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Genetics Final Exam > Lecture 11 / Chapters 13, 14, and 15 - Biological Information Processing > Flashcards

Lecture 11 / Chapters 13, 14, and 15 - Biological Information Processing Flashcards

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1
Q

True or False?

The genetic code is degenerate, meaning that a codon can specify more than one amino acid.

A

False.

(The degeneracy of the genetic code means that an amino acid may be coded for by more than one codon. However, a single codon can only ever specify one amino acid.)

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2
Q

A DNA sequence produces a mutant protein in which several amino acids in the middle of the protein differ from the normal protein. What kind of mutation could have occurred?

  • No mutation
  • An addition and a deletion mutation
  • An addition or deletion mutation
  • A nonsense mutation
A

An addition and a deletion mutation.

(A single addition or deletion would change the reading frame of the protein, but if another mutation occurred to cancel the effects of the first mutation, only those amino acids between the mutations would change.)

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3
Q

Which of the following statements about eukaryotic transcription is false?

  • A polycistronic mRNA may be transcribed if the gene products are used in the same pathway or needed at the same time.
  • Transcription initiation occurs when RNA polymerase binds to a complex of transcription factors at the TATA box.
  • Eukaryotic promoter regions contain a TATA box and a CAAT box.
  • The transcripts produced contain both exons and introns.
A

A polycistronic mRNA may be transcribed if the gene products are used in the same pathway or needed at the same time.

(This statement is false. Polycistronic mRNAs are produced only in prokaryotes. In eukaryotes, a single gene is transcribed at a time.)

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4
Q

Which of the following statements about eukaryotic transcription is false?

  • A polycistronic mRNA may be transcribed if the gene products are used in the same pathway or needed at the same time.
  • Transcription initiation occurs when RNA polymerase binds to a complex of transcription factors at the TATA box.
  • Eukaryotic promoter regions contain a TATA box and a CAAT box.
  • The transcripts produced contain both exons and introns.
A

A polycistronic mRNA may be transcribed if the gene products are used in the same pathway or needed at the same time.

(This statement is false. Polycistronic mRNAs are produced only in prokaryotes. In eukaryotes, a single gene is transcribed at a time.)

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5
Q

True or False?

The code is nonoverlapping, meaning that, assuming “standard translation,” a given base participates in the specification of one and only one amino acid.

A

True

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6
Q

How many different codons code for amino acids?

  • 3
  • 64
  • 20
  • 61
A

61

There are 61 codons that code for amino acids and three stop codons that do not code for an amino acid.

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7
Q

Introns are known to contain termination codons (UAA, UGA, or UAG), yet these codons do not interrupt the coding of a particular protein. Why?

  • Exons are spliced out of mRNA before translation.
  • These triplets cause frameshift mutations, but not termination.
  • UAA, UGA, and UAG are initiator codons, not termination codons.
  • More than one termination codon is needed to stop translation.
  • Introns are removed from mRNA before translation.
A

Introns are removed from mRNA before translation.

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8
Q

What is the initiator triplet in both prokaryotes and eukaryotes? What amino acid is recruited by this triplet?

  • AUG; arginine
  • UAA; no amino acid called in
  • UAA; methionine
  • AUG; methionine
  • UAA or UGA; arginine
A

AUG; methionine

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9
Q

What is the initiator triplet in both prokaryotes and eukaryotes? What amino acid is recruited by this triplet?

  • AUG; arginine
  • UAA; no amino acid called in
  • UAA; methionine
  • AUG; methionine
  • UAA or UGA; arginine
A

AUG; methionine

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10
Q

Which type of mutation helped lead to the understanding that the genetic code is based on triplets?

  • Nonsense
  • Base substitution (substitution of one base for another)
  • Missense mutation within the promoter
  • Frameshift
A

Frameshift

(Insertions or deletions of one or two nucleotides resulted in frameshift mutations. Insertion or deletion of three nucleotides resulted in insertion or deletion of a single amino acid and did not shift the reading frame.)

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11
Q

During transcription in eukaryotes, a type of RNA polymerase called RNA polymerase II moves along the template strand of the DNA in the 3’→5’ direction. However, for any given gene, either strand of the double-stranded DNA may function as the template strand.

