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# Multiple choice question for engineering

## Set 1

1. In the following equation, what does y signifies?

a) Dispersion
b) Displacement
c) Diffusion
d) Distance

Answer: d [Reason:] In single-phase systems, the rate of mass transfer due to molecular diffusion is given by Fick’s law of diffusion, which states that mass flux is proportional to the concentration gradient. JA is the mass flux of component A, N_A is the rate of mass transfer of component A, a is the area across which mass transfer occurs, DAB is the binary diffusion coefficient or diffusivity of component A in a mixture of A and B, C_A is the concentration of component A, and y is distance, (dC_A)/dy is the concentration gradient, or change in concentration of A with distance.

2. Molecular diffusion is which type of motion?
a) Thermal motion
b) Linear motion
c) Non- turbulent motion
d) Rectilinear motion

Answer: a [Reason:] Molecular diffusion, often simply called diffusion, is the thermal motion of all (liquid or gas) particles at temperatures above absolute zero. The rate of this movement is a function of temperature, viscosity of the fluid and the size (mass) of the particles.

3. What is Fick’s law applicable to?
a) Momentum transfer
b) Heat transfer
c) Mass transfer
d) Velocity transfer

Answer: c [Reason:] In single-phase systems, the rate of mass transfer due to molecular diffusion is given by Fick’s law of diffusion, which states that mass flux is proportional to the concentration gradient.

4. Determine the diffusivity of CO2 in a gas mixture having the composition:
CO2: 28.5 %, O2: 15%, N2: 56.5%,
The gas mixture is at 273 k and 1.2 * 105 Pa. The binary diffusivity values are given as: (at 273 K)
D12 P = 1.874 m2 Pa/sec
D13 P = 1.945 m2 Pa/sec
D23 P = 1.834 m2 Pa/sec
a) 1.51×10-5 m2/sec
b) 1.50×10-5 m2/sec
c) 1.61×10-5 m2/sec
d) 1.60×10-5 m2/sec

Answer: c [Reason:] Diffusivity of CO2 in mixture :

5. Refer to Q4, and calculate the diffusivity of O2 in a gas mixture.
a) 1.539×10-5 m2/sec
b) 1.639×10-5 m2/sec
c) 1.530×10-5 m2/sec
d) 1.630×10-5 m2/sec

Answer: a [Reason:] Diffusivity of O2 in the mixture,

6. Methane diffuses at steady state through a tube containing helium. At point 1 the partial pressure of methane is pA1 = 55 kPa and at point 2, 0.03 m apart PA2 = 15 KPa. The total pressure is 101.32 kPa, and the temperature is 298 K. At this pressure and temperature, the value of diffusivity is 6.75 * 10–5 m2/sec.
Calculate the flux of CH4 at steady state for equimolar counter diffusion.
For steady state equimolar counter diffusion, molar flux is given by:

a) 3.666×10-5 kmol/m2 sec
b) 3.866×10-5 kmol/m2 sec
c) 3.833×10-5 kmol/m2 sec
d) 3.633×10-5 kmol/m2 sec

7. Refer to Q6, and calculate the partial pressure at a point 0.02 m apart from point 1.
a) 20.63 kPa
b) 28.33 kPa
c) 28.63 kPa
d) 20.33 kPa

Answer: b [Reason:] Partial pressure at 0.02 m from point 1 is:

pA = 28.33 kPa.

8. In a gas mixture of hydrogen and oxygen, steady state equimolar counter diffusion is occurring at a total pressure of 100 kPa and temperature of 20°C. If the partial pressures of oxygen at two planes 0.01 m apart, and perpendicular to the direction of diffusion are 15 kPa and 5 kPa, respectively and the mass diffusion flux of oxygen in the mixture is 1.6 * 10–5 kmol/m2.sec, calculate the molecular diffusivity for the system.
For equimolar counter current diffusion:

a) 2.898×10-5 m2/sec
b) 3.898×10-5 m2/sec
c) 2.989×10-5 m2/sec
d) 3.989×10-5 m2/sec

Answer: b [Reason:] Substituting these in equation:

9. What is the unit of diffusion flux “J” in the Fick’s law equation?
a) mol m−2 s−1
b) mol m2 s−1
c) mol m2 s1
d) mol m−2 s1

Answer: a [Reason:] J is the “diffusion flux,” of which the dimension is amount of substance per unit area per unit time, so it is expressed in such units as mol m−2 s−1. J measures the amount of substance that will flow through a unit area during a unit time interval.

