Genetics Chap 12, 13, 14, & 15

Front 1

Chap 12 # 1
What is semiconservative replication?

Back 1

In semiconservative replication, the original 2 strands of the double helix serve as templates for new strands of DNA. When replication is complete, two double-stranded DNA molecules will be present. Each will consist of one original template strand and one newly synthesized strand that is complementary to the template.

Front 2

Chap 12 # 2
How did Meselson and Stahl demonstrate that replication in E. Coli takes place in a semiconservative manner?

Back 2

By growing E. Coli cells in a medium containing the heavy isotope of nitrogen (15 N) for several generations. The (15N) was incorporated into the DNA of the E. Coli cells. The E. Coli cells were then switched to a medium containing the common form of nitrogen (14N) and allowed to proceed through a few cycles of cellular generations. Samples of the bacteria were removed at each cellular generation. Using equilibrium density gradient centrifugation, they were able to distinguish DNA's that contained only (15N) from DNA's that contained only (14N) or a mixture of both because DNA's containing the (15N) isotope are heavier. The more (15N) a DNA molecule containes, the further it will sediment during equilibrium density gradient centrifugation. DNA from (15N)produced only a single band at expected position during centrifugation. After one round of replication in the (14N) medium and centrifugation 1 band was present at an intermediate to that of (15N) and after 2 rounds of replication 2 bands were present one at an intermediate to (15N) and one at the expected position for DNA only containing (14N). Results are consistent with predictions of semiconservative replication and incompatible with conservative and dispersive replication.

Front 3

Chap 12 # 3
Draw a molecule of DNA undergoing Theta replication. On your drawing, identify 1) origin, 2) polarity 5' & 3' ends), 3) leading and lagging strands, 4) Okazaki Fragments, and 5) location of primers.

Back 3

Draw this.

Front 4

Chap 12 # 4
Draw a molecule of DNA undergoing rolling-circle replication. On your drawing, identify 1) origin, 2) polarity (5' & 3' ends) of all templae strands and newly synthesized strands, 3) leading and lagging strands, 4) Okazaki fragments, and 5) location of primers.

Back 4

Draw this.

Front 5

Chap 12 # 8
List the different proteins and enzymes taking part in bacterial replication. Give the function of each in the replication process.

Back 5

1) DNA polymerase III is the primary replication polymerase. It elongates a new nucleotide strand from the 3'-OH of the primer.
2) DNA polymerase I removes the RNA nucleotides of the primers and replaces them with DNA molecules.
3) DNA ligase connects Okazaki fragments by sealing nicks in the sugar phosphate backbone.
4) DNA primase synthesizes the RNA primers that provide the 3'-OH group needed for DNA polymerase III to initiate DNA synthesis.
5) DNA helicase unwinds the double helix by breaking the hydrogen bonding between the 2 strands at the replication fork.
6) DNA gyrase reduces DNA supercoiling and torsional strain that is created ahead of the replication fork by making double-stranded breaks in the DNA and passing another segment of the helix through the break before resealing it. Gyrase is also called topoisomerase II.
7) Initiator proteins bind to the replication origin and unwind short regions of DNA
8) Single-stranded binding protein (SSB protein) stabilizes single-stranded DNA prior to replication by binding to it, thus preventing the DNA from pairing with complementary sequences.

Front 6

CHAP 12 # 9
What similarities and differences exist in the enzymatic activities of DNA polymerase I, II, and III? What is the function of each type of DNA polymerase in bacterial cells?

Back 6

Each of the 3 DNA polymerases has a 5' to 3' polymerase activity. They differ in their exonuclease activities. DNA polymerase I has a 3' to 5' as well as a 5' to 3' exonuclease activity. DNA polymerase II & III have only a 3' to 5' exonuclease activity.
1) DNA Polymerase I carries out proofreading. It aslo removes and replaces the RNA primers used to initiate DNA synthesis.
2) DNA polymerase II functions as a DNA repair polymerase. It restarts replication after DNA damage has halted replication. It has proofreading activity.
3) DNA polymerase III is the primary replication enzyme and also has a proofreading function in replication.

