Characterization of human PCa orthotopic mouse models 1. Dissemination pattern of human PCa cells in NSG mice

R ESULTS

1.2. Characterization of human PCa orthotopic mouse models 1. Dissemination pattern of human PCa cells in NSG mice

Chapter 4. Three dimensional structures of proteins

4.1. 3-D stru

4.1. 3-D structure of prcture of proteinsoteins

Generally speaking, hydrophobic amino acids are found buried inside proteins (away from Generally speaking, hydrophobic amino acids are found buried inside proteins (away from water), while polar and charged amino acids are most often found on the surface of proteins. We water), while polar and charged amino acids are most often found on the surface of proteins. We will then get the following distribution for the amino ac

will then get the following distribution for the amino ac ids :ids : Buried inside: Val, Phe, Ileu

Buried inside: Val, Phe, Ileu

On the surface: Glu, Arg, Asn, Lys, Se

On the surface: Glu, Arg, Asn, Lys, Ser, Thrr, Thr S-S S-S

S - S S - S

-- S S S S

--A-C-F-P-K-R-W-C-R-R-V-C A-C-F-P-K-R-W-C-R-R-V-C

C-Y-C-F-C C-Y-C-F-C

*4.2. 3-D structure of proteins

The structure of urea suggests that this molecule denatures proteins by breaking the hydrogen The structure of urea suggests that this molecule denatures proteins by breaking the hydrogen interactions which stabilize the 3-D structure of the macromolecules (i.e. via the interaction of interactions which stabilize the 3-D structure of the macromolecules (i.e. via the interaction of the amino and ketone groups of urea with the amino and ketone groups of the peptide bonds and the amino and ketone groups of urea with the amino and ketone groups of the peptide bonds and side chains).

side chains).

4.3. 3-D stru

4.3. 3-D structure of prcture of proteinsoteins

Even though it is located at the surface of proteins, Phe must avoid contact with water. This can Even though it is located at the surface of proteins, Phe must avoid contact with water. This can be

be accomplished accomplished if if two two or or more pmore protein suburotein subunits interact nits interact via via hydrophobic hydrophobic regions regions (which (which couldcould include Phe), keeping Phe in an hydroiphobic environment.

include Phe), keeping Phe in an hydroiphobic environment.

*4.4. 3-D structure of proteins

The presence of Asp inside proteins is possible if its polar and charged groups are involved in The presence of Asp inside proteins is possible if its polar and charged groups are involved in intermolecular interactions. This is possible if Asp is part of asecondary structure like the intermolecular interactions. This is possible if Asp is part of asecondary structure like the α α--helix.

helix.

4.5. 3-D stru

4.5. 3-D structure of prcture of proteinsoteins

Protein 1 has a high amount of hydrophilic amino acids, and very little hydrophobic residues Protein 1 has a high amount of hydrophilic amino acids, and very little hydrophobic residues (65% hydrophilic/35% hydrophobic). This suggests that a lot of these amino acids will be (65% hydrophilic/35% hydrophobic). This suggests that a lot of these amino acids will be interacting with the solvent, which is the ca

interacting with the solvent, which is the case for rod-shaped proteins (i.e. proteine A).se for rod-shaped proteins (i.e. proteine A).

Protein 3 has the same amount of hydrophobic vs hydrophilic amino acids, while protein 2 has Protein 3 has the same amount of hydrophobic vs hydrophilic amino acids, while protein 2 has many more hydrophobic than hydrophilic residues (protéine 2 = 30%/70%). This suggests that many more hydrophobic than hydrophilic residues (protéine 2 = 30%/70%). This suggests that protein

protein 3 3 would would be be globular, globular, with with several several hydrophobic hydrophobic amino amino acids acids buried buried inside inside and and lots lots ofof hydrophilic amino acids on the surface. Protein 3 would therefore be protein B.

hydrophilic amino acids on the surface. Protein 3 would therefore be protein B.

