Genetics Exam 2

A polypeptide has the following amino acid sequence: Met-Ser-Pro-Arg-Leu-Glu-Gly The amino acid sequence of this polypeptide was determined in series of mutants listed in parts a through c. For each mutant, indicate the type of mutation that occurred in the DNA (single-base substitution, insertion, deletion) and the phenotypic effect of the mutation (nonsense mutation, missense mutation, frameshift, etc.). Mutant 1: Met-Ser-Ser-Arg-Leu-Glu-Gly

A missense mutation has occurred resulting in the substitution of Ser for Pro in the protein. The change is most likely due to a single-base substitution in the Ser codon resulting in the production of a Pro codon. Four of the Ser codons can be changed to Pro codons by a single transition mutation. Pro Ser CCU UCU CCC UCC CCA UCA CCG UCG

If an RNA that consisted only of guanine nucleotides was translated what polypeptide would be made?

A polypeptide consisting only of glycine.

Sense codon

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

A polypeptide has the following amino acid sequence: Met-Ser-Pro-Arg-Leu-Glu-Gly The amino acid sequence of this polypeptide was determined in series of mutants listed in parts a through c. For each mutant, indicate the type of mutation that occurred in the DNA (single-base substitution, insertion, deletion) and the phenotypic effect of the mutation (nonsense mutation, missense mutation, frameshift, etc.). Mutant 2: Met-Ser-Pro

A single-base substitution has occurred in the Arg codon resulting in the formation of a stop codon. Two of the potential codons for Arg can be changed by single substitutions to stop codons. The phenotypic effect is a nonsense mutation and a truncated protein. Arg Stop CGA UGA transition mutation AGA UGA transversion mutation

Would you expect to find more molecules of H2A or more molecules of H3? Explain your reasoning.

Because each nucleosome contains two molecules of H2A and two molecules of H3, eukaryotic cells should have equal amounts of these two histones.

In a typical eukaryotic cell, would you expect to find more molecules of the H1 histone or more molecules of the H2A histone? Explain your reasoning.

Because each nucleosome contains two molecules of histone H2A and only one molecule of histone H1 is associated with each nucleosome, eukaryotic cells will have more H2A than H1.

Give some examples of post-transcriptional control that can affect gene expression.

Capping and polyadenylation of the mRNA as well as intron removal. Control of alternative exon splicing and alternative 3'cleavage. Selective transport of mRNA from the nucleus to the cytoplasm is another method of post-transcriptional control.

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 near the promoter and helps RNA polymerase to bind to the lac promoter, thus resulting in increased levels of transcription from the lac operon. If a mutation prevents CAP from binding near the promoter,RNA polymerase will bind the lac promoter poorly. This will result in significantly lower levels of transcription of the lac structural genes.

How are tRNAs linked to their corresponding amino acids?

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

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

Enhancers are cis-regulatory DNA sequences that are the binding sites of transcriptional activator proteins. Transcription at a distant promoter 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.Enhancers can be upstream (5'), downstream (3'), or even within genes; they can control where and when a gene is transcribed (i.e. tissue and timing of transcription).

Synonymous codon

If two codons specify the same amino acid they are said to be synonymous.

Nonoverlapping code

In a nonoverlapping code, a single nucleotide is part of only one codon. If an overlapping code were present, then a single nucleotide would be found in more than one codon. Because the genetic code is nonoverlapping, codons within the same gene do not overlap.

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. Known exceptions are the termination codons, which in the mitochondria of some organism's code for amino acids. Occasionally, a sense codon is substituted for another sense codon.

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 reduced, 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.

How does bacterial gene regulation differ from eukaryotic gene regulation?

