Gen practice problems test 2

Which of the statements describe how inducers cause changes in gene expression? -Inducers bind to activator proteins and enhance the ability of the activator to bind DNA. -Inducers enhance transcription of all genes. -Inducers are proteins that bind to DNA to enhance transcription of target genes. -Inducers allow transcription of specific genes. -Inducers are ions that are essential for the function of RNA polymerase.

-Inducers bind to activator proteins and enhance the ability of the activator to bind DNA. -Inducers allow transcription of specific genes.

How are proteins regulated after translation? (select all that apply) -active proteins can be inactivated by post‑translational modification -proteins that are no longer required can be transported out of the cell -inactive proteins can be activated by phosphorylation -fewer mRNA molecules can be transcribed to produce fewer proteins -proteins can be tagged with ubiquitin molecules and subsequently degraded

-active proteins can be inactivated by post‑translational modification -inactive proteins can be activated by phosphorylation -proteins can be tagged with ubiquitin molecules and subsequently degraded

Match the description of transcriptional control to the corresponding transcriptional regulator: -Active repressor -Inactive repressor -Active activator -Inactive activator There is one description for each transcriptional regulator: -negative control in a repressible operon -regulator of a positive repressible operon -regulator of a negative inducible operon -regulator of a positive inducible operon

-active repressor- regulator of a negative inducible operon -inactive repressor- negative control in a repressible operon -active activator- regulator of a positive repressible operon -inactive activator- regulator of a positive inducible operon

A mutant strain of E. coli produces β‑galactosidase in the presence and in the absence of lactose. Where in the operon might the mutation in this strain occur, and why? (select all that apply) -in the operator region, where the mutation leads to the failure of the operator to normally bind the repressor -near the lacI gene, where the mutation leads to increased levels of lac repressor being made -in the lacI gene, which leads to an inactive lac repressor -in the promoter region of the operon, where the mutation leads to the failure of the RNA polymerase to bind to the promoter -in the CAP binding site, where the mutation leads to the inefficiency of RNA polymerase activity -in the operator region, which leads to increased binding of the lac repressor to the operator

-in the operator region, where the mutation leads to the failure of the operator to normally bind the repressor -in the lacI gene, which leads to an inactive lac repressor

Determine the nature of the expression of the lacZYA genes, represented by L, and label each genotype as constitutive, inducible, or no transcription in the presence of lactose. Assume that all promoters are wild type. symbols and representations: I−: lacI mutant cannot bind to the operator I^S: lacI mutant always binds to the operator O^C: Operator mutant that prevents repressor binding F': the wild‑type operator, O+, and lacI gene on a plasmid L: lacZYA genes Genotypes to label: 1) I^+ O^+ L^+ 2) I^+ O^c L^+ 3) I^+ O^c L^+ , F' 4) I^- O^+ L^+ , F' 5) I^s O^+ L^+ 6) I^s O^+ L^+ , F'

1) I^+ O^+ L^+ : inducible 2) I^+ O^c L^+ : constitutive 3) I^+ O^c L^+ , F' : constitutive 4) I^- O^+ L^+ , F' : inducible 5) I^s O^+ L^+ : no transcription 6) I^s O^+ L^+ , F' : no transcription

The lac operon in E. coli controls the gene expression of the enzymes that digest lactose in the cell. In the absence of lactose, the lac operon will turn off and gene expression will be inactivated. Place the events of gene regulation by the lac operon in order of their occurrence, from the removal of lactose from the environment to when the cell no longer digests lactose. -Lactose is scarce in the environment. -The cell no longer digests lactose. -The repressor is activated in the absence of lactose. -RNA polymerase is prevented from moving from the promoter. -The repressor binds to the operator. -Lactose enzyme genes are not transcribed.

1) Lactose is scarce in the environment 2) The repressor is activated in the absence of lactose. 3) The repressor binds to the operator. 4) RNA polymerase is prevented from moving from the promoter. 5) Lactose enzyme genes are not transcribed. 6) The cell no longer digests lactose.

