Biology Exam 4

ch. 20 Describe the use of DNA microarray assays.

- A DNA microarray consists of tiny amounts of a large number of single-stranded DNA fragments representing different genes fixed to a glass slide in a tightly spaced array, or grid, of dots (like a DNA (computer) chip). - The mRNAs from the cells being studied are reverse-transcribed into cDNAs, and a fluorescent label is added so the cDNAs can be used as probes on the microarray. - The resulting pattern of colored dots reveals the dots to which each probe was bound and thus the genes that are expressed in the cell samples being tested.

ch. 18 Using the Trp operon as an example, explain the concept of an operon and the function of the operator, repressor, and co-repressor. Understand how the Trp operon is regulated.

- By itself, the trp operon is turned on; that is, RNA polymerase can bind to the promoter and transcribe the genes of the operon. - The trp operon can be switched off by a protein called the trp repressor. A repressor binds to the operator, preventing RNA polymerase from transcribing the genes, often by preventing RNA polymerase from binding. - Only when a tryptophan molecule binds to the trp repressor at an allosteric site does the repressor protein change to the active form that can attach to the operator, turning the operon off. - Tryptophan functions in this system as a corepressor, a small molecule that cooperates with a repressor protein to switch an operon off. As tryptophan accumulates, more tryptophan molecules associate with trp repressor molecules, which can then bind to the trp operator and shut down production of the tryptophan pathway enzymes. - When tryptophan is low, trp repressor proteins would be inactive and release themselves from the operator, turning it on.

ch. 20 What is CRISPR? How does it work?

- CRISPR ("clustered regularly interspaced short palindromic repeats") is a gene editing technique that uses a bacterial enzyme (Cas9) and a guide RNA to direct the enzyme to the location where it needs to cut and modify the DNA sequences. - CRISPR is based on the natural defense mechanism of some bacteria that cut and store the DNA of invading viruses. - CRISPR can be used to disable, repair, or insert new genes in the genome.

ch. 16 What are the roles of DNA polymerase, mismatch repair enzymes, and nuclease in DNA proofreading and repair?

- DNA polymerase synthesizes the new DNA strands by sequencing nucleotides at the 3' end and proofreads newly made DNA, replacing any incorrect nucleotides. - Mismatch repair enzymes remove and replace incorrectly paired nucleotides that have resulted from replication errors. - Nucleotide excision repair - a nuclease (DNA-cutting enzyme) cuts out and replaces damaged stretches of DNA, and the resulting gap is then filled in with nucleotides by DNA polymerase and DNA ligase, using the undamaged strand as a template.

ch. 17 Describe the process of termination in translation and explain which enzymes, protein factors, and energy sources are needed.

- Elongation continues until a stop codon in the mRNA reaches the A site. The nucleotide base triplets UAG, UAA, and UGA act as signals to stop translation. - A release factor, a protein shaped like an aminoacyl tRNA, binds directly to the stop codon in the A site. The release factor causes the addition of a water molecule instead of an amino acid to the polypeptide chain. This reaction hydrolyzes (breaks) the bond between the completed polypeptide and the tRNA in the P site, releasing the polypeptide through the exit tunnel of the ribosome's large subunit. - The remainder of the translation assembly then comes apart in a multistep process, aided by other protein factors. - Breakdown of the translation assembly requires the hydrolysis of two more GTP molecules.

ch. 16 Explain the role of DNA polymerase.

- Enzymes called DNA polymerases catalyze the synthesis and elongation of new DNA at a replication fork by adding nucleotides to the 3′ end of a pre-existing chain. - Most DNA polymerases require a primer and a DNA template strand, along which complementary DNA nucleotides are lined up, one by one.

ch. 20 How is DNA technology used in the forensic sciences?

- If enough blood, semen, or tissue is available, forensic laboratories can determine the blood type or tissue type by using antibodies to detect specific cell-surface proteins. - Genetic markers that vary in the population can be analyzed for a given person to determine that individual's unique set of genetic markers, or genetic profile. - DNA fingerprinting looks at variable sections of DNA called - short tandem repeats (STRs): repeated end-to-end units of two- to five-nucleotide sequences in specific regions of the genome. - PCR is used to amplify particular STRs, using sets of primers that are labeled with colored fluorescent tags; the length of the region, and thus the number of repeats, can then be determined by electrophoresis. - Labs can compare the number of repeats at each of these STRs to a sample taken from a crime scene and calculate the probability that the DNA from a suspect matches that sample.

ch. 20 How can DNA technology have medical applications in such areas as the diagnosis of genetic disease, the development of gene therapy, vaccine production, and the development of pharmaceutical products?

- PCR and labeled nucleic acid probes can detect the DNA of pathogens, even in small amounts. - Medical scientists can now diagnose hundreds of human genetic disorders by using PCR with primers that target the genes associated with these disorders. The amplified DNA product is then sequenced to reveal the presence or absence of the disease-causing mutation. - Gene therapy's aim is to insert a normal allele of the defective gene into the somatic cells of the tissue affected by the disorder.

ch. 17 Define "point mutations." Give an example and note the significance of such a change.

