CHAPTER 6 Multiphase Systems(Elementary princilples of Chemical Engineering 4th Ed)
6.88) Serine (Ser, molecular weight=105.1g/mol) is a non-essential amino acid (seeFootnote16) produced by fermentation. As with many other fermentation products, substantial downstream processing is required to meet specifications on product purity. Crystallization from an aqueous solution is useful in meeting those specifications. The following table shows how serine solubility in water varies with temperature:
|
T(°C) |
5.0 10.0 15.0 20.0 25.0 30.0 35.0 45.0 50.0 |
|
Solubility17 (g Ser/100g H2O) |
18.45 22.71 26.88 30.22 35.91 39.40 44.34 50.77 53.76 |
(a) Prepare a plot of solubility as a function of temperature that can be used for interpolation.
(b) An aqueous solution of serine containing 60 g Ser/100 g H2O is pumped into a batch cooling crystallizer, and the temperature is reduced slowly to 10°C, causing the formation of crystals of the monohydrated salt Ser H2O.Using the given solubility data, estimate the mass of crystals produced per unit mass of feed solution and the fraction of serine fed that is recovered as crystalline product.
(c) The molecular structure of serine makes it much more hydrophilic than other amino acids, and therefore its solubility is about an order of magnitude greater than that of most other amino acids. The addition of methanol to reduce the solubility in the solution has been suggested. Experimental data show that the solubility of Ser as a function of methanol content is given by the correlation S/S0 =exp(-4.8xm)where xm is the mass fraction of methanol in a methanol–water solvent mixture,S0 (g/g solvent) is serine solubility in water at a given temperature, and S is the solubility in the methanol–water solvent. In an alternative to the processing scheme described in Part(b), sufficient methanol is added to the crystallizer after it has reached 10°C to produce a final solution that has a methanol-to-water mass ratio of 55:45, and the resulting system is allowed to come to equilibrium. Estimate the mass of crystals produced per unit mass of feed solution and the fraction of serine fed that is recovered as crystalline product.
In: Chemistry
Chymotrypsin serves as a catalyst in the "Hydrolysis" of pepdite bonds - found in carboxylic groups of amino acids, which consist of aromatic pedand groups and large hydrophobic groups. Chymotrypsin is also known to serve as a catalyst in the Hydrolysis of "ester - and amide bonds" , found in hydrophobic groups.
1. Would you expect the KMof the reactions of Chymotrypsin with the different substrates to have the same value ? explain your reasoning.
2. Would you expect the Vmaxof the reactions of Chymotrypsin with the different substrates to have the same value ? explain your reasoning.
3. Would you expect the " pH-dependence " of the "hydrolysis - reactions" of the different substrates to be the same for all substrates ? explain your reasoning.
In: Chemistry
Given the non-template strand of DNA, draw the template strand with the 5' and 3' ends labeled. Draw the RNA molecule with the proper 5' and 3' ends labeled
Non-Template Strand: 3'- AAT GCT CGT AGC TTC GAT CGG ATC GA-5'
My answers: Template= 5'- TAT CGA GCA TCG AAG CTA GCC TAG CT-3'
RNA Molecule= 3'- AUA GCU CGU UGC UUC GAU CGG AUC GA-5'
Next it asks, how many amino acids would the RNA molecule code for? - how would I figure this out?
In: Biology
Superoxide dismutase is a very unique protein. It is the only protein that can catalyze faster than the rate of diffusion, which is usually a limiting rate for proteins since you cannot catalyze something faster than the amount of time it takes for your substrate to arrive at the active site. However, superoxide dismutase is able to do this. If you look at a diagram of superoxide dismutase, you will find a large cluster of basic amino acids surrounding the active site where the substrate binds. First, state the purpose of superoxide dismutase and what its substrate is. Then discuss how you think superoxide dismutase is able to have a kcat/Km that is faster than the diffusion limit.
