Category: Objective Questions

Bioprocess Engineering test focuses on “General Reaction Kinetics for Biological Systems”.

1. Which order represents Michelis-Menten kinetics?
A. First-Second order
B. Zero-First order
C. Zero-Second order
D. Second order

Answer: B
Explanation: The kinetics of many biological reactions are either zero-order, first-order or a combination of these called Michaelis-Menten kinetics. The reaction order depends on the relative size of the two terms in the denominator. At low substrate concentration [S] <<K_M, under these conditions the reaction rate varies linearly with substrate concentration [S] – First order kinetics. However at higher [S] with [S] >> K_M, the reaction becomes independent of [S] – Zero order kinetics.

2. The relationship between reaction rate and reactant concentration is of which order from the following relation?
r_A = k₁ C_A
A. Zero order
B. First order
C. Zero-First order
D. Second order

Answer: B
Explanation: If a reaction obeys first-order kinetics, the relationship between reaction rate and reactant concentration is as follows:
r_A = k₁ C_A
where r_A is the volumetric rate of reaction and k₁ is the first- order rate constant with dimensions T^-1.

3. The following enzyme reaction is applicable for which order?
A. Zero order
B. Zero-First order
C. Zero-Second order
D. First order

Answer: B
Explanation: The kinetics of most enzyme reactions are reasonably well represented by the Michaelis-Menten equation, where r_A is the volumetric rate of reaction, C_A is the concentration of reactant A, V_max is the maximum rate of reaction at infinite reactant concentration, and K_m is the Michaelis constant for reactant A. V_max has the same dimensions as r_A; K_m has the same dimensions as C_A.

4. Refer to Q3, and estimate the unit of K_m?
A. kg mol m^-2
B. kg mol
C. kg mol m^-1
D. kg mol m^-3

Answer: D
Explanation: K_m is the Michaelis constant for reactant A. K_m has the same dimensions as C_A. Typical units for V_max are kgmol m^-3 s^-I; typical units for K_m are kgmolm^-3.

5. What do you mean by the “Turn- over number”?
A. Occurrence of saturation
B. Conversion of amount of substrate into product
C. Conversion of amount of product into substrate
D. Enzyme unbound with substrate

Answer: B
Explanation: When all enzyme is bound to substrate k_cat, the turnover number, is the maximum number of substrate molecules converted to product per enzyme molecule per second. Further addition of substrate does not increase the rate which is said to be saturated.

6. High substrate concentration follows which order?
A. Zero order
B. First order
C. Zero- Second order
D. Second order

Answer: A
Explanation: v ≈ v_max
Therefore, at high substrate concentrations, the reaction rate approaches a constant value independent of substrate concentration; in this concentration range, the reaction is essentially zero order with respect to the substrate.

7. Convert 35°C into K.
A. 300.08 K
B. 305.18 K
C. 308.15 K
D. 315.18 K

Answer: C
Explanation: Convert temperatures to degrees Kelvin (K) using this equation:
T(K) = T(°C. + 273.15
T = 35°C = 308.15 K.

8. Convert 35 K into °C.
A. 230.18
B. 238.15
C. -230.18
D. -238.15

Answer: D
Explanation: Convert degree Kelvin (K) into temperatures using this equation:
T(°C. = T(K) – 273.15
= -238.15°C.

9. Catalytic efficiency is defined as__________
A. k_cat/K_M
B. K_M/k_cat
C. K_m/k₀
D. K_m/k₁

Answer: A
Explanation: The constant k_cat/K_M (catalytic efficiency) is a measure of how efficiently an enzyme converts a substrate into product.

10. Unbinding of Enzyme- Substrate complex increases the reaction rate.
A. True
B. False

Answer: B
Explanation: The rate of an enzymatic reaction will increase as substrate concentration increases, and that increased unbinding of enzyme-substrate complexes will decrease the reaction rate.