Which of the following initially determines which DNA strand is the template strand, and therefore in which direction RNA polymerase II moves along the DNA?

  • the specific sequence of bases along the DNA strands
  • the location along the chromosome where the double-stranded DNA unwinds
  • which of the two strands of DNA carries the RNA primer
  • the position of the gene’s promoter on the chromosome
  • the location of specific proteins (transcription factors) that bind to the DNA
A

The specific sequence of bases along the DNA strands.

(In eukaryotes, binding of RNA polymerase II to DNA involves several other proteins known as transcription factors. Many of these transcription factors bind to the DNA in the promoter region (shown below in green), located at the 3’ end of the sequence on the template strand. Although some transcription factors bind to both strands of the DNA, others bind specifically to only one of the strands.
Transcription factors do not bind randomly to the DNA. Information about where each transcription factor binds originates in the base sequence to which each transcription factor binds. The positioning of the transcription factors in the promoter region determines how the RNA polymerase II binds to the DNA and in which direction transcription will occur.)

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12
Q

After transcription begins, several steps must be completed before the fully processed mRNA is ready to be used as a template for protein synthesis on the ribosomes.

Which three statements correctly describe the processing that takes place before a mature mRNA exits the nucleus?

  • Coding sequences called exons are spliced out by ribosomes.
  • A translation stop codon is added at the 3’ end of the pre-mRNA.
  • A cap consisting of a modified guanine nucleotide is added to the 5’ end of the pre-mRNA.
  • Noncoding sequences called introns are spliced out by molecular complexes called spliceosomes.
  • A poly-A tail (50-250 adenine nucleotides) is added to the 3’ end of the pre-mRNA.
A
  • A cap consisting of a modified guanine nucleotide is added to the 5’ end of the pre-mRNA.
  • Noncoding sequences called introns are spliced out by molecular complexes called spliceosomes.
  • A poly-A tail (50-250 adenine nucleotides) is added to the 3’ end of the pre-mRNA.

(Once RNA polymerase II is bound to the promoter region of a gene, transcription of the template strand begins. As transcription proceeds, three key steps occur on the RNA transcript:
Early in transcription, when the growing transcript is about 20 to 40 nucleotides long, a modified guanine nucleotide is added to the 5’ end of the transcript, creating a 5’ cap.
Introns are spliced out of the RNA transcript by spliceosomes, and the exons are joined together, producing a continuous coding region.
A poly-A tail (between 50 and 250 adenine nucleotides) is added to the 3’ end of the RNA transcript.
Only after all these steps have taken place is the mRNA complete and capable of exiting the nucleus. Once in the cytoplasm, the mRNA can participate in translation.)

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13
Q

The genetic code is said to be triplet, meaning that they’re ________.

  • are three “nonsense” triplets
  • are three bases in mRNA that code for an amino acid
  • may be three ways in which an amino acid may terminate a chain
  • are three amino acids per base in mRNA
  • None of the answers listed is correct.
A

are three bases in mRNA that code for an amino acid.

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14
Q

Which subunit of RNA polymerase establishes template binding to a promoter in prokaryotes?

  • Beta prime
  • Alpha
  • Beta
  • Sigma
A

Sigma.

(The sigma subunit recognizes the promoter sequence. Different sigma subunits can be employed to regulate the expression of genes at the transcriptional level.)

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15
Q

When considering the initiation of transcription, one often finds consensus sequences located in the region of the DNA where RNA polymerase(s) binds. Which of the following is a common consensus sequence?

  • satellite DNAs
  • GGTTC
  • TATA
  • any trinucleotide repeat
  • TTTTAAAA
A

TATA

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16
Q

What are two main types of posttranscriptional modifications that take place in the mRNA of eukaryotes?

  • The addition of a poly-T sequence at the 5’ end of the gene and the addition of a poly-U tail at the 3’ end.
  • The addition of a poly-A sequence at the 5’ end and the addition of a 7-mG cap at the 3’ end of the RNA transcript.
  • The excision of the introns and the addition of a 7-mG cap to the 3’ end.
  • The addition of a 7-mG cap at the 5’ end of the transcript and the addition of a poly-A sequence at the 3’ end of the message.
A

The addition of a 7-mG cap at the 5’ end of the transcript and the addition of a poly-A sequence at the 3’ end of the message.

