Protein expression and purification
II. Calculation in the molecular
Kód předmětu: Bi8980
Název prezentace v zápatí 1
II. Calculation in the molecular
biosciences
Lubomír Janda, Blanka Pekárová and Radka Dopitová
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.1. Estimation of the results of calculations
Relating the number of amino acids in the polypeptide chain to molecular mass.
1 Dalton (Da) is equal to 1/12 the mass of the 12C isotope of carbon (1.66 x 10-24 g)
1 amino-acid contributes about 110 (112) Da to the protein’s mass.
A: A protein 260 aa long = ? kDa
B: A protein of 40 kDa = ? amino acid ? nt
Strategy A: (260 x 100) + (260 x 10) = 28,600 Da + (2 x 260) = 29,120 Da
The molar concentration of a solution.
Estimate the molarity of a 3.5 mg/ml solution of BSA, whose molecular mass is 66,000 Da
(66 kDa).
Strategy: 3.5/66,000 = 3.5 x 10-5/0.66 ~ 5 x 10-5 = 0.00005 M = 0.05 mM = 50 μM (control
5 x 0.66 = 3.3)
Strategy B: 40,000/100 = 400 aa – 10% (40 aa) = 360 aa
360 aa x 3 = 900 + 180 = 1,080 nt
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.2. Significant figures
•If the molecular mass of a protein is quoted as 30 kDa, this represents only one
significant figure (mass is between 25 and 35 kDa)
•MS might give an answer of 34.503 kDa (5 significant figures).
•If mobility on protein electrophoresis compared with standard proteins of
known molecular mass gives us no more than two significant figures, then
molecular mass also must be quoted to two significant figures, i.e. 35 kDa.molecular mass also must be quoted to two significant figures, i.e. 35 kDa.
A calculator gives the result of a calculation as 4,623.708 Da. Express this result to 1,
2, 3, and 4 significant figures
Answer: The results are 1. 5,000 Da
2. 4,600 Da
3. 4,620 Da
4. 4,624 Da
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.3. Logarithms
The logarithm of a number n is the power to which the reference base
number (usually 10) must be raised to give n.
Thus, 102 = 100, so log 100 = 2; similarly log 100 000 = log 105= 5
10a x 10b = 10a+b, so log (a x b) = log a + log b
10a/10b = 10a-b, so log (a/b) = log a – log b
a2 = a x a, so log (a2) = log a + log a = 2log a; in general log (an) = n log aa2 = a x a, so log (a2) = log a + log a = 2log a; in general log (an) = n log a
An alternative reference base number for logarithms is the Euler number
(e, equal to 2.71828…)
Logarithms to base e are known as natural logarithms and generally
denoted by ln. ln 10 = 2.303, so in general ln x= 2.303 log x
Key properties of logarithms
•The log of 1 = 0 (this is because 100 = 1).
•The log of number between 0 and 1 is negative (pH).
•The log of a number greater than 1 is positive.
•Negative numbers do not have logarithms.
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.3. Logarithms
2.1.3.1. Acid-base behaviour and the pH scale
Acidity is quantitatively defined by the concentration of protons (H+ ions) present in a solution.
Stomach 0.03 M (30 mM)
Duodenum 0.00000001 M (1 x 10-8 M) (10 nM)
Lysosome 0.00003 M (3 x 10-5 M) (30 µM)
pH is defined by the equation: pH = - log [H+]
Lysosome 0.00003 M (3 x 10 M) (30 µM)
Stomach (0.03 M) pH = -log (0.03) = -(-1.52) = 1.52
Duodenum (0.00000001 M) pH = -log (1 x 10-8 M) = -(-8) = 8
Lysosome (0.00003 M) pH = -log (3 x 10-5 M) = -(-4.52) = 4.52
Henderson-Hasselbalch equation pH = pKa + log ([A-]/[HA] (p91)
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.3. Logarithms
2.1.3.2. Variation of reaction rates with temperature
The rates of reactions increase dramatically with temperature as a greater
proportion of the reactants possess the energy necessary to surmount the
activation energy barrier for reaction to occur.
k = Ae-Ea/RT
k rate constant for the reaction
A pre-exponential factor (related to the frequency of
successful collisions between reacting molecules)
Ea activation energy for the reaction
R gas constant (8.31 J/K mol)
T temperature in degrees Kelvin
ln k = ln A – (Ea/RT)
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.3. Logarithms
2.1.3.2. Variation of reaction rates with temperature
ln k = ln A – (Ea/RT)
Arhenius plots show the effect of
temperature on reaction rate.
