Wk 6 Problems

Modifications to chromatin can affect transcriptional activity by changing the accessibility of DNA to the transcription machinery. The given descriptions are examples of various processes that may or may not cause remodeling of chromatin. Match each description to the effect it has on transcriptional activity caused by chromatin remodeling. Activates, Inactivates, Activates and inactivates, No effect Answer Bank: Histone acetyltransferases attach acetyl groups to the N‑terminus of histones. Histone deacetylases remove acetyl groups from the N‑terminus of histones. Histone methylation occurs at different amino acids. RNA polymerase II binds the start site of transcription.

Activates Histone acetyltransferases attach acetyl groups to the N‑terminus of histones. Inactivates Histone deacetylases remove acetyl groups from the N‑terminus of histones. Activates and inactivates Histone methylation occurs at different amino acids. No effect RNA polymerase II binds the start site of transcription. Explanation: Transcriptional activity of a gene can be affected by various molecular processes, including chromatin remodeling. However, not all molecular processes that affect transcriptional activity do so by remodeling chromatin. Chromatin remodeling activates transcription by exposing parts of the gene, such as the promoter. Attachment of acetyl groups to histones causes the DNA strands to associate with histones more loosely. The attachment of acetyl groups is facilitated by enzymes called histone acetyltransferases. Another way by which DNA can be exposed for access by transcriptional proteins is to move the position of nucleosomes. The SWI/SNF enzyme complex translocates the nucleosome so that the promoter of the gene is accessible for transcription. Chromatin remodeling inactivates transcription by allowing DNA and histones to associate more tightly. Histone deacetylases remove the acetyl group from histones, and thus perform the opposite activity of histone acetyltransferases. Transcription can also be altered by chromatin remodeling through direct DNA modification. Methyl groups can be attached to cytosines, especially in regions of DNA where there are many guanines next to cytosine. These regions are called CpG islands. Methylated DNA is less accessible than unmethylated DNA because it is more compact, and because binding sites for transcriptional proteins are obscured. This is why CpG islands are often found near promoters. Methylation of histones can either activate or inactivate transcription, depending on the specific amino acids that are methylated. Proteins that are part of the transcriptional machinery can increase transcriptional activity, but they do not do so by modifying chromatin.

Select the definition of a Barr body. an inactive autosome an inactive X chromosome an active X chromosome an inactive Y chromosome The first cloned cat, CarbonCopy (CC), was tabby, while the cat she was cloned from, Rainbow, was calico. The surrogate mother was a tabby. Select the explanation that best explains why CC would never have been identical in pattern to Rainbow. The coat color genes are on the mitochondrial genome and inherited with the mother's oocyte. The coat color genes are maternal effect genes, so CC looked like the surrogate. The pattern of X-chromosome inactivation is established randomly in a cell lineage. Both X chromosomes remain actively transcribed.

Barr body = an inactive X chromosome The reason CC would never have been identical in pattern to Rainbow is best explained due to : The pattern of X-chromosome inactivation is established randomly in a cell lineage. Explanation: In cells that contain more than one X chromosome, all but one X chromosome must be inactivated to create comparable amounts of gene products between males, who have one X chromosome, and females who have two X chromosomes. The presence of more than one active X chromosome is lethal because the animal has twice the normal gene expression from the X chromosome, which in turn causes imbalances in the expression of all other genes. During early embryogenesis in individuals with two X's, X-inactivation occurs randomly, producing lineages of cells with one X chromosome silenced. Each cell will only express the phenotype associated with the active X chromosome. The inactivated X chromosomes are condensed into structures known as Barr bodies. In each cell, there is one Barr body in XX individuals, one Barr body in XXY individuals, two Barr bodies in XXX individuals, and so on. This random inactivation is responsible for the variation in the coat color of calico cats, as one of the genes controlling coat color is located on the X chromosome. Gametes typically possess a maximum of one X chromosome and have no inactive X chromosomes. However, CC was created by somatic cell nuclear transfer, a form of cloning where the nucleus from an adult cell is transferred into an enucleated oocyte. The adult cell has already had one X chromosome inactivated, so the remaining X chromosome is the active one that dictates the coat color. In this case, the X chromosome with the orange coat color allele was inactivated, and the active X chromosome carries a black tabby allele, which results in CC's black tabby phenotype. There are other processes that might cause the appearance of a clone to differ from the donor, even though they are genetically identical. One process is the maternal effect, wherein the mother supplies mRNA transcripts and proteins to an oocyte. However, the mother's genotype is invariant and supplies the same products to all offspring. Similarly, the mitochondrial genome is inherited from the mother to the oocyte. The mitochondria present in the oocyte are also invariant and are passed on to all offspring.