Front 7

CHAP 12 # 10
Why is primase required for replication?

Back 7

Primase is a DNA-dependent RNA polymerase. Primase synthesizes the short RNA molecules, or primers, that have a free 3'-OH to which DNA polymerase can attach deoxyribonucleotides in replication initiation. The DNA polymerases require a free 3'-OH to which they add nucleotides, and therefore they cannot initiate replication. Primase does not have this requirement.

Front 8

CHAP 12 # 11
What 3 mechanisms ensure the accuracy of replication in bacteria?

Back 8

1) Highly accurate nucleotide selection by the DNA polymerase when pairing bases.
2) the proofreading function of DNA polymerase, which removes incorrectly inserted bases.
3) a mismatch repair apparatus that repairs mistakes after replication is complete.

Front 9

CHAP 12 # 13
In what ways is eukaryotic replication similar to bacterial replication, and in what ways is it different?

Back 9

1) Semiconservative replication
2) replication origins serve as starting points for replication.
3) Short segments of RNA called primers provide a 3'-OH for DNA polymerases to begin synthesis of new strands.
4) Synthesis occurs in 5' to 3' direction.
5) The template strand is read in 3' to 5' direction.
6) Deoxyribonucleoside triphosphates are the substrates
7) Replication is continuous on the leading strand and discontinuous on the lagging strand.
Eukaryotic DNA replication differs from bacterial replication in that:
1) It has multiple origins of replications per chromosome.
2) It has several different DNA polymerases with different functions.
3) Immediately following DNA replication, assembly of nucleosomes takes place.
1) It has mu

Front 10

CHAP 12 # 14
What is the end-of-chromosome problem for replication? Why, in the absence of telomerase, do the ends of chromosomes get progressively shorter each time the DNA is replicated?

Back 10

For DNA polymerases to work, they need the presence of a 3'-OH group to which to add a nucleotide. At the ends of the chromosomes when the RNA primer is removed, ther is no adjacent 3'-OH group to which to add a nucleotide, thus no nucleotides are added leaving a gap at the end of the chromosome. Telomerase can extend the single stranded protruding end by pairing with the overhanging 3' end of the DNA and adding a repeated sequence of nucleotides. In the absence of telomerase, DNA polymerase will be unable to add nucleotides to the end of the strand. After multiple rounds of replication without a functional telomerase the chromosome will become progressively shorter.

Front 11

CHAP 12 # 21
A circular molecule of DNA contains 1 million base pairs. If DNA synthesis at a replication fork occurs at a rate of 100,000 nucleotides per minute, how long will theta replication require to completely replicate a molecule, assuming that theta replication is bi-directional? How long will replication of this circular chromosome take by rolling-circle replication? Ignore replication of the displaced strand in rolling-circle replication.

Back 11

In bi-directional replication there are 2 replication forks, each proceeding at a rate of 100,000 nucleotides per minute. Therefore it would require 5 minutes for circular DNA to be replicated by bi-directional replication because each fork could synthesize 500,000 nucleotides (5 min x 100,000 nucleotides per min) within the time period. Because rolling-circle replication is unidirectional and thus has only one replication fork, 10 minutes will be required to replicate the entire circular molecule.

Front 12

CHAP 12 # 23
The following diagram represents a DNA molecule that is undergoing replication. Draw the strands of the newly synthesized DNA and identify the following:
1) Polarity of newly synthesized strands
2) Leading and Lagging strands
3) RNA primers.

Back 12

Draw This.

Front 13

CHAP 12 # 24
What would be the effect on DNA replication of mutations that destroyed each of the following activities in DNA polymerase I?
a) 3' - 5' exonuclease activity
b) 5' - 3' exonuclease activity
c) 5' - 3' polymerase activity

Back 13

A) The 3' - 5' exonuclease activity is important for proofreading newly synthesized DNA. If the activity is nonfunctional, then the fidelity of replication by DNA polymerase I will decrease, resulting in more misincorporated bases in the DNA
B) Loss of the 5' - 3' exonuclease activity would result in the RNA primers used to initiate replication not being removed by DNA polymerase I.
C) DNA polymerase I would be unable to synthesize new DNA strands if the 5' - 3' polymerase activity was destroyed. RNA primers could be removed by DNA polymerase I using the 5' - 3' exonuclease activity, but new DNA sequencews could not be added in their place by DNA polymerase I.

