genetics exam 2

What is the concept of colinearity? In what way is this concept fulfilled in bacterial and eukaryotic cells?

Colinearity is the concept that the sequence of nucleotides in the DNA of a gene has a direct correspondence with the sequence of amino acids in the protein. In general, bacterial genes and proteins are co-linear with the exception of short sequences at the 5’ and 3’ ends of the genes that do not code for amino acids. Eukaryotic genes often contain large regions of noncoding DNA, called introns, which interrupt the amino acid coding regions.

What are some characteristics of introns?

Introns are intervening sequences that typically do not encode amino acids for proteins. Eukaryotic genes commonly contain introns. More complex organisms have longer introns.

What are the four basic types of introns? In which organisms are they found?

Group I introns are found in some rRNA genes of eukaryotic organisms, and bacterial, with self splicing introns.

Group II introns are found in protein-encoding genes of prokaryotes and eukaryotes. These are self-splicing introns.

Nuclear pre-mRNA introns are found in the protein-encoding genes of the eukaryotic nucleus. These introns are removed by spliceosomes.

Transfer RNA introns are found in tRNA genes of prokaryotes and eukaryotes. These introns are removed by enzymes.

What are the three primary regions of mRNA sequences in bacterial cells?

The 5′ untranslated region which contains the Shine–Dalgarno

sequence

The protein-encoding region, which begins with the start codon and ends with the stop codon

The 3′ untranslated region

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 covalently linked 5′ to 5′ to the mRNA molecule. This nucleotide is methylated at position 7 of the base.

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

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 11–30 nucleotides upstream of the 3′ cleavage site, and to a U-rich site downstream of the cleavage site. Once the complex has formed, the pre-mRNA is cleaved. Other proteins add from 50 to several hundred adenine nucleotides to the 3′ end, creating the poly(A) tail. The presence of the poly(A) tail protects the 3’ end of the mRNA from degradation by endonucleases.

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.

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 two10. 11. 12. transesterification reactions ultimately leading to the joining of the two exons. Initially, sNRP 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 a conformational change of the spliceosome so that the intron loops over, and the branch point and 5′ splice site of the intron are now adjacent to each other. U1 and U4 then disassociate from the spliceosome and base pairing occurs between U6 and U2 and between U6 and the 5’ splice site.

Describe two types of alternative processing pathways. How do these pathways 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 whether the alternative 3′ cleavage site is located before or after the stop codon.

What is RNA editing? Explain the role of guide RNAs in RNA editing.

RNA editing alters the sequence of an RNA molecule after transcription either by the insertion, deletion, or modification of nucleotides within the transcript. The guide RNAs (gRNAs) provide templates for the alteration of nucleotides in some RNA molecules undergoing editing and are partly complementary to regions within the pre-edited RNA molecule.

Summarize the different types of processing that can take place in pre-mRNA

Addition of the 5′ 7-methylguanosine cap to the 5′ end of the pre-mRNA

Cleavage at a site downstream of the AAUAAA consensus sequence at the 3′ end of the pre-mRNA

Addition of the poly(A) tail to the 3′ end of the mRNA immediately following cleavage

Removal of the introns (splicing)

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

A long precursor tRNA may be cleaved to produce one or multiple individual tRNA molecules.

Nucleotides at both the 5′ and 3′ ends of the tRNAs may be removed one at a time, a process called trimming.

For most tRNA molecules, the CCA nucleotides at the 3’ end are not encoded by the gene and are added by an enzyme.

For tRNAs that possess introns, splicing occurs.

Standard bases can be altered by base-modifying enzymes.

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.

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

What are some similarities and differences between siRNAs and miRNAs?

Both silence gene expression of foreign genes or a cell’s own genes after combining with proteins to form an RNA-induced silencing complex (RISC).

• Both are dsRNA molecules 21 to 25 nucleotides in length.

• Both are produced by the action of the enzyme Dicer on larger RNA molecules.

Some differences:

• The miRNAs originate from transcription products of distinct genes, while siRNAs originate from mRNA, RNA transposons, or RNA viruses.

• For miRNAs, Dicer cleaves single-stranded RNAs that form short hairpins, while for siRNAs, Dicer cleaves single-stranded RNAs that form long hairpins or RNA duplexes.

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.

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

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 before it can be translated. 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 eukaryotic pre-mRNA is interrupted by introns.