- Point mutations: chemical changes in one base pair of a gene. - For example, we can trace the genetic basis of sickle-cell disease to the mutation of a single nucleotide pair in the gene that encodes the b-globin polypeptide of hemoglobin. The change of a single nucleotide in the DNA's template strand leads to an altered mRNA and the production of an abnormal protein.

ch. 17 Compare protein synthesis in prokaryotes and eukaryotes.

- Prokaryotes have polycistronic mRNA, which contains the coding sequence of several genes, while eukaryotes have monocistronic mRNA, which contains the coding sequence of only one gene. - Prokaryotes can start protein synthesis before the transcription of mRNA is completed, which is called coupled transcription-translation, while eukaryotes have separate transcription and translation processes. - Prokaryotes do not contain a 5' cap on mRNA, while eukaryotes contain a 5' cap on mRNA. - Eukaryotes have introns or non-coding sequences in their genes, which are removed by splicing, while prokaryotes have only exons or coding sequences.

ch. 18 Explain the role that promoters, enhancers, activators, and repressors play in transcriptional control.

- Promoters are DNA sequences that are located upstream of the gene and serve as the binding site for RNA polymerase, which initiates transcription. - Enhancers are regulatory DNA sequences that can be located far away from the gene they regulate (distal control elements) and can increase the rate of transcription. - Activators are proteins that bind to enhancers and promote transcription by enhancing the interaction of RNA polymerase with the promoter. - Repressors are proteins that bind to DNA sequences and prevent transcription by impeding the progress of RNA polymerase along the DNA strand. They can also block the binding of activating transcription factors.

ch. 20 Describe the process of RT-PCR.

- Reverse transcription polymerase chain reaction (RT-PCR): combines the reverse transcription process with the PCR process. - The sample mRNA is first converted to double-stranded DNA. The enzyme reverse transcriptase is used to synthesize a complementary DNA copy (a reverse transcript) of each mRNA in the sample. The mRNA is then degraded by the addition of a specific enzyme, and a second DNA strand, complementary to the first, is synthesized by DNA polymerase. - complementary DNA (cDNA): the resulting double-stranded DNA from RT-PCR that lacks introns and can be used for protein expression in bacteria. - cDNA can then be thermally broken down into two single-stranded DNAs to serve as a template for PCR amplification of the gene of interest. - the products are run on a gel, in a process called gel electrophoresis, which will reveal amplified DNA products as bands in samples that contain the mRNA of interest.

ch. 18 How does excessive cell division result from mutations in the ras oncogenes?

- The Ras protein, encoded by the ras gene, is a G protein that relays a signal from a growth factor receptor on the plasma membrane to a cascade of protein kinases. - The cellular response at the end of the pathway is the synthesis of a protein that stimulates the cell cycle. - Normally, such a pathway will not operate unless triggered by the appropriate growth factor. But certain mutations in the ras gene can lead to the production of a hyperactive Ras protein that triggers the kinase cascade even in the absence of growth factor, resulting in increased cell division.

ch. 17 Explain what determines the primary structure of a protein and describe how a polypeptide must be modified before it becomes fully functional.

- The primary structure of a protein is determined by the sequence of amino acids in the polypeptide chain. This sequence is encoded in the DNA of the gene that codes for the protein. Thus, a gene determines primary structure, which in turn determines shape. - Before a polypeptide can become a functional protein, it may need to undergo post-translational processing. As the polypeptide is being synthesized, it begins to fold, and the goal is to find the lowest point in the energy landscape, which is generally assumed to be the native or functional state of the polypeptide. It may be necessary for enzymes to cut the polypeptide chain to remove single amino acids or even entire regions of the polypeptide before it can become a functional protein. Once the polypeptide has been correctly folded, it may need to be modified further to become fully functional. This can include the addition of prosthetic groups, such as heme in hemoglobin, or the addition of sugars, lipids, phosphate groups, or others.

ch. 20 How are vaccines made? What is the purpose of vaccines?

- Vaccines are made up of whole bacteria or viruses, or parts of them, often a protein or sugar. These active components of the vaccine, called antigens, are what trigger an immune response when in the body. - Vaccines familiarize your immune system — which makes antibodies to defend your body against harmful invaders — with a certain pathogen so it'll know what to do if you become infected with that pathogen in the future.

ch. 17 Describe the process of elongation in translation and explain which enzymes, protein factors, and energy sources are needed.

- amino acids are added one by one to the previous amino acid at the C-terminus (carboxyl end) of the growing chain; methionine end is called the N-terminus. - each addition involves several proteins called elongation factors and occurs in a three-step cycle. Energy expenditure occurs in the first and third steps. - Codon recognition requires hydrolysis of one molecule of GTP, which increases the accuracy and efficiency of this step. One more GTP is hydrolyzed (broken up) to provide energy for the translocation step; total = 2 GTP molecules. - The mRNA is moved through the ribosome in one direction only, 5′ end first. - elongation cycle repeated as each amino acid is added until the polypeptide is completed.

ch. 16 Explain the "base-pairing rule" and describe its significance.

