BIO 2306 EXAM 3 PRACTICE

In a line of human cells grown in culture, a geneticist isolates a temperature- sensitive mutation at a locus that encodes an acetyltransferase enzyme; at temperatures above 38C the mutant cells produce a nonfunctional form of the enzyme. What would be the most likely effect of this mutation when the cells are grown at 40C?

Acetyltransferase enzymes add acetyl groups to histone proteins preventing the proteins from forming the 30-nm chromatin fiber. Essentially, the chromatin structure is destabilized, which allows for transcription to occur. If the cells are raised to 40C, then the acetyltransferase enzyme would not function and acetyl groups would not be added to the histone proteins that are the target of this enzyme. The result would be that the nucleosomes and the chromatin would remain stabilized and thus block transcriptional activation.

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 posttranslational modification of the synthesized protein

What are some of the modifications in tRNA that take place through processing?

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

what are the three principal elements in mrNA sequences in bacterial cells?

1. the 5' untranslated region, which contains the Shine-Dalgarno sequence 2. the protein-encoding region, which begins with the start codon and ends with the stop codon 3. the 3' untranslated region

What is the difference between a missense mutation and a nonsense mutation? Between a silent mutation and a neutral mutation?

A base substitution that changes the sequence and the meaning of an mRNA codon, resulting in a different amino acid being inserted into a protein, is called a missense mutation. Nonsense mutations occur when a mutation replaces a sense codon with a stop (or nonsense) codon.A nucleotide substitution that changes the sequence of an mRNA codon, but not the meaning, is called a silent mutation. In neutral mutations, the sequence and the meaning of an mRNA codon are changed. However, the amino acid substitution has little or no effect on protein function.

poly(A) tail

A poly(A) tail is added to the 3' end of the pre-mRNA. It affects mRNA stability.

define the following term as it applies to the genetic code: sense codon

A sense codon is a group of three nucleotides that code for an amino acid. In a standard genetic code there are 61 sense codons that code for the 20 amino acids commonly found in proteins.

How is the reading frame of a nucleotide sequence set?

The initiation codon on the mRNA sets the reading frame.

Suppose that a 20-bp deletion occurs in the middle of exon 2 of the gene depicted inFigure 14.12a. What will be the likely effect of this deletion in the proteins produced by alternative splicing?

The proteins produced by the mRNA with exons 1, 2, and 3 will likely contain more amino acids and be larger than the protein produced by the mRNA with exons 1 and 3.

define the following term as it applies to the genetic code: reading frame

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

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

The regulator protein-corepressor complex would normally bind to the operator and inhibit transcription. If a mutation prevented the repressor protein from binding at the operator, then the operon would never be turned off and transcription would occur all the time.

What is the origin of small interfering RNAs, microRNAs, and Piwi-interacting RNAs? What do these RNA molecules do in the cell?

The siRNAs originate from the cleavage of mRNAs, RNA transposons, and RNA viruses by the 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 tranposons. The microRNAs typically silence genes that are different from those from which the microRNAs are transcribed. However, the siRNAs usually silence genes from which they are transcribed.

What makes up the spliceosome? What is the function of the spliceosome?

The spliceosome consists of five small ribonucleoproteins (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 occurs through the action of the spliceosome.

define the following term as it applies to the genetic code: termination codon

The termination codon signals the termination or end of translation and the end of the protein molecule. There are three 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.

What is the difference between a transition and a transversion? Which type of base substitution is usually more common?

Transition mutations are base substitutions in which one purine (A or G) is changed to the other purine, or a pyrimidine (T or C) is changed to the other pyrimidine. Transversions are base substitutions in which a purine is changed to a pyrimidine or vice versa. Although transversions would seem to be statistically favored because there are eight possible transversions and only four possible transitions, about twice as many transition mutations are actually observed in the human genome.

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

Within the operon, the operator region is the 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. Outside of the operon, a mutation in the lacI gene that inactivates the repressor or keeps it from binding to the operator could also lead to constitutive expression of the structural genes.

List some important differences between bacterial and eukaryotic cells that affect the way in which the genes are regulated.

a. Bacterial genes are frequently organized into operons with coordinate regulation, and genes with operons can be transcribed as on a single long mRNA. Eukaryotic genes are not organized into operons and are singly transcribed from their own promoters. In eukaryotic cells, nucleosome structure of the DNA is remodeled prior to transcription occurring. Essentially, the chromatin must assume a more open configuration state, allowing for access by transcription-associated factors. Activator and repressor molecules function in both eukaryotic and bacterial cells. However, in eukaryotic cells activators appear to be more common than in bacterial cells. In bacteria, transcription and translation can occur concurrently. In eukaryotes, the nuclear membrane separates transcription from translation both physically and temporally. This separation results in a greater diversity of regulatory mechanisms that can occur at different points during gene expression.

will the protein produced by the regulator gene be synthesized initially as an active repressor or an inactive repressor? : negative control in an inducible operon

active repressor

will the protein produced by the regulator gene be synthesized initially as an active repressor or an inactive repressor? : negative control in a repressible operon

inactive repressor

At which level of gene regulation shown in Figure 16.1 does attenuation occur?

transcription

Malaria, one of the most pervasive and destructive of all infectious diseases, is caused by protozoan parasites of the genus Plasmodium, which are transmitted from person to person by mosquitoes. Plasmodium parasites are able to evade the host immune system by constantly altering the expression of their var genes, which encode Plasmodium surface antigens (L. H. Freitas-Junior, et al. 2005. Cell 121:25- 36). Individual var genes are expressed when chromatin structure is disrupted by chemical changes in histone proteins. What type of chemical changes in the histone proteins might be responsible for these changes in gene expression?