Whether a gene is transcribed to produce RNA, or not, ab be controlled by the binding of transcription factor proteins to the regulatory region of a gene. The activity of transcription factors themselves can be influenced by signals from the environment. B. Describe how a transcription factor 1 (TF1) and environmental signal A could regulate expression of gene X in bacteria in a manner that is: 1) positive and inducible 2) negative and inducible 3) positive and repressible 4) negative and repressible

1) an activator is turned on by an environmental signal increasing gene expression 2) a repressor is turned off by an environmental signal increasing gene expression 3) an activator is turned off by an environmental signal decreasing gene expression 4) a repressor is turned on by an environmental signal decreasing gene expression

The pez operon codes for 3 structural genes (pezA, pezB, and pezC). The gene pezR codes for a repressor protein that binds the operator region of the pez operon (pezO). Pez operon expression is induced by glucose. For each strain below, indicate if structural gene expression is constitutive, inducible, or if no expression takes place. a+b+c+r+o+ a+b+c+r-o+ a+b+c+r+o^c a+b+c+r-o+ / a-b-c-r+o+ a-b+c+r-o+ / a+b-c+r+o^c a-b-c+r-o^c / a+b-c+r+o+

1) inducible expression a, b, and c 2) constitutive expression a, b, and c 3) constitutive expression a, b, and c 4) inducible expression a, b, and c 5) constitutive expression a and c, inducible expression of b 6) inducible expression of a, no expression of b, constitutive expression of c

RNA interference (RNAi) can affect gene expression in two different ways. 1) if RNAi perfectly matches the RNA it targets, what happens? 2) if RNAi does not perfectly match the RNA it targets, what happens?

1) it degrades the RNA, stopping translation 2) it would inhibit translation, but it would not affect the RNA

Whether a gene is transcribed to produce RNA, or not, ab be controlled by the binding of transcription factor proteins to the regulatory region of a gene. The activity of transcription factors themselves can be influenced by signals from the environment. A. Describe how the following influence gene expression: 1) positive/ negative regulation 2) inducible/ repressible regulation

1) positive regulation is when the binding of a transcription factor increases gene expression. Negative regulation is when the binding of a transcription factor decreases gene expression.

Of the four components of an operon: repressor, promoter sequence, operator, and structural gene. 1) Which components act in cis? What does this mean? 2) Which components act in trans? What does this mean? 3) What is the difference between a supersupressor and a repressor?

1) promoter sequence and operator. This means they cannot jump to another gene to cause it to function. (promoter cannot jump to another gene that does not have a functional promoter; operator cannot jump to another gene that does not have a functional operator) 2) repressor acts in trans. This means it can jump to another gene that does not have a functioning repressor and block expression (if the operator is functional) 3) supersuppressor blocks expression regardless of if there is an inducer present (cannot be blocked, unlike a regular repressor)

RNA interference (RNAi) is a mechanism of gene silencing that is mediated by the presence of double‑stranded RNA. List the seven steps involved in gene silencing by RNAi in order. Antisense RNA pairs with the target RNA Double‑stranded RNA (dsRNA) is introduced into a cell. A gene's expression is silenced. The sense strand is separated from the antisense sense and degraded.. Long dsRNA is cleaved into short dsRNA. RNA-induced silencing complex (RISC) binds to short dsRNA. Target RNA is degraded.

1: Double‑stranded RNA (dsRNA) is introduced into a cell. 2: Long dsRNA is cleaved into short dsRNA. 3: RNA-induced silencing complex (RISC) binds to short dsRNA. 4: The sense strand is separated from the antisense sense and degraded. 5: Antisense RNA pairs with the target RNA. 6: Target RNA is degraded. 7: A gene's expression is silenced.

How does an epigenetic change differ from a mutation? -Mutations involve nucleotide sequences of only one or a few base pairs. An epigenetic change involves large regions of a chromosome, thousands of base pairs or more in length. -Epigenetic changes occur due to environmental effects and only persist as long as the environment that induced them remains constant. Mutations are permanent. -Mutations can be passed on to offspring or to daughter cells. Epigenetic changes cannot be inherited. -A mutation alters the nucleotide sequence in DNA. An epigenetic change does not alter the DNA sequence, but can be inherited by daughter cells.

A mutation alters the nucleotide sequence in DNA. An epigenetic change does not alter the DNA sequence, but can be inherited by daughter cells.