Classify each feature as describing euchromatin, heterochromatin, or both. Answer Bank: has a high level of transcription has a low level of transcription is a state of DNA organization is found in prokaryotes is loosely packed is the major state of most genes Is the major state of the human Y chromosome is tightly packed

Euchromatin is loosely packed has a high level of transcription is found in prokaryotes is the major state of most genes Both is a state of DNA organization Heterochromatin is tightly packed has a low level of transcription is the major state of the human Y chromosome Explanation: Chromatin refers to the state of a particular region of a chromosome and is composed of DNA and interacting proteins. The function of the different forms of chromatin is to organize the structure of DNA and aid in regulating gene expression. Heterochromatin is the more tightly packed form of chromatin. The tight packing of heterochromatin means that it is transcriptionally inactive because the transcription sites are inaccessible to transcription machinery. Heterochromatin contains relatively few genes because the chromosome structural components, such as centromeres and telomeres, lack genes. Additionally, the human genome only contains about 8% heterochromatin. Chromosomes that are largely inactive, such as the human Y chromosome or a female's second X chromosome, are comprised mostly of heterochromatin. Euchromatin is the more loosely packed form of chromatin. This means that it is easily accessible to transcription machinery and is considered the more transcriptionally active form of chromatin. Because the human genome contains over 90% euchromatin, most genes are contained within euchromatin. As prokaryotes do not have different forms of chromatin, all of their genetic material is comprised of active euchromatin. The lack of heterochromatin in prokaryotes suggests it evolved in eukaryotes, possibly as genomes became larger and more complex.

FISH is a technique in molecular biology that refers to ____________________, and this technique allows scientists to _________________________________________. Fluorescence in situ hydridization; locate a specific DNA sequence in a chromosome Fluorescence in vitro hybridization; identify a specific gene in the genome Fluorescence in vivo hybridization; identify a specific mRNA in the cell

Fluorescence in situ hydridization; locate a specific DNA sequence in a chromosome

Circular DNA from SV40 virus was isolated and subjected to gel electrophoresis. The results are shown in lane A (the control) of the adjoining gel patterns. A photo shows three gel panels labeled a through c. Panel A is control and has one wide band at the bottom and one smaller band near the top. Panel B has bands that decrease in width from bottom to top, except the topmost band is slightly wider. Panel C has bands that increase in thickness from bottom to top. Why does the DNA separate in agarose gel electrophoresis? differences in mass differences in charge differences in size differences in base‑pair sequence How does the DNA differ between the two bands in lane A? The bottom band is linear. The bottom band is supercoiled. The top band is circular. The top band is relaxed. The DNA was then incubated with topoisomerase I for 5 minutes and again analyzed by gel electrophoresis with the results shown in lane B. What types of DNA do the various bands in lane B represent? double‑stranded ligase products linear digestion fragments circular molecules with differing linking numbers Okazaki fragments Another sample of DNA was incubated with topoisomerase I for 30 minutes and again analyzed as shown in lane C. For lane C, what is the significance of the fact that more of the DNA is in slower‑moving forms? The DNA is becoming progressively more relaxed. The DNA is becoming progressively more condensed. The DNA is becoming progressively more degraded. The DNA is becoming progressively more elongated.

