Homework 1:1. Which three cellular components are present in both prokaryotes and eukaryotes?A) ribosomes, chloroplasts, mitochondriaB) nucleus,ribosomes, RNAC) RNA, DNA, ribosomesD) endoplasmic reticulum, DNA, RNAE) mitochondria, DNA, RNA- Explanation: pro do not have a nucleus or membrane bound organelles2. Keq can be determined from the change in standard free energy using the equationA) Keq = e^−G°/ RTB) Keq = ln e^−G°/TSC)Keq = e^−H/RTD) Keq = e^−G°/TSE)Keq = log e^−G°/RT3. Sheets composed of two layers of amphipathic molecules arranged with the hydrophilicgroups on the surface and the hydrophobic groups buried in the center that form in water arecalledA) micelles.B) liposomes.C) vacuoles.D) bilayer membranes.E) none of the above- Explanation: in our cell membrane, lipid head are hydrophilic and their tails are hydrophobic4. Consider the reaction A + B <-> C + D. After reaching equilibrium at 25°C, the followingconcentrations of reactants and products were measured: [A] = 10M, [B] = 15M, [C] = 10M, [D] = 10M.Calculate G° for this reaction.A) 1000 J/molB) 10kJ/molC) 1 J/molD) insufficient data to determine answerE) none of the above- Explanation: Keq= products/ reactants : (10*10)/(10*15)= 100/150p- G° = -RTlnKeq: -(0.008314)(298)ln (100/150) = 1000J/mol (multiply by 1000 to get to J)5. The G° for the conversion of glucose 6-phosphate (G6P) to fructose 6-phosphate (F6P) is+1.7kJ/mole. In a particular human cell the concentration G6P is 8.0 μM and the concentrationF6P is 1.0 μM. Calculate the G of the reaction as it occurs in this cell at 37°C?A) −37 kJ/molB) −3.7 kJ/molC) 0 kJ/molD) +7.1 kJ/molE) −0.6 kJ/mol- Explanation: G = G° + RTln(products/reactants)- G = 1700(J/mol) +(8.314)(310)ln(1/8) = -3695 J/ mol - -3.7kJ/mol6. For a reaction with H=23 kJ/mol and S =22 J/K•mol, at 2°C, the reaction is:A) spontaneousB) nonspontaneousC) at equilibriumD) impossible to determine reactivityE) none of the above- Explanation: G = H - TS- G = 23000 (J/mol) - (275 * 22(J/mol)) = 16950- Since G is positive the rxn will be nonspontaneous7. Phosphoric acid is a polyprotic acid, with pK values of 2.14, 6.86, and 12.38. Which ionicform predominates at pH 9.3?A) H3PO4B) H2PO41−C) HPO42−D) PO43−E) none of the above- Explanation: the ionization of phosphoric acid takes place in steps- Triprotic phosphoric acid gives Hydrogens away in 3 steps- From pKa 0-2.12 and from 2.12-7.21 mostly H3PO4 (dihydrogen phosphate)- From pKa 7.21 - 12.67 mostly HPO42- : why it is the answer since it falls w/in that level8. The pK1 of citric acid is 3.09. What is the citric acid: monosodium citrate ratio in a 1.0 Mcitric acid solution with a pH of 2.09?A) 10:1B) 1:1C) 1:10D) 10:11E) 1:11- Explanation: buffer = citric acid + monosodium citrate- pH buffer = 2.09, pKa citric acid = 3.09- Hendersons equation: pH = pKa + log ([salt (citrate)]/[acid (citric acid)])- 2.09 = 3.09 + log (salt/acid) -> -1 = log (salt/ acid) -> 1= -log(salt/acid)- - > 1 = log(citric/acid) - > log10 = log (citric/acid) -> 10/1 = citric/acid9. A graduate student at TAMU wants to measure the activity of a particular enzyme at pH 4.0.To buffer her reaction, she will use a buffer system based on one of the acids listed below, whichacid is most appropriate for the experiment?A) Formic acid (Ka1.78×10−4)B) MES (Ka8.13×10−7)C) PIPES (Ka1.74×10−7)D) Tris (Ka8.32×10−9)E) Piperidine (Ka7.58×10−12)- Explanation: 1st find pKa of all acids and see which has the closest pKa to the given pH- pKa = - logKa- Formic acid: pKa = -log(1.78x10^-4) -> pKa = 3.7496- MES: pKa = -log(8.13x10^-7) -> pKa = 6.0899- PIPES: pKa = -log(1.74x10^-7) -> Pka= 6.7595- Tris: pKa = -log(8.32x10^-7) -> pKa= 8.0799- Piperidine: pKa= -log(7.58x10^-7) -> pKa= 11.1203- Formic acid has the closest pKa value so that is the answer10. The capacity of a buffer to resist changes in pH upon addition of protons or hydroxide ionsdepends on:A) the pKa of the weak acid in the buffer.B) the pH of the buffer.C) the total concentration of the weak acid and its conjugate base in the buffer.D) all of the aboveE) none of the aboveQUIZ 1:1. What are the four most abundant elements in a human body?A) C, N, O, HB) C, N, O, PC) C, S, O, HD) C, Na, O, HE) none of the above2. Using phylogeny all living organisms can be divided into the following domains:A) bacteria, eukarya, and vertebrateB) archaea and eukaryaC) bacteria, eukarya, and archaeaD) eukarya and bacteriaE) none of the above3. A reaction with a ______H and a ______S, will never be spontaneous.A) positive, positiveB) positive, negativeC) negative, positiveD) negative, negativeE) none of the above- Explanation: the reaction is both enthalpically and entropically opposed- It will be nonspontaneous (endergonic) at all temperatures4. Which of the following statements about water is not true?A) It has a high dielectric constant.B) It dissolves salts and polar substances.C) It can form two hydrogen bonds per water molecule.D) It packs in a hexagonal (honeycomb) shaped latticewhen the temperature falls below 0°C.E) In the liquid state it is only 15% less hydrogen bonded than in the solid state at 0°C.5. Amphiphilic moleculesA) have both oxidizing and reducing groups.B) are micelles.C) have chromophores in two different wavelength regions.D) have both acidic and basic groups.E) have both hydrophilic and hydrophobic groups.6. Physical properties that depend on the amounts of various species, rather than theidentities of those species, are calledA) osmotic properties.B) hydrophobic properties.C) London dispersion forces.D) aggregate properties.E) colligative propertiesHOMEWORK 21. Which amino acid does not have a primary alpha-amino group?A) glutamineB) arginineC) lysineD) prolineE) glutamate2. Which of the following amino acids is the least abundant in proteins?A) VB) WC) GD) FE) A3. Which of the following tripeptides carries a net positive charge at pH 7.0?A) Ala─Thr─AsnB) Gln─Val─SerC) Arg─Glu─MetD) Pro─Ile─LeuE) Leu─Lys─Gly- Explanations: Options A,B and D can be ruled out because they do not contain any basic AA- Although option C. (Arg-Glu-Met) has basic amino acid arginine with additionalamine group (+ charge), it gets neutralised by the negative charge conferredby the additional carboxylic acid group of Glutamate. Hence, option C canalso be ruled out.4. The pK1, pK2, and pKR for the