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BIOCHEMISTRY REVIEW: EVALUATION 3 SET D 1 

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

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 Transversion

Purine → purine or 

pyrimidine → pyrimidine 

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 proteinproduct 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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BIOCHEMISTRY REVIEW: EVALUATION 3 SET D 2

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.

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 eIF-

2α 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.

*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.

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 tyrosinyl-

tRNA.

6. _B_ Peptidyl transferase is an intergral part of what

ribosomal sub-unit?

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BIOCHEMISTRY REVIEW: EVALUATION 3 SET D 3

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

Explanation:

Trinucleotide repeat disorders are a set of  geneticdisorders 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 mutationB. gene knockout

C. missense

D. same sense

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).

9. _B_ Which of the following is not involved in translation?

A. Initiation Factor

B. Enhancer

C. ATP

D. GTP

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.

10. _D_ Which of the following can initiate frameshift

mutation?

A. tautomerism of bases

B. alkylating agents

C. base analogs

D. intercalating agents

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

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 three-

dimensional shape of the molecule. These may cause other

mutations in the genetic code but not frameshift mutations.

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.

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.

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.

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

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.

12. _A_ Co-repressor of tryptophan:

A. tryptophan

B. lactose

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BIOCHEMISTRY REVIEW: EVALUATION 3 SET D 4

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 L-tryptophan (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.

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.

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

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.

Below are illustrations of Inosinate base pairing.

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

binding 

30S sub-unit

Streptomycin Initiation, 30S sub-unit

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BIOCHEMISTRY REVIEW: EVALUATION 3 SET D 5

elongation

Tetracycline Aminoacyl tRNA

binding 

A-site

Erythromycin Peptidyl

transferase

50S sub-unit

Lincomycin Peptidyl

transferase

50S sub-unit

Clindamycin Peptidyl

transferase

50S sub-unit

Chloramphenicol Peptidyl

transferase

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 β-

chain 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.

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

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BIOCHEMISTRY REVIEW: EVALUATION 3 SET D 6

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

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-N-

acetylgalactosamine to D-galactose end of H antigen,

producing the A antigen. The B allele encodes

a glycosyltransferase that joins α-D-galactose bonded to D-

galactose 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

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

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.

28. _C_ This eukaryotic initiation factor prevents the re-

association 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 sub-

unit. (Refer to Figure 2 for summary.) 

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

Supplementary figure on the structure of tRNA.

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BIOCHEMISTRY REVIEW: EVALUATION 3 SET D 7

FIGURE 1

FIGURE 2

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BIOCHEMISTRY REVIEW: EVALUATION 3 SET D 8

FIGURE 3