Gel electrophoresis separates DNA by differences in size The DNA difference in the two shown in A: The bottom band is supercoiled. The top band is relaxed. The types of DNA represented by the bands in B are circular molecules with differing linking numbers In C the significance of the fact that more of the DNA is in slower-moving forms is The DNA is becoming progressively more relaxed. Explanation: Differences in size result in the separate bands of SV40 virus DNA in the agarose gel. Although the circular SV40 virus DNA is all of the same sequence and length, topological differences between the DNA molecules affect their hydrodynamic volume. A supercoiled DNA molecule is more compact than a relaxed DNA molecule of the same length. The sample consists of DNA molecules all of the same sequence and length because it is all genomic SV40 virus DNA. Thus, differences in mass or base‑pair sequence are not the reasons why the DNA molecules separate in the gel. Due to the negatively charged phosphate groups in its sugar‑phosphate backbone, DNA has a constant mass‑to‑charge ratio at neutral to basic pH. Therefore, DNA molecules cannot be resolved by their charge using gel electrophoresis. The top band in lane A represents relaxed DNA, whereas the bottom band represents supercoiled DNA. Because it is more compact, supercoiled DNA moves faster than relaxed DNA through the gel matrix during gel electrophoresis. Since all the DNA is closed circular DNA, there are no linear fragments. The topology of a DNA molecule can be quantified by its linking number, 𝐿𝑘. 𝐿𝑘 is equal to the number of times that a strand of DNA winds in the right‑handed direction around the helix axis when the helix axis lies in a plane. The value of 𝐿𝑘 for a DNA molecule does not change as a result of twisting or deforming unless the DNA is cut, wound or unwound, and resealed. DNA molecules of the same length and sequence but differing 𝐿𝑘 values are called topological isomers, or topoisomers. The bands in lane B represent circular DNA molecules with different linking numbers. Topoisomerases catalyze changes in the 𝐿𝑘 of genomic DNA. Therefore, incubation of the SV40 virus DNA with topoisomerase I for five minutes results in a mix of topoisomers. Because topology determines the rate at which a DNA molecule moves through a gel matrix, the topoisomers resolve as multiple bands, as seen in lane B. Topoisomerase I only transiently breaks a single DNA strand during catalysis. Following a single rotation to unwind the DNA helix, the broken strand religates while still in the topoisomerase active site. Thus, linear DNA fragments are not created by incubation with topoisomerase. Okazaki fragments are formed during DNA replication, which is catalyzed by DNA polymerases. Ligation of double‑stranded Okazaki fragments is carried out by ligases. In lane C, more of the DNA is in slower‑moving forms because the DNA is becoming more relaxed with additional incubation time. Topoisomerase I incrementally changes the 𝐿𝑘 of a DNA molecule through a single unwinding rotation per catalytic cycle. Thus, if the SV40 virus DNA is incubated for longer with topoisomerase I, the topoisomerase is able to progressively relax more of the DNA. Because a larger proportion of DNA molecules are less condensed and therefore have a greater hydrodynamic volume, the resulting topoisomers skew toward the slower‑moving end of the gel column. Since topoisomerase I does not catalyze DNA replication or act as a restriction enzyme, the DNA molecules are neither elongated or degraded.

ChIP‑seq, chromatin immunoprecipitation sequencing, and bis‑seq, bisulfite sequencing, data for a particular gene were collected and illustrated as a graph. The graph highlights a region labeled as the active promoter and a region labeled as the active enhancer. One row is labeled as bisulfite sequencing and shows percentages near zero only at the promoter and enhancer. The row labeled H3K4me1 shows relatively high peaks at both the promoter and enhancer. The row labeled H3K4me3 shows a relatively high peak at the promoter and a low level at the enhancer. The row labeled H3K36me3 shows relatively low levels at both the promoter and enhancer. Determine where the histones bind the unmethylated DNA of the gene. H3K4me1 binds the promoter. H3K4me1 binds the enhancer. H3K4me3 binds the promoter. H3K4me3 binds the enhancer. H3K36me3 binds the promoter. H3K36me3 binds the enhancer.

H3K4me1 binds the promoter. H3K4me1 binds the enhancer. H3K4me3 binds the promoter. Explanation: Bisulfite sequencing (bis‑seq) is a method whereby unmethylated DNA can be distinguished from methylated DNA. Upon treatment with bisulfite, any unmethylated cytosine will be converted to uracil. This treatment means that only methylated cytosine will remain and can be distinguished. Chromatin immunoprecipitation sequencing (ChIP‑seq) is a method for determining protein‑DNA interactions and binding sites. In this case, histones were analyzed to determine where they bind in a particular gene. The first row of data indicates the bis-seq data and percentage of DNA methylation. The active promoter region and active enhancer region each show DNA methylation levels near zero. DNA methylation of promoters is typically associated with repression. The histone H3K4me1 row shows high peaks at the promoter and enhancer regions relative to the surrounding regions. These data indicate H3K4me1 binds at both the promoter and enhancer. The histone H3K4me3 row shows a high peak only at the promoter region relative to the surrounding regions. These data indicate H3K4me3 binds at only the promoter. The histone H3K36me3 row shows low levels at the promoter and enhancer regions relative to the surrounding regions. These data indicate H3K36me3 does not bind at the promoter or enhancer.

Methylation on lysine residues can preserve positive charge, which ususally leads to a more condensed chromatin that is incompatible with gene expression. However, methylation in certain context can activate gene expression as well. Which of the following provides a reasonable explantion for that? Methylation leads to the formation of heterochromatin Methylation, especially trimethylation, can create a binding site that could recruit gene activating proteins to that chromatin region. Methylation can facilitate acetylation of the same lysine residue, thus activating gene expression. Methylation serves to recruit more H1 proteins to chromatin

Methylation, especially trimethylation, can create a binding site that could recruit gene activating proteins to that chromatin region.