In: Biology
The role of Voltage Sensitive Na+ Channels (VSSC in neuron signaling) is:
a) Initiation of depolarization
b) propagation of depolarization
c) space from an axonal membrane of a neuron to dendrite membrane of next neuron
d) release of neurotransmitter
ATP production during aerobic respiration is by which mechanism:
a) substrate level phosphorylation
b) oxidative-phosphorylation
c) both substrate-level and oxidative-phosphorylation
d) This depends upon the temperature
e) this depends upon the pH
ATP production during glycolysis is by which mechanism:
a) substrate-level phosphorylation
b) oxidative-phosphorylation
c) both substrate-level and oxidative-phosphorylation
d) this depends upon the temperature
e) this depends upon the pH
Which is not a postulate of cell theory:
a) all life is made of cells
b) heredity is controlled by genes
c) all cells come from pre-existing cells
d) all cells are microscopic
If a molecule has a molecular weight (mw) of 3,685 daltons (Da), then how many grams are required to make 1.0 mole:
a) 3,685 g
b) 368.5 g
c) 36.85 g
d) 3.7 g
In: Biology
Questions 9 to 13 are in reference to the DNA sequence shown in Question 8.
Here is Question 8.
Question 8:
The top strand of the following segment of DNA serves as the template strand:
3’ TACACCTTGGCGACGACT 5’
5’ ATGTGGAACCGCTGCTGA 3’
We will refer to this segment of DNA as the original (or unmutated) sequence.
Please answer the following questions:
(a) What is the mRNA sequence?
The mRNA sequence is 5' 3'.
**Please enter your sequence in the 5' to 3' direction. Deductions will be made if a sequence is inputted in the wrong direction.**
(b) Using the mRNA sequence you determined in part (a) of this question, give the sequence of the protein that would be translated.
The amino acid sequence for this protein is N-terminus C-terminus.
**Please note**
The N-terminus refers to the beginning of the primary sequence for a protein, and the C-terminus refers to the end of the primary sequence for a protein.
i.e. input the amino acids in the order that they would be translated.
If a codon encodes for a stop codon, type STOP.
When inputting your sequence, separate each amino acid with a hyphen (e.g. Ser-Tyr-STOP).
You will need to consult the genetic code to answer this question.
Question 10:
The original (unmutated) DNA sequence (shown above in Question 8) has been mutated to the following (this represents the template strand):
3’ TACGACCTTGGCGACGACT 5’
We will refer to this sequence as mutation #2.
Please note that for simplicity only the template strand for this mutated segment of DNA is shown.
Answer the following questions:
(a) What is the complete mRNA sequence for the mutated segment mutation #2?
The mutated mRNA sequence is 5' 3'.
**Please enter your sequence in the 5' to 3' direction. Deductions will be made if a sequence is inputted in the wrong direction.**
(b) Using the mRNA sequence you determined in part (a) of this question, give the sequence of the protein that would be translated.
The amino acid sequence for this protein is N-terminus C-terminus.
**Please note**
The N-terminus refers to the beginning of the primary sequence for a protein, and the C-terminus refers to the end of the primary sequence for a protein.
i.e. input the amino acids in the order that they would be translated.
If a codon encodes for a stop codon, type STOP.
When inputting your sequence, separate each amino acid with a hyphen (e.g. Ser-Tyr-STOP).
You will need to consult the genetic code to answer this question.
4 points
QUESTION 11
Questions 9 to 13 are in reference to the DNA sequence shown in Question 8.
Question 11:
The original (unmutated) DNA sequence (shown above in Question 8) has been mutated to the following (this represents the template strand):
3’ TACACCTTAGCGACGACT 5’.
We will refer to this sequence as mutation #3.
Please note that for simplicity only the template strand for this mutated segment of DNA is shown.
Answer the following questions:
(a) What is the complete mRNA sequence for the mutated segment mutation #3?
The mutated mRNA sequence is 5' 3'.
