Question

A cell’s immune response to an infection depends on a sequence of chemical mes-
sengers. In researching how such chemical messengers are transported from outside into the cell,

Rutger D. Luteijn and Shivam A. Zaver et al. measured how folic acid (vitamin B9) and other med-
ications inhibit the cell’s uptake (absorption) of immunoreactive chemical messengers [2]. Table 2

summarizes some of their data.

© 2020 Y. Nievergelt MATH 241-001 Problems 1 AND 2 due by midnight PDT on 30 July 2 Example 2 A cell's immune response to an infection depends on a sequence of chemical mes- sengers. In researching how such chemical messengers are transported from outside into the cell, Rutger D. Luteijn and Shivam A. Zaver et al. measured how folic acid (vitamin B9) and other med- ications inhibit the cell's uptake (absorption) of immunoreactive chemical messengers [2]. Table 2 summarizes some of their data. N The slope, By, of the line o least squares through O minimizes U(B) = (4x – Btx)? k=1 Table 2: Cellular uptake of an immunoreactive chemical messenger versus time, with averages (yk) of three measurements computed from [2, Source Data Fig.4, Fig. 4e]. INDEX O ko 2 3 1 TIME [min] ts 0 30 60 150 300 UPTAKE (counts/min)/106 cells y O 266.9 705.1 1772.5 4165.5 Problem 2 The following activities pertain to Table 2 in Example 2. N (2.1) For the data in Table 2, compute the value U (4) = (y2 - 4x)? k=1 N (2.2) With Table 2 and U(B) = (yx - Btk), find the derivative U'(4). 1 (2.3) Find the value of 8 where U (B) = (y2 – Btx)" is minimum. k=1 (2.4) Find the slope, By, of the line through 0 fitted to the data in Table 2. N (2.5) If you have access to a plotting device, plot U(8) = (2x – Btr)?. If you do not have access to a plotting device, then calculate and plot by hand three points on the graph of U. k=1

Answer 1

(2.1) We just evaluate the formula for at
B= 4
,

$U(4) =\sum_{k=1}^4 (y_k - 4 t_k)^2$
$= \left( \frac{266.9}{3} -4\times 30 \right )^2 + \left( \frac{705.1}{3}-4\times 60 \right )^2+ \left(\frac{1772.5}{3} - 4\times 150 \right )^2+ \left( \frac{4165.5}{5}-4\times 300\right )^2$
.

(2.2) We have

Τ' (β) = (Σ yk - βt, k=1

$= \sum_{k=1}^4 \left((y_k - \beta t_k)^2 \right )'$
$= \sum_{k=1}^4 2\times (y_k - \beta t_k)\times (-t_k)$
$=- 2 \sum_{k=1}^4 t_k (y_k - \beta t_k)$

and therefore

$U'(4) = -2 \sum_{k=1}^4 t_k (y_k - 4 t_k) = -10\,789$.

(2.3)
U(B)
is minimized when $U'(\beta) = 0$ ,

$0= U'(\beta)$
$=- 2 \sum_{k=1}^4 t_k (y_k - \beta t_k)$

We can cancel the factor and get

$0 = \sum_{k=1}^4 t_k (y_k -\beta t_k)$
$0 = \sum_{k=1}^4 t_k y_k - \beta\sum_{k=1}^4 t_k^2$

$\beta = \frac{\sum y_k t_k}{\sum t_k^2} = 4.4611$.

(2.4) The value
B= 4.4611
corresponds to the slope $\tilde \beta_y$ of the line passing thought the origin that fits the data the best according to least squares.

(2.5) Here is the plot of vs $\beta$

X106 5 4.5 4 3.5 3 (on 2.5 2 1.5 1 0.5 0 -2 0 2 4 6 8 10 B