Addition or removal of phosphate groups, methyl groups, or acetyl groups to the tails of the histone proteins. For example, acetyltransferase enzymes might be recruited to the promoter of a specific var gene. Acetyltransferase enzymes add acetyl groups to the histone tails, which disrupts chromatin structure and facilitates transcription of the DNA.

Which strand of DNA (upper or lower) in Figure 16.8 is the template strand? Explain your reasoning.

The bottom strand. RNA polymerase binds the promoter and moves downstream, toward the transcription start site, so in this diagram it moves from left to right. The template strand is always read 3 to 5, because the newly synthesized RNA is synthesized 5 to 3. As the polymerase moves left to right, it is the bottom strand that will be read 3 to 5, so it must be the template strand.

How would the deletion of the 5' cap most likely affect a eukaryotic pre-mRNA?

The deletion of the 5′ cap would most likely prevent splicing of the intron that is nearest to the 5′ cap. Ultimately, elimination of the cap will affect the stability of the pre-mRNA as well as its ability to be translated.

Examine Figure 16.7. 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. The result would be that transcription was repressed even in the presence of lactose; thus, no -galactosidase or permease would be produced.

Describe two types of alternative processing pathways. How do they lead to the production of multiple proteins from a single gene?

Alternative processing of pre-mRNA can take the form of either alternative splicing of pre-mRNA introns or the alternative cleavage of 3′ cleavage sites in a pre-mRNA molecule containing two or more cleavage sites for polyadenylation. Alternative splicing results in different exons of the pre-mRNA being ligated to form mature mRNA. Each mRNA formed by an alternative splicing process will yield a different protein. In pre-mRNA molecules with multiple 3′ cleavage sites, cleavage at the different sites will generate mRNA molecules that differ in size. Each alternatively cleaved RNA potentially could code for a different protein depending on the location of the alternative 3′ cleavage site. A single pre-mRNA transcript can undergo both alternative processing steps, thus potentially producing multiple proteins.

define the following term as it applies to the genetic code: initiation codon

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 although rarely.

What is an insulator?

An insulator or boundary element is a sequence of DNA that inhibits the action of regulatory elements called enhancers in a position dependent manner.

What is antisense RNA? How does it control gene expression?

Antisense RNA molecules are small RNA molecules that are complementary to other DNA or RNA sequences and that form RNA-protein complexes. In bacterial cells, antisense RNA molecules can bind to a complementary region in the 5' UTR of a mRNA molecule, blocking the attachment of the ribosome to the mRNA and stopping translation or they pair with specific regions of the mRNA and cleave the mRNA stopping translation.

For the ovalbumin gene shown in Figure 14.3, where would the 5′ untranslated region and 3′ untranslated regions be located in the DNA and in the RNA?

Assuming that alternative splicing is not occurring then the 5′ UTR would be located in exon 1 and the 3′ UTR would be located in exon 8 for both the DNA and RNA molecules.

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 Sproduce a regulatory protein that exhibits positive or negative control? Is this operon inducible or repressible?

Because the blob operon is transcriptionally inactive in the presence of B, gene Smost likely codes for a repressor protein that requires compound B as a corepressor. The data suggest that the blob operon is repressible because it is inactive in the presence of compound B, but active when compound B is absent.

A codon that specifies the amino acid Gly undergoes a single-base substitution to become a nonsense mutation. In accord with the genetic code given in Figure 15.10, is this mutation a transition or a transversion? At which position of the codon does the mutation occur?

By examining the four codons that encode for Gly, GGU, GGC, GGA, and GGG, and the three nonsense codons, UGA, UAA, and UAG, we can determine that only one of the Gly codons, GGA, could be mutated to a nonsense codon by the single substitution of a U for a G at the first position: GGA --> UGA Because uracil is a pyrimidine and guanine is a purine, the mutation is a transversion.

What is the function of the 5′ cap?

CAP binding proteins recognize the 5′ cap and stimulate binding of the ribosome to the 5′ cap and to the mRNA molecule. The 5′ cap may also increase mRNA stability in the cytoplasm. Finally, the 5′ cap is needed for efficient splicing of the intron that is nearest the 5′ end of the pre-mRNA molecule.

X31b is an experimental compound that is taken up by rapidly dividing cells. Research has shown that X31b stimulates the methylation of DNA. Some cancer researchers are interested in testing X31b as a possible drug for treating prostate cancer. Offer a possible explanation for why X31b might be an effective anticancer drug.

