Download Responses to challenges

Responses to challenges:
Summary:
Challenges accepted will result in the following regrading:
Question 6: accept B, C or D
Question 13: all credit
Question 32: accept A or C
Question 3:
I believe that not only is B incorrect but D should be accepted. In Dr. Hardie’s
powerpoint for ch. 12 this is the direct statement: "A transcription factor may turn on
its own gene as well as others. Such a gene is a master regulator of cell differentiative
state." Due to this statement answer D is incorrect because it is stating that the steroid
hormone induces the regulator when his notes directly state it induces itself.
Response: Not accepted.
For the cell to differentiate, first something will have to induce expression of the master
transcription factor(s) governing the new state; then the differentiated state will be maintained for
that cell and its progeny because the master transcription factor(s) induce their own expression.
Question 5:
I believe they should accept all answers to this question because how the syllabus
defines "no punctuation" is very different from how it was stated on the exam. Under
the section "The Genetic Code" pg. 7 the syllabus states: "Once synthesis of a protein
begins there are no PUNCTUATION marks to keep the reading mechanism in the
correct phase. Codons are simply read as consecutive groups of 3 nucleotides" From
this statement it means that there are no mechanisms to prevent the code from being
misread. None of the choices in the question even hint at this statement and answer
choice D most definitely does not.
Response: Not accepted.
If codons all had to start with the same nucleotide, that nucleotide would become a punctuation
mark. The ribosome does not look for punctuation marks, it just translates nucleotides three at a
time starting with the start codon.
6) Key: D, Challenging to accept: B and C
Explanation: Synthetic RNA would not have alternating AT, it would be alternating AU.
RNA has uracil NOT thymine. Therefore, the other correct answers B or C should be
accepted.
From the manual:
Synthetic Trinucleotides Used to Bind Specific tRNAs
Synthetic trinucleotides bind to ribosomes and promotes the binding of specific
tRNA's to the ribosome (UUU causes the tRNA for phenylalanine to bind, etc.). This
provided a direct means for deciphering the code.
The codons UGA, UAA, and UAG do not specify any amino acid. These codons
are recognized by proteins called release factors that cause termination of the
polypeptide chain. Thus UGA, UAA, and UAG are called stop codons or
terminators. AUG is used as the start codon, as will be discussed in the following
chapter. The following is a summary of the genetic code.
Response: Accepted. B,C, or D will be accepted as correct answers.
Exactly as the challenge states, option A should have specified alternating AU, not alternating AT.
#13) Key:A Challenging to accept D
“A” cannot be the answer because it specifically states in the manual under the title
“Stomach” [Chapter 29, page 2] “The low pH environment aids in digestion by killing
bacteria, starting acid hydrolysis of some proteins, carbohydrates and lipids, and by
activating pepsinogen to pepsin.”
Furthermore, “The gastric juices of the stomach contains HCl, the proteolytic enzyme,
pepsin, a lipase, mucus, and intrinsic factor...” [Chapter 29, page 2]
“D” should be accepted as an answer because the other answer choices (A, B, C, and
E) are true statements about fat digestion, and also because answer choice “D” states
that triacylglycerols are rearranged into mixed micelles when actually, it is the degraded
products of triacylglycerols (monoacylglycerol and fatty acid salts) that form mixed
micelles. This is stated in Chapter 29 on page 9. “ the monoacylglycerol and fatty acid
salts form mixed micelles with bile salts and other lipids such as cholesterol and the fatsoluble
vitamins”
Credit will be given for all answers:
The student is correct to note that it was stated in the syllabus that lipid digestion can be
started by acid hydrolysis.
Question 14:
Answer choice C should also be accepted because according to the syllabus proteins
are cleaved by exo and endopeptidases and not unfolded. There was no mention of
protein unfolding in the syllabus or test review. I believe the question "How are we
suppose to answer that question if the information was never given to us?" should be
forwarded to Dr. Hink.
No Credit: The fact that low pH (i.e. HCl) unfolds proteins was in fact mentioned multiple
times:
The lecture notes, page 5, halfway down.
The exam review notes, page 2, point b, near the bottom
This point was also specifically discussed in the lecture.
Question 16:
This question is very misleading because of the mention of hexokinase. Answer choice
C should be accepted because hexokinase catalyzes the reaction from glucose to
glucose-6-P and is an IRREVERSIBLE reaction, not from glucose-6-P to glucose.
The double-headed arrow written in this question is incorrect because it is stating that
the reaction is reversible which is incorrect. Also, answer choice C is stating that the
reaction is catalyzed by glucose-6-phosphatase. Therefore, answer choice C should
be accepted because it is not even involved with the reaction and therefore does not
involve the principle of change.
No credit:
It was stated that the distinction between what we consider as reversible and irreversible
reaction relates to the magnitude of the free energy charge, but in reality all reactions are
reversible.
