Genetics Quiz 7

Name six different levels at which gene expression might be controlled.

(1) Alteration or modification of the gene structure at the DNA level (2) Transcriptional regulation (3) Regulation at the level of mRNA processing (4) Regulation of mRNA stability (5) Regulation of translation (6) Regulation by post-translational modification of the synthesized protein

What are the three steps of cloning a foreign DNA fragment into a plasmid vector using a restriction enzyme that generates sticky ends? Why is ligase needed?

1) Cut foreign dna fragment and plasmid with same restriction enzyme 2) Restriction enzyme makes staggered cuts in dna, producing complementary sticky ends on the foreign dna and the plasmid 3) DNA and plasmid are mixed together, some of their cut ends pairing together. Ligase is used to seal the nicks in the sugar phosphate backbone, creating a recombinant plasmid with the foreign dna fragment in it.

Why must the DNA fragments on the gel, as well as the probe DNA, be denatured prior to hybridization?

Because when in Southern blotting the probe single stranded dna fragments bind to the membrane by completing the complementary sequences on the membrane. If they weren't denatured before hybridization, they would not be able to bind to the membrane.

What antibiotic binds to the 50S subunit of the ribosome and blocks peptide bond formation? Which one binds to the A site? Which one binds to the small subunit and inhibits initiation? Which one blocks translocation?

Chloramphenicol. Tetracyclines. Streptomycin. Erythromycin.

Give signs for B-galactosidase and permease in lactose absent (first parenthesis) and lactose present (second parenthesis) for each of the following. 1. lacI+lacP+lacO+lacZ+lacY+ 2. lacI-lacP+lacO+lacZ+lacY+ 3. lacI+lacP+lacOclacZ+lacY+ 4. lacI-lacP+lacO+lacZ+lacY- 5. lacI-lacP-lacO+lacZ+lacY+ 6. lacI+lacP+lacO+lacZ-lacY+/ lacI-lacP+lacO+lacZ+lacY- 7. lacI-lacP+lacOclacZ+lacY+/ lacI+lacP+lacO+lacZ-lacY- 8. lacI-lacP+lacO+lacZ+lacY-/ lacI+lacP-lacO+lacZ-lacY+ 9. lacI+lacP-lacOclacZ-lacY+/ lacI-lacP+lacO+lacZ+lacY- 10. lacI+lacP+lacO+lacZ+lacY+/ lacI+lacP+lacO+lacZ+lacY+ 11. lacIslacP+lacO+lacZ+lacY-/ lacI+lacP+lacO+lacZ-lacY+ 12. lacIslacP-lacO+lacZ-lacY+/ lacI+lacP+lacO+lacZ+lacY+

1. (-, -) (+,+) 2. (+, +) (+,+) 3. (+, +) (+,+) 4. (+ −) (+ −) 5. (-,-) (-, -) 6. (-, -) (+,+) 7. (+ +) (+ +) 8. (-, -) (+,-) 9. (- -) (+ -) 10. (- -) (+ +) 11.(- -) (- -) 12. (- -) (- -)

A powerful way to detect distinct bands on this gel would be by Southern blotting, describe the 6 steps

1. DNA fragments separated by gel electrophoresis 2. Gel soaked in alkali solution to denature double stranded dna and then placed in a dish of buffer 3. A membrane is positioned above the gel 4. Buffer is drawn to the top layer of blotting paper, carrying dna onto the membrane 5. fixed dna on the membrane is put in a hybridization bottle with a probe 6. Probe binds to complementary dna fragments on the membrane and can now be detected

Examine Figure 12.7. What would be the effect of a drug that altered the structure of allolactose so that it was unable to bind to the regulator protein?

