3.1.1 Dilution

3.1.2 Results

See section 4.1

3.2.1 Mechanism

Glucose itself does not have convenient absorption. Thus, glucose oxidase, ABTS and peroxidase is added to make a green solution (glucose is first oxidised by oxidase to produce gluconic acid and H₂O₂; peroxidase then catalyses the reaction between H₂O₂ and ABTS to produce a green dye).

3.2.2 Determine the volume of enzyme

We want the reaction to ‘virtually’ complete (i.e. > 99% conversion of substrate) in 10 min.

For an enzyme reaction:

\[-\dfrac{ds}{dt}=v-\dfrac{V_\text{max}s}{K_m+S}\]

As the reaction approaches completion, \(s<<K_m\):

\[-\dfrac{ds}{dt}=\dfrac{V_\text{max}s}{K_m}\]

Integrating to determine the total time for a given change in \(s\):

\[ \begin{aligned} -\int_{s_0}^{s_1}{\dfrac{ds}{s}}&=\dfrac{V_\text{max}}{K_m}\int_0^t dt \\ \ln\dfrac{s_0}{s_1}&=\dfrac{V_\text{max}t}{K_m} \end{aligned} \]

For a decrease in \(s\) from 100% to 1% in 10 min:

\[ \begin{aligned} \ln\dfrac{100}{1}&=\dfrac{V_\text{max}\times 10}{K_m} \\ V_\text{max}&=0.4605K_m \text{ (mol L}^{-1}\text{min}^{-1}\text{)} \end{aligned} \]

Our assay volume will be 4.0mL, so the amound of enzyme, in U (\(\mu\text{mol min}^{-1}\)), will be

\[\text{amount of enzyme in U}=0.4605\times K_m \times 4.0 \times 10^{-3}\times10^6=1842K_m\]

where \(K_m\) has unit M.

For GOD, \(K_m=1\text{ mM}\), and the provided enzyme concentration is 100U/mL:

\[ \begin{aligned} K_m&=1\text{ mM} \\ \text{minimum amount of enzyme}&=1842 \times 1 \times 10^{-3}=1.84 \text{ U} \\ \text{minimum volume of enzyme}&=1.84 \div 100=0.0184 \text{ mL} \end{aligned} \]

For POD, \(K_m=100\text{ }\mu\text{M}\), and the provided enzyme concentration is 20U/mL:

\[ \begin{aligned} K_m&=100\text{ }\mu\text{M} \\ \text{minimum amount of enzyme}&=1842 \times 100 \times 10^{-6}=0.184 \text{ U} \\ \text{minimum volume of enzyme}&=0.184 \div 20=0.0092 \text{ mL} \end{aligned} \]

We decided to use an excess amount of enzyme (i.e. 0.1mL for both). This guarantees that the reactions will complete within 10 min, and makes it easier to make up the 4mL total volume for assay (see section 3.2.4.2).

3.2.3 Determining \(\lambda_\text{max}\)

Measuring absorbance near \(\lambda_\text{max}\) reduces error. An 100nmol-glucose-containing (i.e. with 1mL 0.1mM glucose solution, see section 3.2.4.2) assay result is used. The absorbance in the range 350nm to 450nm is measured, and \(\lambda_\text{max} = 418\text{ nm}\). We then used this wavelength for all glucose assays.

3.2.4 Calibration

Calibrate \(A^{418}\) with known glucose concentrations, then we can map \(A^{418}_\text{sample}\) to the calibration curve to determine sample glucose concentration.

3.2.5 Sample Assay

The sample is diluted 1:100 by mixing 1mL sample with 99mL buffer.

4mL assays are made on 1mL 1:100 dilution 3 times, according to the following table (identical to the assays used for calibration except that an unknown concentration is used this time)

3.2.6 Results

See section 4.2

4.1.1 Distinguishing between 4.0mM and 4.5mM ATP

The percentage error (coefficient of variance) of this experiment is:

\[c_v=\dfrac{s_x}{\bar{x}}=\dfrac{0.054}{3.268}=0.0165\]

for \(\bar{x}\) = 4.0 and 4.5, \(s_x=c_v\bar{x}\) = 0.0661 and 0.0744, respectively.