**Please enter your sequence in the 5' to 3' direction. Deductions will be made if a sequence is inputted in the wrong direction.**
(b) Using the mRNA sequence you determined in part (a) of this question, give the sequence of the protein that would be translated.
The amino acid sequence for this protein is N-terminus C-terminus.
**Please note**
The N-terminus refers to the beginning of the primary sequence for a protein, and the C-terminus refers to the end of the primary sequence for a protein.
i.e. input the amino acids in the order that they would be translated.
If a codon encodes for a stop codon, type STOP.
When inputting your sequence, separate each amino acid with a hyphen (e.g. Ser-Tyr-STOP).
You will need to consult the genetic code to answer this question.
4 points
QUESTION 12
Questions 9 to 13 are in reference to the DNA sequence shown in Question 8.
Question 12:
(1 mark each)
In reference to the original sequence (shown in Question 8), classify each type of mutation present from Questions 9 to 11. Choose the best option for each.
|
|
QUESTION 13
Questions 9 to 13 are in reference to the DNA sequence shown in Question 8.
Question 13:
Most mutations have a neutral effect on the phenotype, function or survival of an organism because they do not elicit any noticeable change. Whereas other mutations can have a positive effect on an organism leading to new versions of proteins that help an organism adapt to changes in its environment; while other mutations can have a negative effect on the organism and result in a protein that does not function normally or at all.
Answer the following questions based on the responses you gave above in Questions 8 to 12.
(a) Based on the protein sequences that were produced as a result of mutation #1, mutation #2, or mutation #3, describe the effect, if any, these mutations would likely have on the function of the protein within the cell. Support your answer.
(b) If these mutations occurred within a germline cell and not a somatic cell, how would the effects of these mutations differ?
|
In: Biology

A BLAST search using the amino acid sequence of human parathyroid hormone (PTH) against the protein and DNA sequence databases returned close homologs in many vertebrate species ranging from primates to fish, but no sequence homologs in bacteria, archaea, plants, fungi, slime moulds, arthropods or nematodes. With reference to the tree of life shown above, where do you think PTH evolved? Explain your answer.
In: Biology
Please answer only Question 12 and 13.
Question 12:
In reference to the original sequence (shown in Question 8), classify each type of mutation present from Questions 9 to 11. Choose the best option for each.
|
|
3 points
QUESTION 13
Questions 9 to 13 are in reference to the DNA sequence shown in Question 8.
Question 13:
Most mutations have a neutral effect on the phenotype, function or survival of an organism because they do not elicit any noticeable change. Whereas other mutations can have a positive effect on an organism leading to new versions of proteins that help an organism adapt to changes in its environment; while other mutations can have a negative effect on the organism and result in a protein that does not function normally or at all.
Answer the following questions based on the responses you gave above in Questions 8 to 12.
(a) Based on the protein sequences that were produced as a result of mutation #1, mutation #2, or mutation #3, describe the effect, if any, these mutations would likely have on the function of the protein within the cell. Support your answer.
(b) If these mutations occurred within a germline cell and not a somatic cell, how would the effects of these mutations differ?
Here is Information you need to do Q 12 and 13.
The top strand of the following segment of DNA serves as the template strand:
3’ TACACCTTGGCGACGACT 5’
5’ ATGTGGAACCGCTGCTGA 3’
We will refer to this segment of DNA as the original (or unmutated) sequence.
Please answer the following questions:
(a) What is the mRNA sequence?
The mRNA sequence is 5' 3'.
**Please enter your sequence in the 5' to 3' direction. Deductions will be made if a sequence is inputted in the wrong direction.**
(b) Using the mRNA sequence you determined in part (a) of this question, give the sequence of the protein that would be translated.
The amino acid sequence for this protein is N-terminus C-terminus.
**Please note**
The N-terminus refers to the beginning of the primary sequence for a protein, and the C-terminus refers to the end of the primary sequence for a protein.
i.e. input the amino acids in the order that they would be translated.
If a codon encodes for a stop codon, type STOP.