(These are the two steps in the processing of eukaryotic mRNA.)

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17
Q

It has been recently determined that the gene for Duchenne muscular dystrophy (DMD) is more than 2000 kb (kilobases) in length; however, the mRNA produced by this gene is only about 14 kb long. What is a likely cause of this discrepancy?

  • The exons have been spliced out during mRNA processing.
  • There are more amino acids coded for by the DNA than by the mRNA.
  • The introns have been spliced out during mRNA processing.
  • The DNA represents a double-stranded structure, whereas the RNA is single-stranded.
  • When the mRNA is produced, it is highly folded and therefore less long.
A

The introns have been spliced out during mRNA processing.

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18
Q

It has been recently determined that the gene for Duchenne muscular dystrophy (DMD) is more than 2000 kb (kilobases) in length; however, the mRNA produced by this gene is only about 14 kb long. What is a likely cause of this discrepancy?

  • The exons have been spliced out during mRNA processing.
  • There are more amino acids coded for by the DNA than by the mRNA.
  • The introns have been spliced out during mRNA processing.
  • The DNA represents a double-stranded structure, whereas the RNA is single-stranded.
  • When the mRNA is produced, it is highly folded and therefore less long.
A

The introns have been spliced out during mRNA processing.

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19
Q

True or False?

Heterogeneous nuclear RNA is a primary transcript in eukaryotes that is processed prior to involvement in translation.

A

True.

20
Q

Which of the following lists steps of mRNA production in eukaryotes in the correct order?

  • Transcription, 5’ cap addition, addition of poly-A tail, exon splicing, passage through nuclear membrane
  • 5’ cap addition, addition of poly-A tail, exon splicing, passage through nuclear membrane, transcription
  • Transcription, addition of poly-A tail, 5’ cap addition, exon splicing, passage through nuclear membrane
  • Transcription, 5’ cap addition, addition of poly-A tail, passage through nuclear membrane, exon splicing
A

Transcription, 5’ cap addition, addition of poly-A tail, exon splicing, passage through nuclear membrane.

21
Q

Which of the following is characteristic of transcription in eukaryotes but not in prokaryotes?

  • A single transcript may be transcribed and translated simultaneously.
  • Exon splicing catalyzed by the spliceosome
  • A 5’ untranslated leader sequence
  • A 3’ untranslated trailer sequence
A

Exon splicing catalyzed by the spliceosome.

(Introns must be removed from eukaryotic pre-mRNA and this process may be facilitated by the spliceosome; introns in prokaryotes are rare, small, and self excise.)

22
Q

Which of the following best describe(s) the function of the 5’ mRNA cap?

  • To protect the transcript from degradation
  • Termination of transcription
  • To provide a binding site for poly(A) polymerase
  • It provides a site for ribosome binding in the cytoplasm.
A
  • To protect the transcript from degradation
  • It provides a site for ribosome binding in the cytoplasm.

(The 5’ cap is essential for recognition of the mRNA by ribosomes in the cytoplasm.)

23
Q

Which of the following best describe(s) the function of the 5’ mRNA cap?

  • To protect the transcript from degradation
  • Termination of transcription
  • To provide a binding site for poly(A) polymerase
  • It provides a site for ribosome binding in the cytoplasm.
A
  • To protect the transcript from degradation
  • It provides a site for ribosome binding in the cytoplasm.

(The 5’ cap is essential for recognition of the mRNA by ribosomes in the cytoplasm.)

24
Q

Which of the following is a characteristic of RNA splicing in Eukaryotes?

  • It involves removal of introns from a gene sequence followed by transcription and subsequent splicing of exons.
  • After splicing occurs, the U1, U2, U5, U6 snRNP complex removes remaining exons for degradation.
  • It involves recognition of sequence-specific intron/exon boundary sites by cytoplasmic proteins.
  • Exon/intron boundaries are typically characterized by a 5’ GU splice junction and a 3’ AG splice junction.
A

Exon/intron boundaries are typically characterized by a 5’ GU splice junction and a 3’ AG splice junction.