Allolactose is produced when lactose is present; allolactose normally binds to the repressor protein and makes it inactive, allowing transcription to occur when lactose is present. If a drug altered the structure of allolactose, it would not bind to the repressor and the repressor would continue to bind to the operator, keeping transcription off. Result would be that transcription was repressed even in the presence of lactose; thus, no β-galactosidase or permease would be produced

Briefly describe the lac operon and how it controls the metabolism of lactose

Consists of three structural genes—the lacZ gene, the lacY gene, and lacA gene, which encode β-galactosidase, permease, and thiogalactoside transacetylase. All three genes share a promoter and operator region. Upstream from the lactose operon is the lacI gene that encodes the lac operon repressor, which binds at the operator region and inhibits transcription of the lac operon by preventing RNA polymerase from successfully initiating transcription. When lactose is present in the cell, the enzyme β-galactosidase converts some of it into allolactose, which binds to the lac repressor, altering its shape and reducing the repressor's affinity for the operator. Because the allolactose-bound repressor does not bind to the operator, RNA polymerase can initiate transcription of the lac structural genes from the lac promoter.

Why is gene regulation important for bacterial cells?

Gene regulation allows for biochemical and internal flexibility while maintaining energy efficiency by the bacterial cells

Under which of the following conditions would a lac operon produce the greatest amount of β-galactosidase? The least? Explain your reasoning. Condition 1: Lactose present, glucose not Condition 2: lactose not present, glucose present Condition 3: Lactose and glucose present Condition 4: Lactose and glucose not present

Greatest amount: Condition 1 Max transcription occurs in presence of lactose and absence of glucose. Lactose binds to the lac repressor reducing affinity of the lac repressor to the operator. This results in promoter being accessible to RNA polymerase. Lack of glucose allows for increased synthesis of cAMP, which can complex with CAP. The formation of CAP-cAMP improves the efficiency of RNA polymerase binding to the promoter, which results in higher levels of transcription from the lac operon. Least amount: Condition 2 No lactose present makes the lac repressor active and it will bind to the operator to inhibit transcription. Glucose decreases cAMP levels so a CAP-cAMP complex can't form and rna polymerase cant be stimulated to transcribe the lac operon.

If there were 5 different types of bases in mRNA instead of four, what would be the minimum codon size (# nucleotides) required to specify the following numbers of different amino acid types: a. 4 b. 20 c. 30

Have to determine the # of combinations possible when there are different amounts of bases and codon lengths. Typically you can put the number of different types of bases ^ number of nucleotides in each codon. a. 1 A codon length of 1 nucleotide could specify 4 different amino acids b. 2 A codon length of 2 nucleotides could specify 20 different amino acids c. 3 A codon length of 3 nucleotides could specify 30 different amino acids

Define gene cloning. What is the purpose of making many identical copies of a gene (called gene amplification)? For example, why would a drug company want to clone the gene for insulin?

Make multiple copies of a specific piece of dna by placing the fragment in a bacterial cell and allow the cell to replicate the dna. The purpose of this is to reproduce recombinant dna methods in the cheapest way possible, allowing another organism to do the work.

Give the amino acid sequence of the protein encoded by the mRNA in this sequence: 5'-AUG-CCC-ACG-ACU-GCG-AGC-GUU-CCG-CUA-AGG-UAG-3'

Met-Pro-Thr-Thr-Ala-Ser-Val-Pro-Leu-Arg

Assume we wish to transform a eukaryotic cell with a gene of interest (e.g., we want to perform gene therapy to cure a genetic disease caused by a mutant gene). Can we use an engineered bacterial plasmid to do this? Why or why not?

No. Many eukaryotic proteins are modified after translation, which bacteria do not have the ability to carry out. This means that bacteria do not have the ability to produce a functional protein. However, many cloning vectors have been created to allow the insertion of genes into eukaryotic cells.

What is a constitutive gene?

Not regulated and is expressed continually

A mutation prevents the catabolite activator protein (CAP) from binding to the promoter in the lac operon. What will the effect of this mutation be on the transcription of the operon?

RNA polymerase will bind to the lac promoter poorly, significantly decreasing the transcription of the lac structural genes

An Is mutation results in a defective what?

Regulatory domain of the lac repressor

Assume you digest a sample of human DNA with HindIII and electrophorese the digest on a lane of an agarose gel. After staining the gel with EtBr, you observe an orange smear running down the length of the lane. Why do you not see discrete bands?

The dna fragments must be too small to be seen as distinct bands on an electrophoretic gel.

