BIOCHEM EVALS 3

SUBJECT: BIOCHEMISTRY TOPIC: EVALUATION #3 Set D (Translation, Gene Regulation, Heme Metabolism, Hemostasis) LECTURER: PROF. SHEILA TORRES & DR. MARIA ESPERANZA UY DATE: MARCH 2011

This review will only answer questions 1-28 as Heme Metabolism and Hemostasis are not part of the Comprehensive Exam.  Please read Chapters 37 & 38 in Harper‟s Illustrated Biochemistry (28th Ed) for additional information although most bases are lifted from the book.

REVIEW PROPER 1. _B_ Which of the following is NOT a valid statement about the genetic code? A. Exceptions to the “standard” genetic code have been found in some species and in mitochondria of others. B. There are 64 possible codons, all of which code for amino acids C. The genetic code is redundant with some amino acids represented by as many as six different codons. D. The genetic code is read without punctuation, such that loss of one nucleotide can change the coding for all amino acids downstream from the loss. Explanation: To start, we should define translation. Translation, simply put, is the conversion of genetic information (in form of mRNA) into their protein product that they decode for. Next, we define what a codon is. A codon is a triplet code composed of three nucleotides. B is the answer because through probability, we can get 64 possible codons. Since there are only FOUR nucleotides, and there are only 3 nucleotides in a codon, then we raise 4 to 3. 43 = 64. Next, what makes this statement WRONG? It‟s because of its latter part. NOT ALL of the codons CODE FOR AMINO ACIDS. 3 codons called nonsense codons code for STOP CODONS which terminate the translation process, namely, UAA, UAG, UGA.

Above is a table of the different codons and the amino acids they encode. The codons are written in the 5’→3’ direction. The third base of each codon (in bold type) plays a lesser role in specifying an amino acid than the first two. There are three stop codons and

BIOCHEMISTRY REVIEW: EVALUATION 3 SET D

one initiation codon (AUG; which also decodes for Methionine). All the AAs except met and trp have more than one codon. In most cases, codons that specify the same AA differ only at the third base.

2. _A_ In sickle cell anemia A-T mutation occurred within the β8-globin gene resulting to the change of val to glu amino acid in β-globin. This mutation is known as: A. transversion, missense B. transition, nonsense C. transition, missense D. transversion, nonsense Explanation: Mutations are changes in the DNA sequence. These may happen as single base changes called POINT MUTATIONS and could be of two types: (1) TRANSITION, which is pyrimidine → pyrimidine or purine → purine substitution, and (2) TRANSVERSION, which is pyrimidine → any of the two purines or purine → any of the two pyrimidines.

Again, Transition Purine → purine or pyrimidine → pyrimidine

Transversion Purine → pyrimidine or Pyrimidine → purine

The effects of point mutations may either be:  SILENT MUTATIONs – wherein there is no detectable effect because the codon still decoded for the same amino acid. o Recall the “Wobble Effect” wherein change in the wobble base (the third nucleotide in the codon) had so significant effect on the amino acid produced (i.e. AGA and AGU both decode Arg).  MISSENSE MUTATIONs – when the point mutation results in the production of a different amino acid. o Acceptable missense – when the protein product cannot be distinguished from the normal one o Partially acceptable missense – when protein function of the peptide product is rendered partially abnormal o Unacceptable missense – when protein product molecule is rendered incapable of functioning normally.  NONSENSE MUTATIONs – when the point mutation results in premature termination of the polypeptide

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chain, hence the codon coded for was a STOP CODON. Review: Purines Pyrmidines Adenine Thymidine Guanine Cytosine Since the mutation above was A-T (meaning its purine to pyrimidine) it is a TRANSVERSION and since the protein product decoded a different protein (valine to glutamate), it is a MISSENSE MUTATION. This justifies that the answer is letter A. 3. _A_ How many nucleotides are there in a couple of codons? A. 6 B. 4 C. 3 D. 2 Explanation: Remember, a codon is a triplet code, meaning it is composed of three nucleotides. Having a couple, by this I think Professor Torres was pertaining to a PAIR („cause couple seems so romantic. HAHA), then we would have 2 codons, summing up to 6 nucleotides.

