Q6

How does the amount of antibody influence the number of counts precipitated?

The number of counts increases with increasing antibody concentration until the antigen-binding capacity of antibodies exceed the amount of labelled antigen (when Ab dilution is 10³). In this case, additional unlabelled antigens will just bind free antibodies without interfering with those bound to labelled antigens.

Q7

See Figure 4.1.

Figure 4.1: Q7

Q8

See Figure 4.2

Figure 4.2: Q8

Q9

Calculations see Figure 4.3

Figure 4.3: Q9

Explain the possible causes of the variation, bearing in mind the experiments have been carried out by an experienced operator

• Radioactive decay is a probablistic event, so the number of disintegrations per unit time naturally varies.
• There can also be random errors intrintic to the Geiger counter

Calibration Curve

I extracted experiments with known amount of antigen and calculated the mass of antigen in pg:

known <- raw %>%
  filter(known == 1, # experiments with known antigen amount for calibration
         antibody_dilution == 10000) %>%
  mutate(antigen_amount = 
        antigen_volume / 1000 # \mu L to ml
                       * 10^9  # mg to pg
                       / antigen_dilution) %>%
  filter(antigen_amount<=1000)

First, I plotted count vs antigen amount directly (the blue line is fitted using the LOESS algorithm) in Figure 5.2 . Although this plot is not the most useful for calibration, it shows that the random variation is large when the count is greater than 20000. Therefore I aim at a count around 20000 in the test for the unkown antigen concentration.

known %>%
  ggplot(aes(antigen_amount, count))+
  geom_point()+
  geom_smooth(span=0.9, se=FALSE)+
  labs(x = 'antigen amount (pg)', title = 'Calibration Curve (cpm vs antigen amount)')+
  theme_bw()

Figure 5.2: RIA Calibration Curve

The real calibration curve is shown in Figure 5.3 (The reasoning was described in Q3 and Q4).

known %>% 
  ggplot(aes(antigen_amount, 10000/count))+
  geom_point()+
  geom_smooth(span=0.9)+
  labs(x = 'antigen amount (pg)', title = 'Calibration Curve (10000/cpm vs antigen amount)')+
  theme_bw()

Figure 5.3: RIA Calibration Curve (10000/count vs antigen amount)

The equation of the fitted straight line can be calculated as follows:

with(known, summary(lm(10000/count ~ antigen_amount)))

## 
## Call:
## lm(formula = 10000/count ~ antigen_amount)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.33043 -0.05604  0.00116  0.05493  0.39587 
## 
## Coefficients:
##                 Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    2.591e-01  1.659e-02   15.62   <2e-16 ***
## antigen_amount 2.486e-03  3.819e-05   65.10   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.133 on 134 degrees of freedom
## Multiple R-squared:  0.9693, Adjusted R-squared:  0.9691 
## F-statistic:  4238 on 1 and 134 DF,  p-value: < 2.2e-16

This gives:

\[10000/\text{cpm} = 0.2591 + 2.486\times10^{-3}\text{antigen amount}\]

or:

\[\text{antigen amount (in pg)}=4022526/\text{cpm} - 104.2237\]

Testing for the Unkown

Here is the data when I tried different antigen dilution and volumes to make the count close to 20000. \(1\mu\text{L}\) of \(1\times10^4\) dilution suffices:

unknown_trial <- raw %>%
  filter(`known`== 0,
         `antigen_volume` != 0) %>%
  slice(1:7)
unknown_trial

## # A tibble: 7 x 6
##   antigen_volume antigen_dilution antibody_volume antibody_dilution count known
##            <dbl>            <dbl>           <dbl> <fct>             <dbl> <dbl>
## 1              1              100               1 10000               454     0
## 2              5             1000               1 10000               826     0
## 3              1             1000               1 10000              3857     0
## 4              5            10000               1 10000              8245     0
## 5              1            10000               1 10000             19760     0
## 6              5           100000               1 10000             29922     0
## 7              1           100000               1 10000             33886     0

Then I repeated \(1\mu\text{L}\) of \(1\times10^4\) dilution 276 times (see Section 9.1 for the data). Curiously, the counts do not follow a normal distribution. It is bimodal:

unknown_test <- raw %>%
  filter(`known`== 0,
         `antigen_volume` != 0) %>%
  slice(8:n())
unknown_test %>% 
  ggplot(aes(count))+
  geom_histogram(bins = 40, fill='white', color='black')+
  labs(title = 'Distribution of 276 Radioactivity Count Readings', x='radioactivity count')+
  theme_bw()

This is likely to be a bug in the ImmunoLab application.

