This document contains challenges to exam 1 questions, and response from the faculty member in bold. In summary, question 31 will be regraded to accept b or e as answers. The other challenges were not accepted.
Question #15:
Answers: B & E should be accepted.
H2O
Reason: The glycosidic bond linking two glucose molecules in an α1-4 manner occurs between the anomeric carbon of one glucose and a non-anomeric carbon of a second glucose, such as in Maltose.
This shows two glucose molecules joining together to make a maltose disaccharide, which decreases the reducing power AND results in a loss of OH from a non-anomeric carbon, which happens to be carbon #4 on the second glucose molecule. Two OH groups join together to make a glycosidic bond with water as a byproduct, therefore there is a loss of an OH group from the non-anomeric carbon.
Response: This possibility goes outside of what was covered in class. That is good in the sense that the student has pursued the details. And, yes, a OH is lost. But, the OH lost is from the anomeric C, not the non-anomeric C. Thus, this challenge, while an interesting one, is not accepted.
Q9) Accept C in addition to B
I put c) decrease release of oxygen to tissue because of what was mentioned in class about when the first oxygen BINDS to hemoglobin the second one is bound much more easier and so forth.My understanding of the question was cooperativity referred to how the oxygen binds to hemoglobin and when it is bound it requires other factors to work on it(i.e PH) for it to be released to tissue.Therefore when hemoglobin binds more oxygen we have less supply in the tissue( so release decreases the release to tissue (at the binding level)
Response: (b) and (c) cannot possibly both be correct, because they contradict each other. Cooperativity occurs during both binding and release. In class, we went into detail about how cooperativity resulted in “increased release of O2 to tissues”. Thus, this challenge is not accepted.
Q20) Correct answer B-Permeation is a factor and increases as pH increases, challenge to be D-Permeation is not a factor
Per Ch. 3 page 10, it states "the protonated form of anesthetics is the form that is active in blocking nerve impulses. However the protonated form is too polar (due to its charge) to cross the comparatively non-polar membranes that surround the nerve cells."
It states in the problem that the H+ will leave at pKa of 8. When injecting a range from 4.0 - 9.0 the anesthetic first increases and then decreases.
From pH 4.0-7.9 the H+ would be attached making the anesthetic + charged and would not permeate the cell making the anesthetic incapable of blocking the nerve if permeability was a factor, but since the problem stated that it first increased and then decreased. Meaning the anesthetic is working without crossing the membrane, leaving answer choice "D - permeation is not a factor" a reasonable choice.
Response: As we discussed in detail in class, titration of a H+ ion does not occur in an all-or-nothing fashion. Titration is progressive. A rising, then falling, effect implies that at least two factors are involved. Thus, this challenge is not accepted.
Q24) I think the question should either be thrown out or E should be a possible option based on the reasoning above.
The graph B, while obviously the most "appropriate" of the graphs shown is still somewhat misleading if inspected carefully. The value of y is decreasing at an accelerated rate and it does not appear that it will ever actually actually reach the Vmax value of 10 even if it were to go on forever. Please consider the rough estimates of the coordinates on the line of the graph provided: (2,5), (4,6.5), (6,7.5), (8,8), (10, 8.25). It is fair to say that one could expect the next two points to be (12, 8.375) and (14, 8.438) even continuing on to (with a little bit of calculator work) (30,8.4997). While unlikely, it is possible that the graph will reach the enzyme saturation point or Vmax at 10 s-1but there is nothing on the graph to indicate that it won't flatten out before or after 10 either. Unless the graph specifically show the line flattening out at 10 then I cannot assume that it does, especially when very careful consideration might indicate otherwise.
The main difficulty of this question lies in the fact that E. was even an option at all. Or if E. is to be included then I believe the question should be worded "Which of these plots would be most appropriate.." or "Which of these plots most closely approximates the Km values of 2 and Vmax value of 10s-1?"