Cancer cells are typically rapidly dividing cells with high gene expression and metabolic activity. DNA methylation particularly in genomic regions with many CpG sequences (CpG islands) is associated with transcriptional repression. If the X31b molecules can be uptaken by the rapidly dividing cancer cells, then these molecules can stimulate methylation of DNA sequences in the cancer cells, leading to transcriptional repression of numerous genes. The repression of transcription could decrease the growth of the cancer cells and potentially cause a loss of viability of these highly active cells.

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 transcription of the operon?

Catabolite activator protein binds the CAP site of the lac operon and stimulates RNA polymerase to bind the lac promoter, thus resulting in increased levels of transcription from the lac operon. If a mutation prevents CAP from binding to the site, then RNA polymerase will bind the lac promoter poorly. This will result in significantly lower levels of transcription of the lac structural genes.

exons

Exons are transcribed regions that are not removed in intron processing. They include the 5′ UTR, coding regions that are translated into amino acid sequences, and the 3′ UTR.

Briefly describe expanding nucleotide repeats. How do they account for the phenomenon of anticipation?

Expanding nucleotide repeats occur when DNA insertion mutations result in an increase in the number of copies of a nucleotide repeat sequence. Such an increase in the number of copies of a nucleotide sequence may occur by errors in replication or unequal recombination. Within a given family, a particular type of nucleotide repeat may increase in number from generation to subsequent generation, increasing the severity of the mutation in a process called anticipation.

Listed in parts a through g are some mutations that were found in the 5′ UTR region of the trp operon of E. coli. What will the most likely effect of each of these mutations be on transcription of the trp structural genes?: Deletion of the string of adenine nucleotides that follows region 4 in the 5' UTR

For the attenuator hairpin to function as a terminator, the presence of a string of uracil nucleotides following region 4 in the mRNA 5' UTR is required. The deletion of the string of adenine nucleotides in the DNA will result in no string of uracil nucleotides following region 4 of the mRNA 5' UTR. No termination will occur, and transcription will proceed.

Why is gene regulation important for bacterial cells?

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

define the following term as it applies to the genetic code: overlapping code

If an overlapping code were present, then a single nucleotide would be expected to be included in more than one codon. The result for a sequence of nucleotides within a single gene would be to encode more than one type of polypeptide. However, because the genetic code is nonoverlapping, codons within the same gene do not overlap. Where overlap occurs is in overlapping genes in some viruses, where the same segment of the genome encodes multiple—two or even three (in case of HIV Env gene)—peptides. In such cases, the codons of different peptides are translated in different frames in nonoverlapping fashion.

Listed in parts a through g are some mutations that were found in the 5′ UTR region of the trp operon of E. coli. What will the most likely effect of each of these mutations be on transcription of the trp structural genes?: A mutation that creates a stop codon early in region 1 of the mRNA 5' UTR

If region 1 of the mRNA 5' UTR is free to pair with region 2, then regions 3 and 4 of the mRNA 5' UTR can form the attenuator. An early stop codon will result in the ribosome "falling off" region 1, allowing it to form a hairpin structure with region 2. Transcription will not occur because regions 3 and 4 are now free to form the attenuator.

Listed in parts a through g are some mutations that were found in the 5′ UTR region of the trp operon of E. coli. What will the most likely effect of each of these mutations be on transcription of the trp structural genes?: Deletions in region 2 of the mRNA 5' UTR

If region 2 of the mRNA 5' UTR is deleted, the antiterminator cannot be formed. The attenuator will form and transcription will not occur.

Listed in parts a through g are some mutations that were found in the 5′ UTR region of the trp operon of E. coli. What will the most likely effect of each of these mutations be on transcription of the trp structural genes?: A mutation that prevents the binding of the ribosome to the 5′ end of the mRNA 5′ UTR

If the ribosome does not bind to the 5' end of the mRNA, then region 1 of the mRNA 5' UTR will be free to pair with region 2, thus preventing region 2 from pairing with region 3 of mRNA 5' UTR. Region 3 will be free to pair with region 4, forming the attenuator or termination hairpin. Transcription of the trpstructural genes will be terminated. Essentially, no gene expression will occur.

define the following term as it applies to the genetic code: nonoverlapping code

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

define the following term as it applies to the genetic code: universal code

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 genomes.

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

In an inducible operon, a mutation at the operator site that blocks binding of the repressor would result in constitutive expression and transcription would occur all the time.

How do repressors that bind to silencers in eukaryotes differ from repressors that bind to operators in bacteria?

In bacteria, repressors that bind to the operator block RNA polymerase from binding to the promoter, and thus, directly block transcription. On the other hand, repressors that bind to silencers in eukaryotes block transcriptional activator proteins from binding at an activator site, thus eliminating transcriptional activation.

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

In catabolite repression, the 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.

For E. coli strains with the lac genotypes show 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.

In determining if expression of the β-galactosidase and the permease gene will occur, you should consider several factors. 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. However, the lacI+ gene product or lacrepressor is trans-acting and does not have to be located on the same DNA molecule as β-galactosidase and permease genes to inhibit expression. For the repressor to function, it does require that the cis-acting lac operator be on the same DNA molecule as the functional β-galactosidase and permease genes. Finally, the dominant lacIs gene product is also trans-acting and can inhibit transcription at any functional lac operator region.