If I had to assume it’s a normal distribution, a 95% confidence interval could be calculated:

cpm = unknown_test[['count']]
mass = 4022526/cpm - 104.2237
conc = mass /
          1e-3 / # 1e-3 ml
          1e-4 / # dilution
          1e9  # pg to mg
mean = mean(conc); sd = sd(conc);
t = qt(0.975, length(conc)-1); t_sd = t*sd
c(mean=mean, sd = sd, t = t, t_sd = t_sd, lower = mean-t_sd, upper = mean+t_sd)

##      mean        sd         t      t_sd     lower     upper 
## 0.8361621 0.1897999 1.9686279 0.3736453 0.4625168 1.2098075

(The relationship between cpm and antigen amount has been established previously: \(\text{antigen amount (in pg)}=4022526/\text{cpm} - 104.2237\))

The mean is 0.8362 mg/ml and the confidence interval is \(0.8362 \pm 0.3736\), or [0.4625 1.2098] mg/ml. (This is actually invalid because the distribution was not normal).

Q12

Draw a simple labelled diagram showing the principle of a sandwich ELISA.

See Figure 6.1.

Figure 6.1: Schematic of ELISA

Q13

Give an example of a chromogenic substrate for:

1. phosphatase: BCIP/NBT
2. peroxidase: TMB (3,3’,5,5’-tetramethylbenzidine)

Q14

Measuring the amount of enzyme in a well is technically simpler than measuring radioactivity, but it is much more susceptible to outside influences. List the factors that might affect the accuracy of your estimation of the amount of enzyme (assuming that optical density measurements are completely accurate).

• Variation in temperature and pH
• Aggregation of enzyme proteins (which leads to underestimation)

Q15

What is the Beer-Lambert law and why is it relevant when working out how much labelled antibody to use?

Beer-Lambert Law is:

\[A = \epsilon cl\]

where \(A\) is the absorbance (\(A=\log_{10}\left(\dfrac{I_\text{incident}}{I_\text{transmitted}}\right)\)), \(\epsilon\) is the molar extinction coefficient, \(c\) is the concentration of the light-absorbing substance, \(l\) is the path length.

By measuring absorbance (\(A\)), the concentration of the coloured product, and hence the concentration of the enzyme-conjugated antibodies, can be deduced.

What would happen to the dynamic range if we used more labelled Ab?

Q16

Describe the steps you took to optimise Antibody dilution for the rest of the experiment.

First, I used a wide range of antibody dilutions (but with smaller intervals around \(10^4\), as the optimal concentration is more likely to be around here), these concentrations are \(10^2, 10^3, 4\times10^3, 10^4, 4\times10^4, 8\times10^4, 10^5, 10^6\). The antigen concentrations are evenly distributed, namely \(1\times10^n\) and \(4\times10^n\) for \(n\) in \(\{1,2,3,4,5,6\}\). The results are shown in Figure 8.1. The plot_elisa() function used here were defined in Section 7. I believe the plots I made are a better representation of ELISA results than screenshots or Excel spreadsheets. However, if you wish to view the raw data, you can download them from https://github.com/TianyiShi2001/ox/tree/master/content/lab/2020-03-08-immunoassay-in-silico/.

plot_elisa('./2020-03-08-immunoassay-in-silico/elisa_trial_1.tsv', "ELISA Trial 1 (choosing antibody concentration)")

Figure 8.1: ELISA Trial 1 (choosing antibody concentration) results.

