Draw the structure of the predominant form of pyridoxal-

5-phosphate at pH 7.00

Short Answer

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The structure of pyridoxal-5-phosphate (PLP) at pH=7.00

Step by step solution

01

Step 1: Pyridoxal-5-phosphate (PLP):

Pyridoxal phosphate (PLP, pyridoxal-5-phosphate, P5P), the active form of vitamin B6, is a coenzyme in a variety of enzymatic reactions. The International Union of Biochemistry and Molecular Biology has catalogued more than 140 PLP-dependent activities, corresponding to ~4% of all classified activities. The versatility of PLP arises from its ability to covalently bind the substrate, and then to act as an electrophilic catalyst, thereby stabilizing different types of carbanionic reaction intermediates.

02

Structure of pyridocxal-5-phosphate (PLP):

In this task we will draw the structure of predominant form of pyridoxal-5-phosphate (PLP) at pH=7.00

Only NH has a value of pKa>7.00 so it would not be deprotonated at pH=7.00 and rest of residues would be because their pKa< 7.00

The structure of PLP at pH=7.00 :

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Most popular questions from this chapter

The both [H2A]=[HA]are equal as found to be pH=10.54

Consider the diprotic acidH2AwithK1=1.00×10-4andK2=1.00×10-8.Find the pH and concentrations ofH2A,HA-, andA2-in

(a)role="math" localid="1654926233413" 0.100MH2A;

(b) role="math" localid="1654926403235" 0.100MNaHA ;

(c) 0.100MNa2A .

In this problem, we calculate the pH of the intermediate form of a diprotic acid, taking activities into account.

(a) Including activity coefficients, derive Equation 10 - 11 for potassium hydrogen phthalate (K+HP- in the example following Equation 10 - 12 ).

(b) Calculate the pH of 0.050MKHP , using the results in part (a). Assume that the sizes of both HP- and P2- are 600pm . For comparison, Equation 10 - 11 gives pH=4.18.

Find the pH of 0.002MK+HP-with Equation10-11.

CO2(g)𝆏CO2(aq)KH=[CO2aq]PCO2=10-1.2073molkg-1bar-1at0C=10-1.6048molkg-1bar-1at30CCO2(aq)+H2O𝆏HCO3-+H+Ka1=[HCO3-][H+][CO2aq]=10-6.1004molkg-1at0C=10-5.8008molkg-1bar-1at30CHCO3-𝆏CO32-+H+CaCO3(S,aragonite)𝆏Ca2++CO3-2Ksparg=[Ca2+][CO32-]=10-6.1113mol2kg-2bar-2at0C=10-6.1391mol2kg-2bar-2at30CCaCO3(s,calcite)𝆏Ca2++CO3-2kspcal=[Ca2+][CO32-]=10-6.3652mol2kg-2bar-2at0C=10-6.3713mol2kg-2bar-2at30CEffect of temperature on carbonic acid acidity and the solubility of localid="1654949830957" CaCO3×14Box10-1states that marine life withlocalid="1654949841354" CaCO3shells and skeletons will be threatened with extinction in cold polar waters

before that will happen in warm tropical waters. The following equilibrium constants apply to seawater at0ndlocalid="1654949866125" 30C, when concentrations are measured in moles per kilogram of seawater and pressure is in bars:

localid="1654951136007" CO2g𝆏CO2aqKH=CO2aqPCO2=10-1.2073molkg-1bar-1at0C=10-1.6048molkg-1bar-1at30CCO2aq+H2O𝆏HCO3-+H+Ka1=HCO3-H+CO2aq=10-6.1004molkg-1at0C=10-5.8008molkg-1bar-1at30CHCO3-𝆏CO32-+H+CaCO3S,aragonite𝆏Ca2++CO3-2Ksparg=Ca2+CO32-=10-6.1113mol2kg-2bar-2at0C=10-6.1391mol2kg-2bar-2at30CCaCO3s,calcite𝆏Ca2++CO3-2kspcal=Ca2+CO32-=10-6.3652mol2kg-2bar-2at0C=10-6.3713mol2kg-2bar-2at30C

The first equilibrium constant is calledlocalid="1654949908801" KHfor Henry's law (Problem 10-10). Units are given to remind you what units you must use.

(a) Combine the expressions forlocalid="1654949922835" KH,K21, andlocalid="1654949935065" K22to find an expression forlocalid="1654949944862" [CO3-2]in terms oflocalid="1654949958224" PCO2andlocalid="1654949971486" [H+].

(b) From the result of (a), calculatelocalid="1654949980228" [CO32-](molkg-1)atlocalid="1654950000228" pCO2=800μbar and pH=7.8 at temperatures oflocalid="1654950013597" 0(polar ocean) andlocalid="1654950030147" 30C(tropical ocean). These are conditions that could possibly be reached around the year 2100 .

(c) The concentration oflocalid="1654950042348" Ca2+in the ocean islocalid="1654950053646" 0.010M. Predict whether aragonite and calcite will dissolve under the conditions in (b).

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