Front 14

CHAP 12 # 25
How would DNA replication be affected in a cell that is lacking topoisomerase?

Back 14

Topoisomerase II (gyrase) reduces the positive supercoiling or torsional strain that develops ahead of the replication fork due to the unwinding of the double helix. if the topoisomerase activity was lacking, then the torsional strain would continue to increase, making it more difficult to unwind the double helix. Ultimately, the increasing strain would lead to an inhibition of the replication fork movement.

Front 15

Chap 13 # 2
What are the major classes of cellular RNA? Where would you expect to find each class of RNA within eukaryotic cells?

Back 15

1) Ribosomal RNA, or rRNA, is found in the cytoplasm.
2) Transfer RNA, or tRNA, is found in the cytoplasm.
3) Messenger RNA, or mRNA, is found in the cytoplasm, (However, pre-mRNA is found only in the nucleus.
4) Small nuclear RNA, or smRNA, is found in the nucleus as part of riboproteins called snrps.
5) Small nucleolar RNA, or snoRNA, is found in the nucleus.
6) Small cytoplasmic RNA, or scRNA, is found in the cytoplasm.

Front 16

Chap 13 # 4
What parts of DNA make up a transcription unit? Draw and label a typical transcription unit in a bacterial cell.

Back 16

Draw This.

Front 17

Chap 13 # 6
Describe the structure of RNA polymerase.

Back 17

Bacterial RNA polymerase consists of several polypeptides. The RNA polymerase core enzyme is composed of 4 polypeptide subunits: 2 copies of the alpha subunit, the beta subunit, and the beta prime subunit. The addition of a sigma factor to the core enzyme forms the RNA polymerase holoenzyme.

Front 18

Chap 13 # 8
What are the three basic stages of transcription? Describe what happens at each.

Back 18

1) Initiation: Transcription proteins assemble at the promoter to form the basal transcription apparatus and begins synthesis of RNA.
2) Elongation: RNA polymerase moves along the DNA template in a 3' to 5' direction unwinding the DNA and synthesizing RNA in a 5' to 3' direction.
3) Termination: Synthesis of RNA is terminated, and the RNA molecule separates from the DNA template.

Front 19

Chap 13 # 9
Draw and label a typical bacterial promoter. Include any common consensus sequences.

Back 19

Draw This.

Front 20

CHAP 13 # 11
How is the process of transcription in eukaryotic cells different from that in bacterial cells?

Back 20

Eukaryotic transcription requires the action of 3 RNA polymerases. Each type of polymerase recognizes and transcribes from different types of promoters. Binding to the promoter and initiation from the promoter requires the action of many protein factors; different promoters require different sets of protein factors. The RNA molecule produced by transcription in eukaryotic cells usually requires extensive processing, such as the addition of a 5' cap, a 3' poly(A) tail, and the removal of introns prior to becoming functional. Bacterial promoters tend to be more uniform in composition, and only 1 RNA polymerase does transcription. Bacterial RNAs are typically functional once transcription has taken place.

Front 21

CHAP 13 # 16
The following diagram represents DNA that is part of the RNA-coding sequence of a transcription unit. The bottom strand is the template strand. Give the sequence found on the RNA molecule transcribed from this DNA and label the 5' and 3' ends of the RNA
5'-ATAGGCGATGCCA-3'
3'-TATCCGCTACGGT-5'

Back 21

The RNA molecule would be complementary to the template strand, contain uracil, and be synthesized in an antiparallel fasion.
The RNA strand contains the ssame sequence as the nontemplate DNA strand except that the RNA strand contains uracil instead of thymine.