Typical plot for an enzyme-catalyzed
reaction.
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.3. Logarithms
2.1.3.2. Variation of reaction rates with temperature
Explain the dependence of activity on temperature and determine the activation
energy for the enzyme-catalyzed reaction.
Table: The variation of lactate dehydrogenase activity with temperature
Temperature (°C) 5 15 25 35 45 55 65
Activity (µmol/min.mg) 52.5 108 212 401 488 270 40.3
energy for the enzyme-catalyzed reaction.
The data can be plotted according to the Arrhenius equation to give a straight line
ln k = ln A – (Ea/RT)
T - temperature K = °C + 273
Slope of the graph = -5,810 K = -Ea/R
Ea = 5,810 x 8.31 J/mol = 48.3 kJ/mol
At 45°C, the rate is less than predicted on the basis of this straight line.
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.3. Logarithms
2.1.3.3. First-order processes and bacterial growth
[A]t – [A]0e-kt •The decay of radioactive isotopes
•The decrease in the concentration of drugs in the blood
•Bacterial growth
[A]t concentration at time t
[A]0 concentration at time zero
k is the rate constant for the reaction
Input 100 cells generation time 30 min
after 30 min 200 cells
60 min 400 cells60 min 400 cells
10 h 1.0486 x 108 (220 x 100 cells)
20 h 1.0995 x 1014 (240 x 100 cells)
Q: The number of bacterial cells in a culture increases from 1.2 x 105 to 5.8 x 105
over 120 min. What is the generation time for the bacteria under these
conditions?
ST: The two data points can be used to calculate the slope of the plot of log
against time.
S: The slope of the plot is 0.684/120 = 0.0057 min-1. Thus 0.301/t1/2 = 0.0057
From which t1/2 = 0.301/0.0057 min = 52.8 min
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.3. Logarithms
2.1.3.4. Molecular mass calibration graphs
•Molecular masses of proteins are often estimated by the techniques of gel
filtration and SDS-PAGE.
•Two graphs may be constructed: log molecular mass vs. elution volume and
log molecular mass vs. mobility.
Protein Mobility (cm) Molecular
mass (kDa)
Log (molecular
mass)
35.00 1.54
36.00 1.56
39.00 1.59
mass (kDa) mass)
Phosphorylase b 1.25 97.00 1.99
Bovine serum albumin 2.10 66.00 1.82
Ovalbumin 3.70 43.00 1.63
Carbonic anhydrase 6.10 29.00 1.46
Trypsin inhibitor 8.20 21.50 1.33
α-SNAP 5.30
β-SNAP 5.20
γ-SNAP 4.85
The molecular mass of lysoyzme is 14,300 Da, and its density is
1.4 mg/ml. Avogadro’s number is 6.02 x 1023 mol-1. The volume
of a sphere of radius r is given by (4π/3)r3.
Question 1: Assuming that the lysozyme molecule can be
regarded as spherical, calculate its radius.
I. The molecular principles for understanding proteins – Lubomír Janda
Please solve the problem. 5 points
regarded as spherical, calculate its radius.
Answer: The mass of 1 mol lysozyme is 14,300 g. Hence the
mass of 1 molecule = 14,300/(6.02 x 1023 ) g = 2.375 x 10-20 ml
(cm3). So (4π/3)r3 = 1.697 x 10-20, hence r3 = 4.05 x 10-21, therefore
r = 1.59 x 10-7 cm, or 1.59 nm. Thus, the radius of a lysozyme
molecule is 1.59 nm.
The concentration of the protein required for a nuclear magnetic
resonance experiment is 0.5 mM.
Question 2: If the molecular mass of the protein is 27.5 kDa,
what concentration is required in terms of mg/ml? If the
sample volume is 0.4 ml, what mass of protein is required?
I. The molecular principles for understanding proteins – Lubomír Janda
Please solve the problem. 5 points
sample volume is 0.4 ml, what mass of protein is required?