You use a series of mutations in yeast to study regulation of the GLN-1 gene that codes for an enzyme (GS). Mutations in this gene (gln-1) show no GS enzyme activity. You isolate two additional mutations, gln-2 and gln-3, that reduce but do not eliminate GS enzyme activity. Mating any of these mutations to wild type yeast produce diploids with normal GS activity. Mating haploid gln-2 or gln-3 mutants to each other or to haploid gln-1 mutants produces diploids that show normal GS activity. A) do the gln-2 and gln-3 mutations appear to affect the same or different genes? B) are the gln-2 and gln-3 mutations dominant or recessive? C) do the GLN-2 / GLN-3 gene produces act in cis or in trans to gln-1?

A) GLN-2 And GLN-3 both reduce the expression of GS enzyme but complement each other so they are probably separate genes B) there is no phenotype in GLN-2 and GLN-3 heterozygotes so mutations are probably recessive C) functional GLN-2 and GLN-3 can each restore some GS expression even when located on a different chromosome from functional GLN-1 gene so act in trans. Probably code for activator proteins.

transcriptional regulation of eukaryotic genes often involves multiple proteins. For example, the Gal-1 genes codes for a protein needed to metabolize galactose in yeast. Gal-1 expression is regulated by 3 proteins coded for by the gal-3, gal-4, and gal-80 genes. Explain how each of the following will affect gal-1 expression: A) presence and absence of glucose B) knockout mutations of gal-3, gal-4, gal-80 C) a mutation eliminating the DNA binding domain of gal-4 D) mutations that eliminate the protein interaction domains of gal-3, gal-4, or gal-80 E) mutations that eliminate galactose binding domain of gal-3 F) mutations that alter UAS (response element) sequence

A) Gal-1 expression is promoted when Gal-4 binds to the upstream activating sequence (UAS - enhancer element) and recruits RNA polymerase to the promoter. If galactose is absent, Gal-80 binds to Gal-4 and prevents it from recruiting polymerase blocking expression. If galactose is present, it activates gal-3, which prevents gal-80 from binding to gal-4. This lets gal-4 recruit RNA polymerase to the promoter increasing expression of gal-1. (Look at drawing for illustration) B) knockout of Gal-3: low expression- no response to galactose, can't prevent repressor from interacting with Gal-4 Gal-4: low expression- no response to galactose, no activation Gal-80: constitutive high expression, no repression C) mutation to the Gal-4 DNA binding domain: Most likely low expression- no response to galactose, reduced gal-4 binding to UAS. Alternative: could improve binding to UAS but still regulated by gal-80 D) mutations to the protein interaction domains of Gal-3: low expression- no response to galactose, can't remove repressor Gal-4: two possibilities— 1) can't bind repressor -> constitutive high expression, 2) can't dimerize -> low expression Gal-80: two possibilities— 1) can't bind gal-4 -> constitutive high expression, 2) can't bind gal-3 -> low expression - no response to galactose E) mutation to the glucose binding domain of gal-3: low expression - no response to galactose F) mutation to the UAS (response element): Most likely low expression - no response to galactose, reduced gal-4 binding to UAS. Alternative: could improve gal-4 binding but still regulated by gal-80. Would not affect fgal-1 expression

You isolate a mutant strain of E. coli that produces the enzymes for B-galactosidase and Permease whether lactose is present or not. A) describe two different ways (models) that could explain this behavior. B) you isolate a second mutant that can permease if lactose is present but can never produce B-galactosidase. Describe mutation that could cause this pattern of expression. C) if a partial diploid is made from these two mutants, neither B-galactosidase nor permease are made in the absence of lactose. When lactose is added, the partial diploid makes both enzymes. Which of the two models you described for part A does this observation support?