Listed in parts a through g are some mutations that were found in the 5′ UTR region of the trp operon of E. coli. What will the most likely effect of each of these mutations be on transcription of the trp structural genes?: A mutation that changes the tryptophan codons in region 1 of the mRNA 5'UTR into codons for alanine

In the wild-type trp operon, low levels of tryptophan result in the ribosome pausing in region 1 of the mRNA 5' UTR. The pause permits regions 2 and 3 of the mRNA 5' UTR to form the antiterminator hairpin, allowing transcription of the structural genes to continue. If alanine codons have replaced tryptophan codons, then under conditions of low alanine, the stalling of the ribosome will not occur. The attenuator will form, stopping transcription. The ribosome will stall when alanine is low, so transcription of the structural genes will occur only when alanine is low.

How is the poly(A) tail added to pre-mRNA? What is the purpose of the poly(A) tail?

Initially, a complex consisting of several proteins forms on the 3′ UTR of the pre- mRNA molecule. Proteins necessary for cleavage and polyadenylation bind to the AAUAAA consensus sequence, which is located upstream of the 3′ cleavage site, and to a site downstream of the cleavage site. Once the complex has formed, the pre-mRNA is cleaved. Other proteins add adenine nucleotides to the 3′ end creating the poly(A) tail. The presence of the poly(A) tail increases the stability of the mRNA molecule through the interaction of proteins at the poly(A) tail. The length of the poly(A) tail influences the time in which the transcript remains intact and available for translation. The Chapter Fourteen: RNA Molecules and RNA Processing 301 poly(A) tail also assists with the binding of the ribosome to the mRNA and nuclear export.

How is the 5′ cap added to eukaryotic pre-mRNA?

Initially, the terminal phosphate of the three 5′ phosphates linked to the end of the mRNA molecule is removed. Subsequently, a guanine nucleotide is attached to the 5′ end of the mRNA using a 5′ to 5′ phosphate linkage. Next, a methyl group is attached to position 7 of the guanine base. Ribose sugars of adjacent nucleotides may also be methylated, but at the 2′-OH.

What role do the initiation factors play in protein synthesis?

Initiation factors are proteins that are required for the initiation of translation. In bacteria, there are three initiation factors (IF-1, IF-2, and IF-3). Each one has a different role. IF-1 promotes the disassociation of the large and small ribosomal subunits. IF-3 binds to the small ribosomal subunit and prevents it from associating with the large ribosomal subunit. IF-2 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 RNA helicase activity, which is necessary to resolve secondary structures.

introns

Introns are noncoding sequences of DNA that intervene within coding regions of a gene.

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

Positive transcriptional control requires an activator protein to stimulate transcription at the operon. In negative control, a repressor protein inhibits or turns off transcription at the operon.An inducible operon normally is not transcribed. It requires an inducer molecule to stimulate transcription either by inactivating a repressor protein in a negative inducible operon or by stimulating the activator protein in a positive inducible operon.Transcription normally occurs in a repressible operon. In a repressible operon, transcription is turned off either by the repressor becoming active in a negative repressible operon or by the activator becoming inactive in a positive repressible operon.

Are the 5′ untranslated regions (5′ UTR) of eukaryotic mRNAs encoded by sequences in the promoter, exon, or intron of the gene? Explain your answer.

The 5′ UTR is located in the first exon of the gene. It is not part of the promoter because promoters for RNA polymerase II (which transcribes pre-mRNA) are not normally transcribed, and the 5′ UTR is in the mRNA. It is not located in an intron because introns are removed during processing of pre-mRNA, and the 5′ UTR is part of the mRNA.

5' cap

The 5′ cap functions in the initiation of translation and mRNA stability.

what is the 5' cap?

The 5′ end of eukaryotic mRNA is modified by the addition of the 5′ cap. The cap consists of an extra guanine nucleotide linked 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.

a 5' untranslated region

The 5′ untranslated region lies upstream of the translation start site. The eukaryotic ribosome binds at the 5′ cap of the mRNA molecule and scans to the first methionine codon (AUG). The region 5′ of this start codon is the 5′ UTR.

What role do CRISPR-Cas systems naturally play in bacteria?

The CRISPR-Cas system essentially serves as an adaptive RNA defense system or immune system for bacterial cells and protects cells against foreign DNA genomes from plasmids or infecting bacteriophages. The spacers between the palindromic repeat sequences consist of short nucleotide sequences captured from previous invading DNA molecules and can help to provide adaptive protection by targeting homologous invading foreign DNA molecules for cleavage by the Cas proteins.

An enhancer is surrounded by four genes (A, B, C, and D), as shown in the adjoining diagram. An insulator lies between gene C and gene D. On the basis of the positions of the genes, the enhancer, and the insulator, the transcription of which genes is most likely to be stimulated by the enhancer? Explain your reasoning.

The action of an enhancer is blocked when the insulator is located between the enhancer and the promoter of the gene. It is likely that genes A, B, and C will be stimulated by the enhancer and that gene D will not be stimulated. Insulators block the stimulatory action of enhancers when they are located between the enhancer and the promoter of the gene. In the example from the figure, the insulator is only between gene D and the enhancer. The enhancer's effect on genes A, B, and C is not likely to be affected by the insulator and these genes will be stimulated.

