 0.05 (in fact, p < 0.001)

c) Cohen’s d = (μA - μB) / sqrt((σA² + σB²) / 2) = (0.6767 - 0.4767) / sqrt((0.0252² +

b) Normalized counts:

c) Log2 fold changes and statistical tests:

Assume all exons are 100 bp long.

b) To estimate the relative abundance of each isoform, we can use the junction reads:

observe the following PSI values for a particular exon across three replicates:

d) If you increase the sample size to 10 replicates per condition while maintaining the same

mean and standard deviation, how would this affect the p-value and statistical power?

Solution 2. a) For Condition A: Mean (μA) = (0.65 + 0.70 + 0.68) / 3 = 0.6767 Standard

deviation (σA) = sqrt([(0.65 - 0.6767)² + (0.70 - 0.6767)² + (0.68 - 0.6767)²] / 2) = 0.0252

For Condition B: Mean (μB) = (0.45 + 0.50 + 0.48) / 3 = 0.4767 Standard deviation (σB) =

b) Normalize the read counts using the size factors.

each gene. Assume a negative binomial distribution for read counts.

0.05.

Solution 3. a) To calculate DESeq2 size factors:

1. Calculate geometric mean for each gene: Gene A: (100 * 120 * 500 * 550)^(1/4) ≈ 234.38

2000/1994.99, 50/48.33]) ≈ 1.0025 WT Rep2: median([120/234.38, 2200/1994.99, 60/48.33])

≈ 1.1025 Mut Rep1: median([500/234.38, 1800/1994.99, 40/48.33]) ≈ 0.9023 Mut Rep2:

median([550/234.38, 2000/1994.99, 45/48.33]) ≈ 1.0025

b) Normalized counts:

Gene A: log2FC ≈ 2.43, p-value < 0.001 (significant) Gene B: log2FC ≈ -0.02, p-value > 0.05

Rep2 ≈ 1.0025 b) Normalized counts are as shown in the table above c) Gene A is significantly

differentially expressed (upregulated in Mutant) d) After FDR correction, only Gene A remains

significant at α = 0.05

Solution 1. a) The Percent Spliced In (PSI) value for Exon 2 is calculated as:

PSI = Inclusion reads

Inclusion reads +Skipping reads ×100%

Inclusion reads = (Junction 1-2 + Junction 2-3) / 2 = (250 + 250) / 2 = 250 Skipping reads =

Junction 1-3 = 450

PSI = 250

250 +450 ×100% =35.71%

b) To estimate the relative abundance of each isoform, we can use the junction reads:

Isoform 1 (1-2-3): 250 reads (Junction 1-2 or 2-3) Isoform 2 (1-3): 450 reads (Junction 1-3)

Total: 250 + 450 = 700 reads

above.

observe the following PSI values for a particular exon across three replicates:

Condition A: 0.65, 0.70, 0.68 Condition B: 0.45, 0.50, 0.48

Assume the PSI values follow a normal distribution.

a) Calculate the mean and standard deviation of PSI values for each condition.

b) Perform a t-test to determine if there is a significant difference in PSI values between

conditions. Use α = 0.05.

c) Calculate the effect size (Cohen’s d) for the difference in PSI between conditions.

d) If you increase the sample size to 10 replicates per condition while maintaining the same

mean and standard deviation, how would this affect the p-value and statistical power?

deviation (σA) = sqrt([(0.65 - 0.6767)² + (0.70 - 0.6767)² + (0.68 - 0.6767)²] / 2) = 0.0252

sqrt([(0.45 - 0.4767)² + (0.50 - 0.4767)² + (0.48 - 0.4767)²] / 2) = 0.0252

Degrees of freedom = 2 + 2 = 4

The critical t-value for α = 0.05 and df = 4 is 2.776. Since |10.0| > 2.776, we reject the null

hypothesis.

p-value < 0.05 (in fact, p < 0.001)

c) Cohen’s d = (μA - μB) / sqrt((σA² + σB²) / 2) = (0.6767 - 0.4767) / sqrt((0.0252² +

0.0252²) / 2) = 7.92

This indicates a very large effect size.

d) Increasing the sample size to 10 replicates per condition would: - Decrease the p-value

further - Increase the statistical power - Narrow the confidence intervals

Degrees of freedom = 2 + 2 = 4

The critical t-value for α = 0.05 and df = 4 is 2.776. Since |10.0| > 2.776, we reject the null

hypothesis.

p-value < 0.05 (in fact, p < 0.001)

c) Cohen’s d = (μA - μB) / sqrt((σA² + σB²) / 2) = (0.6767 - 0.4767) / sqrt((0.0252² +

b) Normalized counts:

c) Log2 fold changes and statistical tests:

Assume all exons are 100 bp long.

b) To estimate the relative abundance of each isoform, we can use the junction reads:

observe the following PSI values for a particular exon across three replicates:

d) If you increase the sample size to 10 replicates per condition while maintaining the same

mean and standard deviation, how would this affect the p-value and statistical power?

Solution 2. a) For Condition A: Mean (μA) = (0.65 + 0.70 + 0.68) / 3 = 0.6767 Standard

deviation (σA) = sqrt([(0.65 - 0.6767)² + (0.70 - 0.6767)² + (0.68 - 0.6767)²] / 2) = 0.0252

For Condition B: Mean (μB) = (0.45 + 0.50 + 0.48) / 3 = 0.4767 Standard deviation (σB) =

b) Normalize the read counts using the size factors.

each gene. Assume a negative binomial distribution for read counts.