To distinguish between the two, two sample t test is used. If \(n\) measurements are made on each concentration, then

\[t=\dfrac{\bar{x_1}-\bar{x_2}}{\sqrt{\dfrac{s_1^2}{n}+\dfrac{s_2^2}{n}}}=\dfrac{4.5-4.0}{\sqrt{\dfrac{0.0661^2+0.0744^2}{n}}}=\sqrt{25.26n}\]

then we can get a list of \(p\) values for each \(n\):

n <- 2:8
df <- n - 1
t <- sqrt(25.26 * n)
p <- (1 - pt(t, df)) * 2
names(p) <- paste('n =', n)
round(p, 5)

##   n = 2   n = 3   n = 4   n = 5   n = 6   n = 7   n = 8 
## 0.08898 0.01294 0.00210 0.00036 0.00006 0.00001 0.00000

as n increases, \(p\) decreases. When n = 3, \(p\) is already less than 0.05.

4.2.1 Calibration

The results are shown below:

A1 <- c(0.952, 0.815, 0.709, 0.496, 0.272, 0)
A2 <- c(1.223, 0.969, 0.704, 0.415, 0.222, 0)
A3 <- c(1.194, 0.919, 0.682, 0.452, 0.236, 0)
A <- c(A1, A2, A3)
glucose = c(100, 80, 60, 40, 20, 0)
kable(tibble(`nmol glucose in 4mL assay` = glucose, `$A^{418}_1$` = A1,
             `$A^{418}_2$` = A2, `$A^{418}_3$` = A3))

To fit a straight line, linear regression is used. The coefficients are shown below:

summary(lm(A ~ 0 + rep(glucose, 3)))

## 
## Call:
## lm(formula = A ~ 0 + rep(glucose, 3))
## 
## Residuals:
##       Min        1Q    Median        3Q       Max 
## -0.181273 -0.000982  0.005691  0.039277  0.089727 
## 
## Coefficients:
##                  Estimate Std. Error t value Pr(>|t|)    
## rep(glucose, 3) 0.0113327  0.0002387   47.48   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.06132 on 17 degrees of freedom
## Multiple R-squared:  0.9925, Adjusted R-squared:  0.9921 
## F-statistic:  2254 on 1 and 17 DF,  p-value: < 2.2e-16

which shows there is a linear relationship \(A^{418}=0.01133n_\text{glucose}\) with \(R^2=0.9921\).

Then we can plot it (CI added).

ggplot(data.frame(A, glucose), aes(glucose, A))+
  geom_point()+
  geom_smooth(formula = y ~ 0 + x, method = 'lm')+
  labs(x = 'nmol glucose in 4mL assay', y = expression(A ^ 418),
       title = 'Calibrationn Curve for 4mL essay')

4.2.2 Sample result

Three independent measurements of \(A^{418}_\text{sample}\) gives 0.863, 0.872 and 0.871. According to the linear relationship \(A^{418}=0.01133n_\text{glucose}\) we can calculate the amount (in nmol) of glucose in 1mL 1:100 dilution, then the concentration of the raw sample can be calculated easily:

A <- c(0.863, 0.872, 0.871) # sample absorbance
n <- A/0.01133 # nmol glucose in 1 mL 1:100 dilution
(c_glucose_sample <- n * 1e-9 # converting to mol/mL (1:100 dilution)
       / 1e-3 # converting to M (1:100 dilution)
       * 100 # in M (raw sample)
       * 1e3) # converting to mM (raw sample)

## [1] 7.616946 7.696381 7.687555

Statistical summary:

x <- c_glucose_sample; n <- length(x); df <- n - 1; t = qt(0.975, df)
c(mean=mean(x), sd=sd(x), SE=sd(x)/sqrt(n), t = t, `t*SE` = t * sd(x)/sqrt(n))

##       mean         sd         SE          t       t*SE 
## 7.66696087 0.04353824 0.02513682 4.30265273 0.10815499

Which gives 95% confidence interval: \(\text{[glucose]}=7.667\pm0.057\) mM

4.2.3 Distinguishing between 4.0mM and 4.5mM glucose

Similar to the calculation for ATP shown in section 4.1.1, first calculate the percentage error (coefficient of variance) of this experiment:

\[c_v=\dfrac{s_x}{\bar{x}}=\dfrac{0.057}{7.667}=0.00743\]

for \(\bar{x}\) = 4.0 and 4.5, \(s_x=c_v\bar{x}\) = 0.0297 and 0.0335, respectively.