Consider the following figure. The step that can be facilitated by a topoisomerase is ____________; the circular DNA with the highest linking number is ___________________. Steps B & C; the one in the middle Step A; the one on the left Step A; the one on the right

Step A; the one on the left

Most scientists consider the Human Genome Project (HGP) to be the most significant scientific project of the 21st century. Choose the statements that describe the key findings of the Human Genome Project. There are approximately three billion base pairs in the human genome. The genetic information of a cell is stored in the form of DNA. There are 23 pairs of chromosomes that make up the human genome. DNA exists in a double helical form. The human genome contains approximately 25000 genes.

There are approximately three billion base pairs in the human genome. The human genome contains approximately 25000 genes. Explanation: Completed in 2003, the Human Genome Project (HGP) was an international biological research project with the goal of sequencing and analyzing all of the DNA base pairs that make up the human genome. A genome is the complete set of genetic information stored in the form of DNA. Some of the key new findings of the HGP were the estimates that the human genome has approximately three billion nucleotides and that there are only 25000 total genes. Major discoveries on the identity, form, and packaging of DNA were largely made in the 1950s. In 1944, Oswald Avery, Colin MacLeod, and Maclyn McCarty identified that genetic information is stored in the form of DNA. In 1953, James Watson and Francis Crick deduced that the structure of DNA is a double helix. In 1956, Joe Hin Tjio and Albert Levan experimentally determined that there are 46 chromosomes in the nucleus of a human cell. The findings of the HGP illustrate that the field of genetics is rapidly changing how the genomes of complex organisms are sequenced. Rapid and cost‑effective sequencing is revolutionizing the field of molecular medicine. By rapidly sequencing the genomes of thousands of individuals, scientists can identify the rare genetic variations that increase the risk of diseases such as schizophrenia and autism. Scientists have been working on developing drugs that can be tailored to the patient's genome sequence and treating diseases based on the patient's cellular biochemistry. Before the Human Genome Project, scientists considered it extremely implausible that the human genome, made of billions of nucleotides, could be sequenced in a matter of days. The powerful tools of genome sequencing are quickly clarifying the complex relationships between the information stored in DNA and the structural and functional characteristics of an organism.

Suppose that a mother seeks genetic counseling because she is concerned that her child may have velocardiofacial syndrome, a syndrome that can result in symptoms such as a cleft palate and heart defects. The genetic counselor is aware that this disease is caused only by a small deletion in chromosome 22q11.2, that traditional karyotyping often overlooks. Consequently, the genetic counselor informs the mother that a cost-effective test will be conducted to visually detect the presence or absence of the specific chromosomal change by using velocardiofacial syndrome-specific probes and a sample of the child's DNA. To which technique is the genetic counselor likely referring? restriction mapping sequence tagged site (STS) mapping linkage mapping fluorescence in situ hybridization (FISH)

fluorescence in situ hybridization (FISH) Explanation: Fluorescence in situ hybridization (FISH) is a physical mapping method in which specifically designed fluorescent probes bind to the part of chromosomes with which they share a high degree of similarity. The fluorescence allows the results to be visualized. FISH requires a DNA- or RNA-specific probe that targets a specific DNA sequence and is labeled with a fluorescent molecule. If the DNA sequence of interest is present, the labeled probe will bind to the target DNA. The chromosome region of interest will then fluoresce under fluorescence microscopy. FISH is valuable because it is a relatively rapid approach that is sensitive enough to detect a single gene in a chromosome set, and the fluorescent probes allow easy visualization of the results. This method can be applied in fields such as genetic counseling and medicine to detect chromosomal abnormalities in patients. The disease-specific, fluorescently labeled probes anneal to the portions of the associated chromosomal changes, which can then be visualized. Other mapping methods are useful for identifying the location of a specific gene, but they either do not allow visualization or do not provide high enough resolution to detect the presence or absence of specific DNA sequences, such as genes. Recombination frequency-based mapping, or linkage mapping, uses the positions of genes relative to one another to identify their approximate location on a chromosome. These locations are measured in nonphysical units that are useful for mapping relative positions of genes, but not necessarily detecting specific sequences. Restriction mapping uses restriction enzymes to map unknown segments of DNA. Restriction fragment length polymorphism (RFLP) analysis also uses restriction enzymes for DNA profiling. These two approaches use gel electrophoresis to separated DNA fragments by size, but do not specify chromosomal location. Sequence tagged site (STS) mapping offers high resolution results but is expensive and more commonly used to assemble detailed physical maps of genomes.