- base pairing rule: adenine (A) to thymine (T), as guanine (G) to cytosine (C). - significant because Adenine and guanine are purines, nitrogenous bases with two organic rings, while cytosine and thymine are nitrogenous bases called pyrimidines, which have a single ring. Pairing a purine with a pyrimidine is the only combination that results in a uniform diameter for the double helix. -Although the base-pairing rules dictate the combinations of nitrogenous bases, they do not restrict the sequence of nucleotides along each DNA strand. The linear sequence of the four bases can be varied in countless ways, and each gene has a unique base sequence. - each base has chemical side groups that can form hydrogen bonds with its appropriate partner: Adenine can form two hydrogen bonds with thymine and only thymine; guanine forms three hydrogen bonds with cytosine and only cytosine.

ch. 20 Explain how DNA technology can be used to improve the nutritional value of crops and to develop plants that can produce pharmaceutical products.

- can identify and isolate genes that control the production of specific nutrients, such as vitamins and minerals. These genes can then be introduced into crops to increase their nutritional content. - can develop plants that can produce pharmaceutical products in a process known as molecular farming, which involves introducing genes that code for specific proteins into plants. These proteins can then be harvested and used to produce drugs, vaccines, and other pharmaceutical products.

ch. 20 Describe the safety and ethical questions related to recombinant DNA studies and the biotechnology industry.

- feared hazardous new pathogens might be created. - strains of microorganisms to be used in recombinant DNA experiments are genetically crippled to ensure that they cannot survive outside the laboratory. - concerns about the safety of genetically modified organisms (GMOs), a transgenic (transferred gene) organism that has acquired one or more genes from another species or from another variety of the same species, in food. - fear that transgenic plants might pass their new genes to close relatives in nearby wild areas, creating "super weeds". - other worries include the possibility that transgenic protein products might cause allergic reactions.

ch. 17 Compare where transcription and translation occur in prokaryotes and in eukaryotes.

- in prokaryotes, there is not nucleus, so both processes occur in the cytoplasm. - can occur simultaneously. - in eukaryotes, transcription occurs in the nucleus, and translation occurs in the cytoplasm after mRNA is transported to it. - are separated in space and time.

ch. 18 How does noncoding RNAs play many roles in controlling gene expression?

- microRNAs (miRNAs): small, single-stranded RNA molecules capable of binding to complementary sequences in mRNA molecules. - The miRNA-protein complex degrades the target mRNA or, less often, simply blocks its translation. - small interfering RNAs (siRNAs): small noncoding RNAs, similar in size and function to miRNAs, differing slightly in the structure of their precursors. - can block gene expression by RNA interference (RNAi). - small noncoding RNAs can cause remodeling of chromatin structure. - piwi-interacting RNAs, or piRNAs, induce formation of heterochromatin, blocking the expression of some parasitic DNA elements in the genome known as transposons. - long noncoding RNAs (lncRNAs): range from 200 to hundreds of thousands of nucleotides in length. - One lncRNA, long known to be functional, is responsible for X chromosome inactivation, which prevents expression of genes located on one of the X chromosomes in most female mammals by binding back to and coating that chromosome, leading to condensation of the entire chromosome into heterochromatin. - can act as a scaffold, bringing DNA, proteins, and other RNAs together into complexes, either to condense chromatin or, in some cases, to help bring the enhancer of a gene together with mediator proteins and the gene's promoter, activating gene expression in a more direct fashion.

ch. 18 Explain the difference between positive and negative control. Give examples of each from the lac operon.

- negative control: operons are switched off by the active form of their respective repressor protein. - lac repressor without allolactose binds to operator, blocking transcription. - positive control: operons are switched on by a regulatory protein which interacts directly with the genome to increase transcription. - cAMP binds to CRP protein, attaches to lac promoter, increasing the affinity of RNA polymerase for the lac promoter, and thereby increasing the rate of transcription of the lac operon.

ch. 20 Describe the process of nucleic acid hybridization.

- nucleic acid hybridization: the base pairing of one strand of a nucleic acid to a complementary sequence from another nucleic acid strand, either DNA or RNA (which are called nucleic acid probes that are labeled during synthesis with a fluorescent tag so they can be followed).

ch. 18 Distinguish between proto-oncogenes and oncogenes. Describe three genetic changes that can convert proto-oncogenes to oncogenes.