When inputting your sequence, separate each amino acid with a hyphen (e.g. Ser-Tyr-STOP).
You will need to consult the genetic code to answer this question.
4 points
QUESTION 9
Questions 9 to 13 are in reference to the DNA sequence shown in Question 8.
Question 9:
The original (unmutated) DNA sequence (shown above in Question 8) has been mutated to the following (this represents the template strand):
3’ TACATCTTGGCGACGACT 5’.
We will refer to this sequence as mutation #1.
Please note that for simplicity only the template strand for this mutated segment of DNA is shown.
Answer the following questions:
(a) What is the complete mRNA sequence for the mutated segment mutation #1?
The mutated mRNA sequence is 5' 3'.
**Please enter your sequence in the 5' to 3' direction. Deductions will be made if a sequence is inputted in the wrong direction.**
(b) Using the mRNA sequence you determined in part (a) of this question, give the sequence of the protein that would be translated.
The amino acid sequence for this protein is N-terminus C-terminus.
**Please note**
The N-terminus refers to the beginning of the primary sequence for a protein, and the C-terminus refers to the end of the primary sequence for a protein.
i.e. input the amino acids in the order that they would be translated.
If a codon encodes for a stop codon, type STOP.
When inputting your sequence, separate each amino acid with a hyphen (e.g. Ser-Tyr-STOP).
You will need to consult the genetic code to answer this question.
4 points
QUESTION 10
Questions 9 to 13 are in reference to the DNA sequence shown in Question 8.
Question 10:
The original (unmutated) DNA sequence (shown above in Question 8) has been mutated to the following (this represents the template strand):
3’ TACGACCTTGGCGACGACT 5’
We will refer to this sequence as mutation #2.
Please note that for simplicity only the template strand for this mutated segment of DNA is shown.
Answer the following questions:
(a) What is the complete mRNA sequence for the mutated segment mutation #2?
The mutated mRNA sequence is 5' 3'.
**Please enter your sequence in the 5' to 3' direction. Deductions will be made if a sequence is inputted in the wrong direction.**
(b) Using the mRNA sequence you determined in part (a) of this question, give the sequence of the protein that would be translated.
The amino acid sequence for this protein is N-terminus C-terminus.
**Please note**
The N-terminus refers to the beginning of the primary sequence for a protein, and the C-terminus refers to the end of the primary sequence for a protein.
i.e. input the amino acids in the order that they would be translated.
If a codon encodes for a stop codon, type STOP.
When inputting your sequence, separate each amino acid with a hyphen (e.g. Ser-Tyr-STOP).
You will need to consult the genetic code to answer this question.
4 points
QUESTION 11
Questions 9 to 13 are in reference to the DNA sequence shown in Question 8.
Question 11:
The original (unmutated) DNA sequence (shown above in Question 8) has been mutated to the following (this represents the template strand):
3’ TACACCTTAGCGACGACT 5’.
We will refer to this sequence as mutation #3.
Please note that for simplicity only the template strand for this mutated segment of DNA is shown.
Answer the following questions:
(a) What is the complete mRNA sequence for the mutated segment mutation #3?
The mutated mRNA sequence is 5' 3'.
**Please enter your sequence in the 5' to 3' direction. Deductions will be made if a sequence is inputted in the wrong direction.**
(b) Using the mRNA sequence you determined in part (a) of this question, give the sequence of the protein that would be translated.
The amino acid sequence for this protein is N-terminus C-terminus.
**Please note**
The N-terminus refers to the beginning of the primary sequence for a protein, and the C-terminus refers to the end of the primary sequence for a protein.
i.e. input the amino acids in the order that they would be translated.
If a codon encodes for a stop codon, type STOP.
When inputting your sequence, separate each amino acid with a hyphen (e.g. Ser-Tyr-STOP).
You will need to consult the genetic code to answer this question.