As for protein 2, its high content in hydrophobic amino acids little content in hydrophilic amino As for protein 2, its high content in hydrophobic amino acids little content in hydrophilic amino acids suggest that it could be protein C: the association of several subunits would regions of high acids suggest that it could be protein C: the association of several subunits would regions of high content of hydrophobic amino acids, protecting them from the aquous environment.

content of hydrophobic amino acids, protecting them from the aquous environment.

4.6. 3-D stru

4.6. 3-D structure of prcture of proteinsoteins

The primary structure of this peptide doesn’t have any features expected from a

The primary structure of this peptide doesn’t have any features expected from a ββ-pleated sheet,-pleated sheet, and no Gly and

and no Gly and Pro (which are Pro (which are known to disrupt known to disrupt secondary structures) secondary structures) are present. are present. We canWe can therefore deduce that this peptide would adopt an

therefore deduce that this peptide would adopt an αα-helical structure.-helical structure.

4.7. 3-D stru

4.7. 3-D structure of prcture of proteinsoteins

The primary structure of this peptide is typical for those arranged as

The primary structure of this peptide is typical for those arranged as ββ-pleated sheets.-pleated sheets.

4.8. 3-D stru

4.8. 3-D structure of prcture of proteinsoteins

The presence of several positively charged residues (Arg and Lys) in addition to Gly indicates The presence of several positively charged residues (Arg and Lys) in addition to Gly indicates that this peptide will most likely be a random coil.

that this peptide will most likely be a random coil.

4.9. 3-D stru

4.9. 3-D structure of prcture of proteinsoteins

With the criterias use in the 3 preceding problems, we can deduce that:

Amino acids 2-12:

Amino acids 2-12: αα-helix;-helix;

Amino acids 13-17: random coil;

Amino acids 18-26:

Amino acids 18-26: ββ-pleated sheet-pleated sheet

4.10. 3-D str

4.10. 3-D structure of pucture of proteinsroteins

The high amount of Pro and Hypro indicates that protein C will be a collagen-like triple-helix.

Protien A is rich in in amino acids with small side-chains (Gly, Ser, Ala): it will adopt a Protien A is rich in in amino acids with small side-chains (Gly, Ser, Ala): it will adopt a β β-- pleated sheet type of structure.

pleated sheet type of structure.

Protein B has a lot of amino acids that would be expected to be found in

Protein B has a lot of amino acids that would be expected to be found in αα-helices. However, the-helices. However, the integrity of this helix would be severely perturbed by the presence of Gly and Pro and by the integrity of this helix would be severely perturbed by the presence of Gly and Pro and by the presence of several consecutive acidic or basic amino acids.

presence of several consecutive acidic or basic amino acids.

Chapter 5. Enzymology Chapter 5. Enzymology

5.1. Enzyme

5.1. Enzyme kineticskinetics

a) From the available data, we can notice that the reaction rate doesn’t increase when the a) From the available data, we can notice that the reaction rate doesn’t increase when the substrate concentration is over 2 x 10

substrate concentration is over 2 x 10^-3^-3M. This is the maximal velocity (Vmax) of the enzyme, inM. This is the maximal velocity (Vmax) of the enzyme, in this case 60

this case 60 μμmol/min.mol/min.

b) v is constant at a

b) v is constant at a [S] above 2 x 10[S] above 2 x 10^-3^-3M because the substrates is saturating the enzyme.M because the substrates is saturating the enzyme.

c) Since the maximal velocity is achieved at a [S] of 2 x 10

c) Since the maximal velocity is achieved at a [S] of 2 x 10^-2^-2 M, almost all the enzyme is part of M, almost all the enzyme is part of an enzyme/substrate complex. The amount of enzyme free in solution is then negligible.

an enzyme/substrate complex. The amount of enzyme free in solution is then negligible.