Answer: A 1 M solution of protein contains 27,500 mg/ml.
Thus a 0.5 mM solution contains 0.5 x 10-3 x 27,500 mg/ml = 13.75
mg/ml. In 0.4 ml, there would be 13.75 x 0.4 mg, i.e. 5.5 mg.
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.3. Logarithms
2.1.3.5. Spectrophotometry
Beer–Lambert law A = εεεε x c x l
A is the absorbance at a particular wavelength (without
units; merely a ratio):
A = log (I0/It)
I is the intensity of the incident light (light strikingI0 is the intensity of the incident light (light striking
the cuvette).
It is the intenisty of the transmitted light (the light
leaving the cuvette).
εεεε is the absorption coefficient (degree of absorption) (M-1
cm-1).
l is the path length of the cuvette (cm).
c is the concentration of the solution (M).
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.3. Logarithms
2.1.3.5. Spectrophotometry
Q: What percentage of the incident light is transmitted after
passage through solutions which have absorbance values of (a)
0.05, (b) 0.3, (c) 1.0, and (d) 2.0
ST: This is an application of the equation: A = log (I /I )ST: This is an application of the equation: A = log (I0/It)
10x of 0.05 = 1.122 10x of 0.3 = 1.995
10x of 1 = 10 10x of 2 = 100
S: The value of I0/It at each value of A is multiplied by 100 to give
the percent of the incident light that is transmitted
100/1.122 = 89.1% 100/1.995 = 50.1%
100/10 = 10% 100/100 = 1%
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.3. Logarithms
2.1.3.5. Spectrophotometry
Beer–Lambert law A = εεεε x c x l
A is the absorbance at a particular wavelength (without units; merely a ratio).
εεεε is the absorption coefficient (degree of absorption) (M-1 cm-1).
l is the path length of the cuvette (cm).
c is the concentration of the solution (M).
•The logarithmic nature of A is not always readily appreciated in a digital display of
the absorbance.the absorbance.
•However, it is particularly important to note that you should always aim to measure
absorbance values in the range 0.1–1.0. Despite what the manufacturers of the
instrument may claim, absorbance values significantly above 1.0 are not usually
reliable.
Q: The absorption coefficient of NADH at 340 nm is 6,220 M-1cm-1. You have made up a 2.5 mM
stock solution of NADH. How would you check its concentration spectrophotometrically?
ST: The absorbance to be measured should be in the range for reliable measurements (0.1–1.0).
S: 2.5 mM solution in 1 cm cuvette will have absorbance at 340 nM of
6,220 x 2.5 x 10-3 = 15.6
50-fold diluted solution will have absorbance of 15.6/50 = 0.312.
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.3. Logarithms
2.1.3.6. Energy changes and equilibrium constants of reactions
The standard free energy change in a reaction (∆G0) is related to the equilibrium
constant for the reaction (Keq):
∆∆∆∆G0 = -RT ln Keq
R is the gas constant (8.31 J K-1 mol-1)
T is the absolute temperature.T is the absolute temperature.
Keq will change logarithmically with changes in ∆G0.
∆∆∆∆G= ∆∆∆∆H −−−− T∆∆∆∆S
∆∆∆∆H is the change of enthalpy (change in heat content).
∆∆∆∆S is the change in entropy (change of disorder of a system).
∆∆∆∆G is the change in (Gibbs) free energy.
T is the absolute temperature (degrees Kelvin, K).
Standard state of a substance refers to: solution of 1 M of the solute
pressure of 1 atm at the temperature in question
0 0 0
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.3. Logarithms
2.1.3.6. Energy changes and equilibrium constants of reactions
Q: For the combustion of glucose (C6H12O6 + 6O2 6CO2 + 6H2O), the
values of ∆G0 and ∆H0 at 25°C are -2,872 and -2,822 kJ/mol, respectively. Calculate
the value of ∆S0 and comment on its sign.
∆∆∆∆G= ∆∆∆∆H −−−− T∆∆∆∆S
the value of ∆S0 and comment on its sign.
ST: This is an application of above equation using the standard state values of
the thermodynamic parameters.