What role do the initiation factors play in protein synthesis?

Three initiation factors (IF1, IF2, and IF3). IF1 promotes the disassociation of the large and small ribosomal subunits. IF3 binds to the small ribosomal subunit and prevents it from associating with the large ribosomal subunit. IF2 is responsible for binding GTP and delivering the fMet-tRNAfMet to the initiator codon on the mRNA. In eukaryotes, there are more initiation factors, but many have similar roles. Some of the eukaryotic initiation factors are necessary for recognition of the 5′ cap on the mRNA. Others possess a RNA helicase activity, which is necessary to resolve secondary structures.

The goal of the cloning procedure (in a test tube) is to engineer a recombinant DNA molecule. However, not all the digested plasmids in the test tube will have inserts, these plasmids can ligate back to themselves. To select the desired recombinant plasmids, bacteria are transformed with a sample of DNAs from the test tube, and those bacteria that contain recombinant DNAs are then identified by growth on a selective medium. a. Review the mechanism of transformation; remember Griffith and the transforming principle? b. what is competence?

a. A naked dna fragment is taken up by the recipient cell, a crossover between the bacterial dna and the inserted dna occurs in the bacterium and leads to a recombinant chromosome. Remainder of the naked dna fragment that didnt pair with the bacterial chromosome is degraded. b. Ability to take up DNA from the environment, capable of being transformed

A human male is said to be... for the Y chromosome?

hemizygous

The following diagram illustrates a step in the process of translation. Identify the following elements on the diagram. a. 5′ and 3′ ends of the mRNA 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

pg 189 online book answers

In elongation, the creation of peptide bonds between amino acids is catalyzed by what?

rRNA

During the initiation of translation in bacteria, the small ribosomal unit binds to which consensus sequence?

shine dilarno sequence

For E. coli strains with the lac genotypes given below, use a plus sign (+) to indicate the synthesis of β-galactosidase and permease and a minus sign (−) to indicate no synthesis of the proteins. How would you solve this? (3) lacI+lacP+lacO+lacZ+lacY+ lacI-lacP+lacO+lacZ+lacY+ lacI+lacP+lacOclacZ+lacY+ lacI-lacP+lacO+lacZ+lacY- lacI-lacP-lacO+lacZ+lacY+ lacI+lacP+lacO+lacZ-lacY+/ lacI-lacP+lacO+lacZ+lacY- lacI-lacP+lacOclacZ+lacY+/ lacI+lacP+lacO+lacZ-lacY- lacI-lacP+lacO+lacZ+lacY-/ lacI+lacP-lacO+lacZ-lacY+ lacI+lacP-lacOclacZ-lacY+/ lacI-lacP+lacO+lacZ+lacY- lacI+lacP+lacO+lacZ+lacY+/ lacI+lacP+lacO+lacZ+lacY+ lacIslacP+lacO+lacZ+lacY-/ lacI+lacP+lacO+lacZ-lacY+ lacIslacP-lacO+lacZ-lacY+/ lacI+lacP+lacO+lacZ+lacY+

1. The presence of lacZ+ and lacY+ on the same DNA molecule as a functional promoter (lacP+) is required because the promoter is a cis acting regulatory element. 2. However the lacI+ gene product (repressor) is trans acting and does not have to be located on the same DNA molecule as β-galactosidase and permease genes to inhibit expression. Does require that the cis acting lac operator be on the same DNA molecule as the functional β-galactosidase and permease genes. 3. Dominant lacIs gene product is also trans acting and can inhibit transcription at any functional lac operator region

The fox operon, which has sequences A, B, C, D encodes enzymes 1 and 2. Mutations in sequences A, B, C, D have the following effects, + indicates enzyme is synthesized and - indicates enzyme is not synthesized for no mutation, A, B, C,D accordingly. (Fox absent) Enzyme 1: - - - - + Enzyme 2: - - - - + (Fox present) Enzyme 1: + - - + + Enzyme 2:+ + - - + 1. Is the fox operon inducible or repressible? 2. Indicate which sequence is part of the following components of the operon? Use each sequence once -Regulator gene -Promoter -Structural gene for enzyme 1 -Structural gene for enzyme 2