In eukaryotic cells, gene-coding regions are interrupted by introns, which are generally longer than exons. An individual intron may be much longer than the entire coding region. Gene expression requires the proper splicing of the pre-mRNA to remove these noncoding regions. In prokaryotic cells, gene-coding regions are usually not interrupted. In eukaryotic cells, chromatin structure plays a major role in gene regulation. Chromatin that is condensed inhibits transcription. Therefore, for expression to occur, the chromatin must be altered to allow for changes in structure. Acetylation and phosphorylation of histone proteins and DNA methylation are important in these changes. At the level of transcription initiation, the process is more complex in eukaryotic cells. In eukaryotes, initiation requires a complex machine involving RNA polymerase, general transcription factors, and transcriptional activators. Bacterial RNA polymerase is either blocked or stimulated by the actions of regulatory proteins. Finally, in eukaryotes the action of activator proteins binding to enhancers may take place at a great distance from the promoter and structural gene. These distant enhancers are uncommon in bacterial cells.

With respect to the lac operon, define the term induction.

Induction is the relief of repression of transcription caused by the repressor protein. This occurs after allolactose binds to the repressor causing an allosteric change in its shape.

Assume that the number of different types of bases in RNA is four. What would be the minimum codon size (number of nucleotides) required to specify all amino acids if the number of different types of amino acids in proteins were: 2

1, because 4x1= 4 codons, which is more than enough to specify two different amino acids.

Assume that the number of different types of bases in RNA is four. What would be the minimum codon size (number of nucleotides) required to specify all amino acids if the number of different types of amino acids in proteins were: 17

3

Assume that the number of different types of bases in RNA is four. What would be the minimum codon size (number of nucleotides) required to specify all amino acids if the number of different types of amino acids in proteins were: 75

4

If a tRNA had the anticodon sequence 5' CGA 3', what would be the sequence and polarity of the corresponding codon in the mRNA? In the DNA on the non-template strand?

5' UCG 3' in the mRNA; 5' TCG 3' in the non-template strand of the DNA.

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 one amino acid in the polypeptide to another is called a missense mutation. A base substitution that changes a codon for one amino acid to a stop codon is called a nonsense mutation. A base substitution that changes codon for an amino acid to another codon for the same amino acid is called a silent mutation. Neutral mutations can be either missense or silent mutations. If the neutral mutation is a missense mutation the amino acid substitution had little or no effect on protein function.

Contrast the effects of frameshift, missense and nonsense mutations on a protein amino acid sequence.

A frameshift mutation changes the reading frame and causes the wrong amino acids to be inserted into the polypeptide after the change in the DNA. Typically, frameshift mutations will result in an inactive protein. Missense mutations result in a single amino acid substitution in the protein (the total number of amino acids remains unchanged).The mutation may or may not affect protein function depending on where in the protein it occurs.Nonsense mutations replace a codon specifying an amino acid with a stop codon. The protein is shortened or truncated and usually non-functional.

Give some examples of post-translational control of gene expression.

Phosphorylation and dephosphorylation of proteins may be necessary to make them active/inactive. If a protein needs to be associated with the membrane it may have a lipid group added to it after translation. The lipid group will inset into the lipid bilayer of the membrane and anchor the protein at the membrane. Acetylation, deacetylation, phosphorylation, dephosphorylation and methylation of histones is another form of post-translational control. These changes affect chromatin structure.Finally, some proteins need to associate with ions such as Ca++or Zn++to become active.

Suppose a chemist develops a new drug that neutralizes the positive charges on the tails of histone proteins. What would be the most likely effect of this new drug on chromatin structure? Would this drug have any effect on gene expression? Explain your answers.

Such a drug would disrupt the electrostatic interactions between the histone tails and the phosphate backbone of DNA and thereby cause a loosening of the DNA from the nucleosome. The drug may mimic the effects of histone acetylation, which neutralizes the positively charged lysine residues. Opening of chromatin structure would result from the altered nucleosome-DNA packing. An increase in transcription would result because DNA would be more accessible to transcription factors.

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. Degeneracy in the codon table is often (but not always) because different codons for the same amino acid differ at the third position but are the same at the first two positions.

Explain the wobble hypothesis. In what sense might this be beneficial to the efficiency of translation?

The 5' nucleotide of the anticodon in tRNA can base pair with more than one 3' nucleotide in the mRNA codon. Thus, although each tRNA can be charged with only one specific amino acid, the anticodon can recognize more than one codon for that particular amino acid. This reduces the number of tRNAs needed for translation.

Give an example of translational control that can affect gene expression.