As shown in Figure 8.1, the optimal antibody concentration is between \(4\times10^4\) and \(8\times10^4\). So in the next round of trial, I chosed antibody concentrations of \(3.2\times10^4\), \(4.0\times10^4\), \(6.4\times10^4\), and \(8.0\times10^4\). The antigen concentrations are the same as the previous trial. The experiments on these four antibody concentrations are repeated to fill 96 wells. The results are shown in Figure 8.2.

elisa_trial_2_1 <- plot_elisa('./2020-03-08-immunoassay-in-silico/elisa_trial_2_1.tsv', "ELISA Trial 2-1")
elisa_trial_2_2 <- plot_elisa('./2020-03-08-immunoassay-in-silico/elisa_trial_2_2.tsv', "ELISA Trial 2-2")
ggarrange(elisa_trial_2_1, elisa_trial_2_2, ncol=2, nrow=1, common.legend = TRUE, legend="bottom", labels = c("A", "B"))

Figure 8.2: ELISA Trial 2 (choosing antibody concentration) results.

The plot suggests that an antibody concentration of \(4.0\times10^4\) would be optimal. Its maximum is close to, but does not exceed, 2.0.

Q17

Discuss how you generated data for your calibration curve (you can append titled data sets to the end of this document)

Q18

Plot your results to generate a calibration curve. Ensure axes are labelled, amount (not dilution) of antigen is plotted and that you have provided a descriptive title of the graph.

First, I used the same antigen concentrations as in previous two trials (\(1\times10^n\) and \(4\times10^n\) for \(n\) in \(\{1,2,3,4,5,6\}\).), as shown below (this code reads data into dataframe std1 while printing unique dilutions used).

std1 <- read_tsv('./2020-03-08-immunoassay-in-silico/elisa_std1.tsv') %>%
  janitor::clean_names() %T>%
  # print dilutions used
  (function(d){print(sort(unique(d$ag_dilution)))}) %>%
  # convert antigen dilution into antigen concentration in ng/ml
  # keep only two variables: ag_conc and absorbance
  transmute(ag_conc = 10^6 / ag_dilution, absorbance)

##  [1] 1e+01 4e+01 1e+02 4e+02 1e+03 4e+03 1e+04 4e+04 1e+05 4e+05 1e+06 4e+06

As shown in Figure 8.3, if plotted directly, it can be observed that absorbance increases with increasing antigen concentration as expected, but the mathematical relationship is unclear and so the ‘optimal range’ is also vague.

ggplot(std1, aes(ag_conc, absorbance))+
  geom_point()+
  labs(title = 'ELISA priliminary calibration curve', x = 'antigen concentration (ng/ml)')

Figure 8.3: ELISA priliminary calibration curve

However, as shown in Figure 8.4, a linear relationship can be obtained by a double-log transformation. In this plot, it is clear that the useful range of antigen concentration would be \(1.0<\log_{10}{c}<{3.4}\), or \(10^1 < c < 10^{3.4}\), where \(c\) is the antigen concentration in \(ng/ml\). This corresponds to a range of dilution of \(4\times10^2 < x < 10^5\).

ggplot(std1, aes(log10(ag_conc), log10(absorbance)))+
  geom_point()+
  labs(title = 'ELISA priliminary calibration curve (double log)',
       x = 'log(antigen concentration (ng/ml))')+
  geom_abline(intercept = -3.4, slope = 1)+
  geom_text(x=2,y=-1, label='y = x - 3.4', family="Helvetica", size=4.5)

Figure 8.4: ELISA priliminary calibration curve (double log)

Based on previous knowledge that the dilution should be in the range \(4\times10^2 < x < 10^5\), more experiments are carried out. The dilutions used are shown below.

std2 <- read_tsv('./2020-03-08-immunoassay-in-silico/elisa_std2.tsv') %>%
  janitor::clean_names() %T>%
  # print dilutions used
  (function(d){print(sort(unique(d$ag_dilution)))}) %>%
  # convert antigen dilution into antigen concentration in ng/ml
  # keep only two variables: ag_conc and absorbance
  transmute(ag_conc = 10^6 / ag_dilution, absorbance)

##  [1]   800  1000  2000  4000  8000 10000 20000 32000 40000 64000 80000

The calibration curve is shown in Figure 8.5.

ggplot(std2, aes(log10(ag_conc), log10(absorbance)))+
  geom_point()+
  labs(title = 'ELISA calibration curve',
       x = 'log(antigen concentration (ng/ml))')+
  geom_smooth(method='lm')

Figure 8.5: ELISA calibration curve (double log)

Q19

Briefly comment on the graph:

summary(lm(log10(std2$absorbance) ~ log10(std2$ag_conc)))