*eIF-3 is a key protein since it binds with high affinity to 4G of 4F and links this with 40S sub-unit.

Formation of 80S Initiation complex eIF-5 hydrolyzes the GTP bound to eIF-2 on the 48S initiation complex that leads to release of all the other initiation factors bound to the complex and facilitates rapid binding of the 60S sub-unit to form the 80S ribosome. 5. _D_ A structural analog of tyrosinyl - tRNA that can cause premature termination of translation: A. bromouracil B. rifampicin C. tetracycline D. puromycin Explanation: Puromycin is incorporated via the A site on the ribosome into the carboxyl terminal position of a peptide but causes the premature release of the polypeptide. This inhibitive effect happens to both prokaryotes and eukaryotes. Please refer to page 367 of Harper’s for the comparison of the structures of tyrosinyl – tRNA and puromycin. Below is the mechanism by which puromycin inhibits proteins synthesis.

4. _B_ A phosphorylated eIF-2α that inactivates eIF-2B inhibits protein synthesis by: A. blocking the formation of the 80S complex B. preventing the formation of the 43S complex C. blocking the reduction of secondary structure D. all of the above Explanation: Translation happens in three phases, initiation, elongation, and termination. Initiation involves many processes explained below: (Refer to Figure 1 for the pathway) A. Ribosomal dissociation eIF-3 and eIF-1A bind to the dissociated 40S sub-unit that delays the reassociation with 60S. B. Formation of the 43S Preinitiation Complex GTP binds to eIF-2 and this binary complex binds to met tRNA (that carries the start codon AUG which is methionine). This ternary complex binds to the 40S sub-unit to form the 43S preinitiation complex. *eIF-2 is one of the two control points for initiation of the protein synthesis and is composed of alpha, beta, and gamma sub-units. eIF-2α is phosphorylated by different protein kinases that are activated under conditions of stress, starvation, viral infection, etc. Phosphorylated eIF2α binds tightly to and inactivates the GTP-GDP recycling protein eIF-2B that prevents the formation of the 43S initiation complex. Hence, the answer is B. C. Formation of the 48S Initiation Complex The 5‟ terminal of mRNA is capped with a methyl-guanosyl triphosphate cap that facilitates binding of mRNA to the 43S initiation complex. Cap-binding complex: eIF-4F composed of eIF-4E and eIF-4G - eIF-4A. eIF-4B reduces the complex secondary structure of the 5‟ end of the mRNA through ATPase and ATP-dependent helicase activites. mRNA + 43S Initiation complex = 48S Initiation Complex. BIOCHEMISTRY REVIEW: EVALUATION 3 SET D

Bromouracil is brominated uracil which can act as a base analog for thymine in DNA and can induce DNA mutations. It is a structural analog but NOT of the molecule of interest, hence ruled out. Rifampicin is an antibiotic that inhibits DNA – dependent RNA polymerase in bacterial cells and inhibits the translation to RNA and transcription to proteins. It prevents translation but not in the way specified in the questions, hence ruled out. Tetracycline is an antibiotic that bind to the 30S sub-unit in microbial ribosomes and inhibits protein synthesis by blocking the site of attachment of the charged aminoacyl tRNA that prevents introduction of new amino acids to the growing peptide bond. Yes, it has the same function as stated above but is NOT a structural analog of tyrosinyltRNA. 6. _B_ Peptidyl transferase is an intergral part of what ribosomal sub-unit? 2

A. 30S B. 60S C. 40 S D. 80S Explanation: Peptidyl transeferase is located on the P site of the 60S ribosomal sub-unit. It is involved in the petide bond formation in the elongation step of translation. 7. _C_ Huntington‟s disease is a neurodegenerative disorder which is characterized by 35-120 CAG repeats in the gene‟s coding region. It arose from what kind of mutation? A. transition B. transversion C. expansion D. deletion

multiples of three) will not shift the reading frame since triplets are also added/removed. Insertion or deletion of one or two nucelotides will result into translation of a peptide with garbled peptide chain or probably formation of a stop codon that will prematurely end the translational process. Tautomerism happens when a keto or enol structure is formed. Guanine and thymine can have alternate molecular structures based on different locations of a particular hydrogen atom. A keto structure occurs when the hydrogen atom bonds to a nitrogen atom within the ring. An enol structure occurs when the hydrogen atom bonds to an nearby oxygen atom that sticks out from the ring. Both guanine and thymine can switch easily from one tautomer to another. The change in shape affects the threedimensional shape of the molecule. These may cause other mutations in the genetic code but not frameshift mutations.