Front 22

CHAP 13 # 19
Write the consensus sequence for the following set of nucleotide sequences.
AGGAGTT
AGCTATT
TGCAATA
ACGAAAA
TCCTAAT
TGCAATT

Back 22

The consensus sequence is identified by determining which nucleotide is used most frequently at each position. For the 2 nucleotides that occur at an equal frequency at the first position, both are listed at that position and identified by a slash mark:
T/A GCAATT

Front 23

CHAP 13 # 20
List at least 5 properties that DNA polymerases and RNA polymerases have in common. List at least 3 differences.

Back 23

Similarities:
1) Both use DNA templates
2) DNA templates are read in the 3' - 5' direction.
3) the complementary strand is synthesized in a 5' - 3' direction that is antiparallel to the template.
4) Both use triphosphates as substrates
5) Their actions are enhanced by accessory proteins
Differences:
1) RNA polymerases use ribonucleoside triphosphates as substrates, whereas DNA polymerases use deoxyribonucleoside triphosphates
2) DNA polymerases require a primer that provides an available 3'-OH group where synthesis begins, whereas RNA polymerases do not require primers to begin synthesis
3) RNA polymerases synthesize a copy off only one of the DNA strands, Whereas DNA polymerases can synthesize copies of both strands.

Front 24

CHAP 14 # 2
What are some characteristics of introns?

Back 24

Eukaryotic genes commonly contain introns. However, introns are rare in bacterial genes. The number of introns found in an organism's genome is typically related to complexity - More complex organism's possess more introns. Introns typically do not encode proteins and are usually larger in size than exons.

Front 25

CHAP 14 # 4
What are the 3 principal elements in mRNA sequences in bacterial cells?

Back 25

1) The 5' untranslated region, which contains the Shine-Dalgarno sequence
2) The protein encoding region
3) The 3' untranslated region

Front 26

CHAP 14 # 5
What is the function of the Shine-Delgarno consensus sequence?

Back 26

The Shine-Delgarno consensus sequence functions as the ribosome-binding site on the mRNA molecule.

Front 27

CHAP 14 # 6
What is the 5' cap?

Back 27

The 5' end of eukaryotic mRNA is modified by the addition of the 5' cap. The cap consists of an extra guanine nucleotide linked to 5' to 5' to the mRNA molecule. This nucleotide is methylated at position 7 of the base. The ribose sugars of adjacent bases may be methylated at the 2'-OH.

Front 28

Chap 14 # 7
How is the poly(A) tail added to Pre-mRNA? What is the purpose of the poly(A) Tail?

Back 28

Initially, a complex consisting of several proteins forms on the 3' UTR of the pre-mRNA molecule. Cleavage and polyadenylation specificity factor (CPSF) bind to the AAUAAA consensus sequence, which is located upstream of the 3' cleavage site. Another protein, cleavage stimulation factor (CsTF), binds downstream of the cleavage site. 2 cleavage factors (CFI & CFII) and polyadenylate polymerase (PAP) also become part of the complex. Once the complex has formed, the pre-mRNA is cleaved. CsTF and the 2 cleavage factors leave the complex. PAP adds approximately 10 adenine nucleotides to the 3' end of the pre-mRNA molecule. The addition of the short poly(A) tail allows for the binding of the poly(A) protein (PABII) to the tail. PABII increases the rate of polyadenylation, which subsequently allows for more PABII protein to bind to the tail.

Front 29

Chap 14 # 8
What makes up the spliceosome? What is the function of the spliceosome?

Back 29

The spliceosome consists of 5 small ribonucleotides (snRNPs). Each snRNP is composed of multiple proteins and a single small nuclear RNA molecule or snRNA. The snRNPs are identified by which snRNA (U1, U2, U3, U4, U5, or U6) each contains. Splicing of pre-mRNA nuclear introns takes place within the spliceosome.

Front 30

Chap 14 # 9
Explain the process of pre-mRNA splicing in nuclear genes.