To distinguish between the two, two sample t test is used. If \(n\) measurements are made on each concentration, then

\[t=\dfrac{\bar{x_1}-\bar{x_2}}{\sqrt{\dfrac{s_1^2}{n}+\dfrac{s_2^2}{n}}}=\dfrac{4.5-4.0}{\sqrt{\dfrac{0.0297^2+0.0335^2}{n}}}=\sqrt{124.7n}\]

then we can get a list of \(p\) values for each \(n\):

n <- 2:8
df <- n - 1
t <- sqrt(124.7 * n)
p <- (1 - pt(t, df)) * 2
names(p) <- paste('n =', n)
round(p, 5)

##   n = 2   n = 3   n = 4   n = 5   n = 6   n = 7   n = 8 
## 0.04026 0.00266 0.00020 0.00002 0.00000 0.00000 0.00000

as n increases, \(p\) decreases. When n = 2, \(p\) is already less than 0.05.

5.3.1 Grouping and grouped confidence interval of mean

According to the plots, evidently there are two ATP concentrations, so we need to divide the 21 measurements into two groups, and calculate their mean and CI separately.

ATP1 <- ATP[ATP < 3.5] # less than 3.5mM, around 3.0mM
ATP2 <- ATP[ATP > 3.5] # more than 3.5mM, around 4.0mM

For the ATP around 3.0:

x <- ATP1; n <- length(x); df <- n - 1; t = qt(0.975, df)
c(mean=mean(x), sd=sd(x), SE=sd(x)/sqrt(n), t = t, `t*SE` = t*sd(x)/sqrt(n))

##       mean         sd         SE          t       t*SE 
## 3.07166667 0.12995337 0.03751431 2.20098516 0.08256843

95% confidence interval: \(\text{[ATP]}=3.07\pm0.08\) mM

For the ATP around 4.0:

x <- ATP2; n <- length(x); df <- n - 1; t = qt(0.975, df)
c(mean=mean(x), sd=sd(x), SE=sd(x)/sqrt(n), t = t, `t*SE` = t*sd(x)/sqrt(n))

##       mean         sd         SE          t       t*SE 
## 4.02555556 0.12620530 0.04206843 2.30600414 0.09700998

95% confidence interval: \(\text{[ATP]}=4.03\pm0.10\) mM

5.3.2 Test for significance of the difference in concentration between the two groups

To test there is a significant difference between the two concentrations, t-test² is used:

t.test(ATP1, ATP2)

## 
##  Welch Two Sample t-test
## 
## data:  ATP1 and ATP2
## t = -16.923, df = 17.66, p-value = 2.346e-12
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -1.0724720 -0.8353058
## sample estimates:
## mean of x mean of y 
##  3.071667  4.025556

\(p=2.3\times10^{-12}<<0.05\); the two concentrations are significantly different

5.3.3 Test for diviation from given concentrations

The solutions were actually made up to be 3.0mM and 4.0mM ATP. Were they made up correctly? We need two one-sample t tests using the given concentrations as the population mean, \(\mu\):

3.0mM: (\(H_0\): true concentration is 3.0mM)

t.test(ATP1, mu=3.0)

## 
##  One Sample t-test
## 
## data:  ATP1
## t = 1.9104, df = 11, p-value = 0.08249
## alternative hypothesis: true mean is not equal to 3
## 95 percent confidence interval:
##  2.989098 3.154235
## sample estimates:
## mean of x 
##  3.071667

\(p=0.082>0.05\), failed to reject hypothesis. Solution made up correctly.