- oncogenes: cancer-causing genes. - proto-oncogenes: normal versions of cellular genes that code for proteins that stimulate normal cell growth and division. - an oncogene arises from a genetic change that leads to an increase either in the amount of the proto-oncogene's protein product or in the intrinsic activity of each protein molecule. - The genetic changes are: 1. epigenetic changes: Mutation in a gene for a chromatin-modifying enzyme can lead to loosening of the chromatin and inappropriate expression of a proto-oncogene. 2. translocations: chromosomes can break and rejoin incorrectly, translocating fragments from one chromosome to another. - If a translocated proto-oncogene ends up near an especially active promoter (or other control element), its transcription may increase. 3. gene amplification: increases the number of copies of the proto-oncogene in the cell through repeated gene duplication. 4. point mutations: if either in the promoter or an enhancer that controls a proto-oncogene, can cause an increase in a gene's expression. - if in the coding sequence of the proto-oncogene, can change the gene's product to a protein that is hyperactive or more degradation-resistant than the normal protein.

ch. 18 Explain why a mutation knocking out the p53 gene can lead to excessive cell growth and cancer.

- p53 gene: a tumor-suppressor gene. - it encodes a protein that is a specific transcription factor that promotes the synthesis of cell cycle-inhibiting proteins. - once the p53 protein is activated, it activates several other genes that halt cell cycles and turn on DNA repair or "suicide" genes, so without p53, they would never activate.

ch. 17 Explain in what way the genetic code is redundant and unambiguous.

- redundant because more than one codon may specify a particular amino acid, but not ambiguous because no codon specifies more than one amino acid.

ch. 20 Explain how eukaryotic genes are cloned to avoid the problems associated with introns.

- scientists usually employ an expression vector, a cloning vector that contains a highly active bacterial promoter just upstream of a restriction site where the eukaryotic gene can be inserted in the correct reading frame. - can use a form of the gene that includes only the exons (complementary DNA, or cDNA). - mRNA lacks introns so it can be converted to cDNA using reverse transcriptase. cDNA can then be inserted into the host.

ch. 20 What is a stem cell? Why is it important in cloning plants and animals? What are some problems associated with animal cloning?

- stem cell: a relatively unspecialized cell that can both reproduce itself indefinitely and, under appropriate conditions, differentiate into specialized cells of one or more types. - important because they have the ability to differentiate into any type of cell in the body, meaning that they can be used to create an entire organism from a single cell. - In plants, mature cells can "dedifferentiate" and then give rise to all the specialized cell types of the organism (cells said to be totipotent). - In animals, the older the cell, the less totipotent it is. - problems include various defects such as obesity, pneumonia, liver failure, and premature death. - In the nuclei of fully differentiated cells, a small subset of genes is turned on and expression of the rest of the genes is repressed. - often the result of epigenetic changes in chromatin, such as acetylation of histones (proteins) or methylation of DNA.

ch. 16 How is the lagging strand synthesized even though DNA polymerase can add nucleotides only to the 3' end?

- the DNA polymerase travels away from the replication fork, the same direction as the template strand. - the lagging strand is discontinuous because the polymerase must stop and go back toward the replication fork to continue synthesis, creating a series of short fragments called Okazaki fragments. - each Okazaki fragment is bonded covalently by DNA ligase to the others to create a continuous strand. Each fragment requires a primer. - these small fragments are generated by the polymerase making a small 5' to 3' fragment inside the replication bubble as it opens. Then all the little fragments are joined together.

ch. 17 Describe the process of initiation in translation and explain which enzymes, protein factors, and energy sources are needed.

- the enzyme aminoacyl-tRNA synthetase joins a given amino acid to an appropriate tRNA through covalent bonding, producing a aminoacyl tRNA, or charged tRNA. - the union of mRNA, initiator tRNA, and the small ribosomal subunit is followed by the attachment of a large ribosomal subunit, completing the translation initiation complex. Proteins called initiation factors are required to bring all these components together. The cell also expends energy obtained by hydrolysis of a guanosine triphosphate (GTP) molecule to form the initiation complex.

ch. 16 How long is each helix?

- the helix makes one full turn every 3.4 nm along its length, with ten layers of base pairs in each, stacked just 0.34 nm apart.

ch. 18 How does the lac operon function? Explain the role of the inducer, allolactose.

- the lac operon ("lactose") is an inducible operon, which is usually off but can be stimulated (induced) to be on when a specific small molecule interacts with a different regulatory protein. - The regulatory gene, lacI, located outside the lac operon, codes for an allosteric repressor protein that can switch off the lac operon by binding to the lac operator. The lac repressor is active by itself, binding to the operator and switching the lac operon off. - allolactose is the inducer, the specific small molecule that inactivates the repressor. - In the absence of lactose (and therefore allolactose), the lac repressor is in its active shape and binds to the operator; thus, the genes of the lac operon are silenced. - If lactose is added to the cell's surroundings, allolactose binds to the lac repressor and alters its shape so the repressor can no longer bind to the operator. - Without the lac repressor bound, the lac operon is transcribed into mRNA, and the enzymes for using lactose are made.

ch. 20 Describe the polymerase chain reaction (PCR).