S: We can rearrange the equation to give ∆∆∆∆S0 = (∆∆∆∆H0 – ∆∆∆∆G0)/ T. Putting T = 298 K, we
find that ∆S0 = (-2,822 + 2,872)/298 kJ/K.mol = 0.168 kJ/K.mol, or 168 J/K.mol. The
positive value of ∆∆∆∆S0 reflects the greater number of molecules of products (12)
compared with reactants (7). In this reaction, the enthalpy term makes the
dominant contribution to the free energy change.
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.4. Reciprocals
The reciprocal of a number is 1 divided by that number. Use the 1/x button on the
calculator to calculate reciprocals and explore this function in:
•The Lineweaver-Burk plot of enzyme kinetic data and the subsequent
calculation of the parameters Km and Vmax.
•Calculating the Km or Kd from the slope of an Eadie-Hofstee or a Scatchard plot,
respectively.
•Interconverting association and disocciation constant for binding processes.•Interconverting association and disocciation constant for binding processes.
•In the Arrhenius plot where the x-axis of the plot is 1/T.
Check that you have mastered the key concepts at the start of this section by
attempting the following questions.
Q1:From a graph, 1/Vmax is found to be 0.0235 min/µM. What is the value of
Vmax?
Q2: From the same graph, -1/Km is found to be -0.0065 µM-1. What is the
value of Km?
Q3: The value of Ka for a binding process is 4.53 x 104 M-1. What is the value of
Kd, given that Ka=1/Kd?
S1: Vmax = 42.6 µM/min S2: Km = 154 µM S3: Kd = 22.08 µM
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.5. Testing hypotheses
2.1.5.1. Dependent and independent variables
In a graph, the convention is that the x-axis (abscissa) is used to plot the variable
that the experimenter varies. This is the independent variable.
The y-axis (ordinate) is used to plot the quantity that is then observed. This is the
dependent variable.
y = mx + c
m is the slope (gradient) of the line.m is the slope (gradient) of the line.
c is the intercept of the line on the y-axis.
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.5. Testing hypotheses
2.1.5.1. Dependent and independent variables
y = mx + c
m is the slope (gradient) of the line.
c is the intercept of the line on the y-axis.
2.2
2.3
2.2
2.4
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.5. Testing hypotheses
2.1.5.2. Rearranging equations
Q: Transform equation so as to give the Hanes-Woolf equation for which
the plot is [S]/v versus [S].
ST: The approach is to rearrange the equation so as to be able to separate terms in the
y = mx + c form.
S: By multiplying both sides of equation (Km + [S]), we obtain:S: By multiplying both sides of equation (Km + [S]), we obtain:
vKm + v [S] = Vmax [S]
Rearranging terms: Vmax [S] = v[S] + vKm
Dividing each term on both sides of equation by Vmax
Dividing each term by v: Hanes-Woolf equation
Because Km and Vmax are constants, this equation is of the form y = mx + c, with y =
[S]/v and x = [S]. A plot of [S]/v versus [S] will be a straight line with an intercept on the
y-axis of Km/Vmax and a slope of 1/Vmax. The intercept on the x-axis is -Km.
The molecular mass of lysozyme is 14.3 kDa.
Question 3: What is the molar absorption coefficient of the
protein at 280 nm, given that the A280 of a 1 mg/ml solution in
a cuvette of 1 cm path length is 2.65?
I. The molecular principles for understanding proteins – Lubomír Janda
Please solve the problem. 5 points
280
a cuvette of 1 cm path length is 2.65?
Answer: A 1 M solution of lysozyme would be 14,300
mg/ml. Using the Beer-Lambert law, the A280 of this solution in a
cuvette with a 1-cm path length is 2.65 x 14,300 = 37,900. Thus,
the absorption coefficient of lysozyme is 37,900 M-1 cm-1.
The Tris buffer system has a pKa of 8.1. During a reaction, 5 mM
H+ ions are formed.
Question 4: What are the concentrations of the TrisH+ and
Tris forms at pH 8.3, if the total concentration of Tris species is
50 mM? What is the new pH after adding 5 mM H+ ions?
I. The molecular principles for understanding proteins – Lubomír Janda
Please solve the problem. 5 points
50 mM? What is the new pH after adding 5 mM H ions?