In addition, the reaction indicated under c) is not the reaction catalyzed by glucose-6phosphostase; glucose-6-phosphatase catalyzes instead the following transformation:
Glucose-6-phosphate + H2O -> glucose + Pi
#19) Key A; Challenging to Accept E and A
though the large negative free energy reaction tells us that it is spontaneous and will go
to completion eventually. however this only tells us about the thermodynamics of the
reactions. therefore, yes an enzyme be used to to speed up the process (showing us
the kinetics).
now since we are talking abotu regulating metabolic flux, we are looking to find
an enzyme that will catalize a portion of the reaction that regulates the pathway.
answer choice E talks about enzyme catalyzing a decarboxylation reation. typically
in a decarboxylation reaction, energy is needed to created the breaking of bonds
and separation of molecules. for example, in teh citric acid cycle where isocitrate
dehydrogenase and alpha-ketogluturate catalyze the charging and decarboxlyation
of the respective reactants to their product.. this enzyme is essential to regulating the
metabolic pathway. it should also be noted that both of these pathways are considered
irreversable and the enzymes in these pathways would certainly regulate the forward
flow of the metabolic pathway.
No credit:
Though rate enhancement is critical, all enzymes increase the rate, so this alone does not
distinguish what might be a regulatory enzyme and one that is not.
The argument about decarboxylation reactions cannot be accepted since there is nothing
inherent about a decarboxyltion reaction that would dictate that it necessarily have a large
negative free energy change (though, obviously, they can).
Question 27:
Answer A should also be accepted because although "uniquely" is mentioned
coenzyme Q is required for electron transfer to move the electrons from complex II to
complex III. The wording of the question is confusing because the word "uniquely" could
be interpreted as coenzyme Q's association with complex II versus cytochrome C as an
electron carrier.
No credit:
There is no difference in the way you are interpreting the word uniquely and the way the
word uniquely is used in correct answer. Thus, I am only wiling to give credit for the
answer that is actually correct.
Question 32:
The correct answer is A and not C. According to the syllabus ch. 17 on page 8 and his
powerpoint one equivalent of glucose-6-phosphate produces 2 equivalents of NADPH.
For answer choice C to be correct he would have had to state 1 hexose and not 1
glucose-6-phosphate in order to produce 12 equivalents of NADPH.
Credit will be also given for A:
C in fact is the correct answer when there is a ‘net consumption of one equivalent’ of
glucose-6-phosphate [and its conversion to six equivalents of CO2]. This is explained in
the chapter in the syllabus on page 7, in the lecture notes on page 10, and in the
conference 8 notes on page 27. However, credit is also being given for answer A since the
question implied, but did not explicitly state, that the ‘net consumption of one equivalent’
of glucose-6-phosphate result in the conversion of all six of its carbon atoms to CO2.
Answer A is possible if you consider the ‘‘net consumption of one equivalent’ of glucose6-phosphate to produce 2 equivalents of NADPH, 1 equivalent of ribulose-5-phosphate,
and 1 equivalent of CO2.
#39) key: A; Challenging to Accept B also
In the manual (Chapter 19, Page 6) does shows us a diagram that shows glycogen
phosphorlyase breaking down glycogen to yield Glucose-1-phosphate and the rest of
the branch.
B is essentially still correct since glucose-1-phosphate is still glucose, just
phosphoralated (which means it can not pass membranes). Free glucose is still also
glucose, so I’m sure that the “glucose” meant in the answer might be refering to “free
glucose,” but simply glucose is a bit ambiguous as glucose-1-phosphate is still a
glucose (just phosporylated to an activated state which can enter glycolysis or glycogen
synthesis depending on the need of the body).
maybe the question was trying to point us to show us how answer choice B is incorrect
because “glucose, or free glucose” is generated by the hydrolysis of branched glycogen,
but in the manual it makes the distinction that the glucose is “free” and not just simply
glucose.
No credit:
The question directly asks whether glucose is a product of the reaction catalyzed by
glycogen phosphorylase, so there really is no ambiguity in terms of whats being asked.
Moreover, it was emphasized in multiple places (including my lecture) that the glucose
monomers are removed from glycogen, not as glucose, but as glucose-1-phosphate.
Question 50:
The correct answer should be B and not E. The only enzymes that use NH4+ as a
substrate are carbamoyl phosphate synthetase and glutamine synthetase. Glutamate
dehydrogenase PRODUCES NH4+, it does not use it as a substrate. Ch. 20, page 7 of
the syllabus it directly states this and shows the reaction.
#50) Key: E; Challenging to accept “B”
Glutamate dehydrogenase does not use NH4+ as a substrate. This is stated in
Chapter 20, page 7 of the manual (in equation form) NH4+ is shown as a product (not
as a substrate) of the reaction that is catalyzed by Glutamate dehydrogenase.
Response. No credit.
The reaction catalyzed by glutamate dehydrogenase is reversible. See slide 13 from the lecture
Powerpoint slides. In the cytoplasm the enzyme catalyzes the synthesis of glutamate from ketoglutarate and NH4+.