4.0mM: (\(H_0\): true concentration is 4.0mM)

t.test(ATP2, mu=4.0)

## 
##  One Sample t-test
## 
## data:  ATP2
## t = 0.60748, df = 8, p-value = 0.5604
## alternative hypothesis: true mean is not equal to 4
## 95 percent confidence interval:
##  3.928546 4.122566
## sample estimates:
## mean of x 
##  4.025556

\(p=0.56>0.05\), failed to reject hypothesis. Solution made up correctly.

5.4.1 Grouping and grouped confidence interval of mean

By observing the plots, the boundary between the two concentrations of glucose is 8.0mM, so we can divide them into two groups:

glucose1 <- glucose[glucose < 8.0]
glucose2 <- glucose[glucose > 8.0]

And calculate their 95% CIs separately:

lapply(list(glucose1=glucose1, glucose2=glucose2), function(x){
  n <- length(x); df <- n - 1; t = qt(0.975, df)
  c(mean=mean(x), sd=sd(x), SE=sd(x)/sqrt(n), t = t, `t*SE` = t*sd(x)/sqrt(n))
})

## $glucose1
##      mean        sd        SE         t      t*SE 
## 7.3254545 0.3921317 0.1182322 2.2281389 0.2634377 
## 
## $glucose2
##      mean        sd        SE         t      t*SE 
## 8.6857143 0.2612698 0.0987507 2.4469119 0.2416343

The confidence intervals are: \(7.33\pm0.27\) mM and \(8.69\pm0.22\) mM

5.4.2 Test for significance of the difference in concentration between the two groups

t-test performed to test for significance:

t.test(glucose1, glucose2)  

## 
##  Welch Two Sample t-test
## 
## data:  glucose1 and glucose2
## t = -8.8301, df = 15.912, p-value = 1.573e-07
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -1.686972 -1.033548
## sample estimates:
## mean of x mean of y 
##  7.325455  8.685714

\(p=1.57\times 10^{-7}<<0.05\), which shows the two concentrations are significantly different.

5.4.3 Test for diviation from given concentrations

The solutions were actually made up to be 7.0mM and 9.0mM ATP. Were they made up correctly? We need two one-sample t tests using the given concentrations as the population mean, \(\mu\):

7.0mM: (\(H_0\): true concentration is 7.0mM)

t.test(glucose1, mu=7.0)

## 
##  One Sample t-test
## 
## data:  glucose1
## t = 2.7527, df = 10, p-value = 0.02038
## alternative hypothesis: true mean is not equal to 7
## 95 percent confidence interval:
##  7.062017 7.588892
## sample estimates:
## mean of x 
##  7.325455

\(p=0.020<0.05\), reject null hypothesis. Solution made up incorrectly.

9.0mM: (\(H_0\): true concentration is 9.0mM)

t.test(glucose2, mu=9.0)

## 
##  One Sample t-test
## 
## data:  glucose2
## t = -3.1826, df = 6, p-value = 0.01901
## alternative hypothesis: true mean is not equal to 9
## 95 percent confidence interval:
##  8.444080 8.927349
## sample estimates:
## mean of x 
##  8.685714

\(p=0.019<0.05\), reject null hypothesis. Solution made up incorrectly.

5.5.1 Results of the sample

Counts in the H channel and C channel are shown below:

H <- c(14178.5, 13963, 14065, 14356.5, 13809, 14157.5, 13824.5, 14032, 14094.5, 14014)
C <- c(3411, 3472.5, 3386.5, 3516, 3375.5, 3418.5, 3438, 3464.5, 3513.5, 3438.5)

Statistical summaries are calculated:

n <- 10; df <- n-1; t = qt(0.975, df)
lapply(list(`H Channel`=H, `C Channel`=C), function(i){
  c(mean=mean(i), sd=sd(i), SE=sd(i)/sqrt(n), t = t, `t*SE` = t*sd(i)/sqrt(n))
})

## $`H Channel`
##         mean           sd           SE            t         t*SE 
## 14049.450000   164.091176    51.890186     2.262157   117.383756 
## 
## $`C Channel`
##        mean          sd          SE           t        t*SE 
## 3443.450000   48.359447   15.292600    2.262157   34.594264

