QUESTION NO. 1
Fragile X syndrome is a common form of inherited mental retardation. The mutation in the disease allows the increase of a CGG repeat in a particular gene from a normal of about 30 repeats to 200-1000 repeats. This repeat is normally found in the 5' untranslated region of a gene for the protein FMR1. FMR1 might be involved in the translation of brain-specific mRNAs during brain development. The consequence of the very large number of CGG repeats in the DNA is extensive methylation of the entire promoter region of the FMR1 gene. Methylation of bases in DNA usually

A. facilitates the binding of transcription factors to the DNA.

B. makes a difference in activity only if it occurs in an enhancer region.

C. prevents chromatin from unwinding.

D. inactivates DNA for transcription.

E. results in increased production of the produce of whatever gene is methylated.

QUESTION NO. 2

The best definition of an endonuclease is that it hydrolyzes

A. nucleotide from only the 3' -end of an oligonucleotide .

B. nucleotide from either terminal of an oligonucleotide.

C. phosphodiester bond located in the interior of a polynucleotide .

D. bond only in a specific sequence of nucleotides.

E. bond that is distal co the base chat occupies the 5' position of the bond

QUESTION NO. 3

In eukaryotic transcription

A. RNA polymerase does not require a template.

B. all RNA is synthesiud in the nucleolus.

C. consensus sequences are the only known promoter elements.

D. phosphodiester bond formation is favored because there is pyrophosphate hydrolysis.

E. RNA polymerase requires a primer.

QUESTION NO. 4

Normally, certain kinds of reiterated sequences occur in a chromosome as an interspersion pattern that is

A. highly repetitive DNA sequences.

B. the portion of DNA composed of single-copy DNA.

C. Alu sequences.

D. alcernacing blocks of single-copy DNA and moderately repetitive DNA.

E. alternating blocks of short interspersed repeats and long interspersed repeats.

QUESTION NO. 5

During initiation of protein synthesis,

A. methionyl-tRNA appears at the A site of the 80S initiation complex.

B. eIF3 and the 40S ribosomal subunit participate in forming a preinitiation complex .

C. eIF2 is phosphorylated by GTP .

D. the same methionyl-tRNA is used as is used during elongation .

E. a complex of mRNA, 60S ribosomal subunit, and certain initiation factors is formed.

QUESTION NO. 6

Replication

A. requires that a phosphodiester bond of the incoming dNTP be hydrolyzed in order to be added to the growing chain.

B. uses 5' to 3' polymerase activity to synthesize one strand and 3' to 5' polymerase activity to synthesize the complementary strand.

C. owes its accuracy, in part, to 3' to 5' exonucleolytic activity of DNA polymerases or associated proteins.

D. begins with two dNTPs joining together.

E. requires only proteins with DNA polymerase activity.

QUESTION NO. 7

Streptomycin binds the small subunit of prokaryotic ribosomes and

A. causes premature release of the incomplete peptide.

B. prevents binding of the 40S and 60S subunits .

C. interferes with initiation of protein synthesis.

D. inhibits peptidyl transferase activity.

E. acts as an N-glycosidase .

QUESTION NO. 8

Retroviruses, like HIV which causes AIDS, have their genetic information in the form of RNA. Reverse transcriptase synthesizes a DNA copy of the viral genome. One drug used in treating AIDS is AZT, an analog of deoxythymidine, which has an azido group at the 3' position of the sugar. It can be phosphorylated and competes with dTTP for incorporation into the reverse transcript. Once incorporated, its presence terminates chain elongation. The growing chain is terminated because

A. the analog can not hydrogen bond co RNA.

B. the presence of the AZT analog inhibits the proofreading ability of reverse transcriptase.

C. AZT does not have a free 3' -OH.

D. the analog causes distortion of the growing chain, inhibiting reverse transcriptase.

E. dTTP can no longer be added to the growing chain.

QUESTION NO. 9

Protooncogenes produce products that have specific roles in regulating growth and differentiation of normal cells. Mutations can turn these genes into oncogenes whose products are less responsive to normal control. Unmutated protein p53, a tumor suppressor, is a transcription factor, inhibiting some genes and activating others. P53 inhibits genes with TATA sequences and activates genes for DNA repair. The TATA sequence