(These splice junctions are recognized by the spliceosome so that accurate removal of introns is possible.)

25
Q

A snRNP is best described as _______.

  • small RNAs associated with protein complexes in the nucleus
  • a specific sequence of 7 bases in the mRNA transcript
  • the loop structure that forms when two exons are brought into close proximity
  • short mRNA transcripts combined with nuclear protein complexes
A

Small RNAs associated with protein complexes in the nucleus.

(snRNPs recognize the 5’ and 3’ splice junctions and the branch point sequence, excise the intron, and splice together the exons.)

26
Q

Which of the following is most likely attributable to a base substitution at a 5’ splice junction?

  • A longer than usual final transcript
  • Failure of the U2 snRNP to bind the transcript
  • An amino acid substitution in the final protein product
  • A shorter than usual final transcript
A

A longer than usual final transcript.

Such a mutation could block intron removal, resulting in a longer than usual transcript.

27
Q

Most eukaryotic mRNAs are shorter than the genes that encode them. The reason for this is _______.

  • addition of the poly-A tail to the genomic DNA
  • introns are spliced out of the DNA
  • exons are spliced out of the hnRNA
  • eukaryotic genes contain both introns and exons
A

Eukaryotic genes contain both introns and exons.

After transcription, the introns are spliced out of hnRNA to form mRNA.

28
Q

What is the central dogma of molecular genetics?

A

DNA –> RNA –> protein

Process from DNA to RNA is transcription.

Process from RNA to protein is translation.

29
Q

Which region of a tRNA molecule binds to amino acids?

  • Anticodon loop
  • Codon loop
  • 3′ end
  • Variable loop
A

3′ end.

The 3′ end of a tRNA molecule contains the amino acid binding site.

30
Q

Which of the following statements best describes the function of aminoacyl tRNA synthetase?

  • It attaches a specific amino acid to a tRNA molecule.
  • It synthesizes tRNA molecules.
  • It provides the energy required to attach a specific amino acid to a tRNA molecule.
  • It helps tRNA synthesize proteins.
A

It attaches a specific amino acid to a tRNA molecule.

Aminoacyl tRNA synthetase catalyzes the charging reaction that links a specific amino acid to a tRNA molecule.

31
Q

True or False?

Each aminoacyl tRNA synthetase is specific for one amino acid and a small number of tRNAs.

A

True.

(Each aminoacyl tRNA synthetase enzyme recognizes only one amino acid, but each enzyme can often recognize several tRNAs because there is usually more than one codon for each amino acid.)

32
Q

True or False?

Each aminoacyl tRNA synthetase is specific for one amino acid and a small number of tRNAs.

A

True.

(Each aminoacyl tRNA synthetase enzyme recognizes only one amino acid, but each enzyme can often recognize several tRNAs because there is usually more than one codon for each amino acid.)

33
Q

In bacteria, the methionine that initiates the formation of a polypeptide chain differs from subsequently added methionines in that _______.

  • a formyl group is attached to the initiating methionine
  • its tRNA anticodon is not complementary to the AUG codon
  • incorporation of the initial methionine does not require a tRNA
  • the initiating methionine is not an amino acid
A

a formyl group is attached to the initiating methionine.

This modification is not present on methionine residues added during elongation.

34
Q

In bacteria, the methionine that initiates the formation of a polypeptide chain differs from subsequently added methionines in that _______.

  • a formyl group is attached to the initiating methionine
  • its tRNA anticodon is not complementary to the AUG codon
  • incorporation of the initial methionine does not require a tRNA
  • the initiating methionine is not an amino acid
A

a formyl group is attached to the initiating methionine.

This modification is not present on methionine residues added during elongation.

35
Q

Translation in bacteria is directly dependent on all of the following associations except _______.

  • complementary base pairing between mRNA and rRNA
  • association of the 30S and the 50S ribosomal subunits
  • complementary base pairing between mRNA and tRNA
  • complementary base pairing between mRNA and DNA
A

Complementary base pairing between mRNA and DNA.