11. Linear inhibition is sometimes called as ____________
A. Complete inhibition
B. Partial inhibition
C. Incomplete inhibition
D. Mixed inhibition

Answer: A
Explanation: Linear inhibition is sometimes called complete inhibition, and the contrasting term partial inhibition is sometimes used for a type of non-linear inhibition in which saturation with inhibitor does not decrease the rate to zero. These latter terms are discouraged because they can be misleading, implying, for example, that the rate may indeed be decreased to zero in ‘complete inhibition’ at non-saturating concentrations of inhibitor.

12. The rate limiting step of Michaelis Menten kinetics is _________
A. Complex formation step
B. Non-complex formation step
C. Complex dissociation step
D. Non- complex dissociation step

Answer: C
Explanation: The rate limiting step in the enzyme catalyzed transformation of substrate S into product P is the breakdown of the ES complex. The rate-limiting step is usually the product formation step.

13. For a given enzyme catalyzed reaction, the Michaelis constant is 0.5 mM and the substrate concentration is 3.0 mM. What is the fractional saturation of the enzyme under these conditions?
A. 80.57%
B. 85.5%
C. 85.7%
D. 80.75%

14. In a particular enzyme-catalyzed reaction, V_max = 0.2 mol/sec and K_m = 5 mM. Assume the enzyme shows standard Michaelis-Menten kinetics. What is the rate of the reaction when [S] = 10 mM?
A. 0.133
B. 0.331
C. 0.233
D. 0.332

Answer: A
Explanation: v = V_max[S] / (K_m + [S])
= 0.2×10 / (5+10)
= 0.133.

15. In a particular enzyme-catalyzed reaction, V_max = 0.2 mol/sec and K_m = 6 mM. Assume the enzyme shows standard Michaelis-Menten kinetics. What is the rate of the reaction when [S] = 20 mM?
A. 0.155
B. 0.156
C. 0.153
D. 0.152

Answer: C
Explanation: v = V_max[S] / (K_m + [S])
= 0.2×20 / (6+20)
= 0.153.

Bioprocess Engineering Multiple Choice Questions on “Sterilisation”.

1. What do you mean by “Axenic culture”?
A. Containing single type of organism
B. Containing two types of organism
C. Containing multiple types of organism
D. Not containing any type of organism
Answer: A
Explanation: In biology, axenic describes the state of a culture in which only a single species, variety, or strain of organism is present and entirely free of all other contaminating organisms. Axenic culture is also an important tool for the study of symbiotic and parasitic organisms in a controlled manner.

2. Cell death in solids is more effective than liquid.
A. True
B. False
Answer: B
Explanation: If the liquid contains contaminant particles in the form of flocs or pellets, temperature gradients may develop. Because heat transfer within solid particles is slower than in liquid, the temperature at the centre of the solid will be lower than that in the liquid for some proportion of the sterilising time. As a result, cell death inside the particles is not as effective as in the liquid. Longer holding times are required to treat solid-phase substrates and media containing particles.

3. Hold-up of large volumes of medium for longer periods of time is efficient.
A. True
B. False
Answer: A
Explanation: Sustained elevated temperatures during heating and cooling are damaging to vitamins, proteins and sugars in nutrient solutions and reduce the quality of the medium. Because it is necessary to hold large volumes of medium for longer periods of time. Continuous sterilisation, particularly a high-temperature, short-exposure-time process, can significantly reduce damage to medium ingredients while achieving high levels of cell destruction.

4. Plug flow contains _______________
A. There is mixing
B. There is variation
C. There is no mixing but variation
D. There is neither mixing nor variation
Answer: D
Explanation: The type of flow in pipes where there is neither mixing nor variation in fluid velocity is called Plugflow. Plug flow is an ideal flow pattern; in reality, fluid elements in pipes have a range of different velocities.

5. For an ideal plug flow, Reynolds number should be ____________
A. Low
B. Variably low
C. High
D. Not very high
Answer: C
Explanation: Plug flow is approached in pipes at turbulent Reynolds numbers above about 2 × 10⁴; operation at high Reynolds numbers minimises fluid mixing and velocity variation.

6. Deviation from plug flow behavior is ____________
A. Linear dipersion
B. Axial dispersion
C. Circular dispersion
D. Non-dispersion
Answer: B
Explanation: Deviation from plug-flow behaviour is characterised by the degree of axial dispersion in the system, i.e. the degree to which mixing occurs along the length or axis of the pipe. Axial dispersion is a critical factor affecting design of continuous sterilisers.