A. occurs about 25 bp downstream from the start of transcription.

B. binds directly to RNA polymerase.

C. binds transcription factors which bind RNA polymerase.

D. binds p53.

E. is an enhancer sequence

Question

QUESTION NO. 1
Fragile X syndrome is a common form of inherited mental retardation. The mutation in the disease allows the increase of a CGG repeat in a particular gene from a normal of about 30 repeats to 200-1000 repeats. This repeat is normally found in the 5' untranslated region of a gene for the protein FMR1. FMR1 might be involved in the translation of brain-specific mRNAs during brain development. The consequence of the very large number of CGG repeats in the DNA is extensive methylation of the entire promoter region of the FMR1 gene. Methylation of bases in DNA usually

A. facilitates the binding of transcription factors to the DNA.

B. makes a difference in activity only if it occurs in an enhancer region.

C. prevents chromatin from unwinding.

D. inactivates DNA for transcription.

E. results in increased production of the produce of whatever gene is methylated.

QUESTION NO. 2

The best definition of an endonuclease is that it hydrolyzes

A. nucleotide from only the 3' -end of an oligonucleotide .

B. nucleotide from either terminal of an oligonucleotide.

C. phosphodiester bond located in the interior of a polynucleotide .

D. bond only in a specific sequence of nucleotides.

E. bond that is distal co the base chat occupies the 5' position of the bond

QUESTION NO. 3

In eukaryotic transcription

A. RNA polymerase does not require a template.

B. all RNA is synthesiud in the nucleolus.

C. consensus sequences are the only known promoter elements.

D. phosphodiester bond formation is favored because there is pyrophosphate hydrolysis.

E. RNA polymerase requires a primer.

QUESTION NO. 4

Normally, certain kinds of reiterated sequences occur in a chromosome as an interspersion pattern that is

A. highly repetitive DNA sequences.

B. the portion of DNA composed of single-copy DNA.

C. Alu sequences.

D. alcernacing blocks of single-copy DNA and moderately repetitive DNA.

E. alternating blocks of short interspersed repeats and long interspersed repeats.

QUESTION NO. 5

During initiation of protein synthesis,

A. methionyl-tRNA appears at the A site of the 80S initiation complex.

B. eIF3 and the 40S ribosomal subunit participate in forming a preinitiation complex .

C. eIF2 is phosphorylated by GTP .

D. the same methionyl-tRNA is used as is used during elongation .

E. a complex of mRNA, 60S ribosomal subunit, and certain initiation factors is formed.

QUESTION NO. 6

Replication

A. requires that a phosphodiester bond of the incoming dNTP be hydrolyzed in order to be added to the growing chain.

B. uses 5' to 3' polymerase activity to synthesize one strand and 3' to 5' polymerase activity to synthesize the complementary strand.

C. owes its accuracy, in part, to 3' to 5' exonucleolytic activity of DNA polymerases or associated proteins.

D. begins with two dNTPs joining together.

E. requires only proteins with DNA polymerase activity.

QUESTION NO. 7

Streptomycin binds the small subunit of prokaryotic ribosomes and

A. causes premature release of the incomplete peptide.

B. prevents binding of the 40S and 60S subunits .

C. interferes with initiation of protein synthesis.

D. inhibits peptidyl transferase activity.

E. acts as an N-glycosidase .

QUESTION NO. 8

Retroviruses, like HIV which causes AIDS, have their genetic information in the form of RNA. Reverse transcriptase synthesizes a DNA copy of the viral genome. One drug used in treating AIDS is AZT, an analog of deoxythymidine, which has an azido group at the 3' position of the sugar. It can be phosphorylated and competes with dTTP for incorporation into the reverse transcript. Once incorporated, its presence terminates chain elongation. The growing chain is terminated because

A. the analog can not hydrogen bond co RNA.

B. the presence of the AZT analog inhibits the proofreading ability of reverse transcriptase.

C. AZT does not have a free 3' -OH.

D. the analog causes distortion of the growing chain, inhibiting reverse transcriptase.

E. dTTP can no longer be added to the growing chain.

QUESTION NO. 9

Protooncogenes produce products that have specific roles in regulating growth and differentiation of normal cells. Mutations can turn these genes into oncogenes whose products are less responsive to normal control. Unmutated protein p53, a tumor suppressor, is a transcription factor, inhibiting some genes and activating others. P53 inhibits genes with TATA sequences and activates genes for DNA repair. The TATA sequence

A. occurs about 25 bp downstream from the start of transcription.

B. binds directly to RNA polymerase.

C. binds transcription factors which bind RNA polymerase.

D. binds p53.

E. is an enhancer sequence

Accepted Answer

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