10. The diffusion of gas is ____________
a) Linear process
b) Slow process
c) Spontaneous process
d) Non- spontaneous process

Answer: c [Reason:] Diffusion refers to the process of particles moving from an area of high concentration to one of low concentration. The rate of this movement is a function of temperature, viscosity of the medium, and the size (mass) of the particles. Diffusion results in the gradual mixing of materials, and eventually, it forms a homogeneous mixture.

## Set 2

1. Post-translational modification includes which type of protein modification?
a) Ionic
b) Non-covalent
c) Covalent
d) Metallic

Answer: c [Reason:] Post-translational modification (PTM) refers to the covalent and generally enzymatic modification of proteins following protein biosynthesis. Proteins are synthesized by ribosomes translating mRNA into polypeptide chains, which may then undergo PTM to form the mature protein product. PTMs are important components in cell signaling.

2. What is the full form of GRAS in the regulatory approval of food products?
a) Generally regarded as sacred
b) Generally recognized as safe
c) Generally recommended as standard
d) Generally recognized standard

Answer: b [Reason:] GRAS stands for in the regulatory approval of food products as “Generally recognized as safe”.

3. E.coli is not a perfect host due to __________
c) Retained intercellularly
d) Production at medium levels

Answer: b [Reason:] E. coli is not a perfect host. The major problems result from the fact E. coli does not normally secrete proteins. When proteins are retained intracellularly and produced at high levels, the amount of soluble active protein present is usually limited due to either proteolytic degradation or insolubilization into inclusion bodies.

4. Heat shock method uses ___________
a) Calcium rich environment
b) Magnesium rich environment
c) Potassium rich environment
d) Sodium rich environment

Answer: a [Reason:] In the laboratory, bacterial cells can be made competent and DNA subsequently introduced by a procedure called the heat shock method. Heat shock transformation uses a calcium rich environment provided by calcium chloride to counteract the electrostatic repulsion between the plasmid DNA and bacterial cellular membrane.

5. Inclusion bodies are devoid of proteins.
a) True
b) False

Answer: b [Reason:] Inclusion bodies, sometimes called elementary bodies, are nuclear or cytoplasmic aggregates of stable substances, usually proteins. They typically represent sites of viral multiplication in a bacterium or a eukaryotic cell and usually consist of viral capsid proteins.

6. Resolubilization is generally for the chemical alternations.
a) True
b) False

Answer: b [Reason:] The protein in the inclusion body is misfolded. The misfolded protein has no biological activity and is worthless. If the inclusion bodies are recovered from the culture, the inclusion bodies can be resolubilized and activity restored. When resolubilization is straightforward and recoveries are high, the formation of inclusion bodies can be advantageous, as it simplifies the initial steps of recovery and purification. It is important that during resolubilization the protein be checked by several analytical methods to ensure that no chemical modifications have occurred. Even slight changes in a side group can alter the effectiveness of the product.

7. Endotoxins are intracellular not harmful.
a) True
b) False

Answer: b [Reason:] If the product is retained intracellularly, then the cell must be lysed (broken) during recovery. Lysis usually results in the release of endotoxins (or pyrogens) from E. coli. Endotoxins are lipopolysaccharides (found in the outer membrane) and can result in undesirable side effects (e.g., high fevers) and death. Thus, purification is an important consideration.

8. Pyrogen is __________
a) Antibiotic
b) Non-toxic
c) Drug
d) Fever inducer

Answer: d [Reason:] Pyrogen (fever) is a fever inducing substance.

