Q1: Sumatriptan (Tosymra) type

Transcription

1 Q1: Sumatriptan (Tosymra) type MM = 295 g/mol Corresponds to < 30 nonhydrogen atoms (C,N,O,..) AnBbody or protein contains > 50 amino acids, and each amino acid contains > 3 heavy atoms PepBde or Small molecule? 12 residue pepbde contains at least 4 heavy atoms Bmes 12 = 48 heavy atoms. 48 > 30 hence a small molecule

2 Q14: Kd calcula:on Problem: Drug is being added to 4 μm solubon of its target protein unbl 3:5 molar rabo (total drug:total protein) is achieved. Given that 2.33 μm of 1:1 complexes formed upon equilibrabon, find the dissociabon constant Kd. Solu:on: The molar rabo of 3:5 means that [L tot ] = 2.4 μm Since 2.33 μm of complexes was formed, unbound concentrabons are: [P] = [P tot ] [PL]= 4 μm 2.33 μm = 1.67 μm [L] = [L tot ] [PL]= 2.4 μm 2.33 μm = 0.07 μm K d = [P][L] / [PL] = / μm = 50 nm

3 Q18: K d to ΔG, ΔG = ΔH - TΔS Problem: When performed at 300K, an endothermic 1:1 drug/ target binding reacbon with Kd = 58 nm absorbs the molar heat of 1.4 kcal/mol. What is the molar entropy of this reacbon, ΔS? Solu:on: Find molar ΔG bind : ΔG = RT ln Kd = 0.6 kcal/mol ln ( ) = -10 kcal/ mol Molar ΔH bind of 1.4 kcal/mol is given ΔG = ΔH - TΔS; therefore, ΔS = (ΔH - ΔG) / T = ( kcal/mol) / 300 K = cal/mol / 300 K = 38 cal/(mol K) In this problem, entropy must be posi:ve! Given that ΔH is posibve, this is the only way to achieve negabve ΔG.

4 Week 5 problem solving + equa-ons, + applica-ons Colliga:ve proper:es, Nuclear Medicine

5 Van t Hoff Factor Problem: When the pharmaceubcal formulabon of drug X, [X] 2 Ca 2+ is dissolved in water, 30% of the molecules dissociated into three ions, 30% into two ions, and 40% did not dissociate. Calculate the van t Hoff factor for the solubon. A. 0.3 B. 0.4 C. 1 D. 1.9 E. 2 Solu:on: = = 1.9 OR: Every 100 molecules will produce = 190 parbcles. Answer: the van t Hoff factor is 1.9 (to be used for calculabon of all colligabve properbes)

6 Raoul s Law Problem: The pure water vapor pressure at 47 C is 0.1 bar. EsBmate the vapor pressure when 32 g of NaCl (MW = 58 g/mol) is added to 1 L of water. Assume that the salt is not volable, but it dissociates completely. A bar B bar C bar D bar Solu:on: 32 g = 32/58 ~ 0.55 mol of salt Dissolved AND DISSOCIATED in 1L = 55 mol of water; van t Hoff factor of 2 Molar fracbon of the solute x / 55 = Raoult's Law, ΔP w = x solute P w * = 0.02 * 0.1 = bar P w = P w * ΔP w = = bar Answer: The new vapor pressure is ~ bar.

7 Osmo:c pressure ΔP ~ ΔM 25 [atm L/mol] Problem: The blood glucose level of a diabebc pabent is approximately g/dl (dl = 0.1L). Given that glucose MW = g/mol, calculate the osmobc pressure created by glucose. A. 0.9 mmhg B. 19 mmhg C. 109 mmhg D. 209 mmhg Solu:on: no dissociabon occurs: van t Hoff factor = g/dl = 1.98 g/l ~ 11 mm = osmolarity ΔP osm = 11 mosm/l 25 L atm/mol = atm = 209 mmhg Answer: 209 mmhg Note that glucose does permeate membranes; the actual ΔP osm due to glucose between plasma and intersbbum is lower

8 Week 5 problem solving + equa-ons, + applica-ons EM radia:on, fluorescence, radioac:vity

9 EM bands & photon energies λ = m, ν = Hz h ev s h J s (Planck constant) 1 ev = J Energy vs Frequency relabon: Photon E = hν ν λ = c c m/s (speed of light)

10 Range λ ν Photon E Radio 1 m-10 5 km 3 Hz-300 MHz 12.4 fev-1.24 µev Microwa ve 1 mm-1 m GHz 1.24 µev-1.24 mev Far IR µm THz mev IR µm THz mev Near IR nm THz ev VIS nm THz ev Near UV nm PHz ev UV nm PHz ev Far UV nm PHz ev X-ray nm 30 PHz-30 EHz kev Gamma <10 pm >30 EHz >124 kev

11 Radia:on: wavelength to frequency T is Tera (12) P is Peta (15)

12 EM radia:on: frequency to photon energy The longest wavelength of EM radiabon that is needed to break a C=C bond is 200 nm Calculate the frequency of this radiabon in Hz Calculate the energy of a single photon in ev A. 47 mev B. 6.2 ev C. D. 3.1 kev 9.3 MeV Solu-on: Have just found that frequency ν = 1.5 PHz E photon = hν = ( ev s) ( Hz) = 6.2 ev Answer: 6.2 ev

13 Ac:vity of radiopharmaceu:cals CounBng the rate of radionuclei disintegra?on 1 Bq (Becquerel) = 1 event per second = 1 s -1 1 Ci (Curie) = 37 GBq Energy released depends on the specific nuclei and disintegrabon mode an average value may be associated with each radionuclide and decay mode AcBvity goes down as the radionuclide decays provided values are at?me of calibra?on Radiopharmaceu:cals are prescribed as ac-vity in Bq Source: FDA

14 Dosimetry-related quan::es Exposure: a measure of ionizabon v Electric charge freed by radiabon per kg of air v 1 R (roentgen) = C/kg (Coulomb per kg) Absorbed dose (energy/body mass): v 1 Gy (gray) = 1 J/kg v 1 rad = 0.01 Gy v Depends on the type of ma{er that absorbs the radiabon, e.g. for an exposure of 1 roentgen by 1 MeV γ-rays: the dose in air = rad the dose in water = rad the dose in averaged human Bssue = rad Dose equivalent: v Different radiabon types have different biological effects for the same deposited energy v W R is a correcbve radia?on weigh?ng factor: dependent on radiabon type converts the absorbed dose into an esbmate of Bssue damage v 1 Sv (Sievert) = W R Gy = W R (J/kg) v 1 rem (roentgen equivalent man) = 0.01 Sv Ionizing radia:on is prescribed as absorbed dose in Gy

15 Ac:vity of radiopharmaceu:cals Problem: 131 I decay releases energy in the form of β- (~90%) and γ- (~10%) radiabon. The average energy release per nucleus is 192 kev for β and 364 kev for γ. Given a sample of 131 I with total acbvity of 5 mci, how much energy does it emit per second in the form of β-radiabon? Give your answer in J/s. A. 57 kj/s B J/s C mj/s D µj/s Solu:on: 1 Bq = 1 nucleus per second = 1 s -1 ; 1 Ci = 37 GBq 5 mci = GBq = 185 MBq ( nuclei disintegrate per second) : converbng Curie units to Bq units. For β: ( ev) ( Bq) ( J/eV) = J/s