7. Peclet number has no units.
A. True
B. False
Answer: A
Explanation: The Peclet number is a dimensionless number used in calculations involving convective heat transfer. It is the ratio of the thermal energy convected to the fluid to the thermal energy conducted within the fluid.

8. Performance of sterilizer is directly proportional to the Peclet number.
A. True
B. False
Answer: A
Explanation: At any given sterilisation temperature defining the value of k_d and Da, performance of the steriliser declines significantly as the Peclet number decreases.

9. The pores of filter for filtration should be between in diameter is _________
A. 0.1 – 0.55 μm
B. 0.2 – 0.45 μm
C. 0.3 – 0.45 μm
D. 0.3 – 0.50 μm
Answer: B
Explanation: Membranes used for filter sterilisation of liquids are made of cellulose esters or other polymers and have pores between 0.2 and 0.45 μm in diameter. The membranes themselves must be sterilised before use, usually by steam.

10. Heat sterilization is more effective than filtration for liquid medium.
A. True
B. False
Answer: A
Explanation: Liquid filtration is generally not as effective or reliable as heat sterilisation. Viruses and mycoplasma are able to pass through membrane filters; care must also be taken to prevent holes or tears in the membrane. Usually, filter-sterilised medium is incubated for a period of time before use to test its sterility.

11. Membrane filters and depth filters are same.
A. True
B. False
Answer: B
Explanation: A Depth Filter is a filter consisting of either multiple layers or a single layer of a medium having depth, which captures contaminants within its structure, as opposed to on the surface. Depth filters typically have nominal pore size ratings, Whereas, Membrane filters or “membranes” are microporous plastic films with specific pore size ratings. Also known as screen, sieve or microporous filters, membranes retain particles or microorganisms larger than their pore size primarily by surface capture.

12. Membrane filters are more efficient than depth filters.
A. True
B. False
Answer: A
Explanation: Depth filters do not perform well if there are large fluctuations in flow rate or if the air is wet; liquid condensing in the filter increases the pressure drop, causes channelling of the gas flow, and provides a pathway for organisms to grow through the bed, Whereas, Membrane filter cartridges typically contain a pleated, hydrophobic filter with small and uniformly-sized pores 0.45 pm or less in diameter. The hydrophobic nature of the surface minimises problems with filter wetting while the pleated configuration allows a high filtration area to be packed into a small cartridge volume.

13. Deep bed filters are used for filtration of?
A. Liquid
B. Solid
C. Gas
D. Non-fluid
Answer: A
Explanation: Deep bed filtration is a rapid and efficient method for removing small particles from liquids. Such dispersions of particles in liquids are common in a wide range of industries.

14. 20-micron filter is more efficient than 5-micron filter.
A. True
B. False
Answer: B
Explanation: The average size of the openings between pieces of the filter media are represented in microns. For example, a 20-micron filter has larger openings than a 5-micron filter. Consequently, the 20-micron filter element will let larger particles pass through the filter than the 5-micron media would.

15. What is an MPR rating on air filters?
A. Magnitude performance rating
B. Micro-particle performance rating
C. Macro-particle performance rating
D. Moles per rate
Answer: B
Explanation: MPR (Micro-Particle Performance Rating) MPR Rating is a rating system developed by 3M. It rates the manufacturer’s filters and their ability to capture airborne particles smaller than 1 micron.

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Bioprocess Engineering online quiz focuses on “Major Classes of Interactions in Mixed Cultures”.

1. Which of the following defines the given statement – “An association between two organisms in which one benefits and the other derives neither benefit nor harm”?
A. Neutralism
B. Mutualism
C. Commensalism
D. Amensalism
Answer: C
Explanation: Commensalism, in biology, a relationship between individuals of two species in which one species obtains food or other benefits from the other without either harming or benefiting the latter.