The binding of miRNAs to the 3'UTR of an mRNA can prevent translation of the mRNA.The miRNA may interfere with binding of proteins to the 3'UTR and poly(A) tail thus preventing the interaction of these proteins with proteins at the 5'cap structure. This in turn will prevent the small subunit of the ribosome from recognizing the cap structure.

A polypeptide has the following amino acid sequence: Met-Ser-Pro-Arg-Leu-Glu-Gly The amino acid sequence of this polypeptide was determined in series of mutants listed in parts a through c. For each mutant, indicate the type of mutation that occurred in the DNA (single-base substitution, insertion, deletion) and the phenotypic effect of the mutation (nonsense mutation, missense mutation, frameshift, etc.). Mutant 3: Met-Ser-Pro-Asp-Trp-Arg-Asp-Lys

The deletion of a single nucleotide at the first position in the Arg codon (most likely CGA) has resulted in a frameshift mutation in which the mRNA is read in a different frame, producing a different amino acid sequence for the protein.

The following amino acid sequence isfound in a tripeptide: Met-Trp-His. How many different nucleotide sequences could encode this tripeptide?

The potential mRNA nucleotide sequences encoding for the tripeptide Met-Trp-His can be determined by using the codon table. From the table, we can see that the amino acid His has two codons, while the amino acids Met and Trp each have only one codon.In addition, there are three stop codons.Therefore, there are six different mRNA nucleotide sequences that could encode the tripeptide(1 x 1 x 2 x 3).

What events bring about the termination of translation?

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

What is the histone code?

The tails of histones in nucleosomes are chemically modified and different combinations of these modifications are observed in different regions of chromatin.Many combinations of different modifications are possible and each combination has a meaning that must be interpreted by the cell. Essentially the histone code is a language based on chemical modifications of the histone proteins that have meaning to the cell.

Termination codon

The termination codon signals the termination or end of translation and the end of the polypeptide. 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?

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.

For each of the following sequences indicate the process most immediately affected by deleting the sequence. Choose only one process for each sequence. DNA Replication Transcription RNA Processing Translation start codon

Translation The start codon is necessary for translation initiation.

How does the term allosteric transition apply to the regulation of the lac operon?

When lactose binds to the repressor, the repressor undergoes a conformational change (allosteric transition). The altered repressor is unable to bind to the operator and inhibit transcription by RNA polymerase.

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 one possible 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.

What are isoaccepting tRNAs?

are tRNA molecules that have different anticodon sequences but accept the same amino acids. For example, two tRNAs that carry the amino acid serine but have different anticodons are isoaccepting tRNAs.

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.

For each of the following sequences indicate the process most immediately affected by deleting the sequence. Choose only one process for each sequence. DNA Replication Transcription RNA Processing Translation -10 consensus sequence in bacteria

transcription In bacteria, the -10 sequence is an important component of the promoter. Deletion of the -10 sequence will prevent transcription initiation from occurring.

For each of the following sequences indicate the process most immediately affected by deleting the sequence. Choose only one process for each sequence. DNA Replication Transcription RNA Processing Translation Shine-Dalgarno sequence

translation The Shine-Dalgarno sequence or ribosome-binding site is recognized by the 16S rRNA in the 30s ribosomal subunit during the initiation of translation.If the sequence is deleted,the ribosome will not bind to the mRNA molecule and translation will not occur.

For each of the following sequences indicate the process most immediately affected by deleting the sequence. Choose only one process for each sequence. DNA Replication Transcription RNA Processing Translation poly(A) tail

translation The poly(A) tail is involved in mRNA stability and initiation of translation. If the tail is missing, then the mRNA will be degraded more rapidly thus affecting translation.

Gene X has an enhancer that must bind proteins A, B, and C to activate transcription of gene X. Proteins A, B and Care produced in the following organs and times in human development. -the gene X enhancer binding protein A is made in the heart, lungs and kidney; the protein first appears 13 days into development -the gene X enhancer binding protein B is made in the heart, liver and kidney; the protein first appears 17 days into development -the gene X enhancer binding protein C is made in the heart, lungs and kidney; the protein first appears 15 days into development In which organs(s) will gene X be expressed? At what time in development will gene X expressed at a high level?