A) constitutive expression of B-gal and permease could be caused in 2 ways: - a mutation in regulator gene (i+ to i-) that produces either no repressor or a defective repressor protein that cannot bind to the operator. - a mutation in the operator (o+ to o^c) that prevents the repressor protein from recognizing/binding to the operator. B) permease expression in the second mutant is inducible so lack, lacO, and lacY must be normal. The second bacteria strain must have a mutation in the lacZ gene in order for no B-gal to be produced and its genotype must be i+o*z-y+ C) B-gal is inducible in the partial diploid. This means that regulation of the functional copy of lacZ in the first strain of bacteria has been restored. Only an i+ gene could act in trans in the diploid to make mutant 1 inducible for B-gal production. Mutant 1 must be a mutation from i+ to i- rather than an operator mutation from o+ to o^c. The partial diploid be i-o+z+y+ / i+o+z-y+

Chromosomes consist of DNA and proteins in a complex known as chromatin. A) describe the difference between euchromatin and heterochromatin to a non-expert. B) explain how: i. Acetylation/ deacetylation of lysine amino acids on histone proteins can affect chromatin structure and gene expression ii. The Addition/ removal of methyl groups to cytosine nucleotides can alter gene expression and chromatin structure iii. How insulator sequences and the insulator binding protein CTCF can promote or inhibit the expression of a gene.

A) heterochromatin - DNA is tightly associated with histone proteins into a compact structure in which little of the DNA is exposed and available to interact with other proteins B) chromatin modifications i. Lysine amino acids have a (+) charged R group that can interact strongly with (-) charged nucleotides promoting the tight association of DNA with histones (heterochromatin). This association can prevent the binding of transcription factors and RNA polymerase to DNA needed for transcription and gene expression. Adding acetyl (CH3 O-) functional groups to the R-group of lysine neutralizes this charge and weakens the interaction between histones and DNA promoting the formation of euchromatin. The looser association between DNA and histones in euchromatin allow transcription factors and RNA Polymerase to access and transcribe genes increasing gene expression rates. ii) Gene expression is promoted by the binding of transcription factors and RNA polymerase to DNA to initiate transcription. Methyl groups added to DNA can interfere with these interactions reducing transcription. Regions of methylated cytosine nucleotides can also be recognized and bound by other proteins that remove acetyl groups from histones and promote the formation of heterochromatin. iii) Insulator sequences help to mark the boundaries between DNA regions. CTCF can bind to insulator sequences and (along with other proteins) can organize DNA in to loops. These loops can bring enhancer sequences close to the promoters of some genes promoting their expression and away from other genes, inhibiting the expression of these genes. Insulator sequences also help to limit the presence of heterochromatin and euchromatin to specific DNA regions influencing the expression of genes in these regions.

Planaria have an RNA interference (RNAi) pathway but the genes associated with the pathway have not been isolated. You discover a gene (rdeA) in planaria that codes for a protein very similar to one involved in RNAi in other eukaryotes to determine if rdeA is part of an RNAi mechanism, you expose planaria to two types of double stranded RNA (dsRNA) molecules. One dsRNA (myo) is complementary to the myosin gene. A second dsRNA (scr) has the same nucleotide composition (% A, U, C, and G) as the myo dsRNA but the order of the nucleotides has been scrambled. You feed these dsRNAs to rdeA(+) and rdeA(-) planaria and then examine expression of both myosin and actin genes in these groups using gel electrophoresis. A) predict the results you might expect to see if the rdeA protein is part of an RNAi mechanism and if it is not. B) how would your experiment be affected or the dsRNA produced in the myo bacterial strain is not perfectly complementary to the myosin gene? C) What is the purpose of the scrambled dsRNA sequence? Why did you look at expression of the actin gene?

A) if rdeA protein is involved in RNAi, rdeA(+) planaria that eat the myo dsRNA should show a reduction in expression of the myosin gene. Planaria that eat the scr dsRNA and rdeA(-) that eat either dsRNA should show normal levels of myosin gene expression. B) Planaria exposed to myo dsRNA that is perfectly complementary to the myosin gene will show reduced myosin mRNA and protein because the RISC complex will cleave the myosin mRNA. If the myo dsRNA is imperfectly complementary, the RISC complex will bind to the myosin mRNA blocking translation but will not cleave the mRNA. As a result myosin mRNA levels will be normal. C) actin RNA helps determine if equal amounts of RNA were included in the samples (all samples should show equal levels of actin RNA). The scrambled dsRNA is a (-) control. If changes in myosin mRNA expression are due to RNAi, it should only occur using dsRNA sequences complementary to the myosin RNA (myo) and not with the scrambled (scr) sequence.