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 an exon, usually part of 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

Know a typical eukaryotic gene and the pre-mRNA and mRNA derived from it. Assume that the gene contains three exons. Identify the following items and, for each item, give a brief description of its function:

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.

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

The AAUAAA consensus sequence lies near the 3′ end of the pre-mRNA. It determines the location of the 3′ cleavage and poly(A) tail addition to the pre-mRNA molecule.

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 of the promoter.

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

A poly(A) tail is added to the 3′ end of the pre-mRNA. It affects mRNA stability, nuclear export, and interaction with the ribosome.

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

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

What is the one gene, one enzyme hypothesis? Why was this hypothesis an important advance in our understanding of genetics

The one gene, one enzyme hypothesis proposed by Beadle and Tatum states that each gene encodes a single, separate enzyme of a biochemical pathway.

What different methods were used to help break the genetic code? What did each method reveal, and what were the advantages and disadvantages of each one?

Marshall Nirenberg and Johann Heinrich Matthaei used the enzyme polynucleotide kinase to create homopolymers of synthetic RNAs. Using a cell-free protein synthesizing system, they were able to determine the amino acid coded by each homopolymer. By this method,the meanings for the amino acids specified by the codons UUU, AAA, and CCC were determined.

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.

How is the reading frame of a nucleotide sequence set?

The initiation codon, typically the first AUG, on the mRNA sets the reading frame.

How are tRNAs linked to their corresponding amino acids?

Each of the 20 different amino acids that are commonly found in proteins has a8. 9. 10. corresponding aminoacyl-tRNA synthetase, an enzyme that covalently links the amino acid to the correct tRNA molecule.

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

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 (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.

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 attaches to a tRNA charged with an amino acid. The charged tRNA is delivered to the ribosome at the A site.

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.

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 and have several differences in protein synthesis. First, bacteria and eukaryotes share the universal genetic code. However, the initiation codon, AUG, in eukaryotic cells codes for methionine, whereas in bacteria, the AUG initiation codon codes for N-formylmethionine. In bacterial cells, transcription and translation are coupled and occur nearly simultaneously. By contrast, in eukaryotes, transcription and translation are kept temporally and spatially separate.

Arrange the following components of translation in the approximate order in which they would appear or be used in prokaryotic protein synthesis:

1) Initiation factor 3

2) fMet-tRNAfMet

3) 30S initiation complex

4) 70S initiation complex

5) Elongation factor Tu

6) Elongation factor G

7) Release factor 1

What do you think would be the potential effect of a mutation in the part of the tRNA gene that encodes: (a) the acceptor stem; (b) the anticodon; and (c) one of the red-colored nucleotides?

(a) The tRNA would not attach to the amino acid. (b) The tRNA would potentially pairwith the wrong codon. (c) Potentially, the tRNA would be charged with the wrong amino

acid.

In what ways are spliceosomes (see Chapter 14) and ribosomes similar? In what ways are they different? Can you suggest some possible reasons for their similarities?

Spliceosomes and ribosomes are both large complexes that are composed of several different RNA and protein molecules. In essence, the spliceosome and ribosome areRNA-based enzymes or ribozymes. The RNA molecules in both structures are necessary for catalysis

Why is gene regulation important for bacterial cells?

Gene regulation allows for biochemical flexibility and rapid responses to changes in the environment while maintaining energy efficiency by the bacterial cells.

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

(1) Alteration of DNA or chromatin structure.

(2) Transcription

(3) mRNA processing

(4) mRNA stability

(5) Translation

(6) Posttranslational modification of the synthesized protein

What is the 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.5. 6. In negative control, a repressor protein inhibits or turns off transcription.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

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

The lac operon, shown in Figure 16.7, 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 lacZ gene, 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 (lacP) and operato

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. Glucose is the preferred sugar

because it can enter glycolysis without modification and therefore requires less cellular

energy for its metabolism. Because the gene expression necessary for using other

sugars is turned off, or down to very low levels, only enzymes involved in the

metabolism of glucose will be synthesized.

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 hairpin structure or attenuator in the mRNA.

Two types of secondary structures can be formed by the 5′ UTR of the mRNA transcript

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 continuation of

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.