All three proteins are made in the heart and kidney so gene X will be expressed in these organs.The last protein to be made is protein B at 17 days so a high level of transcription will be observed at this time.

Referring to the codon table in your slides, state the polypeptide specified by the following bacterial mRNA sequence.Begin translation at the first start codon and show the amino and carboxy terminal ends. 5'-AUGUUUAAAUUUAAAUUUUGA-3 ́

Amino-fMet-Phe-Lys-Phe-Lys-Phe-CarboxylThemRNA sequence begins with the three nucleotides AUG. This indicates the start point for translation and allows for a reading frame to be set. In bacteria, the AUG initiation codon codes for N-formylmethionine. Also, a stop codon is present either at the end of the translatable sequence. The amino terminal refers to the end of the protein with a free amino group and will be the first amino acid in the chain. The carboxyl terminal refers to the end of the protein with a free carboxyl group and is the last amino acid in the chain.

Initiation codon

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

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 important differences in protein synthesis. While, bacteria and eukaryotes share the universal genetic code, 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 translation takes place in the cytoplasm. Therefore, transcription and translation in eukaryotes are uncoupled. 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 by base pairing with the 16S rRNA. In eukaryotic mRNAs, the small ribosomal subunit binds the 5' cap of the mRNA and scans downstream until it encounters the first AUG codon in a Kozak consensus. Finally, elongation and termination in bacterial and eukaryotic cells are functionally similar, although different elongation and termination factors are used.

How is bacterial gene regulation similar to eukaryotic gene regulation?

Bacterial and eukaryotic gene regulation involves the action of protein repressors and protein activators. Regulation of gene expression at the transcriptional level is also common in both types of cells. Both have coordinated expression, although, through different mechanisms. Bacterial genes are often clustered in operons and are coordinately expressed through the synthesis of a single polycistronic mRNA. Eukaryotic genes are typically separate, with each containing its own promoter and transcribed on individual mRNAs. Coordinate expression of multiple genes is accomplished through the presence of common [shared] response elements(consensus sequences). Genes sharing the same response element will be regulated by the same regulatory factors.

How does the process of translation 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.Translation initiation in eukaryotes also requires more initiation factors than in bacteria.

Which of the following amino acid changes could result from a mutation that changed a single base? For each change that could result from the alteration of a single base, determine which position of the codon (first, second, or third nucleotide) in the mRNA must be altered for the change to result. Phe->Ser

Both Phe codons (UUU and UUC) could be mutated at the second position to produce Ser codons: UUU (Phe) -Change the second position to C to produce UCU (Ser).UUC (Phe) -Change the second position to C to produce UCC (Ser).

A codon that specifies the amino acid Gly undergoes a single-base substitution to become a nonsense mutation. Using the codon table determine if this mutation is a transition or a transversion .At which position of the codon does the mutation occur?

By examining the four codons that encode 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 àUGA Because uracil is a pyrimidine and guanine is a purine, the mutation is a transversion.

How is chromatin structure regulated and why is this important for eukaryotic gene regulation?

Changes in chromatin structure can result in repression or stimulation of gene expression. Acetylation of histone tails by acetyltransferase proteins results in the destabilization of the nucleosome structure and increases transcription. The reverse reaction by deacetylases stabilizes nucleosome structure. Similarly, phosphorylation of histone tails opens chromatin whereas dephosphorylation close chromatin.DNA cytosine methylation is associated with closed chromatin and 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 is related to the acetylation of the histone proteins.

Under which of the following conditions would a lac operon produce the greatest amount of β-galactosidase? The least?Explain your reasoning. L (lactose) G (glucose) P (present) Condition 1: LP-Yes GP-No Condition 2: LP- No GP-Yes Condition 3: LP-Yes GP-Yes Condition 4: LP-No GP-No

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 preventing bindingof thelac repressor to the operator. This results in the promoter being accessible to RNA polymerase. The lack of glucose allows for normal levels ofsynthesis 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. Conditions 2 and 4 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.

For each of the following sequences indicate the process most immediately affected by deleting the sequence. Choose only one process for each sequence. DNA Replication Transcription RNA Processing Translation ori site

DNA replication The ori site or origin of replication is necessary for the initiation of replication.