1. When no mutations are present, enzymes 1 and 2 are produced in the presence of Fox but not in its absence, indicating that the operon is inducible and that Fox is the inducer 2. Mutation A allows the production of enzyme 2 in presence of Fox, but enzyme 1 is not produced in presence/absence of Fox so A must have a mutation in strucutral gene for enzyme 1. With mutation in B, neither enzyme is produced so mutations occurs in promotor and prevent rna polymerase from binding. Mutation in C affects only enzyme 2, which isn't produced in presence or absence of Fox. Enzyme 1 is produced normally in presence of fox so mutation in C occurs in structural gene for enzyme 2. Mutation in D is constitutive, producing enzymes 1 and 2 with or w/o Fox. Occurs in regulator gene, producing a defective repressor that is unable to bind to the operator under any conditions. Regulator gene: D Promoter: B Structural gene for enzyme 1: A Structural gene for enzyme 2: C

What are some types of posttranslational modification of proteins? (6)

1. amino terminal methionine may be removed. Sometimes, in bacteria only the formyl group is cleaved from the N-formylmethionine, leaving a methionine at the amino terminal. 2. Precursor proteins are cleaved and trimmed by protease enzymes to produce a functional protein. 3. Glycoproteins are produced by the attachment of carbohydrates to newly synthesized proteins. 4. Molecular chaperones are needed by many proteins to ensure that the proteins are folded correctly. 5. 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. 6. Acetylation of amino acids in the amino terminal of some eukaryotic proteins

What events bring about the termination of translation?

1. termination codon enters "A" site in ribosome 2. Release factors (RF-1, RF-2, and RF-3) to bind the ribosome. RF-1 recognizes and interacts with the stop codons UAA and UAG, while RF-2 can interact with UAA and UGA. A RF-3 GTP complex binds to the ribosome. 3. The GTP is hydrolyzed to GDP, cleaving the polypeptide chain from the tRNA located at the "P" site.

Amino acids bind to which part of tRNA?

3' end

Draw a picture illustrating the general structure of an operon and identify its parts

5' regulator protein gene then operon: [promoter, operator, structural genes 3']

Define cloning vector. What is a plasmid? Give three reasons why a plasmid is an ideal cloning vector. 5 Plasmid vectors have been engineered to contain a variety of Multiple Cloning Site (MCS) casettes, each of which has a battery of 20 or so unique restriction sites (ones that appear only once in the plasmid sequence). Why is it essential to have a unique site for cloning?

A cloning vector is a stable, replicating dna molecule into which a foreign dna fragment can be inserted for introduction into a cell. A plasmid is a circular dna molecule that exist in bacteria Good for cloning because they have origin of replication (able to replicate independently of the bacterial chromosome), has one or more selectable markers (Traits that enable cells containing the vector to be selected or identified), and has several recognition sites for 1+ restriction enzymes A vector must have one restriction site for cloning for each restriction enzyme used or else there will be multiple restriction sites, which will generate several pieces of dna that are extremely difficult to piece back together in the correct order

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

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 the elongation of the peptide by preventing peptide-bond formation.

True or false A copy of the F factor is transferred from an F- cell to a F+ cell during the process of bacterial transformation

False

How are tRNAs linked to their corresponding amino acids?

For each of the 20 different amino acids commonly found in proteins, a corresponding aminoacyl-tRNA synthetase covalently links the amino acid to the correct tRNA molecule.

For e. coli strains with the following lac genotypes make a table and use a + to indicate synthesis of B-galactosidase and permease and a - to indicate no synthesis of the protein when lactose is absent and when it is present. a. lacI+ lacP+ lacO+ lacZ+ lacY+ b.lacI+ lacP+ lacOc lacZ- lacY+ c. lacI+ lacP- lacO+ lacZ+ lacY- d. lacI+ lacP+ lacO+ lacZ- lacY-/ lacI- lacP+ lacO+ lacZ+ lacY+