Back 30

Removal of the intron from the pre-mRNA requires the assembly of the spliceosome complex on the pre-mRNA, cleavage at both the 5' & 3' splice sites of the intron, and 2 transesterification reactions ultimately leading to the joining of the 2 exons. Initially, snRNP UI binds to the 5' splice site through complementary base pairing of the UI snRNA. Next, snRNP U2 binds to the branch point within the intron. The U5 and U4-U6 complex joins the spliceosome, resulting in the looping of the intron so that the bracnch point and 5' splice site of the intron are now adjacent to each other. UI and U4 now disassociate from the spliceosome, and the spliceosome is activated. The pre-mRNA is then cleaved at the 5' splice site, producing an exon with a 3'-OH. The 5' end of the intron folds back and forms 5'-2' phosphodiester linkage through the first transesterification reaction with the adenine nucleotide at the branch point of the intron. This looped structure is called the lariat. Next, the 3' splice site is cleaved and then immediately ligated to the 3;-OH of the first exon through the second transesterification reaction. Thus, the exons are now joined and the intron has been excised.

Front 31

CHAP 14 # 13
What are some of the modifications in tRNA that occur through processing?

Back 31

1) The precursor RNA may be cleaved into smaller molecules.
2) Nucleotides at both 5' & 3' ends of the tRNAs may be removed or trimmed.
3) Standard bases can be altered by base-modifying enzymes.

Front 32

CHAP 14 # 15
Explain how rRNA is processed.

Back 32

Most rRNAs are synthesized as large precursor RNAs that are processed by methylation, cleavage, and trimming to produce the mature mRNA molecules. In E. Coli, methylation occurs to specific bases and the 2'-OH of the ribose sugars of the 30S rRNA precursor. The 30S precursor is cleaved and trimmed to produce the 16S rRNA, 23S rRNA, and the 5S rRNA. In eukaryotes, a similar process occurs. However, small nucleolar RNAs help to cleave and modify the precursor rRNAs.

Front 33

CHAP 14 # 16
What is the origin of siRNAs and microRNAs? What do these RNA molecules do in the cell?

Back 33

The siRNAs originate from the cleavage of mRNAs, RNA transposons, and RNA viruses by enzyme Dicer. Dicer may produce multiple siRNAs from a single double-stranded RNA molecule. The double-stranded RNA molecule may occur due to the formation of hairpins or by duplexes between different RNA molecules. The miRNAs arise from the cleavage of individual RNA molecules that are distinct from other genes. The enzyme Dicer cleaves these RNA molecules that have formed small hairpins. A single miRNA is produced from a single RNA molecule.
Both siRNAs and microRNAs silence gene expression through a process called RNA interference. Both function by shutting off gene expression of a cell's own genes or to shut off expression of genes from the invading foreign genes of viruses or transposons. The micrRNAs typically silence genes that are different from those which tha microRNAs are transcribed. However, the siRNAs usually silence genes from which they are transcribed.

Front 34

CHAP 14 # 20
How do the mRNA of bacterial cells and the pre-mRNA of eukaryotic cells differ?
How do the mature mRNAs of bacterial cells and eukaryotic cells differ?

Back 34

Bacterial mRNA is translated immediately upon being transcribed. Eukaryotic pre-mRNA must be processed. Bacterial mRNA and eukaryotic pre-mRNA have similarities in structure. Each has a 5' untranslated region as well as a 3' untranslated region. Both also have protein-coding regions. However, the protein-coding region of the pre-mRNA is disrupted by introns. The eukaryotic pre-mRNA must be processed to produce the mature mRNA. Eukaryotic mRNA has a 5' cap and a poly(A) tail, unlike bacterial mRNAs. Bacterial mRNA also contains the Shine-Dalgarno consensus sequence. Eukaryotic mRNA does not have the equivalent.

Front 35

CHAP 14 # 21
Draw a typical eukaryotic gene and the pre-mRNA and mRNA derived from it. Assume that the gene contains 3 exons. Identify the following items and, for each item, give a brief description of its function.
a) 5' untranslated region
b) Promoter
c) AAUAAA consensus sequence
d) Transcription start site.
e) 3' untranslated region
f) Introns
g) Exons
h) Poly(A) tail
i) 5' Cap

Back 35

Draw This.

Front 36

MY CARDS
What is the function of
DNA Polymerase III?

Back 36

It is the primary replication polymerase. It elongates a new nucleotide strand from the 3'-OH of the primer.

