CHEM-642   Biochemistry
Comprehensive Final Examination
Friday, 26 May 2000
Part I 125 Points (Dr. White)
Drs. White and Chen Instructors
 

There are 11 pages to this part of the examination. Write your name on each new page. Read every question so that you understand what is being asked. If you feel any question is unclear or ambiguous, clearly explain your answer or interpretation. Please call our attention to any errors you encounter.

You will be given separate pages on which you will find many metabolic pathways and the genetic code.

This examination will assess your learning, problem-solving skills, and ability to communicate clearly. It is intended to be challenging even to the best students in the class. Some of the questions will deal with material you have not seen before and is not in your text; however, they can be answered by applying basic principles discussed in the course.

Do not expose your answers to the scrutiny of your neighbors. Please fold under each page before you go on to the next.

Breakdown of this part of the examination by sections:
 

Short Answer           25 Points
Multiple Choice        21 Points
Problems                 69 Points
Essay                      10 Points
Total                     125 Points
Exam Statistics:    N=50    Range = 20.5 to 106        Mean±SD   62.23 ± 19.97


Part I - Short Answer Questions (25 Points)

Pyruvate is one of the common intermediates of metabolism. A number of compounds that can generate or be derived from pyruvate in one step are drawn below. Identify each compound, indicate the type of reaction to or from pyruvate, and indicate any other substrates or cofactors required in the reaction.
 
Metabolite one-step from pyruvate Name of compound 

(half point each)

Reaction type

(1 point each)

Other substrates and cofactors required 
a. _____________ _________________ _________________
b. _____________ _________________ _________________
c. _____________ _________________ _________________
d. _____________ _________________ _________________
e. _____________ ________________  _________________
f. _____________ _________________ _________________

________________ 1. fMet-tRNAfMet is to protein synthesis as this compound is to fatty acid synthesis.

________________ 2. Coenzyme source of the methyl groups on the 5' Cap structure of mRNA.

________________ 3. Guanine is to guanosine as uracil is to ___.

________________ 4. Functions as a dehydrating agent in metabolism.

________________ 5. Inositol is to PI as ___ is to PC and Serine is to PS..

________________ 6. Radioactive isotope of hydrogen.

________________ 7. Cellular compartment where fatty acid synthesis occurs.

________________ 8. Self mutilation is associated with this human genetic disease involving the purine salvage pathway.

________________ 9. Five carbon precursor of cholesterol biosynthesis.

________________ 10. Compound that accumulates in gout.
 



Part II - Multiple Choice Questions (3 points a piece)

The following questions are based on concepts and understanding, not memorization.

_____ 1. Which of the following functional group or group of atoms is not represented in the structure of acetyl Coenzyme A (AcCoA) shown above?

A. Thioester             B. Methylene group             C. Amide

D. Glycosidic bond     E. Acid anhydride               F. Phosphodiester
 

_____ 2. If AcCoA were hydrolyzed to completion, which of the following compounds would not be among the products?
A. Acetate               B. Phosphate               C. AMP

D. Ribose               E. Adenine                   F. Beta-Alanine


_____ 3. From the structure of AcCoA one could predict that it would

A. absorb UV light around 260 nm.

B. bind tightly to a negatively charged ion exchange resin.

C. be quite soluble in an organic solvent.

D. diffuse readily through biological membranes.

E. have GTP as a biosynthetic precursor.


_____ 4. Which one of the following compounds is synthesized by both humans and E. coli ?

A. Thiamin           B. Tryptophan           C. Immunoglobulins

D. Thymine          E. Riboflavin              F. Creatine


_____ 5. Methotrexate, a potent anticancer drug, inhibits dihydrofolate reductase, the NADP-dependent enzyme that converts dihydrofolate to tetrahydrofolate. Tumor cells are especially sensitive to methotrexate because:

A. It interferes with dTTP synthesis thereby inhibiting replication.

B. It interferes with purine biosynthesis and thereby inhibits replication and transcription.

C. It disrupts the NAD/NADP redox balance in turn all reactions coupled to NAD and NADP.

D. The accumulation of dihydrofolate interferes with replication by intercalating in DNA.

E. Serine and glycine cannot be metabolized.

_____ 6. Threonine aminoacyl tRNA synthetase catalyzes the ATP-dependent activation of threonine which becomes covalently bonded to the 3' end of its cognate tRNAThr as depicted below.