9. Secretion and excretion are two different terms.
a) True
b) False

Answer: a [Reason:] Secretion is defined here as the translocation of a protein across the inner membrane of E. coli. Excretion is defined as release of the protein into the extracellular compartment.

10. B. subtilis has a number of problems that have hindered its commercial adoption mainly due to:
a) Increased levels of proteins
b) Decreased levels of proteins
c) Increased levels of proteases
d) Decreased levels of proteases

Answer: c [Reason:] A primary concern has been that B. subtilis produces a large amount and variety of proteases. These proteases can degrade the product very rapidly. Mutants with greatly reduced protease activity have become available, but even these mutants may have sufficient amounts of minor proteases to be troublesome. B. subtilis is also much more difficult to manipulate genetically than E. coli because of a limited range of vectors and promoters.

11. What do you mean by glycosylation?
c) Lysis of sugar moieties
d) Blockage of sugar molecules

Answer: a [Reason:] Glycosylation is a critical function of the biosynthetic-secretory pathway in the endoplasmic reticulum (ER) and Golgi apparatus. Approximately half of all proteins typically expressed in a cell undergo this modification, which entails the covalent addition of sugar moieties to specific amino acids.

12. What is the full form of AOX in Pichia pastoris expression system?
a) Artificial oxidase
b) Acetone oxidase
c) Alcohol oxidase
d) Acetyl oxidase

Answer: c [Reason:] Pichia pastoris has two alcohol oxidase genes, Aox1 and Aox2, which have a strongly inducible promoter. These genes allow Pichia to use methanol as a carbon and energy source. The AOX promoters are induced by methanol and are repressed by e.g. glucose. Usually the gene for the desired protein is introduced under the control of the AOX1 promoter, which means that protein production can be induced by the addition of methanol.

13. Transformed cell lines are immortal.
a) True
b) False

Answer: a [Reason:] Cells that can be propagated indefinitely are called continuous, immortal, or transformed cell lines. Cancer cells are naturally immortal. All cancerous cell lines are transformed, although it is not clear whether all transformed cell lines are cancerous.

14. Baculovirus infects insect cell lines and are also pathogenic to humans.
a) True
b) False

Answer: b [Reason:] The baculovirus that infects insect cells is an ideal vector for genetic engineering, because it is nonpathogenic to humans and has a very strong promoter that encodes for a protein that is not essential for virus production in cell culture. The insertion of a gene under the control of this promoter can lead to high expression levels (40% of the total protein as the target protein).

15. Plant cell cultures are better than animal cultures.
a) True
b) False

Answer: a [Reason:] Plant cell cultures, compared to animal cell cultures, grow to very high cell density, use defined media, and are intrinsically safer.

## Set 3

1. Prophylactic is also an antibiotic.
a) True
b) False

Answer: a [Reason:] A preventive measure. The word comes from the Greek for “an advance guard,” an apt term for a measure taken to fend off a disease or another unwanted consequence. A prophylactic is a medication or a treatment designed and used to prevent a disease from occurring. For example, prophylactic antibiotics may be used after a bout of rheumatic fever to prevent the subsequent development of Sydenham’s chorea. A prophylactic is also a drug or device, particularly a condom, for preventing pregnancy.

2. Hybridoma cells have an application to produce:
a) Antigens
b) Antibodies
c) Cancer cells
d) Cell lines

Answer: b [Reason:] Hybridoma cells are obtained by fusing lymphocytes (normal blood cells that make antibodies) with myeloma (cancer) cells. Lymphocytes producing antibodies grow slowly and are mortal. After fusion with myeloma cells, hybridomas become immortal, can reproduce indefinitely, and produce antibodies. Using hybridoma cells, highly specific, monoclonal (originating from one cell) antibodies can be produced against specific antigens.

3. Monoclonal antibodies are also used for chromatographic separations.
a) True
b) False

Answer: a [Reason:] MAb’s are also used for chromatographic separations to purify protein molecules. Purification of interferon by affinity chromatography is an example of the use of MAb’s for protein purification purposes.