What is attenuation? What is the mechanism by which the attenuator forms when tryptophan levels are high and the antiterminator forms when tryptophan levels are low?

Attenuation is the termination of transcription prior to the structural genes of an operon. It is a result of the formation of a termination hairpin structure or attenuator in the mRNA. Two types of secondary structures can be formed by the mRNA 5′ UTR of the trp operon. If the 5′ UTR forms two hairpin structures from the base pairing of region 1 with region 2 and the pairing of region 3 with region 4, then transcription of the structural genes will not occur. The hairpin structure formed by the pairing of region 3 with region 4 results in a terminator being formed that stops transcription. When region 2 pairs with region 3, the resulting hairpin acts as an antiterminator allowing for transcription to proceed. Region 1 of the 5' UTR also encodes a small protein and has two adjacent tryptophan codons (UGG). Tryptophan levels affect transcription due to the coupling of translation with transcription in bacterial cells. When tryptophan levels are high, the ribosome quickly moves through region 1 and into region 2, thus preventing region 2 from pairing with region 3. Therefore, region 3 is available to form the attenuator hairpin structure with region 4, stopping transcription. When tryptophan levels are low, the ribosome stalls or stutters at the adjacent tryptophan codons in region 1. Region 2 now becomes available to base pair with region 3, forming the antiterminator hairpin. Transcription can now proceed through the structural genes.

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.

Bacterial and eukaryotic cells share several similarities as well as have several differences in protein synthesis. Initially, bacteria and eukaryotes share the universal genetic code. However, the initiation codon, AUG, in eukaryotic cells codes for methionine, whereas in bacteria the AUG codon codes for N-formylmethionine. In eukaryotes, transcription takes place within the nucleus, whereas most translation takes place in the cytoplasm (although some translation does take place within the nucleus). Therefore, transcription and translation in eukaryotes are kept temporally and spatially separate. However, in bacterial cells transcription and translation are coupled and occur nearly simultaneously.Stability of mRNA in eukaryotic cells and bacterial cells is also different. Bacterial mRNA is typically short-lived, lasting only a few minutes. Eukaryotic mRNA may last hours or even days. Charging of the tRNAs with amino acids is essentially the same in both bacteria and eukaryotes. The ribosomes of bacteria and eukaryotes are different as well. Bacteria and eukaryotes 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.During translation initiation, the bacterial small ribosomal subunit recognizes the Shine-Dalgarno 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. Finally, elongation and termination in bacterial and eukaryotic cells are functionally similar, although different elongation and termination factors are used.

How does the process of initiation differ in bacterial and eukaryotic cells?

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 the 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 AUG 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.

How do the mRNAs of bacterial cells and the pre-mRNAs of eukaryotic cells differ? How do the mature mRNAs of bacterial and eukaryotic cells differ?

Bacterial mRNA is translated immediately upon being transcribed. Eukaryotic pre- mRNA must be processed and exported from the nucleus. 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.

How do base analogs lead to mutations?

Base analogs have structures similar to the nucleotides and, if present, may be incorporated into the DNA during replication. Many analogs have an increased tendency for mispairing, which can lead to mutations. DNA replication is required for the base analog-induced mutations to be incorporated into the DNA.

Suppose that a mutation occurs in the middle of a large intron of a gene encoding a protein. What will the most likely effect of the mutation be on the amino acid sequence of that protein? Explain your answer.

Because introns are removed prior to translation, an intron mutation would have little effect on a protein's amino acid sequence unless the mutation occurred within the 5′splice site, the 3′ splice site, or the branch point. If mutations within these sequences altered splicing such that a new exon is recognized by the spliceosome, then the mature mRNA would be altered, thus altering the amino acid sequence of the protein. The result could be a protein with additional amino acid sequence. Or, possibly, the inclusion of the new exon that was previously intron could introduce a stop codon that stops translation prematurely. If a mutation in the new exon or inclusion of the new exon induced a frameshift, the reading frame and the amino acid sequence would be altered from that point onward in the protein.

What changes take place in chromatin structure and what role do these changes play in eukaryotic gene regulation?

Changes in chromatin structure can result in repression or stimulation of gene expression. As genes become more transcriptionally active, DNA shows increased sensitivity to DNase I digestion, suggesting that the chromatin structure is more open. Acetylation of histone proteins by acteyltransferase proteins results in the destabilization of the nucleosome structure and increases transcription as well as hypersensitivity to DNase I. The reverse reaction by deacetylases stabilizes nucleosome structure and lessens DNase I sensitivity. Other transcription factors and regulatory proteins, called chromatin remodeling complexes, bind directly to the DNA-altering chromatin structure without acetylating histone proteins. The chromatin remodeling complexes allow for transcription to be initiated by increasing accessibility to the promoters by transcription factors.DNA methylation is also associated with decreased transcription. Methylated DNA sequences stimulate histone deacetylases to remove acetyl groups from the histone proteins, thus stabilizing the nucleosome and repressing transcription. Demethylation of DNA sequences is often followed by increased transcription, which may be related to the deacetylation of the histone proteins.