0.05.

Solution 3. a) To calculate DESeq2 size factors:

1. Calculate geometric mean for each gene: Gene A: (100 * 120 * 500 * 550)^(1/4) ≈ 234.38

2000/1994.99, 50/48.33]) ≈ 1.0025 WT Rep2: median([120/234.38, 2200/1994.99, 60/48.33])

≈ 1.1025 Mut Rep1: median([500/234.38, 1800/1994.99, 40/48.33]) ≈ 0.9023 Mut Rep2:

median([550/234.38, 2000/1994.99, 45/48.33]) ≈ 1.0025

b) Normalized counts:

Gene A: log2FC ≈ 2.43, p-value < 0.001 (significant) Gene B: log2FC ≈ -0.02, p-value > 0.05

Rep2 ≈ 1.0025 b) Normalized counts are as shown in the table above c) Gene A is significantly

differentially expressed (upregulated in Mutant) d) After FDR correction, only Gene A remains

significant at α = 0.05

Solution 1. a) The Percent Spliced In (PSI) value for Exon 2 is calculated as:

PSI = Inclusion reads

Inclusion reads +Skipping reads ×100%

Inclusion reads = (Junction 1-2 + Junction 2-3) / 2 = (250 + 250) / 2 = 250 Skipping reads =

Junction 1-3 = 450

PSI = 250

250 +450 ×100% =35.71%

b) To estimate the relative abundance of each isoform, we can use the junction reads:

Isoform 1 (1-2-3): 250 reads (Junction 1-2 or 2-3) Isoform 2 (1-3): 450 reads (Junction 1-3)

Total: 250 + 450 = 700 reads

above.

observe the following PSI values for a particular exon across three replicates:

Condition A: 0.65, 0.70, 0.68 Condition B: 0.45, 0.50, 0.48

Assume the PSI values follow a normal distribution.

a) Calculate the mean and standard deviation of PSI values for each condition.

b) Perform a t-test to determine if there is a significant difference in PSI values between

conditions. Use α = 0.05.

c) Calculate the effect size (Cohen’s d) for the difference in PSI between conditions.

d) If you increase the sample size to 10 replicates per condition while maintaining the same

mean and standard deviation, how would this affect the p-value and statistical power?

deviation (σA) = sqrt([(0.65 - 0.6767)² + (0.70 - 0.6767)² + (0.68 - 0.6767)²] / 2) = 0.0252

sqrt([(0.45 - 0.4767)² + (0.50 - 0.4767)² + (0.48 - 0.4767)²] / 2) = 0.0252

Degrees of freedom = 2 + 2 = 4

The critical t-value for α = 0.05 and df = 4 is 2.776. Since |10.0| > 2.776, we reject the null

hypothesis.

p-value < 0.05 (in fact, p < 0.001)

c) Cohen’s d = (μA - μB) / sqrt((σA² + σB²) / 2) = (0.6767 - 0.4767) / sqrt((0.0252² +

0.0252²) / 2) = 7.92

This indicates a very large effect size.

d) Increasing the sample size to 10 replicates per condition would: - Decrease the p-value

further - Increase the statistical power - Narrow the confidence intervals

(Wild-type and Mutant), you observe the following read counts for three genes across two

replicates:

a) Calculate the DESeq2 size factors for each sample.

b) Normalize the read counts using the size factors.

c) Estimate the log2 fold change and perform a statistical test for differential expression for

each gene. Assume a negative binomial distribution for read counts.

d) Calculate the False Discovery Rate (FDR) using the Benjamini-Hochberg procedure with α =

0.05.

Solution 3. a) To calculate DESeq2 size factors:

1. Calculate geometric mean for each gene: Gene A: (100 * 120 * 500 * 550)^(1/4) ≈ 234.38

Gene B: (2000 * 2200 * 1800 * 2000)^(1/4) ≈ 1994.99 Gene C: (50 * 60 * 40 * 45)^(1/4) ≈

c) Log2 fold changes and statistical tests:

Gene A: log2FC ≈ 2.43, p-value < 0.001 (significant) Gene B: log2FC ≈ -0.02, p-value > 0.05

(not significant) Gene C: log2FC ≈ -0.28, p-value > 0.05 (not significant)

d) Benjamini-Hochberg procedure:

Rank p-values: 1. Gene A: < 0.001 2. Gene B: > 0.05 3. Gene C: > 0.05

Therefore: a) Size factors: WT Rep1 ≈ 1.0025, WT Rep2 ≈ 1.1025, Mut Rep1 ≈ 0.9023, Mut

In an RNA-seq experiment, you observe the following read counts:

Assume all exons are 100 bp long.

a) Calculate the Percent Spliced In (PSI) value for Exon 2.

b) Estimate the relative abundance of each isoform.

c) Calculate the expected number of reads for each exon and junction if the total number of

reads is doubled, assuming the same relative abundances.

Solution 1. a) The Percent Spliced In (PSI) value for Exon 2 is calculated as:

PSI = Inclusion reads

Inclusion reads +Skipping reads ×100%