Answer: From the Henderson-Hasselbalch equation, the
ratio [Tris]/[TrisH+] = 100.2 = 1.58. Thus, [Tris] = 61.2 mM and
[TrisH+] = 38.8 mM. Upon addition of 5 mM H+, [Tris] will be
reduced by 5 mM to 56.2 mM and [TrisH+] will be raised by 5 mM
to 43.8 mM. Application of the Henderson-Hasselbalch equation
shows that the new pH would be 8.21. (Note: If no buffer had
been present, 5 mM H+ would give a pH of 2.3.)
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.6. Some basic statistics
2.1.6.1. Distributions of variables
arithmetic mean ( )
10.2, 10.7, 11.0, 11.1, 11.5, 11.6, 11.9, 12.2 median
The mode is valuable in describing a distribution of
11.3
mode
The mode is valuable in describing a distribution of
variables which cannot be ranked and distribution
which might show two peaks.
The mode is the most
frequently occurring
measurement in a data
set or distribution.
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.6. Some basic statistics
2.1.6.2. The normal distribution
The population mean is defined by:
Where, x1, x2, …… are the individual values
of the property and n is the number of
values in the population.
68.2%
95.4%
99.7%values in the population.
x is an individual value of the property.
µµµµ is the population mean.
n is the number of values in the population.
Introduction of the term n-1 rather than n into
the expression for standard deviation
In terms of experimentally derived values, this would indicate the degree of
confidence we had in stating the value of the given parameter.
99.7%
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.6. Some basic statistics
2.1.6.2. The normal distribution
Clearly, the larger the sample size n,
the smaller the value of the SEM.
Q: The operation of a pipette was checked by repeatedly dispensing and
weighing volumes of water. The volume on the pipette was set at 1 ml, and the
following volumes (mL) were dispensed in succession: 0.932, 0.927, 0.948,following volumes (mL) were dispensed in succession: 0.932, 0.927, 0.948,
0.937, 0.918, 0.929, 0.940, and 0.942. What is the mean and standard
deviation of these values?
ST: We use equation and
S: The mean value is 0.934 ml and the standard deviation is 0.0096 ml. From
the properties of the normal distribution, 99.7% of the values would be within
the range 0.905 to 0.963 ml, which is significantly different from the nominal
value of 1.000 ml. Thus we can conclude that the pipette is precise, but it is
not accurate. If the experiment had given a mean of 1.002 mL with a standard
deviation of 0.0096 mL the pipette would be precise and accurate.
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.6. Some basic statistics
2.1.6.3. Testing the difference between two means
Student’s t function – used to determine
the significance of the differences (more
appropriate for smaller sample sizes).
|| means ‘irrespective of sign’.
The probability (p) that the two sample means are
identical can be deduced from the properties of the t
function for the appropriate number of degrees of
freedom (equal to n1 + n2 -2).
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.6. Some basic statistics
2.1.6.3. Testing the difference between two means
Q: In a small-scale drug trial, the diastolic blood pressures of the drug and placeboQ: In a small-scale drug trial, the diastolic blood pressures of the drug and placebo
were 122.5 and 110.3 mm, respectively. There were 20 patients in each group. The
standard deviations for the two groups were 20.5 and 18.1 mm, respectively. Do the
data show (at >95% confidence) that the drug has an effect on the blood pressure?
ST: We calculate the standard error of the difference (equation above), and from
that the value of t.
S: The value of SEd = 6.11 mm. The value of = 12.2 mm. Hence t = 1.997.
Reference to the appendix shows that t is below the entry value (2.02) for 95%
confidence. Hence, we cannot reject the null hypothesis and must conclude that the
drug has not been shown to have an effect. Since t is quite close to the entry value, it
would probably be worthwhile extending the test to include more patients.
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.6. Some basic statistics
2.1.6.4. The correlation coefficient and linear regression
The term “correlation” refers to how strongly two variables are related.
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.6. Some basic statistics
2.1.6.4. The correlation coefficient and linear regression
r = +1, perfect positive correlation
r = -1, perfect negative correlation
95% confidence level
r ≥ 0.754 (n=5)
r ≥ 0.576 (n=10)
The determination of the best straight line
is known as linear regression.
r = -1, perfect negative correlation
is known as linear regression.