Explanation: Trinucleotide repeat disorders are a set of genetic disorders caused by trinucleotide repeat expansion, a kind of mutation where trinucleotide repeats in certain genes exceeding the normal, stable, threshold, which differs per gene. During protein synthesis, the expanded CAG repeats are translated into a series of uninterrupted glutamine residues forming what is known as a polyglutamine tract ("polyQ"). Such polyglutamine tracts may be subject to increased aggregation. (Wikipedia) 8. _B_ A manipulated mutation that results to the elimination or loss of function of the gene product: A. neural mutation B. gene knockout C. missense D. same sense

Alkylating agents do not cause frameshift mutations because they only add alkyl groups (alkanes, alkenes, alkynes) to the molecule. They do not change the reading frame.

Explanation: Gene knockout is a genetic technique (meaning done by man or by machine and is not natural) in which one of an organism‟s genes is made inoperative (“knocked out” of the organism).

Base analogs are molecules that are structurally similar to the nucleotide in the DNA chain but cannot cause a frameshift mutation because they do not change the reading frame. They may cause a mutation like a point mutation due to confusion in reading the analog but not a frameshift mutation.

9. _B_ Which of the following is not involved in translation? A. Initiation Factor B. Enhancer C. ATP D. GTP

Intercalating agents are ligands that are small enough, polycyclic, aromatic, and planar that can fit in between base pairs of the DNA strand. Since they resemble insertions, they can cause shifting of the reading frame and therefore, frameshift mutations.

Explanation: The important “ingredients” needed for translation are: tRNA, rRNA, mRNA, ribosomes, eIFs (or initiation factors in general), ATP, GTP, and amino acids. Enhancers are involved in TRANSCRIPTION, not translation.

11. _B_ Photolyases directly repair what kind of damage? A. base locked in its enol-form B. thymine dimers C. apurinic site D. apyrimidinic site

10. _D_ Which of the following can initiate frameshift mutation? A. tautomerism of bases B. alkylating agents C. base analogs D. intercalating agents

Explanation: Photolyases are DNA-linked enzymes that repair DNA damage caused by exposure to UV rays. Specifically, bind complementary DNA strands and break ceratin pyrimidine dimers that form when thymine or cytosine bases covalently link on the same strand and form bulges on the chain.

Explanation: Frameshift mutations occur when the reading frame of the codon is shifted either due to an insertion or deletion of a nucleotide. Deletion or insertion however of three (or BIOCHEMISTRY REVIEW: EVALUATION 3 SET D

12. _A_ Co-repressor of tryptophan: A. tryptophan B. lactose 3

C. glucose D. cAMP Explanation: Many regulatory proteins are allosteric, and bind small metabolic intermediates that modify their activities, allowing the intracellular pattern of gene expression to respond to physiological change. For example, the Escherichia coli tryptophan aporepressor (TrpR) protein binds DNA poorly. However, when complexed with Ltryptophan (corepressor), it binds three sites (operators) on the E. coli genome to repress transcription of the trp (BENNETT and YANOFSKY, 1978), aroH (ZURAWSKI et al., 1981 ), and trpR operons (GUNSALUS and YANOFSKY, 1980). Thus, Trp aporepressor mediates the homeostatic control of tryptophan biosynthesis.

D. all of the choices Explanation: Inosine is often found at the 5‟ wobble position and can form hydrogen bonds with adenine, cytosine, or uracil. The anticodon with inosine can recognize more than one synonymous codon.

13. _C_ The structural genes of Lac operon are: A. repressible B. constitutive C. inducible D. all of the choices Explanation: The lac operon is normally repressed and needs to be induced to displace the repressor and thus activated for transcription of the structural genes to occur.

Below are illustrations of Inosinate base pairing.