95% confidence intervals:

• H channel: \(14049.5\pm 117.38\) cpm
• C channel: \(3443.5\pm 34.59\) cpm

5.5.2 Correction for overlap

Let counts due to ³H = \(X\) and counts due to ¹⁴C = \(Y\). Let \(\alpha\) be the fraction in H channel of all counts due to ³H and \(\beta\) be the fraction in C channel of all counts due to ¹⁴C, so that for a sample with mixed isotopes:

\[H = \alpha X + (1-\beta)Y\] \[C = (1-\alpha)X+\beta Y\]

\(\alpha\) and \(\beta\) can be determined by counting pure ³H and ¹⁴C, respectively. Here are the results:

c(H_in_H = mean(c(464814.5, 462463.5, 464775.5)),
  H_in_C = mean(c(14825.5, 15057.5, 14210.0)),
  C_in_H = mean(c(10295, 9842.5, 9922.5)),
  C_in_C = mean(c(47989, 48868.5, 49133.5)))

##    H_in_H    H_in_C    C_in_H    C_in_C 
## 464017.83  14697.67  10020.00  48663.67

which gives:

\[ \begin{aligned} \alpha &= 464017.83/(464017.83+14697.67)=0.9693 \\ \beta &= 48663.67/(48663.67+10020.00)=0.8293 \end{aligned} \]

Plugging in sample results:

\[ \begin{aligned} H &= \alpha X + (1-\beta)Y = 0.9693X+(1-0.8293)Y=14049.5(\pm117.38) \\ C &= (1-\alpha)X+\beta Y = (1-0.9693)X+0.8293Y = 3443.5(\pm34.59) \end{aligned} \]

Simplifying…

\[ \begin{aligned} 0.9693X+0.1707Y&=14049.5(\pm117.38) \\ 0.0307X+0.8293Y &= 3443.5(\pm34.59) \end{aligned} \]

Solving for X: (note that uncertainties are added)

\[ \begin{aligned} 4.7075X+0.8293Y &=60721.7(\pm507.31) \\ 0.0307X+0.8293Y &= 3443.5(\pm34.59) \\ 4.6769X &= 57278.2(\pm541.9) \\ X &= 12247.2(\pm115.9) \end{aligned} \]

Solving for Y:

\[ \begin{aligned} 0.9693X+0.1707Y&=14049.5(\pm117.38) \\ 0.9693X+26.18Y &= 108722.6(\pm1058.66) \\ 26.01Y &= 94673.1(\pm1176.04) \\ Y &= 3639.4(\pm45.2) \end{aligned} \]

which gives:

\[ \begin{aligned} \text{actual H (ATP) counts in 0.2mL sample}&=12247.2(\pm115.9) \\ \text{actual C (glucose) counts in 0.2mL sample} &= 3639.4(\pm45.2) \end{aligned} \]

5.5.3 Specific Radioactivity

The radioactivity counting were done on 0.2mL sample. Our ATP concentration (section 4.1) is \(\text{[ATP]}=3.268\pm0.054\) mM and glucose concentration (section 4.2.2) is \(\text{[glucose]}=7.667\pm0.057\) mM. Thus, the specific radioactivity of ATP and glucose are:

\[ \begin{aligned} \text{ATP: }12247.2(\pm115.9)/[0.2\times10^{-3}\times3.268\times 10^{-3}(\pm0.054\times 10^{-3})]&= 1.87(\pm0.05)\times 10^{10} \text{ cpm mol}^{-1}\\ \text{glucose: } 3639.4(\pm45.2)/[0.2\times10^{-3}\times7.667\times 10^{-3}(\pm0.057\times 10^{-3})]&= 2.37(\pm0.05) \times 10^{9} \text{ cpm mol}^{-1} \end{aligned} \]

(the \(\pm0.05\) comes from the maximal possible numerator (e.g. \(12247.2 + 115.9\) for ATP) over the least possible denominator (e.g. \((3.268-0.054)\times10^{-3}\) for ATP))