Transcription, not translation, is dependent on this association.

36
Q

Which of the following best describes the first step in the formation of the translation initiation complex?

  • The small ribosomal subunit binds to an mRNA sequence near the 3’ end of the transcript.
  • The large ribosomal subunit binds to the small ribosomal subunit.
  • The small ribosomal subunit binds to an mRNA sequence near the 5’ end of the transcript
  • The large ribosomal subunit binds to an mRNA sequence near the 5’ end of the transcript
A

The small ribosomal subunit binds to an mRNA sequence near the 5’ end of the transcript.

37
Q

At which site does the charged initiator tRNA bind during protein synthesis?

  • A site
  • E site
  • T site
  • P site
A

P site

(The initiator tRNAfmet binds to the mRNA codon in the P site of the ribosome. The initiator tRNA is the only one that binds in the P site; all other tRNAs bind the ribosome in the A site.)

38
Q

True or False?

The enzyme EF‑Tu catalyzes the formation of a peptide bond between the amino acid held by the tRNA in the A site and the elongating amino acid chain held by the tRNA in the P site.

A

False.

(Peptidyl transferase is the enzyme that catalyzes the formation of peptide bonds during translation. EF‑Tu is an elongation factor that facilitates the entry of charged tRNAs into the A site.)

39
Q

What event occurs during translocation?

  • The two ribosomal subunits join to form a complex.
  • mRNA shifts in the 5’ direction along the ribosome.
  • Amino acids are added to the polypeptide chain.
  • The polypeptide is cleaved from the terminal tRNA.
A

mRNA shifts in the 5’ direction along the ribosome.

(Translocation is the process by which mRNA shifts by 3 bases in the 5’ direction along the ribosome to bring another codon into the A site.)

40
Q

True or False?

Different sets of human hemoglobins are found at different times in development.

A

True.

(During embryonic and fetal development, the set of polypeptides found in hemoglobin is completely different from that found in the hemoglobin of adults.)

41
Q

True or False?

Different sets of human hemoglobins are found at different times in development.

A

True.

(During embryonic and fetal development, the set of polypeptides found in hemoglobin is completely different from that found in the hemoglobin of adults.)

42
Q

The β chain of adult hemoglobin is composed of 146 amino acids of a known sequence. In comparing the normal β chain with the β chain in sickle cell hemoglobin, what alteration is one likely to find?

  • glutamic acid replacing valine in the first position
  • frameshift substitutions
  • valine instead of glutamic acid in the sixth position
  • extensive amino acid substitutions
  • trinucleotide repeats
A

Valine instead of glutamic acid in the sixth position.

43
Q

The β chain of adult hemoglobin is composed of 146 amino acids of a known sequence. In comparing the normal β chain with the β chain in sickle cell hemoglobin, what alteration is one likely to find?

  • glutamic acid replacing valine in the first position
  • frameshift substitutions
  • valine instead of glutamic acid in the sixth position
  • extensive amino acid substitutions
  • trinucleotide repeats
A

Valine instead of glutamic acid in the sixth position.

44
Q

The primary structure of a protein is determined by ________.

  • hydrogen bonds formed between the components of the peptide linkage
  • pleated sheets
  • the sequence of amino acids
  • covalent bonds formed between fibroin residues
  • a series of helical domains
A

The sequence of amino acids.

45
Q

The primary structure of a protein is determined by ________.

  • hydrogen bonds formed between the components of the peptide linkage
  • pleated sheets
  • the sequence of amino acids
  • covalent bonds formed between fibroin residues
  • a series of helical domains
A

The sequence of amino acids.

46
Q

The secondary structure of a protein includes _____.

  • alpha and gamma
  • disulfide bridges
  • hydrophobic clusters
  • gamma and delta
  • α-helix and β-pleated sheet
A

α-helix and β-pleated sheet

47
Q

One form of posttranslational modification of a protein includes ________.

  • removal of introns
  • self-splicing
  • removal or modification of terminal amino acids
  • removal of exons
  • shuffling of exons
A

Removal or modification of terminal amino acids

Genetics Final Exam (5 decks)

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