2. Which of the following defines the given statement – “Any relationship between organisms of different species in which one organism is inhibited or destroyed while the other organism remains unaffected”?
A. Neutralism
B. Mutualism
C. Commensalism
D. Amensalism
Answer: D
Explanation: One type of relationship that has been classified by biologists and ecologists is amensalism. Amensalism is any relationship between organisms of different species in which one organism is inhibited or destroyed while the other organism remains unaffected.

3. Which of the following defines the given statement – “Two organisms of different species exist in a relationship in which each individual benefits from the activity of the other”?
A. Neutralism
B. Mutualism
C. Commensalism
D. Amensalism
Answer: B
Explanation: Mutualism is the way two organisms of different species exist in a relationship in which each individual benefits from the activity of the other. Similar interactions within a species are known as co-operation.

4. Symbiosis is a Short term interaction.
A. True
B. False
Answer: B
Explanation: Biological interactions are effects that organisms have on each other, whether
▪ Short-term interactions, especially predation
▪ Long-term interactions, now called symbiosis, which includes commensalism, parasitism, and mutualism.

5. Which of the following is an endosymbiotic in nature?
A. Nucleus
B. Vacuole
C. Chloroplast
D. Ribosome
Answer: C
Explanation: Endosymbiotic theory that attempts to explain the origins of eukaryotic cell organelles such as mitochondria in animals and fungi and chloroplasts in plants was greatly advanced by the seminal work of biologist Lynn Margulis in the 1960s.

6. Mutualistic symbiosis is a type of mutualism.
A. True
B. False
Answer: A
Explanation: Mutualistic Symbiosis is a type of mutualism in which individuals interact physically, or even live within the body of the other mutualist. Frequently, the relationship is essential for the survival of at least one member.

7. Competition is having in between two organism:
A. Negative effects on both
B. Positive effects on both
C. On one positive and on other negative
D. No one is either positively or negatively affected
Answer: A
Explanation: Competition is an indirect interaction between two populations that has negative effects on both. In competition, each population competes for the same substrate. Two populations or microorganisms with similar nutrient requirements usually compete for a number of common, required nutrients when grown together.

8. Which of the following is correct according to an exclusion principle?
A. Both the organism competing for the resource exist
B. The organism with slowest growth rate will displace the other
C. The organism with fastest growth rate will displace the other
D. Both organism will equally survive
Answer: C
Explanation: The competitive exclusion principle, sometimes referred to as Gause’s Law of competitive exclusion or just Gause’s Law, states that two species that compete for the exact same resources cannot stably coexist. The organisms with the fastest growth rate will displace the others from the culture. This is known as the exclusion principle.

9. For Protocooperation species the interaction is not necessary.
A. True
B. False
Answer: A
Explanation: Protocooperation is where two species interact with each other beneficially; they have no need to interact with each other – they interact purely for the gain that they receive from doing this. It is not at all necessary for protocooperation to occur; growth and survival is possible in the absence of the interaction.

10. Symbiosis and mutualism are same.
A. True
B. False
Answer: B
Explanation: Symbiosis and mutualism are not the same. Symbiosis implies a relationship when two organisms live together. A symbiotic relationship may be mutualistic, but it may also be neutralistic, parasitic, commensalistic, and so on.

11. Competition and commensalism can occur at the same time.
A. True
B. False
Answer: A
Explanation: Note that these interactions can, and often do, exist in combination. For example, A and B may compete for glucose as a nutrient, but A requires a growth factor from B to grow. In such a case, both competition and commensalism would be present.

12. Lichens are ____________
A. A fungal-Bacterial association
B. A fungal-yeast association
C. A Bacterial-yeast association
D. A fungal-algal association
Answer: D
Explanation: Lichens are a fungal-algal symbiosis (that frequently includes a third member, a cyanobacterium). The mass of fungal hyphae provides a protected habitat for the algae, and takes up water and nutrients for the algae. In return, the algae (and cynaobacteriA. provide carbohydrates as a source of energy for the fungus.

13. Which of the following is not an example of Mutualistic Symbiosis?
A. Fungal-algal symbiosis
B. Algae and the corals
C. Mushroom and fly
D. Racoon and poison ivy
Answer: B
Explanation: Resource-resource mutualism -A relationship where one resource is traded for another. Corals and the symbiotic algae .The algae get inorganic nutrients from the corals, and the corals get sugars that are by-products of photosynthesis from the algae. When a coral ‘bleaches’ it is actually kicking out the zooxanthellae that live in it, so all you see is the coral’s skeleton, which is white.

