Biochemistry Mid-Semester Exam — 2019 (Past OmpathStudy

Biochemistry Mid-Semester Exam — 2019 (Past Paper) University of Nairobi Level 1 Biochemistry mid-semester exam, 2019. Full text transcription of the original scanned paper — questions only. --- Surname: COSPETER Other names FAITH NAIPUTA Reg. No. I29/1966/2018 Group. B2 UNIVERSITY OF NAIROBI DEPARTMENT OF BIOCHEMISTRY MID OF FIRST SEMESTER CONTINUOUS ASSESSMENT TEST FOR YEAR-I MBCHB, BPHARM & BDS (2018/2019) DATE: Tuesday, January 22, 2019 TIME: 0900AM - 1200 NOON SECTION A (30 Marks): Answer ALL questions in the spaces provided 1. (a) State:- Beer's Law When a ray of monochromatic light passes through an absorbing medium, its absorbance is directly proportional to the concentration of the solution. Lamberts Law When a ray of monochromatic light passes through an absorbing medium, its absorbance is directly proportional to the length of the medium light path. (b) Write the formula/expression for Beer- Lamberts Law and show how Absorbance and Transmittance are related Transmittance = Absorbance x Extinction coefficient Extinction coefficient Absorbance (A) = log(T) Absorbance = I0/I :---------------------- :---------------- log I/I = kcl I0 - intensity of incident light I - final intensity of light (c) With illustration, explain the relationship between absorbance and concentration of a given absorbing medium? The higher the concentration of a given solution the higher the absorbance this is due to higher number of particles per unit volume hence more light is absorbed by the particles. e.g. 1 mole of hydrochloric acid in 500ml of water & 1 mole of hydrochloric acid in 1000ml of water The solution A obtained has higher absorbance as compared to the solution B, this is because the number of hydrochloric particles per unit volume of water in solution A is higher than in solution B thus greater Absorbance of solution A, when a ray of monochromatic light is passed through it. Page 1 of 14 (c) A solution Y of 29.3 g/litre has an absorbance of 0.25 at 260nm. If the light path is 1 cm and the molecular weight is 586, calculate:- The molar extinction coefficient E = Absorbance / length of light path = 0.25 / 1 = 0.25 The transmittance of 10 µmole/litre of solution Y T = 0.25 x 10 = 2.5 (d) What is the importance of Beer-Lambert Law in biochemistry/medicine? used in pharmaceutical in the manufacture of drugs, in determination of the drug's concentration. used in diagnosis and prognosis of a certain disease, by use of biochemistry knowledge. Page 2 of 14 2. (a) Identify the three bonding theories (i) Lewis Bonding theory (ii) Valence Bonding theory (iii) Molecular orbital theory (b) In reference to NO (i) How many electrons are in NO molecule? 15 electrons (ii) Draw a clearly labeled molecular orbital diagram of NO (Molecular orbital diagram of NO with energy levels, atomic orbitals (2s, 2p) and molecular orbitals (sigma, pi) for Nitrogen and Oxygen, showing electron filling) State (i) The magnetic properties Paramagnetic (ii) The Bond Order 2.5 and, comment on the existence of the NO? It exists. Justify your answer Its bond order is more than zero Page 3 of 14 3. The compound, 2-methyl-pent-2-ene can undergo an acid catalyzed addition reaction with hydrogen halides following Markovnikov addition reaction rule. (a) Explain the meaning of Markovnikov's rule (1 mark) When Hx (hydrogen halide) is added to an alkene compound the major product is formed when the H is bonded to a carbon atom with most number of hydrogen atoms while X is bonded to the carbon atom with number of hydrogen atoms. (b) Using curly arrows to show the movement of electron pairs, describe the stepwise mechanism of an electrophilic addition reaction between 2-methyl-pent-2-ene and hydrogen bromide (4 marks) (Chemical structures showing the reaction mechanism of 2-methyl-pent-2-ene with HBr, forming a carbocation intermediate and then the final product) When 2-methyl pent-2-ene and hydrogen bromide undergo additional reaction we would have an electron rich area hence reacts with hydrogen ion electron deficiency at point of carbon atom. The carbon atom is electron positively charged with less number of hydrogen bromide ion likely carbocations formed in the reaction mechanism. (c) Draw the structure of the two most likely carbocations formed in the reaction above (1 mark) (Chemical structures of two carbocations) (d) Give the structural formula and the IUPAC names of the major and minor products formed in the above addition reaction (4 marks) (Chemical structures of major and minor products) Major product 2-Bromo-2-methylpentane Minor product 2-Bromo-3-methylpentane SECTION B (30 Marks) : Answer all questions in the spaces provided 1. Write the Lewis structures of the following molecule/ions and indicate dative bonds and formal charges (on the respective atoms). Atomic number: C=6, N=7, O=8, H=1 Page 4 of 14 1. (i) N₂O N₂O HCO₃⁻ anion :---------------- :------------------------------ 8. The following elements are paramagnetic except; A. N, 7 1S² 2S² 2P