                                Thr + tRNAThr + ATP Thr~tRNAThr + AMP + PPi

The chemical nature of the covalent bond between threonine and its tRNA is:
 

A. N-glycoside between the alpha-amino group of threonine and the anomeric carbon of the 3' ribose.

B. Phosphoanhydride between the alpha-carboxyl group of threonine and a tRNAThr phosphoryl group

C. Amide between the alpha-amino group of threonine and a tRNAThr carboxyl group.

D. Thioester between the alpha-carboxyl group of threonine and a sulfur-modified base at the 3' terminus of tRNAThr.

E. Ester between the alpha-carboxyl group of threonine and a tRNAThr hydroxyl group.


_____ 7. When used as an artificial message, a synthetic RNA with a repeating nucleotide sequence encoded an alternating polypeptide containing a basic and an acidic amino acid. What is the sequence of the synthetic mRNA?

A. (GAA)n   B. (CGC)n     C. (AG)n   D. (CG)n     E. (AAG)n   F. (AC)n


Part III Problems and short response.
1.    (6 points) When humans eat the glycogen in turkey breast meat, each glucose unit gives us a different  average yield of ATP in glycolysis than that obtained when we use our own liver glycogen. Explain clearly and briefly why this is so and what is the ATP yield per glucose in both situations.

         Dietary glycogen ________                        Liver glycogen ________
 

Explanation:
 

Give two important and distinct reasons why glucose units are stored as glycogen in the liver.
    A.______________________________________________________________________

    B. ______________________________________________________________________
 

2. The following diagram outlines the biosynthetic pathway for threonine and relates it to the biosynthesis of other amino acids derived from aspartic acid.

2A. (10 Points) Two intermediates (A & B) on the pathway to threonine are omitted along with coenzymes associated with reactions C, D, E, & F. Based on similarity to reactions portrayed on the metabolic handout sheets, predict the structures of the missing intermediates and identify the cofactors.
                                                                                        Coenzymes for Reactions:
 
C =

D =

E =

F =

Compound A             Compound B
2B. (4 Points) Consider reactions C, D, E, & F. What can you say about the flux of metabolites through each reaction in relation to the others?
3A. (4 Points) Threonine, an amino acid with two chiral centers, is the precursor of isoleucine, the only other common amino acid with two chiral centers, as is shown below. Put a circle around the carbon atoms in isoleucine that are derived from pyruvate.

3B. (4 Points) Interestingly, each of the enzymes above has dual substrate specificity such that a methyl group can replace the ethyl group. Thus, in contrast to isoleucine, all of the carbon atoms of valine synthesized by this pathway come from pyruvate. Acetolactate synthase, the TPP-dependent enzyme, is the target for the potent sulfonylurea herbicides, Oust® and Glean®, manufactured by Dupont [Trends in Biotech.2(6), 158-161 (1984)]. In cells inhibited by these compounds, alpha-ketobutyrate accumulates but pyruvate does not. How do you explain this?

3C. (4 Points) While Oust® and Glean®, with Ki values in the nM range, are toxic to plants at a few grams per hectare, they have low toxicity to animals. What is a reasonable explanation for this large difference in toxicity?

3D. (8 Points) In addition to transamination to yield valine, alpha-ketovalerate also is a precursor to leucine. Examine the chemical relationship between valine and leucine and the corresponding alpha-ketoacids that get transaminated to form each. Now consider the metabolic steps that relate another more familiar pair of alpha-ketoacids bearing the same relationship. Using these hints, write out the sequence of biosynthetic reactions that convert alpha-ketovalerate to leucine.


4. Regulation of Amino Acid Metabolism by Attenuation.
This is a multipart question that will take time.

A bacterial cell must be sensitive to nutrients in its environment so that it does not waste energy     synthesizing those compounds when they are available. For example, E. coli can synthesize all of the amino acids found in proteins. If one or more of these amino acids becomes available, the biosynthetic pathways for them are shut down by end product inhibition and the synthesis of the enzymes of the pathways is repressed. As a result of DNA sequence and RNA sequence analysis combined with the physiologic response of various regulatory mutants, the fine control of the repression process is fairly well understood.