Lactose present Glucose present Condition 1 YES NO Condition 2 NO YES Condition 3 YES YES Condition 4 NO NO Under which of the following conditions would a lac operon produce the greatest amount of β-galactosidase? The least? Explain your reasoning.

Condition 1 will result in the production of the maximum amount of β- galactosidase. For maximum transcription, the presence of lactose and the absence of glucose are required. Lactose (or allolactose) binds to the lac repressor reducing the affinity of the lac repressor to the operator. This decreased affinity results in the promoter being accessible to RNA polymerase. The lack of glucose allows for increased synthesis of cAMP, which can complex with CAP. The formation of CAP-cAMP complexes improves the efficiency of RNA polymerase binding to the promoter, which results in higher levels of transcription from the lac operon. Condition 2 will result in the production of the least amount of β-galactosidase. With no lactose present, the lac repressor is active and binds to the operator, inhibiting transcription. The presence of glucose results in a decrease of cAMP levels. A CAP-cAMP complex does not form, and RNA polymerase will not be stimulated to transcribe the lac operon.

Where are DNase I hypersensitivity sites found and what do they indicate about the nature of chromatin?

DNase I hypersensitivity sites are typically found near regulatory promoter elements, approximately 1000 nucleotides upstream of a transcription start site, and occasionally, at the 3ʹ end of the gene as well. This sensitivity to DNase I suggests a more relaxed or open configuration state for the chromatin, allowing for access to the DNA of transcriptional regulatory proteins as well as access by DNase I. Essentially the DNA is more exposed due to the lack of nucleosome structure.

Listed in parts a through g are some mutations that were found in the 5′ UTR region of the trp operon of E. coli. What will the most likely effect of each of these mutations be on transcription of the trp structural genes?: Deletions in region 4 of the mRNA 5' UTR

Deletions in region 4 will prevent formation of the attenuator by the 5' UTR mRNA. Transcription will proceed.

What would be the effect of moving the insulator shown in Figure 17.8 to a position between Enhancer II and the promoter for gene B?

Enhancers I and II would now stimulate gene A. Neither enhancer would stimulate gene B.

What is an enhancer? How does it affect transcription of distant genes?

Enhancers are DNA sequences that are the binding sites of transcriptional activator proteins. Transcription at a distant gene is affected when the DNA sequence located between the gene's promoter and the enhancer is looped out, allowing for the interaction of the enhancer-bound proteins with proteins needed at the promoter, which in turn stimulates transcription. Additionally, the transcription of short enhancer (e)RNA molecules from an enhancer template may be involved in transcriptional activation, but a precise mechanism for such activation has not been determined.

Transformation is a process in which bacteria take up new DNA released by dead cells and integrate it into their own genomes (see p. 265 in Chapter 9). InStreptococcus pneumoniae (which causes many cases of pneumonia, inner-ear infections, and meningitis) the ability to carry out transformation requires from 105- 124 genes, collectively termed the com regulon. The com regulon is activated in response to a protein called competence-stimulating peptide (CSP), which is produced by bacteria and is exported into the surrounding medium. When enough CSP accumulates, it attaches to a receptor protein that stimulates the transcription of genes within the com regulon and sets in motion a series of reactions that ultimately results in transformation. Does the com regulon in Streptococcus pneumoniaeexhibit positive or negative control? Explain your answer.

It is regulated through positive control. Increased levels of CSP induce the receptor to stimulate transcriptional activator functions, which allow gene expression to occur similarly to what occurs in positive inducible regulation.

define the following term as it applies to the genetic code: non universal codons

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 known exceptions are the termination codons, which in some organisms do code for amino acids. Occasionally, a sense codon is substituted for another sense codon.

Explain how rRNA is processed.

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.

Some mutations in the trp 5' UTR region increase termination by the attenuator. Where might these mutations occur and how might they affect the attenuator?

Mutations that disrupt the formation of the antiterminator will increase termination by the attenuator. Such disruptions could be caused by a deletion in region 2 that prevents region 2 from pairing with region 3. Mutations in region 1 could also affect the antiterminator if the mutations prevented the ribosome from stalling at the adjacent tryptophan codons within region 1. For example, any mutation that blocks translation initiation or stops translation early within region 1 would not allow the ribosome to migrate on the trp operon mRNA. Another type of mutation affecting antiterminator formation in region 1 is one that eliminates or replaces the two adjacent tryptophan codons in the small protein. Elimination of these codons would prevent the ribosome from stalling in region 1, thus increasing the rate of the terminator formation.

define the following term as it applies to the genetic code: nonsense codon

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

How would the deletion of the poly(A) tail most likely affect a eukaryotic pre-mRNA?

Polyadenylation increases the stability of the mRNA. If eliminated from the pre- mRNA, then the mRNA would be degraded quickly by nucleases in the cytoplasm.

Explain the process of pre-mRNA splicing in nuclear genes.