The nuc operon in bacteria produces enzymes needed to convert the amino acid aspartate into precursors for nucleotide synthesis. The operon has 4 structural genes (nucB, nucC, nucD, and nucE). In addition, the nucR gene codes for a repressor protein and the nucA gene codes for an activator protein that each interact with the nuc operon. Expression of the operon requires high levels of aspartate and low levels of ATP. Aspartate binds to the nucA protein and ATP binds to the nucR protein. A) describe a model for how the nuc operon is regulated. Does the operon display characteristics of positive / negative and inducible / repressible regulation? B) how would nuc expression be affected by mutations that eliminate nucA gene function? nucR? C) how do puke expression of the operon be affected by mutations that can result in constitutively active forms of the proteins coded for by the nucA gene? Of the nucR gene?

A) the operon displays the following regulation: - negative repressible: ATP binding activates repressor (nucR) and blocks expression - positive inducible: High aspartate activates activator (nucA) and promotes expression B) nucA (-) eliminates activator function: reduces expression under all conditions nucR (-) eliminates repressor function: expression when high aspartate regardless of ATP levels C) nucA^c -activator always bound - expression with low ATP regardless of aspartate levels nucR^c -repressor always bound - reduces expression under all conditions

Activator proteins increase gene expression, whereas repressor proteins inhibit gene expression. Which statement concerning activator and repressor proteins in eukaryotes is true? -Repressors are more common than activators in eukaryotes. -Activators are more common than repressors in eukaryotes. -Activators and repressors are equally common in eukaryotes. -Eukaryotes use transcription factors instead of activators or repressors.

Activators are more common than repressors in eukaryotes.

How do cells regulate gene expression using alternative RNA splicing? -Alternative RNA splicing determines which proteins are produced from each gene. -Alternative RNA splicing determines which genes are underexpressed. -Alternative RNA splicing determines which genes are transcribed to mRNA. -Alternative RNA splicing determines how fast certain proteins are degraded.

Alternative RNA splicing determines which proteins are produced from each gene.

Place the steps of chromatin immuno‑precipitation (ChIP) in order from first to last. .A-Reverse cross linking of DNA and protein B-Shear protein-bound DNA into fragments C-crosslink DNA-binding proteins to DNA D-purify and sequence DNA E-immunoprecipitate target protein

C, B, E, A, D

Place the steps of DNA methylation maintenance in order from first to last. A-Fully methylated DNA is produced. B-The replicated DNA strand is synthesized unmethylated. C-Prior to replication, CpGs are fully methylated D-Any unmethylated DNA is methylated E-Methyltransferase binds to DNA methyl groups

C, B, E, D, A

All of the cells in the human body contain the same genes. How do cells have different morphologies and functions when they contain the same genetic information? -Different cell types obtain different types of nutrients from their surroundings. -Different cell types express particular genes at different levels. -Different cell types receive different endocrine signals. -Different cell types replicate at a different rate.

Different cell types express particular genes at different levels.

Epigenetic changes to chromatin can persist in a cell for many cellular generations. Describe a mechanism that can accomplish this considering that, after each cell division, half of each DNA molecule and half of the chromatin proteins are newly synthesized and unmodified.

For methylated DNA, after replication only one strand will be methylated. However, methylated nucleotides at CpG sites can recruit methyltransferase enzymes that add methyl groups to cytosines on the complementary strand. For histones, DNA methylation may promote modification of associated histone proteins. Also, as with methylated DNA modified histones may be associated with unmodified partners in newly assembled histones after replication. The modified histones may recruit enzymes that add similar marks to the unmodified proteins.

Classify the characteristics of eukaryotic gene regulation

Genes are located on different chromosomes. A 5' cap and 3' poly-A tail are added to mRNA Transcription occurs in the nucleus, whereas translation occurs in the cytoplasm.

Classify the characteristics of prokaryotic gene regulation

Genes are located on one chromosome. Some genes are organized into operons, and mRNA transcripts often specify more than one protein. mRNA can be transcribed from DNA and translated into protein at the same time

You perform an experiment to examine the 3D organization of the DNA associated with a gene you are studying and find that the gene is physically near another sequence of DNA located far away on the same chromosome in cell type A. However, in cell by B the gene is not located near this other DNA sequence. You also know that your gene is expressed in cell type A but not B. How might you interpret this result?