14. Which one is the correct interaction sign for Amensalism?
A. +/+
B. 0/0
C. 0/+
D. 0/-
Answer: D
Explanation: Type of Interaction Sign Effects Mutualism +/+ both species benefit from interaction Obligate Mutualism +/+ obligatory; both populations benefit Commensalism +/0 one species benefits, one unaffected Neutralism 0/0 populations do not affect one another Amensalism 0/- One species is disadvantaged/one species unaffected.

15. Which one is the correct interaction sign for Commensalism?
A. +/+
B. 0/0
C. +/0
D. 0/-
Answer: C
Explanation: Type of Interaction Sign Effects Mutualism +/+ both species benefit from interaction Obligate Mutualism +/+ obligatory; both populations benefit Commensalism +/0 one species benefits, one unaffected Neutralism 0/0 populations do not affect one another Amensalism 0/- One species is disadvantaged/one species unaffected.

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Chemical Reaction Engineering Multiple Choice Questions & Answers on “Design Equations for Flow Reactors”.

1. The differential form of the plug flow reactor design equation is ______
A. FA0(frac{dX}{dV}) = -r_A
B. (frac{dX}{dV}) = -r_A
C. V(frac{dX}{dV}) = -r_A
D. (frac{dX}{dV}) = -r_A F_A0

Answer: A
Explanation: Molar flowrate and conversion are related in the mole balance equation to get the above equation.

2. The design equation for a plug flow reactor is ___________
A. V = (int_0^Xfrac{dX}{-rA} )
B. V = F_A0 (int_0^Xfrac{dX}{-rA} )
C. (frac{V}{-rA} = int_0^Xfrac{dX}{FA0} )
D. F_A0 = -r_A(int_0^Xfrac{dX}{V} )

Answer: B
Explanation: To determine the volume of PFR required to obtain a certain conversion, the mole balance equation in terms of molar flowrate and conversion is written and integrated to get the above equation.

3. The design equation of PBR to determine the weight of the catalyst is ________
A. (frac{W}{FA0} = int_0^X )-rA dX
B. (frac{W}{FA0} = int_0^X frac{dX}{dV} )
C. W = F_A0 (int_0^X frac{dX}{-rA} )
D. W = (int_0^X )FA0 dX

Answer: C
Explanation: The general mole balance for flow reactors is modified to get the above equation. The reaction rate is expressed in mol/s.g of catalyst in this equation.

4. In CSTR, the rate of reaction is evaluated at the exit conditions.
A. True
B. False

Answer: A
Explanation: The composition at exit is identical to the composition in the reactor. Hence, reaction rate is measured at the exit conditions.

5. A space velocity of 7 hr^-1 means ___________
A. 7 reactor volumes of feed is fed into the reactor per hour
B. Feed is fed at 7 hrs interval
C. Takes 7 hours to treat one reactor volume of feed
D. Reaction time is 7 hours

Answer: A
Explanation: Number of reactor volumes of feed that can be treated in unit time is called space velocity.

6. A space time of 10 minutes means __________
A. 10 reactor volumes of feed is fed into the reactor per minute
B. Feed is fed at 10 mins interval
C. Takes 10 mins to treat one reactor volume of feed
D. Reaction time is 10 hrs

Answer: C
Explanation: Time required to process one reactor volume of feed is called Space time.

7. Space time in terms of molar federate and initial concentration is _________
A. τ = C_A0 V₀
B. τ = (frac{CA0}{FA0} )
C. τ = C_A0 V
D. τ = (frac{VCA0}{FA0} )

Answer: D
Explanation: By definition, Space time = (frac{Volume , of , the , reactor}{Volumetric , feed , rate} = frac{V}{V0} ) and F_A0 = C_A0 V₀,
Therefore τ = (frac{VCA0}{FA0}. )

8. The performance equation for CSTR is_________
A. (frac{τ}{CA0} = frac{XA}{-rA} )
B. τ = (frac{XA}{-rA} )
C. τ = C_A0 -r_A
D. τ = (frac{XA}{CA0(-rA.} )

Answer: A
Explanation: General design equation of CSTR is modified and written in terms of initial concentration and space time to get the above equation.