4. Monoclonal antibodies are referred as ___________
a) Magic bullets
b) Magic gun
c) Magic shots
d) Magic bomb

Answer: a [Reason:] In the early 1900s, German Nobel Laureate Paul Ehrlich imagined an ideal therapy for disease, a drug precisely targeted to an invader, which if linked to a toxic chemical would act like a missile, carrying a destructive payload directly to the disease. Ehrlich said the drug would be a ‘Magische Kugel,’ which in English means ‘Magic Bullet.’ Such a therapy, he theorized, would be ideal for countless diseases, including cancer.

5. Antibody fragments lack fc domain.
a) True
b) False

Answer: a [Reason:] There are a range of applications in which Fc mediated effects are not required and are even undesirable. A common solution for applications where the antibody is only being used to block a signalling molecule or receptor is the use of antibody fragments that lack the Fc domain. This also helps to reduce the other main failure of therapeutic antibodies, namely the lack of delivery, which is especially true for anti-cancer antibodies.

6. Antibody fragments are advantageous than Monoclonal antibodies.
a) True
b) False

Answer: a [Reason:] The large size of MAb’s has limited their ability to penetrate some tumors. Antibody fragments can be used instead; these products can be made in nonmammalian cells. The use of smaller fragments enables deeper penetration with the affinity of the antibody also being critical and if it is too high this will restrict its ability to penetrate a tumour.

7. Interferon is a virus.
a) True
b) False

Answer: b [Reason:] Interferon (an anticancer glycoprotein secreted by animal cells upon exposure to cancer causing agents) is an example of an immunoregulator produced by mammalian cells. Interferon can be produced by either animal cells or recombinant (genetically engineered) bacteria.

8. Lymphokines are produced both by T-cell and B-cell.
a) True
b) False

Answer: b [Reason:] Lymphokines are a subset of cytokines that are produced by a type of immune cell known as a lymphocyte. They are protein mediators typically produced by T cells to direct the immune system response by signaling between its cells. Lymphokines aid B cells to produce antibodies.

9. Thymosins are:
a) Small proteins
b) Medium proteins
c) Large proteins
d) Globular proteins

Answer: a [Reason:] Thymosins are small proteins present in many animal tissues. They are named thymosins because they were originally isolated from the thymus, but most are now known to be present in many other tissues. Thymosins have diverse biological activities, and two in particular, thymosins α1 and β4, have potentially important uses in medicine, some of which have already progressed from the laboratory to the clinic. In relation to diseases, thymosins have been categorized as biological response modifiers.

10. Subunit vaccines from Virus contain viral DNA.
a) True
b) False

Answer: b [Reason:] A vaccine which, through chemical extraction, is free from viral nucleic acid and contains only specific protein subunits of a given virus; such vaccines are relatively free of the adverse reactions (for example, influenza virus) associated with vaccines containing the whole virion.

11. The poor antigen in a conjugate vaccine is:
a) Strong protein
b) Weak protein
c) A Polysaccharide
d) Non-polysaccharide

Answer: c [Reason:] A conjugate vaccine is created by covalently attaching a poor antigen to a strong antigen thereby eliciting a stronger immunological response to the poor antigen. Most commonly, the poor antigen is a polysaccharide that is attached to strong protein antigen.

12. Eicosanoids is a type of ________________
a) Hormone
b) Antibiotic
c) Vaccine
d) Antigen

Answer: a [Reason:] Eicosanoids are lipid hormones – hormones made from lipids, kinds of fats.

13. Amines is a type of hormone.
a) True
b) False

Answer: a [Reason:] There are two major chemical classes of hormones, peptides (proteins) and steroid hormones. Protein based hormones can be divided into three categories: proteins, peptides and amines.

14. Baculovirus infects insect cell lines and are also pathogenic to humans.
a) True
b) False

Answer: b [Reason:] The baculovirus that infects insect cells is an ideal vector for genetic engineering, because it is nonpathogenic to humans and has a very strong promoter that encodes for a protein that is not essential for virus production in cell culture. The insertion of a gene under the control of this promoter can lead to high expression levels (40% of the total protein as the target protein).