Removal of an intron from the pre-mRNA requires the assembly of the spliceosome complex on the pre-mRNA, cleavage at both the 5′ and 3′ splice sites of the intron, and two transesterification reactions ultimately leading to the joining of the two exons. Initially, snRNP U1 binds to the 5′ splice site through complementary base pairing of the U1 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 branch point and 5′ splice site of the intron are now adjacent to each other. U1 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.

Describe the basic structure of ribosomes in bacterial and eukaryotic cells.

Ribosomes in both eukaryotes and bacteria consist of a complex of protein and RNA molecules. A functional ribosome is composed of a large and a small subunit. The bacterial 70S ribosome consists of a 30S small subunit and a 50S large subunit. Within the small subunit are a single 16S RNA molecule and 21 proteins. The 23S and the 5S RNA molecules, along with 31 proteins, are found in the large bacterial subunit. The eukaryotic 80S ribosome is comprised of a 60S large subunit and a 40S small subunit. Three RNA molecules, the 28S RNA, the 5.8S RNA, and the 5S RNA, are located in the large subunit as well as 49 proteins. The eukaryotic small subunit contains only a single 18S RNA molecule and 33 proteins.

What are riboswitches? How do they control gene expression? How do riboswitches differ from RNA-mediated repression?

Riboswitches are regulatory sequences in RNA molecules. Most can fold into compact secondary structures consisting of a base stem and several branching hairpins. At riboswitches, regulatory molecules bind and influence gene expression by affecting the formation of secondary structures within the mRNA molecule. The binding of the regulatory molecule to a riboswitch sequence may result in repression or induction. Some regulatory molecules bind the riboswitch sequence and stabilize a terminator structure in the mRNA, which results in premature termination of the mRNA molecule. Other regulatory molecules bind riboswitch sequences resulting in the formation of secondary structures that block the ribosome binding sites of the mRNA molecules, thus preventing translation initiation. In induction, the regulatory molecule acts as an inducer, stimulating the formation of a secondary structure in the mRNA that allows for transcription or translation to occur.

How do insertions and deletions arise?

Strand slippage that occurs during DNA replication and unequal crossover events due to misalignment at repetitive sequences have been shown to cause deletions and additions of nucleotides to DNA molecules. Strand slippage results from the formation of small loops on either the template or the newly synthesized strand. If the loop forms on the template strand, then a deletion occurs. Loops formed on the newly synthesized strand result in insertions. If, during crossing over, a misalignment of the two strands at repetitive sequence occurs, then the resolution of the crossover will result in one DNA molecule containing an insertion and the other molecule containing a deletion.

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

Synonymous codons code for the same amino acid, or, in other words, they have the same meaning. A nucleotide at the third position of a codon pairs with a nucleotide in the first 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 the "wobble," or nonstandard base pairing with the anticodons, affects the third nucleotide position, the redundancy of codons ensures that the correct amino acid is inserted in the protein when nonstandard pairing occurs.

3' untranslated region

The 3′ untranslated region is a sequence of nucleotides at the 3′ end of the mRNA that is not translated into proteins. However, it does affect the translation of the mRNA molecule as well as the stability of the mRNA.

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

The lac operon consists of three structural genes involved in lactose metabolism, the lacZ gene, the lacY gene, and the lacA gene. Each of these three genes has a different role in the metabolism of lactose. The lacZ gene codes for the enzyme β- galactosidase, which breaks the disaccharide lactose into galactose and glucose, and converts lactose into allolactose. The lacY gene, located downstream of the lacZgene, codes for lactose permease. Permease is necessary for the passage of lactose through the E. coli cell membrane. The lacA gene, located downstream of lacY, encodes the enzyme thiogalactoside transacetylase whose function in lactose metabolism has not yet been determined. All of these genes share a common overlapping promoter and operator region. Upstream from the lactose operon is thelacI gene that encodes the lac operon repressor. The repressor 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. Allolactose binds to the lac repressor, altering its shape and reducing the repressor's affinity for the operator. Since this allolactose-bound repressor does not occupy the operator, RNA polymerase can initiate transcription of the lac structural genes from the lac promoter.

Explain why mutations in the lacI gene are trans in their effects, but mutations in the lacOgene 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 different molecules of DNA. The lacO gene encodes the operator. The binding of the lacrepressor to the operator affects the binding of RNA polymerase to the DNA, and therefore affects only the expression of genes on the same molecule of DNA.

Duchenne muscular dystrophy is caused by a mutation in a gene that comprises 2.5 million base pairs and specifies a protein called dystrophin. However, less than 1% of the gene actually encodes the amino acids in the dystrophin protein. On the basis of what you now know about gene structure and RNA processing in eukaryotic cells, provide a possible explanation for the large size of the dystrophin gene.

The large size of the dystrophin gene is likely due to the presence of many intervening sequences or introns within the coding region of the gene. Excision of the introns through RNA splicing yields the mature mRNA that encodes the dystrophin protein.

A geneticist discovers that two different proteins are encoded by the same gene. One protein has 56 amino acids and the other 82 amino acids. Provide a possible explanation for how the same gene could encode both of these proteins.