How to solve by strain: a. All genes have normal sequences so the lac gene functions normally; When lactose is absent, regulator protein binds to the operator and inhibits the transcription of the structural genes, so B-galactosidase and permease aren't produced. When lactose is present, some of it is inverted into allolactose which binds to the repressor and makes it inactive. So the structural genes are transcribed and B-galactosidase and permease are produced. b. lacZ gene is mutated, so B-galactosidase won't be produced. lacO gene has a constitutive mutation, so repressor can't bind to it and transcription will take place. Therefore, permease will be produced with and without lactose. c. Promoter is mutated, so RNA polymerase can't bind and transcription doesn't take place. B-galactosidase and permease are not produced. d. This is a partial diploid strain with two copies of the lac operon, one on bacterial chromosome and one on the plasmid. First lac operon has mutations in both lacZ and lacY genes, so it cant encode B-galactosidase or permease. Second lac operon has a defective regulator gene, but normal regulator gene in first operon can diffuse to other molecules so that is binds to the second operon in the absence of lactose, inhibiting transcription. Therefore, No B-galactosidase and permease is produced because of the absence of lactose. In the presence of lactose, repressor can't bind to the operator so the second operator is transcribed and B-galactosidase and permease are produced. (Lactose absent) B-galactosidase: - - - - permease: - + - - (Lactose present) B-galactosidase: + - - + permease: + + - +

What is the difference between positive and negative control? What is the difference between inducible and repressible operons?

Positive: activator protein Negative: a repressor protein Inducible operon: transcription is off and goal is to turn on Repressible: Transcription is on and goal is to turn off

Arrange the following components of translation in the approximate order in which they would appear or be used in prokaryotic protein synthesis: Elongation factor Tu fMet-tRNAfMet Elongation factor G Release factor 1 30S initiation complex Initiation factor 3 70S initiation complex

Potential exception is initiation factor 3. Initiation factor 3 could possibly be listed first because it is necessary to prevent the 30S ribosome from associating with the 50S ribosome. It binds to the 30S subunit prior to the formation of the 30S initiation complex. However, during translation events the release of initiation factor 3 allows the 70S initiation complex to form, a key step in translation. Initiation factor 3 fMet-tRNAfMet 30S initiation complex 70S initiation complex Elongation factor Tu Elongation factor G Release factor 1

What is catabolite repression? How does it allow a bacterial cell to use glucose in preference to other sugars?

Presence of glucose inhibits or represses the transcription of genes involved in the metabolism of other sugars. Because the gene expression necessary for utilizing other sugars is turned off, only enzymes involved in the metabolism of glucose will be synthesized. Operons that exhibit catabolite repression are under the positive control of catabolic activator protein (CAP). For CAP to be active, it must form a complex with cAMP. Glucose affects the level of cAMP. The levels of glucose and cAMP are inversely proportional—as glucose levels increase, the level of cAMP decreases. Thus, CAP is not activated.

Define probe. Assume you want to identify a HindIII restriction fragment on the gel in #1 that contains the human gene for actin, a muscle protein. What will be your probe?

Probe is a dna or rna molecule with a base sequence complementary to a sequence in the gene of interest. The probe would be 3'-TTCGAA-5'

Compare and contrast the process of protein synthesis in bacterial and eukaryotic cells, giving similarities and differences in the process of translation in these two types of cells

Similarities: 1. share the universal genetic code. 2. Charging of the tRNAs with amino acids Differences: 1. Initiation codon, AUG, in eukaryotic cells codes for methionine, whereas in bacteria the AUG codon codes for N-formyl methionine. 2. In eukaryotes, transcription takes place within the nucleus, whereas most translation takes place in the cytoplasm. Bacterial cells transcription and translation occur nearly simultaneously. 3. Stability of mRNA; Bacterial mRNA is typically short-lived, lasting only a few minutes. Eukaryotic mRNA may last hours or even days. 4. Both have large and small ribosomal subunits, but they differ in size and composition. The bacterial large ribosomal consists of two ribosomal RNAs, while the eukaryotic large ribosomal subunit consists of three. 5. During translation initiation, the bacterial small ribosomal subunit recognizes the ShineDalgarno consensus sequence in the 5′ UTR of the mRNA and to regions of the 16S rRNA. In most eukaryotic mRNAs, the small subunit binds the 5′ cap of the mRNA and scans downstream until it encounters the first AUG codon. 6. Use different elongation and termination factors.