Front 37

MY CARDS
What is the function of
DNA Polymerase I?

Back 37

It removes the RNA nucleotides of the primers and replaces them with DNA molecules.

Front 38

MY CARDS
What is the function of
DNA Ligase?

Back 38

It connects the Okazaki fragments by sealing nicks in the sugar phosphate backbone.

Front 39

MY CARDS
What is the function of
DNA Primase?

Back 39

It synthesizes the RNA primers that provide the 3'OH group needed for DNA polymerase III to initiate synthesis.

Front 40

MY CARDS
What is the function of
DNA Helicase?

Back 40

It unwinds the double helix by breaking the hydrogen bonding between the 2 strands at the replication fork.

Front 41

MY CARDS
What is the function of
DNA Gyrase?

Back 41

It reduces the supercoiling and torsional strain that is created ahead of the replication fork by making double-stranded breaks in the DNA and passing another segment of the helix through the break before resealing it. Gyrase also called topoismerase.

Front 42

MY CARDS
What is the function of
Initiator Proteins?

Back 42

They bind to the replication origin and unwind short regions of DNA.

Front 43

MY CARDS
What is the function of
Single-stranded binding protein (SSB)?

Back 43

It stabilizes single-stranded DNA prior to replication by binding to it, thus preventing the DNA from pairing with complementary sequences.

Front 44

MY CARDS
Describe function of
DNA polymerase I.

Back 44

Carries out proofreading. It also removes and replaces the RNA primers used to initiate DNA synthesis.

Front 45

MY CARDS
Describe function of
DNA polymerase II.

Back 45

Functions as a DNA repair polymerase. It restarts replication after DNA damage has halted replication. It has proofreading activity.

Front 46

MY CARDS
Describe function of
DNA polymerase III.

Back 46

It is the primary replication enzyme and also has a proofreading function in replication.

Front 47

MY CARDS
What is the similarities of the 3 DNA polymerases?

Back 47

All have 5' to 3' polymerase activity.

Front 48

MY CARDS
What are the differences between the 3 DNA polymerases?

Back 48

they differ in exonuclease activities.
DNA polymerase I & III have only a 3' to 5' exonuclease activity.
DNA polymerase II has both 3' to 5' and 5' to 3' exonuclease activity.

Front 49

MY CARDS
Name the major classes of RNA.

Back 49

Ribosomal (rRNA)
Transfer (tRNA)
Messenger (mRNA)
Small Nuclear (smRNA)
Small Nucleolar (snoRNA)
Smal Cytoplasmic (scRNA)

Front 50

MY CARDS
Where is Ribosomal RNA found?

Back 50

In the cytoplasm.

Front 51

MY CARDS
Where is Transfer RNA found?

Back 51

In the Cytoplasm.

Front 52

MY CARDS
Where is Messenger RNA found?

Back 52

In the Cytoplasm. (However, pre-mRNA is only found in the nucleus)

Front 53

MY CARDS
Where is pre-mRNA found?

Back 53

In the Nucleus.

Front 54

MY CARDS
Where is Small nuclear RNA found?

Back 54

In the nucleus (as part of riboproteins called snrps.)

Front 55

MY CARDS
Where is Small Nucleolar RNA found?

Back 55

In the Nucleus.

Front 56

MY CARDS
Where is Small CytoplasmicRNA found?

Back 56

In the Nucleus.

Front 57

MY CARDS
What are snrps?

Back 57

small nuclear ribosomal proteins.

Front 58

MY CARDS
Describe Initiation.

Back 58

Transcription proteins assemble at the promoter to form the basal transcription apparatus and begins synthesis of RNA.

Front 59

MY CARDS
Describe Elongation.

Back 59

RNA polymerase moves along the DNA template in a 3' to 5' direction unwinding the DNA and synthesizing RNA in a 5' to 3' direction.

Front 60

MY CARDS
Describe Termination.

Back 60

Synthesis of RNA is terminated, and the RNA molecule seperates from the DNA template.

Front 61

CHAP 15 # 3
What are isoaccepting tRNAs?