- the polymerase chain reaction, or PCR, is what biologists use to obtain the foreign DNA to be inserted. - has a three-step cycle (repeated 30-40 times) in which the reaction mixture is... 1. heated to high temperatures to denature (separate) the strands of the double-stranded DNA. 2. cooled to allow annealing (hydrogen bonding) of short, single-stranded DNA primers complementary to sequences on opposite strands at each end of the target sequence. 3. a special DNA polymerase extends the primers in the 5' to 3' direction. - Taq polymerase is the special DNA polymerase used because its heat-stabling quality allows the enzyme to function at temperatures up to 95°C. - Pfu polymerase, which is more accurate and stable, is also used but is more expensive. -With each cycle, the number of target segment molecules of the correct length doubles, so the number of molecules equals 2^n, where n is the number of cycles.

ch. 17 What is the significance of the reading frame during translation?

- the reading frame is a way of dividing the sequence of nucleotides in a nucleic acid (DNA or RNA) molecule into a set of consecutive, non-overlapping triplets (codons) that can be read as three letter words. - codons must be read in the correct reading frame (correct groupings and direction) in order for the specified polypeptide to be produced.

ch. 17 Explain why polypeptides begin with methionine when they are synthesized.

- the start codon on the mRNA is AUG, which signals the protein-synthesizing machinery to begin translating the mRNA at that location, and it also codes for the amino acid methionine. In most eukaryotic proteins, the methionine is cleaved off the polypeptide before it becomes a functional protein.

ch. 16 Define "antiparallel" and explain why continuous synthesis of both DNA strands is not possible.

- the two strands of DNA in a double helix are antiparallel, meaning that they are oriented in opposite directions to each other. - 5'→ 3' direction of one strand runs counter to 3' → 5' direction of other strand. - Continuous synthesis of both stands is not possible because the structure allows DNA polymerases to add nucleotides only to the free 3' end of a primer or growing DNA strand, never the 5' end.

ch. 17 What is the function of tRNA?

- transfer RNA (tRNA) is a translator that transfers an amino acid from the cytoplasmic pool of amino acids to a growing polypeptide in a ribosome. - a tRNA bears a specific amino acid at one end of its three-dimensional structure, while at the other end is a nucleotide triplet (anticodon) that can base-pair with the complementary codon on mRNA. - is a translator in the sense that, in the context of the ribosome, it can read a nucleic acid word (the mRNA codon) and interpret it as a protein word (the amino acid).

ch. 16 How did Watson and Crick deduce the structure of DNA? Describe the evidence they used.

- used an X-ray diffraction picture of DNA discovered by Maurice Wilkins and Rosalind Franklin, which established that the DNA molecule existed in a helical conformation.

ch. 16 Describe the semiconservative model of replication.

- when a double helix replicates, each of the two daughter molecules will have one old strand, from the parental molecule, and one newly made strand. - each strand serves as a template for ordering nucleotides on complementary strands. - nucleotides line up along template strands according to base-pairing rules. - nucleotides are linked to form new strands.

ch. 16 What are Chargaff's rules?

1. DNA base composition varies between species. 2. For each species, the percentages of A and T bases are roughly equal, as are those of G and C bases.

ch. 20 What are three techniques used to aggressively introduce recombinant DNA into eukaryotic cells?

1. Electroporation: involves applying an electric field to the cells, creating temporary pores in the cell membrane. The DNA is then able to enter the cell through these pores. 2. Microinjection: a fine needle is used to directly inject the recombinant DNA into the nucleus of the cell. 3. Viral vectors: use modified viruses to deliver the recombinant DNA into the cells. The virus is modified so that it can't cause disease, but it still has the ability to enter the cell and deliver the DNA.

ch. 20 Explain the advantages and limitations of PCR.

Advantages: Highly specific: PCR can distinguish DNA sequences by just one nucleotide, making it very accurate. Sensitive: it allows the detection of even a single copy of DNA. Rapid: PCR can amplify DNA sequences in a matter of hours. Versatile: PCR can be used to amplify DNA from a variety of sources, including blood, saliva, and hair. Quantitative: Real-time PCR can be used to quantify the amount of DNA present in a sample. Limitations: Contamination: highly sensitive to contamination, which can lead to false-positive results. Limited length: PCR can only amplify DNA fragments up to a certain length, typically less than 10 kb. Errors: PCR can introduce errors into the amplified DNA sequence, such as mutations or deletions. Inhibitors: PCR can be inhibited by substances present in the sample.

ch. 20 What are the procedures used for cloning a eukaryotic gene into a bacterial plasmid?

DNA cloning: identical copies of a gene. plasmid: small, circular DNA molecule with only a small number of genes. cloning vector: a DNA molecule that can carry foreign DNA into a host cell and be replicated there. - Researchers insert DNA they want to study ("foreign" DNA) into the plasmid. The resulting plasmid is now a recombinant DNA molecule. The plasmid is then returned to a bacterial cell, producing a recombinant bacterium. This single cell reproduces through repeated cell divisions to form a clone of cells, a population of genetically identical cells.

ch. 20 Explain how advances in recombinant DNA technology have helped scientists study the eukaryotic genome.