14. _D_ The inducer or repressor molecule binds to what part of the operon? A. regulatory B. promoter C. structural D. operator Explanation: Refer to figure 3. 15. _A_ Codons: GUU, GUC, GUG, GUA, all code for valine, this feature of the genetic code is known as: A. degeneracy B. unambiguous C. non-overlapping D. universal Explanation: Since 61 codons (minus three from original 64 since they are stop codons) decode for 20 amino acids, multiple codons must decode a specific amino acid. This explains degeneracy as a feature of the genetic code. The genetic code as unambiguous means that a specific codon may only decode for a single amino acid. Non-overlaping refers to the reading frame of the RNA. Since there are no punctuations between codons, the mRNA sequence is read in a continuing sequence of nucleotide triplets. Universal means that for almost all the species, the codons decode for the same amino acid (except for some exceptions of course). Since in the question, four codons decode for valine, it exemplifies multiple codons decode for a single amino acid, hence DEGENERACY is the answer. 16. _D_ The wobble base I (inosine) of an anticodon can pair with what base/s of the codon: A. U B. C C. A

BIOCHEMISTRY REVIEW: EVALUATION 3 SET D

17. _A_ Which of the following antibiotics inhibit peptide bond formation in prokaryotes? A. clindamycin B. tetracycline C. puromycin D. cycloheximide Explanation: Below is a table of Protein Synthesis inhibitors. Recall that prokaryotes have 30S and 50S ribosomal sub-units. Since Clindamycin affects reactions catalyzed by peptidyl transferase (which is peptide bond formation), it is the answer for this number. Inhibitor Process Affected Site of Action Kasugamycin Intiator tRNA 30S sub-unit binding Streptomycin Initiation, 30S sub-unit

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Tetracycline Erythromycin Lincomycin Clindamycin Chloramphenicol

elongation Aminoacyl tRNA binding Peptidyl transferase Peptidyl transferase Peptidyl transferase Peptidyl transferase

A-site 50S sub-unit 50S sub-unit 50S sub-unit 50S sub-unit

18. _D_ A eukaryotic mRNA can form how many polypeptide after one round of translation? A. four B. three C. two D. one Explanation: Translation of mRNA starts at the 5' terminal where the amino terminal is formed. The sequence is read from 5' to 3' and ends with the formation of the carboxy terminal. Recall that transcription of DNA into RNA will form the 5' end of the RNA first, hence in prokaryotes (which have no compartmentalization since they have a nucleiod region instead of a membrane-bound nucleus), translation can readily proceed transcription and occur simultaneously. In eukaryotes, since they have a membrane bound nucleus where transcription occurs, the mRNA has to be brought out into the cytoplasm first where translation will occur. Hence, after once round of translation, only ONE polypeptide chain can be formed from eukaryotic mRNA.

19. _D_ Which of the following nucleic acid base is not present in codons? A. adenine B. cytosine C. guanine D. thymine Explanation: If you remember in RNA synthesis, when RNA is transcribed from DNA, the complementary base for adenine is NOT thymine, but uracil. Hence, in codons (which are found in RNA sequences) there is NO THYMINE, just uracils. 20. _B_ In one form of Thalassemia, codon 17 of the βchain is changed from UGG to UGA. This kind of mutation is: A. same sense B. nonsense C. missense D. gene knockout Explanation: As explained earlier, point mutations are mutations that happen on single base nucleotides. In this case, there was a point mutation on the third nucleotide in the codon. Also, codons can also encode for non-amino acid products like the STOP CODONS, which are UAA, UAG, and UGA. Since the point mutation resulted in UGA, the translation process is prematurely stopped and this kind of mutation is called a NONSENSE mutation. BIOCHEMISTRY REVIEW: EVALUATION 3 SET D