The lac operon is transcribed at high levels when lactose is the only carbon source. However, transcription of the operon is greatly reduced in the presence of lactose and glucose. Why?

Glucose lowers the levels of cAMP. cAMP binds to CAP and together they help RNA polymerase to bind to the promoter. If cAMP levels are low, CAP and RNA polymerase have difficulty binding to the promoter to initiate transcription.

Why does E. coli prefer to use glucose as a carbon source even if lactose is also present?

Lactose must be cleaved to generate glucose and galactose. If glucose is available the cell does not need to make b-galactosidase and permease to utilize lactose. This would be an inefficient use of cellular energy resources.

For the lac operon contrast negative and positive control of transcription.

Negative control occurs when the repressor binds to the operator in the absence of lactose. The operon is not transcribed. Positive control occurs in the presence of lactose when CAP-cAMP help RNA polymerase to bind to the promoter and activate transcription.

Nonsense codon

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

Which of the following amino acid changes could result from a mutation that changed a single base? For each change that could result from the alteration of a single base, determine which position of the codon (first, second, or third nucleotide) in the mRNA must be altered for the change to result. Leu->Gln

Of the six codons that encode for Leu, only two could be mutated by the alteration of a single base to produce the codons for Gln: CUA (Leu) -Change the second position to A to produce CAA (Gln). CUG (Leu) -Change the second position to A to produce CAG (Gln).

Which types of chemical modifications to histone proteins or DNA would be involved in closing chromatin?

On the histone tails, deacetylation of lysine, dephosphorylation of serine or histidine and methylation of some lysine R groups.Also, cytosine methylation of DNA.

For each of the following sequences indicate the process most immediately affected by deleting the sequence. Choose only one process for each sequence. DNA Replication Transcription RNA Processing Translation 3' splice site consensus sequence

RNA processing The 3' splice site is necessary for proper excision of the intron. Therefore, RNA processing events will be affected.

Degenerate code

The genetic code is degenerate because most amino acids are specified by more than one codon.For example, arginine is specified by 6 different codons.

Unambiguous code

The genetic code is unambiguous because each codon specifies only one amino acid. For example, the codon UUU always specifies phenylalanine.

How is the reading frame of a nucleotide sequence set?

The initiation codon on the mRNA sets the reading frame.

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

The lac operon consists of three structural genes, of which two areinvolved in lactose metabolism, the lacZ gene and the lacY gene. The lacZ gene encodesthe enzyme β-galactosidase, which cleavesthe disaccharide lactose into galactose and glucose, and converts lactose into allolactose. The lacY gene, located downstream of the lacZ gene, encodes apermease. Permease is necessary for the passage of lactose through the E. coli cell wall. The genes share a common promoter and operator. In another location on the bacterial chromosomeis the lacI gene that encodes the lac repressor. In the absence of lactose, the repressor binds at the operator region and inhibits transcription of the lac operon by preventing RNA polymerase from binding to the promoter.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 so that the repressor can no longer bind to 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 lacO are cis in their effects.

The lacI gene encodes the lac repressor protein, which can diffuse within the cell and bind to any operator. It can therefore affect the expression of genes on the same or different molecules of DNA. lacO is the operator. Because lacO does not make a product it can only affect the transcription of structural genes found physically on the same piece of DNA.

Describe the composition and structure of the nucleosome.

The nucleosome core particle contains two molecules each of histones H2A,H2B, H3, and H4, which form a protein core withDNA wound twice around the core. For each nucleosome, one molecule of a fifth histone, H1, binds to DNA entering and exiting the nucleosome core to clamp the DNA around the nucleosome.

Name 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, however we focus on two,EF-Tu and EF-G. EF-Tubinds toGTP and then to a tRNA charged with an amino acid. The charged tRNA is delivered to the ribosome at the A site. The elongation factor EF-G also binds GTP and is necessary for the translocation or movement of the ribosome along the mRNA during translation.

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 polycistronic mRNA. Eukaryotic genes are not organized into operons and are singly transcribed from their own promoters. b. 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. c. 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.