DNA technology: techniques used to manipulate DNA. genetic engineering: the direct manipulation of genes for practical purposes. recombinant DNA molecule: a molecule containing DNA from two different sources. - key advances include the discovery of enzymes that modify DNA molecules in ways that enable them to be joined together in new combinations, the demonstration that DNA molecules can be cloned, propagated, and expressed in bacteria, the development of methods for chemically synthesizing and sequencing DNA molecules, and the development of the polymerase chain reaction method for amplifying DNA in vitro. - advances have allowed scientists to overcome the difficulties associated with studying the eukaryotic genome, such as the length of DNA, the fact that genes make up a small fraction of the genome and that genes are nearly identical to the noncoding sequences.

ch. 17 How is RNA different from DNA?

DNA: - contains deoxyribose sugar - has nitrogenous base A, G, C, and T - double-stranded RNA: - contains ribose sugar - has nitrogenous base A, G, C, and U (uracil instead of thymine) - single-stranded

ch. 17 How is RNA modified after transcription in eukaryotic cells?

During RNA processing, both ends of the primary transcript are altered, and sometimes interior sections of the RNA molecule are cut out and the remaining parts spliced together (RNA splicing). - the 3′ end, which is synthesized first, receives a 5′ cap, a modified form of a guanine (G) nucleotide added onto the 5′ end after transcription of the first 20-40 nucleotides have been transcribed - at the 3′ end, an enzyme then adds 50-250 more adenine (A) nucleotides, forming a poly-A tail.

ch. 20 Define and distinguish between genomic libraries using plasmids, phages, and cDNA.

Genomic libraries are collections of cloned DNA fragments that represent the entire genome of an organism. - Plasmid libraries contain thousands of randomly cut pieces of genomic DNA that are inserted into plasmids. Plasmids are naturally occurring small circles of DNA that can be specifically cut, rearranged, and sealed with enzymes. After transformation, the bacteria copy the plasmid, resulting in gene cloning. - Phage libraries contain thousands of randomly cut pieces of genomic DNA that are inserted into phages. Phages are viruses that infect bacteria and can be used to clone genomic DNA fragments. Phage genomes are bigger than plasmids and can be engineered to remove large amounts of DNA that are not needed for infection and replication in host cells. - cDNA libraries contain only cloned copies of mRNAs. The process of creating a cDNA library involves isolating mRNA and "transcribing" it with reverse transcriptase. The resulting cDNA is then inserted into plasmids or phages.

ch. 18 How are repressible and inducible enzymes different? How do those differences reflect differences in the pathways they control?

Inducible enzymes: - their synthesis is induced by a chemical signal. - usually function in catabolic pathways, which break down a nutrient to simpler molecules. - By producing the appropriate enzymes only when the nutrient is available, the cell avoids wasting energy and precursors making proteins that are not needed. Repressible enzymes: - usually function in anabolic pathways, which synthesize essential end products from raw materials (precursors). - By suspending production of an end product when it is already present in sufficient quantity, the cell can allocate its organic precursors and energy for other uses.

ch. 16 What is the difference between the leading strand and the lagging strand of DNA?

Leading Strand: - template strand in the 3' to 5' direction. - synthesis occurs toward replication fork. - continuously synthesized (DNA polymerase synthesizes the copied strand continuously by moving in the complementary 5' to 3' direction). Lagging Strand: - template strand in the 5' to 3' direction. - synthesis occurs away from replication fork. - discontinuously synthesized.

ch. 18 Explain how cyclic AMP and the cyclic AMP receptor protein are affected by glucose concentration.

Only when lactose is present and glucose is in short supply does E. coli use lactose as an energy source, and only then does it synthesize appreciable quantities of the enzymes for lactose breakdown. - the mechanism depends on the interaction of an allosteric regulatory protein with a small organic molecule, cyclic AMP (cAMP) in this case, which accumulates when glucose is scarce. - The regulatory protein, called cAMP receptor protein (CRP), is an activator, a protein that binds to DNA and stimulates transcription of a gene. (CRP is also called catabolite activator protein, or CAP.) - When cAMP binds to this regulatory protein, CRP assumes its active shape and can attach to a specific site at the upstream end of the lac promoter. - the rate of transcription is controlled by whether CRP has cAMP bound to it: With bound cAMP, the rate is high; without it, the rate is low (like a volume control). - Amount of glucose increases→ cAMP concentration falls, and without cAMP, CAP detaches from operon.

ch. 17 Describe the role of the promoter, the terminator, and the transcription unit.