The blob operon produces enzymes that convert compound A into compound B. The operon is controlled by a regulatory gene S. Normally, the enzymes are synthesized only in the absence of compound B. If gene S is mutated, the enzymes are synthesized in the presence and in the absence of compound B. Does gene S produce a repressor or an activator? Is this operon inducible or repressible?

Since the operon is inactive in the presence of B, gene S codes for a repressor protein requiring B as a corepressor. Data suggests the blob operon is repressible because its inactive in presence of B, but active when B is absent

What is the difference between a structural gene and a regulator gene?

Structural genes encode proteins, regulator genes control the transcription of structural genes

Explain why mutations in the lacI gene are trans in their effects, but mutations in the lacO gene are cis in their effects.

The lacI gene encodes the lac repressor protein, which can diffuse within the cell and attach to any operator. It can therefore affect the expression of genes on the same or on a different molecule of DNA. The lacO gene encodes the operator. It affects the binding of RNA polymerase to the DNA, and therefore affects the expression of genes only on the same molecule of DNA.

A mutant strain of E. coli produces β-galactosidase in the presence and in the absence of lactose. Where in the operon might the mutation in this strain be located?

The operator region is most probable location of the mutation. If the mutation prevents the lac repressor protein from binding to the operator, then transcription of the lac structural genes will not be inhibited. Expression will be constitutive.

Why is transcription a particularly important level of gene regulation in both bacteria and eukaryotes?

Transcription is the first step in the process of information transfer from dna to protein. For cellular efficiency, gene expression is often regulated early in the process of protein production

In a negative repressible operon, the regulator protein is synthesized as a. an active activator b. an inactive activator c. an active repressor d. an inactive repressor

d. an inactive repressor

Radioactive labeling of a probe: A DNA probe is labeled in vitro (in a test tube) by DNA polymerase using, as one of the four nucleotides in the reaction mixture, one that contains a radioactive isotope of phosphorus (usually 32P-ATP). This reaction gives rise to new strands of DNA with "hot" phosphorus in their phosphodiester bonds. Hybridization can be detected using X-ray film; radioactivity in the hybrid DNA (on the filter) will expose the film, and complementary DNA sequences in the lane will appear as a dark band. Assume you see multiple bands on the Southern blot, using an actin probe. What can you conclude?

You can conclude that there are several specific DNA fragments in the genome of the hybrid dna??

Walk through Figure 14-6. A common way of selecting recombinants relies on the use of plasmid vectors that contain the lacZ gene (for -galactosidase, which converts lactose to galactose and glucose, as in the lac operon). This gene has unique restriction sites for cloning. In the lacZ selection scheme, the lacZ gene on the host bacterial genome is mutated such that it cannot make - gal (designated lacZ-). Hence, the ability of these cells to undergo the -gal reaction depends on the presence of a functional lacZ gene on the plasmid (designated lacZ+). Conversely, if the plasmid lacZ is interrupted by a foreign gene, gal will not be produced. a. What is the substrate (substance acted upon by an enzyme) of -gal in this selection scheme? What color is the substrate? What color is the product? b. Why is this type of selection also known as blue-white selection? Hint: what color are lacZ- cells that contain or lack plasmids with recombinant DNAs?

a. Bacteria with a recombinant plasmid, which is white. The product remains white because the B-galactosidase would cleave X-gal and make the colonies blue if the substrate was a nonrecombinant plasmid. b. Called blue-white selection because if foreign dna is inserted into the restriction site of the lacZ gene, no B-galactosidase will be produced and therefore the chemical X-gal will not be cleaved to produce a blue color, it will remain white. So lacZ- cells without plasmids with recombinant DNA will be blue, cells with recombinant dna are white.