5.2. Enzyme

5.2. Enzyme kineticskinetics a) Vmax = 0,25

a) Vmax = 0,25 μμmol/min;mol/min;

b) Km can be determined using the Michaelis-Menten equation:

v = [S] Vmax v = [S] Vmax [S] + Km [S] + Km v[S] + vKm = [S]Vmax v[S] + vKm = [S]Vmax vKm = [S]Vmax - v[S]

vKm = [S]Vmax - v[S]

vKm = [S] (Vmax - v) vKm = [S] (Vmax - v) Km = [S] (Vmax - v) Km = [S] (Vmax - v)

v v

Using the data for a [S] of

Using the data for a [S] of 5 x 105 x 10^-6^-6::

Km = 5 x 10

Km = 5 x 10^-6^-6 M x (0.25 μ M x (0.25μmol/min – 0.071mol/min – 0.071 μμmol/min)mol/min) 0.071

0.071 μμmol/minmol/min Km = 1.26 X 10 Km = 1.26 X 10^-5^-5 M M Note

Note: Similar data would be obtained if : Similar data would be obtained if a different [S] is chosen, as long a different [S] is chosen, as long as v < Vmax.as v < Vmax.

c) The initial velocity can be obtained using the Michaelis-Menten equation:

- for [S] = 1 x 10 - for [S] = 1 x 10^-6^-6 M: M:

v = [S] Vmax v = [S] Vmax [S] + Km [S] + Km v = 1 x 10

v = 1 x 10^-6^-6 M x 0.25 M x 0.25 μμmol/minmol/min 1 x 10

1 x 10^-6^-6M + 1.26 x 10M + 1.26 x 10^-5^-5MM v = 0.0184

v = 0.0184 μμmol/minmol/min - for [S] = 1 x 10

- for [S] = 1 x 10^-1^-1M: v = 0.25 μM: v = 0.25μmol/min (saturating [S]: v = Vmax).mol/min (saturating [S]: v = Vmax).

d) For a [S] of 2 x 10

d) For a [S] of 2 x 10^-3^-3 M, the initial velocity will be M, the initial velocity will be equal to Vmax. We therefore get:equal to Vmax. We therefore get:

v = Vmax = 0.25

v = Vmax = 0.25 μμmol/minmol/min After 5 min, we get: 0.25

After 5 min, we get: 0.25 μμmol/min x 5 min. = 1.25mol/min x 5 min. = 1.25 μμmol of product.mol of product.

- For a [S] of 2 x 10

- For a [S] of 2 x 10^-6^-6 M, we must first find the initial velocity using the Michaelis-Menten M, we must first find the initial velocity using the Michaelis-Menten equation:

equation:

v = [S] Vmax v = [S] Vmax [S] + Km [S] + Km v = 2 x 10

v = 2 x 10^-6^-6 M x 0.25 M x 0.25 μμmol/minmol/min 2 x 10

2 x 10^-6^-6 M + 1.25 x 10 M + 1.25 x 10^-5^-5 M M v = 0.035

v = 0.035 μμmol/minmol/min After 5 minutes of reaction, we get:

After 5 minutes of reaction, we get:

v = 0.035

v = 0.035 μμmol/min x 5 min. = 0mol/min x 5 min. = 0.175 μ.175 μmol of product.mol of product.

e) Since Km is independent of enzyme concentration, it will not be affected by the increase in e) Since Km is independent of enzyme concentration, it will not be affected by the increase in [E] and will remain equal to 1.25 x 10

[E] and will remain equal to 1.25 x 10^-5^-5 M. M.

However, Vmax will be changed because: Vmax = k

However, Vmax will be changed because: Vmax = k _cat_cat x x [E[E_tt]. Therefore, since k ]. Therefore, since k _cat_cat is a constant, is a constant, the 4 fold increase in [E] will also multiply the

the 4 fold increase in [E] will also multiply the Vmax by 4. Therefore, Vmax = Vmax by 4. Therefore, Vmax = 11 μμmol/min.mol/min.

