text: user9709: I have these Ct data from qPCR for a series of samples A-H, I want to quantify the RNA copies of my samples using R:
  Sample_ID   Replicate1_Ct Replicate2_Ct
  A           22.92         24.21           
  B           24.29         22.16           
  C           24.29         22.16           
  D           27.91         23.76           
  E           21.44         22.81           
  F           16.50         18.31           
  G           17.79         16.56                 
  H           17.34         19.09 
  
  In addition, I have measured a standard (STD) curve of copy number of the RNA in question, that I want to use as a reference:
  STD_Ct      STD_copies      Log10(STD_copies)
  33          4.00E+03        3
  28          4.00E+04        4
  24          4.00E+05        5
  20          4.00E+06        6
  15          4.00E+07        7
  12          4.00E+08        8
  
  
  It seems
  This the original data form:
  
  
  user2108: Now that we've straightened out the data, it seems like the simplest solution is to just linearly interpolate the copy number from your standard curve.
  When I read your standard curve (std) and your sample data (ct) into R, I see that the log10(Ct) fits the copy number fairly well, whereas the linear is a bit harder:
  
  So we can fit a model to predict copy number based on Ct on the STD data:
  > fit = lm(std$Log10.STD_copies. ~ std$STD_Ct)
  > fit
  
  Call:
  lm(formula = std$Log10.STD_copies. ~ std$STD_Ct)
  
  Coefficients:
  (Intercept)   std$STD_Ct  
      10.6847      -0.2357  
  
  # we plug those numbers into the linear equation,
  # taking the mean of the two replicates for a best estimate.
  
  > rowMeans(ct[,2:3]) * -.2357 + 10.6847
  [1] 5.130429 5.210567 5.210567 4.595390 5.469837 6.582341
  [7] 6.636553 6.391424
  
  
  So those are log10(copy number) estimates, of a sort. You can get much more elaborate, of course.
  
original_text: 9709: <p>I have these Ct data from qPCR for a series of samples A-H, I want to quantify the RNA copies of my samples using R:</p>
  <pre><code>Sample_ID   Replicate1_Ct Replicate2_Ct
  A           22.92         24.21           
  B           24.29         22.16           
  C           24.29         22.16           
  D           27.91         23.76           
  E           21.44         22.81           
  F           16.50         18.31           
  G           17.79         16.56                 
  H           17.34         19.09 
  </code></pre>
  <p>In addition, I have measured a standard (STD) curve of copy number of the RNA in question, that I want to use as a reference:</p>
  <pre><code>STD_Ct      STD_copies      Log10(STD_copies)
  33          4.00E+03        3
  28          4.00E+04        4
  24          4.00E+05        5
  20          4.00E+06        6
  15          4.00E+07        7
  12          4.00E+08        8
  
  </code></pre>
  <p>It seems</p>
  <p>This the original data form:
  <a href="https://i.stack.imgur.com/amPJv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/amPJv.png" alt="enter image description here" /></a></p>
   2108: <p>Now that we've straightened out the data, it seems like the simplest solution is to just linearly interpolate the copy number from your standard curve.</p>
  <p>When I read your standard curve (<code>std</code>) and your sample data (<code>ct</code>) into R, I see that the log10(Ct) fits the copy number fairly well, whereas the linear is a bit harder:</p>
  <p><a href="https://i.stack.imgur.com/2dhUM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2dhUM.png" alt="enter image description here" /></a></p>
  <p>So we can fit a model to predict copy number based on Ct on the STD data:</p>
  <pre><code>&gt; fit = lm(std<span class="math-container">$Log10.STD_copies. ~ std$</span>STD_Ct)
  &gt; fit
  
  Call:
  lm(formula = std<span class="math-container">$Log10.STD_copies. ~ std$</span>STD_Ct)
  
  Coefficients:
  (Intercept)   std$STD_Ct  
      10.6847      -0.2357  
  
  # we plug those numbers into the linear equation,
  # taking the mean of the two replicates for a best estimate.
  
  &gt; rowMeans(ct[,2:3]) * -.2357 + 10.6847
  [1] 5.130429 5.210567 5.210567 4.595390 5.469837 6.582341
  [7] 6.636553 6.391424
  
  </code></pre>
  <p>So those are log10(copy number) estimates, of a sort. You can get much more elaborate, of course.</p>
  