Genome organization. The distant sequence may contain an enhancer sequence that is involved in regulating the expression of your gene. The DNA appears to have formed a loop in cell type A that has brought the sequence close to your gene within the same sub-domain. Different cells express different genes and this is reflected in the organization of the DNA in the nucleus. The DNA is organized differently in cell type B and a loop is not formed. As a result, the gene is not expressed.

Easily detectable proteins used for reporter constructs

Green fluorescent protein (GFP) - glows under UV light B-galactosidase (B-gal) - enzyme produces blue pigment from a colorless compound Luciferase - produces light in presence of added substrate

The restriction enzymes Hpa-I and Msp-I cut the same DNA sequence (5'CCGG 3'). However, Hpa-I is sensitive to methylated DNA methylated while Msp-I is not. The ANL gene is genomically imprinted with the maternal copy being methylated. You use Msp-I and Hpa-I to examine the promoter region for the ANL gene in adult animals and in gametes. a) What "genomic imprinting" refer to? b) Describe the results you would expect if you looked at adult cells in males and females. c) Describe the results you would expect to see in sperm and eggs. d) Which copy of the ANL gene (maternal / paternal) in an adult cell would you expect to see expressed and why? e) There are two alleles for the ANL gene (ANL and anl) with the ANL allele displaying complete dominance. If you crossed two individuals, one of which was homozygous recessive, the other heterozygous for the ANL gene, what kind of offspring would you expect to see and in what proportions?

Imprinting a) Genomic imprinting refers to patterns of DNA methylation in an embryo inherited from the parents. Usually, a stretch of DNA inherited from one gender is methylated but not the other. E.g. a region of DNA inherited from mom may be methylated but the copy from dad is not. The reverse may be true for a different region of DNA. These methylations patters influence whether the maternal or paternal copy of a gene will be expressed in the embryo. b) Both copies of the ANL gene can be cut by Msp-I. However, one copy of the gene will be methylated and resistant to cleavage by Hpa-I c) The ANL gene in sperm should not be methylated and can be cut by both Hpa-I and Msp-I. ANL gene in eggs should be methylated and can only be cut by Msp-I. d) Methylation is usually associated with gene silencing. The ANL gene should be active on the paternal copy of a chromosome. e) The ANL allele that came from the mother is expected to be methylated and inactive. The copy of the gene that came from the father should not be methylated and should be active. If the father is homozygous recessive, all of the offspring will show the recessive phenotype. If the father is heterozygous, half of the offspring will receive the recessive allele and display a recessive phenotype.

You have identified a region of DNA that you believe contains a previously unidentified insulator sequence. a) What are insulator sequences? b) How could you test this? c) What would be suitable control(s) for your experiment?

Insulator A) Insulator sequences are short stretches of DNA that can influence interactions between neighboring DNA regions. They can separate regions of hetero- and euchromatin, organize DNA into loops, and prevent enhancer sequences from influencing the expression of nearby genes. b) When placed between an activator sequence and a promoter, an insulator sequence should prevent activation of gene expression. Prepare a reporter construct in which the potential insulator sequence has been placed between an activator response element and a promoter driving expression of a visible product (b-gal or GFP). If this reduces expression, this would suggest that the sequence being tested could act as an insulator. Alternatively, you could remove the suspected insulator sequence from the region of DNA being examined and look at expression of neighboring genes. If neighboring genes not normally expressed show increased expression after the deletions, this could also suggest the action of an insulator. c) Two potentially good controls i) A reporter construct containing a different but equal sized piece of DNA between the activator response element and promoter to see if simply inserting any DNA between the promoter and response element affects gene expression (negative control). ii) Place a DNA sequence previously shown to act as an insulator between the activator response element and promoter to demonstrate that an insulator can interfere with expression of the gene (positive control). iii) Place a DNA sequence previously shown to act as an insulator between the activator response element and promoter to demonstrate that an insulator can interfere with expression of the gene (positive control).