5.3. Enzyme

5.3. Enzyme kineticskinetics

The Lineweaver-Burke plot is graph of the reciprocal of the initial velocity (1/v) as a function of The Lineweaver-Burke plot is graph of the reciprocal of the initial velocity (1/v) as a function of the reciprocal of the substrate concentration (1/[S]). From the a

the reciprocal of the substrate concentration (1/[S]). From the a vailable data, we get:vailable data, we get:

1/[S] M

The Lineweaver-Burke plot is shown below. The Km is easily found as the reciprocal value of The Lineweaver-Burke plot is shown below. The Km is easily found as the reciprocal value of the x-intercept.

the x-intercept. In this In this case: Km case: Km = 2.7 x = 2.7 x 1010^-5^-5 M. As for Vmax, we can obtain its value by the M. As for Vmax, we can obtain its value by the reciprocal of the intersection with the y axis. Here,Vmax = 67

reciprocal of the intersection with the y axis. Here,Vmax = 67 μ μmol/min.mol/min.

-1/Km = 36 750

5.4. Enzyme

5.4. Enzyme kineticskinetics

To determine the affinity of the enzyme for its substrate, we must find the value of the To determine the affinity of the enzyme for its substrate, we must find the value of the Michaelis-Menten constant. This is easily done using the Lineweave

Michaelis-Menten constant. This is easily done using the Lineweave r-Burke plot:r-Burke plot:

1/ [S] (M

1/ [S] (M^-1^-1 x 10 x 10^-4^-4)) 1/v (1/v (

(pH 7.6) (pH 7.6)

1/v ( 1/v (

(pH 9.0) (pH 9.0) 5.74

5.74 13.51 13.51 29.4129.41 3.75

3.75 11.76 11.76 21.2821.28 1.90

1.90 10.20 10.20 13.3313.33 0.60

0.60 8.77 8.77 7.817.81

0.25

0.25 - - 5.995.99

The Lineweaver-Burke plot is shown below.

The intersection with the x axis gives us -1/Km. From the graph, we can see that the smaller the The intersection with the x axis gives us -1/Km. From the graph, we can see that the smaller the value of -1/Km, the greater Km will be. Km is a measure of the affinity of the enzyme for its value of -1/Km, the greater Km will be. Km is a measure of the affinity of the enzyme for its substrate, and is equivalent to the amount of substrate required to reach 1/2 Vmax. Therefore, substrate, and is equivalent to the amount of substrate required to reach 1/2 Vmax. Therefore, when Km is high, the enzyme has a low affinity for its substrate: a lot of substrate is needed to when Km is high, the enzyme has a low affinity for its substrate: a lot of substrate is needed to acheive 1/2 Vmax.

acheive 1/2 Vmax.

From the graph, we can easily determine that the value of Km is greater at pH 9 than pH 7.6.

Therefore, the enzyme will have

Therefore, the enzyme will have a greater affinity for its substrate at pH 7.6.a greater affinity for its substrate at pH 7.6.

pH = 9 pH = 9

0 0 5 5 10 10 15 15 20 20 25 25 30 30 35 35

--1100 --88 --66 --44 --22 00 22 44 66 88

1/[S]

1 1 / / V V

pH = 7.6 pH = 7.6

mol

mol^-1^-1 x min.) x min.) molmol^-1^-1 x min.) x min.)

5.5. Enzyme

5.5. Enzyme kineticskinetics

To solve this problem, we must first draw a Lineweav

To solve this problem, we must first draw a Lineweav er-Burke plot:er-Burke plot:

1/[S] (M

1/[S] (M^-1^-1)) 1/v (mmol1/v (mmol^-1^-1 x min.) x min.) No inhibitor No inhibitor

1/v (mmol

1/v (mmol^-1^-1 x min.) x min.) With inhibitor With inhibitor 10 000

10 000

0.0357 0.0588

6 666.67 6 666.67

0.0277 0.0435

5 000 5 000

0.0233 0.0345

2 000 2 000

0.0154 0.0200

1 333.33 1 333.33

0.0135 0.0164

The Lineweaver-Burke is shown below.

a) Vmax is given by the

a) Vmax is given by the reciprocal of the intersection on the y reciprocal of the intersection on the y axis:axis:

1/Vmax = 0.0101

1/Vmax = 0.0101 μμmolmol^-1^-1 x min. x min.