9. Calculate the volume of CSTR in litre required to achieve 70% conversion. The feed contains 10 molA/litre with feed rate of 3 lit/min and rate constant 0.15 for the reaction -r_A = k(C_A)^1.5.
A. 3
B. 9
C. 18
D. 27

10. For a gaseous reaction carried out in 25 litre PFR with 5molA/lit and space time 3 mins. Calculate the conversion attained if reaction kinetics is given as follows
A → 2.5P, -r_A = (2 min^-1) C_A .
A. 0.45
B. 0.65
C. 0.95
D. 0.98

Answer: C
Explanation: For gaseous phase reaction
ε_A = (frac{2.5-1}{1}) = 1.5
For first order reaction in PFR by integral method
kτ = (1+ ε_A) ln((frac{1}{1-XA} )) – ε_AX_A
2*3 = (1+1.5)ln((frac{1}{1-XA} )) – 1.5X_A
6 = 2.5ln ((frac{1}{1-XA} )) – 1.5X_A
X_A = 0.95.

Chemical Reaction Engineering Multiple Choice Questions & Answers on “Design of Fluid Particle Reactors”.

1. State true or false.
The fluidized bed reactor is used for mixed flow of solids.
A. True
B. False
Answer: A
Explanation: Fluidised bed provides high agitation of solids by high fluid velocity. The gas flow is more complicated than mixed flow.

2. Which of the following does not provide plug flow of Solid – fluids?
A. Fluidised bed reactor
B. Countercurrent flow in blast furnaces
C. Crossflow in moving belt
D. Concurrent flow in driers
Answer: A
Explanation: Blast furnaces and cement kilns provide plug flow in countercurrent operation. Moving belt feeders are used in furnaces. Cocurrent flow is employed in polymer driers.

3. Which of the following does not control the design of fluid – solid reactor?
A. Reaction kinetics for single particles
B. Density of fluid being treated
C. Size distribution of solids
D. Flow patterns of solids and fluid
Answer: B
Explanation: The complexity of the reaction determines the use of flow reactor or plug flow reactor. The variation of temperature conditions determines the flow patterns of fluid by affecting the viscosity. The range of sizes determines the operational factors.

4. If (overline{X_{Ri}}) is the mean conversion of a reactant of particle size R_i, R_m is the particle of maximum size in the feed and F(R_i) is the fraction of R_i fed to the reactor, then the mean conversion of solids of a particular size ‘i’ leaving a plug flow reactor converting a mixture of particles of varying sizes is ____
A. (overline{X_{(B.Ri}} = ∑_{R(τ)}^{Rm})[1- X_(B.Ri](frac{F(Ri)}{F} )
B. 1-(overline{X_{(B.Ri}} = ∑_{R(τ)}^{Rm})[ X_Ri](frac{F(Ri)}{F} )
C. 1-(overline{X_{(B.Ri}} = ∑_{R(τ)}^{Rm})[1-X_(B.Ri] F (Ri)
D. 1-(overline{X_{(B.Ri}} = ∑_{R(τ)}^{Rm})[1-X_(B.Ri] (frac{F(Ri)}{F} )
Answer: D
Explanation: Mean value for fraction of ‘i’ unconverted is the summation of the product of fraction of the reactant unconverted in the particle size Ri and the fraction of feed in the size Ri.
1-(overline{X_{(B.Ri}}) = ∑(fraction of the reactant unconverted in the particle size Ri × the fraction of feed in the size Ri).