Were the partial diploid strains of E. coli used by Jacob and Monod in their analysis of

the lac operon F+

, F-, or F' cells? Explain your answer. (See

They were F′ cells, in which the F plasmid contains part of the bacterial chromosome.

The partial diploid cells used by Jacob and Monod contained a copy of the lac operon

from the bacterial chromosome.

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.

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 and24. 25. 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

lac repressor 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.

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

Transcription

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 or base substitution 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.

Would you expect to see attenuation in the lac operon and other operons that control the

metabolism of sugars? Why or why not?

No, attenuation of the lac operon and other operons that control the metabolism of

sugars is not likely to occur because the products of these operons do not affect the

process of translation. Attenuation depends on the products of the operon affecting

translation, which in turn affect termination of transcription

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. This sensitivity to

DNase I suggests a more relaxed or open configuration state for the chromatin, allowing

transcriptional regulatory proteins access to binding sites on the DNA

What changes take place in chromatin structure, and what role do these changes play in

eukaryotic gene regulation?

Changes in chromatin structure include (1) chromatin remodeling, (2) histonemodification, and (3) DNA methylation. These changes can result in repression or

stimulation of gene expression. Chromatin remodeling complexes, which include

transcription factors and regulatory proteins that bind directly to the DNA, allow for

transcription to be initiated by increasing accessibility to the promoters by the transcription

apparatus.

What is the histone code?

The histone code refers to modifications to the tails of histone proteins. These

modifications include the addition or removal of phosphate groups, acetyl groups, methyl

groups, or ubiquitin. As a consequence of these modifications, chromatin structure is

altered and the pattern of gene expression changes.

5. How is chromatin immunoprecipitation used to determine t

Briefly explain how transcription factors affect the level of transcription of eukaryotic

genes.

Some transcription factors (called general transcription factors) stimulate and stabilize thebasal transcription apparatus at the core promoter either directly through interaction with

the basal transcription apparatus or indirectly through interaction with coactivator or with

chromatin modifying proteins. Other transcription factors stimulate transcription by

binding DNA at specific base sequences within an enhancer or at consensus sequences of

the regulatory promoter and attracting or stabilizing the basal transcription factor

apparatus.

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

S

Enhancers are DNA sequences that are the binding sites of transcription factors.

Transcription at a distant gene (thousands to millions of base pairs away) 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.

What is a response element? How do response elements bring about the coordinated

expression of eukaryotic genes?

Response elements are regulatory DNA sequences consisting of short consensus sequences

located at various distances, either in the regulatory promoter or an enhancer, from the

genes that they regulate. If the same response element sequence is located in the control

regions of different genes, these genes will be activated by the same stimuli, thusproducing a coordinated response.

8. What is an insulator?

An insulator, also called a boundary element, is a sequence of DNA that inhibits the action

of regulatory elements called enhancers in a position-dependent manner. Specific proteins

bind to insulators and play a role in their blocking activity. Insulators bind proteins that

allow chromatin to form topologically associated domains or TADs (see Chapter 11).

TADs are large, spatially interacting regions of chromatin that are separated physically

from other TADs.

How does bacterial gene regulation differ from eukaryotic gene regulation? How are they

similar?

Similarities: Bacterial and eukaryotic gene regulation involves the action of protein

repressors and protein activators; that is, there is both negative and positive regulation.

Cascades of gene regulation in which the activation of one set of genes affects another set

of genes take place in both eukaryotes and bacteria. Regulation of gene expression at the

transcriptional level is also common in both types of cells.In bacterial cells, transcription and translation occur concurrently. Translation

of a particular mRNA can affect the continuation of transcription taking place at the same

time. The presence of the nuclear membrane in eukaryotic cells separates transcription and

translation in time and space.

DNase I hypersensitivity provides information about changes in chromatin structure (see

Chapter 11). At which genes and states of development shown in Figure 11.7 is chromatin

in an open configuration?

Sites of DNase I hypersensitivity correspond to regions where chromatin is in a more open

configuration: the embryonic globin gene U in erythroblasts at 5 days, and the aD and aA

globin genes in erythroblasts at 14 days

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. The blockage of RNA polymerase is also

position dependent; the operator must be adjacent to or overlapping with the promoter. On

the other hand, repressors that bind to silencers in eukaryotes block transcription factors

from binding at an activator site, thus eliminating transcriptional activation. The repression

of transcription by a silencer and the repressor protein is position-independent