Response: Plot B was generated using the Michaelis-Menten equation with a Km value of 2 mM and a Vmax value of 10 s-1. This is easily demonstrated if the data are converted to double-reciprocal form. The point that the Vmax value is only reached at very high concentrations of the substrate was emphasized several times in class.
Q27) I believe they should accept answers B and C.
Answer choice B states that it is an "inactivation of chymotrypsinogen" which can be interpreted to mean that when the peptide bond is cleaved that zymogen losses its function (irreversible). The wording of that question is confusing and vague and for that reason I believe they should accept B and C.
Response: Inactivation means loss of activity. Cleavage of chymotrypsin results in activation of the protein.
Question #28:
Answer: ALL answers should be given credit OR question thrown out.
Reason: When talking about the control of enzyme activity, allosteric enzymes, isozymes, zymogens, and phosphorylation we covered. The question asks which of the following would be most useful for a cell to rapidly modulate the responses of enzyme to changes in intracellular environment. In our lecture manual, the word “intracellular” was specifically used in the allosteric enzymes such and the covalent modifications (zymogens and phosphorylation) section. Because there are both negative and positive allosteric enzymes, it is not a guarantee that allosteric enzymes can rapidly modulate the responses. With negative cooperativity, “the velocity increases less slowly with an increase in substrate concentration,” which could be easily interpreted as not all allosteric enzymes rapidly modulate responses in the cell.
Since “intracellular” modulation was also used in the zymogen and phosphorylation section, it seems logical that zymogens and phosphorylation could rapidly modulate responses intracellularly. Therefore, ALL answers should be given credit OR question should be thrown out.
Response:
The question was not which ones could or could not be used, but which was “most useful.” Zymogen activation occurs outside of cells. Phosphorylation is generally in response to signals from outside the cell. The point that allostery allows a cell to adjust its metabolism rapidly was made both in lecture and in the review session.
Question 28 (again):
Given below, I feel that since 2/4 answers were correct in this case, the answer of (e) should be accepted or the question should be thrown out entirely.
I believe not only allosteric enzymes but also phosphorylation is useful for a cell to rapidly module the responses of enzymes to changes in the intracellular environment. On Pg.18 of Chapter 6, it reads, "Once the target enzyme is phosphorylated, its activity is changed. The activity may be either increased or decreased." Therefore, since I saw that both (b) allosteric enzymes and (d) phosphorylation could modulate the responses of enzymes to changes in the intracellular environment, I chose (e) all of the above (even though I did not completely agree with the other two answers).
Response: See above.
Question #30:
Answers: C & B should be given credit.
Reason: Part of what made the enzymes chapter confusing was the fact that Dr. Fitzpatrick told us in class that much of the information in the manual was “outdated” terminology and that we should go by what he has in the powerpoints. With that in mind, on slide #42 in his lecture it says, “The Vmax/Km value is an additional value that can give insight into enzyme function.” It also on that same slide says, “The Vmax/Km value describes the reaction between the free enzyme and substrate free in solution.” Both of these statements do not clearly state the same information as answer choice C: “The Vmax/Km value for an enzyme is used as a measure of the substrate specificity”.
In terms of answer choice B: “The kcat value for an enzyme tells you how the enzyme increases the rate of the reaction”, in my biochemistry book from college (Principles of Biochemistry 5th Edition – David L Nelson) it says that kcat is used to “describe the rate of any enzyme-catalyzed reaction at substrate saturation”. Because enzymes can only increase the rate of a reaction by lowering the activation energy, kcat clearly can “describe”/tell you how the enzyme increases the rate of the reaction. Therefore, answer choice B should also be counted correct because neither answer choice was clearly taught in lecture, and the manual was said be to incorrect in many places by Dr. Fitzpatrick
Response: See slide 43 for substrate specificity and Vmax/Km.
Q31) Which one of the following statements about subcellular organelles is FALSE?
B) In prokaryotes, ATP is synthesized in the mitochondrion.
E) Human phagocytes have degradative organelle called LYSOZYME*.