5. For mixed flow of particles containing a single unchanging size and uniform gas composition, the fraction unconverted for film resistance controlling is ____
A. (int_0^τ)(-(frac{t}{τ}))(frac{e}{t}^frac{-t}{t}) dt
B. (int_0^τ)(1 – (frac{t}{τ}))(frac{e}{t}^frac{-t}{t}) dt
C. (int_0^τ)(1 –(frac{t}{τ}))dt
D. (int_0^τ)(1 – (frac{t}{τ}))(frac{e}{t}^frac{-t}{t}) dt
Answer: B
Explanation: For a mixed flow reactor, the mean residence time, t in the reactor is, E = (frac{e}{t}^frac{-t}{t}). For a single size of particles converted in time τ, 1-(overline{X_{(B.}}) = ∫₀^τ(1 – X_B)(frac{e}{t}^frac{-t}{t})dt for an individual particle. For film resistance controlling, X_B = (frac{t}{τ}.)

6. For mixed flow of particles containing a single unchanging size and uniform gas composition, the fraction unconverted for chemical reaction controlling is ____
A. (int_0^τ)(1-(frac{t}{τ}))³(frac{e}{t}^frac{-t}{t}) dt
B. (int_0^τ)(1-(frac{t}{τ}))²(frac{e}{t}^frac{-t}{t}) dt
C. (int_0^τ)(1-(frac{t}{τ}))(frac{e}{t}^frac{-t}{t}) dt
D. (int_0^τ)(1-(frac{t}{τ}))^0.5(frac{e}{t}^frac{-t}{t}) dt
Answer: A
Explanation: For chemical reaction controlling, (frac{t}{τ}) = 1-(1- X_B)(^frac{1}{3}). Hence, 1-(overline{X_{(B.}}) = ∫₀^τ(1-(frac{t}{τ}))³(frac{e}{t}^frac{-t}{t}) dt.

7. If (frac{τ}{t}) = 0.5, average conversion for a particle B of constant size in a mixed flow reactor for film resistance controlling is ____
A. 0.45
B. 0.38
C. 0.536
D. 0.743
Answer: C
Explanation: 1-(overline{X_{(B.}}) = 0.5((frac{τ}{overline{t}}) – frac{1}{3!}(frac{τ}{overline{t}})^2+frac{1}{4!} (frac{τ}{overline{t}}))³-…, as obtained by expansion.
1-(overline{X_{(B.}}) = 0.4635 and (overline{X_{(B.}}) = 0.536

8. If (frac{τ}{t}=frac{1}{3},) average conversion for a particle B of constant size in a mixed flow reactor for chemical reaction controlling is ____
A. 0.92
B. 0.98
C. 0.75
D. 0.76
Answer: A
Explanation: (overline{X_{(B.}} = 3(frac{overline{t}}{τ}) -6 (frac{overline{t}}{τ})^2+ 6(frac{overline{t}}{τ})^3(1-e^frac{τ}{overline{t}}, 1-overline{X_{(B.}}) = 0.078 and (overline{X_{(B.}}) = 0.92

9. For mixed flow of a mixture of sizes of unchanging size particles, the mean residence time of a material of a given size is ____
A. t = (frac{Weight , of , all , solids , in , the , reactor}{Feed , rate , of , solids , to , the , reactor} )
B. t = (frac{Feed , rate , of , solids , to , the , reactor}{Weight of all solids in the reactor} )
C. t = (frac{Weight , of , all , solids , in , the , reactor}{Reactant , concentration} )
D. t = (frac{Reactant , concentration}{Weight , of , all , solids , in , the , reactor} )
Answer: A
Explanation: The mean residence time of a material of a given size is equal to mean residence time of solids in the bed. It is the fraction of weight of all solids in the reactor to the feed rate of solids in the reactor.

10. If weight of all solids in the reactor is 300kg and the feed rate of solids is (frac{50kg}{hr},) then the mean residence time(in hrs) of a material of a given size for mixed flow of a mixture of sizes of unchanging size particles is ____
A. 7
B. 6
C. 4
D. 3
Answer: B
Explanation: t = (frac{W}{F} = frac{300}{50}) = 6 hrs.

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Molecular Biology Multiple Choice Questions on “Double Stranded DNA Forms a Double Helix”.

1. The first X-ray diffraction patterns of DNA were taken in 1938 by ___________
A. William Asbury
B. Rosalind Franklin
C. Francis H. Crick
D. Linus Pauling

Answer: A
Explanation: The first X–ray diffraction patterns of DNA of low quality taken in 1983 by William Asbury. He used using DNA supplied by Ola Hammarsten and Torbjorn Caspersson.

