Question: Trouble understanding (2^Delta CT). Why ZEROES? Also, using them as a feature.
gravatar for galapacheco
6 months ago by
galapacheco0 wrote:

Good evening my fellow bioinformatics/staticians!

I have some data of a RNA expression in an excel, a very few little samples where we could get the $2^-Delta CT$ values for each one of my target RNA.

I want to use these values as a feature in a multidimensional analysis, probably PCA using NIPAL, i havent decide it yet since i have just about 6 controls and 7 cases for one of my RNA, and 17Control 21 Cases for another; and i have stumbled upon some intriguin values:


I have yet to ask the person who created that data, cause i dont see how $2^-Delta CT$ can be zero. ¿Any ideas? I have some theories.

  1. Its an Error, and its the DeltaCT =Ct_target - Ct_reference= 0; so 2^0 = 1. There is no Diference between the expresion of the target and the house reference gene.
  2. We have no expression of the target gene; so its 0; and there is no $2^-Delta CT$ applied, its just 0.
  3. ¿Maybe we just have to use the zero? Because. Dont know why tho.

Cause if we got 2^-Delta CT; lower the value, more the expresion of our target RNA. ¿Right?

Example: TargetC1T = 50 ; RefCT = 10; Delta CT =50-10=40 ; -DeltaCT = -40.

2^-40 = 9.094947e-13

TargetC1T = 40 ; RefCT = 10; Delta CT =40-10=30 ; -DeltaCT = -30.

2^-30 = 9.313226e-10

So 9.09 e-13 < 9.31 e-10 ; its inverse to the CT ; CT1 > CT2.

¿So, what we do when we got that 0? It would mean that there is an infinite concentration? Because i cant quite understand if a zero makes sense.

So, what do you think? ¿Maybe i should use pseudocounts like $log_2(1+^2^{-Delta CT})$ so i have no 0 (therefore no Inf) values?

Once i solve that, i pretend to use these values as a feature (gene expression) in a multidimensional data with a lot, and i mean a lot, of clinical variables. Yet i have to sort wich ones whould i use, and if a logistic multinomial model could be applied (Since the low n, and the n

Thank you for your time.

rna-seq deltact R • 275 views
ADD COMMENTlink modified 6 months ago • written 6 months ago by galapacheco0
gravatar for Kevin Blighe
6 months ago by
Kevin Blighe69k
Republic of Ireland
Kevin Blighe69k wrote:

I think that a situation like this can occur when a gene is set to the equivalent of NA, which may be cycle number 40 depending on which lab you're in. Re-using my example from here ( A: How to report and plot qPCR data ), I set the Ct values for my target gene, Gene1, to 40 across my 3 replicates:

                  Sample1_Rep1 Sample1_Rep2 Sample1_Rep3 | Control_DNA1 Control_DNA2 Control_DNA3
Gene1             40           40           40           | 21           21           20
HousekeeperGene   22           22           22           | 21.5         21           22

[1] Gene1, Delta Ct in Sample1:

((40+40+40) / 3) - ((22+22+22) / 3) = 18

[2] Gene1, Delta Ct in Control DNA:

((21+21+20) / 3] - [(21.5+21+22) / 3) = -0.8333

[3] Gene1, Delta Delta Ct:

18 - (-0.8333) = 18.8333

[4] fold increase / decrease:

2 ^ (- ddCt) = 2 ^ (- 18.8333) = 0.000002140976

If you round that up, it's virtually 0 (zero).

I would confirm with the person who did the calculations, though.


ADD COMMENTlink modified 6 months ago • written 6 months ago by Kevin Blighe69k

That would make sense, but i have really low values like 0,00000001. Although we could just have more cycles, therefore the data should be able to have such lower values. When i got the answer from the person who did the calculations im going to post it here. Thanks, Kevin Blighe

Btw, with that explanation, i understood the CT backwards, right? The lower the CT the lower the quantity (more cicles are needed to get a signal?)?

ADD REPLYlink modified 6 months ago • written 6 months ago by galapacheco0

A lower Ct value actually indicates higher quantity of starting DNA target template. If the starting amount is so small, however, such that it never passes the threshold, then it is either assigned NA or the last amplification cycle on the instrument (~40)

ADD REPLYlink written 6 months ago by Kevin Blighe69k

¿Ct or Delta Ct or Delta Delta Ct? Because if we assign the last amplification cycle on the instrument, we obtain the lowest value, therefore the max concentration posible. (Thats what i understand)

ADD REPLYlink written 6 months ago by galapacheco0

No, if a sample is assigned the last Ct because it never reached the detection threshold, it can be inferred that the DNA template is small or non-exstent [or allude to some error].

Each DNA template is amplified via PCR in each cycle - the amount of DNA increases exponentially wth each cycle. So, the earlier the cycle threshold [Ct] at which a sample reaches detection threshold, the greater the initial starting DNA.

ADD REPLYlink written 6 months ago by Kevin Blighe69k

Allright, So 2^-30 , has more DNA that 2^-40; wich has a direct relation to its value 2^-30 = 9.3132257e-10 ; 2^-40 = 9.9.094947e-13; 2^-30 >> 2^-40 by roughly 10^3 times more. Now i dont know why i had so much trouble understanding this, i think my brain saw the negative exponent and tough "negative values" instead of fractions :D. ¡Thanks a lot Kevin!

ADD REPLYlink written 6 months ago by galapacheco0

No hay de que, pero, solo para esclarecer, el numero de Ct 30 sí representa más ADN que Ct 40. Entonces, el delta Ct, dependiendo del punto da vista, poderla ser -10 o +10.

                     Cyclic threshold [Ct]
0 .... 5 .... 10 .... 15 .... 20 .... 25 .... 30 .... 35 .... 40
            <--- Más ADN     ||||     Menos ADN --->
ADD REPLYlink written 6 months ago by Kevin Blighe69k
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