Suppose an operon has the following characteristics: (1) The operon codes for structural proteins that convert compound Q to compound B. (2) The operon is controlled by a constitutively expressed regulatory gene called reg. (3) In wild‑type individuals, the operon is transcribed in the absence of compound B but not in the presence of compound B. (4) In reg− mutants, the operon is constitutively transcribed. Is this operon inducible or repressible? Why? -It cannot be determined because the number of structural genes is unknown. -It is repressible because wild‑type transcription is repressed in the presence of compound B. -It is inducible because the operator in reg− mutants can still bind to compound B. -It is repressible because reg− transcription only occurs in the presence of compound B. -It is inducible because wild‑type transcription does not occur in the presence of compound B.

It is repressible because wild‑type transcription is repressed in the presence of compound B.

Certain behavioral traits in humans are believed to be encoded by the X chromosome. Examples of such traits include verbal ability and seeking social interactions. For these traits, research has shown that there is less variability between identical male twins than between identical female twins. Which phenomenon best explains these results? -Male twins both get the same mitochondria from their mother. -Females have double the amount of X expression so there's more room for error. -Males have a Y chromosome, which is much smaller than the X. -One X chromosome is randomly inactivated in females.

One X chromosome is randomly inactivated in females.

Explain what a consensus sequence is

The "average" nucleotide sequence in a stretch of DNA/RNA with a particular function, often a region recognized by a binding protein Slightly different DNA/RNA sequences may have the same function (bound by the same protein) and may influence the function of the region (how tightly the protein binds)

Arrange the steps of the regulation of the trp operon in order of occurrence. -The level of tryptophan is low -Trytophan binds to the trp repressor and induces a conformational change. -The trp repressor protein binds to the operator. -Sufficient quantities of tryptophan make further synthesis unnecessary. -Products of trp genes synthesize tryptophan. -The trp repressor blocks RNA polymerase from binding to the promoter. -RNA polymerase binds to promoter, allowing transcriptions of trp genes to proceed. -Transcription of gene stops.

The level of tryptophan is low: I. RNA polymerase binds to promoter, allowing transcriptions of trp genes to proceed. II. Products of trp genes synthesize tryptophan. III. Sufficient quantities of tryptophan make further synthesis unnecessary. IV. Trytophan binds to the trp repressor and induces a conformational change. V. The trp repressor protein binds to the operator. VI. The trp repressor blocks RNA polymerase from binding to the promoter. VII. Transcription of gene stops.

A mutation at the operator site of an operon prevents the repressor from binding. What effect will this mutation have on transcription in a repressible operon? -There will be no change in the operon's activity. -It will be impossible to turn on the transcription of the structural genes. -The operon will always be transcriptionally active. -There will be a significant decrease in the operon's activity.

The operon will always be transcriptionally active.

A mutation at the operator site of an operon prevents the repressor from binding. What effect will this mutation have on transcription in an inducible operon? -It will be impossible to turn on the transcription of the structural genes. -There will be a significant decrease in the operon's activity. -There will be no change in the operon's activity. -The operon will always be transcriptionally active.

The operon will always be transcriptionally active.

What would be required to prove that a phenotype is caused by an epigenetic change? -The phenotype must be reversible in the presence of mutagenic compounds. -The phenotype must be passed to progeny only through the maternal DNA. -The phenotype must be unexplained by known genetic mechanisms. -The phenotype must be heritable with modified chromatin and an unaltered nucleotide sequence. -The phenotype must be the result of a nucleotide change that is present in one gender but not the other.

The phenotype must be heritable with modified chromatin and an unaltered nucleotide sequence.

What is the function of the promoter in the E. coli lac operon? -The promoter is a protein that prevents the transcription of the lactose genes by binding to the beginning of the lac operon. -The promoter is a region of DNA at the start of the lac operon, where RNA polymerase binds to begin transcription of the lactose genes. -The promoter is a region of DNA at the start of the lac operon that switches lactose gene expression on or off through protein binding. -The promoter is a region of DNA outside of the lac operon that expresses a protein that inactivates the transcription of the lactose genes.

The promoter is a region of DNA at the start of the lac operon, where RNA polymerase binds to begin transcription of the lactose genes.

What are sigma factors and how might gene expression in a prokaryote be affected by a mutation that blocks production of a particular sigma factor?