15. What do you mean by glycosylation?
c) Lysis of sugar moieties
d) Blockage of sugar molecules

Answer: a [Reason:] Glycosylation is a critical function of the biosynthetic-secretory pathway in the endoplasmic reticulum (ER) and Golgi apparatus. Approximately half of all proteins typically expressed in a cell undergo this modification, which entails the covalent addition of sugar moieties to specific amino acids.

## Set 4

1. A compound dissolves in water at a rate proportional to the product of the amount of undissolved solid and the difference between the concentration in a saturated solution and the actual solution; i.e., Csat – C(t). A saturated solution of this compound contains 40 g per 100 g of water. In a test run starting with 20 kg of undissolved compound in 100 kg of water, it was found that 5 kg dissolved in 3 hr. if the test continues, how many kg of compound will remain undissolved after 7 hr? Assume that the system is isothermal.
a) 11.56 kg
b) 10.72 kg
c) 11.76 kg
d) 10.52 kg

Answer: b [Reason:] Definitions of variables used: C = kg dissolved compound/ kg water Cs = saturated m = mass of undissolved solid mo = initial mass of solid W = kg of water C7= 0.093 and m: mo – C7.100 kg

m = 10.72 kg.

2. What is the first – order decay?
a) The rate of loss of the pollutant is constant
b) The rate of loss of the reactant is constant
c) The rate of loss of the pollutant is directly proportional to its concentration
d) The rate of loss of the reactant is directly proportional to its concentration

Answer: c [Reason:] First – order decay is the rate of loss of the pollutant is directly proportional to its concentration:

| dc/dt| reaction = -kC

For such a pollutant, mreaction = – V kC.

3. The water level in a municipal reservoir has been decreasing steadily during a dry spell, and there is a concern that the drought could continue for another 60 days. The local water company estimates that the consumption rate in the city is approximately 107 L/day. The state conservation service estimates that rainfall and stream drainage into the reservoir coupled with evaporation from the reservoir should yield a net water input rate of 106 exp (-t/100) L/day, where t is the time in days from the beginning of the drought, at which time the reservoir contained an estimated 109 liters of water.
Integrate the balance to calculate the reservoir volume at the end of the 60 days of continued drought.

a) 4.50 × 109 L
b) 4.50 × 108 L
c) 4.45 × 108 L
d) 4.45 × 109 L

Answer: c [Reason:] We now separate variables and integrate the differential balance equation from t=0 to t= 60 days.

4. Which one process is not a transient process?
a) Fed- batch
b) Batch
c) Continuous
d) Semi- batch

Answer: a [Reason:] Batch, semi-batch and continuous are transient as when they are start up, shut down or become transient at other times due to planned or unexpected changes in operating conditions.

5. Which of the following is the steady state condition based on the water tank concept?
a) Qin ≠ Qout
b) Qin = Qout
c) Qin > Qout
d) Qin < Qout

Answer: b [Reason:] Water flows in = Water flows out; If Qin = Qout: Steady state (No accumulation) No increase in level, mass, volume etc. In time, there is no change in the system.

6. Determine the volume required for a PFR to obtain the degree of pollutant reduction, Assume that the flow rate and first-order decay rate constant are unchanged (Q = 50 m3/day, k = 0.216 day-1), Cout / Cin = 32/100 = 0.32.
a) 204 m3
b) 234 m3
c) 254 m3
d) 264 m3

7. In a batch process, the reaction takes place in the presence of an acid medium. The acid is drained from the reaction vessel at the rate of 15ml/s as a result of the density difference of the acid from the reacting component. To avoid wastage of acid, it is recycled to an acid tank which has 1000 L capacity. The acid drained from the reaction vessel, picks up 50 g/L solids from the reactor. Acid is fed once again to the process from acid tank. When the process is started, the acid is almost pure in the tank as a result of filtration. As the reaction proceeds, acid in the tank gets more and more contaminated with the solids. The concentration of the solids should not exceed 100 g/L from the process point of view. The batch time is 16h. Estimate whether the concentration of the solids will exceed 100g/L during the batch reaction.
a) 37.04 h
b) 30.05 h
c) 36.04 h
d) 32.05 h

Answer: a [Reason:] Input of the solids to the tank = 15 (c+50) t∆g/1000

Output of solids from tank = 15 Δt c/ 1000g

Accumulation in the tank = 1000 (c+ Δc) – 1000c

= 1000Δc g Input – output = Accumulation

1/1000 [15 (c+50) Δt – 15cΔt] = 1000Δt

750 Δt = 1000000 Δc

Converting into differential form ,

dt = 1333.33dc, when t=0 and c=0.