2. In early 1950s high quality X-ray diffraction photographs of DNA suggesting the DNA being double helix and composed of two nucleotide strands. Who took those photographs?
A. Rosalind Franklin
B. William Asbury
C. Francis H. Crick and James D. Watson
D. Rosalind Franklin and Maurice Wilkins

Answer: D
Explanation: The high quality X-ray diffraction photographs were taken by Rosalind Franklin and Maurice Wilkins. It suggested that DNA structure was helical and might be composed of more than one nucleotide strand in early 1950s.

3. In 1952, an unambiguous established work in the laboratory of Alexander Todd led to the discovery of _____________
A. Chemical nature of DNA
B. X-ray diffraction structure of DNA
C. 3’-5’ phosphodiester linkage regularly links the nucleotides of DNA
D. Nucleic acid strands are held together by hydrogen bonds

Answer: C
Explanation: Chemical nature of DNA was discovered by Watson and Crick in 1952. They also found that nucleic acid strands are held together by hydrogen bonds. Again the first X-ray diffraction structure was discovered by William Asbury in 1938.

4. What should be the complementary strand of 3’….ATGGCTTGA….5’?
A. 3’….TACCGAACT….5’
B. 5’….TACCGAACT….3’
C. 3’….TAGGCAAGT….5’
D. 5’….TAGGCAAGT….3’

Answer: B
Explanation: The complementary of A is T and for G is C, therefore the complementary sequence should be ….TACCGAACT…. And the two strands are antiparallel as the given strand is in 3’ to 5’ direction the complementary strand should be in 5’ to 3’ direction.

5. The hydrogen bonds formed during A and T bonding occurs between C₆NH₂ of A and __________
A. C4O of T
B. PO3^–
C. C6O of A
D. C1O of T

Answer: A
Explanation: As the sugar molecule is connected to N1 in T the C4O and C1O are free. Also the C1O is farthest from C₆NH₂ thus the bond forms between C₆NH₂ and C4O. This stabilizes the model because of being in preferred tautomeric state.

6. In a DNA double helix the bases are held together by hydrogen bonds. These hydrogen bonds are _________
A. Covalent bonds
B. Non-covalent bonds
C. Ionic bonds
D. Van der Waals forces

Answer: B
Explanation: Hydrogen bond is a weak non covalent bond. This occurs between two molecules resulting from an electrostatic attraction between a proton and an electronegative atom of the other molecule. No share of electron takes place which is the criteria for non-covalent bonding.

7. It is easy to break the bond between A and T than in between G and C.
A. True
B. False

Answer: A
Explanation: It is easy to break the bond between A and T than in between G and C. This is because A and T has a double bond whereas G and C has triple bond connecting them. Thus, the energy required to break a double bond is lower compared to a triple bond.

8. With respect to the importance of hydrogen bonding and DNA double helix stability, which of the following statements is false?
A. Favorable tautomeric form of nucleotide bases
B. Contributes to the thermodynamic stability
C. Decreases the entropy
D. Specificity of base pairing

Answer: C
Explanation: When the two complementary strands of DNA comes together in an aqueous solution the water molecules attached to the polynucleotide strands are displaced from the bases. This creates disorder and this increases the entropy. Therefore, it finally increases the stability of the DNA duplex.

9. In the late 1970s non double helical form of DNA was also suggested but was discarded on the basis of certain factors. Which of the following is not a factor responsible?
A. X-ray crystallography
B. Nuclein
C. Topoisomerase
D. Nucleosome core particle

Answer: B
Explanation: Nuclein, later named as nucleic acid and eventually DNA was first discovered by Swiss physiological chemist Friedrich Miescher inside the nuclei of human WBCs in 1869. Thus nuclein is not a term related to the structural orientation and chemical nature of DNA.

10. Levene investigated and found that the nucleic acid is composed of poly-nucleotides and each nucleotide is composed of one base, a sugar molecule and a phosphate. This was performed on the genome of________________
A. Bacteria
B. Fungi
C. WBCs
D. Yeast

Answer: D