21. _C_ The enzyme that is involved during tRNA charging and amino acid activation: A. translocase B. peptidase C. aminoacyl synthetase D. DNA glycosylase Explanation: Charging or recognition and attachment of amino acids is catalyzed by AMINOACYL tRNA SYNTHETASES that catalyze esterification of the AA to the tRNA. This step is energy requiring (specifically ATP). 22. _D_ Which of the following events of translation is/are GTP/ATP requiring? A. charging of tRNA B. entry of aminoacyl tRNa into the A site C. formation of peptide bond D. all of the above Explanation: As discussed in previous questions, all these processes require energy either in the form of ATP or GTP. 23. _A_ Which of the following congenital defect/s could not carry out excision repair? A. Xeroderma pigmentosum B. Fanconi‟s syndrome C. Bloom‟s syndrome D. all of the choices Explanation: Xeroderma Pigementosum is an autosomal recessive genetic disorder of DNA repair in which the ability to repair damage caused by ultraviolet (UV) light is deficient. The most common defect in xeroderma pigmentosum is an autosomal recessive genetic defect in which nucleotide excision repair (NER) enzymes are mutated, leading to a reduction in or elimination of Nucleotide Excision Repair. If left unchecked, damage caused by UV light can cause mutations in individual cell's DNA. Fanconi‟s disease is a disease of the proximal renal tubules of the kidney in which glucose, amino acids, uric acid, phosphate and bicarbonate are excreted in the urine rather than absorbed. It has no relation with base excision repair. Bloom‟s syndrome is a rare autosomal recessive chromosomal disorder characterized by a high frequency of breaks and rearrangements in an affected person‟s chromosomes. 24. _B_ An enzyme that removes damaged base during base excision repair: A. Dam methylase B. DNA glycosylase C. AP endonucease D. ABC exonuclease Explanation: Base Excision Repair is initiated by DNA glycosylases, which recognize and remove specific damaged or inappropriate bases, forming AP sites. These are then cleaved by an AP 5

endonuclease. The resulting single-strand break can then be processed by either short-patch (where a single nucleotide is replaced) or long-patch Base Excision Repair (where 2-10 new nucleotides are synthesized). (Wikipedia) 25. _A_ Which of the following gene/alleles of blood arose from frameshift mutation in the glycoslytransferase gene? A. O B. B C. A D. all of the choices

28. _C_ This eukaryotic initiation factor prevents the reassociation of the 60S to 40S ribosomal sub-units: A. eIF1 B. eIF2 c. eIF3 d. eIF5 Explanation: eIF1A and eIF3 bind to the 40S sub-unit and delays the reassociation of 40S and 60S sub-units. eIF-3 plays a more important role since it also has high affinity for the 4F subunit. (Refer to Figure 2 for summary.)

Explanation: The ABO blood group system is determined by what type of glucosyltransferases are expressed in the body. The ABO gene locus expressing the glucosyltransferases has three main alleleic forms: A, B, and O. The A allele encodes a glycosyltransferase that bonds α-Nacetylgalactosamine to D-galactose end of H antigen, producing the A antigen. The B allele encodes a glycosyltransferase that joins α-D-galactose bonded to Dgalactose end of H antigen, creating the B antigen. In case of O allele the exon 6 contains a deletion that results in a loss of enzymatic activity. The O allele differs slightly from the A allele by deletion of a single nucleotide -Guanine at position 261. The deletion causes a frameshift and results in translation of an almost entirely different protein that lacks enzymatic activity. This results in H antigen remaining unchanged in case of O groups. The combination of glucosyltransferases by both alleles present in each person determines whether there is a AB, A, B or O blood type. (Wikipedia) 26. _B_ Spontaneous deamination of cytosine converts this base into: A. adenine B. uracil C. thymine D. guanine

-------------------------------END OF TRANSCRIPTION----------------------------

Explanation: Spontaneous deamination of cytosine happens through action of the enzyme cytosine deaminase that hydrolyzes cytosine into uracil and liberates ammonia as a by-product. Reaction is seen below: (Cytosine + H2O → Uracil + NH3)

27. _B_ The following are some components of 48s intiation complex EXCEPT: A. mRNA B. 60S C. 40S D. charged tRNA

Supplementary figure on the structure of tRNA.

Explanation: Recall as initiation of translation was discussed earlier: the 48S complex is formed from the capped mRNA and the 43S preinitiation complex (ternary complex [met tRNA] + 40S sub-unit with bound eIF-3 and eIF-1A). Hence, the 60S sub-unit is not part of the 48S complex.

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FIGURE 1

FIGURE 2 BIOCHEMISTRY REVIEW: EVALUATION 3 SET D

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FIGURE 3

BIOCHEMISTRY REVIEW: EVALUATION 3 SET D

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