Integration therefore yields,

t = 1333.33c C = 100g/L is the limit

t= 100 × 1333.33 = 133.33 s = 37.04 h.

8. Which of the following is not an example of batch process?
a) Cooking
b) Specialty chemicals
c) Refining
d) Brewing

Answer: c [Reason:] Refining is a continuous process as feed and products flow continuously through process. System is open, and usually modeled as steady flow. Examples: Petroleum refining (except coking, blending).

9. Due to algal growth in a water supply reservoir in a neighborhood, the water has acquired an unpleasant taste. The compound is non-degradable, and the water utility has decided that the best way to deal with the situation is simply to flush the contaminant out of the system. The reservoir contains 11,000 m3 of water, and the proposal is to pump 300 m3 /h of clean water through the system and discharge the effluent to the sewer system until 95% of the offending compound has been removed. If the reservoir is intensely mixed, how much flushing water will be required?
a) 30, 000 m3
b) 33, 000 m3
c) 35, 000 m3
d) 39, 000 m3

Answer: b [Reason:] The compound is non-reactive and therefore behaves as a tracer. The contaminant concentration in the flushing water is zero. When 95% of the contaminant has been flushed out, the concentration will be 5% of the initial concentration, so:

10.” A balloon is filled with air at a steady rate.” Which process can be correct for this condition?
a) Semi – batch
b) Batch
c) Fed – batch
d) Continuous

Answer: a [Reason:] Semibatch Process is neither batch nor continuous process. A process is operating at a steady state if the values of all the variables in the process do not change with time, except for small fluctuations about a constant mean values. A transient or unsteady-state exists when the process variables change with time. Furthermore, batch and semibatch processes are transient operations, and continuous process can be steady-state or transient.

## Set 5

1.” Bacteria have slightly higher nitrogen content than fungi” is this statement true or false?
a) True
b) False

Answer: a [Reason:] Bacteria tend to have slightly higher nitrogen contents (11-14%) than fungi (6.3-9.0%).

2. Estimate the degree of reduction of Methane, Glucose and Ethanol?

3. Calculate the stoichiometric coefficients of the following biological reaction:
Hexadecane: C16H34 + a O2 + b NH3 = c (C4.4H7.3N0.86O1.2) + d H2O + e CO2
a) a= 12.427, b= 2.085, c= 2.42, d= 12.43, e= 5.33
b) a= 12.345, b= 3.456, c= 2.42, d= 12.46, e= 5.44
c) a= 12.594, b= 2.345, c= 3.42, d= 12.49, e= 5.66
d) a= 12.345, b = 3.560, c= 3.42, d= 12.46, e= 5.44

Answer: a [Reason:] Amount of carbon in 1 mole of substrate = 16 (12) =192 g Amount of carbon converted to biomass = 192 (2/3) = 128 g Then, 128 = c (4.4)(12); c = 2.42 Amount of carbon converted to CO2 = 192 – 128 = 64 g 64 = e (12) e = 5.33 The nitrogen balance yields 14b = c (0.86)(14) b = (2.42)(0.86) = 2.085 The hydrogen balance is 34 (1) + 3b = 7.3c + 2d d = 12.43 The oxygen balance yields 2a(16) = 1.2c(16) + 2e(16) + d(16) a =12.427.