Back 61

Isoaccepting tRNAs are tRNA molecules that have different anticodon sequences but accept the same amino acids.

Front 62

CHAP 15 # 4
What is the significance of the fact that many synonymous codons differ only in the third nucleotide position?

Back 62

Synonymous codons code for the same amino acid, or, in other words, have the same meaning. A nucleotide at the 3rd position of a codon pairs with a nucleotide in the 1st position of the anticodon. Unlike the other nucleotide positions involved in the codon-anticodon pairing, this pairing is often weak or wobbles, and nonstandard pairings can occur. Because of the wobble, or nonstandard base-pairing with the anticodons, affects the 3rd nucleotide position, the redundency of codons ensures that the correct amino acid is inserted in the protein when nonstandard pairings occur.

Front 63

CHAP 15 # 5A
Define:
Reading Frame.

Back 63

The reading frame refers to how the nucleotides in a nucleic acid molecule are grouped into codons containing 3 nucleotides. Each sequence of nucleotides has 3 possible sets of codons, or reading frames.

Front 64

CHAP 15 # 5B
Define:
Overlapping code.

Back 64

If an overlapping code is present, then a single nucleotide is included in more than 1 codon. The result for a sequence of nucleotides is that more than one type of polypeptide can be encoded within that sequence.

Front 65

CHAP 15 # 5C
Define:
Nonoverlapping code.

Back 65

In a nonoverlapping code, a single nucleotide is part of only 1 codon. This results inthe production of a single type of polypeptide from one polynucleotide sequence.

Front 66

CHAP 15 # 5D
Define:
Initiation codon:

Back 66

An initiation codon establishes the appropriate reading frame and specifies the first amino acid of the protein chain. Typically, the initiation codon is AUG; however, GUG and UUG can also serve as initiation codons.

Front 67

CHAP 15 # 5E
Define:
Termination codon:

Back 67

The termination codon signals the termination or end of translation and the end of the protein molecule. There are 3 types of termination codons - UAA, UAG, and UGA - which can also be referred to as stop codons or nonsense codons. These codons do not code for amino acids.

Front 68

CHAP 15 # 5F
Define:
Sense codon:

Back 68

A sense codon is a group of 3 nucleotides that code for an amino acid. There are 61 sense codons that code for 20 amino acids commonly found in proteins.

Front 69

CHAP 15 # 5G
Define:
Nonsense codon:

Back 69

Nonsense codons or termination codons signal the end of translation. These codons do not code for amino acids.

Front 70

CHAP 15 # 5H
Define:
Universal code:

Back 70

In a universal code, each codon specifies, or codes, for the same amino acid in all organisms. The genetic code is nearly universal, but not completely. Most of the exceptions occur in mitochondrial genes.

Front 71

CHAP 15 # 5I
Define:
Nonuniversal codons:

Back 71

Most codons are universal (or nearly universal) in that they specify the same amino acids in almost all organisms. However, there are exceptions where a codon has different meanings in different organisms. Most of the know exceptions are the termination codons, which in some organisms do code for amino acids. Occasionally, a sense codon is substituted for another sense codon.

Front 72

CHAP 15 # 7
How are tRNAs linked to their corresponding amino acids?

Back 72

Each of the 20 different amino acids that are commonly found in proteins has a corresponding aminoacyl-tRNA synthetase that covalently links tha amino acid to the correct tRNA molecule.

Front 73

CHAP 15 # 9
How does the process of initiation differ in bacterial cells and eukaryotic cells?

Back 73

Bacterial initiation of translation requires that sequences in the
16S rRna of the small ribosomal subunit bind to the mRNA at the ribosome binding site or Shine-Dalgarno sequence. The Shine-Dalgarno sequence is essential in placing the ribosome over the start codon (typically AUG). In eukaryotes, there is no Shine-Dalgarno sequence. The small ribosomal subunit recognizes the 5' cap of the eukaryotic mRNA with the assistance of initiation factors. Next, the ribosomal small subunit migrates along the mRNA scanning for the AG start codon. In eukaryotes, the start codon is located with a consensus sequence called the Kozak sequence (5' - ACCAUGG - 3'). transcription in eukaryotes also requires more initiation factors.