This question will focus on the exquisite fine control known as attenuation which couples transcription to translation. The following are some of the features of the attenuation mechanism quoted or paraphrased from Keller & Calvo [PNAS, 76, 6186-90 (1979)].

 
i.   Most transcription initiated at the relevant promotor terminates before the structural genes of the operon are reached, resulting in the synthesis of a leader RNA of about 150 nucleotides.

ii.   The site at which termination occurs ("attenuator") is similar to previously identified transcription termination sites (rho-independent termination sites). It is a palindromic G-C rich region followed by a series of adenosines on the template strand. The corresponding region of the leader RNA, which has a potential stem and loop structure followed by a series of uridines, is called the "terminator."

iii.    Each of the known leader RNAs contains a second potential stem-and-loop structure     proximal to the terminator and overlapping with it in such a way that pairing with one region precludes pairing with the other.

iv.    Within each leader RNA, translational start and stop signals are positioned so that a peptide of 14-28 amino acids might be synthesized.

v.     Each leader peptide contains in high frequency the amino acid corresponding to the particular operon.

vi.     The derepression of the operon requires the transcriptional read-through of the attenuator. This occurs only if a ribosome initiates the synthesis of the leader peptide and is retarded in its progress by lower than normal amounts of a specific amino acyl tRNA. This favors a different conformation of the leader RNA and signals transcriptional read through.


The DNA sequence on the next page corresponds to the control region of a particular amino acid operon [PNAS 76(4),1706-1710 (1979)]. Like mRNA, it is complementary to the template strand. A genetic code sheet is provided for you assistance in some of the questions that follow.
 

5'. . .ACAGATAAAAATTACAGAGTACACAACATCCATGAAACGCATTAGCACCACC
 

ATTACCACCACCATCACCATTACCACAGGTAACGGTGCGGGCTGACGCGTACA
 

GGAAACACAGAAAAAAGCCCGCACCTGACAGTGCGGGCTTTTTTTTTCGACCA
 

AAGGTAACGAGGTAACAACCATGCGAGTGTTGAAGTTCGGCGGTACATCA...3'
 

4A.   (1 Point) Underline the transcriptional termination region.

4B.   (2 Points) This DNA sequence (sense strand) corresponds to that of the relevant leader RNA and mRNA. Circle the initiation and termination codons of the leader peptide.

4C.    (2 Point) Put a box around the initiation codon for the first structural gene.

4D.    (5 Points) Write the predicted amino acid sequence of the leader peptide above the appropriate codons. (3 point bonus if you can use the correct one letter representations for the amino acids.)

4E.    (3 Points) This sequence is derived from the control region of the operon for what amino acid?

4F.    (3 Points) Identify the palindromic region corresponding to the terminator with opposing arrows drawn above the sequences. (Note that the palindromic properties refer to double-stranded DNA of which only one strand is shown. Each strand in these regions have the potential to form stem-loop structures.)

4G.    (3 Points) Identify the second palindromic region which overlaps the terminator with opposing arrows drawn under the sequences.

4H.    (6 Points) Depict in a general way the alternative base-paired structures possible for the leader RNA corresponding to the above sequence of DNA. Which conformation would be favored by low levels of the relevant amino acid?

4I.   (4 Points) Attenuation is an extremely elegant and finely tuned mechanism for controlling the expression of amino acid operons. One must consider that even higher order regulation is involved such as that between operons. Consider the metabolism of the amino acid you have identified in Part E and the sequence of the leader peptide from Part D. What other amino acid seems to be important in controlling the expression of this operon? Does this make metabolic sense? If so, explain.



Part IV Essay (10 Points)

Defend or refute the following proposition: It is only a matter of organizational convenience that biochemistry courses and textbooks treat metabolism and molecular biology as separate topics. Support your position with a reasoned argument supported by relevant examples. Your answer will be graded on its quality and not on the position you take.


Genetic Code Chart

UUU PHE UCU SER UAU TYR UGU CYS
UUC UCC UAC UGC
UUA
 
 

LEU

UCA UAA End UGA End
UUG UCG UAG UGG TRP
CUU CCU PRO CAU HIS CGU ARG
CUC CCC CAC CGC
CUA CCA CAA GLN CGA
CUG CCG CAG CGG
AUU ILE ACU THR AAU ASN AGU SER
AUC ACC AAC AGC
AUA ACA AAA LYS AGA ARG
AUG MET ACG AAG AGG
GUU VAL GCU ALA GAU ASP GGU GLY
GUC GCC GAC GGC
GUA GCA GAA GLU GGA
GUG GCG GAG GGG



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Last updated 2 June 2000 by Hal White.
Copyright 2000, Harold B. White, Department of Chemistry and Biochemistry, University of Delaware, Newark, DE 19716