In: Biology
| SALARY | EDUC | EXPER | TIME |
| 39000 | 12 | 0 | 1 |
| 40200 | 10 | 44 | 7 |
| 42900 | 12 | 5 | 30 |
| 43800 | 8 | 6 | 7 |
| 43800 | 8 | 8 | 6 |
| 43800 | 12 | 0 | 7 |
| 43800 | 12 | 0 | 10 |
| 43800 | 12 | 5 | 6 |
| 44400 | 15 | 75 | 2 |
| 45000 | 8 | 52 | 3 |
| 45000 | 12 | 8 | 19 |
| 46200 | 12 | 52 | 3 |
| 48000 | 8 | 70 | 20 |
| 48000 | 12 | 6 | 23 |
| 48000 | 12 | 11 | 12 |
| 48000 | 12 | 11 | 17 |
| 48000 | 12 | 63 | 22 |
| 48000 | 12 | 144 | 24 |
| 48000 | 12 | 163 | 12 |
| 48000 | 12 | 228 | 26 |
| 48000 | 12 | 381 | 1 |
| 48000 | 16 | 214 | 15 |
| 49800 | 8 | 318 | 25 |
| 51000 | 8 | 96 | 33 |
| 51000 | 12 | 36 | 15 |
| 51000 | 12 | 59 | 14 |
| 51000 | 15 | 115 | 1 |
| 51000 | 15 | 165 | 4 |
| 51000 | 16 | 123 | 12 |
| 51600 | 12 | 18 | 12 |
| 52200 | 8 | 102 | 29 |
| 52200 | 12 | 127 | 29 |
| 52800 | 8 | 90 | 11 |
| 52800 | 8 | 190 | 1 |
| 52800 | 12 | 107 | 11 |
| 54000 | 8 | 173 | 34 |
| 54000 | 8 | 228 | 33 |
| 54000 | 12 | 26 | 11 |
| 54000 | 12 | 36 | 33 |
| 54000 | 12 | 38 | 22 |
| 54000 | 12 | 82 | 29 |
| 54000 | 12 | 169 | 27 |
| 54000 | 12 | 244 | 1 |
| 54000 | 15 | 24 | 13 |
| 54000 | 15 | 49 | 27 |
| 54000 | 15 | 51 | 21 |
| 54000 | 15 | 122 | 33 |
| 55200 | 12 | 97 | 17 |
| 55200 | 12 | 196 | 32 |
| 55800 | 12 | 133 | 30 |
| 56400 | 12 | 55 | 9 |
| 57000 | 12 | 90 | 23 |
| 57000 | 12 | 117 | 25 |
| 57000 | 15 | 51 | 17 |
| 57000 | 15 | 61 | 11 |
| 57000 | 15 | 241 | 34 |
| 60000 | 12 | 121 | 30 |
| 60000 | 15 | 79 | 13 |
| 61200 | 12 | 209 | 21 |
| 63000 | 12 | 87 | 33 |
| 63000 | 15 | 231 | 15 |
| 46200 | 12 | 12 | 22 |
| 50400 | 15 | 14 | 3 |
| 51000 | 12 | 180 | 15 |
| 51000 | 12 | 315 | 2 |
| 52200 | 12 | 29 | 14 |
| 54000 | 12 | 7 | 21 |
| 54000 | 12 | 38 | 11 |
| 54000 | 12 | 113 | 3 |
| 54000 | 15 | 18 | 8 |
| 54000 | 15 | 359 | 11 |
| 57000 | 15 | 36 | 5 |
| 60000 | 8 | 320 | 21 |
| 60000 | 12 | 24 | 2 |
| 60000 | 12 | 32 | 17 |
| 60000 | 12 | 49 | 8 |
| 60000 | 12 | 56 | 33 |
| 60000 | 12 | 252 | 11 |
| 60000 | 12 | 272 | 19 |
| 60000 | 15 | 25 | 13 |
| 60000 | 15 | 36 | 32 |
| 60000 | 15 | 56 | 12 |
| 60000 | 15 | 64 | 33 |
| 60000 | 15 | 108 | 16 |
| 60000 | 16 | 46 | 3 |
| 63000 | 15 | 72 | 17 |
| 66000 | 15 | 64 | 16 |
| 66000 | 15 | 84 | 33 |
| 66000 | 15 | 216 | 16 |
| 68400 | 15 | 42 | 7 |
| 69000 | 12 | 175 | 10 |
| 69000 | 15 | 132 | 24 |
| 81000 | 16 | 55 | 33 |
This data set was obtained by collecting information on a randomly selected sample of 93 employees working at a bank.