Vmax = 99

Vmax = 99 μμmol/minmol/min

Since the intersection on the y axis is the same in the presence than in the absence of the Since the intersection on the y axis is the same in the presence than in the absence of the inhibitor, we can conclude that we are dealing with a competitive inhibition.

inhibitor, we can conclude that we are dealing with a competitive inhibition.

b) Km Km is is given given by by the the negative negative value value of of the the reciprocal reciprocal of of the the intersection intersection on on the the x x axis, axis, in in thisthis case:

case:

- In the absence of inhibitor:

-1/Km = 3 800 M -1/Km = 3 800 M^-1^-1 Km

Km = = 2.63 x 2.63 x 1010^-4^-4 M M - In the presence of the

- In the presence of the inhibitor:inhibitor:

-1/Km

-1/Km_app_app = 2000 M = 2000 M^-1^-1 Km

Km_app_app = 5 x 10 = 5 x 10^-4^-4 M M

c) Ki can be found with the values determined above and the equation:

Km_app_app = Km + Km [I] = Km + Km [I]

Ki Ki

Ki = Km [I]

a) To find Km and Vmax, we must first draw a Lineweaver-Burke plot (shown below):Vmax, we must first draw a Lineweaver-Burke plot (shown below):

1/[S] (M

The value of Vmax is given by the reciprocal of the intersection on the y axis:

Vmax

Vmax = = 2.362.36 μμmol/min.mol/min.

Similarly, Km is given by the negative value of the reciprocal of the intersection on the x Similarly, Km is given by the negative value of the reciprocal of the intersection on the x axis:

axis:

Km = 4.4 x 10 Km = 4.4 x 10^-4^-4 M. M.

b) From the Lineweaver-Burke plot, inhibitor A leads to a Vmax

b) From the Lineweaver-Burke plot, inhibitor A leads to a Vmax_app_app of 2.36 mmol/min and of 2.36 mmol/min and a

a KmKm_app_app of 8.3 x 10 of 8.3 x 10^-4^-4 M. This is expected from a competitive inhibitor (for example, a M. This is expected from a competitive inhibitor (for example, a substrate analog).

substrate analog).

As for inhibitor B, Vmax

As for inhibitor B, Vmax_app_app is 0,889 is 0,889 μμmol/min Kmmol/min Km_app_app is 4,4 x 10is 4,4 x 10^-4^-4 M. Since the Km is M. Since the Km is identical to the Km observed in the absence of the inhibitor, we are dealing with a identical to the Km observed in the absence of the inhibitor, we are dealing with a non-competitive inhibition. This type of inhibition is observed when the inhibitor does not competitive inhibition. This type of inhibition is observed when the inhibitor does not interfere with the enzyme-sunstrate interaction. This is expected from alkylating agents, interfere with the enzyme-sunstrate interaction. This is expected from alkylating agents, which would ever so slightly modify the structure of the enzyme, leading to alterations to which would ever so slightly modify the structure of the enzyme, leading to alterations to the active site and a less effective enzyme.

the active site and a less effective enzyme.

-1 -1 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7

--4400000 0 --2200000 0 0 0 2200000 0 4400000 0 6600000 0 8800000 0 110000000 0 1122000000

1/[S]

1 1 / / v v

Sans inhibiteur Sans inhibiteur Inhibiteur A Inhibiteur A Inhibiteur B Inhibiteur B No inhibitor No inhibitor Inhibitor A Inhibitor A Inhibitor B Inhibitor B

c) The inhibition constant for inhibitor A can be determined as follows:

Km_app_app = Km + Km [I] = Km + Km [I]

Ki Ki Ki = Km [I]

Ki = Km [I]

(Km

(Km_app_app-Km)-Km)

Ki = 4.25 x 10

Ki = 4.25 x 10^-4^-4 M x 5 x 10 M x 5 x 10^-4^-4MM (1.11 x 10

(1.11 x 10^-3^-3 M – 4.25 x 10 M – 4.25 x 10^-4^-4 M) M) Ki = 5.64 x 10

Ki = 5.64 x 10^-4^-4 M M

The inhibition constant for inhibitor B can be found as follows:

Vmaxapp = Vmax Vmaxapp = Vmax

(1 + [I]/Ki) (1 + [I]/Ki) Ki = Vmax

Ki = Vmax_app_app [I] [I]

Vmax- Vmax Vmax- Vmax_app_app

Ki = 0.889

Ki = 0.889 μμmol/min (3.2 x 10mol/min (3.2 x 10^-6^-6M)M) (2.36

(2.36 μμmol/min – 0.889 mmol/min)mol/min – 0.889 mmol/min) Ki = 1.93 x 10

Ki = 1.93 x 10^-6^-6MM

d) The initial velocity can be determined by a modification of the Michaelis-Menten d) The initial velocity can be determined by a modification of the Michaelis-Menten equation for competitive inhibitors:

equation for competitive inhibitors:

v =

v = Vmax Vmax x [S]x [S]

[S] + Km [S] + Km_app_app and Km

and Km_app_app = Km + Km [I] = Km + Km [I]

Ki Ki

Km_app_app = 4.4 x 10 = 4.4 x 10^-4^-4 M M + + 4.4 4.4 x x 1010^-4^-4 M x 2 x 10 M x 2 x 10^-5^-5 M M 5.64 x 10

5.64 x 10^-4^-4 M M Km

Km_app_app = 4.56 x 10 = 4.56 x 10^-4^-4 M M

Therefore:

v =

v = Vmax Vmax x [S]x [S]

[S] + Km [S] + Km_app_app

v = 2.36

v = 2.36 μμmol/min x 3 x 10mol/min x 3 x 10^-4^-4 M M 3 x 10

3 x 10^-4^-4 M + 4.56 x 10 M + 4.56 x 10^-4^-4 M M v = 0.936

v = 0.936 μμmol/minmol/min

5.7. Enzym

5.7. Enzyme catalysise catalysis

The fact that the optimal pH is 8 suggests that the (de)protonation of specific amino acids The fact that the optimal pH is 8 suggests that the (de)protonation of specific amino acids is important for enzyme activity. For example, at pH below 8, protonation of the side is important for enzyme activity. For example, at pH below 8, protonation of the side chain of His (pKaR = 6) could alter the active site either directly or indirectly (by chain of His (pKaR = 6) could alter the active site either directly or indirectly (by inducing a conformational change in the protein). At pH above 8, the deprotonation of inducing a conformational change in the protein). At pH above 8, the deprotonation of other side chains (e.g Tyr or Lys) co

other side chains (e.g Tyr or Lys) could have samilar effects.uld have samilar effects.

5.8. Enzym

5.8. Enzyme catalysise catalysis

a) For an enzyme whose optimal pH is 4, we can propose that the ascending part of the a) For an enzyme whose optimal pH is 4, we can propose that the ascending part of the curve would be attributable to the ionization of the lateral group of Asp or the C-terminal curve would be attributable to the ionization of the lateral group of Asp or the C-terminal carboxyl group of the protein. For the descending part of the curve, it could be caused by carboxyl group of the protein. For the descending part of the curve, it could be caused by the deprotonation of His or the ionization

the deprotonation of His or the ionization of Glu.of Glu.

b) For

b) For an ean enzyme whose nzyme whose optimal pH optimal pH is 11, is 11, the athe ascending pascending part of rt of the curve the curve might be might be duedue to the deprotonation of the lateral group of Lys. For the descending portion of the curve, to the deprotonation of the lateral group of Lys. For the descending portion of the curve, it could be due to the deprotonation of Arg.

it could be due to the deprotonation of Arg.

5.9. Enzym

5.9. Enzyme catalysise catalysis To be able to

To be able to answer this question, we must first draw the graph answer this question, we must first draw the graph of the rate of the of the rate of the reactionreaction

In document Docetaxel response characterization and identification of predictive biomarkers in metastatic castration resistant prostate cancer (página 79-85)