Least-squares method is the most widely
used method.
The value of the y-axis intercept, c, can
be calculated from the equation above.
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.6. Some basic statistics
2.1.6.4. The correlation coefficient and linear regression
sm = standard errors in the estimates of the slope
sc = y-axis intercept
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.6. Some basic statistics
2.1.6.4. The correlation coefficient and linear regression
Q: The rate (v, in units of µµµµM/min) of an enzyme-catalyzed reaction was studied as a
function of substrate concentration ([S], in units of µµµµM). The data was analyzed by the
Hanes-Woolf plot in which [S]/v is plotted against [S]. The following values were
obtained:
[S]/v 13.2 18.0 18.2 22.9 25.3 27.0 30.7
[S] 10 20 30 40 50 60 70
Calculate the correlation coefficient for the plot of [S]/v against [S], and use linear
regression to calculate the best straight line.
[S] 10 20 30 40 50 60 70
ST:
S: [S]/v is designated as y and [S] as x. The values of y and x are 22.19 and 40, respectively.
The values of ∑ (y-y)2 and ∑ (x-x)2 are 220.03 and 2,800 respectively. The value of ∑(x-x)(yy)
is 776. From this, using the first equation r = 0.9886; this is highly significant correlation
(p < 0.001 for 5 degrees of freedom, i.e. the number of (x , y) data points (7) -2). The slope
(m) and y-axis intercept (c) of the least-squares line are 0.227 and 11.1, respectively.
Further analysis using the equations for sm and sc above gives 0.029 and 1.3, respectively.
II. Calculation in the molecular biosciences – Lubomír Janda
2.1. The key mathematical tools
2.1.6. Some basic statistics
2.1.6.5. Non-linear regression
NRMSD – the normalized root mean square deviation
yobs and ycal are the observed and calculated values of y, at each specified value of x.
The following values of blood cholesterol (mM) were found
in a sample of eight healthy females: 4.3, 4.1, 5.8, 5.0, 3.9,
5.5, 5.2, and 4.9.
Question 5: What is the median, mean and standard
deviation of these values? What is the 95% confidence limit for
I. The molecular principles for understanding proteins – Lubomír Janda
Please solve the problem. 5 points
deviation of these values? What is the 95% confidence limit for
the population mean (µµµµ).
Answer: The values are: median, 4.95 mM; mean, 4.84 mM;
standard deviation, 0.68 mM; 95% confidence limit for µ,
4.37–5.31 mM.
A group of 21 healthy students undertook a glucose tolerance test in which
they fasted overnight and then ingested 75 g glucose. Before taking the
glucose, the blood glucose levels of the group had a mean value of 4.73 mM
(SD = 0.48 mM). 30 min after taking the glucose, the blood glucose levels had
a mean value of 7.95 mM (SD = 1.63 mM). After another 90 min, the levels
had a mean value of 5.25 mM (SD = 1.53 mM).
I. The molecular principles for understanding proteins – Lubomír Janda
Please solve the problem.
Question 6: Are the levels at 30 and 120 min significantly different
5 points
Question 6: Are the levels at 30 and 120 min significantly different
from that at the start?
Answer: Comparing 30 min and start values, t = 8.68; this gives p < 0.01, i.e.
the null hypothesis can be rejected with >99% confidence. Comparing 120
min and start values, t = 1.49; this gives a p value between 0.2 and 0.1 (0.2
> p > 0.1); i.e. the null hypothesis cannot be rejected with at least 95%
confidence. Thus, the 30 min value is significantly higher than the start
value, but the 120 values is not.
A Hanes-Woolf plot used to analyze a set of enzyme kinetic
data obtained at eight values of substrate concentration
showed a correlation coefficient (r) of 0.669.
Question 7: What would you recommend to the investigator
who produced the data?
I. The molecular principles for understanding proteins – Lubomír Janda
Please solve the problem. 5 points
who produced the data?
Answer: The value of r is below the value required for 95%
confidence of a positive correlation. It would not, therefore, be
appropriate to use the plot to try to obtain reliable values of the
kinetic parameters (Km and Vmax) for the enzyme. It would be
sensible to try to improve the experimental technique and to
obtain more data points.