A synthetic mRNA added to a cell-free protein-synthesizing system produces a peptide with the following amino acid sequence: Met-Pro-Ile-Ser-Ala. What would be the effect on translation if the following components were omitted from the cell-free proteinsynthesizing system? What, if any, type of protein would be produced? a. Initiation factor 3 b. Initiation factor 2 c. Elongation factor Tu d. Elongation factor G e. Release factors RF-1, RF-2, and RF-3 f. ATP g. GTP

a. Decrease the amount of protein synthesized. IF-1 promotes the disassociation of the large and small ribosomal subunits. Would reduce the rate of initiation because more of the small ribosomal subunits would remain bound to the large ribosomal subunits. b.No translation would occur. The lack of IF-2 would prevent fMet-tRNAfMet from being delivered to the small ribosomal subunit, thus blocking translation. c. No other amino acids would be delivered to the ribosome. EF-Tu, which binds to GTP and the charged tRNA, is necessary for elongation. This three-part complex enters the A site of the ribosome. If EF-Tu is not present, the charged tRNA will not enter the A site, thus stopping translation. d. Necessary for the translocation (movement) of the ribosome along the mRNA in a 5′ to 3′ direction. When a peptide bond has formed between the Met and Pro, the lack of EF-G would prevent the movement of the ribosome along the mRNA, and so no new codons would be read. The formation of the dipeptide Met-Pro does not require EF-G. e. Would prevent termination. These recognize the stop codons and bind to the ribosome at the A site. They then interact with RF-3 to promote the cleavage of the peptide from the tRNA at the P site. f. Without ATP, the charging of tRNAs by aminoacyl-tRNA synthetases would not take place, and no amino acids will be available for protein synthesis. g. If GTP is absent, protein synthesis will not take place since its required for initiation, elongation, and termination of translation.

For each of the following types of transcriptional control, indicate whether the protein produced by the regulator gene will be synthesized initially as an active repressor, inactive repressor, active activator, or inactive activator. a. Negative control in a repressible operon b. Positive control in a represisble operon c. Negative control in an inducible operon d. Positive control in an inducible operon

a. Inactive repressor b. Active activator c. Active repressor d. Inactive activator

Refer to the diagram in the last problem (22) to answer the following questions. a. What will be the anticodon of the next tRNA added to the A site of the ribosome? b. What will be the next amino acid added to the growing polypeptide chain?

a. The anticodon 5′-CGU-3′ is complementary to the codon 5′-ACG-3′, which is located at the A site of the ribosome. b.The codon 5′-ACG-3′ encodes the amino acid threonine.

A mutation at the operator prevents the regulator protein from binding. What effect will this mutation have in the following types of operons? a. Regulator protein is a repressor in a repressible operon b. Regulator protein is a repressor in an inducible operon

a. The operon would never be turned off, and transcription will take place all the time. b. The result will be constitutive expression, and the transcription will take place all the time.

Identifying bacterial cells that have recombinant plasmids a. Define selectable marker b. Antibiotic resistance as a selectable marker. Assume a bacterium contains a plasmid DNA with a gene whose product confers resistance to the antibiotic ampicillin. This bacterium is designated ampR . (for ampicillin resistance), and it will grow and divide on an agar plate supplemented with ampicillin, giving rise to a colony. Bacteria lacking the plasmid will die on an amp-containing plate. .

a. Traits that enable cells containing the vector to be selected or identified

For each of the sequences in the following table, indicate the process most immediately affected by deleting the sequence. replication, rna processing, translation, or transcription a. ori site b. 3' splice site consensus c. polyA tail d. Terminator e. Start codon f. -10 consensus g. Shine Dalgarno

a. replication b. rna processing c. translation d. transcription e. translation f. transcription g. translation

In a polyribosome the polypeptide associated with which ribosome will be the longest? a. those at 5' end mrna b. those at 3' end mrna c. those in the middle of mrna

b

Which one of the steps below occurs before the other four? a. 70S initiation complex is formed b. IF-3 binds to the 30S ribosomal subunit c. fMet-tRNA binds to the initiation codon d. a charged tRNA enters the A site e. Release factor 1 (RF-1) binds to stop codon

b

In the presence of allolactose, the lac operon repressor a. binds to the operator b. binds to the promoter c. cannot bind to the operator d. binds to the regulator gene

c. cannot bind to the operator