4. Calculate the stoichiometric coefficients of the following biological reaction:
Glucose: C6H12O6 + aO2 + bNH3 = c (C4.4H7.3N0.86O1.2) + d H2O + e CO2
a) a= 1.573, b= 0.685, c= 0.470, d= 2.564, e= 2
b) a= 2.789, b= 1.896, c= 0.438, d= 1.395, e= 1
c) a= 1.473, b= 0.782, c= 0.909, d= 3.854, e= 2
d) a= 2.390, b= 1.295, c= 0.943, d= 2. 564, e= 1

Answer: c [Reason:] Amount of carbon in 1 mole of substrate = 72 g Amount of carbon converted to biomass = 72(2/3) = 48 g Then, 48 = 4.4c (12); c= 0.909 Amount of carbon to CO2 = 72-48 = 24 g 24 = 12e e = 2 The nitrogen balance yields 14b = 0.86c (14) b= 0.782 The hydrogen balance is 12 + 3b = 7.3c + 2d d= 3.854 The oxygen balance yields 6(16) + 2(16)a = 1.2(16)c + 2(16)e +16d a = 1.473.

5. H10O5) and ammonium hydroxide (NH4OH) with a respiratory quotient of 1.4. Estimate the stoichiometric coefficient of the equation:
aC5H10O5 + bO2 + cNH4OH → CH1.66N0.13O0.40 + dCO2 + eH2O
a) a = 0.2823, b = 0.2938, c = 0.13, d = 0.4113, e = 0.9065
b) a = 0.2893, b = 0.3638, c = 0.13, d = 0.4321, e = 0.9478
c) a = 0.2843, b = 0.2590, c = 0.13, d = 0.4321, e = 0.9576
d) a= 0.2823, b = 0.2130, c = 0.13, d= 0.4113, e = 0.8743

Answer: a [Reason:] Equations for coefficients: C atom balance: 5a = 1 + d (1) H atom balance: 10a + 5c = 1.66 + 2e (2) O atom balance: 5a + 2b + c = 0.40 + 2d + e (3) N atom balance: c = 0.13 (4) Respiratory quotient: RQ = 1.4 = d/b (5)

Therefore, eq.2× eq.(3) – eq.(2), we get, b = 0.2938 From eq.(5), we get, d = 0.4113 From eq.(1), we get, a = 0.2823 From eq.(3), we get, e = 0.9065.

6. Estimating Yield from Stoichiometry:
3C6H12O6 + 2 + 2NH3 -> 2C5H7O2N + 8CO2 +14H2O
Given: 3(180) 8(32) 2(17) 2(113)
Calculate the yield of glucose.
a) 0.53 g cells/ g glucose used
b) 0.27 g cells/ g glucose used
c) 0.42 g cells/ g glucose used
d) 0.51 g cells/ g glucose used

7. The growth of S.cerevisiae on glucose under anaerobic conditions can be described by the following reaction:
C₆H₁₂O₆ + aNH₃ → 0.59 CH₁.₆₄N₀.₁₆O₀.₅₂ (biomass) + 0.43 C₃H₈O₃ + 1.54CO₂ + 1.3C₂H₅OH +0.036H₂O
Determine the biomass (MW= 23.74) yield coefficient Y x/s
a) 0.078 g. g⁻¹
b) 0.070 g. g⁻¹
c) 0.068 g. g⁻¹
d) 0.060 g. g⁻¹

Answer: a [Reason:] Yx/s= 0.59xMW biomass/1xMWglucose = 0.59×23.74/1×180 = 0.078 g. g⁻¹.

8. Refer Q7, and determine the product yield coefficient Y etoh/s and determine coefficient a.
a) 0.57 g. g⁻¹, 0.078
b) 0.33 g. g⁻¹, 0.094
c) 0.45 g. g⁻¹, 0.086
d) 0.44 g. g⁻¹, 0.056

Answer: b [Reason:] Yetoh/s = 1.3 x MW EtOH / 1x MW glucose = 0.59 x 46/ 1×180 = 0.33 g.g⁻¹ N in NH₃ = N in biomass (CH₁.₆₄N₀.₁₆O₀.₅₂) ∴ a = (0.59)(0.16) = 0.094.

9. What is the degree of reduction of glucose?
a) 4
b) 3
c) 12
d) 24