They help RNA polymerase recognize the promoter and initiate mRNA transcription. different sigma factors recognize different promoter sequences and regulate expression of whole classes of proteins in response to environmental conditions (heat shock, nitrogen metabolism, spore formation, etc.) If a sigma factor were not produced, this might reduce or prevent expression of one class of proteins while others would not be affected

How do reporter constructs help you study the regulatory region of a gene?

They make it easier to study gene expression by modifying the gene to include DNA that codes for a different, more easily detectable protein. The new coding region can be added to the existing gene creating a "fusion protein" that has both the original and new amino acids or can completely replace the old coding region.

In eukaryotes, transcription factors and enhancer sequences are used to regulate transcription. Classify the statements as true or false. Transcription factors bind to the entire enhancer sequence. Enhancer sequences are composed of DNA base pairs Enhancer sequences can be located thousands of base pairs downstream from the transcription start site. Transcription factors always increase transcription levels. Enhancer sequences directly alter transcription levels

True: Enhancer sequences can be located thousands of base pairs upstream/downstream from the transcription start site. Enhancer sequences are composed of DNA base pairs False: Enhancer sequences directly alter transcription levels Transcription factors always increase transcription levels. Transcription factors bind to the entire enhancer sequence.

Suppose you are studying a gene of interest. How would you be able to determine if a histone with a particular modification binds to your gene of interest? -Do not unlink the DNA-protein complex when sequencing. -Use fluorescent in situ hybridization of the gene of interest. -Use an antibody specific to the modified histone. -Flourescently label and visualize the modified histone.

Use an antibody specific to the modified histone.

Monozygotic twins have identical DNA sequences, yet twins can differ in some physical or behavioral traits. What evidence suggests that epigenetic effects may cause these phenotypic differences in monozygotic twins? -Younger twins increase DNA methylation, whereas older twins increase histone acetylation. -Older twins increase DNA methylation, whereas younger twins increase histone acetylation. -Younger twins have more similar DNA methylation and histone acetylation patterns than older twins. -Older twins have more similar DNA methylation and histone acetylation patterns than younger twins.

Younger twins have more similar DNA methylation and histone acetylation patterns than older twins.

which statement best describes an operon? -protein modifications such as the addition of a functional group, or alternate folding of the protein -mRNA modifications, such as the addition of a 5′5′‑cap and 3′3′ poly‑A tail and the removal of introns -the processing of exons in mRNA that results in a single gene coding for multiple proteins -a gene cluster controlled by a single promoter that transcribes to a single mRNA strand -heritable changes in gene expression that occur without altering the DNA sequence

a gene cluster controlled by a single promoter that transcribes to a single mRNA strand

constitutive expression

continuously expressed under normal cellular conditions (genes always on)

DNA can undergo many types of modifications. One of these is methylation. Which base(s) are methylated more often in human DNA?

cytosine

The yeast Saccharomyces cerevisiae has several genes encoding enzymes that function in the importation and metabolism of galactose. The genes are located on several chromosomes, and are transcribed separately. The GAL genes have similar promoters, and gene transcription is under regulation by the proteins Gal3p, Gal4p, and Gal80p. Place the statements about GAL gene regulation in the correct order, starting from conditions with an absence of galactose through conditions with abundant galactose. Only five statements will be placed.

glucose absent, galactose absent: 1) Gal4p, in complex with Gal80p, is bound to DNA. glactose becomes abundant 2) galactose binds Gal3p 3) Gal3p associates with Gal80p 4) Gal80p inhibition of Gal4p is abated 5) Gal4p is freed to function as an activator at GAL promoters gene transcription

What are the key differences between euchromatin and heterochromatin?

heterochromatin- tightly packed and is less likely to contain an actively expressed gene, methylated DNA euchromatin- loosely packed and is more likely to contain an actively expressed gene, unmethylated DNA

What is methylated DNA?

only occurs in cytosine In heterochromatin, Cytosine on the template strand becomes Uracil, so when it is copied, it pairs with Adenine instead of Guanine

Choose the examples of epigenetic modifications. ubiquitylation lipidation acetylation methylation polyadenylation

ubiquitylation acetylation methylation