## 
## Call:
## lm(formula = log10(std2$absorbance) ~ log10(std2$ag_conc))
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.09315 -0.03981  0.00023  0.04625  0.08439 
## 
## Coefficients:
##                      Estimate Std. Error t value Pr(>|t|)    
## (Intercept)         -3.397722   0.015712  -216.3   <2e-16 ***
## log10(std2$ag_conc)  0.996370   0.007425   134.2   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.04711 on 94 degrees of freedom
## Multiple R-squared:  0.9948, Adjusted R-squared:  0.9948 
## F-statistic: 1.801e+04 on 1 and 94 DF,  p-value: < 2.2e-16

As shown in Figure 8.5 and in the linear model summary above, when the antigen concentration is in the range \(10^1 < c < 2.5\times10^3\) (i.e. the absorbance is in the range \(0.004 < A < 1.000\)), there is a significant correlation between the antigen concentration and the absorbance, which can be described by this equation:

\[\log_{10}{A} = \log_{10}{c} - 3.40\]

where \(A\) is the absorbance and \(c\) is the antigen concentration in \(ng/ml\).

As shown in Figure 8.5, when log(antigen concentration) is less than 1.7, the variation within each sets of absorbance readings is slightly larger, so the optimal range should be narrow down to \(10^{1.7} < c < 10^{3.4}\) for antigen concentration, or \(0.020 < A < 1.000\) for absorbance.

Q20

Using appropriate dilutions and number of replicates (discuss briefly below). Calculate the concentration of antigen in the original (unknown) solution, stating your calculation process. Where relevant, you can append titled data sets to the end of this document. Esnure your answer is statistically valid (think back to the stats test used in RiaLab)

First, I used a broad range of dilutions (again, \(1\times10^n\) and \(4\times10^n\) for \(n\) in \(\{1,2,3,4,5,6\}\)).

df <- read_tsv('./2020-03-08-immunoassay-in-silico/elisa_exp1.tsv') %>%
  janitor::clean_names() %>%
  transmute(ag_dilution = factor(ag_dilution), absorbance)
ggplot(df, aes(ag_dilution, absorbance))+
    geom_boxplot()+
    labs(title = "ELISA Experiment for the Unknown")+
    geom_hline(yintercept = 1.000, color='red', size = 0.6)+
    geom_hline(yintercept = 0.020, color='red', size = 0.6)

Figure 8.6: Finding the appropriate dilution of the unknown so that the absorbance fall within the linear range.

    # scale_x_discrete(limits = rev(levels(df$ag_dilution)))+

As learnt from Q19, the linear range is where the absorbance lies between 0.020 and 1.000. According to Figure 8.6, a \(10^3\) dilution would be the most appropriate. Therefore, I repeat 480 times with this dilution.

elisa_absorbance <- read_tsv('./2020-03-08-immunoassay-in-silico/elisa_exp_fin.tsv') %>%
  janitor::clean_names() %>%
  extract2('absorbance')

Curiously, again, the distribution is bimodal (the data is shown in Section 9.2:

qplot(elisa_absorbance, geom = 'histogram', bins = I(40), fill=I('white'), color=I('black'),
      main = "distribution of 480 absorbance readings for the unknown")

Figure 8.7: Distribution of 480 absorbance readings for the ELISA assay with 1:1000 unknown sample and 1:40000 antibody

If I had to assume it’s a normal distribution, a 95% confidence interval could be calculated:

conc <- 10^(log10(elisa_absorbance) + 3.40) *
                10^3 / # dilution
                10^6 # ng/ml to mg/ml
mean <- mean(conc); sd = sd(conc);
t <- qt(0.975, length(conc)-1); t_sd <- t*sd
c(mean=mean, sd = sd, t = t, t_sd = t_sd, lower = mean-t_sd, upper = mean+t_sd)

##       mean         sd          t       t_sd      lower      upper 
## 0.80949203 0.08313395 1.96492886 0.16335230 0.64613973 0.97284433

(Used the relationship between antigen concentration and absorbance established previously: \(\log_{10}{A} = \log_{10}{c} - 3.40\))

The mean is 0.8095 mg/ml and the confidence interval is \(0.8094\pm0.1634\), or [0.6461 0.9728] mg/ml. (This is actually invalid because the distribution was not normal).