Which description applies to epigenetic gene regulation? a gene cluster controlled by a single promoter that transcribes to a single mRNA strand processing of exons in mRNA that results in a single gene coding for multiple proteins mRNA modifications such as additions of a 5′ cap and 3′ poly‑A tail and removal of introns heritable changes in gene expression that occur without altering the DNA sequence protein modifications such as addition of a functional group, or structural changes such as folding

heritable changes in gene expression that occur without altering the DNA sequence Explanation: Gene regulation is a tightly controlled process that can occur at several different stages within a cell. Operons and epigenetic changes control whether transcription occurs. Following transcription, RNA molecules can be altered through additions of molecules and alternative splicing of exons. Proteins can also be altered after translation. Epigenetic gene regulation influences whether and how often genes are transcribed. Unlike mutations, epigenetic changes alter gene expression by means other than direct changes to the DNA sequence. Some of these changes have been shown to pass from generation to generation. Two of the most widely studied sources of epigenetic change are DNA methylation and histone modification, which both appear to influence how tightly wound the DNA molecule is and how accessible it is to cellular transcription machinery. Operons also control whether transcription occurs. An operon is a cluster of genes along a stretch of DNA that operate under the coordinated control of a single promoter. The genes in an operon turn on or off together and are controlled by environmental influences, such as the availability of certain nutrients. Following transcription, messenger RNA (mRNA) undergoes modifications such as the addition of a 5′ cap and 3′ poly‑A tail, the removal of introns, and the splicing of exons. Without such modifications, an mRNA molecule will not be transcribed properly. Furthermore, exons transcribed from a gene may not always get spliced together the same way. Alternative splicing of exons in mRNA is a regulated process that allows a single gene to code for multiple proteins. Finally, post‑translational modifications can occur after proteins are made. These modifications can include the addition of functional groups and peptides, the alteration of the chemical nature of the amino acids, and structural changes such as folding.

Select the post‑translational modifications of histones that are most commonly associated with changes in transcription levels in eukaryotes. methylation and acetylation phosphorylation and glycosylation nitrosylation and ubiquitination acetylation and lipidation Which of the post‑translational modifications generally targets a protein for degradation in eukaryotes? carboxylation methylation ubiquitination lipidation acetylation

methylation and acetylation are the most common post-translational modification of histones associated with changes in eukaryotic transcription levels. ubiquitination is the eukaryotic post-translational modification that generally targets proteins for degradation. Explanation: There are many types of post‑translational modifications of proteins that can affect transcription levels in eukaryotes. Histones are proteins that package DNA into tightly‑wound chromatin. Methylation and acetylation are common post‑translational modifications to histones that affect transcription levels in eukaryotes. The methylation of histones affects the availability of DNA for transcription, and can affect levels of transcription by silencing genes or increasing transcription, depending on which amino acids are methylated. Acetylation decreases the interaction between the histone and DNA and relaxes the DNA, allowing greater levels of transcription. Lipidation is a post‑translational modification that targets a protein to associate with membranes, such as the cell membrane or the organelles membrane. Phosphorylation is a very common post‑translational modification and has many roles, such as activating or inactivating a protein, and aiding in cellular signal transduction. Glycosylation is another common post‑translational modification that can affect protein folding and stability within the cell. Ubiquitination is a post‑translational modification that attaches multiple copies of the protein ubiquitin to the side‑chain lysine residues using a multi‑step process. The addition of ubiquitin targets the protein for degradation by the proteasome.

Histone proteins increase the transcription rate for a particular gene. regulate gene expression by reducing the rate of transcription. stabilize condensed chromosomal DNA during transcription. bind tightly to DNA and strengthen hydrogen bonds between bases.

regulate gene expression by reducing the rate of transcription. Explanation: The DNA‑histone protein complex is the core association in chromatin. DNA is wrapped around histone proteins like thread on a spool. Histone proteins regulate gene expression by reducing the rate of transcription. Keep in mind that transcription is the synthesis of an mRNA strand from DNA. Gene expression involves the transcription of a gene on DNA to mRNA and the subsequent translation of the mRNA to the encoded protein. Histone proteins are one way of regulating gene expression. Histones can be modified to either bind more tightly or less tightly to a DNA segment, blocking or allowing access to DNA transcription machinery and therefore slowing or assisting transcription. There are likely additional ways that histone proteins affect DNA transcription that have not yet been discovered. Although histone proteins play a role in compacting DNA, condensed chromosomal DNA is visible during mitosis, not‑transcription.