The pre-mRNA molecules transcribed from the gene are likely processed by alternative processing pathways. Two possible mechanisms that could have produced the two different proteins from the same pre-mRNA are alternative splicing or multiple 3′ cleavage sites in the pre-mRNA. The cleavage of the pre-mRNA molecule at different 3′ cleavage sites would produce alternatively processed mRNA molecules that differ in size. Translation from each of the alternative mRNAs would produce proteins containing different numbers of amino acids.Alternative splicing of the pre-mRNA could produce different mature mRNAs, each containing a different number of exons and thus the mRNAs differ in size. Again, translation from each alternatively spliced mRNA would generate proteins that differ in the number of amino acids contained.

What events bring about the termination of translation?

The process of termination begins when a ribosome encounters a termination codon. Because the termination codon would be located at the A site, no corresponding tRNA will enter the ribosome. This allows for the 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. Termination of protein synthesis is complete when the polypeptide chain is cleaved from the tRNA located at the P site. During this process, the GTP is hydrolyzed to GDP.

promoter

The promoter is the DNA sequence that the transcription apparatus recognizes and binds to initiate transcription.

A geneticist induces a mutation in a cell line growing in the laboratory. The mutation occurs in a gene that encodes a protein that participates in the cleavage and polyadenylation of eukaryotic mRNA. What will be the immediate effect of this mutation on RNA molecules in the cultured cells?

The protein is needed as part of the process for cleavage of the 3′ UTR and for polyadenylation. A nonfunctional protein would result in mRNA lacking a poly(A) tail, and the mRNA would be degraded more quickly in the cytoplasm by nucleases.

What role does RNA stability play in gene regulation? What controls RNA stability in eukaryotic cells?

The total amount of protein synthesized is dependent on how much mRNA is available for translation. The amount of mRNA present is dependent on the rates of mRNA synthesis and degradation. Less-stable mRNAs will be degraded faster so there will be fewer copies available to serve as templates for translation.The presence of the 5′ cap, 3' poly(A) tail, the 5' UTR, 3' UTR, and the coding region in the mRNA molecule are features that can affect the stability of the mRNA molecule. Poly(A) binding proteins (PABP) bind at the 3' poly(A) tail. These proteins contribute to the stability of the tail and protect the 5' cap through direct interaction. Once a critical number of adenine nucleotides have been removed from the tail, the protection is lost and the 5' cap is removed. The removal of the 5' cap allows for 5' to 3' nucleases to degrade the mRNA. AU-rich sequence elements in the 3ʹ UTR can also increase degradation of the mRNA.

transcription start site

The transcription start site is the location of the first transcribed nucleotide of the mRNA and is located 25 to 30 nucleotides downstream of the TATA box.

Listed in parts a through g are some mutations that were found in the 5′ UTR region of the trp operon of E. coli. What will the most likely effect of each of these mutations be on transcription of the trp structural genes?: Deletions in region 3 of the mRNA 5' UTR

The trp operon mRNA 5' UTR will be unable to form the attenuator if region 3 contains a deletion. Attenuation or termination of transcription will not occur, resulting in continued transcription of the trp structural genes.

Hemoglobin is a complex protein that contains four polypeptide chains. The normal hemoglobin found in adults—called adult hemoglobin—consists of two alpha and two beta polypeptide chains, which are encoded by different loci. Sickle-cell hemoglobin, which causes sickle-cell anemia, arises from a mutation in the beta chain of adult hemoglobin. Adult hemoglobin and sickle-cell hemoglobin differ in a single amino acid: the sixth amino acid from one end in adult hemoglobin is glutamic acid, whereas sickle-cell hemoglobin has valine at this position. After consulting the genetic code provided in Figure 15.10, indicate the type and location of the mutation that gave rise to sickle-cell anemia.

There are two possible codons for glutamic acid, GAA and GAG. Single-base substitutions at the second position in both codons can produce codons that encode valine: GAA--------> GUA (Val) GAG--------> GUG (Val) Both substitutions are transversions. However, in the gene encoding the β chain of hemoglobin, the GAG codon is the wild-type codon and the mutated GUG codon results in the sickle-cell phenotype.

Give the elongation factors used in bacterial translation and explain the role played by each factor in translation.

Three elongation factors have been identified in bacteria: EF-TU, EF-TS, and EF-G. EF-TU joins with GTP and then to a tRNA charged with an amino acid. The charged tRNA is delivered to the ribosome at the A site. During the process of delivery, the GTP joined to EF-TU is cleaved to form an EF-TU-GDP complex. EF-TS is necessary to regenerate EF-TU-GTP. The elongation factor EF-G binds GTP and is necessary for the translocation or movement of the ribosome along the mRNA during translation.

Briefly explain how transcriptional activator proteins and repressors affect the level of transcription of eukaryotic genes.

Transcriptional activator proteins stimulate transcription by binding DNA at specific base sequences such as an enhancer or regulatory promoter and attracting or stabilizing the basal transcription factor apparatus. Repressor proteins bind to silencer sequences or promoter regulator sequences. These proteins may inhibit transcription by either blocking access to the enhancer sequence by the activator protein, preventing the activator from interacting with the basal transcription apparatus, or preventing the basal transcription factor from being assembled.