M201PS5Answers2008 - MICB 201 Chapter 5 Answers Problem#1(a Glucose and vitamins(b There are sources of C(CO2 from air N S P and trace elements

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MICB 201 Chapter 5 Answers Problem #1 (a) Glucose and vitamins (b) There are sources of C (CO 2 from air), N, S, P and trace elements that all bacteria (i) to (vi) can use, so this is not an issue. (i) Require NH 4 + as an energy and electron source and use O 2 as TEA, both are present so these organisms will grow. (ii) Require NO 2 - as an energy/electron source and O 2 as TEA; because NO 2 - not present these organisms will not grow. (iii) Require S or H 2 S as energy/electron sources and use O 2 as a TEA; because S or H 2 S not present these organisms will not grow. (iv) Require H 2 as energy/electron sources and CO 2 as a TEA; because H 2 not present these organisms will not grow. Note : these organisms are obligate anaerobes so will be killed by air. (v) Require light as an energy source and H 2 O as an electron source, no nutrient is required as an TEA. Since all their nutritional requirements are satisfied, these organisms will grow. (vi) Require light as an energy source and H 2 S or S as electron sources, no nutrient is required as an TEA. Since H 2 S or S are not present these organisms will not grow. Note : these organisms are obligate anaerobes so will be killed by air even if H 2 S or S were present. (c) (i) Yes, if the organism could use N 2 from the air as an N-source. (ii) No, there is no other source of S. (iii) No, there is no other source of P. Problem #2 (a) NADPH (b) Ribose + NADPH + 1 H + ---------> Deoxyribose + NADP + + H 2 O Problem #3 (a) An enzyme (b) In the cell, the energy that would have been released as heat is trapped in ATP. Problem# 4 (a) If a lipid bliayer was not impermeable to H + , a H + gradient could not be established across a membrane. (b) pH = 7 corresponds to a [H + ] = 1 x 10 -7 M = 1 x 10 -7 mol/L Volume of a sphere = 4/3 π r 3 Volume of cell = 4/3 π( 0.5 x 10 -4 cm) 3 = 4/3 π( 0.125 x 10 -12 ) cm 3 = 4/3 π( 0.125 x 10 -12 )mL = 4/3 π (0.125 x 10 -15 ) L = 0.523 x 10 -15 L 1 x 10 -7 mol/L x 0.523 x 10 -15 L x 6.02 x 10 23 H + /moL = 31.5 H +
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Problem #5 Electron Source IGNORE NH 4 + to N 2 : This related to the "Annamox reaction" (WebCT Box 5-3) which will not be discussing this term. + 2H + + NO 2 - NO 3 - H 2 O 2 e - NADP + + H + NADPH H 2 S S o + 2H + Electron Source + O 2 + H + NO 2 - 5 H + + NH 4 + 2 e - NADP + + H + NADPH + 2H 2 O SO 4 -2 + 4H + S o + O 2 2 e - NADP + + H + NADPH 2 e - NADP + + H + NADPH SO 4 + 4H + H 2 S + 1.5 O 2 + H 2 O 2 e - NADP + + H + NADPH Note if the bacterium was using S o or H 2 S as an electron source with SO 4
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This note was uploaded on 01/25/2010 for the course MICB MICB201 taught by Professor Kion during the Spring '10 term at The University of British Columbia.

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M201PS5Answers2008 - MICB 201 Chapter 5 Answers Problem#1(a Glucose and vitamins(b There are sources of C(CO2 from air N S P and trace elements

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