Front 74

CHAP 15 # 14
What are some types of posttranslational modification of proteins?

Back 74

Several different modifications can occur to a protein following translation. Frequently, the amino terminal methionine may be removed. Sometimes, in bacteria only the formyl group is cleaved from the N-formyl methionine, leaving a methionine at the amino terminal. More extensive modification occurs in some proteins that are originally synthesized as precursor proteins. These precursor proteins are cleaved and trimmed by protease enzymes to produce a functionalprotein. Glycoproteins are produced by the attachment of carbohydrates to newly synthesized proteins. Molecular chaperones are needed by many proteins to ensure that the proteins are folded correctly. Secreted proteins that are targeted for the membrane or other cellular locations frequently have 15 to 30 amino acids, called the signal sequence, removed from the amino terminal. Finally acetylation of amino acids in the amino terminal of some eukaryotic proteins also occurs.

Front 75

CHAP 15 # 15
Explain how some antibiotics work by affecting the process of protein synthesis.

Back 75

A number of antibiotics bind the ribosome and inhibit protein synthesis at different steps in translation. Some antibiotics, such as streptomycin, bind to the small subunit and inhibit translation initiation. Other antibiotics, such as chloramphenicol bind to the large subunit and block elongation of the peptide by preventing peptide bond formation. Antibiotics such as tetracycline and neomycin, bind the ribosome near the "A" site yet have different effects. Tetracyclines block entry of charged tRNAs to the "A" site, while neomycin induces translation errors. Finally, some antibiotics such as erythromycin, block the translation of the ribosome along the mRNA.

Front 76

What 3 different methods were used to help break the genetic code? What did each reveal and what were the advantages and disadvantages?

Back 76

Marshal Nirenberg and Johann Heinrich Matthaei used the enzyme polynucleotide kinase to create homopolymers of synthetic RNAs. Using a cell-free protein synthesizing system they were able to determine the amino acid coded by each homopolymer. By this method, the meanings for the amino acids specified by the codons UUU, AAA, CCC, and GGG were determined. The disadvantage is that the meanings for only 4 codons could be determined.
The same system was also used to create copolymers that contained random mixtures of 2 nucleotides in a known ratio. Different amino acids in the protein depended on the ratio of the 2 nucleotides. To determine or predict the composition of the codons, the frequency of amino acids produced using the copolymer was compared with the theoretical frequencies expected for the codons. A disadvantage of this procedure is that it depended on random incorporation of nucleotides, which did not always happen. A further problem was that the base sequence of the codon could not be determined - only the bases contained within the codon. The redundancy of the code also provided difficulties because several different codons could specify the same amino acid. To solve these problems, they mixed ribosomes bound to short RNAs of known sequences with charged tRNAs. The mixture was passed through a nitrocellulose filter to which the tRNAs paired to ribosome-mRNA stuck. They next determined the amino acids attached to the bound tRNAs. More then 50 codons were identified by this method. The difficulty is that not all tRNAs and codons could be identified with this method.
Gobind Khorana and his colleagues used a 3rd method. They synthesized RNA molecules of known repeating sequences. Using a cell-free protein synthesizing system they produced proteins of alternating amino acids. However, this procedure could not specify which codon encodes which amino acid.

Front 77

CHAP 15 # 27
Arrange the following components of translation in the appropriate order in which they would appear or be used during protein synthesis.
70S initiation complex
release factor 1
peptidyl transferase
elongation factor G
30S initiation complex
elongation factor TU
initiation factor 3
fMet-tRNA

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fMet-tRNA
30S initiation complex
Initiation factor 3
70S initiation complex
Elongation factor Tu
Peptidyl transferase
elongation factor G
Release factor 1

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The following diagram illustrates a step in the process of translation.
Identify the following:
a) 5' and 3' ends of the RNA
b) A,P, and E sites
c) Start codon
d) Stop codon
e) Amino and carboxyl ends of the newly synthesized polypeptide chain.
f) Approximate location of the next peptide bond that will be formed
g) Place on the ribosome where release factor 1 will bind

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