Promoter: the DNA sequence where RNA polymerase attaches and initiates transcription. (in eukaryotes) Transcription factors: a collection of proteins that help guide the binding of RNA polymerase and the initiation of transcription. Transcription initiation complex: the whole complex of transcription factors and RNA polymerase bound to the promoter. TATA box: promoter DNA sequence. (in bacteria) Terminator: the coding sequence that signals for RNA polymerase the end of transcription. Transcription unit: the stretch of DNA that is transcribed into an RNA molecule.

ch. 17 Explain how RNA polymerase recognizes where transcription should begin.

RNA polymerase: an enzyme that joins together RNA nucleotides complementary to the DNA template strand in the 5' to 3' direction, thus elongating the RNA polynucleotide. - don't need to add the first nucleotide onto a pre-existing primer. - it slides along the chromosome, feeling for sections that are not tightly wound on the histones or turned off by methylation. When it runs along such "open" sections, it feels for a certain DNA sequence, (...TGNTATAAT... in prokaryotes) called the promoter and binds to it.

ch. 18 Explain eukaryotic processing of pre-mRNA.

RNA processing includes enzymatic addition of a 5′ cap and a poly-A tail, as well as splicing out of introns, to yield a mature mRNA.

ch. 17 What is the general process of transcription? Know the steps of initiation, elongation, and termination.

Transcription is the process in which RNA is synthesized under the direction of DNA. It consists of three steps: 1. Initiation: once transcription factors are attached to the promoter DNA and the polymerase is bound to them in the correct orientation on the DNA, the enzyme unwinds the two DNA strands and begins transcribing the template strand at the start point. 2. Elongation: As RNA polymerase moves along the DNA, it untwists the double helix, exposing about 10-20 DNA nucleotides at a time for pairing with RNA nucleotides. The enzyme adds nucleotides to the 3′ end of the growing RNA molecule as it moves along the double helix. The newly synthesized RNA molecule behind the RNA polymerase peels away from its DNA template, and the DNA double helix re-forms. 3. Termination: In bacteria, transcription proceeds through a terminator sequence in the DNA. The transcribed terminator (an RNA sequence) functions as the termination signal, causing the polymerase to detach from the DNA and release the transcript. In eukaryotes, RNA polymerase II transcribes a sequence on the DNA called the polyadenylation signal sequence (AAUAAA), which proteins find and then cut the RNA transcript free from the polymerase at a point about 10-35 nucleotides downstream, releasing the pre-mRNA, which then undergoes processing.

ch. 17 Distinguish between transcription and translation.

Transcription: the synthesis (production) of RNA using information in the DNA. - the two nucleic acids are written in different forms of the same language, and the information is simply transcribed, or "rewritten," from DNA to RNA. - a DNA strand can serve as a template for assembling a complementary sequence of RNA nucleotides. - produces mRNA. messenger RNA (mRNA): carries a genetic message from the DNA to the protein-synthesizing machinery of the cell. Translation: the synthesis of a polypeptide using the information in the mRNA in ribosomes. - there is a change in language: The cell must translate the nucleotide sequence of an mRNA molecule into the amino acid sequence of a polypeptide.

ch. 18 Describe the process of alternative splicing.

alternative RNA splicing: different mRNA molecules are produced from the same primary transcript, depending on which RNA segments are treated as exons and which as introns.

ch. 16 What is the structure of DNA?

double helix: two stranded. antiparallel: subunits run in opposite directions. - double, anti-parallel, right-handed (up-right direction) helix with two outer sugar-phosphate backbones, with the nitrogenous bases paired in the molecule's interior, held together by hydrogen bonds. - each DNA nucleotide monomer consists of the sugar deoxyribose (blue) attached to both a nitrogenous base (A, T, G, or C), and a phosphate group (yellow). The phosphate group of one nucleotide is attached to the sugar of the next by a covalent bond, forming a "backbone" of alternating phosphates and sugars from which the bases project. A polynucleotide strand has directionality, from the 5′ end (with the phosphate group) to the 3′ end (with the ¬OH group of the sugar). 5′ and 3′ refer to the numbers assigned to the carbons in the sugar ring (see magenta numbers).

ch. 18 Explain the regulation of chromatin structure.

histone acetylation: the addition of an acetyl group to an amino acid in a histone tail—appears to promote transcription by opening up chromatin structure. DNA methylation: the addition of methyl groups to histones can lead to the condensation of chromatin and reduced transcription.

ch. 17 Describe RNA splicing.

introns (intervening sequences): the noncoding segments of nucleic acid that lie between coding regions. exons: the other regions that are eventually expressed, usually by being translated into amino acid sequences. The removal of introns is accomplished by a large complex made of proteins and small RNAs called a spliceosome. This complex binds to several short nucleotide sequences along an intron, including key sequences at each end. The intron is then released (and rapidly degraded), and the spliceosome joins together the two exons that flanked the intron. so, during the process of RNA splicing, introns are removed and exons joined to form a contiguous coding sequence.

ch. 17 Define base-pair insertions and deletions. Give examples and note the significance of such a change.