B & E should be accepted because in E, the organelle is a lysosome, not a lysozyme.
Response: This challenge is accepted. The incorrect word usage in E was inadvertant.
Question #32:
Answers: Both A & C should be given credit.
Reason: The question states, “A biochemist devises a way to measure the accumulation of product of an enzyme, and uses that to measure how much of the enzyme if present in various fraction during purification of the enzyme. This measurement is called .”
On page 10 in chapter 1 of the manual, it says “Classical purification of enzymes and other proteins generally proceeds by the procedure of multiple chromatographic fractionations.” The question is somewhat vague in what is asking for. “Various fractions” refer to chromatography which was taught to us as a “classical assay”.
Even on the page in the “updated manual” with the large diagraph of chromatography, the title of the page reads “Classical purification of enzymes and proteins by Chromatographic Fractionations.
Based upon my understand, I believed from the information provided that chromatography was a type of biochemical assay because as the bottom of page 10 it also reads “The assay could be: a. detection of the enzymatic activity by a classical assay (chromatography)”. Therefore, if you are measuring enzyme related things, a classical assay could be used, specifically, chromatography (a type of enzyme assay). Therefore, I believe both A and C should be counted correct due to ambiguity of information provided to us.
Response: This challenge is not accepted. The objective is to understand that the assay is the fundamental tool that biochemists use to determine how much activity is present. The assay is used in many contexts. In combination with a fractionation method (in this case column chromatography) the assay reveals which fractions have the activity and are therefore a more purified preparation of the enzyme than it was before column chromatography. In combination with affinity purification, the assay reveals if the protein has properly folded. In combination with kinetic analysis, the assay can reveal if the enzyme is allosterically regulated, or the characteristics of various inhibitors. The assay could be employed on cell lysates of different individuals to discover if any of them were missing this enzyme, perhaps due to genetic disease.
Q34) A and B should be accepted in addition to C
I was wondering if you could add Q34 to be challenged also. I know there is a chart in Ch.8 page 5 that blatantly says Type 2 collagen is found in cartilage.
However, Type 1 cartilage is the predominant form in Fibrocartilage. So "B" could also be right?
But purelysynthesizingcollagen does not create cartilage. Synthesizing collagen is the beginning steps of cartilage formation. However, it is my understanding that fully functional cartilage is produced once collagen fibers cross-link each other. That is why I selected answer "A".
So my take home message: Cartilage is not formed as soon as nascent collagen fibers are synthesized, because these fibers must be modified in a variety of ways before it can form cartilage. Cross-linking of collagen fibers is the last step of cartilage formation, and the best given answer choice to specifically indicate cartilage formation.
Type 1collagenis the predominant form in Fibrocartilage. So "B" could also be right?
Response: This challenge is not accepted. All of the options except synthesis of type II collagen also occur during synthesis of non cartilagenous connective tissues. Among these options, only synthesis of type II collagen exclusively occurs during synthesis of cartilage, and hence specifically indicates that cartilage is being formed.
Q40) Answer A should be accepted in addition to answer B; Or the question should be thrown out because all the information that was provided to us in the manual about Cytokines are as follows: ( page 4 of Chapter 10)
"There are many protein signals passed between cellswhich are variously called cytokines, lymphokines, interleukins (IL), growth factors or colony forming factors. These proteins bind to cell surface receptors on their target cells and cause a variety of changes in cell behavior.The changes may only last while the factor is present, may last until some other signal cancels the effect, or may be permanenet. Some of the major recognized signals are tabulated below."
How is a student suppose to know that a cytokine is a "secreted protein" if it was never stated, and that it coordinates responses "locally and systemically" if only local responses were mentioned.
Answer A states that cytokines are cell surface proteins mediating signals which is true. From the information in the manual, it seems that cytokines can mediate signals between cells that are in physical contact as well as cells that are not physically in contact. Therefore, it is not incorrect to say that a cytokine is "a cell surface protein mediating signals between cells in physical contact."