SALARY- starting annual salary at the time of hire
EDUC - number of years of schooling at the time of the hire
EXPER - number of months of previous work experience at the time of hire
TIME - number of months that the employee has been working at the bank until now
2. Use the least squares method to fit a simple linear model that relates the salary (dependent variable) toeducation (independent variable).
a) What is your model? State the hypothesis that is to be tested, the decision rule, the test statistic, and your decision, usinga level of significance of 5%.
b) What percentage of the variation in salary has been explained by the regression?
c) Provide a 95% confidence interval estimate for the true slope value.
d) Based on your model, what is the expected salary of a new hire with 12 years of education
e ) What is the 95% prediction interval for the salary of a new hire with 12 years of education? Use the fact that the distance value = 0.011286
In: Statistics and Probability
1. In the complex reactions that make up cellular respiration, energy is transferred by
A. phosphorylation reactions
B. redox reactions
C. neither 1 or 2
D. both 1 and 2
2. Respiration of carbohydrate substrates provides more energy than proteins substrates because
A. proteins have no C-H bonds
B. proteins cannot be processed by glycolysis
C. proteins compete with NADH reduction reactions
D. proteins release NH3
3. Which of the 4 stages of respiration is anaerobic, and can continue without oxygen?
A. citric acid cycle
B. glycolysis
C. electron transport and oxidative phosphorylation
D. pyruvate processing
4. ATP is a regulatory molecule for the allosteric enzyme phospho-fructose kinase. This regulatory mechanism of the activity of glycolysis is called:
A. interference inhibition
B. competitive inhibition
C. feedback inhibition
D. active site inhibition
5. The NET output of glycolysis of 1 glucose molecule is
A. 4 ATP, 2 NADH
B. 2ATP, 2 FADH2
C. 2 ATP, 2 NADH
D. 2 ATP, 2 FADH2
6. Which of the four stages of respiration does NOT take place in mitochondria
A. pyruvate processing
B. glycolysis
C. Citric acid cycle
D. electron transport and oxidative phosphorylation
7. The citric acid cycle is a set of sequential reactions where 2 carbon atoms from Acetyl CoA (that derives from a glucose molecule) are coupled to a 4 carbon molecule. The carbon atoms from Acetyl CoA are released as CO2. Redox reactions and phosphorylation reactions are used to harvest energy in the form of ATP, NADH and FADH2. The product of these reactions is again a 4 carbon molecule that can be coupled with 2 carbon atoms from another Acetyl CoA to repeat the whole process
A. true, this is a correct description of the citric acid cycle
B. false, the citric acid cycle starts with pyruvate as substrate
C. false, the citric acid cycle employs only redox reactions
D. false, the citric acid cycle does not yield NADH
8. The net total output of energy rich molecules from respiration of a molecule of glucose after the citric acid cycle is
A. 2 ATP, 50 NADH, 4 FADH2
B. 2 ATP, 10 NADH, 2 FADH2
C. 4 ATP, 10 NADH, 2 FADH2
D. 4 ATP, 2 NADH, 10 FADH2
9. In the absence of oxygen, cellular respiration switches to fermentation. Fermentation is a set of alternative reactions to complete glycolysis and produce 2 ATP for a glucose molecule, by making sure that the following reagent is available
A. NAD (oxidized electron carrier)
B. EtOH (alcohol)
C. NADH (reduced electron carrier)
D. Acetyl CoA (substrate for citric acid cycle)
In: Biology