nucleotide-pair insertions or deletions: additions or losses of nucleotide pairs in a gene. - may alter the reading frame of the genetic message (codons), causing a frameshift mutation (when the number of nucleotides inserted or deleted is not a multiple of three) and extensive missense or nonsense mutation that leads to premature termination. - Unless the frameshift is very near the end of the gene, the protein is almost certain to be nonfunctional.

ch. 18 What is the adaptive advantage of genes grouped into an operon?

operon: a unit of genetic function consisting of the operator, the promoter, and the genes they control. operator: segment of DNA that behaves as an on-off switch. - a single "on-off switch" can control the whole cluster of functionally related genes (genes are coordinately controlled). - allows bacteria to respond quickly to changes in their environment. - no expenditure of extra energy and raw material.

ch. 16 What is the process of DNA replication? Note the structure of the many origins of replication and replication forks.

origins of replication: short stretches of DNA that have a specific sequence of nucleotides. replication fork: a Y-shaped region where the parental strands of DNA are being unwound. - Replication begins at sites called origins of replication, where the two DNA strands are separated by proteins that attach to them, opening up a replication "bubble". - Replication proceeds in both directions from each origin, until the entire molecule is copied. - At the end of each replication bubble is a replication fork.

ch. 20 What is the natural function of restriction enzymes?

restriction enzymes (restriction endonucleases): enzymes that cut DNA molecules at a limited number of specific locations. - created by bacteria to cut up foreign DNA from other organisms or phages. - recognize a particular short DNA sequence, or restriction site, and cut both DNA strands at precise points within this restriction site. - Host DNA protected by methylation, the addition of methyl groups (—CH3) to adenines or cytosines at recognition sites.

ch. 20 How is the creation of sticky ends by restriction enzymes useful in producing a recombinant DNA molecule?

restriction fragments: the pieces left after a restriction enzyme makes many cuts in DNA molecules with 4 to 6 nucleotide pairs. sticky end: the single-stranded end of a double-stranded fragment. - These short extensions can form hydrogen-bonded base pairs with complementary sticky ends on any other DNA molecules cut with the same restriction enzyme. - associations can be made permanent by DNA ligase, an enzyme that catalyzes the formation of covalent bonds that close up the sugar-phosphate backbones of DNA strands, forming a stable recombinant plasmid containing foreign DNA.

ch. 17 Define base-pair substitutions. Give examples and note the significance of such a change.

single nucleotide-pair substitutions: the replacement of one nucleotide and its partner with another pair of nucleotides. - the wrong base pair is present but the overall number of base pairs is correct. - If the change is in the 3rd base of a codon, because of redundancy in the genetic code, there will be no change in the amino acid translated and the protein will be the same (silent mutation: no observable effect on the phenotype). - If the change is in the 1st or 2nd spot, the amino acid translated will be different (missense mutation: when an incorrect amino acid is coded for; little change occurs). - If the change causes a premature Stop codon, this is a nonsense mutation, resulting in a polypeptide that will be shorter than the polypeptide encoded by the normal gene, usually leading to nonfunctional proteins.

ch. 16 Describe the functions of telomeres. Explain the significance of telomerase to healthy and cancerous cells.

telomeres: special, short nucleotide sequences at the ends of eukaryotic chromosomal DNA molecules. - protect genes from being eroded through multiple rounds of DNA replication (postpones but does not prevent). - telomeres decrease in length each replication in mature cell - an enzyme called telomerase catalyzes the lengthening of telomeres in eukaryotic germ cells, thus restoring their original length and compensating for the shortening that occurs during DNA replication. - it contains its own RNA molecule that it uses as a template to artificially "extend" the leading strand, allowing the lagging strand to maintain a given length. Telomerase is not active in most human somatic cells, but its activity varies from tissue to tissue. - normal shortening of telomeres may protect organisms from cancer by limiting the number of divisions that somatic cells can undergo. - in healthy cells, telomerase activity is low, and telomeres shorten with each cell division until they reach a critical length, leading to apoptosis. In contrast, most cancer cells have high telomerase activity, which allows them to maintain telomere length and avoid cell death.

ch. 17 Define "codon". What is the relationship between the linear sequence of codons on mRNA and the linear sequence of amino acids in a polypeptide?

triplet code: the genetic instructions for a polypeptide chain are written in the DNA as a series of nonoverlapping, three-nucleotide words. template strand: the DNA strand that provides the pattern, or template, for the sequence of nucleotides in an RNA transcript. coding strand: the non-template DNA strand. - codon: a sequence of three nucleotides which together form a unit of genetic code in a DNA or mRNA molecule, written in the 5' to 3' direction. - the linear sequence of the codon on the mRNA strand will code for one of the 20 amino acids that will be incorporated at the corresponding position along a polypeptide to create a protein.

ch. 18 How do mutations in tumor-suppressor genes contribute to cancer?

tumor-suppressor genes: encode proteins that help prevent uncontrolled cell growth. - Any mutation that decreases the normal activity of a tumor-suppressor protein may contribute to the onset of cancer, in effect stimulating growth through the absence of suppression.